Kepler's second and third Law are the same for all two-body systems with a potential energy that depends on the distance between them $V(r)$. The Orbital Equation is the differential equation that gives the possible orbits, which is $\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} + u = - \frac{\mu}{L_z^2} \frac{f(u)}{u^2}$, where $u = \frac{1}{r}$, $\mu$ is the reduced mass of both bodies, $L_z$ is the angular momentum perpendicular to the plane in which the bodies move and $f(u)=-\frac{\mathrm{d}V}{\mathrm{d}r}|_{r=\frac{1}{u}}$. If you use $V(r)=-\frac{V_0}{r}$ then you get the solutions to the orbits of two bodies due to gravity or of two charges due to (only) the electric field. You could try using $V(r) = -\frac{V_0}{2r^2}$ and see which orbits you get.

Kepler's Second Law is the same for all two-body systems with a potential energy that depends on the distance between them $V(r)$. The Orbital Equation is the differential equation that gives the possible orbits, which is $\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} + u = - \frac{\mu}{L_z^2} \frac{f(u)}{u^2}$, where $u = \frac{1}{r}$, $\mu$ is the reduced mass of both bodies, $L_z$ is the angular momentum perpendicular to the plane in which the bodies move and $f(u)=-\frac{\mathrm{d}V}{\mathrm{d}r}|_{r=\frac{1}{u}}$. If you use $V(r)=-\frac{V_0}{r}$ then you get the solutions to the orbits of two bodies due to gravity or of two charges due to (only) the electric field. You could try using $V(r) = -\frac{V_0}{2r^2}$ and see which orbits you get.