Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 4 / 10B (Dynamics and Relativity)

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.

The radial energy equation is ${\dot{r}}^2=2(E-U(r))$, so motion requires $U(r)\le E$. For $Q>0$ and $h>0$, the effective potential minimum and its limiting value zero at infinity implyAt the lower endpoint the particle stays at $r=h^2/Q$, giving a circular orbit. Strictly between that endpoint and zero, two positive turning points confine the radial motion. For $E$ below the minimum there is no allowed motion, while $E\ge 0$ does not give a radially bounded orbit.

Write $\epsilon =|E|$ and $D=Q^2-2\epsilon h^2\ge 0$. At each turning point $E=U(r)$, soThe smaller and larger roots give the periapsis distance and apoapsis distance:They coincide for the circular orbit. Their sum and product are

The speed satisfies $v^2=2(E+Q/r)$, which decreases strictly with $r$ because $Q>0$. Thus its maximum is at the periapsis and its minimum at the apoapsis. At either endpoint the radial velocity vanishes, so $v=h/r$. Therefore the apsidal speed sum for an attractive inverse-square orbit isHence $\boxed{v_{\min }+v_{\max }=2Q/h}$. Individually, $v_{\min }=(Q-\sqrt{D})/h$ and $v_{\max }=(Q+\sqrt{D})/h$, also valid when the orbit is circular.

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.