Lagrange Triangle

The setup

Three equal masses sit at the corners of an equilateral triangle inscribed in the unit circle, so every side is √3 ≈ 1.732. Each velocity has the same length, 0.7598, turned a quarter turn from its own radius. That sets the whole figure spinning about the centre without changing its shape.

Why it turns

Take one corner. Its two neighbours are each at distance √3, pulling with 1/3 apiece, and the two pulls are 60° apart. They add to 2 · (1/3) · cos 30° = 1/√3, aimed straight at the centre. A circle of radius 1 needs v² for that, which fixes v = √(1/√3) ≈ 0.7598. One turn then takes 2π/v ≈ 8.27 seconds. The same sum works for a triangle of any size.

History

Lagrange found the equilateral configuration in 1772, one of the few exact solutions of the three-body problem. The Trojan asteroids, sixty degrees ahead of and behind Jupiter along its orbit, are the case nature runs.

Why it does not last

Exact is not the same as stable. The equilateral solution survives a nudge only when one mass dominates the other two, and equal masses fail that test badly. Here the sides hold to within a percent for some thirteen seconds, a turn and a half. By twenty-one seconds the triangle is unrecognisable. Two bodies then pair off into a tight binary and the third is flung away.

Try

• Top in the deck: the triangle flat, turning as one piece for the first two turns.
• Carry more digits, y=0.8660254 and speeds from vy=0.7598357 on down. The figure lasts three and a half turns instead of one and a half. Digits buy turns, never stability.
• Delete every vx and vy. With nothing said the speeds start at zero, so the three fall straight through the middle and are slung back out.
• Raise Span once the triangle has broken, to follow the body that leaves.

Read more

• Lagrange point
• Trojan (celestial body)
• Central configuration
• Three-body problem