The orbital velocity of an satellite in a circular orbit just above the earth's surface is `v`. For a satellite orbiting at an altitude of half of the earth's radius, the orbital velocity is

Options

A. `3/2v`
B. `sqrt(3/2)v`
C. `sqrt(2/3)v`
D. `2/3v`

Answer

C

Explanation

`v=sqrt((GM)/(R+h))`
For first satellite `h=0, v_(1)=sqrt((GM)/R)`
For second satellite `h=R/2, v_(2)=sqrt((2GM)/(3R))`
`v_(2)sqrt(2/3)v_(1) =sqrt(2/3)v`