If the gravitational acceleration at surface | vedclass

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Question

$\mathrm{U}_{\mathrm{i}}=-\frac{\mathrm{GMm}}{\mathrm{R}}, \mathrm{U}_{\mathrm{f}}=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{R} / 2}$

$\mathrm{KE}_{\

Vedclass · Accepted Answer

$\frac{{mgR}}{3}$

Vedclass · Answer

$\mathrm{U}_{\mathrm{i}}=-\frac{\mathrm{GMm}}{\mathrm{R}}, \mathrm{U}_{\mathrm{f}}=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{R} / 2}$

$\mathrm{KE}_{\mathrm{i}}=\mathrm{KE}_{\mathrm{f}}=0$

$\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=-\frac{2 \mathrm{GMm}}{3 \mathrm{R}}+\frac{\mathrm{GMm}}{\mathrm{R}}$

$\Delta \mathrm{U}=\frac{\mathrm{GMm}}{3 \mathrm{R}} \quad$ As $\frac{\mathrm{GM}}{\mathrm{R}^{2}}=\mathrm{g}$

$\Delta \mathrm{U}=\frac{\mathrm{mgR}}{3}$

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