Exercises from "Fundamentals of Astrodynamics" Part 5

02 Feb 2020

Moving on to the next problem in chapter 1 of “Fundamentals of Astrodynamics” by Bate, Mueller, and White, we arrive at a simple proof.

Exercise 1.7 Prove that $r_{apoapsis}=a(1 + e)$.

Solution:

By inspection of the figure in my previous post:

$$r(\pi) = r_{apoapsis}.$$

In simple language, apoapsis (i.e. maximum radial distance) happens 180 ^o, or $\pi$ radians from periapsis (i.e. minimum radial distance). To show that $r_{apoapsis}=a(1 + e)$, we’ll revisit equation 1.5-4 from the book, a form of the trajectory equation called the polar equation of a conic section:

$$r(\nu) = \frac{p}{1+e\cos{\nu}},$$

where $r$, is the radial distance from the focus as a function of the true anomaly, $\nu$, the semi-latus rectum, $p$, and the eccentricity, $e$. We can substitute equation 1.5-6 for the semi-latus rectum and complete the proof:

$$\displaylines{ p = a \left (1 - e^2 \right ) \\ \therefore r(\nu) = \frac{a \left (1 - e^2 \right )}{1+e\cos{\nu}} \\ }$$ $$\begin{align*} r(\pi) &= \frac{a \left (1 - e^2 \right )}{1 - e} \\ &= \frac{a(1 + e)(1 - e)}{1 - e} \\ &= \frac{a(1 + e) \cancel{(1 - e)}}{\cancel{1 - e}} \\ &= a(1 + e) \\ \end{align*}$$ $$\displaylines{ \because r_{apoapsis}=r(\pi) \\ r_{apoapsis}=a(1 + e). \blacksquare }$$ eccentricity true anomaly semi-latus rectum apoapsis trajectory equation