Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 6
Problem 6.1 A large spacecraft has a mass of 125,000 kg. Its orbital maneuvering engines produce a
thrust of 50 kN. The spacecraft is in a 400 km circular earth orbit. A delta-v maneuver transfers the spacecraft to a coplanar 300 km by 400 km elliptical orbit. Neglecting propellant loss and using elementary
physics (linear impulse equals change in linear momentum, distance equals speed times time), estimate
(a) the time required for the v burn and
(b) the distance traveled by the spacecraft during the burn.
(c) Calculate the ratio of your answer for (b) to the circumference of the initial circular orbit.
(d) What percent of the initial mass was expelled as combustion products?
Solution
Orbit 1 circle :
r1  6378  400  6778km
v1 

r1

398,600
 7.6686km s
6778
Orbit 2 ellipse :
ra  r1  6778km
2
rp  6378  300  6678km
2
h2  2
ra rp
2
2
ra  rp
2
2
 2  398,600
6778  6678
 51,784 km 2 s
6778  6678
h
51,784
va  2 
 7.6401km s
2
ra
6778
2
v  v a  v1  7.6401  7.6686  0.028548km s  2.8548m s
2
(a)
Thrust  t  mv
50000 t  125000 2.8548
t  71.371s
(b)
v  v1  v 
v
0.028548
vavg  1
 v1 
 7.6686 
 7.6829km s
2
2
2
s  vavg t  7.682971.371  548.34km
(c)
s
548.34

 0.012875 or 1.29%
orbit circumference 2  6778
(d)
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
v
Chapter 6
0.028548

I g
m
 1  e sp 0  1  e 3000.00981  0.00965353
m
m
 0.965%
m
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Chapter 6
Problem 6.2 A satellite traveling 8 km/s at a perigee altitude of 500 km fires a retrorocket. What deltav is necessary to reach a minimum altitude of 200 km during the next orbit?
Solution
v perigee  8.0 km s
1
rperigee  6378  500  6878 km
1
rapogee  rperigee  6878 km
2
1
rperigee  6378  200  6578km
2
h2  2
v apogee 
2
rperigee rapogee
2
2
rperigee  rapogee
2
2
h  2  398600
6578  6878
 51773 km 2 s
6578  6878
h2
51773

 7.52734 km s
rapogee
6878
2
v  vapogee  v perigee  7.52734  8.0
2
1
v  0.47266km s
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Chapter 6
Problem 6.3 A spacecraft is in a 500 km altitude circular earth orbit. Neglecting the atmosphere, find
the delta-v required at A in order to impact the earth at (a) point B; (b) point C.
Solution
Orbit 1 is the original circular orbit and orbit 2 is the impact trajectory.

vA  
1
r
398 600
 7.613 km s
6378  500

h 2
rapogee  2
1
 1  e2
2
6378  500 
h 2
rB  2
2
h2 2
1
 1  e2

 6878 1  e2 
1
1
 1  e2 cos 60

h2 2
 1  e2 cos B
h 2
6378  2
h22

h 2
 2



h22

 6378 1  0.5e2 
6378 1  0.5e2   6878 1  e2 

e2  0.04967
 h2  6878 1  e2   6878  398 600  1  0.04967   51040 km 2 s
(a)
h
51040
vA   2 
 7.421 km s
2 r
6878
A
v  v A   v A   7.421  7.613  0.1915 km s
2
1
(b) To fall through the point directly below, we must remove completely the transverse component of
velocity:
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Chapter 6
v  0  v A   7.613 km s
1
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Chapter 6
Problem 6.4 A satellite is in a circular orbit at an altitude of 250 km above the earth’s surface. If an
onboard rocket provides a delta-v of 200 m/s in the direction of the satellite’s motion, calculate the altitude of the new orbit’s apogee.
Solution
r1  6378  250  6628km
v1 

r1

398,600
 7.75492 km s
6628
v perigee  v1  v  7.75492  0.2  7.95492 km s
2
rperigee  r1  6628km
2
h2  rperigee v perigee  66287.95492  52725.2 km 2 s
2
2
h 2
rperigee  2
2
6628 
1
 1  e2
52725.22 1
398,600 1  e2
e2  0.0522453
h 2
rapogee  2
2
1
52725.2 2
1

 7358.74 km
 1  e2
398,600 1  0.0522453
zapogee  7358.74  6378  980.741km
2
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Chapter 6
Problem 6.5 A spacecraft S is in a geocentric hyperbolic trajectory with a perigee radius of 7000 km and
a perigee speed of 1.3v esc . At perigee, the spacecraft releases a projectile B with a speed of 7.1 km/s parallel to the spacecraft’s velocity. How far d from the earth’s surface is S at the instant B impacts the earth?
Neglect the atmosphere.
Solution
To determine where the projectile B impacts the earth we need its orbital elements.
rapogee
B
v apogee
B
 7000 km
 7.1 km s
hB  rapogee B v apogee B  7000 7.1  49700 km
2
rapogee
B
7000 
h
 B
2
s
1
 1  eB
497002 1
 eB  0.1147
398600 1  eB
2   hB
TB  2 
2
 
 1  eB
3
 49700 3

2

  4952 s period of B' s orbit 

 398 6002 
 1  0.1147 2 

At impact, rB  Rearth .
Rearth 
6378 
hB2
1
 1  eB cos  impact
At impact,
rB  Rearth 
49700 2
1
  impact  104.3 from perigee of B' s elliptical orbit
398 600 1  0.1147 cos  impact

Determine the time of flight ( tof ) to impact by first finding t impact , the time from perigee to B’s impact
point.
 impact
1  eB
1  0.1147
104.3
tan

tan
 Eimpact  1.708 rad
2
1  eB
2
1  0.1147
2
M impact  E impact  eB sin Eimpact  1.708  0.1147 sin 1.708  1.594 rad
tan
E impact
t impact  TB

M impact
2
 4952
1.594
 1257 s (from impact point to perigee)
2
Then
T
4952
tof  B  t impact 
1257  1220 s
2
2
Find the orbital elements of spacecraft S trajectory.
rperigee  7000 km
S
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v perigee  1.3vesc  1.3
S
2
rperigee
 1.3
S
2  398 600
 13.97 km s
7000
hS  rperigee S v perigee S  7000 13.97  97 110 km
h
rperigee  S
2
s
2
1
 1  eS
S
7000 
Chapter 6
97110 2 1
 eS  2.38
398 600 1  eS
Location of S on its hyperbolic trajectory when B impacts the earth:
2
e 2 1
3 S
Mh 
3/ 2
398600 2
 2.382 1
3/ 2
1220  2.131 rad
hS
97 110
eS sinh F  F  M h
2.38sinh F  F  2.131  F  1.118 Algorihm 3.2 
tan
S
2

tof 
3
eS  1
F
2.38  1
1.118
tanh

tanh
  S  76.87
eS 1
2
2.38  1
2
h 2
rS  S
971102
1
1

 15 360 km
 1  eS cos s 398 600 1  2.38 cos 76.87
d  rS  6378  8978 km
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Orbital Mechanics for Engineering Students Third Edition
Chapter 6
Problem 6.6 A spacecraft is in a 200 km circular earth orbit. At t = 0, it fires a projectile in the direction
opposite to the spacecraft’s motion. Thirty minutes after leaving the spacecraft, the projectile impacts the
earth. What delta-v was imparted to the projectile? Neglect the atmosphere.
Solution
A is apogee of impact trajectory, I is the impact point, a is the semimajor axis of the projectile’s orbit.
rA  a1  e
r
e  A  1
a
(1)
From Equation 3.25,
rI  a1  ecos E
 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 
t
t
t
M  2 2
   2    2
 
3/
2 3/ 2
T
T
a 2
a

Let f E  M  E  esin E . Then Kepler’s equation is f E  0 .
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Chapter 6
f E  M  E  e sin E
t
f E   
3/ 2
 E  esin E   
a
t
 rI  rA cos E 


1  cos E 
3/ 2
E
rA  rI
sin E
rA cos E  rI
Setting rA  6578 km , rI  6378 km and t  30  60  1800s ,
f E   
1136 400
 6378  6578 cos E 


1  cos E
3/ 2
E
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  0.01975
a  6451 km
True anomaly of the impact point:


1 e
1  0.01975
5.319
tan I 
tan I 
tan
  I  303.8 123.8 ccw from apogee 
2
1 e
2
1  0.01975
2


h   a1  e2   398 600  6451  1  0.019752  50700 km 2 s
vA 
vc 
h 50 700

 7.707 km s velocity of projectile at apogee.
rA
6578

r

398 600
 7.784 km s velocity of spacecraft in circular orbit.
6578
v  v A  vc  7.707  7.784  0.07725 km s
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Chapter 6
Problem 6.7 A spacecraft is in a circular orbit of radius r and speed v around an unspecified planet. A
rocket on the spacecraft is fired, instantaneously increasing the speed in the direction of motion by the
amount v   v , where   0 . Calculate the eccentricity of the new orbit.
Solution
v

Circular orbit velocity
r
The maneuver point is periapsis of the new orbit, so its angular momentum is
h  r v   v   rv 1     r

r
1    
r 1   
The periapsis radius of the new orbit is
 r 1   
1
1
2 1

r

 r 1   
 1 e

1 e
1 e
2
h2
It follows that
1  e  1   
2
or
e     2
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Chapter 6
Problem 6.8 A spacecraft is in a 300 km circular earth orbit. Calculate
(a) the total delta-v required for a Hohmann transfer to a 3000 km coplanar circular earth orbit and
(b) the transfer orbit time.
Solution
(a)
Orbit 1 (circle):

v1 
r

398 600
 7.726 km s
6378  300
Orbit 2 (transfer ellipse):
e2 
rapogee  rperigee
2
2
rapogee  rperigee
2
2
h 2
rperigee  2
2
6678 
2
v apogee 
2
6378  3000  6378  300
 0.1682
6378  3000  6378  300
1
 1  e2
h2 2
1
398 600 1  0.1682
v perigee 

h2
rperigee
rapogee
2
h2  55 760 km 2 s

55 760
 8.35 km s
6678

55 760
 5.946 km s
9378
2
h2

Orbit 3 (circle):
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v3 

r

Orbital Mechanics for Engineering Students Third Edition
Chapter 6
398 600
 6.519 km s
6378  3000
v1  v perigee  v1  8.350  7.726  0.6244 km s
2
v 2  v 3  vapogee  6.519  5.946  0.5734 km s
2
v total  v1  v 2  1.198 km s
(b)


a2 
1
1
r
r
 6678  9378  8028 km s
2 perigee 2 apogee 2
2
T2 
2 3/ 2
a


2
398 600
80283/ 2  7159 s
period of tranfer ellipse 
T
t perigee to apogee  2  3579 s  59.65 min
2
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Chapter 6
Problem 6.9 A space vehicle in a circular orbit at an altitude of 500 km above the earth executes a
Hohmann transfer to a 1000 km circular orbit. Calculate the total delta-v requirement.
Solution
rA  6878 km
vA  
1

rA
h2  2

rB  7378 km
398 600
 7.6127 km s
6878
rA rB
6878  7378
 2  398 600
 53 270.3 km 2 s
rA  rB
6878  7378
h
53 270
vA   2 
 7.7450 km s
2 r
6878
A
h
53 270
vB   2 
 7.2202 km s
2 r
7378
B
Alternatively, the energy equation, v 2 2   r    2a , implies
 2 1
v   
 r a
so that, since a  rA  rB  2  7128 km ,
1 
 2 1
 2
vA   
   

  398 600  7.7450 km s
2

6878 7128 
 rA a
1 
 2 1
 2
vB        

 398 600  7.2202 km s
2
 7378 7128 
 rB a
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vB  
3

rB
Orbital Mechanics for Engineering Students Third Edition

Chapter 6
398 600
 7.3502 km s
7378
v  v A   v A   v B   v B   0.13235  0.13005
2
1
3
2
v  0.2624 km s
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Chapter 6
Problem 6.10 Assuming the orbits of earth and Mars are circular and coplanar, calculate
(a) the time required for a Hohmann transfer from earth orbit to Mars orbit and
(b) the initial position of Mars (  ) in its orbit relative to earth for interception to occur.
Solution
(a) For the transfer ellipse
1
1
rMars  rearth   227.9  149.6   106  188.8  106 km
2
2
3/ 2
2 3/ 2
2
188.8  106   44.73  106 s = 517.7 days
T 
a


132.7  10 9
T
tof   258.8 days time of fligh from earth to Mars 
2
a
(b) Period of Mars in its orbit,
2
TMars 
rMars 3/ 2 

2
227.9  10 6 3/ 2  59.34  106
132.7  10 9
180  
tof
258.8



 0.7537
TMars 343.4
180
2
s  686.8 days
  44.33
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Chapter 6
Problem 6.11 Calculate the total delta-v required for a Hohmann transfer from the smaller circular orbit to the larger one.
3r
3
2
B
A
1
r
Solution
rA  r
rB  3r
v A   v1 
1
h2  2 

rA


r
rA rB
r 3r 
 2
 1.225  r
rA  rB
r  3r
h
1.225  r

vA   2 
 1.225
2 r
r
r
A
(Alternatively, use the energy equation.)
h
1.225  r

vB   2 
 0.4082
2 r
3r
r
B
vB   v 3 
3

rB


3r
 0.5774

r



v  v A   v A   v B   v B   0.2247
 0.1691
 0.3938
 0.3938v1
2
1
3
2
r
r
r
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Chapter 6
Problem 6.12 With a v A of 1.500 km/s, a spacecraft in the circular 6700 km geocentric orbit 1 initiates
a Hohmann transfer to the larger circular orbit 3. Calculate  v B at apogee of the Hohmann transfer ellipse 2.
Solution

vA  
1
rA

398 600
 7.7131 km s
6700
h2  rA  v A   v A   6700 7.7131  1.5  61728 km 2 s


1
h 2
rA  2
1
 1  e2
6700 
617282 1
398 600 1  e2

e2  0.42677
h 2
rB  2
617282
1
1

 16 676 km
 1  e2 398 600 1  0.42677
h
61728
vB   2 
 3.7016 km s
2 r
B 16 676
vB  
3

rB

398 600
 4.8890 km s
16 676
vB  vB   vB   4.8890  3.7016
3
2
vB  1.187 km s
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Chapter 6
Problem 6.13 Two geocentric elliptical orbits have common apse lines and their perigees are on the
same side of the earth. The first orbit has a perigee radius of rp  7000 km and e  0.3 , whereas for the
second orbit rp  32000 km and e  0.5 .
(a) Find the minimum total delta-v and the time of flight for a transfer from the perigee of the inner orbit to the apogee of the outer orbit.
(b) Do part (a) for a transfer from the apogee of the inner orbit to the perigee of the outer orbit.
e = 0.5
2
3
Earth
e = 0.3
1
B
D
A
C
4
7000 km
32 000 km
Solution
rA  7000 km
rC  32000 km
e1  0.3
r r
e1  B A
rB  rA
r  7000
0.3  B
rB  7000

rB  13 000 km
e2  0.5
r r
e2  D C
rD  rC
r  32 000
0.5  D
rD  32 000
h1  2
Howard D. Curtis

rD  96 000 km
rA rB
7000 13 000
 2  398 600
 60 230 km 2 s
rA  r B
7000  13 000
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h2  2
rC rD
32 000  96 000
 2  398 600
 138 300 km 2 s
rC  r D
32 000  96 000
h3  2
rA rD
7000  96 000
 2  398 600
 72120 km 2 s
rA  r D
7000  96 000
h4  2
rBrC
13 000  32 000
 2  398 600
 85 850 km 2 s
rB  r C
13 000  32 000
h
60 230
vA   1 
 8.604 km s
1 r
7000
A
h
72120
vA   3 
 10.3 km s
3 r
7000
A
h
60 230
vB   1 
 4.633 km s
1 r
13
000
B
h
85 850
vB   4 
 6.604 km s
4 r
13
000
B
h
138 300
vC   2 
 4.323 km s
2 r
32 000
C
h
85 850
vC   4 
 2.683 km s
4 r
32 000
C
h
138 300
vD   2 
 1.441 km s
2 r
96 000
D
h
72120
vD   3 
 0.7512 km s
3 r
D 96 000
T3 
2  rA  rD 


 2 
T4 
2  rB  rC 


 2 
3/ 2
2
 7000  96 000 


2
398 600 

3/ 2

Chapter 6
3/ 2
2
 13 000  32 000 


2
398 600 
 116 300 s  32.31 h
3/ 2
 33 590 s  9.33 h
(a) Transfer orbit 3:
v D  v D   v D   1.441  0.7512  0.6896 km s
2
3
v A  v A   v A   10.3  8.604  1.699 km s
3
1
v total  v A  v D  2.389 km s
T
32.32
tof3  3 
 16.15 h
2
2
(b) Transfer orbit 4:
vB  vB   vB   6.604  4.633  1.971 km s
4
1
vC  vC   vC   4.323  2.683  1.64 km s
2
4
v total  vB  vC  3.611 km s
T
9.33
tof 4  4 
 4.665 h
2
2
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Chapter 6
Problem 6.14 The Space Shuttle was launched on a fifteen-day mission. There were four orbits after
injection, all of them at 39° inclination.
Orbit 1: 302 by 296 km.
Orbit 2 (day 11): 291 by 259 km.
Orbit 3 (day 12): 259 km circular.
Orbit 4 (day 13): 255 by 194 km.
Calculate the total delta-v, which should be as small as possible, assuming Hohmann transfers.
Solution
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
rapogee rperigee
h1  2
1
1
rapogee  rperigee
1
1
v apogee 
1
4
 2  398 600
6680  6674
 51 590 km 2 s
6680  6674
h1
51 590

 7.723 km s
rapogee
6680
1
v perigee 
1
h1

rperigee
1
rapogee rperigee
h2  2
2
2
v perigee 
2

r3
2
rapogee  rperigee
2
2
v apogee 
v3 
51 590
 7.730 km s
6674

h4  2
v apogee 
4
h2
rapogee

51 500
 7.722 km s
6669

51 500
 7.759 km s
6637
2
h2
rperigee
 2  398 600
2
6669  6637
 51 500 km 2 s
6669  6637
398 600
 7.750 km s
6637
rapogee rperigee
4
4
rapogee  rperigee
4
4
 2  398 600
6633  6572
 51 300 km 2 s
6633  6572
h4
51 300

 7.734 km s
rapogee
6633
4
v perigee 
4
Howard D. Curtis
h4
rperigee

4
51 300
 7.806 km s
6572
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Orbital Mechanics for Engineering Students Third Edition
Chapter 6
Apogee of orbit 1 to perigee of orbit 2:
h12  2
v12 
rapogee rperigee
1
2
rapogee  rperigee
1
2
h12
 vapogee 
rapogee
1
1
 2  398 600
h12
rperigee
6680  6637
 51 520 km 2 s
6680  6637
 v perigee
2
 7.712  7.723  7.762  7.759  0.01393 km s
2
Perigee of orbit 2 to orbit 3 (tangent):
v 23  v perigee  v 3  7.759  7.750  0.009313 km s
2
Orbit 3 to perigee of orbit 4:
h34  2
r3 rperigee4
r3  rperigee4
 2  398 600
6572  6637
 51 310 km 2 s
6572  6637
h
h34
v 34  34  v 3 
 v perigee  7.731  7.75  7.807  7.806  0.02026 km s
4
r3
rperigee
4
v total  v12  v 23  v 34  0.01393  0.009313  0.02026  0.04351km s
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Chapter 6
Problem 6.15 Calculate the total delta-v required for a Hohmann transfer from a circular orbit of radius r to a circular orbit of radius 12r.
3
2
1
B
r
A
12r
Solution

vA  
1
r
r 12r 
 1.359 r
r  12r
h2  2
h

v A   2  1.359
2
r
r
vB  
2
h2

 0.1132
12r
r
Alternatively, using the energy equation,
a2 
r  12r
 6.5r
2


2 1 
v A        1.846  1.359
2
r
a
r
r

2
1


 2
vB   

  0.01282  0.1132
2
r
r
 12r a2 
vB  
3

12r
 0.2887

r



v  v A   v A   v B   vB   0.3587
 0.1754
 0.5342
2
1
3
2
r
r
r
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Orbital Mechanics for Engineering Students Third Edition
Chapter 6
Problem 6.16 A spacecraft in circular orbit 1 of radius r leaves for infinity on parabolic trajectory 2 and
returns from infinity on a parabolic trajectory 3 to a circular orbit 4 of radius 12r. Find the total delta-v
required for this non-Hohmann orbit change maneuver.
Solution
vA  

vB  
2

 0.4082
12r
r
1
3
r
2

 1.414
r
r
vA  
2
vB  
4

12r
v  v A   v A   v B   vB   0.4142
2
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1
4
3
 0.2887

r

r
 0.1196
256

r
 0.5338

r
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 6
Problem 6.17 A spacecraft is in a 300 km circular earth orbit. Calculate
(a) the total delta-v required for the bielliptical transfer to a 3000 km altitude coplanar circular orbit shown, and
(b) the total transfer time.
4
3000 km
e = 0.3
2
vA
300 km
vB
B
A
C vC
1
3
Solution
rA  6678 km
rC  9378 km
(a)
Orbit 1:
vA  
1

rA

398 600
 7.726 km s
6678
Orbit 2:
rB  rA
 e2
rB  rA
rB  6678
 0.3  rB  12 402 km
rB  6678
h2  2
rA rB
6678 12 402
 2  398 600
 58 830 km 2 s
rA  rB
6678  12 402
h
58 830
vA   2 
 8.809 km s
2 r
6678
A
h
58 830
vB   2 
 4.743 km s
2 r
12
402
B
Orbit 3:
h3  2
rBrC
12 402  9378
 2  398 600
 65 250 km 2 s
rB  rC
12 402  9378
h
65 250
vB   3 
 5.261 km s
3 r
B 12 402
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Chapter 6
h
65 250
vC   3 
 6.957 km s
3 r
9378
C
Orbit 4:
vC  
4

rC

398 600
 6.519 km s
9 378
v total  v A   v A   vB   vB   vC   vC   1.083  0.5177  0.4379  2.039 km s
2
1
3
2
4
3
(b)
2  rA  rB 


 2 
3/ 2
T2 
2  rB  rC 


 2 
3/ 2
T3 
t total 


2
3/ 2
2
3/ 2
 6678  12 402 


2
398 600 
 12 402  9378 


2
398 600 
 9273 s
 11 310 s
1
T  T2  10 290 s  2.859 hr
2 1
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Chapter 6
Problem 6.18 Verify Equations 6.4.
rB
rC
2
rA
5
4
1
D
B
A
C
F
3
Solution
Angular momenta of the five orbits:
h1  rA
h2  2
rA rB
rA  rB
h3  2
rBrC
rB  rC
h4  rC
h5  2
rA rC
rA  rC
Let
rC   rA
rB   rA
v0 

rA
Bielliptical transfer:
h
vA   1 
1 r
A
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
rA
 v0
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Chapter 6
Orbital Mechanics for Engineering Students Third Edition
h

vA   2  2
2 r
rA
A
rB

 2v 0
rA  rB
1 
h
2
vB   2 
2 r
r
B
B
rA rB
 rA
 2
rA  rB
rA rB
rA  rB
h
2
vB   3 
3 r
rB
B
rBrC

 2
rB  rC
rA
rA
rB
rC
2 

v0
rB  rC
 
h
2
vC   3 
3 r
rC
C
rBrC

 2
rB  rC
rA
rA
rC
rB
2 

v0
rB  rC
 
h
vC   4 
4 r
C

rC
rA
rC


rA
1


1
2   1 

vC  v C   v C  
3
4
2 
 
v0 
2 
 
v0 
1 
2
 1 
1

v0 
v0 
2   
 2     1  
v bielliptic  





2

1     
   1  
  
1 

v0
v0
2   
v0 
2   1 
v bielliptic  v A  v B  v C 
 1 
v0
v A  v A   v A   2v 0
 v0 
2
1
1 
v B  vB   vB  
3
2
2

 
2

v
0
v0
1     
   1  
v
0
v0

2
1    v 0

1   

Hohmann transfer:
h
2
vA   5 
5 r
rA
A
rA rC

 2
rA  rC
rA
rC
2 

v0
rA  rC
1
h
2
vD   5 
5 r
r
C
C
rA rC

 2
rA  rC
rA
rA
rC
Howard D. Curtis
rA
2

v0
rA  rC
1 
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v D   vC  
4
4
1

v0
v A  v A   v A  
5
1
Chapter 6
2 
1
v0  v0 
2   1
1
v0
1
2
1  2
v D  v D   v D  
v0 
v0 
v0
4
5

1 
1 
v Hohmann  v A  v D 
2   1
1
v0 
1  2
1 
v0
 1
2 1    
v Hohmann  

 1 v0
 


1






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Chapter 6
Problem 6.19 The space station and spacecraft A and B are all in the same circular earth orbit of 350 km
altitude. Spacecraft A is 600 km behind the space station and Spacecraft B is 600 km ahead of the space
station. At the same instant, both spacecraft apply a  v  so as to arrive at the space station in one revolution of their phasing orbits.
(a) Calculate the times required for each spacecraft to reach the space station.
(b) Calculate the total delta-v requirement for each spacecraft.
Solution
For the circular orbit of the space station,
r  6728 km
vc 

r

398 600
2 3/ 2
 7.697 km s Tc 
r

6728

2
398 600
67283/ 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
2  6728  600
2 r  600
 5492
 5414 s = 90.2 min
2 r
2  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
2  6728  600
2 r  600
 5492
 5570 s = 92.8 min
2 r
2  6728
(b)
The period of spacecraft A’s phasing orbit, is t SA , which determines the semimajor axis of that orbit:
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2
398 600
Chapter 6
aA 3/ 2  aA  6664 km
Spacecraft A is at the apogee of its phasing orbit. From the energy equation
1 
2 1 
 2
vA    
 398 600 

  7.660 km s


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 A 
 2 v A  0.07388 km s
total
The period of spacecraft B’s phasing orbit, is t BS , which determines the semimajor axis of that orbit:
5570 
2
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
v B       398 600 

  7.733 km s

6728 6791 
 r aB 
The prograde delta-v required to enter the phasing orbit is
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
vB 
total
 2 vB  0.07153 km s
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Chapter 6
Problem 6.20 Satellites A and B are in the same circular orbit of radius r. B is 180° ahead of A. Calculate
the semimajor axis of a phasing orbit in which A will rendezvous with B after just one revolution in the
phasing orbit.
r
B
F
A
Solution
T
T is the period of the original circular orbit.
2
2 3/ 2 1 2 3/ 2
a

r
2 

Tphasing 
a3/ 2 
1 3/ 2
r
2
1

a   r 3/ 2 
2

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Chapter 6
Problem 6.21 Two spacecraft are in the same elliptical earth orbit with perigee radius 8000 km and apogee radius 13 000 km. Spacecraft 1 is at perigee and spacecraft 2 is 30° ahead. Calculate the total delta-v
required for spacecraft 1 to intercept and rendezvous with spacecraft 2 when spacecraft 2 has traveled
60°.
D intercept
2
C Spacecraft 2
60°
P Spacecraft 1
30°
A
1
13 000 km
8000 km
Solution
rA  13000 km
rP  8000 km
Orbit 1:
r r
a1  A P  10 500 km
2
rA  rP
e1 
 0.2381
rA  rP
h1   1  e rP  398 600 1  0.2381  8000  62 830 km 2 s
T1 
2 3/ 2
a


2
398 600
10 5003/ 2  10 710 s
Time of flight from P to C:
 1  e1
 1  0.2381
 
30 
EC  tan 1 
tan C   tan 1 
tan
  0.4144 rad

2 
1  0.2381
2 
 1  e1
M C  EC  e1 sin EC  0.4144  0.2381sin 0.4144  0.3185 rad
M
0.3185
tC  C T 
10 710  542.8 s
2
2
Time of flight from P to D:
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 1  e1
 1  0.2381
 
90 
ED  tan 1 
tan D   tan 1 
tan
  1.330 rad

2 
1  0.2381
2 
 1  e1
M D  ED  e1 sin Ed  1.330  0.2381sin 1.330  1.099 rad
M
1.099
tD  D T 
10 710  1873 s
2
2
Time of flight from C to D:
t CD  tD  t C  1873  542.8  1330 s
To determine the trajectory from P to D is Lambert’s problem. Note that
rD 
h12
62 8302
1
1

 9905 km
 1  e1 cos  D 398 600 1  0.2381cos 90
so that in perifocal coordinates
rP  8000p̂ km
rD  9905q̂ km
Note as well, that on orbit 1,
vP  
1

398 600
  sin  Pp̂  e  cos  P q̂  
  sin 0p̂  0.2381  cos 0 q̂ 
h1
62 830
 7.854q̂ km s 
vD  
1
398 600

  sin  D p̂  e  cos  D q̂  
  sin 90p̂  0.2381  cos 90q̂ 
h1 
62 830
 6.344p̂  1.510q̂ km s 
The following MATLAB script calls upon Algorithm 5.2, implemented as the M-function lambert.m in
Appendix D.25, to solve Lamberts’ problem for the velocities on orbit 2 at P and D ( v D  and v P  ). The
2
2
output to the MATLAB Command Window is listed afterwards.
%
%
%
%
%
%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Problem 6_21
~~~~~~~~~~~~
%
%
%
%
%
%
%
%
%
%
deg
pi
mu
r1, r2
dt
dtheta
R1, R2
string
This program uses Algorithm 5.2 to solve Lambert's problem for the
data of in Problem 6.21.
-
factor for converting between degrees and radians
3.1415926...
gravitational parameter (km^3/s^2)
initial and final radii (km)
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
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Chapter 6
% ----------------------------------------------------------clear
global mu
mu
= 398600;
deg
= pi/180;
r1
r2
dt
dtheta
=
=
=
=
8000;
9905;
1330;
90;
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 6.21: Lambert''s Problem\n')
fprintf('\n Input data:\n');
fprintf('\n
Gravitational parameter (km^3/s^2) = %g\n', mu)
fprintf('\n
Radius 1 (km)
= %g', r1)
fprintf('\n
Position vector R1 (km)
= [%g %g %g]\n',...
R1(1), R1(2), R1(3))
fprintf('\n
Radius 2 (km)
= %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 6.21: Lambert's Problem
Input data:
Gravitational parameter (km^3/s^2) = 398600
Radius 1 (km)
Position vector R1 (km)
= 8000
= [8000
Radius 2 (km)
Position vector R2 (km)
= 9905
= [6.06506e-13
0
0]
9905
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]
-----------------------------------------------------
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Chapter 6
v P   2.532p̂  9.576q̂ km s 
2
v D   7.734p̂  4.373q̂ km s 
2
v P  v P   v P   2.532p̂  9.576q̂  7.854q̂  2.532p̂  1.722q̂ km s 
2
1
2
v P  2.532  1.7222  3.062 km s 
v D  v D   v D   7.734p̂  4.373q̂  6.344p̂  1.510q̂   1.391p̂  2.863q̂ km s 
2
1
2
v D  1.391  2.8632  3.183 km s 
v total  v P  v D  3.062  3.183  6.245 km s
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Chapter 6
Problem 6.22 At the instant shown, spacecraft S1 is at point A of circular orbit 1 and spacecraft S2 is at
point B of circular orbit 2. At that instant, S1 executes a Hohmann transfer so as to arrive at point C of
orbit 2. After arriving at C, S1 immediately executes a phasing maneuver in order to rendezvous with S2
after one revolution of its phasing orbit. What is the total delta-v requirement?
Solution
h1  rA  398 600  8000  56 469 km 2 s
h
v A   1  7.0587 km s
1 r
A
h3  2
rA rB
8000 14 000
 2  398 600
 63 706 km 2 s
rA  rB
8000  14 000
h
63 706
vA   3 
 7.9633 km s
3 r
8000
A
To put S1 on the Hohmann transfer ellipse at A requires
v A  v A   v A   7.9633  7.0587  0.90459 km s
3
1
h
63 706
vC   3 
 4.5504 km s
3 r
C 14 000
h2  rB  398 600 14 000  74 702 km 2 s
h
74 702
vC   2 
 5.3359 km s
2 r
14
000
C
To boost S1 from orbit 3 to S 2 s orbit (orbit 2) at C requires
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Chapter 6
vC  vC   vC   5.3359  4.5504  0.78542 km s
2
3
Total delta-v to transfer S1 from orbit 1 to orbit 2 is
v Hohmann  v A  vC  0.90459  0.78542  1.6900 km s
Period of orbit 2:
T2 
2

3
rB 2 
2
398 600
3
14 000 2  16 486 s
Semimajor axis of orbit 3:
r r
8000  14 000
a3  A C 
 11000 km
2
2
Period of orbit 3:
T3 
2

3
a3 2 
2
398 600
3
11000 2  11 482 s
T3 2 is S1 s flight time from A to C. After S1 s delta-v maneuver at C, the true anomaly of S 2 on orbit 2
is
S 
2
2 T3
2 11 482

 2.1880 rad
T2 2 16 486 2
S1 is ahead of S 2 on orbit 2 by
    2.1880  0.95360 rad
Time for the target vehicle S 2 to reach C and then complete one orbit is
t

0.95360
T  T2 
16 486  16 486  18 988 s
2 2
2
This is the period T4 of S1 s phasing orbit.
T4 
2

3
a4 2
From this we obtain the semimajor axis of the phasing orbit
2
2
 T   3  18 988 398 600  3
a4   4
  15 383 km
 
2
 2 


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Chapter 6
Now we can find the apogee ra  of orbit 4,
4
a4 
rC  ra 
15 383 
4
2
14 000  ra 
4

2
ra   16 766 km
4
Then
h4  2
rC ra 
4
rC  ra 
4
 2  398 600
14 000 16 766
 77 988 km 2 s
14 000  16 766
It follows that
h
77 988
vC   4 
 5.5706 km s
4 r
C 14 000
The delta-v to boost S1 from orbit 2 to the phasing orbit and then return it orbit 2 after one revolution is
v phasing  2  vC   vC    2 5.5706  5.3359  0.46944 km s

4
2
Therefore, the total delta-v required for S1 to encounter S 2 in the prescribed fashion is
v total  v Hohmann  v phasing  1.6900  0.46944
v total  2.1594 km s
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Chapter 6
Problem 6.23 Spacecraft B and C, which are in the same elliptical earth orbit 1, are located at the true
anomalies shown. At this instant, spacecraft B executes a phasing maneuver so as to rendezvous with
spacecraft C after one revolution of its phasing orbit 2. Calculate the total delta-v required. Note that the
apse line of orbit 2 is at 45° to that of orbit 1.
1
C
Ap se line
of orbit 2
2
150°
B
45°
Phasing
orbit
Ap se line
of orbit 1
Earth
18 900 km
8100 km
Solution
For orbit 1:
rp ra
h1  2
e1 
a1 
T1 
1
1
rp  ra
1
1
ra  rp
1
1
ra  rp
1
1

 2
8100 18 900
 67 232 km 2 s
8100  18 900
18 900  8100
 0.4000
18 900  8100
rp  ra
8100  18 900
1
1

 13 500 km
2
2
2

2
3
a1 2 
3
398 600
13 500 2  15 610 s
At point B on orbit 1:
h2
rB  1
67 2322
1
1

 8839.7 km
 1  e1 cos B 398 600 1  0.4 cos 45
v
vr
1  rB1  8839.7  7.6056 km s
h
B
B
67 232

1  h1 e1 sin B 
2
398 600
0.4 sin 45  1.6769 km s
67 232
2
vB   v 
 vr
 7.60562  1.67692  7.7883 km s
1
B 1
B 1
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 B 1  tan 1
Chapter 6
1  tan 1 1.6769  12.434
v 
7.6056
1
vr
B
B
The period of orbit 2 is the time tCB it takes to fly from C to B on orbit 1.
E

1  e1
1  0.4
150
tan C 
tan C 
tan

2
1  e1
2
1  0.4
2
MC  EC  e1 sin EC  2.3646  0.4 sin 2.3646 
tC 
MC  2.0841
MC
2.0841
T 
15 610  5178 s
2 1
2
E
1  e1

1  0.4
45
tan B 
tan B 
tan
2
1  e1
2
1  0.4
2
MB  EB  e1 sin EB  0.52960  0.4 sin 0.52960
tB 
EC  2.3646

EB  0.52960

MB  0.32752
MB
0.32752
T 
15 610  813.72 s
2 1
2
tCB  T1  tC  tB  15 610  5178  813.72  11 246 s
T2  11 246 s
Now we can find the semimajor axis of orbit 2:
2
2
 T µ  3  11 246 398 600  3
a2   2
  10 849 km
 
2
 2 



Since a2  rB  ra2
 2 , the apogee of orbit 2 is
ra  2a2  rB  2 10 849  8839.7  12 858 km
2
The angular momentum of orbit 2 is
h2  2
rBra
2
rB  ra
 2  398 600
2
8839.7 12 858
 64 623 km 2 s
8839.7  12 858
Since B is the perigee of orbit 2,
h
64 623
vB   2 
 7.3105 km s
2 r
8839.7
B
and
 B 2  0
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Chapter 6
The delta-v required to transfer from orbit 1 to orbit 2 at B is
2
2
vB  vB   vB   2 vB  vB  cos  B    B  

1
2
1
2
2
1
vB  7.78832  7.31052  2  7.7883  7.3105cos 0  12.434  1.7027 km s
The delta-v required to transfer from orbit 2 back to orbit 1 at the end of the phasing maneuver is the
same. Therefore,
v total  2vB  3.3054 km s
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Chapter 6
Problem 6.24 (a) With a single delta-v maneuver, the earth orbit of a satellite is to be changed from a
circle of radius 15 000 km to a collinear ellipse with perigee altitude of 500 km and apogee radius of 22000
km. Calculate the magnitude of the required delta-v and the change in the flight path angle  .
(b) What is the minimum total delta-v if the orbit change is accomplished instead by a Hohmann transfer?
3
vA
v
2
B
A
2
vA
15 000 km
1
2
1
C
D
E
Common apse
line
Earth
4
22 000 km
6878 km
Solution
rA  rC  rE  15000 km
rB  22000 km
rD  6878 km
(a)
Orbit 1:
vA 
1

rA

398 600
 5.155 km s
15 000
 A1  0
Orbit 2:
r r
22 000  6878
e2  B D 
 0.5237
rB  rD 22 000  6878
h2  2
rB rD
22 000  6878
 2  398 600
 64 630 km 2 s
rB  rD
22 000  6878
At the maneuvering point A:
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Chapter 6
h 2
rA  2
1
 1  e2 cos  A
15 000 
64 6302
1
  A  125.1
398 600 1  0.5237 cos  A
  rA2  15 000  4.309 km s
h
vA
2
vA
2
64 630

r  h2 e2 sin A 
2
398 600
0.5237 sin 125.1  2.641 km s
64 630
2
vA  vA
 vA
 4.3092  2.6412  5.054 km s
2
2 
2 r
vA
2.641
2 r
 A  tan 1
 tan 1
 0.5499   A  31.51
2
2
vA
4.309
2


 A   A   A  31.51  0  31.51
2
1
v A  v A 2  v A 2  2v A v A cos  A  5.1552  5.054 2  25.1555.054 cos  A  2.773 km s T
1
2
1
2
(b)
Try Hohmann transfer (orbit 3) from point E on orbit 1 to point B on orbit 2.
h3  2
rErB
15 000  22 000
 2  398 600
 84 320 km 2 s
rE  rB
15 000  22 000
v E  v A  5.155 km s
1
1
h
84 320
vE  3 
 5.621 km s
3
rE 15 000
h
84 320
vB  3 
 3.833 km s
3
rB 22 000
h
64 630
vB  2 
 2.938 km s
2
rB 22 000
v total  v E  v E  v B  v B  0.4665  0.985  1.362 km s
3
1
2
3
Try Hohmann transfer (orbit 4) from point C on orbit 1 to point D on orbit 2.
h4  2 
Howard D. Curtis
15000  6878
rC rD
 2  398600
 61310 km 2 s
rC  rD
15000  6878
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Chapter 6
v C  v A  5.155 km s
1
1
h4 61 310
vC 

 4.088 km s
4
rC 15 000
h
61 310
vD  4 
 8.914 km s
4
rD
6878
h
64 630
vD  2 
 9.397 km s
4
rD
6878
v total  v C  v C  v D  v D  1.067  0.4824  1.55 km s
4
1
2
4
This is larger than the total computed above; thus for minimum Hohmann transfer
v  1.362 km s
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Orbital Mechanics for Engineering Students Third Edition
Chapter 6
Problem 6.25 An earth satellite has a perigee altitude of 1270 km and a perigee speed of 9 km/s. It is
required to change its orbital eccentricity to 0.4, without rotating the apse line, by a delta-v maneuver at
  100 . Calculate the magnitude of the required v and the change in flight path angle  .
Solution
Orbit 1:
rperigee  6378  1270  7648 km
1
v perigee  9 km s
1
h1  rperigee v perigee  7648  9  68 832 km 2 s
1
1
h12
rperigee 
1
 1  e1
1
2
7648 
68 832
1
 e1  0.5542
398 600 1  e1
At the maneuver point,   100 .
68832 2
1
1

 13150 km
 1  e1 cos  398600 1  0.5542cos 100
68 832
h
v1   1 
 5.234 km s
r
13150

398600
v1 r  e1 sin  
0.5542 sin 100  3.16 km s
h1
68 832
r
h12
2
2
2
2
v1  v 1   v 1r  5.234  3.16  6.114 km s
v1
3.16
 1  tan 1 r  tan 1
 31.13
v1 
5.234
Orbit 2:
e2  0.4
r
h2 2
1
 1  e 2 cos 
h22
1
 h2  69 840 km 2 s
398 600 1  0.4cos 100
69 840
h
v2   2 
 5.311 km s
r
13150

398600
v2 r 
e2 sin  
0.4 sin 100  2.248 km s
h2
69 840
13150 
2
2
2
2
v 2  v 2   v 2 r  5.311  2.248  5.767 km s
v2
2.248
 2  tan 1 r  tan 1
 22.94
v2 
5.767
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Orbital Mechanics for Engineering Students Third Edition
Chapter 6
   2   1  22.94  31.13  8.181
v  v12  v 2 2  2v1 v 2 cos   6.114 2  5.767 2  2  6.114  5.767 cos 8.181  0.9155 km s
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Orbital Mechanics for Engineering Students Third Edition
Chapter 6
Problem 6.26 The velocities at points A and B on orbits 1, 2 and 3, respectively, are (relative to the
perifocal frame)
v A   3.7730p̂  6.5351q̂ (km/ s)
1
v A   3.2675p̂  8.1749q̂ (km/ s)
2
v B   3.2675p̂  3.1442q̂ (km/ s)
2
v B   2.6679p̂  4.6210q̂ (km/ s)
3
Calculate the total v for a transfer from orbit 1 to orbit 3 by means of orbit 2.
Solution
v A  v A   v A   3.2675p̂  8.1749q̂  3.7730p̂  6.5351q̂ 
2
1
v A  0.50549p̂  1.6398q̂ (km/ s)
v A  v A  1.716 km/ s
v B  v B   v B   2.6679p̂  4.6210q̂  3.2675p̂  3.1442q̂ 
3
2
v B  0.5996p̂  1.4768q̂ (km/ s)
vB  v B  1.5939 km/ s
v total  v A  v B
v total  3.3098 km/ s
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Orbital Mechanics for Engineering Students Third Edition
Chapter 6
Problem 6.27 Trajectories 1 and 2 are ellipses with eccentricity 0.4 and the same angular momentum h.
Their speed at B is v. Calculate, in terms of v, the v required at B to transfer from orbit 1 to orbit 2.
Solution
Orbit 1:
rB 
h2
1
h2

 1  ecos 90 
1  h esin 90  h e
vB   h  h  
1 rB h2  h
vB
r

 B  tan 1
1
vB
r
vB

1  tan 1 e h  tan 1 e
 h
1
Orbit 2:
2  h esin 90  h e
vB   h  h  
2 rB h2  h
vB
r

 B  tan 1
2
vB
r
vB

2  tan 1  e h  tan 1 e   tan 1 e
 
 h
2
vB  vB 2  vB 2  2vB vB cos  B  v 2  v 2  2v  v cos  B  2v 2 1  cos  B 
1
2
1
2
 B   B   B  2 tan 1 e
2
1
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
Chapter 6

cos  B  cos 2 tan 1 e  cos 2 tan 1 e sin 2 tan 1 e
2
2
e 

 1 
1  e2




2 
 1e 
 1  e2 
1  e2

1  e2 
2e
 v B  2v 2  1 
v
 
2

1  e 
1  e2
v B 
2  0.4
1  0.4 2
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v  0.7428v
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Chapter 6
Problem 6.28 A satellite is in a circular earth orbit of altitude 400 km. Determine the new perigee and
apogee altitudes if the satellite’s onboard rocket
(a) Provides a delta-v in the tangential direction of 240 m/s.
(b) Provides a delta-v in the radial (outward) direction of 240 m/s.
Solution
r0  6378  400  6778 km
v0 

r0
398 600
 7.6686 km s
6778

(a)
rp  r0  6778 km
 

h  rv   r0 v 0  v    r0 
 v  
 r0

rp 
r0 
ra 
h2
1
 1 e
h2
1
 1 e

e
1
h2

 1 e 
1
h2
r0
ra 
2
h2
1
 r0
 h2

1
 1
  r0


2
 
 r0

v
 r

 0


1

r0
2  r0
h2
r0
2  r0

1
  

 v   
 r0 

  r0
6778
398 600
1
2 6778
 398 600

 0.24 

 6778

2
2
1
 7698.3 km
z a  7698.3  6378  1320.3 km
(b)
h  rv   r0 v0
r 2  r0 
r 2v 2
1
1
1
1
 0 0
 0
 r0
 1  ecos 
 1  ecos 

1  ecos 
1  ecos 
cos   0
   90
r0 
h2
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
vr 
h
esin  
Chapter 6


e 1  cos 2  
e
r
r0
0
v
e 
rp 
Orbital Mechanics for Engineering Students Third Edition
 r0
h2
1
 1 e

r0

1
1
v

 r0
1
r0
v
 r0
6778
0.24

1
 6572.3 km
398 600 6778
z p  6572.3  6378  194.31km
ra 
h2
1

 1 e
1
r0
v
 r0

1
6778
0.24
 6997.0 km
398 600 6778
z a  6997.0  6378  618.98 km
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284
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Orbital Mechanics for Engineering Students Third Edition
Chapter 6
Problem 6.29 At point A on its earth orbit, the radius, speed and flight path angle of a satellite are
rA  12 756 km , v A  6.5992 km/ s and  A  20 . At point B, at which the true anomaly is 150°, an impulsive maneuver causes v   0.75820 km/ s and v r  0 .
(a) What is the time of flight from A to B?
(b) What is the rotation of the apse line as a result of this maneuver?
2
B

P2
A
150°

P1
1
Solution
Orbit 1:
vA

 v A cos  A  6.5992 cos 20  6.20122 km s
 h1  rA v A  12 756  6.20122  79102.8 km 2 s

v A  v A sin  A  6.5992  sin 20  2.25706 km s
r
vA 
r

e sin  A
h1 1
2.25706 
398 600
e sin  A
79102.8 1
h2
rA  1
e1 sin  A  0.447917

1
 1  e1 cos  A
12 756 
79102.8 2
1
398 600 1  e1 cos  A


e1 cos  A  0.230641

e12 sin 2  A  cos 2  A  0.447917 2  0.2306412
e12  0.253825
sin  A 
Howard D. Curtis

e1  0.50381
0.447917
 0.889 058
0.50381

 A  62.755 2 or  A  117.235
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Chapter 6
Since cos  A  0 ,  A  62.7552 .
The period of orbit 1 is
T1 

h1
2 
2
  1  e12

2

 398 6002


79102.8

 1  0.503812

  30 368.2 s

The eccentric anomaly at point A of orbit 1 is
 1 e

 1  0.50381
62.7552 
1 tan  A  2 tan 1
EA  2 tan 1 
tan


  0.67392
2 
2
 1  0.50381

 1  e1
From Kepler’s equation, the corresponding mean anomaly is
M A  EA  e1 sin EA  0.67392  0.50381sin 0.67392  0.35951
Therefore, the time since perigee passage is
tA 
MA
0.35951
T1 
30 368  1737.6 s
2
2
At point B the eccentric anomaly is
 1 e

 1  0.50381
150 
1 tan B  2 tan 1
EB  2 tan 1 
tan


  2.2687
2
2 
 1  0.50381
 1  e1
Thus
M B  EB  e1 sin EB  2.2687  0.50381sin 2.2687  1.8826
and
tB 
MB
1.8826
T1 
30 368  9099.2 s
2
2
It follows that the time of flight from A to B is
tof  tB  t A  9099.2  1737.6  7361.6 s  2.045 hr
(b)
h2
rB  1
79102.82
1
1

 27 848.9 km
 1  e1 cos B
398 600 1  0.50381cos150
(1)
1
vB


h
79102.8
 1
 2.84043 km s
1 rB 27 848.9
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(2)
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vB
Orbital Mechanics for Engineering Students Third Edition

r
1  h1 e1 sin B
1

Chapter 6
398 600
 0.50381  sin 150  1.26945 km s
79102.8
(3)
vB  0.75820 km s
(4)
vBr  0
(5)
 vBr   vBr   vBr  1.26945  0  1.26945 km s
2
(6)
1
According to Equation 6.18b
 
tan B 
2



v


 B 1   v B   v Br 1   v Br 

2

2
vB
 1
  rB
v
  v B  e1 cos B   2 v B
 vB
 B 1

 
1
 1
 


Substituting Equations (1) through (5) above yields
2.8404  0.75821.2694  0
2.8404 2
2.8404  0.75822  0.50381  cos150  2  2.8404  0.7582 398 600 27 849
 
tan   3.3521
tan B
2

B2

B  73.389 or 106.612°
2
According to Equation (6) above, the spacecraft is flying away from perigee on orbit 2, so 0  B  180 .
2
Therfore,
B  106.612
2
This means
  150  106.612  43.3877
That is, the apse line is rotated 43.387 7° ccw from that of orbit 1.
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Chapter 6
Problem 6.30 A satellite is in elliptical orbit 1. Calculate the true anomaly  (relative to the apse line of
orbit 1) of an impulsive maneuver that rotates the apse line an angle  counterclockwise but leaves the
eccentricity and the angular momentum unchanged.

1

Original apse
line
2
Solution
r1  r2
h2
1
h2
1

 1  ecos   1  ecos    
cos   cos    
       2     
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
2
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Chapter 6
Problem 6.31 A satellite in orbit 1 undergoes a delta-v maneuver at perigee P1 such that the new orbit
2 has the same eccentricity e , but its apse line is rotated 90° clockwise from the original one. Calculate the
specific angular momentum of orbit 2 in terms of that of orbit 1 and the eccentricity e .
2
1
F
P2
P1
Solution
Orbit 1:
2
h
rP1  1
1
 1e
Orbit 2:
2
2
h
rP1  2
1
h
 2
 1  ecos 90

rP  rP
1

1
h2 2

h2 

h12
1
 1 e
h1
1 e
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Chapter 6
Problem 6.32 Calculate the delta-v required at A in orbit 1 for a single impulsive maneuver to rotate
the apse line 180° counterclockwise (to become orbit 2), but keep the eccentricity e and the angular momentum h the same.
Solution
At A:
r
h
2

v r1 

h
esin 90 


h
e

esin 90   e
h
h
h
v1  v 2 
r
 v   0
vr 2 

v r  v r2  v r1  2 e
h
 v  v r 
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h
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Chapter 6
Problem 6.33 Spacecraft A and B are in concentric, coplanar circular orbits 1 and 2, respectively. At the
instant shown, spacecraft A executes an impulsive delta-v maneuver to embark on orbit 3 in order to intercept and rendezvous with spacecraft B in a time equal to the period of orbit 1. Calculate the total deltav required.
Solution
rA  8000p̂ km 
The time of flight from A to C is
t AC
3  T1 
T2 
2

2

3
R1 2 
2
398 600
3
 8000 2  7121.09 s
3
R2 2
The true anomaly of C on orbit 2 is
C 2  2
t AC
3
T2
3
3
R 2
 8 000  2
 2  1   2 
 2.71408 rad  155.505
 14 000 
 R2 
Thus, the perifocal position vector of C is
rC  R 2 cos Bp̂  sin Bq̂  14 0000 cos155.505p̂  sin 155.505q̂ 
 12 470p̂  5804.51q̂ km 
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
With rA , rC and t AC , we use Algorithm 5.2 to determine orbit 3. The following MATLAB code calls
3
upon the M-file lambert.m in Appendix D.25 to do so.
% ~~~~~~~~~~~~
% Problem 6_33
% -----------clear
global mu
mu
= 398600;
%grav. parameter (km^3/s^2)
R1
= 8000;
%radius of orbit 1 (km)
R2
= 14000;
%radius of orbit 2 (km)
r1
= R1*[1 0 0];
%position vector of A
dtheta = 2*pi*(R1/R2)^(3/2);
%true anomaly of C (rad)
r2
= R2*[cos(dtheta) sin(dtheta) 0]; %position vector of C
dt
= 2*pi/sqrt(mu)*R1^(3/2);
%flight time from A to C (s)
string = 'pro';
[v1 v2] = lambert(r1, r2, dt, string);
%Algorithm 5.2
fprintf('\n\n Solution of Lambert''s Problem:\n')
fprintf('\n Velocity vector v1 (km/s) = [%12.5e %12.5e %12.5e]',...
v1(1), v1(2), v1(3))
fprintf('\n Velocity vector v2 (km/s) = [%12.5e %12.5e %12.5e]\n',...
v2(1), v2(2), v2(3))
The output to the MATLAB command window is
Solution of Lambert's Problem:
Velocity vector v1 (km/s) = [ 2.06162e+00 7.73163e+00 0.00000e+00]
Velocity vector v2 (km/s) = [-6.10235e-01 -4.57700e+00 -0.00000e+00]
Therefore
v A   2.06162p̂  7.73163q̂ km s 
3
vC
3  0.610235p̂  4.57699q̂ km s 
Because orbits 1 and 2 are circular, Equation 2.125 yields
vA  
1
vC
2 

R1

R2
q̂ 
398 600
q̂  7.05868q̂ km s 
8000
 sin Bp̂  cos Bq̂ 
398 600
 sin 155.505p̂  cos155.505q̂ 
14 000
 2.21229p̂  4.85564q̂ km s 
v A  v A 3  v A 1  2.06162p̂  7.73163q̂  7.05868q̂  2.06162p̂  0.672947q̂ km s 
v A  v A  2.16866 km s
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Chapter 6
v C  v C 2  v C 3  2.21229p̂  4.85564q̂  0.610235p̂  4.57699q̂ 
 1.60205p̂  0.278646q̂ km s 
vC  v C  1.62611 km s
v total  v A  vC  3.79478 km s
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Chapter 6
Problem 6.34 Spacecraft A is in orbit 1, a 10 000 km radius equatorial earth orbit. Spacecraft B is in elliptical polar orbit 2, having eccentricity 0.5 and perigee radius 16 000 km. At the instant shown, both
spacecraft are in the equatorial plane and B is at its perigee. At that instant, spacecraft A executes an impulsive delta-v maneuver to intercept spacecraft B one hour later at point C. Calculate the delta-v required for A to switch to the intercept trajectory 3.
Solution
rB  16000Ĵ km 
h 2
rB  2
1
 1  e2

h2  rB 1  e2 
h2  398 600 16000  1  0.5  97 808.0 km 2 s
vB 
h2
97 808.0
K̂ =
K̂  6.11300K̂ km s 
rB
16 000

Knowing rB , v B and tBC  1 hr , we use Algorithm 3.4 to find rC and v C . The following MATLAB
2
script calls upon the M-file rv_from_r0v0.m in Appendix D.16 to do so.
% ~~~~~~~~~~~~
% Problem 6.34a
% ~~~~~~~~~~~~
global mu
mu = 398600;
%gravitational parameter (km^3/s^2)
r0 = [0 16000 0];
%initial position vector (km)
v0 = [0 0 6.11300]; %initial velocity vector (km/s)
t = 3600;
%elapsed time (s)
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[r v] = rv_from_r0v0(r0, v0, t); %Algorithm 3.4
fprintf('\n Final position vector:')
fprintf('\n
r = [%12.5e, %12.5e, %12.5e] (km)\n', r(1), r(2), r(3))
fprintf('\n Final velocity vector:')
fprintf('\n
v = [%12.5e, %12.5e, %12.5e] (km/s)\n', v(1), v(2), v(3))
The output to the MATLAB command window is
Final position vector:
r = [ 0.00000e+00, 7.82906e+03,
1.84968e+04] (km)
Final velocity vector:
v = [ 0.00000e+00, -3.75299e+00,
3.62618e+00] (km/s)
That is,
rC  7829.06Ĵ 18 496.8K̂ km 
vC
2  3.75299Ĵ 3.62618K̂ km s 
Clearly,
rA  10000Î km 

Knowing rA , rC and the flight time t AC  1 hr , we use Algorithm 5.2 to find v A  and v C . The fol3
3
lowing MATLAB code calls upon the M-file lambert.m in Appendix D.25 to do so.
% ~~~~~~~~~~~~~
% Problem 6_34b
% ------------clear
global mu
mu
rA
rC
tAC
=
=
=
=
398600;
[10000 0 0];
[0 7829.06 18496.8];
3600;
%gravitational parameter (km^3/s^2)
%position vector of A (km)
%position vector of B (km)
%time of flight from A to C (s)
string = 'pro';
[vA vC] = lambert(rA, rC, tAC, string); %Algorithm 5.2
fprintf('\n\n Solution of Lambert''s Problem:\n')
fprintf('\n Velocity vector vA (km/s) = [%12.5e %12.5e %12.5e]',...
vA(1), vA(2), vA(3))
fprintf('\n Velocity vector vC (km/s) = [%12.5e %12.5e %12.5e]\n',...
vC(1), vC(2), vC(3))
The output to the MATLAB command window is
Solution of Lambert's Problem:
Velocity vector vA (km/s) = [ 9.48539e-01
Velocity vector vC (km/s) = [-4.00570e+00
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3.13609e+00 7.40926e+00]
1.20499e+00 2.84688e+00]
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Chapter 6
That is,
v A   0.948539Î  3.13609Ĵ 7.40926K̂ km s 
3
vC
3  4.00570Î  1.20499Ĵ 2.84688K̂ km s 
Therefore,


v A  v A 3  v A 1  0.948539Î  3.13609Ĵ 7.40926K̂  6.31348Ĵ
 0.948539Î  3.17739Ĵ 7.40926K̂ km s 
v A  v A  8.11743 km s
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Chapter 6
Problem 6.35 Spacecraft B and C are in the same elliptical orbit 1, characterized by a perigee radius of
7000 km and an apogee radius of 10 000 km. The spacecraft are in the positions shown when B executes
an impulsive transfer to orbit 2 in order to catch and rendezvous with C when C arrives at apogee A. Find
the total delta-v requirement.
q̂
B
C
1
2
120°
A
p̂
Earth
10000 km
7000 km
Solution
Parameters of orbit 1:
e1 
ra  rp
1
1
ra  rp
1
1

h2
rp  1
1
1
 1  e1
10 000  7000
 0.176471
10 000  7000

h1  rp 1  e1 
1
h1  398 600  7000  1  0.176471  57 293.9 km 2 s
a1 
T1 
ra  rp
10 000  7000
1
1

 8500 km
2
2
2

3
a1 2 
2
398 600
3
8500 2  7799.01 s
Time since perigee passage at C ( t C ):
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Chapter 6
 1 e

 1  0.176471
120 
1 tan C  tan 1
EC  tan 1 
tan


  1.93354 rad
2 
2 
 1  0.176471
 1  e1
M C  EC  esin EC  1.93354  0.176471 sin 1.93354  1.76855 rad
tC 
MC
1.76855
T 
 7799.01  2195.22 s
2 1
2
Time of flight from C to A on orbit 1:
T
7799.01
tCA  1  tC 
 2195.22  1704.290 s
1
2
2
Time of flight from B to A on orbit 2:
tBA  tCA  1704.290 s
2
1
State vector of point A on orbit 1:
rA   10 000p̂ km 
h
57 293.9
v A    1 q̂  
q̂  5.72939q̂ km s 
1
rA
10 000
State vector of point B on orbit 1:
h2
rB  1
1
cos Bp̂  sin Bq̂ 
 1  e1 cos B

57 293.9 2
1
cos 90p̂  sin 90q̂ 
398 600 1  0.176471cos 90
rB  8235.29q̂ km 
vB  
1
398 600

  sin Bp̂  e1  cos B q̂  


 57 293.9   sin 90p̂  0.176471  cos 90q̂ 
h1 
v B   6.95711p̂  1.22773q̂ km s 
1
Knowing rA , rB and t BA , we can use Algorithm 5.2 to determine orbit 2. The following MATLAB script
2
calls upon the M-file lambert.m in Appendix D.25 to do so.
% ~~~~~~~~~~~~
% Problem 6_35
% -----------clear
global mu
mu
rA
= 398600;
= [-10000 0 0];
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%gravitational parameter (km^3/s^2)
%position vector of A (km)
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rB = [0 8235.29 0]; %position vector of B (km)
tBA = 1704.290;
%time of flight from B to A (s)
string = 'pro';
[vB vA] = lambert(rB, rA, tBA, string); %Algorithm 5.2
fprintf('\n\n Solution of Lambert''s Problem:\n')
fprintf('\n Velocity vector vA (km/s) = [%12.5e %12.5e %12.5e]',...
vA(1), vA(2), vA(3))
fprintf('\n Velocity vector vB (km/s) = [%12.5e %12.5e %12.5e]\n',...
vB(1), vB(2), vB(3))
The output to the MATLAB command window is
Solution of Lambert's Problem:
Velocity vector vA (km/s) = [-2.35674e+00 -6.78141e+00
Velocity vector vB (km/s) = [-8.23457e+00 -9.03578e-01
0.00000e+00]
0.00000e+00]
That is,
v A   2.35674p̂  6.78141q̂ km s 
2
v B   8.23457p̂  0.903578q̂ km s 
2
Thus
v A  v A 1  v A 2
 5.72939q̂  2.35674p̂  6.78141q̂ 
 2.35674p̂  1.05202q̂ km s 
 v A  v A  2.58089 km s
v B  v B 2  v B 1
 8.23457p̂  0.903578q̂  6.95711p̂  1.22773q̂ 
 1.27746p̂  2.13130q̂ km s 
 v B  v B  2.48482 km s
v total  2.58089  2.48482  5.06571 km s
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Chapter 6
Orbital Mechanics for Engineering Students Third Edition
Problem 6.36 At time t = 0, manned spacecraft a and unmanned spacecraft b are at the positions shown
 
 
in circular earth orbits 1 and 2, respectively. For assigned values of 0a and 0b , design a series of impulsive maneuvers by means of which spacecraft a transfers from orbit 1 to orbit 2 so as to rendezvous
with spacecraft b (i.e., occupy the same position in space). The total time and total delta-v required for the
transfer should be as small as possible. Consider earth’s gravity only.
Solution
Design problem.
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Chapter 6
Problem 6.37 What must the launch azimuth be if the satellite in Example 4.10 is launched from (a)
Kennedy Space Center (latitude = 28.5°N); (b) Vandenburgh AFB (latitude = 34.5°N) (c) Kourou, French
Guiana (latitude 5.5°N).
Solution
 cos i 
A  sin 1 

cos  
(a)
 cos116.57 
A  sin 1 
 sin 1 0.5088  210.6° or 329.4
 cos 28.5 
(b)
 cos116.57 
A  sin 1 
 sin 1 0.5427   212.9 or 327.1
 cos 34.5 
(c)
 cos116.57 
A  sin 1 
 sin 1 0.4494   206.7 or 333.3
 cos 5.5 
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Chapter 6
Problem 6.38 The state vector of a spacecraft in the geocentric equatorial frame is r  rÎ and v  v Ĵ . At
that instant an impulsive maneuver produces the velocity change v  0.5v Î  0.5v K̂ . What is the inclination of the new orbit?
Solution
h  r  v  v  
Î
r
0.5v
Ĵ K̂
0
0  0.5rv Ĵ rv K̂
v 0.5v
h  1.11803rv
h 
rv


i  cos 1  Z   cos 1 
 cos 1 0.894427 
 1.11803rv 
 h 
i  26.5651
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Chapter 6
Problem 6.39 An earth satellite has the following orbital elements: a  15000 km , e  0.5 ,   45 ,
  30 , i  10 . What minimum delta-v is required to reduce the inclination to zero?
Solution
h  a1  e
2

398600 15000  1  0.5
2
2
rascending node 
h
1
 1  e cos  
rdescending node 

  66960
km
2
s
66 9602
1
 7851km
398600 1  0.5cos 30
h2
66 9602
1
1

 19 840 km
 1  e cos     398 600 1  0.5cos 30  180
Rotate the orbital plane 10 degrees around the node line. That means hold v r fixed and rotate v  10°. For
minimum delta-v, do this maneuver at the furthest distance from the focus (at the descending node, rather than the ascending node).
v 
66 960
h

 3.375 km s
rd escending nod e 19 840
v  2v  sin
i
10
 2  3.375  sin
 0.5883 km s
2
2
(Note: if the maneuver is done at the ascending node,
v 
66 960
h

 8.53 km s
rascend ing node
7851
v  2v  sin
i
10
 2  8.53  sin
 1.487 km s
2
2
Over twice the delta-v requirement.)

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
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Chapter 6
Problem 6.40 With a single impulsive maneuver, an earth satellite changes from a 400 km circular orbit
inclined at 60° to an elliptical orbit of eccentricity e  0.5 with an inclination of 40°. Calculate the minimum required delta-v.
Solution
For the circular orbit

v1 
398 600
 7.668 km s
6778

r
Assume the maneuver is done at apogee of the ellipse (orbit 2).
h 2
r 2
1
 1  e2
6778 
h2 2
1
 h2  36 750 km 2 s
398 600 1  0.5
Then
rperigee 
h2 2
367502 1
1

 2259 km
 1  e2 398600 1  0.5
which is inside the earth. So the maneuver cannot occur at apogee. Assume it occurs at perigee.
h2
r 2
1
 1  e2
6778 
v2 
h22
1
 h2  63 660 km 2 s
398 600 1  0.5
h2 63660

 9.392 km s
r
6778
v  v12  v 22  2v1v 2 cos   7.6682  9.3922  2  7.668  9.392cos   3.414 km s
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Chapter 6
Problem 6.41 An earth satellite is in an elliptical orbit of eccentricity 0.3 and angular momentum
60000 km 2 s . Find the delta-v required for a 90° change in inclination at apogee (no change in speed).
Solution
60 000 2 1
1

 12 900 km
 1  e 398 600 1  0.3
60 000
h
v apogee 

 4.65 km s
rapogee 12 900
rapogee 
h2
v  2v apogee sin
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
2
 2  4.65 sin
90
 6.577 km s
2
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Chapter 6
Problem 6.42 A spacecraft is in a circular, equatorial orbit (1) of radius ro about a planet. At point B it
impulsively transfers to polar orbit (2), whose eccentricity is 0.25 and whose perigee is directly over the
North Pole. Calculate the minimum delta-v required at B for this maneuver.
N
Orbit 1 shown edge-on
ro
B
1
2
S
Solution

v B1 
ro
h 2
ro  2
h 2
h 2
1
1
 2
 2  h2   ro
 1  ecos 
 1  0.25 cos 90

 ro
h
v B2  2 
ro
v Br2 
v 
ro


ro



 0.25  sin 90   0.25
r
 ro
o
e sin  
h2 2
vBr2  vBr1 2  vB12  vB22  2vB1vB2 cos 
2
2
2





 
v   0.25
 0 

2
cos 90
ro
ro
ro
ro ro


v  0.0625
v  2.0625
v  1.436
Howard D. Curtis

ro


ro


ro
0

ro

ro
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Chapter 6
Problem 6.43 A spacecraft is in a circular, equatorial orbit (1) of radius r0 and speed v 0 about an unknown planet (   398 600 km 3 s2 ). At point C it impulsively transfers to orbit (2), for which the ascending node is point C, the eccentricity is 0.1, the inclination is 30° and the argument of periapsis is 60°. Calculate, in terms of v 0 , the single delta-v required at C for this maneuver.
Z
Periapsis
2
e =0.1
30°
1 e= 0

X
Y
C
Solution
At C on circular orbit 1:
vr  0
1
v   v0 
1

r0
The true anomaly of C on orbit 2 is
C   2  60
2
Thus
h 2
r0  2
1
 1  e2 cos C

h2 
1  e2 cos  2 r0
2
so that
h2 
1  0.1cos 60r0  1.0247
r0
Then at C on orbit 2,
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Chapter 6
h

v   2  1.0247
 1.0247v 0
2
r0
r0
and
vr 
2

h2
vC 

e2 sin C 
2
v r
2
 vr
1

1.0247 r0
 0.1  sin 60  0.084515v 0
2  v  2  v  2  2v  v 
1
2
1
2
cos 
0.084515v0  02  v02  1.0247v0 2  2v0 1.0247v0 cos 30
 0.0071429v 0 2  v 0 2  1.05v 0 2  1.7748v 0 2
vC  0.53134v0
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Chapter 6
Problem 6.44 A spacecraft is in a 300 km circular parking orbit. It is desired to increase the altitude to
600 km and change the inclination by 20°. Find the total delta-v required if
(a) The plane change is made after insertion into the 600 km orbit (so that there are a total of three
delta-v burns).
(b) If the plane change and insertion into the 600 km orbit are accomplished simultaneously (so
that the total number of delta-v burns is two).
(c) The plane change is made upon departing the lower orbit (so that the total number of delta-v
burns is two).
Solution
The initial and target orbits are “1” and “2”, respectively, and “3” is the transfer orbit.
r1  6678 km

v1 
r1

398 600
 7.726 km s
6678
r2  6978 km

v2 
r2

398 600
 7.558 km s
6978
r r
6678  6978
a3  1 2 
 6828 km
2
2
1 
2 1
 2
v perigee       398 600 

  7.810 km s

3
6678 6828 
 r1 a3 
1
1 
 2
 2
v apogee       398 600 

  7.474 km s

3
r
a
6978
6828
 2
3
(a)
i
2
20
 7.810  7.726   7.558  7.474   2  7.558 sin
2
 0.0844  0.083 48  2.625  2.793 km s


v  v perigee  v1  v 2  v apogee
3
3
 2  v 2 sin
(b)


 v  v perigee  v1  v apogee
3
2
3
 v 2 2  2v apogee v 2 cos  i
3
 7.810  7.726   7.474 2  7.5882  2  7.474  7.558 cos 20
 0.0844  2.612  2.696 km s
(c)
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
v  v perigee 2  v12  2v perigee v1 cos i  v 2  v apogee
3
3
3
Chapter 6

 7.812  7.7262  2  7.81  7.726 cos 20  7.558  7.474 
 2.699  0.083 48  2.783 km s
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Chapter 6
Problem 6.45 Calculate the total propellant expenditure for Problem 6.3 using finite-time delta-v maneuvers. The initial spacecraft mass is 4000 kg. The propulsion system has a thrust of 30 kN and a specific
impulse of 280 s.
Solution: Computer project.
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Chapter 6
Problem 6.46 Calculate the total propellant expenditure for Problem 6.3 using finite-time delta-v maneuvers. The initial spacecraft mass is 4000 kg. The propulsion system has a thrust of 30 kN and a specific
impulse of 280 s.
Solution: Computer project.
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Chapter 6
Problem 6.47 At a given instant t 0 , a 1000 kg earth-orbiting satellite has the inertial position and velocity vectors r0  436î  6083 ĵ  2529k̂ km  and v 0  7.340î  0.5125 ĵ  2.497k̂ km s . 89 minutes later a
rocket motor with Isp  300s and 10 kN thrust aligned with the velocity vector ignites and burns for 120
seconds. Use numerical integration to find the maximum altitude reached by the satellite and the time it
occurs.
Solution
The solution is implemented in the following MATLAB function file. The command window output is at
the end.
% ~~~~~~~~~~~~~~~~~~~
function Problem_6_47
% ~~~~~~~~~~~~~~~~~~~
clear all
hours
= 3600; % converts hours to seconds
minutes = 60;
% converts minutes to seconds
mu = 398600; % earth's gravitational paramater (km^3/s^2)
g0 = 9.807; % earth's sea-level gravitational acceleration (m/s)
R = 6378;
% earth's radius (km)
m0
= 2000;
r0
v0
% initial spacecraft mass
= [436 6083 2529];
% initial spacecraft position vector (km)
= [-7.34 -0.5126 2.497]; % initial spacecraft velocity vector (km/s)
t0
= 0;
% initial time (s)
tf
= 4.5*hours; % final time (s)
tspan = [t0 tf];
% time interval for the solution (s)
thrust
Isp
tig
tb
=
=
=
=
10;
300;
t0 + 89*minutes;
4*minutes;
y0
= [r0 v0 m0]';
%
%
%
%
onboard rocket thrust (kN)
specific impulse (s)
rocket ignition time (s)
rocket burn time (s)
% column vector of initial conditions
%... Use MATLAB's ode45 to integrate the equations of motion (listed
%
in the subfunction 'rates') over the time interval 'tspan' with
%
initial conditions 'y0'. The solution is delivered in the columns
%
of the matrix y whose rows correspond to the discrete times returned
%
in the column vector 'times'.
[times,y] = ode45(@rates, tspan, y0);
%... Extract the discrete time histories of the position vector components:
X = y(:,1); Y = y(:,2); Z = y(:,3);
%... Find the radius at each time:
for i = 1:length(times)
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Chapter 6
r(i) = norm([X(i) Y(i) Z(i)]);
end
%...Find the maximum radius 'rmax' and the time step 'imax' at which it
%
occurs:
[rmax imax] = max(r);
%...Output to the command window:
fprintf('\n\n--------------------------------------------------------\n')
fprintf('\n Initial mass = %g kg. Final mass = %g kg.\n', m0, y(end,7))
fprintf('\n For a constant thrust of %g kN for %g s starting at %g m,',...
thrust, tb, tig/minutes)
fprintf('\n and a flight time of %g hours:\n', tf/hours)
fprintf('\n The maximum altitude is %g km at time = %g h.',...
rmax-R, times(imax)/hours)
fprintf('\n--------------------------------------------------------\n\n')
% ~~~~~~~~~~~~~~~~~~~~~~~~
function dydt = rates(t,f)
% ~~~~~~~~~~~~~~~~~~~~~~~~
%{
This function calculates the acceleration vector using Equation 2.22
with the addition of a thrust T in the direction of the velocity
vector
t
f
- time
- column vector containing the position vector and the
velocity vector at time t
x, y, z
- components of the position vector r
r
- the magnitude of the the position vector
vx, vy, vz - components of the velocity vector v
ax, ay, az - components of the acceleration vector a
dydt
- column vector containing the velocity and acceleration
components
%}
% -----------------------x
= f(1);
y
= f(2);
z
= f(3);
vx
= f(4);
vy
= f(5);
vz
= f(6);
m
= f(7);
r
v
= norm([x y z]);
= norm([vx vy vz]);
if t < tig
T = 0;
elseif t >= tig & t <= tig + tb
T = thrust;
else
T = 0;
end
ax
ay
= -mu*x/r^3 + T/m*vx/v;
= -mu*y/r^3 + T/m*vy/v;
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Chapter 6
az
= -mu*z/r^3 + T/m*vz/v;
mdot = -T*1.e3/g0/Isp;
dydt = [vx vy vz ax ay az mdot]';
end %rates
end %Problem_6_47
Output to the MATLAB command window:
-------------------------------------------------------Initial mass = 2000 kg. Final mass = 1171.11 kg.
For a constant thrust of 10 kN for 240 s starting at 89 m,
and a flight time of 4.5 hours:
The maximum altitude is 10562.2 km at time = 3.2419 h.
--------------------------------------------------------
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Chapter 6
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