Answers (1)
H Harsh

If you do the math (equating \frac{GMm}{R^{2}} to m\omega R^{2} and using M(Mass of the earth) =M=\frac{4\pi}{3}*\rho *R^{3} and \omega =2\pi f), it follows that the size of the object is entirely irrelevant and that only the density ρ of the sphere enters into the equation for f, the number of revolutions per unit time:   f^{2}=\frac{1}{3\pi}G\rho

For ρ=5.5×10^3 kilogram per cubic meter (the density of planet earth) it follows that f=0.197×10^(−3) revolutions per second, corresponding to a revolution period of 50705070seconds (1 hour and 24 minutes).