The Moon's orbit is tilted 5 degrees to the ecliptic so as it orbits the
Earth it alternately dips below and rises above said ecliptic. The two
places where the orbit crosses the ecliptic are called "nodes", an
ascending node and a descending node. Precession of the Moon's orbit causes
the orbital plane (and the nodes) to rotate counter to the Moon's rotation
once every 18.6+ years or somewhere around 19.4 degrees per year. The
rotation of the orbital plane is separate from the nodes lining up twice a
year due to the Earth/Moon combo orbiting the Sun which will cause the said
nodes to line, once with a node between the Sun and the Earth and the other
behind the Earth and once just the oposite. All of the before info is
readily available from Astronomy books or through Google but what is not
available (I cant find it anyway) is a chart to show where the Moons
orbital plane is at any given time relative to it's 19.4 degree per
year/18.6 year cycle. Thanks in advance for any help.

So, call me redundant but from the nodes alone you can determine the position of the orbital plane. In other words, 'behind' the descending node is the part above the ecliptic and 'behind' the ascending node is below the ecliptic, with 'behind' determined relative to the direction of orbit. I can't draw pretty pictures, either with pad and paper or with Java, but that's a start.

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The Moon's ascending node is given here as 125.08° on 2000 Jan. 1.50 TT. Its ascending node at any other time can be estimated by multiplying the number of elapsed years by -19.4.

Actually the Moon doesn't orbit about the Earth, it orbits about the sun, but since the Earth and Moon's orbits are so close they perturb each other's orbits in such a way as to appear to orbit about a common centre of mass. Technically we are in a binary Planet system.

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phantom, I'd buy that if the center of mass wasn't firmly inside the earths surface. As such, the earth isn't even perturbed by 1 earth radii. Therefore I'd call it a moon.

If you call us a binary planet, what's to stop mars and it's moons being a trinary? or jupiter and its moons...

Now, if that center of mass was outside the earth...then I'd agree whole heartedly.

"Actually"? That's just two different ways of describing the situation. They're equivalent.
Technically, that's false. Technically, there are only nine planets.

Actually Milli, in some sort of way, he's right. The sun attracts the moon over twice as strongly as the Earth does. It is the only major (over 1000 km in diameter) and even most of the minor satelllites of which this can be said. The only other satellites that are in this situation are minor (and very minor) satellites of the gas giants, which are considered captured asteroids. If you could draw the orbits of the Earth and the Moon to scale, you would see that the moon's orbit (like the Earth's and the rest of the planets) is always concave toward the sun. All the other satellites (except for those minor ones) are, at some point in their orbit, concave toward their primary.

Yeah, close enough not to really matter.

While I don't think the moon is, and agree that there are only, nine planets (no comments about Pluto ), don't you find it interesting that the moon's motion carries it through the ecliptic (tilted by 5 degrees or so) , and not, like almost all other satellites, through the plane (or thereabouts) of their primarie's equator.

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No, I know all about the specifics, that's not what I was objecting to. I was objecting to the use of the word "actually." But it's all relative.

RAAN = Right Ascension of Ascending Node
http://www.amsat.org/amsat/keps/kepmodel.html#raan
"Draw a line from the center of the earth to the point where our satellite crosses the equator (going from south to north). If this line points directly at the vernal equinox, then RAAN = 0 degrees."

On this page (thanks to cityboy916)
http://ssd.jpl.nasa.gov/sat_elem.html
It is stated that for Epoch 2000 Jan. 1.50 TT the RAAN is 125.08 degrees.


Edit : Thanks to all, I think I have it now.
This is a great place, I read it first every morning.

Not in the slightest. The ecliptic is, by definition, the plane of the Earth's orbit. As such, it bisects Earth through its center of mass (well, OK, the Earth-Moon system's COM). Anything orbiting that same COM is either going to (a) cross the plane twice per orbit (b) orbit in that plane. If you compute Jupiter's ecliptic ("jocliptic"?) all its moons will cross that plane twice per orbit as well.

And unless my understanding of physics is woefully insufficient, the Moon also crosses the plane of Earth's equator twice per orbit.

Or did I misunderstand that statement completely?

The inner moons of the major planets all orbit within a few degrees of the equator. The equators of the major planets are inclined by tens of degrees to their orbital planes. Saturn is inclined 27 degrees (I think), and Uranus is inclined 98 degrees. I forget the inclinations of Jupiter and Neptune, but I think they're between 10 and 30 degrees. Corrections will be gratefully accepted.

Now the Earth's equator is inclined 23.5 degrees to the ecliptic. The Moon's orbit is inclined roughly 5 degrees to the ecliptic, not the equator. The fact that the moon's inclination puts it closer to the ecliptic than the equator is taken as an argument in favor of the Moon being made from the debris of a Mars-sized protoplanet that also orbited in a plane near the ecliptic. Satellites wih inclinations near their primary's equator are believed to have formed from the material that went into forming the primary.

The Moon's orbit has an inclination to the equator that varies from 18 to 28 degrees over a period of 18.6 years. It just so happens that we are at the part of the cycle where the inclination is near maximum, resulting in very low full moons in summer and very high full moons in winter. This may be the reason behind some of the woo-woos claiming that the moon is "out of its orbit". There is nothing wrong with the Moon's orbit.

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I get it now. Thanks for explaining, and that is an interesting little factoid. But in my defense, I don't think Tensor's comment said quite what s/he meant it to

What?

The tidal pull of the Moon on Earth is 4 times stronger than the Sun pull so how could the Sun attracts tthe Moon over twice as strongly as the Earth does?

Moon orbits the Earth and it is a satellite of Earth. The center of mass is 2000 km below Earth surface.

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Well, gravity varies at an inverse square with distance and tidal force varies at an inverse cube with distance. Also, the moon's tidal force on the Earth is only twice as much as the sun's IIRC.

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Just to throw some numbers out there:

M_sun = 2x10^30kg
M_earth = 6x10^24kg
1AU = 150,000,000,000m
Earth-Moon distance = 350,000,000m
(from Google)

F_g = GMm/r^2

F_g(sun)/F_g(earth) = (M_sun)*(r^2)/(M_earth)*(R^2)
= (2x10^30)*(1.2x10^17)/6x10^24)*(2.2x10^22)
= 2.4x10^41/1.32x10^47
= 2.4/1.32
= 1.82

(Huzzah for showing all your work! God I'm bored.)

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Google said 350,000,000m? I don't think the moon ever gets that close--and right now it's 400,000,000m away.
Here's an extra-credit question then. How come Tensor says the sun attracts the moon over twice as strongly as the Earth does, but your number is 1.82?

NASA Reference Publication 1349
Twelve Year Planetary Ephemeris: 1995 - 2006

http://sunearth.gsfc.nasa.gov/eclips...n4.html#mo2004

Moon on August 14, 2004 is at 402,100 kilometers
Total range 356,000 min. to 405,000 max. kilometers

Let's try this with the 400,000 km value:

M_s = 2e30 kg
M_e = 6e24 kg
r_s = 1.5e8 km
r_e = 4e5 km

F~M/r^2 =>
F_s/F_e = (M_s*r_e^2)/(M_e*r_s^2)
= (2e30*16e10)/(6e24*2.25e16)
= 38e40/13.5e40
= 38/13.5
F_s/F_e = 2.81 > 2

Now, just to check: 2.81*(3.5/4)^2 = 2.81*0.766 = 2.15. Hmm. What about: (2.81/1.82)^(1/2) = (1.54)^(1/2) = 1.24; 4e5/1.24 = 3.2e5. Hmm again. I can't get my value to turn into Ut's, but at least it's bigger than 2.

Jupiter has an obliquity of 3.13° and Neptune's is 28.32°.

Thank you!

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