The orbit here can be taken as the limit as $r_0 \to a$ of the case where the orbit is an eccentric circle with radius $a$ and center a distance $r_0$ from the origin. I solved for the potential and the force law for this general case in this answer. In the limit of $r_0 \to a$ the results simply become
$$
U(r) = -\frac{ k}{r^4}, \quad F(r) = - \frac{4k}{r^5} \hat{r}.
$$
So this orbit could arise if the force law was proportional to $r^{-5}$ rather than $r^{-2}$ as in the Kepler problem. There are no known two-body forces with this behavior, but one could contrive such a force law by imagining a mass (or charge) distribution spread out through some region of space acting on a massive (or charged) test particle. Alternately, one would expect a $r^{-5}$ dependence for gravity in a Universe with 6 spatial dimensions. (Whether this seems more or less contrived than the previous example is a matter of taste.)
As noted in my previous answer, only particles with special initial conditions (namely $L^2/\mu = k/(2a^2)$) will actually describe these clean circular paths. The general paths for orbits in this potential will be much more complicated. In fact, for a general $r^{-6}$ potential, most orbits will either crash into the origin (as these do) or fly off to infinity.