I am curious about a system of gravitating bodies with particular characteristics. Before describing the characteristics of the system, let me say what I want to know about it:

  1. Can this be a solution?
  2. Is it a stable solution?
  3. Can the orbiting objects all lie on the same elliptical orbit?

Here are the characteristics of the system: It is a system of 3 bodies of negligible mass in closed orbits around a central massive object. The three bodies of negligible mass orbit such that their distances to the central massive object and to each other are always changing, BUT the sum of their inverse distances (i.e. the sum of their individual potential energies) is constant. In this system energy is conserved, therefore it follows that the sum of the objects' squared velocities (i.e. their kinetic energies) is also always constant, even though the individual velocities are always changing.

  • $\begingroup$ it depends, do your negligible objects gravitationally interact with each other? $\endgroup$ Nov 11 '14 at 15:25
  • $\begingroup$ @julianfernandez No, they have negligible mass so there are no lines of force between them. The only lines of force in this system are between these objects and the central massive object. $\endgroup$
    – ben
    Nov 11 '14 at 15:30
  • $\begingroup$ @ben but would this not basically make it a multiple two body problem, since you could just leave out two of the three bodies of negligible mass when looking at the remaining one, which would yield a Kepler orbit. $\endgroup$
    – fibonatic
    Nov 12 '14 at 4:16
  • $\begingroup$ @fibonatic exactly, but in an elliptical orbit potential and kinetic energies are exchanged all the time, what he asks is if having three bodies you can make the total kinetic energy (and potential) constant at all times (so one one body gains the other's compensate) $\endgroup$ Nov 12 '14 at 5:01

The answer to the 3 questions is yes. To see this consider that both the T (kinetic energy) and V (potential energies) must be constant at all times.

$T=\Sigma T_i=\Sigma \frac{1}{2}m_iv_i^2$


$V=\Sigma V_i=-GM\Sigma \frac{m_i}{r_i}$

For an elliptical orbit: $v^2=2\frac{m}{r}-\frac{1}{a}$

Replacing this on $T$ and defining a new variable $x_i=1/r_i$, we can rearrange the above conditions as:

$ \Sigma {m_i}{x_i}=C_1$


$\Sigma {m_i^2}{x_i}=C_2$

Any solution must be an the intersection of these two planes, and you can make them non-parallel by having different $m_i$'s. thus there is an infinite number of solutions, even if the three masses are on the same orbit. The same conclusion is true for any finite number of bodies.

If the masses are the same, you can make the planes identical by using $mC_1=C_2$, this is possible if $\Sigma v_{0i}^2=GM-\Sigma \frac{1}{a}_i$

or equivalently: $ GM=2m\Sigma \frac{1}{r_{0i}}$

Disclaimer: Ben, you should re-check the equations to make sure I dint make any algebraic mistake.

  • $\begingroup$ Thanks. Note that the $m_i$ are negligible (relative to the central mass), so I'm not sure you can make them different. $\endgroup$
    – ben
    Nov 11 '14 at 22:52
  • $\begingroup$ Another possibility is to make $C1/m=C2/m^2$ so the two planes are the same. I am gonna check if that is possible. $\endgroup$ Nov 12 '14 at 2:19
  • $\begingroup$ But in any case that is not an issue. if m1 is negligible 2*m2 is negligible too. Other wise you should also say that the kinetic energies are the same (negligible) and thus in that case the answer would be yes by definition. So you are using the concept of negligibility different in both cases. $\endgroup$ Nov 12 '14 at 2:30
  • $\begingroup$ I updated the answer $\endgroup$ Nov 12 '14 at 3:15

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