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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.

First adopt the uniform-angular-motion approximation implicit in the requested formula. Set the line of sight to longitude zero, the $b$ exoplanet transit to $t=0$, and the next $c$ exoplanet transit to $t=\tau$, with $0\le \tau <P_c$. Write $n_b=2\pi /P_b$, $n_c=2\pi /P_c$, with $P_c>P_b$. The next conjunction satisfiesThe common longitude is $n_{b}t$ modulo $2\pi$. Subtracting one $2\pi$ gives the conjunction longitude from a transit time lagSubstitution of the period ratio yieldsIt is exact in this uniform-angle model; no first-order expansion in $\delta$ is necessary here. For other choices of which exoplanet transit is used, choose the appropriate whole-turn branch.

For the finite-eccentricity orbit specified earlier, this is an approximation. Let $f_L$ and $M_L$ be the true anomaly and mean anomaly at the line of sight. The eccentric planet has $M_c(t)=M_L+n_c(t-\tau )$ and true angular advance $n_c(t-\tau )+{\Delta }_e(t)$, whereTrue conjunction therefore requires $(n_b-n_c)t+n_c\tau -{\Delta }_e(t)=2\pi$. The PDF omits this term, which is generally $O(e_c)$.

For a concrete counterexample take $p=q=1$, $\delta =0$, $e_c=1/5$, $\tau =P_b/4$, and a line of sight along $c$'s periapsis. The uniform model predicts $t=7P_b/4$ and longitude $3\pi /2$. At that time $M_c=3\pi /2$, but Kepler's equation gives $f_c\ne 3\pi /2$, so the planets are not truly aligned. Accurate finite-eccentricity conjunctions must be found from the corrected equation.

A simultaneous transit in a resonant chain requires both conjunction patterns to admit the observer's longitude at the same time. In the leading mean-conjunction geometry put $x=\Lambda -{\varpi }_c$. The necessary phase compatibility at symmetric centres isLet $d=\gcd (q,n)$. Eliminating $x$, and using that $q\ell -nk$ ranges over multiples of $d$, gives the exact modular compatibility criterionHere $\gcd$ denotes the greatest common divisor. For the orders $q,n\in \{1,2,3\}$ above, the two phase sets intersect only whenThe common directions are the two quadratures relative to periapsis. In reduced resonances this requires $p$ and $m$ odd; otherwise the apparent second-order ratios reduce to different resonance orders. There is no additional restriction on those odd values from angular compatibility alone. A shared direction and suitable temporal phase can be chosen, and the rational period ratios allow its recurrence.

This is a potential low-amplitude configuration, not a guarantee of an observed triple exoplanet transit. The observer must lie in the common orbital plane and near an allowed direction. Finite stellar radii, nonzero libration amplitudes of a resonant argument, apsidal motion and asymmetric centres broaden or change the possibilities. At finite $e_c$, true simultaneous alignment obeys the more precise equations $(p+q)y-px={\varphi }_{bc}$ and $(m+n)x-my={\varphi }_{cd}$, with $y=M_c(x)$. Thus the order-only conclusion is not a universal necessary condition for every eccentric resonant chain.

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.

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.

If the proper eccentricity is negligible and the semi-major axis ranges over $a\pm \Delta a$, particles lie on nearly aligned forced Kepler orbits. For a locally constant forced vector, these are geometrically similar ellipses, represented to first order by circles of radius $a^{\prime }$ centered at $-a^{\prime }{\mathbf{e}}_f$.

The boundaries of the aligned eccentric ring are thereforeIt is narrower at forced periapsis and wider at forced apoapsis, unlike the constant-width proper-eccentricity annulus. Its centers shift slightly between the two edges, as shown in the right panel of the figure. The condition $e_p/e_f\ll \Delta a/a$ makes the proper radial excursion negligible compared with the variation in forced center location.

The forced vector need not actually be constant across the interval. More generallyFor aligned apsides this becomes $2\Delta a[1-(e_f+a{e}_f^{\prime })\cos \theta ]$, provided the orbits remain nested. The radial gradient of forced eccentricity matters for the density comparison even in a narrow ring.

Measure angles from periapsis in the direction of motion. The true anomaly $f$ is the angle at the stellar focus between the pericentre ray and the star-particle radius vector. Draw the auxiliary circle of radius $a$ around the ellipse's center; the point on that circle with the same projected major-axis coordinate as the particle defines the eccentric anomaly $E$ at the center.

The mean anomaly $M=n(t-\tau )$ is a uniformly advancing time phase, with $M=0$ at pericentre passage $t=\tau$. It is the angle swept by a fictitious uniformly moving circular phase marker, not generally the actual position angle. The mean motion is the angular frequencyThe second formula is Kepler's equation, which converts the uniform time phase into geometric position.

For the $5:3$ exterior mean-motion resonance, $p=3$, $q=2$. The truncated dissipative resonant eccentricity equilibrium givesFor weak drag and $C>0$, take the resonant-argument libration centre ${\varphi }_0\simeq \pi$. The pericentre geometry of an exterior resonance then places the three pericentres at angles $\pi /3$, $\pi$, and $5\pi /3$ relative to the planet. The trajectory closes after three particle orbits and five planet orbits. It moves retrograde in the rotating frame on average but develops a prograde loop near each periapsis, because the periapsis angular speed exceeds the planet's angular speed.

For a finite drag offset ${\varphi }_0=\pi +\arcsin S$, rotate the whole particle pattern through $\arcsin S/3$. The sketch uses the eccentricity requested by the second-order model; $e\simeq 0.555$ is not small, so the curve is an illustration of that approximation rather than a quantitatively accurate integration of the full dissipative resonance.

The specific angular momentum is $h=\sqrt{G{M}_{\star }a(1-e^2)}$, so the periapsis angular speed isAt the exterior mean-motion resonance, neglecting radiation pressure, $n/n_p=p/(p+q)$. Thus at the limiting orbital eccentricity $e=\sqrt{2q/(5p-q)}$,The condition for overtaking the planet in angle isFor the formal second-order equilibrium, it holds for every physical bound value $0<q<5p/3$. To see this, rewrite $q/p=5e^2/(2+e^2)$, so $p/(p+q)=(2+e^2)/(2+6e^2)$. For $0<e<1$, $\sqrt{1+e}/(1-e{)}^{3/2}>1+2e$, whileThe instantaneous angular motion is therefore prograde relative to the planet near periapsis, although its mean relative motion is retrograde. The range $q\ll p$ is the regime where the eccentricity expansion is most reliable.

At periapsis, the particle's mean longitude equals its longitude of periapsis: $\lambda =\varpi$. Hence the pericentre geometry of an exterior resonance givesThe resonant argument therefore fixes the angular separation from the planet whenever the particle reaches periapsis. With $r_{\mathrm{peri}}=a(1-e)$ and planetary radius $a_p$, the actual separation isThus a restricted libration phase can prevent close encounters even when the radial ranges of the two orbits overlap: this is resonance protection.

At the periapsis $r_1$ and apoapsis $r_2$, the given Kepler orbit satisfiesHenceTherefore

The limiting trajectory grazes the planet at periapsis $R_p$. Equating the asymptotic and periapsis values of specific angular momentum and using the vis-viva equation givesHence the comet avoids the planet when

For some orbital phase of the moon to permit an impact, the comet's planetocentric hyperbolic Kepler orbit must at least cross the moon's circular orbit. At the limiting case $a_m$ is the comet's periapsis, so the radial velocity in part (iv) vanishes. ConsequentlyThis is gravitational focusing by the planet; an actual collision additionally requires the moon to occupy the crossing point at the right time.

At periapsis, $r=a(1-e)=a\delta$. The grain's energy from part (ii) is nonnegative whensoThe strict inequality gives a hyperbolic Kepler orbit, while equality gives a parabolic Kepler orbit. The printed eccentricity formula yields $\beta >\delta /(2+\delta )$, which has the same requested lowest-order limit $\beta >\delta /2$.

At periapsis the release velocity is tangential. A grain with $\beta =1$ feels no net stellar inverse-square force and therefore moves on the tangent line. A grain with $\beta <1$ retains an inward acceleration and bends toward the star; a grain with $\beta >1$ feels a net outward acceleration and bends away from it. At the common observation time, joining these positions in increasing $\beta$ gives the synchrone shown below. It starts near the low-$\beta$ orbital trajectory, passes through the force-free $\beta =1$ position, and extends outward through the repelled high-$\beta$ grains.

The sudden potential change leaves the star's position and velocity unchanged. Its new energy is thereforeThus $E_1>0$ precisely when $r<a$. Parametrize the orbit by eccentric anomaly $\eta$:A phase-mixed orbit is uniform in the mean anomaly ${\theta }_r$. For $e>0$, the condition $r<a$ is $-\pi /2<\eta <\pi /2$ modulo one period. The corresponding mean-anomaly interval has lengthHenceThe energy increase $G(M_0-M_1)/r$ is largest near periapsis, so stars are more readily unbound there than near apoapsis. A perfectly circular orbit is the measure-zero marginal case $E_1=0$ at every phase.

At periapsis, $\lambda =\varpi$. If the planet has longitude ${\lambda }_0$ there, thenso ${\varphi }_1/p$ is the angular displacement of periapsis from the planet, modulo $2\pi /p$. At an astronomical conjunction, $\lambda ={\lambda }_1={\lambda }_c$, andso ${\varphi }_1/q$ is the conjunction longitude measured from periapsis, with the $q$ possible branches differing by $2\pi /q$.

The planet's mean motion is $n_1=(p+q)n/p$. Hence the synodic period, or mean interval between conjunctions, iswhere $T=2\pi /n$ is the planetesimal's orbital period.