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A small asteroid is orbiting around the sun in a circular orbit of radius $r_0$ with speed $v_0$. A rocket is launched from the asteroid with speed $v=\alpha v_0$, where $v$ 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 eff ects of other planets)
$\sqrt{2}$
$2$
$\sqrt{3}$
$1$
Solution
(d)
Total energy of rocket at instant of launch = Potential energy due to its position on asteroid + Kinetic energy due to motion of asteroid + Kinetic energy given for launch $=-\frac{G M m}{r_0}+\frac{1}{2} m V_0^2+\frac{1}{2} m\left(\alpha V_0\right)^2$
Rocket remains bounded to the solar system, if this energy is less than or equal to zero.
$\text { i.e. } \quad \frac{-G M m}{r_0}+\frac{1}{2} m V_0^2\left(1+\alpha^2\right) \leq 0$
$\Rightarrow \quad \frac{1}{2} m V_0^2\left(1+\alpha^2\right)=\frac{G M m}{r_0}$
$\Rightarrow \quad \frac{1}{2} m \frac{G M}{r_0}\left(1+\alpha^2\right)=\frac{G M m}{r_0}$
$\Rightarrow \quad \alpha^2=1 \Rightarrow \alpha=1$
Similar Questions
Match list-$I$ with list-$II$:
| List-$i$ | List-$2$ |
| $(A)$Kinetic energy of plant | $(1)$ $-\frac{\mathrm{GMm}}{\mathrm{a}}$ |
| $(B)$ Gravaitatioin potentiyal energy of sun -plant system | $(2)$ $\frac{\mathrm{GMm}}{2 \mathrm{a}}$ |
| $(C)$Total mecaniacal energy of palnt | $(3)$ $\frac{\mathrm{Gm}}{\mathrm{r}}$ |
| $(D)$Escap energyat the surface of plant for unit mass object | $(4)$ $-\frac{\mathrm{GMm}}{2 \mathrm{a}}$ |
(Where $\mathrm{a}=$ radius of planet orbit, $\mathrm{r}=$ radius of planet, $M=$ mass of Sun, $m=$ mass of planet)
Choose the correct answer from the options given below:
