 Solution

For the inverse-square potential $V(r)=-Q/r$, the central force is $-V^{\prime }(r)\hat{r}=-Q\hat{r}/r^2$. The transverse component of acceleration in polar coordinates is therefore zero:Thus $\boxed{h=r^2\dot{\theta }\text{ is constant}}$; for unit mass it is the signed angular momentum perpendicular to the orbital plane. The radial component givesConsequentlyThis is the mechanical energy, since the full kinetic energy is $\frac{1}{2}({\dot{r}}^2+r^2{\dot{\theta }}^2)$ and $r^2{\dot{\theta }}^2=h^2/r^2$.

For $h\ne 0$ and $Q>0$, the effective potential goes to $+\infty$ as $r\downarrow 0$ and to zero from below as $r\to \infty$. Its derivative is $U^{\prime }(r)=(Qr-h^2)/r^3$, so it decreases up to $r_0=h^2/Q$ and then increases. Its zero and minimum areFor $Q<0$, $U=h^2/(2r^2)+|Q|/r$ is positive and strictly decreasing, with limits $+\infty$ at zero and $0^{+}$ at infinity. It has no minimum at a finite positive radius.

If $h=0$, the centrifugal term disappears. For $Q>0$ the graph is instead $-Q/r$, increasing from $-\infty$ to $0^{-}$ without a minimum; for $Q<0$ it remains positive and decreasing. The bounded-orbit calculation below uses the stipulated $h>0$, which excludes this radial collision case.