The initial velocity v_{i} required to projec | vedclass

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(C) Using the law of conservation of energy between the surface of the earth and the maximum height $h = 10 R$:Total energy at surface = Total energy

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$10$

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(C) Using the law of conservation of energy between the surface of the earth and the maximum height $h = 10 R$:Total energy at surface = Total energy at height $h$.$-\frac{GMm}{R} + \frac{1}{2}mv_{i}^{2} = -\frac{GMm}{R+h} + 0$.Since $h = 10R$,the total distance from the center is $R + 10R = 11R$.$-\frac{GMm}{R} + \frac{1}{2}mv_{i}^{2} = -\frac{GMm}{11R}$.$\frac{1}{2}mv_{i}^{2} = GMm \left( \frac{1}{R} - \frac{1}{11R} ight) = GMm \left( \frac{10}{11R} ight)$.$v_{i}^{2} = \frac{20GM}{11R}$.We know the escape velocity $v_{e} = \sqrt{\frac{2GM}{R}}$,so $v_{e}^{2} = \frac{2GM}{R}$.Substituting this into the equation for $v_{i}^{2}$:$v_{i}^{2} = \frac{10}{11} 	imes \left( \frac{2GM}{R} ight) = \frac{10}{11} v_{e}^{2}$.Comparing this with $v_{i} = \sqrt{\frac{x}{y}} v_{e}$,we get $\frac{x}{y} = \frac{10}{11}$.Thus,$x = 10$.

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