The fastest possible of rotation of a planet is that for which the gravitational force on material at the equtor barely provides the centripetal force needed for the rotation. Show then that the corresponding shortest period of rotation is given by `T=sqrt((3pi)/(Grho))` where `rho` is the density of the planet, assumed to be homogeneous. Evalute the rotation period assuming a density of `3.0 gm//cm^(2)`, typical of many planets, satellites, and asteroids. No such object is found to be spinning with a period shorter than found by this analysis.

Answer

(ii). 1.9 h

Explanation

(i). At equator
`momega^(2)R=(GMm)/(R^(2))`
`implies(2pi)/(T)=omega=sqrt((Grho)/(R^(3)).(4piR^(3))/(3))impliesT=sqrt((3pi)/(Grho))`
(ii). `T=sqrt((3pi)/(6.63xx10^(-11)xx3000))=1.9hr`