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Validity and the complete angular equation. The Lagrange planetary equations here describe a planar, circular-planet, small-orbital eccentricity, weakly perturbed exterior mean-motion resonance. Require $M_{\mathrm{pl}}/M_{\star }\ll 1$, $e\ll 1$, one isolated slowly varying resonant argument, perturbation times long compared with an orbital period, and no close planetary encounters. Other resonant harmonics and short-period terms must be negligible; coefficients may be frozen only across a narrow range of semi-major axis. Take $p>0$, $q=1$ or $2$ in a reduced integer ratio. Exactly $e=0$ makes the longitude of periapsis undefined, so the displayed angular variables must then be replaced by nonsingular eccentricity vector components.

Neglecting the perturbation to the accumulated mean longitude gives $\dot{\lambda }\simeq n(a)$; this is an accumulated-phase approximation, not differentiation of a fictitious expression $n(t)t$ while discarding $t\dot{n}$. With $n^2{a}^3=G{M}_{\star }$,Differentiating, using $\dot{n}=-3n\dot{a}/(2a)$ and the supplied orbital element equations, gives the full result within this constant-coefficient model:An equivalent formula replaces the last two terms in brackets by $q^2{C}_r{e}^{q-2}[(p+q)n-p{n}_{\mathrm{pl}}-2q{C}_s]-2q^3{C}_r^2{e}^{2q-4}\cos \varphi$. No derivatives of $C_r,C_s$ are included because they were specified as constants.

A full fixed point requires $\sin \varphi =0$ and $\dot{\varphi }=0$. Hence ${\varphi }_{\ast }=0$ or $\pi$, with its corresponding detuned semi-major axis fixed byAt either point $\dot{a}=\dot{e}=0$. The longitude of periapsis still precesses, so the fixed point refers to the reduced resonant dynamics.

Linear stability analysis and encounter geometry. Linearizing the full equation about a fixed point givesFor $q=2$, $\pi$ is a center equilibrium and zero is a saddle equilibrium. For $q=1$, $\pi$ is always a center equilibrium, while zero is a saddle equilibrium ifAt sufficiently small orbital eccentricity the full truncated equations instead admit a center at zero as well; equality is a degenerate case requiring higher-order analysis. Thus an unconditional instability claim at zero does not follow from the full equation. In the usual fixed-eccentricity weak-resonance regime the inequality holds and the stable libration center is $\pi$ for either order.

The physical explanation is resonance protection. At equal mean longitudes, the conjunction direction relative to periapsis obeys $q(\Lambda -\varpi )=\varphi$ modulo $2\pi$. For $q=1$, $\varphi =0$ places conjunction near periapsis, where the exterior particle comes closest to the planet; $\varphi =\pi$ places it near apoapsis. For $q=2$, zero includes both apsidal conjunctions, one of them near periapsis, whereas $\pi$ places the two conjunction branches near quadrature. The protected arrangements give restoring kicks in the positive-$C_r$ leading-harmonic model. This geometric explanation is conditional on its non-crossing, small-orbital eccentricity approximation; it does not override the low-$e$ term retained in the full stability calculation.

Small resonant-argument librations. Define the nominal resonant semi-major axis and the displaced center byThis follows by expanding $n=n_r[1-3(a-a_r)/(2a_r)]$ and taking $\cos (\pi +\delta )=-1+O({\delta }^2)$. It gives the requested initial semi-major axis to first order in the precession-induced detuning; $e_0=e(0)$ and $\delta =\Delta \varphi$.

For the pendulum approximation of a mean-motion resonance, also require fractional changes in $e$ small across a libration, and precession terms small compared with $n$. In particular $C_r/(n_r{e}_0^3)\ll 1$ for $q=1$ makes the extra full-equation curvature negligible; for $q=2$, small fractional eccentricity changes require $C_r/(n_r{e}_0^2)\ll 1$. Freeze $e=e_0$ in the leading restoring coefficient and setThen $u=\varphi -\pi$ obeys simple harmonic motion. Using $u(0)=\delta$, $\dot{u}(0)=0$ and integrating the remaining Lagrange planetary equations givesThe omitted apsidal modulation comes from higher perturbative orders and $u^2$. All expressions are leading resonant approximations, valid while the omitted fractional changes remain small. The small-libration phase portrait iswith period $2\pi /{\omega }_0$. The semi-major axis half-width is $2a_r{\omega }_0|\delta |/(3p{n}_{\mathrm{pl}})$ and the orbital eccentricity half-width is $q{C}_r{e}_0^{q-1}|\delta |/{\omega }_0$.

Dividing the original $\dot{e}$ and $\dot{a}$ equations gives the exact invariant of their retained terms,Thus the $e$–$a$ plot is a segment of this increasing curve traversed back and forth, not a closed ellipse. Locally its slope is $q/[2(p+q)e_0{a}_c]$.

Finite amplitude and the separatrix. Keep the same weak-resonance approximation but allow $\delta$ to be finite. The leading angular equation is $\ddot{\varphi }={\omega }_0^2\sin \varphi$, or $\ddot{u}=-{\omega }_0^2\sin u$. Multiplication by $\dot{u}$ givesThis resonant pendulum energy is constant in the pendulum approximation of a mean-motion resonance. It is not an exact integral of the earlier full equation when $e$ and its restoring coefficient vary: the $\dot{\varphi }\sin \varphi$ term alone prevents that conclusion in general.

For $|\delta |<\pi$ the phase portrait has closed libration curves around $\pi$, withwhere $K$ is the complete elliptic integral of the first kind. The largest speed is $2{\omega }_0\sin (|\delta |/2)$ and the leading semi-major axis half-width is $4a_r{\omega }_0\sin (|\delta |/2)/(3p{n}_{\mathrm{pl}})$. As $|\delta |\to \pi$, the energy approaches $2{\omega }_0^2$, the separatrix through the saddles at zero and $2\pi$; the particle spends increasingly long intervals near those saddles. With $ϵ=\pi -|\delta |$,At exactly $\delta =\pi$, $\dot{\varphi }=0$, the particle stays at the unstable fixed point in the ideal model. Nonstationary separatrix trajectories approach the saddle only in infinite time; above separatrix energy, the resonant argument circulates and resonance protection is lost. These distinctions matter when describing the limiting motion.

Let $\mu =M_p/M_{\star }$, $q=\sqrt{1-\beta }$, and restrict first to $0\le \beta <1$. The radiation-pressure coefficient reduces the grain's central gravitational parameter to $G{M}_{\star }{q}^2$, whereas the planet's is approximately $G{M}_{\star }$. Their mean motions obey $n/n_p=q(1+ϵ{)}^{-3/2}$. Setting this ratio to $j/(j+1)$ gives the radiation-pressure shift of a mean-motion resonance:It lies interior to the planet exactly when $q(j+1)<j$, or $j>q/(1-q)$. Equality puts the frequency commensurability at the planet's radius. For $\beta =0$ every such commensurability is exterior.

There is a frame qualification in the printed encounter formula. The circular inertial speed of the grain at $a_p(1+{ϵ}_1)$ is $v_1=q{v}_p(1+{ϵ}_1{)}^{-1/2}$. Subtracting the planet's inertial velocity at conjunction gives the local planet-relative inertial speedThis is the velocity used in the question's two-body gravitational assist calculation. In a frame actually rotating about the star at $n_p$, the grain's tangential velocity is instead $v_1-n_p{a}_p(1+{ϵ}_1)$, with magnitude $v_p[1+{ϵ}_1-q(1-{ϵ}_1/2)]$. In particular its nonradiating shear is $3v_p{ϵ}_1/2$, rather than $v_p{ϵ}_1/2$. Thus the requested expression uses a local translating, inertial encounter convention despite the source's rotating-frame wording. The following derivation consistently uses that convention, and its coefficients are an encounter estimate rather than an exact rotating-frame result.

Take impact parameter $b\simeq a_p{ϵ}_1$ and a small gravitational scattering angle $\theta$. Expanding the supplied hyperbolic scattering relation gives $\theta \simeq 2G{M}_p/(b{v}_{\infty }^2)$. Initially the relative tangential velocity is $-v_{\infty }$; after the deflection it has tangential component $-v_{\infty }\cos \theta$. The relative speed is unchanged in the local hyperbolic Kepler orbit, but the stellar-frame kinetic energy changes. Taking the dot product with the planet's velocity givesThe specific orbital energy of the grain is $-G{M}_{\star }(1-\beta )/(2a)$. Comparing it immediately before and after the encounter at the same position givesConsequentlyThe encounter increases the grain's semi-major axis within this geometry. These steps assume ${ϵ}_1\ll 1$, $\theta \ll 1$, and an orbital kick small enough for the subsequent linearization.

Choose the initial conjunction longitude as zero. The synodic period is $2\pi /(n_p-n)$, and the planet's longitude accumulated by the next conjunction isThese are unwrapped longitudes, counting complete revolutions. The second uses the new mean orbital period while neglecting the encounter's immediate phase offset and the difference between true and mean conjunction. A comparison reduced modulo $2\pi$ could not implement the stated one-turn sensitivity test.

For $\beta =0$, $v_{\infty }\simeq v_p{ϵ}_1/2$, so $\delta ϵ\simeq 32{\mu }^2{ϵ}_1^{-5}$ and ${\lambda }_c\simeq 4\pi /(3ϵ)$. Conjunction-longitude sensitivity to an encounter givesRequiring this to exceed $2\pi$ gives ${ϵ}_1^7<64{\mu }^2/3$. At a high-index first-order mean-motion resonance, ${ϵ}_j=(1+1/j{)}^{2/3}-1\simeq 2/(3j)$, henceThe interpretation of the resonance-overlap encounter estimate is loss of coherent encounter phases: a small kick changes the next synodic encounter time by more than one planetary revolution. Resonant protection by repeatedly meeting at controlled orbital phases can then fail, allowing irregular kicks and diffusion. This sensitivity criterion motivates an unstable region; it is not a proof that every orbit or resonant phase in it must escape.

For a quantitative sketch with radiation, put $h=1-q(1-ϵ/2)$ and $d=1-q(1+ϵ{)}^{-3/2}$. Differentiating the conjunction formula and substituting the kick givesEvaluate this at the positive ${ϵ}_j$ and take $\mathcal{S}=1$ as the critical-index curve. At small $\beta$, increased encounter speed weakens planetary scattering, moving the threshold to larger $j$. As $\beta$ rises further, the radiation-pressure shift of a mean-motion resonance brings each commensurability closer to the planet, strengthening the close encounters. The curve turns over and approaches the exterior-limit boundary $j=q/(1-q)$. Larger planetary mass produces stronger kicks and a lower critical index. These competing effects are shown below; the curves retain the factors in $\mathcal{S}$, so their $\beta =0$ intercepts differ slightly from the leading high-$j$ formula. The boundary for disappearing exterior commensurabilities is geometric; divergence of the encounter approximation near $ϵ=0$ should not be read as a precise physical scattering rate.

Write $R_p$ for the physical planet radius and $R_c=\eta R_{H1}$ for the debris-cloud radius; the source uses $R_1$ for both even though their stated size orderings are opposite. A central transit duration of a small planet on a nearly circular Kepler orbit is $D_p\simeq 2R_{\star }/(n_1{a}_1)=P_1{R}_{\star }/(\pi a_1)$. Kepler's third law, with $M_{\star }=4\pi {\rho }_{\star }{R}_{\star }^3/3$, givesA noncentral planetary exoplanet transit is shorter, so within the small-radius and nearly circular approximations, a duration requiringis too long for the planet alone and favors extended debris. For a central cloud transit with $R_c\gg R_{\star }$, the path length is approximately $2R_c$. Using the Hill radius $R_{H1}=a_1(M_1/(3M_{\star }){)}^{1/3}$ gives $D_1/P_1\simeq R_c/(\pi a_1)$, hence the Hill-cloud transit mass estimateThis equality assumes a central cloud chord. An orbit described merely as nearly edge-on need not have central cloud crossings. If its projected cloud impact distance is $s$, the measured half-chord is $\sqrt{R_c^2-s^2}$, so the boxed mass is multiplied by $[1-(s/R_c{)}^2{]}^{-3/2}$. It is a lower estimate when that geometry is unknown.

For the dynamical calculation use ${\mu }_2=M_2/M_{\star }$, $A=G{M}_2/a_2$, and $\mathcal{R}=Af(\alpha )e_1\cos {\varphi }_1$. The derivatives in the supplied Lagrange planetary equations hold the other osculating orbital elements fixed:Since $A/(n_1{a}_1^2)=n_1{\mu }_2\alpha$, the physical mean-longitude equation for a first-order resonant term isThe source's epoch notation needs care when $n_1$ varies. Literally differentiating ${\lambda }_1=n_1(t)t+{ϵ}_1(t)$ adds $t{\dot{n}}_1$; one cannot retain that definition and also identify the printed ${\dot{ϵ}}_1$ with ${\dot{\lambda }}_1-n_1$. A consistent version writes ${\lambda }_1={\int }_0^t{n}_1(s)ds+{ϵ}_{\mathrm{int}}$ and assigns the displayed correction to ${\dot{ϵ}}_{\mathrm{int}}$. Equivalently, the canonical mean-longitude equation for a first-order resonant term follows from the disturbing function directly. The result for $g$ is the intended physical equation under that convention.

Away from resonant-argument libration, neglect changes in $e_1,\alpha ,{\varpi }_1$ on one slow cycle to this order. With $n_0=2\pi /P_1$ the reference mean motion, the signed slow frequency isThis is the near-resonant transit-timing superperiod. Choose ${\varphi }_1(0)=0$. Differentiating $n_1\propto a_1^{-3/2}$ with the semi-major axis equation givesLet $C=3j{\mu }_2{n}_0^2\alpha f{e}_1$ and $n_s=n_1(0)$. Direct integration giveswith zero initial mean longitude. The constant and linear terms are absorbed into the fitted transit ephemeris: the measured mean frequency is $\bar{n}=n_s-C/\omega$. This distinction avoids mistaking an arbitrary initial frequency for the long-term fitted period.

For $|\Delta |\ll 1$, the double-integrated frequency term dominates the direct $g$ term provided $g/f$ stays finite. Relative to the fitted linear ephemeris, ${\lambda }_1(t_k)=2\pi k$ givesHere $P_1$ denotes the fitted mean period at the retained order. The signed-frequency convention fixes the phase sign. Replacing $\omega$ by $2\pi /P_t>0$ changes the sign of the signed $A_t$ when needed; a nonnegative amplitude hasUsing a signed $P_t=2\pi /\omega$ instead gives the source's signed proportionality $A_t/P_t\simeq 3{\mu }_2\alpha f{e}_1/(2\pi \Delta )$. The leading implicit transit equation can equally use $P_{1}k$ inside the sine, since the difference is higher order in the perturbation. This is the eccentricity-enhanced near-resonant transit-timing variation; its derivation assumes small longitude oscillations, ${\mu }_2{e}_1\ll {\Delta }^2$ at fixed resonance coefficients, and the effectively fixed eccentricity used in this truncated model. Dominance over forced-eccentricity timing terms typically also requires $e_1\gg |\Delta |$ at fixed $j$. A full transit calculation also includes forced orbital eccentricity and the conversion from mean longitude to the actual sky-crossing angle. Those are lower order in this eccentricity-enhanced regime, rather than identically absent; broader near-resonant formulas are derived by Lithwick, Xie and Wu.

Inside a mean-motion resonance, ${\varphi }_1$ undergoes resonant-argument libration and its evolution must be solved together with $a_1,e_1,{\varpi }_1$. The constant circulation frequency $\omega$ and the associated ${\Delta }^{-2}$ expansion then fail. At approximately fixed eccentricity, the resonant resonant pendulum approximation has libration frequency of order $j{n}_0\sqrt{{\mu }_2\alpha |f|e_1}$, up to numerical factors. The transit-timing variation follows that libration period and depends on its libration amplitude of a resonant argument. Exactly at a resonant equilibrium there need be no libration signal; near a separatrix the period grows, and eccentricity dynamics can modify the simple estimate. Resonant amplitudes remain controlled by the bounded resonant motion rather than diverging as the circulating detuning tends to zero.

In the center of mass frame, the velocities are ${\mathbf{v}}_1=(M_2/M)\mathbf{v}$ and ${\mathbf{v}}_2=-(M_1/M)\mathbf{v}$. Hence the kinetic energy is $T=\frac{1}{2}\mu v^2$, while the Newtonian gravitational potential energy is $U=-G{M}_1{M}_2/r=-GM\mu /r$. The two-body orbital energy is thereforeThe relative equation of Newtonian gravity is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. It is a central force, so conservation of angular momentum confines the motion to a plane and preserves the specific angular momentum $h=|\mathbf{r}\times \mathbf{v}|=r^2\dot{\theta }$.

For completeness, derive the polar equation of a Kepler orbit. Let $w=1/r$ and use primes for $\theta$ derivatives. Then $\dot{r}=-h{w}^{\prime }$ and $\ddot{r}-r{\dot{\theta }}^2=-h^2{w}^2(w^{\prime \prime }+w)$. The radial equation gives the Binet equationChoose the angular origin at closest approach. Its solution is $w=(GM/h^2)(1+e\cos \theta )$, soFor the ellipse, the orbital eccentricity satisfies $0\le e<1$. Its extreme separations are $r_p=l/(1+e)$ and $r_a=l/(1-e)$. The semi-major axis is half their sum, giving $a=l/(1-e^2)$ and thus $l=a(1-e^2)$. The true anomaly $\theta$ runs through a full $2\pi$; the endpoints are the same position, and for a circular Kepler orbit the angular origin is arbitrary.

The radial and transverse relative speeds are $\dot{r}=(GM/h)e\sin \theta$ and $r\dot{\theta }=h/r$. Substituting these into the specific orbital energy givesTherefore the conserved orbital energy isThe negative specific orbital energy and fixed angular momentum characterize the bound Kepler orbit; neither should be assumed unchanged through a mass-ejecting explosion.

Initially the circular Kepler orbit has $r=a$ and $v^2=GM/a$. Treat the supernova kick in a binary star as impulsive: the relative position does not change, the companion's velocity is unchanged during the impulse, and the new neutron star receives $\mathbf{u}$ with $|\mathbf{u}|=\alpha v$. Thus ${\mathbf{v}}^{\prime }=\mathbf{v}+\mathbf{u}$, and with $\psi$ the angle between these vectors before the kick,Write $M^{\prime }=M_1^{\prime }+M_2$, $\beta =M^{\prime }/M$ and ${\mu }^{\prime }=M_1^{\prime }{M}_2/M^{\prime }$. The post-explosion specific orbital energy isUsing ${\epsilon }^{\prime }=-G{M}^{\prime }/(2a^{\prime })$ givesHere $a^{\prime }>0$ for a bound Kepler orbit. For a hyperbolic Kepler orbit, this formula uses the signed energy parameter $a^{\prime }<0$, rather than the positive geometric magnitude of the hyperbola's semi-major axis. At the parabolic Kepler orbit boundary, $1/a^{\prime }=0$.

The kick-direction binary survival criterion is bound if $S<2\beta$ and unbound with positive asymptotic speed if $S>2\beta$. To prove the printed sufficient disruption condition, choose a perpendicular kick, $\cos \psi =0$: then $S=1+{\alpha }^2$, so $M^{\prime }<\frac{1}{2}(1+{\alpha }^2)M$ produces positive specific orbital energy. This is sufficient, not the sharp existence threshold. Sincethe complete kick-direction binary survival criterion isThe second inequality follows by taking a kick directly opposite to $\mathbf{v}$. All directions remain bound if $2\beta >(1+\alpha {)}^2$, while every direction has positive escape energy if $2\beta <(1-\alpha {)}^2$. For $\alpha >0$, the marginal direction obeys $\cos \psi =(2\beta -1-{\alpha }^2)/(2\alpha )$ when the right-hand side lies in $[-1,1]$. Equalities give marginal parabolic Kepler orbits for the relevant extreme direction. With zero kick, losing more than half the original total mass unbinds the circular Kepler orbit.

For an escaping pair, the Newtonian gravitational potential energy approaches zero at infinite separation. Conservation of the post-explosion two-body orbital energy then gives $\frac{1}{2}{\mu }^{\prime }{V}^2={\mu }^{\prime }{\epsilon }^{\prime }$, or the asymptotic relative speed of a disrupted binaryThis is the relative recession speed, not the velocity of the new center of mass or either star's individual velocity in the original inertial frame. A marginal parabolic Kepler orbit separates with $V=0$ at infinity.

Use the separation vector $\mathbf{r}$ from the star to the comet, and let $m_{\mathrm{red}}=m{M}_{\star }/(m+M_{\star })$. The two-body problem separates into uniform centre of mass motion andTaking a cross product with $\mathbf{r}$ proves conservation of angular momentum; taking the scalar product with $\dot{\mathbf{r}}$ proves conservation of energy. The relative constants areThe total relative angular momentum and energy are $m_{\mathrm{red}}\mathbf{h}$ and $m_{\mathrm{red}}\epsilon$, respectively. The conserved eccentricity vector is $\mathbf{e}=\dot{\mathbf{r}}\times \mathbf{h}/\mu -\hat{\mathbf{r}}$. Its scalar product with $\hat{\mathbf{r}}$ gives $r=h^2/[\mu (1+e\cos f)]$, with $f$ the true anomaly. The periapsis and apoapsis distances are therefore $h^2/[\mu (1+e)]$ and $h^2/[\mu (1-e)]$. Their sum is twice the semi-major axis, soAt either apsis $v=h/r$; substituting an apsidal radius gives $\epsilon =-\mu /(2a)$. Thus the specific orbital energy implies the vis-viva equationThese are exact relative-coordinate formulas for an elliptic orbit, $0\le e<1$. In the barycentric frame the comet's speed is $M_{\star }/(M_{\star }+m)$ times the relative speed; the printed identification with the comet's stellar orbital speed uses $M_{\star }\gg m$.

In the weak, non-crossing, leading-eccentricity model the exterior resonant coefficient has positive sign: write its disturbing function as $R_{\mathrm{res}}=B_q{e}_c^q\cos {\varphi }_{bc}$, with $B_q>0$ for the orders considered. The physical feedback can be seen for first order. Before an outer conjunction near apoapsis, the faster inner planet pulls the outer one backwards; after passing it pulls forwards. On the outward side, the nearer, stronger earlier pull wins, reducing the outer semi-major axis and increasing its mean motion. Thus ${\varphi }_{bc}<\pi$ is driven upward. On the inward side, the later forward pull wins and drives ${\varphi }_{bc}>\pi$ downward. The paired impulses cancel at the symmetric centre.

For higher order, sum the feedback over all $q$ conjunction phases. The first surviving angular Fourier harmonic is $\sin {\varphi }_{bc}$; the lower Fourier harmonics cancel over the equally spaced directions. A local Hamiltonian reduction with canonical momentum $J$ hasThe negative curvature of the Kepler orbit Hamiltonian gives the displayed sign. Linear stability analysis at $\pi$ gives a restoring acceleration. The symmetric libration centres of an eccentricity resonance are thereforeThe corresponding conjunction phases relative to periapsis are $\{\pi \}$, $\{\pi /2,3\pi /2\}$ and $\{\pi /3,\pi ,5\pi /3\}$. They suppress repeatedly close passages and provide resonance protection.

These are expected centres of the specified leading-harmonic model, not centres fixed solely by the integer order for arbitrary orbital eccentricity and masses. Asymmetric resonant-argument libration can occur when higher Fourier harmonics matter. For example, if $R=A\cos \varphi +B\cos 2\varphi$, $A>0$, $B>A/4$, the symmetric $\pi$ equilibrium loses stability and stable centres satisfy $\cos \varphi =-A/(4B)$. Thus further dynamical information is needed beyond the PDF's unrestricted eccentric orbit to assert unique centres. Exterior asymmetric branches are documented by Winter and Murray, “Resonance and chaos. II. Exterior resonances and asymmetric libration”.

Now the eccentric planet $c$ is inside the circular perturber $d$. The interior eccentricity-type mean-motion resonance hasIn the mean-conjunction approximation, ${\varphi }_{cd}=n(\Lambda -{\varpi }_c)$, so ${\Lambda }_j={\varpi }_c+({\varphi }_{cd}+2\pi j)/n$. Successive conjunctions advance by $2\pi m/n$; coprime $m,n$ visit $n$ distinct phases.

The near-conjunction tangential pulls are reversed for an eccentric body inside its perturber. Just before conjunction the inner body is pulled forwards and afterwards backwards. Near periapsis, an inward-moving inner body experiences the stronger earlier forward pull; its semi-major axis grows, its mean motion falls, and ${\varphi }_{cd}$ increases because its coefficient of ${\lambda }_c$ is negative. On the outward side the feedback reverses, restoring the first-order centre zero. Higher-order feedback is strongest on the side closest to the external perturber, near apoapsis; the leading coefficient alternates in sign with order.

Writing $R_{\mathrm{res}}=B_n{e}_c^n\cos {\varphi }_{cd}$ gives $B_n<0$ for odd $n$ and $B_n>0$ for even $n$ in the leading non-crossing model. The same negative-curvature Hamiltonian argument therefore givesConjunction phases relative to periapsis are $\{0\}$, $\{\pi /2,3\pi /2\}$ and $\{0,2\pi /3,4\pi /3\}$, respectively, as shown in the lower row of the figure. As before, these are small-eccentricity symmetric predictions; additional resonant Fourier harmonics and coupled secular dynamics can change the stable branches.

For actual alignment of the eccentric body at true longitude $\Lambda$, the exact phase relation is instead ${\varphi }_{cd}=(m+n)x-my$, where $x=\Lambda -{\varpi }_c$ and $y=M_c(x)$. The simple equally spaced angles replace $y$ by $x$ at leading order.

Let $\chi =v_{\mathrm{esc}}/v_k$, with $v_k=\sqrt{G{M}_{\star }/a_p}$. The Safronov number is $\Theta ={\chi }^2/2$. In the instantaneous planet frame let the asymptotic vectors be ${\mathbf{u}}_{\mathrm{in}}$ and ${\mathbf{u}}_{\mathrm{out}}$; their magnitudes agree. In the stellar frame the corresponding specific orbital energy change during a short encounter isThe quadratic relative-speed terms cancel. A comet with semi-major axis comparable to $a_p$ has stellar binding energy per unit mass of order $v_k^2/2$. Consequently $\chi$ measures whether a single strong encounter can change a substantial fraction of that binding.

Large escape-to-orbital-speed ratios favour ejection; small ratios favour collision or accretion during repeated encounters. For $\chi \gg 1$, a suitably oriented gravitational assist can eject the comet before it strikes the planet. For $\chi \ll 1$, most individual kicks are too weak; many close passages may be needed, providing repeated chances of collision.

The crossover is of order unity, not a sharp theorem about each orbit. The initial binding energy, encounter speed and geometry matter: an already nearly parabolic comet may be ejected by a low-$\chi$ planet, while an unfavourably directed kick can make a comet more tightly bound even when $\chi$ is large.

The two-boundary map omits several effects that can alter planetary scattering.

The initial semi-major axis, orbital eccentricity, orbital inclination and orbital phase determine whether encounters occur and their relative velocities. A strongly bound comet needs more energy to escape; a nearly parabolic one needs less. The planetary radius and mass density, the finite comet radius, and gravitational focusing determine collision probabilities. Tidal disruption, atmospheric gas drag, sublimation and physical fragmentation can destroy a body before a nominal point-particle scattering sequence is completed.

Other planets can hand a comet from one scatterer to another, eject it, or lift its periapsis clear of the original scatterer's orbit. Mean-motion resonances and secular perturbations can protect objects from encounters or correlate kicks, contradicting the independent random walk assumption. Planetary migration changes the encounter geometry over time.

Stellar flybys and a galactic tide can change distant comet periapses, allowing new encounters or detaching an object from the planetary region. Stellar mass evolution changes both binding and planetary orbits. Finally, the age and the supply rate of new comets determine whether the observed population is a residual one or continuously replenished. Thus the mass-radius map is a useful conditional classification, not a complete survival law.

Write $x=1/a$ for the comet's inverse semi-major axis. Its specific orbital energy is $-G{M}_{\star }x/2$, so escape corresponds to $x=0$. The printed quantityhas dimensions of inverse length. It cannot, literally, be a standard variance-per-time diffusion coefficient. Interpret it as the characteristic root mean square step in $x$ per periapsis passage, as in a discrete comet energy diffusion model.

If successive kicks are unbiased and uncorrelated, after $N$ passages the root mean square displacement is $s_x\sqrt{N}$. Starting at $a_0$ gives $x_0=1/a_0$, hence $N_{\mathrm{ej}}\sim (x_0/s_x{)}^2$. Taking the characteristic interval to be the initial orbital period $P(a_0)$ givesHere $P_p=2\pi \sqrt{a_p^3/(G{M}_{\star })}$. In particular, for $a_0\sim a_p$ this is $P_p(M_{\star }/M_p{)}^2/100$. Without specifying the initial semi-major axis and the time per statistically independent encounter, the PDF cannot determine a unique time.

With the convention $⟨(\Delta x{)}^2⟩=2D_{x}t$, the actual diffusion coefficient would be $D_x=s_x^2/[2P(a_0)]$. The estimate requires $N_{\mathrm{ej}}\gg 1$, orbit crossing, and sufficient phase decorrelation. Near escape the orbital period grows, and resonant or secular correlations invalidate the simple constant-step clock. This is a characteristic diffusion scale, not an exact mean first-passage time: even an unbiased Brownian motion on an unbounded half-line has an infinite mean time to reach its absorbing endpoint.

Set $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $M=M_1+M_2$ and $\mu =M_1{M}_2/M$, the reduced mass. Newton's law of universal gravitation gives ${\ddot{\mathbf{x}}}_1=-G{M}_2\mathbf{r}/r^3$ and ${\ddot{\mathbf{x}}}_2=G{M}_1\mathbf{r}/r^3$. Subtraction establishesFor fixed masses, the orbital energy and angular momentum areTaking derivatives gives $\dot{E}=\mu \mathbf{v}\cdot (\ddot{\mathbf{r}}+GM\mathbf{r}/r^3)=0$ and $\dot{\mathbf{J}}=\mu \mathbf{r}\times \ddot{\mathbf{r}}=0$. These are conservation of energy and conservation of angular momentum in the unperturbed Kepler orbit. To see directly that the orbital eccentricity is fixed, put $\mathbf{h}=\mathbf{r}\times \mathbf{v}$ and $\mathbf{n}=\mathbf{r}/r$. The eccentricity vector isSince $\dot{\mathbf{h}}=0$, the cross product identity $\mathbf{r}\times (\mathbf{r}\times \mathbf{v})=\mathbf{r}(\mathbf{r}\cdot \mathbf{v})-r^2\mathbf{v}$ gives $\ddot{\mathbf{r}}\times \mathbf{h}=GM\dot{\mathbf{n}}$, so $\dot{\mathbf{e}}=0$. Its magnitude $e$ is the orbital eccentricity. Equivalently,Thus the semi-major axis and orbital eccentricity of a bound noncollision Kepler orbit are constant, including $e=0$.

Take the centre of mass as origin. The two positions are ${\mathbf{x}}_1=(M_2/M)\mathbf{r}$ and ${\mathbf{x}}_2=-(M_1/M)\mathbf{r}$. For $x\gg r$, a Taylor expansion givesThe mass dipole vanishes because ${\sum }_A{M}_A{\mathbf{x}}_A=0$, and the second mass moment tensor is ${\sum }_A{M}_A{x}_{Ai}{x}_{Aj}=\mu r_i{r}_j$. The gravitational quadrupole potential of a point-mass binary is consequentlyBoth printed tensors are trace-free. Contracting them with the Kronecker delta summed over three spatial dimensions givesThis proves the requested expression through second order. In particular the printed convention is $q_{ij}=\frac{3}{2}{Q}_{ij}$, where $Q_{ij}=\mu (r_i{r}_j-r^2{\delta }_{ij}/3)$ is the standard mass quadrupole moment; the radiation coefficient must be adjusted accordingly. No mass dipole term may be retained in centre of mass coordinates.