If g is the acceleration due to gravity at the Earth's surface, the gain of the potential energy of an object of mass m raised from the surface of the Earth to the height equal to the radius R of

Question

If g is the acceleration due to gravity at the Earth's surface, the gain of the potential energy of an object of mass m raised from the surface of the Earth to the height equal to the radius R of the Earth is

TestHub · Accepted Answer

Potential energy of an object at the surface of the Earth is

$U_{1} = - \frac{G M m}{R}$ ..... (i)

Potential energy of the object at a height, $h = R$ from the surface of the Earth.

$U_{2} = - \frac{G M m}{R + h} = - \frac{G M m}{R + R}$

Gain in potential energy

$∆ U = U_{2} - U_{1} = (- \frac{G M m}{2 R}) - (- \frac{G M m}{R}) = \frac{1}{2} \frac{G M m}{R}$

Also, $G M = g R^{2}$

$∆ U = \frac{1}{2} \frac{g R^{2} m}{R} = \frac{m g R}{2}$

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