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(B) According to the law of conservation of mechanical energy,the loss in gravitational potential energy equals the gain in kinetic energy.Initial pot

Vedclass · Accepted Answer

$\sqrt{g R}$

Vedclass · Answer

(B) According to the law of conservation of mechanical energy,the loss in gravitational potential energy equals the gain in kinetic energy.Initial potential energy at height $h = R$ is $U_i = -\frac{GMm}{R+R} = -\frac{GMm}{2R}$.Final potential energy at the earth's surface $(r = R)$ is $U_f = -\frac{GMm}{R}$.Loss in potential energy = $U_i - U_f = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{2R}$.Gain in kinetic energy = $\frac{1}{2}mv^2$.Equating the two: $\frac{1}{2}mv^2 = \frac{GMm}{2R}$.Using $GM = gR^2$,we get $\frac{1}{2}v^2 = \frac{gR^2}{2R} = \frac{gR}{2}$.Therefore,$v^2 = gR$,which gives $v = \sqrt{gR}$.

An object is allowed to fall from a height $R$ above the earth,where $R$ is the radius of the earth. Its velocity when it strikes the earth's surface,ignoring air resistance,will be:

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