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(C) Let $R$ be the radius of the earth and $M$ be its mass. The escape velocity at the surface is $v_e = \sqrt{\frac{2GM}{R}} = 11.2 \ km/s$.When the

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$10$

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(C) Let $R$ be the radius of the earth and $M$ be its mass. The escape velocity at the surface is $v_e = \sqrt{\frac{2GM}{R}} = 11.2 \ km/s$.When the rocket is at an altitude $h = \frac{R}{4}$,its distance from the center of the earth is $r = R + h = R + \frac{R}{4} = \frac{5R}{4}$.To escape the earth's gravitational pull from this point,the rocket must have a velocity $v'$ such that its total energy is at least zero.Using the conservation of energy: $\frac{1}{2}mv'^2 - \frac{GMm}{r} = 0$.$v' = \sqrt{\frac{2GM}{r}} = \sqrt{\frac{2GM}{5R/4}} = \sqrt{\frac{8GM}{5R}} = \sqrt{\frac{4}{5}} 	imes \sqrt{\frac{2GM}{R}}$.Substituting $v_e = 11.2 \ km/s$:$v' = \sqrt{0.8} 	imes 11.2 \approx 0.8944 	imes 11.2 \approx 10.02 \ km/s$.Thus,the minimum velocity is approximately $10 \ km/s$.

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