The initial velocity $v_{i}$ required to project a body vertically upward from the surface of the earth to reach a height of 10 R, where R is the radius of the earth, may be described in terms of escape velocity $v_{e}$ such that $v_{i} = \sqrt{\frac{x}{y}} \times v_{e}$ . The value of x will be——

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answer is 10.

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Detailed Solution

$Conservation of energy,$

$\begin{array}{l} U_{i} + K_{i} = U_{f} + K_{f} \\ \Rightarrow - \frac{G M m}{R} + \frac{1}{2} m v_{i}^{2} = - \frac{G M m}{11 R} + 0 \\ If v_{i} = k v \\ \Rightarrow \frac{1}{2} k^{2} v_{e}^{2} = \frac{10}{11} \frac{G M}{R} \end{array}$

$\begin{array}{l} Put, v_{e} = \sqrt{\frac{2 G M}{R}} \\ \Rightarrow \frac{1}{2} k^{2} \frac{2 G M}{R} = \frac{10}{11} \frac{G M}{R} \\ \Rightarrow k^{2} = \frac{10}{11} \\ As per question, k = \sqrt{\frac{x}{y}} \\ \Rightarrow k^{2} = \frac{10}{11} = \frac{x}{y} \\ So, x = 10 \end{array}$