Solutions Manual for Interplanetary Astrodynamics, 1e by David Spencer, Davide Conte (Selective Chapters, 2-6)

MEDSTUDY.COOM

Solutions Manual for Interplanetary Astrodynamics, 1e by David
Spencer, Davide Conte (Selective Chapters, 2-6)


Interplanetary Astrodynamics

Chapter 2 Problem Solutions

For all numerical problems, use 𝜇 = 398, 600 km3/s2 as the gravitational parameter of the Earth.


Problem 1
Starting with the unperturbed two-body equations of motion, Equation (2.9), derive its state space
M
form in spherical coordinates.
Solution
ED
Consider the Cartesian (𝑥, 𝑦, and 𝑧) formulation of the equations of motion for the two-body
problem:
𝜇𝑥
𝑥=−
𝑟3
𝜇𝑦
ST
𝑦=−
𝑟3
𝜇𝑧
𝑧=−
𝑟3
In order to convert between Cartesian and spherical coordinates, we use the following relationships
U
𝑥 = 𝜌 sin 𝜙 cos 𝜃
𝑦 = 𝜌 sin 𝜙 sin 𝜃
D
𝑧 = 𝜌 cos 𝜙

where 𝜌, 𝜙, and 𝜃 are the spherical coordinates.
Y

Taking one time-derivative of the above equations for the 𝑥, 𝑦, and 𝑧 coordinates expressed in
terms of 𝜌, 𝜙, and 𝜃 gives

𝑥 = 𝜌 cos 𝜃 sin 𝜙 + 𝜌𝜙 cos 𝜙 cos 𝜃 − 𝜌𝜃 sin 𝜙 sin 𝜃
𝑦 = 𝜌 sin 𝜙 sin 𝜃 + 𝜌𝜙 cos 𝜙 sin 𝜃 + 𝜌𝜃 cos 𝜃 sin 𝜃
𝑧 = 𝜌 cos 𝜙 − 𝜌𝜙 sin 𝜙




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Taking another time-derivative:

𝑥 = 𝜌 cos 𝜃 sin 𝜙 − 𝜌𝜙2 cos 𝜃 sin 𝜙 − 𝜃2 cos 𝜃 sin 𝜙 + 𝜌𝜙 cos 𝜙 cos 𝜃+
— 𝜃𝜌 sin 𝜙 sin 𝜃 + 2𝜌𝜙 cos 𝜙 cos 𝜃 − 2𝜌𝜃 sin 𝜙 sin 𝜃 − 2𝜌𝜙𝜃 cos 𝜙 sin 𝜃
𝑦 = 𝜌 sin 𝜙 sin 𝜃 − 𝜌𝜙2 sin 𝜙 sin 𝜃 − 𝜌𝜃2 sin 𝜙 sin 𝜃 + 𝜌𝜙 cos 𝜙 sin 𝜃+
+ 𝜌𝜃 cos 𝜃 sin 𝜙 + 2𝜌𝜙 cos 𝜙 sin 𝜃 + 2𝜌𝜃 cos 𝜃 sin 𝜙 + 2𝜌𝜃𝜙 cos 𝜙 cos 𝜃
𝑧 = 𝜌 cos 𝜙 − 2𝜌𝜙 sin 𝜙 − 𝜌𝜙 sin 𝜙 − 𝜌𝜙2 cos 𝜙

Equating each 𝑥, 𝑦, and 𝑧 acceleration expressed in spherical coordinates with its respective
acceleration terms gives us the equations of motion for the two-body problem in terms of spherical
coordinates 𝜌, 𝜙, and 𝜃

𝜌 cos 𝜃 sin 𝜙 − 𝜌𝜙2 cos 𝜃 sin 𝜙 − 𝜃2 cos 𝜃 sin 𝜙 + 𝜌𝜙 cos 𝜙 cos 𝜃+
M
— 𝜃𝜌 sin 𝜙 sin 𝜃 + 2𝜌𝜙 cos 𝜙 cos 𝜃 − 2𝜌𝜃 sin 𝜙 sin 𝜃 − 2𝜌𝜙𝜃 cos 𝜙 sin 𝜃+
𝜇 sin 𝜙 cos 𝜃
+ =0
𝜌2
ED
𝜌 sin 𝜙 sin 𝜃 − 𝜌𝜙2 sin 𝜙 sin 𝜃 − 𝜌𝜃2 sin 𝜙 sin 𝜃 + 𝜌𝜙 cos 𝜙 sin 𝜃+
+ 𝜌𝜃 cos 𝜃 sin 𝜙 + 2𝜌𝜙 cos 𝜙 sin 𝜃 + 2𝜌𝜃 cos 𝜃 sin 𝜙 + 2𝜌𝜃𝜙 cos 𝜙 cos 𝜃
𝜇 sin 𝜙 sin 𝜃
+ =0
𝜌2
𝜇 cos 𝜙
ST
𝜌 cos 𝜙 − 2𝜌𝜙 sin 𝜙 − 𝜌𝜙 sin 𝜙 − 𝜌𝜙2 cos 𝜙 + =0
𝜌2
√
where we used the fact that 𝜌 = 𝑟 = 𝑥2 + 𝑦2 + 𝑧2.
U
D
Y



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Problem 2
Prove that for the unperturbed two-body problem, orbital energy is constant.
Solution
2
Start with the vis-viva equation, Equation (2.50): 𝐸 = 𝑣 − 𝜇
2 𝑟

To prove that energy is constant, we need to take its time derivative and show that it is equal to
zero:

𝑑𝐸 𝑑 𝐯⋅𝐯 𝑑 𝜇
= )−
𝑑𝑡 𝑑𝑡 ( 2 𝑑𝑡 [(𝐫 ⋅ 𝐫)1/2 ]

𝐯⋅𝐯+𝐯⋅𝐯 1
= — 𝜇 — 𝐫−3 (2𝐫 ⋅ 𝐫)
M
( 2 ) [ 2 )

Recall that 𝐯 = 𝐫 = −𝜇𝐫
𝑟3 and 𝐫 = 𝐯, so
ED
𝑑𝐸 −𝜇𝐫 𝜇𝐫
=𝐯⋅( )+ ⋅𝐯
𝑑𝑡 𝑟3 𝑟3
𝜇𝐫 𝜇𝐫
= −𝐯 ⋅ ( 3)+𝐯⋅ =0
𝑟 𝑟3
ST
Thus, 𝑑𝐸
𝑑𝑡
= 0 which means that orbital energy is constant.
U
D
Y



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Problem 3
Prove that the angular momentum vector and eccentricity vector are orthogonal to each other.
Solution
In order to prove that two vectors are orthogonal, or perpendicular, to each other, one must show
that their dot product is zero.
Starting with the definitions of eccentricity, Equation (2.27),
𝐯×𝐡 𝐫
𝐞= —
𝜇 𝑟
and angular momentum, Equation (2.41),

𝐡= 𝐫×𝐯
M

we take the dot product between angular momentum and eccentricity,
ED
𝐯×𝐡 𝐫
𝐡⋅𝐞 = 𝐡⋅ −
( 𝜇 𝑟)
1 1
= 𝐡 ⋅ (𝐯 × 𝐡) − (𝐫 × 𝐯) ⋅ 𝐫
𝜇 𝑟
ST
where we used the definition of angular momentum for the second term. We then use the scalar
triple product on the above equation, which, for three generic vectors 𝐀, 𝐁, and 𝐂 is

𝐀 ⋅ (𝐁 × 𝐂) = 𝐁 ⋅ (𝐂 × 𝐀) = 𝐂 ⋅ (𝐀 × 𝐁)

This helps us simplify the first term as
U

1 1 1
𝐡 ⋅ (𝐯 × 𝐡) = 𝐯 ⋅ (𝐡 × 𝐡) = 𝐯 ⋅ 𝟎 = 0
𝜇 𝜇 𝜇
D

and the second term as
1 1 1
Y
— 𝐫 ⋅ (𝐯 × 𝐫) = − 𝐯 ⋅ (𝐫 × 𝐫) = − 𝐯 ⋅ 𝟎 = 0
𝑟 𝑟 𝑟
which proves that 𝐡 ⋅ 𝐞 = 0 and thus 𝐡 ⟂ 𝐞 = 0.




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