Three Body

Math · chapter 7 of 13

The two ways three rocks can hold a shape

Three bodies can keep the same shape while they spin, and there are exactly two shapes that allow it: a straight line and an equilateral triangle. Euler found the first in 1767, Lagrange the second in 1772.

Forget the general case for a minute and ask a smaller question. Can three bodies move so that the triangle they make never changes shape — only turns and scales? That is a shape you could write down.

Euler, 1767. Put all three on a line, spinning about the balance point like a baton. It works, but only at one spacing, and finding it means solving a fifth-degree equation. There is no formula for a general fifth-degree equation — that is Abel's theorem, a different piece of mathematics arriving in the same century — so even this tidy case ends in a number you have to hunt for rather than write. It has exactly one answer for any three weights you choose.

Lagrange, 1772. Put the three at the corners of an equilateral triangle and spin. This one works for any three weights, with no equation to solve: the triangle holds, each body running an ellipse of its own around the balance point. A heavy body and a light one at two corners is the Trojan asteroid case from the last chapter.

Those two are the only ones. For three bodies there are five such arrangements in total — three collinear ones, depending on which body sits in the middle, and the two mirror-image triangles — and that is the complete list. Nobody has proved the equivalent for every number of bodies; whether the count is always finite is Smale's sixth problem, still open.

This is worth sitting with. The general three-body problem has no formula, and yet within it sit these exact, eternal, writable solutions. A problem being unsolvable in general does not mean it is dark everywhere.

r5 − (3−μ)r4 + (3−2μ)r3μr2 + 2μrμ = 0

Reading it: Where the inner balance point sits: the one distance that satisfies this fifth-degree equation.

This is the L1 case of Euler's quintic. Solved by hunting, not by formula. For the Sun and the Earth it puts L1 about 1.5 million km sunward of us — the figure printed on the page is this equation solved at build time.

ω2 = G(m1+m2+m3)a3

Reading it: An equilateral triangle of any three weights, sides of length a, spins at this rate and holds its shape.

Compare Kepler's third law for two bodies — same form, with all three weights added. Lagrange's triangle is the closest thing the three-body problem has to a napkin answer.

3 collinear + 2 triangles = 5

Reading it: Five arrangements that keep their shape, for any three weights. That is all of them.

Called central configurations. Five for three bodies; fifty for five bodies with generic weights; no general count is proved.

Try it. Press the triangle in the demo and then drag one weight up. The shape holds — the spin rate changes and the balance point moves, and the triangle stays equilateral. Now press the line and drag: the spacing has to be re-solved every time, and you can watch the solver hunt.

This one runs in the browser, with JavaScript on. The words above it stand on their own.

r = 0.0099700.20.40.60.81value of the quinticdistance from the small body

Euler's quintic for the Sun and the Earth, drawn and then solved by bisection at build time: the root is r = 0.009970 in units of the Earth's distance, which puts L1 1,491,539 km sunward of us. No formula exists for a general fifth-degree equation, so this is a hunt, and it takes about forty steps.

L1L2L3L4L5the heavy onethe light onemass ratio μ = 0.2, L1 at x = 0.4381, L2 at 1.2710, L3 at -1.0828

The five places, computed for a mass ratio of 0.2. Three on the line through the pair, from the quintics; two at the corners of equilateral triangles, from Lagrange.

Sources

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