Warren's proof would seem to prove that the equilateral triangle configuration is unstable. My impression is that it's at least neutrally stable (for some mass ratios). I haven't done the details, so I'll leave this as a question, rather than an assertion of fact. [One possible out: the rotation might be non-uniform?] Rich ---- Quoting meekerdb <meekerdb@verizon.net>:
On 1/10/2013 3:53 PM, Warren Smith wrote:
Suppose you have N planets moving under Newton's laws of gravity& motion (treat as point masses). Further, suppose that in an appropriate -- rotating about the center of mass -- reference frame, all the planets are stationary. Finally, let the configuration be STABLE against small perturbations.
MY CLAIM: If N>2 then no such configuration exists.
Why not? I.The pseudo-potential from the centrifugal force from the rotation, is proportional to -r, where r=distance to rotation axis.
II. The (actual) potential cased by gravitational attractions behaves proportionally to -1/r where r is the distance to the attractor.
III. All (pseudo and actual) potentials sum.
IV. The Laplacian LI of the psuedo-potential (I) is LI = -2/r. The Laplacian of the actual potential (II) is LII = 0. Thus upon summing we see that every body experiences a total potential whose Laplacian is NEGATIVE. Since Laplacian is a sum of second derivatives in orthogonal directions, that means at least one such second derivative is NEGATIVE. In other words, each planet can be moved infinitesimally in a way which makes its potential LOWER. Hence that planet was not in a (which would have been stable) potential-minimum.
Escape hatch: if there are only 2 (or 1) planets then you cannot move any planet at all (relative to the center of mass) without violating conservation of angular momentum and without keeping center of mass fixed (except for perturbations which increase energy), so the 2-body and 1-body problems are stable. But with 3 or more planets it is easy to see that some linear combination of the unstable shifts must exist which preserves both total angular momentum and center of mass, since you have>=3 degrees of freedom but only 2 constraints in a (linear, since infinitesimal perturbations) system... that is we have a 3-dimensional (or more) space of perturbations and within it there must exist a 2D (or less) subspace preserving these two invariants.
Right. Perturbing a mass at an L4 or L5 causes it to make a small orbit around the Lagrange point. Of course this also causes a small perturbation in the motion of the two larger bodies.
Brent Meeker
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