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,
$$
J_r+L=\sqrt{GMa}.
$$
Consequently $Ma$ and $e$ remain constant. When the central mass halves, every semimajor axis doubles while each eccentricity is unchanged:
$$
\boxed{a_1=2a_0,
\qquad e_1=e_0}.
$$
No orbit becomes unbound as long as the final mass remains positive, so
$$
\boxed{f_{\rm unbound}=0}.
$$
Because the action mapping changes no eccentricity and introduces no orientation preference, the initially isotropic distribution remains isotropic rather than becoming radially or tangentially anisotropic.

Solution

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

When a gravitational potential changes slowly relative to an orbital period, the orbital actions remain constant provided no separatrix is crossed. In a spherical potential, angular momentum is conserved exactly and the radial action is adiabatically conserved.


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