Assuming $g_{(m o o n)} = (\frac{1}{6}) g_{e a r t h}$ and $D_{(m o o n)} = (\frac{1}{4}) D_{e a r t h}$ where $g$ and $D$ are the acceleration due to gravity and diameter respectively, the escape velocity from the moon is:

\frac{11.2}{24} {k m s}^{- 1}

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11.2 \times \sqrt{24} {k m s}^{- 1}

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\frac{11.2}{\sqrt{24}} {k m s}^{- 1}

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11.2 \times 24 {k m s}^{- 1}

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Solution

The correct option is C $\frac{11.2}{\sqrt{24}} {k m s}^{- 1}$
Escape velocity, $v_{e} = \sqrt{2 g R}$

$\Rightarrow \frac{v_{m}}{v_{e}} = \sqrt{\frac{\frac{1}{6} g_{e} \times \frac{1}{4} R_{e}}{g_{e} \times R_{e}}} = \frac{1}{\sqrt{24}}$

$\Rightarrow v_{m} = \frac{11.2}{\sqrt{24}} {k m s}^{- 1}$

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Assuming g(moon)=(16)gearth and D(moon)=(14)Dearth where g and D are the acceleration due to gravity and diameter respectively, the escape velocity from the moon is:

The correct option is C 11.2√24 kms−1Escape velocity, ve=√2gR⇒vmve=  ⎷16ge×14Rege×Re=1√24⇒vm=11.2√24 kms−1

Assuming $g_{(m o o n)} = (\frac{1}{6}) g_{e a r t h}$ and $D_{(m o o n)} = (\frac{1}{4}) D_{e a r t h}$ where $g$ and $D$ are the acceleration due to gravity and diameter respectively, the escape velocity from the moon is:

The correct option is C $\frac{11.2}{\sqrt{24}} {k m s}^{- 1}$
Escape velocity, $v_{e} = \sqrt{2 g R}$

$\Rightarrow \frac{v_{m}}{v_{e}} = \sqrt{\frac{\frac{1}{6} g_{e} \times \frac{1}{4} R_{e}}{g_{e} \times R_{e}}} = \frac{1}{\sqrt{24}}$

$\Rightarrow v_{m} = \frac{11.2}{\sqrt{24}} {k m s}^{- 1}$