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An orbit about a planet. Stellar tides, radiation pressure and other perturbations can limit its stability; a sufficiently small semi-major axis compared with the Hill sphere permits a local approximately planet-centred Kepler orbit.

Assume the physical coefficients $Q_a,Q_b$ are positive. Differentiate the catastrophic disruption threshold:The bracket is strictly increasing from a negative value to positive infinity, so there is one critical point and it is the global minimum. In fact $Q_D^{\ast }$ tends to infinity at both endpoints $D\to 0$ and $D\to \infty$. The strength-gravity disruption transition is thereforeAt this diameter, $b{Q}_b{D}_w^b=a{Q}_a{D}_w^{-a}$, so $Q_w^{\ast }=[(a+b)/b]Q_a{D}_w^{-a}$. This minimizes the full catastrophic disruption threshold; it need not be where its two contributions are equal.

For equal-density spherical planetesimals, the specific impact energy per unit target mass is $\frac{1}{2}(D_{\mathrm{im}}/D_w{)}^3{v}_{\mathrm{rel}}^2$. Setting this to $Q_w^{\ast }$ gives the catastrophic projectile diameterHere $v_{\mathrm{rel}}$ is the impact speed, the threshold is defined per target mass, and energy-coupling changes are not included. If projectile and target densities differ, multiply the diameter by $({\rho }_{\mathrm{target}}/{\rho }_{\mathrm{projectile}}{)}^{1/3}$. For $a=1/2$, $b=3/2$,so collecting the powers givesThe result presumes that this projectile diameter is available in the population.

For the dust, $n(D)dD$ is the total number in a diameter interval, not a number per unit volume. The mass normalization of a power-law size distribution with dust mass $M/2$ givesThe approximation uses $(D_{\min }/D_{\max }{)}^{4-\alpha }\ll 1$. Merely having a small diameter ratio does not guarantee this accuracy if $\alpha$ approaches four, so retain the exact denominator in that case.

With $\gamma >0$, the stated Poynting–Robertson drag law integrates directly:Thus dust arrives in order of increasing diameter. This is burst transport under inverse-size radial drag, assuming an instantaneous release at $r_0$ and no intervening destruction or changes of diameter.

Dust that reaches the planet's orbital radius has several possible fates. It can pass without a close encounter and continue spiralling towards the star; undergo resonant trapping of dust in an exterior mean-motion resonance, where resonant torque temporarily balances the drag; be scattered onto a different stellar Kepler orbit or ejected; or collide with the planet and be accreted. Resonant trapping of dust can increase orbital eccentricity until escape or an encounter becomes possible. Temporary planet-bound capture can become permanent if drag or another dissipative process removes sufficient energy. Arrival at a radial circle alone does not guarantee a collision with the planet at the correct longitude.

A more massive planet has a larger Hill sphere, stronger resonant perturbations and stronger planetary scattering. Capture into a mean-motion resonance is favoured when drift across it is slow compared with its libration timescale. Since $|\dot{r}|\propto D^{-1}$ here, larger grains drift more slowly and are generally easier to trap; small fast-drifting grains more readily cross a resonance. Grain radiation pressure can also move resonance locations, though the prescribed nearly circular drift model omits that additional change. For direct impact, a larger physical radius increases the geometric cross-section, and gravitational focusing gives, in the two-body encounter approximation,A compact massive planet can nevertheless scatter or eject grains rather than hit them: its encounter region can be much larger than its physical radius. Slower dust migration permits more repeated encounter opportunities. These competing effects mean neither planet mass nor grain diameter alone specifies an accretion probability.

For the idealized complete-accretion calculation, put $T=t_p(D_{\max })$ and $\eta =D_{\min }/D_{\max }$. At time $t$, the arriving diameter is $D_t=D_{\max }t/T$ and $d{D}_t/dt=D_{\max }/T$. Hence the dust-burst accretion pulse follows from the arriving mass per diameter:Dropping the small lower-cutoff normalization correction gives the requested expressionThe rate is zero before the smallest grains arrive and after the largest arrive. Integrating the exact expression over its support returns $M/2$, as required by mass conservation. The approximate expression has that integral to the same lower-cutoff accuracy. This pulse assumes no resonant delay or scattering loss before accretion.

For the surface density of a size-sorted dust burst, take $0<r<r_0$ and writeAt fixed $t>0$, the size-radius map is monotone and $d{D}_r/dr=2r{D}_r/H_r$. Define $\Sigma (r,t)$ as the azimuthally averaged surface density, so the mass in an annulus is $2\pi r\Sigma dr$. Transforming the power-law size distribution into radial mass givesIt follows thatThe requested leading approximation omits $1-{\eta }^{4-\alpha }$ from the denominator. The surface density is zero where the inferred $D_r$ lies outside the original size interval. Equivalently, one can obtain it from the instantaneous inward mass flux divided by $2\pi r|\dot{r}(D_r)|$. At $t=0$ the idealized initial distribution is a delta-function ring at $r_0$; the formula above describes the subsequent radial spread. No uniform initial azimuthal distribution is required for this azimuthal average, but the local density of a clump would require its angular evolution. If the complete-absorption planet assumption is retained, this expression holds exterior to $r_p$ and $\Sigma =0$ interior to the planet.

Synchronous radius and its limits. Kepler's third law for a circumplanetary orbit givesHere $G$ is the gravitational constant. A planet-synchronous orbit repeats after one spin period. A planet-stationary orbit additionally requires a circular orbit, zero orbital inclination to the equator and prograde motion. Then an antenna fixed on the ground points continuously at the same satellite; merely matching the orbital period does not give this property.

Write the planetary radius as $R_p=(3M/(4\pi \rho ){)}^{1/3}$. Requiring the semi-major axis to exceed $R_p$ gives $P^2>3\pi /(G\rho )$. The stellar tidal force restricts the circumplanetary orbit to the Hill sphere, of radius $r_H=a(M/(3M_{\star }){)}^{1/3}$. ThusThese are the surface and Hill-radius constraints in the idealized spherical, small-$M/M_{\star }$ model. Long-term prograde stability generally requires a radius appreciably inside the Hill sphere; for nonzero orbital eccentricity, the surface constraint applies to the periapsis, not just to the semi-major axis.

Collision time and the launch population. A phase-mixed isotropic swarm occupies a shell of radial scale $e{a}_s$. Its number density scales as $N_s/(e{a}_s^3)$, its relative speed as $a_s/P$, and its geometric collision cross-section as $r^2$. Consequently the total collision rate scales as $N_s^2{r}^2/(Pe{a}_s^2)$. To give the numerical normalization used here, adopt an effective shell volume $V=4\pi e{a}_s^3$. At a fixed position the velocity directions are uniformly distributed in the tangent plane, so their mean relative speed is $4v_K/\pi =8a_s/P$. For an unordered pair, the collision cross-section is $4\pi r^2$, givingThus the mean interval between collisions is $A/N_s^2$ in this large-population kinetic model. The effective shell width fixes an order-one coefficient: small orbital eccentricity and random planes alone do not specify a unique radial probability density. Phase mixing, negligible gravitational focusing, $r\ll e{a}_s$, and uncorrelated encounters are implicit in this estimate. Exactly identical orbital periods with perfectly fixed phases do not themselves produce a memoryless collision process.

For nearly planet-stationary orbits with $e\ll I\ll 1$, the swarm volume is smaller by a factor of order $I$, while the relative speed is smaller by the same factor, since vertical motion of scale $I{v}_K$ dominates the eccentric motion. The two changes cancel in the rate $n\sigma v$: there is no parametric factor $I$ in the collision time within the same phase-mixed kinetic approximation. Numerical factors and phase correlations can differ. This is not an argument that bringing all satellites into one nearly circular plane makes their phases random.

With $N_s=Rt$, an Inhomogeneous Poisson process has cumulative hazard functionThe survival function of the first collision is $\exp [-H(t)]$. Setting the expected number of collisions to one givesThis is a characteristic first-event population, with probability $1-e^{-1}$ of an earlier event. It is not a median: the median has an additional factor $(\log 2{)}^{1/3}$.

Which population collides next? Immediately after the first disruption let $S=N_{s0}-2$ and $F=2x^{-3}$. The latter follows from mass conservation for equal-density spherical pieces. Using the same unordered-pair counting and geometric collision cross-sections as above, define $b=(1+x{)}^2/2$; thenHere $S^2,F^2$ stand for the usual large-population approximations to $S(S-1),F(F-1)$. The factor two distinguishing identical and different species is essential. The probability that the next collision is fragment–fragment isIt is the most likely type if $S<4/[x(1+x{)}^2]$ and $S<2/x^2$. It has probability greater than one half ifFor $x\ll 1$, a strongly fragment-dominated next event therefore requires $Sx\ll 4$; the fragment–satellite comparison is more restrictive than the satellite–satellite comparison. If $S=0$, the next event must be fragment–fragment, provided fragments remain.

Population equations and the normalization discrepancy. Every satellite–satellite event produces $2x^{-3}$ fragments and destroys two satellites. Every satellite–fragment event produces a net $x^{-3}-1$ fragments and destroys one satellite; every fragment–fragment event destroys two fragments. Consistent collision counting therefore givesFor $x\ll 1$, $b\simeq 1/2$ and $x^{-3}-1\simeq x^{-3}$. The last two terms in $A{\dot{N}}_f$ are then twice those in the printed equation. This is a genuine factor-of-two inconsistency: equal-size fragment pairs have a collision cross-section smaller by $x^2$, so their event rate must be $x^2{N}_f^2/A$ if the satellite event rate is $N_s^2/A$; destroying both fragments necessarily gives the sink $-2x^2{N}_f^2/A$.

If the printed equation is taken as a prescribed approximate rate model instead, its implicit event rates are $N_s^2/A$, $N_f{N}_s/(4A)$ and $x^2{N}_f^2/(2A)$. Under precisely that mixed normalization, its corresponding satellite equation isThe printed source term also neglects the consumed fragment in a satellite–fragment event, a legitimate relative $O(x^3)$ approximation. That approximation does not repair the pair-counting discrepancy.

The ensuing cascade. This is a two-size fragmentation cascade. First use the printed approximate model, dropping satellite–satellite events as requested. Put $S=N_s$, $F=N_f$, $k=4x^2$ and take $x<1/2$. ThenIntegrating this linear differential equation, with $F_0=2x^{-3}$ and $S_0=N_{s0}-2>0$, givesInitially the collisional cascade grows if $S_0>8x^2$, with approximate early exponential growth time $4A{x}^3/S_0$ when the satellite population is nearly fixed. The fragment population reaches its maximum atAfterwards satellites are depleted and fragment–fragment losses dominate. For $S_0>0$ the continuum solution has $S\to 0$, $F\to 0$, withWhen no satellites remain initially, $F(t)=F_0/[1+x^2{F}_{0}t/A]$ directly. Small integer populations eventually invalidate these deterministic differential equations.

The consistently counted model has, to leading order in $x$, exactly the same curve $F(S)$ and peak, but both retained time derivatives are twice as large. Its growth time is $2A{x}^3/S_0$, and $F(t)\sim A/(2x^{2}t)$. Keeping $(1+x{)}^2$ and the consumed fragment replaces $k$ by $4x^2/(1+x{)}^2$ and $x^{-3}$ in the curve by $x^{-3}-1$. This explicitly separates the physical collision bookkeeping from the printed normalization while giving the evolution under both conventions.

Let ${\mu }_p=G{M}_p$, $A=\beta G{M}_{\star }/c$ and $\kappa =A/a_p^2$. A tightly bound circumplanetary orbit lies well inside the Hill sphere, so $a/a_p\ll 1$ and its orbital period is short compared with the planet's year. Treat the stellar flux and stellar direction $\hat{\mathbf{R}}$ as constant during one dust orbit. With stellar-frame velocity ${\mathbf{V}}_p+\mathbf{u}$, the velocity-dependent acceleration isThe terms independent of $\mathbf{u}$, including the leading static radiation pressure, do no net work on an unperturbed closed circular orbit. HenceFor a coplanar circular orbit, $u^2={\mu }_p/a$ and the component along the stellar direction has mean square $u^2/2$. Thus $\dot{a}=2a^2{\dot{\epsilon }}_p/{\mu }_p$ yields the Poynting–Robertson decay of a circumplanetary orbitThe coefficient three assumes coplanarity, which the PDF does not state. For orbital normal $\hat{\mathbf{k}}$, the general circular covariance is $⟨u_i{u}_j⟩=(u^2/2)({\delta }_{ij}-k_i{k}_j)$, givingAn orbital plane normal to the stellar radial direction has coefficient two during that orbit, a counterexample to an orientation-independent coefficient three. Averaging also over the planet's circular stellar orbit, with fixed dust-plane orbital inclination $I$ to it, gives coefficient $3-\frac{1}{2}{\sin }^{2}I$. The conservative forces must be weak enough for the assumed approximately circular, planet-bound orbit to persist; the small planet-to-star mass ratio alone does not ensure this for arbitrary grain $\beta$.

Radiative drag is only one of several loss mechanisms. Around a star, radiation-pressure blowout can eject small fragments; its $\beta =1/2$ threshold applies specifically to zero-kick release from a circular parent orbit. Stellar-wind drag and gas drag can drive planetary migration, while sublimation destroys grains approaching high-temperature regions. Collisional cascades destroy or fragment grains and can feed the unbound size range. Planetary scattering can cause ejection, collision with a planet, or a stellar impact; resonant trapping of dust can instead delay planetary migration.

For circumplanetary orbits, collisions with the planet or its satellites, disruption in collisions, and escape under stellar tidal forces are additional losses. Orbits near or outside the Hill sphere need not remain planet-bound. Radiation pressure on circumplanetary dust can excite planetocentric orbital eccentricity or unbind very small grains; it need not act only through slow Poynting–Robertson drag. For charged grains, the Lorentz force in stellar or planetary magnetic fields can alter or destabilize an orbit. Shadowing of circumplanetary dust changes the radiation-force average and can reduce the quoted decay rate. Which mechanism dominates depends on grain size and composition, environment, orbit orientation and the available collision or gas density.

Let $r_0$ be the separation of the two centres at release. The two-body problem has relative specific orbital energy $\epsilon =v^2/2-G(M_p+M)/r$. Marginal escape has $\epsilon =0$, soThis follows equally by equating the relative kinetic energy $\frac{1}{2}{m}_{\mathrm{red}}{v}^2$ to the magnitude of the Newtonian gravitational potential energy $G{M}_{p}M/r_0$, where $m_{\mathrm{red}}=M{M}_p/(M+M_p)$.

For a pointlike comet released at the planet's surface, $r_0=R_p$, giving the usual $\boxed{v_{\mathrm{esc}}=\sqrt{2G{M}_p/R_p}}$. If the finite comet touches that surface, its centre is at $r_0=R_p+R$; the same formula uses this larger separation. The assumption $M_p\gg M$ by itself does not imply $R\ll R_p$. Subsequent formulas written in terms of the planet-only escape velocity use the additional small-comet approximation.

This is local escape from an isolated planet. In a stellar Newtonian gravitational field, reaching the edge of the Hill sphere and escaping the stellar system are different energy thresholds.

Consider a circular circumplanetary orbit well inside the Hill sphere, with dust speed $u^2=G{M}_p/a$ and unit plane normal $\hat{\mathbf{k}}$. During one dust orbital period, take the stellar direction $\hat{\mathbf{R}}$ and flux at the planet's stellar radius $a_p$ as constant. The velocity-dependent Poynting–Robertson drag has relative workThe constant terms have zero work over the closed circular orbit. Since $⟨u_i{u}_j⟩=(u^2/2)({\delta }_{ij}-k_i{k}_j)$,The coefficient is three for coplanar stellar and dust orbits; it is two for a dust plane normal to the instantaneous stellar direction. Averaging over a circular planetary year at fixed dust orbital inclination $I$ gives $3-{\sin }^{2}I/2$. Conservative perturbations must remain small enough for this circular-orbit average.

The leading uniform stellar radiation pressure is a differential acceleration relative to a planet, of scale $\beta G{M}_{\star }/a_p^2$. Relative to planetocentric gravity at dust radius $a$, its magnitude isThus lying inside a Hill sphere alone does not guarantee small radiative perturbations. Approximately circular tightly bound dust requires this ratio sufficiently small, as well as weak tides; the uniform radiative force can otherwise excite orbital eccentricity or unbind grains.