A body of mass m is lifted up from surface of earth to a height three times radius of earth. change

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7.Gravitation

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A body of mass $m$ is lifted up from the surface of the earth to a height three times the radius of the earth. The change in potential energy of the body is

where $g$ is acceleration due to gravity at the surface of earth.

A

$3mgR$

B

$\frac{3}{4} mgR$

C

$\frac{1}{3} mgR$

D

$\frac{2}{3} mgR$

Solution

$\Delta U=\frac{m g h}{1+\frac{h}{R}}$

$h=3 R$

hence.

$\Delta U=\frac{m g(3 R)}{1+\frac{3 R}{R}}=\frac{3 m g R}{4}$

Standard 11

Physics

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