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{\bf Question}

A space ship approaches a planet of mass $M$ along the path
(relative to the planet) $$ \frac{l}{r} = 1 + 2\cos \phi$$
\begin{description}
\item[(a)] Show that the closest approach (if there is no
collision) is to $R = \frac{1}{3}l$ at $\phi = 0$.
\item[(b)] Differentiate to find $\ddot r$ in terms of $l$ and
$\dot \phi^2$ at the point of closest approach.  Use this result,
and the fact that the spaceship's acceleration toward the planet
is known in terms of $G$ and $M$ at that position, to find the
speed at the closest approach in terms of $G, \, M$ and $l$.
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
Stationary point of $ r(\phi)$ is when $r'(\phi)=0$; $\ds
\frac{l}{r} = 1 + 2 \cos \phi$

i.e. $\ds \frac{dr}{d \phi} = -\frac{l}{(1 + 2\cos \phi)}\times -2
\sin \phi = 0 \Rightarrow \phi = 0, \pi$

Inspection gives that $\phi = 0$ is the minimum value i.e. $R =
\frac{l}{3}$

\item[(b)]
$\ds r = \frac{l}{1 + 2 \cos \phi} \Rightarrow \dot r =
-\frac{l}{(1 + 2 \cos \phi)^2} \times -2 \sin \phi \, \dot \phi =
\frac{2l\sin\phi \, \dot \phi}{(1+2\cos\phi)^2}$

Therefore

$\ds \ddot r = 2l \left\{\frac{\cos \phi \,  \dot \phi^2}{(1+2
\cos \phi)^2} + \frac{\sin \phi \, \ddot \phi}{(1 + 2 \cos
\phi)^2} + \frac{4 \sin \phi \, \dot \phi^2}{(1 + 2 \cos \phi)^3}
\right\}$

Therefore $\ds \ddot r = \frac{2l\dot \phi}{9}$ at the closest
approach $\phi = 0$

Using Newton's 2nd law; radial component: \begin{eqnarray*}
m(\ddot r - r \dot \phi^2) & = & -\frac{\mu}{r^2} \\ m
\left(\frac{2l}{g} \dot \phi^2 - r \dot \phi^2\ \right) & = &
-\frac{\mu}{r^2} \\ m\left(\frac{2l}{9} = \frac{l}{3}\right) \dot
\phi^2 & = & -\frac{\mu}{l^2}9 \\ \dot \phi^2 & = & \frac{81
\mu}{l^3m}
\end{eqnarray*}

The speed is purely tangential at closest approach (as $\dot r =
0$)

so $\ds v = r \dot \phi = \frac{l}{3} \sqrt{\frac{81\mu}{l^3m}}=
3\sqrt{\frac{\mu}{ml}}$

Now $\mu = GMm$ Therefore $\ds v = 3\sqrt{\frac{GM}{l}}$
\end{description}


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