A body of mass m is placed on the earth's surface. It is taken from the earth's surface to a height h=3R. The change in gravitational energy of the body is

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Solution

The correct option is B

$△ U = \frac{m g h}{1 + \frac{h}{R}} = \frac{m g \times 3 R}{1 + \frac{3 R}{R}} = \frac{3}{4} m g R$

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A body of mass m is placed on the earth's surface. It is taken from the earth's surface to a height h=3R. The change in gravitational energy of the body is

The correct option is B △U=mgh1+hR=mg×3R1+3RR=34mgR

The correct option is B

$△ U = \frac{m g h}{1 + \frac{h}{R}} = \frac{m g \times 3 R}{1 + \frac{3 R}{R}} = \frac{3}{4} m g R$