[canuck_conservative]

Sorry about the late reply.

This is actually easy to do, just use Kepler’s famous Third Law:

P squared / R cubed = a constant for any planet or sun.

where P = orbital period
and R = orbital diameter.

Example - Mars’ orbit is 1.52x Earth’s; So if you cube 1.52, then take the square root, you get that Mars’ orbital period is 1.87x Earth’s, or 684 days.

So in this case, cube 0.3333, then take the square root, and multiply it by the Moon’s current period (about 29 days), and you get - just under 6 days for 1 orbit back then.

[Verginius Rufus]

Thanks for satisfying my idle curiosity. Not sure what I can do with the knowledge, but it's something interesting to know.

I was aware of Kepler's 3rd law as it applies to the planets (Jupiter's period squared is equal to its distance cubed, etc.) but it had not occurred to me to use it to find an unknown value.