Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 316 / 2

Write $k=\mu (1-\beta )$. Taking the cross product of the equation of motion with $\mathbf{r}$ givesSince $\mathbf{r}\cdot \mathbf{h}=0$ at every time, the orbit lies in the fixed plane perpendicular to the conserved specific angular momentum $\mathbf{h}$. Taking the inner product with $\dot{\mathbf{r}}$ gives the conserved specific orbital energyFor $\beta >1$, the effective inverse-square force is repulsive and the potential term is positive.

In polar coordinates in the orbital plane, $h=r^2\dot{\theta }$. Setting $u=1/r$, the Binet equation for the effective inverse-square force isIts general solution isand henceThis is a branch of a hyperbolic Kepler orbit; because $1-\beta <0$, its physical branch has a negative denominator.

The parent body's circular Kepler orbit has speed $v_K=\sqrt{\mu /a_p}$. Release without a velocity impulse therefore givesThe release point is the dust orbit's point of closest approach, where $\theta -\varpi =\pi$. Substituting $r=a_p$ in the orbit equation givesThe longitude of periapsis $\varpi$ specifies the symmetry axis of the conic. With this signed repulsive-force convention, it points opposite the release direction, so the closest point is at $\theta =\varpi +\pi$.

At release,At large radius the potential vanishes, soThe asymptote satisfies $1+e\cos f_{\infty }=0$, where $f=\theta -\varpi$. If the angular displacement from the release direction is $\Delta =\pi -f_{\infty }$, thenFor $\beta \gg 1$, $-\cos \Delta \simeq -1+{\Delta }^2/2$, and therefore

Resolving the conserved energy into radial and azimuthal parts givesUsing the release values of $C$ and $h$, and choosing the outward root,Equivalently, with $x=r/a_p$,

In a steady axisymmetric outflow, mass conservation makes $2\pi r\Sigma (r)v_r(r)$ independent of $r$. The steady radial dust outflow therefore hasIt has an integrable $(r-a_p{)}^{-1/2}$ pile-up just outside the source ring, where the radial speed starts from zero, and approaches $Sigma\propto r^{-1}$ at large radius.

The large-distance profile is proportional to $[\sqrt{2\beta -1},x{]}^{-1}$. At $x=2$, the exact profile is proportional to $(4\beta -1{)}^{-1/2}$. Thusas $\beta \to \infty$. The enhancement is the finite-radius remnant of the source-ring pile-up.