A hexagonal Klemperer rosette

A Klemperer rosette is a system of an equal number of two kinds of bodies arranged in a regular alternating pattern around the center of mass. These systems were first described by Klemperer in 1962.
The figure below shows a hexagonal rosette. The distances of the two kinds of bodies from the center are in general different.
Klemperer also described rhombic and octagonal rosettes.

To construct this simulation we have used some rather complicated formulas to calculate the velocities of the bodies after choosing their distances from the center of the polygon and the mass of the lightest bodies (1, 3 and 5). The mass of bodies 2, 4 and 6 must then have a particular value, as shown in the right column.
The parameters are calculated to high precision.

Exercise

  1. Start the simulation. The orbits seem stable at first, but what happens after a while?
    To make the orbits stable for many revolutions, the initial positions and velocities must have even higher precision.
    All rosettes are unstable. Any deviation along a tangent from the "correct" position brings a body closer to one neighbor and further from another. Then the gravitational force becomes greater for the closer neighbor and less for the farther neighbor, pulling the perturbed body further towards its closer neighbor and amplifying the deviation rather than reducing it. An inward radial deviation causes the perturbed body to get closer to all other bodies, increasing the force on the body and increasing its orbital velocity, which leads indirectly to a tangential deviation.
  2. Looking at the graph in the right column, it is apparent that there exists another possible domain for
    μ less than 0.41. The maximum of μ is 0.31891 for ρ = 0.02801.
    Construct a simulation using these values.

Here we derive equations (2) and (5) in the paper by Klemperer (which is not done in the paper).
We then calculate the velocities in a particular example of a hexagonal rosette.

(Adapted from fig. 2 in Klemperer.)

We use the law of cosines to find the distances d45 and d46:

To calculate the y-components of the gravitational forces on body 4 we need cos α and cos β. The law of cosines gives

From the figure we see that β = 30o so

Because the x-components of the force cancel pairwise, the total force on body 4 (and on 2 and 6) is

To find the force on the body 1, 3 and 5 we just change the indices:

We now postulate that all six bodies may rotate with the same angular velocity ω. Then

These two equations can only be true simultaneously if there is a certain relationship between the radii and the masses. Solving the first equation with respect to ω2 and substituting in the second equation, we get:

Introducing new variables

this can be written as

Simplifying we get

Solving for μ:

which is equation 5 in the paper by Klemperer.
This is the relevant part of the graph showing μ as a function of ρ:

If we assume that m1 is the the largest mass, μ must be less than 1, and then we see that ρ must be less than 1.62 and greater than 1. If ρ = 1 the rosette reduces to the system of six equal bodies described in the simulation "Hexagonal system".

If we choose ρ = 1.4 then μ = 0.76168. With m1 = 1·1030 kg and r1 = 1·108 km the angular velocity becomes

The velocities are