Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 316 / 2

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.

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 $1$.

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 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.

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.

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.

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.