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.

Assumes The two-body problem.

Two bodies under gravity are solved completely. The orbit is a conic, the timing follows from Kepler’s equation, and every question has an answer in closed form.

Add a third body and all of that ends. There is no general solution — not a hard one, not an ugly one, none. Poincaré proved in 1890 that no set of algebraic integrals exists to reduce the problem, and in the course of proving it discovered orbits that never repeat and never settle, which is where chaos theory begins.

And yet, in a restricted version of the problem, five points can be located exactly. Three of them require solving a quintic numerically. The other two are the vertices of equilateral triangles, obtainable by inspection, and they are the only clean result in the whole subject.

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.
Fig. 1 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; the two off it form exact equilateral triangles with both bodies.

What “unsolvable” means

The claim needs stating carefully, because it is often overstated.

Three-body trajectories can be computed. Numerical integration gives them to whatever precision the arithmetic supports, and space missions are flown on exactly such computations. What does not exist is a formula — an expression giving the positions at time tt without stepping through the intervening motion.

The reason is a shortage of conserved quantities. The two-body problem has enough — energy, angular momentum, the centre-of-mass motion, and the extra vector that makes the orbit close — to reduce it to something integrable. Three bodies have the same conserved quantities and far more degrees of freedom, and Poincaré showed no further independent ones exist.

The practical consequence is worse than inconvenience. Trajectories in the three-body problem can be sensitively dependent on initial conditions: two starts differing in the twelfth decimal place diverge completely after enough time. The solar system’s long-term stability is, for this reason, still an open question — the best integrations show Mercury’s eccentricity wandering enough over billions of years that a small fraction of simulations lose the planet entirely.

Restricting it until something works

The circular restricted three-body problem makes three simplifications. The third body is massless, so it does not disturb the other two. The two massive bodies move on circles about their common centre. And everything stays in one plane.

That last restriction is not what makes it tractable; the first two are. With the primaries on fixed circles, there is a frame that rotates with them in which both are stationary, and in that frame the problem becomes a question about a potential.

Effective potential along the line of centres, mass fraction 0.12. The combined gravitational and centrifugal potential along the line joining two bodies in the frame that rotates with them. Its three stationary points are the collinear Lagrange points, and all three are maxima along this line.
Fig. 2 The combined gravitational and centrifugal potential along the line joining two bodies, in the frame that rotates with them. Its three stationary points are the collinear Lagrange points, and every one of them is a maximum along this line.

The rotating frame introduces two fictitious forces. Centrifugal force depends only on position and can be folded into an effective potential; Coriolis force depends on velocity and cannot, which turns out to matter enormously. Ignoring Coriolis for a moment, equilibrium requires the effective potential to be stationary, and the figure shows exactly where along the line that happens.

The three collinear points come out of a fifth-degree polynomial. Quintics have no general solution in radicals — Abel proved that in 1824 — so the positions of L1, L2 and L3 are irreducibly numerical. They are found by iteration, in the figures on this page as everywhere else.

The two that were guessed

L4 and L5 are different in kind. Lagrange found in 1772 that a body forming an equilateral triangle with the two primaries stays in equilibrium, exactly, for any mass ratio whatsoever.

The reason is a small geometric miracle. At the third vertex of an equilateral triangle, the two gravitational pulls — one from each primary, in proportion to their masses and both at the same distance — add to a resultant that points exactly at the barycentre, with exactly the magnitude the centrifugal term requires. The equal distances are what make it work, and they make it work for every mass ratio at once.

No quintic, no iteration, no numerical anything. Two exact solutions, in a problem that has none.

The five Lagrange points at mass fraction 0.03. 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.
Fig. 3 The same construction at a mass ratio nearer the Sun–Jupiter case. The triangular points have not moved relative to the two bodies at all — they are equilateral whatever the masses — while the collinear three have crept toward the smaller body.

Comparing the two figures makes the difference plain. The collinear points depend on the mass ratio and move as it changes. The triangular points do not move at all, because “equilateral” contains no mass.

Stable, unstable, and the force that saves it

All five points are stationary. Only two are stable, and the reason is the term that was set aside.

The collinear points are saddles of the effective potential: stable across the line of centres and unstable along it. A body left at L1 with a small displacement toward either primary accelerates away, and the timescale is short — for the Sun–Earth L1 it is about 23 days. Nothing stays there without help.

The triangular points look worse on the potential surface, because they are maxima — a ball on a hilltop. Yet for a mass ratio below about 1:25 they are stable, and the rescuer is the Coriolis force. A body drifting away from L4 acquires a velocity, and Coriolis deflects that velocity sideways, curving the drift into a closed loop around the point instead of a departure from it. Stability from a velocity-dependent force acting on an object sitting on a hill, which is not a mechanism anybody would have guessed.

The mass-ratio condition is real and observable. The Sun–Jupiter ratio is 1:1047, comfortably stable, and Jupiter’s L4 and L5 hold over ten thousand catalogued asteroids — the Trojans, which have sat there for most of the age of the solar system. Neptune, Mars and even the Earth have their own. Where the ratio fails, the points are empty.

What the points are used for

Each collinear point suits a different job, and every one of them is occupied.

Sun–Earth L1 sits a million and a half kilometres sunward, with an uninterrupted view of the Sun and no possibility of eclipse. SOHO has watched from there since 1996; DSCOVR gives the continuous full-disc view of the daylit Earth.

Sun–Earth L2, the same distance in the anti-sunward direction, has Earth, Moon and Sun all in the same part of the sky — an alignment of the kind that also produces eclipses — so a single shield blocks all three. That is why JWST is there and why Planck and Gaia were. A telescope operating at 40 kelvin needs its heat sources in one direction, and L2 is the only place in the inner solar system that offers it.

Earth–Moon L2, beyond the far side, sees the lunar farside and the Earth at once, and is where a relay satellite sits to talk to landers that have no line of sight home.

None of these are stationary in practice. Because the collinear points are unstable, spacecraft fly halo orbits around them — large loops, hundreds of thousands of kilometres across, that are themselves unstable and require a station-keeping burn every few weeks. JWST spends a few metres per second a year staying put, and the propellant budget for that is what sets the mission’s lifetime.

The residuals that found a planet

The three-body problem’s most famous product is not a point but a planet.

By the 1840s Uranus was measurably off the path an ellipse about the Sun predicted — by about two arcminutes, which does not sound like much and was far outside the errors. Two people independently assumed the discrepancy was a further planet and solved the inverse problem: given the residuals, where must the perturber be?

Le Verrier’s prediction reached Berlin on 23 September 1846, and Neptune was found that night, within a degree of the position given. It is the most spectacular result perturbation theory has produced, and it worked because Uranus’s deviation was a small perturbation on a two-body orbit — the regime where the mathematics is reliable.

Le Verrier then applied the same method to Mercury’s anomalous precession — a residual left over after every Newtonian effect had been subtracted — predicted a planet inside its orbit, and named it Vulcan. It does not exist. The residual was real and its cause was not another body but a defect in the theory of gravity, which is a good illustration of how the same technique can be brilliantly right and completely wrong.

The same picture at a different ratio

The mass fraction changes the collinear points and leaves the triangular ones alone, and comparing two cases makes the split visible.

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 0.04. The two are the same line to the width of the stroke across the whole planetary range — the cube-root formula runs 0.34% short of the exact L₁ distance at the Earth's mass ratio and 7.6% 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.
Fig. 4 The same quintic asked a different question. A Hill radius is not a separate piece of mechanics from L₁ — it is the L₁ distance, up to a factor the cube-root formula throws away — and drawing a(μ/3)1/3a(\mu/3)^{1/3} against the exact root shows how nearly they coincide and where they stop. Below a mass ratio of 10310^{-3}, which is every planet in the solar system, the two agree to better than 2.5 per cent; at the mass ratios of a close binary they part company entirely. The satellites plotted on it are real ones, each at its own fraction of its own planet’s radius, and none is outside the stability limit the figure draws.

For the Sun–Earth pair the mass fraction is 3 × 10⁻⁶, and L₁ and L₂ sit 1.5 million kilometres either side of the Earth — a hundredth of the way to the Sun. The distance scales as the cube root of the mass fraction, which is why it is such a convenient number for so wide a range of pairs. Getting to L₂ is therefore cheap — a few hundred metres per second beyond escape from low orbit — which is very unlike a planetary transfer. Staying is the expensive part, and it is expensive because the point is a saddle rather than a minimum.

What was actually observed there

The five points are a result about a model. What makes them more than that is that four of the five have been checked against things sitting in them, and the checks are quantitative.

The Trojans are the largest test. Jupiter’s L4 and L5 hold more than twelve thousand catalogued asteroids, and the population is estimated at around a million larger than a kilometre — comparable to the entire main asteroid belt. They are not at the points. Each one librates around its point on a tadpole-shaped path taking 150 to 200 years to complete, with excursions of tens of degrees in longitude; the point is the centre of an oscillation, not an address. Their inclinations reach 40°, far outside the plane the model assumes, and they are stable anyway.

There is an asymmetry nobody predicted and nobody has fully explained: the leading swarm at L4 contains roughly 1.4 times as many objects as the trailing one at L5. Proposed causes include the geometry of Jupiter’s early migration and a collisional history that differed between the two clouds. The Lucy mission, launched in 2021, is visiting members of both.

The five Lagrange points at mass fraction 0.02. 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.
Fig. 5 A more lopsided pair, at a mass fraction of 0.02. L₁ and L₂ have closed in on the smaller body — their distance from it scales as the cube root of the mass fraction — while L₄ and L₅ have not moved at all. The Sun–Jupiter fraction is smaller still, at 0.00095, and the Sun–Earth smaller again by a factor of three hundred.

Smaller bodies have Trojans too. Neptune has about thirty known, Mars has several, Uranus has one, and the Earth’s first confirmed Trojan — 2010 TK7 — was found in 2010 at L4, librating in an enormous tadpole that carries it from near the Earth out to the far side of the Sun’s direction and back over 400 years. Its existence had been predicted for decades and repeatedly not found, because from the Earth it is always seen close to the Sun in a bright twilight sky.

The collinear points are occupied only by machines. Nothing natural stays there, which is the strongest available confirmation that they are unstable. The instability timescale for Sun–Earth L1 and L2 is about 23 days, meaning a displacement grows by a factor of ee in that time; a body left there with no control would be a million kilometres away within a year.

The number that measures the instability directly is a station-keeping budget. JWST performs a small burn roughly every three weeks and spends of order two to four metres per second per year holding its halo orbit — a quantity so small that the mission’s twenty-year propellant estimate rests on it, and so unavoidable that no design can remove it. The burns are also deliberately one-sided, always pushing outward from the Sun, because a burn in the other direction would send the telescope through L2 onto a trajectory it could not recover from with its sunshield in the way.

There is a third kind of co-orbital motion the five points do not cover, and it exists because L₄ and L₅ are not separate basins.

A body librating about L₄ with enough amplitude does not stay near it. Its excursion carries it past L₃ — the unstable point on the far side of the primary — and into the neighbourhood of L₅, where it turns round and comes back. Seen in the rotating frame the path is a horseshoe: an enormous arc spanning nearly the whole orbit, closed at neither end, with the secondary sitting in the gap.

Saturn’s moons Janus and Epimetheus are the clearest case. Their orbits differ in radius by 50 kilometres — less than either moon’s own diameter — and every four years the inner one catches the outer, they exchange a little angular momentum without ever coming within 10,000 km of each other, and they swap orbits. Each has been doing this for as long as anyone has watched, and neither has ever passed the other.

The Earth has one too. Asteroid 3753 Cruithne is on a horseshoe with respect to the Earth, taking 770 years to traverse it.

The five points exist at every mass ratio and their arrangement changes with it, so the geometry is worth drawing at two values far below the one the essay uses to make the picture legible.

The five Lagrange points at mass fraction 0.001. 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.
Fig. 6 The five points at a mass fraction of a thousandth, which is roughly Jupiter’s. The three collinear points have crowded in close to the secondary and the two triangular ones have not moved at all — they are at the vertices of an equilateral triangle for any mass ratio whatever.
The five Lagrange points at mass fraction 0.05. 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.
Fig. 7 And at a twentieth, which is above nothing in the solar system and below the stability threshold of 0.0385 — so these two triangular points exist and hold nothing. The threshold is a statement about stability rather than about existence, and the difference is easiest to see where both apply.

The mass ratio the stability depends on

The condition for the triangular points to be stable is a single inequality, and it is worth writing down because it is sharp and because the solar system sits nowhere near it.

Linearising the motion about L4 and expanding gives a quartic in the oscillation frequency, and its roots are real — which is to say the motion is bounded — only when 27μ(1μ)<127\mu(1-\mu) < 1, with μ\mu the smaller body’s fraction of the total mass. The root of that is μ0.0385\mu \approx 0.0385, a mass ratio of about one to twenty-five.

Above the threshold the equilibrium is unstable and a body near L4 leaves. Below it, the Coriolis deflection wins and the body loops indefinitely.

The solar system’s pairs are all far below. The Sun and Jupiter give μ=0.00095\mu = 0.00095, twenty times inside the limit, which is why the Trojan swarms have survived for billions of years. The Earth and Moon give 0.0121, which is inside it by a factor of three — so the Earth–Moon triangular points are linearly stable, and yet no permanent population sits there, because the Sun’s perturbation is not in the model that produced the criterion and is large enough to eject anything over long times. Faint concentrations of dust have been reported at those points repeatedly since the 1960s and the observations remain contested.

Pluto and Charon are the solar system’s nearest approach to the boundary, at μ0.11\mu \approx 0.11, comfortably above it — so that pair has no stable triangular points at all, and its small moons sit elsewhere.

A criterion derived in a model with three bodies and circular orbits gives the right answer for Jupiter and the wrong answer for the Earth and Moon, and the difference is entirely what the model left out.

Where the model stops

Circular primaries. Real orbits are eccentric, and even a small eccentricity breaks the construction: the elliptic restricted problem has no fixed points at all, and the equilibria become periodic paths.

A massless third body. Trojans are massless to excellent approximation; a third star is not, and the general problem returns.

One plane. Real Trojans have inclinations, and their motion about L4 is a three-dimensional libration.

Only three bodies. Jupiter’s Trojans are also perturbed by Saturn, and the long-term stability of the swarm is a numerical question rather than an analytic one — the same difficulty that makes the orbits of the planets themselves only provisionally stable.

The figures have a specific and important limitation: they are drawn in the rotating frame, where the primaries stand still. In the inertial frame nothing in the picture is stationary — the Lagrange points sweep round the primary once per orbital period, and a Trojan asteroid is on an ordinary heliocentric orbit with Jupiter’s period, sixty degrees ahead. The word “point” makes them sound like places, and they are better thought of as phase relationships.

And the Hill radius at a mass ratio between the two, since it is the quantity the collinear points are placed by.

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 0.008. The two are the same line to the width of the stroke across the whole planetary range — the cube-root formula runs 0.34% short of the exact L₁ distance at the Earth's mass ratio and 4.6% 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.
Fig. 8 The Hill radius against mass ratio with the marker near eight thousandths. The cube-root dependence is shallow enough that a factor of a thousand in mass moves the radius by only a factor of ten, which is why every planet in the solar system has a Hill sphere within an order of magnitude of a hundredth of its orbital radius.

One more reading covers the mass ratio the solar system’s most-used pair of points actually has.

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 10⁻⁷ to 3·10⁻⁶. 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 0.3% 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.
Fig. 9 The Hill radius against mass ratio with the marker at the Earth–Sun value of three parts in a million. The radius is one and a half million kilometres, which is where the Sun-observing and infrared observatories are parked — a distance derived from a cube root of a number nobody chose.

The ladder from here

Later rungs: the effective potential in two dimensions, and the zero-velocity curves that bound where a body can go. Jacobi’s integral, the one conserved quantity the restricted problem retains. The quintic for the collinear points. Coriolis stabilisation worked through. Halo orbits and their station-keeping. Trojan populations across the solar system. Horseshoe orbits, where a body swaps between L4 and L5 by way of L3. Poincaré’s discovery of homoclinic tangles. And the low-energy transfer networks that thread between Lagrange points, which is how a spacecraft can travel across the solar system on almost no fuel and a great deal of patience.

Lagrange published the triangular solutions in a prize essay on the motion of three bodies, and regarded them as a curiosity of no possible application. The first Trojan asteroid was found 134 years later, and the first spacecraft arrived 206 years later.

What this makes readable

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About the same objects

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Angular momentumEccentricityEffective potentialLagrange pointsMass ratioOrbital stabilityRotating frameThree-body problem