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Interpret projection from very far away as the asymptotic incoming state at $r\to \infty$. For $v>0$ and impact parameter $b>0$, conservation of energy and conservation of angular momentum give $E=v^2/2$ and $|h|=bv$. At the closest approach the radial velocity vanishes. With $k=-\kappa <0$, the central-force radial turning point equation isOnly the positive root is physical. Put $q=\kappa /v^2$. The pericentre distance and the purely tangential speed there areThus $p<b$ and $u>v$: attraction bends the path inward and increases the speed. With positive energy, the Kepler orbit is a hyperbola. Measured from the pericentre direction, its polar coordinates satisfy $r=h^2/[\kappa (1+e\cos \varphi )]$ with $e=\sqrt{1+b^2{v}^4/{\kappa }^2}>1$.

The hyperbola shown has a nonzero impact parameter. For $b=0$, the attractive radial orbit instead reaches the singular origin; there is no regular turning point with a finite closest-approach speed.

Treat $a,e$ as instantaneous osculating orbital elements of the fixed-central-parameter Kepler orbit. The perturbing work changes the specific orbital energy according toThe specific angular momentum and conic equation giveSubstituting proves the Gauss planetary equation for the semi-major axis:Here $f$ denotes true anomaly, not the fractional luminosity used in Question 1. The equation follows from work alone; a normal perturbing acceleration has no instantaneous contribution because the velocity lies in the orbital plane. For slowly averaged evolution, the perturbation must be weak over one orbital period.

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”.

For a spherical planet of mass density ${\rho }_p$, $R_p=(3M_p/4\pi {\rho }_p{)}^{1/3}$. The point-comet surface escape velocity and circular Kepler orbit speed therefore giveRaise this equation to the power $3/2$ and solve for $M_p$:Thus the constant of proportionality is $\boxed{\sqrt{3/(32\pi )}}$, and the gravitational constant cancels. Dimensional checking gives mass on the right: $M_{\star }^{3/2}{a}_p^{-3/2}{\rho }_p^{-1/2}$ has that dimension.

This uses the planet's own escape velocity. If instead $\chi$ is defined at finite-comet contact, an additional factor $(1+R/R_p{)}^{3/2}$ appears on the right, and $R_p$ itself depends on $M_p$; the quoted pure-planet scaling is then only the $R/R_p\to 0$ limit.

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.

To evaluate the quadrupole formula without assuming a circular orbit, put $k=GM$, $u=\dot{r}=\mathbf{n}\cdot \mathbf{v}$ and ${\mathbf{v}}_{\perp }=\mathbf{v}-u\mathbf{n}$. Differentiating the gravitational acceleration gives the kinematic jerkFor $S_{ij}=r_i{r}_j$, the product rule then givesInserting these into the printed mass quadrupole moment convention yieldsThe two tensors in brackets are orthogonal in their tensor contraction, because $\mathbf{n}\cdot {\mathbf{v}}_{\perp }=0$. Their squared norms are $6$ and $2v_{\perp }^2$, respectively. HenceThe instantaneous quadrupole luminosity of a Kepler binary is thereforeThe radial derivative $\dot{r}$ is not the vector speed. The bracket equals $v_{\perp }^2+u^2/12$, so the radiated luminosity is nonnegative for every instantaneous velocity.

For a slowly evolving circular orbit, there is no preferred orbital phase at which to excite a persistent eccentricity vector. More precisely, the rotating mass quadrupole moment emits at twice the orbital frequency, with angular harmonic number two; its gravitational-wave energy and angular-momentum balance is $\dot{E}=\Omega \dot{J}$. The circular Kepler orbit sequence has $J=\mu \sqrt{GMa}$ and $dE/dJ=\Omega$, so this loss is tangent to the sequence. At leading adiabatic order it preserves zero secular orbital eccentricity. This is a circular gravitational-wave inspiral, with a small radial drift rather than an exactly fixed-radius Newtonian circle. The energy flux alone would not establish circularity without this symmetry and angular momentum balance.

To leading radiation order use $r=a$, $u=0$ and ${\mathbf{v}}^2=GM/a$ in the loss formula. Combining it with $E=-GM\mu /(2a)$ givesKeeping the fixed masses and leading quadrupole formula, integration givesThis formal coalescence time displays the strong $a_0^4$ dependence: gravitational-wave emission from a binary system matters far more for close binary stars than for wide ones. The orbital period decreases as $a^{3/2}$ and the gravitational-wave frequency increases, producing a chirp. A detached binary can be driven into Roche-lobe overflow; further evolution then depends on mass-transfer stability. A sufficiently close compact pair may merge. The weak-field, slow-motion, point-mass approximation ceases to apply before literal $a=0$; finite stellar radii or strong relativistic effects determine the final interaction.

The radial velocity of a Kepler orbit follows from the vis-viva equation after subtracting the tangential component:Each radial interval is crossed twice per orbital period, so its fraction of time isThis is the phase-mixed radial probability of a Kepler orbit, normalized to one between $q$ and $Q$. Dividing by the annular area $2\pi rdr$ gives the phase-mixed comet surface density per comet:For $Q\gg a_p>q$,The printed density has the powers of $Q$ and $a_p$ interchanged. Its expression is too large by $Q/a_p$ in this limit. The normalized residence-time derivation fixes the corrected expression, which is also required for the later $Q^{1/2}$ ejection-time scaling. For a population of $N_c$ independent identical comets, multiply ${\Sigma }_c$ by $N_c$ to obtain a number surface density of a disk.

For an unperturbed Kepler orbit, successive revolutions advance the planet-relative orientation by $-n_p{T}_c$ modulo $2\pi$, where $T_c$ is the comet's orbital period. An irrational period ratio makes these orientations dense; sufficiently distant mean-motion resonance and adequate observation time justify the corresponding phase mixing approximation.

At each radius between its pericentre distance and apocentre distance, the comet eventually visits every azimuth in the rotating reference frame. The spatial projection is thereforeIt is an annulus, with a nonuniform radial residence probability, rather than a uniformly filled area. Strictly, the full position-velocity phase space is not this annulus: at each radius the energy and angular momentum constrain the velocity, with inward and outward branches. The annulus is its position-space projection. This description assumes the orbital elements have not yet been substantially changed by planetary scattering.

Suppose the inclined population samples its arguments of periapsis and nodal orientations, or that sufficiently slow precession supplies this sampling. Near the planet the vertical amplitude is $H\simeq a_{p}I$. For uniformly distributed vertical phase, $z=H\sin \psi$ has probability density function $[\pi \sqrt{H^2-z^2}{]}^{-1}$, so the midplane volume probability is ${\Sigma }_c/(\pi H)$.

In the genuinely three-dimensional regime $R_p\ll H\ll a_p$, the geometric collision cross-section is $\pi R_p^2$. The relative velocity differs from the coplanar value only at order $I^2$ for the specified high-speed encounters. Therefore the inclined comet collision time isAt fixed ${\rho }_p,M_{\star },I$ the leading scaling is $Q^{3/2}{M}_p^{-2/3}{a}_p^2$, with no leading $q$ dependence when $q\ll a_p$. The displayed factor retains the first dependence on $q/a_p$. Also $t_{\mathrm{col},\mathrm{I}}/t_{\mathrm{col}}\simeq 2a_{p}I/R_p$ in this averaging convention.

Inclination alone does not specify a collision time for one fixed orbit. If its orbital nodes miss the planet's orbit, a tilted Kepler orbit can have no collisions at all. The finite result is an orientation-averaged population result, or requires precession or scattering to sample those orientations. When $a_{p}I\lesssim R_p$, the three-dimensional estimate crosses over to the coplanar result instead of tending to zero.

For a circular Kepler orbit, $e=0$ remains an exact solution of the averaged Poynting–Robertson drag equations, and $\dot{a}=-2A/a$. Integrating gives $a^2(t)=a_0^2-4At$, so the point-star inspiral time under Poynting–Robertson drag isFor a star of radius $R_{\star }$, contact occurs instead at $(a_0^2-R_{\star }^2)/(4A)$ within this model. Sublimation or other forces can remove a real grain earlier.

For one isolated perturbing planet, its Kepler orbit has fixed longitude of periapsis, so $z_1$ is constant and its secular forcing frequency is zero. A constant forced eccentricity satisfies $0=iA{z}_f+i{A}_1{z}_1$. Since $A=-A_{1}f({\alpha }_1)$,This assumes $A\ne 0$. If an external process made the planet precess at frequency $g$, the particular solution would instead be $z_f=A_1{z}_1/(g-A)$; the stated formula relies on the isolated single-planet case.

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.

With the ellipse center as origin and its major axis horizontal, the auxiliary-circle construction gives particle coordinates $X=a\cos E$, $Y=a\sqrt{1-e^2}\sin E$, while the stellar focus is at $(ae,0)$. ThereforeSquaring and adding gives $r^2=a^2(1-e\cos E{)}^2$. The Kepler orbit radius and the anomaly conversion are consequentlyEquivalently,The sine and cosine formulas, or a quadrant-aware angle function, fix the correct branch when using the half-angle relation.

Write $ϵ={\mu }^{1/3}$ and ${\delta }_j=(a_j-a_0)/a_0$. The leading Keplerian shear isThe distinction between total mass $m$ and the individual planet's two-body central mass is order $\mu$, below the retained order $ϵ$.

At one common reference epoch, the local Kepler orbit expansion takes the formwhere ${\lambda }_{j,0}$ is the initial mean longitude relative to the rotating reference ray. Comparing its sine and cosine coefficients with the free solution of Hill equations givesThe required orbital elements from Hill coordinates are thereforeThese equalities have the first-order accuracy of the local approximation. If $e_j=0$, the longitude of periapsis is undefined. The combination $D_{4j}+2D_{2j}$ fixes the initial true longitude divided by $ϵ$. The special initial alignment imposed in part v would require $D_{4j}=-2D_{2j}$, but the general two-planet solution need not impose it.

Substitution also verifies ${\ddot{\xi }}_j-2n_0{\dot{\eta }}_j-3n_0^2{\xi }_j=0$ and ${\ddot{\eta }}_j+2n_0{\dot{\xi }}_j=0$: these are the unforced Hill equations. Close encounters add mutual-gravity forcing and can change the constants. Replacing $n_j$ by $n_0$ in the oscillation is consistent at leading order over local orbital times; accumulated phase differences must be retained when following much longer evolution.

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

Let $h=r^2\dot{f}$ be the specific angular momentum. The radial Kepler orbit equation has semi-latus rectum $p=h^2/\mu$, so comparison withgives $h^2=\mu a(1-e^2)$ for an elliptic orbit and $h^2=\mu a(e^2-1)$ for a hyperbolic Kepler orbit. At either apsis, $\dot{r}=0$ and $v=h/r$. Substituting the apsidal radius and angular momentum into part (i), or equivalently using the vis-viva equation, givesThus the signs are $C=\mp \mu /(2a)$ in the convention of the question.

A cometary dust tail is a superposition of synchrones and syndynes, rather than a material line emitted in one fixed direction. For grains larger than the stellar wavelength, $\beta \propto D^{-1}$: small grains have large $\beta$, are displaced rapidly in the anti-stellar direction, and make a relatively straight broad tail. Large grains have small $\beta$, remain closer to the comet's Kepler orbit, and make a more strongly curved dust trail. The observed position angle consequently depends on grain size, release time, orbital phase, and projection onto the plane of the sky.

Equating the satellite's mean density inside $r_t$ with the host's mean density inside orbital radius $R$ givesThus a measured tidal radius and independently estimated satellite mass infer $M(<R)\simeq m(R/r_t{)}^3$. Measurements from satellites over a range of $R$ trace the host's enclosed-mass profile and hence its gravitational potential. Globular clusters are more often tidally limited: their stellar extent can approach the Jacobi boundary, whereas dwarf galaxies commonly occupy extended dark-matter haloes and their observed stars may substantially underfill it. Orbital eccentricity, mass loss, and nonequilibrium structure complicate this inference.

Let the separation vector point from the host to the satellite. Subtracting the two Newton's second law equations givesThe relative orbit is therefore a one-body Kepler orbit with gravitational parameter $G(M+m)$. A circular orbit of separation $R_0$ hasThe centre-of-mass condition gives $x_m=M{R}_0/(M+m)$ for the satellite's distance from the barycentre.

Consider the inner collinear equilibrium a distance $r_J$ toward the host from the satellite. Taking the positive axis from the host toward the satellite, differentiation of the gravitational plus centrifugal effective potential givesSince $r_J\ll R_0$,The constant host-gravity term cancels ${\Omega }^2{x}_m=GM/R_0^2$. The remaining equation isFor a low-mass satellite, the term proportional to $m{r}_J$ is negligible compared with $3M{r}_J$, and the Jacobi tidal radius isThe outer collinear point gives the same leading result.

Stars that cross the two nearby Lagrange points are no longer bound to the satellite. Small energy and angular-momentum offsets place them on slightly different host orbits, producing one leading and one trailing tidal tail. Differential orbital frequency stretches these streams around the host, while epicyclic motion can create density clumps. The tail's position and velocity structure retain information about the satellite orbit and host potential.

During half a radial oscillation,Useand the two root productsThe stated elementary integrals then give the azimuthal advance over one radial period:Consequently the frequency ratio of the spherical isochrone model isor equivalently $T_{\varphi }/T_r=2/(1+L/\sqrt{L^2+4GMb})$.

Deep in the constant-density core, $L^2\ll 4GMb$ and ${\Omega }_{\varphi }/{\Omega }_r\to 1/2$, as for an isotropic harmonic oscillator: the radius completes two oscillations per revolution. Far outside the core, $L^2\gg 4GMb$ for the corresponding circular scale and the ratio tends to one, recovering the closed Kepler orbit. Intermediate orbits undergo apsidal precession because the ratio is generally not rational.