How would a large a mass be stable at the Earth Sun L4 or L5 point?

Question

I've heard about the Trojan asteroids and there is the famous idea of putting a space colony at one of these points, but the explanations I see for how something is stable at those points it they are 'insignificant mass.' What does that mean? Insignificant to Earth could be pretty big, insignificant to the sun could be downright huge. What are we talking about here, the mass of a large asteroid? The moon? The Earth? Surely it couldn't be bigger than one of the two bodies in the system?

Answer 1

The L4 and L5 points rotate around the barycenter of the Earth-Sun system. A "negligable" mass would be one that doesn't materially affect the location of this barycenter. Obviously, with the Earth being much lighter than the Sun, the L4 point is almost equidistant to the Sun and the Earth, and thus will influence the Earth far more. That means the mass should be negligable to the Earth's mass, not the Suns.

You'd obviously be able to balance the mass in L4 and L5. That's a first-order approach, though, as both masses will actually be in an orbit around L4 and L5 and (due to other planets) are quite unlikely to have the same exact orbits.

See also http://wmap.gsfc.nasa.gov/media/ContentMedia/lagrange.pdf

[edit] I found this discussion which has a closed-form solution on possible masses for a stable three-body configuration rotating around their common barycenter : (m₁+m₂+m₃)² -27*(m₁*m₃+m₃*m₂+m₁*m₂) >= 0 (quoted from Volume 5 of of "What's Happening in the Mathematical Sciences" by Barry Cipra, apparently.)

[edit2] The sun weighs 3.3E5 times more than the earth. That means we can reduce the formula to

(3.3E5 m_earth+m_L4)² >= 0. In practical terms, since the sun is far, far heavier than the other bodies, every term but the mass of the sun squared is negligable. That in turn means that even a second earth would be stable in Earth-Sun's L4.