Why doesn't the moon leave earth's orbit?

Actually, within the solar system's reference frame the moon actually orbits the sun - the Earth just disturbs the Moon's orbit slightly because they're kind of tied at the hip. We of course, see the Moon as orbiting the Earth - Big is beautiful!<img src=icon_smile_wink.gif border=0 align=middle>

heheh - I thought that's what I said, but never mind.

I appreciate both replies, but makis is closer to what I was asking. I am wishing to know if there is a relatively uncluttered mathematical explanation as to why the moon should remain gravitationally bound to the earth, in the face of the fact that the sun pulls on it twice as strongly as earth(and thus it would seem that the moon should depart from earth orbit and orbit the sun instead).

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The gravitational pull of the sun on the moon is slightly more than twice the earth's pull on the moon(Newton's gravitational Law). So the moon is constantly accelerating towards the sun(in the order of several millimeters per second squared).
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This is an interesting observation from the point of view of the Moon. But, considering again, from the point of view of a double planet pair (Earth and Moon) sharing the same orbit around the sun, and taking the simplistic viewpoint of comparing the Sun's forces acting on the Earth and the Moon.

If we calculate the force between the Earth and the Sun -

Force of gravity = G * (gravitational constant 6.67E-11 Nm^2/Kg^2) * ( mass of Sun ) * ( mass of Earth ) / ( distance ^2 )
=6.67E-11*2E+30*6E+24/2.25E+22
.56E+22g

The force between the Moon and the Sun is -

Force = G * ( mass of Sun ) * ( mass of the Moon ) / ( distance ^2 )
=6.67E-11*2E+30*7.35E+22/2.25E+22
=4.36E+20g

So we actually have an attractive force between the Earth and the Sun about eighty times stronger than that between the Moon and the Sun, so if anything, it would be the Earth that the Sun would suck away from the Moon!!

I won't clutter up this thread with maths, but there's a bit more than simple gravitational attraction going on here.

makis makes a good point about the receding Moon. This is a phenomenon which can be considered in terms of the Earth/Moon system alone. The tidal effects are non-linear over time (the creationists cite a linear regression of the Moon's recession as evidence of a younger Earth), and as the continents move around in the oceans due to tectonic activity, the tidal response function of the oceans changes over geological time periods. The ocean response function is pretty complex, consisting of a whole spectrum of resonance states depending on the distribution of continental masses, and the current distribution has the oceans in resonance. The net result of this is that the current recession rate of the Moon from the Earth is anomalously high.

Ever increasing complexity - there's never a straightforward solution...

Well, makis I didn't forget the Moon calculation, it's just not relevant to my point so I excluded it, but thanks for including it anyway - as you can see the gravitational force between the Earth and the Moon is rather small when compared to the force exerted by the Sun on both bodies.

The Moon's orbit about the Earth is actually slightly elliptical, with a small ratio of difference between its major and minor axes in the region of 5%, this results in about a 19,200km variation in distance. Close satellites like the Moon and the Jovian satellites tend to have their orbits "smoothed out" to circular ones due to the tidal effects alluded to above.

PS - Sorry for the calculations, but I thought it better to actually demonstrate from basic principles the magnitude of the forces per se. I'll keep the squiggles to a minimum from now - promise!
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oops - forgot to add that although the Moon orbits the Earth from our reference frame, the Earth and Moon orbit about a mutual centre of mass, or barycenter - which in turn orbits the Sun, with the two bodies wobbling about this barycenter. This was the point I made in my first post, i.e. the Moon doesn't have to escape the Earth; it's already in an elliptical orbit around the sun, but two lines probably wasn't enough.

Thanks for the discussion, Atko and makis. My concern all along has been how to account for the continual acceleration of the moon towards the sun due to the net force pulling it towards the sun. If there were no accounting for it somehow, then the moon should indeed depart earth orbit, and rather quickly, as calculations point out.

We can calculate the moon's acceleration towards the sun 2 ways:

1) via net force/mass of moon (using an average net force)
2) via v^2/r (centripetal acceleration of objects in circular motion)

If the 2 calculations equate, then that would support the idea that the moon's orbits around the sun(as a partner of earth), where acceleration may be calculated the 2nd way, accounts for the acceleration due to the "tug of war", calculated by the 1st method.

So, Method 1:

Force on moon towards sun when moon is behind earth from sun("outer syzygy") = (4.36+1.99)E+20 = 6.35E+20 N
Force on moon at "inner syzygy"(moon between earth and sun) = (4.36-1.99)E+20= 2.37E+20 N
"Average" force on moon, directed towards sun: 4.36E+20 N
"Average" acceleration of moon towards sun, as it orbits earth, is average force / mass of moon: 4.36E+20/7.36E+22 = .006m/s^2(about 6mm per second squared)

My concern here is, if nothing accounts for this acceleration, then after a couple of weeks(about 1.2 million sec),the velocity resulting from this acceleration compares to the moon's orbital speed.

Calculating the acceleration via method 2:
r=1.50E+11 m
2*pi*r= 9.43E+11m (the circumference of earth's orbit)
Orbital speed V = circumference/year(in seconds)
= 29886 m/s
acceleration = v^2/r =.006m/s^2 (about 6mm per second squared)

So, the centripetal acceleration accounts for the "tug of war" acceleration. So we can lay this problem to rest.

Of course in all this we haven't considered tidal effects, or the ever widening earth-moon distance because of tidal effects. A topic for a future discussion?