Orbital period of the Moon?

[Bob B.]

The orbital period of the Moon can be calculated from the formula,

P^2 = 4pi^2*(R-c)R^2/GM

where P is the period, R is the Earth-Moon distance, c is the distance between the Earth and the earth-moon center-of-gravity, G is the constant of gravitation, and M is the mass of Earth. We know,

R = 384,403,000 m
GM = 3.986005E+14 m^3/s^2

We can calculate c from the formula,

c = mR/(M+m)

The earth/moon mass ratio is 81.3, thus,

c = 1*384,403,000/(81.3+1)
c = 4,671,000 m

Therefore the orbital period is,

P^2 = 4pi^2*(384,403,000-4,671,000)*384,403,000^2/3.986005E+14
P^2 = 5.5574E+12 s^2
P = 2,357,400 s
P = 27.285 days

The published figure for the Moon’s orbital period is 27.3217 days. Why the difference?

The solution I show above assumes only two bodies, Earth and the Moon. My guess is the difference between the calculated period and the actual period is due to third body perturbations, specifically, the Sun. Can anyone confirm this? Is there another explanation for the discrepancy? Thanks in advance.

[antoniseb]

I haven't checked your numbers, but the first place I'd look would be at "R". The moon moves in an elliptical orbit around the Earth. Your formula seems to be about a circular orbit. Also, the moon is perturbed in that orbit somewhat, and has its perigee angle move around the ecliptic once every 18 or 19 years... and I believe because of the Sun's interaction, the moon does not follow a strictly elliptical orbit, but one that is squared out a little bit from that. Your calcualtion only deviates by 0.13% from the observed value.

You've chosen an interesting but complex topic. Hopefully someone with more depth of knowledge can tell you more specifically.

[Bob B.]

I haven't checked your numbers, but the first place I'd look would be at "R". The moon moves in an elliptical orbit around the Earth. Your formula seems to be about a circular orbit.

The formula should work for an elliptical orbit if "R" is equal to the semi-major axis. All the sources I've seen give this as 384,403 km (or some rounded off value). If I back-calculate the semi-major axis from the actual orbital period I get 384,748 km, which doesn't match any figure I've seen.

[grant hutchison]

If I back-calculate the semi-major axis from the actual orbital period I get 384,748 km, which doesn't match any figure I've seen.

This is the precise value of one of several numbers Jean Meeus discusses in his chapter What is the mean value of the Earth-Moon distance? in Mathematical Astronomy Morsels, ISBN 0-943396-51-4.
He derives it from a formula that is mathematically interchangeable with yours but goes on:

But this is not yet the end of the story. The gravitational attraction of the Sun not only strongly perturbs the motion of the Moon (whence the many periodic terms in the expressions for the Moon's longitude, latitude and distance), but it also has as consequence that [this formula] is not 'exact'. Due to the presence of the Sun, the sidereal revolution period P of the Moon is not equal to what might be deduced from the value of a by means of [this formula]. According to A. Danjon (Astronomie Générale, Paris, page 275 of the edition of 1959), all is going on as if the presence of the Sun would decrease the attraction Earth-Moon by the factor F = 1.002723.

Trickling through your formula, that translates into multiplying P^2 by 1.002723, or P by sqrt(1.002723) = 1.0013606, which does the trick to the three decimal places you provide for your calculated P.

Grant Hutchison

PS: In a handwaving sort of way, I can imagine this is dynamically related to the way in which the oblateness of the Earth slightly shortens the period of orbiting satellites.

[tony873004]

Can anyone confirm this?

Another way to write the formula is:

p = Sqr(4 * Pi ^ 2 * a ^ 3 / (G * (M1 + M2)))
Here's some values that are more accurate. These are the ones JPL uses. But they don't make a huge difference. The values you used were accurate enough:
M Earth = 5.97369125232006E+24 kg
M Moon = 7.34766310628124E+22
G = 6.6725985E-11

I just simulated this three times in Gravity Simulator.

The first simulation included the Sun, Mercury, Venus, Earth, Moon, Mars, Ceres, Pallas, Vesta, Jupiter, Io, Europa, Ganymede, Callisto, Saturn, Titan, Uranus, Neptune, Triton, Pluto, Charon. It took the Moon 27 days, 10 hours, 19 minutes to complete 1 orbit (27.4299 days). This was done in January when the Earth is closest to the Sun.

The second simulation included the above mentioned objects, but was performed in July, when Earth is furthest from the Sun. It took 27 days, 9 hours 46 minutes to complete 1 orbit. (27.407 days). I was expecting these numbers would avarage 27.323 from your original post of the published value, but they did not.

The 3rd simulation was 2-body only. Since the Moon's SMA varies over the course of an orbit, I had to wait until it was exactly 384,403 before deleting the other objects. Once I did this, the Moon's SMA remained locked at 384,403. It took the Moon 27 days 6 hours 50 minutes (27.285 days) This is correct to within 3 places to the right of the decimal with the computed answers.

If I back-calculate the semi-major axis from the actual orbital period I get 384,748 km, which doesn't match any figure I've seen.

The Moon's instantainous SMA varies between ~379,800 - 386,300 km in a single orbit. Back-calculating it for different days will give you different answers. (inclination and eccentricity too).

Historical records on the Moon's position go back centuries. I imagine the "published" value for period is simply time / orbits using historical values.

In a handwaving sort of way, I can imagine this is dynamically related to the way in which the oblateness of the Earth slightly shortens the period of orbiting satellites.

So this is where the great Grant hangs out now! I always learn something when you post.

The simulation I did uses point mass, and got the answer correct as a 2-body problem. I wouldn't have expected the Earth's oblateness to make a difference at the distance of the Moon. I think I'm correct in assuming that as distance approaches infinity, any object's gravity field, no matter what its shape, approaches what it would be if it were a point mass.

[grant hutchison]

I think I'm correct in assuming that as distance approaches infinity, any object's gravity field, no matter what its shape, approaches what it would be if it were a point mass.

You're right, the Moon's far enough away that I doubt the Earth's oblateness has much effect. I was referring to the effects of the equatorial bulge on closer-in artificial satellites, which have a predictable precession of nodes and periapse, and a non-Keplerian mean motion, all induced by the fact that they "see" additional mass in the equatorial plane.
The presence of the Sun (and other solar system bodies) induces similar movements in satellites that are distant from their parent planet - the satellite "sees" additional mass concentrated in the ecliptic plane. (These effects are still present, obviously, for close-in orbiters, but are swamped by the overwhelming influence of the nearby equatorial bulge of the parent planet.) The influence of the Sun is complicated, too, because its apparent position moves around as the planet orbits - so it has both periodic effects (those that oscillate around some mean value during the course of a year) and secular effects (those that accumulate over time). The alteration to the Moon's average period is obviously a secular effect, measurable only over centuries as you say, and any given lunar orbit will differ from that because of the periodic effects. So I think you might struggle to reproduce all the complexities within your gravity simulator in any reasonable length of time.

Grant Hutchison

[tony873004]

So I think you might struggle to reproduce all the complexities within your gravity simulator in any reasonable length of time.

The simulator doesn't make any effort to reproduce any of the complexities. But it still reproduces most of them, as they're a natural consequence bodies in motion being accelerated. The simulator simply runs through its list of objects with each object asking:

*Where am I (x,y,z)?
*How fast am I going (x,y,z)?
*How much am I being accelerated in the current time step due to the other objects' gravitational forces (x,y,z)?
*What's my new velocity (x,y,z) based on this acceleration?
*What's my new position based upon my new velocity applied in the current time step?
*Do it again and again.....

So it's easy to simulate things like the Moon's 18-year precession of nodes (Saros cycle), Cruithne's co-orbital relationship with Earth, Janus & Epimetheus' horseshoe orbits around Saturn, etc.

Only the complexities due to forces created by things other than point mass gravity are ignored. So it can't do things like Sun-synchronous orbits, the precession of Mercury due to GR, etc.

If I grab JPL numbers for the positions and velocities of solar system objects, simulate the numbers forward for 50 years at a slow time step (<16 seconds), then grab JPL numbers for positions and velocities 50 years in the future and compare them to my results, the Earth and Moon are still within 1 Earth diameter and 1 Moon diameter of where JPL's numbers say they should be. I can almost predict eclipses with them.

But the same is not true for asteroid 2004 MN4 (the asteroid that is going to make a close pass of Earth in 2029.) After only 7 years, my simulated position and JPL's numbers differ by ~90,000 kilometers, or ~7 Earth Radii. This is only an error in position of ~0.0016% considering the distance the asteroid traveled during this 7-year period, but it's much larger than the Earth or Moon's error.

I wonder why? The asteroid doesn't pass close enough to any body during this period for non-spherical shapes to have an effect. Solar radiation pressure and solar wind will have an easier time pushing around a 600m asteroid than a planet or moon. But are either of these forces strong enough to noticably change its orbit over a 7 year period? Or are other non-point mass forces more dominant? Or maybe Earth & Moon's error windows are larger but they just coincidentally are near the middle of their larger error windows giving the appearance of better precision.

[grant hutchison]

The simulator doesn't make any effort to reproduce any of the complexities. But it still reproduces most of them, as they're a natural consequence bodies in motion being accelerated.

Yes indeed.
My comment was just a response to your post:

I was expecting these numbers would avarage 27.323 from your original post of the published value, but they did not.

Given how these things work, I'd predict you'd need a long continuous run over several centuries (and some means of cumulatively averaging the Moon's sidereal period of revolution), rather than a couple of snapshots at orbital extremes.

Grant

[Bob B.]

Thanks, folks. I think you've answered my question.

[tony873004]

I did another simulation and found that the "published" value seems to be tied to the Saros cycle. Instead of taking 2 "snapshots" 6 months apart, I took them ~+18 years apart, the length of 1 saros cycle. During this run, it took the Moon 568877861 seconds to complete 241 orbits:

period = time / orbits

period = 2360489.04979253 seconds
or 27.3204751133395

This is only 1 minute 45 seconds different from the "Published" value.

Then as the simulation runs longer, the simulated period gets a little worse compared to the "published" value, until I reach 36 years (2 saros cycles) where the answer seems to get almost exact again, missing again but this time by 1 minute 44 seconds

So this leads me to believe that the "published" value is simply the average value over 1 complete Saros cycle.

Thinking about it, it doesn't make sense to go back centuries to get a good average. The Moon's period may have been slightly different a few centuries ago, and may not reflect today's value if it is continually increasing. For example, we could go back billions of years when the Moon orbited in about 1 day, average it with today's value and call the Moon's period ~13 days.

[grant hutchison]

Thinking about it, it doesn't make sense to go back centuries to get a good average. The Moon's period may have been slightly different a few centuries ago, and may not reflect today's value if it is continually increasing. For example, we could go back billions of years when the Moon orbited in about 1 day, average it with today's value and call the Moon's period ~13 days.

Sorry, I wasn't actually suggesting that the average was worked out from centuries of data - just that to estimate it precisely using your gravity simulator, you'd need to do centuries of simulation. The Saros cycles aren't exact repetitions of each other, as you know, nor is the position of Jupiter exactly the same from one Saros to the next.
I suspect the published mean period is derived mathematically, given that the theory of the Moon is now very precise. A mean motion can be calculated by average out all the periodic variations, and taking an inverse would give you a mean period.

Grant