Determining this horseshoe/co-orbital moon's possible orbital periods

Question

I've been working on a system with a habitable moon (Moon A) for a story, and I'm now trying to populate the other moons around the parent planet. I'm considering placing one (Moon B) in a co-orbit, horseshoe configuration, with my habitable moon.

While I understand the concept of these orbits, the actual calculations of the orbital times is beyond my mathematical capabilities.

I understand that on a horseshoe orbit there is an inside section and outside section where Moon B is catching up to, or falling behind, Moon A, respectively.

I understand that if the 'inside ring' is nearer the parent body (the 'planet' in the case of co-orbital moons) than the L1 Lagrange point of Moon A, then Moon B will escape the horseshoe orbit and just start orbiting the planet directly. And, similarly, if the 'outside' ring is outside of L2 Lagrange point of Moon A, Moon B will escape and orbit the planet directly. And if either ring is too near the actual orbit of Moon A, it will be a tadpole orbit(never passing the L3 Lagrange point of Moon A before returning to Moon A from the same direction it left it), instead of a full horseshoe.

What I can't figure out (again, math skills limitation) is How long it would take (longest and shortest possible times) for Moon B to make one complete cycle of the horseshoe? And so I can't decide if I should use this type of orbit or not. For example, 3753 Cruithne takes about 770 years to complete it's horseshoe cycle relative to Earth, far too long to be useful for my story. But I don't know how to calculate ho long my co-orbit scenario would take. I'm confident it could be made significantly shorter than 770 years, but exactly how short it would be is still a key factor in the decision.

I would like to know how often the two bodies would approach each other at the two extremes of possibilities, the longest possible time between complete cycles, and the shortest possible time between cycles.

For this question: Planet mass is 477 Earth masses, Moon A's mass is 0.11 Earth masses, Moon B's mass is 0.01 Earth masses. Moon A's semi major axis is 4 million kilometers. (please let me know if any other variables or details are needed)

To re-word the original question: By adjusting the semi major axes of the inner and outer rings of Moon B's horseshoe orbit, either closer to or farther from the semi major axis of Moon A, what are the longest cycle time, and shortest cycle time possible for a horseshoe orbit in this system?

Here is a visual representation of the types of orbit changes I'm referring to. The contour lines inside the highlighted one (nearer L3 L4 and L5) are what I'm referring to when I mentioned adjusting the horseshoe orbit Axes nearer to the semi major axis of Moon A. And the contour lines outside of the highlighted one are what I'm referring to when I mention moving those axes farther from the semi major axis of Moon A. When I refer to a complete 'cycle', I mean the time it takes for Moon B to go from Point A on that image, through Points B, C, D, and E and back to A.

Answer 1

The direct calculation of the orbital times is beyond just about everybody's mathematical capabilities. This is an example of the infamous three-body problem, for which there is no analytic solution in the general case (and it's not one of the few special cases that do have analytic solutions, either).

In order to get accurate results, therefore, you will have to resort to numeric simulation, and just sweep over a range of orbital parameters to simulate until you find the inner and outer edges where the system no longer displays horseshoe behavior.

To get a very rough approximation, though, we can just compare the keplerian periods of each orbit. Orbital period $T$ is proportional to $R^\frac{3}{2}$. The period of a horseshoe cycle will be the time it take the small world to lap the large one on the inner track, and vice-versa when the small world is on the outer track. The lap time $T_L$ is given by $T_L = \frac{T_O T_I}{T_O - T_I}$, where $T_O$ and $T_I$ are the orbital periods of the outer and inner moons, respectively. We can also make the simplifying assumptions that the large world will have its orbit changed only insignificantly, and the small world's orbit will differ from the larger world's orbit by the same amount in either direction.

Putting those all together, we first get the cycle time in terms of individual orbital periods to be $C = \frac{T_M T_i}{T_M - T_i} + \frac{T_M T_o}{T_o - T_M}$, where $T_M$ is the period of the larger moon, and $T_i$ and $T_o$ are the inner and outer orbital periods of the smaller moon, respectively. If we then sub in $R^\frac{3}{2}$ for $T$, using $r$ for the difference in orbital radius between the large and small moons, we get

$C = \frac{(R (R - r))^\frac{3}{2}}{R^\frac{3}{2} - (R - r)^\frac{3}{2}} + \frac{(R (R + r))^\frac{3}{2}}{(R + r)^\frac{3}{2} - R^\frac{3}{2}}$

This is obviously not a good approximation for the longest cycle time, because it tends towards infinity where $r$ approaches zero (and you end up with a fixed-separation Trojan orbit), by which time you will long have left the horseshoe regime behind. It also won't tell you how much separation you can have before the worlds are too far apart to exhibit co-orbital behavior anymore, but if you can establish that independently, it should be reasonably accurate (within an order of magnitude, anyway) for the fastest possible cycle times.