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(C) According to the law of conservation of energy,the total energy at the surface of the Earth equals the total energy at the maximum height $h$. At

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$4$

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(C) According to the law of conservation of energy,the total energy at the surface of the Earth equals the total energy at the maximum height $h$. At the surface: $E_i = \frac{1}{2}mv^2 - \frac{GMm}{R}$. At maximum height $h$: $E_f = 0 - \frac{GMm}{R+h}$. Equating $E_i = E_f$: $\frac{1}{2}mv^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$. Using $V_e = \sqrt{\frac{2GM}{R}}$,we have $GM = \frac{V_e^2 R}{2}$. Substituting this: $\frac{v^2}{2} - \frac{V_e^2}{2} = - \frac{V_e^2 R}{2(R+h)}$. Rearranging gives: $\frac{1}{R+h} = \frac{1}{R} \left(1 - \frac{v^2}{V_e^2}\right)$. Given $v = 10 \ km/s$ and $V_e = 11.2 \ km/s$: $h = \frac{R}{\left(\frac{V_e^2}{v^2} - 1\right)} = \frac{R}{\left(\left(\frac{11.2}{10}\right)^2 - 1\right)} = \frac{R}{(1.2544 - 1)} = \frac{R}{0.2544} \approx 3.93R$. Thus,the maximum height attained is approximately $4R$.

$A$ rocket is launched with velocity $10 \ km/s$. If the radius of the Earth is $R$,then the maximum height attained by it will be: (in $R$)

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