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


{\bf Question}

For a planetary orbit the apses are defined as those points where
$\ds \frac{dr}{d\phi}=0.$
\begin{description}
\item[(a)] Show that $\ds \frac{du}{d\phi} = 0$ at the apse of the
orbit.
\item[(b)] At what points of the orbit is $\ds \frac {d^2u}{d\phi^2} =
0?$  Show that the acceleration term $\ds \ddot r = 0$ there, and
that at these points $\ds \dot \phi =
\left(\frac{\mu}{r^3}\right)^{\frac{1}{2}}$
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
$\ds \frac{du}{d \phi} = -\frac{1}{r^2} \frac{dr}{d \phi}
\Rightarrow \frac{du}{d \phi} = 0$ at an apse
\item[(b)]
$\ds \frac{l}{r} = 1 + e \cos \phi.$\ \  Therefore $\ds
\frac{d^2u}{d\phi^2} = -\frac{e}{l} \cos \phi $

Therefore $\ds \frac{d^2u}{d\phi^2} = 0$ at $\ds \phi =
\frac{\pi}{2}, \frac{3\pi}{2}$

Now $\ds \dot r = -\frac{1}{u^2}\frac{du}{d\phi}\frac{d\phi}{dt}.$

Angular momentum $\ds r^2 \dot \phi = y \Rightarrow \dot \phi =
hu^2$

Therefore $\ds \dot r = -h\frac{du}{d\phi} \Rightarrow \ddot r =
-h\frac{d^2u}{d\phi^2} \dot \phi = -h^2u^2 \frac{d^2u}{d\phi^2}$

Thus $\ds \frac{d^2u}{d\phi^2} \Leftrightarrow \ddot r = 0$

Using Newton's 2nd law; radial component: $\ds \ddot r = r \dot
\phi = -\frac{\mu}{r^2}$

At these points $\ds \ddot r  = 0 \Rightarrow \dot \phi^2 =
\frac{\mu}{r^3}$
\end{description}



\end{document}