A "rubble pile" asteroid can be modeled as a sphere of uniform density ρ\rhoρ. The asteroid rotates about its central axis. If the asteroid rotates too quickly, loose material at its equator will lose contact with the surface and fly off into space.
Show that the critical period of rotation TTT for which loose material at the equator is on the verge of losing contact with the surface is given by:
T=3πGρ T = \sqrt{\frac{3\pi}{G\rho}} T=Gρ3πwhere GGG is the gravitational constant.
A particular asteroid has a uniform density of 2.6×103 kg m−32.6 \times 10^3 \text{ kg m}^{-3}2.6×103 kg m−3. Calculate this minimum period of rotation. Give your answer in hours to an appropriate number of significant figures. Gravitational constant G=6.67×10−11 N m2 kg−2\text{Gravitational constant } G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}Gravitational constant G=6.67×10−11 N m2 kg−2.