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——

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 10.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$Conservation of energy,$

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

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