A body of mass m is taken from Earth’s surface to the height h equal to twice the radius of Earth , the increase in potential energy will be: (g = acceleration due to gravity on the surface of Earth) [2023]

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Solution

Q.
A body of mass m is taken from Earth’s surface to the height h equal to twice the radius of Earth $(R_{e})$ , the increase in potential energy will be: (g = acceleration due to gravity on the surface of Earth) [2023]

1 $\frac{1}{2} m g R_{e}$

2 $3 m g R_{e}$

3 $\frac{2}{3} m g R_{e}$

4 $\frac{1}{3} m g R_{e}$

Ans.
(3)

$U = \frac{- G M_{e} m}{r} and U_{i} = \frac{- G M_{e} m}{R_{e}}$

$U_{f} = \frac{- G M_{e} m}{(R_{e} + h)} = \frac{- G M_{e} m}{R_{e} + 2 R_{e}} = \frac{- G M_{e} m}{3 R_{e}}$

$Increase in internal energy Δ U = U_{f} - U_{i}$

$= \frac{2}{3} \frac{G M_{e} m}{R_{e}}$

$= \frac{2}{3} \frac{G M_{e}}{R_{e}^{2}} m R_{e} = \frac{2}{3} m g R_{e}$

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