# How to resolve the Moon's observed angular velocity with that given by Kepler's Third Law

According to the sidereal period of the Moon's orbit around the Earth of 27.32166 earth days of 86400 seconds we get an angular velocity of $$\frac{2\pi}{27.32166×86400} = 2.6617×10^{-6}$$. (NASA's factsheet has the Moon's sidereal period at 655.728 hrs.)

If we use the NASA factsheet figures for the Earth mass of $$5.9724×10^{24}$$ kg, the Moon mass of $$7.346×10^{22}$$ kg, the semimajor axis of 384,400,000 m, and using a figure for 'Big G' of $$6.6743×10^{-11}$$, then we get an angular velocity of $$2.6654×10^{-6}$$. Kepler's angular velocity implies that the combined mass of the Earth and the Moon are too high.

Assuming that the figures are all accurate to the last significant figure, as stated, then there are possible errors of 0.0015% in Big G, 0.0017% in the Earth mass, 0.014% in the Moon mass, and probably as small as 0.0003% in the semimajor axis, but the error in omega is 0.139%.

How to reconcile ... what have I missed ? Or is it something to do with a time-weighted average separation that is somewhat closer to 385,000,000 m ?

The widely cited figure of 384400 km is the Moon's mean inverse sine parallax value. This is not the same as the semi-major axis length, which is closer to 385000 km. An even better value is 384748 km, which is the value specified in The lunar ephemeris ELP2000. I generally use 385000 km unless I need a more precise figure.

A nice easily memorizable value for the mass of the Moon is 0.012305 Earth masses. Think of 0.012345 and remember to change the 4 to a 0. The Earth's gravitational parameter is $$3.986004418\times10^{14}\,\text{m}^3/\text{s}^2$$. Applying Kepler's laws, this results in an orbital period of $$\frac{2\pi}{\sqrt{(3.986004418\times10^{14}\,\text{m}^3/\text{s}^2) * (1.012305) / (384748\,\text{km})^3}} = 655.72\,\text{hours}$$

• Solving for the semi-major axis length using a sidereal rotation period of 655.728 hours yields 384749 km as opposed to 384748 km. This is akin to arguing about the number of angels that can dance on the point of a pin. Jul 30 at 10:34
• Yes, I worked backwards and got that figure, which mysteriously appears on a Wikipedia page, but with no explanation. I also eventually found some notes on Geophysical Gravity from eps McGill, CA, at eps.mcgill.ca/~courses/c510/gravity.pdf, which confirmed the figure. Aug 1 at 7:52
• Where do you get your figures for G' and the relative mass ? The G' figure is mildly different from the one in the ELP2000 paper and both are different from those in another paper I found at proba2.sidc.be/aux/data/spice/docs/…. Not that I need such accuracy, but I do like to get a consistent set of data. Aug 2 at 5:06
• @JerryFrog Don't use G, don't use M. Instead use the standard gravitational parameter mu, which is the product of the two. Aug 2 at 7:21
• agreed, that was why it was G' instead of G. Aug 2 at 9:48

For the general case of two masses interacting according to Newton's Law of Universal Gravitation, the two masses actually orbit about the center of mass of the system, not necessarily the center of the more massive body. Recall the equation for center of mass. For a two mass system, we will refer to the separation of the two masses as a = r1 + r2 , where r1 is the distance of mass m1 from the center, and r2 is the distance of mass m2 from the center. Consider the case when the two masses are in circular orbits. During their motion, the two planets must be acted on by a centripetal force. Now, by Newton's third law, these two forces must be equal in magnitude (and opposite in direction), which means m1r1 = m2r2 . This actually proves that the center of the circular motion is the center of mass. PP = [4pipi /G(m1+ m2)]aa*a, which is the expression corresponding to Kepler's third law.

• This doesn't answer the OP's question, where the commonly cited value of 384400 km as the semi-major axis length results in a conflict with the Moon's sidereal rotation rate. Jul 30 at 19:56