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  4. Given mass of the moon is (1/81) of the mass of the earth and corresponding radius is (1/4) of the earth. If escape velocity on the earth surface is 11.2 km/s, the value of same on the surface of the moon is

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Q. Given mass of the moon is $ \frac{1}{81} $ of the mass of the earth and corresponding radius is $ \frac{1}{4} $ of the earth. If escape velocity on the earth surface is 11.2 km/s, the value of same on the surface of the moon is

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Solution:

On earth $ {{v}_{e}}=\sqrt{\frac{2GM}{R}} $ $ =11.2\,km/s $ On moon $ {{v}_{m}}=\sqrt{\frac{2GM\times 4}{81\times R}} $ $ =\frac{2}{9}\sqrt{\frac{2GM}{R}} $ $ =\frac{2}{9}\times 11.2=2.5\,km/s $