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{\bf Question}

Show that the equation of the elliptical orbit of a particle
moving under the inverse square force central force $-\mu
r^{-2}{\bf e}_r$ is $$ \frac{(x + ae)^2}{a^2} + \frac{y^2}{b^2} =
1 $$ where $\ds a = -\frac{\mu}{2E}$ and $\ds b = \frac{h}{\sqrt
{\frac{-2E}{m}}}.$

\vspace{.25in}

{\bf Answer}

$\ds \frac{l}{r} = 1 + e \cos \phi.$\ \ \  $0<e<1$ for an ellipse

\begin{eqnarray*} l = r + e \cos \phi & = & r + ex \\ r^2 & = &
(l+ex)^2 \\ \Rightarrow x^2 + y^2 & = & l^2 - 2elx + e^2 x^2 \\
x^2(1 - e)^2 + 2elx + y^2 & = & l^2 \\ (1 - e^2)\left[ \left( x +
\frac{el}{1-e^2}\right)^2 - \frac{e^2l^2}{(1-e^2)^2}\right] + y^2
& = & l^2 \\ \left(x + \frac{el}{1-e^2}\right)^2 + \frac{y^2}{1 -
e^2} & = & \frac{l^2}{(1 - e^2)} \left[1 + \frac{e^2}{1-e^2}
\right] \\ & = & \frac{l^2}{(1-e^2)^2} \end{eqnarray*}

Put $\ds a = \frac{l}{1 - e^2}$ then $\ds \frac{(x + al)^2}{a^2} +
\frac{y^2}{(1-e^2)a^2} = 1$

Therefore $b = a\sqrt{1 - e^2}$

Now from orbit theory   $\ds l = \frac{mh^2}{\mu} \hspace{.2in} e
= \sqrt{1 + \frac{2mh^2E}{\mu^2}}$

Therefore $\ds 1 - e^2 = -\frac{2mh^2E}{\mu^2}$ Therefore $\ds a =
\frac{mh^2}{\mu}\, \frac{\mu}{-2mh^2E} = -\frac{\mu}{2E}$

So $\ds b = -\frac{\mu}{2E} \sqrt{\frac{-2mh^2E}{\mu^2}} =
\frac{h}{\sqrt{\frac{-2E}{m}}}$



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