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7.Gravitation
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If the gravitational acceleration at surface of Earth is $g$, then increase in potential energy in lifting an object of mass $m$ to a height equal to half of radius of earth from surface will be
A
$\frac{{mgR}}{2}$
B
$\frac{{2mgR}}{3}$
C
$\frac{{mgR}}{4}$
D
$\frac{{mgR}}{3}$
Solution
$\mathrm{U}_{\mathrm{i}}=-\frac{\mathrm{GMm}}{\mathrm{R}}, \mathrm{U}_{\mathrm{f}}=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{R} / 2}$
$\mathrm{KE}_{\mathrm{i}}=\mathrm{KE}_{\mathrm{f}}=0$
$\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=-\frac{2 \mathrm{GMm}}{3 \mathrm{R}}+\frac{\mathrm{GMm}}{\mathrm{R}}$
$\Delta \mathrm{U}=\frac{\mathrm{GMm}}{3 \mathrm{R}} \quad$ As $\frac{\mathrm{GM}}{\mathrm{R}^{2}}=\mathrm{g}$
$\Delta \mathrm{U}=\frac{\mathrm{mg} \mathrm{R}}{3}$
Standard 11
Physics
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