14 Jun '13 16:24

So I was wondering what kind of acceleration you would feel if you were on a space elevator at geo synchronous orbit, call it 22,000 miles up or 35,200,000 meters high. That plus the radius of the Earth, call it 4000 miles or 6,400,000 meters (these numbers are close enough for my purposes) and we have about 9.8 meters per second squared sitting on the Earth. But on a space elevator, my thoughts went like this:

You would have the opposite forces of gravitational attraction at that altitude Vs the centripetal acceleration away from Earth, so using the numbers provided by google, GM is 3.985617248E14 and dividing that by r squared, using 4000 miles or 6,400,000 meters I get 9.73 m/sec^2. Close enough to 9.8 for me for now. So extending that as if you were sitting on a rod 22,000 miles out from the surface of the Earth, you get an altitude of 41,600,000 meters from the center of the Earth. So GM/r^2 up there gives you 0.23 meters/sec^2.

So the next step to do is to calculate the centripetal acceleration using V^2/r.

I got the circumference of 261,380,508 meters using my original numbers.

So divide that by 86,400 and I get a velocity of that orbit of 3025 meters per second. Squared=9152061 and I get the centripetal acceleration of 0.22 meters per second squared.

So the 0.23 minus 0.22 almost cancels each other out and you seem to be darn close to weightless if you were floating around in a space elevator. Does this sound correct?

And is there a better way to do this calculation?

You would have the opposite forces of gravitational attraction at that altitude Vs the centripetal acceleration away from Earth, so using the numbers provided by google, GM is 3.985617248E14 and dividing that by r squared, using 4000 miles or 6,400,000 meters I get 9.73 m/sec^2. Close enough to 9.8 for me for now. So extending that as if you were sitting on a rod 22,000 miles out from the surface of the Earth, you get an altitude of 41,600,000 meters from the center of the Earth. So GM/r^2 up there gives you 0.23 meters/sec^2.

So the next step to do is to calculate the centripetal acceleration using V^2/r.

I got the circumference of 261,380,508 meters using my original numbers.

So divide that by 86,400 and I get a velocity of that orbit of 3025 meters per second. Squared=9152061 and I get the centripetal acceleration of 0.22 meters per second squared.

So the 0.23 minus 0.22 almost cancels each other out and you seem to be darn close to weightless if you were floating around in a space elevator. Does this sound correct?

And is there a better way to do this calculation?