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

On Tue, 10 Jul 2007 23:16:26 -0400, "J.J. O'Shea"
<try.not.to@xxxxxxxxxxx> enriched this group when s/he wrote:

On Tue, 10 Jul 2007 18:13:09 -0400, Terry wrote
(in article <v1189318noq8fivjci92qo2ab1kj744hg8@xxxxxxx>):

http://sciencehack.com/

The vid on the link shows that the moon is moving exactly the right
speed to prevent it from shooting off into space or be pulled to
earth.

It seems like something besides speed has to be keeping the moon where
it is.


Way, way, way back in the 17th century, a very bright (but not very nice) man
named Newton had a thought or two on that very subject.

f = G * M * m /(r^2)

where 'f' is the force due to gravity between two items

G is Newton's gravitational constant (experimentally discovered to be 6.67 x
10^-11 Nm^2kg^-2, in SI units. If you use some heathen Yankee units,
substitute those.)

M is the mass of the large item; the mass of the Earth is on the order of 6 x
10^24 kg.

m is the mass of the small item; the mass of the Moon is on the order of 7.35
x 10^22 kg.

and r is the distance between the centre of mass of the two items; the
distance between the relative centres of mass is about 3.85 x 10^8 metres.

Are you with me so far? Good.

Now, Mr. Newton also had a thought or two on that type of mathematics known
as 'calculus'. (This was one place where he worked hard to maintain his
reputation as not being a very nice man.) Those of us who are familiar with
calculus can use it to derive orbital and escape velocities from the above.
I'll spare you the math, but

vsubo = sqrt (G * M / r)
vsube = sqrt (2 * G * M / r)

where G, M, and r are as above, and vsubo is the velocity the smaller item
has in its orbit around the larger item. and vsube is the minimum velocity
required to escape orbit. Note that 'm' is no longer a factor, and that the
smaller 'r' is, the larger the velocity.

Determining the approximate orbital velocity of an object orbiting at the
Moon's distance from the Earth is left as a exercise for the student. Note
that this is only approximate, as the Moon's orbit is not a true circle, but
rather an eclipse, and that there are other bodies which act on _both_ the
Earth and the Moon, notably the Sun, and whose actions affect the Moon's
orbit. (There are also relativistic effects, but as those tend to have a lot
of sqrt (1 - (v^2)/(c^2)) factors, where c is the speed of light in vacuum
and equals approximately 3 x 10^8 m/s, they can be ignored for the purpose of
quick-and-dirty calculations.) Note that if you change vsubo, _by definition
_ r must always change... and it must always change the same way. The larger
the velocity, the smaller the distance.

r = G * M / (vsubo^2)

This is something which has been known since the 17th century.

You might want to look up 'Newton', and 'Kepler' and 'Brahe' and 'gravity'.

Perhaps now you know what keeps the Moon in its orbit.

Perhaps you also know just how much of an idiot you have made yourself
appear.

There is a very good article on how the messure the distance to the
Moon at http://physics.ucsd.edu/~tmurphy/apollo/basics.html

--
Bob.

.

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