# Time period of revolution of a satellite around a planet of radius R is T. period of revolution around another planet, whose radius is 3R but having same density is?

1
Log in to add a comment

Log in to add a comment

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

Uniform circular motion:

We know that centripetal acceleration for a satellite of mass m going around around a planet of mass M, in a circular orbit of radius R , with a uniform velocity v :

m v² / R = G M m / R²

=> v = √ [GM/R] = R ω

=> ω = √[ GM / R³ ]

=> T = 2π √[ R³/GM ]

= K √[ R³/M ] , K is a constant

mass = M = d 4π/3 * R³

=> R³/M = 3/(4πd)

=> T = K ' √ [ 1/d ]

So the time period of a satellite going around a planet is inversely proportional to the square root of the density of the planet.

Hence, the time period remains same.... as T.

We know that centripetal acceleration for a satellite of mass m going around around a planet of mass M, in a circular orbit of radius R , with a uniform velocity v :

m v² / R = G M m / R²

=> v = √ [GM/R] = R ω

=> ω = √[ GM / R³ ]

=> T = 2π √[ R³/GM ]

= K √[ R³/M ] , K is a constant

mass = M = d 4π/3 * R³

=> R³/M = 3/(4πd)

=> T = K ' √ [ 1/d ]

So the time period of a satellite going around a planet is inversely proportional to the square root of the density of the planet.

Hence, the time period remains same.... as T.