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 eqn i by eqn ii `(v^(2))/(v^('2))=(R')/R`
`implies (v')/v=sqrt(R/R')sqrt(2/3) [:'R'=(3R)/2]`
`impliesv'=sqrt(2/3)v`