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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.

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.

A circumplanetary orbit whose orbital period equals the central planet's spin period. Its semi-major axis follows from Kepler's third law. Matching periods alone does not fix the satellite above one ground location; that additionally requires a planet-stationary orbit.

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.

A planet can block stellar radiation during part of a circumplanetary orbit. Then the full-orbit constant-flux averages for radiation pressure and Poynting–Robertson drag require modification. The detailed shadow duty cycle depends on the orbital inclination, stellar angular extent and dust radius.

Differential gravitational acceleration, multiplied by the affected mass, across a spatially extended body. In the linear approximation the acceleration difference is $T\Delta \mathbf{r}$, where $T$ is the tidal tensor. Stellar tidal forces limit the domain of a circumplanetary orbit; sufficiently strong ones can produce tidal disruption.