Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 316 / 2 / vii / Solution

For the bound release orbits, use $e_0=\beta /(1-\beta )$ andThe inspiral time under Poynting–Robertson drag is thenThe limiting branches areThus the formal bound-grain curve, shown in the right panel of the preceding figure, diverges at both ends and has a minimum at an intermediate $\beta$. At $\beta =0$ there is no Poynting–Robertson drag and the circular grain survives indefinitely in this collision-free model.

Initially unbound grains do not have a bound-orbit inspiral lifetime. For $\beta \ge 1/2$ the release calculation gives outward trajectories; their residence time in a specified finite observation region is a dynamical escape time, not the averaged $t_{\mathrm{PR}}$ above. A numerical escape lifetime cannot be assigned without specifying the outer boundary. Extremely close to $1/2$, energy lost on the first passage can alter even this bound/unbound classification: orbital averaging requires roughly $1-2\beta \gg v_b/c$. Accordingly the divergence is a formal limit of the supplied secular equations, not an exact prediction for marginal grains.