A satellite is revolving round the earth with orbital speed `v_(o)`. If it stops suddenly, the speed with which it will strike the surface of earth would be (`v_(e)` = escape velocity of a particle on earth's surface)

Options

A. `v_e^2/v_0`
B. `v_0`
C. `sqrt(v_e^2 -v_0^2)`
D. `sqrt(v_e^2 -2v_0^2)`

Answer

D

Explanation

Let r be the radius of the satellite. Then `v_0^2=(GM)/r`
Applying conservation of mechanical energy between points A and B we have:
`1/2mv^2 =-(GMm)/r-(-(GMm)/R)`
`v^2=(2GM)/R-(2Gm)/r=v_e^2-2v_0^2`
`therefore v=sqrt(v_e^2-2v_0^2)` .