 Solution

For the power law central potential $V=\beta r^k$, differentiate the effective potential:The circular orbit condition is $m\beta k{r}_0^{k+2}=J^2$. Thus, when $J\ne 0$, $\beta k>0$ and $k\ne -2$,The sign condition $\beta k>0$ expresses attraction. At $k=0$ the force vanishes and there is no nontrivial circular orbit. At $k=-2$, attraction requires $\beta <0$ and the condition reduces to $J^2=-2m\beta$. In that exceptional case the effective potential is identically zero and every $r_0>0$ is a possible circular radius for the tuned $J$; there is no isolated preferred radius. A small radial velocity then gives $\ddot{r}=0$, rather than an oscillation.