= Solution
For the <inverse-square potential> $V(r)=-Q/r$, the <central force> is $-V'(r)\hat r=-Q\hat r/r^2$. The transverse component of <acceleration in polar coordinates> is therefore zero:
$$
r\ddot\theta+2\dot r\dot\theta=0,
\qquad
\frac{d}{dt}(r^2\dot\theta)=0.
$$
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 gives
$$
\ddot r-r\dot\theta^2=-\frac Q{r^2},\qquad
\ddot r=\frac{h^2}{r^3}-\frac Q{r^2}=-U'(r),
\qquad U(r)=\frac{h^2}{2r^2}-\frac Qr.
$$
Consequently
$$
\frac{d}{dt}\left(\frac12\dot r^2+U(r)\right)
=\dot r\bigl(\ddot r+U'(r)\bigr)=0,
\qquad
\boxed{E=\frac12\dot r^2+U(r)\text{ is constant}.}
$$
This is the <mechanical energy>, since the full <kinetic energy> is $\tfrac12(\dot r^2+r^2\dot\theta^2)$ and $r^2\dot\theta^2=h^2/r^2$.
For $h\ne0$ and $Q>0$, the <effective potential> goes to $+\infty$ as $r\downarrow0$ and to zero from below as $r\to\infty$. Its derivative is $U'(r)=(Qr-h^2)/r^3$, so it decreases up to $r_0=h^2/Q$ and then increases. Its zero and minimum are
$$
\boxed{U\left(\frac{h^2}{2Q}\right)=0,\qquad
r_0=\frac{h^2}{Q},\qquad U(r_0)=-\frac{Q^2}{2h^2}.}
$$
For $Q<0$, $U=h^2/(2r^2)+|Q|/r$ is positive and strictly decreasing, with limits $+\infty$ at zero and $0^+$ at infinity. It has no minimum at a finite positive radius.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-4-effective-potential.png]
{title=Attractive and repulsive inverse-square effective potentials and bounded-orbit turning points}
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.
Back to article page
