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

Answer

C

Explanation

`(mv^(2))/R=(GMm)/(R^(2))......(i)`
and `(mv'^(2))/(R')=(GMm)/(R'^(2))......(ii)`
Dividing (i) by (ii)
`(v^(2))/(v'^(2))=(R')/R`
`(v'^(2))/(v^(2))=(R')/R`
`(v')/v=sqrt(R/(R'))` as `R'=(3R)/2`
`=sqrt(2/3)implies v'=sqrt(2/3).v`