Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 4 / 10B / b / Solution

At infinity the potential energy tends to zero, so the conserved mechanical energy is $E=\frac{1}{2}{v}_{\infty }^2$. The impact parameter $b$ is the perpendicular distance from the origin to the incoming asymptotic line. Conservation of angular momentum gives $|h|=b{v}_{\infty }$.

For $b>0$, a turning point of the radial motion satisfiesOnly the positive root is physical, givingFor $Q=0$ this reduces to the straight-line closest distance $b$.

For nonzero $Q$, put $\eta =b{v}_{\infty }^2/|Q|\ll 1$. If $Q<0$, the numerator is $|Q|(\sqrt{1+{\eta }^2}+1)$, soIf $Q>0$, rationalize the numerator instead:Thus the requested leading distances areRepulsion gives a finite barrier even for zero impact parameter; attraction allows much closer passage. At $b=0$, the attractive formula has the limiting closest distance zero, corresponding to collision with the singular centre rather than a positive turning point.