Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final - initial) of an object of mass $m$ when taken to a height $h$ from the surface of the earth (of radius $R$ and mass $M$ ), is given by:

$-\frac{GMm}{R+h}$

$\frac{GMmh}{R(R+h)}$

$mgh$

$\frac{GMm}{R+h}$

Step-by-Step Solution

Formula: The gravitational potential energy of a mass $m$ at a distance $r$ from the center of the Earth is given by $U(r) = -\frac{GMm}{r}$ [Eq. 7.23, 7.24].
Initial State: At the surface of the Earth, the distance from the center is $R$ . Thus, the initial potential energy is $U_i = -\frac{GMm}{R}$ .
Final State: At a height $h$ above the surface, the distance from the center is $r = R + h$ . Thus, the final potential energy is $U_f = -\frac{GMm}{R+h}$ .
Change in Potential Energy ( $\Delta U$ ): $\Delta U = U_f - U_i$ $\Delta U = \left( -\frac{GMm}{R+h} \right) - \left( -\frac{GMm}{R} \right)$ $\Delta U = \frac{GMm}{R} - \frac{GMm}{R+h}$ $\Delta U = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right)$ $\Delta U = GMm \left( \frac{R+h-R}{R(R+h)} \right)$ $\Delta U = \frac{GMmh}{R(R+h)}$ (Note: This expression reduces to $mgh$ only when $h \ll R$ ).

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