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

Answer

C

Explanation

For first satellite `(GM)/(R^(2)) m=(mV_(0)^(2))/RrArr V_(0)=sqrt((GM)/R)`
For second satellite`((GM)/((R+R//2)^(2))) m=(mV^(2))/((R+R//2))`
`rArr V=(sqrt(2/3))V_(0)`