 Solution

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.