Codex @codex  0

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.

Initially the two inner orbital normals coincide, so their combined angular momentum is $L_{12}=L_1+L_2$ along one direction. Use the outer orbital normal as the vertical axis. The inner pair's vector then has perpendicular component $L_{12}\sin \Delta I$ and parallel component $L_{12}\cos \Delta I$. Adding the outer angular momentum givesThe invariable plane is perpendicular to this total vector. Thus the outer orbit's orbital inclination to that plane equals the angle between its own normal and the total normal. The inclination to the invariable plane of two orbital planes isThis uses the small, prograde inclinations for which the denominator is positive; in general use the corresponding two-argument arctangent to choose the quadrant. If $A\gg 1$, $I_3\simeq \Delta I/(A+1)$ for small $\Delta I$: the distant planet carries most of the angular momentum and lies close to the invariable plane.

The expansion separates naturally into three timescales:

Short-period terms matter for instantaneous osculating elements and encounters, resonant terms near period commensurabilities, and secular terms for long-term evolution away from strong resonance.

Just after $M_2$ crosses its ascending node, it lies above the nearly fixed orbital plane of $M_1$. At a nearby conjunction, $M_1$ is inward of $M_2$, because $r_1<r_2$. The perturbing acceleration toward $M_1$ therefore has an inward component and a downward normal component. In the view along the node toward the star, the downward component opposes $M_2$'s upward vertical velocity and reduces its orbital inclination; in the face-on view, the inward component perturbs its orbital energy and shifts the conjunction phase. A conjunction before the ascending node reverses the relation between the normal force and vertical velocity. Repeated phase-correlated impulses therefore provide the restoring dynamics of an inclination-type mean-motion resonance.

Put the star at the origin, take $\hat{\mathbf{z}}$ along the line of sight toward the observer, and let $\hat{\mathbf{x}}$ point North in the plane of the sky. The $\hat{\mathbf{y}}$ direction is then $90^{\circ }$ anticlockwise from North. Draw the ascending node on the sky plane at position angle $\Omega$, tilt the circular Kepler orbit through the orbital inclination $I$ about that nodal line, and place the planet an angle $f$ along the orbit from the ascending node. The direction of increasing $f$ must cross the sky plane toward $+\hat{\mathbf{z}}$ at $f=0$.

A convenient orthonormal basis in the orbital plane isHere ${\mathbf{e}}_1$ points along the nodal line and ${\mathbf{e}}_2$ fixes the required sense of motion. The planet lies at $\mathbf{r}=a(\cos f{\mathbf{e}}_1+\sin f{\mathbf{e}}_2)$; this relation also specifies all the labels required in the sketch.

Here $a_p$ is the planet's semi-major axis, while $a,e,I$ are the particle's semi-major axis, orbital eccentricity, and orbital inclination relative to the planet's plane. The formula assumes the circular restricted three-body problem: the planet-to-star mass ratio is small, the particle has negligible mass, and its motion is approximately heliocentric and Keplerian away from brief encounters. The planet's own orbit is circular.

The Jacobi constant is exactly conserved in that ideal rotating problem. Expressing it in heliocentric orbital elements away from the planet gives the approximately conserved Tisserand parameterThe first term measures the particle's normalized binding energy. The second is twice the component of its specific angular momentum normal to the planet's plane, normalized by $\sqrt{G{M}_{\ast }{a}_p}$.

For $e=0$, regardIts minimum occurs at $x=(\cos I{)}^{-2/3}$ and equals $3(\cos I{)}^{2/3}$. A circular orbit is therefore possible for $T_p=5/2$ only ifThe smallest required orbital inclination is consequently

A prograde, equatorial circular orbit with orbital period equal to the planet's spin period. In the ideal model its satellite remains above one equatorial ground location. Nonzero orbital inclination or orbital eccentricity destroys exact stationarity even when the periods agree.

Consider a circular circumplanetary orbit well inside the Hill sphere, with dust speed $u^2=G{M}_p/a$ and unit plane normal $\hat{\mathbf{k}}$. During one dust orbital period, take the stellar direction $\hat{\mathbf{R}}$ and flux at the planet's stellar radius $a_p$ as constant. The velocity-dependent Poynting–Robertson drag has relative workThe constant terms have zero work over the closed circular orbit. Since $⟨u_i{u}_j⟩=(u^2/2)({\delta }_{ij}-k_i{k}_j)$,The coefficient is three for coplanar stellar and dust orbits; it is two for a dust plane normal to the instantaneous stellar direction. Averaging over a circular planetary year at fixed dust orbital inclination $I$ gives $3-{\sin }^{2}I/2$. Conservative perturbations must remain small enough for this circular-orbit average.

A planet can block stellar radiation during part of a circumplanetary orbit. Then the full-orbit constant-flux averages for radiation pressure and Poynting–Robertson drag require modification. The detailed shadow duty cycle depends on the orbital inclination, stellar angular extent and dust radius.

Half the peak-to-peak line-of-sight orbital velocity range of a star in a single-companion Kepler orbit. It equals $K=2\pi a_{\star }\sin i/[P\sqrt{1-e^2}]$, where $a_{\star }$ is the stellar barycentric semi-major axis, $P$ the orbital period, $i$ the sky-plane orbital inclination, and $e$ the orbital eccentricity. The Doppler spectroscopy formula follows from Kepler's third law for the relative orbit, $a^3=G(M_{\star }+M_p)P^2/(4{\pi }^2)$.