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\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
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{\bf Question}

There are many reasons why $\ds {\bf F} = -\frac{GmM}{r^2}{\bf
e}_r$ is not exactly correct for a planet's orbit.  Suppose that
the correct force is $${\bf F} = -\frac{GmM}{r^2}\left(1 +
\frac{k}{r}\right){\bf e}_r,$$ with $\frac{k}{r} <\!\! <1$ (so
that inverse square gravity is very nearly correct).
\begin{description}
\item[(a)] Show that the equation for the orbit is
$$\frac{d^2u}{d\theta^2} + u = \left(
\frac{\mu}{mh^2}\right)\left(1 + ku \right).$$
\item[(b)] Solve this (linear) equation, and show that any bounded
resulting orbit has apses separated by an angle $\ds\pi +
\frac{1}{2}\frac{\pi k \mu}{mh^2}$ approximately.
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] book work: plug in for E

\item[(b)]
Solve ordinary differential equation. $$u = \frac{\mu}{mh^2\left(1
- \frac{k\mu}{mh^2}\right)} + A \cos \left\{\phi \sqrt{1 -
\frac{\mu k}{mh^2}}\right\}$$

An apse occur when $\ds \frac{du}{d \phi} = 0$

i.e. \begin{eqnarray*}\ds \sin\left( \phi \sqrt{1 -
\frac{k\mu}{mh^2}}\right) & = & 0 \\ \phi & = & 0 \, , \,
\frac{\pi}{\sqrt{1 - \frac{\mu k}{mh^2}}}\ \ {\rm etc.}
\end{eqnarray*}

Now the angle between the apses is $\ds \frac{\pi}{\sqrt{1 -
\frac{\mu k}{mh^2}}} \approx \pi + \frac{1}{2} \frac{\mu k}{mh^2}
+ ...$

\end{description}



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