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(A) The escape velocity $V_e$ is given by $V_e = \sqrt{\frac{2GM}{R}}$.Applying the law of conservation of energy between the Earth's surface and infi

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$\sqrt{3} V_e$

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(A) The escape velocity $V_e$ is given by $V_e = \sqrt{\frac{2GM}{R}}$.Applying the law of conservation of energy between the Earth's surface and infinity:Total energy at surface = Total energy at infinity$-\frac{GMm}{R} + \frac{1}{2}m(2V_e)^2 = 0 + \frac{1}{2}mV^2$Here,$V$ is the final velocity at infinity.Substituting $V_e^2 = \frac{2GM}{R}$,we get:$-\frac{GM}{R} + \frac{1}{2}(4 \cdot \frac{2GM}{R}) = \frac{1}{2}V^2$$-\frac{GM}{R} + \frac{4GM}{R} = \frac{1}{2}V^2$$\frac{3GM}{R} = \frac{1}{2}V^2$$V^2 = \frac{6GM}{R} = 3 \left( \frac{2GM}{R} \right) = 3V_e^2$$V = \sqrt{3}V_e$

$A$ body is projected vertically upwards from the Earth's surface with velocity $2 V_e$,where $V_e$ is the escape velocity from the Earth's surface. The velocity when the body escapes the gravitational pull is

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