Past exam of the physics 1 paper of the University of Oxford / 2019 / 8 / b / Solution

The effective potential includes the angular part of the kinetic energy. Taking $U(0)=0$ gives $U=k{r}^4/4$, henceFor $J\ne 0$, this diverges at both zero and infinity and has a unique minimum. Finite-energy motion therefore remains between two radial turning points, or at the minimum itself. For $J=0$, the centrifugal barrier is absent, but $U\to \infty$ still bounds every finite-energy trajectory; radial motion can pass through the force centre.

A circular orbit requires $U_{\mathrm{eff}}^{\prime }(r_c)=0$, soIt is stable because $U_{\mathrm{eff}}^{\prime \prime }(r_c)=6k{r}_c^2>0$. These are nonzero-radius orbits for $J\ne 0$; $J=0$ also permits the stationary particle at the origin. Boundedness does not mean that every noncircular orbit is closed.