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

Write $q=|\dot{\mathbf{r}}|=\sqrt{{\dot{r}}^2+r^2{\dot{\theta }}^2}$. Differentiating the angular momentum and applying Newton's second law givesThe central force contributes no torque, because it is parallel to $\mathbf{r}$. The angular momentum direction is fixed whenever it is nonzero; the motion is consequently planar. In the plane's polar coordinates, $\mathbf{L}=mh{\mathbf{e}}_z$, soFor the mechanical energy, use the chain rule, Newton's second law, and $V^{\prime }(r)=-F_c(r)$:ThusFor a trajectory on a time interval, both $h$ and $E$ are each a constant of motion precisely when $\alpha =0$ or the velocity vanishes throughout that interval. Under the stated strictly attractive central force at a regular point $r>0$, a permanently stationary trajectory is impossible by Newton's second law. For the stated attractive motion, both are conserved if and only if $\alpha =0$. An isolated instant of zero speed only makes the derivatives vanish at that instant. A radial trajectory has $h=0$ even with quadratic drag, but its mechanical energy still decreases whenever it moves.

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.