Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 11
Problem 11.1 A two stage, solid-propellant sounding rocket has the following properties.
m0  249.5 kg
m f  170.1 kg
&e  10.61 kg s Isp  235 s
m
Second stage: m0  113.4 kg
m f  58.97 kg
&e  4.053 kg s Isp  235 s
m
First stage:
Delay time between burnout of first stage and ignition of second stage: 3 seconds.
As a preliminary estimate, neglect drag and the variation of earth’s gravity with altitude to calculate the
maximum height reached by the second stage after burnout.
Solution
First stage:
c  Isp go  235  9.81  2943 m s


 m  m0  m f go
 249.5  249.5  170.1  9.81
v bo  c ln  0  
 2943 ln 

 1127  73.38  1054 m s
 170.1 
m&e
10.61
 mf 
2
h bo
 1  m0  m f 
c   mf 
 ln 
 m f  m0  m f   
 go
m&e   m0 
 2  m&e 
2
 1 249.5  170.1 
2943  170.1 
 170.1  249.5  170.1 
 9.81
ln 


10.61  249.5 
 2  10.61
 3947  274.4
 3673 m

After 3 second staging delay:
v  v bo  gt s  1054  9.81 3  1024 m s
h  hbo  v bot s 
1
1
gt s 2  3673  1054  3   9.81  32  3673  3117  6790 m
2
2
Second stage:
v 0  1024 m s
h0  6790 m
c  Isp go  235  9.81  2305 m s


m0  m f g o
m 
v bo  v 0  c ln  0  
m&e
mf 
 113.4  113.4  58.97   9.81
 1024  2305 ln 

 58.97 
4.0573
 1024  1508  131.7
 2400 m s
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 m0  m f 
c
h bo h0  v 0 

 m&e  m&e
  mf 
 ln 
 m f  m0  m f
  m0 
Chapter 11
2
 1  m0  m f 
 
 go
 2  m&e 
2
 1  113.4  58.97 
 113.4  59.97  2305   59.97 
 6790  1024 

ln 
 58.97  113.4  58.97   


  9.81


 4.053   113.4 
4.053
4.063
 2
 6790  13760  9028  884.7
 28 690 m
Coast to apogee:
v 0  2400 m s
h0  28 690 m
v
2400
0  v0  gt max  t max  0 
 244.7 s
g
9.81
1
1
hmax  h0  v0t max  gt max 2  28 690  2400  244.7   9.81  244.7 2  322 300 m
2
2
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Chapter 11
Problem 11.2 A two-stage launch vehicle has the following properties:
First stage: two solid propellant rockets. Each one has a total mass of 525 000 kg, 450 000 kg of
which is propellant. Isp  290 s .
Second stage: two liquid rockets with Isp  450 s . Dry mass = 30 000 kg, propellant mass = 600 000
kg.
Calculate the payload mass to a 300 km orbit if launched due east from KSC. Let the total gravity and
drag loss be 2 km/s.
Solution
v 0   earth Rearth cos   7.29210 5  6378 cos 28  0.4107 km s
398 600
 2  0.4107  9.315 km s
6678
vbo  v  9.315 km s
v bo  v bo1  v bo2
 m01 
m 02 
v bo  I sp go ln 
 I sp go ln 

1
m f 
2
m f 

 1 
 2 
v 
2  525 000  30 000  600 000  m PL


9315  290  9.81  ln 
 2 (525 000-450 000)+30 000+600 000+m PL 
 30 000  600 000  m PL 
 450  9.81  ln 

30 000  m PL

 1 680 000  m PL 
 630 000  mPL 
9315  2845  ln 
 4414  ln 

 780 000+m PL 
 30 000  m PL 
To find the value of m PL satisfying this equation, graph the function
 1 680 000  m PL 
 630 000  m PL 
f  9315  2845  ln 
 4414  ln 

780
000+m

 30 000  m PL 
PL 
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Chapter 11
f  0 when mPL  110 800 kg
Problem 11.3 Suppose a spacecraft in permanent orbit around the earth is to be used for delivering
payloads from low earth orbit (LEO) to geostationary equatorial orbit (GEO). Before each flight from
LEO, the spacecraft is refueled with propellant, which it uses up in its round trip to GEO. The outbound
leg requires four times as much propellant as the inbound return leg. The delta-v for transfer from LEO to
GEO is 4.22 km/s. The specific impulse of the propulsion system is 430 s. If the payload mass is 3500 kg,
calculate the empty mass of the vehicle.
Solution
m p  m p out  m p in  m p out 
mp
out
4

5
m
4 pout
Outbound leg:

me  m p  m PL


 me  m p  m pout  m PL 
v  Isp go ln 
5
5




me  m pout  3500
me  m pout  3500 



4
4
4220  430  9.81  ln 
  430  9.81  ln 

5
1
 me  m pout  m pout  3500 
 me  m pout  3500 




4
4
5
m
 3500
4 pout

 2.719
1
me  m pout  3500
4
me 
0.5702m p
out
 1.719m e  6018
(1)
Return from GEO to LEO:

1

m e  4 m p out  3500 
v  I sp go ln 

me



1

m e  4 m pout 
4220  430 9.81 ln 

me


1
me  mp
4 out  2.719
me
m p  6.876m e
out
(2)
Substitute (2) into(1):
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Chapter 11
0.5702 6.876me  1.719me  6018
me  2733 kg
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Chapter 11
Problem 11.4 Consider a rocket comprising three similar stages (i.e., each stage has the same specific
impulse, structural ratio and payload ratio). The common specific impulse is 310 s. The total mass of the
vehicle is 150 000 kg, the total structural mass (empty mass) is 20 000 kg and the payload mass is 10 000
kg. Calculate
(a) The mass ratio n and the total v for the three-stage rocket.
(b) m p1 , m p2 , and m p3 .
(c) mE1 , mE2 and mE .
3
(d) mo1 , mo2 and mo3 .
Solution
 PL 

m PL
10 000

 0.06667
m0
150 000
 PL1/ 3
1   PL1/ 3
 0.682

mE
20 000

 0.1429
m0  m PL 150 000  10 000
n
1 
1  0.682

 2.039
   0.1429  0.682
(a)
 v  Isp go ln n 3  310  0.009 81  ln 2.0393  6.5 km s
(b)
m p1 
1   PL1/ 3 1    m
m p2 
1   PL1/ 3 1    m
PL 
m p3 
1   PL1/ 3 1    m
PL 
mE1 
1   PL1/ 3  m
mE2 
1   PL1/ 3  m
PL 
mE3 
1   PL1/ 3  m
PL 
(c)
 PL
PL 
 PL 2/ 3
 PL1/ 3
 PL
 PL 2/ 3
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 PL1/ 3
PL 
1  0.066671/ 3 1  0.1429
0.06667
1  0.066671/ 3 1  0.1429
0.06667 2/ 3
1  0.066671/ 3 1  0.1429
0.066671/ 3
1  0.066671/ 3  0.1429
0.06667
1  0.066671/ 3  0.1429
0.06667 2/ 3
1  0.066671/ 3  0.1429
0.066671/ 3
456
10 000  76 440 kg
10 000  30 990 kg
10 000  12 570 kg
10 000  12 740 kg
10 000  5166 kg
10 000  2095 kg
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Chapter 11
(d)
m03  mE3  m p3  m PL  2095  12 570  10 000  24 660 kg
m02  mE2  m p2  m03  5166  30 990  24 660  60 820 kg
m01  mE1  m p1  m02  12 740  76 440  60 820  150 000 kg
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Chapter 11
Problem 11.5 Find the extrema of the function z  x  y 2 subject to x and y lying on the circle
x  12  y 2  1 .
Solution
z  x 2  y 2  2xy
(1)
g  x 2  2x  y 2
(2)

h  z   g  x 2  y 2  2xy   x 2  2x  y 2

(3)
Local extrema are found from the equations
h
0

h
 2x  2y   2x  2  0
x
h
 2y  2x   2y   0
y

x 2  2x  y 2  0
(4)

  1 x  y  
(5)

x    1 y  0
(6)
From (5) and (6)
1   x   
  1
 
    
1

 1  y   0 

(7)
For   0 , (7) yields
y  x
(8)
which, when substituted into (4) gives x x  1  0 , which means x  0 and x  1 . Then (8) implies that
y  0 and y  1 , respectively. In both cases z  0 [from (1)].
  0 : Local extrema are
x1  0
y1  0
z1  0
x2  1
y 2  1
z2  0
If   0 , then (7) yields
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Chapter 11
   1 
1
x    1
1   
  1 1      2 
1
   
  


   
  1 
y   1
 0     2  1   10   1 
   2 
Substituting these values of x and y into (4), we get
  12
 1
1
2

0
2
  2   2 2
  2 
2
Multiply through by   2 to get
  12  2  1  2  1  0
or
2  4   2  0
The two roots are 0.5858 and 3.414 .
  0.5858: The extremum is
x3 
  1 0.5858  1

 0.2929
  2 0.5858  2
y3  
1
1

 0.7071
2
0.585  2
z 3  x 32  y 32  2x 3 y 3  0.29292  0.7071  2  0.2929 0.7071  0.1716
2
  3.414: The extremum is
x4 
  1 3.414  1

 1.707
  2 3.414  2
y4  
1
1

 0.7071
2
3.414  2
z 4  x 4 2  y 4 2  2x 4 y 4  1.707 2  0.70712  2 1.707  0.7071  5.828
To determine whether the above four extrema are local minima or local maxima, we first note that
  2z
  2z
  2z
2g 2
2g 
2g  2
d2h  

dx  2 

dxdy  



 dy
  y 2
 x y 
  x y
 x2
x2 
 y 2 
d 2 h  2    2 dx 2  2 2    0 dxdy  2    2 dy 2


d 2 h  2   1 dx 2  dy 2  4dxdy
(9)
From (4) we get
2xdx  2dx  2ydy  0
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Chapter 11
or
x  1dx  ydy  0
(10)
  0:
From (9),


d 2 h  2 dx 2  dy 2  4dxdy
(11)
At extremum 1, (10) becomes
x1  1dx  y1dy  0
0  1dx  0  dy  0
 dx  0
so that (11) becomes


d 2 h1  2 02  dy 2  4  0  dy  2dy 2
d 2 h1  0

Extremum 1 is a local minimum
At extremum 2, (10) becomes
x 2  1dx  y 2dy  0
1  1dx  1dy  0
 dy  0
so that (11) becomes
d 2 h1  2 dx 2  02  4dx  0  2dx 2
 d 2 h1  0

Extremum 2 is a local minimum
  0.5858 :
From (9),




d 2 h3  2 0.5858  1 dx 2  dy 2  4dxdy  0.8284 dx 2  dy 2  4dxdy
(12)
At extremum 3, (10) becomes
x 3  1dx  y 3dy  0
0.2929  1dx  0.7071dy  0
0.7071dx  0.7071dy  0
 dy  dx
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Chapter 11
so that (12) becomes
d 2 h3  0.8284 dx 2  dx 2  4dx dx 
d 2 h3  2.343dx 2
 d 2 h3  0

Extremum 3 is a local maximum
  3.414 :
From (9),




d 2 h4  2 3.414  1 dx 2  dy 2  4dxdy  4.828 dx 2  dy 2  4dxdy
(13)
At extremum 4, (10) becomes
x 4  1dx  y 4 dy  0
1.7071  1dx  0.7071dy  0
 dy  dx
so that (13) becomes
d 2 h4  4.828 dx 2  dx 2  4dx dx 
d 2 h4  13.66dx 2
 d 2 h4  0 
Extremum 4 is a local maximum
The constrained function and its four extrema are shown in the figure below.
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Chapter 11
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Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 11
Problem 11.6 A small two-stage vehicle is to propel a 10 kg payload to a speed of 6.2 km/s. The properties of the stage are:
First stage:
Isp  300 s and   0.2 .
Second stage: Isp  235 s and   0.3 .
Estimate the optimum mass of the vehicle.
Solution
c1  Isp1 go  300  0.00981  2.943 km s
c2  Isp2 go  235  0.00981  2.305 km s
 1 0.2
2  0.3
vbo  6.2 km s
v bo 
2
c   1
 c   1
c  1 
i1
i i
1 1
2 2
 c i ln  ci     c1 ln  c1     c 2 ln  c2   
 2.943 1 
 2.305  1 
6.2  2.943 ln 
  2.305 ln 

2.943 0.2 
2.305 0.3 
2.943  1 
2.305  1 
6.2  2.943 ln 
  2.305ln 

0.5886



 0.6915 
To find  , graph the function
2.943 1 
2.305 1 
f  2.943 ln 
  2.305 ln 
  6.2
 0.5886 
 0.6915 
As shown below, f  0 when   1.726 .
c1 1 2.943 1.726  1

 4.016
c11 2.943 0.2 1.726
c2 1 2.305 1.726 1
n2 

 2.496
c 2 2 2.305 0.3 1.726
n1 
n2  1
2.496  1
m 
10  59.53 kg
1  2 n 2 PL 1  0.3  2.496
n1 1
4.016 1
59.53  10  1065 kg
m1 
m2  m PL  
1  1n1
1  0.2 4.016
m2 
M  m1  m2  1065  59.53  1124 kg
Howard D. Curtis
463
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Chapter 11
f
0
-1
1.5
2
2.5

1.726
Howard D. Curtis
464
Copyright © 2013, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Third Edition
Howard D. Curtis
465
Chapter 11
Copyright © 2013, Elsevier, Inc.