[B-Chan]

You're wrong. The time it takes a satellite in orbit to travel once around the Earth is called its orbital period (P). P can be calculated using the following formula:

P=2p*sqrt(a³/m)

where P = the orbital period, a = the semi-major axis of the orbit (same as the radius of the circle for a circular orbit), and m) = the gravitational parameter (~398601 km³ / Sec² for Earth).

Where P=24 hours, a=35,786 kilometers.

NASA has a handy online calculator that allows one to calculate the period of any circular orbit with an altitude >185 km. Try it!

[DuncanWaring]

Everything you've said is correct for a satellite that's orbiting "freely".

The problem is that for a satellite that's above the geosynchronous altitude, as would be the top of the space elevator ribbon, it's natural period is more than a day.

The period of the moon, for example, is 28 days (but you knew that ;-) )

The anchor at the top of this 100,000 km ribbon would have to be orbiting the earth at a rate of once a day, faster than a free satellite at the same altitude, which is what gives it the ability to "lift" the ribbon. If the ribbon were to disconnect, the satellite would immediately go to a higher altitude orbit, in the manner of a rock being swung on a string if you let go of the string.

Even ignoring the tether, going back to the geosynchronous satellite, it travels about 150,000 miles in 24 hours to keep up with the point on the surface of the earth it's above, whereas that point on the earth only travels 25,000 miles in the same 24 hours.

[wastedyears]

I think I’ll leave the thread, the numbers and equations are making me feel funny.