# How many celestial bodies could be in stable orbit at roughly the same distance from a star?

How many planet-like celestial bodies of roughly the same mass (say within 50%) could orbit a star at roughly (say within 10%) the same distance from the star and be in stable orbits. By stable I mean they would not require tethers or significant energy to stay there.

I feel sure the answer is at least 2, Earth plus another body at Lagrangian Point L3. I have heard it's not 3 -- that putting Earth-size bodies at Earth's L4 and L5 would destabilize. Is that true? Please explain concisely and cite reputably. I'm asking in the interests of Earthling living space within the solar system, in the far future of course. (Or other stellar systems in the uber far future.)

Thanks to @dgh for steering the terminology away from "planet" which is defined by the IAU as having cleared its neighborhood (i.e. orbital radius) of other bodies.

## Research Possibilities

• Klemerer Rosette (theoretically unlimited, but practically no more stable than a ball on a hill)
• Iridium Satellites (66 are active in stable, Low Earth Orbit, in six planes)

EDIT - Here's an excellent tool for playing around with this idea. (Though more visually than mathematically - Kids don't try this on your home planet.) Stefano Meschiari's Super Planet Crash game:

Note the untimely and unfortunate exit stage left of Planet 1, at what appears to be chillingly in excess of solar escape velocity. In the above simulation, 5 planets of varying sizes started out in the habitable zone, and stayed quite orderly for almost 25 years: they varied in distance from each other but stayed remarkably equidistant from the sun. The rerun seems to play out differently each run, due to butterfly effects I suppose.

On the other hand, the following is an 11-planet simulation that was stable for over 500 years:

I just clicked out the planets at the same distance to the sun, and they stayed there. This apparently defies the quote above, that the Klemerer Rosette is "no more stable than a ball on a hill". It may not sustain forever, but some dynamic could be observed keeping these 11 bodies apart.

• en.wikipedia.org/wiki/Klemperer_rosette , though of course such a thing would have to be artificial. Mar 1, 2014 at 3:55
• Very cool, @dmckee! Artificiality is no constraint. Alas, Klemperer admitted this formation is unstable, from wikipedia: "any motion away from the perfect geometric configuration causes an oscillation, eventually leading to the disruption of the system." Mar 1, 2014 at 4:07
• Regarding your edit ("this defies the instablity of the Klemerer Rosette"): 500 years is an eyeblink in terms of planetary evolution. Consider the possibility that the semimajor axis of Neptune doubled after its formation, and that 500 years is only three of Neptune's current orbits.
– rob
Jan 13, 2015 at 2:07
• @rob well agreed. Klemerer's exaggeration of the instability of his Rosette could lead to false hopes. Jan 13, 2015 at 16:06

This is trivially the case if you allow them to occupy the two stable Lagrange points of a Jupiter-sized planet. The theory behind the Largrange points assumes a stiff mass hierarchy for the three bodies, $M_1 \gg M_2 \gg m$.
• Here is a proof that orbits at $L_4$ and $L_5$ are unstable unless $M_1/M_2 \geq ({25 + \sqrt{625-4}})/{2} \approx 25:1.$
• Here is an answer at astronomy.SE claiming that stability requires $M_2/m \gtrsim 10$ for the case where $M_1/M_2 = M_\text{sun}/M_\text{earth}$;
The horseshoe orbit seems to be a stable configuration. I suppose it's possible to put four equal-mass objects into horseshoe-type orbits, if they alternate distances: the first and third objects could be on the outer track, and the second and fourth objects on the inner track. The horseshoe exchanges would take place on opposite sides of the central mass — initially, at least. However, I suspect that this configuration would be unstable for the same reason that $L_3$ is unstable; eventually you would wind up with three equal-mass objects quite near each other, and you'd probably have either a collision or an ejection.