# Kepler Orbits: Small Perturbations->Elliptical Orbits

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!

• What do you mean by a radial perturbation? A perturbing force in the radial direction, or apply a radial perturbation once (impulse)? – fibonatic Jun 9 '16 at 3:07
• theory.tifr.res.in/~sgupta/courses/cm2011/hand12.pdf – user83548 Jun 9 '16 at 3:24
• farside.ph.utexas.edu/teaching/336k/Newtonhtml/node45.html – user83548 Jun 9 '16 at 3:24
• Please can you show your working so that we can see where you are getting stuck and advise what to do? – sammy gerbil Jun 9 '16 at 22:53
• Is there really anything to prove here? The general Kepler orbit is elliptical. This includes the circular orbit (e=0). Any deviation from e=0 is no longer circular but remains an elliptical orbit. – sammy gerbil Jun 10 '16 at 15:17

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.

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 $\omega^2\gt 0\;.$)

$\ddagger$ 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 $r~=~r_0\;.$

You may have mis-interpreted the question slightly. The question means to imply an instantaneous perturbation of a circular orbit (an impulse -- momenum kick -- for example) gives an elliptical orbit. Your second approach implies you are adding a simple harmonic motion to the ellipse, or perhaps that you are applying a time-dependent SHM external force to the particle, and neither of of those cases necessarily ends up with an elliptical orbit.

In classical mechanics, the best way to do this problem is by the effective potenial method. An object of mass $m$ and at an instantaneous radius $r$ orbiting with angular momentum $J$ behaves as if in addition to the attractive potential there is another potential $$V_{\mbox{eff}}= \frac{J^2}{2mr^2}$$ Now $J$ is a constant of the motion (conservation of angular momentum. Then the sum of this and the $-\frac{GmM}{r}$ gravitational potential gives a total "potential" for the radial motion that rises to infinity near $r=0$, approaches zero from below for large $r$ and has a minimum at some radius $r_0$. The angular motion can be computed once you know the radial motion since $$\dot{\phi} = \frac{J}{mr^2}$$

A circular orbit starts and forever stays at $r=r_0, \dot{r} = 0$. If you instead start at $r = r_0, \dot{r} \neq 0$ you get oscillation movement in $r$, and thus a non-circular orbit.

But that is not at all the same as proving the orbit is an ellipse. In fact, the reason the orbit is a closed curve is that the $1/r$ form of the potential gives an oscillation period which exactly matches the angular period. Thus there is no advance of the perihelion in the 2-body problem in classical gravitation.

If the potential were, say, $r^{-1.1}$ the orbits would not be elliptical. And in fact, in general relativity, a body orbiting in a Schartzchild-metric does not follow an elliptical orbit. But the orbit for any one angular period is very very close to an ellipse, and if you consider each orbit as an ellipse, then because there is not an exact match in radial and angular periods, you get the small advance in the perihelion.

## protected by Qmechanic♦Jul 20 '16 at 10:00

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