Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The work done to raise a mass $m$ from the surface of the earth to a height $h$ , which is equal to the radius of the earth, is:

$\frac{3}{2}mgR$

$mgR$

$2mgR$

$\frac{1}{2}mgR$

Concept: The work done to lift a body against gravity is equal to the change in its gravitational potential energy. $W = \Delta U = U_{\text{final}} - U_{\text{initial}}$
Formulas:

Gravitational potential energy at a distance $r$ from the center of the Earth: $U = -\frac{GMm}{r}$ [Eq. 7.23, 7.24].
Relation between $G$ and $g$ : $g = \frac{GM}{R^2} \implies GM = gR^2$ .

Initial State: At the surface of the Earth, $r_1 = R$ . $U_i = -\frac{GMm}{R}$
Final State: At a height $h = R$ above the surface, the distance from the center is $r_2 = R + h = R + R = 2R$ . $U_f = -\frac{GMm}{2R}$
Calculation: $W = -\frac{GMm}{2R} - \left( -\frac{GMm}{R} \right)$ $W = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$ Substitute $GM = gR^2$ : $W = \frac{(gR^2)m}{2R} = \frac{1}{2}mgR$

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