Interplanetary Astrodynamics Solution Manual | Spencer & Conte (1st Edition)

SOLUTION MANUAL
Interplanetary Astrodynamics
DAVID B. SPENCER, DAVIDE CONTE
1st Edition

, 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 form in spherical coordinates.
Solution
Consider the Cartesian (𝑥, 𝑦, and 𝑧) formulation of the equations of motion for the
two-body problem:
𝜇𝑥
𝑥̈ = − 3
𝑟
𝜇𝑦
𝑦̈ = − 3
𝑟
𝜇𝑧
𝑧̈ = − 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




1

,Taking another time-derivative:

𝑥̈ = 𝜌̈ cos 𝜃 sin 𝜙 − 𝜌𝜙2 cos 𝜃 sin 𝜙 − 𝜃2 cos 𝜃 sin 𝜙 + 𝜌𝜙̈ cos 𝜙 cos 𝜃+
− 𝜃̈𝜌 sin 𝜙 sin 𝜃 + 2𝜌𝜙 cos 𝜙 cos 𝜃 − 2𝜌𝜃 sin 𝜙 sin 𝜃 − 2𝜌𝜙𝜃 cos 𝜙 sin 𝜃
𝑦̈ = 𝜌̈ sin 𝜙 sin 𝜃 − 𝜌𝜙2 sin 𝜙 sin 𝜃 − 𝜌𝜃2 sin 𝜙 sin 𝜃 + 𝜌𝜙̈ cos 𝜙 sin 𝜃+
+ 𝜌𝜃̈ cos 𝜃 sin 𝜙 + 2𝜌𝜙 cos 𝜙 sin 𝜃 + 2𝜌𝜃 cos 𝜃 sin 𝜙 + 2𝜌𝜃𝜙 cos 𝜙 cos 𝜃
𝑧̈ = 𝜌̈ cos 𝜙 − 2𝜌𝜙 sin 𝜙 − 𝜌𝜙̈ 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

𝜌̈ cos 𝜃 sin 𝜙 − 𝜌𝜙2 cos 𝜃 sin 𝜙 − 𝜃2 cos 𝜃 sin 𝜙 + 𝜌𝜙̈ cos 𝜙 cos 𝜃+
− 𝜃̈𝜌 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
𝜌̈ sin 𝜙 sin 𝜃 − 𝜌𝜙2 sin 𝜙 sin 𝜃 − 𝜌𝜃2 sin 𝜙 sin 𝜃 + 𝜌𝜙̈ cos 𝜙 sin 𝜃+
+ 𝜌𝜃̈ 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
𝜌̈ cos 𝜙 − 2𝜌𝜙 sin 𝜙 − 𝜌𝜙̈ sin 𝜙 − 𝜌𝜙2 cos 𝜙 + =0
𝜌2
√
where we used the fact that 𝜌 = 𝑟 = 𝑥2 + 𝑦2 + 𝑧2.




2

, 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⋅ 1
𝐯 − 𝜇 − −3
𝐫 (2𝐫 ⋅ 𝐫)
=
( 2 ) [ 2 )

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

𝑑𝐸 −𝜇 𝜇𝐫
𝐫
=𝐯⋅( 3 )+ 3 ⋅𝐯
𝑑𝑡 𝑟 𝑟

𝜇𝐫 𝜇𝐫
= −𝐯 ⋅ ( 3 ) + 𝐯 ⋅ 3 = 0
𝑟 𝑟

Thus, 𝑑𝑡𝑑𝐸 = 0 which means that orbital energy is constant.




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