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Assuming, $g_{\text {(Moon) }}=\left(\frac{1}{6}\right) g_{\text {Earth }}$ and $D_{\text {(Moon) }}=\left(\frac{1}{4}\right) D_{\text {Earth }}$, where $g$ and $D$ are the acceleration due to gravity and diameter respectively, the escape velocity from the Moon is
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The correct answer is:
$\frac{11.2}{\sqrt{24}} \mathrm{kms}^{-1}$
As we know, escape velocity, $v_{e}=\sqrt{2 g R}$
Let $v_{m}$ be escape velocity from moon and $v_{e}$ be the escape velocity from Earth. Then,
$\frac{v_{m}}{v_{e}}=\sqrt{\frac{2 \times g_{m} \times R_{m}}{2 \times g_{e} \times R_{e}}}$
$=\sqrt{\frac{g_{m} \times D_{m}}{g_{e} \times D_{e}}} \quad {(\because D=2 R)}$
$=\sqrt{\frac{\frac{g_{e}}{6} \times \frac{D_{e}}{4}}{g_{e} D_{e}}}=\frac{1}{\sqrt{24}}$
$\therefore \quad v_{m}=\frac{v_{e}}{\sqrt{24}}=\frac{11.2}{\sqrt{24}} \mathrm{~km} / \mathrm{s} \quad\left(\because v_{e}=11.2 \mathrm{~km} / \mathrm{s}\right)$
Let $v_{m}$ be escape velocity from moon and $v_{e}$ be the escape velocity from Earth. Then,
$\frac{v_{m}}{v_{e}}=\sqrt{\frac{2 \times g_{m} \times R_{m}}{2 \times g_{e} \times R_{e}}}$
$=\sqrt{\frac{g_{m} \times D_{m}}{g_{e} \times D_{e}}} \quad {(\because D=2 R)}$
$=\sqrt{\frac{\frac{g_{e}}{6} \times \frac{D_{e}}{4}}{g_{e} D_{e}}}=\frac{1}{\sqrt{24}}$
$\therefore \quad v_{m}=\frac{v_{e}}{\sqrt{24}}=\frac{11.2}{\sqrt{24}} \mathrm{~km} / \mathrm{s} \quad\left(\because v_{e}=11.2 \mathrm{~km} / \mathrm{s}\right)$
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