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{\bf Question}

For an  Earth satellite near the end of its life suppose $$r^2
\dot \phi - h_0e^{-k\phi}$$ for some constants $h_0$ and $k$. Find
the corresponding differential equation for $u(\phi),$ assuming
inverse square gravity.

\vspace{.25in}

{\bf Answer}

Newton's 2nd law: $\ds m\{\ddot r - r \dot \phi^2\} =
-\frac{\mu}{r^2} \hspace{.2in} r^2 \dot \phi = h_0 e^{-k\phi}$

Lets put $u = \frac{1}{r},$ so $\phi = h_0 u^2 e^{-k \phi}$

\begin{eqnarray*} \dot r & = & \dot \phi \frac{du}{d \phi}{dr}{du}
\\ & = & h_0 u^2 e^{-k\phi} - u^{-2} \frac{du}{d \phi} \\ & = &
-h_0 e^{-k\phi} \frac{du}{d\phi} \\ \\ \ddot r & = & - \dot h_0
\left\{ -ke^{-k\phi} \frac{du}{d \phi} +
e^{-k\phi}\frac{d^2u}{d\phi^2} \right\}\dot \phi \\ & = & -h_0^2
e^{-2k\phi}u^2\left\{ -k\frac{du}{d\phi} + \frac{d^2u}{d\phi^2}
\right\}\end{eqnarray*}

Hence the radial component of Newton's 2nd law becomes

\begin{eqnarray*}
-h_0^2e^{-2k\phi}u^2\left\{-k\frac{du}{d\phi} +
\frac{d^2u}{d\phi^2} \right\} - h_0^2u^3e^{-2k\phi} & = &
-\frac{\mu u^2}{m} \\  \frac{d^2u}{d\phi^2}- k \frac{du}{d\phi} +
u^2 & = & \frac{\mu e^{2k\phi}}{mh_0^2} \end{eqnarray*}



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