Past exam of the physics 1 paper of the University of Oxford / 2022 / 4

For motion in a central potential, conservation of angular momentum gives $L=m{r}^2\dot{\theta }$. In polar coordinates, the conserved mechanical energy isConsequently, for $L\ne 0$,The sign in $\dot{r}$ distinguishes inward and outward motion. Choosing the orientation so $L>0$, it is also the sign of $dr/d\theta$. The expression under the square root must be nonnegative. If $L=0$, the motion is radial and $\theta$ is constant, so $r$ cannot be parametrized by $\theta$ this way.

Take $\beta >0$. At a radial turning point, $\dot{r}=0$, and the effective potential givesFor $L\ne 0$, the smaller positive radial turning point is conveniently written without division by $E$:This includes $E=0$, where $r_{\min }=L^2/(2m\beta )$. For $E<0$ there is also an outer radial turning point; the two coincide for the circular orbit at $E=-m{\beta }^2/(2L^2)$. Smaller energies are inaccessible at that $L$. For $E>0$ there is just one positive radial turning point. If $L=0$, there is no centrifugal barrier and an inward trajectory can reach the singular centre; its limiting closest distance is zero.

The requested omission of $\beta$ is not possible. For example, at $E=0$ and fixed nonzero $L,m$, doubling $\beta$ halves $r_{\min }$. Thus $E,L,m$ alone cannot determine closest approach.