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Mrjnicoll · May 12, 2010, 07:14 PM

If I tied a ball to the end of an indistructible string and tethered the string to the earth, how long would the string have to be to keep the ball from falling to the earth?

Would the mass of the string or the ball be a factor?

Could I climb up the string?

cdad · May 12, 2010, 07:31 PM

Ref:

Geostationary orbit - Wikipedia, the free encyclopedia

ballengerb1 · May 12, 2010, 07:33 PM

What is the purpose of the string if the ball is in a low earth orbit? I sthis an actual question that's been presented to you?

ebaines · May 13, 2010, 06:55 AM

Geostationary orbit is a good answer; however, it assumes that the string is weightless. If the string has mass (and hence weight), or if you tried to climb the string, the satellite would have to be a bit further away than a normal geostationary orbit.

The math works like this: for a satelliite in circular orbit the outward force on it due to its centripetal acceleration is balanced by the inward force due to gravity:

$ mr \omega ^2 = \frac {GMm} {r^2} \\ r^3 = \frac {GM} {\omega ^2} $

Where $r$ is the distance from the center of the earth to the satellite, and $M$ is the mass of the earth. So given that $\omega$ is set by the rotational speed of the earth, you can calculate $r$ , which is the radius of a geostationary orbit.

Now, if you add the weight of the string or a person climbing on it, there's an additional force component that you have to include:

$ mr \omega ^2 = \frac {GMm} {r^2} + W $

This means the $r$ must be a bit larger to counteract the extra weight.