The orbital velocity of a satellite in a circular orbit just above the earth's surface is `v_(0)`. The orbital velocity for a satellite orbiting in a circular orbit at an altitude of half of earth's radius is:

Options

A. `sqrt((3)/(2))v_(0)`
B. `sqrt((2)/(3)) v_(0)`
C. `(3)/(2)v_(0)`
D. `((2)/(3))v_(0)`

Answer

B

Explanation

`v_(0)=sqrt(gR)`
and orbital speed at a height h above the surface of earth is
`v=sqrt((GM)/(R+h))=sqrt((gR^(2))/(R+(R )/(2)))=sqrt((2)/(3)gR)`
`v= sqrt((2)/(3))*v_(0)`
So correct choice is (b).