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

{\bf Question}

A particle of mass $m$ moves under the influence of the
gravitational force of a particle of mass $M$ fixed at the origin.
Find its equations of motion in spherical polar coordinates.  Show
that $r^2 \sin^2 \theta \dot \phi$ is constant during the motion.

\vspace{.25in}

{\bf Answer}

\bigskip

$\ds m \ddot{\bf  r} =- \frac{GMm}{r^2}{\bf e}_r$

\begin{center}
\setlength{\unitlength}{.5in}
\begin{picture}(3,1)

\put(-.1,-.1){$\bullet$}

\put(1.5,1,5){$\bullet$}

\put(0,0){\line(1,1){1.5}}

\put(0.2,0){\vector(1,1){1.5}}

\put(0.5,0.3){\vector(-1,-1){0.3}}

\put(0,0.3){\makebox(0,0){$m$}}

\put(1.9,1.5){\makebox(0,0){$m$}}

\put(1.5,1.5){\vector(1,1){.5}}

\put(2.1,2.1){\makebox(0,0){${\bf e}_r$}}

\put(1,0.6){\makebox(0,0){$r$}}
\end{picture}
\end{center}


Therefore using ${\bf a}$ in spherical polar coordinates
\begin{eqnarray*} {\bf a} & = & (\ddot r - r \dot \theta^2 - r
\sin^2 \theta \dot \phi^2){\bf r}_r + \left[ \frac{1}{r}
\frac{d}{dt}(r^2 \dot \theta) - r \sin \theta \cos \theta \dot
\phi^2 \right] {\bf e}_\theta \\ & & + \frac{1}{r \sin
\theta}\frac{d}{dt}(r^2 \sin^2 \theta \dot \phi){\bf e}_\phi
\end{eqnarray*} we find on equating components  that

$\begin{array}{lrcl} ({\bf e}_r) & \ddot  r - r \dot \theta ^2 -
r\sin \theta \dot \phi^2 & = &\ds -\frac{GM}{r^2} \\ ({\bf
e}_\theta) &\ds \frac{1}{r} \frac{d}{dt}(r^2 \dot \theta) - r\sin
\theta \cos \theta \dot \phi^2 & = & 0 \\ ({\bf e}_\phi ) &
\ds\frac{1}{r \sin \theta}\frac{d}{dt} (r^2 \sin ^2 \theta \dot
\phi) & = & 0
\end{array}$

By integrating the ${\bf e}_\phi$ equation we obtain $\ds r^2
\sin^2 \theta \dot \phi = $ constant.



\end{document}
