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

For polar coordinates, the rotating unit vectors satisfy ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$ and ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$. Applying the product rule twice to $\mathbf{r}=r{\mathbf{e}}_r$ gives the velocity and accelerationNewton's second law for a central force now givesMultiplying the azimuthal force equation by $r/m$ yieldsThis is conservation of specific angular momentum. Here $F(r)$ is the signed radial component of the force; a negative value points inward.

When $h\ne 0$, $\theta$ is a valid local parameter along the trajectory. Put $u=1/r$ and let primes denote derivatives with respect to $\theta$. The chain rule yieldsSubstituting into the radial Newton's second law equation and dividing by $-m{h}^2{u}^2$ proves the Binet equation:

Here $m{h}^2=b$, so the Binet equation reduces toThe initial conditions becomeMultiply the differential equation by $2u^{\prime }$ and use the chain rule:The initial conditions give $C=0$. Since $e^{-u}>0$, $u^{\prime }$ cannot change sign at a finite value of $u$. Thus $u^{\prime }=e^{-u/2}$ on the physical branch. Separation of variables givesThereforeThe natural logarithm must be positive because $r>0$. For the forward trajectory, $\theta \ge 0$ and $h=\sqrt{b/m}>0$, so $\dot{\theta }=h/r^2>0$: the particle travels counterclockwise while $r$ strictly decreases.

For $a=1$, it starts at $(x,y)=(1,0)$, with tangent vector proportional to $(-e^{-1/2},1)$. Its successive crossings of the positive $x$ coordinate axis have radiiIt makes infinitely many inward turns, approaching the origin as $\theta \to \infty$. There is no finite-time arrival at the origin: $dt/d\theta =r^2/h$ behaves as $1/(4h{\log }^2\theta )$, whose improper integral diverges. If extended backwards, it comes from infinity as $\theta \downarrow 2(1-\sqrt{e})$.