A geostationary satellite is revolving around the earth. To make it escape from gravitational field of earth its velocity must be increase ….....

Options

A. `100%`
B. `41.4%`
C. `50%`
D. `59.6%`

Answer

B

Explanation

`implies` Centripetal force on satellite = Gravitional force between satellite and earth.
`(mv_0^2)/r = (GM_em)/r^2`
(r = distance of satellite from the centre of earth)
`v_0^2=(GM_e)/r`
`:.v_0 = (GM_e)/r " "...(1)`
Escape speed for satellite
`:. v_e = sqrt(2GM_e)/r " "...(2)`
`:. v_e =sqrt2 sqrt((GM_e)/r)`
`:. v_e = sqrt2 v_0`
`:. v_e =1.414 v_0`
`:. v_e/v_0 = 1.414`
`:. (v_e-v_0)/v_0 =(1.414-1)/1 = 0.414`
`:. (v_e-v_0)/v_0xx100% =41.4%`