Circular orbit + small radial oscillation about circular orbit

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The potential energy of a particle of mass $m$ is $U(r)= k/r + c/3r^3$ where $k<0$ and $c$ is very small. Find the angular velocity $\omega$ in a circular orbit about this orbit and the angular frequency $\omega'$ of small radial oscillation about this circular orbit. Hence show that a nearly circular orbit is approximately an ellipse whose axes precess at an angular velocity $\Omega \simeq (c/\vert k \vert a^2) \omega$

\begin{equation}L= \frac{m}{2}(\dot{r}^2 + r^2 \dot{\theta}^2) - U
\end{equation}

\begin{equation}\frac{\partial L }{\partial r} = mr\theta^2 - \frac{\partial U}{\partial r}
\end{equation}

\begin{equation}\frac{d}{dt} \frac{\partial L }{\partial \dot{r}} = m\ddot{r}
\end{equation}

\begin{equation} m\ddot{r} - mr \dot{\theta} ^2 = - \frac{\partial U}{\partial r} \tag{1} \end{equation}

$\ddot{r} = 0$ at $r=a$.
Thus \begin{equation}-ma \omega ^2 = -\frac{\partial U}{\partial r}\big\vert_a = \frac{k}{a^2}+\frac{c}{a^4}\end{equation}

\begin{equation} \omega^2 = \frac{(ka^2+c)}{m a^5}\end{equation}

\begin{equation}\omega = \sqrt{\frac{(ka^2+c)}{m a^5}} \end{equation}

$l= mr^2 \dot{\theta}$ is constant.


\begin{equation}V(r) = U(r) + \frac{l^2}{2m r^2}
\end{equation}

\begin{equation} \ddot{r} - r \dot{\theta}^2 = -\frac{1}{m} \frac{\partial U}{\partial r}= \ddot{r} - \frac{l^2}{m^2 r^3} \tag{2}
\end{equation}
Consider small oscillation where $r=a$ changes to $r=a+x$.
Then $\ddot{r}=\ddot{x}$.
Thus (2) becomes \begin{equation}\ddot{x} - \frac{l^2}{m^2 (a+x)^3} = -\frac{1}{m}\frac{\partial U}{\partial r}\bigg\vert_{a+x} \end{equation}

\begin{equation} \ddot{x} - \frac{l^2}{m^2 a^3} (1+ \frac{x}{a})^{-3} \simeq \ddot{x} - \frac{l^2}{m^2 a^3} (1- \frac{3x}{a}) \tag{3}\end{equation}



\begin{equation}-\frac{1}{m}\frac{\partial U}{\partial r}\bigg\vert_{a+x} \simeq -\frac{1}{m}[\frac{\partial U}{\partial r}\bigg\vert_{a} + \frac{\partial ^2 U}{\partial r ^2}\bigg\vert_a x ]
\tag{4}\end{equation}
(2) at $r=a$ is
\begin{equation}-\frac{1}{m}\frac{\partial U}{\partial r}\big\vert_{a}= -\frac{l^2}{m^2 a^3}
\end{equation}
Thus (3)=(4) yields
\begin{equation} \ddot{x} + \frac{1}{m}(\frac{a}{3}\frac{\partial U}{\partial r}\big\vert_{a} +\frac{\partial ^2 U}{\partial r ^2}\big\vert_a )x = 0\end{equation}
Thus \begin{equation} \omega'^2 = \frac{1}{m}(\frac{a}{3}\frac{\partial U}{\partial r}\big\vert_{a} +\frac{\partial ^2 U}{\partial r ^2}\big\vert_a ) =- \frac{k}{ma^3}+\frac{c}{ma^5} \end{equation}

\begin{equation}
\omega^2= - \frac{k}{ma^3}-\frac{c}{ma^5}
\end{equation}
\begin{equation}
\omega'^2 = - \frac{k}{ma^3}+\frac{c}{ma^5}
\end{equation}
\begin{equation}\frac{\omega'^2}{\omega^2} = \frac{-k a^2 + c}{k a^2 + c}
= 1 + \frac{2c}{\vert k \vert a^2 -c} \simeq 1 + \frac{2c}{\vert k \vert a^2} \end{equation}

\begin{equation} \omega^2 - \omega'^2 \simeq \frac{2c \omega^2}{\vert k \vert a^2}
\end{equation}
\begin{equation}
\omega - \omega' \simeq \frac{2c \omega^2}{\vert k \vert a^2 (\omega+ \omega ')} \simeq \frac{2c \omega^2}{\vert k \vert a^2 (2\omega)}=\frac{c \omega^2}{\vert k \vert a^2 \omega}\end{equation}

How should I show that circular orbit with small oscillation is approximately an ellipse with precession? Also why is angular velocity of prececssion $\Omega = \omega - \omega'$?
 

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