The orbital speed of a satellite moving in circular orbit around the earth (Radius = R) is equal to 50% of the escape velocity on the surface of earth. Then height of the satellite above the surface of earth is

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1.5R

answer is C.

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Detailed Solution

V=50% of Ve∴ GMr=122GMR⇒r=2R∴ h=r−R=2R−R=R

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