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Q.

A satellite revolving around earth at a height 2R above earth surface. If the satellite is stopped and thenreleased, then velocity of body on reaching to earth’s surface (g is acceleration due to gravity on surface R is radius of earth)

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a

$\sqrt{\frac{2 g R}{3}}$

b

$\sqrt{4 g R}$

c

$\sqrt{\frac{4 g R}{3}}$

d

$\sqrt{2 g R}$

answer is B.

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Detailed Solution

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$\frac{1}{2} m v^{2} - \frac{G M m}{R} = - \frac{G M m}{R + 2 R}$

$\frac{1}{2} m v^{2} = \frac{2}{3} \frac{G M m}{R}$

$V = \sqrt{\frac{4}{3} g R}$

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A satellite revolving around earth at a height 2R above earth surface. If the satellite is stopped and thenreleased, then velocity of body on reaching to earth’s surface (g is acceleration due to gravity on surface R is radius of earth)