Physics hanging around the Earth

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SUMMARY

The discussion focuses on calculating the necessary length (L) of a rope with density 0.33 kg/m that hangs above the Earth, considering the Earth's mass (5.98 x 1024 kg) and radius (6.37 x 106 m). The relevant equation is T2 = [(4π2)/(Re2 * g)] (Re + L/2), which incorporates the angular velocity of the Earth (ω). The user expresses confusion over the omission of the Earth's mass and the rope's density in their initial calculations, indicating a need for a more comprehensive approach to the problem.

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Homework Statement


Someone once considered hanging a rope, (of density p), above the Earth so that it hangs slightly above the ground. How long does the rope need to be (length L)? (mass of the Earth = 5.98 * 10^24, Radius of the Earth = 6.37 * 10^6, angular velocity of Earth equals omega, and density p = .33 kg/m) The rope has both ends free.


Homework Equations


T^2 = [(4(pi)^2)/(Re^2 * g)] (Re + L/2)


The Attempt at a Solution


I thought you could just isolate L from the above equations (one of keplers laws). However, that equation doesn't use the mass of the Earth or the density of the rope. So there must be something I'm missing. Some help would be greatly appreciated.
 
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