Let's have a bit of fun.
Imagine we stick a straight, rigid line into the equator, extending far into space, far beyond geosynchronous orbit. The line has a uniform density D and a constant cross-section S. We wanna know how much this thing would weigh and how far into space it would have to extend to hold itself up.
Keep in mind, anything under geosynchronous altitude effectively acts as weight, because force of gravity is larger then "centrifugal" force. Anything over geosynchronous altitude acts like an anchor, because force of gravity is lower then centrifugal force.
First we need to figure out how much force we need to keep the line up, if it only extended from Earth's surface to geosynchronous altitude.
F marks force, m marks mass, M marks the mass of Earth, G is gravitational constant, w is angular velocity and r is radius.
dF = dFg - dFc
dF = dm * g[r] - dm * w^2 * r
dF = dm * (g[r] - w^2 * r)
dm = S * D * dr
g[r] = M * G / r^2
dF = S * D * dr * (M * G / r^2 - w^2 * r)
dF = S * D * (M * G * dr / r^2 - w^2 * r * dr)
Now we integrate from the surface of the planet, marked as R0, to some high altitude marked as R:
F = S * D * (M * G * Integrate[dr/r^2, {r, R0, R}] - w^2 * Integrate[r * dr, {r, R0, R}]
The result is a bit ugly, but don't worry...
F = -D * S * (M * G * (1/R - 1/Ro) + (w^2)/2 * (R^2 - Ro^2))
Now, in order for the line to stay up there, the net force has to be 0. If not, it'll go flying off into space or come crashing down into Earth - or at the very least, end up in an elliptical orbit...
So:
0 = -D * S * (M * G * (1/R - 1/Ro) + (w^2)/2 * (R^2 - Ro^2))
The first thing that's noticeable is that density and cross-section don't matter (in this case).
You turn the equation around and you get this:
R * (2 * M * G + w^2 * Ro^3) = Ro * (2 * M * G + w^2 * R^3)
The first solution is obviously R = R0. But that would mean the length of the line is 0.
Of the two remaining solutions, one is negative. The line could stretch into the planet for the equation to be true.
The one we're interested in is this:
R = (Sqrt[R0^3 * w^2 + 8 * G * M] / (2 * Sqrt[R0] * w)) - R0 / 2
Obviously the whole thing fails if the length of the line is 0, or if the planet doesn't rotate... but we don't need to worry about that..................
---------- Post added at 14:36 ---------- Previous post was at 14:25 ----------
And some values:
For Earth, such a line would need to extend to about 1.5*10^8 m out (that's radius measured from the center of Earth - but not that it makes much difference).
The Moon, for comparison is 3.6 * 10^8 m out at it's closest, so assuming I did the calculation correctly, the line for the space elevator would have to extend out to almost half the Moon's orbital radius. A bit more, if we wanna keep the line straight and haul up some weight.
I hope someone can go over my calculations and check for mistakes...