Knowing that the mass of the moon is $1 / 81$ times that of earth and its radius is $1 / 4$ the radius of earth. If the escape velocity at the surface of the earth is $11.2 k m / s e c$ , then the value of escape velocity at the surface of the moon is

2.5 k m / s e c

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0.14 k m / s e c

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5 k m / s e c

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8 k m / s e c

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Solution

The correct option is A $2.5 k m / s e c$
For escape velocity, $v_{e} = \sqrt{\frac{2 G M}{R}}$
$v_{e}^{'} = \sqrt{\frac{2 G M / 81}{R / 4}} = \sqrt{\frac{4 \times 2 G M}{81 R}} = \frac{2}{9} \sqrt{\frac{G M}{R}}$
$v_{e}^{'} = \frac{2}{9} \times 11.2 = 2.5 k m / s e c$

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Knowing that the mass of the moon is 1/81 times that of earth and its radius is 1/4 the radius of earth. If the escape velocity at the surface of the earth is 11.2km/sec, then the value of escape velocity at the surface of the moon is

The correct option is A 2.5km/secFor escape velocity, ve=√2GMRv′e=√2GM/81R/4=√4×2GM81R=29√GMRv′e=29×11.2=2.5km/sec

Knowing that the mass of the moon is $1 / 81$ times that of earth and its radius is $1 / 4$ the radius of earth. If the escape velocity at the surface of the earth is $11.2 k m / s e c$ , then the value of escape velocity at the surface of the moon is

The correct option is A $2.5 k m / s e c$
For escape velocity, $v_{e} = \sqrt{\frac{2 G M}{R}}$
$v_{e}^{'} = \sqrt{\frac{2 G M / 81}{R / 4}} = \sqrt{\frac{4 \times 2 G M}{81 R}} = \frac{2}{9} \sqrt{\frac{G M}{R}}$
$v_{e}^{'} = \frac{2}{9} \times 11.2 = 2.5 k m / s e c$