The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just nearly provides the centripetal force needed for rotation. The corresponding shortest period of rotation is
(If `rho` is the density of the earth)

Options

A. `sqrt((3pi)/(Grho))`
B. `sqrt((3pirho)/(G))`
C. `sqrt((3piG)/(rho))`
D. `sqrt((Grho)/(3pi))`

Answer

a

Explanation

`mRomega^(2)=(GMm)/(R^(2))`
`omega^(2)=(GM)/(R^(3))`
`omega=sqrt((GM)/(R^(3)))=sqrt((Grho)/(K^(3)))(4pi)/(3)R^(3)=sqrt((4pi)/(3)Grho)`
`T=(2pi)/(omega)=(2pi)/(sqrt((4pi G rho)/(3)))=sqrt((3xx4pi^(2))/(4piG rho))=sqrt((3pi)/(Grho)).`