# How to show circular orbit as a precessing ellipse

A uniform distribution of dust in the solar system adds to the gravitational attraction of the sun on the planet an additional force linear in $r$ where m is the mass of the planet, k is a constant (proportional to the gravitational constant and the density of the dust) and ⃗r is the radius vector from the sun to the planet. This additional force is very small compared to the direct sun-planet gravitational force.

Show that nearly circular orbits can be approximated by a precessing ellipse and find the precession frequency. Is the precession in the same or opposite direction to the orbital angular velocity?`

I can calculate the precession frequency by noting the difference between the small oscillation frequency and the frequency of circular motion. But I have little idea about how to prove the first part. Hints would be appreciated.

My own work is as follows :-

Euler Lagrange EOM:-

$ma_{r}=\frac{l^2}{mr^3}-\frac{GM_{s}m}{r^3}-mkr$ Linearizing this gives frequency of small oscillations about circular motion $w_{osc}=w_{0}+\frac{3k}{w_{0}}$ where $w_{0}$ is the frequency of circular motion. The difference between the two frequencies $w_{osc}-w_{0}$ can be understood as the precession frequency. I am unsure how this is the precession of the ellipse and how the precessing ellipse can be proved to give a nearly circular orbit. Will showing that the angular velocity $w_{0}$ being nearly constant work ?

• Your question is in danger of being closed as homework with insufficient work, closed as unclear, or unanswered. You need to show some work, and it would help to say your educational background. Hint: If you know what the term "Lagrange's planetary equations" means you should be using those. If that sounds like a foreign language, you'll need to use something else. – David Hammen Oct 10 '14 at 6:33
• @DavidHammen : made the changes. – user56199 Oct 10 '14 at 7:28
• Not a physicist but a find the question as quite valid in an empirical sense. Since we have the Law of Large Bodies AND "that which is in motion tends to stay in motion" I would argue some form of "interstellar surface tension" is indeed hard at work trying...and only ultimately succeeding...in keeping the Planets in a "mostly circular" orbit. I still find the fact that Mercury is not tidally locked to the Sun (a very recent discovery) whereas Earth's truly massive Moon is to Earth very odd. Perhaps we have vastly overstated the Sun's mass while vastly understating our own? – Doctor Zhivago Nov 25 '16 at 4:09

I am unsure how this is the precession of the ellipse and how the precessing ellipse can be proved to give a nearly circular orbit. Will showing that the angular velocity $w_0$ being nearly constant work?
Consider the figure below showing the effective potential (continuous line) of a particle of mass $m$ and angular momentum $L$ under the gravitational force $-GMm/r^2$,
The circular orbit of radius $r_0$ correspond to the minimum mechanical energy $-\epsilon<0$. To perturb this orbit means to add a small mechanical energy so that the orbit is limited between $r_{min}$ and $r_{max}$. As long as this increment is small (the total mechanical energy is lesser than zero) the resulting orbit is an ellipse. The perturbation may be due to the gravitational attraction of the dust distribution you mentioned.
In this post of mine I show step by step the precession rate for a perturbed orbit around the circular orbit. This rate is $$\Omega=\frac{-\sqrt{F(r_0)}-\sqrt{-3F(r_0)-r_0F'(r_0)}}{\sqrt{mr_0}},$$ where $r_0$ is the radius of the circular orbit and the force $F$ is decomposed into the main contribution $F_0$ and the perturbation $F_p$, $$F=F_0+F_p.$$ If $\Omega>0$ then the ellipse axis precesses counterclockwise. In your case, $$F_0=-\frac{GMm}{r^2},\quad F_p=-mkr.$$ Plugging into $\Omega$ one gets $$\Omega=\frac{\sqrt{-F_0(r_0)-F_p(r_0)}-\sqrt{-F_0(r_0)-4F_p(r_0)}}{\sqrt{mr_0}}.$$ Taylor expanding up to first order around $F_p/F_0$ gives $$\Omega=-\frac{3}{2}\frac{F_p(r_0)}{\sqrt{-mr_0F_0(r_0)}}=\frac{3}{2}\sqrt{\frac{m^2k^2r_0^2}{GMm^2/r_0}}=\frac{3k}{2}\sqrt{\frac{r_0^3}{GM}}.$$