 Solution

For a circular Kepler orbit, $e=0$ remains an exact solution of the averaged Poynting–Robertson drag equations, and $\dot{a}=-2A/a$. Integrating gives $a^2(t)=a_0^2-4At$, so the point-star inspiral time under Poynting–Robertson drag isFor a star of radius $R_{\star }$, contact occurs instead at $(a_0^2-R_{\star }^2)/(4A)$ within this model. Sublimation or other forces can remove a real grain earlier.