Q65 / 90 · KVPY 2012 (SB/SX) · Physics
A small asteroid is orbiting around the sun in a circular orbit of radius $\mathrm{r}_{0}$ with speed $\mathrm{V}_{0}$. A rocket is launched from the asteroid with speed $\mathrm{V}=\alpha \mathrm{V}_{0}$, where $\mathrm{V}$ is the speed relative to the sun. The highest value of $\alpha$ for which the rocket will remain bound to the solar system is (ignoring gravity due to the asteroid and effects of other planets) -
- A$\sqrt{2}$
- B2
- C$\sqrt{3}$
- Correct answerD1
Solution
$\begin{array}{l}
\text { B.E }=\frac{G m M}{2 R} \\
\text { B.E }=\frac{1}{2} m v_{0}^{2} \\
\text { so, } \alpha=1
\end{array}$
Hence, the correct answer is option: 'D'.