Re: OT? Why Doesn't the Moon Fall Down?

On Thu, 12 Jul 2007 07:42:35 -0400, the following appeared
in talk.origins, posted by "J.J. O'Shea"
<try.not.to@xxxxxxxxxxx>:

On Wed, 11 Jul 2007 21:54:07 -0400, Timberwoof wrote
(in article
<timberwoof.spam-31CECE.18540711072007@xxxxxxxxxxxxxxxxxxxxxxxx>):

Since the moon departing is coupled with the Earth's rotation slowing
down, no. Once the Earth and moon face each other continuously, that
orbit will, by itself, be stable (except for the loss of energy by
General Relativity). Then, as someone else pointed out, the sun's effect
to try to "stabilize" the orbits will start to take hold. That's where
Lagrangian orbital analyses come in. How far away is the moon supposed
to be when the Earth's rotation stops, and where are those near/far
Earth Lagrange pints?

There are five Lagrange points, or special-case solutions to the three-body
problem. For this to work, one of the three bodies must be _much_ smaller
than the other two. One of the large bodies must be smaller than the other
one, and must be in orbit around that larger one.

L1 is on the straight line linking the centres of mass of the two large
bodies, and is inbetween the two large masses, at approximately the point
where the gravitational pull of each body is equal.

L2 is on the straight line linking the centres of mass of the two large
bodies, is on the _other side_ of the smaller of the two large bodies.

L3 is in the same orbit as the smaller of the two large bodies, but at a
separation of 180 degrees.

L4 is in the same orbit as the smaller of the two large bodies, but at
separation of 60 degrees ahead. It is one point of an equilateral triangle,
with the larger of the two large bodies and the smaller of the two large
bodies at the other two points. In the Earth-Moon subsystem, that would mean
that all three are approximately 3.85 x 10^8 metres away from each other.

L5 is in the same orbit as the smaller of the two large bodies, but at
separation of 60 degrees behind. It is one point of an equilateral triangle,
with the larger of the two large bodies and the smaller of the two large
bodies at the other two points. In the Earth-Moon subsystem, that would mean
that all three are approximately 3.85 x 10^8 metres away from each other.

L4 and L5 are stable Lagrange points; objects in those positions will tend to
stay there. The others are unstable points; objects there will tend to move.
L4 and L5 are two _different_ equilateral triangles. As the orbit of the Moon
is not a perfect circle, the exact size of the equilateral triangles will
vary depending on where the Moon is in its orbit. L1, L2, and L3 are
effectively points; L4 and L5 are volumes. Bodies not exactly at the L1, L2
or L3 points will tend to move away from those points. Bodies at L4 and L5
will tend to move around inside a roughly kidney-shaped volume.

r = cuberoot (m / (M * R))

r is the distance from the smaller large body to either L1 or L2

R is the distance between the centres of mass of the two large bodies

m is the mass of the smaller of the two large bodies

M is the mass of the larger of the two large bodies.

Substituting in the masses of the Earth and the Moon and the semi-major axis
of the Moon's orbit gives L1 and L2 for the Earth-Moon subsystem of 6.15 x
10^4 metres from the Moon.

L3 is, of course, exactly opposite the position of the smaller of the two
large bodies.

Thanks! "I used to know all that stuff." ;-)

There are Lagrange points for the Sun-Earth subsystem, in the same relative
positions as for the Earth-Moon subsystem; in fact there are Lagrange points
for all the planets. One of the things which lead to Lagrange working this
stuff out was the observation of the leading and trailing Trojan asteroids,
60 degrees ahead and behind Jupiter.

*That*, I remembered.
--

Bob C.

"Evidence confirming an observation is
evidence that the observation is wrong."
- McNameless

.

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