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A body is released from a height equal to the radius $$(R)$$ of the earth. The velocity of the body when it strikes the surface of the earth will be: (Given $$g$$ = acceleration due to gravity on the earth.)
A body is released from height $$h = R$$ above the Earth's surface. Using conservation of energy:
$$-\frac{GMm}{R+R} = \frac{1}{2}mv^2 - \frac{GMm}{R}$$
$$\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$$
$$v^2 = \frac{GM}{R}$$
Since $$g = \frac{GM}{R^2}$$, we have $$GM = gR^2$$:
$$v^2 = \frac{gR^2}{R} = gR$$
$$v = \sqrt{gR}$$
This matches option 2.
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