 Solution

Put $r=r_0+\eta$, where $|\eta |\ll r_0$. The Taylor expansion at a circular orbit isThe constant $V_{\mathrm{eff}}(r_0)$ has no effect on the force. The paper's displayed harmonic potential should be read as this expansion after subtracting that constant and neglecting higher orders; it is not an equality for the absolute effective potential in general. For example, when $m=J=\beta =1$ and $k=2$, $r_0=2^{-1/4}$ and $V_{\mathrm{eff}}(r_0)=\sqrt{2}$, whereas the printed quadratic expression is zero at $r_0$.

Differentiating again and then using $\beta k{r}_0^{k-2}=J^2/(m{r}_0^4)$ givesThe linearized radial equation of motion is $m\ddot{\eta }=-V_{\mathrm{eff}}^{\prime \prime }(r_0)\eta$, so the radial frequency in a power-law central potential isFor $k<-2$ the circular orbit is unstable and radial disturbances grow exponentially, with growth rate $\sqrt{-(k+2)}|J|/(m{r}_0^2)$. At $k=-2$ the linear restoring force vanishes, as described in part (c). Thus an attractive central force alone does not guarantee small oscillations.