\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\begin{document}
\parindent=0pt


{\bf Question}

A satellite is initially near a planet, at a distance $r_0$ with
perpendicular velocity $v_0$, where $r_0v_0^2 < \frac{\mu}{m}$.
Show that $\ddot r_0 < 0$ so that the orbit must have the form
$$\frac{l}{r} = 1 - e \cos \phi$$ if $\phi = 0$ is the initial
line.  If the planet has radius $R$ what is the condition on $R$
for the satellite to hit the planet?  Where is the impact?

\vspace{.25in}

{\bf Answer}

Newton's 2nd law: $\ds\ddot r - r \dot \phi^2 = -\frac{\mu}{mr^2}
\hspace{.2in} r^2\dot \phi = h$

$\ds h = r_0v_0$ (as initially ${\bf v} = v_0 {\bf e}_\phi, r =
r_0$)

$\ds \ddot r = r \frac{h^2}{r^4} - \frac{\mu}{mr^2} =
\frac{r_0^2v_0^2}{r_0^3} - \frac{\mu}{mr_0^2} = \frac{mr_0^2v_0^2
- \mu}{mr_0^2} < 0$ as $mr_0v_0^2 < \mu$

$\ds \frac{l}{r} = 1 - e\cos \phi$ therefore $\ds \frac{l}{r_0}= 1
- e \cos 0 \Rightarrow \frac{l}{r_0} = 1 - e$

The planet has radius $R,$ thus the satellite hits at $\phi =
\phi_0,$

when $\ds \frac{l}{r} = 1 - e \cos \phi_0.$

This only has a solution if $\frac{l}{r} \leq 1 + e,$  therefore
$\ds R \geq \frac{l}{1+e} = \frac{r_0(1 - e)}{1+e}$.

Impact is at $\ds \cos\phi_0 = \frac{1}{e} \left( 1 -
\frac{l}{R}\right)$

i.e., $\ds \cos \phi_0 = \frac{1}{e} \left( 1 - \frac{r_0( 1 -
e)}{R} \right)$


\end{document}