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(A) According to the law of conservation of energy, the total energy at the surface of the earth must equal the total energy at infinity.Let $m$ be th

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$2 \,V_{e}$

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(A) According to the law of conservation of energy, the total energy at the surface of the earth must equal the total energy at infinity.Let $m$ be the mass of the body and $M$ be the mass of the earth.The initial speed is $v = \sqrt{5} V_{e}$, where $V_{e} = \sqrt{\frac{2GM}{R}}$.The total initial energy $E_{i} = \frac{1}{2}mv^2 - \frac{GMm}{R}$.Substituting $v = \sqrt{5} V_{e}$, we get $E_{i} = \frac{1}{2}m(5 V_{e}^2) - \frac{GMm}{R} = \frac{5}{2}m(\frac{2GM}{R}) - \frac{GMm}{R} = \frac{5GMm}{R} - \frac{GMm}{R} = \frac{4GMm}{R}$.At infinity, the potential energy is $0$. Let the final speed be $v_{f}$.The total final energy $E_{f} = \frac{1}{2}mv_{f}^2 + 0$.Equating $E_{i} = E_{f}$, we get $\frac{4GMm}{R} = \frac{1}{2}mv_{f}^2$.$v_{f}^2 = \frac{8GM}{R} = 4 	imes (\frac{2GM}{R}) = 4 V_{e}^2$.Therefore, $v_{f} = 2 V_{e}$.

$A$ body is projected from the earth's surface with a speed $\sqrt{5}$ times the escape speed $(V_{e})$. The speed of the body when it escapes from the gravitational influence of the earth is

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