You are exploring a newly discovered planet. The radius of the planet is 7.00 Ti

Question

You are exploring a newly discovered planet. The radius of the planet is 7.00 Times 10^7 m. You suspended a lead weight from the lower end of a light string that is 4.00 long and has mass 0.0280 kg. You measure that it takes 0.0675 s for a transverse pulse to travel from the lower end to the upper end of the string. On earth, for the same string and lead weight, it takes 0.0320 s for a transverse pulse to travel the length of the string. The weight of the string is small enough that its effect on the tension in the string can be neglected. Assuming that the mass of the planet is distributed with spherical symmetry, what is the mass? Express your answer with the appropriate units.

Explanation / Answer

That is a really silly way to work out the mass, when you could just let the mass swing as a pnedulum on the end of the string.

v = d/t = 4 / 0.0675 = 59.25925926 m/s

V = T / (m/L)
v^2 . m / L = T = 3511.659808 . 0.028 / 4 = 24.58161866 N

On Earth
v = 4/0.032 = 125 m/s
v^2 . m / L = T = 109.375 N

g(planet) = g(earth) . 24.58161866 / 109.4 = 2.247462277 m/s/s

2.247462277 = GM/R^2 = 6.67 . 10^–11 . M / 49 . 10^14

M = 16.51059244 * 10^25 kg

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