Exercises from "Fundamentals of Astrodynamics" Part 1

23 Jul 2019

Over the coming months, years, decades, no one really knows, I will be working through the exercises from “Fundamentals of Astrodynamics” by Bate, Mueller, and White. The intent is to not only provide myself with a refresher, but to provide a resource for enterprising students.

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Exercise 1.1 The position, $\vec{r}$, and velocity, $\vec{v}$, of a satellite at a given instant in canonical units: Distance Units (DU) and Distance Units per Time Unit (TU), respectively, are described by

$$\displaylines{ \vec{r} = 2\vec{I} + 2\vec{J} + 2\vec{K} \quad \textrm{DU} \\ \vec{v} = -.4\vec{I} + .2\vec{J} + .4\vec{K} \quad \textrm{TU} }$$

Find the specific angular momentum, $\vec{h}$, and specific mechanical energy, $\mathcal{E}$, of the system.

Solution:

Equation for specific angular momentum:

$$\vec{h} = \vec{r} \times \vec{v} \\$$

An easy way to set up the crossproduct is using Sarrus’ scheme:

This yields:

$$\displaylines{ 2 \cdot 0.4 \cdot \vec{I} - 2 \cdot 0.2 \cdot \vec{I} = 0.4 \cdot \vec{I} \\ 2 \cdot (-0.4) \cdot \vec{J} - 2 \cdot 0.4 \cdot \vec{J} = -1.6 \cdot \vec{J} \\ 2 \cdot 0.2 \cdot \vec{K} - (-0.4) \cdot 0.2 \cdot \vec{K} = 1.2 \cdot \vec{K} \\ \\~\\ \therefore \vec{h} = 0.4 \vec{I} - 1.6 \vec{J} + 1.2 \vec{K} \quad \frac{\textrm{DU}^2}{\textrm{TU}} }$$

Equation for specific mechanical energy:

$$\displaylines{ \mathcal{E} = \frac{v^2}{2} - \frac{\mu}{r} \\ v^2 = |\vec{v}|^{2} \\ \therefore v^2 = (-0.4)^2 + 0.2^2 + 0.4^2 = 0.36 \quad \frac{\textrm{DU}^2}{\textrm{TU}^2} \\ r = |\vec{r}| = \left( 2^2 + 2^2 +2^2 \right)^{\frac{1}{2}} = 12^{\frac{1}{2}} \quad \textrm{DU} \\ }$$

Since we’re using canonical units, let $\mu = 1$:

$$\therefore \mathcal{E} = \frac{0.36}{2} - \frac{1}{12^{\frac{1}{2}}} = -0.1087 \quad \frac{\textrm{DU}^2}{\textrm{TU}^2}$$