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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:   $f^{2}=\frac{1}{3\pi}G\rho$