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Question

A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth R; h<<R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to (Neglect the effect of atmosphere.)

A
2gR
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B
gR
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C
gR/2
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D
gR(21)
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Solution

The correct option is D gR(21)
For a satellite in orbit, the velocity can be evaluated from the following equation
GMmr2=mv2or
vo=GMr
Here, r=R+h

For satellite to escape, total energy of satellite must be zero. Let escape velocity of satellite at height h be ve.
E=PE+KE=0
GMmr+12mv2e=0
ve=2GMr

Difference in velocities is given by
Δv=vevo
Δv=GMr(21)
ΔvGMR(21) as h<<R

But g=GMR2

ΔvgR(21)

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