The region a planet may keep a moon in

Nothing in the sky is a moon because it goes round a planet. Every planet is itself going round the Sun, and a body loosely accompanying one at a great distance is not held by it at all — it is on a solar orbit that happens, for the moment, to pass nearby. Being a moon is a statement about a region rather than about a motion.

The region has an edge and the edge is a cube root. Outside it the Sun’s pull, differenced across the planet’s neighbourhood, exceeds what the planet can supply; inside it the planet keeps what it has. The formula fits on one line, and all the interesting content is in the two things it does not say: whether the edge is where anything actually stops, and whether the answer depends on which way the satellite travels.

It is not, and it does. The figure puts the formal boundary and the observed one on one pair of axes, and the second lies between a half and seven tenths of the way in.

One distance, two questions #

The Hill radius is no new piece of mechanics. It is, to within the factor the cube root discards, the distance from the planet to $L_1$ — the innermost of the five points at which a small body keeps station with two larger ones. That essay solves the quintic; this one does not re-solve it, and the hero figure draws both curves so that the coincidence is visible rather than asserted.

What differs is the question. Asked as an equilibrium, the collinear point is where the net rotating-frame force vanishes and the answer is a set of coordinates. Asked as a boundary, the same distance is where a satellite stops being retained and the answer is a region with an inside. The two are the same number and not the same claim: one is where a body may sit still, the other where a body may not leave.

Where the three in the cube root comes from #

The derivation is four lines and belongs in full, because the factor of 3 is the part copied without being understood.

Work in the frame rotating with the planet, on the line joining planet and Sun, at $r \ll a$ from the planet. Three accelerations act on a test satellite there. The planet pulls it inward at $Gm/r^2$. The Sun’s field, differenced against its value at the planet, pulls it outward at $2GMr/a^3$ — the same difference that raises two tides rather than one. And the frame’s rotation adds $\omega^2 r$ outward, with $\omega^2 = G(M+m)/a^3 \approx GM/a^3$ by Kepler’s third law. Setting the inward against the sum of the outward,

$$\frac{Gm}{r^2} = \frac{2GMr}{a^3} + \frac{GMr}{a^3} = \frac{3GMr}{a^3},$$

$$r^3 = \frac{m\,a^3}{3M}, \qquad R_H = a\left(\frac{m}{3M}\right)^{1/3} = a\left(\frac{\mu}{3}\right)^{1/3}.$$

Two of the three come from the tide and one from the rotation, so a reader who has met only the tide places the boundary 14 per cent too far out. The mass enters at the one-third power, which is why the boundary is so insensitive: the Earth and Neptune are a factor of 17 apart in mass and two and a half in $\mu^{1/3}$.

The Hill fraction is a period in disguise #

A second reading of the same fraction turns a distance nobody can perceive into a duration everybody can.

A satellite at radius $r = f R_H$ has mean motion $n_{\text{sat}}^2 = Gm/r^3$. Substituting $r^3 = f^3 a^3 m/3M$ cancels the planet’s mass entirely and leaves

$$\frac{T_{\text{sat}}}{T_{\text{orb}}} = \frac{f^{3/2}}{\sqrt{3}},$$

a relation with no masses in it at all: the Hill fraction and the period ratio are the same number twice.

Put the hero figure’s four fractions through it. The Moon at 0.257 $R_H$ should take 0.0752 of a year, which is 27.5 days; the sidereal month is 27.32. Sinope at 0.451 should take 0.175 of Jupiter’s 11.86-year orbit, or 758 days, which is Sinope’s period. Phoebe at 0.198 gives 547 days against Saturn’s 29.46 years, and Neso at 0.424 gives 0.159 of Neptune’s 164.8 — 26 years, the longest satellite period known, from a Hill fraction and nothing else.

The agreement is not a coincidence but a warning. At the stability limits the period ratio is 0.20 prograde and 0.34 retrograde, so an outer satellite makes three to five orbits in one of its planet’s years. There is no separation of timescales out there, and nothing at that edge is a small perturbation on a two-body orbit.

Allowed is not the same as occupied #

The Hill radius is a boundary in the strong sense: the zero-velocity surface of the restricted problem closes around a planet below a critical value of the Jacobi constant, and a body inside a closed surface cannot cross it whatever it does. It forbids escape; it does not promise a persisting orbit. A trajectory may stay inside for ever while striking the planet or being flung about by encounters that keep it bounded without keeping it in orbit.

The gap is not small: the two limits enclose between an eighth and a third of the volume the boundary permits, and the rest is empty. Every statement in this essay about where satellites actually stop comes from numerical integration and from nothing derivable above. Both are the outcome of following orbits inside a Hill sphere for thousands of planetary years and recording which survive. They have no closed form, they are quoted at one significant figure because that is how well they are known, and no rearrangement of the algebra above produces them.

Four outer edges, and every one of them the wrong way round #

The outer satellites of the giant planets are irregular: large orbits, high inclinations, often high eccentricities, and no plausible history of forming where they are. They are captured bodies. And the furthest out of them, at every planet that has any, goes round the wrong way. Sinope at Jupiter, Phoebe at Saturn and Neso at Neptune are all retrograde; the one prograde satellite among the hero figure’s four is the Moon, which was not captured. No equilibrium argument predicts this, and the equilibrium picture has almost nothing to say about it. The effective potential of the restricted problem is a function of position. It contains no velocity, so it cannot distinguish a satellite going one way from the same satellite going the other, and the Hill radius derived above is identical for both. The observed edge is not: 0.451 for the outermost retrograde satellite against 0.257 for the prograde one, a ratio of 1.8 where the two integrated limits stand at only 1.4 — so the sense of the motion is doing more here than moving a ceiling.

Why a backwards orbit survives further out #

The mechanism is the one that rescues the triangular Lagrange points, and that connection is what this essay exists to make.

$L_4$ and $L_5$ are maxima of the effective potential, so a position-only argument declares them unstable, and they are stable anyway for a lopsided enough pair. What saves them is the Coriolis force, which depends on velocity and cannot appear in a potential at all. Stability decided by a term no potential can hold is the normal case here rather than an exotic one, and the retrograde satellites are that effect measured on real objects instead of drawn on a contour map. The specific route is a rate. A satellite at mean motion $n$ meets the solar perturbation at frequency $n - n_{\text{orb}}$ if prograde and $n + n_{\text{orb}}$ if retrograde. The retrograde case is the faster, so the perturbation averages away more completely over an orbit and the resonances that break prograde orbits fall further in. A difference of two frequencies against a sum of them also decides where a resonance clears a gap and where it locks a moon, and here it puts the retrograde edge at 0.7 rather than 0.5.

What is actually measured, and what it took #

Four quantities go into a Hill fraction, with four different provenances.

The satellite’s semi-major axis comes from astrometry: repeated images giving two angles apiece, fitted over an arc long enough to constrain an orbit, and for the outer irregulars that arc is measured in years. Phoebe was found on photographic plates in 1899 — the first satellite discovered that way — and its retrograde motion took years more plates to establish, because at 547 days per revolution a season’s images barely bend.

The planet’s mass comes from spacecraft tracking, as a Doppler shift on a radio carrier, and what tracking measures is $GM$ rather than $M$. That matters less than it looks: the boundary depends only on $\mu = m/(M+m)$, and a ratio of masses needs no value of $G$ whatever — fortunate, since $G$ is the worst-determined constant in the argument while the mass ratios are known to nine figures. The planet’s semi-major axis enters linearly and is the least troublesome number here. The stability limits come from none of these; they are integrations.

The inner edge of the same region #

A region bounded on one side is half an argument. The other edge is also a cube root of a ratio of densities and lies orders of magnitude nearer the planet: inside the distance at which the tide across a satellite beats the satellite’s own surface gravity a body held together by weight alone comes apart. A moon’s permissible home is therefore an annulus — outside the Roche limit, inside about half the Hill radius — and both edges are tidal, neither an equilibrium. The same differenced field that removes a satellite at large radius tears it apart at small radius, and the two limits are the two ways one mechanism ends a moon.

What the picture cannot show #

One satellite per planet is a survey result, not a dynamical one. The hero figure marks the outermost known satellite of four planets. Phoebe held that position at Saturn for a century only because it was bright enough for a plate in 1899; fainter retrograde irregulars found since sit further out, so 0.198 is a lower bound on that system’s edge rather than a measurement of it. The ordering survives — every later discovery has been retrograde too — but the fractions will move.

Nothing here is three-dimensional, and the Moon is the wrong kind of object. Every curve lies in the orbital plane, and the irregular satellites are precisely the population that does not: inclinations beyond 150 degrees are routine, the limit depends strongly on inclination, and the two numbers drawn average over a range no planar figure can represent. The one prograde point, meanwhile, sits at 0.257 $R_H$ because of a giant impact rather than a capture, so a fair test of that limit needs a captured prograde irregular instead.

Eccentricity is missing twice over. The planet’s own modulates its Hill radius through the year, and the satellite’s carries it towards the boundary at apoapsis, so a satellite whose semi-major axis is inside 0.5 $R_H$ can spend part of every orbit outside it. Semi-major axis is what the figure plots, and it drifts: the Moon’s distance is measured to be increasing by centimetres a year, so its Hill fraction is rising.

The same cube root around a cluster and around a hot Jupiter #

The construction transfers wherever a small mass holds something against a large one, and two destinations are hardly recognisable as the same problem.

A globular cluster orbiting a galaxy has a tidal radius given by the same cube root of a mass ratio, measured the same way — by where the counted star density stops falling smoothly and simply ends. Neither body is a point mass, so the coefficient changes; the $\mu^{1/3}$ does not.

The second is a prediction of absence. The Hill radius is proportional to $a$, so a planet moved inward loses its region in proportion: a hot Jupiter at 0.05 AU has a twentieth of the same planet’s region at 1 AU, and half of that, outside the Roche limit, leaves very little. A planet that migrated to where one cannot form arrives with nowhere to keep a moon whatever it set out with, and no exomoon has been confirmed at short period — though detection limits remain the better explanation.

There is a third destination for the same cube root, and it is the one that decides how large the planet itself could become. A growing body sweeps up whatever crosses its orbit within a few Hill radii — outside that, the solar tide pulls a passing planetesimal away before the encounter can complete. So the mass available to a planet is the surface density of the disc times the area of an annulus a few Hill radii wide, and since the Hill radius itself depends on the mass being accumulated, the two close into an equation whose solution is the isolation mass: the largest body that region can make without something bringing it more material. The same factor of three appears in it, and the answer at one astronomical unit is a fraction of an Earth mass — which is why the terrestrial planets require a stage of mutual collisions after the isolation masses have formed, and why a feeding zone sets the spacing of the bodies that emerge from it.

One conflation is worth correcting on the way out. The Hill radius is not the sphere of influence a patched-conic trajectory is stitched at, which goes as $\mu^{2/5}$: that is an accuracy criterion rather than a statement about escape, and a spacecraft is routinely outside it while inside this one.

Looking for one somewhere else #

The region has been mapped for four planets by finding what is in it. For a planet around another star the region is computable and its contents are not, and the search is one of the harder measurements anyone has attempted.

Three signatures are available, and all are second-order effects on a transit.

The moon’s own transit. A satellite crossing the star produces its own dip, smaller than the planet’s by the square of their radius ratio and arriving at a different time in each transit because the moon is at a different point of its orbit. Averaging many transits therefore smears it out rather than building it up, which removes the usual defence against noise.

Transit timing variations. The planet and moon orbit their common barycentre, and what follows a Keplerian path around the star is that barycentre rather than the planet. So the planet transits early or late by an amount proportional to the moon’s mass and its distance, up to minutes for a large satellite.

Transit duration variations. The same wobble gives the planet a velocity component along its orbit that varies from transit to transit, so the crossing takes slightly longer or shorter. The two effects are ninety degrees out of phase, and detecting both is what distinguishes a moon from a second planet.

The region this essay defines is what makes the search finite. A satellite must lie outside the Roche limit and inside about half a Hill radius, and for a planet close enough to its star to transit repeatedly that annulus is narrow — which bounds the period of any moon, and therefore how many transits are needed before the pattern repeats.

The candidates so far are contested. The best known involves a Neptune-sized object around a Jupiter-sized planet on a long-period orbit, where the Hill radius is generous and only a handful of transits exist; reanalyses disagree about whether the signal survives the detrending. What the searches have established firmly is an absence at short period, which is what the shrinking of the region predicts, and which is also what the sensitivity limits predict — so the two explanations are not yet separated.

A confirmed exomoon would also do something the solar system cannot: measure the region’s outer limit around a planet whose formation history is known independently, since a captured satellite and a co-formed one sit in different parts of the annulus.

The Hill radius is one distance and the five Lagrange points are five places, and it is worth seeing them drawn for a real system rather than for the exaggerated mass ratios the geometry needs to be legible.

That last fact is the one worth carrying, because it is the opposite of the Hill radius’s behaviour. The collinear points are at a distance that shrinks as the cube root of the mass fraction, so a lighter planet keeps a smaller region; the triangular points do not move at all. They are at the vertices of an equilateral triangle for any mass ratio whatever, and their stability — not their position — is what depends on the masses.

The stability condition is a single inequality: the triangular points are stable when the mass fraction is below about 0.0385, which every planet in the solar system satisfies by a wide margin. Above that the two points still exist and are no longer stable, so a binary star of comparable masses has an L4 and an L5 that hold nothing.

That is why the Trojans exist in the numbers they do and why they are found around planets rather than around stars in equal-mass binaries. The region a planet may keep a moon in shrinks with the planet’s mass; the region it may keep a Trojan in does not, and the two are answers to different questions that happen to be asked of the same three-body problem. One is about a region in which a body’s motion is dominated by the planet rather than by the star; the other is about a point at which a body’s motion is dominated by neither, and is held in place by the combination. The cube root appears in the first and not in the second, and the reason is that only the first involves a competition between two attractions of very different sizes.

Where the ladder goes next #

Later rungs: the Hill radius derived from the Jacobi constant rather than from a force balance along a line, which makes it a theorem rather than an estimate. The inclination dependence of the stability limit, the largest thing this rung leaves out. Capture — how a body on a solar orbit gets inside a Hill sphere at all, given that a purely gravitational encounter cannot leave it. And the Hill radius in planet formation, where it sets an embryo’s feeding zone and so the spacing of planets.

Phoebe’s retrograde motion was reported in 1905 as a curiosity about one satellite. It is now the first entry in a pattern that holds at every giant planet, and that pattern is the only evidence anybody has that the region a planet may keep a moon in has a shape the equations of equilibrium cannot see.