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

Options

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

Answer

A

Explanation

`v= sqrt(((GM)/(r)))` where r is radius of the orbit of the satellite.
Here r = `R_(e) + h = R_(e) + (R_(e))/(2) = (3)/(2) R_(e)`
So, v = `sqrt((2GM)/(3R_(e))) = v_(0) sqrt((2)/(3))` ,
where `v_(0)` is the orbital velocity of the satellite , which is moving in circular orbit of radius , r = `R_(e)`