# Orbital Mechanics for Engineering Students ISM

```SOLUTIONS MANUAL
to accompany
ORBITAL MECHANICS FOR ENGINEERING STUDENTS
Howard D. Curtis
Embry-Riddle Aeronautical University
Daytona Beach, Florida
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 1
Problem 1.1
(a)
(
)(
A ⋅ A = Ax iˆ + Ay ˆj + Az kˆ ⋅ Ax iˆ + Ay ˆj + Az kˆ
(
)
)
(
)
(
= Ax iˆ ⋅ Ax iˆ + Ay ˆj + Az kˆ + Ay ˆj ⋅ Ax iˆ + Ay ˆj + Az kˆ + Az kˆ ⋅ Ax iˆ + Ay ˆj + Az kˆ
)
=  Ax 2 ( iˆ ⋅ iˆ ) + Ax Ay iˆ ⋅ ˆj + Ax Az ( iˆ ⋅ kˆ )  +  Ay Ax ˆj ⋅ iˆ + Ay 2 ˆj ⋅ ˆj + Ay Az ˆj ⋅ kˆ 

 

+  Az Ax ( kˆ ⋅ iˆ ) + Az Ay kˆ ⋅ ˆj + Az 2 ( kˆ ⋅ kˆ ) 


2
=  Ax (1) + Ax Ay ( 0 ) + Ax Az ( 0 ) +  Ay Ax ( 0 ) + Ay 2 (1) + Ay Az ( 0 ) +  Az Ax ( 0 ) + Az Ay ( 0 ) + Az 2 (1)

 
 

( )
( )
( )
( )
( )
= Ax 2 + Ay 2 + Az 2
But, according to the Pythagorean Theorem, Ax2 + Ay 2 + Az 2 = A2 , where A = A , the magnitude of
the vector A . Thus A ⋅ A = A2 .
(b)
iˆ
A ⋅ ( B &times; C ) = A ⋅ Bx
ˆj
By
kˆ
Bz
Cx
Cy
Cz
(
) (
= Ax iˆ + Ay ˆj + Az kˆ ⋅  iˆ By Cz − Bz C y − ˆj ( Bx Cz − Bz Cx ) + kˆ Bx C y − By Cx 


= Ax By Cz − Bz C y − Ay ( Bx Cz − Bz Cx ) + Az Bx C y − By Cx
(
)
)
(
)
(
)
or
A ⋅ ( B &times; C) = AxBy C z + Ay Bz C x + Az BxC y − AxBz C y − Ay BxC z − Az By C x
(1)
Note that ( A &times; B) ⋅ C = C ⋅ ( A &times; B) , and according to (1)
C ⋅ ( A &times; B) = C x Ay Bz + C y Az Bx + C z AxBy − C x Az By − C y AxBz − C z Ay Bx
(2)
The right hand sides of (1) and (2) are identical. Hence A ⋅ ( B &times; C) = ( A &times; B) ⋅ C .
(c)
iˆ
A &times; ( B &times; C ) = Ax iˆ + Ay ˆj + Az kˆ &times; Bx
ˆj
By
kˆ
Bz =
Cx
Cy
Cz
(
)
iˆ
Ax
ˆj
Ay
kˆ
Az
By Cz − Bz C y
Bz Cx − Bx C y
Bx C y − By Cx
=  Ay Bx C y − By Cx − Az ( Bz Cx − Bx Cz )  î +  Az By Cz − Bz C y − Ax Bx C y − By Cx  ˆj




+  Ax ( Bz Cx − Bx Cz ) − Ay By Cz − Bz C y  k̂


= Ay Bx C y + Az Bx Cz − Ay By Cx − Az Bz Cx iˆ + Ax By Cx + Az By Cz − Ax Bx C y − Az Bz C y ˆj
(
)
(
(
)
(
)
)
(
) (
)
+ ( Ax Bz Cx + Ay Bz C y − Ax Bx Cz − Ay By Cz ) k̂
= Bx ( Ay C y + Az Cz ) − Cx ( Ay By + Az Bz )  î + By ( Ax Cx + Az Cz ) − C y ( Ax Bx + Az Bz )  ĵ




+ Bz ( Ax Cx + Ay C y ) − Cz ( Ax Bx + Ay By )  k̂


Add and subtract the underlined terms to get
1
Solutions Manual
Orbital Mechanics for Engineering Students
(
)
(
)
A &times; ( B &times; C ) = Bx Ay C y + Az Cz + Ax Cx − Cx Ay By + Az Bz + Ax Bx  î


+ By Ax Cx + Az Cz + Ay C y − C y Ax Bx + Az Bz + Ay By  ĵ


+ Bz Ax Cx + Ay C y + Az Cz − Cz Ax Bx + Ay By + Az Bz  kˆ


ˆ
ˆ
ˆ
ˆ
ˆ
= Bx i + By j + Bz k Ax Cx + Ay C y + Az Cz − Cx i + C y j + Cz kˆ Ax Bx + Ay By + Az Bz
(
)
(
)
(
)
(
)
(
Chapter 1
)(
) (
)(
)
or
A &times; ( B &times; C) = B( A ⋅ C) − C( A ⋅ B)
Problem 1.2 Using the interchange of Dot and Cross we get
[( A &times; B) &times; C] ⋅ D
( A &times; B) ⋅ (C &times; D) =
But
[( A &times; B) &times; C] ⋅ D = − [C &times; ( A &times; B)] ⋅ D
(1)
Using the bac – cab rule on the right, yields
[( A &times; B) &times; C] ⋅ D = −[ A(C ⋅ B) − B(C ⋅ A)] ⋅ D
or
[( A &times; B) &times; C] ⋅ D = −( A ⋅ D)(C ⋅ B) + ( B ⋅ D)(C ⋅ A)
(2)
Substituting (2) into (1) we get
[( A &times; B) &times; C] ⋅ D = ( A ⋅ C)( B ⋅ D) − ( A ⋅ D)( B ⋅ C)
Problem 1.3
Velocity analysis
From Equation 1.38,
v = v o + Ω &times; rrel + v rel .
(1)
From the given information we have
ˆ
v o = −10Iˆ + 30 Jˆ − 50K
(2)
(
) (
)
ˆ − 300Iˆ + 200 Jˆ + 100K
ˆ = −150Iˆ − 400 Jˆ + 200K
ˆ
rrel = r − ro = 150Iˆ − 200 Jˆ + 300K
Iˆ
Jˆ
(3)
ˆ
K
ˆ
Ω &times; rrel = 0.6 −0.4 1.0 = 320Iˆ − 270
0 Jˆ − 300K
−150 −400 200
2
(4)
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 1
v rel = −20iˆ + 25 ˆj + 70kˆ
(
ˆ
= −20 0.57735Iˆ + 0.57735 Jˆ + 0.57735K
)
(
ˆ
+ 25 −0.74296Iˆ + 0.66475 Jˆ + 0.078206K
)
(
ˆ
+ 70 −0.33864Iˆ − 0.47410 Jˆ + 0.81274K
)
so that
ˆ ( m s)
v rel = −53.826Iˆ − 28.115 Jˆ + 47.300K
(5)
Substituting (2), (3), (4) and (5) into (1) yields
(
) (
) (
ˆ + 320Iˆ − 270 Jˆ − 300K
ˆ + −53.826Iˆ − 28.115 Jˆ + 47.300K
ˆ
v = −10Iˆ + 30 Jˆ − 50K
)
ˆ
v = 256.17Iˆ − 268.12 Jˆ − 302.7K
= 478.68 0.53516Iˆ − 0.56011Jˆ − 0.63236K ( m s )
(
)
Acceleration analysis
From Equation 1.42,
a = a O + Ω &times; rrel + Ω &times; ( Ω &times; rrel ) + 2Ω &times; v rel + a rel
(6)
Using the given data together with (4) and (5) we obtain
ˆ
ao = 25Iˆ + 40 Jˆ − 15K
Iˆ
(7)
Jˆ
ˆ
K
ˆ
−1.0 = −340Iˆ + 230 Jˆ + 205K
Ω &times; rrel = −0.4 0.3
−150 −400
Iˆ
(8)
200
Jˆ
ˆ
K
ˆ
Ω &times; ( Ω &times; rrel ) = 0.6 −0.4 1.0 = 390IIˆ + 500 Jˆ − 34K
320 −270 −300
(9)
ˆ
Iˆ
Jˆ
K
ˆ
2Ω &times; v rel = 2 0.6
−0.4
1.0 = 2 9.151Iˆ − 82.206 Jˆ − 38.399K
−53.826 −28.115 47.300
(
)
(10)
a rel = 7.5iˆ − 8.5 ˆj + 6.0kˆ
(
ˆ
= 7.5 0.57735Iˆ + 0.57735 Jˆ + 0.57735K
)
(
ˆ
− 8.5 −0.74296Iˆ + 0.66475 Jˆ + 0.078206K
)
(
+ 6.0 −0.33864Iˆ − 0.47410 Jˆ + 0.81274K̂
)
ˆ
a rel = 8.6134Iˆ − 4.1649 Jˆ + 8.5418K
(11)
Substituting (7), (8), (9), (10) and (11) into (6) yields
(
) (
) (
ˆ + −340Iˆ + 230 Jˆ + 205K
ˆ + 390Iˆ + 500 Jˆ − 34K
ˆ
a = 25Iˆ + 40 Jˆ − 15K
(
) (
)
ˆ
ˆ  + 8.6134Iˆ − 4.1649 Jˆ + 8.5418K
+  2 9.151Iˆ − 82.206 Jˆ − 38.399K

3
)
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 1
ˆ
a = 102Iˆ + 601.42 Jˆ + 87.743K
)(
(
ˆ m s2
= 616.29 0.16551Iˆ + 0.97588 Jˆ + 0.14327K
)
Problem 1.4 From Example 2.8, we have
F = ω &times; F + 2ω &times; ( ω &times; F ) + ω &times; ω &times; F + ω &times; ( ω &times; F ) 
Substituting the given values for the quantities on the right hand side,
ω &times; F = 0 &times; 10iˆ = 0
2ω &times; ( ω &times; F ) = 2 ( −2kˆ ) &times; ( 3kˆ ) &times; (10iˆ )  = 2 ( −2kˆ ) &times; 30 ˆj = 120iˆ
( )
ω &times; ( ω &times; F ) = ( 3kˆ ) &times; ( −2kˆ ) &times; (10iˆ )  = ( 3kˆ ) &times; ( −20 ĵj) = 60iˆ
ω &times; ω &times; ( ω &times; F )  = ( 3kˆ ) &times;
{( 3kˆ ) &times; ( 3kˆ ) &times; (10iˆ )} = ( 3kˆ ) &times; ( 3kˆ ) &times; ( 30ˆj) = ( 3kˆ ) &times; ( −90iˆ ) = −270 ĵ
(
)
Thus, F = 0 + 120iˆ + 60iˆ − 270 ˆj = 120iˆ − 270 ˆj N s3 .
Problem 1.5
iˆ = sin θ Iˆ + cos θ Jˆ
ˆj = − cos θ Iˆ + sin θ Jˆ
ˆ
kˆ = K
(1)
Velocity analysis
The absolute velocity of the airplane is
v = vIˆ
(2)
The absolute velocity of the origin of the moving frame is
vo = 0
(3)
The position of the airplane relative to the moving frame is
rrel =
sin θ ˆ
h ˆ
h
i=
sin θ Iˆ + cos θ Jˆ = h
I + hJˆ
cos θ
cos θ
cos θ
(
)
(4)
The angular velocity of the moving frame is
Ω = −θ K̂
(5)
The velocity of the airplane relative to the moving frame is, making use of (1)
(
)
v rel = vrel iˆ = vrel sin θ Iˆ + cos θ Jˆ = vrel sin θ Iˆ + vrel cos θ Jˆ
According to Equation 1.38, v = v o + Ω &times; rrel + v rel . Substituting (2), (3), (4), (5) and (6) yields
ˆ ) &times;  h sin θ Iˆ + hJˆ  + v sin θ Iˆ + v cos θ Jˆ
vIˆ = 0 + ( −θ K
rel
rel
 cos θ

(
or
4
)
(6)
Solutions Manual
Orbital Mechanics for Engineering Students

 sin θ  ˆ 
vIˆ = 0 + ( hθ ) Iˆ −  hθ
J + vrell sin θ Iˆ + vrel cos θ Jˆ
 cos θ  

(
Chapter 1
)
Collecting terms,
sin θ 

vIˆ = hθ + vrel sin θ Iˆ +  vrel cos θ − hθ
 Ĵ

cos θ 
(
)
Equate the Î and Ĵ components on each side to obtain
hθ + vrel sin θ = v
− hθ
sin θ
+ vrel cos θ = 0
cos θ
Solving these two equations for θ and vrel yields
v
cos2 θ
h
(7)
vrel = v sin θ
(8)
θ=
Acceleration analysis
The absolute acceleration of the airplane, the absolute acceleration of the origin of the moving frame,
and the angular acceleration of the moving frame are, respectively,
a=0
ˆ
Ω = −θ K
ao = 0
(9)
The acceleration of the airplane relative to the moving frame is, making use of (1),
(
)
a rel = arel iˆ = arel sin θ Iˆ + cos θ Jˆ = arel sin θ Iˆ + arel cos θ Jˆ
(10)
Substituting (7) into (5), the angular velocity of the moving frame becomes
ˆ = − v cos2 θ K
ˆ
Ω = −θ K
h
(11)
Substituting (8) into (6) yields
(
)
v rel = vrel iˆ = v sin θ sin θ Iˆ + cos θ Jˆ = v sin 2 θ Iˆ + v sin θ cos θ Jˆ
(12)
From (4) and (9) we find
ˆ ) &times;  h sin θ Iˆ + hJˆ  = hθ Iˆ − hθ sin θ Jˆ
Ω &times; rrel = ( −θ K

 cos θ
cos θ
(13)
Using (5) and (7) we get
sin θ ˆ
v

v
 sin θ ˆ
Ω &times; rrel = hθ Iˆ − hθ
J = h  cos2 θ  Iˆ − h  cos2 θ 
J = v cos2 θ Iˆ − v sin θ cos θ Jˆ
h

h
 cos θ
cos θ
From (11) and (14) we have
5
(14)
Solutions Manual
Orbital Mechanics for Engineering Students
Iˆ
Jˆ
0
0
v cos2 θ
− v sin θ cos θ
Ω &times; ( Ω &times; rrel ) =
Chapter 1
ˆ
K
−
v2
v2
v
cos2 θ = −
sin θ cos3 θ Iˆ −
cos4 θ Jˆ
h
h
h
(15)
0
From (11) and (12),
2Ω &times; v rel = 2
Iˆ
Jˆ
ˆ
K
0
0
v sin 2 θ
v sin θ cos θ
−
v
v2
v2
cos2 θ = 2
sin θ cos3 θ Iˆ − 2
sin 2 θ cos2 θ Jˆ
h
h
h
(16)
0
According to Equation 1.42, a = a o + Ω &times; rrel + Ω &times; ( Ω &times; rrel ) + 2Ω &times; v rel + a rel . Substituting (9), (10),
(13), (15) and (16) yields

sin θ ˆ   v 2
v2

0 = 0 +  hθ Iˆ − hθ
J +  −
sin θ cos3 θ Iˆ −
cos4 θ Jˆ 


cos θ   h
h
 v2

v2
+2
sin θ cos3 θ Iˆ − 2
sin 2 θ cos2 θ Jˆ  + arel sin θ Iˆ + arel cos θ Jˆ
 h

h
(
)
Collecting terms


v2
v2
0 =  hθ −
sin θ cos3 θ + 2
sin θ cos3 θ + arel sin θ  Iˆ


h
h


sin θ v 2
v2
+  − hθ
−
cos4 θ − 2
sin 2 θ cos2 θ + arel cos θ  Jˆ


cos θ
h
h
or




v2
sin θ v 2
0 =  hθ +
sin θ cos3 θ + arel sin θ  Iˆ +  − hθ
−
cos2 θ (1 + sin 2 θ ) + arel cos θ  Jˆ




h
h
cos θ
Equate the Î and J√ components on each side to obtain
hθ + arel sin θ = −
− hθ
v2
sin θ cos3 θ
h
v2
sin θ
cos 2 θ (1 + sin 2 θ )
+ arel cos θ =
h
cos θ
Solving these two equations for θ&laquo;&laquo;and arel yields
θ = −2
v2
h2
cos3 θ sin θ
arel =
v2
cos3 θ
h
Problem 1.6 From Equation 2.58b with z = 0 we have
 y2

a = −2Ω y sin φ iˆ + Ω 2 RE sin φ cos φ ˆj − 
+ Ω 2 RE cos2 φ  k̂
 RE

where
6
(1)
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 1
RE = 6378 &times; 103 m
φ = 30&deg;
1000 &times; 103
= 27.78 m s
3600
2π
Ω=
= 7.2921 &times; 10−5 rad s
23.934 &times; 3600
y=
Substituting these numbers into (1), we find
a = −0.0020256iˆ + 0.014686 ˆj − 0.025557kˆ ( m s )
From F = ma , with m = 1000 kg , we obtain the net force on the car,
F=-2.0256 iˆ + 14.686 ˆj − 25.557kˆ ( N )
Flateral = Fx = −2.0256 N = −0.4554 lb , that is
Flateral = 0.4554 lb to the west
The normal force N of the road on the car is given by N = Fz + mg , so that
N = −25.557 + 1000 &times; 9.81 = 9784 N
Problem 1.7 From Equation 1.61b, with z = 0 ,
z
a = Ω 2 RE sin l cos lˆj − Ω 2 RE cos2 lkˆ
∑ Fy = ma y
From
we get
T sin θ = mΩ 2 RE sin φ cos φ
T=
mΩ 2 RE sin φ cos φ
sin θ
∑ Fz = ma z
From
L = 30 m
g
we obtain
T cos θ − mg = − mΩ 2 RE cos2 φ
y
North
2
mΩ RE sin φ cos φ
cos θ − mg = − mΩ 2 RE cos2 φ
sin θ
tan θ =
Ω 2 RE sin φ cos φ
g − Ω 2 RE cos2 φ
Since d = L tan θ , we deduce
d=L
Ω 2 RE sin φ cos φ
g − Ω 2 RE cos2 φ
Setting
7
d
Solutions Manual
Orbital Mechanics for Engineering Students
L = 30 m
RE = 6378 &times; 1000 = 6.378 &times; 10 6 m
φ = 29&deg;
yields
g = 9.81 m s2
2π
= 7.2921 &times; 10 −5 rad s
Ω=
23.9344 &times; 3600
d = 44.1 mm ( to the south )
8
Chapter 1
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 2
Problem 2.1
ˆ
r = 3t 4 Iˆ + 2t 3 Jˆ + 9t 2K
( 3t 4Iˆ + 2t3 Jˆ + 9t 2Kˆ ) ⋅ ( 3t 4Iˆ + 2t3 Jˆ + 9t 2Kˆ ) =
r =
r=
9t 8 + 4t 6 + 81t 4
d r
36t7 + 12t 5 + 162t 3
=
dt
9t 8 + 4t 6 + 81t 4
At t = 2 sec ,
r&laquo; =
4608 + 384 + 1296
2304 + 256 + 1296
= 101.3 m s
ˆ
r =12t 3Iˆ + 6t 2 Jˆ + 18tK
r =
(12t3Iˆ + 6t 2 Jˆ + 18tKˆ ) ⋅ (12t3Iˆ + 6t 2 Jˆ + 18tKˆ ) =
144t 6 + 36t 4 + 324t 2
At t = 2 sec ,
r&laquo; = 9216 + 576 + 1296 = 105.3 m s
Problem 2.2
uˆ r ⋅ uˆ r = 1 ⇒
duˆ
duˆ
duˆ
d
(uˆ ⋅ uˆ ) = 0 ⇒ dtr ⋅ uˆ r + uˆ r ⋅ dtr = 0 ⇒ uˆ r ⋅ dtr = 0
dt r r
Or,
dr
dr
−r
r
duˆ r
d  r
dt
dt = rr − rr
=  =
2
dt
dt  r 
r
r2
duˆ
r rr − rr r ⋅ r rr
uˆ r ⋅ r = ⋅
=
−
dt
r
r2
r2 r2
duˆ
But according to Equation 2.25, r ⋅ r = rr . Hence uˆ r ⋅ r = 0
dt
Problem 2.3 Both particles rotate with a constant angular velocity
around the center of mass c.m., which lies midway along the line
joining the two masses. Let û be the unit vector drawn from one of
the masses to c.m., which is the origin of an inertial frame. The only
force on m is that of mutual gravitational attraction,
Fg = G
m2
d2
uˆ
m
The absolute acceleration of m is normal to its circular path around c.m.,
a = ω2
d
uˆ
2
From Newton’s second law, Fg = ma , so that G
m2
d
2
uˆ = mω 2
9
d
uˆ , or
2
m
d
Fg
c.m.
Solutions Manual
ω =
Orbital Mechanics for Engineering Students
2Gm
d3
Problem 2.4 The center of mass of the three equal masses lies
at the centroid of the equilateral triangle, whose altitude h is
given by h = d o sin 60&deg; . The distance r of each mass from the
center of mass is, therefore
r=
Chapter 2
3
m
Fg
2
2
h = d o sin 60&deg;
3
3
1 3
m
x
2
h
3
Fg
1
2 2
ω d sin 60&deg;
3 o o
o
c.m.
30&deg;
30&deg;
Relative to an inertial frame with the center of mass as its
origin, the acceleration of each particle is
a = ω o2 r =
y
do
1 2
m
2
do
and this acceleration is directed toward the center of mass. The net force on each particle is the vector
sum of the gravitational force of attraction of its two neighbors. This net force is directed towards the
center of mass, so that its magnitude, focusing on particle 1 in the figure, is
Fnet = Fg
1−2
cos 30&deg; + Fg
1−3
cos 30&deg; = G
m ⋅m
d o2
cos 30&deg; + G
m ⋅m
d o2
cos 30&deg; = 2
Gm2
d o2
cos 30&deg;
Setting Fnet = ma , we get
2
Gm2
d o2
2
cos 30&deg; = m ω o2 d o sin 60&deg;
3
3Gm cos 30&deg; 3Gm
=
d o3 sin 60&deg;
d o3
3Gm
ωo =
d o3
ω o2 =
Problem 2.5
(a)
(b)
&micro;
398 , 600
=
= 7.697 km s
r
6378 + 350
3
2π 32
2π
(6378 + 350) 2 = 5492 sec = 91 min 32 s
T=
r =
&micro;
398 , 600
v=
Problem 2.6 The mass of the moon is 7.348 &times; 10 22 kg . Therefore, for a satellite orbiting the moon,

km 3 
km 3
22
&micro; = Gm moon =  6.67259 &times; 10 −20
7
.
348
&times;
10
kg
=
4903

kg − s2 
s2

(
)
The radius of the moon as 1738 km. Hence
&micro;
4903
=
= 1.642 km s
r
1738 + 80
3
2π 32
2π
(1738 + 80) 2 = 6956 sec = 1 hr 56 min
T=
r =
&micro;
4903
v=
10
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 2
Problem 2.7 The time between successive crossings of the equator equals the period of the orbit.
That is
d
2π
=
( REarth + z)3/2
Ω Earth REarth
&micro;
where d = 3000 km is the distance between ground tracks, z is the altitude of the orbit,
REarth = 6378 km and Ω Earth = 2π (23.934 ⋅ 3600) = 7.2921 &times; 10 −5 rad s . Thus
3000
(7.2921 &times; 10 )(6378)
−5
=
2π
398 600
(6378 + z)3/2
so that
z = 1440.7 km
Problem 2.8 From Example 2.3 we know that vGEO = 3.0747 km s . From Equation 2.82 we know
that
vesc = 2 vcircular . Hence ∆v = ( 2 − 1) vGEO = 0.41421 ⋅ 3.0747 = 1.2736 km s .
Problem 2.9
&micro;sun = 1.3271 &times; 1011 km 3 s2
rearth = 149.6 &times; 10 6 km
vearth =
&micro;sun
1.3271 &times; 1011
=
= 29.784 km s
rearth
149.6 &times; 10 6
vesc = 2 ⋅ 29.784 = 42.121 km s
vrelative = 42.121 − 29.784 = 12.337 km s
Problem 2.10
A
πab
=
T 3
T
πab
A=
= 1.0472 ab
3
Problem 2.11
&micro;
e sin θ
h
h
v⊥ = = 2
r h
vr =
h
1
&micro; 1 + e cos θ
v = v r2 + v ⊥ 2
&micro; 2( 2
=
e cos θ + sin 2 θ ) + 2e cos θ + 1
h
&micro; 2
v=
e + 2e cos θ + 1
h
Problem 2.12 For the ellipse, according to Problem 2.11,
11
Solutions Manual
vellipse2 =
Orbital Mechanics for Engineering Students
&micro;2
(e 2 + 2e cos θ + 1)
h2
For the circle, at the point of intersection with the ellipse,
vcircle2 =
&micro;
&micro;
&micro;2
= 2
= 2 (1 + e cos θ )
r h
1
h
&micro; 1 + e cos θ
Setting vcircle2 = vellipse2 ,
&micro;2
h2
2
(1 + e cos θ ) = &micro; (e 2 + 2e cos θ + 1)
2
h
yields e cos θ = −e 2 , or θ = cos −1 ( −e ) .
Problem 2.13 From Equation 3.42
tan γ =
e sin θ
1 + e cos θ
From Problem 2.12 θ = cos −1 ( −e ) . Hence
tan γ =
[
e sin cos −1 ( −e )
1 −e
[
]
2
]
But sin cos −1 ( −e ) = 1 − e 2 . Therefore,
tan γ =
e 1 − e2
1 −e
2
e
=
1 − e2
Problem 2.14
(a)
e=
rapogee − rperigee
rapogee + rperigee
=
70 000 − 7000
= 0.81818 (ellipse)
70 000 + 7000
=
77 000
= 38 500 km
2
(b)
a=
rapogee + rperigee
2
(c)
T=
2π
&micro;
a 3/2 =
2π
398 600
(38 500)3/2 = 75 180 s
(d)
ε=−
398 600
&micro;
=−
= −5.1766 km 2 s2
2a
2 ⋅ 38 500
(e) From Equation 3.62,
38 500(1 − 0.81818 2 )
1 + 0.81818 cos θ
cos θ = 0.88615 ⇒ θ = 27.607&deg;
6378 + 1000 =
12
(20.88 h )
Chapter 2
Solutions Manual
Orbital Mechanics for Engineering Students
(f) From Equation 3.40
h = &micro;(1 + e )rperigee = 398 600 ⋅ (1 + 0.81818) ⋅ 7000 = 71 226 km 2 s
Then
h 71 226
=
= 9.6538 km s
r
7378
398 600
&micro;
v r = e sin θ =
⋅ 0.81818 ⋅ sin 27.607&deg; = 2.1218 km s
h
71 226
v⊥ =
(g)
h
vperigee =
rperigee
h
vapogee =
rapogee
=
71 226
= 10.175 km s
7000
=
71 226
= 1.0175 km s
70 000
Problem 2.15
rperigee = 6378 + 250 = 6628 km
rapogee = 6378 + 300 = 6678 km
a=
T=
rperigee + rapogee
2π
&micro;
2
a
3/2
=
=
2π
6628 + 6678
= 6653 km
2
398 600
tperigee to apogee =
⋅ 6653 3/2 = 5400.5 s (90.009 m )
T
= 45.005 m
2
Problem 2.16
(a)
e=
rapogee − rperigee
rapogee + rperigee
=
(6378 + 1600) − (6378 + 600)
= 0.066863
(6378 + 1600) + (6378 + 600)
(b)
h = &micro;(1 + e )rperigee = 398 600 ⋅ (1 + 0.066863) ⋅ (6378 + 600) = 54 474 km 2 s
vperigee =
54 474
h
=
= 7.8065 km s
rperigee 6378 + 600
vapogee =
54 474
h
=
= 6.8280 km s
rapogee 6378 + 1600
(c)
3
3
T=
54 474
2π 
h 
2π


=
= 6435.6 s = 107.26 m
2 

2 
2
2
&micro;  1 −e 
398 600  1 − 0.066863 
Problem 2.17
h = rperigee vperigee = (6378 + 1270) ⋅ 9 = 68 832 km 2 s
rperigee =
h2
1
1
&micro; +e
2
13
Chapter 2
Solutions Manual
Orbital Mechanics for Engineering Students
68 832 2
1
⇒ e = 0.55416
1
+
e
398 600
e sin θ
0.55416 ⋅ sin 100&deg;
tan γ =
=
= 0.60385 ⇒ γ = 31.13&deg;
1 + e cos θ 1 + 0.55416 cos 100&deg;
6378 + 1270 =
z + Rearth =
z + 6378 =
2
h2
1
&micro; 1 + e cos θ
68 832 2
1
⇒ z = 6773.8 km
398 600 1 + 0.55416 cos 100&deg;
h = rperigee vperigee = (6378 + 1270) ⋅ 9 = 68 832 km 2 s
h2 1
rperigee = 2
&micro; 1+e
68 832 2
1
⇒ e = 0.55416
1
+e
398 600
e sin θ
0.55416 ⋅ sin 100&deg;
=
= 0.60385 ⇒ γ = 31.13&deg;
tan γ =
1 + e cos θ 1 + 0.55416 cos 100&deg;
6378 + 1270 =
z + Rearth =
z + 6378 =
2
h2
1
&micro; 1 + e cos θ
68 832 2
1
⇒ z = 6773.8 km
398 600 1 + 0.55416 cos 100&deg;
Problem 2.18
v r = v sin γ = 9.2 ⋅ sin 10&deg; = 1.5976 km s
v ⊥ = v cos γ = 9.2 ⋅ cos 10&deg; = 9.0602 km s
∴ h = rv ⊥ = (6378 + 640) ⋅ 9.0602 = 63 585 km 2 s
r=
h2
1
&micro; 1 + e cos θ
63 5852
1
398 600 1 + e cos θ
e cos θ = 0.445 29
6378 + 640 =
&micro;
e sin θ
h
398 600
e sin θ
1.5976 =
63 585
e sin θ = 0.254 84
sin θ 0.254 84
tan θ =
=
= 0.572 31 ⇒ θ = 29.783&deg;
cos θ 0.445 29
vr =
e sin 29.783&deg; = 0.254 84 ⇒ e = 0.51306
3
3
63 585
2π 
h 
2π


T= 2
=
= 16 075 s = 4.4654 h

2 
2
2
&micro;  1 −e 
398 600  1 − 0.51306 
14
Chapter 2
Solutions Manual
Orbital Mechanics for Engineering Students
Problem 2.19
rperigee + rapogee (6378 + 250) + (6378 + 42 000)
=
= 27 503 km
2
2
2π 3/2
2π
T=
a
=
27 503 3/2 = 45 392 s = 12.61 h
&micro;
398 600
a=
e=
rapogee − rperigee (6378 + 42 000) − (6378 + 250)
=
= 0.75901
rapogee − rperigee (6378 + 42 000) + (6378 + 250)
h = &micro;(1 + e )rperigee = 398 600 ⋅ (1 + 0.75901) ⋅ (6378 + 250) = 68 170 km 2 s
h
vperigee =
rperigee
68 170
= 10.285 km s
6378 + 250
=
Problem 2.20
h = rperigee vperigee = (6378 + 640) ⋅ 8 = 56 144 km s
rperigee =
7018 =
h2 1
&micro; 1+e
56 144 2 1
⇒ e = 0.126 82
398 600 1 + e
56 144 2
h2 1
1
=
= 9056.6 km
&micro; 1 − e 398 600 1 − 0.126 82
= 9056.6 − 6378 = 2678.6 km
rapogee =
zapogee
a=
T=
rperigee + rapogee
=
2
2π
&micro;
(7018) + (9056.6)
2
2π
a 3/2 =
398 600
= 8037.3 km
8037.3 3/2 = 7171 s = 1.992 h
Problem 2.21
T=
2π
&micro;
a 3/2
2π
2 ⋅ 3600 =
398 600
a 3/2 ⇒ a = 8059 km
Using the energy equation
vperigee2
2
−
&micro;
rperigee
=−
&micro;
2a
2
398 600
398 600
8
−
=−
⇒ rperigee = 7026.2 km
2
rperigee
2 ⋅ 8059
zperigee = 7026.2 − 6378 = 648.25 km
15
Chapter 2
Solutions Manual
Orbital Mechanics for Engineering Students
Problem 2.22
T=
2π
a 3/2
&micro;
2π
90 ⋅ 60 =
&micro;
a 3/2 ⇒ a = 6652.6 km
rperigee + rapogee = 2 a
(6378 + 150) + rapogee = 2 ⋅ 6652.6 ⇒ rapogee = 6777.1 km
e=
rapogee − rperigee
rapogee + rperigee
=
6777.1 − 6528
= 0.018 723
6777.1 + 6528
Problem 2.23
(a)
vesc = 2
v∞ =
398 600
&micro;
= 2
= 10.926 km s
r
6378 + 300
vperigee2 − vesc 2 = 152 − 10.926 2 = 10.277 km s
(b)
h = rperigee vperigee = 6678 ⋅ 15 = 100 170 km 2 s
rperigee =
6678 =
r=
h2 1
&micro; 1+e
100 170 2 1
⇒ e = 2.7696
398 600 1 + e
100 170 2
h2
1
1
=
= 48 497 km s
&micro; 1 + e cos θ
398 600 1 + 2.7696 cos 100&deg;
(c)
398 600
&micro;
e sin θ =
⋅ 2.7696 ⋅ sin 100&deg; = 10.853 km s
h
100 170
h 100 170
v⊥ = =
= 2.0655 km s
r 48 497
vr =
Problem 2.24
(a) From Equation 3.47
ε=
v 2 &micro; 2.23 2 398 600
− =
−
= 1.4949 km 2 s2
2
2
402 000
r
From Equation 3.50
h2 = −
2
1 &micro;2 (
1 398 600 (
1 − e2 ) = −
1 − e2 )
2 ε
2 1.4949
h2 = (5.3141e 2 − 1) &times; 1010
From the orbit equation
16
Chapter 2
Solutions Manual
Orbital Mechanics for Engineering Students
h2 = &micro;r (1 + e cos θ ) = 398 600 ⋅ 402 000 ⋅ (1 + e cos 150&deg;)
h2 = (16.024 − 13.877 e ) &times; 1010
Equating the two expressions for h2 ,
(5.3141e 2 − 1) &times; 1010 = (16.024 − 13.877 e) &times; 1010
yields
e 2 + 22.6113e − 4.0153 = 0
which has the positive root
e = 1.086
(b) Using this value of the eccentricity we find
h2 = (16.024 − 13.877 ⋅ 1.086) &times; 1010 = 9.5334 &times; 10 9 km 4 s2
so that
h2 1
9.5334 &times; 10 9
1
=
= 11 466 km
&micro; 1+e
398 600 1 + 1.086
= 11 466 − 6378 = 5087.6 km
rperigee =
zperigee
(c)
vperigee =
h
rperigee
=
9.5334 &times; 10 9
= 8.5158 km s
11 466
Problem 2.25 From the energy equation
v ∞ 2 + vesc 2 = v 2
2&micro;
2
= (1.1v ∞ )
r
&micro;
r = 9.5238 2
v∞
v ∞2 +
Substituting Equation 3.105
&micro;
r = 9.5238
&micro; 2

e −1
h

2
= 9.5238
h2 1
&micro; e2 − 1
Using Equation 3.40
[
r = 9.5238 rperigee (1 + e )
1
] e 2 − 1 = 9.5238
rperigee
e −1
Problem 2.26
v∞ =
398 600 2
&micro; 2
e −1 =
3 − 1 = 10.737 km s
h
105 000
17
Chapter 2
Solutions Manual
Orbital Mechanics for Engineering Students
Problem 2.27
(a)
r1 =
1
h2
&micro; 1 + e cos θ1
6378 + 1700 =
h2
1
398 600 1 + e cos 130&deg;
h2 = (3.2199 − 2.0697 e ) &times; 10 9
r2 =
1
h2
&micro; 1 + e cos θ 2
6378 + 500 =
h2
1
398 600 1 + e cos 50&deg;
h2 = (2.7416 + 1.7622e ) &times; 10 9
(3.2199 − 2.0697 e ) &times; 10 9 = (2.7416 + 1.7622e ) &times; 10 9
3.832e = 0.478 32
e = 0.124 82
(b)
h2 = (2.7416 + 1.7622 ⋅ 0.124 82) &times; 10 9 = 2.9615 &times; 10 9 km 4 s2
2.9615 &times; 10 9
1
h2 1
=
= 6605.4 km
&micro; 1+e
398 600 1 + 0.124 82
zperigee = 6605.4 − 6378 = 227.35 km
rperigee =
(c) From Equation 3.63,
a=
rperigee
1 −e
=
6605.4
= 7547.5 km
1 − 0.124 82
Problem 2.28
(a)
vr
= tan γ = tan 15&deg; = 0.26795 ⇒ v r = 0.26795v ⊥
v⊥
v2 = v r2 + v ⊥ 2
2
7 2 = (0.26795v ⊥ ) + v ⊥ 2
∴ v ⊥ = 6.7615 km s
v r = 1.8117 km s
h = rv ⊥ = 9000 ⋅ 6.7615 = 60 853 km 2 s
&micro;
e sin θ
h
398 600
1.8117 =
e sin θ
60 853
e sin θ = 0.27659
vr =
18
Chapter 2
Solutions Manual
r=
Orbital Mechanics for Engineering Students
1
h2
&micro; 1 + e cos θ
60 853 2
1
398 600 1 + e cos θ
e cos θ = 0.032259
9000 =
tan θ =
0.27659
e sin θ
=
= 8.574 ⇒ θ = 83.348&deg;
e cos θ 0.032259
(b)
e cos 83.348&deg; = 0.032259 ⇒ e = 0.27847
Problem 2.29 From Equation 3.50,
1 &micro;2 (
1 − e2 )
2 h2
1 398 600 2 (
−20 = −
1 − e 2 ) ⇒ e = 0.30605
2 60 000 2
ε=−
60 000 2
h2 1
1
=
= 6915.2 km
&micro; 1 + e 398 600 1 + 0.30605
= 6915.2 − 6378 = 537.21 km
rperigee =
zperigee
60 000 2
h2 1
1
=
= 13 015 km
&micro; 1 − e 398 600 1 − 0.30605
= 13 015 − 6378 = 6636.8 km
rapogee =
zapogee
Problem 2.30
(a)
v ⊥ = v cos γ = 8.85 ⋅ cos 6&deg; = 8.8015 km s
h = rv ⊥ = (6378 + 550) ⋅ 8.8015 = 60 977 km 2 s
r=
h2
1
&micro; 1 + e cos θ
6928 =
60 977 2
1
⇒ e cos θ = 0.34644
398 600 1 + e cos θ
v r = v sin γ = 8.85 ⋅ sin 6&deg; = 0.92508 km s
&micro;
v r = e sin θ
h
398 600
0.92508 =
e sin θ ⇒ e sin θ = 0.14152
60 977
e sin θ
= 0.40849 ⇒ θ = 22.22&deg;
e cos θ
e sin 22.22&deg; = 0.14152 ⇒ e = 0.37423
∴ tan θ =
(b)
3
3
T=
60 977
2π 
h 
2π


=
= 11 243 s = 187.39 m
2 

2

&micro;  1 − e2 
398 600  1 − 0.37423 2 
19
Chapter 2
Solutions Manual
Orbital Mechanics for Engineering Students
Problem 2.31
v ⊥ = v cos γ = 10 ⋅ cos 30&deg; = 8.6603 km s
h = rv ⊥ = (10 000) ⋅ 8.6603 = 86 603 km 2 s
r=
10 000 =
h2
1
&micro; 1 + e cos θ
86 603 2
1
⇒ e cos θ = 0.88159
398 600 1 + e cos θ
v r = v sin γ = 10 ⋅ sin 30&deg; = 5 km s
&micro;
e sin θ
h
398 600
5=
e sin θ ⇒ e sin θ = 1.0863
86 603
vr =
∴ tan θ =
e sin θ
= 1.2323 ⇒ θ = 50.94&deg;
e cos θ
Problem 2.32
v ⊥ = v cos γ = 10 ⋅ cos 20&deg; = 9.3969 km s
h = rv ⊥ = (15 000) ⋅ 9.3969 = 140 950 km 2 s
r=
15 000 =
h2
1
&micro; 1 + e cos θ
140 950 2
1
⇒ e cos θ = 2.323
398 600 1 + e cos θ
v r = v sin γ = 10 ⋅ sin 20&deg; = 3.4202 km s
&micro;
v r = e sin θ
h
398 600
3.4202 =
e sin θ ⇒ e sin θ = 1.2095
140 950
∴ tan θ =
e sin θ
= 0.52065 ⇒ θ = 27.504&deg;
e cos θ
Problem 2.33 Using the orbit equation r =
h2
1
we find
&micro; 1 + e cos θ
h2
= 10 000(1 + e cos 30&deg;) = 10 000 + 8660.3e
&micro;
h2
= 30 000(1 + e cos 105&deg;) = 30 000 − 7764.6e
&micro;
∴ 10 000 + 8660.3e = 30 000 − 7764.6e
16 425e = 20 000 ⇒ e = 1.2177
Problem 2.34
v1 =
398 600
&micro;
=
= 7.6127 km s
r
6378 + 500
20
Chapter 2
Solutions Manual
Orbital Mechanics for Engineering Students
v
v2 = v1 + 1 = 11.419 km s
2
2
v∞
&micro; v 2 &micro;
− = 2 −
∞
r
2
2
2
2
398 600
11.419
v∞
=
−
= 7.2441 ⇒ v ∞ = 3.8062 km s
2
2
6878
Problem 2.35
398 600
&micro;
=
= 7.7143 km s
r
6378 + 320
= v1 + 0.5 = 8.2143 km s
v1 =
vperigee
h = rperigee vperigee = 6698 ⋅ 8.2143 = 55 019
rperigee =
6698 =
h2 1
&micro; 1+e
55 019 2 1
⇒ e = 0.13383
398 600 1 + e
55 019 2
h2 1
1
=
= 8767.8 km
&micro; 1 − e 398 600 1 − 0.13383
= 8767.8 − 6378 = 2389.8 km
rapogee =
zapogee
Problem 2.36
&micro;
v=
rperigee
vperigee = v + αv = (1 + α )
&micro;
rperigee
h = rperigee vperigee = rperigee (1 + α )
&micro;
= (1 + α ) &micro;rperigee
rperigee
2
rperigee
(1 + α ) &micro;rperigee 1
h2 1
=
=
&micro; 1+e
&micro;
1+e
∴ 1 + e = (1 + α )
2
1 + e = 1 + 2α + α 2 ⇒ e = α (α + 2)
Problem 2.37
(a)
v1 =
v⊥2
398 600
&micro;
=
= 7.6686 km s
r
6378 + 400
= v1 + 0.24 = 7.9086 km s
h2 = rv ⊥ 2 = 6778 ⋅ 7.9086 = 53 605 km 2 s
r=
h2 2 1
&micro; 1+e
53 6052 1
⇒ e = 0.063572
398 600 1 + e
zperigee = 400 km
6778 =
2
21
Chapter 2
Solutions Manual
Orbital Mechanics for Engineering Students
6378 + zapogee =
2
Chapter 2
53 6052
h2 2 1
1
=
⇒ zapogee = 1320.3 km
2
&micro; 1 − e 398 600 1 − 0.063572
(b)
398 600
&micro;
=
= 7.6686 km s
6378 + 400
r
v⊥2 =
h2 = rv ⊥ 2 = 6778 ⋅ 7.6686 = 51 978 km 2 s
&micro;
e sin θ
h2 2
398 600
0.24 =
e sin θ ⇒ e2 sin θ = 0.031296
51 978 2
v r2 =
1
h2
r= 2
&micro; 1 + e2 cos θ
6778 =
51 978 2
1
⇒ e2 cos θ = 0 ⇒ θ = 90&deg; (since e cannot be zero if v r ≠ 0)
398 600 1 + e2 cos θ
e2 sin 90&deg; = 0.031296 ⇒ e2 = 0.031296
51 978 2
h2 1
1
6378 + zperigee = 2
=
⇒ zperigee = 196.49 km
2
2
&micro; 1 + e 398 600 1 + 0.031296
51 978 2
h2 1
1
6378 + zapogee = 2
=
⇒ zapogee = 631.3 km
2
2
&micro; 1 − e 398 600 1 − 0.031296
Problem 2.38 In Figure 3.30
m1 = msun = 1.989 &times; 10 30 kg
m2 = mearth = 5.974 &times; 10 24 kg
r12 = 149.6 &times; 10 6 km
From Equation 3.169
π2 =
m2
= 3.0035 &times; 10 −6
m1 + m2
Substitute π 2 into Equation 3.195,
f (ξ ) =
1 − π2
ξ + π2
3
(ξ + π 2 ) +
π2
ξ + π2 − 1
3
(ξ + π 2 − 1) − ξ
The graph of f (ξ ) is similar to Figure 3.33, with the two crossings on the right much more closely
spaced. Zeroing in on the regions where f (ξ ) = 0 , with the aid of a computer, reveals
ξ1 = 0.990 026 6
ξ2 = 1.010 034
ξ3 = −1.000 001
Then,
x1 = ξ1r12 = 148.108 &times; 10 6 km
x2 = ξ2 r12 = 151.101 &times; 10 6 km
x3 = ξ3 r12 = −149.600 &times; 10 6 km
22
Solutions Manual
Orbital Mechanics for Engineering Students
These are the locations of L1 , L2 and L3 relative to the center of mass of the sun-earth system
(essentially the center of the sun).
23
Chapter 2
Solutions Manual
Orbital Mechanics for Engineering Students
24
Chapter 2
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 3
Problem 3.1 Graph the function
f = x2 − 5x + 4 − 10e sin x
f(x)
to get an idea where the roots lie.
5
f ′ = 2 x − 5 − 10 cos xe
xi+1 = xi −
sin x
0
f ( xi )
x
-5
f ′( xi )
x 2 − 5xi + 4 − 10e sin xi
xi+1 = xi − i
2 xi − 5 − 10 cos xie sin xi
5
6
7
First root:
x0 = 4 (Estimate)
x1 = 4.7733482
x2 = 4.9509264
x3 = 4.9577657
Second root:
x0 = 7 (Estimate)
x1 = 6.7673080
x2 = 6.7732223
x3 = 6.7732128
Third root:
x0 = 8 (Estimate)
x1 = 7.9259101
x2 = 7.9198260
x3 = 7.9197836
x4 = 4.9577768
x5 = 4.9577768
x5 = 6.7732128
x4 = 7.9197836
Problem 3.2 Graph the function
f(x)
1
f = tan x − tanh x
to get an idea where the roots lie.
Clearly, the first root is x = 0 .
0
x
2
-1
4
6
8
f ′ = sec 2 x − sech 2 x
f ( xi )
x i +1 = x i −
f ′ ( xi )
tan xi − tanh xi
x i +1 = x i −
sec 2 xi − sech 2 xi
Second root:
x0 = 4 (Estimate)
x1 = 3.9322455
x2 = 3.9266343
x3 = 3.9266023
x4 = 3.9266023
Third root:
x0 = 7 (Estimate)
x1 = 7.0730641
x2 = 7.0686029
x3 = 7.0685827
x5 = 7.0685827
Problem 3.3
1
1
rapogee + rperigee = (6978 + 6578) = 6778 km
2
2
2π 3/2
2π
T=
a
=
6778 3/2 = 5553.5 s
&micro;
398 600
rapogee − rperigee 6978 − 6578
e=
=
= 0.029 507
rapogee + rperigee 6978 + 6578
a=
(
8
)
25
Fourth root:
x0 = 10 (Estimate)
x1 = 10.247568
x2 = 10.211608
x3 = 10.210178
x4 = 10.210176
x5 = 10.210176
10
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 3
Let B denote the point where the satellite flies through 400 km altitude on the way to apogee.
rB =
6378 + 400 =
E
tan B =
2
a (1 − e 2 )
1 + e cos θ B
(
6778 1 − 0.029 507 2
)
1 + 0.029 507 cos θ B
⇒ θ B = 91.691&deg;
1 − 0.029 507
θ
1 −e
91.691&deg;
tan B =
tan
1+e
2
1 + 0.029 507
2
MB = EB − e sin EB = 1.5708 − 0.029 507 sin 1.5708 = 1.5413 rad
M T 1.5413 ⋅ 5553.5
tB = B =
= 1362.3 s
2π
2π
tB is the time after perigee at which the spacecraft goes above 400 km. Let C denote the point at
which the satellite flies downward through 400 km altitude on its way to perigee. The time of flight
tBC from B to C is
tBC = T − 2tB = 5553.5 − 2 ⋅ 1362.3 = 2828.9 s = 47.148 m
Problem 3.4
(a)
1
1
r
+r
= (10 000 + 7000) = 8500 km
2 apogee perigee
2
2π 3/2
2π
T=
a
=
8500 3/2 = 7799 s
&micro;
398 600
rapogee − rperigee 10 000 − 7000
e=
=
= 0.176 47
rapogee + rperigee 10 000 + 7000
a=
(
)
t1 = 0.5 ⋅ 3600 = 1800 s
2πt1 2π ⋅ 1800
M1 =
=
T
7799
E1 − e sin E1 = M1
E1 − 0.176 47 sin E1 = 1.4501
E1 = 1.6263 rad (Algorithm 3.1)
∴ tan
1 + 0.176 47
θ1
E
1+e
1.6263
=
tan 1 =
tan
⇒ θ1 = 103.28&deg;
2
1 −e
2
1 − 0.176 47
2
t2 = 1.5 ⋅ 3600 = 5400 s
2πt2 2π ⋅ 5400
M2 =
=
T
7799
E2 − e sin E2 = M2
E2 − 0.176 47 sin E2 = 4.3504
E2 = 4.1969 rad (Algorithm 3.1)
1 + 0.176 47
θ2
1+e
E
4.1969
tan 2 =
tan
=
⇒ θ 2 = 231.99&deg;
2
1 −e
2
1 − 0.176 47
2
∆θ = θ 2 − θ1 = 128.7&deg;
tan
(b)
h = &micro; rperigee (1 + e ) = 398 600 ⋅ 7000 ⋅ (1 + 0.176 47 ) = 57 294 km 2 s
∆A h
=
∆t 2
∆t = 3600 s
26
Solutions Manual
∴ ∆A =
Orbital Mechanics for Engineering Students
1
1
h∆t = 57 294 ⋅ 3600 = 103.13 &times; 106 km 2
2
2
Problem 3.5
(a)
T=
15.743 ⋅ 3600 =
2π
&micro;
a 3/2
2π
398 600
a 3/2 ⇒ a = 31890 km
1
r
+r
2 perigee apogee
1
31890 = 12756 + rapogee ⇒ rapogee = 51024 km
2
rapogee − rperigee 51024 − 12756
=
= 0.6000
e=
rapogee + rperigee 51024 + 12756
a=
(
)
(
M = 2π
)
10
t
= 2π
15.743
T
E − e sin E = M
E − 0.6 sin E = 3.9911
E = 3.6823 rad (Algorithm 3.1)
tan
θ
E
1+e
1 + 0.6
3.6823
=
tan =
tan
⇒ θ = 195.78&deg;
2
1 −e
2
1 − 0.6
2
r=
31890(1 − 0.6 2 )
a (1 − e 2 )
=
= 48 924 km
1 + e cos θ 1 + 0.6 cos 195.78&deg;
(b)
v2 &micro;
&micro;
− =−
2
r
2a
398 600
v 2 398 600
−
=−
⇒ v = 2.0019 km s
2
48 924
2 ⋅ 31890
(c)
h = &micro;a (1 − e 2 ) = 398 600 ⋅ 31 890 ⋅ (1 − 0.6 2 ) = 90 196 km 2 s
vr =
398 600
&micro;
e sin θ =
⋅ 0.6 ⋅ sin 195.78&deg; = −0.720 97 km s
h
90 196
Problem 3.6
(a)
MPB = E − e sin E
2πtPB π
π
= − e sin
T
2
2
π
T


tPB =
−e
2
 2π
1 e
tDPB = 2tPB =
− T
2 π
(b)
T
−t
2 PB
T π
e 
 T 1
tBA = −
−e
=
+
T



2
2
2π
4 2π 
tBA =
27
Chapter 3
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 3
1 e
+
T
2 π
Problem 3.7
tan
EB
θ
π
1 −e
1 − 0.3
tan B =
tan = 0.733 80 ⇒ EB = 1.2661 rad
=
2
1+e
2
1 + 0.3
4
MB = EB − e sin EB = 1.2661 − 0.6 sin 1.2661 = 0.97992 rad
M
0.97992
tB = B T =
T = 0.48996T
2π
2π
Problem 3.8
a=
e=
T=
rapogee + rperigee
rapogee
2
− rperigee
rapogee + rperigee
2π
&micro;
a 3/2 =
14 000 + 7000
= 10 500 km
2
14 000 − 7000
=
= 0.33333
14 000 + 7000
=
2π
398 600
10 500 3/2 = 10708 s
E
1 −e
1 − 0.33333
 60&deg; 
tan θ = 60&deg; =
tan
=
tan (30&deg;) = 0.40825 ⇒ Eθ = 60&deg; = 0.77519 rad
 2 
2
1+e
1 + 0.33333
Mθ = 60&deg; = 0.77519 − 0.33333 sin (0.77519) = 0.54191 rad
M
0.54191
tθ = 60&deg; = θ = 60&deg; T =
10708 = 923.51 s
2π
2π
tθ = tθ = 60&deg; + 30 ⋅ 60 = 2723.5 s
t
2723.5
Mθ = 2π θ = 2π
T
10708
Eθ − e sin Eθ = Mθ
Eθ − 0.33333 sin Eθ = 1.5981 ⇒ Eθ = 1.9122 rad (Algorithm 3.1)
tan
θ
1+e
E
1 + 0.33333
1.9122
=
tan θ =
tan
⇒ θ = 126.95&deg;
2
1 −e
2
1 − 0.33333
2
Problem 3.9
a cos θ + b sin θ = c
b
c
cos θ + sin θ =
a
a
b
sin φ
≡ tan φ =
a
cos φ
sin φ
c
cos θ +
sin θ =
cos φ
a
cos θ cos φ + sin θ sin φ =
c
cos φ
a
c
cos φ
a
c

θ − φ = &plusmn; cos −1 cos φ
a

cos(θ − φ ) =
θ = φ &plusmn; cos −1
c

cos φ
a

28
Solutions Manual
Orbital Mechanics for Engineering Students
Problem 3.10
MB = EB − e sin EB
π
π
t
2π B = − e sin
T
2
2
π
π
− e sin
2 T = (0.25 − 0.15915e )Τ
tB = 2
2π
Problem 3.11

1
h2

rperigee (1 + e )
&micro; 1 + e cos θ 
 ⇒ r=
2
1 + e cos θ
1 
h
rperigee =

&micro; 1+e
r=
2rperigee =
rperigee (1 + 0.5)
1 + 0.5 cos θ B
cos θ B = −0.5 ⇒ θ B = 120&deg;
θ
1 −e
1 − 0.5
120&deg;
E
π
tan B =
tan B =
tan
2
1+e
2
1 + .5
2
2
π
π
MB = EB − e sin EB = − 0.5 sin = 1.0708 rad
2
2
1.0708
MB
tB =
T=
T = 0.17042T
2π
2π
Problem 3.12 From Example 3.3 we have
e = 0.24649
T = 8679.1 s
θ c = 143.36
Thus
tan
1 −e
1 − 0.24649
143.36&deg;
Ec
θ
=
tan c =
tan
2
1+e
2
1 + 0.24649
2
Mc = Ec − e sin Ec = 2.3364 − 0.24649 ⋅ sin 2.3364 = 2.1587 rad
2.1587
M
⋅ 8679.1 = 2981.8 s
tc = c T =
2π
2π
Problem 3.13
rSOI = 925 000 km
rperigee (1 + e )
r=
1 + e cos θ
(6378 + 500)(1 + 1)
925 000 =
⇒ θ = 170.11&deg;
1 + 1 ⋅ cos θ
θ 1
θ 1
1
170.11&deg; 1
170.11&deg;
M p = tan + tan 3 = tan
+ tan 3
= 262.82
2
2 6
2 2
2
6
2
h2
rperigee =
⇒ h = 2&micro;rperigee = 2 ⋅ 398 600 ⋅ 6878 = 74 048 km 2 s
2&micro;
29
Chapter 3
Solutions Manual
t=
Orbital Mechanics for Engineering Students
Chapter 3
74 048 3
h3
Mp =
⋅ 262.82 = 671 630 s = 7d 18h 34m
398 600
&micro;
Problem 3.14
(a)
h = &micro;rperigee (1 + e ) = 398 600 ⋅ 7500 ⋅ (1 + 1) = 77 324 km 2 s
1
90&deg; 1
90&deg;
Mp
= tan
+ tan 3
= 0.66667
θ = 90&deg;
2
2
6
2
77 324 3
h3
0.66667 = 1939.9 s
tθ = 90&deg; = 2 M p
=
θ = 90&deg; 389 600 2
&micro;
t −90&deg; to +90&deg; = 2 ⋅ 1939.9 = 3879.8 s = 1.0777 h
)
)
(b)
Mp =
&micro;2 t
=
h3
398 600 3 ⋅ (24 ⋅ 3600)
77 324 3
= 29.692
1/3
−1/3
θ 



2
2
= 3 M p + (3 M p ) + 1  − 3 M p + (3 M p ) + 1 




2
1/3
θ
2
2
tan = 3 ⋅ 29.692 + (3 ⋅ 29.692) + 1
− 3 ⋅ 29.692 + (3 ⋅ 29.692) + 1
2
θ = 159.2&deg;
tan
[
r=
] [
77 324 2
1
1
h2
=
= 230 200 km
&micro; 1 + cos θ 398 600 1 + cos 159.2&deg;
Problem 3.15
(a)
vperigee = 1.1
2 ⋅ 398 600
2&micro;
= 1.1
= 11.341 km s
rperigee
7500
h = rperigee vperigee = 7500 ⋅ 11.341 = 85056 km 2 s
rperigee =
7500 =
h2 1
&micro; 1+e
85 056 2 1
⇒ e = 1.4200
398 600 1 + e
90&deg;
1.42 − 1
90&deg;
F90&deg;
e −1
=
tan
=
tan
⇒ F90&deg; = 0.887 14
2
2
1.42 + 1
2
e +1
M h )90&deg; = e sinh F90&deg; − F90&deg; = 1.42 ⋅ sinh 0.88714 − 0.88714 = 0.544 46
tanh
M h )90&deg; =
&micro;2
0.544 46 =
t -90&deg; to 90&deg;
3
(e 2 − 1)3/2 t90&deg;
h
398 600 2
3
(1.42 2 − 1)3/2 t90&deg;
85056
= 2t90&deg; = 4115.7 s = 1.1433 h
⇒ t90&deg; = 2057.9 s
(b)
Mh =
&micro;2
2
(e 2 − 1)3/2 t = 398 6003 (1.42 2 − 1)3/2 ⋅ 24 ⋅ 3600 = 22.859
3
h
85056
e sinh F − F = M h
1.42 sinh F − F = 22.859 ⇒ F = 3.6196 (Algorithm 3.2)
30
]
−1/3
= 5.4492
Solutions Manual
Orbital Mechanics for Engineering Students
tan
e +1
F
1.42 + 1
3.6196
θ
=
⇒ θ = 132.55&deg;
tanh =
tanh
2
e −1
2
1.42 − 1
2
r=
85 056 2
h2
1
1
=
= 455 660 km
&micro; 1 + e cos θ 398 600 1 + 1.42 ⋅ cos 132.55&deg;
Chapter 3
Problem 3.16
h = rperigee vperigee = (6378 + 300) ⋅ 11.5 = 76 797 km 2 s
rperigee =
e=
a=
h2 1
&micro; 1+e
h2
&micro;rperigee
−1 =
76 797 2
− 1 = 1.2157 ( hyperbola)
398 600 ⋅ 6878
76 797 2
h2 1
1
=
= 30 964 km
2
&micro; e − 1 398 600 1.2157 2 − 1
At 6 AM:
v2 &micro;
&micro;
− =
2
r 2a
2
398
600
398 600
10
−
=
⇒ r = 9149.9 km
2
r
2 ⋅ 30 964
r=
9149.9 =
1
h2
&micro; 1 + e cos θ
76 797 2
1
⇒ θ = −59.494&deg; ( flying towards earth )
398 600 1 + 1.2157 cos θ
1.2157 − 1
θ
F
e −1
−59.494&deg;
tan =
tan
=
= −0.10384 ⇒ F = −0.360 45
2
2
1.2157 + 1
2
e +1
M h = e sinh F − F = 1.2157 sinh (−0.360 45) − (−0.360 45) = −0.087 287
tanh
3/2
&micro;2
M h = 3 (e 2 − 1) t
h
398 600 2
(1.2157 2 − 1)3/2 t ⇒ t = −753.3 s
−0.087 287 =
3
76 797
( negative means time until perigee)
At 11 AM:
t = 5 ⋅ 3600 − −753.3 = 17 247 s
Mh =
&micro;2
3
( time since perigee at 11 AM)
2
(e 2 − 1)3/2 t2 = 398 6003 (1.2157 2 − 1)3/2 ⋅ 17 247 = 1.9984
h
e sinh F − F = M h
76 797
1.2157 sinh F − F = 1.9984 ⇒ F = 1.8760 (Algorithm 3.2)
tanh
θ
=
2
e +1
F
1.8760
1.2157 + 1
tan =
tan
⇒ θ = 133.96&deg;
e −1
1.2157 − 1
2
2
76 797 2
1
1
h2
=
= 94 771 km
&micro; 1 + e cos θ 398 600 1 + 1.2157 ⋅ cos 133.96&deg;
z = r − 6378 = 88 393 km
r=
31
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 3
Problem 3.17
v ⊥ = v cos γ = 8 ⋅ cos( −65&deg;) = 3.3809 km s
h = rv ⊥ = (37 000 + 6378) ⋅ 3.3809 = 146 660 km 2 s
r=
43 378 =
h2
1
&micro; 1 + e cos θ
146 660 2
1
⇒ e cos θ = 0.24397
398 600 1 + e cos θ
v r = v sin γ = 8 ⋅ sin ( −65&deg;) = −7.2505 km s
&micro;
v r = e sin θ
h
398 600
−7.2505 =
e sin θ ⇒ e sin θ = −2.6677
146 660
e sin θ −2.6677
tan θ =
=
= −10.935 ⇒ θ = −84.775&deg;
e cos θ 0.24397
e sin ( −84.775&deg;) = −2.6677 ⇒ e = 2.6788 ( hyperbola)
rperigee =
146 660 2
1
h2 1
=
= 14 668 km ( no impact )
&micro; 1 + e 398 600 1 + 2.6788
F
e −1
θ
2.6788 − 1
 −84.775&deg; 
=
tan =
tan
⇒ F = −1.4389


e +1
2
2
2.6788 + 1
2
M h = e sinh F − F = 2.6788 sinh ( −1.4389) − ( −1.4389) = −3.8906
tanh
3/2
&micro;2
M h = 3 (e 2 − 1) t
h
398 600 2
(2.6788 2 − 1)3/2 t ⇒ t = −5032.5 s = −1.3979 h
−3.8906 =
3
46 660
1.3979 hours until perigee passage.
Problem 3.18 Write the following MATLAB script to use M-function kepler_U to implement
Algorithm 3.3.
clear
global mu
mu = 398600;
ro
vro
a
dt
=
=
=
=
7200;
1;
10000;
3600;
x
= kepler_U(dt, ro, vro, 1/a);
fprintf('\n\n-----------------------------------------------------\n')
fprintf('\n Initial radial coordinate = %g',ro)
= %g',vro)
fprintf('\n Elapsed time
= %g',dt)
fprintf('\n Semimajor axis
= %g\n',a)
fprintf('\n Universal anomaly
= %g',x)
fprintf('\n\n-----------------------------------------------------\n\n')
Running this program produces the following output in the MATLAB Command Window:
32
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 3
= 1
Elapsed time
= 3600
Semimajor axis
= 10000
Universal anomaly
= 229.341
----------------------------------------------------That is, χ = 229.34 km1/2 . To check this with Equation 3.55, proceed as follows.
h2
1
&micro; 1 + e cos θ
r=
7200 =
h2
1
398 600 1 + e cos θ
∴ cos θ =
3.844 &times; 10 −10 h2 − 1
e
(1)
&micro;
e sin θ
h
398 600
1=
e sin θ
h
1
h
∴ sin θ =
398 600 e
vr =
(2)
sin 2 θ + cos2 θ = 1
Substitute (1) and (2):
1

 398 600

2
2 
h
3.844 &times; 10 −10 h2 − 1 
 =1
+



e
e
(3)
h2 1
&micro; 1 − e2
h2
1
10 000 =
398 600 1 − e 2
a=
∴ h = 3.986 &times; 10 9 (1 − e 2 )
(4)
Substitute (4) into (3):

1

 398 600
2
2

 3.844 &times; 10 −10 3.986 &times; 10 9 (1 − e 2 ) − 1 
3.986 &times; 10 9 (1 − e 2 )
 =1
 +
e
e



[
Expanding and collecting terms yields
1
1993e
2
(3844.5e 4 − 4195.9e 2 + 351.41) = 0
or
3844.5e 4 − 4195.9e 2 + 351.41 = 0
33
]
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 3
The positive roorts of this equation are e = 1.0000 and e = 0.30233 . Obviously, we choose the latter.
e = 0.30233
(5)
Substituting (5) into (4), we get
h = 60180 km 2 s
(6)
Substituting (5) and (6) into (1) or (2) yields
θ1 = 29.959&deg;
(7)
Compute the time at this initial true anomaly as follows:
tan
E1
=
2
1 −e
1 − 0.30233
29.959&deg;
θ
= 0.19584
tan 1 =
tan
1+e
2
1 + 0.30233
2
(8)
M1 = E1 − e sin E1 = 0.38678 − 0.30233 ⋅ sin 0.38678 = 0.27273 rad
M
0.27273
t1 = 1 T =
⋅ 9952.0 = 431.99 s
2π
2π
Obtain E one hour later.
t2 = t1 + 3600 = 4032 s
t
4032
M2 = 2 π 2 = 2 π
T
9952.0
E2 − e sin E2 = M2
E2 − 0.30233 sin E2 = 2.5456
Using Algorithm 3.1,
(9)
According to Equation 3.55,
χ = a (E2 − E1 ) = 10 000 (2.6802 − 0.38678) = 229.34 km1/2
This is the same as the value obtained via Algorithm 3.3
Problem 3.19 Write the following MATLAB script to use the M-function rv_from_r0v0 to
execute Algorithm 3.4.
clear
global mu
mu = 398600;
R0
V0
t
= [20000 -105000 -19000];
= [ 0.9
-3.4
-1.5];
=2*3600;
[R V] = rv_from_r0v0(R0, V0, t);
fprintf('-----------------------------------------------------')
fprintf('\n Initial position vector (km):')
fprintf('\n
r0 = (%g, %g, %g)\n', R0(1), R0(2), R0(3))
fprintf('\n Initial velocity vector (km/s):')
34
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 3
fprintf('\n
v0 = (%g, %g, %g)', V0(1), V0(2), V0(3))
fprintf('\n\n Elapsed time = %g s\n',t)
fprintf('\n Final position vector (km):')
fprintf('\n
r = (%g, %g, %g)\n', R(1), R(2), R(3))
fprintf('\n Final velocity vector (km/s):')
fprintf('\n
v = (%g, %g, %g)', V(1), V(2), V(3))
fprintf('\n-----------------------------------------------------\n')
The output to the MATLAB Command Window is as follows:
----------------------------------------------------Initial position vector (km):
r0 = (20000, -105000, -19000)
Initial velocity vector (km/s):
v0 = (0.9, -3.4, -1.5)
Elapsed time = 7200 s
Final position vector (km):
r = (26337.8, -128752, -29655.9)
Final velocity vector (km/s):
v = (0.862796, -3.2116, -1.46129)
-----------------------------------------------------
35
Solutions Manual
Orbital Mechanics for Engineering Students
36
Chapter 3
Solutions Manual
Orbital Mechanics for Engineering Students
Problem 4.1 Algorithm 4.1 (MATLAB M-function coe_from_sv in Appendix D.8):
(1) r = r = 16 850 km
( 2 ) v = v = 5.7415 km s
( 3 ) vr = r ⋅ v = 0.001 885 6 km s ( &gt; 0 )
r
( 4 ) h = r &times; v = 82 234Iˆ − 23 035Jˆ + 41 876Kˆ ( km 2 s)
( 5 ) h = h = 95 360 km 2 s
( 6 ) i = cos−1  hZ  = cos−1  41 876  = 63.952&deg;
 
h
 96 350 
( 7 ) N = Kˆ &times; h = 24 035Iˆ + 82 234 Jˆ ( km 2 s)
( NY &gt; 0 )
2
( 8 ) N = N = 85 674 km s
( 9 ) Ω = cos−1  N X  = cos−1  24 035  = 73.707&deg;


 85 674 
N
(10 ) e =
1
 5.7415 2 − 398 600 16 850 2615Iˆ + 15 881Jˆ + 3980K
ˆ
398 600 
(
)(
(
)
)
ˆ 
− (16 580 ) ( 0.001 855 6 ) −2.767Iˆ − 0.7905 Jˆ + 4.98K

ˆ ( e &gt; 0)
= 0.059 521Iˆ + 0.360 32 Jˆ + 0.08988K
Z
(11) e = e = 0.37602
(12) ω = cos −1
(13) θ = cos −1
 N ⋅e
= 15.43&deg;
 Ne 
 e⋅r
= 0.067 42&deg;
 er 
Problem 4.2 Algorithm 4.1 (MATLAB M-function coe_from_sv in Appendix D.8):
(1) r = r = 12 670 km
( 2 ) v = v = 3.9538 km s
( 3 ) vr = r ⋅ v = −0.7905 km s ( &lt; 0 )
r
( 4 ) h = r &times; v = 49 084Iˆ ( km 2 s)
( 5 ) h = h = 49 084 km 2 s
( 6 ) i = cos−1  hZ  = cos−1 
 
0 
= 90&deg;
49
 084 
h
( 7 ) N = Kˆ &times; h = 49 084 Jˆ ( km 2 s)
( NY &gt; 0 )
( 8 ) N = N = 49 084 km 2 s
( 9 ) Ω = cos−1  N X  = cos−1 


N
(10 ) e =
0 
= 90&deg;
 49 084 
1
 3.9538 2 − 398 600 12 670 12 670K
ˆ
398 600 
(
)(
(
)
)
ˆ 
− (12 670 ) ( −0.7905 ) −3.8744 Jˆ − 0.7905K

ˆ
ˆ
= −0.097342 J − 0.52296K ( eZ &lt; 0 )
(11) e = e = 0.53194
37
Chapter 4
Solutions Manual
Orbital Mechanics for Engineering Students
(12) ω = 360&deg; − cos −1
(13) θ = 360&deg; − cos −1
Chapter 4
 N ⋅e
= 360&deg; − 100.54&deg; = 259.46&deg;
 Ne 
 e⋅r
= 360&deg; − 169.46&deg; = 190.54&deg;
 er 
Problem 4.3 Algorithm 4.1 (MATLAB M-function coe_from_sv in Appendix D.8):
(1) r = 10189 km
( 2 ) v = 5.8805 km s
( 3 ) vr = r ⋅ v = 1.2874 km/s ( &gt; 0 )
r
( 4 ) h = r &times; v = 31 509Iˆ + 11 468Jˆ + 47 888Kˆ ( km 2 s)
( 5 ) h = h = 58 461 km 2 s
( 6 ) i = cos−1  hZ  = cos−1  47 888  = 35&deg;
 
h
 58 461 
( 7 ) N = Kˆ &times; h = −11 468Iˆ + 31 509Jˆ ( km 2 s)
( NY &gt; 0 )
( 8 ) N = N = 33 532 km 2 s
( 9 ) Ω = cos−1  N X  = cos−1  −11 468  = 110&deg;


N
(10 ) e =
 33 532 
1
 5.8805 2 − 398 600 10189 6472.7Iˆ − 7470.8 Jˆ − 2469.8K
ˆ
398 600 
(
)(
(
)
ˆ 
− (10 189 )(1.2874 ) 3.9914Iˆ + 2.7916 Jˆ − 3.2948K

ˆ
ˆ
ˆ
= −0.2051I − 0.006 738 2 J + 0.13657K ( eZ &gt; 0 )
(11) e = e = 0.2465
(12) ω = cos −1
(13) θ = cos −1
 N ⋅e
= 74.996&deg;
 Ne 
 e⋅r
= 130&deg;
 er 
Problem 4.4
v⋅r
= −2.30454 km s (Flying towards perigee)
r
 e⋅r
θ = 360&deg; − cos −1
= 360&deg; − 30&deg; = 330&deg;
 er 
vr =
Problem 4.5
ˆ
r&times;e
−3353.1Iˆ + 6361.8 Jˆ + 2718.1K
=
7687.9
r&times;e
ˆ
ˆ
ˆ
w = −0.43616I + 0.827 51Jˆ + 0.353 55K
ˆ =
w
ˆ ) = cos−1 ( 0.353 55 ) = 69.295&deg;
ˆ ⋅K
i = cos−1 ( w
Problem 4.6
(a)
→
AB = ( 4 − 1) iˆ + ( 6 − 2 ) ˆj + ( 5 − 3 ) kˆ = 3iˆ + 4 ˆj + 2kˆ
→
AC = ( 3 − 1) iˆ + ( 9 − 2 ) ˆj + ( −2 − 3 ) kˆ = 2iˆ + 7 ˆj − 5kˆ
38
)
Solutions Manual
Orbital Mechanics for Engineering Students
→
X ′ = AB = 3iˆ + 4 ˆj + 2kˆ
→
Z′ = X ′ &times; AC = −34 iˆ + 19 ˆj + 13kˆ
Y ′ = Z′ &times; X ′ = −14 iˆ + 107 ˆj − 193kˆ
(b)
X′
iˆ′ =
= 0.557 09iˆ + 0.742 78 ˆj + 0.371 39kˆ
X′
ˆj′ = Y ′ = −0.063 314 iˆ + 0.483 90 ˆj − 0.872 83kˆ
Y′
Z′
kˆ ′ =
= −0.828 04 iˆ + 0.462 73 ˆj + 0.316 60kˆ
Z′
  iˆ′  

  0.5557 09 0.742 78 0.371 39 
ˆ
[Q] =   j′   =  −0.063 314 0.483 90 −0.872 83 

 

 kˆ ′    −0.828 04 0.462 73 0.316 60 
0.557 09 −0.063 314 −0.828 04   2 
 
0.483 90
0.462 73  −1 
 0.371 39 −0.872 83 0.316 60   3 
{v} = [Q ]T {v ′} =  0.742 78
−1.3066 
{v} =  2.3898 
 2.5654 


( v = −1.3066iˆ + 2.3898ˆj + 2.5654k̂ˆ )
Problem 4.7
 0.267 26 0.534 52 0.801 78  −50 
{V}u = [Q ]xu {V}x =  −0.443 76 0.806 84 −0.389 97  100 
 −0.855 36 −0.251 58 0.452 84   75 
100.22 
{V}u = 73.624  ( V = 100.22uˆ + 73.624 vˆ + 51.573wˆ )
51.573 


Problem 4.8
1
0
0 

[R1 ] = 0 cos 40&deg; sin 40&deg; 
0 − sin 40&deg; cos 40&deg;
cos 25&deg; 0 − sin 25&deg;

0
[R2 ] =  0 1

 sin 25&deg; 0 cos 25&deg; 
cos 25&deg; sin 40&deg; sin 25&deg; − cos 40&deg; sin 25&deg;


cos 40&deg;
sin 40&deg;
[Q ] = [R2 ][R1 ] =  0

 sin 25&deg; − sin 40&deg; cos 25&deg; cos 40&deg; cos 25&deg; 
0.906 31 0.27165 −0.32374 

0.766 04
0.64279
[Q ] =  0

0.422 62 −0.582 56 0.694 27 
Problem 4.9 ro = −5102Iˆ − 8228Jˆ − 2106Kˆ ( km ) v o = −4.348Iˆ + 3.478Jˆ − 2.846Kˆ ( km s)
Method 1
39
Chapter 4
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 4
Use Algorithm 3.4 (MATLAB M-function rv_from_r0v0 in Appendix D.7):
(1a )
ro = 9907.6 km
vo = 6.2531 km s
(1b) vr o = −0.044 678 km s
(1c ) α = 103.77 &times; 10−6 km-1
( 2 ) χ = 195 km1/2
( 3 ) f = −0.365 39 g = 1394.4 s-1
( 4 ) r = ( −0.36 539 ) ( −5102Iˆ − 8228Jˆ − 2106Kˆ ) + 1394.4 ( −4.348Iˆ + 3.478Jˆ − 2.846Kˆ )
ˆ ( km )
= −4198.4Iˆ + 7856.1Jˆ − 3199.2K
( 5 ) f = −6.0467 &times; 10−4 s-1 g = −0.429 31
( 6 ) v = ( −6.0467 &times; 10−4 ) ( −5102Iˆ − 8228Jˆ − 2106Kˆ ) + ( −0.429 31) ( −4.348Iˆ + 3.478Jˆ − 2.846Kˆ )
ˆ ( km s )
= 4.9517Iˆ + 3.4821Jˆ + 2.4946K
Method 2
Compute the orbital elements using Algorithm 4.1 (MATLAB M-function coe_from_sv in Appendix
D.8):
(1) r = r = 9907.6 km
( 2 ) v = v = 6.2531 km s
( 3 ) vr = r ⋅ v = −0.044 678 km s ( &lt; 0 )
r
( 4 ) h = r &times; v = 30 738Iˆ − 5367.8Jˆ − 53 520Kˆ ( km 2 s)
( 5 ) h = h = 61 952 km 2 s
( 6 ) i = cos−1  hZ  = cos−1  −53 520  = 149.76&deg;


 
h
61 952
ˆ
ˆ
( 7 ) N = K &times; h = 5367.8I + 30 738Jˆ km 2 s
(
) ( NY &gt; 0 )
( 8 ) N = N = 31 203 km 2 s
( 9 ) Ω = cos−1  N X  = cos−1  5367.8  = 80.094&deg;


N
(10 ) e =
 31 203 
1
 5.7415 2 − 398 600 9907.6 −5102Iˆ − 8228 Jˆ − 2106K
ˆ
398 600 
(
)(
(
)
ˆ 
− ( 9907.6 ) ( −0.044 678 ) −4.348Iˆ + 3.478 Jˆ − 2.846K

ˆ
ˆ
ˆ
= 0.009 638 51I + 0.027193 J + 0.002 808 3K ( e &gt; 0 )
Z
(11) e = e = 0.028 987
(12) ω = cos −1
 N ⋅e
= 11.09&deg;
 Ne 
(13) θ = 360&deg; − cos −1
 e⋅r
= 360&deg; − 166.14&deg; = 193.86&deg;
 er 
3
2π 
h 
T= 2
= 9414.9 s
&micro;  1 − e 2 
Determine the time since perigee passage at true anomaly θ o = 193.86&deg; :
40
)
Solutions Manual
E
tan o =
2
Orbital Mechanics for Engineering Students
Chapter 4
1 − 0.028 987
1 −e
193.86&deg;
θ
⇒ E o = −2.8926 rad
tan o =
tan
1+e
2
1 + 0.028 987
2
M o = E o − e sin E o = −2.8926 − 0.028 987 ⋅ sin ( −2.8926) = −2.8855 rad
M
−2.8855
9414.9 = −4323.7 s ( minus means time until perigee passage)
to = o T =
2π
2π
Update the true anomaly of the spacecraft. t = t o + 50 ⋅ 60 = −1323.7 s .
M = 2π
t
−1323.7
= 2π
9414.9
T
E − e sin E = M
E − 0.028 987 sin E = −0.883 41 ⇒ E = −0.90623 rad (Algorithm 3.1)
tan
1 + 0.028 987
1+e
−0.90623
E
θ
=
⇒ θ = −53.242&deg;
tan =
tan
2
1 −e
2
1 − 0.028 987
2
Algorithm 4.2 (MATLAB M-function sv_from_coe in Appendix D.9):
(1)
2
1
{r } x = h
&micro; 1 + e cos θ
cos( −53.242&deg;) 
cos θ 
2
1



 61 952
sin ( −53.242&deg;) 
sin θ  =
( −53.242&deg;) 
398
600
1
+
0
.
028
987
cos

 0 
0




 5663.9 


=  −7582.8  ( km )
 0



 − sin θ 
(2) {v} x = &micro; e + cos θ  =
(3)
[Q]xX
  5.1548 

− sin ( −53.242&deg;)
398 600 
 

(
)
0.028 987 + cos −53.242&deg;  = 4.0368  (km s)
61 952 
  0 
0

 

h

 0

cos Ω cos ω − sin Ω sin ω cos i

= sin Ω cos ω + cos Ω cos i sin ω

sin i sin ω
− cos Ω sin ω − sin Ω cos i cos ω
− sin Ω sin ω + cos Ω cos i cos ω
sin i cos ω
0.33251 0.80204
0.49616 


= 0.93811 −0.33532 −0.086644 
0.09688 0.49426
−0.8639 
−4198.4 

( 4 ) {r}X = [Q ]xX {r}x =  7856.1  ( km )
or
 −3199.2 


sin Ω sin i 

− cos Ω sin i 

cos i
ˆ ( km )
r = −4198.4Iˆ + 7856.1Jˆ − 3199.2K
 4.9517 
{v}X = [Q ]xX {v}x =  3.4821  ( km s) or v = 4.9517Iˆ + 3.4821Jˆ + 2.4946Kˆ ( km s)
 2.4946 


Problem 4.10 e = 1.5 Ω = 130&deg; i = 35&deg; ω = 115&deg; θ = 0&deg;
rperigee = 6678 km
h = &micro;(1 + e )rperigee = 398 600 ⋅ (1 + 1.5) ⋅ 6678 = 81 576 km 2 s
Algorithm 4.2 (MATLAB M-function sv_from_coe in Appendix D.9):
41
Solutions Manual
2
1
(1) {r}x = h
&micro; 1 + e cos θ
Orbital Mechanics for Engineering Students
Chapter 4
cos ( 0 ) 6678 
cos θ 
2

 
1
 81 576


 sin ( 0 )  =  0  ( km )
 sin θ  =
(
)
398
600
.
cos
+
1
1
5
0
 0   0 
 0 




 
r = 6678pˆ ( km )
 − sin ( 0 )   0 
 − sin θ 
398 600 
 
&micro;



( 2 ) {v}x = e + cos θ  =
1.5 + cos ( 0 ) = 12.216  ( km s )
h
  0 
 81 576 
0

 
 0


v = 12.216qˆ ( km s )
(3)
cos Ω cos ω − sin Ω sin ω cos i − cos Ω sin ω − sin Ω cos i cos ω
[Q]xX = sin Ω cos ω + cos Ω cos i sin ω − sin Ω sin ω + cos Ω cos i cos ω

sin i sin ω
sin i cos ω
 −0.297 06 0.847 76 0.439 39 


=  −0.800 95 −0.47175 0.368 69 
 0.519 84
−0.2424 0.81915 
sin Ω sin i 

− cos Ω sin i 

cos i
−1983.8 
( 4 ) {r}X = [Q ]xX {r}x = −5348.8  ( km )
or
r = −1983.8Iˆ − 5348.8 Jˆ + 3471.5K̂ ( km )
 3471.5 


 10.356 
{v}X = [Q ]xX {v}x = −5.7627  ( km s) or v = 10.356Iˆ − 5.7627 Jˆ − 2.9611Kˆ ( km s)
 −2.9611 


Problem 4.11 h = 81 576 km 2 s
e = 1.5
Ω = 130&deg;
i = 35&deg; ω = 115&deg; .
t = 7200 s
Mh =
&micro;2
2
(e 2 − 1)3/2 t = 398 6003 (1.52 − 1)3/2 ⋅ 7200 = 2.945
3
h
81 576
e sinh F − F = M h
1.5 sinh F − F = 2.945 ⇒ F = 1.886 ( Algorithm 3.2)
tan
θ
=
2
e +1
F
1.5 + 1
1.886
⇒ θ = 117.47&deg;
tanh =
tan
e −1
2
1.5 − 1
2
Algorithm 4.2 (MATLAB M-function sv_from_coe in Appendix D.9):
cos (117.47&deg; ) −25 007 
cos θ 
81 576 2

 
1
1
h2



(1) {r}x =
 siin (117.47&deg; )  =  48 093  ( km )
 sin θ  =
(
)
&micro; 1 + e cos θ 
  0 
 398 600 1 + 1.5 cos 117.47&deg; 
0
 0 

 

r = −25 007pˆ + 48 093qˆ ( km )
 − sin (117.47&deg; )  −4.3352 
 − sin θ 
398 600 
 
&micro;



( 2 ) {v}x = e + cos θ  =
1.5 + cos (117.47&deg; ) =  5.0752  ( km s )
8
h
1
576

  0 

0



 0
 
ˆ
ˆ
v = −4.3352p + 5.0752q ( km s )
42
Solutions Manual
(3)
[Q]xX
Orbital Mechanics for Engineering Students
 −0.297 06 0.847 76 0.439 39 


=  −0.800 95 −0.47175 0.368 69 
 0.519 84
−0.2424 0.81915 
Chapter 4
(Problem 4.10)
 48 200 
ˆ ( km )
( 4 ) {r}X = [Q ]xX {r}x =  −2658  ( km )
or r = 48 200Iˆ − 2658 Jˆ − 24 658K
−24 658 


 5.5903 
{v}X = [Q ]xX {v}x =  1.0781  ( km s) or v = 5.5903Iˆ + 1.0781Jˆ − 3.4838Kˆ ( km s)
−3..4838 


Problem 4.12 ro = 6472.7Iˆ − 7470.8Jˆ − 2469.8Kˆ ( km ) vo = 3.9914Iˆ + 2.7916Jˆ − 3.2948Kˆ ( km s2 )
Method 1
Use Algorithm 3.4 (MATLAB M-function rv_from_r0v0 in Appendix D.7):
(1a )
ro = 10189 km
vo = 5.805 km s
(1b) vr o = 1.2874 km s
(1c ) α = 109.54 &times; 10−6 km-1
( 2 ) χ = 171.31 km1/2
( 3 ) f = −0.093 379
g = 1870.6 s-1
( 4 ) r = ( −0.093 379 ) ( 6472.7Iˆ − 7470.8Jˆ − 2469.8Kˆ ) + 1870.6 ( 3.9914Iˆ + 2.7916Jˆ − 3.2948Kˆ )
ˆ ( km )
= 6861.9Iˆ + 5919.6 Jˆ − 5932.7K
( 5 ) f = −5.3316 &times; 10−4 s-1 g = −0.028 475
( 6 ) v = ( −5.3316 &times; 10−4 ) ( 6472.7Iˆ − 7470.8Jˆ − 2469.8Kˆ ) + ( −0.028 475 ) ( 3.9914Iˆ + 2.7916Jˆ − 3.2948Kˆ )
ˆ ( km s )
= −3.5647Iˆ + 3.9037 Jˆ + 1.4106K
Method 2
From Problem 4.3 h = 58 461 km 2 s , e = 0.2465, i = 35&deg;, Ω = 110&deg;, ω = 74.996&deg;, θ = 130&deg; .
3
3
T=
tan
58 461 
2π 
2π
h 

=
= 8680.3 s
2 
2


398 600  1 − 0.24652 
&micro;  1 − e2 
1 −e
1 − 0.2465
130&deg;
Eo
θ
=
tan o =
tan
⇒ E o = 2.0612 rad
2
1+e
2
1 + 0.2465
2
M o = E o − e sin E o = 2.0612 − 0.2465 ⋅ sin (2.0612) = 1.8437 rad
M
1.8437
to = o T =
8680.3 = 2547.1 s ( minus means time until perigee passage)
2π
2π
Update the true anomaly: t = t o + 50 ⋅ 60 = 5547.1 s
43
Solutions Manual
Orbital Mechanics for Engineering Students
M = 2π
Chapter 4
t
5547.1
= 2π
T
8680.3
E − e sin E = M
E − 0.2465 sin E = 4.0153 ⇒ E = 3.8541 rad (Algorithm 3.1)
tan
3.8541
θ
1+e
E
1 + 0.2465
tan =
tan
=
⇒ θ = −147.73&deg;
2
1 −e
2
1 − 0.2465
2
Algorithm 4.2 (MATLAB M-function sv_from_coe in Appendix D.9):
(1)
2
1
{r } x = h
&micro; 1 + e cos θ
cos( −147.73&deg;) 
cos θ 
2
1



 58 461
sin ( −147.73&deg;) 
sin θ  =
( −147.73&deg;) 
398
600
1
+
0
.
2465
cos

 0 
0




 −9158.1 


=  −5783.8  ( km )
 0



(2)
(3)
  3.6408 

− sin ( −147.73&deg;)
 − sin θ 
398 600 
&micro;
 


{v} x = e + cos θ  =
0.2465 + cos( −147.73&deg;)  =  −4.0841 (km s)
h
8 461 
  0 

0
 0


 

cos Ω cos ω − sin Ω sin ω cos i − cos Ω sin ω − sin Ω cos i cos ω sin Ω sin i 
[Q]xX = sin Ω cos ω + cos Ω cos i sin ω − sin Ω sin ω + cos Ω cos i cos ω − cos Ω sin i 


sin i sin ω
sin i cos ω
cos i
 −0.8321 0.13078 0.538 99 


=  −0.026 991 −0.9802 0.19617 
 0.553 97 0.148 69 0.81915 
 6864 
( 4 ) {r}X = [Q ]xX {r}x =  5916.5  ( km )
or
−5933.3 


ˆ ( km )
r = 6864Iˆ + 5916.5 Jˆ − 5933.3K
−3.5636 
{v}X = [Q ]xX {v}x =  3.905  ( km s) or v = −3.5636Iˆ + 3.905Jˆ + 1.4096Kˆ ( km s)
 1.40996 


Problem 4.13 e = 1.2 Ω = 75&deg; i = 50&deg; ω = 80&deg; θ = 0&deg;
rperigee = 6578 km
h = &micro;(1 + e )rperigee = 398 600 ⋅ (1 + 1.2) ⋅ 6578 = 79 950 km 2 s
Algorithm 4.2 (MATLAB M-function sv_from_coe in Appendix D.9):
cos ( 0 ) 6578 
cos θ 
79 950 2

 
1
1
h2



(1) {r}x =
 sin ( 0 )  =  0  ( km )
 sin θ  =
(
)
398
600
&micro; 1 + e cos θ 
.
cos
1
1
2
0
+
 0   0 


 

 0 
r = 6578pˆ ( km )
 − sin ( 0 )   0 
 − sin θ 
398 600 
 
&micro;



( 2 ) {v}x = e + cos θ  =
1.2 + cos ( 0 ) = 11.546  ( km s )
h
9
950
7
  0 


0
 

 0


v = 11.546qˆ ( km s )
44
Solutions Manual
(3)
Orbital Mechanics for Engineering Students
cos Ω cos ω − sin Ω sin ω cos i
[Q]xX = sin Ω cos ω + cos Ω cos i sin ω

sin i sin ω
− cos Ω sin ω − sin Ω cos i cos ω
− sin Ω sin ω + cos Ω cos i cos ω
sin i cos ω
Chapter 4
sin Ω sin i 

− cos Ω sin i 

cos i
 −0.56 561 −0.3627
0.739 94 


=  0.331 57 −0.92236 −0.198 27 
 0.754 41 0.133 02
0.64279 
−3726.5 
( 4 ) {r}X = [Q ]xX {r}x =  2181.1  ( km )
 4962.5 


−4.1878 
{v}X = [Q ]xX {v}x =  −10.65 
 1.5359 


Problem 4.14 h = 75950 km 2 s
or
ˆ ( km s )
or v = −4.1878Iˆ + 10.65 Jˆ + 1.5359K
( km s)
e = 1.2
ˆ ( km )
r = −3726.5Iˆ + 2181.1Jˆ + 49962.5K
Ω = 130&deg;
i = 50&deg; ω = 80&deg;
t = 7200 s
Mh =
2
&micro;2
(e 2 − 1)3/2 t = 398 6003 (1.2 2 − 1)3/2 ⋅ 7200 = 0.762 09
3
h
e sinh F − F = M h
75950
1.2 sinh F − F = 0.762 09 ⇒ F = 1.3174 ( Algorithm 3.2)
tan
θ
=
2
0.762 09
e +1
F
1.2 + 1
tanh =
tan
86&deg;
⇒ θ = 124.86
e −1
2
1.2 − 1
2
Algorithm 4.2 (MATLAB M-function sv_from_coe in Appendix D.9):
cos (124.86&deg; ) −26 336 
cos θ 
 

h2
1
1
75950 2



(1) {r}x =
 siin (124.86&deg; )  =  37 806  ( km )
 sin θ  =
&micro; 1 + e cos θ 
  0 
 398 600 1 + 1.2 cos (124.86&deg; ) 
0
 


 0 
r = −26 336pˆ + 37 806qˆ ( km )
 − sin (124.86&deg; )  −4.3064 
 − sin θ 
398 600 
 
&micro;



( 2 ) {v}x = e + cos θ  =
1.5 + cos (124.86&deg; ) =  3.298  ( km s )
h
1
576
8
  0 


0


 0
 

ˆ
ˆ
v = −4.3064p + 3.298q ( km s )
(3)
[Q]xX
 −0.56 561 −0.3627
0.739 94 


=  0.331 57 −0.92236 −0.198 27 
 0.754 41 0.133 02
0.64279 
 1207.2 
( 4 ) {r}X = [Q ]xX {r}x =  −43 603  ( km )
−14 839 


 1.2434 
{v}X = [Q ]xX {v}x = −4.4698 
−2.8100 


or
( km s)
45
(Problem 4.13)
ˆ ( km )
r = 1207.2Iˆ − 43 603 Jˆ − 14 8399K
ˆ ( km s )
or v = 1.2434Iˆ − 4.4698 Jˆ − 2.8100K
Solutions Manual
Orbital Mechanics for Engineering Students
Problem 4.15 h = 75000 km 2 s
e = 0.7
Chapter 4
θ = 25&deg;
 − sin (25&deg;)   −2.2461
398 600 
 

0.7 + cos(25&deg;)  =  8.537  (km s)
h
  0 
 75000 
0
 0


 

 −0.83204 0.02741 0.55403   −2.2461  2.1028 
 



= [Q ]xX {v} x =  -0.13114 -0.98019 -0.14845  8.537  =  −8.0733  (km s)
 0.53899 -0.19617 0.81915   0   −2.8853 
 − sin θ 
{v} x = &micro; e + cos θ  =
{v} X
or
ˆ ( km s )
v = 2.1028Iˆ − 8.0733 Jˆ − 2.8853K
Problem 4.16
Ω = 60&deg; ω = 0
B
Z
i = 90&deg;
perigee
apogee
 −3.208 
{v} x =  −0.8288  (km s)
 0



Y
60&deg;
X
cos Ω cos ω − sin Ω sin ω cos i − cos Ω sin ω − sin Ω cos i cos ω
[Q]xX = sin Ω cos ω + cos Ω cos i sin ω − sin Ω sin ω + cos Ω cos i cos ω

sin i sin ω
sin i cos ω
 0.5
0 0.86603 


= 0.86603 0
−0.5 
 0
1
0 
 −1.604 
{v}X = Q  {v}x = −2.7782  ( km s)
xX
−0.8288 


or
ˆ ( km s )
v = −1.604Iˆ − 2.7782 Jˆ − 0.8288K
Problem 4.17 a = 7016 km
( 2)
{r } x = a 1 − e
1 + e cos θ
e = 0.05
i = 45&deg;
Ω =0
ω = 20&deg;
θ = 10&deg;
cos θ 
2 cos 10&deg;  6568.7 

 7016(1 − 05 ) 
 

sin θ  =
sin 10&deg;  = 1158.2  ( km )
+
.
cos
&deg;
1
0
05
10
 0 
 0   0 



 

46
sin Ω sin i 

− cos Ω sin i 

cos i
Solutions Manual
Q  xX
Orbital Mechanics for Engineering Students
cos Ω cos ω − sin Ω sin ω cos i − cos Ω sin ω − sin Ω cos i cos ω

= sin Ω cos ω + cos Ω cos i sin ω − sin Ω sin ω + cos Ω cos i cos ω

sin i sin ω
sin i cos ω
0
 0.93969 -0.34202



= 0.24184 0.66446 -0.70711 
0.24184 0.66446 0.70711 
Chapter 4
sin Ω sin i 

− cos Ω sin i 

cos i
5776.4 
{r}X = Q  {r}x =  2358.2  ( km )
xX
 2358.2 


or
ˆ ( km )
r = 5776.4Iˆ + 2358.2 Jˆ + 2358.2K
Problem 4.18
Ω=−
2
3 &micro; J 2 Rearth
cos i
2
2(
1 − e 2 ) a7/2
1
1
r
+r
= ( 6878 + 7378 ) = 7128 km
2 perigee apogee
2
rapogee − rperigee 7378 − 6878
= 0.035 073
e=
=
rapogee + rperigee 7378 + 68788
(
a=
Ω=
2π
= 1.991 &times; 10−7 raad s
365.26 ⋅ 24 ⋅ 3600
1.991 &times; 10−7 = −
1.991 &times; 10
−7
)
3
2
398 600 ⋅ 0.001082 6 ⋅ 6378 2
(1 − 0.035 073 )
2
= −1.367 &times; 10
−6
2
7128
cos i
7/2
cos i ⇒ i = 98.372&deg;
Problem 4.19
T=
Ω =-
ω earth
2π
&micro;
3
2
r 3/2 =
2π
398 600
&micro; J 2 Rearth 2
( 6378 + 180 )3/2 = 5285.3 s
cos i = -
3
2
398 600 ⋅ 0.001082 6 ⋅ 6378 2
r 7/2
( 6378 + 180 )7/2
2π
2π +
.26 = 7.2921 &times; 10−5 rad s
365
=
24 ⋅ 3600
cos 30&deg; = −1.5814 &times; 10−6 rad s
Change in east longitude of the ascending node after 1 orbit of the satellite:
∆λ = ω earth − Ω T = 7.2921 &times; 10−5 − ( −1.5814 &times; 10−6 )  ⋅ 5285.3 = 0.393 77 rad
(
)
Spacing s,
s = Rearth ∆λ = 6378 ⋅ 0.393 77 = 2511.4 km
Problem 4.20
The change in east longitude λ of the ascending node of a satellite after n s orbits is
47
Solutions Manual
Orbital Mechanics for Engineering Students
(
Chapter 4
)
∆λ = ω earth − Ω nsT
If ∆λ is an integral multiple n e of earth rotations ( 2π ) then the ground track will close on itself,
(
)
2π ne = ω earth − Ω nsT
(
)
Let ν = n s n e . Then 2π = ω earth − Ω ν T , or
T=
1
2π
ν ω earth − Ω
But T = 2πr 3/2
(1)
&micro; , where r is the radius of the orbit. Thus
2π 3/2 1
2π
r
=
ν ω earth − Ω
&micro;
or

&micro;
r=
 ν ω earth − Ω
(
)



2/3
(2)
For a circular orbit
Ω= −
3
2
&micro; J 2 Rearth 2
r 7/2
cos i
or
cos i = −
2
3
r 7/2
&micro; J 2 Rearth 2
Ω
Substituting (2) we get


&micro;


2  ν ω earth − Ω 
cos i = −
3
&micro; J 2 Rearth 2
(
7/3
)
(3)
Ω
Substituting
ω earth = 7.2921 &times; 10−5 rad
J 2 = 0.001082 6
Ω = 1.997 &times; 10−7 rad s
Rearth = 6378 km
&micro; =398 600 km 3 s2
into (1), (2) and (3) we get
 73.948 
i = cos −1  − 7/3 
 ν

T=
86 400
ν
z=
42 241
ν 2/3
From these we obtain the following table of scenarios:
48
− 6378
Solutions Manual
ν
17
16
15
14
13
12
11
10
9
8
Orbital Mechanics for Engineering Students
z ( km ) i (deg)
T (h)
11.042
95.711 1.4118
274.55
567.03
96.583
97.658
1.5000
1.6000
893.93
99.006
1.7143
1262.2
100.72
2.000
1681.0
102.96
105.95
2.1818
2162.6
2722.6
110.07
2.4000
2.6667
3384.8
116.03
3.0000
4182.3
125.29
3.4286
Problem 4.21
5

ω = − f  sin 2 i − 2
2

ω
7
f =−
=−
= 7.2384 deg day
5
5
2
2
sin i − 2
sin 40&deg; − 2
2
2
Ω = − f cos i = −7.2384 ⋅ cos 40&deg; = −5.545 deg day
Problem 4.22 From
ˆ ( km )
ro = −2429.1Iˆ + 4555.1Jˆ + 4577.0K
ˆ ( km s )
v o = −4.7689Iˆ − 5.6113 Jˆ + 3.0535K
we obtain the orbital elements by means of Algorithm 4.1:
(
h = 55000 km 2 s
)
e = 0.1
Ω o = 70&deg;
i = 50&deg;
ω o = 60&deg;
θo = 0
55 000 2
1
h2 1
=
= 7665.8 km
&micro; 1 − e 2 398 600 1 − 0.12
2π 3/2
2π
7665.8 3/2 = 6679.5 s
T=
a
=
&micro;
398 600
a=
The satellite is at perigee ( θ o = 0 ) so t o = 0 . After 72 hours, t f = t o + 72 ⋅ 3600 = 259 000 s .
t f T = 38.805 , so t f is in orbit 39. The time since perigee in orbit 39 is
t39 = (38.805 − 38)T = 5378.3 s
2π
2π
t =
T 39 6679.5
= M39
∴ M39 =
E39 − e sin E39
E39 − 0.1 sin E39 = 5.0952 ⇒ E39 = 4.9623 rad ( Algorithm 3.1)
tan
1+e
1 + 0.1
4.9623
θ 39
E
=
tan 39 =
tan
⇒ θ 39 = 278.68&deg;
2
1 −e
2
1 − 0.1
2
The perifocal state vector is
49
Chapter 4
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 4
cos(278.68&deg;)   1128.9 
cos θ 
55000 2
h2
1
1

 



{r } x =
sin (278.68&deg;)  =  −7390.5 ( km )
sin θ  =
(
)
&micro; 1 + e cos θ 
  0 
 398 600 1 + 0.1 cos 278.68&deg; 
0
 0 

 

 − sin (278.68&deg;)  7.1642 
 − sin θ 
398 600 
 

{v} x = &micro; e + cos θ  =
0.1 + cos(278.68&deg;)  = 1.8191  (km s)
h
  0 
 55000 
0
 0


 

Update Ω and ω :
Ω=−
2
3 &micro; J 2 Rearth
3
cos i = −
2
2(
2
1 − e 2 ) a7/2
3998 600 ⋅ 0.0010826 ⋅ 6378 2
(1 − 0.12 )
2
7665.87/2
cos 50&deg; = −1.0789 &times; 10−6 cos 50&deg;
= −6.9352 &times; 10−7 rad s = −3.9736 &times; 10−5 deg s
∴ Ω39 = Ω1 + Ω∆t = 70&deg; − 3.9736 &times; 10−5 ⋅ 259 200 = 59.701&deg;
ω=−
2
3 &micro; J 2 Rearth
2
2(
1 − e 2 ) a7/2
5


2
−6  5
2
 sin i − 2 = −1.0789 &times; 10  sin 50&deg; − 2
2
2
= 5.7599 &times; 10−7 rad s=3.2945 &times; 10-5 deg s
∴ ω 39 = ω1 + ω∆t = 60&deg; + 3.2945 &times; 10-5 ⋅ 259 200 = 68.539&deg;
Update the transformation matrix between perifocal and geocentric equatorial coordinates:
cos Ω39 cos ω 39 − sin Ω39 sin ω 39 cos i
[Q]xX = sin Ω39 cos ω 39 + cos Ω39 cos i sin ω 39

sin i sin ω
− cos Ω39 sin ω 39 − sin Ω39 cos i cos ω 39
− sin Ω39 sin ω 39 + cos Ω39 cos i cos ω 39
sin i cos ω 39
sin Ω39 sin i 

− cos Ω39 sin i 

cos i
 −0.33192 −0.672 58 0.66141 


=  0.6177
−0.684 89 −0.386 48 
 0.712 94
0.280 27
0.64278 
Compute the geocentric equatorial state vector at t f :
 4596 
ˆ ( km )
{r}X = [Q ]xX {r}x =  5759  ( km )
or r = 4596Iˆ + 5759 Jˆ − 1266.5K
−1266.5 


−3.6014 
{v}X = [Q ]xX {v}x =  3.1794  ( km s) or v = −3.6014Iˆ + 3.1794 Jˆ + 5.6174K̂ˆ ( km s)
 5.6174 


50
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 5
Problem 5.1
The following MATLAB script uses the given data to compute v 2 by means of Algorithm 5.1, which is
implemented as the M-function gibbs in Appendix D.10. The output to the MATLAB Command
Window is listed afterwards.
% ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
% Problem_5_01
% ~~~~~~~~~~~~
%
% This program uses Algorithm 5.1 (Gibbs method) to obtain the state
% vector from the three coplanar position vectors provided in
% Problem 5.1.
%
% mu
- gravitational parameter (km^3/s^2)
% r1, r2, r3 - three coplanar geocentric position vectors (km)
% ierr
- 0 if r1, r2, r3 are found to be coplanar
%
1 otherwise
% v2
- the velocity corresponding to r2 (km/s)
%
% User M-function required: gibbs
% -------------------------------------------------------------------clear
global mu
mu = 398600;
r1 = [5887 -3520 -1204];
r2 = [5572 -3457 -2376];
r3 = [5088 -3289 -3480];
%...Echo the input data to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.1: Gibbs Method\n')
fprintf('\n Input data:\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n r1 (km) = [%g %g %g]', r1(1), r1(2), r1(3))
fprintf('\n r2 (km) = [%g %g %g]', r2(1), r2(2), r2(3))
fprintf('\n r3 (km) = [%g %g %g]', r3(1), r3(2), r3(3))
fprintf('\n\n');
%...Algorithm 5.1:
[v2, ierr] = gibbs(r1, r2, r3);
%...If the vectors r1, r2, r3, are not coplanar, abort:
if ierr == 1
fprintf('\n These vectors are not coplanar.\n\n')
return
end
%...Output the results to the command window:
fprintf(' Solution:')
fprintf('\n');
fprintf('\n v2 (km/s) = [%g %g %g]', v2(1), v2(2), v2(3))
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.1: Gibbs Method
Input data:
Gravitational parameter (km^3/s^2)
51
= 398600
Solutions Manual
r1 (km) = [5887
r2 (km) = [5572
r3 (km) = [5088
Orbital Mechanics for Engineering Students
-3520
-3457
-3289
Chapter 5
-1204]
-2376]
-3480]
Solution:
v2 (km/s) = [-2.50254 0.723248 -7.13125]
-----------------------------------------------------
ˆ
v 2 = −2.5025Iˆ + 0.72325 Jˆ − 7.1312K
( km/s )
Problem 5.2
The following MATLAB script uses r2 and v 2 from Problem 5.1 to compute the orbital elements by
means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8. The
output to the MATLAB Command Window is listed afterwards.
%
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%
%
%
%
%
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%
%
%
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_02
~~~~~~~~~~~~
This program uses Algorithm 4.1 to obtain the orbital
elements from the state vector obtained in Problem 5.1.
pi
deg
mu
r
v
coe
T
-
3.1415926...
factor for converting between degrees and radians
gravitational parameter (km^3/s^2)
position vector (km) in the geocentric equatorial frame
velocity vector (km/s) in the geocentric equatorial frame
orbital elements [h e RA incl w TA a]
where h
= angular momentum (km^2/s)
e
= eccentricity
RA
= right ascension of the ascending node (rad)
w
TA
a
= semimajor axis (km)
- Period of an elliptic orbit (s)
User M-function required: coe_from_sv
--------------------------------------------------------------------
clear
global mu
deg = pi/180;
mu = 398600;
%...Data declaration for Problem 5.2:
r = [
5572
-3457
-2376];
v = [-2.50254 0.723248 -7.13125];
%...
%...Algorithm 4.1:
coe = coe_from_sv(r,v);
%...Echo the input data and output results to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.2: Orbital elements from state vector\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n State vector:\n')
fprintf('\n r (km)
= [%g %g %g]', ...
52
Solutions Manual
Orbital Mechanics for Engineering Students
r(1), r(2), r(3))
= [%g %g %g]', ...
v(1), v(2), v(3))
fprintf('\n v (km/s)
disp(' ')
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
Chapter 5
Angular momentum (km^2/s)
Eccentricity
Right ascension (deg)
Inclination (deg)
Argument of perigee (deg)
True anomaly (deg)
Semimajor axis (km):
=
=
=
=
=
=
=
%g',
%g',
%g',
%g',
%g',
%g',
%g',
coe(1))
coe(2))
coe(3)/deg)
coe(4)/deg)
coe(5)/deg)
coe(6)/deg)
coe(7))
%...if the orbit is an ellipse, output its period (Equation 2.73):
if coe(2)&lt;1
T = 2*pi/sqrt(mu)*coe(7)^1.5;
fprintf('\n Period:')
fprintf('\n
Seconds
= %g', T)
fprintf('\n
Minutes
= %g', T/60)
fprintf('\n
Hours
= %g', T/3600)
fprintf('\n
Days
= %g', T/24/3600)
end
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.2: Orbital elements from state vector
Gravitational parameter (km^3/s^2) = 398600
State vector:
r (km)
v (km/s)
= [5572 -3457 -2376]
= [-2.50254 0.723248 -7.13125]
Angular momentum (km^2/s) = 52948.9
Eccentricity
= 0.0127382
Right ascension (deg)
= 150.003
Inclination (deg)
= 95.0071
Argument of perigee (deg) = 151.688
True anomaly (deg)
= 48.3093
Semimajor axis (km):
= 7034.71
Period:
Seconds
= 5871.93
Minutes
= 97.8655
Hours
= 1.63109
Days
= 0.0679622
-----------------------------------------------------
52 949 2
h2 1
1
=
= 6945.1 km
&micro; 1 + e 398 600 1 + 0.012 738
= 6945.1 − 6378 = 567.11 km
rperigee =
zperigee
Problem 5.3
(
)
As in Example 5.3, we set r1 = r1 iˆ and r2 = r2 cos ∆θ iˆ + sin ∆θ ˆj , where r1 = 6978 km , r2 = 6678 km
and ∆θ = 60&deg; . The following MATLAB script uses this data to compute v1 and v 2 by means of
Algorithm 5.2, which is implemented as the M-function lambert in Appendix D.11. The output to the
MATLAB Command Window is listed afterwards.
% ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
% Problem 5_03a
53
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 5
% ~~~~~~~~~~~~~
%
% This program uses Algorithm 5.2 to solve Lambert's problem for the
% data provided in Problem 5.3.
%
%
%
%
%
%
%
%
%
%
deg
pi
mu
r1, r2
dt
dtheta
R1, R2
string
-
factor for converting between degrees and radians
3.1415926...
gravitational parameter (km^3/s^2)
time between r1 and r2 (s)
change in true anomaly during dt (degrees)
initial and final position vectors (km)
= 'pro' if the orbit is prograde
= 'retro if the orbit is retrograde
V1, V2 - initial and final velocity vectors (km/s)
% User M-function required: lambert
% ----------------------------------------------------------clear
global mu
mu
= 398600;
%km^3/s^2
deg
= pi/180;
r1
r2
dt
dtheta
=
=
=
=
6378 + 600;
6378 + 300;
15*60;
60;
%km
%km
%sec
%degrees
R1 = [r1 0 0];
R2 = [r2*cos(dtheta*deg)
r2*sin(dtheta*deg)
0];
%...Algorithm 5.2:
string = 'pro';
[V1 V2] = lambert(R1, R2, dt, string);
%...Echo the input data and output results to the command window:
fprintf('\n-----------------------------------------------------')
fprintf('\n Problem 5.3: Lambert''s Problem\n')
fprintf('\n Input data:\n');
fprintf('\n
Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n
= %g', r1)
fprintf('\n
Position vector R1 (km)
= [%g %g %g]\n',...
R1(1), R1(2), R1(3))
fprintf('\n
= %g', r2)
fprintf('\n
Position vector R2 (km)
= [%g %g %g]\n',...
R2(1), R2(2), R2(3))
fprintf('\n
Elapsed time (s)
= %g', dt)
fprintf('\n
Change in true anomaly (deg) = %g', dtheta)
fprintf('\n\n Solution:\n')
fprintf('\n Velocity vector V1 (km/s) = [%g %g %g]',...
V1(1), V1(2), V1(3))
fprintf('\n Velocity vector V2 (km/s) = [%g %g %g]',...
V2(1), V2(2), V2(3))
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.3: Lambert's Problem
Input data:
54
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 5
Gravitational parameter (km^3/s^2) = 398600
Position vector R1 (km)
= 6978
= [6978
0
Position vector R2 (km)
= 6678
= [3339
5783.32
0]
0]
Elapsed time (s)
= 900
Change in true anomaly (deg) = 60
Solution:
Velocity vector V1 (km/s) = [-0.544135 7.68498 0]
Velocity vector V2 (km/s) = [-6.98129 3.96849 0]
----------------------------------------------------To find the perigee altitude, we need the orbital elements. The following MATLAB script uses
r = 6978iˆ ( km ) and v = −0.544135iˆ + 7.68498 ˆj ( km/s ) from above to compute the orbital elements
1
1
by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8.
The output to the MATLAB Command Window is listed afterwards.
%
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%
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_03b
~~~~~~~~~~~~
This program employs Algorithm 4.1 to obtain the orbital
elements from the state vector found from the solution
of Lambert's problem using the data given in Problem 5.3.
pi
deg
mu
r
v
coe
T
-
3.1415926...
factor for converting between degrees and radians
gravitational parameter (km^3/s^2)
position vector (km) in the geocentric equatorial frame
velocity vector (km/s) in the geocentric equatorial frame
orbital elements [h e RA incl w TA a]
where h
= angular momentum (km^2/s)
e
= eccentricity
RA
= right ascension of the ascending node (rad)
w
TA
a
= semimajor axis (km)
- Period of an elliptic orbit (s)
User M-function required: coe_from_sv
--------------------------------------------------------------------
clear
global mu
deg = pi/180;
mu = 398600;
%...Data declaration for Problem 5.3:
r = [6978 0 0];
v = [-0.544135 7.68498 0];
%...
%...Algorithm 4.1:
coe = coe_from_sv(r,v);
%...Echo the input data and output results to the command window:
55
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 5
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.3: Orbital elements from state vector\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n State vector:\n')
fprintf('\n r (km)
= [%g %g %g]', ...
r(1), r(2), r(3))
fprintf('\n v (km/s)
= [%g %g %g]', ...
v(1), v(2), v(3))
disp(' ')
fprintf('\n Angular momentum (km^2/s) = %g', coe(1))
fprintf('\n Eccentricity
= %g', coe(2))
fprintf('\n Right ascension (deg)
= %g', coe(3)/deg)
fprintf('\n Inclination (deg)
= %g', coe(4)/deg)
fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg)
fprintf('\n True anomaly (deg)
= %g', coe(6)/deg)
fprintf('\n Semimajor axis (km):
= %g', coe(7))
%...if the orbit is an ellipse, output its period (Equation 2.73):
if coe(2)&lt;1
T = 2*pi/sqrt(mu)*coe(7)^1.5;
fprintf('\n Period:')
fprintf('\n
Seconds
= %g', T)
fprintf('\n
Minutes
= %g', T/60)
fprintf('\n
Hours
= %g', T/3600)
fprintf('\n
Days
= %g', T/24/3600)
end
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.3: Orbital elements from state vector
Gravitational parameter (km^3/s^2) = 398600
State vector:
r (km)
v (km/s)
= [6978 0 0]
= [-0.544135 7.68498
0]
Angular momentum (km^2/s) = 53625.8
Eccentricity
= 0.0806743
Right ascension (deg)
= 0
Inclination (deg)
= 0
Argument of perigee (deg) = 0
True anomaly (deg)
= 294.849
Semimajor axis (km):
= 7261.83
Period:
Seconds
= 6158.57
Minutes
= 102.643
Hours
= 1.71071
Days
= 0.0712798
-----------------------------------------------------
53 626 2
h2 1
1
=
= 6676 km
&micro; 1 + e 398 600 1 + 0.0806743
= 6676 − 6378 = 298 km
rperigee =
zperigee
Problem 5.4 The following MATLAB script uses r1 , r2 and ∆t to compute v1 and v 2 by means of
Algorithm 5.2, which is implemented as the M-function lambert in Appendix D.11. The output to the
MATLAB Command Window is listed afterwards.
56
Solutions Manual
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
Orbital Mechanics for Engineering Students
Chapter 5
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_04
~~~~~~~~~~~~
This program uses Algorithm 5.2 to solve Lambert's problem for the
data provided in Problem 5.4.
mu
r1, r2
dt
string
-
gravitational parameter (km^3/s^2)
initial and final position vectors (km)
time between r1 and r2 (s)
= 'pro' if the orbit is prograde
= 'retro if the orbit is retrograde
v1, v2 - initial and final velocity vectors (km/s)
coe
- orbital elements [h e RA incl w TA a]
User M-function required: lambert
-----------------------------------------------------------
clear
global mu
deg = pi/180;
%...Data
mu
=
r1
=
r2
=
dt
=
string =
%...
declaration for Problem 5.4:
398600;
[-3600
3600
5100];
[-5500 -6240
-520];
30*60;
'pro';
%...Algorithm 5.2:
[v1, v2] = lambert(r1, r2, dt, string);
%...Echo the input data and output the results to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.4: Lambert''s Problem\n')
fprintf('\n Input data:\n');
fprintf('\n
Gravitational parameter (km^3/s^2) = %g\n', mu);
fprintf('\n
r1 (km)
= [%g %g %g]', ...
r1(1), r1(2), r1(3))
fprintf('\n
r2 (km)
= [%g %g %g]', ...
r2(1), r2(2), r2(3))
fprintf('\n
Elapsed time (s) = %g', dt);
fprintf('\n\n Solution:\n')
fprintf('\n
v1 (km/s)
= [%g %g %g]', ...
v1(1), v1(2), v1(3))
fprintf('\n
v2 (km/s)
= [%g %g %g]', ...
v2(1), v2(2), v2(3))
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.4: Lambert's Problem
Input data:
Gravitational parameter (km^3/s^2) = 398600
r1 (km)
= [-3600
r2 (km)
= [-5500
Elapsed time (s) = 1800
3600 5100]
-6240 -520]
57
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 5
Solution:
v1 (km/s)
= [-5.68521 -5.19833 0.348733]
v2 (km/s)
= [3.42204 -3.24131 -4.71994]
-----------------------------------------------------
ˆ ( km/s ) v = 3.4220Iˆ − 3.2413 Jˆ − 4.7199K
ˆ ( km/s )
v1 = −5.6852Iˆ − 5.1983 Jˆ + 0.348 73K
2
Problem 5.5 The following MATLAB script uses r1 = −3600Iˆ + 3600Jˆ + 5100Kˆ ( km ) and
ˆ ( km/s ) from Problem 5.4 to compute the orbital elements by
v1 = −5.6852Iˆ − 5.1983 Jˆ + 0.348 73K
means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8. The
output to the MATLAB Command Window is listed afterwards.
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_05
~~~~~~~~~~~~
This program employs Algorithm 4.1 to obtain the orbital
elements from the state vector found in Problem 5.4.
pi
deg
mu
r
v
coe
T
-
3.1415926...
factor for converting between degrees and radians
gravitational parameter (km^3/s^2)
position vector (km) in the geocentric equatorial frame
velocity vector (km/s) in the geocentric equatorial frame
orbital elements [h e RA incl w TA a]
where h
= angular momentum (km^2/s)
e
= eccentricity
RA
= right ascension of the ascending node (rad)
w
TA
a
= semimajor axis (km)
- Period of an elliptic orbit (s)
User M-function required: coe_from_sv
--------------------------------------------------------------------
clear
global mu
deg = pi/180;
mu = 398600;
%...Data declaration for Problem 5.5:
r = [
-3600
3600
5100];
v = [-5.68521 -5.19833 0.348733];
%...
%...Algorithm 4.1:
coe = coe_from_sv(r,v);
%...Echo the input data and output results to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.5: Orbital elements from state vector\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n State vector:\n')
fprintf('\n r (km)
= [%g %g %g]', ...
r(1), r(2), r(3))
fprintf('\n v (km/s)
= [%g %g %g]', ...
58
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 5
v(1), v(2), v(3))
disp(' ')
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
Angular momentum (km^2/s)
Eccentricity
Right ascension (deg)
Inclination (deg)
Argument of perigee (deg)
True anomaly (deg)
Semimajor axis (km):
=
=
=
=
=
=
=
%g',
%g',
%g',
%g',
%g',
%g',
%g',
coe(1))
coe(2))
coe(3)/deg)
coe(4)/deg)
coe(5)/deg)
coe(6)/deg)
coe(7))
%...if the orbit is an ellipse, output its period (Equation 2.73):
if coe(2)&lt;1
T = 2*pi/sqrt(mu)*coe(7)^1.5;
fprintf('\n Period:')
fprintf('\n
Seconds
= %g', T)
fprintf('\n
Minutes
= %g', T/60)
fprintf('\n
Hours
= %g', T/3600)
fprintf('\n
Days
= %g', T/24/3600)
end
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.5: Orbital elements from state vector
Gravitational parameter (km^3/s^2) = 398600
State vector:
r (km)
v (km/s)
= [-3600 3600 5100]
= [-5.68521 -5.19833
0.348733]
Angular momentum (km^2/s) = 55458
Eccentricity
= 0.0982445
Right ascension (deg)
= 45.0287
Inclination (deg)
= 45.0497
Argument of perigee (deg) = 46.034
True anomaly (deg)
= 43.9458
Semimajor axis (km):
= 7791.19
Period:
Seconds
= 6844.1
Minutes
= 114.068
Hours
= 1.90114
Days
= 0.0792142
-----------------------------------------------------
55 458 2
h2 1
1
=
= 7025.7 km
&micro; 1 + e 398 600 1 + 0.098 244
= 7025.7 − 6378 = 647.74 km
rperigee =
zperigee
Problem 5.6 The following MATLAB script uses r1 , r2 and ∆t to compute v1 and v 2 by means of
Algorithm 5.2, which is implemented as the M-function lambert in Appendix D.11. The output to the
MATLAB Command Window is listed afterwards.
(a)
% ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
% Problem_5_06
% ~~~~~~~~~~~~
%
% This program uses Algorithm 5.2 to solve Lambert's problem for the
% data provided in Problem 5.6.
%
59
Solutions Manual
%
%
%
%
%
%
%
%
%
%
mu
r1, r2
dt
string
Orbital Mechanics for Engineering Students
Chapter 5
-
gravitational parameter (km^3/s^2)
initial and final position vectors (km)
time between r1 and r2 (s)
= 'pro' if the orbit is prograde
= 'retro if the orbit is retrograde
v1, v2 - initial and final velocity vectors (km/s)
coe
- orbital elements [h e RA incl w TA a]
User M-function required: lambert
-----------------------------------------------------------
clear
global mu
deg = pi/180;
%...Data
mu
=
r1
=
r2
=
dt
=
string =
%...
declaration for Problem 5.6:
398600;
[ 5644 -2830 -4170];
[-2240
7320 -4980];
20*60;
'pro';
%...Algorithm 5.2:
[v1, v2] = lambert(r1, r2, dt, string);
%...Echo the input data and output the results to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.6: Lambert''s Problem\n')
fprintf('\n Input data:\n');
fprintf('\n
Gravitational parameter (km^3/s^2) = %g\n', mu);
fprintf('\n
r1 (km)
= [%g %g %g]', ...
r1(1), r1(2), r1(3))
fprintf('\n
r2 (km)
= [%g %g %g]', ...
r2(1), r2(2), r2(3))
fprintf('\n
Elapsed time (s) = %g', dt);
fprintf('\n\n Solution:\n')
fprintf('\n
v1 (km/s)
= [%g %g %g]', ...
v1(1), v1(2), v1(3))
fprintf('\n
v2 (km/s)
= [%g %g %g]', ...
v2(1), v2(2), v2(3))
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.6: Lambert's Problem
Input data:
Gravitational parameter (km^3/s^2) = 398600
r1 (km)
= [5644 -2830
r2 (km)
= [-2240 7320
Elapsed time (s) = 1200
-4170]
-4980]
Solution:
v1 (km/s)
= [-4.13223 9.01237 -4.3781]
v2 (km/s)
= [-7.28524 6.31978 2.5272]
-----------------------------------------------------
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ˆ ( km/s ) v = −7.2852Iˆ + 6.3198 Jˆ + 2.5272K
ˆ ( km/s )
v1 = −4.1322Iˆ + 9.0124 Jˆ − 4.3781K
2
Problem 5.7 The following MATLAB script uses r1 = 5644Iˆ − 2830Jˆ − 4170Kˆ ( km ) and
ˆ ( km/s ) from Problem 5.4 to compute the orbital elements by means
v1 = −4.1322Iˆ + 9.0124 Jˆ − 4.3781K
of Algorithm 4.1, which is implemented as the M-function coe_from_sv in Appendix D.8. The
output to the MATLAB Command Window is listed afterwards.
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_07
~~~~~~~~~~~~
This program employs Algorithm 4.1 to obtain the orbital
elements from the state vector found in Problem 5.6.
pi
deg
mu
r
v
coe
T
-
3.1415926...
factor for converting between degrees and radians
gravitational parameter (km^3/s^2)
position vector (km) in the geocentric equatorial frame
velocity vector (km/s) in the geocentric equatorial frame
orbital elements [h e RA incl w TA a]
where h
= angular momentum (km^2/s)
e
= eccentricity
RA
= right ascension of the ascending node (rad)
w
TA
a
= semimajor axis (km)
- Period of an elliptic orbit (s)
User M-function required: coe_from_sv
--------------------------------------------------------------------
clear
global mu
deg = pi/180;
mu = 398600;
%...Data declaration for Problem 5.7:
r = [
5644
-2830
-4170];
v = [-4.13223 9.01237 -4.3781];
%...
%...Algorithm 4.1:
coe = coe_from_sv(r,v);
%...Echo the input data and output results to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.7: Orbital elements from state vector\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n State vector:\n')
fprintf('\n r (km)
= [%g %g %g]', ...
r(1), r(2), r(3))
fprintf('\n v (km/s)
= [%g %g %g]', ...
v(1), v(2), v(3))
disp(' ')
fprintf('\n Angular momentum (km^2/s) = %g', coe(1))
fprintf('\n Eccentricity
= %g', coe(2))
fprintf('\n Right ascension (deg)
= %g', coe(3)/deg)
fprintf('\n Inclination (deg)
= %g', coe(4)/deg)
fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg)
fprintf('\n True anomaly (deg)
= %g', coe(6)/deg)
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fprintf('\n Semimajor axis (km):
Chapter 5
= %g', coe(7))
%...if the orbit is an ellipse, output its period (Equation 2.73):
if coe(2)&lt;1
T = 2*pi/sqrt(mu)*coe(7)^1.5;
fprintf('\n Period:')
fprintf('\n
Seconds
= %g', T)
fprintf('\n
Minutes
= %g', T/60)
fprintf('\n
Hours
= %g', T/3600)
fprintf('\n
Days
= %g', T/24/3600)
end
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.7: Orbital elements from state vector
Gravitational parameter (km^3/s^2) = 398600
State vector:
r (km)
v (km/s)
= [5644 -2830 -4170]
= [-4.13223 9.01237 -4.3781]
Angular momentum (km^2/s) = 76096.4
Eccentricity
= 1.20053
Right ascension (deg)
= 130.007
Inclination (deg)
= 59.0184
Argument of perigee (deg) = 259.98
True anomaly (deg)
= 320.023
Semimajor axis (km):
= -32922.3
-----------------------------------------------------
76 096 2
h2 1
1
=
= 6601.8 km
&micro; 1 + e 398 600 1 + 1.2005
= 6601.8 − 6378 = 223.82 km
rperigee =
zperigee
Problem 5.8 The following MATLAB script uses the M-function J0 in Appendix D.12 to compute
the Julian day for the date given in part (a) of Problem 5.8. The output to the MATLAB Command
Window is listed afterwards, as are the results for the dates (b) through (c).
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_08a
~~~~~~~~~~~~~
This program computes J0 and the Julian day number using the data
in Problem 5.8.
year
month
day
hour
minute
second
ut
j0
jd
-
range: 1901 - 2099
range: 1 - 12
range: 1 - 31
range: 0 - 23 (Universal Time)
rage: 0 - 60
range: 0 - 60
universal time (hr)
Julian day number at 0 hr UT
Julian day number at specified UT
User M-function required: J0
--------------------------------------------------------------------
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clear
%...Data
year
=
month =
day
=
declaration for Problem 5.8a:
1914;
8;
14;
hour
= 5;
minute = 30;
second = 00;
%...
ut = hour + minute/60 + second/3600;
%...Equation 5.46:
j0 = J0(year, month, day);
%...Equation 5.47:
jd = j0 + ut/24;
%...Echo the input data and output the results to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Example 5.8a: Julian day calculation\n')
fprintf('\n Input data:\n');
fprintf('\n
Year
= %g',
year)
fprintf('\n
Month
= %g',
month)
fprintf('\n
Day
= %g',
day)
fprintf('\n
Hour
= %g',
hour)
fprintf('\n
Minute
= %g',
minute)
fprintf('\n
Second
= %g\n', second)
fprintf('\n Julian day number = %11.3f', jd);
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Example 5.8a: Julian day calculation
Input data:
Year
Month
Day
Hour
Minute
Second
=
=
=
=
=
=
1914
8
14
5
30
0
Julian day number = 2420358.729
----------------------------------------------------Example 5.8b: Julian day calculation
Input data:
Year
Month
Day
Hour
Minute
Second
=
=
=
=
=
=
1946
4
18
14
0
0
Julian day number = 2431929.083
----------------------------------------------------Example 5.8c: Julian day calculation
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Chapter 5
Input data:
Year
Month
Day
Hour
Minute
Second
=
=
=
=
=
=
2010
9
1
0
0
0
Julian day number = 2455440.500
----------------------------------------------------Example 5.8d: Julian day calculation
Input data:
Year
Month
Day
Hour
Minute
Second
=
=
=
=
=
=
2007
10
16
12
0
0
Julian day number = 2454390.000
-----------------------------------------------------
Problem 5.9 This is similar to Example 5.5. The MATLAB script listed in the solution to Problem
5.8 can be used to obtain the Julian day numbers.
Problem 5.10 The following MATLAB script uses Algorithm 5.3, which is implemented in MATLAB
by the M-function LST in Appendix D.13, to compute the local sidereal time for the data given in part
(a) of Problem 5.10. The output to the MATLAB Command Window is listed afterwards, as are the
results for the data in (b) through (e).
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_10a
~~~~~~~~~~~~~
This program uses Algorithm 5.3 to obtain the local sidereal
time from the data provided in Problem 5.10.
lst
EL
- local sidereal time (degrees)
- east longitude of the site (west longitude is negative):
degrees (0 - 360)
minutes (0 - 60)
seconds (0 - 60)
WL
- west longitude
year - range: 1901 - 2099
month - range: 1 - 12
day
- range: 1 - 31
ut
- universal time
hour (0 - 23)
minute (0 - 60)
second (0 - 60)
User M-function required: LST
--------------------------------------------------------------------
clear
%...Data declaration for Problem 5.10a:
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Chapter 5
% East longitude:
degrees = 18;
minutes = 3;
seconds = 0;
% Date:
year
= 2008;
month
= 1;
day
= 1;
% Universal time:
hour
= 12;
minute = 0;
second = 0;
%...
%...Convert negative (west) longitude to east longitude:
if degrees &lt; 0
degrees = degrees + 360;
end
%...Express the longitudes as decimal numbers:
EL = degrees + minutes/60 + seconds/3600;
WL = 360 - EL;
%...Express universal time as a decimal number:
ut = hour + minute/60 + second/3600;
%...Algorithm 5.3:
lst = LST(year, month, day, ut, EL);
%...Echo the input data and output the results to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.10a: Local sidereal time calculation\n')
fprintf('\n Input data:\n');
fprintf('\n
Year
= %g', year)
fprintf('\n
Month
= %g', month)
fprintf('\n
Day
= %g', day)
fprintf('\n
UT (hr)
= %g', ut)
fprintf('\n
West Longitude (deg)
= %g', WL)
fprintf('\n
East Longitude (deg)
= %g', EL)
fprintf('\n\n');
fprintf(' Solution:')
fprintf('\n');
fprintf('\n Local Sidereal Time (deg) = %g', lst)
fprintf('\n Local Sidereal Time (hr) = %g', lst/15)
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.10a: Local sidereal time calculation
Input data:
Year
Month
Day
UT (hr)
West Longitude (deg)
East Longitude (deg)
=
=
=
=
=
=
2008
1
1
12
341.95
18.05
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Solution:
Local Sidereal Time (deg) = 298.572
Local Sidereal Time (hr) = 19.9048
----------------------------------------------------Problem 5.10b: Local sidereal time calculation
Input data:
Year
Month
Day
UT (hr)
West Longitude (deg)
East Longitude (deg)
=
=
=
=
=
=
2007
12
21
10
215.033
144.967
Solution:
Local Sidereal Time (deg) = 24.5646
Local Sidereal Time (hr) = 1.63764
----------------------------------------------------Problem 5.10c: Local sidereal time calculation
Input data:
Year
Month
Day
UT (hr)
West Longitude (deg)
East Longitude (deg)
=
=
=
=
=
=
2005
7
4
20
118.25
241.75
Solution:
Local Sidereal Time (deg) = 104.676
Local Sidereal Time (hr) = 6.9784
----------------------------------------------------Problem 5.10d: Local sidereal time calculation
Input data:
Year
Month
Day
UT (hr)
West Longitude (deg)
East Longitude (deg)
=
=
=
=
=
=
2006
2
15
3
43.1
316.9
Solution:
Local Sidereal Time (deg) = 146.884
Local Sidereal Time (hr) = 9.79228
----------------------------------------------------Problem 5.10e: Local sidereal time calculation
Input data:
Year
Month
Day
UT (hr)
West Longitude (deg)
East Longitude (deg)
=
=
=
=
=
=
2006
3
21
8
228.067
131.933
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Chapter 5
Solution:
Local Sidereal Time (deg) = 70.6348
Local Sidereal Time (hr) = 4.70899
-----------------------------------------------------
Problem 5.11. θ = 117&deg;
φ = 51&deg;
Α = 28&deg;
a = 68&deg; .
From Equation 5.83a,
δ = sin −1 (cos φ cos A cos a + sin φ sin a )= sin −1 (cos 51&deg; cos 28&deg; cos 68&deg; + sin 117&deg; sin 68&deg;)
δ = 68.235&deg;
From Equation 5.83b, since A &lt; 180&deg; ,
h = 360&deg; − cos −1
= 360&deg; − cos −1
 cos φ sin a − sin φ cos A cos a 


cos δ
 cos 51&deg; sin 68&deg; − sin 51&deg; cos 28&deg; cos 68&deg; 


cos 68.235&deg;
= 331.69&deg;
From Equation 5.83c
α = θ − h = 117&deg; − 331.69&deg; = −214.69
Placing this within the range 0 ≤ θ ≤ 360&deg; ,
α = 145.31&deg;
Problem 5.12 The following MATLAB script uses Algorithm 5.4, which is implemented in MATLAB
by the M-function rv_from_observe in Appendix D.14, to compute the state vector of a space
object from the data given in Problem 5.12. The output to the MATLAB Command Window is listed
afterwards.
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_12
~~~~~~~~~~~~
This program uses Algorithm 5.4 to obtain the state
vector from the observational data provided in Problem 5.12.
deg
pi
mu
- conversion factor between degrees and radians
- 3.1415926...
- gravitational parameter (km^3/s^2)
Re
f
wE
omega
-
equatorial radius of the earth (km)
earth's flattening factor
angular velocity of the earth (rad/s)
earth's angular velocity vector (rad/s) in the
geocentric equatorial frame
-
slant range of object (km)
range rate (km/s)
azimuth (deg) of object relative to observation site
time rate of change of azimuth (deg/s)
elevation angle (deg) of object relative to observation
% rho
% rhodot
% A
% a
site
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- time rate of change of elevation angle (degrees/s)
% theta
% phi
% H
- local sidereal time (deg) of tracking site
- geodetic latitude (deg) of site
- elevation of site (km)
% r
% v
- geocentric equatorial position vector of object (km)
- geocentric equatorial velocity vector of object (km)
%
%
%
%
%
%
%
%
%
%
%
%
%
%
coe
Chapter 5
- orbital elements [h e RA incl w TA a]
where
h
= angular momentum (km^2/s)
e
= eccentricity
RA
= right ascension of the ascending node (rad)
incl = inclination of the orbit (rad)
w
TA
a
= semimajor axis (km)
- period of elliptical orbit (s)
rp
T
User M-function required: rv_from_observe
--------------------------------------------------------------------
clear
global
deg
f
Re
wE
mu
f Re wE mu
=
=
=
=
=
%...Data
rho
=
rhodot =
A
=
=
a
=
=
theta =
phi
=
H
=
%...
pi/180;
0.0033528;
6378;
7.2921e-5;
398600;
declaration for Problem 5.12:
988;
4.86;
36;
0.59;
36.6;
-0.263;
40;
35;
0;
%...Algorithm 5.4:
%...Echo the input data and output the solution to
%
the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.12')
fprintf('\n\n Input data:\n');
fprintf('\n Slant range (km)
= %g', rho);
fprintf('\n Slant range rate (km/s)
= %g', rhodot);
fprintf('\n Azimuth (deg)
= %g', A);
fprintf('\n Azimuth rate (deg/s)
fprintf('\n Elevation (deg)
= %g', a);
fprintf('\n Elevation rate (deg/s)
fprintf('\n Local sidereal time (deg)
= %g', theta);
fprintf('\n Latitude (deg)
= %g', phi);
fprintf('\n Altitude above sea level (km) = %g', H);
fprintf('\n\n');
68
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fprintf(' Solution:')
fprintf('\n\n State vector:\n');
fprintf('\n r (km)
= [%g, %g, %g]', ...
r(1), r(2), r(3));
fprintf('\n v (km/s)
= [%g, %g, %g]', ...
v(1), v(2), v(3));
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.12
Input data:
Slant range (km)
Slant range rate (km/s)
Azimuth (deg)
Azimuth rate (deg/s)
Elevation (deg)
Elevation rate (deg/s)
Local sidereal time (deg)
Latitude (deg)
Altitude above sea level (km)
=
=
=
=
=
=
=
=
=
988
4.86
36
0.59
36.6
-0.263
40
35
0
Solution:
State vector:
r (km)
= [3794.66, 3792.71, 4501.31]
v (km/s)
= [-7.72483, 7.72134, 0.0186586]
-----------------------------------------------------
ˆ ( km )
r = 3794.7Iˆ + 3792.7 Jˆ + 4501.3K
ˆ ( km/s )
v = −7.7248Iˆ + 7.72134 Jˆ + 0.018 659K
Problem 5.13 The following MATLAB script uses the state vector found in Problem 5.12 to
compute the orbital elements by means of Algorithm 4.1, which is implemented as the M-function
coe_from_sv in Appendix D.8. The output to the MATLAB Command Window is listed afterwards.
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_13
~~~~~~~~~~~~
This program employs Algorithm 4.1 to obtain the orbital
elements from the state vector found in Problem 5.12.
pi
deg
mu
r
v
coe
T
-
3.1415926...
factor for converting between degrees and radians
gravitational parameter (km^3/s^2)
position vector (km) in the geocentric equatorial frame
velocity vector (km/s) in the geocentric equatorial frame
orbital elements [h e RA incl w TA a]
where h
= angular momentum (km^2/s)
e
= eccentricity
RA
= right ascension of the ascending node (rad)
w
TA
a
= semimajor axis (km)
- Period of an elliptic orbit (s)
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% User M-function required: coe_from_sv
% -------------------------------------------------------------------clear
global mu
deg = pi/180;
mu = 398600;
%...Data declaration for Problem 5.13:
r = [ 3794.66, 3792.71,
4501.31];
v = [-7.72483, 7.72134, 0.0186586];
%...
%...Algorithm 4.1:
coe = coe_from_sv(r,v);
%...Echo the input data and output results to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.13: Orbital elements from state vector\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n State vector:\n')
fprintf('\n r (km)
= [%g %g %g]', ...
r(1), r(2), r(3))
fprintf('\n v (km/s)
= [%g %g %g]', ...
v(1), v(2), v(3))
disp(' ')
fprintf('\n Angular momentum (km^2/s) = %g', coe(1))
fprintf('\n Eccentricity
= %g', coe(2))
fprintf('\n Right ascension (deg)
= %g', coe(3)/deg)
fprintf('\n Inclination (deg)
= %g', coe(4)/deg)
fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg)
fprintf('\n True anomaly (deg)
= %g', coe(6)/deg)
fprintf('\n Semimajor axis (km):
= %g', coe(7))
%...if the orbit is an ellipse, output its period (Equation 2.73):
if coe(2)&lt;1
T = 2*pi/sqrt(mu)*coe(7)^1.5;
fprintf('\n Period:')
fprintf('\n
Seconds
= %g', T)
fprintf('\n
Minutes
= %g', T/60)
fprintf('\n
Hours
= %g', T/3600)
fprintf('\n
Days
= %g', T/24/3600)
end
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.13: Orbital elements from state vector
Gravitational parameter (km^3/s^2) = 398600
State vector:
r (km)
v (km/s)
= [3794.66 3792.71 4501.31]
= [-7.72483 7.72134 0.0186586]
Angular momentum (km^2/s) = 76490.5
Eccentricity
= 1.09593
Right ascension (deg)
= 315.13
Inclination (deg)
= 39.9968
Argument of perigee (deg) = 89.8097
True anomaly (deg)
= 0.0797759
Semimajor axis (km):
= -73003.5
-----------------------------------------------------
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Chapter 5
Problem 5.14 The local sidereal time θ , azimuth A, angular elevation a and slant range ρ are
provided at three observation times. The rates are not provided, but we can still use Algorithm 5.4,
implemented in MATLAB as rv_from_observe in Appendix D.14, to find just the position vectors
at each of the times. The following MATLAB script carries out this procedure, passing zeros to
rv_from_observe as values for the rates. The output to the MATLAB Command Window follows.
%
%
%
%
%
%
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_10a
~~~~~~~~~~~~~
This program uses Algorithm 5.4 to find the geocentric position
vectors corresponding to the three sets of azimuth, elevation
and slant range data given in Problem 5.14
% deg
% pi
- conversion factor between degrees and radians
- 3.1415926...
% Re
% f
% wE
%
- equatorial radius of the earth (km)
- earth's flattening factor
- angular velocity of the earth (rad/s) (not required in
this problem)
%
%
%
%
%
%
%
- vector of three observation times (min)
- vector of slant ranges (km) of the object at the three
observation times
- vector of azimuths (deg) of the object relative to the
observation site at the three observation times
- vector of elevation angles (deg) of the object relative to
the observation site at the three observation times
t
rho
az
el
% theta
at
%
% phi
% H
- vector of local sidereal times (deg) of the tracking site
% r
% v
%
given)
- geocentric equatorial position vector of object (km)
- geocentric equatorial velocity vector of object (km)
(not computed since the rates of rho, az and el are not
the three observation times
- geodetic latitude (deg) of site
- elevation of site (km)
% User M-function required: rv_from_observe
% -------------------------------------------------------------------clear
global f Re wE
deg
Re
f
wE
=
=
=
=
pi/180;
6378;
0.0033528;
7.292115e-5;
%...Data declaration for Problem 5.14:
phi
= -20;
H
= 0.5;
t
= [0 2 4];
theta = [60.0
60.5014 61.0027];
az
= [165.931 145.967 2.40962];
el
= [9.53549 45.7711 21.8825];
rho
= [1214.89 421.441 732.079];
%...
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%...Echo the input data to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.14')
fprintf('\n\n Input data (angles in degrees):\n');
fprintf('\n Time')
fprintf('\n (min)
Azimuth
Elevation
Slant range\n')
for i = 1:3
fprintf('\n %5.1f%15.5e%15.5e%15.5e',t(i), az(i), el(i), rho(i))
end
%...Output the solution to the command window:
fprintf('\n\n Solution:')
fprintf('\n\n Time')
fprintf('\n (min)
Geocentric position vector (km)\n')
for i = 1:3
%...Algorithm 5.4:
[r,v] = rv_from_observe(rho(i), 0, az(i), 0, el(i), ...
0, theta(i), phi, H);
fprintf('\n %5.1f
[%g %g %g]',t(i), r(1), r(2), r(3))
end
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.14
Input data (angles in degrees):
Time
(min)
0.0
2.0
4.0
Azimuth
1.65931e+02
1.45967e+02
2.40962e+00
Elevation
Slant range
9.53549e+00
4.57711e+01
2.18825e+01
1.21489e+03
4.21441e+02
7.32079e+02
Solution:
Time
(min)
Geocentric position vector (km)
0.0
[2641.68 5158.02 -3328.73]
2.0
[2908.04 5474.36 -2500.03]
4.0
[3118.6 5685.65 -1623.34]
-----------------------------------------------------
ˆ ( km )
r1 = 2641.7Iˆ + 5158.0 Jˆ − 3328.7K
ˆ ( km )
r = 2908.0Iˆ + 5474.4 Jˆ − 2500.0K
2
ˆ ( km )
r3 = 3118.6Iˆ + 5685.6 Jˆ − 1623.3K
Using these three position vectors we employ Gibbs’ method, Algrorithm 5.1, which is implemented
in MATLAB as the M-function gibbs in Appendix D.10. The following MATLAB script calls upon
gibbs to find the velocity vector v 2 corresponding to the position vector r2 . The output is listed
afterward.
%
%
%
%
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_14b
~~~~~~~~~~~~~
This program uses Algorithm 5.1 (Gibbs method) to obtain the state
72
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Chapter 5
% vector from the three coplanar position vectors found in the first
% part of Problem 5.14.
%
% mu
- gravitational parameter (km^3/s^2)
% r1, r2, r3 - three coplanar geocentric position vectors (km)
% ierr
- 0 if r1, r2, r3 are found to be coplanar
%
1 otherwise
% v2
- the velocity corresponding to r2 (km/s)
%
% User M-function required: gibbs
% -------------------------------------------------------------------clear
global mu
mu = 398600;
r1 = [2641.68 5158.02 -3328.73];
r2 = [2908.04 5474.36 -2500.03];
r3 = [3118.6 5685.65 -1623.34];
%...Echo the input data to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.14: Gibbs Method\n')
fprintf('\n Input data:\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n r1 (km) = [%g %g %g]', r1(1), r1(2), r1(3))
fprintf('\n r2 (km) = [%g %g %g]', r2(1), r2(2), r2(3))
fprintf('\n r3 (km) = [%g %g %g]', r3(1), r3(2), r3(3))
fprintf('\n\n');
%...Algorithm 5.1:
[v2, ierr] = gibbs(r1, r2, r3);
%...If the vectors r1, r2, r3, are not coplanar, abort:
if ierr == 1
fprintf('\n These vectors are not coplanar.\n\n')
return
end
%...Output the results to the command window:
fprintf(' Solution:')
fprintf('\n');
fprintf('\n v2 (km/s) = [%g %g %g]', v2(1), v2(2), v2(3))
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.14: Gibbs Method
Input data:
Gravitational parameter (km^3/s^2)
= 398600
r1 (km) = [2641.68 5158.02 -3328.73]
r2 (km) = [2908.04 5474.36 -2500.03]
r3 (km) = [3118.6 5685.65 -1623.34]
Solution:
v2 (km/s) = [1.99357 2.20552 7.12881]
-----------------------------------------------------
ˆ ( km/s )
v 2 = 1.9936Iˆ + 2.2055Jˆ + 7.1288K
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Chapter 5
Problem 5.15 The following MATLAB script uses r2 and v 2 from Problem 5.14 to compute the
orbital elements by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in
Appendix D.8. The output to the MATLAB Command Window is listed afterwards.
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%
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_15
~~~~~~~~~~~~
This program uses Algorithm 4.1 to obtain the orbital
elements from the state vector obtained in Problem 5.14.
pi
deg
mu
r
v
coe
T
-
3.1415926...
factor for converting between degrees and radians
gravitational parameter (km^3/s^2)
position vector (km) in the geocentric equatorial frame
velocity vector (km/s) in the geocentric equatorial frame
orbital elements [h e RA incl w TA a]
where h
= angular momentum (km^2/s)
e
= eccentricity
RA
= right ascension of the ascending node (rad)
w
TA
a
= semimajor axis (km)
- Period of an elliptic orbit (s)
User M-function required: coe_from_sv
--------------------------------------------------------------------
clear
global mu
deg = pi/180;
mu = 398600;
%...Data declaration for Problem 5.15:
r = [2908.04 5474.36 -2500.03];
v = [1.99357 2.20552
7.12881];
%...
%...Algorithm 4.1:
coe = coe_from_sv(r,v);
%...Echo the input data and output results to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.15: Orbital elements from state vector\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n State vector:\n')
fprintf('\n r (km)
= [%g %g %g]', ...
r(1), r(2), r(3))
fprintf('\n v (km/s)
= [%g %g %g]', ...
v(1), v(2), v(3))
disp(' ')
fprintf('\n Angular momentum (km^2/s) = %g', coe(1))
fprintf('\n Eccentricity
= %g', coe(2))
fprintf('\n Right ascension (deg)
= %g', coe(3)/deg)
fprintf('\n Inclination (deg)
= %g', coe(4)/deg)
fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg)
fprintf('\n True anomaly (deg)
= %g', coe(6)/deg)
fprintf('\n Semimajor axis (km):
= %g', coe(7))
%...if the orbit is an ellipse, output its period (Equation 2.73):
if coe(2)&lt;1
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T = 2*pi/sqrt(mu)*coe(7)^1.5;
fprintf('\n Period:')
fprintf('\n
Seconds
fprintf('\n
Minutes
fprintf('\n
Hours
fprintf('\n
Days
=
=
=
=
%g',
%g',
%g',
%g',
Chapter 5
T)
T/60)
T/3600)
T/24/3600)
end
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.15: Orbital elements from state vector
Gravitational parameter (km^3/s^2) = 398600
State vector:
r (km)
v (km/s)
= [2908.04
= [1.99357
5474.36
2.20552
-2500.03]
7.12881]
Angular momentum (km^2/s) = 51626.3
Eccentricity
= 0.00102595
Right ascension (deg)
= 60.0001
Inclination (deg)
= 95.0003
Argument of perigee (deg) = 270.34
True anomaly (deg)
= 67.6075
Semimajor axis (km):
= 6686.59
Period:
Seconds
= 5441.5
Minutes
= 90.6916
Hours
= 1.51153
Days
= 0.0629803
-----------------------------------------------------
Problems 5.16 and 5.17 The following MATLAB script uses Equations 5.55 and 5.56 to convert
the data given in Problem 5.16 into three tracking site position vectors ( R1 , R2 , R3 ) and three space
object direction cosine vectors ( ρ̂
ρ1 , ρ̂ρ2 , ρ̂ρ3 ). These vectors together with the three observation times are
then handed off to the M-function gauss (Appendix D.15). gauss implements both the Gauss
Algorithm 5.5 to compute an approximation of the state vector ( r , v) and Algorithm 5.6. which
iteratively improves it. The output to the MATLAB Command Window is listed afterwards.
%
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%
%
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_16
~~~~~~~~~~~~
This program uses Algorithms 5.5 and 5.6 (Gauss's method) to compute
the state vector from the angles only data provided in Problem 5.16.
deg
pi
mu
Re
f
H
phi
t
ra
dec
theta
R
-
factor for converting between degrees and radians
3.1415926...
gravitational parameter (km^3/s^2)
earth's flattening factor
elevation of observation site (km)
latitude of site (deg)
vector of observation times t1, t2, t3 (s)
vector of topocentric equatorial right ascensions
at t1, t2, t3 (deg)
- vector of topocentric equatorial right declinations
at t1, t2, t3 (deg)
- vector of local sidereal times for t1, t2, t3 (deg)
- matrix of site position vectors at t1, t2, t3 (km)
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% rho
- matrix of direction cosine vectors at t1, t2, t3
% fac1, fac2
- common factors
% r_old, v_old - the state vector without iterative improvement (km,
km/s)
% r, v
- the state vector with iterative improvement (km,
km/s)
%
% User M-function required: gauss
% -------------------------------------------------------------------clear
global mu
deg
mu
Re
f
=
=
=
=
pi/180;
398600;
6378;
1/298.26;
%...Data declaration for Problen 5.16:
H
= 0;
phi
= 29*deg;
t
= [
0
60
120
];
ra
= [
0
6.59279e+01
7.98500e+01]*deg;
dec
= [5.15110e+01 2.79911e+01
1.46609e+01]*deg;
theta = [
0
2.50684e-01
5.01369e-01]*deg;
%...
%...Equations 5.56, 5.57:
fac1 = Re/sqrt(1-(2*f - f*f)*sin(phi)^2);
fac2 = (Re*(1-f)^2/sqrt(1-(2*f - f*f)*sin(phi)^2) + H)*sin(phi);
for i = 1:3
R(i,1) = (fac1 + H)*cos(phi)*cos(theta(i));
R(i,2) = (fac1 + H)*cos(phi)*sin(theta(i));
R(i,3) = fac2;
rho(i,1) = cos(dec(i))*cos(ra(i));
rho(i,2) = cos(dec(i))*sin(ra(i));
rho(i,3) = sin(dec(i));
end
%...Algorithms 5.5 and 5.6:
[r, v, r_old, v_old] = gauss(rho(1,:), rho(2,:), rho(3,:), ...
R(1,:),
R(2,:), R(3,:), ...
t(1),
t(2),
t(3));
%...Echo the input data and output the solution to
%
the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problems 5.16 and 5.17: Orbit determination')
fprintf('\n
by the Gauss method\n')
= %g', Re)
fprintf('\n Flattening factor
= %g', f)
fprintf('\n Gravitational parameter (km^3/s^2) = %g', mu)
fprintf('\n\n Input data:\n');
fprintf('\n Latitude (deg) of tracking site = %g', phi/deg);
fprintf('\n Altitude (km) above sea level
= %g', H);
fprintf('\n\n Observations:')
fprintf('\n
Right')
fprintf('
Local')
fprintf('\n
Time (s)
Ascension (deg)
Declination (deg)')
fprintf('
Sidereal time (deg)')
for i = 1:3
fprintf('\n %9.4g %11.4f %19.4f %20.4f', ...
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Chapter 5
t(i), ra(i)/deg, dec(i)/deg, theta(i)/deg)
end
fprintf('\n\n Solution:\n')
fprintf('\n Without iterative improvement (Problem 5.16)...\n')
fprintf('\n r (km)
= [%g, %g, %g]', r_old(1), r_old(2), r_old(3))
fprintf('\n v (km/s) = [%g, %g, %g]', v_old(1), v_old(2), v_old(3))
fprintf('\n');
fprintf('\n\n With iterative improvement (Problem 5.17)...\n')
fprintf('\n r (km)
= [%g, %g, %g]', r(1), r(2), r(3))
fprintf('\n v (km/s) = [%g, %g, %g]', v(1), v(2), v(3))
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problems 5.16 and 5.17: Orbit determination
by the Gauss method
= 6378
Flattening factor
= 0.00335278
Gravitational parameter (km^3/s^2) = 398600
Input data:
Latitude (deg) of tracking site = 29
Altitude (km) above sea level
= 0
Observations:
Time (s)
(deg)
0
60
120
Right
Ascension (deg)
Declination (deg)
0.0000
65.9279
79.8500
Local
Sidereal time
51.5110
27.9911
14.6609
0.0000
0.2507
0.5014
Solution:
Without iterative improvement (Problem 5.16)...
r (km)
= [5788.09, 484.257, 3341.52]
v (km/s) = [-0.460072, 8.05816, -0.265618]
With iterative improvement (Problem 5.17)...
r (km)
= [5788.42, 485.007, 3341.96]
v (km/s) = [-0.460926, 8.0706, -0.266112]
-----------------------------------------------------
Problem 5.18 The following MATLAB script uses r2 and v 2 from Problem 5.17 to compute the
orbital elements by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in
Appendix D.8. The output to the MATLAB Command Window is listed afterwards.
%
%
%
%
%
%
%
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_18
~~~~~~~~~~~~
This program uses Algorithm 4.1 to obtain the orbital
elements from the state vector obtained in Problem 5.17.
pi
- 3.1415926...
77
Solutions Manual
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
deg
mu
r
v
coe
T
Orbital Mechanics for Engineering Students
Chapter 5
-
factor for converting between degrees and radians
gravitational parameter (km^3/s^2)
position vector (km) in the geocentric equatorial frame
velocity vector (km/s) in the geocentric equatorial frame
orbital elements [h e RA incl w TA a]
where h
= angular momentum (km^2/s)
e
= eccentricity
RA
= right ascension of the ascending node (rad)
w
TA
a
= semimajor axis (km)
- Period of an elliptic orbit (s)
User M-function required: coe_from_sv
--------------------------------------------------------------------
clear
global mu
deg = pi/180;
mu = 398600;
%...Data declaration for Problem 5.18:
r = [ 5788.42, 485.007,
3341.96];
v = [-0.460926, 8.0706, -0.266112];
%...
%...Algorithm 4.1:
coe = coe_from_sv(r,v);
%...Echo the input data and output results to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.18: Orbital elements from state vector\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n State vector:\n')
fprintf('\n r (km)
= [%g %g %g]', ...
r(1), r(2), r(3))
fprintf('\n v (km/s)
= [%g %g %g]', ...
v(1), v(2), v(3))
disp(' ')
fprintf('\n Angular momentum (km^2/s) = %g', coe(1))
fprintf('\n Eccentricity
= %g', coe(2))
fprintf('\n Right ascension (deg)
= %g', coe(3)/deg)
fprintf('\n Inclination (deg)
= %g', coe(4)/deg)
fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg)
fprintf('\n True anomaly (deg)
= %g', coe(6)/deg)
fprintf('\n Semimajor axis (km):
= %g', coe(7))
%...if the orbit is an ellipse, output its period (Equation 2.73):
if coe(2)&lt;1
T = 2*pi/sqrt(mu)*coe(7)^1.5;
fprintf('\n Period:')
fprintf('\n
Seconds
= %g', T)
fprintf('\n
Minutes
= %g', T/60)
fprintf('\n
Hours
= %g', T/3600)
fprintf('\n
Days
= %g', T/24/3600)
end
fprintf('\n-----------------------------------------------------\n')
Problem 5.18: Orbital elements from state vector
Gravitational parameter (km^3/s^2) = 398600
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Chapter 5
State vector:
r (km)
v (km/s)
= [5788.42 485.007 3341.96]
= [-0.460926 8.0706 -0.266112]
Angular momentum (km^2/s) = 54201.2
Eccentricity
= 0.100054
Right ascension (deg)
= 270
Inclination (deg)
= 30.0001
Argument of perigee (deg) = 89.9993
True anomaly (deg)
= 4.15098
Semimajor axis (km):
= 7444.75
Period:
Seconds
= 6392.73
Minutes
= 106.546
Hours
= 1.77576
Days
= 0.07399
-----------------------------------------------------
Problems 5.19 The following MATLAB script uses Equations 5.55 and 5.56 to convert the data
given in Problem 5.19 into three tracking site position vectors ( R1 , R2 , R3 ) and three space object
direction cosine vectors ( ρˆ1 ,ρˆ2 ,ρˆ3 ) . These vectors together with the three observation times are then
handed off to the M-function gauss (Appendix D.15). gauss implements both the Gauss Algorithm
5.5 to compute an approximation of the state vector ( r , v) and Algorithm 5.6. which iteratively
improves it. The output to the MATLAB Command Window is listed afterwards.
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
% Problem_5_19
% ~~~~~~~~~~~~
%
% This program uses Algorithms 5.5 and 5.6 (Gauss's method) to
compute
% the state vector from the angles only data provided in Problem
5.16.
%
% deg
- factor for converting between degrees and radians
% pi
- 3.1415926...
% mu
- gravitational parameter (km^3/s^2)
% Re
% f
- earth's flattening factor
% H
- elevation of observation site (km)
% phi
- latitude of site (deg)
% t
- vector of observation times t1, t2, t3 (s)
% ra
- vector of topocentric equatorial right ascensions
%
at t1, t2, t3 (deg)
% dec
- vector of topocentric equatorial right declinations
%
at t1, t2, t3 (deg)
% theta
- vector of local sidereal times for t1, t2, t3 (deg)
% R
- matrix of site position vectors at t1, t2, t3 (km)
% rho
- matrix of direction cosine vectors at t1, t2, t3
% fac1, fac2
- common factors
% r_old, v_old - the state vector without iterative improvement (km,
km/s)
% r, v
- the state vector with iterative improvement (km,
km/s)
%
% User M-function required: gauss
% -------------------------------------------------------------------
79
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Chapter 5
clear
global mu
deg
mu
Re
f
=
=
=
=
pi/180;
398600;
6378;
1/298.26;
%...Data declaration for Problem 5.19:
H
= 0;
phi
= 29*deg;
t
= [
0
60
120];
ra
= [15.0394 25.7539
48.6055]*deg;
dec
= [20.7487 30.1410
43.8910]*deg;
theta = [
90 90.2507
90.5014]*deg;
%...
%...Equations 5.56, 5.57:
fac1 = Re/sqrt(1-(2*f - f*f)*sin(phi)^2);
fac2 = (Re*(1-f)^2/sqrt(1-(2*f - f*f)*sin(phi)^2) + H)*sin(phi);
for i = 1:3
R(i,1) = (fac1 + H)*cos(phi)*cos(theta(i));
R(i,2) = (fac1 + H)*cos(phi)*sin(theta(i));
R(i,3) = fac2;
rho(i,1) = cos(dec(i))*cos(ra(i));
rho(i,2) = cos(dec(i))*sin(ra(i));
rho(i,3) = sin(dec(i));
end
%...Algorithms 5.5 and 5.6:
[r, v, r_old, v_old] = gauss(rho(1,:), rho(2,:), rho(3,:), ...
R(1,:),
R(2,:), R(3,:), ...
t(1),
t(2),
t(3));
%...Echo the input data and output the solution to
%
the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problems 5.19 and 5.20: Orbit determination')
fprintf('\n
by the Gauss method\n')
= %g', Re)
fprintf('\n Flattening factor
= %g', f)
fprintf('\n Gravitational parameter (km^3/s^2) = %g', mu)
fprintf('\n\n Input data:\n');
fprintf('\n Latitude (deg) of tracking site = %g', phi/deg);
fprintf('\n Altitude (km) above sea level
= %g', H);
fprintf('\n\n Observations:')
fprintf('\n
Right')
fprintf('
Local')
fprintf('\n
Time (s)
Ascension (deg)
Declination (deg)')
fprintf('
Sidereal time (deg)')
for i = 1:3
fprintf('\n %9.4g %11.4f %19.4f %20.4f', ...
t(i), ra(i)/deg, dec(i)/deg, theta(i)/deg)
end
fprintf('\n\n Solution:\n')
fprintf('\n Without iterative improvement (Problem 5.19)...\n')
fprintf('\n r (km)
= [%g, %g, %g]', r_old(1), r_old(2), r_old(3))
fprintf('\n v (km/s) = [%g, %g, %g]', v_old(1), v_old(2), v_old(3))
fprintf('\n');
80
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Chapter 5
fprintf('\n\n With iterative improvement (Problem 5.20)...\n')
fprintf('\n r (km)
= [%g, %g, %g]', r(1), r(2), r(3))
fprintf('\n v (km/s) = [%g, %g, %g]', v(1), v(2), v(3))
fprintf('\n-----------------------------------------------------\n')
Problems 5.19 and 5.20: Orbit determination
by the Gauss method
= 6378
Flattening factor
= 0.00335278
Gravitational parameter (km^3/s^2) = 398600
Input data:
Latitude (deg) of tracking site = 29
Altitude (km) above sea level
= 0
Observations:
Time (s)
(deg)
0
60
120
Right
Ascension (deg)
Declination (deg)
15.0394
25.7539
48.6055
Local
Sidereal time
20.7487
30.1410
43.8910
90.0000
90.2507
90.5014
Solution:
Without iterative improvement (Problem 5.19)...
r (km)
= [765.19, 5963.6, 3582.88]
v (km/s) = [-7.50919, 0.705265, 0.423729]
With iterative improvement (Problem 5.20)...
r (km)
= [766.265, 5964.12, 3583.57]
v (km/s) = [-7.51882, 0.706233, 0.42431]
----------------------------------------------------Approximate state vector:
ˆ ( km )
r = 765.19Iˆ + 5963.60Jˆ + 3582.88K
ˆ ( km/s )
v = −7.50919Iˆ + 0.705265Jˆ + 0.423729K
Problem 5.20 From the MATLAB output in the previous problem solution, the refined state vector
is
ˆ ( km )
r = 766.265Iˆ + 5964.12Jˆ + 3583.57K
ˆ ( km/s )
v = −7.518 82Iˆ + 0.706 233Jˆ + 0.424 31K
Problem 5.21 The following MATLAB script uses r and v from Problem 5.20 to compute the
orbital elements by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in
Appendix D.8. The output to the MATLAB Command Window is listed afterwards.
%
%
%
%
%
%
%
%
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_21
~~~~~~~~~~~~
This program uses Algorithm 4.1 to obtain the orbital
elements from the state vector obtained in Problem 5.20.
pi
deg
- 3.1415926...
- factor for converting between degrees and radians
81
Solutions Manual
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
mu
r
v
coe
T
Orbital Mechanics for Engineering Students
Chapter 5
-
gravitational parameter (km^3/s^2)
position vector (km) in the geocentric equatorial frame
velocity vector (km/s) in the geocentric equatorial frame
orbital elements [h e RA incl w TA a]
where h
= angular momentum (km^2/s)
e
= eccentricity
RA
= right ascension of the ascending node (rad)
w
TA
a
= semimajor axis (km)
- Period of an elliptic orbit (s)
User M-function required: coe_from_sv
--------------------------------------------------------------------
clear
global mu
deg = pi/180;
mu = 398600;
%...Data declaration for Problem 5.21:
r = [ 766.265,
5964.12, 3583.57];
v = [-7.51882, 0.706233, 0.42431];
%...
%...Algorithm 4.1:
coe = coe_from_sv(r,v);
%...Echo the input data and output results to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.21: Orbital elements from state vector\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n State vector:\n')
fprintf('\n r (km)
= [%g %g %g]', ...
r(1), r(2), r(3))
fprintf('\n v (km/s)
= [%g %g %g]', ...
v(1), v(2), v(3))
disp(' ')
fprintf('\n Angular momentum (km^2/s) = %g', coe(1))
fprintf('\n Eccentricity
= %g', coe(2))
fprintf('\n Right ascension (deg)
= %g', coe(3)/deg)
fprintf('\n Inclination (deg)
= %g', coe(4)/deg)
fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg)
fprintf('\n True anomaly (deg)
= %g', coe(6)/deg)
fprintf('\n Semimajor axis (km):
= %g', coe(7))
%...if the orbit is an ellipse, output its period (Equation 2.73):
if coe(2)&lt;1
T = 2*pi/sqrt(mu)*coe(7)^1.5;
fprintf('\n Period:')
fprintf('\n
Seconds
= %g', T)
fprintf('\n
Minutes
= %g', T/60)
fprintf('\n
Hours
= %g', T/3600)
fprintf('\n
Days
= %g', T/24/3600)
end
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.21: Orbital elements from state vector
Gravitational parameter (km^3/s^2) = 398600
82
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Chapter 5
State vector:
r (km)
v (km/s)
= [766.265 5964.12 3583.57]
= [-7.51882 0.706233 0.42431]
Angular momentum (km^2/s) = 52946.7
Eccentricity
= 0.00474691
Right ascension (deg)
= 360
Inclination (deg)
= 30.9997
Argument of perigee (deg) = 90.3282
True anomaly (deg)
= 353.388
Semimajor axis (km):
= 7033.16
Period:
Seconds
= 5869.98
Minutes
= 97.833
Hours
= 1.63055
Days
= 0.0679396
-----------------------------------------------------
Problem 5.22 The following MATLAB script uses the given three tracking site position vectors
( R1 , R2 , R3 ) and three space object direction cosine vectors ( ρˆ1 , ρˆ2 , ρˆ3 ) together with the three
observation times to find the state vector ( r , v) by means of Algorithm 5.5 and then iteratively
improve it using Algorithm 5.6. Both algorithms are implemented in the MATLAB M-function gauss
in Appendix D.15. The output to the MATLAB Command Window is listed afterwards.
% ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
% Problem_5_22
% ~~~~~~~~~~~~
%
% This program uses Algorithms 5.5 and 5.6 (Gauss's method) to compute
% the state vector from the angles only data provided in Problem 5.16.
%
% deg
- factor for converting between degrees and radians
% pi
- 3.1415926...
% mu
- gravitational parameter (km^3/s^2)
% t
- vector of observation times t1, t2, t3 (s)
% theta
- vector of local sidereal times for t1, t2, t3 (deg)
% R
- matrix of site position vectors at t1, t2, t3 (km)
% rho
- matrix of direction cosine vectors at t1, t2, t3
% r_old, v_old - the state vector without iterative improvement (km,
km/s)
% r, v
- the state vector with iterative improvement (km,
km/s)
%
% User M-function required: gauss
% -------------------------------------------------------------------clear
global mu
deg = pi/180;
mu = 398600;
%...Data declaration for Problem 5.22:
t
= [
0
60
120];
R
= [ -1825.96
-1841.63
-1857.25
3583.66
3575.63
3567.54
rho
= [-0.301687
-0.793090
0.200673
-0.210324
4933.54
4933.54
4933.54];
0.932049
0.571640
83
Solutions Manual
Orbital Mechanics for Engineering Students
-0.873085
-0.362969
Chapter 5
0.325539];
%...
%...Algorithms 5.5 and 5.6:
[r, v, r_old, v_old] = gauss(rho(1,:), rho(2,:), rho(3,:), ...
R(1,:),
R(2,:),
R(3,:), ...
t(1),
t(2),
t(3));
%...Echo the input data and output the solution to
%
the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problems 5.22 and 5.23: Orbit determination')
fprintf('\n
by the Gauss method\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g', mu)
fprintf('\n\n Input data:\n');
fprintf('\n Site position vector (R) and space object')
fprintf('\n direction cosine vector (rho) at three times:\n')
for i = 1:3
fprintf('\n t = %g s:\n',t(i))
fprintf('\n
R
= [%g %g %g]', R(i,1), R(i,2), R(i,3))
fprintf('\n
rho = [%g %g %g]', rho(i,1), rho(i,2), rho(i,3))
disp(' ')
end
fprintf('\n\n Solution:\n')
fprintf('\n Without iterative improvement (Problem 5.22)...\n')
fprintf('\n r (km)
= [%g, %g, %g]', r_old(1), r_old(2), r_old(3))
fprintf('\n v (km/s) = [%g, %g, %g]', v_old(1), v_old(2), v_old(3))
fprintf('\n');
fprintf('\n\n With iterative improvement (Problem 5.23)...\n')
fprintf('\n r (km)
= [%g, %g, %g]', r(1), r(2), r(3))
fprintf('\n v (km/s) = [%g, %g, %g]', v(1), v(2), v(3))
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problems 5.22 and 5.23: Orbit determination
by the Gauss method
Gravitational parameter (km^3/s^2) = 398600
Input data:
Site position vector (R) and space object
direction cosine vector (rho) at three times:
t = 0 s:
R
= [-1825.96 3583.66 4933.54]
rho = [-0.301687 0.200673 0.932049]
t = 60 s:
R
= [-1841.63
rho = [-0.79309
3575.63 4933.54]
-0.210324 0.57164]
t = 120 s:
R
= [-1857.25 3567.54 4933.54]
rho = [-0.873085 -0.362969 0.325539]
Solution:
84
Solutions Manual
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Chapter 5
Without iterative improvement (Problem 5.22)...
r (km)
= [-2350.74, 3440.62, 5300.49]
v (km/s) = [-6.61345, -3.88226, -0.413321]
With iterative improvement (Problem 5.23)...
r (km)
= [-2351.59, 3440.39, 5301.1]
v (km/s) = [-6.62403, -3.8885, -0.414013]
----------------------------------------------------Approximate state vector:
ˆ ( km )
r = −2350.74Iˆ + 3440.62Jˆ + 5300.49K
ˆ ( km/s )
v = −6.613 45Iˆ − 3.882 26Jˆ − 0.413 321K
Problem 5.23 From the MATLAB output listed in the previous problem solution, the iteratively
improved state vector is
ˆ ( km )
r = −2351.59Iˆ + 3440.39Jˆ + 5301.1K
ˆ ( km/s )
v = −6.624 03Iˆ − 3.8885Jˆ − 0.414 013K
Problem 5.24 The following MATLAB script uses r and v from Problem 5.23 to compute the
orbital elements by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in
Appendix D.8. The output to the MATLAB Command Window is listed afterwards.
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
% Problem_5_24
% ~~~~~~~~~~~~
%
% This program uses Algorithm 4.1 to obtain the orbital
% elements from the state vector obtained in Problem 5.23.
%
% pi
- 3.1415926...
% deg - factor for converting between degrees and radians
% mu
- gravitational parameter (km^3/s^2)
% r
- position vector (km) in the geocentric equatorial frame
% v
- velocity vector (km/s) in the geocentric equatorial frame
% coe - orbital elements [h e RA incl w TA a]
%
where h
= angular momentum (km^2/s)
%
e
= eccentricity
%
RA
= right ascension of the ascending node (rad)
%
%
w
%
TA
%
a
= semimajor axis (km)
% T
- Period of an elliptic orbit (s)
%
% User M-function required: coe_from_sv
% ------------------------------------------------------------------clear
global mu
deg = pi/180;
mu = 398600;
%...Data declaration for Problem 5.24:
r = [-2351.59, 3440.39,
5301.1];
v = [-6.62403, -3.8885, -0.414013];
85
Solutions Manual
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Chapter 5
%...
%...Algorithm 4.1:
coe = coe_from_sv(r,v);
%...Echo the input data and output results to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.24: Orbital elements from state vector\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n State vector:\n')
fprintf('\n r (km)
= [%g %g %g]', ...
r(1), r(2), r(3))
fprintf('\n v (km/s)
= [%g %g %g]', ...
v(1), v(2), v(3))
disp(' ')
fprintf('\n Angular momentum (km^2/s) = %g', coe(1))
fprintf('\n Eccentricity
= %g', coe(2))
fprintf('\n Right ascension (deg)
= %g', coe(3)/deg)
fprintf('\n Inclination (deg)
= %g', coe(4)/deg)
fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg)
fprintf('\n True anomaly (deg)
= %g', coe(6)/deg)
fprintf('\n Semimajor axis (km):
= %g', coe(7))
%...if the orbit is an ellipse, output its period (Equation 2.73):
if coe(2)&lt;1
T = 2*pi/sqrt(mu)*coe(7)^1.5;
fprintf('\n Period:')
fprintf('\n
Seconds
= %g', T)
fprintf('\n
Minutes
= %g', T/60)
fprintf('\n
Hours
= %g', T/3600)
fprintf('\n
Days
= %g', T/24/3600)
end
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.24: Orbital elements from state vector
Gravitational parameter (km^3/s^2) = 398600
State vector:
r (km)
v (km/s)
= [-2351.59
= [-6.62403
3440.39
-3.8885
5301.1]
-0.414013]
Angular momentum (km^2/s) = 51868.3
Eccentricity
= 0.000957299
Right ascension (deg)
= 28.0006
Inclination (deg)
= 51.9999
Argument of perigee (deg) = 88.9231
True anomaly (deg)
= 4.99817
Semimajor axis (km):
= 6749.43
Period:
Seconds
= 5518.38
Minutes
= 91.973
Hours
= 1.53288
Days
= 0.0638701
-----------------------------------------------------
Problems 5.25 The following MATLAB script uses Equations 5.55 and 5.56 to convert the data
given in Problem 5.25 into three tracking site position vectors ( R1 , R2 , R3 ) and three space object
direction cosine vectors ( ρˆ1 , ρˆ2 , ρˆ3 ) . These vectors together with the three observation times are then
86
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 5
handed off to the M-function gauss (Appendix D.15). gauss implements both the Gauss Algorithm
5.5 to compute an approximation of the state vector ( r , v) and Algorithm 5.6. which iteratively
improves it. The output to the MATLAB Command Window is listed afterwards.
% ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
% Problem_5_25
% ~~~~~~~~~~~~
%
% This program uses Algorithms 5.5 and 5.6 (Gauss's method) to compute
% the state vector from the angles only data provided in Problem 5.25.
%
% deg
- factor for converting between degrees and radians
% pi
- 3.1415926...
% mu
- gravitational parameter (km^3/s^2)
% Re
% f
- earth's flattening factor
% H
- elevation of observation site (km)
% phi
- latitude of site (deg)
% t
- vector of observation times t1, t2, t3 (s)
% ra
- vector of topocentric equatorial right ascensions
%
at t1, t2, t3 (deg)
% dec
- vector of topocentric equatorial right declinations
%
at t1, t2, t3 (deg)
% theta
- vector of local sidereal times for t1, t2, t3 (deg)
% R
- matrix of site position vectors at t1, t2, t3 (km)
% rho
- matrix of direction cosine vectors at t1, t2, t3
% fac1, fac2
- common factors
% r_old, v_old - the state vector without iterative improvement (km,
km/s)
% r, v
- the state vector with iterative improvement (km,
km/s)
%
% User M-function required: gauss
% -------------------------------------------------------------------clear
global mu
deg
mu
Re
f
=
=
=
=
pi/180;
398600;
6378;
1/298.26;
%...Data declaration for Problem 5.25:
H
= 0.5;
phi
= 60*deg;
t
= [
0
300
600];
ra
= [157.783 159.221
160.526]*deg;
dec
= [24.2403 27.2993
29.8982]*deg;
theta = [
150 151.253
152.507]*deg;
%...
%...Equations 5.56, 5.57:
fac1 = Re/sqrt(1-(2*f - f*f)*sin(phi)^2);
fac2 = (Re*(1-f)^2/sqrt(1-(2*f - f*f)*sin(phi)^2) + H)*sin(phi);
for i = 1:3
R(i,1) = (fac1 + H)*cos(phi)*cos(theta(i));
R(i,2) = (fac1 + H)*cos(phi)*sin(theta(i));
R(i,3) = fac2;
rho(i,1) = cos(dec(i))*cos(ra(i));
87
Solutions Manual
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Chapter 5
rho(i,2) = cos(dec(i))*sin(ra(i));
rho(i,3) = sin(dec(i));
end
%...Algorithms 5.5 and 5.6:
[r, v, r_old, v_old] = gauss(rho(1,:), rho(2,:), rho(3,:), ...
R(1,:),
R(2,:), R(3,:), ...
t(1),
t(2),
t(3));
%...Echo the input data and output the solution to
%
the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problems 5.25 and 5.26: Orbit determination')
fprintf('\n
by the Gauss method\n')
= %g', Re)
fprintf('\n Flattening factor
= %g', f)
fprintf('\n Gravitational parameter (km^3/s^2) = %g', mu)
fprintf('\n\n Input data:\n');
fprintf('\n Latitude (deg) of tracking site = %g', phi/deg);
fprintf('\n Altitude (km) above sea level
= %g', H);
fprintf('\n\n Observations:')
fprintf('\n
Right')
fprintf('
Local')
fprintf('\n
Time (s)
Ascension (deg)
Declination (deg)')
fprintf('
Sidereal time (deg)')
for i = 1:3
fprintf('\n %9.4g %11.4f %19.4f %20.4f', ...
t(i), ra(i)/deg, dec(i)/deg, theta(i)/deg)
end
fprintf('\n\n Solution:\n')
fprintf('\n Without iterative improvement (Problem 5.25)...\n')
fprintf('\n r (km)
= [%g, %g, %g]', r_old(1), r_old(2), r_old(3))
fprintf('\n v (km/s) = [%g, %g, %g]', v_old(1), v_old(2), v_old(3))
fprintf('\n');
fprintf('\n\n With iterative improvement (Problem 5.26)...\n')
fprintf('\n r (km)
= [%g, %g, %g]', r(1), r(2), r(3))
fprintf('\n v (km/s) = [%g, %g, %g]', v(1), v(2), v(3))
fprintf('\n-----------------------------------------------------\n'
----------------------------------------------------Problems 5.25 and 5.26: Orbit determination
by the Gauss method
= 6378
Flattening factor
= 0.00335278
Gravitational parameter (km^3/s^2) = 398600
Input data:
Latitude (deg) of tracking site = 60
Altitude (km) above sea level
= 0.5
Observations:
Time (s)
(deg)
0
300
600
Right
Ascension (deg)
Declination (deg)
157.7830
159.2210
160.5260
24.2403
27.2993
29.8982
88
Local
Sidereal time
150.0000
151.2530
152.5070
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 5
Solution:
Without iterative improvement (Problem 5.25)...
r (km)
= [-19050.2, 7702.56, 14469.6]
v (km/s) = [-3.27477, -0.482844, 5.07464]
With iterative improvement (Problem 5.26)...
r (km)
= [-19081, 7714.25, 14486.6]
v (km/s) = [-3.27846, -0.484358, 5.08206]
----------------------------------------------------Approximate state vector:
ˆ ( km )
r = -19050.2Iˆ + 7702.56Jˆ + 14469.6K
ˆ ( km/s )
v = −3.27477Iˆ − 0.482844Jˆ + 5.07464K
Problem 5.26 From the MATLAB output listed in the previous problem solution, the iteratively
improved state vector is
ˆ ( km )
r = −19081Iˆ + 7714.25Jˆ + 14486.6K
ˆ ( km/s )
v = −3.27846Iˆ − 0.484 358Jˆ + 5.082 06K
Problem 5.27 The following MATLAB script uses r and v from Problem 5.26 to compute the
orbital elements by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in
Appendix D.8. The output to the MATLAB Command Window is listed afterwards.
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_5_27
~~~~~~~~~~~~
This program uses Algorithm 4.1 to obtain the orbital
elements from the state vector obtained in Problem 5.26.
pi
deg
mu
r
v
coe
T
-
3.1415926...
factor for converting between degrees and radians
gravitational parameter (km^3/s^2)
position vector (km) in the geocentric equatorial frame
velocity vector (km/s) in the geocentric equatorial frame
orbital elements [h e RA incl w TA a]
where h
= angular momentum (km^2/s)
e
= eccentricity
RA
= right ascension of the ascending node (rad)
w
TA
a
= semimajor axis (km)
- Period of an elliptic orbit (s)
User M-function required: coe_from_sv
--------------------------------------------------------------------
clear
global mu
deg = pi/180;
mu = 398600;
%...Data declaration for Problem 5.27:
r = [ -19081,
7714.25, 14486.6];
v = [-3.27846, -0.484358, 5.08206];
%...
89
Solutions Manual
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Chapter 5
%...Algorithm 4.1:
coe = coe_from_sv(r,v);
%...Echo the input data and output results to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.27: Orbital elements from state vector\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n State vector:\n')
fprintf('\n r (km)
= [%g %g %g]', ...
r(1), r(2), r(3))
fprintf('\n v (km/s)
= [%g %g %g]', ...
v(1), v(2), v(3))
disp(' ')
fprintf('\n Angular momentum (km^2/s) = %g', coe(1))
fprintf('\n Eccentricity
= %g', coe(2))
fprintf('\n Right ascension (deg)
= %g', coe(3)/deg)
fprintf('\n Inclination (deg)
= %g', coe(4)/deg)
fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg)
fprintf('\n True anomaly (deg)
= %g', coe(6)/deg)
fprintf('\n Semimajor axis (km):
= %g', coe(7))
%...if the orbit is an ellipse, output its period (Equation 2.73):
if coe(2)&lt;1
T = 2*pi/sqrt(mu)*coe(7)^1.5;
fprintf('\n Period:')
fprintf('\n
Seconds
= %g', T)
fprintf('\n
Minutes
= %g', T/60)
fprintf('\n
Hours
= %g', T/3600)
fprintf('\n
Days
= %g', T/24/3600)
end
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.27: Orbital elements from state vector
Gravitational parameter (km^3/s^2) = 398600
State vector:
r (km)
v (km/s)
= [-19081 7714.25 14486.6]
= [-3.27846 -0.484358 5.08206]
Angular momentum (km^2/s)
Eccentricity
Right ascension (deg)
Inclination (deg)
Argument of perigee (deg)
True anomaly (deg)
Semimajor axis (km):
=
=
=
=
=
=
=
76005.8
1.08937
136.949
62.9772
287.335
112.915
-77612.3
Problem 5.28 The following MATLAB script uses the given three tracking site position vectors
( R1 , R2 , R3 ) and three space object direction cosine vectors ( ρˆ1 , ρˆ2 , ρˆ3 ) together with the three
observation times to find the state vector ( r , v) by means of Algorithm 5.5 and then iteratively
improve it using Algorithm 5.6. Both algorithms are implemented in the MATLAB M-function gauss
in Appendix D.15. The output to the MATLAB Command Window is listed afterwards.
% ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
% Problem_5_28
% ~~~~~~~~~~~~
%
90
Solutions Manual
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Chapter 5
% This program uses Algorithms 5.5 and 5.6 (Gauss's method) to compute
% the state vector from the angles only data provided in Problem 5.28.
%
% deg
- factor for converting between degrees and radians
% pi
- 3.1415926...
% mu
- gravitational parameter (km^3/s^2)
% t
- vector of observation times t1, t2, t3 (s)
% theta
- vector of local sidereal times for t1, t2, t3 (deg)
% R
- matrix of site position vectors at t1, t2, t3 (km)
% rho
- matrix of direction cosine vectors at t1, t2, t3
% r_old, v_old - the state vector without iterative improvement (km,
km/s)
% r, v
- the state vector with iterative improvement (km,
km/s)
%
% User M-function required: gauss
% -------------------------------------------------------------------clear
global mu
deg = pi/180;
mu = 398600;
%...Data declaration for Problem 5.28:
t
= [
0
300
600];
R
= [ 5582.84
5581.50
5577.50
0
122.122
244.186
3073.90
3073.90
3073.90];
rho
= [ 0.846428,
0.749290,
0.529447,
0,
0.463023,
0.777163,
0.532504
0.473470
0.340152];
%...
%...Algorithms 5.5 and 5.6:
[r, v, r_old, v_old] = gauss(rho(1,:), rho(2,:), rho(3,:), ...
R(1,:),
R(2,:),
R(3,:), ...
t(1),
t(2),
t(3));
%...Echo the input data and output the solution to
%
the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problems 5.28 and 5.29: Orbit determination')
fprintf('\n
by the Gauss method\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g', mu)
fprintf('\n\n Input data:\n');
fprintf('\n Site position vector (R) and space object')
fprintf('\n direction cosine vector (rho) at three times:\n')
for i = 1:3
fprintf('\n t = %g s:\n',t(i))
fprintf('\n
R
= [%g %g %g]', R(i,1), R(i,2), R(i,3))
fprintf('\n
rho = [%g %g %g]', rho(i,1), rho(i,2), rho(i,3))
disp(' ')
end
fprintf('\n\n Solution:\n')
fprintf('\n Without iterative improvement (Problem 5.28)...\n')
fprintf('\n r (km)
= [%g, %g, %g]', r_old(1), r_old(2), r_old(3))
fprintf('\n v (km/s) = [%g, %g, %g]', v_old(1), v_old(2), v_old(3))
fprintf('\n');
fprintf('\n\n With iterative improvement (Problem 5.29)...\n')
fprintf('\n r (km)
= [%g, %g, %g]', r(1), r(2), r(3))
fprintf('\n v (km/s) = [%g, %g, %g]', v(1), v(2), v(3))
91
Solutions Manual
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Chapter 5
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problems 5.28 and 5.29: Orbit determination
by the Gauss method
Gravitational parameter (km^3/s^2) = 398600
Input data:
Site position vector (R) and space object
direction cosine vector (rho) at three times:
t = 0 s:
R
= [5582.84 0 3073.9]
rho = [0.846428 0 0.532504]
t = 300 s:
R
= [5581.5 122.122 3073.9]
rho = [0.74929 0.463023 0.47347]
t = 600 s:
R
= [5577.5 244.186 3073.9]
rho = [0.529447 0.777163 0.340152]
Solution:
Without iterative improvement (Problem 5.28)...
r (km)
= [8282.6, 1791.26, 4780.7]
v (km/s) = [-1.07108, 5.89508, -0.618321]
With iterative improvement (Problem 5.29)...
r (km)
= [8306.27, 1805.89, 4795.66]
v (km/s) = [-1.07872, 5.94219, -0.622807]
----------------------------------------------------Approximate state vector
ˆ ( km )
r = 8282.6Iˆ + 1791.26Jˆ + 4780.7K
ˆ ( km/s )
v = -1.07108Iˆ + 5.895 08Jˆ − 0.618 321K
Problem 5.29 From the MATLAB output listed in the previous problem solution, the iteratively
improved state vector is
ˆ ( km )
r = 8306.27Iˆ + 1805.89Jˆ + 4795.66K
ˆ ( km/s )
v = −1.078 72Iˆ + 5.94219Jˆ − 0.622 807K
Problem 5.30 The following MATLAB script uses r and v from Problem 5.29 to compute the
orbital elements by means of Algorithm 4.1, which is implemented as the M-function coe_from_sv in
Appendix D.8. The output to the MATLAB Command Window is listed afterwards.
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
% Problem_5_30
% ~~~~~~~~~~~~
92
Solutions Manual
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
-
Orbital Mechanics for Engineering Students
Chapter 5
This program uses Algorithm 4.1 to obtain the orbital
elements from the state vector obtained in Problem 5.29.
pi
deg
mu
r
v
coe
T
-
3.1415926...
factor for converting between degrees and radians
gravitational parameter (km^3/s^2)
position vector (km) in the geocentric equatorial frame
velocity vector (km/s) in the geocentric equatorial frame
orbital elements [h e RA incl w TA a]
where h
= angular momentum (km^2/s)
e
= eccentricity
RA
= right ascension of the ascending node (rad)
w
TA
a
= semimajor axis (km)
- Period of an elliptic orbit (s)
User M-function required: coe_from_sv
-------------------------------------------------------------------
clear
global mu
deg = pi/180;
mu = 398600;
%...Data declaration for Problem 5.30:
r = [ 8306.27, 1805.89,
4795.66];
v = [-1.07872, 5.94219, -0.622807];
%...
%...Algorithm 4.1:
coe = coe_from_sv(r,v);
%...Echo the input data and output results to the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 5.30: Orbital elements from state vector\n')
fprintf('\n Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n State vector:\n')
fprintf('\n r (km)
= [%g %g %g]', ...
r(1), r(2), r(3))
fprintf('\n v (km/s)
= [%g %g %g]', ...
v(1), v(2), v(3))
disp(' ')
fprintf('\n Angular momentum (km^2/s) = %g', coe(1))
fprintf('\n Eccentricity
= %g', coe(2))
fprintf('\n Right ascension (deg)
= %g', coe(3)/deg)
fprintf('\n Inclination (deg)
= %g', coe(4)/deg)
fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg)
fprintf('\n True anomaly (deg)
= %g', coe(6)/deg)
fprintf('\n Semimajor axis (km):
= %g', coe(7))
%...if the orbit is an ellipse, output its period (Equation 2.73):
if coe(2)&lt;1
T = 2*pi/sqrt(mu)*coe(7)^1.5;
fprintf('\n Period:')
fprintf('\n
Seconds
= %g', T)
fprintf('\n
Minutes
= %g', T/60)
fprintf('\n
Hours
= %g', T/3600)
fprintf('\n
Days
= %g', T/24/3600)
end
93
Solutions Manual
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Chapter 5
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 5.30: Orbital elements from state vector
Gravitational parameter (km^3/s^2) = 398600
State vector:
r (km)
v (km/s)
= [8306.27 1805.89 4795.66]
= [-1.07872 5.94219 -0.622807]
Angular momentum (km^2/s) = 59242.6
Eccentricity
= 0.0995646
Right ascension (deg)
= 270
Inclination (deg)
= 30.0002
Argument of perigee (deg) = 269.945
True anomaly (deg)
= 190.718
Semimajor axis (km):
= 8893.18
Period:
Seconds
= 8346.36
Minutes
= 139.106
Hours
= 2.31843
Days
= 0.0966014
-----------------------------------------------------
94
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 6
Problem 6.1
Orbit 1 (circle):
398 600
&micro;
=
= 7.726 km /s
r
6378 + 300
vc =
Orbit 2 (ellipse):
e=
rapogee =
6378 + 300 =
vapogee =
rapogee − rperigee
rapogee + rperigee
=
(6378 + 300) − (6378 + 200)
= 0.003758
(6378 + 300) − (6378 + 200)
h2 1
&micro; 1 −e
h2
1
⇒ h = 51 500 km 2 s
398 600 1 − 0.003758
h
rapogee
=
51 500
= 7.711 km s
6678
∆v = 7.726 − 7.711 = 0.01453 km s = 14.53 m s
Thrust ⋅ ∆t = m∆v
53 400 ⋅ ∆t = 125000 ⋅ 14.53 ⇒ ∆t = 34.01 s
(a)
vcircle + ( vcircle + ∆v)
∆v
14.53
= vcircle +
= 7.726 +
= 7.733 m s
2
2
2
∆s = v avg ∆t = 7.733 ⋅ 34.01 = 263 km
v avg =
(b)
∆s
263
=
= 0.006 268 or 0.63%
orbit circumference 2π ⋅ 6600
(c)
Problem 6.2
vperigee = 8.2 km s
1
rperigee = 6378 + 480 = 6858 km
1
rapogee = rperigee = 6858 km
2
1
rperigee = 6378 + 160 = 6538 km
2
e2 =
rapogee − rperigee
2
2
rapogee + rperigee
2
2
h2 2 1
rapogee =
2
&micro; 1 − e2
=
6858 − 6538
= 0.023 89
6858 + 6538
h2 2
1
⇒ h2 = 51 660 km 2 s
398 600 1 − 0.023 89
51 660
h2
=
=
= 7.532 km s
rapogee
6858
6858 =
vapogee
2
2
∆v = vapogee − vperigee = 7.532 − 8.2 = −0.6678 km s
2
1
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Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 6
Problem 6.3
(a)
Orbit 1 (circle):
v1 =
398 600
&micro;
=
= 7.726 km s
r
6378 + 300
Orbit 2 (transfer ellipse):
e2 =
rapogee − rperigee
2
rapogee + rperigee
2
rperigee
2
2
2
=
(6378 + 3000) − (6378 + 300)
= 0.1682
(6378 + 3000) + (6378 + 300)
h2 1
= 2
&micro; 1 + e2
h2 2
1
⇒ h2 = 55760 km 2 s
398 600 1 + 0.1682
55760
h2
=
=
= 8.35 km s
rperigee
6678
6678 =
vperigee
2
vapogee =
2
2
h2
rapogee
=
2
55760
= 5.946 km s
9378
Orbit 3 (circle):
v3 =
398 600
&micro;
=
= 6.519 km s
r
6378 + 3000
∆v1 = vperigee − v1 = 8.350 − 7.726 = 0.6244 km s
2
∆v2 = v3 − vapogee = 6.519 − 5.946 = 0.5734 km s
2
∆vtotal = ∆v1 + ∆v2 = 1.198 km s
(b)
1
1
r
= (6678 + 9378) = 8028 km s
+r
2 perigee2 apogee2
2
2π 3/2
2π
T2 =
a
=
8028 3/2 = 7159 s ( period of tranfer ellipse)
&micro;
398 600
a2 =
tperigee to apogee =
(
)
T2
= 3579 s = 59.65 m
2
Problem 6.4 To determine where the projectile B impacts the earth we need the orbital elements.
rapogee = 7000 km
B
vapogee = 7.1 km s
B
hB = rapogee vapogee = 7000 ⋅ 7.1 = 49 700 km 2 s
B
rapogee =
B
7000 =
2
B
hB
1
&micro; 1 + eB
49 700 2 1
⇒ eB = 0.1147
398 600 1 + eB
hB
2π 
TB =
2 
&micro;  1 − e 2
B
3
3

49 700


2π
 =
 = 4952 s ( period of B's orbit )
2 
2
398 600  1 − 0.1147 

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Solutions Manual
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Chapter 6
At impact, rB = Rearth .
Rearth =
hB2
1
&micro; 1 + eB cos θ impact
6378 =
(At impact, rB = Rearth )
49 700 2
1
⇒ θ impaact = 104.3&deg; ( from perigee of B's elliptical orbit )
398 600 1 + 0.1147 cos θ impact
Determine the time of flight ( tof ) to impact by first finding t impact , the time from perigee to B’s
impact point.
E impact
θ impact
1 − eB
1 − 0.1147
104.3&deg;
tan
tan
=
⇒ E impact = 1.708 rad
2
1 + eB
2
1 + 0.1147
2
M impact = E impact − eB sin E impact = 1.708 − 0.1147 sin 1.708 = 1.594 rad
tan
=
t impact = TB
M impact
= 4952
2π
1.594
= 1257 s (from impact point to perigee)
2π
Then
T
4952
tof = B − t impact =
− 1257 = 1220 s
2
2
Find the orbital elements of spacecraft S trajectory.
rperigee = 7000 km
S
vperigee = 1.3 v esc = 1.3
S
2&micro;
rperigee
= 1.3
S
2 ⋅ 398 600
= 13.97 km s
7000
hS = rperigee vperigee = 7000 ⋅ 13.97 = 97 110 km 2 s
S
rperigee =
S
7000 =
2
S
hS
1
&micro; 1 + eS
97 110 2 1
⇒ eS = 2.38
398 600 1 + eS
Location of S on its hyperbolic trajectory when B impacts the earth:
Mh =
&micro;2
2
(e 2 − 1)3/2 tof = 398 6003 ⋅ (2.382 − 1)3/2 ⋅1220 = 2.131 rad
3 S
97 110
hS
eS sinh F − F = M h
2.38 sinh F − F = 2.131 ⇒ F = 1.118 (Algorihm 3.2)
θ
e +1
F
2.38 + 1
1.118
⇒ θ S = 76.87&deg;
tan S = S
tanh =
tanh
2
eS − 1
2
2.38 − 1
2
rS =
97 110 2
hS2
1
1
=
= 15 360 km
&micro; 1 + eS cos θ s 398 600 1 + 0.28 cos 76.87&deg;
distance = rS − 6378 = 8978 km
Problem 6.5
(a) For the transfer ellipse
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Solutions Manual
Orbital Mechanics for Engineering Students
1
1
(r + r ) = 2 (227.9 + 149.6) &times; 10 6 = 188.8 &times; 10 6 km
2 Mars earth
2π 3/2
2π
(188.8 &times; 10 6 )3/2 = 44.73 &times; 10 6 s = 517.7 days
T=
a
=
9
&micro;
132.7 &times; 10
T
tof = = 258.8 days ( time of flight from earth to Mars )
2
a=
(b) Period of Mars in its orbit,
2π
2π
(227.9 &times; 10 6 )3/2 = 59.34 &times; 10 6 s = 686.8 days
TMars =
rMars3/2 =
&micro;
132.7 &times; 10 9
180&deg; − α
258.8
tof
∴
=
=
= 0.7537
T
180&deg;
343.4
Mars
2
α = 44.33&deg;
Problem 6.6
rA = 7000 km rC = 32 000 km
e1 = 0.3
r −r
e1 = B A
rB + rA
r − 7000
0.3 = B
⇒ rB = 13 000 km
rB + 7000
e2 = 0.5
r −r
e2 = D C
rD + rC
r − 32 000
0.5 = D
⇒ rD = 96 000 km
rD + 32 000
h1 = 2&micro;
7000 ⋅ 13 000
rArB
= 2 ⋅ 398 600
= 60 230 km 2 s
rA + r B
7000 + 13 000
h2 = 2&micro;
32 000 ⋅ 96 000
rC rD
= 2 ⋅ 398 600
= 138 300 km 2 s
rC + r D
32 000 + 96 000
h3 = 2&micro;
7000 ⋅ 96 000
rArD
= 2 ⋅ 398 600
= 72120 km 2 s
rA + r D
7000 + 96 000
h4 = 2&micro;
13 000 ⋅ 32 000
rBrC
= 2 ⋅ 398 600
= 85850 km 2 s
13 000 + 32 000
rB + r C
60 230
h
v A1 = 1 =
= 8.604 km s
rA
7000
60 230
h
vB1 = 1 =
= 4.633 km s
rB 13 000
138 300
h
vC 2 = 2 =
= 4.323 km s
rC
32 000
138 300
h
vD2 = 2 =
= 1.441 km s
rD
96 000
72120
h
v A3 = 3 =
= 10.3 km s
rA
7000
85850
h
v B4 = 4 =
= 6.604 km s
rB 13 000
85850
h
vC 4 = 4 =
= 2.683 km s
rC 32 000
72120
h
vD3 = 3 =
= 0.7512 km s
rD 96 000
98
Chapter 6
Solutions Manual
Orbital Mechanics for Engineering Students
2π  rA + rD 
T3 =
&micro; 2 
2π  rB + rC 
T4 =
&micro; 2 
3/2
3/2
 7000 + 96 000 


=

2
398 600 
2π
Chapter 6
3/2
 13 000 + 32 000 


=

2
398 600 
2π
= 116 300 s = 32.31 h
3/2
= 33 590 s = 9.33 h
(a) Transfer orbit 3:
∆v A = v A3 − v A1 = 10.3 − 8.604 = 1.699 km s
∆vD = vD2 − vD3 = 1.441 − 0.7512 = 0.6896 km s
∆vtotal = ∆v A + ∆vD = 2.389 km s
tof 3 =
T3 32.32
=
= 16.15 h
2
2
(b) Transfer orbit 4:
∆vB = vB4 − vB1 = 6.604 − 4.633 = 1.971 km s
∆vC = vC 2 − vC 4 = 4.323 − 2.683 = 1.64 km s
∆vtotal = ∆vB + ∆vC = 3.611 km s
T
9.33
tof 4 = 4 =
= 4.665 h
2
2
Problem 6.7 Orbit 1 is the original circular orbit and orbit 2 is the impact trajectory.
v A1 =
398 600
&micro;
=
= 7.613 km s
r
6378 + 500
h2 1
rapogee = 2
2
&micro; 1 − e2
h2 1
h2
6378 + 500 = 2
⇒ 2 = 6878(1 − e2 )
&micro; 1 − e2
&micro;
1
h2
rB2 = 2
&micro; 1 + e2 cos θ B
1
h2
h2
6378 = 2
⇒ 2 = 6378(1 + 0.5e2 )
&micro; 1 + e2 cos 60&deg;
&micro;
h2 2 h2 2
=
⇒ 6378(1 + 0.5e2 ) = 6878(1 − e2 ) ⇒ e2 = 0.04967
&micro;
&micro;
∴ h2 = 6878&micro;(1 − e2 ) = 6878 ⋅ 398 600 ⋅ (1 − 0.04967 ) = 51 040 km 2 s
(a)
51 040
h
v A2 = 2 =
= 7.421 km s
rA
6878
∆v = v A2 − v A1 = 7.421 − 7.613 = −0.1915 km s
(b) To fall through the point directly below, we must remove completely the transverse component of
velocity:
∆v = 0 − v A1 = −7.613 km s
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Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 6
Problem 6.8 A is apogee of impact trajectory, I is the
impact point, a is the semimajpr axis.
rA = a (1 + e )
r
∴e = A −1
a
(1)
A
spacecraft
6578 km
From Equation 3.22,
I
rI = a (1 − e cos E)
projectile
 r


= a 1 − A − 1 cos E 


a


= (1 + cos E) a − rA cos E
r + r cos E
∴a = I A
1 + cos E
(2)
Substitute (2) into (1) to get
e=
rA
rA − rI
−1 =
rI + rA cos E
rA cos E + rI
1 + cos E
(3)
Mean anomaly of the impact point (measured ccw from perigee) is
T

+t
2

&micro;t
t
t
M = 2π
= π + 2π = π + 2π
= π + 3/2
2π 3/2
T
T
a
a
&micro;
Let f (E) = M − E + e sin E . Then Kepler’s equation is f (E) = 0 .
f (E) = M − E + e sin E
f (E) = π +
&micro;t
a
3/2
&micro;t
− E + e sin E = π +
3/2
 ri + ra cos E 
 1 + cos E 
Setting rA = 6578 km , rI = 6378 km , t = 30 ⋅ 60 = 1800 s ,
f (E) = π +
1136 400
 6378 + 6578 cos E 


1 + cos E
3/2
−E +
−E +
ra − ri
sin E
ra cos E + ri
200
sin E
6578 cos E + 6378
Graphing f (E) reveals that f (E) = 0 at E = 5.319 rad . Substituting this into (1) and (2) yields
e = 0.019 75
a = 6451 km
True anomaly of the impact point:
tan
1 + 0.019 75
θI
1+e
Ε
5.319
=
⇒ θ = 303.8&deg;
tan I =
tan
2
1 −e
2
1 − 0.019 75
2
(
)
h = &micro;a (1 − e 2 ) = 398 600 ⋅ 6451 ⋅ 1 − 0.019752 = 50700 km 2 s
100
(123.8&deg; cw from apogee)
Solutions Manual
vA =
vc =
Orbital Mechanics for Engineering Students
Chapter 6
h 50700
=
= 7.707 km s velocity of projectile at apogee.
rA
6578
398 600
&micro;
=
= 7.784 km s
r
6578
velocity of spacecraft in circular orbit.
∆v = v a − vc = 7.707 − 7.784 = −0.07725 km s
Problem 6.9
rapogee = 6378 + 302 = 6680 km
rperigee = 6378 + 296 = 6674 km
rapogee = 6378 + 291 = 6669 km
rperigee = 6378 + 259 = 6637 km
r3 = 6378 + 259 = 6637 km
rapogee = 6378 + 255 = 6633 km
rperigee = 6378 + 194 = 6572 km
1
1
2
2
4
h1 = 2&micro;
4
rapogee rperigee
1
1
rapogee + rperigee
1
vapogee =
1
vperigee =
1
1
1
h1
=
rperigee
1
51 590
= 7.730 km s
6674
rapogee rperigee
2
2
rapogee + rperigee
2
2
vperigee =
2
6680 ⋅ 6674
= 51 590 km 2 s
6680 + 6674
51 590
h1
=
= 7.723 km s
rapogee
6680
h2 = 2&micro;
vapogee =
= 2 ⋅ 398 600
h2
rapogee
2
h1
rperigee
= 2 ⋅ 398 600
2
6669 ⋅ 6637
= 51 500 km 2 s
6669 + 6637
51 500
=
= 7.722 km s
6669
=
2
51 500
= 7.759 km s
6637
398 600
&micro;
=
= 7.750 km s
r3
6637
v3 =
h4 = 2&micro;
rapogee rperigee
4
rapogee + rperigee
4
vapogee =
4
vperigee =
4
4
= 2 ⋅ 398 600
4
6633 ⋅ 6572
= 51 300 km 2 s
6633 + 6572
51 300
h4
=
= 7.734 km s
rapogee
6633
4
h4
rperigee
=
4
51 300
= 7.806 km s
6572
Apogee of orbit 1 to perigee of orbit 2:
h12 = 2&micro;
rapogee rperigee
1
rapogee + rperigee
1
∆v12 =
2
= 2 ⋅ 398 600
2
6680 ⋅ 6637
= 51 520 km 2 s
6680 + 6637
h12
h12
− vapogee +
− vperigee = 7.712 − 7.723 + 7.762 − 7.759 = 0.013 93 km s
1
2
rapogee
rperigee
1
2
Perigee of orbit 2 to orbit 3 (tangent):
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∆v23 = vperigee − v3 = 7.759 − 7.750 = 0.009 313 km s
2
Orbit 3 to perigee of orbit 4:
h34 = 2&micro;
∆v34 =
r3 rperigee
4
r3 + rperigee
= 2 ⋅ 398 600
4
6572 ⋅ 6637
= 51310 km 2 s
6572 + 6637
h34
h34
− v3 +
− vperigee = 7.731 − 7.75 + 7.807 − 7.806 = 0.020 26 km s
4
r3
rperigee
4
∆vtotal = ∆v12 + ∆v23 + ∆v34 = 0.013 93 + 0.009 313 + 0.020 26 = 0.04351 km s
Problem 6.10 rA = 6878 km
v A1 =
rB = 7378 km
398 600
&micro;
=
= 7.613 km s
rA
6878
h2 = 2&micro;
h
v A2 = 2
rA
h
v B2 = 2
rB
rArB
6878 ⋅ 7378
= 2 ⋅ 398 600
= 53 270 km 2 s
rA + rB
6878 + 7378
53 270
=
= 7.745 km s
6878
53 270
=
= 7.220 km s
7378
Alternatively, the energy equation, v 2 2 − &micro; r = − &micro; (2 a ) , implies
v=
 2 1
− &micro;
 r a
so that, since a = (rA + rB ) 2 = 7128 km ,
1 
 2 1
 2
v A2 = 
− &micro; =
−
⋅ 398 600 = 7.745 km s
 6878 7128 
 rA a 
1 
 2 1
 2
v B2 =  −  &micro; =
−
⋅ 398 600 = 7.220 km s

7378 7128 
 rB a 
v B3 =
398 600
&micro;
=
= 7.35 km s
rB
7378
∆v = v A2 − v A1 + vB3 − vB2 = 0.1323 + 0.1300 = 0.2624 km s
Problem 6.11
v A1 =
&micro;
r
r (12r )
= 1.359 &micro;r
r + 12r
h
&micro;
= 2 = 1.359
r
r
h2 = 2&micro;
v A2
102
Chapter 6
Solutions Manual
v B2 =
Orbital Mechanics for Engineering Students
&micro;
h2
= 0.1132
12r
r
Alternatively, using the energy equation,
a2 =
r + 12r
= 6.5r
2
&micro;
&micro;
2 1 
v A2 =  −  &micro; = 1.846 = 1.359
r
r
 r a2 
&micro;
&micro;
1
 2
v B2 = 
−  &micro; = 0.01282 = 0.1132
r
r
 12r a2 
v B3 =
&micro;
&micro;
= 0.2887
r
12r
∆v = v A2 − v A1 + vB3 − vB2 = 0.3587
&micro;
&micro;
&micro;
+ 0.1754
= 0.5342
r
r
r
Problem 6.12
&micro;
2&micro;
&micro;
v A2 =
= 1.414
r
r
r
2&micro;
&micro;
&micro;
&micro;
v B3 =
v B4 =
= 0.4082
= 0.2887
r
r
12r
12r
&micro;
&micro;
&micro;
∆v = v A2 − v A1 + vB4 − vB3 = 0.4142
+ 0.1196
= 0.5338
r
r
r
v A1 =
Problem 6.13 rA = r
v A1 = v1 =
h2 = 2&micro;
rB = 3r
&micro;
&micro;
=
rA
r
rArB
r (3r )
= 1.225 &micro;r
= 2&micro;
rA + rB
r + 3r
1.225 &micro;r
h
&micro;
= 1.225
v A2 = 2 =
rA
r
r
(Alternatively, use the energy equation.)
1.225 &micro;r
h
&micro;
= 0.4082
v B2 = 2 =
rB
3r
r
v B3 = v 3 =
&micro;
&micro;
&micro;
=
= 0.5774
3r
rB
r
∆v = v A2 − v A1 + vB3 − vB2 = 0.2247
Problem 6.14 rA = 6678 km
(a)
Orbit 1:
v A1 =
&micro;
&micro;
&micro;
+ 0.1691
= 0.3938
= 0.3938 v1
r
r
r
rC = 9378 km
398 600
&micro;
=
= 7.726 km s
rA
6678
Orbit 2:
103
Chapter 6
Solutions Manual
Orbital Mechanics for Engineering Students
rB − rA
= e2
rB + rA
rB − 6678
= 0.3 ⇒ rB = 12 402 km
rB + 6678
h2 = 2&micro;
6678 ⋅ 12 402
rArB
= 2 ⋅ 398 600
= 58 830 km 2 s
rA + rB
6678 + 12 402
58 830
h
v A2 = 2 =
= 8.809 km s
rA
6678
58 830
h
v B2 = 2 =
= 4.743 km s
rB 12 402
Orbit 3:
h3 = 2&micro;
12 402 ⋅ 9378
rBrC
= 2 ⋅ 398 600
= 65 250 km 2 s
rB + rC
12 402 + 9378
65 250
h
v B3 = 3 =
= 5.261 km s
rB 12 402
65 250
h
vC 3 = 3 =
= 6.957 km s
rC
9378
Orbit 4:
398 600
&micro;
=
= 6.519 km s
vC 4 =
9 378
rC
∆vtotal = v A2 − v A1 + vB3 − vB2 + vC 4 − vC 3 = 1.083 + 0.5177 + 0.4379 = 2.039 km s
(b)
2π  rA + rB 
&micro; 2 
3/2
T2 =
2π  rB + rC 
&micro; 2 
3/2
T3 =
ttotal =
 6678 + 12 402 



2
398 600 
3/2
=
 12 402 + 9378 



2
398 600 
3/2
=
2π
2π
= 9273 s
= 11 310 s
1
(T + T2 ) = 10 290 s = 2.859 hr
2 1
Problem 6.15 rA = rC = rE = 15 000 km
(a)
Orbit 1:
v A )1 =
rB = 22 000 km
rD = 6878 km
398 600
&micro;
=
= 5.155 km s
rA
15000
γ A1 =
Orbit 2:
22 000 − 6878
r −r
= 0.5237
e2 = B D =
rB + rD 22 000 + 6878
h2 = 2&micro;
22 000 ⋅ 6878
rBrD
= 2 ⋅ 398 600
= 64 630 km 2 s
22 000 + 6878
rB + rD
At the maneuvering point A:
104
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Chapter 6
1
h2
rA = 2
&micro; 1 + e2 cos θ A
64 630 2
1
⇒ θ A = 125.1&deg;
398 600 1 + 0.5237 cos θ A
64 630
h
v A2 ) = 2 =
= 4.309 km s
⊥
rA 15000
398 600
&micro;
v A2 ) =
e sin θ A =
0.5237 sin 125.1&deg; = 2.641 km s
r
h2 2
64 630
15 000 =
v A2 = v A2 )
γ A2 = tan −1
2
⊥
+ v A2 )
v A2 )
v A2 )
r
2
r
= 4.309 2 + 2.6412 = 5.054 km s
= tan −1
⊥
2.641
= 0.5499 ⇒ γ A2 = 31.51&deg;
4.309
∆γ A = γ A2 − γ A1 = 31.51&deg; − 0 = 31.51&deg;
∆v A = v A12 + v A2 2 − 2 v A1 v A2 cos ∆γ A = 5.1552 + 5.054 2 − 25.1555.054 cos ∆γ A = 2.773 km s T
(b)
Try Hohmann transfer (orbit 3) from point E on orbit 1 to point B on orbit 2.
h3 = 2&micro;
15 000 ⋅ 22 000
rErB
= 2 ⋅ 398 600
= 84 320 km 2 s
rE + rB
15 000 + 22 000
vE1 = v A1 = 5.155 km s
h3 84 320
=
= 5.621 km s
rE 15000
84 320
h
= 3 =
= 3.833 km s
rB 22 000
64 630
h
= 2 =
= 2.938 km s
rB 22 000
v E3 =
v B3
v B2
∆vtotal = vE3 − vE1 + vB2 − vB3 = 0.4665 + 0.985 = 1.362 km s
Try Hohmann transfer (orbit 4) from point C on orbit 1 to point D on orbit 2.
h4 = 2&micro;
15 000 ⋅ 6878
rC rD
= 2 ⋅ 398 600
= 61 310 km 2 s
rC + rD
15 000 + 6878
vC1 = v A1 = 5.155 km s
61 310
h
vC 4 = 4 =
= 4.088 km s
rC 15 000
61 310
h
vD4 = 4 =
= 8.914 km s
rD
6878
64 630
h
vD4 = 2 =
= 9.397 km s
rD
6878
∆vtotal = vC 4 − vC1 + vD2 − vD4 = 1.067 + 0.4824 = 1.55 km s
This is larger than the total computed above; thus for minimum Hohmann transfer
∆v = 1.362 km s
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Chapter 6
Problem 6.16
Orbit 1:
rperigee = 6378 + 1270 = 7648 km
1
vperigee = 9 km s
1
h1 = rperigee vperigee = 7648 ⋅ 9 = 68 832 km 2 s
1
1
2
rperigee =
1
7648 =
h1
1
&micro; 1 + e1
68 832 2 1
⇒ e1 = 0.5542
398 600 1 + e1
At the maneuver point, θ = 100&deg; .
68 832 2
h12
1
1
=
= 13 150 km
&micro; 1 + e1 cos θ 398 600 1 + 0.5542 cos 100&deg;
68 832
h
v1 ⊥ = 1 =
= 5.234 km s
r
13150
r=
v1 r =
398 600
&micro;
e sin θ =
0.5542 sin 100&deg; = 3.16 km s
h1 1
68 832
v1 = v1 ⊥ 2 + v1 r 2 = 5.234 2 + 3.16 2 = 6.114 km s
v1
3.16
= 31.13
γ 1 = tan −1 r = tan −1
v1 ⊥
5.234
Orbit 2:
e2 = 0.4
h2
1
r= 2
&micro; 1 + e2 cos θ
h2 2
1
⇒ h2 = 69 840 km 2 s
398 600 1 + 0.4 cos 100&deg;
69 840
h
v2 ⊥ = 2 =
= 5.311 km s
13150
r
13150 =
v2 r =
398 600
&micro;
0.4 sin 100&deg; = 2.248 km s
e2 sin θ =
69 840
h2
v2 = v2 ⊥ 2 + v2 r 2 = 5.3112 + 2.248 2 = 5.767 km s
v2
2.248
= 22.94&deg;
γ 2 = tan −1 r = tan −1
5.767
v2 ⊥
∆γ = γ 2 − γ 1 = 22.94&deg; − 31.13&deg; = −8.181&deg;
∆v = v12 + v2 2 − 2 v1 v2 cos ∆γ = 6.114 2 + 5.767 2 − 2 ⋅ 6.114 ⋅ 5.767 cos( −8.181) = 0.9155 km s
Problem 6.17
rA = 12 756 km
v A = 6.5992 km
γ A = 20&deg;
Orbit 1:
v A ⊥ = v A cos γ A = 6.5992 cos 20&deg; = 6.20122 km s
∴ h1 = rA v A ⊥ = 12756 ⋅ 6.20122 = 79102.8 km 2 s
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Solutions Manual
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v A r = v A sin γ A = 6.5992 ⋅ sin 20&deg; = 2.25706 km s
&micro;
e sin θ A
h1 1
398 600
2.25706 =
e sin θ A ⇒ e1 sin θ A = 0.447 917
79 102.8 1
vAr =
2
B
79 102.8 2
1
⇒ e1 cos θ A = 0.230 641
398 600 1 + e1 cos θ A
(
)
∴ e12 sin 2 θ A + cos2 θ A = 0.447 917 2 + 0.230 6412
e12 = 0.253 825 ⇒ e1 = 0.503 81
∴ sin θ A =
0.447 917
= 0.889 058 ⇒ θ A = 62.7552&deg; or θ A = 117.235&deg;
0.503 81
Since cos θ A &gt; 0 , θ A = 62.755 2&deg; .
79 102.8 2
h2
1
1
rB = 1
=
= 27 848.9 km
&micro; 1 + e1 cos θ B1
398 600 1 + 0.503 81 cos 150&deg;
79 102.8
h
= 2.840 43 km s
vB ⊥ ) = 1 =
1
rB 27 848.9
398 600
&micro;
⋅ 0.503 81 ⋅ sin 150&deg; = 1.269 45 km s
vB r ) =
e sin θ B1 =
1
h1 1
79 102.8
Orbit 2:
∆vB ⊥ = 0.758 20 km s
∴ vB ⊥ ) = vB ⊥ ) + ∆vB ⊥ = 2.840 43 + 0.758 20 = 3.598 63 km s
2
1
h2 = rB vB ⊥ ) = 27 848.9 ⋅ 3.598 63 = 100 218 km 2 s
2
∆vB r = 0
∴ vB r ) = vB r ) + ∆vB ⊥ = 1.269 45 + 0 = 1.269 45 km s
2
1
&micro;
e sin θ B2
h2 2
398 600
e sin θ B2 ⇒ e2 sin θ B2 = 0.319 146
1.269 45 =
100 218 2
vB r ) =
2
rB =
27 848.9 =
A
150&deg;
1
1
h2
rA = 1
&micro; 1 + e1 cos θ A
12 756 =
Chapter 6
h2 2
1
&micro; 1 + e2 cos θ B2
100 218 2
1
⇒ e2 cos θ B2 = −0.095216 6
398 600 1 + e2 cos θ B2
107
Solutions Manual
(
Orbital Mechanics for Engineering Students
)
2
e2 2 sin 2 θ B2 + cos2 θ B2 = 0.319146 2 + (−0.095216 6)
e2 2 = 0.110 921
e2 = 0.333 048
0.319146
= 0.958 261 ⇒ θ B2 = 73.387 7&deg; or 106.612&deg;
0.333 048
&lt; 0 , θ B2 = 106.612&deg; .
∴ sin θ B2 =
Since cos θ B2
∆θ = 150 − 106.612&deg; = 43.3877&deg;
That is, the apse line is rotated 43.387 7&deg; ccw from that of orbit 1.
Problem 6.18
r1 = r2
h2
h2
1
1
=
&micro; 1 + e cos θ
&micro; 1 + e cos(η − θ )
cos θ = cos(η − θ )
∴ θ = η − θ ⇒ 2θ = η ⇒ θ =
η
2
Problem 6.19
Orbit 1:
rP1 =
h12 1
&micro; 1+e
Orbit 2:
rP1 =
∴
h2 2
1
h2
= 2
&micro; 1 + e cos 90&deg;
&micro;
h2 2 h12 1
h1
=
⇒ h2 =
&micro;
&micro; 1+e
1+e
Problem 6.20
At A:
r=
h2
&micro;
&micro;
&micro;
e sin 90&deg; = e
h
h
&micro;
&micro;
v r 2 = e sin ( −90&deg;) = − e
h
h
h
v ⊥1 = v ⊥ 2 =
r
∴ ∆v ⊥ = 0
&micro;
∆v r = v r 2 − v r1 = −2 e
h
v r1 =
108
Chapter 6
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 6
2&micro;e
∴ ∆v = ∆v r =
h
Problem 6.21 For the circular orbit of the space station,
r = 6728 km
vc =
398 600
&micro;
=
= 7.697 km s
r
6728
Tc =
2π
&micro;
r 3/2 =
2π
398 600
6728 3/2 = 5492 s = 91.54 m
(a) The time required for spacecraft A to reach the space station is the time it takes for the space
station to fly around to the original position of spacecraft A.
tSA = Tc
2π ⋅ 6728 − 600
2πr − 600
= 5492
= 5414 s = 90.2 min
2πr
2π ⋅ 6728
The time required for spacecraft B to reach the space station is the time it takes for the space station to
fly around to the original position of spacecraft B.
tBS = Tc
2π ⋅ 6728 + 600
2πr + 600
= 5492
= 5570 s = 92.8 min
2πr
2π ⋅ 6728
(b)
The period of spacecraft A’s phasing orbit, is tSA , which determines the semimajor axis of that orbit:
5414 =
2π
398 600
a A3/2 ⇒ a A = 6664 km
Spacecraft A is at the apogee of its phasing orbit. From the energy equation
1 
1 
 2
2
v A = &micro; −
−
= 7.660 km s
 = 398 600
6728 6664 
 r aA 
The delta-v required to drop into the phasing orbit is
∆v A = v A − vc = 7.660 − 7.697 = −0.036 94 km s
Spacecraft A must therefore slow down in order to speed up (i.e., catch the space station). After one
circuit of its phasing orbit, this delta-v must be added in order to rejoin the circular orbit. Thus
∆v Atotal = 2 ∆v A = 0.073 88 km s
orbit:
The period of spacecraft B’s phasing orbit, is tBS , which determines the semimajor axis of that
5570 =
2π
398 600
aB3/2 ⇒ aB = 6791 km
Spacecraft B is at the perigee of its phasing orbit. From the energy equation
1 
2 1 
 2
vB = &micro; −  = 398 600
−
= 7.733 km s
 6728 6791 
 r aB 
The prograde delta-v required to enter the phasing orbit is
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Chapter 6
∆vB = vB − vc = 7.733 − 7.697 = +0.03576 km s
Spacecraft B must therefore speed up in order to slow down (i.e, allow the space station to catch up).
After one circuit of its phasing orbit, this delta-v must be subtracted in order to rejoin the circular
orbit. Thus
∆vBtotal = 2 ∆vB = 0.071 53 km s
Problem 6.22
T
2
2π 3/2 1 2π 3/2
a
=
r
2 &micro;
&micro;
1
a 3/2 = r 3/2
2
Tphasing =
a=
Problem 6.23
 1 3/2 
r
2

2/3
⇒ a = 0.63r
rA = 13 000 km
rP = 8000 km
Orbit 1:
rA + rP
= 10 500 km
2
r −r
e1 = A P = 0.2381
rA + rP
a1 =
h1 = &micro;(1 + e )rP = 398 600(1 + 0.2381) ⋅ 8000 = 62 830 km 2 s
T1 =
2π
&micro;
a 3/2 =
2π
398 600
10 500 3/2 = 10 710 s
Time of flight from P to C:
 1 − e1
 1 − 0.2381
θ 
30&deg; 
EC = tan −1 
tan C  = tan −1 
tan
 1 + 0.2381
2 
2 
 1 + e1
MC = EC − e1 sin EC = 0.4144 − 0.2381 sin 0.4144 = 0.3185 rad
0.3185
M
⋅ 10 710 = 542.8 s
tC = C T =
2π
2π
Time of flight from P to D:
 1 − e1
 1 − 0.2381
θ 
90&deg; 
ED = tan −1 
tan D  = tan −1 
tan

2 
1 + 0.2381
2 
 1 + e1
MD = ED − e1 sin Ed = 1.330 − 0.2381 sin 1.330 = 1.099 rad
1.099
M
tD = D T =
⋅ 10 710 = 1873 s
2π
2π
Time of flight from C to D:
tCD = tD − tC = 1873 − 542.8 = 1330 s
To determine the trajectory from P to D is Lambert’s problem. Note that
rD =
62 830 2
1
1
h12
=
= 9905 km
&micro; 1 + e1 cos θ D 398 600 1 + 0.2381 cos 90&deg;
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Solutions Manual
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Chapter 6
so that in perifocal coordinates
rP = 8000pˆ km
rD = 9905qˆ km
Note as well, that on orbit 1,
v P1 =
398 600
&micro;
 − sin θ P pˆ + ( e + cos θ P ) qˆ  =
 − sin 0pˆ + ( 0.2381 + cos 0 ) qˆ 
h1 
62 830 
= 7.854qˆ ( km s )
v D11 =
398 600
&micro;
 − sin θ Dpˆ + ( e + cos θ D ) qˆ  =
 − sin 90&deg;pˆ + ( 0.2381 + cos 90&deg; ) qˆ 
h1 
62 830 
= −6.344pˆ + 1.510q̂ ( km s )
The following MATLAB script calls upon Algorithm 5.2, implemented as the M-function lambert in
Appendix D.11, to solve Lamberts’ problem for the velocities on orbit 2 at P and D. The output to the
MATLAB Command Window is listed afterwards.
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
% Problem 6_23
% ~~~~~~~~~~~~
%
% This program uses Algorithm 5.2 to solve Lambert's problem for the
% data of in Problem 6.23.
%
%
%
%
%
%
%
%
%
%
deg
pi
mu
r1, r2
dt
dtheta
R1, R2
string
-
factor for converting between degrees and radians
3.1415926...
gravitational parameter (km^3/s^2)
time between r1 and r2 (s)
change in true anomaly during dt (degrees)
initial and final position vectors (km)
= 'pro' if the orbit is prograde
= 'retro if the orbit is retrograde
V1, V2 - initial and final velocity vectors (km/s)
% User M-function required: lambert
% ----------------------------------------------------------clear
global mu
mu
= 398600;
%km^3/s^2
deg
= pi/180;
r1
r2
dt
dtheta
=
=
=
=
8000;
9905;
1330;
90;
%km
%km
%sec
%degrees
R1 = [r1 0 0];
R2 = [r2*cos(dtheta*deg)
r2*sin(dtheta*deg)
0];
%...Algorithm 5.2:
string = 'pro';
[V1 V2] = lambert(R1, R2, dt, string);
%...Echo the input data and output results to the command window:
fprintf('\n-----------------------------------------------------')
fprintf('\n Problem 6.23: Lambert''s Problem\n')
fprintf('\n Input data:\n');
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Solutions Manual
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
Orbital Mechanics for Engineering Students
Chapter 6
Gravitational parameter (km^3/s^2) = %g\n', mu)
= %g', r1)
Position vector R1 (km)
= [%g %g %g]\n',...
R1(1), R1(2), R1(3))
= %g', r2)
Position vector R2 (km)
= [%g %g %g]\n',...
R2(1), R2(2), R2(3))
Elapsed time (s)
= %g', dt)
Change in true anomaly (deg) = %g', dtheta)
fprintf('\n\n Solution:\n')
fprintf('\n Velocity vector V1 (km/s) = [%g %g %g]',...
V1(1), V1(2), V1(3))
fprintf('\n Velocity vector V2 (km/s) = [%g %g %g]',...
V2(1), V2(2), V2(3))
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 6.23: Lambert's Problem
Input data:
Gravitational parameter (km^3/s^2) = 398600
Position vector R1 (km)
= 8000
= [8000
Position vector R2 (km)
= 9905
= [-0 9905
0
0]
0]
Elapsed time (s)
= 1330
Change in true anomaly (deg) = 90
Solution:
Velocity vector V1 (km/s) = [-2.53168 9.57638 0]
Velocity vector V2 (km/s) = [-7.73458 4.37347 0]
-----------------------------------------------------
v P 2 = −2.532pˆ + 9.576qˆ ( km s )
v D 2 = −7.734pˆ + 4.373qˆ ( km s )
∆v P = v P 2 − v P1 = ( −2.532pˆ + 9.576qˆ ) − 7.854qˆ = −2.532pˆ + 1.722qˆ ( km s )
2
∆v P = ( −2.532 ) + 1.722 2 = 3.062 ( km s )
∆v D = v D 2 − v D1 = ( −7.734pˆ + 4.373qˆ ) − ( −6.344pˆ + 1.510qˆ ) = −1.391pˆ + 2.863qˆ ( km s )
2
∆v D = ( −1.391) + 2.863 2 = 3.183 ( km s )
∆vtotal = ∆v P + ∆v D = 3.062 + 3.183 = 6.245 km s
Problem 6.24
h = &micro;a (1 − e 2 ) = 398 600 ⋅ 15 000 ⋅ (1 − 0.52 ) = 66 960 km 2 s
rascending node =
66 960 2
h2
1
1
=
= 7851 km
&micro; 1 + e cos( −ω ) 398 600 1 + 0.5 cos( −30&deg;)
112
Solutions Manual
Orbital Mechanics for Engineering Students
rdescending node =
Chapter 6
66 960 2
h2
1
1
=
= 19 840 km
&micro; 1 + e cos( −ω + π ) 398 600 1 + 0.5 cos( −30&deg; + 180&deg;)
Rotate the orbital plane 10 degrees around the node line. That means hold v r fixed and rotate v ⊥ 10
degrees. For minimum delta-v, do this maneuver at the furthest distance from the focus (at the
descending node, rather than the ascending node).
v⊥ =
h
rdescending node
∆v = 2 v ⊥ sin
=
66 960
= 3.375 km s
19 840
∆i
10&deg;
= 2 ⋅ 3.375 ⋅ sin
= 0.5883 km s
2
2
(Note: if the maneuver is done at the ascending node,
v⊥ =
h
rascending node
∆v = 2 v ⊥ sin
=
66 960
= 8.53 km s
7851
10&deg;
∆i
= 2 ⋅ 8.53 ⋅ sin
= 1.487 km s
2
2
Over twice the delta-v requirement.)
Problem 6.25 For the circular orbit
v1 =
398 600
&micro;
=
= 7.668 km s
6778
r
Assume the maneuver is done at apogee of the ellipse (orbit 2).
h2 1
r= 2
&micro; 1 − e2
6778 =
1
h2 2
⇒ h2 = 36 750 km 2 s
398 600 1 − 0.5
Then
rperigee =
36 750 2
h2 2 1
1
=
= 2259 km
&micro; 1 + e2 398 600 1 + 0.5
which is inside the earth. So the maneuver cannot occur at apogee. Assume it occurs at perigee.
h2 1
r= 2
&micro; 1 + e2
6778 =
1
h2 2
⇒ h2 = 63 660 km 2 s
398 600 1 + 0.5
63 660
h
= 9.392 km s
v2 = 2 =
6778
r
∆v = v12 + v2 2 − 2 v1 v2 cos δ = 7.668 2 + 9.392 2 − 2 ⋅ 7.668 ⋅ 9.392 cos δ = 3.414 km s
113
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 6
Problem 6.26
60 000 2 1
h2 1
=
= 12 900 km
&micro; 1 − e 398 600 1 − 0.3
60 000
h
vapogee =
=
= 4.65 km s
rapogee 12 900
rapogee =
∆v = 2 vapogee sin
δ
90&deg;
= 2 ⋅ 4.65 sin
= 6.577 km s
2
2
Problem 6.27
&micro;
ro
vB1 =
h2
h2
h2
1
1
ro = 2
= 2
= 2 ⇒ h2 = &micro;ro
&micro; 1 + e cos θ
&micro; 1 + 0.25 cos( −90&deg;)
&micro;
vB ⊥ =
&micro;ro
&micro;
h2
=
=
ro
ro
ro
vB r =
&micro;
&micro;
&micro;
e sin θ =
⋅ 0.25 ⋅ sin ( −90&deg;) = −0.25
h2 2
r
&micro;ro
o
2
2
∆v =
(vB r2 − vB r1 )2 + vB ⊥12 + vB ⊥ 22 − 2 vB ⊥1vB ⊥ 2 cos δ
2


&micro;
&micro;
∆v =  −0.25
− 0 +
r
r


o
o
∆v = 0.0625
&micro; &micro; &micro;
+ + −0
ro ro ro
∆v = 2.0625
&micro;
ro
∆v = 1.436
2
&micro;
+
ro
2
−2
&micro;
ro
&micro;
cos( −90&deg;)
ro
&micro;
ro
Problem 6.28 The initial and target orbits are “1” and “2”, respectively, and “3” is the transfer
orbit.
r1 = 6678 km
398 600
&micro;
=
= 7.726 km s
6678
r1
r2 = 6978 km
v1 =
398 600
&micro;
=
= 7.558 km s
6978
r2
6678 + 6978
r +r
= 6828 km
a3 = 1 2 =
2
2
1 
2 1 
 2
−
= 7.810 km s
vperigee = &micro; −  = 398 600


3
6678
6828
r
a
 1
3
v2 =
1
1 
2
 2
vapogee = &micro; −  = 398 600
−
= 7.474 km s

3
6978 6828 
 r2 a3 
114
Solutions Manual
Orbital Mechanics for Engineering Students
(a)
(
) (
∆v = vperigee − v1 + v2 − vapogee
3
∆i
3
) + 2 ⋅ v2 sin 2
= (7.810 − 7.726) + (7.558 − 7.474) + 2 ⋅ 7.558 sin
20&deg;
2
= 0.0844 + 0.083 48 + 2.625 = 2.793 km s
(b)
(
)
∆v = vperigee − v1 + vapogee
3
2
3
+ v2 2 − 2 vapogee v2 cos ∆i
3
= (7.810 − 7.726) + 7.474 2 + 7.588 2 − 2 ⋅ 7.474 ⋅ 7.558 cos 20&deg;
= 0.0844 + 2.612 = 2.696 km s
(c)
∆v = vperigee
2
3
+ v12 − 2 vperigee v1 cos ∆i + v2 − vapogee
3
(
3
)
= 7.812 + 7.726 2 − 2 ⋅ 7.81 ⋅ 7.726 cos 20&deg; + (7.558 − 7.474)
= 2.699 + 0.083 48 = 2.783 km s
Problem 6.29 Design problem.
Problem 6.30
 cos i 
A = sin −1 

 cos φ 
(a)
A = sin −1
 cos 116.57&deg; 
= sin −1 ( −0.5088) = 329.4&deg;
 cos 28.5&deg; 
(b)
A = sin −1
 cos 116.57&deg; 
= sin −1 ( −0.5427 ) = 327.1&deg;
 cos 34.5&deg; 
(c)
A = sin −1
 cos 116.57&deg; 
= sin −1 ( −0.4493) = 333.3&deg;
 cos 34.5&deg; 
115
Chapter 6
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 7
Problem 7.1
ˆ
iˆ = 0Iˆ + Jˆ + 0K
ˆj = 0Iˆ + 0 Jˆ + K
ˆ
ˆ
kˆ = Iˆ + 0 Jˆ + 0K
0 1 0 


∴ [Q ]Xx = 0 0 1 
1 0 0 
ˆ = 6628K
ˆ ( km )
rB = ( 6378 + 250 ) K
398 600 ˆ
&micro;ˆ
J=−
J = −7.754 92 Jˆ ( km s )
rB
6628
vB = −
vB 2 ˆ
7.755 2 ˆ
ˆ km 2 s
K = −0.009 073 45K
K=−
rB
6628
(
aB = −
)
ˆ = 6678 Jˆ ( km )
rA = ( 6378 + 300 ) K
vA =
398 600 ˆ
&micro; ˆ
ˆ ( km s )
K=−
K = 7.725 84K
rA
6678
aA = −
vA 2 ˆ
7.755 2 ˆ
J = −0.008 938 08 Jˆ km 2 s
J=−
6628
rA
(
)
v
7.726 ˆ
Ωxyz = A Iˆ =
I = 0.001 156 91Iˆ ( rad s )
rA
6678
Ωxyz = 0
ˆ ( km )
rrel = rB − rA = −6678 Jˆ + 6628K
{rrel }xyz = [Q]Xx {rrel }XYZ
0 1 0   0  −6678 
 



= 0 0 1  −6678  =  6628 
1 0 0   6628   0 
rrel = −6678iˆ + 6628 ˆj ( km )
v B = v A + Ωxyz &times; rrel + v rel
ˆ + 0.0001 156 91Iˆ &times; −6678 Jˆ + 6628K
ˆ +v
−7.754 92 Jˆ = 7.725 84K
rel
ˆ
ˆ
ˆ
ˆ
−7.754 92 J = 7.725 84K + −7.667 99 J − 7.725 84K + v
) (
(
(
)
)
rel
v reel = −0.086 931 6 Ĵ ( km s )
0
 −0.086 931 6 
0 1 0  



0
{v rel }xyz = [Q]Xx {v rel }XYZ = 0 0 1  −0.086 931 6  = 

 

1 0 0  
0
0
 

ˆ
v rel = −0.086 931 6i ( km s )
(
)
aB = a A + Ωxyz &times; rrel + Ωxyz &times; Ωxyz &times; rrel + 2Ωxyz &times; v rel + a rel
(
) (
) (
)
) (
)
ˆ = −0.008 983 08 Jˆ + 0 + 0.001 156 91Iˆ &times;  0.001 156 91Iˆ &times; −6678 Jˆ + 6628K
ˆ 
−0.009 073 45K


ˆ


+ 2  0.001 156 91I &times; −0.086 931 6 Ĵ  + a rel
(
116
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 7
ˆ = −8.938 08 (10−3 ) Jˆ + 0 + 1.156 91 (10−3 ) Iˆ  &times; −7.667 99 Jˆ − 7.725 84K
ˆ
−0.009 073 45K


(
)
ˆ +a
+ 2  −0.000 100 572K
 reel
ˆ = −8.938 08 (10−3 ) Jˆ + 0 + 0.008 938 08 Jˆ − 0.008 87116K
ˆ
−0.009 073 45K
(
(
)
)
ˆ +a
+ 2 −0.000100 572K
rel
ˆ = −0.009 072 31K
ˆ +a
−0.009 073 45K
rel
(
ˆ km s2
a rel = −1.140 18 &times; 10−6 K
)
0

0
 
0 1 0  


 
−6 
0
{arel }xyz = [Q]Xx {arel }XYZ = 0 0 1  
 = −1.140 18 &times; 10 


1 0 0  −1.140 18 &times; 10−6  
0



a rel = −1.140 18 &times; 10−6 ˆj km s2
(
)
Problem 7.2
vA =
398 600 ˆ
&micro;ˆ
j=−
j = 7.058 68 ˆj ( km s )
8000
rA
vB =
398 600 ˆ
&micro;ˆ
j=−
j = 7.546 05 ˆj ( km s )
rB
7000
v
7.058 68 ˆ
k = 0.000 882 335kˆ ( rad s )
Ωxyz = A kˆ =
8000
rA
Ωxyz = 0
rrel = rB − rA = −1000iˆ ( km )
v B = v A + Ωxyz &times; rrel + v rel
(
) (
)
7.546 05 ˆj = 7.058 68 ˆj + 0.000 882 335kˆ &times; −1000iˆ + v rel
7.546 05 ˆj = 7.058 68 ˆj − 0.882 335 ˆj + v
rel
v rel = 1.369 70 ˆj ( km s )
aA = −
vA 2 ˆ
7.058 68 2 ˆ
i=−
i = −0.006 228 12iˆ km
m s2
8000
rA
aB = −
vB 2 ˆ
7.546 05 2 ˆ
i=−
i = −0.008 134 69iˆ km s2
7000
rB
(
(
(
)
)
)
aB = a A + Ωxyz &times; rrel + Ωxyz &times; Ωxyz &times; rrel + 2Ωxyz &times; v rel + a rel
(
) (
)
−0.008 134 69iˆ = −0.006 228 12iˆ + 0 + 0.000 882 335kˆ &times;  0.000 882 335kˆ &times; ( −1000iˆ ) 
+ 2 0.000 882 335kˆ &times; 1.369 70 ĵ + a
(
) (
)
117
rel
Solutions Manual
Orbital Mechanics for Engineering Students
(
) (
−0.008 134 69iˆ = −0.006 228 12iˆ + 0.000 882 335kˆ &times; −0.882 335 ˆj
(
)
)
+ 2 −0.001 208 54 î + a rel
0.008 134 69iˆ = −0.006 228 12iˆ + 0.000 778 516kˆ
− 0.002 417 07iˆ + a
rell
−0.008 134 69iˆ = −0.007 866 68iˆ + a rel
(
a rel = −0.000 268 012iˆ km s2
)
Problem 7.3 r = ro + δr
(a)
1/2
r = [( ro + δr ) ⋅ ( ro + δr ) ]
1/2
= ( ro ⋅ ro + 2 ro ⋅ δr + δr ⋅ δr )
)1/2
(
= ro2 + 2 ro ⋅ δr + δr 2
1/2
 2 r ⋅ δr  δr  2 
= ro 1 + o 2 +   
 ro  
ro

Keep only terms of the first order in δr :
2 r ⋅ δr 

r = ro  1 + o 2 
ro 

1/2
2 r ⋅ δr 

∴ r = ro  1 + o 2 
ro 

1/4
By means of the binomial theorem,
1 2 ro ⋅ δr 
1 ro ⋅ δr

r = ro  1 +
= ro +

2
4 ro 
2 ro3/2

1 ro ⋅ δr
∴ O(δr ) =
2 ro3/2
(b)
ro = 3iˆ + 4 ˆj + 5kˆ
δ r = 0.01iˆ − 0.01ˆj + 0.03kˆ
1/2
ro = (3 2 + 4 2 + 52 )
(δ r = 0.033 166 2)
= 7.071 07
ro = 2.659 15
O (δ r ) =
(
)(
)
1 ro ⋅ δ r 1 3iˆ + 4 ˆj + 5kˆ ⋅ 0.01iˆ − 0.01ˆj + 0.03kˆ
1 0.140 000
= 0.003 722 81
=
=
3/2
2 ro3/2
2
2 18.803 00
7.071 07
r = 3.01iˆ + 3.99 ˆj + 5.03kˆ
1/2
r = (3.012 + 3.99 2 + 5.03 2 )
= 7.090 92
r = 2.662 88
r − ro = 0.003 729 58
118
Chapter 7
Solutions Manual
(
Orbital Mechanics for Engineering Students
)
r − ro − O(δr )
r − ro
⋅ 100 =
Thus, O(δr ) is within 0.2% of
(c) δ r = iˆ − ˆj + 3kˆ
O (δ r ) =
0.003 729 58 − 0.003 722 81
⋅ 100 = 0.181 565%
0.003 729 58
r − ro ,
(δ r = 3.316 62,
δ r ro = 4.69 &times; 10−1
(
)(
)
)
k
1 ro ⋅ δ r 1 3iˆ + 4 ˆj + 5kˆ ⋅ iˆ − ˆj + 3k̂
1
14
=
=
= 0.372 281
3
/
2
3
/
2
2 r
2
2 18.803 00
7.071 07
o
r = ro + δ r = 4 iˆ + 3 ˆj + 8kˆ
1/2
r = (4 2 + 3 2 + 8 2 )
= 9.433 98
r = 3.071 48
r − ro = 0.412 331
(
)
r − ro − O(δr )
r − ro
⋅ 100 =
0.412 331 − 0.372 281
⋅ 100 = 9.713 08%
0.412 331
In this case O(δr ) is a poor approximation, exceeding
Problem 7.4 For e &lt;&lt; 1 :
r=
a (1 − e 2 )
1 + e cos θ
≈ a (1 − e 2 )(1 − e cos θ )
= a (1 − e cos θ ) − ae 2 + ae 3 cos θ
≈ a (1 − e cos θ )
Problem 7.5 x + 9x = 10
10
9
x = 3 A cos 3t − 3B sin 3t
x = A sin 3t + B cos 3t +
At t = 0 , x=5:
5 = A sin (0) + B cos(0) +
5 = B+
10
9
10
9
35
9
At t = 0 , x = −3 :
∴B =
−3 = 3 A cos(0) − 3 ⋅
35
sin (0)
9
−3 = 3 A
A = −1
119
r − ro by almost 10 percent.
Chapter 7
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 7
Thus
35
10
cos 3t +
9
9
35
x = −3 cos 3t −
sin 3t
3
x = − sin 3t +
At t = 1.2 :
35
10
cos ( 3 ⋅ 1.2 ) +
= −1.934
9
9
35
x = −3 cos ( 3 ⋅ 1.2 ) −
sin ( 3 ⋅ 1.2 ) = 7.853
3
x = − sin ( 3 ⋅ 1.2 ) +
Problem 7.6
x + 10 x + 2 y = 0
y + 3x = 0
(1)
(2)
Initial conditions (at t = 0 ):
x0 = 1
(3)
x0 = −3
y0 = 2
y0 = 4
(4)
(5)
(6)
From (2)
d
( y + 3x ) = 0
dt
y + 3 x = const
y + 3 x = y0 + 3 x0 = 4 + 3 ⋅ 1 = 7
y = 7 − 3x
(7)
Substitute (7) into (1):
x + 10 x + 2 ( 7 − 3 x ) = 0
x + 4 x = −14
General solution:
7
2
x = 2 A cos 2t − 2B sin 2t
(8)
x = A sin 2t + B cos 2t −
(9)
Evaluating (8) at t = 0 and using (3),
1 = A sin (0) + B cos(0) −
1 =B−
B=
9
2
7
2
7
2
(10)
Evaluating (9) at t = 0 and using (4),
120
Solutions Manual
Orbital Mechanics for Engineering Students
−3 = 2 A cos(0) − 2B sin (0)
3
A=−
2
Chapter 7
(11)
With (10) and (11), (8) becomes
3
9
7
x = − sin 2t + cos 2t −
2
2
2
(12)
Substituting (12) into (7) yields
9
 3
y = 7 − 3  − sin 2t + cos 2t −
 2
2
9
27
35
y = sin 2t −
cos 2t +
2
2
2
7

2
(13)
Then
9
27
35
y = − cos 2t −
sin 2t +
t +C
4
4
2
(14)
Evaluating (14) at t = 0 and using (5), we get
9
27
35
(0) + C
2 = − cos(0) −
sin (0) +
4
4
2
2=−
C=
9
+C
4
17
4
(15)
Substituting this into (14) yields
9
27
35
17
y = − cos 2t −
sin 2t +
t+
4
4
2
4
(16)
At t = 5 , (12) and (14) yield, respectively,
3
9
7
x = − sin (2 ⋅ 5) + cos(2 ⋅ 5) − = −6.460
2
2
2
9
27
35
17
y = − cos(2 ⋅ 5) −
sin (2 ⋅ 5) +
⋅5 +
= 97.31
4
4
2
4
Problem 7.7
2π
= 0.001163 6 s -1
90 ⋅ 60
t = 15 ⋅ 60 = 900 s
n=
[Φrr (t)]
 4 − 3 cos nt 0
0   2.5
0 0 

 

= 6(sin nt − nt ) 1
0  =  −1.087 1 0 

0
0 cos nt   0
0 0.5
121
Solutions Manual
[Φrv (t)]
Orbital Mechanics for Engineering Students
 1
 n sin nt
2
=  (cos nt − 1)
n

0

1 
{δr0 } = 0  ( km )
0 
 
2
(1 − cos nt )
n

 

  744.29 859.44 0 
 =  −859.44
1
0 
0
 
0
0
0
.5
1

sin nt 

n
0
1
(4 sin nt − 3 nt )
n
0
 0 
{δv 0 } = 0.01 (km s)
 0 


{δr } = [Φrr (t ) ]{δr0 } + [Φrv (t ) ]{δv 0 }
{δr } = [Φrr (t ) ]{δr0 } + [Φrv (t ) ]{δv 0 }
 2.5
0 0  1   744.29 859.44 0   0 


  

1
0  0.01
=  −1.087 1 0  0  +  −859.44
 0
0 0.5 0   0
0
0.5  0 
 2.5  8.5944 


 
=  −1.087  + 2.7718 
 0   0 

 

11.094 


= 1.6847 
 0 


δr = 11.222 km
Problem 7.8
398 600
&micro;
=
= 7.7714 km s
r
6600
v 7.7714
n= =
= 0.001177 5 s -1
r
6600
2π
T=
= 5446.1 s
n
T
t = = 1778.7 s
3
 4 − 3 cos nt 0
0   5.5
0
0 

 
[Φrr (t)] = 6(sin nt − nt) 1 0  = −7.3702 1 0 

0
0 cos nt   0
0 −0.5
v=
[Φrv (t)]
[Φvr (t)]
[Φvv (t)]
 1
2

0
(1 − cos nt )
 n sin nt
 
n
2547.8
0 
2
  735.49
1

 =  −2547.8 −2394.2
0 
=  (cos nt − 1)
0
(4 sin nt − 3 nt )
n
n
 
0
735.49 
1
 0

sin nt 
0
0


n

 3 n sin nt

0
0
0

0.003 059 2 0

 

= 6 n(cos nt − 1) 0
0
0
.
010
597
0
0
=
−
 


0
0 − n sin nt  
0
0 −0.0010197 
 cos nt
2 sin nt
0   −0.5
1.7321
0 

 

=  −2 sin nt 4 cos nt − 3
0  =  −1.7321
0 
−5

0
0
cos nt   0
0
−0.5
122
Chapter 7
Solutions Manual
Orbital Mechanics for Engineering Students
{δr f } = [Φrr ]{δr0 } + [Φrv ]{δv0 + }
0
0  1  735.49
2547.8
0 
0   5.5
  
  

0  1 +  −2547.8 −2394.2
0  δv 0 +
0  =  −7.3702 1
0   0
0 −0.5 1  0
0
735.49 
  
 735.49
 5.5
2547.8
0 
0
0  1  −5.5 




  
+
0  δv 0 = −  −7.3702 1
0  1 = 6.3702 
 −2547.8 −2394.2
 0
 0
0
735.49 
0 −0.5 1  0.5 
−1
 735.49
2547.8
0   −5.5   −0.000 647 34 
 


 
0  6.3702  =  −0.001 971 8  (km s)
δv 0 + =  −2547.8 −2394.2
 0
0
735.49   0.5   0.000 679 82 
{
{
{
}
}
}
 −0.000 647 34   0   −0.000 647 34 
{∆v1} = δv 0 − δv 0 =  −0.0019718  −  0  =  −0.0019718  (km s)
 0.000 679 82  0.005  −0.004 320 2 

 
 

∆v1 = ∆v1 = 0.004792 8 km s
{
+
} {
−
}
{δv f − } = [Φvr ]{δr0 } + [Φvv ]{δv0 + }
0.003 059 2 0
0
1.7321
0   −0.000 647 34 
 1  −0.5

  



δv f =  −0.010 597 0
0
1 + −1.7321
−5
0   −0.0019718 
  

0
0 −0.0010197  1  0
0
−0.5  0.000 679 82 

 0.003 059 2   −0.003 0917 

 

δv f − =  −0.010 597  +  0.010 98 
 −0.0010197   −0.000 339 91

 

 −0.000 032 48 


δv f − =  0.000 38312  (km s)
 −0.001359 6 


0 
 
δv f + = 0 
0 
 
0   −0.000 032 48   0.000 032 48 
+
−
{∆v 2 } = δv f − δv f = 0  −  0.000 38312  = −0.000 38312  (km s)
0   −0.001359 6   0.001359 6 
  
 

∆v2 = ∆v 2 = 0.001413 km s
{
−
}
{
}
{
}
{
}
{
} {
}
vtotal = ∆v1 + ∆v2 = 0.004792 8 + 0.001413 = 0.006 2058 km s = 6.206 m s
Problem 7.9
T0
2
2π
nt =
=π
T0
 4 − 3 cos nt 0
0   7

[Φrr (t)] = 6(sin nt − nt) 1 0  = −6π

0
0 cos nt   0
t=
0 0

1 0
0 −1
123
Chapter 7
Solutions Manual
(a)
Orbital Mechanics for Engineering Students
 1
 
2
0
(1 − cos nt )
 n sin nt
  0
n
2
 
[Φrv (t)] =  n (cos nt − 1) 1n (4 sin nt − 3 nt) 0  = − 4n

 
1
0

0
0
sin nt  

 
n
 3 n sin nt
  0
0
0
0 0

 

0
[Φvr (t)] = 6 n(cos nt − 1) 0
 =  −12 n 0 0 

0
0 − n sin nt   0
0 0 
 cos nt
2 sin nt
0   −1 0
0

 
[Φvv (t)] = −2 sin nt 4 cos nt − 3 0  =  0 −7 0 

0
0
0 −1
cos nt   0
{δr f } = [Φrr ]{δr0 } + [Φrv ]{δv0 + }
0   7
  
0  =  −6π
0   0
  

 0

0 0  δρ  
4


1 0  δy0  +  −
 n
0 −1  0   0

4
n
3π
−
n
0

0
  0 
0  δv0 


0  0 

 4δv0 


7δρ
0  
  n

  
  3πδv0 
+
−
=
−
+
0
6
πδρ
δ
y
  
0 
n 
0  
 
0

  
  0

4δv0


7δρ +


n
 0 

 −6πδρ − 3πδv0 + δ  = 0 
y0   

n

 0 
0


4δv0
7
7δρ +
= 0 ⇒ δv0 = − nδρ
n
4
3π  7
 3
δy0 = 6πδρ +
− nδρ = πδρ
 4
n  4
 δρ 
 0 
3

 7

+
∴ {δr0 } =  πδρ 
δv 0 =  − vδρ 
4
4


 0 
 0 


{
}
(b)
{δv f − } = [Φvr ]{δr0 } + [Φvv ]{δv0 + }
{δv f − }
{δv f − }
 0
0 0   δρ   −1 0
0  0 
 

 3

 7
=  −12 n 0 0   πδρ  + 0 −7 0   − vδρ 


4
4
 0
0 0   0   0
0 −1  0 




 0   0   0   0 

  49
  1 πδρ 
 1
=  −12 nδρ  +  nδρ  =  nδρ  = 

 0  4
  4 0   2 T0 
0

 
  0 
 
124
4
n
3π
−
n
0
Chapter 7

0

0

0

Solutions Manual
Orbital Mechanics for Engineering Students
x
3π
δρ
4
t=0
0
y
Problem 7.10
nt = 2π
2π
T
 4 − 3 cos nt

= 6(sin nt − nt )

0
 1
 n sin nt
2
=  (cos nt − 1)
n

0

 3 n sin nt

= 6 n(cos nt − 1)

0
n=
[Φrr (t)]
[Φrv (t)]
[Φvr (t)]
[Φvv (t)]
0
0
  1
 
1
0  =  −12π
0 cos nt   0
2
(1 − cos nt )
n
1
(4 sin nt − 3 nt )
n
0 0

1 0
0 1 

 0
0
 
6π
 = 0 −
0
n
 
1
0
0
0
sin nt  

n
 0 0 0 
0
0
 

0
0
 = 0 0 0 
0 −n sin nt  0 0 0 
 cos nt
2 sin nt
0  1 0 0 

 

0  = 0 1 0 
=  −2 sin nt 4 cos nt − 3

0
0
cos nt  0 0 1 
0
{δr f } = [Φrr ]{δr0 } + [Φrv ]{δv0 + }
0
0  δu0 + 
0 0   0  0



6π


1 0  δy0  + 0 −
0  δv0 + 
n


0 1   0  0
0
0   0 
0

0   0  

6πδv0 + 
  


0  = δy0  +  −
n 
0   0  
0
  
 

0

0  
nδy0
6πδv0 + 
  
+
 ⇒ δv0 =
0  = δy0 −
n
6π

0  
0
  

0   1
  
0  =  −12π
0   0
  
Thus, assuming a Hohmann transfer, δu0 + = 0 ,
{δv0 }
+
 0 
 nδy0 
=

 60π 


125
0

0
0 
Chapter 7
Solutions Manual
{∆v1} = {δv 0
Orbital Mechanics for Engineering Students
+
∆v1 = ∆v1 =
} − {δv0 }
−
 0  0   0 
 nδy0     nδy0 
=
 − 0  = 

 60π  0   60π 

   

nδy0
6π
{δv f − } = [Φvr ]{δr0 } + [Φvv ]{δv0 + }
0 0 0   0  1 0 0   0 
 


  nδy0 
δv f = 0 0 0  δy0  + 0 1 0  

6π
0 0 0   0  0 0 1   0 


 0 
 nδy0 
δv f − = 

 60π 


0   0   0 
nδy
nδy
+
−
{∆v 2 } = δv f − δv f = 0  −  6π 0  = − 6π 0 
0  
 

   0   0 
nδy0
∆v2 = ∆v 2 =
6π
{
{
−
}
}
{
} {
}
∆vtotal = ∆v1 + ∆v2 =
nδy0 nδy0 nδy0 2δy0
+
=
=
6π
6π
3π
3T
x
y0
y
t=0
y0
Problem 7.11
2π
= 0.000 872 66
2 ⋅ 3600
t = 30 ⋅ 60 = 1800 s
π
nt =
2
 4 − 3 cos nt
[Φrr (t)] = 6(sin nt − nt)

0
 1
 n sin nt

[Φrv (t)] =  n2 (cos nt − 1)


0

 3 n sin nt

[Φvr (t)] = 6 n(cos nt − 1)

0
n=
s -1
0
0   4
0 0
 

1
0  =  −3.425 1 0 
0 cos nt   0
0 0 
2

0
(1 − cos nt )
 
n
2292
0 
  1146
1


=  −2292 −816.3
0
0 
(4 sin nt − 3 nt )
n
 
0
1146 
1
 0
sin nt 
0

n
  0.002 618 0

0
0
0
 

0
0
0
 =  −0.005236 0

0 − n sin nt  
0
0 −0.000 8727 
126
Chapter 7
Solutions Manual
Orbital Mechanics for Engineering Students
 cos nt
2 sin nt
0  0
2 0

 
[Φvv (t)] = −2 sin nt 4 cos nt − 3 0  = −2 −3 0 

cos nt   0
0
0
0 0 
{δr f } = [Φrr ]{δr0 } + [Φrv ] δv0 +
{
}
 4
0 0  0   1146
2292
0  0 


  

0   −0.003 
δr f =  −3.425 1 0  6  +  −2292 −816.3
 0
0 0  0   0
0
1146   0 
0   −6.875  −6.875
 

  
δr f = 6  +  2.449  =  8.449  ( km )
0   0   0 
  
 

δr f = 10.89 km
{ }
{ }
{δv f } = [Φvr ]{δr0 } + [Φvv ]{δv0 + }
 0.002 618 0
 0   0
0
2 0  0 


  

δv f =  −0.005236 0
0
 6  +  −2 −3 0   −0.003 

0
0 −0.000 8727  0   0
0 0   0 
0   −0.006  0   −0.006 
   

  
δv f = 0  +  0.009  = 0  +  0.009  (km s)
0   0  0   0 
  
   

δv f = 0.010 82 km s
{
}
{
}
x (km)
y (km)
t=0
8
4
6
2
0
-2
-4
-6
t = 30 min
Problem 7.12 Use C-W frame attached to original location of satellite in GEO.
2π
365.26 = 7.292 &times; 10 −5 s -1
24 ⋅ 3600
2π +
n=
First determine the relative position and velocity after two hours.
t = 2 ⋅ 3600 = 7200 s
nt = 0.1671π
127
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 4 − 3 cos nt 0   1.404
0
=
=


6(sin nt − nt ) 1   −0.1427 1 
 1
2

(1 − cos nt )   6874 3694 
 n sin nt
n
[Φrv (t)] =  2

=
1
 (cos nt − 1)
(4 sin nt − 3 nt )   −3694 5895
n

n
 3 n sin nt
0   10.97 &times; 10 −5 0 
[Φvr (t)] = 6 n(cos nt − 1) 0  = −5.893 &times; 10 −5 0 


 
 cos nt
2 sin nt   0.8653 1.002 
[Φvv (t)] = −2 sin nt 4 cos nt − 3  = −1.002 0.4612 

 

[Φrr (t)]
{δr2 } = [Φrr ]{δr0 } + [Φrv ]{δv 0 }
0  0   6874 3694 
 −10   1.404

=
  + 
{δv 0 }
 10   −0.1427 1  0   −3694 5895
 6874 3694 
 −10 
 −3694 5895{δv 0 } =  10 




 6874
{δv 0 } =  −3694

−1
3694    −10   −0.00177 

 =
 (km s)
5895   10  0.000 587 
{δv2 − } = [Φvr ]{δr0 } + [Φvv ]{δv0 }
 97 &times; 10 −5 0  0   0.8653
{δv2 − } = −105..893
  + 
−1.002
&times; 10 −5 0 0
{δv2 − }

  
 −0.000 943 4 
=
 (km s)
 0.002 045 

1.002   −0.00177 


0.4612  0.000 587 
Initiate rendezvous with the origin.
t = 6 ⋅ 3600 = 21600 s
nt = 0.5014π
{δr6 } = [Φrr ]{δr2 } + [Φrv ]{δv 2 + }
0   4.013 0   −10   13710 27 540 
+
 =
+

 δv 2
0   −3.451 1   10   −27 540 −9947 
 13710 27 540 
 4.013 0   −10 
+

−
 δv 2 = − 

−
27
540
9947


 −3.451 1   10 
{
{
}
}
 13710 27 540 
 40.13 
+
 −27 540 −9947  δv 2 =  −44.51




{
{δv2 + } = −13710
27 540

}
27 540 
−9947 
−1
 40.13   0.001329 

=
 (km s)
 −44.51 0.0007954 
{δv6 − } = [Φvr ]{δr0 } + [Φvv ]{δv2 + }
218 8 0   −10   −0.004 3
{δv6 − } = −00.000

+
.000 439 4 0  10
−2
2   0.001329 


−3.017  0.0007954 


 
−
0
.
002188
0
.
001
585   −0.000 602 5

 
δv 6 − = 
+
=
 (km s)
 0.004 394   −0.005057   −0.000 663 
{
}
128
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 0.001 329   −0.000 943 4   0.002 272 
=
 (km s)

  0.002 045   −0.001 25
{∆v 2 } = {δv 2 + } − {δv 2 − } = 0.000 795 4  − 
∆v2 = ∆v 2 = 0.002 593 km s
0   −0.000 602 5 0.000 602 5
=
 (km s)
663   0.000 663 
  
{∆v 6 } = {δv 6 + } − {δv 6 − } = 0  −  −0.000
∆v6 = ∆v 6 = 0.000 8958 km s
∆vtotal = ∆v2 + ∆v6 = 0.002 593 + 0.000 8958 = 0.003 489 km s = 3.489 m s
x (km)
t=0
0
y (km)
10
8
6
4
2
-2
Return
-4
Outbound
-6
-8
t = 2 hr
-10
Problem 7.13 Design problem.
Problem 7.14
398 600
6600
n=
= 0.001 177 5 s -1
6600
3
3
δv = − nδx = − ⋅ 0.001 177 5 ⋅ 5 = 0.008 8311 km s = 8.8311 m s
2
2
Problem 7.15
nt =
π
2
0
 4 − 3 cos nt 0   4
=
=


6(sin nt − nt ) 1   −3.425 1 
 1
2
  1
(1 − cos nt )  
 n sin nt
n
n
[Φrv (t)] =  2
= 2
1
 (cos nt − 1)
(4 sin nt − 3 nt )   −
n
  n
n
δr f = [Φrr ]{δr0 } + [Φrv ]{δv 0 }
[Φrr (t)]

2

n
0.7124 

−
n 
{ }
 1
  nπδr 
2
 4
0   δr   n
  16 
n
δr f = 

+ 2

0.7124   7 nδr 
 −3.425 1  πδr   −

 −
−
 n
4 
n 
 4δr
  −3.304δr   0.6963δr 
δr f = 
+
=

 −0.2832δr   0.854δr  0.85708δr 
{ }
{ }
129
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d = δr f = 0.9004δr
Problem 7.16
nt = π nt = π
0
 4 − 3 cos nt 0   7
[Φrr (t)] = 6(sin nt − nt) 1  = −18.85 1 

 

 1
2
 
(1 − cos nt )   0
 n sin nt
n
[Φrv (t)] =  2
= 4
1
 (cos nt − 1)
(4 sin nt − 3 nt )   −
n
  n
n
4 
n 
9.425 

−
n 
{δr f } = [Φrr ]{δr0 } + [Φrv ]{δv0 }

4 
0  δx0   0
0   7
n {δv }
 =
 0  +  4
9
.
425  0
0
18
85
1
−
.
  −
  

−
 n
n 

4 
 0
n {δv } =  −7δx0 


 4
9.425  0
18.85δx0 

−
−
 n
n 
or
4 

 0

{δv 0 } =  4 9n.425 
−

−
n 
 n
 −0.589 nδx 
{δv 0 } =  −1.75nδx 0 
0 

−1
 −7δx0 


18.85δx0 
δ v 0 = −0.589nδ x0 iˆ − 1.75nδ x0 ˆj ( km s )
Problem 7.17
nt =
π
4
[Φrr (t)]
[Φrv (t)]
0
 4 − 3 cos nt 0   1.879
=
=


6(sin nt − nt ) 1   −0.4697 1 
2
  0.7071
 1
(1 − cos nt )  
 n sin nt
n
n
=
 =  0.5858
2
1
 (cos nt − 1)
(4 sin nt − 3 nt )   −
 
n
n
n
{δr f } = [Φrr ]{δr0 } + [Φrv ]{δv0 }
 0.7071 0.5858 
0   −d  
n
n  0 


+

1  δy0   0.5858 0.4722  δv0 
−


n
n 
 0.5858δv0 
0   −1.879 d  

n
 =
 +  0.4722δv 
0
0  0.4697 d + δy0  
n


0.5858δv0


− 1.879 d

 0 
n
 0.4722δv
= 
0
0

+ 0.4697 d + δy0   
n


0   1.879
 =
0   −0.4697
130
0.5858 
n 
0.4722 

n 
Chapter 7
Solutions Manual

0

1

Orbital Mechanics for Engineering Students
0.5858 
n  δy0  =  1.879 d 
0.4722  δv0   −0.4697 d 

n 
−1
0.5858 

δy0  0
n   1.879 d 

=
  −0.4697 d 
.
0
4722
δv0  1

 
n 

δy0   −0.8062 1   1.879 d   −1.984 d 

=
=


δv0   1.707 n 0   −0.4697 d  3.207 nd 
δy0 = −1.984 d
131
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132
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Solutions Manual
Orbital Mechanics for Engineering Students
Problem 8.1
&micro;sun = 132.7 &times; 10 9 km 3 s2
&micro;earth = 398 600 km 3 s2
Rearth = 147.4 &times; 10 6 km
rearth = 6378 km
a=
1
1
+ R2 ) = (147.4 &times; 10 6 + 120 &times; 10 6 ) = 138.7 &times; 10 6 km
(R
2 earth
2
Heliocentric spacecraft velocity at earth’s sphere of influence:
V
( v)
1
2
1
 2


= &micro;sun 
−  = 132.7 &times; 10 9 
−
 = 28.43 km s
6
6
 147.4 &times; 10
 Rearth a 
138.7 &times; 10 
Heliocentric velocity of earth:
Vearth =
&micro;sun
132.7 &times; 10 9
=
= 30.06 km s
Rearth
149.6 &times; 10 6
∴ v ∞ = Vearth − V
( v)
= 30.06 − 28.43 = 1.579 km s
Geocentric spacecraft velocity of spacecraft at perigee of departure hyperbola:
v p = v ∞2 +
2 ⋅ 398 600
2&micro;earth
= 1.579 2 +
= 11.12 km s
6378 + 200
rp
Geocentric spacecraft velocity in its circular parking orbit:
vc =
398 600
&micro;earth
=
= 7.784 km s
r
6378 + 200
∴ ∆v = v p − vc = 11.12 − 7.784 = 3.337 km s
Problem 8.2
&micro;sun = 132.7 &times; 10 9 km 3 s2
&micro;earth = 398 600 km 3 s2
&micro;Mercury = 22 930 km 3 s2
Rearth = 149.6 &times; 10 6 km
RMercury = 57.91 &times; 10 6 km
rearth = 6378 km
rMercury = 2440 km
Vearth =
&micro;sun
132.7 &times; 10 9
=
= 29.78 km s
Rearth
149.6 &times; 10 6
VMercury =
&micro;sun
RMercury
=
132.7 &times; 10 9
57.91 &times; 10 6
= 47.87 km s
133
Chapter 8
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Semimajor axis of Hohmann transfer ellipse:
a=
1
1
Rearth + RMercury = (149.6 &times; 10 6 + 57.91 &times; 10 6 ) = 103.8 &times; 10 6 km
2
2
(
)
Departure from earth:
Spacecraft heliocentric velocity:
V
( v)
1
2
1
 2


= &micro;sun 
−  = 132.7 &times; 10 9 
−
 = 22.25 km s
6
6
 149.6 &times; 10
 Rearth a 
103.8 &times; 10 
∴ v ∞ = Vearth − V
( v)
= 29.78 − 22.25 = 7.532 km s
Spacecraft geocentric velocity in circular parking orbit:
vc =
398 600
&micro;earth
=
= 7.814 km s
rp
6378 + 150
Spacecraft geocentric velocity at perigee of departure hyperbola:
v p = v ∞2 +
2 ⋅ 398 600
2&micro;earth
= 7.532 2 +
= 13.37 km s
rp
6378 + 150
∴ ∆v1 = v p − vc = 13.37 − 7.814 = 9.611 km s
Arrival at Mercury:
Spacecraft heliocentric velocity:
V
( v)
2
1
2
1



= &micro;sun 
−  = 132.7 &times; 10 9 
−
= 57.48 km s
6
6 

R
a
57.91 &times; 10
103.8 &times; 10
 Mercury

( )
∴ v ∞ = V v − VMercury = 57.48 − 47.87 = 9.611 km s
Spacecraft velocity relative to Mercury at periapse of approach hyperbola:
v p = v ∞2 +
2&micro;Mercury
rp
= 9.6112 +
2 ⋅ 22 930
= 10.49 km s
2440 + 150
Spacecraft parking orbit speed relative to Mercury:
vc =
&micro;Mercury
rp
=
22 930
= 2.975 km s
2440 + 150
∴ ∆v2 = v p − vc = 10.49 − 2.975 = 7.516 km s
∆vtotal = ∆v1 + ∆v2 = 9.611 + 7.516 = 15.03 km s
Problem 8.3
 mp 
rSOI = R

 msun 
msun = 1.989 &times; 10 30 kg
Mercury:
134
Chapter 8
Solutions Manual
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R = 57.91 &times; 10 6 km
m p = 3.302 &times; 10 23 kg
 3.302 &times; 10 23 
5
rSOI = 57.91 &times; 10 6 
 = 1.124 &times; 10 km
 1.989 &times; 10 30 
Venus:
R = 108.2 &times; 10 6 km
m p = 4.869 &times; 10 24 kg
 4.869 &times; 10 24 
5
rSOI = 108.2 &times; 10 6 
 = 6.162 &times; 10 km
 1.989 &times; 10 30 
Mars:
R = 227.9 &times; 10 6 km
m p = 6.419 &times; 10 23 kg
 6.419 &times; 10 23 
5
rSOI = 227.9 &times; 10 6 
 = 5.771 &times; 10 km
 1.989 &times; 10 30 
Jupiter:
R = 778.6 &times; 10 6 km
m p = 1.899 &times; 10 27 kg
 1.899 &times; 10 27 
7
rSOI = 778.6 &times; 10 6 
 = 4.882 &times; 10 km
 1.989 &times; 10 30 
Problem 8.4
 mp 
rSOI = R

 msun 
msun = 1.989 &times; 10 30 kg
Saturn:
R = 1433 &times; 10 6 km
m p = 5.685 &times; 10 26 kg
 5.685 &times; 10 26 
7
rSOI = 1433 &times; 10 6 
 = 5.479 &times; 10 km
 1.989 &times; 10 30 
Uranus:
R = 2872 &times; 10 6 km
m p = 8.683 &times; 10 25 kg
 8.683 &times; 10 25 
7
rSOI = 2872 &times; 10 6 
 = 5.178 &times; 10 km
 1.989 &times; 10 30 
Neptune:
R = 4495 &times; 10 6 km
m p = 1.024 &times; 10 26 kg
 1.024 &times; 10 26 
7
rSOI = 1.024 &times; 10 26 
 = 8.658 &times; 10 km
 1.989 &times; 10 30 
135
Chapter 8
Solutions Manual
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Chapter 8
Pluto:
R = 5870 &times; 10 6 km
m p = 1.25 &times; 10 22 kg
 1.25 &times; 10 22 
6
rSOI = 5870 &times; 10 6 
 = 3.076 &times; 10 km
 1.989 &times; 10 30 
Problem 8.5
&micro;sun = 132.7 &times; 10 9 km 3 s2
&micro;Jupiter = 126.7 &times; 10 6 km 3 s2
Rearth = 149.6 &times; 10 6 km
RJupiter = 778.6 &times; 10 6 km
rJupiter = 71490 km
Semimajor axis of Hohmann transfer ellipse:
a1 =
1
1
Rearth + RJupiter = (149.6 &times; 10 6 + 778.6 &times; 10 6 ) = 464.1 &times; 10 6 km
2
2
(
)
VJupiter =
&micro;sun
132.7 &times; 10 9
=
= 13.06 km s
RJupiter
778.6 &times; 10 6
Use the energy equation to obtain the spacecraft’s velocity upon arrival at Jupiter’s sphere of
influence:
2
1

( )
V1 v = &micro;sun 
−  = 132.7 &times; 10 9
R
a
 Jupiter
1
2
1


−

 = 7.412 km s
 778.6 &times; 10 6 464.1 &times; 10 6 
( )
v ∞ = VJupiter − V1 v = 13.06 − 7.412 = 5.643 km s
Sun
û S
VJupiter = 13.06 km/s
V1 v
778.6(106) km
7.412 km s
149.6(106) km
ûV
Eccentricity of hyperbolic swing by trajectory:
e =1 +
rp v ∞ 2
&micro;Jupiter
=1 +
271490 ⋅ 5.643 2
126.7 &times; 10 6
= 1.068
Turn angle:
 1
 1 
= 2 sin −1
= 138.8&deg;
 e
 1.068 
Angle between VJupiter and v ∞ at inbound crossing: φ1 = 180&deg; . At the outbound crossing,
δ = 2 sin −1
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Chapter 8
φ2 = 180&deg; + δ = 318.8&deg;
At the outbound crossing
( )
V2v = VJupiter + v ∞ 2
= VJupiter uˆ V + ( v∞ cos φ 2uˆ V + v∞ sin φ 2 ) uˆ S
= 13.06uˆ V + ( 5.643 ⋅ cos 318.8&deg;uˆ V + 5.643 sin 318.8&deg; ) uˆ S
p
û S
= 17.30uˆ V − 3.716uˆ S ( km s )
( v)
∆V
( )
318.8&deg;
v
( )
= V2v − V1 v
2
ûV
= (17.30uˆ V − 3.716uˆ S ) − 7.412uˆ V
= 9.890uˆ V − 3.716uˆ S ( km s )
∴ ∆V
( v)
= ∆V
( v)
V
= 10.57 km s
v
v
v
1
v
1
2
138.8&deg;
δ
( )
Alternatively, ∆V v = 2 v ∞ sin = 2 ⋅ 5.643 ⋅ sin
= 10.57 km s .
2
2
Angular momentum of the new orbit after flyby:
( )
h2 = RJupiterV⊥v = 778.6 &times; 10 6 ⋅ 17.30 = 13.47 &times; 10 9 km 2 s
Obtain the semimajor axis a2 from the energy equation:
( )2
V2 v
2
−
&micro;sun
&micro;
= − sun
RJupiter
2 a2
17.70 2 132.7 &times; 10 9
132.7 &times; 10 9
−
=
−
⇒ a2 = 4.791 &times; 10 9 km
2
2 a2
778.6 &times; 10 6
Observe that a2 a1 = 10.32 . The new orbit dwarfs the original one in size and, therefore, energy.
Use Equation 3.61 to find the eccentricity:
a2 =
1
h2 2
&micro;sun 1 − e2 2
9
4.791 &times; 10 =
13.47 &times; 10 9
132.7 &times; 10
2
9
1
1 − e2 2
⇒ e2 = 0.8453
Use the orbit equation to find the true anomaly in the new orbit:
RJupiter =
h2 2
1
&micro;sun 1 + e2 cos θ 2
778.6 &times; 10 6 =
13.47 &times; 10 9
132.7 &times; 10
2
9
1
⇒ θ 2 = 26.5&deg;
1 + 0.8453 cos θ 2
The new and original heliocentric orbits are illustrated below.
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Periapse of
orbit 2
Chapter 8
1
26.5&deg;
Sun
Jupiter
Earth at launch
2
Problem 8.6
Algorithm 8.1, which makes use of the data in Table 8.1, is implemented in MATAB as the M-function
planet_elements_and_sv in Appendix D.17. The MATLAB script Example_8_07, which also
appears in Appendix D.17, calls upon planet_elements_and_sv to calculate the orbital elements
of earth on the date specified in Example 8.7. The output to the Command Window is also listed in
Appendix D.17 for comparison with the results presented in Example 8.7. By changing planet_id to
4, the following Command Window output is obtained for Mars.
----------------------------------------------------Problem 8.6
Input data:
Planet:
Year :
Month :
Day
:
Hour :
Minute:
Second:
Mars
2003
August
27
12
0
0
Julian day: 2452879.000
Orbital elements:
Angular momentum (km^2/s)
= 5.47595e+09
Eccentricity
= 0.0934167
Right ascension of the ascending node (deg) = 49.5682
Inclination to the ecliptic (deg)
= 1.85035
Argument of perihelion (deg)
= 286.488
True anomaly (deg)
= 358.131
Semimajor axis (km)
= 2.27936e+08
Longitude of perihelion (deg)
Mean longitude (deg)
Mean anomaly (deg)
Eccentric anomaly (deg)
=
=
=
=
138
336.057
334.513
358.457
358.298
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 8
State vector:
Position vector (km) = [1.85954e+08
Magnitude
= 2.06653e+08
Velocity (km/s)
= [11.4744
Magnitude
= 26.4984
-----------------------------------------------------
-8.99155e+07
23.8842
-6.45661e+06]
0.218255]
Problem 8.7 The following MATLAB script calls upon Algorithm 8.1, implemented as the
MATLAB M-function planet_elements_and_sv in Appendix D.17, to compute the distance of the
earth from the sun on the first day of each month of the year 2005, at 12:00:00 UT. The output to the
MATLAB command window is listed afterwards.
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_8_07a
~~~~~~~~~~~~~
This program uses Algorithm 8.1 to compute the orbital elements
and state vector of Mars at the date and time specified
in Example 8.7.
mu
- gravitational parameter of the sun (km^3/s^2)
coe
- vector
[h e
where
h
e
RA
incl
w
TA
a
w_hat
L
M
E
r
v
of heliocentric orbital elements
RA incl w TA a w_hat L M E],
=
=
=
=
=
=
=
=
=
=
=
angular momentum
(km^2/s)
eccentricity
right ascension
(deg)
inclination
(deg)
argument of perihelion
(deg)
true anomaly
(deg)
semimajor axis
(km)
longitude of perihelion ( = RA + w) (deg)
mean longitude ( = w_hat + M)
(deg)
mean anomaly
(deg)
eccentric anomaly
(deg)
- heliocentric position vector (km)
- heliocentric velocity vector (km/s)
planet_id - planet identifier:
1 = Mercury
2 = Venus
3 = Earth
4 = Mars
5 = Jupiter
6 = Saturn
7 = Uranus
8 = Neptune
9 = Pluto
year
month
day
hour
minute
second
-
range:
range:
range:
range:
range:
range:
1901 - 2099
1 - 12
1 - 31
0 - 23
0 - 60
0 - 60
User M-functions required: planet_elements_and_sv,
month_planet_names
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Chapter 8
% -------------------------------------------------------------------global mu
mu = 1.327124e11;
deg = pi/180;
%...Data declaration for Problem 8.6 :
planet_id = 3;
year
= 2005;
day
= 1;
hour
= 12;
minute
= 0;
second
= 0;
%...
fprintf('\n-----------------------------------------------------\n')
fprintf(' Problem 8.7a: Determine the month\n')
fprintf('\n Year = %g Time = 12:00:00 UT\n', year)
for month = 1:12
%...Algorithm 8.1:
[coe, r, v, jd] = planet_elements_and_sv ...
(planet_id, year, month, day, hour, minute, second);
%...Convert the month numbers into names for output:
[month_name, planet_name] = month_planet_names(month, planet_id);
fprintf('\n %10s 1st:
Distance = %11.5e (km)', month_name,
norm(r))
end
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 8.7a: Determine the month
Year = 2005
Time = 12:00:00 UT
January
1st:
Distance = 1.47100e+08 (km)
February 1st:
Distance = 1.47417e+08 (km)
March
1st:
Distance = 1.48243e+08 (km)
April
1st:
Distance = 1.49505e+08 (km)
May
1st:
Distance = 1.50745e+08 (km)
June
1st:
Distance = 1.51707e+08 (km)
July
1st:
Distance = 1.52093e+08 (km)
August
1st:
Distance = 1.51830e+08 (km)
September 1st:
Distance = 1.50963e+08 (km)
October
1st:
Distance = 1.49758e+08 (km)
November 1st:
Distance = 1.48464e+08 (km)
December 1st:
Distance = 1.47505e+08 (km)
----------------------------------------------------This list reveals that the greatest distance occurs in the month of July. We can modify the above
MATLAB script to loop through the days of July:
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_8_07b
~~~~~~~~~~~~~
This program uses Algorithm 8.1 to determine the distance of the
earth from the sun, according to Problem 8.7
User M-functions required: planet_elements_and_sv,
month_planet_names
--------------------------------------------------------------------
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Chapter 8
global mu
mu = 1.327124e11;
%...Data declaration for Problem 8.7 :
planet_id = 3;
year
= 2005;
month
= 7;
hour
= 12;
minute
= 0;
second
= 0;
%...
fprintf('\n-----------------------------------------------------\n')
fprintf(' Problem 8.7b: Determine the day\n')
fprintf('\n Year = %g Time = 12:00:00 UT\n', year)
%...Convert the planet_id and month numbers into names for output:
[month_name, planet_name] = month_planet_names(month, planet_id);
for day = 1:31
%...Algorithm 8.1:
[coe, r, v, jd] = planet_elements_and_sv ...
(planet_id, year, month, day, hour, minute, second);
%...Convert the planet_id and month numbers into names for output:
[month_name, planet_name] = month_planet_names(month, planet_id);
fprintf('\n %5s %4g:
Distance = %14.7e (km)',...
month_name, day, norm(r))
end
fprintf('\n-----------------------------------------------------\n')
The output to the MATLAB command window is:
----------------------------------------------------Problem 8.7b: Determine the day
Year = 2005
July
July
July
July
July
July
July
July
July
July
July
July
July
July
July
July
July
July
July
July
July
July
July
July
July
July
Time = 12:00:00 UT
1:
2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
12:
13:
14:
15:
16:
17:
18:
19:
20:
21:
22:
23:
24:
25:
26:
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
Distance
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
1.5209314e+08
1.5209524e+08
1.5209664e+08
1.5209732e+08
1.5209728e+08
1.5209653e+08
1.5209506e+08
1.5209287e+08
1.5208998e+08
1.5208637e+08
1.5208204e+08
1.5207701e+08
1.5207126e+08
1.5206481e+08
1.5205765e+08
1.5204978e+08
1.5204122e+08
1.5203195e+08
1.5202198e+08
1.5201132e+08
1.5199996e+08
1.5198792e+08
1.5197519e+08
1.5196178e+08
1.5194769e+08
1.5193292e+08
141
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(km)
(km)
(km)
(km)
(km)
(km)
(km)
(km)
(km)
(km)
(km)
(km)
(km)
(km)
(km)
(km)
(km)
(km)
(km)
(km)
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Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 8
July
27:
Distance = 1.5191748e+08 (km)
July
28:
Distance = 1.5190138e+08 (km)
July
29:
Distance = 1.5188461e+08 (km)
July
30:
Distance = 1.5186718e+08 (km)
July
31:
Distance = 1.5184910e+08 (km)
----------------------------------------------------The furthest distance occurs on July 4.
Problem 8.8 For the data given in this problem, the following MATLAB script invokes Algorithm
8.2, which is implemented as the MATLAB M-function interplanetary in Appendix D.18.
interplanetary uses Algorithms 8.1 and 5.2 to compute v ∞ at the home and target planets. The
output to the MATLAB Command Window is listed afterwards.
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_8_08
~~~~~~~~~~~~
This program uses Algorithm 8.2 to obtain v-infinities
in Problem 8.8.
mu
deg
pi
- gravitational parameter of the sun (km^3/s^2)
- conversion factor between degrees and radians
- 3.1415926...
planet_id
- planet identifier:
1 = Mercury
2 = Venus
3 = Earth
4 = Mars
5 = Jupiter
6 = Saturn
7 = Uranus
8 = Neptune
9 = Pluto
year
month
day
hour
minute
second
-
depart
- [planet_id, year, month, day, hour, minute, second]
at departure
- [planet_id, year, month, day, hour, minute, second]
at arrival
arrive
range:
range:
range:
range:
range:
range:
1901 - 2099
1 - 12
1 - 31
0 - 23
0 - 60
0 - 60
planet1
planet2
trajectory
- [Rp1, Vp1, jd1]
- [Rp2, Vp2, jd2]
- [V1, V2]
coe
- orbital elements [h e RA incl w TA]
where
h
= angular momentum (km^2/s)
e
= eccentricity
RA
= right ascension of the ascending
incl = inclination of the orbit (rad)
w
TA
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Orbital Mechanics for Engineering Students
a
Chapter 8
= semimajor axis (km)
jd1, jd2
tof
- Julian day numbers at departure and arrival
- time of flight from planet 1 to planet 2 (days)
Rp1, Vp1
Rp2, Vp2
R1, V1
- state vector of planet 1 at departure (km, km/s)
- state vector of planet 2 at arrival (km, km/s)
- heliocentric state vector of spacecraft at
departure (km, km/s)
- heliocentric state vector of spacecraft at
arrival (km, km/s)
R2, V2
vinf1, vinf2 - hyperbolic excess velocities at departure
and arrival (km/s)
User M-functions required: interplanetary, coe_from_sv,
month_planet_names
--------------------------------------------------------------------
clear
global mu
mu = 1.327124e11;
deg = pi/180;
%...Data declaration for Problem 8.8:
%...Departure
planet_id = 3;
year
= 2005;
month
= 12;
day
= 1;
hour
= 0;
minute
= 0;
second
= 0;
depart = [planet_id
year
month
day
hour
minute
second];
%...Arrival
planet_id = 2;
year
= 2006;
month
= 4;
day
= 1;
hour
= 0;
minute
= 0;
second
= 0;
arrive = [planet_id
year
month
day
hour
minute
second];
%...
%...Algorithm 8.2:
[planet1, planet2, trajectory] = interplanetary(depart, arrive);
R1 = planet1(1,1:3);
Vp1 = planet1(1,4:6);
jd1 = planet1(1,7);
R2 = planet2(1,1:3);
Vp2 = planet2(1,4:6);
jd2 = planet2(1,7);
V1
V2
= trajectory(1,1:3);
= trajectory(1,4:6);
tof = jd2 - jd1;
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Chapter 8
%...Use Algorithm 5.1 to find the orbital elements of the
%
spacecraft trajectory based on [Rp1, V1]...
coe = coe_from_sv(R1, V1);
%
... and [R2, V2]
coe2 = coe_from_sv(R2, V2);
%...Equations 8.102 and 8.103:
vinf1 = V1 - Vp1;
vinf2 = V2 - Vp2;
%...Echo the input data and output the solution to
%
the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 8.8: Earth to Venus')
fprintf('\n\n Departure:\n');
fprintf('\n
Planet: %s', planet_name(depart(1)))
fprintf('\n
Year : %g', depart(2))
fprintf('\n
Month : %s', month_name(depart(3)))
fprintf('\n
Day
: %g', depart(4))
fprintf('\n
Hour : %g', depart(5))
fprintf('\n
Minute: %g', depart(6))
fprintf('\n
Second: %g', depart(7))
fprintf('\n\n
Julian day: %11.3f\n', jd1)
fprintf('\n
Planet position vector (km)
= [%g %g %g]', ...
R1(1),R1(2), R1(3))
fprintf('\n
Magnitude
= %g\n', norm(R1))
fprintf('\n
Planet velocity (km/s)
= [%g %g %g]', ...
Vp1(1), Vp1(2), Vp1(3))
fprintf('\n
Magnitude
= %g\n', norm(Vp1))
fprintf('\n
Spacecraft velocity (km/s)
= [%g %g %g]', ...
V1(1), V1(2), V1(3))
fprintf('\n
Magnitude
= %g\n', norm(V1))
fprintf('\n
v-infinity at departure (km/s) = [%g %g %g]', ...
vinf1(1), vinf1(2), vinf1(3))
fprintf('\n
Magnitude
= %g\n', norm(vinf1))
fprintf('\n\n Time of flight = %g days\n', tof)
fprintf('\n\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n\n
fprintf('\n
Arrival:\n');
Planet: %s', planet_name(arrive(1)))
Year : %g', arrive(2))
Month : %s', month_name(arrive(3)))
Day
: %g', arrive(4))
Hour : %g', arrive(5))
Minute: %g', arrive(6))
Second: %g', arrive(7))
Julian day: %11.3f\n', jd2)
Planet position vector (km)
= [%g %g %g]', ...
R2(1), R2(2), R2(3))
fprintf('\n
Magnitude
fprintf('\n
Planet velocity (km/s)
= [%g %g %g]', ...
Vp2(1), Vp2(2), Vp2(3))
= %g\n', norm(R1))
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Chapter 8
fprintf('\n
Magnitude
= %g\n', norm(Vp2))
fprintf('\n
Spacecraft Velocity (km/s)
= [%g %g %g]', ...
V2(1), V2(2), V2(3))
fprintf('\n
Magnitude
= %g\n', norm(V2))
fprintf('\n
v-infinity at arrival (km/s) = [%g %g %g]', ...
vinf2(1), vinf2(2), vinf2(3))
fprintf('\n
Magnitude
= %g', norm(vinf2))
fprintf('\n\n\n Orbital elements of flight trajectory:\n')
fprintf('\n
Angular momentum (km^2/s)
= %g',...
coe(1))
fprintf('\n Eccentricity
= %g',...
coe(2))
fprintf('\n Right ascension of the ascending node (deg) = %g',...
coe(3)/deg)
fprintf('\n Inclination to the ecliptic (deg)
= %g',...
coe(4)/deg)
fprintf('\n Argument of perihelion (deg)
= %g',...
coe(5)/deg)
fprintf('\n True anomaly at departure (deg)
= %g',...
coe(6)/deg)
fprintf('\n True anomaly at arrival (deg)
= %g\n',...
coe2(6)/deg)
fprintf('\n Semimajor axis (km)
= %g',...
coe(7))
% If the orbit is an ellipse, output the period:
if coe(2) &lt; 1
fprintf('\n Period (days)
= %g',...
2*pi/sqrt(mu)*coe(7)^1.5/24/3600)
end
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 8.8: Earth to Venus
Departure:
Planet:
Year :
Month :
Day
:
Hour :
Minute:
Second:
Earth
2005
December
1
0
0
0
Julian day: 2453705.500
Planet position vector (km)
Magnitude
= [5.33243e+07
= 1.47517e+08
Planet velocity (km/s)
Magnitude
= [-28.2595
= 30.202
10.6564
-6.01367e-05]
Spacecraft velocity (km/s)
Magnitude
= [-27.0436
= 27.9729
6.58196
2.7931]
v-infinity at departure (km/s) = [1.2159
145
1.37541e+08
-4.07444
2.79316]
-1830.84]
Solutions Manual
Orbital Mechanics for Engineering Students
Magnitude
Chapter 8
= 5.08735
Time of flight = 121 days
Arrival:
Planet:
Year :
Month :
Day
:
Hour :
Minute:
Second:
Venus
2006
April
1
0
0
0
Julian day: 2453826.500
Planet position vector (km)
Magnitude
= [-5.74135e+07
= 1.47517e+08
-9.1938e+07
Planet velocity (km/s)
Magnitude
= [29.4613
= 34.955
-18.7099
-1.95644]
Spacecraft Velocity (km/s)
Magnitude
= [30.3918
= 37.8351
-22.2324
-3.68154]
v-infinity at arrival (km/s)
Magnitude
= [0.930541
= 4.0311
-3.52248
2.05581e+06]
-1.7251]
Orbital elements of flight trajectory:
Angular momentum (km^2/s)
= 4.0914e+09
Eccentricity
= 0.183291
Right ascension of the ascending node (deg) = 68.8159
Inclination to the ecliptic (deg)
= 5.77973
Argument of perihelion (deg)
= 142.255
True anomaly at departure (deg)
= 217.738
True anomaly at arrival (deg)
= 26.8913
Semimajor axis (km)
= 1.30519e+08
Period (days)
= 297.66
----------------------------------------------------The output shows that at departure from earth, v ∞ = 5.087 km s . Hence, the spacecraft velocity at
perigee of the departure hyperbola is
v p = v ∞2 +
2&micro;earth
rp2
= 5.087 2 +
2 ⋅ 398 600
(6378 + 180)2
= 12.14 km s
The spacecraft velocity in its circular 180 km parking orbit is
vc =
398 600
&micro;earth
=
= 7.796 km s
rp
6378 + 180
Hence, the delta-v requirement at earth is
∆v1 = v p − vc = 12.14 − 7.796 = 4.346 km s
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Chapter 8
At Venus ( &micro;Venus = 324 900 km 3 s2 , rVenus = 6052 km ) the above output shows that
v ∞ = 4.031 km s . The speed at the 300 km altitude periapse on the arrival hyperbola is therefore
vp
hyperbola
= v ∞2 +
2&micro;Venus
rp2
= 4.0312 +
2 ⋅ 324 900
(6052 + 300)2
= 10.89 km s
The semimajor axis of the elliptical capture orbit is
a=
1
+ 300) + (rVenus + 9000) ] = 10 702
[(r
2 Venus
Therefore the velocity at periapse on the ellipse is, using the energy equation,
vp
ellipse
2
1 
 2 1

= &micro;Venus  −  = 324 900
−
 = 8.482 km s
 6052 + 300 10702 
 rp a 
It follows that the delta-v requirement at Venus is
∆v2 = v p
hyperbola
− vp
ellipse
= 10.89 − 8.482 = 2.406 km s
The total delta-v requirement is
∆vtotal = ∆v1 + ∆v2 = 4.346 + 2.406 = 6.753 km s
Problem 8.9 For the data given in this problem, the following MATLAB script invokes Algorithm
8.2, which is implemented as the MATLAB M-function interplanetary in Appendix D.18.
interplanetary uses Algorithms 8.1 and 5.2 to compute v ∞ at the home and target planets. The
output to the MATLAB Command Window is listed afterwards.
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem_8_09
~~~~~~~~~~~~
This program uses Algorithm 8.2 to obtain v-infinities
in Problem 8.9.
mu
deg
pi
- gravitational parameter of the sun (km^3/s^2)
- conversion factor between degrees and radians
- 3.1415926...
planet_id
- planet identifier:
1 = Mercury
2 = Venus
3 = Earth
4 = Mars
5 = Jupiter
6 = Saturn
7 = Uranus
8 = Neptune
9 = Pluto
year
month
day
hour
minute
second
-
range:
range:
range:
range:
range:
range:
1901 - 2099
1 - 12
1 - 31
0 - 23
0 - 60
0 - 60
147
Solutions Manual
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depart
arrive
Orbital Mechanics for Engineering Students
Chapter 8
- [planet_id, year, month, day, hour, minute, second]
at departure
- [planet_id, year, month, day, hour, minute, second]
at arrival
planet1
planet2
trajectory
- [Rp1, Vp1, jd1]
- [Rp2, Vp2, jd2]
- [V1, V2]
coe
- orbital elements [h e RA incl w TA]
where
h
= angular momentum (km^2/s)
e
= eccentricity
RA
= right ascension of the ascending
incl = inclination of the orbit (rad)
w
TA
a
= semimajor axis (km)
jd1, jd2
tof
- Julian day numbers at departure and arrival
- time of flight from planet 1 to planet 2 (days)
Rp1, Vp1
Rp2, Vp2
R1, V1
- state vector of planet 1 at departure (km, km/s)
- state vector of planet 2 at arrival (km, km/s)
- heliocentric state vector of spacecraft at
departure (km, km/s)
- heliocentric state vector of spacecraft at
arrival (km, km/s)
R2, V2
vinf1, vinf2 - hyperbolic excess velocities at departure
and arrival (km/s)
User M-functions required: interplanetary, coe_from_sv,
month_planet_names
--------------------------------------------------------------------
clear
global mu
mu = 1.327124e11;
deg = pi/180;
%...Data declaration for Problem 8.9:
%...Departure
planet_id = 3;
year
= 2005;
month
= 8;
day
= 15;
hour
= 0;
minute
= 0;
second
= 0;
depart = [planet_id
%...Arrival
planet_id =
year
=
month
=
day
=
hour
=
minute
=
second
=
year
month
day
4;
2006;
3;
15;
0;
0;
0;
148
hour
minute
second];
Solutions Manual
Orbital Mechanics for Engineering Students
arrive = [planet_id
year
month
day
hour
minute
Chapter 8
second];
%...
%...Algorithm 8.2:
[planet1, planet2, trajectory] = interplanetary(depart, arrive);
R1 = planet1(1,1:3);
Vp1 = planet1(1,4:6);
jd1 = planet1(1,7);
R2 = planet2(1,1:3);
Vp2 = planet2(1,4:6);
jd2 = planet2(1,7);
V1
V2
= trajectory(1,1:3);
= trajectory(1,4:6);
tof = jd2 - jd1;
%...Use Algorithm 5.1 to find the orbital elements of the
%
spacecraft trajectory based on [Rp1, V1]...
coe = coe_from_sv(R1, V1);
%
... and [R2, V2]
coe2 = coe_from_sv(R2, V2);
%...Equations 8.102 and 8.103:
vinf1 = V1 - Vp1;
vinf2 = V2 - Vp2;
%...Echo the input data and output the solution to
%
the command window:
fprintf('-----------------------------------------------------')
fprintf('\n Problem 8.9: Earth to Mars')
fprintf('\n\n Departure:\n');
fprintf('\n
Planet: %s', planet_name(depart(1)))
fprintf('\n
Year : %g', depart(2))
fprintf('\n
Month : %s', month_name(depart(3)))
fprintf('\n
Day
: %g', depart(4))
fprintf('\n
Hour : %g', depart(5))
fprintf('\n
Minute: %g', depart(6))
fprintf('\n
Second: %g', depart(7))
fprintf('\n\n
Julian day: %11.3f\n', jd1)
fprintf('\n
Planet position vector (km)
= [%g %g %g]', ...
R1(1),R1(2), R1(3))
fprintf('\n
Magnitude
= %g\n', norm(R1))
fprintf('\n
Planet velocity (km/s)
= [%g %g %g]', ...
Vp1(1), Vp1(2), Vp1(3))
fprintf('\n
Magnitude
= %g\n', norm(Vp1))
fprintf('\n
Spacecraft velocity (km/s)
= [%g %g %g]', ...
V1(1), V1(2), V1(3))
fprintf('\n
Magnitude
= %g\n', norm(V1))
fprintf('\n
v-infinity at departure (km/s) = [%g %g %g]', ...
vinf1(1), vinf1(2), vinf1(3))
fprintf('\n
Magnitude
= %g\n', norm(vinf1))
149
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 8
fprintf('\n\n Time of flight = %g days\n', tof)
fprintf('\n\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n
fprintf('\n\n
fprintf('\n
Arrival:\n');
Planet: %s', planet_name(arrive(1)))
Year : %g', arrive(2))
Month : %s', month_name(arrive(3)))
Day
: %g', arrive(4))
Hour : %g', arrive(5))
Minute: %g', arrive(6))
Second: %g', arrive(7))
Julian day: %11.3f\n', jd2)
Planet position vector (km)
= [%g %g %g]', ...
R2(1), R2(2), R2(3))
fprintf('\n
Magnitude
fprintf('\n
Planet velocity (km/s)
= [%g %g %g]', ...
Vp2(1), Vp2(2), Vp2(3))
fprintf('\n
Magnitude
= %g\n', norm(Vp2))
fprintf('\n
Spacecraft Velocity (km/s)
= [%g %g %g]', ...
V2(1), V2(2), V2(3))
fprintf('\n
Magnitude
= %g\n', norm(V2))
fprintf('\n
v-infinity at arrival (km/s) = [%g %g %g]', ...
vinf2(1), vinf2(2), vinf2(3))
fprintf('\n
Magnitude
= %g\n', norm(R1))
= %g', norm(vinf2))
fprintf('\n\n\n Orbital elements of flight trajectory:\n')
fprintf('\n
Angular momentum (km^2/s)
= %g',...
coe(1))
fprintf('\n Eccentricity
= %g',...
coe(2))
fprintf('\n Right ascension of the ascending node (deg) = %g',...
coe(3)/deg)
fprintf('\n Inclination to the ecliptic (deg)
= %g',...
coe(4)/deg)
fprintf('\n Argument of perihelion (deg)
= %g',...
coe(5)/deg)
fprintf('\n True anomaly at departure (deg)
= %g',...
coe(6)/deg)
fprintf('\n True anomaly at arrival (deg)
= %g\n',...
coe2(6)/deg)
fprintf('\n Semimajor axis (km)
= %g',...
coe(7))
% If the orbit is an ellipse, output the period:
if coe(2) &lt; 1
fprintf('\n Period (days)
= %g',...
2*pi/sqrt(mu)*coe(7)^1.5/24/3600)
end
fprintf('\n-----------------------------------------------------\n')
----------------------------------------------------Problem 8.9: Earth to Mars
Departure:
Planet: Earth
150
Solutions Manual
Year :
Month :
Day
:
Hour :
Minute:
Second:
Orbital Mechanics for Engineering Students
Chapter 8
2005
August
15
0
0
0
Julian day: 2453597.500
Planet position vector (km)
Magnitude
= [1.19728e+08
= 1.51517e+08
-9.28572e+07
798.533]
Planet velocity (km/s)
Magnitude
= [17.7711
= 29.405
23.4274
-0.000315884]
Spacecraft velocity (km/s)
Magnitude
= [20.6107
= 33.0431
25.7677
1.75181]
v-infinity at departure (km/s) = [2.83967
Magnitude
= 4.07557
2.34021
1.75212]
Time of flight = 212 days
Arrival:
Planet:
Year :
Month :
Day
:
Hour :
Minute:
Second:
Mars
2006
March
15
0
0
0
Julian day: 2453809.500
Planet position vector (km)
Magnitude
= [-8.33472e+07
= 1.51517e+08
2.26736e+08
Planet velocity (km/s)
Magnitude
= [-21.8221
= 22.7173
-6.30169
0.40447]
Spacecraft Velocity (km/s)
Magnitude
= [-20.4825
= 20.9374
-4.25753
-0.845206]
v-infinity at arrival (km/s)
Magnitude
= [1.33959
= 2.74496
2.04416
-1.24968]
Orbital elements of flight trajectory:
Angular momentum (km^2/s)
= 5.00602e+09
Eccentricity
= 0.246978
Right ascension of the ascending node (deg) = 322.198
Inclination to the ecliptic (deg)
= 3.03935
Argument of perihelion (deg)
= 355.671
True anomaly at departure (deg)
= 4.33479
True anomaly at arrival (deg)
= 152.278
Semimajor axis (km)
= 2.01098e+08
Period (days)
= 569.273
-----------------------------------------------------
151
6.79991e+06]
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 8
The output shows that at departure from earth, v ∞ = 4.076 km s . Hence, the spacecraft velocity at
perigee of the departure hyperbola is
v p = v ∞2 +
2&micro;earth
rp2
= 4.076 2 +
2 ⋅ 398 600
(6378 + 190)2
= 11.75 km s
The spacecraft velocity in its circular 190 km parking orbit is
vc =
398 600
&micro;earth
=
= 7.790 km s
rp
6378 + 190
Hence, the delta-v requirement at earth is
∆v1 = v p − vc = 11.75 − 7.790 = 3.957 km s
At Mars ( &micro;Mars = 42 830 km 3 s2 , rMars = 3396 km ) the above output shows that v ∞ = 2.745 km s .
The speed at the 300 km altitude periapse on the arrival hyperbola is therefore
vp
hyperbola
= v ∞2 +
2&micro;Mars
= 2.745 2 +
rp2
2 ⋅ 42 830
(3396 + 300)2
= 5.542 km s
The semimajor axis of the capture ellipse is found from the required 35 hour period.
T=
2π
&micro;Mars
35 ⋅ 3600 =
a 3/2
2π
42 830
a 3/2 ⇒ a = 25 830 km
Therefore the velocity at periapse on the ellipse is, using the energy equation,
vp
ellipse
2
1 

 2 1
= &micro;Mars  −  = 42 830
−
 = 4.639 km s
3396
300
25830
r
a
+


 p

It follows that the delta-v requirement at Mars is
∆v2 = v p
hyperbola
− vp
ellipse
= 5.542 − 4.639 = 0.9030 km s
The total delta-v requirement is
∆vtotal = ∆v1 + ∆v2 = 3.957 + 0.9030 = 4.860 km s
Problem 8.10
Rearth = 149.6 &times; 10 6 km
RSaturn = 1433 &times; 10 6 km
Semimajor axis of Hohmann transfer ellipse:
a=
1
1
Rearth + RSaturn ) = 149.6 &times; 10 6 + 1433 &times; 10 6 = 791.3 &times; 10 6 km
(
2
2
(
)
Period of Hohmann transfer ellipse:
152
Solutions Manual
T=
2π
&micro;sun
Orbital Mechanics for Engineering Students
a 3/2 =
2π
132.7 &times; 10 9
Chapter 8
(791.3 &times; 10 6 )3/2 = 383.9 &times; 10 6 s = 12.17 y
Therefore, the time of flight to Saturn’s orbit is T 2 = 6.083 y . Cassini departed on 15 October 1997
(Julian day 2 450 736.5) and arrived on 1 July 2004 (Julian day 2 453 187.5). The number of years for
Cassinni’s flight was
2 453 187.5 - 2 450 736.5
= 6.71 y
365.25
Cassini, with its several flyby maneuvers, required a flight time only about 10 percent longer than the
Hohmann transfer.
The velocity of the spacecraft at the outbound crossing of the earth’s sphere of influence is
V
( v)
1
2
1
 2


= &micro;sun 
−  = 132.7 &times; 10 9 
−
 = 40.08 km s
 149.6 &times; 10 6 791.3 &times; 10 6 
 Rearth a 
The velocity of earth in its (assumed) circular orbit is
Vearth =
&micro;sun
132.7 &times; 10 9
=
= 29.78 km s
Rearth
149.6 &times; 10 6
Thus
v∞ = V
( v)
− Vearth = 40.08 − 29.78 = 10.3 km s
The spacecraft velocity at the 180 km altitude perigee of the departure hyperbola is
v p = v ∞2 +
2 ⋅ 398 600
2&micro;earth
= 10.3 2 +
= 15.09 km s
rp
6378 + 180
The velocity in the circular parking orbit is
vc =
398 600
&micro;earth
=
= 7.796 km s
rp
6378 + 180
Hence,
∆v = v p − vc = 15.09 − 7.796 = 7.289 km s
From Equation 6.1


7.289
 ∆v  


∆m = m o 1 − exp −
= m o 1 − exp −
 = 0.916m o


−
3
 300 ⋅ 9.81 &times; 10  
 I sp go  


But ∆m equals the mass m p of propellant expended, and the initial mass m o equals m p plus the mass
of the spacecraft (2000 kg). Thus,
(
m p = 0.916 2000 + m p
)
⇒ m p = 21810 kg
153
Solutions Manual
Orbital Mechanics for Engineering Students
154
Chapter 8
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 9
Problem 9.1
(
)
ˆ as the basis.
Method 1: Use Iˆ , Jˆ , K
ˆ
iˆ = sin θ sin φ Iˆ − sin θ cos φ Jˆ + cos θ K
ˆj = cos φ Iˆ + sin φ Jˆ
(1)
ˆ
kˆ = − cos θ sin φ Iˆ + cos θ cos φ Jˆ + sin θ K
(3)
ˆ + 3 ˆj + 4 iˆ
ω = 2K
(4)
(2)
α = ω = 3 ˆj + 4 iˆ
(
)
ˆj = ( 2K
ˆ ) &times; ˆj = ( 2K
ˆ ) &times; cos φ Iˆ + sin φ Jˆ = −2 sin φ Iˆ + 2 cos φ Jˆ
(
)
(
) (
ˆ + 3 ˆj &times; iˆ = 2K
ˆ + 3 cos φ Iˆ + 3 sin φ Jˆ &times; sin θ sin φIˆ − sin θ cos φ Jˆ + cos θ K
ˆ
iˆ = 2K
ˆ
= ( 3 cos θ sin φ + 2 sin θ cos φ ) Iˆ + ( 2 sin θ sin φ − 3 cos θ cos φ ) Jˆ − 3 sin θ K
(
)
)
ˆ
α = 3 −2 sin φ Iˆ + 2 cos φ Jˆ + 4 ( 3 cos θ sin φ + 2 sin θ cos φ ) Iˆ + ( 2 sin θ sin φ − 3 cos θ cos φ ) Jˆ − 3 sin θ K

ˆ
ˆ
ˆ
= ( −6 sin φ + 12 cos θ sin φ + 8 sin θ cos φ ) I + ( 6 cos φ − 12 cos θ cos φ + 8 sin θ sin φ ) J − 12 sin θ K
α = α ⋅α
=
( −6 sin φ + 12 cos θ sin φ + 8 sin θ cos φ )2 + ( 6 cos φ − 12 cos θ cos φ + 8 sin θ sin φ )2 + 144 sin 2 θ
=
(sin 2 φ + cos2 φ ) ( 36 + 144 cos2 θ − 144 cosθ + 64 sin 2 θ ) + 144 sin 2 θ
= 36 + 144 ( cos2 θ + sin 2 θ ) − 144 cos θ + 64 sin 2 θ
= 36 + 144 − 144 cos θ + 64 sin 2 θ
α = 180 − 144 cos θ + 64 sin 2 θ
(
)
Method 2: Use iˆ , ˆj, kˆ as the basis. Multiply (1) through by cos θ and (2) by sin θ to obtain
ˆ = cos θ iˆ
sin θ cos θ sin φ Iˆ − sin θ cos θ cos φ Jˆ + cos2 θ K
ˆ = sin θ kˆ
− sin θ cos θ sin φ Iˆ + sin θ cos θ cos φ Jˆ + sin 2 θ K
ˆ = cos θ iˆ + sin θ kˆ
K
Then (4) can be written
ω = 2 ( cos θ iˆ + sin θ kˆ ) + 3 ˆj + 4 iˆ = ( 4 + 2 cos θ ) iˆ + 3 ˆj + 2 sin θ kˆ
α=
dω 
 +Ω&times;ω
dt  rel
dω 
 = −2θ sin θ iˆ + 2θ cos θ kˆ = −6 sin θ iˆ + 6 cos θ kˆ
dt  rel
155
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Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 9
ˆ + 3 ˆj = 2 ( cos θ iˆ + sin θ kˆ ) + 3 ˆj
Ω = 2K
iˆ
Ω &times; ω = 2 cos θ
4 + 2 cos θ
ˆj
kˆ
3 2 sin θ = 8 sin θ ˆj − 12k̂
3 sin θ
∴ α = ( −6 sin θ iˆ + 6 cos θ kˆ ) + 8 sin θ ˆj − 12kˆ = −6 sin θ iˆ + 8 sin θ ˆj + ( 6 cos θ − 12 ) kˆ
α = α ⋅ α = 36 sin 2 θ + 64 sin 2 θ + ( 36 cos2 θ − 144 cos θ + 144 )
= 36 ( sin 2 θ + cos2 θ ) + 64 sin 2 θ − 144 cos θ + 144
= 36 + 64 sin 2 θ − 144 cos θ + 144
α = 180 + 64 sin 2 θ − 144 cos θ
Problem 9.2
(a)
{
}
ˆ + ψ nˆ
ω plate = (θ kˆ ) + φ ˆj  + ν m
(1)
ˆ = sin φ iˆ + cos φ kˆ
ˆ &times; ˆj = − cos φ iˆ + sin φ kˆ
m
pˆ = m
nˆ = cos ν ˆj + sin ν pˆ = − cos φ sin ν iˆ + cos ν ˆj + sin φ sin ν kˆ
Substituting m̂ and n̂ into (1) yields the result,
ω plate = (ν sin φ − ψ cos φ sin ν ) iˆ + (φ + ψ cos ν ) ˆj + θ + ν cos φ + ψ sin φ sin ν k̂
(
)
(2)
(b)
α plate =
dω plate
dt
=
dω plate 
 + Ω &times; ω plate
dt  rel
(3)
where Ω is the angular velocity of the xyz frame, which is θ k̂ That is,
Ω = θ k̂
(4)
From (2)
dω plate 
 d
 d

d
θ + ν cos φ + ψ sin φ sin ν  kˆ
 =  (ν sin φ − ψ cos φ sin ν )  iˆ +  (φ + ψ cos ν )  ˆj + 
dt
  dt
  dt

rel  dt
(
)
Taking the derivatives, bearing in mind that θ , φ and ψ are all constant, we get
dω plate 
os φ cos ν  iˆ
 = φ (ν cos φ + ψ sin φ sin ν ) − ψν co
dt  rel 
− ψν sin ν ˆj + φ ( −ν sin φ + ψ cos φ sin ν ) + ψν cos ν sin φ  k̂
From (2) and (4),
156
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Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 9
Ω &times; ω plate = θ kˆ &times; (ν sin φ − ψ cos φ sin ν ) iˆ

+ (φ + ψ cos ν ) ˆj + θ + ν cos φ + ψ sin φ sin ν kˆ 

ˆ
= −θ (ψ cos ν + φ ) i + θ (ν sin φ − ψ cos φ sin ν ) ˆj
(
)
(6)
Substituting (5) and (6) into (3) and collecting terms yields the result,
α plate = ν φ cos φ − ψ cos φ cos ν + ψφ sin φ sin ν − ψθ cos ν − φθ  iˆ


ˆ


+ ν θ sin φ − ψ sin ν − ψθ cos φ sin ν j


+ ψν cos ν sin φ + ψφ cos φ sin ν − φν sin φ  kˆ
(
(
)
)
(7)
(c)
(
aC = a B + α BC &times; rC B + ω BC &times; ω BC &times; rC B
)
(8)
ˆ = l sin φ iˆ + cos γ kˆ
rC B = lm
ω
= θ kˆ + φ ˆj
(9)
(10)
BC
α BC =
dω BC dω BC 
=
 + Ω &times; ω BC = 0 + θ kˆ &times; θ kˆ + φ ˆj = −θφ iˆ
dt
dt  rel
(
a B = a O + α AB &times; rB
O
(
+ ω AB &times; ω AB &times; rB
O
)
(11)
)
(12)
= 0 + 0 &times; 1.25ljˆj + θ kˆ &times; θ kˆ &times; 1.25lˆj = −1.25θ 2 lˆj
(
)
(
)
Substituting (9), (10), (11) and (12) into (8) yields
(
) (
aC = −1.25θ 2 lˆj + −θφ iˆ &times; l sin φ iˆ + cos γ kˆ + θ kˆ + φ ˆj &times;  θ kˆ + φ ˆj &times; l sin φ iˆ + cos γ kˆ 
) (
) (
) (
Upon expanding and collecting terms, we get the result
(
)
(
)
aC = − l φ 2 + θ 2 sin φ iˆ + 2lφθ cos φ − 1.25lθ 2 ĵj − lφ 2 cos φ kˆ
Problem 9.3
dv 
 +ω&times;v
dt  rel
d 3ˆ
=
t i + 4 ˆj + ( 2t 2kˆ ) &times; t 3 iî + 4 ˆj
dt
= 3t 2 iˆ + −8t 2 iˆ + 2t 5 ˆj
aG =
(
)
(
(
)
)
= −5t 2 iˆ + 2t 5 ˆj
At t = 2 s
(
aG = −20iˆ + 64 ˆj m 2 s
)
Problem 9.4
α=
dω 
 +Ω&times;ω
dt  rel
α=
d
ω iˆ + ω y ˆj + ω z kˆ + ω x iˆ + ω y ˆj &times; ω x iˆ + ω y ˆj + ω z k̂
dt x
(
) (
) (
157
)
) (
)
Solutions Manual
Orbital Mechanics for Engineering Students
iˆ
α = 0 + ωx
ˆj
kˆ
ωy
0
Chapter 9
ωx ω y ωz
α = ω yω z iˆ − ω xω z ˆj
Problem 9.5
y 2 + z 2
i
 i
[I i ] = mi  −xi y i
 −x z
i i

 20 −10
[I1 ] =  −10 20
 −10 −10
 256 128
I
=
[ 3 ]  128 256
 −128 128
216 108

[I 5 ] = 108 216
108 −108
− xi y i
2
xi + z i
2
− y i zi
− xi z i 

− y i zi 
xi2 + y i2 
−10 
−10  kg - m 2

20 
−128 

128  kg - m 2
256 
108 
−108  kg - m 2

216 
(
)
(
)
(
)
 20
[I 2 ] =  −10
 −10
 64

I
=
[ 4 ]  32
 −32
216

[I 6 ] = 108
108
792 356 36 
[I O ] = ∑ [I i ] = 356 792 −76  kg - m 2
i =1
 36 −76 792 
6
(
−10 −10 
20 −10  kg - m 2

−10 20 
32 −32 

64 32  kg - m 2
32 64 
108
108 
216 −108  kg - m 2
−108 216 
(
)
(
)
(
)
)
6
m=
∑ mi = 60
kg
i =1
6
∑ m i xi
xG = i =1
m
=
16
= 0.2667 m
60
=
−16
= −0.2667 m
60
=
16
= 0.2667 m
60
6
∑ mi y i
y G = i =1
m
6
∑ m i zi
zG =
[
i =1
m
y 2 + z 2
G
 G
Ι mP = m  − xG yG
 −x z
G G

]
[I G ] = [I P ] − [ Ι mP ]
− xG yG
2
xG + zG
− yG zG
2
− xG zG   8.533

− yG zG  =  4.267

xG2 + yG2   −4.267
792 356 36   8.533 4.267
= 356 792 −76  −  4.267 8.533

 
 36 −76 792   −4.267 4.267
158
4.267
8.533
4.267
−4.267 
4.267 

8.533 
−4.267 
4.267  kg - m 2

8.533 
(
)
Solutions Manual
783.5
[I G ] = 351.7
40.27
Orbital Mechanics for Engineering Students
351.7
783.5
−80.27
40.27 
−80.27  kg - m 2
783.5 
(
)
Problem 9.6
From the previous problem
792 356 36 

[I O ] = 356 792 −76  kg ⋅ m 2
 36 −76 792 
(
iˆ′ =
iˆ + 2 ˆj + 2kˆ
12 + 2 2 + 2 2
)
= 0.3333iˆ + 0.6667 ˆj + 0.6667kˆ
ˆ  = 0.3333 0.6667 0.6667 
 i′
 792 356 36 
T




I x′ =  î′   356 792 −76   iˆ′ 
 36 −76 792 
 7922 356 36  0.3333 



= 0.3333 0.6667 0.6667   356 792 −76  0.6667 
 36 −76 792  0.6667 
 535.3 


= 0.3333 0.6667 0.6667   596 
 489.3 


I x′ = 898.7 kg ⋅ m 2
Problem 9.7
 b2 + c 2
m
[I G ] = 12  −ab
 − ac

− ab
a + c2
− bc
2
− ac 

− bc 
a 2 + b2 
a = l cos θ
b = l sin θ
c =0
 ( l sin θ )2 + (0)2 −( l cos θ )( l sin θ )

−( l cos θ )(0)

m
2
2
−( l sin θ )(0)
[I G ] = 12  −( l cos θ )( l sin θ ) ( l cos θ ) + (0)

2
2
 −( l cos θ )(0)
(
)(
)
(
)
(
)
l
l
l
sin
θ
cos
θ
sin
θ
0
−
+



2
 sin θ
ml2  sin 2θ
[I G ] = 12  − 2

 0
−
sin 2θ
2
cos2 θ
0

0

0

1 
159
Chapter 9
Solutions Manual
Orbital Mechanics for Engineering Students
Problem 9.8
(a)
 b2 + c 2
0
0 

m
2
2
a +c
0 
[I G ] = 12  0
 0
a 2 + b2 
0

2 2 + 12
0
0 

1000 
2
2
=
3 +1
0 
 0
12 
0
3 2 + 2 2 
 0
416.7
0
0 

=
0
833.3
0  kg ⋅ m 2


 0
0
1083 
(
y 2 + z 2
G
 G
I mO = m  − xG yG
 −x z
G G

[
]
xG =
a
= 1.5 m
2
[
∴ I mO
]
− xG yG
xG2 + zG2
− yG zG
yG =
)
− xG zG 

− yG zG 
xG2 + yG2 
b
=1 m
2
zG =
c
= 0.5 m
2
 1250 −1500 −750 


=  −1500 2500 −500  kg ⋅ m 2
 −750 −500 3250 
(
)
0
0   1250 −1500 −750 
416.7
 0
833.3
0  +  −1500 2500 −500 
[I O ] = [I G ] + I mO = 
 0
0
1083   −750 −500 3250 
 1667 −1500 −750 
[I O ] =  −1500 3333 −500  kg ⋅ m 2
 −750 −500 4333 
[
]
(
)
(b)
1667 − λ
−1500
−1500
−750
3333 − λ
−500
−750
−500 = 0
4333 − λ
λ3 − 9333λ2 + 24.16(10 6 )λ − 10.91(10 9 ) = 0
λ1 = 568.9 kg ⋅ m 2
λ2 = 4209 kg ⋅ m 2
λ3 = 4556 kg ⋅ m 2
Principal direction 1:
( )
1667 − 568.9
−1500
−750
  v x1  0 

 −1500
 v (1)  0 
3333 − 568.9
−500

 y  =  

−750
−500
4333 − 568.9   v (z1)  0 


 1098 −1500 −750   1  0 
 −1500 2764 −500   v (1)  = 0 

 y   
 −750 −500 3764   v (z1)  0 


(1 ) 

 2764 −500   v y  1500 
 −500 3764   (1)  =  750 

  v z  

160
Chapter 9
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 9
 v (1)   2764 −500  −1 1500  0.5929 
y
=

 (1)  = 
 
 v z   −500 3764   750   0.278 
iˆ + 0.5829 ˆj + 0.278kˆ
( )
vˆ 1 =
= 0.8366iˆ + 0.496 ˆj + 0.2326kˆ
2
2
2
1 + 0.5829 + 0.278
Principal direction 2:
( )
1667 − 4209
−1500
−750
  v x2  0 

 −1500
 v (2 )  0 
3333
−
4209
−
500

 y  =  
 −750
−500
4333 − 4209   v (z2 )  0 


 −2542 −1500 −750   1  0 
 −1500 −875.5 −500   v (2 )  = 0 

 y   
 −750
−500 124.5  v (z2 )  0 


 −875.5 −500   v (y2 )  1500 
 −500 124.5  (2 )  =  750 

  v z  

−
1
(
)
2
 v   −875.5 −500  1500   −1.565 
y
 (2 )  = 
=

 
 v z   −500 124.5  750   −0.2601
iˆ − 1.565 ˆj − 0.2601kˆ
( )
= 0.5333iˆ − 0.8345 ˆj − 0.1387kˆ
vˆ 2 =
2
2
2 (
1 + −1.565 ) + ( −0.2601)
Principal direction 3:
( )
1667 − 4556
−1500
−750
  v x3  0 

 −1500
 (3 )  0 
3333 − 4556
−500

 v y  =  
 −750
−500
4333 − 4556   v (z3 )  0 


 −2889 −1500 −750   1  0 
 −1500 −1222 −500   v (3 )  = 0 

 y   
 
( )
 −750 −500 −222.3   v z3  0 
 −1222 −500   v (y3 )  1500 
 −500 −222.3   (3 )  =  750 

  v z  

 v (3 )   −1222 −500  −1 1500   1.916 
y
 (3 )  = 

=

−500 −222.3   750   −7.685
v
 z  
iˆ + 1.916 ˆj − 7.685kˆ
( )
vˆ 3 =
= 0.1253iˆ + 0.2401ˆj − 0.9626kˆ
2
12 + 1.916 2 + ( −7.685 )
The following MATLAB script uses the built-in function eig to obtain these results, as shown in the
Command Window output which follows.
fprintf('\n---------------------------------------------------\n')
Matrix = [ 1666.7
-1500
-750
-1500 3333.3
-500
-750
-500 4333.3]
[eigvector, eigvalue] = eig(I);
fprintf('\n Eigenvalue
for i = 1:3
Eigenvector')
161
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 9
fprintf('\n %g
[%g %g %g]', eigvalue(i,i), eigvector(1,i),
eigvector(2,i), eigvector(3,i))
end
fprintf('\n---------------------------------------------------\n')
--------------------------------------------------Matrix =
1666.7
-1500
-750
-1500
3333.3
-500
-750
-500
4333.3
Eigenvalue
Eigenvector
4208.83
[-0.533304 0.834478 0.138683]
4555.59
[0.125278 0.240046 -0.962644]
568.881
[0.836596 0.496008 0.232559]
--------------------------------------------------(c)
iˆ′ =
3iˆ + 2 ˆj + kˆ
2
2
3 + 2 +1
2
= 0.8018iˆ + 0.5345 ˆj + 0.2673k̂
 iˆ′  = 0.8018 0.5345 0.2673 


Ix′
 1667 −1500 −750  0.8018 
T



ˆ
ˆ






I
=  i′   O   i′  = 0.8018 0.5345 0..2673   −1500 3333 −500   0.5345 
 −750 −500 4333   0.2673 
334.1 


I x′ = 0.8018 0.5345 0.2673  445.4 
239.5 


I x′ = 583.3 kg ⋅ m 2
Problem 9.9
1


ml2ω

0
0
 ω  
12
1

ml2 
sin 2θ     1
2
2
{H C } = [I C ]{ω } = 12 0 sin θ − 2   0  = − 24 ml Ω sin 2θ 


 Ω   1
sin 2θ
2
2


Ω
θ
ml
cos2 θ 
cos
0 −
 12



2
1
1
1
HC =
ml 2ω iˆ −
ml 2Ω sin 2θ ˆj + ml 2Ω cos2 θ kˆ
12
24
12
H P = H C + rC P &times; mv C
rC P = d iˆ
v C = Ω kˆ &times; d iˆ = Ω dˆj
H P = HC + d iˆ &times; mΩ dˆj = HC + mΩ d 2kˆ
1
1

 1
H P =  ml 2ω iˆ −
ml 2Ω sin 2θ ˆj + ml 2Ω cos2 θ kˆ  + mΩ d 2kˆ
 12

24
12
1
1
 1

ml 2ω iˆ −
ml 2Ω sin 2θ ˆj +  ml 2 cos2 θ + md 2  Ω kˆ
HP =


12
24
12
162
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 9
Problem 9.10
1000
0
−300  ω x 
ω x 
 

  =
1000 500  ω y  1000 ω y 
 0
ω 
 −300 500 1000  ω z 
 z
1000ω x − 300ω z

 1000ω x 


 
1000
ω
+
500
ω

y
z
 = 1000ω y 
 −300ω x + 500ω y + 1000ω z  1000ω z 


 
−300ω z

 0 

  
500ω z

 = 0 
 −300ω + 500ω  0 
x
y  

ωz =0
−300ω x + 500ω y = 0 ⇒ ω y =
3
ω
5
 ωx 
3

∴ω =  ω x 
5

 0 
ω = 1.166ω x = 20 ⇒ ω x = 17.15
17.15 


∴ ω = 10.29  or ω = 17.15iˆ + 10.29jˆ ( s-1 )
 0 


Problem 9.11
ω = 2t 2 iˆ + 4 ˆj + 3tkˆ
dω 
dω 
+ω &times;ω =

dt rel
dt  rel
α = 4tiˆ + 3kˆ
t = 3:
ω = 18iˆ + 4 ˆj + 9kˆ
α=
α = 12iˆ + 3kˆ
{M} = [I G ]{α } + {ω } &times; [I G ]{ω }
10
0
0  12  18  10 0 0  18 
   
 
20 0   0  +  4  &times;  0 20 0   4 




 0 0 30   3   9   0 0 30   9 
120  18  180 
{M} =  0  +  4  &times;  80 
 90   9  270 

   

{M} =  0
ˆj
iˆ
kˆ
M = (120iˆ + 90kˆ ) + 18 4
9 = (120iˆ + 90k̂
k ) + 360iˆ − 3240 ˆj + 720kˆ
180 80 270
(
M = 480iˆ − 3240 ˆj + 810kˆ
2
M = M = 480 2 + ( −3240) + 810 2 = 3374 N ⋅ m
163
)
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 9
Problem 9.12
{M G } = [I G ]{α } + {ω } &times; [I G ]{ω }
{ω } = 0
ˆ  &times; 100Iˆ = 34.29 ˆj + 25.36kˆ ( N ⋅ m )
MG = r &times; F = ( 0 − 0.075 ) Iˆ + ( 0 − 0.2536 ) Jˆ + ( 0 − 0.05714 ) K

−0.03975 0.0120  α X 
 0   0.1522
 

 
34.29  =  −0.03975 0.07177 0.040 57  αY 
25.36   0.0120
0.040 57
0.1569  α Z 

 
−1
−0.039 75 0.0120   0  143.9 
α X   0.1522
  

 


2
αY  =  −0.039 75 0.07177 0.040 57  34.29  =  553.1 m s
α   0.0120
0.040 57
0.1569  25.36   7.61 
 Z 
(
)
Problem 9.13
(a)
∑ Fx = maG x
 v2

cos θ 
mg sin θ = m
 R

tan θ =
v2
gR
(b)
iˆ
M = ω &times; H = ωx
Aω x
ˆj
ωy
kˆ
ω z = ( C − B ) ω yω z iˆ + ( A − C ) ω xω z ˆj + ( B − A ) ω xω y kˆ
Bω y
Cω z
v
v
sin θ
ωy =0
ω z = cos θ
R
R
 v
 v

M = ( A − C )  − sin θ   cos θ  ˆj
 R
R

ωx = −
M y = (C − A)
v2
R2
sin θ cos θ = (C − A)
v2
2 R2
sin 2θ
Problem 9.14
iˆ
M = ω &times; H = ωx
Aω x
ωx =0
ˆj
kˆ
ωy
ω z = ( C − B ) ω yω z iˆ + ( A − C ) ω xω z ˆj + ( B − A ) ω xω y kˆ
Bω y
Cω z
ω y = ω Z sin α
ω z = ω Z cos α
M = ( C − B ) (ω Z sin α )(ω Z cos α ) iˆ
Mx =
1
(C − B)ω Z2 sin 2α
2
My = 0
Mz = 0
G (–l[1 + cos ]/12, lsin /12, 0)
Problem 9.15
l
 l l
xG1 = − − cos = − cos θ
 2 3
6
l
yG1 = sin θ
6
zG1 = 0
y
2l/
2
3
1
C (0,0,0) B(l/3, 0, 0)
x
A(–2l/3, 0, 0)
l/
3
164
2l/3
l/3
Solutions Manual
xG2 = −
Orbital Mechanics for Engineering Students
Chapter 9
l
6
yG2 = zG2 = 0
xG =
yG =
mxG1 + mxG2
l
l
1 l
− cos θ −
= − (1 + cos θ )
2 6
6
12
l
1 l

sin θ = sin θ
=


2 6
12
=
2m
myG1 + myG2
2m
zG = 0
Free-body diagram (no bearing couples and no thrust components of bearing force):
y
A
Az
Ay
z
Bz
x
By
l/3
2l/3
∑ Fx = 2maG x :
B
C
0 =0
∑ Fy = 2maG y :
Ay + By = −
∑ Fz = 2maG z :
Az + Bz = 0
mω 2 l
sin θ
6
(1)
(2)
Moments of inertia about G1 (inferred from results of Exercise 9.7):

2
 sin θ
ml2  1
( )
 sin 2θ
I G1 =
1
12  2
 0
[ ]
1
sin 2θ
2
cos2 θ
0

0

0

1 
From Equation 9.60:
[I( ) ]
1
mC
y 2 + z 2
G1
 G1
= m  − xG1 yG1

 − xG1 zG1
− xG1 yG1
xG12 + zG12
− yG1 zG1
2

l


sin θ
6


  l
 l
= m  − − cos θ
sin θ
 6
6


0


− xG1 zG1 

− yG1 zG1 

xG12 + yG12 
 l
 l
− − cos θ
sin θ
 6
6
2
 l

− cos θ
 6

0
165




0


2
2
 l

l
 
− cos θ +
sin θ 
 6

6
 
0
Solutions Manual
Orbital Mechanics for Engineering Students
1
2
 3 sin θ
ml2  1
( )
 sin 2θ
I m1 C =
12  6
 0

[
1
sin 2θ
6
1
cos2 θ
3
]
[I (C1) ] = [I (G1) ] + [I (m1)C ]
1

0

0

1
3 
0
[ ]
2
sin 2θ
3
4
cos2 θ
3
0
0 0 0 
ml2 
IG =
0 1 0


2
12
0 0 1 
y 2 + z 2
G2
 G2
( )
I m2 C = m  − xG2 yG2

 − xG2 zG2

1
2
0
 3 sin θ
 ml2  1
 sin 2θ
0 +
 12  6
1 
 0

1
sin 2θ
2

2
 sin θ
ml2  1
 sin 2θ
=
12  2
 0
4
2
 3 sin θ
ml2  2
( )
 sin 2θ
I C1 =
12  3
 0

Chapter 9
cos2 θ
0
1
sin 2θ
6
1
cos2 θ
3
0

0

0

1
3 

0

0

4
3 
[ ]
(2 )
[
]
[I ]
(2 )
mC
0
ml2 
=
0
12 

0

0
1
3
0

ml 0
( )
I C2 =
12 
0

[ ]
2
xG2 2 + zG2 2
− yG2 zG2
0
0
− xG2 zG2 
  l 2

− yG2 zG2  = m 0 −
  6 

xG2 2 + yG2 2 

0
0

0

0
1

3
0
0
0 0 0 
2 
ml 
ml 0
( )

I m2 C =
0 1 0  +
12 
12 
0 0 1 
0

0 0
4

0
3
4
0

3
[I (C2 ) ] = [I (G2 ) ] + [
2
− xG2 yG2
]
4
2
 3 sin θ
ml2  2
[I C ] = I C(1) + I C(2 ) = 12  3 sin 2θ

 0

[ ] [ ]
0
1
3
2
2
sin 2θ
3
4
cos2 θ
3
0
0
0

0
1

3

0
0
2
 ml 0

0+

12

0
4


3
166
0
4
3
0
0

0
4

3
0



0

2
 l  
−
 6  
Solutions Manual
Orbital Mechanics for Engineering Students
 ml2
sin 2 θ

9
 2
ml
[I C ] =  18 sin 2θ


0

ml2
sin 2θ
18
2
ml (
1 + cos2 θ )
9

0 


0 

2ml2 
9 
0
 ml2
sin 2 θ

9
 2
ml
{
}
{H C } = [I C ] ω =  18 sin 2θ


0



 ml2ω
sin 2 θ 
0 

 ω   92


    ml ω
sin 2θ 
0  0  = 
18

  
0
2ml2   0  

9 
ml2
sin 2θ
18
ml2 (
1 + cos2 θ )
9
0
iˆ
ω
ˆj
0
ml 2ω
sin 2 θ
9
ml 2ω
sin 2θ
18
MC = ω &times; H C =
kˆ
ml 2ω 2
sin 2θ k̂
0 =
18
0
MC x = 0
MC y = 0
MC z =
Chapter 9
(3)
(4)
ml2ω 2
sin 2θ
18
(5)
Calculate the moments of the bearing reactions in the above free body diagram:
1
1

 2

 2 
l 
2
MC =  − liˆ  &times; Ay ˆj + Az kˆ +  iˆ  &times; By ˆj + Bz kˆ =  Az l − Bz l  ˆj +  − Ay l + By l  ĵ

 3

 3 
3 
3
3
3
(
)
(
)
MC x = 0
2
1
MC y = Az l − Bz l
3
3
2
1
MC z = − Ay l + By l
3
3
(6)
(7)
(8)
From (4) and (7)
2 Az − Bz = 0
(9)
From (5) and (8)
2
1
ml2ω 2
− Ay l + By l =
sin 2θ
3
3
18
(10)
From (1) we have
By = −
mω 2 l
sin θ − Ay
6
(11)
Substituting this into (10):
−

2
1  mω 2 l
ml2ω 2
ml2ω 2
Ay l +  −
sin θ − Ay  l =
sin 2θ ⇒ Ay = −
sin θ (1 + 2 cos θ )

3
3
6
18
18
167
(12)
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 9
Therefore, from (11),
By = −
 ml2ω 2

mω 2 l
mω 2 l
sin θ −  −
sin θ (1 + 2 cos θ )  = −
sin θ (1 − cos θ )


6
18
9
(13)
From (2) we have
Bz = − Az
Substituting this into (9)
2 Az − ( − Az ) = 0 ⇒ Az = 0
Therefore, Bz = 0 .
The only reactions at each bearing are in the plane of the rod and shaft, normal to the shaft, as given
by Equations (12) and (13).
H = m l2sin /9
y
A
B
l/3
2l/3
mw2lsin (1 + 2cos )/18
x
m l2sin (1 – cos )/9
Problem 9.16
MG = HG
)rel + ω &times; H
iˆ
MG = Aω x iˆ + Bω y ˆj + Cω z kˆ + ω x
Aω x
ˆj
ωy
kˆ
ωz
Bω y
Cω z
ˆj
iˆ
k̂
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
)
(
)
600i + MG y j + MGz k = 5ω x i + 5 0 j + 10 0 k + 0
0.5
100
5 ( 0 ) 5 ⋅ 0.5 10 ⋅ 100
600iˆ + MGx ˆj + MG y kˆ = ( 5ω x + 250 ) iˆ
5ω x + 250 = 600 ⇒ ω x = 70 rad s2
Problem 9.17
MO = IOxω x iˆ + IO yω y ˆj + IOzω z kˆ +
iˆ
ωx
ˆj
ωy
kˆ
ωz
IO xω x
IO yω y
IOzω z
2
1
mL2
 L
mL2 + m
=
IOy = 0
 6
12
9
ω&laquo;x = ω&laquo;y = ω&laquo;z = 0
ωx =0
ω y = −ω cos θ
ω z = ω sin θ
IOx = IOz =
168
Solutions Manual
Orbital Mechanics for Engineering Students
ˆj
iˆ
MO = 0 −ω cos θ
0
kˆ
ω sin θ
=−
mL2
ω sin θ
9
0
Chapter 9
1
mω 2 L2 sin θ cos θ iˆ
9
1
∴ MOx = − mω 2 L2 sin θ cos θ
9
Moment of the weight vector about O:
L
MOx = −mg sin θ
6
1
L
3
g
∴ − mω 2 L2 sin θ cos θ = −mg sin θ ⇒ ω =
9
6
2 L sin θ
Problem 9.18
)rel + Ω &times; H
MG = HG
MG = IGxα x iˆ + IG yα y ˆj + IGzα z kˆ +
iˆ
10 ⋅ 9.81 ⋅ 0.25iˆ = 0 + 0
iˆ
Ωx
ˆj
Ωy
kˆ
Ωz
IGxω x
IG yω y
IGzω z
ˆj
ωp
0 0.014 06ω p
kˆ
630
0.02812 ⋅ 630
24.52iˆ = 17.72ω p iˆ ⇒ ω p = 1.384 rad s
Or, using Equation 9.96,
ωp =
mgd 10 ⋅ 9.81 ⋅ 0.25
=
= ω p = 1.384 rad s
I zω s 0.02812 ⋅ 630
Problem 9.19
ω wheel =
v
=
r
î
1000
0.3
130
k̂
MB
I wheel = 25 ⋅ 0.2 2 = 1 kg ⋅ m 2
M = ω p &times; H s = 0.8 ˆj &times; [1 ⋅ 120.4kˆ ] = 96.3iˆ ( N ⋅ m )
ĵ
169
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 9
Problem 9.20
I rotor = 4 ⋅ 0.07 2 = 0.0196 kg ⋅ m 2
2π ˆ 

k  = 41.05 ˆj ( N ⋅ m )
M = ω p &times; H s = 2iˆ &times;  0.0196 ⋅ 10 000 ⋅

60 
By = F
Ay = − F
0.04 F = 41.05
F = 1026 N
Problem 9.21
i
G
C = 12.5 kg - m 2
j
k
MG
I rotor = 200 ⋅ 0.25 2 = 12.5 kg ⋅ m 2
2π
60
650 000
v
ω p = = 3600 = 0.072 22 rad s
r
2500
M = ω p &times; H s = 0.072 22iˆ &times; 12.5 ( −1571kˆ )  = 1418iˆ ( N ⋅ m )
ω s = 15 000 ⋅
The moment reaction on the airframe is clockwise, pitching the nose down.
Problem 9.22
MG = ω p &times; H s
ω p = 20î ( rad s )
1
⋅ 10 ⋅ 0.05 2 ⋅ 200kˆ = 2.5kˆ kg ⋅ m s2
2
MG = 20iˆ &times; 2.5kˆ = −50 ˆj ( N ⋅ m )
(
Hs =
0.6RB ˆj = −50 ˆj
RB = −83.33 N
)
RA = − RB = 83.33 N
Problem 9.23
lx = cosψ cos φ − sin φ sin ψ cos θ
= cos 70&deg; cos 50&deg; − sin 50&deg; sin 70&deg; cos 25&deg;
= 0.6428 ⋅ 0.3420 − 0.7660 ⋅ 0.9397 ⋅ 0.9063
= −0.4326
α xX = cos −1 ( −0.4326) = 115.6&deg;
170
forward
Solutions Manual
Orbital Mechanics for Engineering Students
Problem 9.24
 20
{H } = [I ]{ω } =  −10

 0
−10 0  10   0 

  
30 0  20  =  500 
0 40  30  1500 
10 
1
1
 
T = H ⋅ ω = 0 500 1500  20  = 23 000 J
2
2
30 
 
171
( J ⋅ s)
Chapter 9
Solutions Manual
Orbital Mechanics for Engineering Students
172
Chapter 9
Solutions Manual
Orbital Mechanics for Engineering Students
Problem 10.1
ωp =
ωs
C
1200
6
=
A − C cos θ 2600 − 1200 cos 6&deg;
H = Aω p = 2600 ⋅ 5.171 = 13 450 kg ⋅ m 2 s
Problem 10.2
C
500
6
ωs
=
A − C cos θ 300 − 500 cos 10&deg;
2π
2π
T=
=
= 0.4173 s
ω p 15.06
ωp =
Problem 10.3
C
mr 2
ωs
ωs
ωs
=
= −2
A − C cos θ 1 2
θ
θ
cos
cos
mr − mr 2
2
θ2
cos θ = 1 −
2
ωp =
−1
 θ2 
θ2
1
= 1 −  = 1 +
2 
2
cos θ 
2

θ 

∴ ω p = −2ω s  1 +

2 
(θ &lt;&lt; 0)
Problem 10.4
H = Aω p = 1000 ⋅ 2 = 2000 kg ⋅ m 2 s
Problem 10.5
HG
)rel + ω &times; HG = 0
[I G ]{α } + ω &times; H G = 0
385.4

 0
 0
385.4

 0
 0
385.4

 0
 0
385.4

 0
 0
0
416.7
0
0
416.7
0
0
416.7
0
0
416.7
0
0 
0
0   0.01  0 
 0.01  385.4
 
  
 {α } 

+  −0.03  &times;  0
0 
416.7
0   −0.03  = 0 
 0.02   0
52.08 
0
52.08   0.02  0 

 
0 
 0.01   3.854  0 
 
  
 {α } 
+  −0.03  &times;  −12.50  = 0 
0 
 0.02   1.042  0 
52.08 

 
  
0 
 0.2188  0 
  


0  {α } +  0.066 6  = 0 
 −0.009 39  0 
52.08 

  
0 
 0.2188  0 
  


0  {α } +  0.066 6  = 0 
 −0.009 39  0 
52.08 

  
173
Chapter 10
Solutions Manual
Orbital Mechanics for Engineering Students
385.4
0
0 
{α } = − 0
416
.
7
0 

 0
0
52.08 
2
α = 0.000 6167 m s
−1
 0.2188   −0.000 567 6 

 

 0.066 6  =  −0.000 16 
 −0.009 39   0.000 180 3 

 

Chapter 10
( m s2 )
Problem 10.6
C
0.72
30
ωs
=
C − A cos θ 0.72 − 0.36 cos 15&deg;
ωn = 0
ωp =
ω x = ω p sin θ sin ψ + ω n cosψ = −62.12 sin 15&deg; sin ψ + 0 ⋅ cosψ = −16.08 sin ψ
ω y = ω p sin θ cosψ − ω n cosψ = −62.12 sin 15&deg; cosψ − 0 ⋅ cosψ = −16.08 cosψ
ω z = ω s + ω p cos θ = 30 + ( −62.12) cos15&deg; = −30
1
Aω x2 + Bω y 2 + Cω z 2
2
1
2
2
2
= 0.36(−16.08 sin ψ ) + 0.36(−16.08 cosψ ) + 0.72( −30)
2
1
= 93.05 sin 2 ψ + cos2 ψ + 648
2
1
= (93.05 + 648)
2
= 370.5 J
TR =
(
[
[
)
(
)
]
]
Or,
H = Aω p = 0.36( −62.12) = −22.36 kg ⋅ m 2 / s
TR =
1 H2 
C−A

1+
sin 2 θ


2 C
A
2
1 ( −22.36) 
0.72 − 0.36

1+
sin 2 15&deg;

2 0.72 
0.36
1
= ⋅ 694.5 ⋅ (1 + 0.066 99)
2
= 370.5 J
=
Problem 10.7
1
 12 m

[I G ] = 



[l2 + (2 l)2 ]
0
0
1
2
m l2 + (3 l)
12
0
0
5
2
 12 ml

{H 0 } = [I G ]{ω } =  0

 0

2
H 0 = 1.121ml ω 0
[
0
5 2
ml
6
0
 5
2
  12 ml
 
= 0
0
 
1
2
2
m (2 l) + (3 l)   0
 
12
0
]
[
]


2


 1.5ω 0   0.625ml ω 0 
0  0.8ω 0  = 0.6667 ml2ω 0 


2
 
13 2  0.6ω 0   0.65ml ω 0 
ml

12
0
174
0
5 2
ml
6
0



0 

13 2 
ml

12
0
Solutions Manual
T0 =
Orbital Mechanics for Engineering Students
1
1
ω ⋅ H 0 = 1.5ω 0
2
2
0.8ω 0
(a)
Chapter 10
 0.625ml2ω 0 


0.6ω 0  0.6667 ml2ω 0  = 0.9305ml2ω 0 2
 0.65ml2ω 
0 

1
1 13 2 2
Cω 2 =
ml ω = 0.5417 ml2ω 2
2
2 12
T = T0
T=
0.5417 ml2ω 2 = 0.9305ml2ω 0 2
ω = 1.718ω 0 = 1.311ω 0
(b)
H = Cω =
13 2
ml ω
12
H = H0
13 2
ml ω = 1.121ml2ω 0
12
ω = 1.035ω 0
Problem 10.8
H initial = 1000 ⋅ 6 = 6000 kg ⋅ m 2 s
1
Tinitial = ⋅ 1000 ⋅ 6 2 = 18 000 kg ⋅ m 2 s2
2
H final = 5000ω final
H final = H initial
5000ω final = 6000
ω final = 1.2 rad s
1
1
⋅ 5000 ⋅ ω final2 = ⋅ 5000 ⋅ 1.2 2 = 3600 kg ⋅ m 2 s2
2
2
∆T = T final − Tinitial = 3600 − 18 000 = −14 400 J
T final =
Problem 10.9
 0.1522
−0.03975 0.012  10   1.522 

{H G0 } = −0.03975 0.07177 0.04057  0  = −0.3975 kg ⋅ m 2 s
 0.012
0.04057
0.1569   0   0.1200 
2
H G0 = H G0 = 1.5776 kg ⋅ m s
(
T0 =
10 
1
1
 
H G0 ⋅ ω 0 = 1.522 −0.3975 0.1200   0  = 7.610 J
2
2
0 
 
I max = 0.1747 kg ⋅ m 2
H G f = I maxω
f
( from Example 10.11)
= 0.1747ω
f
H G f = H G0
0.1747ω f = 1.578
ω f = 9.03 rad s
175
)
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 10
1
1
⋅I
ω 2 = ⋅ 0.1747 ⋅ 9.03 2 = 7.123 J
2 max f
2
∆T = T f − T0 = 7.123 − 7.610 = −0.4867 J
Tf =
Problem 10.10
(
( )
( )
( p)
)
( )
(
( p)
( )
C rω 2r + C p ω 2r + ω rel 2 = C rω 1 r + C p ω 1 r + ω rel 1
(r)
(r)
ω 2 = ω1 +
( )
ω 2r = 3 +
(
( p)
( p)
C p ω rel 1 − ω rel 2
)
)
Cp + Cr
500(1 − 0.5)
500 + 1000
Problem 10.11
ωs
C
1000
0.1
=
A − C cos θ 5000 − 1000 cos 20&deg;
π
π
t=
=
= 118.1 s
ω p 0.0266
ωp =
Problem 10.12
A
cos γ =
A2 + C 2 tan 2
φ
2
m ( 2 2 ) 500 (
1
1
3r + l =
3 ⋅ 0.52 + 2 2 ) = 197.9 kg ⋅ m 2
C = mr 2 = 500 ⋅ 0.52 = 62.5 kg ⋅ m 2
12
12
2
2


2


2
  A  − A2 
  197.9  − 197.9 2 
  cos γ 


−1   cos 5&deg; 
φ = 2 tan −1 
 = 2 tan 
 = 30.97&deg;
2
C
62.52




A=
Problem 10.13
Npd = I (ω 2 − ω 1 )
Npd
ω 2 = ω1 +
I
30 ⋅ 15 ⋅ 1.5
ω 2 = 0.01 ⋅ 2π +
= 0.4003 rad s = 0.0637 rev s
2000
Problem 10.14
HG 0 = Aω 0 x iˆ + Bω 0 y ˆj + Cω 0 z kˆ
= 2000 ⋅ 0.1iˆ + 4000 ⋅ 0.3 ˆj + 6000 ⋅ 0.5kˆ
= 200iˆ + 1200 ˆj + 3000kˆ
H G = H G 0 + ∆H G
(
= 200iˆ + 1200 ˆj + 3000kˆ + 50iˆ − 100 ˆj + 300kˆ
= 250iˆ + 1100 ˆj + 3300kˆ
)
176
Solutions Manual
Orbital Mechanics for Engineering Students
Aω x iˆ + Bω y ˆj + Cω z kˆ = 250iˆ + 1100 ˆj + 3300kˆ
250
250
=
A
2000
1100 1100
=
=
B
4000
3300 3300
=
=
C
6000
ωx =
ωy
ωz
ω = ω x2 + ω y 2 + ω z 2 = 0.6275 rad s
Problem 10.15
1 2 1
1
1
mr +
ml2 = 300 ⋅ 1.52 +
300 ⋅ 1.52 = 225 kg ⋅ m 2
4
12
4
12
1
1
C = mr 2 = 300 ⋅ 1.52 = 337.5 kg ⋅ m 2
2
2
A=
 2π ˆ 
HG = Cω1kˆ = 337.5 ⋅  1 ⋅
k = 35.34kˆ kg ⋅ m 2
1
 60 
(
HG = HG + I
2
1
M
(
HG = 35.34kˆ + −I M ˆj
2
)
H G2 = 35.34 2 + I M 2 = 1249 + I M 2
H G2 = Aω p
1249 + I M 2 = 225 ⋅ (0.1 ⋅ 2π ) = 141.4
I M 2 = 18740
I M = 136.9 N ⋅ m ⋅ s
Problem 10.16
K =1 +
(a)
C
2mR
lf = R K
t=
2
=1 +
ω0 −ω
f
ω0 + ω
f
300
2 ⋅ 3 ⋅ 1.52
= 23.22
= 1.5 23.22 ⋅
5 −1
= 5.902 m
5 +1
K ω0 −ω f
23.22 5 − 1
=
= 0.7869 s
2 ω +ω
52 5 + 1
ω0
f
0
(b)
lf = R K
t=
ω0 −ω
f
ω0 + ω
f
= 1.5 23.22 ⋅
5 −0
= 7.228 m
5+0
K ω0 −ω f
23.22 5 − 0
=
= 0.9636 s
2 ω +ω
52 5 + 0
ω0
0
f
177
)
Chapter 10
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 10
Problem 10.17
T0 =
1
1
1
1
A ω x2 + ω y 2 + Cω z 2 = AΩ 2 + Cω 0 2
2
2
2
2
(
)
C
60
2
ωs
=
A − C cos θ 30 − 60 cos 15&deg;
30
A
( −4.141) cos 15&deg; = −2 rad s
ω 0 = ω p cos θ =
C
60
Ω = ω p sin θ = −4.141 sin 15&deg; = −1.072 rad s
ωp =
∴ T0 =
(a)
1
1
2
2
⋅ 30( −1.072) + ⋅ 60( −2) = 137.2 J
2
2
H 0 = Aω p = 30 ⋅ 4.141 = 124.2 kg ⋅ m 2 s
H f = H0
60ω f = 124.2 ⇒ ω
f
(b)
1
1
Cω f 2 = ⋅ 60 ⋅ 2.0712 = 128.6 J
2
2
∆T = T f − T0 = 128.6 − 137.2 = −8.616 J
Tf =
(c)
lf = R 1+
C
2
2mR
=1⋅ 1 +
60
= 2.299 m
2 ⋅ 7 ⋅ 12
Problem 10.18
A 0 0  0   0 
{H G } = [I G ]{ω } =  0 A 0   n  =  nA 
 0 0 C   −ω s   −ω sC 
ˆj
iˆ
kˆ
M =Ω&times;H = 0 n
0 = − Cω niˆ
G
G
s
0 nA −ω sC
2π
60
2π
2π +
365.26 = 7.292 &times; 10 −5 rad /s
n=
24 ⋅ 3600
C = 550 kg ⋅ m 2
ωs =1⋅
∴ MG x = −550 ⋅ 0.1047 ⋅ 7.292 &times; 10 −5 = −0.004 2 N ⋅ m
Problem 10.19 ω 0 and ω f are the initial and final angular velocities of the spacecraft. ω is the
angular velocity of the flywheel relative to the vehicle.
[H (Gv) + H (Gv) ]0 = [H (Gv) + H ( w) ] f
( Aω iˆ + Aω ˆj + Cω kˆ ) + (I ω iˆ + I ω ˆj + I ω kˆ )
= ( Aω iˆ + Aω ˆj + Cω kˆ ) +  I (ω + ω ) iˆ + I ω ˆj + I ω kˆ 


0x
0y
x
0z
y
x 0x
z
x
y 0y
x
z 0z
y
178
y
z z
Solutions Manual
Orbital Mechanics for Engineering Students
( A + I z )ω z = ( A + I z )ω 0 z
(
⇒ ω z = ω0z = 0
A + I y ω y = A + I y ω 0 y ⇒ ω y = ω 0 y = 0.05 rad s
)
(
)
( A + I x )ω x + I xω = ( A + I x )ω 0 x
A

⇒ ω =  1 +  (ω 0 x − ω x )
Ix 

1000 

(0.1 − 0.003) = 4.947 rad s
∴ω = 1 +

20 
Problem 10.20
Given: I 3 &gt; I 2 &gt; I1 .
Figure 10.29, Stable region I: I roll &gt; I yaw &gt; I pitch :
I1 axis in pitch direction (normal to orbital plane)
I 2 axis in yaw direction (radial)
I 3 axis in roll direction (local horizon)
Figure 10.29, Stable region II (preferred): I pitch &gt; I roll &gt; I yaw :
I1 axis in yaw direction (radial)
I 2 axis in roll direction (local horizon)
I 3 axis in pitch direction (normal to orbital plane)
179
Chapter 10
Solutions Manual
Orbital Mechanics for Engineering Students
180
Chapter 10
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 11
Problem 11.1
mp = mp
out
+ mp
in
= mp
out
+
mp
out
4
=
5
m
4 p out
Outbound leg:
m e + m p + mPL


∆v = I sp go ln 

 m e + m p − m p out + mPL 
5


m e + m p + 3500


out
4
4220 = 430 ⋅ 9.81 ⋅ ln 

5
 m e + m p − m p + 3500 


out
out
4
5


+ 3500
m + m
 e 4 p out

= 430 ⋅ 9.81 ⋅ ln 

1
 m e + m p + 3500 


out
4
5
m e + m p + 3500
out
4
= 2.719
1
m e + m p + 3500
out
4
0.5702m p − 1.719m e = 6018
out
(1)
Return from GEO tp LEO:
1


m + m
+ 3500
 e 4 p out

∆v = I sp go ln 

me


1


m + m
 e 4 p out 
4220 = 430 ⋅ 9.81 ⋅ ln 

me


1
me + mp
out
4
= 2.719
me
m p = 6.876m e
(2)
out
Substitute (2) into(1):
0.5702(6.876m e ) − 1.719m e = 6018
m e = 2733 kg
Problem 11.2
First stage:
c = I sp go = 235 ⋅ 9.81 = 2943 m s
(
)
 m  m0 − m f go
 249.5  (249.5 − 170.1) ⋅ 9.81
= 1127 − 73.38 = 1054 m s
v bo = c ln  0  −
= 2943 ln
−
 170.1 
&laquo;
me
10.61
mf 
 mf 
 1  m0 − m f  2
 ln 
 m f + m0 − m f  − 
 go
  m0 
 2  m&laquo;e 
 1  249.5 − 170.1  2
2943   170.1 
=
ln
⋅
170
.
1
+
249
.
5
−
170
.
1
9.81
−2

10.61   249.5 
10.61

c
h bo =
&laquo;
me
181
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 11
= 3947 − 274.4
= 3673 m
After 3 second staging delay:
v = v bo − g∆t s = 1054 − 9.81 ⋅ 3 = 1024 m s
1
1
h = hbo + v bo ∆t s − g∆t s2 = 3673 + 1054 ⋅ 3 − ⋅ 9.81 ⋅ 3 2 = 3673 + 3117 = 6790 m
2
2
Second stage:
v0 = 1024 m s
h0 = 6790 m
c = I sp go = 235 ⋅ 9.81 = 2305 m s
(
)
 m  m0 − m f go
v bo = v0 + c ln  0  −
m&laquo;e
mf 
 113.4  (113.4 − 58.97 ) ⋅ 9.81
−
 58.97 
4.0573
= 1024 + 1508 − 131.7
= 1024 + 2305 ln
= 2400 m s
 1  m0 − m f  2
 m0 − m f 
c  mf 
h bo = h0 + v0 
 ln 
 go
+
 m f + m0 − m f  − 
 m&laquo;e  m&laquo;e   m0 
 2  m&laquo;e 
 1  113.4 − 58.97  2
 113.4 − 59.97  2305   59.97 
+
ln
⋅ 58.97 + 113.4 − 58.97  −
⋅ 9.81


 4.053   113.4 

4.053
4.063
 2
= 6790 + 13 760 + 9028 − 884.7
= 28 690 m
= 6790 + 1024
Coast to apogee:
v0 = 2400 m s
h0 = 28 690 m
v
2400
0 = v0 − gt max ⇒ t max = 0 =
= 244.7 s
g
9.81
1
1
hmax = h0 + v0 t max − gt max 2 = 28 690 + 2400 ⋅ 244.7 − ⋅ 9.81 ⋅ 244.7 2 = 322 300 m
2
2
Problem 11.3
v0 = ω earth Rearth cos φ = 7.292(10 −5 ) ⋅ 6378 ⋅ cos 28&deg; = 0.4107 km s
398 600
+ 2 − 0.4107 = 9.315 km s
6678
= ∆v = 9.315 km s
= v bo1 + v bo2
∆v =
v bo
v bo
 m0 
 m0 
v bo = I sp go ln  1  + I sp go ln  2 
1
2
 m f1 
 m f2 
182
Solutions Manual
Orbital Mechanics for Engineering Students
2 ⋅ 525 000 + 30 000 + 600 000 + mPL


9315 = 290 ⋅ 9.81 ⋅ ln 

 2 ⋅ (525000 - 450000) + 3000 + 600000 + mPL 
 30 000 + 600 000 + mPL 
+ 450 ⋅ 9.81 ⋅ ln 

30 000 + mPL


 1 680 000 + mPL 
 630 000 + mPL 
9315 = 2845 ⋅ ln 
 + 4414 ⋅ ln 

 753 000 + mPL 
 30 000 + mPL 
To find the value of mPL satisfying this equation, graph the function
 1 680 000 + mPL 
 630 000 + mPL 
f = 9315 − 2845 ⋅ ln 
 − 4414 ⋅ ln 

 753 000 + mPL 
 30 000 + mPL 
f
500
0
-500
100 000
120 000
mPL (kg)
110 800
f = 0 when mPL = 110 800 kg
Problem 11.4
π PL =
π PL1/3
λ=
ε=
10 000
mPL
=
= 0.06667
150 000
m0
1 − π PL1/3
= 0.682
20 000
mE
=
= 0.1429
m0 − mPL 150 000 − 10 000
(a)
1+λ
1 + 0.682
=
= 2.039
ε + λ 0.1429 + 0.682
∆v = I sp go ln n 3 = 310 ⋅ 0.009 81 ⋅ ln 2.039 3 = 6.5 km s
n=
(b)
mp =
(1 − πPL1/3 )(1 − ε) m
mp =
(1 − πPL1/3 )(1 − ε) m
=
mp =
(1 − πPL1/3 )(1 − ε) m
=
1
2
3
π PL
π PL2/3
π PL1/3
PL =
PL
PL
(1 − 0.06667 1/3 )(1 − 0.1429)
0.06667
(1 − 0.06667 1/3 )(1 − 0.1429)
0.06667 2/3
(1 − 0.06667 1/3 )(1 − 0.1429)
0.06667 1/3
183
⋅ 10 000 = 76 440 kg
⋅ 10 000 = 30 990 kg
⋅ 10 000 = 12 570 kg
Chapter 11
Solutions Manual
Orbital Mechanics for Engineering Students
(c)
mE1 =
(1 − πPL1/3 )ε m
mE2 =
(1 − πPL1/3 )ε m
=
mE3 =
(1 − πPL1/3 )ε m
=
PL
π PL
π PL2/3
π PL1/3
PL
PL
=
(1 − 0.06667 1/3 ) ⋅ 0.1429
0.06667
(1 − 0.06667 1/3 ) ⋅ 0.1429
0.06667 2/3
(1 − 0.06667 1/3 ) ⋅ 0.1429
0.06667 1/3
10 000 = 12740 kg
10 000 = 5166 kg
10 000 = 2095 kg
(d)
m0 3 = mE3 + m p + mPL = 2095 + 12 570 + 10 000 = 24 660 kg
3
m0 2 = mE2 + m p + m0 3 = 5166 + 30 990 + 24 660 = 60 820 kg
2
m01 = mE1 + m p + m0 2 = 12740 + 76 440 + 60 820 = 150 000 kg
1
Problem 11.5
c1 = I sp go = 300 ⋅ 0.009 81 = 2.943 km s
1
c2 = I sp go = 235 ⋅ 0.009 81 = 2.305 km s
2
ε 1 = 0.2
ε2 = 0.3
v bo = 6.2 km s
2
 c iη − 1 
 c1η − 1 
 c 2η − 1 

 = c1 ln 
 + c2 ln 
c iε iη 
 c1ε1η 
 c2 ε2η 
i =1
 2.943η − 1 
 2.305η − 1 
6.2 = 2.943 ln 
 + 2.305 ln 

 2.943 ⋅ 0.2η 
 2.305 ⋅ 0.3η 
v bo =
∑ c i ln 
 2.943η − 1 
 2.305η − 1 
6.2 = 2.943 ln 
 + 2.305 ln 

 0.5886η 
 0.6915η 
To find η , graph the function
 2.943η − 1 
 2.305η − 1 
f = 2.943 ln 
 + 2.305 ln 
 − 6.2
 0.5886η 
 0.6915η 
As shown below, f = 0 when η = 1.726 .
c η − 1 2.943 ⋅ 1.726 − 1
=
= 4.016
n1 = 1
2.943 ⋅ 0.2 ⋅ 1.726
c1ε1η
c η − 1 2.305 ⋅ 1.726 − 1
=
= 2.496
n2 = 2
c2 ε2η 2.305 ⋅ 0.3 ⋅ 1.726
2.496 − 1
n2 − 1
⋅ 10 = 59.53 kg
mPL =
1 − ε2 n2
1 − 0.3 ⋅ 2.496
4.016 − 1
n −1
m1 = 1
(m + mPL ) = 1 − 0.2 ⋅ 4.016 ( 59.53 + 10) = 1065 kg
1 − ε1 n1 2
m2 =
M = m1 + m2 = 1065 + 59.53 = 1124 kg
184
Chapter 11
Solutions Manual
Orbital Mechanics for Engineering Students
Chapter 11
f
0
-1
1.5
2
2.5
1.726
Problem 11.6
z = x2 + y 2 + 2 xy
g = x2 − 2 x + y 2
(
h = z + λg = x2 + y 2 + 2 xy + λ x2 − 2 x + y 2
)
∂h
= 2 x + 2 y + λ (2 x − 2) = 0 ⇒ (λ + 1) x + y = λ
∂x
∂h
= 2 y + 2 x + λ (2 y) = 0 ⇒ x + (λ + 1) y = 0
∂y
λ + 1
1   x  λ 
∴
 = 
λ + 1  y   0 
 1
1 
 x  λ + 1
 =
λ + 1
y  1
−1
 λ +1 
λ + 1 −1  λ   λ + 2 
λ 
1
 = (


  = 
 0  λ λ + 2)  −1 λ + 1  0   − 1 
 λ +2
x2 − 2 x + y 2 = 0
(λ + 1)2
(λ + 2)2
−2
1
λ +1
+
=0
λ + 2 (λ + 2)2
2
Multiply through by (λ + 2) (this is okay since λ + 2 = 0 clearly does not correspond to a local
extremum). Then
(λ + 1)2 − 2(λ + 1)(λ + 2) + 1 = 0
or
λ2 + 4λ + 2 = 0
The two roots are −0.5858 and − 3.414 .
λ = −0.5858:
λ + 1 −0.5858 + 1
=
= 0.2929
λ + 2 −0.5858 + 2
1
1
y=−
=−
= −0.7071
λ +2
−0.585 + 2
x=
185
Solutions Manual
Orbital Mechanics for Engineering Students
2
z1 = x2 + y 2 + 2 xy = 0.2929 2 + ( −0.7071) + 2 ⋅ 0.2929( −0.7071) = 0.1716
λ = −3.414:
λ + 1 −3.414 + 1
=
= 1.707
λ + 2 −3.414 + 2
1
1
y=−
=−
= 0.7071
−3.414 + 2
λ +2
z2 = x2 + y 2 + 2 xy = 1.707 2 + 0.70712 + 2 ⋅ 1.707 ⋅ 0.7071 = 5.828
x=
Note that
 ∂2z
 ∂2z
 ∂2z
∂ 2 g
∂2 g 
∂ 2 g 2
d 2 h =  2 + λ 2  dx2 + 2
+λ
 dxdy +  2 + λ 2  dy
 ∂x
∂x∂y 
 ∂x∂y
∂x 
∂y 
 ∂y
d 2 h = (2 + λ ⋅ 2) dx2 + 2(2 + λ ⋅ 0) dxdy + (2 + λ ⋅ 2) dy 2
(
)
(
)
d 2 h = 2(λ + 1) dx2 + dy 2 + 4 dxdy
For λ = −0.5858 ,
(
)
d 2 h = 2(λ + 1) dx2 + dy 2 + 4 dxdy = 2( −0.5858 + 1) dx2 + dy 2 + 4 dxdy
(
)
= 0.8284 dx2 + dy 2 + 4 dxdy
Since d 2 h &gt; 0 , z1 = zmin .
For λ = −3.414 ,
(
)
(
)
d 2 h = 2(λ + 1) dx2 + dy 2 + 4 dxdy = 2( −3.414 + 1) dx2 + dy 2 + 4 dxdy
(
)
= −4.828 dx2 + dy 2 + 4 dxdy
Since d 2 h &lt; 0 , z2 = zmax .
186
Chapter 11
Solutions Manual
Orbital Mechanics for Engineering Students
187
Chapter 11
Solutions Manual
Orbital Mechanics for Engineering Students
188
Chapter 11
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