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If release imparts no velocity kick, a grain's initial Kepler orbit under reduced stellar attraction has $q=r_b$, $Q=r_b/(1-2\beta )$, and $e=\beta /(1-\beta )$ for $0\le \beta <1/2$. Its pericentre distance remains at the dust birth ring, while its apocentre distance grows with the radiation-pressure coefficient.

For an unperturbed Kepler orbit, successive revolutions advance the planet-relative orientation by $-n_p{T}_c$ modulo $2\pi$, where $T_c$ is the comet's orbital period. An irrational period ratio makes these orientations dense; sufficiently distant mean-motion resonance and adequate observation time justify the corresponding phase mixing approximation.

At each radius between its pericentre distance and apocentre distance, the comet eventually visits every azimuth in the rotating reference frame. The spatial projection is thereforeIt is an annulus, with a nonuniform radial residence probability, rather than a uniformly filled area. Strictly, the full position-velocity phase space is not this annulus: at each radius the energy and angular momentum constrain the velocity, with inward and outward branches. The annulus is its position-space projection. This description assumes the orbital elements have not yet been substantially changed by planetary scattering.

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.

Using $q=a(1-e)$ and $Q=a(1+e)$, differentiate each apsidal distance. The apsidal evolution under Poynting–Robertson drag isBoth the pericentre distance and apocentre distance decrease. Since $a=(Q+q)/2$ and $e=(Q-q)/(Q+q)$, the same rates areTheir ratio isIn particular $Q\gg q$ gives $dQ/dq\simeq (5/3)(Q/q{)}^2$, whereas the circular limit gives slope $1$.

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.

For a Kepler orbit with pericentre distance $q$, apocentre distance $Q$, and semi-major axis $a=(q+Q)/2$, a uniform mean anomaly givesThe two radial passages per orbital period are included, and ${\int }_q^Q{p}_r(r)dr=1$.