The Rubble-Pile Spin Limit

Using the fact that a uniform spherical shell exerts no net gravitational force on a mass inside it, show that the gravitational field strength at distance $r$ from the centre ($r \le R$) is $g(r) = \tfrac{4}{3}\pi G\rho r$.

By considering the forces acting on a loose boulder resting on the surface at the equator, show that the shortest rotation period the asteroid can have without the boulder lifting off is $T_c = \sqrt{\dfrac{3\pi}{G\rho}}$, and calculate $T_c$ in hours.

An asteroid of this density rotates with period $3.50\ \text{h}$. Calculate the ratio of the apparent weight of a boulder at the equator to its apparent weight at a pole.

Surveys find that almost no asteroids larger than about 300 m rotate faster than about 2.2 h, while many smaller asteroids spin in minutes. Use parts (a) and (b) to explain why the critical period does not depend on the asteroid's size, why at $T_c$ the whole body (not just its surface) is on the verge of flying apart, and why small asteroids can beat the limit. Comment on one assumption of the model.