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 Earth-sun 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.
(In reply to
Problem is oversimplified? by Kenny M)
KM wrote: "I accept the three previous solutions, as well as the 'official' solution."
Hmm - well, there is no 'official' solution here, except perhaps the published one in the Brannen paper. But, this is also the one I give in my post. Note - the solution says that for the Moon to make loops with a 27 day period (as the problem specifies) it must orbit beyond 6.9 M miles, not within. When you leave the orbit period fixed, as the problem asks, then it is the more distant Moon that orbits faster and make loops. A stable orbit depends on the central mass, and so changing the radius of the orbit by a factor of f while leaving its period at 27 days requires changing the Moon's speed by f as well, and also changing the Earth mass by f^3. For the solution, the factor f is ~ 6.9 M miles/0.24 M miles = 28.4. For the other part of the problem - the requirement for convexity, see below.
In the sense that keeping the orbit at 27 days is making an oversimplified problem, I am in complete agreement with KM. Likewise, considering a Moon orbit at a different radius but still taking 27 day is a wacky violation of Kepler's 2nd (& Newton's) Laws. But, showing that the actual orbit is reliably convex, even taking into account its non-circular aspects, is the point. I have more on this below.
That brings us to the solutions of t and C. t's intuitive solution for the minimum distance for loops to occur is correct: it is the distance beyond which the Moon speed exceeds the Earth speed. Using units of time as 1 year, and distance units as 1 Earth-Moon distance, the distance to the Sun d, and the period of the Moon as 27 days, we have:
speed_M > speed_E
circ_M/period_M > circ_E/period_E
2 pi (1) /(27/365)yr > 2 pi d/1yr
where circ = orbit circumference.
This Earth-Moon distance is > 1/13.5 d, where d is the distance to the Sun. So the Moon orbit radius is > 93/13.5 M miles or 6.9 M Miles.
This, in essence, is t's calculation and it agrees exactly with the curvature solution for loops. The "always-convex" requirement is more stringent and much more difficult to calculate: the Moon must orbit more slowly and closer than 6.9 M miles, with a distance < d/p^2 or < 93/13.5^2 = 0.51 M miles, which it does. It's at about 0.24 M Miles.
Charlie, on the other-hand, solved a different problem by finding a 0 total centripetal force on the new moon from it's Earth and Sun orbits by asking that their centripetal accelerations cancel. To make them cancel, an R_em (Earth-Moon) distance was be found so that v_em^2/R_em = M_m v_sm^2/R_sm (where M_m drops out.) But he also noted the violation of Kepler's law to move the Moon outward while leaving its period alone. He essentially found a Lagrangian Point that would apply if the Earth were 28.6^3 more massive (in order to support this 27 day orbit with the needed centripetal force). The 28.6^3 comes from Newton's version of Kepler's 3rd law relating period, radius and mass.
Finally, onto KM's comments, with my comments interspersed:
"However I submit that the constant 27 day lunar orbit period is not at all negligible (as implied by Steve Lord). [I was suggesting that inserting a more the exact period for the Moon will not affect the result at all significantly] In fact, in Steve's solution, the parameter "p" (ratio of Earth period to Moon period) actually varies (using appropriate physics) proportionally to the lunar orbit radius to the -3/2 power. This means the rate of change of "p" can be faster than that of "d". I *think* this flips the solution space on it's head, meaning, (for Earth/Moon masses):
[Ah, I am not sure what's being suggested here. It's the distance to period ratio that's at stake, not minor changes of either, nor their "rates of change". Here is how I view things: The problem as posed asks to consider a 27 day sidereal period. It is completely true the Moon's sidereal period, synodic period and particularly its anomalistic period (as the apsidal precession completes in 8.89 years) all vary as the
complex elliptical orbit does its thing. But these "lunar month" variations are on the order of 7 hours each 27 day sidereal or 29 day synodic orbit and do not really enter into the problem. The problem solution is aimed at pointing out how well into the "always convex" zone the Moon is. This result does not change by adding 7 hours more of less to its 27 day sidereal period.]
Loops only occur for very small orbital radii. [Ah, make that high velocity orbits.] As the lunar orbit radius increases, you then transition to wiggles, then convex. [Not if you keep the period fixed - then it's the opposite. What you are pointing out is that a smaller radius lunar orbit while keeping the same Earth mass will result in higher orbit velocities and, when fast enough, loops. I don't contest this, but it is not the problem as stated.] I think the transition points are: Loops below approx. 453 km lunar radius, Loop to Wiggles transition - about 453km, Wiggles to Convex - about 260600km. Convex above 260600km lunar radius."
Edited on May 25, 2021, 6:31 am
re: Problem is oversimplified?
