If earth has a mass nine times and radius twice to the of a planet $$P$$. Then $$\frac{v_e}{3}\sqrt{x}$$ ms$$^{-1}$$ will be the minimum velocity required by a rocket to pull out of gravitational force of $$P$$, where $$v_e$$ is escape velocity on earth. The value of $$x$$ is

We are given that Earth has mass $$M_e = 9M_p$$ and radius $$R_e = 2R_p$$, where $$M_p$$ and $$R_p$$ are the mass and radius of planet $$P$$.

The escape velocity on Earth is $$v_e = \sqrt{\frac{2GM_e}{R_e}}$$, and on planet $$P$$ it is $$v_p = \sqrt{\frac{2GM_p}{R_p}}$$.

From the given relations, $$M_p = \frac{M_e}{9}$$ and $$R_p = \frac{R_e}{2}$$. Substituting these in:

$$v_p = \sqrt{\frac{2G \cdot \frac{M_e}{9}}{\frac{R_e}{2}}} = \sqrt{\frac{2GM_e \cdot 2}{9R_e}} = \sqrt{\frac{2}{9}} \cdot \sqrt{\frac{2GM_e}{R_e}}$$

So $$v_p = \frac{\sqrt{2}}{3} \cdot v_e = \frac{v_e}{3}\sqrt{2}$$.

Comparing with the given form $$\frac{v_e}{3}\sqrt{x}$$, we get $$x = 2$$. Hence, the correct answer is $$2$$.