A body of mass m is situated at a distance eq | vedclass

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(B) The distance of the body from the center of the Earth is $r = R + 2R = 3R$.The gravitational potential energy $(PE)$ of the body at this distance

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$\frac{mgR}{3}$

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(B) The distance of the body from the center of the Earth is $r = R + 2R = 3R$.The gravitational potential energy $(PE)$ of the body at this distance is given by $PE = -\frac{GMm}{r} = -\frac{GMm}{3R}$.Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting this into the potential energy expression: $PE = -\frac{(gR^2)m}{3R} = -\frac{mgR}{3}$.The energy required to escape the gravitational field is equal to the magnitude of the potential energy,which is the binding energy: $E_{escape} = -PE = \frac{mgR}{3}$.

$A$ body of mass $m$ is situated at a distance equal to $2R$ ($R$ is the radius of the Earth) from the Earth's surface. The minimum energy required to be given to the body so that it may escape out of the Earth's gravitational field is:

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