[Physics] Kepler Orbits: Small Perturbations->Elliptical Orbits

Question

A group of my physics classmates and I have been stuck on this problem. We have tried a few approaches. The problem is to show that a body following a circular orbit, when given a small radial perturbation, will oscillate about its original orbit with simple harmonic motion producing an elliptical orbit. The task is to show the new orbit is an ellipse.

We have tried two directions, one that starts from forces and inputs a $r=r_0+\Delta r$, to retrieve a force of the form $F=-kx.$

The second approach is to start with the kepler orbit equation $r(\phi)=1/(1+\epsilon\cdot \cos(\phi))$ where $\epsilon$ is the eccentricity. However, once we take this radial equation and add a SHM perturbation in the form of $A\sin(\phi)$ to the RHS, we don't see a way to manipulate this into the form of an ellipse. We are also not sure if the result embodies the perihelion advance, or if it is just a simple non-advancing ellipse. Any pointers on how to tackle this problem would be greatly appreciated!

Answer 1

Short answer: Use the radial acceleration component $a_\mathrm r$ of the total $\bf a$ by the central force and then deduce the SHM equation using the properties of circular orbital motion (effective potential energy curves can also be used) that $\dot r= 0$ and keeping in mind the fact that SHM would only occur when the circular orbit is stable.

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Here the approach is elaborately discussed:

The radial component of the central force $\bf F$ is given by

$$\mathbf F\cdot \mathbf e_\mathrm r~=~ m\left[\ddot r - r(\dot \theta)^2\right]\;.\tag 1 $$

Due to the small perturbation, the radius of the circular orbit changed from $r_0$ to $r(t) ~=~r_0 + \rho(t)~~_; ~~~\rho(0)\ll r_0\;.$

Using $\mathit l= mr^2\dot \theta=\textrm{const}$, $(1)$ becomes

$$\ddot\rho(t)- \frac{l^2}{m^2(r_0+ \rho(t))^3}~=~ \frac{F_\mathrm r(r_0+ \rho(t))}{m}~\tag {1-a} $$

Now, \begin{align}(r_0+ \rho(t))^{-3}&=r_0^{-3}\left(1+ \frac{\rho}{r_0}\right)^{-3}\\ &\approx r_0^{-3}\left(1-3\frac{\rho}{r_0}\right)\\ F_\mathrm r(r_0+\rho)&= F_\mathrm r(r_0) +\frac{\mathrm dF_\mathrm r}{\mathrm dr}\bigg|_{r=r_0}~\rho+ \ldots \\ &\approx F_\mathrm r(r_0) +\frac{\mathrm dF_\mathrm r}{\mathrm dr}\bigg|_{r=r_0}~\rho\end{align}

So,

$$\ddot\rho- \frac{l^2}{m^2r_0^3}\left(1-3\frac{\rho}{r_0}\right)= \frac{F_\mathrm r(r_0)}m +\frac{\mathrm dF_\mathrm r}{\mathrm dr}\bigg|_{r=r_0}~\cdot \frac{\rho}m\tag{1-b} $$

From the orbit-equation$^\ddagger$, we get

$$\mathit l^2=- mr_0^3 F_\mathrm r(r_0)_; $$

So,

$$\ddot \rho+ \left[\frac{3\mathit l^2}{m^2r_0^4}- \frac{1}{m}\frac{\mathrm dF_\mathrm r}{\mathrm dt}\bigg |_{r=r_o}\right]\rho~=~0\;.$$

This can be written as

$$\ddot \rho + \omega^2 \rho ~=~0\;.\tag 2$$

Now, this is a second-order differential equation for simple harmonic oscillator with frequency $\omega\;,$ where $$\omega^2 ~=~ \frac{3\mathit l^2}{m^2r_0^4}- \frac{1}{m}\frac{\mathrm dF_\mathrm r}{\mathrm dt}\bigg |_{r=r_o}\;.$$

Unless, the circular orbit is unstable, a simple harmonic radial oscillation about $r=r_0$ would ensue (i.e. SHM would occur if $\omega^2\gt 0\;.$)

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$\ddagger$ We would derive the orbit-equation from $(1)$ expressing $r$ and its derivatives in terms of $\theta\;.$

\begin{align}\dot r &= \frac{\mathrm dr}{\mathrm d\theta}\frac{\mathrm d\theta}{\mathrm dt}\\ \ddot r&=\frac{\mathrm d}{\mathrm dt}\left(\frac{\mathrm dr}{\mathrm d\theta}\dot \theta\right)\\&=\frac{\mathrm d^2r}{\mathrm d\theta^2}~\dot\theta ^2 + \frac{\mathrm dr}{\mathrm d\theta}~\ddot \theta \end{align}

Let

$$\mathbf H \stackrel{\text{def}}{=} \mathbf r\times \dot{\mathbf r} =\frac{\mathbf L}m \;.$$

Therefore

\begin{align}\dot \theta &= \frac{|\mathbf H|}{r^2}\\ \implies \ddot \theta &=-\frac{2|\mathbf H|}{r^3} \, \dot r\;.\end{align}

Putting all these in $(1),$ we get

\begin{align}\frac{\mathrm d^2r}{\mathrm d\theta^2}~\dot\theta ^2 + \frac{\mathrm dr}{\mathrm d\theta}~\ddot \theta- r(\dot \theta)^2& = \frac{F_\mathrm r}{m} \\ \implies\frac{\mathrm d^2r}{\mathrm d\theta^2}~\left(\frac{|\mathbf H|}{r^2}\right)^2 +\frac{\mathrm dr}{\mathrm d\theta}~\left(-\frac{2|\mathbf H|}{r^3}\dot r\right) - r ~\left(\frac{|\mathbf H|}{r^2}\right)^2 &= \frac{F_\mathrm r}{m}\\ \implies \left(\frac{|\mathbf H|}{r^2}\right)^2\left[\frac{\mathrm d^2r}{\mathrm d\theta^2} - 2\frac{\left(\frac{\mathrm dr}{\mathrm d\theta}\right)^2}{r}-r\right]&=\frac{\mathrm F(r)}{m}\\ \implies \frac{\mathrm d^2r}{\mathrm d\theta^2} - 2\frac{\left(\frac{\mathrm dr}{\mathrm d\theta}\right)^2}{r}-r&= \frac{r^4 F_\mathrm r}{|\mathbf H|^2 m}\;.\tag{i}\end{align}

$(\rm i)$ is the orbital equation.

Now, applying the facts that for circular orbits, $\dot r,~\ddot r~=~0\,,$ we get the value of $$|\mathbf H|~=~\sqrt{ -\frac{r^3F_\mathrm r}{m}}\;,$$ where $r~=~r_0\;.$