You are exploring a newly discovered planet. The radius of the planet is 7.00×107m . You suspend a lead weight from the lower end of a light string that is 4.00 {rm m} long and has mass 0.0280 {rm kg}. You measure that it takes 0.0600s 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 0.0320s 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 its mass in kilograms?
1 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 / 0.06 = 66.67 m/s
V = √ T / (m/L)
v^2 . m / L = T = 4444 . 0.028 / 4 = 31.1 N
On Earth
v = 4/0.032 = 125 m/s
v^2 . m / L = T = 109.4 N
g(planet) = g(earth) . 31.1 / 109.4 = 2.787 m/s/s
2.787 = GM/R^2 = 6.67 . 10^–11 . M / 49 . 10^14
M = 20.47 . 10^25 kg