It comes as a surprise to many that the orbit of the moon around the sun has no loops in it. Indeed, it is a convex curve not very different from the orbit of Earth around the sun.
How far away from Earth would our moon have to be for the moon's orbit around the sun to have a loop? How far away for it to be nonconvex?
Assume all orbits are circular and all lie in the same plane (so that "loop" and "convex" have clear planar meanings), the Earthsun distance is 93 million miles, the Earth's orbit requires 365 days, and the moon's orbit around Earth takes 27 days (and that is constant in this problem). Using such approximations has negligible impact on the problem.
Note that the moon's orbit is "prograde": in the same direction as Earth moves around the sun. Both motions are counterclockwise, viewed from our north pole.
Initially I thought that the boundary between a convex lunar path and a concave lunar path would be at the point where the fullmoon position (opposite the sun) would form the vertices of an almostdodecagon (nonclosed, as there aren't an integral number of months in a year), where the newmoon positions (between earth and sun) were at the midpoints of the pseudododecagon.
However I realized that at full moon, the moon would not be headed directly toward the new moon position, so it would extend outside the dodecagon and then approach it asymptotically, implying it's already concave, so this is not the boundary condition.
Thinking further leads me to conclude that we want the inward acceleration to equal the outward acceleration at new moon. The outward (towards earth) acceleration, from earth's gravity is v^2/r where v is the moon's velocity relative to the earth and r is its distance from the earth.
The inward (towards the sun) acceleration would be the same formula but with v relative to the solar system and r the distance from the sun.
As the earthmoon system moves 2*pi*93,000,000 miles in 365 days, either of these is moving 1,600,000 mi/day around the sun.
Let r represent the lunar distance from earth. Its speed is then 2*pi*r/27 ~= .116*r mi/da.
Setting
(1.60x10^6)^2/(93x10^6) = .116^2 * r
r = (1.60x10^6)^2/(93x10^6) / .116^2
This is over 2 million miles, over 8 times its actual distance.
This ignores the fact that the speed of the moon would be considerably less if it were that far away from the earth, but I believe the spirit of the problem is more mathematical that in accord with physical reality, as the relation between distance and velocity is not mentioned in the puzzle.
In fact I think a balance between the gravitation of the sun and of the earth (resulting in equal accelerations in opposite directions), using the appropriate relations of speed to distance, would actually place the moon at one of the Lagrangian points of the earth's orbit, where the moon kept station between the earth and the sun, rather than going around the earth.

Posted by Charlie
on 20210520 11:22:45 