It is the first time I post here, but I used to post in the Mathematics forum (I finished my Math degree two years ago and decided to study Physics now).
The following question is from an exam I took two weeks ago and it gave me headaches:
A particle moves in a circular orbit of radius $a$ under a central potential which attracts it to a point $O$, which is not in the center of the circumference.
Let $v_1$ and $v_2$ be the maximum and minimum speed. Compute the kinetic moment, areal velocity and orbit period. Could this central potential be a gravitational or electrostatic potential?
How could this situation happen? I don't understand how it can be a circular orbit while the attracting point $O$ is not in the center.
When I was in Mathematics, I studied a course on Celestial Mechanics and I am familiar with gravitational orbits, but I think maybe I am taking things for granted that do not happen in other kinds of orbital motion.
Best Answer
We can actually find the force law which produces any curve $r(\theta)$ using the Binet equation, as outlined in my previous answer. In this case, we have $$ r = r_0 \cos \theta + \sqrt{a^2 - r_0^2 \sin^2 \theta} $$ where $r_0$ is the distance from the origin to the center of the circle (displaced along the $x$-axis), and $a > r_0$ is the radius of the circle. After a significant amount of algebra (confession: I used Mathematica), we find that we must have $$ F(r) = \frac{8 a^2 L^2 r}{\mu (a^2 - r_0^2 + r^2)^3}, $$ which implies a potential of $$ U(r) = - \frac{2 a^2 L^2}{\mu (a^2 - r_0^2 + r^2)^2}. $$
It may look a bit odd to have the properties of a particular particle's orbit in a potential that's supposed to apply to all particles. A better way to look at this is that such trajectories are possible if the potential is of the form $$ U(r) = - \frac{k}{(b^2 + r^2)^2}, $$ But there are constraints on the shapes of the eccentric-circle trajectories; for a fixed potential, only eccentric circles with $a^2 - r_0^2 = b^2$ are possible. Moreover (as was pointed out in the comments), most trajectories in this potential will not be eccentric circles, and in fact will not even be closed orbits. The eccentric-circle trajectories arise only for special initial conditions; in particular, a particle describing an eccentric-circle trajectory with parameters $a$ and $r_0$ must have $L^2/\mu = k/2 a^2$.
For those who are interested, I have provided Mathematica code to do numerical integration of this potential. It certainly looks like an eccentric circle.