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Put $k=p+1$, $\mu =G{M}_{\star }$ and $n_{\mathrm{pl}}=\sqrt{\mu /a_{\mathrm{pl}}^3}$. To leading order the mean-motion resonance fixes $a=a_{\mathrm{pl}}{k}^{-2/3}$. The specific angular momentum isThe asteroid's instantaneous angular speed is $h/r^2$. Thus the high-eccentricity angular-speed crossover satisfies $h/r_x^2=n_{\mathrm{pl}}$, givingHere $q\ll r_x\ll a$, so a parabolic Kepler orbit gives the local motion accurately. Its polar equation is $r=2q/(1+\cos f)$. Write $f_x=\pi -{ϵ}_x$. Since $1+\cos f_x\simeq {ϵ}_x^2/2$,This is the outward crossing; the inward crossing has true anomaly $-f_x$ modulo $2\pi$.

For the pericentre-to-crossover flight time use $D=\tan (f/2)$. The parabolic Kepler orbit has $r=q(1+D^2)$ and $dt=r^{2}df/h$, whence direct integration gives the Barker equationAt the crossover $D_x\simeq \sqrt{r_x/q}\gg 1$. Consequently the planet's angular displacement isThe error tends to zero as $q/a_{\mathrm{pl}}\to 0$ at fixed $k$; it includes the finite binding energy of the elliptic Kepler orbit as well as the subleading term in the Barker equation.

At the $j$th pericentre passage the asteroid's unwrapped mean longitude is $\lambda =\varpi +2\pi j$. The resonant argument therefore givesThus $\varphi /k$ specifies the planet's direction relative to the asteroid's longitude of pericentre when the asteroid passes pericentre, with $k$ branches separated by $2\pi /k$. It is not the instantaneous true-longitude separation throughout the orbit. At high orbital eccentricity the asteroid quickly sweeps to a direction nearly opposite pericentre, while the planet scarcely moves. During the long outer excursion the asteroid's direction changes slowly and the planet advances substantially. This makes an astronomical conjunction during the outer excursion much more dangerous than an astronomical conjunction close to pericentre.

At the following apocentre, the planet has advanced $\pi /k$. The asteroid's direction in the rotating reference frame is consequentlyThese $k$ apocentre directions are equally spaced. Phase protection of interior integer resonances places the planet in the largest gap between them, rather than at an apocentre direction. For $k=2$, $\varphi =\pi$ puts an apocentre toward the planet, whereas $\varphi =0$ puts the apocentres at $\pm \pi /2$. For $k=3$, $\varphi =0$ puts an apocentre toward the planet, whereas $\varphi =\pi$ puts the apocentres at $\pi /3,\pi ,5\pi /3$. Hence the geometrically favored resonant-argument libration centres in this regime of high orbital eccentricity areThis is an encounter-avoidance argument for likely stability, not a calculation of a resonant Hamiltonian or a claim that every orbit near either centre is stable at arbitrary orbital eccentricity.

The following rotating-frame drawing uses $a_{\mathrm{pl}}=1$, $a=3^{-2/3}$, $q=0.002$ and $\varphi =\pi$. It follows three successive Kepler orbits, which fill one planet orbital period and close the pattern. The blue segment is just the first outward half-orbit. The marked crossover is a maximum of its rotating-frame polar angle, since $d\psi /dt=h/r^2-n_{\mathrm{pl}}$ changes from positive to negative there. The planet stays at $(1,0)$; the three apocentres avoid that direction.

To estimate angular-momentum kicks to nearly radial orbits, note that $q\simeq h^2/(2\mu )$, so an order-one fractional change in $q$ requires $|\Delta h|$ of order $h$. A characteristic distant encounter at radii of order $a_{\mathrm{pl}}$ has tangential acceleration of order $G{M}_{\mathrm{pl}}/a_{\mathrm{pl}}^2$ and duration of order $n_{\mathrm{pl}}^{-1}$. Its torque per unit asteroid mass therefore producesWith the most favorable coherent signs, the crude encounter count isOrder-one factors depend on $p$ and encounter geometry. Uncorrelated signs instead give a random walk count of order $(M_{\star }/M_{\mathrm{pl}}{)}^{2}q/a_{\mathrm{pl}}$ when this is large. A librating phase-protected orbit can suppress the kicks even further.

There is an important limit to calling this a minimum. The question does not specify an encounter impact parameter. For a weak close encounter with impact parameter $b$ and relative speed $u\sim \sqrt{G{M}_{\star }/a_{\mathrm{pl}}}$, the impulse estimate gives $|\delta h|\sim G{M}_{\mathrm{pl}}{a}_{\mathrm{pl}}/(bu)$ and hence $N\sim (M_{\star }/M_{\mathrm{pl}})(b/a_{\mathrm{pl}})\sqrt{q/a_{\mathrm{pl}}}$, capped below by one. In the 2:1 case the Kepler orbit can cross the planet's orbit, so one suitably close encounter can suffice: there is no impact-parameter-independent large lower bound. For $k\ge 3$, $Q\simeq 2a_{\mathrm{pl}}{k}^{-2/3}<a_{\mathrm{pl}}$ bounds the separation below by $a_{\mathrm{pl}}-Q$, and supplies a geometric factor at fixed $k$. The boxed count is the usual orbital-scale, maximally coherent estimate; the assumptions are essential.

The central equation must be read as $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$; its printed left-hand expression lacks $=0$. Initially $\mu$ is constant. Dotting this equation with the velocity givesTaking its cross product with $\mathbf{r}$ gives $d(\mathbf{r}\times \mathbf{v})/dt=0$. In the fixed orbital plane the magnitude is $h=r^2\dot{\theta }$. These are conservation of specific orbital energy and specific angular momentum.

Let $f=\theta -\varpi$. Differentiating the polar equation of the Kepler orbit, with its elements initially constant, givesSubstitution into the specific orbital energy yields

For isotropic stellar mass loss with no recoil or drag, the force is still central, so $\mathbf{h}$ remains exactly constant, although $C$ does not. NowThe same algebraic relation between $C,h,e$ holds for the instantaneous osculating orbital elements. Differentiating it and using $h^2/(\mu r)=1+e\cos f$ givesand, for $e>0$,A particularly useful nonsingular description uses the eccentricity vectorThe vector equation remains meaningful at a circular orbit, where a longitude of pericentre is undefined. Its components along and perpendicular to $\mathbf{e}$ give the scalar formula above and

The osculating pericentre is $q=h^2/[\mu (1+e)]$. ThereforeFor mass loss, $\dot{\mu }<0$, this is nonnegative: pericentre does not decrease. It is positive except at pericentre itself, where $\cos f=1$ and $\dot{q}=0$. Thus a strictly opposite sign at every instant is not literally true; a pericentre passage with nonzero mass-loss rate is a counterexample to the strict wording. The intended monotonicity follows exactly, without requiring slow mass loss.

For adiabatic orbital expansion under isotropic mass loss, average over the unperturbed Kepler orbit and hold $\dot{\mu }/\mu$ constant to leading order during that orbit. If $E$ is the eccentric anomaly and $n=\sqrt{\mu /a^3}$, thenIt follows immediately thatAveraging with uniform true anomaly instead of uniform time would give the wrong result. Since $\mu a(1-e^2)=h^2$, conserved $h$ and constant secular orbital eccentricity giveThe pericentre and apocentre expand in the same secular proportion. These are leading adiabatic invariants, rather than exact invariants for arbitrary time-dependent $\mu$.

For the instantaneous precession fraction write $\theta =f+\varpi$ and $\dot{\theta }=h/r^2$. The adiabatic apsidal condition is controlled byIt is a signed fraction: apsidal advance contributes positively on the outward leg during mass loss, and negatively on the inward leg. Let ${\tau }_{\mu }=|\mu /\dot{\mu }|$. Ordinary orbit averaging requires ${\tau }_{\mu }\gg P$, together with negligible variation of the mass-loss law over one orbit. If the pericentre direction is also to vary negligibly relative to the instantaneous azimuthal motion, the maximum absolute fraction must be small. A precise condition for $0<e<1$ isThe maximizing cosine is $c=(1-\sqrt{1+8e^2})/(2e)$: differentiating the last factor gives $c+2e-e{c}^2=0$. Thus one can evaluate $F(e)$ explicitly by putting $\sqrt{1-c^2}/(1+ec{)}^2$ into the formula. Its useful limits areFor orbital eccentricity of order unity, mass loss much slower than an orbital period suffices; near a circular orbit, keeping the pericentre direction slowly varying additionally requires $n{\tau }_{\mu }e\gg 1$. The latter divergence is a singularity of the pericentre coordinate, not a divergence of the eccentricity vector dynamics. Indeed, at frozen $\dot{\mu }/\mu$, integrating the scalar evolution to first order gives $\delta e=-(\dot{\mu }/\mu )(1-e^2)\sin E/n$ up to an integration constant. Its absolute amplitude stays small for slow mass loss even when the scalar longitude of pericentre becomes unsuitable.

Use rotating barycentric coordinates $\mathbf{R}=(x,y,z)$, rotating velocity $\mathbf{w}$, and $\mathbf{\Omega }=\hat{\mathbf{z}}$. Multiplying the three equations by the corresponding velocity components and adding cancels the Coriolis force terms:Thus the Jacobi constant is $C=2U-|\mathbf{w}{|}^2$. The inertial barycentric velocity is $\mathbf{V}=\mathbf{w}+\mathbf{\Omega }\times \mathbf{R}$. If $H_z=(\mathbf{R}\times \mathbf{V}{)}_z$ and $E_b=|\mathbf{V}{|}^2/2-{\mu }_1/r_1-{\mu }_2/r_2$, thenThe inertial energy $E_b$ and inertial angular momentum $H_z$ need not separately be constant. Their combination is constant because the two gravitational sources rotate steadily at unit angular speed.

For the Tisserand parameter take ${\mu }_2\ll {\mu }_1\simeq 1$, and compare the osculating orbital elements well outside a close encounter, where ${\mu }_2/r_2$ and barycentre-to-primary corrections are negligible. ThenThis requires a circular secondary orbit, the restricted three-body problem approximation, and no dissipative force or other perturber. $T$ is conserved to leading order between encounter episodes; the osculating $T$ during a close encounter need not be constant. The exact invariant is $C$.

For clarity the exact two-body energy relation is $a={\mu }_1/(2{\mu }_1/r_1-v_1^2)$, rather than the printed formula without the numerator ${\mu }_1$. At the leading order ${\mu }_1=1$ used for $T$, the two agree; retaining finite ${\mu }_1$ in just part of that formula is inconsistent.

At a close encounter put the secondary radius and speed equal to one at leading order, and denote the particle's incoming relative speed outside the secondary's strong-deflection region by $u$. Conservation of relative speed in the short gravitational encounter givesAn incoming orbit with $a=1$ has $|{\mathbf{v}}_1|=1$ at $r_1=1$ by the vis-viva equation; escape there requires an outgoing speed exceeding $\sqrt{2}$. The triangle inequality therefore gives the necessary single-encounter escape velocity boundEquality only gives a marginal parabolic limit. This argument deliberately gives a weak necessary bound, not a sufficient scattering criterion.

The planet-encounter relative velocity is related to the Tisserand parameter byHere $v_{1,\mathrm{tangential}}=h_z$ at the encounter radius, and the formula also applies to inclined passages at the planet's orbital plane. For the intended prograde coplanar incoming orbit, $I=0$ and $a=1$, so $u^2=2-2\sqrt{1-e^2}$. Applying the preceding necessary bound givesCoplanar alone also allows $I=\pi$. For that retrograde case $u^2=2+2\sqrt{1-e^2}$, and even a circular incoming orbit can be scattered onto an escaping orbit; the orbital eccentricity restriction therefore presumes prograde motion. Also, retaining the outgoing relative-velocity geometry gives the sharper necessary condition $|{\mathbf{v}}_{1,\mathrm{out}}|\le 1+u$, hence $u>\sqrt{2}-1$ and $e>\sqrt{4\sqrt{2}-5}/2$ for the same prograde incoming orbit. This stronger bound is consistent with, and implies, the weaker bound above.

For the Tisserand periapsis bound, first writeA bound orbit capable of another close encounter must intersect the secondary's circular radius, so $q\le 1\le Q$. For fixed $q<1$, the largest possible $T$ occurs at $I=0$. The remaining expression decreases with $Q\ge 1$, sinceThus every such orbit obeysOn $0<q<1$, $g(q)$ increases monotonically from $2$ to $3$. For $2<T<3$ the least accessible pericentre is the unique root of $T=g(q)$: the limiting orbit is prograde and coplanar, with apocentre just at the planet. Introduce $s=\sqrt{3-T}$. Its tangential speed at this apocentre is $1-s$, and its vis-viva equation gives the particularly well-behaved answerEquivalently, setting $w=\sqrt{2q/(1+q)}$ gives $T=2-w^2+2w$ and $w=1-s$, with $q=w^2/(2-w^2)$. Rationalizing the previous result yieldsThe apparent singularity at $T=\sqrt{8}$ is removable; the first form gives $q_{\min }=(\sqrt{2}-1)/2$ there. The limits are $q_{\min }\to 0$ as $T\downarrow 2$ and $q_{\min }\to 1$ as $T\uparrow 3$. The same lower envelope also excludes a smaller pericentre on an escaping trajectory when $2<T<3$. To see this without introducing an apocentre, use the planet-frame velocity sphere: $u=\sqrt{3-T}<1$ forces the tangential component at an encounter to be at least $w=1-u>0$, so total specific angular momentum obeys $h\ge w$. At pericentre the stellar energy obeys $h^2=2q+2E{q}^2$, while $E=h_z-T/2\le h-T/2$. ThusIf $q<q_{\min }=w^2/(2-w^2)$, then $q^2<w\le h$, and the left side is increasing with $h$. Hence $w^2\le 2q+(2w-T)q^2$. But $2w-T=w^2-2$, and the right side is smaller than $w^2$ for $0\le q<q_{\min }$, a contradiction. This establishes the barrier for all Kepler trajectories that reach the encounter radius, including unbound ones.

For $0<T\le 2$, bound encounter-crossing orbits can approach $q=0$ (take $Q\to 2/T$), so the positive root extrapolated from the displayed expression is not a universal barrier. For $T>3$ there is no trajectory reaching a close encounter in this approximation, since $u^2=3-T$ would be negative. Multiple encounters can approach the derived lower envelope by changing energy, angular momentum and inclination while preserving the Tisserand parameter; the invariant alone does not guarantee a particular encounter history reaches it.

At each fixed semimajor axis, let $B=b_{3/2}^2(\alpha )/b_{3/2}^1(\alpha )$ and $Z_f=B{e}_{\mathrm{pl}}$. The planet's longitude of pericentre is the zero of longitude. The secular perturbation equation becomes $\dot{z}=iA(z-Z_f)$. Integrating it with the initial condition $z(0)=0$ givesThe forced eccentricity vector is the constant real $Z_f$; the free eccentricity vector is $-Z_f{e}^{iAt}$. Consequently the complex eccentricity follows a circle in the Argand plane centered at $Z_f$, with radius $Z_f$, starting at the origin and moving counterclockwise. Its initial velocity is downward, $\dot{z}(0)=-iA{Z}_f$. The secular period is $2\pi /A$, andThe longitude of pericentre is undefined at the origin, but the complex eccentricity passes smoothly through it. At $At=\pi$, $z=2Z_f$ and the orbit is maximally eccentric and aligned with the planet.

Using the small-$\alpha$ Laplace coefficient expansion givesAlso $n=n_{\mathrm{pl}}{\alpha }^{-3/2}$ to leading order in the planet-to-star mass ratio. The secular forcing by a distant outer planet is thereforeA relative $O({\alpha }^2)$ correction to $A$ can accumulate a phase error at very long times; the leading-frequency formula is not a uniform-in-time expansion.

If $a_2=a_1(1+\delta )$ with $\delta =\delta a/a_1\ll 1$, then $Z_{f,2}/Z_{f,1}=1+\delta +O({\delta }^2)$ and $A_2/A_1=1+3\delta /2+O({\delta }^2)$. The outer planetesimal therefore has a slightly larger forced eccentricity circle and a shorter secular period. Its accumulated secular phase lead is $\Delta (At)\simeq 3A_{1}t\delta /2$. Initially both circles start at the origin, but their eccentricity vectors progressively lose alignment. Linearizing the exponential in $\delta$ additionally requires $|A_{1}t\delta |\ll 1$.

The orbit crossing is caused by this accumulating differential secular phase, even though individual orbital eccentricities remain small. To first order in orbital eccentricity, the radius of an orbit at fixed inertial longitude $\theta$ isFor an infinitesimally neighboring orbit,This derivative is evaluated within the first-order orbital-shape model; derivatives of the neglected quadratic terms give relative $O(Z_f)$ corrections near crossing. The least separation over longitude is consequently $\delta a[1-|z+a{\partial }_{a}z|]$. The differential secular orbit-crossing criterion is $|z+a{\partial }_{a}z|=1$. Since $Z_f\propto a$ and $A\propto a^{3/2}$,The first term stays bounded by $4Z_f\ll 1$, whereas the second grows in proportion to time. To leading order in $Z_f$, crossing begins when $(3/2)Z_{f}At=1$. ThusMore precisely, in this first-order orbital-shape approximation the exact first crossing lies within a fractional $O(Z_f)$ window of this value, since the magnitude differs from $(3/2)Z_{f}At$ by at most $4Z_f$. Higher-order Laplace coefficients give additional corrections. The local derivation takes the neighboring-orbit limit before the long-time limit; for a finite separation it requires $\delta a/a_1\ll Z_f$ near crossing. A broad continuous disk contains such neighboring orbits. The crossings first appear in directions selected by the phase of $z+a{\partial }_{a}z$, not simultaneously at every longitude.

For the collision energy estimate adopt the phase-averaged mean orbital eccentricity $\overline{e}=4Z_f/\pi \simeq 5\alpha e_{\mathrm{pl}}/\pi$. The specified velocity estimate givesUsing the projectile kinetic energy per target mass convention for specific impact energy, equal-size and equal-density bodies have equal masses and $Q=\frac{1}{2}{v}_{\mathrm{col}}^2$. A catastrophic planetesimal collision requires $Q\gtrsim Q_D^{\ast }$, soThe numerical coefficient is only illustrative because the collision speed was specified only to order of magnitude. If the disruption threshold is instead defined using the centre of mass kinetic energy per combined mass, equal masses give $Q_R=v_{\mathrm{col}}^2/8$ and the coefficient becomes $8{\pi }^2/25$ for a threshold expressed in that convention. The robust result is a lower bound of order $Q_D^{\ast }{a}_{\mathrm{pl}}/(G{M}_{\star }{e}_{\mathrm{pl}}^2)$.

Within the inner-disk approximation, catastrophic collisions require sufficiently large $a_1/a_{\mathrm{pl}}$, not arbitrarily small radii. Although the Keplerian speed rises inward, the secularly induced orbital eccentricity falls faster: $v_{\mathrm{col}}^2\propto \alpha$. The lower bound must be much smaller than one to leave a domain compatible with $\alpha \ll 1$. A small planet orbital eccentricity, a large disruption threshold, or a distant planet can eliminate that domain. The planet mass controls the time to crossing but cancels from the eventual forced eccentricity amplitude and this collision-energy criterion. The threshold alone does not ensure that crossing has occurred within the disk's available lifetime.

Take $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $r=|\mathbf{r}|$ and $M=M_1+M_2$. Newton's law of universal gravitation gives ${\ddot{\mathbf{x}}}_1=-G{M}_2\mathbf{r}/r^3$ and ${\ddot{\mathbf{x}}}_2=G{M}_1\mathbf{r}/r^3$. Subtracting yields the relative two-body equationThe reduced mass is $\mu =M_1{M}_2/M$. The orbital energy and angular momentum in the center of mass frame areDifferentiation givesThus both are conserved. The eccentricity vector isFor $\mathbf{h}=\mathbf{J}/\mu$, one has $\ddot{\mathbf{r}}\times \mathbf{h}/(GM)=\dot{\mathbf{r}}/r-\mathbf{r}(\mathbf{r}\cdot \dot{\mathbf{r}})/r^3=d(\mathbf{r}/r)/dt$, so $\dot{\mathbf{e}}=0$. In particular the orbital eccentricity is constant, withFor a bound Kepler orbit, $E=-GM\mu /(2a)$ and $J^2=GM{\mu }^{2}a(1-e^2)$.

For the far-field multipole expansion, set the center of mass to zero. Then ${\mathbf{x}}_1=(M_2/M)\mathbf{r}$ and ${\mathbf{x}}_2=-(M_1/M)\mathbf{r}$. At $x=|\mathbf{x}|\gg r$, expand each point-mass Newtonian gravitational potential:Repeated indices are summed. The dipole vanishes since ${\sum }_A{M}_A{\mathbf{x}}_A=0$, and ${\sum }_A{M}_A{x}_{Ai}{x}_{Aj}=\mu r_i{r}_j$. The second-order gravitational quadrupole potential of a point-mass binary is thereforeBoth tensors specified in the question are trace-free. Direct contraction givesso to the requested orderFor equal masses the cubic multipole also vanishes, but it need not vanish for unequal masses.

To evaluate the quadrupole formula, use $k=GM$, $\mathbf{v}=\dot{\mathbf{r}}$, $\mathbf{n}=\mathbf{r}/r$, $u=\dot{r}=\mathbf{n}\cdot \mathbf{v}$, and $\mathbf{w}=\mathbf{v}-u\mathbf{n}$. The acceleration and its derivative areThe third derivative of $r_i{r}_j$ is ${\dot{b}}_i{r}_j+r_i{\dot{b}}_j+3b_i{v}_j+3v_i{b}_j$. Its trace is $d^3(r^2)/d{t}^3=-2ku/r^2$. Substituting in $q_{ij}=\mu (3r_i{r}_j-r^2{\delta }_{ij})/2$ givesBecause $\mathbf{n}\cdot \mathbf{w}=0$, the cross tensor contraction is zero. The squared norms of the two bracketed pieces are $6u^2$ and $72w^2$, respectively. Thusand the instantaneous energy loss isThe parenthesis is $w^2+u^2/12\ge 0$, as required for positive radiated power. This is the instantaneous quadrupole luminosity of a Kepler binary with loss sign for the orbital energy.

A slowly evolving initially circular binary is expected to follow a quasi-circular inspiral at leading radiation order. One needs angular-momentum loss as well as energy loss for this conclusion: the rotating quadrupole has frequency $2\Omega$ and azimuthal harmonic number two, so its fluxes satisfy the gravitational-wave energy and angular-momentum balance $\dot{E}=\Omega \dot{J}$. Along the circular family,The flux relation is precisely the tangent condition for remaining on that family; equivalently it keeps $e^2=1+2E{J}^2/(G^2{M}^2{\mu }^3)$ equal to zero at leading secular order. The shrinking orbit is not an exact fixed-radius circle: its small radial drift is neglected at this adiabatic order.

For a circular orbit of instantaneous separation $a$, $u=0$ and $v^2=GM/a$. ThereforeDifferentiating $E=-GM\mu /(2a)$ at fixed masses now givesWriting $C=64G^3{M}^2\mu /(5c^5)$, the circular gravitational-wave inspiral obeys $a^4=a_0^4-4Ct$. The formal point-mass coalescence time is $a_0^4/(4C)$, and $\dot{P}/P=(3/2)\dot{a}/a$ gives a decreasing orbital period and increasing gravitational-wave frequency.

Gravitational radiation tightens a detached close binary and becomes more effective at small separation. It can bring a stellar donor into Roche-lobe overflow or drive compact objects toward merger. Finite radii, tides and mass transfer can then alter the evolution; the fixed-mass inspiral law is not the entire evolution law once these matter. Near merger the weak-field, slow-motion quadrupole formula also ceases to be sufficient.

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.