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For coplanar circular dust, constant $A=\beta G{M}_{\star }/c$, fixed stellar radius $a_p$ and initial planetocentric radius $a_0>R_p$, integrating $\dot{a}=-3Aa/a_p^2$ givesCompare the inspiral time under Poynting–Robertson drag for a circular circumstellar orbit, $t_{\mathrm{cs}}=(a_p^2-R_{\star }^2)/(4A)$. Planetary arrival is not automatically faster; in the point-star limit it is faster only for $a_0/R_p<e^{3/4}$. A different constant orientation coefficient replaces three.

Keep $A=\beta G{M}_{\star }/c$ constant. A circular circumstellar orbit stays circular in the secular drag approximation, and $da/dt=-2A/a$ integrates to $a^2=a_p^2-4At$. The inspiral time under Poynting–Robertson drag to the stellar surface isThe final expression treats the star as a point, or assumes $R_{\star }\ll a_p$.

For the coplanar circumplanetary orbit used in the preceding result, $da/dt=-3Aa/a_p^2$, so $a=a_0\exp (-3At/a_p^2)$. The inspiral time of circumplanetary dust to the planet's surface, for $a_0>R_p$, isThus the timescales have the same dependence on stellar flux and $\beta$, but planetary arrival contains a logarithm of the initial planetocentric radius. It is not always shorter: for a point star, $t_{\mathrm{cp}}<t_{\mathrm{cs}}$ only if $a_0/R_p<e^{3/4}$. A constant tilted-orbit average replaces three by its appropriate orientation coefficient. Sublimation or other removal can terminate the evolution before either idealized arrival time.

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.

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.

Assume negligible fragment release speed relative to a circular parent planetesimal. Its initial speed is $v_b=\sqrt{G{M}_{\star }/r_b}$, but radiation pressure reduces the grain's gravitational parameter to $(1-\beta )G{M}_{\star }$. Its specific orbital energy and specific angular momentum areFor $0\le \beta <1/2$, the resulting dust orbit released from a circular parent ring hasRelease is at periapsis: the inherited tangential speed exceeds the new circular speed. Low-$\beta$ grains remain near the dust birth ring; higher bound $\beta$ gives increasingly eccentric orbits extending well outside it. Uniform release longitude produces an approximately axisymmetric dust halo, with long residence times near apoapsis. At $\beta =1/2$ the conservative release orbit is parabolic; above this radiation-pressure blowout threshold it is initially unbound. For $\beta \ge 1$, zero or repulsive effective gravity also prevents a bound ellipse.

The left panel shows one release longitude; rotating it around the ring fills the halo. At longer times Poynting–Robertson drag also carries bound grains inward, so an initially outward-spanning release distribution does not imply an empty inner region in a continuously supplied, collision-free population.

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.