Concept

Linear stability — where it appears

The behaviour of small disturbances to a solution when the equations are approximated to first order in the disturbance. A linearly stable orbit has disturbances that oscillate rather than grow, which is necessary but not sufficient for lasting stability.

Named by 2 essays across 2 fields — each of them below, with the objects they name alongside it.

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.

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.

spaceflight · Station-keeping

The figure-eight with equal masses, and with one mass changed. The paths of three bodies in the figure-eight choreography over 100 periods, with equal masses (left) and with the third body's mass changed by 0.0001 (right). With equal masses the three paths lie on one another: each body runs round the same figure-of-eight, a third of a period behind the next. With the mass changed the three bodies stay bound but leave the single curve, filling a broad band round where it was. The choreography is an exact solution only for exactly equal masses, and the island of stable orbits near it is narrow.

A periodic orbit that the sky cannot keep

Three equal masses can chase one another for ever round a single figure of eight. The orbit exists, it is stable, and it will almost certainly never be seen — because its island of stability is a hundredth of a per cent wide in the masses, and nothing in nature makes three stars equal to that precision.

gravitation · The three-body problem

Named alongside it

The objects these essays reach for when they reach for this one.

Choreography Collinear points Δv E-folding time Figure-eight orbit Halo orbit Hill sphere Invariant manifold Island of stability KAM theory Lagrange points Lunisolar perturbation

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