Is there anything interesting about an n-body system whose potential energy is constant?

Question

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.

Answer 1

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.

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

and

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

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$

and

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

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.