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For the Poynting–Robertson drag invariant $C$ and $A=\beta G{M}_{\star }/c$, the orbit-averaged time to the point-star limit isThis tends to $a_0^2/(4A)$ for a circular orbit and to $4Q_0^{1/2}{q}_0^{3/2}/(5A)$ for a highly eccentric orbit. Stellar radius, sublimation, collisions, and failure of orbital averaging can terminate the evolution earlier.

Parameterize the trajectory by the decreasing orbital eccentricity. The Poynting–Robertson drag invariant givesFor $Q_0\gg q_0$, $C\simeq 2q_0$. Initially $e\simeq 1$, and the orbit moves almost vertically down a plot of $Q/q_0$ against $q/q_0$: the apocentre distance shrinks rapidly while the pericentre distance changes little. Integrating the high-eccentricity slope givesOnce the orbit has moderate orbital eccentricity, both distances change appreciably. For example $e=1/2$ gives $q\simeq 0.766q_0$ and $Q\simeq 2.30q_0$ in the $Q_0/q_0\to \infty$ limit. Eventually $e\ll 1$, $q\simeq Q\simeq C{e}^{4/5}$, and the trajectory approaches the diagonal $Q=q$ before reaching the origin. Thus the two approximate phases are apocentre contraction at nearly fixed pericentre, followed by nearly circular inward migration. They are a smooth crossover, not two separate exact solutions.

For $0<e<1$, differentiate the logarithm of the proposed Poynting–Robertson drag invariant:Substitution of the orbit-averaged Poynting–Robertson drag rates gives, over the common denominator $a^2(1-e^2{)}^{3/2}$, the numerator $A[-(2+3e^2)+5e^2+2(1-e^2)]=0$. HenceEquivalently, eliminating time gives $d\log a/de=(2/5)(2+3e^2)/[e(1-e^2)]$, whose integral is $\log a=(4/5)\log e-\log (1-e^2)+\text{constant}$. An exactly circular orbit remains circular; the expression for $C$ is singular at $e=0$ and is replaced by that separate solution.

Eliminate $a$ using the Poynting–Robertson drag invariant, $a=C{e}^{4/5}/(1-e^2)$. The eccentricity rate becomesSince the point-star limit corresponds to $a\to 0$ and $e\to 0$, the inspiral time under Poynting–Robertson drag isFor $e_0\to 0$, the integral is $(5/8)e_0^{8/5}+o(e_0^{8/5})$, recovering $a_0^2/(4A)$ despite the apparent singularity of $C$.

For $e_0\to 1$, use $C\simeq 2q_0$ and $1-e_0=2q_0/(Q_0+q_0)\simeq 2q_0/Q_0$. The endpoint asymptotic integral is $[2(1-e_0){]}^{-1/2}$, givingMost of this long lifetime is spent near the initially large apocentre distance, before substantial circularization. The result assumes weak drag and a valid orbital average; near-parabolic release can violate that assumption.