A body of mass m falls from a height R above | vedclass

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(B) According to the law of conservation of energy,the total mechanical energy at height $R$ above the surface (distance $2R$ from the center) must eq

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$\sqrt{gR}$

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(B) According to the law of conservation of energy,the total mechanical energy at height $R$ above the surface (distance $2R$ from the center) must equal the total mechanical energy at the surface (distance $R$ from the center).Initial energy at height $R$ (where $r = 2R$): $E_i = -\frac{GMm}{2R} + 0$Final energy at the surface (where $r = R$): $E_f = -\frac{GMm}{R} + \frac{1}{2}mv^2$Equating $E_i = E_f$:$-\frac{GMm}{2R} = -\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}$Since the acceleration due to gravity on the surface is $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting $GM = gR^2$ into the equation for $v^2$:$v^2 = \frac{gR^2}{R} = gR$Therefore,$v = \sqrt{gR}$.

$A$ body of mass $m$ falls from a height $R$ above the surface of the earth,where $R$ is the radius of the earth. What is the velocity attained by the body on reaching the ground? (Acceleration due to gravity on the surface of the earth is $g$)

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