Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 320 / 1

For a bound Kepler orbit about mass $M$, the specific orbital energy and specific angular momentum areThe apsidal radii areequivalently $a=(r_{-}+r_{+})/2$ and $e=(r_{+}-r_{-})/(r_{+}+r_{-})$. Combining the formulas also gives

At fixed energy, the circular orbit has $L_c^2=GMa$, whileAn isotropic galactic distribution function has conditional angular-momentum density $n(L)dL\propto LdL$. ThereforeNormalization on $0\le e<1$ gives the thermal eccentricity distribution

The sudden potential change leaves the star's position and velocity unchanged. Its new energy is thereforeThus $E_1>0$ precisely when $r<a$. Parametrize the orbit by eccentric anomaly $\eta$:A phase-mixed orbit is uniform in the mean anomaly ${\theta }_r$. For $e>0$, the condition $r<a$ is $-\pi /2<\eta <\pi /2$ modulo one period. The corresponding mean-anomaly interval has lengthHenceThe energy increase $G(M_0-M_1)/r$ is largest near periapsis, so stars are more readily unbound there than near apoapsis. A perfectly circular orbit is the measure-zero marginal case $E_1=0$ at every phase.

Average the phase probability over the thermal eccentricity distribution:Therefore

For slow mass loss, the adiabatic invariance of an orbital action applies. Spherical symmetry conserves $L$ exactly, and adiabatic evolution conserves the radial action. For a Kepler orbit,Consequently $Ma$ and $e$ remain constant. When the central mass halves, every semimajor axis doubles while each eccentricity is unchanged:No orbit becomes unbound as long as the final mass remains positive, soBecause the action mapping changes no eccentricity and introduces no orientation preference, the initially isotropic distribution remains isotropic rather than becoming radially or tangentially anisotropic.