The orbital velocity of an artifical satellite in a cirular orbit above the earth's surface at a distance equal to radiu of earth is v. For a satellite orbiting at an altitude half of earth's radius, orbital velocity is

Options

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

Answer

C

Explanation

The satellite is moving at an attitude above the surface equal to the radius of the earth R.
`therefore (mv_(0)^(2))/(2R)=G(Mm)/(4R^(2))`
`implies v_(0)=(sqrt((GM)/(2R)))`
Now if the satellite moves at an attitude equal to R/2 then,
`(mv'_(0)^(2))/(3R//2)=G(Mm)/(9xx(R//2)^(2))`
`implies v_(0)^(2)=(2)/(3)*(GM)/(R)`
`implies v'_(0)=(sqrt(2))/(sqrt(3))sqrt((GM)/(R))=(sqrt(2))/(sqrt(3))v_(0)sqrt(2)`
`=(sqrt(4))/(sqrt(3))v_(0)(2)/(sqrt(3))v_(0)`