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(B) According to the law of conservation of mechanical energy,the total energy at the initial height must equal the total energy at the surface of the

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$\left[ \frac{GM}{R} \right]^{1/2}$

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(B) According to the law of conservation of mechanical energy,the total energy at the initial height must equal the total energy at the surface of the Earth.Initial height $h = R$. The distance from the center of the Earth is $r_i = R + h = 2R$.Initial potential energy $U_i = -\frac{GMm}{2R}$ and initial kinetic energy $K_i = 0$.Final potential energy at the surface $U_f = -\frac{GMm}{R}$ and final kinetic energy $K_f = \frac{1}{2}mv^2$.Applying conservation of energy: $U_i + K_i = U_f + K_f$.$-\frac{GMm}{2R} + 0 = -\frac{GMm}{R} + \frac{1}{2}mv^2$.$\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$.$v^2 = \frac{GM}{R}$.$v = \sqrt{\frac{GM}{R}} = \left[ \frac{GM}{R} \right]^{1/2}$.

Given that the mass of the 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 velocity will be:

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