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( 16 comments — Leave a comment )
elven_ranger
Jan. 28th, 2009 01:53 pm (UTC)
I love the link about the second "moon"
fayanora
Jan. 28th, 2009 10:14 pm (UTC)
Me too! :-)
(Deleted comment)
fayanora
Jan. 28th, 2009 10:14 pm (UTC)
Silly people!
arinwolfe
Jan. 28th, 2009 04:09 pm (UTC)
you always post the coolest/ oddest stuff
fayanora
Jan. 28th, 2009 10:16 pm (UTC)
Thanks!
kengr
Jan. 28th, 2009 09:55 pm (UTC)
Actually, the article itself mentions another moon (other than Luna). So that makes *three*. Add in the fact that I'd heard of two (other than Luna) *before* this, and the total is at least *four*.

fayanora
Jan. 28th, 2009 10:10 pm (UTC)
WOW!
kengr
Jan. 28th, 2009 11:55 pm (UTC)
Ah. You are awake. Getcher butt in gear and get over here. :-)
christinaathena
Jan. 29th, 2009 04:37 am (UTC)
Well, none of them are actually "moons". They're simply objects in a 1:1 orbital resonance with the Earth. They orbit the Sun, not the Earth, and are therefore not true moons. Several of them are likely to become are quasi-satellites at some point, but it will only be a temporary situation before returning to a normal solar orbit.

EDIT: Wikipedia's inconsistent. The quasi-satellite article lists three quasi-satellites, but the article for Cruithne says it's in a horseshoe orbit, and the article for 2003 YN107 says it was in a quasi-satellite orbit until 2006.

Edited at 2009-01-29 04:43 am (UTC)
kengr
Jan. 29th, 2009 07:28 am (UTC)
Luna orbits the sun rather than earth too.

Sol exerts 4 times the force on Luna that Earth does.

christinaathena
Jan. 29th, 2009 02:54 pm (UTC)
The Moon orbits the Earth, not the Sun.

http://blogs.discovermagazine.com/badastronomy/2008/09/29/the-moon-that-went-up-a-hill-but-came-down-a-planet/ (Edited for a better explanation)

the fact that the Sun’s effect of gravity on the Moon is stronger than the Earth’s is interesting, but not relevant. Why not? Because the Sun is pulling on both the Earth and the Moon! Since both the Earth and Moon are roughly the same distance from the Sun (the Sun is 400 times farther away from us than the Moon is), the acceleration due to the Sun’s gravity on the Moon and the Earth is about the same. That means the Sun’s gravity is important in keeping the Earth and Moon together orbiting the Sun, but doesn’t affect the Earth and Moon as a system. The Moon is still gravitationally bound to the Earth.

Edited at 2009-01-29 03:00 pm (UTC)
kengr
Jan. 29th, 2009 08:33 pm (UTC)
And the exact same thing applies to the other bodies that you were saying weren't moons.

They are also gravitationally bound to the earth.

And as I said in the first place, work out the acceleration the Earth applies to the moon, and the acceleration that the sun does. It's quite simple math. And you'll see that the sun exerts 4 times the force the earth does.

Earth & Luna are co-orbitals. More tightly bound than most, but still...
christinaathena
Jan. 29th, 2009 09:12 pm (UTC)
Actually, no. The relevant issue isn't the relative gravitational strength of the Sun on the Moon, but the DIFFERENCE between the force the Sun exerts on the Moon and the force it exerts on teh Earth, which is minimal. It's like saying that the Sun is the main influence on the tides because the Sun's gravitational force on the Earth is stronger than the Moon's. But, that would be a false conclusion, because the relevant issue is the difference in the gravity between the facing and opposite sides (which is why the high tides are the points immediately underneath the Moon and the antipodal point, while the relative positions of the Sun and the Moon determine the strength of the tides). Likewise, the relevant issue isn't the Sun's gravitational force on the Moon, but the *relative attraction*. Earth and the Moon are pulled towards the Sun with only minimal difference. Even at its greatest, when the Moon is at new or full, the difference is tiny.

Viewed from a geocentric perspective, the Moon's path deviates only slightly from what you'd predict ignoring the Sun's gravity. Viewed from a heliocentric perspective, the difference is far greater between the actual path and the Earth-ignoring calculations.

For objects such as Cruithne, the path from a geocentric orbit doesn't even circle the Earth, but an imaginary point further away. It's orbit is predominantly determined by the Sun's gravity with influence from the Earth's.

Earth and Cruithne's orbits around the Sun: http://en.wikipedia.org/wiki/File:Orbits_of_Cruithne_and_Earth.gif

Cruithne's orbit from a geocentric frame of reference:
http://en.wikipedia.org/wiki/File:Horseshoe_orbit_of_Cruithne_from_the_perspective_of_Earth.gif

As you can see, Cruithne's path relative to the Sun is a more-or-less standard ellipse. Relative to the Earth, it is a weird bean-shaped path.

On the other hand, the Moon's path relative to the Sun is a wavy path, while relative to the Earth it is an ellipse.

In as far as it makes sense to describe it as orbiting only one object, the Moon orbits the Earth with significant influence from the Sun, while Cruithne orbits the Sun with significant influence from the Earth.

Edited at 2009-01-29 09:14 pm (UTC)
kengr
Jan. 29th, 2009 11:09 pm (UTC)
I should have dropped the reference to relative gravitational force but I didn't feel like going back and editing.

Note, btw, that *tidal* forces work on an inverse *cube* law, and the forces on earth from the sun and the moon are quite similar.

A/L = GM/r^3

A = acceleration
L = distance from center of body that is having tides
G = gravitational constant
M = mass of body exerting tidal forces
r. Distance between centers of body exerting tidal forces and body tides are being induced in.


As for orbital shape, please note that Luna's path around the sun is at all points *concave* towards the un, while the moons of other planets are convex at many points.

So it's just as valid to say that Luna is orbiting the sun with its orbit strongly influnced by earth.
christinaathena
Jan. 30th, 2009 01:51 am (UTC)
Right. And the *reason* that tidal forces work on a cube rule is that the relevant issue is the relative difference between the near and far side. When the moon was closer, the ratio between the Earth's radius and the Moon's orbital radius was larger. Wikipedia has a good explanation of how the formula is derived. http://en.wikipedia.org/wiki/Tidal_force

The moon*'s orbit around the Sun is far more accurately described as a combination of an elliptical orbit around the Earth and the Earth's elliptical orbit around the Sun. Thus, it is far more reasonable to describe the Earth as its primary. Yes, it can be said to orbit both, but the Earth is more relevant to its path than the Sun. (Of course, if one ignores the Sun in one's calculations, there will be significant error)

Cruithne, etc., however, have paths that are primarily determined by the Sun. Cruithne's orbit is an ellipse with the Sun at one focus, just like any other planet or asteroid. The Moon's orbit is an ellipse with the Earth at one focus, just like any other satellite. They are very different things. If Cruithne and the like are satellites of the Earth, then you'd have to call all of the Trojan asteroids satellites of Jupiter, which no astronomer does, and would stretch the meaning of "satellite" to a preposterous extent. So, one COULD argue that the Earth has no "true" moon (although I would not argue so), but any number greater than two is incorrect under any definition of the word "moon" in use by astronomers.


*Incidentally, why do you keep using "Luna", but still use Earth and Sun? It seems to me that it's more consistent to either use the English names for all three or the Latin names for all three. I personally prefer the English names, but that's more of an esthetic issue.
kengr
Jan. 30th, 2009 02:34 am (UTC)
Actually. tidal forces are a lot more complex than that. You get forces twice as strong trying to *compress* the body at right angles to the forces trying to stretch it.

There's a good diagram and explanation in the late Dr. Forward's book "Indistinguishable from Magic".

And I used Luna, because that's unambigous. "moon" in a discussion involving Luna, Tor, etc is more than a bit ambiguous. :-)
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