Home
Class 11
PHYSICS
The fastest possible of rotation of a pl...

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.

Text Solution

Verified by Experts

The correct Answer is:
`T=sqrt((3pi)/(Grho))`
`g'=g-Romega^(2)implies0g-Romega^(2)`
`omega=sqrt(g//R)implies(2pi)/T=sqrt(g//R)`
`T=2pisqrt(R//g)impliesT=2pisqrt(R/(GM//R^(2)))`
`T=2pisqrt((R^(3))/(GM))impliesT=2pisqrt((R^(3))/(Gxx4/3piR^(3)rho))`
`T=sqrt((3pi)/(Grho))`
Promotional Banner

Similar Questions

Explore conceptually related problems

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)

If rho is the density of the planet , the time period of nearby satellite is given by

Can centripetal force produe rotation ?

If rho is the density of its material, then its rotational K.E. is given by

The angular velocity of rotation of a planet of mass M and radius R, at which the matter start to escape from its equator is

A planet revolves around the sun under the effect of gravitational force exerted by the sun. Why is the torque on the planet due to the gravitational force zero ?

For uniform circular motion, does the direction of centripetal force depends on the sense of rotation?