standard elliptical orbit - Moonstick Information Site

The Standard Elliptical Orbit

First let me propose a orbital path. Then I will prove that it satisfies Kepler's First Law (the orbits of the planets are ellipses with the sun at one focus), Kepler's Second Law (a line connecting the planet and the sun sweeps out equal areas in equal times), "conservation of angular momentum" (which is actually the same as Kepler's second law), and "conservation of energy". (The last two proofs are actually sufficient by themselves; the first two proofs are just conceptual conveniences.)

mathematical orbital path solution with proof

As you can see, this solution checks on all four points and is indeed the correct solution.

Now lets put Kepler's laws to work on the orbit of the moon. In the above equations, e is the eccentricity. The greater the eccentricity, the more elliptical the orbit. For the moon's orbit, this is about .055.

As you can see, the main effect of the eccentricity of the moon's orbit is that the moon will periodically be 6.3° ahead or behind of where it is expected to be. This changes the times of moonrise and moonset by about 25 minutes.

When the moon is at is closest approach to the Earth (on the left side of the diagram, called perigee), it is at its expected longitude. About one week later (on the bottom of the diagram), the moon is about 6.3° ahead (to the east of) where it would be if it circled the Earth uniformly. By the time it reaches apogee, it is back in step with the fictitious "mean moon". Halfway back to perigee (as depicted in the diagram), the moon is 6.3° behind causing moonrise and moonset to be about 25 minutes early everywhere on the Earth.