Circular orbit + small radial oscillation about circular orbit

[unitdisk]

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 $\Omega = \omega - \omega'$?

 

[eljose79]

To show that a nearly circular orbit is approximately an ellipse with precession, we can use the fact that the angular frequency of small radial oscillations $\omega'$ is related to the angular velocity of precession $\Omega$ by $\Omega = \omega - \omega'$. This means that the difference between the angular frequencies of the circular orbit and small oscillations is equal to the angular velocity of precession.

We can then use this relation to show that the orbit is approximately an ellipse by considering the change in the angular frequency of small oscillations as the orbit precesses. As the orbit precesses, the angular frequency of small oscillations changes by a small amount, which can be approximated by $\Delta \omega' = \Omega \Delta t$, where $\Delta t$ is the time it takes for the orbit to precess by a small angle.

Using the relation between $\omega'$ and $\omega$, we can rewrite this as $\Delta \omega' = (\omega - \omega') \Delta t$. We can then substitute in the expression for $\Delta \omega'$ from equation (14) to get $\Delta \omega' = \frac{c \omega^2}{\vert k \vert a^2 \omega} \Delta t$.

Next, we can use the fact that the angular frequency of small oscillations is related to the potential energy by $\omega'^2 = - \frac{1}{m} \frac{\partial U}{\partial r} \big\vert_{a}$, which we can rewrite as $\omega'^2 = \frac{k}{ma^3} + \frac{c}{ma^5}$. Substituting this into our equation for $\Delta \omega'$, we get $\Delta \omega' = \frac{c}{\vert k \vert a^2 \omega} (\frac{k}{ma^3} + \frac{c}{ma^5}) \Delta t$.

Finally, we can use the fact that the angular velocity of precession is related to the potential energy by $\Omega = \frac{1}{m} \frac{\partial U}{\partial r} \big\vert_{a}$, which we can rewrite as $\Omega = \frac{k}{ma^3} + \frac{c}{ma^5}$. Substituting this into our equation for $\Delta \omega'$, we get $\Delta \omega