Question

The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for the rotation. Calculate the minimum rotation period assuming a density of 3.70 g/cm3.

Answer 1

m ω^2 R = centripetal force

G M m / R^2 = gravitational force

ω^2 = G M / R^3 for the forces to be equal

d (density) = M / V = M / (4/3 R^3) = 3/4 M / R^3

or M / R^3 = 4 d / 3

Then

ω^2 = G 4 d / 3

d = 3.7 g/cm^2 = 3700 kg/m*3 converting

ω^2 = 6.67E-11 * 3700 * 4 / 3 = 32.9E-8

ω = 5.74E-4

P = 1 / f = 2 pi / ω = 6.28 / 5.74E-4 = 1.09E4 sec

10900 / 3600 = 3.04 hrs for the minimum rotation period