An object is taken to height 2 R above surface of earth, increase in potential energy is [R is radiu

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7.Gravitation

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An object is taken to height $2 R$ above the surface of earth, the increase in potential energy is $[R$ is radius of earth]

A

$\frac{m g R}{2}$

B

$\frac{m g R}{3}$

C

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

D

$2 m g R$

Solution

(c)

Potential energy at surface $=-\frac{G M m}{R}$

Potential energy at height, $2 R=-\frac{G M m}{3 R}$

Change in potential energy $=-\frac{G M m}{3 R}+\frac{G M m}{R}$

$=\frac{G M m}{R}\left(\frac{-1+3}{3}\right)$

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

$=\frac{2}{3}\left(\frac{G M}{R^2}\right) m R$

$=\frac{2}{3} m g R$

Standard 11

Physics

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