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Standard XII

Physics

Escape Speed

The mass of t...

Question

The mass of the moon is $1 / 81$ of earth's mass and its radius $1 / 4$ that of the earth. If the escape velocity from the earth's surface is $11.2$ km/sec. its value from the surface of the moon will be

0.14 k m s^{- 1}

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0.5 k m s^{- 1}

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2.5 k m s^{- 1}

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5.0 k m s^{- 1}

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Solution

The correct option is C $2.5 k m s^{- 1}$
$v_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}}$
$v_{m} = \sqrt{\frac{2 G \frac{M_{e}}{81}}{\frac{R_{e}}{4}}} = \frac{2}{9} v_{e}$
$= 2 / 9 \times 11.2 k m s^{- 1} = 2.5 k m s^{- 1}$

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The mass of the moon is 1/81 of earth's mass and its radius 1/4 that of earth. If the escape velocity from the earth's surface is 11.2 km/s ,its value for the moon...

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The mass of the moon is 1/81 of earth's mass and its radius 1/4 that of the earth. If the escape velocity from the earth's surface is 11.2 km/sec. its value from the surface of the moon will be

The correct option is C 2.5kms−1ve=√2GMeRevm=    ⎷2GMe81Re4=29ve=2/9×11.2kms−1=2.5kms−1

The mass of the moon is $1 / 81$ of earth's mass and its radius $1 / 4$ that of the earth. If the escape velocity from the earth's surface is $11.2$ km/sec. its value from the surface of the moon will be

The correct option is C $2.5 k m s^{- 1}$
$v_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}}$
$v_{m} = \sqrt{\frac{2 G \frac{M_{e}}{81}}{\frac{R_{e}}{4}}} = \frac{2}{9} v_{e}$
$= 2 / 9 \times 11.2 k m s^{- 1} = 2.5 k m s^{- 1}$