Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics IV / 4 / b

With $\alpha =0$, a gives constant $h$, and the polar coordinates basis satisfies ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$, ${\mathbf{e}}_r\times {\mathbf{e}}_z=-{\mathbf{e}}_{\theta }$. Also $\ddot{\mathbf{r}}=(F_c/m){\mathbf{e}}_r$. Differentiating the proposed Laplace-Runge-Lenz vector therefore givesFor this to vanish for arbitrary $h$, including nonzero $h$, the central force must satisfyAttraction requires $k>0$. The condition “for any value of $h$” matters: with $h=0$ alone, the Laplace-Runge-Lenz vector is constant on each radial ray for any central force, so that special case would not determine the force law.

Assume $h\ne 0$. The velocity in polar coordinates is $\dot{\mathbf{r}}=\dot{r}{\mathbf{e}}_r+r\dot{\theta }{\mathbf{e}}_{\theta }$. HenceTaking the dot product of the conserved Laplace-Runge-Lenz vector with $\mathbf{r}$ givesConsequently the polar equation of a conic isOnly angles for which the denominator is positive describe the physical trajectory. Its eccentricity is $A/(mk)$, giving an ellipse, parabola, or hyperbola when this ratio is respectively less than, equal to, or greater than $1$.

If $A=0$, the angle $\gamma$ is undefined, but the same dot product identity gives the circular orbit $\boxed{r=m{h}^2/k}$. If $h=0$, the trajectory is radial and the displayed division degenerates; for $k>0$, $\mathbf{A}=-mk{\mathbf{e}}_r$ and $\gamma =\pi$ on a fixed ray, so the identity does not determine $r$ as a function of a varying angle.