Codex @codex  0

For an opaque or detectably extinguishing spherical debris cloud of radius $R_c=\eta R_H$ much larger than the stellar radius, a central exoplanet transit lasts approximately $2R_c/(na)$. Combining this with the Hill radius gives the displayed mass estimate. For cloud impact distance $s$, replace $R_c$ by $\sqrt{R_c^2-s^2}$: the central formula then gives a lower mass estimate unless the chord geometry is known.

The collinear Lagrange point beyond the smaller primary, away from the larger one, is $L_2$. For a small mass ratio it lies approximately one Hill radius from the secondary and has a growing linear mode.

Let $M=M_1+M_2$ and $\alpha =M_2/M$, with $0<\alpha <1$. In the centre of mass frame the two primary distances are $r_1=\alpha R$ and $r_2=(1-\alpha )R$. Applying Newton's law of universal gravitation and circular centripetal acceleration to the first primary gives $M_1{\omega }^2\alpha R=G{M}_1{M}_2/R^2$, and henceThe same result follows from the second primary or from the relative two-body equation of motion.

A rotating-frame equilibrium has zero rotating-frame velocity, so its Coriolis force vanishes. Above the orbital plane both primaries have downward gravitational components; below it both have upward components. The centrifugal force has no $z$ component. More explicitly,which vanishes only at $z=0$ for finite nonsingular points. Thus all equilibria lie in the orbital plane.

The equation of motion in a rotating frame for the dwarf isOn the orbital plane the centrifugal contribution is $m{\omega }^2\mathbf{r}$. Thus the rotating-frame effective potential is $E=U-m{\omega }^2(x^2+y^2)/2$; equilibria are exactly its stationary points. The Coriolis force affects motion about those equilibria but not their positions. This is the circular restricted three-body problem, with the dwarf treated as a test mass.

The printed energy unit has the wrong dimensions. The correct total energy unit is $mGM/R$, or $GM/R$ for specific energy; $GM/R^2$ is an acceleration. Dividing $E$ by $mGM/R$ and measuring lengths in units of $R$ gives the dimensionless effective potentialThe primary coordinates are $(-\alpha ,0)$ and $(1-\alpha ,0)$. Restricting to $y=0$ gives $F(x)=\mathcal{E}(x,0)$ andThe absolute values are essential; dropping them would reverse the attraction on the left of a primary. On each of the intervals separated by the two primaries, $F^{\prime }$ decreases strictly from $+\infty$ to $-\infty$. There is exactly one Collinear Lagrange point on each interval, a maximum along the $x$ direction.

For the inner Lagrange point, write $x=1-\alpha -\delta$ with $\delta >0$. For the L2 Lagrange point, write $x=1-\alpha +\delta$. The respective equations areBalancing the leading terms gives $\delta =(\alpha /3{)}^{1/3}+O({\alpha }^{2/3})$. Therefore the small-mass-ratio collinear Lagrange points obeyThe omitted $-\alpha$ displacement of the secondary is smaller than the stated error. The leading distance from the secondary is the dimensionless Hill radius.

For the L3 Lagrange point put $x=-1+c\alpha +O({\alpha }^2)$. It lies to the left of both primaries, soExpanding the two positive inverse-square factors gives $(1-\alpha )/(x+\alpha {)}^2=1+(2c+1)\alpha +O({\alpha }^2)$ and $\alpha /(x-1+\alpha {)}^2=\alpha /4+O({\alpha }^2)$. Thus $F^{\prime }=-(3c+5/4)\alpha +O({\alpha }^2)$, soAll three have $y=z=0$. The small-$\alpha$ equalities in the paper are asymptotic expressions, not exact coordinates for finite mass ratio.

For the remaining Lagrange points, introduce plane polar coordinates about the first primary: $x=-\alpha +s\cos \theta$, $y=s\sin \theta$. Its distance to the second primary is $d=(s^2+1-2s\cos \theta {)}^{1/2}$, andThe angular stationary condition is ${\partial }_{\theta }\mathcal{E}=\alpha s\sin \theta (d^{-3}-1)=0$. For an off-axis point $s>0$ and $\sin \theta \ne 0$, so $d=1$. The radial stationary condition becomesHence $s=1$, and $d=1$ implies $\cos \theta =1/2$. The Triangular Lagrange points are exactlyBoth triangles are equilateral. The collinear and off-axis calculations together exhaust the stationary points.

For the requested sketch, $\mathcal{E}$ tends to $-\infty$ at either primary and at large in-plane radius. Each Collinear Lagrange point is a saddle of the planar effective potential: its $x$ curvature is negative and its $y$ curvature is positive. To see the latter, let $d_1=|x+\alpha |$, $d_2=|x-1+\alpha |$ and $D=(1-\alpha )/d_1^3+\alpha /d_2^3$. At an inner point both distances are below one, so $D>1$. At an exterior stationary point, use barycentric positions $a_1=-\alpha$, $a_2=1-\alpha$ and weights $1-\alpha ,\alpha$: the equation implies $x(D-1)=\sum w_i{a}_i/|x-a_i{|}^3$. To the right, the weighted sum is positive because the positive primary is closer; to the left it is negative because the negative primary is closer. Thus $D>1$ there as well, and ${\mathcal{E}}_{yy}=D-1>0$.

At either Triangular Lagrange point, the planar Hessian isIts trace is $-3$ and determinant $27\alpha (1-\alpha )/4>0$, so these are local maxima. Their height is $\mathcal{E}=-3/2+\alpha (1-\alpha )/2$. The sketch below uses a finite mass ratio to separate the five points clearly; its singular wells are clipped only for display. These curvature labels concern the effective potential, not a stability test that ignores the Coriolis force.

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.

The normalization of the circular restricted three-body problem implies ${\mu }_1+{\mu }_2=1$, by Kepler's third law. The two bodies have barycentric positions $(-{\mu }_2,0)$ and $({\mu }_1,0)$. ConsequentlyDifferentiate the effective potential directly:Since $x={\mu }_1(x+{\mu }_2)+{\mu }_2(x-{\mu }_1)$ and $y=({\mu }_1+{\mu }_2)y$, these becomeOn $y=0$, centrifugal acceleration is the straight line $a_{\mathrm{cf}}=x$. Each gravitational term points towards its source, diverges at that source, and tends to zero far away. Let $f(x)=U_x(x,0)$; away from the sources,Thus the net acceleration is strictly increasing on each of the three intervals separated by the bodies. On the left interval $f$ runs from $-\infty$ to $+\infty$; on the middle interval it runs from $-\infty$ just right of $M_1$ to $+\infty$ just left of $M_2$; on the right interval it runs from $-\infty$ to $+\infty$. The intermediate value theorem and strict monotonicity give exactly three collinear Lagrange points, $L_3,L_1,L_2$ in increasing $x$ order. The sketch plots gravity and centrifugal acceleration separately; their intersections with $a_{\mathrm{grav}}=-x$ locate the equilibria.

Write $x_{L2}={\mu }_1+\delta$, so $r_{2,L2}=\delta$ and $r_{1,L2}=1+\delta$. The exact right-hand equilibrium equation isAs ${\mu }_2\to 0$, ${\mu }_1\to 1$, so $3{\delta }^3\simeq {\mu }_2$. Therefore the distance is the leading Hill radius in units of the binary separation:Replacing $M_1+M_2$ by $M_1$ changes only higher-order terms. In dimensional units the right-hand side is multiplied by the binary separation.

Use the displacement vector $\mathbf{q}=(X,Y,\dot{X},\dot{Y}{)}^T$, where $X=x-x_{L2}$ and $Y=y$. The absolute coordinate $x$ cannot be the first entry of a homogeneous linear system about $L_2$; it must be translated. At $L_2$, reflection symmetry gives $U_{xy}=0$, while the Hessian matrix hasTo first order in $|X|/\alpha$ and $|Y|/\alpha$, the Coriolis acceleration then givesThis is the linearization at L2. The expression for $B$ is exact at the true equilibrium; in the small-secondary limit, ${\mu }_2/{\delta }^3\simeq 3$ and ${\mu }_1/(1+\delta {)}^3\simeq 1$, so $B\simeq 4$.

For an eigenmode proportional to $e^{\lambda t}$, eliminate the velocity coordinates. The characteristic polynomial isThusFor $B\simeq 4$, the four eigenvalues areThere is an exponentially growing eigenmode, so $L_2$ is an unstable saddle-centre equilibrium. A generic displacement has an unstable component whose distance grows as $e^{\sqrt{1+2\sqrt{7}}t}$. The L2 escape e-folding time isThis is a local exponential timescale, not the time to reach a prescribed distance from an unspecified initial offset. Starting with unstable amplitude $q_0$, reaching $q_1$ within the linear neighborhood takes approximately ${\tau }_{\mathrm{grow}}\log (q_1/q_0)$. Initial data exactly on the centre-stable subspace need not grow forward at this linear order; the existence of the growing eigenmode already proves instability.

Exclude collisions from the ejection count. The effective collision width is $2R_p$, while the favorable noncolliding ejection width is $(b_{\mathrm{ej}}-R_p{)}_{+}$. Define the first flyby-ejection threshold and the equal-rate crossover byThere are no direct noncolliding ejections in this weak-deflection model for $Q\le Q_{\mathrm{hit}}$. Above that threshold, comparison of the two rates gives collision is more likely for $Q<Q_{\mathrm{crit}}$, while ejection is more likely for $Q>Q_{\mathrm{crit}}$. In the deep-crossing limit $Q_{\mathrm{crit}}\simeq [9/(2\sqrt{2})](M_{\star }/M_p)R_p$, so its scaling at fixed density is $M_{\star }{M}_p^{-2/3}{\rho }_p^{-1/3}$. Order-one factors depend on encounter geometry; the robust condition compares $(M_p/M_{\star })Q$ with $R_p$.

Adding the mutually exclusive rare-event loss rates gives the collision-ejection lifetime of a comet:For $Q\le Q_{\mathrm{hit}}$ this is $t_{\mathrm{col}}\propto Q^{3/2}$. For $Q\gg Q_{\mathrm{hit}}$ it approaches $t_{\mathrm{ej}}\propto Q^{1/2}$. A logarithmic sketch therefore bends from slope $3/2$ to slope $1/2$, with no maximum. At the equal-rate crossover the lifetime is half either individual loss time.

These asymptotic branches use the corrected residence density from part iii and require $Q\gg a_p$ and a local encounter scale below the Hill radius. They cannot be extrapolated indefinitely to arbitrarily distant weakly bound orbits.

The two halo centres orbit their centre of mass with separation $R$. Their relative coordinate obeys $\ddot{\mathbf{R}}=-G(M_1+M_2)\mathbf{R}/R^3$, hence circular motion requiresReflection symmetry about the orbital plane makes the vertical force point back toward $z=0$, while the centrifugal force has no vertical component. An equilibrium away from that plane is therefore impossible.

In the uniformly rotating frame, an equilibrium is a stationary point of the gravitational plus centrifugal effective potentialSet $R=1$, divide by $G(M_1+M_2)/R$, and write $\alpha =M_2/(M_1+M_2)$. The primary and secondary lie at $x=-\alpha$ and $x=1-\alpha$, so on their line

For the two roots near the secondary, put $x=1-\alpha \mp d$ in $F^{\prime }(x)=0$. Dominant balance gives $3d\simeq \alpha /d^2$, so $d=(\alpha /3{)}^{1/3}$. Expanding the root beyond the primary directly in powers of $\alpha$ givesto the requested orders. The other equilibria are the two Triangular Lagrange points

The distance from $H_2$ to either nearby collinear point is the Hill radius. Since $\alpha \simeq M_2/M_1$,Inside this tidal radius, the subhalo's gravity dominates the host's differential gravitational field; outside it, material can escape through the neighborhoods of $L_1$ and $L_2$. For an extended spherical host, $M_1$ is replaced by enclosed mass and the coefficient becomes $3-d\log M_1/d\log R$, giving the Jacobi tidal radius. An extended subhalo requires the bound mass inside $r_t$ to be found self-consistently. On an eccentric orbit there is no time-independent rotating potential or exact tidal boundary; stripping is strongest near pericentre and the instantaneous radius varies around the orbit.

Because the dark component is more extended, tidal stripping first sends dark matter through both $L_1$ and $L_2$, producing leading and trailing dark-matter tidal tails. The compact stellar component is stripped more deeply and also produces a leading and a trailing stellar tail. The two constituents therefore give four tails distinguished by composition, with the dark tails broader and more extended.

For each point $(\Delta {\varphi }_1,e)$, sample $\delta \in [-\Delta {\varphi }_1,\Delta {\varphi }_1]$ and the three conjunction branchesFor every branch, solve Kepler's equation for the eccentric anomaly, evaluate $r=a(1-e\cos E)$, and minimize the planet-planetesimal separation over $\delta$ and $k$. Mark the point as encounter-capable when this minimum is below a chosen $R_{\mathrm{enc}}$, naturally the Hill radius for strong scattering. Repeating this calculation on a grid traces the boundary in the $\Delta {\varphi }_1$--$e$ plane.

Direct integrations of the circular restricted three-body problem can then refine the geometric map by allowing the resonant argument, eccentricity, and conjunction kicks to evolve self-consistently. The integrations distinguish merely orbit-crossing initial data from trajectories that actually enter the encounter region, and reveal chaotic layers near the boundary.

At conjunction, the planetesimal's mean anomaly isIf ${\varphi }_1=\pi +\delta$ with $|\delta |\le \Delta {\varphi }_1\le \pi$, the conjunction branch that can come closest to periapsis hasLet $E_{min}$ solve Kepler's equationAt that phase the orbital radius is $r_{min,c}=a(1-e\cos E_{min})$. A geometrical close encounter is possible only ifwhere $R_{\mathrm{enc}}$ may be chosen as the planet's Hill radius or another encounter distance. With a point planet, set $R_{\mathrm{enc}}=0$. This implicit inequality is the requested eccentricity constraint as a function of $\Delta {\varphi }_1$.

The weaker necessary condition that the orbits cross isIt becomes sufficient for phase access only as $\Delta {\varphi }_1\to \pi$, when a conjunction can approach periapsis. Smaller libration amplitude keeps conjunctions farther from periapsis and requires a larger eccentricity than this orbit-crossing bound.

In barycentric coordinates with unit primary separation, a small secondary mass fraction $\alpha$ gives distances $(\alpha /3{)}^{1/3}$ from that secondary to the two neighboring Lagrange points at leading order. This is the dimensionless Hill radius. The opposite exterior L3 Lagrange point has coordinate $-1-5\alpha /12+O({\alpha }^2)$.