Given that mass of earth is M and its radius is R. A body is dropped from a height equal to the radius of the earth above the surface of the earth. When it reaches the ground, its speed will be

\frac{G M}{R}

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\sqrt{\frac{G M}{R}}

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\sqrt{\frac{2 G M}{R}}

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\frac{2 G M}{R}

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Solution

The correct option is B $\sqrt{\frac{G M}{R}}$
$k i + u i = k f + u f$
$0 + \frac{- G M m}{(R + R)} = \frac{1}{2} m v^{2} + (\frac{- G M m}{R})$
$\frac{1}{2} m v^{2} = \frac{G M m}{2 R}$
$v = \sqrt{\frac{G M}{R}}$
Hence, Option B is correct answer

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Given that mass of earth is M and its radius is R. A body is dropped from a height equal to the radius of the earth above the surface of the earth. When it reaches the ground, its speed will be

The correct option is B √GMRki+ui=kf+uf0+−GMm(R+R)=12mv2+(−GMmR)12mv2=GMm2Rv=√GMRHence, Option B is correct answer

The correct option is B $\sqrt{\frac{G M}{R}}$
$k i + u i = k f + u f$
$0 + \frac{- G M m}{(R + R)} = \frac{1}{2} m v^{2} + (\frac{- G M m}{R})$
$\frac{1}{2} m v^{2} = \frac{G M m}{2 R}$
$v = \sqrt{\frac{G M}{R}}$
Hence, Option B is correct answer