= 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> is
$$
\boxed{t_{\rm circ}=\frac{a_0^2}{4A}=\frac{ca_0^2}{4\beta GM_\star}.}
$$
For 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.
Back to article page
