• question_answer The escape velocity of a body on the Earth's surface is ${{v}_{e}}$. A body is thrown up with a speed $\sqrt{5{{v}_{e}}}.$ Assuming that the Sun and planets do not influence the motion of the body, the velocity of the body at infinite distance is ${{v}_{\infty }}$. Then, the value of $\frac{{{v}_{\infty }}}{{{v}_{e}}}$ is A) Zero B) 1 C) 2 D) 3

 Applying conservation of mechanical energy. Energy at initial position = Energy at Infinity $\Rightarrow$$\frac{1}{2}mv_{i}^{2}-\frac{GmM}{R}=\frac{1}{2}mv_{\infty }^{2}+GPE$ at infinity Or $\frac{1}{2}m{{\left( \sqrt{5}{{V}_{e}} \right)}^{1}}-\left( \frac{GM}{{{R}^{2}}} \right)mR=\frac{1}{2}mv_{\infty }^{2}+0$ Or $\frac{1}{2}m\left( 5\operatorname{v}_{e}^{2} \right)-mgR=\frac{1}{2}mv_{\infty }^{2}$$\Rightarrow$$\frac{5v_{e}^{2}}{2}-gR=\frac{1}{2}v_{\infty }^{2}$$\Rightarrow$$\frac{5v_{e}^{2}}{2}-\frac{2gR}{2}=\frac{1}{2}v_{\infty }^{2}$ $\frac{5v_{e}^{2}}{2}-\frac{v_{e}^{2}}{2}=\frac{1}{2}v_{\infty }^{2}\left( \because v_{e}^{2}=2gR \right)$$\Rightarrow$$\frac{1}{2}\left( 4v_{e}^{2} \right)=\frac{1}{2}v_{\infty }^{2}$$\Rightarrow$${{\left( \frac{{{v}_{\infty }}}{{{v}_{e}}} \right)}^{2}}=4\Rightarrow \frac{{{v}_{\infty }}}{{{v}_{e}}}=2$