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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.

For a central potential, angular momentum is conserved and the motion lies in a plane. With $L=m{r}^2\dot{\theta }$, radial energy conservation becomesAllowed radii have $\mathcal{E}\ge V_{\mathrm{eff}}$; equality gives radial turning points. A well with two finite turning points gives regular radially bound motion. Access to arbitrarily large radii gives an unbound branch, while access to the origin can lead to collision in a singular potential. Stationary points are circular-orbit candidates; a strict local minimum gives radial stability. For a potential tending to zero at infinity, energy relative to that asymptote is useful, but a singular inner potential and centrifugal barrier must still be examined.

Integrating the angular-momentum relation with the origin of angle at $t=0$ givesFor $V=-A/r+B/r^2$ at $L=L_0$, the effective potential is $-A/r+(L_0^2+2m{B}_0)/(2m{r}^2)$. It is exactly the Kepler effective potential with $L_{\mathrm{K}}=\sqrt{L_0^2+2m{B}_0}$, provided that expression is real. Equal radial energy and initial radial phase therefore give identical $r(t)$, although the angular motion differs. Writing $\psi$ for the Kepler angle, $\dot{\theta }/\dot{\psi }=L_0/L_{\mathrm{K}}=k$. With matched periapsis origins,Here $r_0=L_{\mathrm{K}}^2/(mA)$ and the eccentricity is fixed by radial energy. For $B_0>0$, $k<1$ and successive periapses are separated by $2\pi k$, so there is retrograde apsidal motion relative to a closed Kepler ellipse. The radial equivalence is the relevant content of the Newton's theorem of revolving orbits.

For the pure inverse-cube central force set $A=0$. The free-particle reference orbit at $B=0$ is a straight line, $r=d/\cos (\psi -{\psi }_0)$, not a finite-$A$ conic with its old $r_0$ held fixed. To cover attractive as well as repulsive cases completely, putwhere primes denote angle derivatives and $L\ne 0$. The inverse-cube central orbit classification gives the following physically positive-radius branches.

For $\lambda >0$, write $u=C\cos [\sqrt{\lambda }(\theta -{\theta }_0)]$ with $C>0$, so $r=C^{-1}\sec [\sqrt{\lambda }(\theta -{\theta }_0)]$. These are scattering orbits with positive energy and an inner turning radius $1/C$; the angular interval between the two infinite-radius ends is $\pi /\sqrt{\lambda }$. Repulsion ($B>0$) gives an interval smaller than $\pi$; weak attraction ($-L^2/(2m)<B<0$) gives a larger interval and can cause winding. The $B=0$ limit is the straight line.

For $\lambda =0$, $u=C+D(\theta -{\theta }_0)$. If $D\ne 0$, reciprocal-linear spirals connect an infinite-radius end to collision at infinite accumulated angle. If $D=0$ and $C>0$, every radius admits a circular solution at zero energy: attraction exactly balances centrifugal acceleration. Each is a marginal circular orbit under an inverse-cube force, with no radial restoring force.

For $\lambda <0$, put $\beta =\sqrt{-\lambda }$ and $u=C_{+}{e}^{\beta \theta }+C_{-}{e}^{-\beta \theta }$. Up to angle shifts and reversal, the three energy classes arewith positive-radius domains understood. The negative-energy branch is radially bounded but reaches the singular centre on either side of its outer turning point; it is not a regular closed orbit. Collision requires additional physics to continue through $r=0$. Zero angular momentum gives radial trajectories and is excluded from the angular parametrization above.