Astronomy & Celestial Mechanics
the sky, computed
Concept

Lagrange points — where it appears

The five positions in a two-body rotating frame at which a third small body feels no net force. Three are collinear and unstable, two are triangular and stable for a mass ratio below one in 25, and the stable pair holds the Trojan asteroids.

Named by 14 essays across 6 fields — each of them below, with the objects they name alongside it.

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.

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.

gravitation · Lagrange points 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.

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.

gravitation · The three-body problem Zero-velocity curves, mass fraction 0.15. Level sets of the Jacobi constant in the frame that rotates with the two bodies. A particle with a given value of C is confined to the side of its own curve where the kinetic energy would be positive; as C falls the curves open, first at the inner point between the bodies, then behind the smaller one, and finally around the whole system.

The curve that says where a body cannot go

The restricted three-body problem has no solution and one conserved quantity. That quantity is enough to draw a boundary the body can never cross — without integrating anything, without knowing where it started, and for all time.

gravitation · The three-body problem The surface a star stops at. The equipotential through L₁ — the Roche lobe — at mass fractions 0.50, 0.20, 0.05, in the frame that rotates with the pair. Each is a level set of exactly the same function the zero-velocity curves are level sets of, at exactly the critical value, so nothing here is a new construction: the Roche lobe is the last closed equipotential, and it is closed only because the two lobes touch at a single point. Material that reaches that point is no longer bound to the star it came from, and it leaves through an opening of zero area. The lobes are drawn in the orbital plane; in three dimensions each is a teardrop, and its volume-equivalent radius is what "the size of a Roche lobe" means. As the mass ratio becomes extreme the smaller star's lobe shrinks towards it, which is why a white dwarf accreting from a companion has a lobe smaller than the Sun.

The surface a star stops at

Around each star of a close pair there is a last closed equipotential, and the two touch at a single point. A star that swells to reach it hands its outer layers to its companion through an opening of zero area — and the transfer, once started, makes itself worse.

gravitation · Lagrange points 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.

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.

gravitation · Hill sphere A Hohmann transfer, 2.6 to 1 in radius. Two circular orbits and the ellipse that touches both. The first burn raises the far point to the outer orbit; the second, half an orbit later, circularises. The costs are computed from the vis-viva relation.

The cheapest way between two orbits, and why it is so slow

Two burns and a long coast is the least fuel that will move a spacecraft between two circular orbits. It is also, for anything beyond the Moon, an unreasonably long wait.

spaceflight · Orbital transfer Two responses to the same transfer, turning round at q = 0.79 and q = 1. What conservative mass transfer does to the orbit and to the lobe, plotted against the mass ratio of donor to accretor on a logarithmic axis. Both curves are logarithmic derivatives with respect to the donor's mass, so a positive value means the quantity shrinks as the donor loses mass and a negative one means it grows. The orbit's response is exactly twice the mass ratio less one, which follows from holding the total mass and the total angular momentum fixed and nothing else, and it crosses zero at equal masses: transfer from the heavier star draws the orbit in, transfer from the lighter one pushes it out. The lobe's response adds to that the change in the lobe's shape, and it crosses zero earlier, at a mass ratio of 0.788. Between those two crossings the orbit is still widening while the lobe is already closing. To the right of both, a donor that loses mass finds its lobe shrinking around it, which is the runaway the essay is about: the transfer narrows the valve it is flowing through.

The flow that narrows its own channel

Two stars close enough share a surface, and the point where that surface pinches is a valve. What comes through it changes the orbit, and the orbit changes the valve — with a sign that reverses at equal masses, which is why some binaries transfer quietly for a hundred million years and others tear themselves apart in a thousand.

stars · Mass transfer A stream that is not the orbit it came from. 456 stars released in pairs from the two saddles of a 10⁵ solar-mass cluster over 4.0 billion years, integrated in a halo whose circular speed is 220 kilometres a second, drawn with the progenitor's own orbit. The cluster runs between 10 and 25 kiloparsecs and the orbit is the thin closed-looking curve; the stars are everything else. The point of the figure is the discrepancy. Stars leaving through the inner saddle are on slightly smaller orbits, turning round at a median of 24.77 kiloparsecs rather than the progenitor's 25.00, and therefore running ahead; stars leaving through the outer saddle reach 25.22 and fall behind. The whole spread is 1.8 per cent of the apocentre, which is the number worth carrying: an offset far too small to see in this drawing builds the entire stream, because it acts for four billion years. The two arms are therefore not merely displaced along the orbit, they are on different orbits, and the track a survey measures is a family of them rather than any single one. Fitting a Galactic potential by demanding that a stream lie along an orbit is wrong by exactly this much, and the size of the error grows with the mass of the progenitor, because the mass is what sets the distance between the two doors.

A stream is not the orbit it came from

A cluster torn apart by a galaxy leaves a thin trail of stars across the sky, and the obvious thing to do with it is fit an orbit. That is wrong by a knowable amount, because stars leave through two doors with a small energy offset and end up on a family of orbits rather than on one.

galaxies · Stellar streams A 10 Earth-mass Trojan started 10° from its point, seen in the planet's transits. A Jupiter-mass planet on a 4-day orbit about a 1 solar-mass star, with a 10 Earth-mass companion started 10 degrees beyond the leading Lagrange point of the same orbit, integrated for 160 days. Above, the angle between the companion and the planet as seen from the star: it swings about 60 degrees, between 51.1 and 70.3, with a period of 48.5 days measured off the curve, against 49.8 from the small-amplitude formula P/√(27μ/4). Below, the planet's own transit times minus a straight line: they swing by ±300.0 seconds at the same period, because the planet and the companion orbit their common centre of mass and the companion's libration moves that centre along the orbit. The companion shares the planet's period exactly, so it produces no timing signal at any orbital period of its own and would not transit on a schedule distinct from the planet's; the libration period is the only clock it has.

A companion on the same orbit, seen in the planet's clock

A body sharing a planet's orbit at one of its Lagrange points has the planet's period exactly, so no search for periodic dips or wobbles can find it at a period of its own. It shows instead in two ways the planet's own signals carry — a slow swing of the transit times at the companion's libration period, and a fixed offset between the planet's transit and its star's velocity curve.

exoplanets · Transit timing 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 When a target can be observed from a telescope that cannot look near the Sun. The days of the year (across) on which a target at ecliptic longitude 0° and a given ecliptic latitude (up) can be observed by a telescope allowed to point only between 85° and 135° from the Sun, the constraint of a large infrared telescope at the Sun–Earth L2 point. The Sun moves along the ecliptic once a year, and the allowed band sweeps round with it. A target on the ecliptic is inside the band twice a year, for about 51 days each time, a little before and a little after its opposition — never at it, since opposition is 180° from the Sun. The two windows widen with latitude and merge into one above 45°, where the elongation at opposition no longer exceeds 135°; within 5° of either ecliptic pole, where the elongation at conjunction no longer falls below 85°, the target is never outside the band at all, which is the continuous viewing zone. The pattern depends on ecliptic, not equatorial, coordinates: the season of a space telescope is set by the Sun's path, and the Earth's axis has no part in it.

A season without a night

A telescope in space has no horizon and no night, and it can still only look at most of the sky for part of the year. Its season is a band of solar elongation swept round the ecliptic by the Sun, so an object on the ecliptic can be seen for two stretches of about seven weeks and never at opposition, while the few degrees round the ecliptic poles can be watched all year. From low orbit the Earth takes the night's place, and the zones it never hides are carried round the sky every eight weeks by the planet's equatorial bulge.

sky · Sidereal time How fast an error in the departure grows, near a neck and on a conic. How much a small error in a spacecraft's velocity is magnified in its later position — the largest singular value of the block of the state-transition matrix that maps velocity errors to position errors, in the Earth–Moon system's units — against time in days, integrated from the variational equations of the restricted three-body problem. A trajectory held at the L1 point between the two bodies (blue) amplifies errors exponentially, by a factor of e every 1.48 days — the rate the linearised dynamics there predict, 2.932 in the system's units, which the integration reproduces to within two per cent. A two-body circular orbit at half the separation (green), integrated the same way with the second body's mass switched off, amplifies them only secularly — an error in speed is an error in period, and that grows along the track in proportion to time. Over 14 days the neck has multiplied an error by 2,359; the conic by 9.

A trajectory with no good first guess

A trajectory designed from conics is corrected onto the real one by Newton's method, and the method works because the arrival depends on the departure only polynomially. Near a Lagrange point the dependence is exponential: every day and a half near the Earth–Moon neck multiplies an error by e. The first guess has to be exponentially good, the digits of the departure run out in weeks, and the only cure is to stop treating the trajectory as one piece.

spaceflight · Patched conics An arrival held without a burn, and released. A body approaching a planet from a heliocentric orbit just outside the planet's, drawn in the planet's rotating frame (the Sun far to the left, the planet's orbital motion upward), in units in which the Hill radius is 0.69. It drifts in along the shear of the differential rotation, passes through the neck at the Lagrange point L2 into the region the planet controls, loops round the planet 3.3 times and leaves through a neck again. No burn is drawn: the capture and the release are both the Sun's tide working against the planet's pull. For Mars one unit of distance is 1.56 million km and an orbit round the planet near the Hill radius takes about a year.

An arrival the Sun hands over

A spacecraft that reaches a planet with no speed to spare can be captured with no burn at all, passing through the neck between the planet's gravity and the Sun's into a loose, looping orbit. The capture is temporary and chaotic — a few revolutions, and arrivals a few hundred kilometres apart end differently — but while it lasts the spacecraft is already bound, and the burn that keeps it is taken from there.

spaceflight · Escape A loop round the Earth that the Earth does not make. The path over one year, in the frame that turns with the Earth (Earth at the centre, the Sun far to the left, the Earth's motion upward), of bodies on heliocentric orbits with the Earth's period and eccentricities of 0.05, 0.1, 0.15, started at the Earth's mean longitude. Each traces a retrograde loop round the Earth, about 2ae long in the direction of motion and ae wide towards the Sun: for e = 0.1, 0.395 by 0.199 AU. The small circle is the Earth's Hill sphere, 0.010 AU in radius, inside which the Earth's gravity dominates the Sun's tide. Every loop lies far outside it. The loop is a difference between two ellipses with one period, and would be traced the same way if the Earth had no mass at all.

A moon the Earth does not hold

Seen from the Earth, the small asteroid Kamoʻoalewa circles the planet once a year, backwards, between forty and a hundred times as far away as the Moon. It is not a moon. It orbits the Sun, with the Earth's period and a slightly different eccentricity, and the loop is the difference between two ellipses — one the Earth would trace even with no mass at all. What the Earth's gravity does is subtler: it keeps the loop from drifting away.

gravitation · Lagrange points

Named alongside it

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

Hill sphereRestricted three-body problemJacobi constantMass ratioRotating frameThree-body problemAngular momentumMass transferRoche lobeAccretionCo-orbitalΔv

All concepts