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(A) The gravitational potential energy at the surface of the Earth is $U_i = -\frac{GMm}{R}$.The object is taken to a height $h = 3R$. The final dista

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$48$

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(A) The gravitational potential energy at the surface of the Earth is $U_i = -\frac{GMm}{R}$.The object is taken to a height $h = 3R$. The final distance from the center of the Earth is $r = R + h = R + 3R = 4R$.The gravitational potential energy at this height is $U_f = -\frac{GMm}{4R}$.The gain in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{4R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{4R} = \frac{3GMm}{4R}$.Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$. Substituting this into the equation:$\Delta U = \frac{3(gR^2)m}{4R} = \frac{3}{4} mgR$.Given $m = 1 \, kg$,$g = 10 \, m/s^2$,and $R = 6400 \, km = 6.4 	imes 10^6 \, m$:$\Delta U = \frac{3}{4} 	imes 1 	imes 10 	imes 6.4 	imes 10^6 = 3 	imes 2.5 	imes 6.4 	imes 10^6 = 48 	imes 10^6 \, J = 48 \, MJ$.

An object of mass $1 \, kg$ is taken to a height from the surface of Earth which is equal to three times the radius of Earth. The gain in potential energy of the object will be $.... \, MJ$ [Given: $g = 10 \, m/s^2$ and radius of Earth $R = 6400 \, km$].

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