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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?

Mark Eichenlaub
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Scott
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1 Answers1

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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 mearth+mL4)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.

MSalters
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  • I was going to write this answer, but I got a queasy feeling that since the Lagrange points are at equilateral triangles, as you vary the mass, you might have stability for any introduced masses, even enormous masses, just with Jupiter and the sun now orbiting the new massive object. Is this false? – Ron Maimon Nov 21 '11 at 11:20
  • For the Lagrange Point L3 a so called "Anti Earth" is discussed sometimes by ScFi-conspirationists. – Georg Nov 21 '11 at 11:53
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    @Georg: see the first link. It quantisizes the stability of Earth-Sun's L3: ~150 years. – MSalters Nov 21 '11 at 15:26
  • Ok. So it's not so 'generic' as SF stories would lead you to believe. Putting a huge mass in L4 might be possible, but it would move that point closer to the Earth (if I'm picturing that right) so that the 'earth' pull would be stronger. That makes sense, but is there a limit to the size of the third body? Is there a fraction of the smaller body that above which you just can't get the three forces - sun, earth, circular momentum - to balance? – Scott Nov 21 '11 at 22:30
  • @Scott: Well, two of the three terms in that formula are known. You wouldn't care about tidal forces, either (i.e. the object internally would not need to be extremely rigid). The L4 point is the third point of an equilateral triangle. Therefore, the tidal forces between the 3 bodies depend purely on mass, and again the sun dominates the tidal forces. On earth, the tidal force of the Sun is half that of the moon: easily measurable, but it mostly affects oceans - those are obviously not rigid. – MSalters Nov 22 '11 at 08:25
  • Ok. Very interesting. I know Jupiter had a huge amount of mass there. Thanks so much for taking the time to answer. – Scott Nov 24 '11 at 10:51
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    +1 for the formula you found from Barry Cipra's article. However, I'm not at all sure your intuition that you can add stability by balancing the masses at L4 and L5 is correct. Orbital dynamics is truly non-intuitive—although as long as they're both small with respect to the Earth and the sun I expect it should be stable. – Peter Shor Dec 22 '11 at 14:38