Generator

The lagrange-points generator

The five Lagrange points at mass fraction 0.12

The five points at which a small body can keep station with two larger ones. The three on the line of centres are roots of a quintic and are unstable; the two forming equilateral triangles are stable for a sufficiently lopsided mass ratio.

6 essays call lagrange-points. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.

Where it is called

Every figure listed here is the same construction drawn at different numbers, so a correction to one is a correction to all of them.

$The five Lagrange points at mass fraction 0.12. The five points at which a small body can keep station with two larger ones. The three on the line of centres are roots of a quintic and are unstable; the two forming equilateral triangles are stable for a sufficiently lopsided mass ratio.$ Gravitation

Five places that keep station, in a problem with no solution

Three bodies under gravity cannot be solved. Restrict the problem slightly and five exact answers fall out anyway — three of them roots of a quintic, two of them perfect equilateral triangles.

Three bodies, integrated. Three equal masses integrated forward under mutual gravity: the figure-eight choreography — all three bodies on one closed curve. Every point is a step of the equations of motion, and the total energy is conserved to 2.5e-11 across the run.

Gravitation

Three bodies, and what "no solution" actually means

The three-body problem is routinely called unsolvable. Trajectories are computed for it every day, exact periodic solutions are known, and both statements are true — the word is doing more work than it looks.

$The Hill radius against the mass ratio, and where four planets keep their outermost moon. The Hill radius R_H = a(μ/3)^⅓ and the exact L₁ distance from the same quintic, both in units of the planet's own semi-major axis, against the mass ratio μ = m/(M+m) from 10⁻⁸ to 0.12. The two are the same line to the width of the stroke across the whole planetary range — the cube-root formula runs 0.33% short of the exact L₁ distance at the Earth's mass ratio and 10.0% short at the right-hand edge — so the departure between them is drawn on its own percentage scale at the right, which is the only place it can be seen. Below the curves are the numerically determined stability limits, 0.5 R_H for a prograde satellite and 0.7 R_H for a retrograde one, and the outermost known satellite of each of four planets, each at a Hill fraction computed from that planet's own mass and orbit: Moon 0.257, Sinope 0.451, Phoebe 0.198, Neso 0.424. Of the four, three are retrograde — Sinope, Phoebe, Neso — and they reach 0.451 of their planet's Hill radius against 0.257 for the one prograde satellite: 1.8 times as far out, in a diagram whose two stability limits stand in the ratio 0.7 to 0.5. No equilibrium argument predicts that, and it is what the drawing makes plain.$ Gravitation

The region a planet may keep a moon in

A satellite is not held by orbiting but by orbiting inside the radius at which the Sun's tidal field would take it away. That radius is the same distance the innermost Lagrange point sits at, asked a different question — and the boundary the sky actually respects is smaller, by a factor that depends on which way the moon is going round.

At the Sun–Earth L₂ the cheapest correction is every 23 days, and it costs e σ per e-folding. The annual station-keeping cost at the Sun–Earth L₂ point, against the interval between corrections, on logarithmic axes, for velocity errors of 0.5 cm/s, 2.0 cm/s, 5.0 cm/s along the unstable direction at each correction. Correcting often costs a lot because every correction carries its own error σ; correcting rarely costs a lot because the error has grown by e^(T/τ) in between, with an e-folding time τ = 23.4 days set by the point's growth rate of 2.484 times the orbital mean motion. The product (365.25/T) σ e^(T/τ) has its minimum at exactly T = τ, where the annual cost is 365.25 e σ/τ: 0.21 m/s a year for σ = 0.5 cm/s, 0.85 m/s a year for σ = 2.0 cm/s, 2.12 m/s a year for σ = 5.0 cm/s. The minimum is broad, so an operator can correct at a convenient interval near the e-folding time for little penalty, and the cost scales linearly with how well the spacecraft's velocity is known and executed. This is a one-dimensional caricature: a real halo orbit's correction also removes a stable component it need not, and solar radiation pressure on a large sunshield is a steady error source of its own. The figure's claim is the structure — an unstable equilibrium is cheap to hold if the instability is caught while it is still small, and its cost is a navigation budget rather than a force budget.

Spaceflight

An unstable point that costs less to hold than a stable orbit

A spacecraft at the Sun–Earth L₂ point sits on an equilibrium that throws it away, doubling any error every sixteen days. It holds station for a few metres per second a year — a twentieth of what a geostationary satellite pays to stay on an orbit that is stable. The difference is what is being paid for — an instability caught small costs a navigation budget, and a steady torque costs a force budget.

$The quantity a capture has to change. The Jacobi constant along three trajectories through the same encounter with the secondary, integrated in the rotating frame at mass fraction 0.0009543. The undissipated one holds its value to 3e-11 over the whole passage — that is not an approximation but an identity of the equations of motion, and the residual is the integrator's. A body that enters through the neck at L₁ with the neck open leaves through it, because C has not changed and the neck is still open; a purely gravitational two-body encounter is time-reversible and cannot end in a bound orbit. The other two runs remove energy in the rotating frame. A drag proportional to velocity raises C continuously at a rate 2k v², and a single impulse against the motion raises it in a step. Both finish above C(L₁) = 3.039, which is the line at which the neck shuts and the body is trapped in the secondary's lobe. Which mechanism operated for a given satellite is not recoverable from its orbit today — but it is partly recoverable from the population, because the three leave different distributions behind.$ Gravitation

A moon that two bodies could not have caught

A body that falls into a planet's Hill sphere under gravity alone leaves it again, because the quantity the rotating frame conserves is the same on the way out as on the way in. Every captured moon is therefore a record of something that removed energy — and which of the three candidates did it is written in the inclinations rather than in any one orbit.

A limit with a notch cut in the middle of it. The stability limit for a satellite, in units of its planet's Hill radius, against the inclination of its orbit to the planet's. The two flat stretches are the coplanar answers — 0.5 prograde and 0.7 retrograde, the difference between them being the Coriolis asymmetry that a rotating frame imposes and no potential can hold. Between them the limit is not an interpolation. Inside the shaded band, from 39.23° to 140.77°, the Sun's averaged perturbation exchanges the orbit's inclination for its eccentricity, and a satellite whose semi-major axis is comfortably inside the lobe has an apocentre that is not — so the boundary is set by a(1 + eₘₐₓ) rather than by a, and it falls to exactly half the coplanar value at ninety degrees. The band edges are cos²i = 3/5 exactly, with no fitted quantity anywhere in them, and the notch is the reason a quoted stability limit in Hill radii is a statement about a coplanar orbit and not about a satellite.

Gravitation

A stability limit with a notch cut in it

The two numbers usually quoted for how far out a moon may orbit are coplanar numbers. A tilted orbit is not somewhere between them: inside a band of inclination fifty-six degrees wide the Sun converts the tilt into an eccentricity, the apocentre rather than the semi-major axis has to stay inside the lobe, and the limit falls to half. The band is empty in the sky.

The whole library · All essays