SOLUTIONS MANUAL
to accompany
ORBITAL MECHANICS FOR ENGINEERING STUDENTS
Third Edition
Howard D. Curtis
Embry-Riddle Aeronautical University
Daytona Beach, Florida
Solutions Manual
Chapter 1
Orbital Mechanics for Engineering Students Third Edition
Problem 1.1 Given the three vectors A  A x î  A y ĵ  A z k̂ , B  Bx î  By ĵ  Bz k̂ and C  Cx î  Cy ĵ  Cz k̂ ,
show analytically that
(a) A  A  A 2
(b) A  B  C   A  BC
(c) A  B  C  B A C  C A  B
Solution


A  A  A x î  A y ĵ  A z k̂  A x î  A y ĵ  A z k̂






 A x î  A x î  A y ĵ  A z k̂  A y ĵ  A x î  A y ĵ  A z k̂  A z k̂  A x î  A y ĵ  A z k̂
 
 
 
 

 
  A x 2 î  î   A x A y î  ĵ  A x A z î  k̂    A y A x ĵ  î  A y 2 ĵ  ĵ  A y A z ĵ  k̂ 

 

2
  A z A x k̂  î   A z A y k̂  ĵ  A z k̂  k̂ 


  A x 2 1  A x A y 0  A x A z 0   A y A x 0  A y 2 1  A y A z 0   A z A x 0  A z A y 0  A z 2 1

 
 

 Ax2  A y2  Az2
But, according to the Pythagorean Theorem, A x2  A y 2  A z 2  A 2 , where A  A , the magnitude of the
vector A . Thus A  A  A 2 .
(b)
î
A  B  C   A  Bx
ĵ
By
k̂
Bz
Cx
Cy
Cz

 




 A x î  A y ĵ  A z k̂   î By Cz  Bz Cy  ĵ Bx Cz  Bz Cx  k̂ Bx Cy  By Cx 


 A x By Cz  Bz Cy  A y Bx Cz  Bz Cx  A z Bx Cy  By Cx



or
A  B  C   A x By Cz  A y Bz Cx  A z Bx Cy  A x Bz Cy  A y Bx Cz  A z By Cx
(1)
Note that A  B C  C  A  B , and according to (1)
C  A  B  Cx A y Bz  Cy A z Bx  Cz A x By  Cx A z By  Cy A x Bz  Cz A y Bx
(2)
The right hand sides of (1) and (2) are identical. Hence A  B  C   A  B C .
(c)
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Chapter 1
Orbital Mechanics for Engineering Students Third Edition
A  B  C   A x î  A y ĵ  A z k̂  Bx
î
ĵ
By
k̂
Bz 
Cx
Cy
Cz




î
Ax
ĵ
Ay
k̂
Az
By Cz  Bz Cy
Bz Cx  Bx Cy
Bx Cy  By Cx




  A y Bx Cy  By Cx  A z Bz Cx  Bx Cz  î   A z By Cz  Bz Cy  A x Bx Cy  By Cx  ĵ

 

  A x Bz Cx  Bx Cz  A y By Cz  Bz Cy  k̂




 A y Bx Cy  A z Bx Cz  A y By Cx  A z Bz Cx î  A x By Cx  A z By Cz  A x Bx Cy  A z Bz Cy ĵ
 A x Bz Cx  A y Bz Cy  A x Bx Cz  A y By Cz k̂
 Bx A y Cy  A z Cz  Cx A y By  A z Bz  î  By A x Cx  A z Cz  Cy A x Bx  A z Bz  ĵ


 
 Bz A x Cx  A y Cy  Cz A x Bx  A y By  k̂


Add and subtract the underlined terms to get








A  B  C   Bx A y C y  A z Cz  A x Cx  Cx A y By  A z Bz  A x Bx  î



 By A x Cx  A z Cz  A y C y  Cy A x Bx  A z Bz  A y By  ĵ


 Bz A x Cx  A y C y  A z Cz  Cz A x Bx  A y By  A z Bz  k̂



 




 Bx î  By ĵ  Bz k̂ A C   Cx î  Cy ĵ  Cz k̂ A  B

 Bx î  By ĵ  Bz k̂ A x Cx  A y Cy  A z Cz  Cx î  Cy ĵ  Cz k̂ A x Bx  A y By  A z Bz

Or,
A  B  C   B A C   C A  B
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Chapter 1
Problem 1.2 Use just the vector identities in Problem 1.1 to show that
A  B C  D  A CB  D  A  D B C
Solution
From Problem 1.1(b)
A  B  C  D   A  B  C D
(1)
But
A  B  C D   C  A  B D
Using Problem 1.1(c) on the right yields
A  B  C D   A C  B  B C  A  D
or
A  B  C D   A  D C  B  B  D C  A 
(2)
Substituting (2) into (1) we get
A  B C  D   A CB  D   A  D B C
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Chapter 1
Problem 1.3 Let A  8î  9 ĵ  12k̂ , B  9î  6 ĵ  k̂ and C  3î  5 ĵ  10k̂ . Calculate the (scalar) projection
of CAB of C onto the plane of A and B .
Solution
The unit normal û n to the plane of A and B is
û n 
A  B 63î  100 ĵ  33k̂

 0.34115î  0.54141 ĵ  0.17870k̂
A B
17 10.863
The projection Cn of C in the direction of the unit normal to the plane is



Cn  C  û n  3î  5 ĵ  10k̂  0.34115î  0.54141ĵ  0.17870k̂  0.10289
C is the hypotenuse of the right triangle whose other two sides are of length Cn and CAB , where CAB
lies in the plane of A and B. Therefore, the square of the length of C is
C
2
 Cn 2  CAB2
That is
32  52  102  0.10289  CAB2
2
which means
CAB  11.575
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Chapter 1
Problem 1.4 Since û t and û n are perpendicular and û t  û n  û b , use the bac-cab rule to show that
û b  û t  û n and û n  û b  û t , thereby verifying Equation 1.29.
Solution
bac-cab rule: A  B  C  B A C  C A  B
û b  û t  û t  û n  û t
 û t  û t  û n 
  û t û t  û n  û n û t  û t 
  û t 0 û n 1
 û b  û t  û n
û n  û b  û n  û t  û n 
 û t û n  û n  û n û n  û t 
 û t 1 û n 0
 û n  û b  û t
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Chapter 1
Problem 1.5 The x, y and z coordinates (in meters) of a particle as a function of time (in seconds) are
x  sin 3t , y  cost and z  sin 2t . At t  3s determine:
(a) The velocity v, in Cartesian coordinates.
(b) The speed v.
(c) The unit tangent û t .
(d) The angles  x ,  y and  z that v makes with the x, y and z axes.
(e) The acceleration a in Cartesian coordinates.
(f) The unit binormal vector û b .
(g) The unit normal vector û n .
(h) The angles  x ,  y and  z that a makes with the x, y and z axes.
(i) The tangential component at of the acceleration.
(j) The normal component an of the acceleration.
(k) The radius of curvature of the path of P.
(l) The Cartesian coordinates of the center of curvature of the path.
Solution
(a)
v


dr 
d


sin 3t î  cost ĵ  sin 2tk̂   3cos 3t î  sin t ĵ  2cos 2tk̂

dt  t3 dt
t3
t3
v  2.2791î  0.95892 ĵ  1.6781k̂ ( m s)
(b)
v v 
2.27912  0.958922  1.67812
v  2.9883 m s
(c)
û t 
v 2.2791î  0.95892 ĵ  1.6781k̂

v
2.9883
û t  0.76267 î  0.32089 ĵ  0.56157k̂
(d)
 
 x  cos 1 û t gî  cos 1 0.76267 
 x  139.70
 
 y  cos 1 û t gĵ  cos 1 0.32089 
 y  71.283


 z  cos 1 û t gk̂  cos 1 0.56157 
 z  124.16
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Chapter 1
Orbital Mechanics for Engineering Students Third Edition
(e)
a


dv 
d


3cos 3t î  sin t ĵ  2cos 2tk̂   9sin 3t î  cost ĵ  4sin 2tk̂
dt  t3 dt
t3
t3
 
a  5.8526î  0.28366 ĵ  2.1761k̂ m s 2
(f)
û b 





2.2791î  0.95892 ĵ  1.6781k̂  5.8526î  0.28366 ĵ  2.1761k̂
v a

v a
2.2791î  0.95892 ĵ  1.6781k̂  5.8526î  0.28366 ĵ  2.1761k̂
1.6107 î  14.781 ĵ  6.2587k̂
1.6107 î  14.781 ĵ  6.2587k̂



1.6107 î  14.781 ĵ  6.2587k̂
16.132
û b  0.099844î  0.91625 ĵ  0.38797k̂
(g)


û n  û b  û t  0.099844 î  0.91625 ĵ  0.38797k̂  0.76267 î  0.32089 ĵ  0.56157k̂

û n  0.63904 î  0.23982 ĵ  0.73083k̂
(h)
a a 
5.85262  0.283662  2.17612  6.2505 m
s2
 ax 
 5.8526 
 cos 1 
 6.2505 
 a 
x  cos 1 
x  159.45
 ay 
1  0.28366 
  cos 
a
6.2505 
 
 y  cos 1 
 y  92.601
 az 
 2.1761 
 cos 1 

 6.2505 
 a
z  cos 1 
z  69.626
(i)


at  a  û t  5.8526î  0.28366 ĵ  2.1761k̂  0.76267 î  0.32089 ĵ  0.56157k̂

at  3.1505 m s 2
(j)
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Chapter 1
Orbital Mechanics for Engineering Students Third Edition


an  a  û n  5.8526î  0.28366 ĵ  2.1761k̂  0.63904 î  0.23982 ĵ  0.73083k̂

an  5.3984 m s 2
(k)

v 2 2.98832

an
5.3984
  1.6542 m
(l)
rC  r  û n



 0.65029î  0.28366 ĵ  0.54402k̂  1.6542 0.63904î  0.23982 ĵ  0.73083k̂

rC  0.40678î  0.11304 ĵ  0.66489k̂ m 
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Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.6 An 80 kg man and 50 kg woman stand 0.5 meter from each other. What is the force of
gravitational attraction between the couple?
Solution
F
Gm1m 2
r2

6.6742  1011  80  50
0.52
F  1.0679  106 N
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Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.7 If a person’s weight is W on the surface of the earth, calculate the earth’s gravitational pull
on that person at a distance equal to the moon’s orbit.
Solution
Let m be the person’s mass, ME the mass of the earth, and RE the radius of the earth. Then
W 
GmM E
RE 2
or
GmM E  WRE2
(1)
If rmoon is the moon’s orbital radius, then the gravitational pull of the earth at that distance is
F
GmM E
(2)
rmoon 2
Substituting (1) into (2) yields
2
 R 
F E  W
 rmoon 
Since RE  6378 km and rmoon  384 400 km , we find
F  0.0002753W
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Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.8 If a person’s weight is W on the surface of the earth, calculate what it would be, in terms
of W, at the surface of
(a) the moon
(b) Mars
(c) Jupiter
Solution
The force of gravity on mass m at the earth’s surface is
W 
GmM E
RE 2
so that
Gm 
WRE 2
ME
(1)
At the surface of planet P, the force of gravity is. using (1),
WP 
GmM P
RP2
2
 R  MP
 E
W
 RP  M E
(2)
(a)
2
2
 RE  M moon
 6378  7.348  1022
W moon  
W 
W


 1737  597.4  1022
ME
 R moon 
W moon  0.1658W
(b)
2
2
 RE  M Mars
 6378  64.19  1022
W Mars  
W

W
 3396 
ME
 R Mars 
597.4  1022
W Mars  0.3790W
(c)
2
2
 R
 M Jupiter
 6378  189 900  1022
E
W Jupiter  
W

W

 71 490 
ME
597.4  10 22
 R Jupiter 
W Jupiter  2.530W
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Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.9 A satellite of mass m is in a circular orbit around the earth, whose mass is M. The orbital
radius from the center of the earth is r. Use Newton’s Second Law of motion, together with Equations
1.25 and 1.40, to calculate the speed v of the satellite in terns of M, r and the gravitational constant G.
Solution
Writing Newton’s Second Law in the direction normal to the circular orbital path,
Fn  man
(1)
From Equation 1.25
an 
v2
r
From Equation 1,40m
Fn 
GmM
r2
Therefore, (1) becomes
mv 2 GmM

r
r2
so that
v
GM
r
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Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.10 If the earth takes 365.25 days to complete its circular orbit of radius 149.6  106 km
around the sun, use the result of Problem 1.9 to calculate the mass of the sun.
Solution
From Problem 1.9
GM
r
v
The constant speed v equals the distance 2 r traveled in one period T,
v
2 r
T
Thus
GM 2 r

r
T
or
M
4 2 r 3
G T2
Thus
M
4 2
1.496  10 
8
3
6.6742  1020 365.25  24  3600 2
M  1.9886  1030 kg
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Chapter 1
Orbital Mechanics for Engineering Students Third Edition
Problem 1.11 F is a force vector of fixed magnitude embedded on a rigid body in plane motion (in the
&  5k̂ rad s 2 , 
&& 3k̂ rad s 3 and F  15î  10 ĵ N . At that
xy plane). At a given instant   2k̂ rad s , 
instant calculate &
F&&.
Solution
From Example 1.12, we have
&
&& 
&& F  2&   F   
& F      F
F
Substituting the given values for the quantities on the right hand side,



&& F  3k̂  15î  10 ĵ  30î  45 ĵ N s 3





 


&   F 2 5k̂  2k̂  15î  10 ĵ   2 5k̂  20î  30 ĵ  2 150î  100 ĵ  300î  200 ĵ N s 3
2





& F  2k̂  5k̂  15î  10 ĵ   2k̂  50î  75 ĵ  150î  100 ĵ N s 3
  








       F  2k̂  2k̂  2k̂  15î  10 ĵ   2k̂  2k̂  20î  30 ĵ   2k̂  60î  40 ĵ

 80î  120 ĵ N s 3


Thus,




&
F&& 30î  45 ĵ  300î  200 ĵ  150î  100 ĵ  80î  120 ĵ

&
F&& 500î  225 ĵ N s 3
Howard D. Curtis


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Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.12 The absolute position, velocity and acceleration of O are
rO  16Î  84 Ĵ 59K̂ (m)
v O  7Î  9 Ĵ 4K̂ ( m s)
aO  3Î  7 Ĵ 4K̂ ( m s 2 )
The angular velocity and acceleration of the moving frame are
  0.8Î  0.7 Ĵ 0.4K̂ ( rad s)
&  0.4Î  0.9Ĵ 1.0K̂ ( rad s 2 )

The unit vectors of the moving frame are
î  0.15617Î  0.31235Ĵ 0.93704K̂
ĵ  0.12940I  0.94698Ĵ 0.29409K̂
k̂  0.97922I  0.075324 Ĵ 0.18831K̂
The absolute position of P is
r  51Î  45Ĵ 36K̂ (m)
The velocity and acceleration of P relative to the moving frame are
v rel  31î  68 ĵ  77k̂ ( m s)
arel  2î  6 ĵ  5k̂ ( m s)
Calculate the absolute velocity v P and acceleration a P of P.
Solution
Velocity analysis
From Equation 1.66,
v P  v O    rrel  v rel .
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Orbital Mechanics for Engineering Students Third Edition
Chapter 1
From the given information we have
v O  7Î  9Ĵ 4K̂
(2)
rrel  r  rO  51Î  45Ĵ 36K̂ 16Î  84 Ĵ 59K̂ 67Î  129Ĵ 77K̂
(3)
Î
Ĵ
K̂
  rrel  0.6 0.7 0.4  35.5Î  8.4 Ĵ 56.3K̂
67 129 23
(4)
v rel  31î  68 ĵ  77k̂

 31 0.15617Î  0.31235 Ĵ 0.93704K̂


 68 0.12940Î  0.94698 Ĵ 0.29409K̂

 77 0.97922Î  0.075324 Ĵ 0.18831K̂ 
so that
v rel  79.358Î  68.277 Ĵ 23.550K̂ m s 
(5)
Substituting (2), (3), (4) and (5) into (1) yields
Î
Ĵ
K̂
v P  7Î  9Ĵ 4K̂  0.8 0.7 0.4  79.358Î  68.277 Ĵ 23.550K̂
67 129 77








v P  7Î  9Ĵ 4K̂  35.5Î  8.4 Ĵ 56.3K̂  79.358Î  68.277 Ĵ 23.550K̂
v P  121.86Î  50.877 Ĵ 83.85K̂ m s 
v P  156.42 0.77902Î  0.32525Ĵ 0.53605K̂m s 
Acceleration analysis
From Equation 1.70
&  r      r  2  v  a
a  aO  
rel
rel
rel
rel
(6)
Using the given data together with (4) and (5) we obtain
2
aO  3Î  7 Ĵ 4K̂ ( m s )
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Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Î
Ĵ
K̂
&  r  0.4 0.9 1.0  149.7Î  76.2Ĵ 8.7K̂

rel
67 129 23
(8)
Î
Ĵ
K̂
    rrel  0.8 0.7 0.4  36.05Î  59.24 Ĵ 31.57K̂
35.5 8.4 56.3
(9)
2  v rel  2
Î
Ĵ
K̂
0.8
0.7
0.4  87.592Î  101.17 Ĵ 1.857K̂
79.358 68.277 23.55
(10)
a rel  2î  6 ĵ  5k̂


 6 0.12940Î  0.94698 Ĵ 0.29409K̂ 
 5 0.97922Î  0.075324 Ĵ 0.18831K̂ 
 2 0.15617Î  0.31235 Ĵ 0.93704K̂
arel  4.432Î  6.6832Ĵ 0.83202K̂
(11)
Substituting (7), (8), (9), (10) and (11) into (6) yields
r
&
r
a

4 4 4 48 6 4 4 4 44 7 4rel4 4 4 48
6 44 70 4 48 6 4 4 4 4 7 rel
a  3Î  7 Ĵ 4K̂  149.7Î  76.2Ĵ 8.7K̂  36.05Î  59.24 Ĵ 31.57K̂ 


a
2v
6 4 4 4 4 4 7 4rel 4 4 4 48 6 4 4 4 4 44 7rel4 4 4 4 4 48
 87.592Î  101.17 Ĵ 1.857K̂  4.432Î  6.6832Ĵ 0.83202K̂ 



a  85.129 0.32292Î  0.82842Ĵ 0.45765K̂ m s 2
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
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.13 An airplane in level flight at an altitude h and a uniform speed v passes directly over a
radar tracking station A. Calculate the angular velocity  and angular acceleration  of the radar
antenna as well as the rate r& at which the airplane is moving away from the antenna. Use the equations
of this chapter (rather than polar coordinates, which you can use to check your work). Attach the inertial
frame of reference to the ground and assume a non-rotating earth. Attach the moving frame to the
antenna, with the x axis pointing always from the antenna towards the airplane,
Solution
î  sin Î  cos  Ĵ
ĵ   cos Î  sin  Ĵ
k̂  K̂ (1)
Velocity analysis
The absolute velocity of the airplane is
v  v Î
(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 
h
h
sin 
sin Î  cos  Ĵ h cos
î 
Î  hĴ
cos 
cos 

(4)
The angular velocity of the moving frame is
 = &K̂
(5)
The velocity of the airplane relative to the moving frame is, making use of (1)
v rel  v rel î  v rel sin Î  cos  Ĵ v rel sin Î  v rel cos  Ĵ
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18
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Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
According to Equation 1.66, v  v O    rrel  v rel . Substituting (2), (3), (4), (5) and (6) yields
 sin 

v Î  0  &K̂   h
Î  hĴ  v rel sin  Î  v rel cos  Ĵ
 cos 



or

 sin   
v Î  0  h&Î   h&
Ĵ  v rel sin  Î  v rel cos  Ĵ
 cos   



Collecting terms,
sin  

v Î  h& v rel sin  Î   v rel cos   h&
 Ĵ

cos  
Equate the Iˆ and ˆJ components on each side to obtain
h& v rel sin   v
sin 
h&
 v rel cos   0
cos 
Solving these two equations for Ýand v rel yields
v
cos 2 
h
(7)
v rel  v sin 
(8)
&
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
aO  0
&  &&K̂

(9)
The acceleration of the airplane relative to the moving frame is, making use of (1),
arel  arel î  arel sin Î  cos  Ĵ arel sin Î  arel cos  Ĵ
(10)
Substituting (7) into (5), the angular velocity of the moving frame becomes
v
  &K̂   cos 2  K̂
h
(11)
Substituting (8) into (6) yields
v rel  v rel î  v sin  sin Î  cos  Ĵ v sin 2 Î  v sin  cos  Ĵ
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Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
From (4) and (9) we find
&  r  &&K̂    h sin  Î  hĴ  h&&Î  h&&sin  Ĵ

rel
 cos 

cos 
(13)
Using (5) and (7) we get
sin  ˆ
v
v
2 
2  sin  ˆ
2
  rrel  hÝ
Iˆ  hÝ
J  h cos  Iˆ  h cos  
J  v cos Iˆ  v sin  cos  ˆJ
 h

 h
 cos 
cos 
(14)
From (11) and (14) we have
Î
Ĵ
0
0
v cos 2 
v sin  cos 
    rrel  
K̂
v
v2
v2
 cos 2   
sin  cos 3  Î 
cos 4  Ĵ
h
h
h
(15)
0
From (11) and (12),
2  v rel  2
Î
Ĵ
0
0
v sin 2 
v sin  cos 
K̂
v
v2
v2
 cos 2   2 sin  cos 3  Î  2 sin 2  cos 2  Ĵ
h
h
h
(16)
0
&  r      r  2  v  a . Substituting (9), (10), (13),
According to Equation 1.70, a  aO  
rel
rel
rel
rel
(15) and (16) yields
sin    v 2

0  0   h&&Î  h&&
Ĵ   
sin  cos 3  Î 

cos    h
 v2

v2
  2 sin  cos 3  Î  2 sin 2  cos 2  Ĵ 
 h

h

v2
cos 4  Ĵ

h
arel sin Î  arel cos  Ĵ
Collecting terms


v2
v2
0   h&&
sin  cos 3   2 sin  cos 3   arel sin   Î


h
h


sin  v 2
v2
  h&&

cos 4   2 sin 2  cos 2   arel cos   Ĵ


cos  h
h
or
2
2
 Ý
 Ý
Ý v sin  cos 3   arel sin  
Ýsin   v cos 2  1 sin 2    arel cos  
Iˆ  h
ˆJ
0  h

 

h
cos 
h
Equate the Iˆ and ˆJ components on each side to obtain
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Orbital Mechanics for Engineering Students Third Edition
h&& arel sin   
Chapter 1
v2
sin  cos 3 
h
sin 
v2
h&&
 arel cos  
cos 2  1  sin 2  
cos 
h
ÝÝand a yields
Solving these two equations for 
rel
&& 2
v2
cos 3  sin 
h2
Howard D. Curtis
arel 
v2
cos 3 
h
21
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.14 At 30° north latitude, a 1000 kg (2205 lb) car travels due north at a constant speed of 100
km/h (62 mph) on a level road at sea level. Taking into account the earth’s rotation, calculate the lateral
(sideways) force of the road on the car and the normal force of the road on the car.
Solution
From Equation 1.86b with z  0 we have
 y&2

a  2 y&sin  î   2 RE sin  cos  ĵ  
  2 RE cos 2   k̂
 RE

(1)
where
RE  6378  103 m
  30
100  103
 27.78 m s
3600
2

 7.2921  105 rad s
23.934  3600
y&
Substituting these numbers into (1), we find
a  0.0020256î  0.014686 ĵ  0.025557k̂ m s 
From F  ma , with m  1000 kg , we obtain the net force on the car,
F  2.0256î  14.686 ĵ  25.557k̂ 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  9.81
N  9784 N 2205 lb 
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Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.15 At 29° north latitude, what is the deviation d from the vertical of a plumb bob at the end
of a 30 m string, due to the earth’s rotation?
Solution
From Equation 1.90b, with z  0 ,
a   2RE sin  cos  ĵ   2RE cos 2 k̂
From
 Fy  may
we get
T sin   m 2RE sin  cos 
or
T
From
m 2 RE sin  cos 
sin 
 Fz  maz
(1)
we obtain
T cos   mg  m 2RE cos 2 
Substituting (1),
m 2 RE sin  cos 
cos   mg  m 2 RE cos 2 
sin 
from which we find
tan  
 2 RE sin  cos 
g   2 RE cos 2 
Since d  L tan  , we deduce
dL
 2 RE sin  cos 
(2)
g   2 RE cos 2 
Setting
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Orbital Mechanics for Engineering Students Third Edition
Chapter 1
L  30 m
RE  6378  1000  6.378  106 m
  29
g  9.81 m s 2
2

 7.2921  105 rad s
23.9344  3600
yields
d  44.1 mm to the south 
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Copyright © 2013, Elsevier, Inc.
Solutions Manual
Chapter 1
Orbital Mechanics for Engineering Students Third Edition
Problem 1.16 Verify by substitution that Equation 1.114a is the solution of Equation 1.113.
Solution
Equation 1.113:
x&& 2 n x&  n 2 x 
F0
sin  t
m
(1)
Equation 1.114a:
x  e n t A sin  d t  Bcos  d t 


F0 m

2
n
2

  2   2 sin  t  2  cos  t 
n
 n

 2 n  

(2)
where
 d  n 1   2
(3)
xc  en t A sin  dt  Bcos  dt 
(4)
x&c  en t A d  B n cos  dt  B d  A n sin  dt 
(5)
Let
Then




  B  2A     2
A d  B n  B n 2 


x&&c  e n t   n 2 2   d 2 B  2A d n  cos  dt    n 2 2   d 2 A  2B d n  sin  dt




(6)
Using (4), (5) and (6) we get


 cos  t 

x&&c  2 n x&c   n 2 x c  e n t sin  d t   n 2 2   d 2 A  2B d n  2 n B d  A n  A  n 2 


d

2 2
n
2
d
d n


 cos  t B 

n

   A 2 
 e n t sin  d t  A  n 2 2   d 2  2 2 n 2   n 2  B 2 d n  2 n d 


d

   d 2  2 2 n 2
2 2
n

2
n
n d


 2 d n 





 e n t sin  dt  A  n 2   n 2 2   d 2  B 0  cos  dt B  n 2   n 2 2   d 2  A 0




 



 e n t A  1   2  n 2   d 2   B  1   2  n 2   d 2 




Howard D. Curtis
25
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Chapter 1
Orbital Mechanics for Engineering Students Third Edition
Making use of (3), we obtain
x&&c  2 n x&c   n 2 xc  0
(7)
Let
xp 

F0 m
2
n

2
  2  
2
2


(8)


(9)
  2   2 sin  t  2  cos  t 
n
 n

n
Then
x&p 
x&&p 
F0 m

n  

2
2

 2 n 
2
2
F0 m
2
n

2
  2  
2
2
  2   2  cos  t  2 2  sin  t 
n
 n



   2   2  2 sin  t  2 3  cos  t 
n
n


(10)
n
Using (8), (9) and (10) yields
x&&p  2 n x&p   n 2 x p 

sin  t   


 2  
F0 m
2
n 
2

2
2
2
n



  2  2  4 2 2 n 2   n 2  n 2   2 

n


 cos  t  2 3 n  2 n  n 2   2    n 2 2 n 




F0 m
2
n

2
  2  
2
2

sin  t  2 2   4  4 2 2 2   4   2 2
n
n
n
n

n




 cos  t 2 3 n  2 n 3  2 n  3  2 n 3 




F0 m
2
n

2
  2  
2
2

sin  t  4  2 n 2 2   n 4  4 2 2 n 2

n


2

2
2

  2  2 n   sin  t

n
2
2

 n 2   2  2 n  

F0 m

so that
x&&p  2 n x&p   n 2 x p 
Howard D. Curtis
F0
sin  t
m
(11)
26
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
It follows from (7) and (11) that x  x c  x p satisfies Equation (1).
Howard D. Curtis
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Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.17 Verify that Equation 1.114b is valid
Solution
Equation 1.114b comprises

x&
A   n x0  0 
d
d


 2  2 2  1  n 2
 F0
d m
n 2   2   2n 2
2
(1)
and
B  x0 
n
2 n
2

2
  2n 
2
2
F0
m
(2)
From Equation 1.114a,
x  e n t A sin  d t  Bcos  d t 

F0 m
2
n

2
  2  
2
2


  2   2 sin  t  2  cos  t 
n
 n

(3)
n
At t = 0, x  x 0 . Evaluating (3) at t = 0, we find
x0  B 

2 n
2
n
2
  2  
2
2
F0
m
n
Solving for B yields (2).
Differentiating (3) yields
x& e n t   n A sin  d t  B cos  d t   d A cos  d t  B sin  d t 
F0 m
  2   2  cos  t  2 2  sin  t 

n
 n
2

2
 n 2   2  2 n 




(4)
At t = 0, x& x&0 . Evaluating (4) at t = 0, we get
x&0   n B   d A 


2
n
2
n
2

2 
  2  
2
2
F0
m
n
Solving for A,
Howard D. Curtis
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Solutions Manual

A
 n B
d d
x&0
Orbital Mechanics for Engineering Students Third Edition


2
n
n 2   2

2
2

 2 n 
2
 F0
d m
Chapter 1
(5)
Substituting (2) into (5) yields




n 2   2
x& 
2 n 2 2
 F0 
 F0
A  0  n x0  

2
2


d d

m
2

m
2
2
2
 2 n  d   n 2   2  2 n  d
 n  




Collecting terms gives (1).
Howard D. Curtis
29
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.18 Numerically solve the fourth-order differential equation &&
y&& 2&
y& y  0 for y at t = 20, if
the initial conditions are y  1 , y& &
y& &
y&& 0 at t = 0.
Solution Using MATLAB.
%~~~~~~~~~~~~~~~~~~~~
function problem_1_18
%~~~~~~~~~~~~~~~~~~~~
% Numerically integrate the differential equation D4y + 2*D2y + y = 0
% from t = 0 to t = 20. Print out the value of y at t = 20 if the initial
% conditions are y(0) = 1, Dy(0) = D2y(0) = D3y(0) = 0.
%-------------------%...Initial values of y and its first three time derivatives:
y0 = 1; Dy0 = 0; D2y0 = 0; D3y0 = 0;
%...Place the initial values in a column vector:
f0 = [y0; Dy0; D2y0; D3y0];
%...Initial and final times:
t0 = 0; tf = 20;
%...Invoke MATLAB's ode45 variable time step numerical integrator to
%
solve for y, Dy, D2y and D3y at discrete times over the interval
%
[t0 tf]. The times are returned in the column vector 't'.
%
The values of y and its derivatives at each time are returned in
%
the columns of 'f'. The subfunction 'rates' below furnishes the
%
time derivatives.
[t, f] = ode45(@rates, [t0 tf], f0);
output
return
function
y
Dy
D2y
D3y
D4y
dydt
end
dydt = rates(t, f)
= f(1);
= f(2);
= f(3);
= f(4);
= -y - 2*D2y;
= [Dy; D2y; D3y; D4y];
function output
fprintf('\n---------------------------------------------------\n')
fprintf('\n Numerical solution of the differential equation')
fprintf('\n D4y + 2*D2y + y = 0.\n')
fprintf('\n y(%g) = %g\n', t(end), f(end,1))
fprintf('\n---------------------------------------------------\n\n')
end %output
end %problem_1_18
Howard D. Curtis
30
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Command Window session:
>> problem_1_18
--------------------------------------------------Numerical solution of the differential equation
D4y + 2*D2y + y = 0.
y(20) = 9.54524
-------------------------------------------------->>
Howard D. Curtis
31
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.19 Numerically solve the differential equation &y&& 3&
y& 4 y& 12y  te2t for y at t = 3 if, at t =
0, y  y& &
y& 0 .
Solution Using MATLAB
%~~~~~~~~~~~~~~~~~~~~
function problem_1_19
%~~~~~~~~~~~~~~~~~~~~
% Numerically integrate the differential equation
% D3y + 2*D2y - 4*Dy - 12*y = t*exp(2*t)
% from t = 0 to t = 3. Print out the value of y at t = 3 if the initial
% conditions are y(0) = Dy(0) = D2y(0) = 0.
%-------------------%...Initial values of y and its first three time derivatives:
y0 = 0; Dy0 = 0; D2y0 = 0;
%...Place the initial values in a column vector:
f0 = [y0; Dy0; D2y0];
%...Initial and final times:
t0 = 0; tf = 3;
%...Invoke MATLAB's ode45 variable time step numerical integrator to
%
solve for y, Dy, and D2y at discrete times over the interval
%
[t0 tf]. The times are returned in the column vector 't'.
%
The values of y and its derivatives at each time are returned in
%
the columns of 'f'. The subfunction 'rates' below furnishes the
%
time derivatives.
[t, f] = ode45(@rates, [t0 tf], f0);
output
return
function
y
Dy
D2y
D3y
dydt
end
dydt = rates(t, f)
= f(1);
= f(2);
= f(3);
= t*exp(2*t) - 3*D2y + 4*Dy + 12*y;
= [Dy; D2y; D3y];
function output
fprintf('\n---------------------------------------------------\n')
fprintf('\n Numerical solution of the differential equation')
fprintf('\n D3y + 3*D2y - 4*Dy - 12*y = t*exp(2*t).\n')
fprintf('\n y(%g) = %g\n', t(end), f(end,1))
fprintf('\n---------------------------------------------------\n\n')
end %output
end %problem_1_19
Howard D. Curtis
32
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Command Window session:
>> problem_1_19
--------------------------------------------------Numerical solution of the differential equation
D3y + 3*D2y - 4*Dy - 12*y = t*exp(2*t).
y(3) = 66.6189
-------------------------------------------------->>
Howard D. Curtis
33
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.20 Numerically solve the differential equation t&
y& t 2 y& 2y  0 to obtain y at t = 4 if the
initial conditions are y = 0 and y& 1 at t = 1.
Solution Using MATLAB
%~~~~~~~~~~~~~~~~~~~~
function problem_1_20
%~~~~~~~~~~~~~~~~~~~~
% Numerically integrate the differential equation
% t*D2y + t^2*Dy - 2*y = 0
% from t = 1 to t = 4. Print out the value of y at t = 4 if the initial
% conditions are y(1) = 0 and Dy(1) = 1.
%-------------------%...Initial values of y and its first three time derivatives:
y0 = 0; Dy0 = 1;
%...Place the initial values in a column vector:
f0 = [y0; Dy0];
%...Initial and final times:
t0 = 1; tf = 4;
%...Invoke MATLAB's ode45 variable time step numerical integrator to
%
solve for y and Dy at discrete times over the interval
%
[t0 tf]. The times are returned in the column vector 't'.
%
The values of y and its derivatives at each time are returned in
%
the columns of 'f'. The subfunction 'rates' below furnishes the
%
time derivatives.
[t, f] = ode45(@rates, [t0 tf], f0);
output
return
function
y
Dy
D2y
dydt
end
dydt = rates(t, f)
= f(1);
= f(2);
= (-t^2*Dy + 2*y)/t;
= [Dy; D2y];
function output
fprintf('\n---------------------------------------------------\n')
fprintf('\n Numerical solution of the differential equation')
fprintf('\n t*D2y + t^2*Dy - 2*y = 0.\n')
fprintf('\n y(%g) = %g\n', t(end), f(end,1))
fprintf('\n---------------------------------------------------\n\n')
end %output
end %problem_1_20
Howard D. Curtis
34
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Command Window session:
>> problem_1_20
--------------------------------------------------Numerical solution of the differential equation
t*D2y + t^2*Dy - 2*y = 0.
y(4) = 1.28873
-------------------------------------------------->>
Howard D. Curtis
35
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.21 Numerically solve the system
x&

y
2

x
2
x
2

y&


y


2
z
2
z
2
z&
 0
 0
 0
to obtain x, y and z at t = 20. The initial conditions are x = 1 and y = z = 0 at t = 0.
Solution
%~~~~~~~~~~~~~~~~~~~~
function problem_1_21
%~~~~~~~~~~~~~~~~~~~~
% Numerically integrate the system of differential equations
%
%
Dx
+ y/2
- z/2
= 0
%
-x/2 + Dy
+ z/sqrt(2) = 0
%
x/2 - y/sqrt(2) + Dz
= 0
%
% from t = 0 to t = 20. Print out the values of x, y and z at t = 20
% if the initial conditions are x = 1 and y = z = 0 at t = 0.
%-------------------%...Initial values of x y and z:
x0 = 1; y0 = 0; z0 = 0;
%...Place the initial values in a column vector:
f0 = [x0; y0; z0];
%...Initial and final times:
t0 = 0; tf = 20;
%...Invoke MATLAB's ode45 variable time step numerical integrator to
%
solve for x, y and z at discrete times over the interval
%
[t0 tf]. The times are returned in the column vector 't'.
%
The values of x, y and z at each time are returned in
%
the columns of 'f'. The subfunction 'rates' below furnishes the
%
time derivatives.
[t, f] = ode45(@rates, [t0 tf], f0);
output
return
function
x =
y =
z =
Dx =
dydt = rates(t, f)
f(1);
f(2);
f(3);
-y/2 + z/2;
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Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Dy = x/2 - z/sqrt(2);
Dz = -x/2 + y/sqrt(2);
dydt = [Dx; Dy; Dz];
end
function output
fprintf('\n---------------------------------------------------\n')
fprintf('\n Numerical solution of the system of differential')
fprintf('\n equations\n')
fprintf('\n
Dx
+ y/2
- z/2
= 0')
fprintf('\n
-x/2 + Dy
+ z/sqrt(2) = 0')
fprintf('\n
x/2 - y/sqrt(2) + Dz
= 0\n')
fprintf('\n x(%g) = %g\n', t(end), f(end,1))
fprintf('\n y(%g) = %g\n', t(end), f(end,2))
fprintf('\n z(%g) = %g\n', t(end), f(end,3))
fprintf('\n---------------------------------------------------\n\n')
end %output
end %problem_1_21
Command Window session:
>> problem_1_21
--------------------------------------------------Numerical solution of the system of differential
equations
Dx
+ y/2
- z/2
= 0
-x/2 + Dy
+ z/sqrt(2) = 0
x/2 - y/sqrt(2) + Dz
= 0
x(20) = 0.703137
y(20) = 0.666531
z(20) = -0.246703
-------------------------------------------------->>
Howard D. Curtis
37
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.22 Use one of the numerical methods discussed in Section 1.8 to solve Equation 1.127
x&&
g0 RE 2
x2
0
(1)
for the time required for the moon to fall to the earth if it were somehow stopped in its orbit while the
earth remained fixed in space. Compare your answer with the analytical solution,
t

r0
1  r0  2r  
 r0  r r0  r   sin 

2
 r0  
2g0 RE  4
r0
(2)
2
Solution
Note that at impact the distance between the center of the earth and the moon would be
Rearth  Rmoon  8115 km .
Using MATLAB:
% ~~~~~~~~~~~~~~~~~~~
function problem_1_22
% ~~~~~~~~~~~~~~~~~~~
% Rectilinear orbit problem. Numerically integrate the equation
%
% D2x + g0*RE^2/x^2 = 0
%
% from t = 0 to the time required for x to reach a specified value.
%-------------------clear all
RE
= 6378;
% Earth's radius
g0
= 9.807e-3; % Sea-level gravitational acceleration
days = 24*3600; % Conversion from days to seconds
%...Initial values of x (km) and Dx (km/s):
x0 = 384400; Dx0 = 0;
%...Place the initial values in a column vector:
f0 = [x0; Dx0];
%...Initial and final times:
t0 = 0; tf = 4.844453*days;
%...Invoke MATLAB's ode45 variable time step numerical integrator to
%
solve for x and Dx at discrete times over the interval
%
[t0 tf]. The times are returned in the column vector 't'.
%
The values of x and Dx at each time are returned in
%
the columns of 'f'. The subfunction 'rates' below furnishes the
%
time derivatives.
[t, f] = ode45(@rates, [t0 tf], f0);
Howard D. Curtis
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Copyright © 2013, Elsevier, Inc.
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Orbital Mechanics for Engineering Students Third Edition
Chapter 1
output
return
function dydt = rates(t, f)
x
= f(1);
Dx
= f(2);
D2x = -g0*RE^2/x^2;
dydt = [Dx; D2x];
end %rates
function output
fprintf('\n---------------------------------------------------\n')
fprintf('\n Numerical solution of the differential equation\n')
fprintf('\n D2x + g0*RE^2/x^2 = 0\n')
fprintf('\n x(%.8g days) = %g km\n', t(end)/days, f(end,1))
fprintf('\n---------------------------------------------------\n\n')
end %output
end %problem_1_22
% ~~~~~~~~~~~~~~~~~~~
Command Window session:
>> problem_1_22
--------------------------------------------------Numerical solution of the differential equation
D2x + g0*RE^2/x^2 = 0
x(4.844453 days) = 8115.6 km
-------------------------------------------------->>
From (2) we have
t
384 400
2  9.807  10
3

 384 400  2  8115  
384 400
sin 1 
  384 400  8115 384 400  8115 
 
2
384 400

 6378  4

2
t  418 560 s  4.84445 days
Howard D. Curtis
39
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
Problem 1.23 Use a Runge-Kutta solver to solve the nonlinear Lorenz equations
x&  y  x 
y& x   z  y
z& xy   z
Use   10 ,   8 3 and   28 and the initial conditions x = 0, y = 1 and z = 0 at t = 0. Let t range to a
value of 20 or higher. Plot the phase trajectory x = x(t), y= y(t), z = z(t).
Solution Using MATLAB
% ~~~~~~~~~~~~~~~~~~~
function problem_1_23
% ~~~~~~~~~~~~~~~~~~~
% Numerical solution of the Lorenz equations:
%
% Dx = sigma*(y - x)
% Dy = x*(rho - z) - y
% Dz = x*y - beta*z
%
% ------------------clear all; close all; clc
%...Values for the constants:
beta = 8/3;
sigma = 10;
rho
= 28;
%...Initial values of x, y and z:
x0 = 0; y0 = 1; z0 = 0;
%...Place the initial values in a column vector:
f0 = [x0; y0; z0];
%...Initial and final times:
t0 = 0; tf = 30.2;
%...Invoke MATLAB's ode45 variable time step numerical integrator to
%
solve for x, y and z at discrete times over the interval
%
[t0 tf]. The times are returned in the column vector 't'.
%
The values of x, y and z at each time are returned in
%
the columns of 'f'. The subfunction 'rates' below furnishes the
%
time derivatives.
[t, f] = ode45(@rates, [t0 tf], f0);
plotit
return
Howard D. Curtis
40
Copyright © 2013, Elsevier, Inc.
Solutions Manual
function
x =
y =
z =
Dx =
Dy =
Dz =
dfdt
end
Orbital Mechanics for Engineering Students Third Edition
Chapter 1
dfdt = rates(t,f)
f(1);
f(2);
f(3);
sigma*(y - x);
x*(rho - z) - y;
x*y - beta*z;
= [Dx; Dy; Dz];
function plotit
figure(1)
plot3(f(:,1), f(:,2), f(:,3))
grid on
xlabel('x'); ylabel('y'); zlabel('z')
text(f(1,1), f(1,2), f(1,3),'o Start')
text(f(end,1), f(end,2), f(end,3), 'o Stop')
axis tight
view([1 -1 .5])
end %plotit
end %problem_1_23
% ~~~~~~~~~~~~~~~~~~~
The MATLAB plot is shown below
Howard D. Curtis
41
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Howard D. Curtis
42
Chapter 1
Copyright © 2013, Elsevier, Inc.