Suppose I have a body orbiting in a (unstable, as shown by the effective potential of the system) circular orbit around another one in accordance to a inverse law of the form $$-kmm_1\frac{1}{r^4}$$ Could I have the same orbit with and a three body system composed of the two masses mentioned above plus a third one identical with the first positioned in the Lagrange point L3 of the system (as shown in figure) and obeying a Newtonian inverse square law of the form: $$-Kmm_1(\frac{1}{r^2})$$? My idea would be that the two forces can be equated with each other (regardless of how fine tuned both configurations would be to have such a scenario) in order to "mimick" in a Newtonian force field the inverse 4 attractive law through the additional body. Would this imply that a circular orbit leaves the central force generating it undetermined under very specific symmetry conditions without specifying the number and positions of all other eventually-interacting bodies?
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$\begingroup$ I'm not sure it's clear what you mean, do you mean the first situation has the first force and the second has the standard inverse square force? Shouldn't you also show that the three body's have the same orbit in the inverse four situation? $\endgroup$– TriatticusCommented Dec 4 at 20:51
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$\begingroup$ So yes, in the first case I have the "exotic" ˜ 1/(r^4) force law, in the second the standard inverse square one. But in the first case I have n=2 bodies, and in the second n=3. My question is if it is possible to arrange the 3 bodies affected by the standard force in such a way that the orbit of one of them equals the one it would have in case n=2 and F=K*1/(r^4). Sorry for not having been clear $\endgroup$– CraterusCommented Dec 4 at 21:05
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$\begingroup$ What I mean is that you should have four situations demonstrating this, why use different numbers of masses for the inverse four vs inverse square and not make all four situations. Two masses in both potentials and three masses in both. $\endgroup$– TriatticusCommented Dec 4 at 22:10
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