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


{\bf Question}

The spaceship Enterprise finds itself spiralling towards a strange
planet along the path (relative to the planet) $$r = Ae^{-k\phi},
\hspace{.2in} \dot \phi = Be^{2k\phi}.$$
\begin{description}
\item[(a)] What force is acting on the Enterprise, in terms of
$r$, its mass $m$, and the constants $A$ and $B$.
\item[(b)] Find $r(t)$ for this orbit, and determine when $r = 0$
given that $\phi = 0$ when $t = 0$.
\end{description}

\vspace{.25in}

{\bf Answer}

$\ds  r^2 \dot \phi = A^2B =$ constant, therefore it is a central
force as the angular momentum is constant.

\begin{eqnarray*} \dot r & = & -Ake^{-k\phi}\dot\phi \\ & = & -kr
\dot \phi \\ & = & -k Ae^{-k\phi} \cdot Be^{2k\phi} \\ & = &
-kABe^{k \phi} \\ \\ \ddot r & = & -kABe^{k \phi}Be^{2k\phi} \\ &
= & -k^2 AB^2e^{3k\phi}\\ & = & -k^2a^4b^2r^{-3} \\ \\ {\rm
Radial\ compont\ of\ force} & = & m(\ddot r - r \dot \phi ^2) \\ &
= & m\left[-k^2a^2b^2r^{-3} - AB^2e^{3k\phi}\right] \\ & = &
-mA^4b^2(1+k^2)r^{-3}
\end{eqnarray*}
Therefore there is an inverse cube radial force .

Now $\ds \dot \phi = Be^{2k\phi}.$ Integrating gives $\ds \int
\frac{d\phi}{e^{2k\phi}} = Bt + const$

Therefore $\ds \frac{-1}{2k}e^{-2k\phi} = Bt + C$

Now $\ds \phi = 0$ at $\ds t = 0 \Rightarrow c = \frac{-1}{2k}
\Rightarrow Bt = \frac{1}{2k}\left(1 - e^{-2k\phi}\right),$

whence $r = Ae^{-k \phi} = A \sqrt{1 - 2kBt}$

So $r=0$ when $\ds t=\frac{1}{2kB}$



\end{document}
