1.

The orbital period of sa satellite near the surface of the planet of radius Ris given by (rho is the density of the planet):

Answer»

`(3PI)/(rho G)^(1//2)`
`(4pi)/(3rho G)^(1//2)`
`(4pi)/(rho G)^(1//2)`
`(3pi)/(4rhoG)^(1//2)`

Solution :`(Mv^(2))/(r )=(GMm)/(r^(2))`
`&v=(2pir)/(T)`
`therefore T=(2pir^(3//2))/(GM)^(1//2)`
`For r =R & M = (4)/(3) pi R^(3) rho`
`T=((3pi)/(rhoG))`


Discussion

No Comment Found

Related InterviewSolutions