Solution Manual for Interplanetary Astrodynamics 1st Edition by David Spencer & Davide Conte – Complete Exam Prep & Answers

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Solutions Manual
Interplanetary Astrodynamics

1st Edition

by David Spencer, Davide Conte




Table of Contents
1. Introduction
2. Kinematics, Dynamics, and Astrodynamics
3. N-Body Problem
4. Coordinate Frames, Time, and Planetary Ephemerides
5. Trajectory Design
6. Navigation and Targeting




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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
form in spherical coordinates.
Solution
Consider the Cartesian (𝑥, 𝑦, and 𝑧) formulation of the equations of motion for the two-body
problem:
𝜇𝑥
𝑥¨ = − 3
𝑟
𝜇𝑦
𝑦¨ = − 3
𝑟
𝜇𝑧
𝑧¨ = − 3
𝑟
In order to convert between Cartesian and spherical coordinates, we use the following relationships

𝑥 = 𝜌 sin 𝜙 cos 𝜃
𝑦 = 𝜌 sin 𝜙 sin 𝜃
𝑧 = 𝜌 cos 𝜙

where 𝜌, 𝜙, and 𝜃 are the spherical coordinates.
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 − 𝜌𝜙˙2 cos 𝜃 sin 𝜙 − 𝜃˙2 cos 𝜃 sin 𝜙 + 𝜌𝜙¨ cos 𝜙 cos 𝜃+
— 𝜃¨𝜌 sin 𝜙 sin 𝜃 + 2𝜌˙𝜙˙ cos 𝜙 cos 𝜃 − 2𝜌˙𝜃˙ sin 𝜙 sin 𝜃 − 2𝜌𝜙˙𝜃˙ cos 𝜙 sin 𝜃
𝑦¨ = 𝜌¨ sin 𝜙 sin 𝜃 − 𝜌𝜙˙2 sin 𝜙 sin 𝜃 − 𝜌𝜃˙2 sin 𝜙 sin 𝜃 + 𝜌𝜙¨ cos 𝜙 sin 𝜃+
+ 𝜌𝜃¨ cos 𝜃 sin 𝜙 + 2𝜌˙𝜙˙ cos 𝜙 sin 𝜃 + 2𝜌˙𝜃˙ cos 𝜃 sin 𝜙 + 2𝜌𝜃˙𝜙˙ cos 𝜙 cos 𝜃
𝑧¨ = 𝜌¨ cos 𝜙 − 2𝜌˙𝜙˙ sin 𝜙 − 𝜌𝜙¨ sin 𝜙 − 𝜌𝜙˙2 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 − 𝜌𝜙˙2 cos 𝜃 sin 𝜙 − 𝜃˙2 cos 𝜃 sin 𝜙 + 𝜌𝜙¨ cos 𝜙 cos 𝜃+
— 𝜃¨𝜌 sin 𝜙 sin 𝜃 + 2𝜌˙𝜙˙ cos 𝜙 cos 𝜃 − 2𝜌˙𝜃˙ sin 𝜙 sin 𝜃 − 2𝜌𝜙˙𝜃˙ cos 𝜙 sin 𝜃+
𝜇 sin 𝜙 cos 𝜃
+ =0
𝜌2
𝜌¨ sin 𝜙 sin 𝜃 − 𝜌𝜙˙2 sin 𝜙 sin 𝜃 − 𝜌𝜃˙2 sin 𝜙 sin 𝜃 + 𝜌𝜙¨ cos 𝜙 sin 𝜃+
+ 𝜌𝜃¨ cos 𝜃 sin 𝜙 + 2𝜌˙𝜙˙ cos 𝜙 sin 𝜃 + 2𝜌˙𝜃˙ cos 𝜃 sin 𝜙 + 2𝜌𝜃˙𝜙˙ cos 𝜙 cos 𝜃
𝜇 sin 𝜙 sin 𝜃
+ =0
𝜌2
𝜇 cos 𝜙
𝜌¨ cos 𝜙 − 2𝜌˙𝜙˙ sin 𝜙 − 𝜌𝜙¨ sin 𝜙 − 𝜌𝜙˙2 cos 𝜙 + =0
𝜌2
√
where we used the fact that 𝜌 = 𝑟 = 𝑥2 + 𝑦2 + 𝑧2.




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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˙)
( 2 ) [ 2 )

Recall that 𝐯 = 𝐫 = −𝜇𝐫
𝑟3 and 𝐫 = 𝐯, so

𝑑𝐸 −𝜇𝐫 𝜇𝐫
=𝐯⋅( )+ ⋅𝐯
𝑑𝑡 𝑟3 𝑟3
𝜇𝐫 𝜇𝐫
= −𝐯 ⋅ ( + 𝐯 ⋅ =0
𝑟3 ) 𝑟3
Thus, 𝑑𝐸
𝑑𝑡
= 0 which means that orbital energy is constant.




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