View Full Version : Why not an equatorial moon

Cheap Astronomy
2008-Nov-02, 04:57 AM
Probably showing my ignorance of some basic astronomy here, but welcome edification.

I understand that by conservation of angular momentum, moons generally orbit a planet's equator (thinking of Uranus on it's side).

Anyway, our Moon is generally within 5 of the ecliptic whereas I would have thought it should out by around 23 at solstices and 0 at equinoxes because it's be roughly following the equator (with Earth axis tilted at 23). Wikipedia tells me it's between 18.29 and 28.58 to Earth's equator.

Am I expecting too much from our satellite companion?

2008-Nov-02, 06:49 AM
Too much from the Earth. The moon to some degree isnt a moon but a co-orbital planet.

To put it simply, the effect of the Sun on the Moon is large enough to force the moon to act a bit like a planet, and so stay more on the ecliptic, than a moon, and stay equatorial.

2008-Nov-02, 08:38 AM
I looked at the Wiki article and they obviously meant the celestial equator for that figure, not the Earths equator.

2008-Nov-02, 09:41 AM
Laplace plane (http://en.wikipedia.org/wiki/Laplace_plane).

Ken G
2008-Nov-02, 02:41 PM
I think that the key point here is, the Moon is thought to have broken off from the Earth in a huge collision with a third body. The angular momentum of the Earth and that other planet is then largely unconstrained, and the orbit of the Moon is also not forced to follow the spin of the Earth, though I suspect it would be unlikely for them to be vastly different.

Cheap Astronomy
2008-Nov-03, 10:58 AM
Awesome - thanks all. Particularly the Laplace plane - which is nearly as interesting as a Lagrange point.

This leaves me thinking that after the big whack (http://en.wikipedia.org/wiki/Big_Whack), the Moon probably was in an equatorial orbit, but has moved out through a Laplace plane to where it is now - increasingly lining up with the solar system disk due to the Sun's gravitational influence.