Past exam of the physics B course of the University of Cambridge / 2012 / Paper 1 / Section C / C11

For an attractive inverse-square force, write $V(r)=-\kappa /r$ with $\kappa >0$, and take nonzero conserved angular momentum $J$. The conic section orbit hasThe physically allowed cases are ellipse for $-m{\kappa }^2/(2J^2)\le E<0$, parabola for $E=0$, and hyperbola for $E>0$. The minimum allowed energy gives the circular special case $e=0$. Here $E$ is measured relative to infinity. Zero angular momentum instead gives radial collision or escape trajectories, interpreted as degenerate limits rather than regular closed ellipses.

For a repulsive inverse-square force, $V(r)=+\kappa /r$. The effective potential is positive and decreases from infinity to zero, so a trajectory reaching a finite radius has $E>0$. For nonzero angular momentum, it is the accessible branch of a hyperbola, with $e^2=1+2E{J}^2/(m{\kappa }^2)>1$. There are no bound ellipses or finite-distance zero-energy parabolas. At zero angular momentum, the motion is straight inward followed by straight outward after reflection at a radial turning point.

For the scattering calculation, take the heavy target as fixed and define $\kappa =2Z{e}^2/(4\pi {ϵ}_0)>0$. The incoming kinetic energy and angular momentum are $E=m{v}^2/2$ and $J=mvb$. In the original sketch, the horizontal dashed incoming asymptote has perpendicular distance $b$ from the nucleus, while $d$ is the radius from the nucleus to the point where $\dot{r}=0$.

At closest approach, the velocity is tangential and $v_t=J/(md)=vb/d$. conservation of energy givesChoose the positive root and put $s=\kappa /(m{v}^2)$:Head-on incidence gives $d_{\mathrm{rep}}=2s$, and repulsion always gives $d_{\mathrm{rep}}>b$. With an attractive Coulomb force, change the sign of the potential term:For $b>0$, the attractive and repulsive inverse-square closest approaches obey $d_{\mathrm{att}}{d}_{\mathrm{rep}}=b^2$. For $b=0$ attraction causes a collision with the point centre, so $d=0$ is a singular endpoint, not a regular scattering turning point. Finite nuclear size is excluded by the stated point-force model.

For the repulsive inverse-cube central force $\mathbf{F}={\kappa }_3\hat{\mathbf{r}}/r^3$, the torque vanishes, $\mathbf{r}\times \mathbf{F}=0$, and angular momentum is still conserved. Its potential energy is $+{\kappa }_3/(2r^2)$, since $-dV/dr=+{\kappa }_3/r^3$. ThusThe new coefficient is denoted ${\kappa }_3$ because its units are $\mathrm{N}{\mathrm{m}}^3$, unlike the $\mathrm{N}{\mathrm{m}}^2$ units of the inverse-square coefficient. Conserving angular momentum depends on centrality, not on a particular radial power law.