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?
Answers
Answered by
47
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.
kvnmurty:
click on thanks blue button above please
Answered by
14
See the 23rd one in the image below
Attachments:
Similar questions