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Question

# An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth. The height (h) of the satellite above the earth's surface is (Take radius of earth as Re).

A
h=R2e
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B
h=Re
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C
h=2Re
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D
h=4Re
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Solution

## The correct option is C h=ReThe escape velocity from earth is given byve=√2gRe.....(i)The orbital velocity of a satellite revolving around earth is given byv0=√GMe(Re+h)where, Me= mass of earth, Re= radius of earth, h= height of satellite from surface of earth.By the relation GMe=gR2eSo, v0=√gR2e(Re+h).....(ii)Dividing equation (i) by (ii), we getvev0=√2(Re+h)(Re)Given, v0=ve22veve=√2(Re+h)ReSquaring on both side, we get4=2(Re+h)Re or Re+h=2Re.⟹ h=Re

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