Solutions Manual
Orbital Mechanics for Engineering Students Fourth Edition
Chapter 2
Problem 2.16 A spacecraft is in a circular orbit of Mars at an altitude of 200 km. Calculate its speed and
its period.
Solution
For Mars, µ = 42,828km 3 s 2
The radius of Mars is 3396 km. Hence
µ
42,828
=
= 3.4511 km s
r
3396 + 200
3
2π 32
2π
( 3396 + 200 ) 2 = 6547 sec
T=
r =
µ
42,828
v=
T = 1 h 49 min 7 s
Howard D. Curtis
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Copyright 2020, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Fourth Edition
Chapter 2
Problem 2.22 The altitude of a satellite in an elliptical orbit around the earth is 2000 km at apogee and
500 km at perigee. Determine (a) the eccentricity of the orbit; (b) the orbital speeds at perigee and apogee;
(c) the period of the orbit.
Solution
(a)
e=
rapogee − rperigee
rapogee + rperigee
=
( 6378 + 500 ) − ( 6378 + 2000 )
= 0.098322
( 6378 + 500 ) + ( 6378 + 2000 )
(b)
h = µ (1+ e ) rperigee = 398,600 ⋅ (1+ 0.098322 ) ⋅ ( 6378 + 500 ) = 54,874 km 2 s
v perigee =
v apogee =
h
rperigee
=
54,874
= 7.9782 km s
6378 + 500
h
54,874
=
= 6.5497 km s
rapogee 6378 + 2000
(c)
3
T=
3
2π ⎛ h ⎞
2π
54,874
⎛
⎞
=
= 6630.2 s = 110.5 m
µ 2 ⎜⎝ 1− e 2 ⎟⎠
398,600 2 ⎜⎝ 1− 0.098322 2 ⎟⎠
Howard D. Curtis
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Copyright 2020, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Fourth Edition
Chapter 2
Problem 2.35 What velocity, relative to the earth, is required to escape the solar system on a parabolic
path from earth’s orbit?
Solution
µsun = 1.3271× 1011 km 3 s 2
rearth = 149.6 × 106 km
v earth =
µsun
1.3271× 1011
=
= 29.784 km s
rearth
149.6 × 106
v esc = 2 ⋅ 29.784 = 42.121 km s
v relative = 42.121− 29.784 = 12.337 km s
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Copyright 2020, Elsevier, Inc.
Solutions Manual
Orbital Mechanics for Engineering Students Fourth Edition
Chapter 2
Problem 2.36 A hyperbolic earth departure trajectory has a perigee altitude of 250 km and a perigee
speed of 11 km/s.
(a) Calculate the hyperbolic excess speed (km/s).
(b) Find the radius (km) when the true anomaly is 100 degrees.
(c) Find v r and v ⊥ (km/s) when the true anomaly is 100 degrees.
Solution
(a)
v esc = 2
µ
398,600
= 2
= 10.967 km s
r
6378 + 250
v ∞ = v perigee
2
− v esc
2
= 112 − 10.967 2 = 0.84994 km s
(b)
h = rperigee v perigee = 6628 ⋅11 = 72,908 km 2 s
rperigee =
6628 =
r=
h2 1
µ 1+ e
72,908 2 1
⇒ e = 1.1012
398,600 1+ e
h2
1
72,908 2
1
=
= 16,179 km
µ 1+ e cosθ 398,600 1+ 1.1012 cos100°
(c)
µ
398,600
e sin θ =
⋅1.1012 ⋅sin100° = 5.4488 km s
h
72,908
h 72,908
v⊥ = =
= 4.5064 km s
r 16,179
vr =
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Copyright 2020, Elsevier, Inc.
Solutions Manual
Chapter 2
Orbital Mechanics for Engineering Students Fourth Edition
Problem 2.43 At a given instant, a spacecraft has the position and velocity vectors r0 = 7000 î ( km ) and
v 0 = 7 î + 7 ĵ ( km s ) relative to an earth-centered non-rotating frame. (a) What is the position vector after
the true anomaly increases by 90°? (b) What is the true anomaly of the initial point?
Solution
(a) Algorithm 2.2
1.
a.
r0 = 7000î = 7000 km
v 0 = 7 î + 7 ĵ = 9.8995 km s
b,
(
)
7000î ⋅ 7 î + 7 ĵ
r ⋅v
49 000
v r0 = 0 0 =
=
= 7 km s
r0
7000
7000
c.
h = r0v ⊥0 = r0 v 0 2 − v r0 2 = 7000 9.8995 2 − 7 2 = 7000 ⋅7 = 49 000 km 2 s
d.
r=
=
h2
µ
1
⎛h
⎞
hv
1+ ⎜
− 1⎟ cos Δθ − r0 sin Δθ
µ
⎝ µr0 ⎠
2
49 000 2
398600
1
⎛ 49 000 2
⎞
49 000 ⋅7
1+ ⎜
− 1⎟ cos 90° −
sin 90°
398600
⋅7000
398600
⎝
⎠
= 43183 km
e.
f = 1−
µr
h
2
(1− cos Δθ ) = 1− 398600 ⋅ 43183
(1− cos 90°) = −6.1691
2
49 000
rr
43183 ⋅7000
g = 0 sin Δθ =
sin 90° = 6169.1 s
h
49 000
2.
(
)
(
)
r = fr0 + gv 0 = −6.1691 7000î + 6169.1 7 î + 7 ĵ = 43183 ĵ ( km )
(b)
r0 =
h2
1
h2
49 000 2
⇒ e cosθ 0 =
−1=
− 1 = −0.13949
µ 1+ e cosθ 0
µr0
398600 ⋅7000
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Solutions Manual
v r0 =
Orbital Mechanics for Engineering Students Fourth Edition
Chapter 2
hv
µ
49 000 ⋅7
e sin θ 0 ⇒ e sin θ 0 = r0 =
= 0.89501
h
µ
398600
e sin θ 0 −0.13949
=
= −6.1691
e cosθ 0
0.89501
θ 0 = tan −1 ( −6.1691) ⇒ θ 0 = 99.208° or θ 0 = 279.21°
Since sin θ 0 > 0 and cosθ 0 < 0 , θ 0 must lie in the second quadrant.
∴ θ 0 = 99.208°
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