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The mass of the moon is $\frac{1}{81}$ of earth’s mass and its radius $\frac{1}{4} th$ that of the earth. If the escape velocity from the earth’s surface is $11.2 k m s^{- 1}$ , its value for the moon will be..(up to one decimal place)

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answer is 2.5.

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Detailed Solution

$\frac{g_{m}}{g_{e}} = \frac{M_{m}}{M_{e}} \times (\frac{R_{e}}{R_{m}}) = \frac{1}{81} \times (4)^{2} = \frac{16}{81}$

$g_{m} = \frac{16}{81} g_{e}$

$∴ v_{e} = \sqrt{2 g_{e} R_{e}} = \sqrt{2 \times 9.8 \times (6400 \times 1000)}$

$\approx 11.2 {kms}^{- 1}$

$v_{m} = \sqrt{2 g_{m} R_{m}} = \sqrt{2 \times \frac{16}{81} g_{e} \times \frac{1}{4} R_{e}}$

$= \frac{2}{9} \sqrt{2 g_{e} R_{e}} = \frac{2}{9} \times 11 \approx 2.5 {kms}^{- 1}$

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