If earth has a mass nine times and radius twice to that of a planet $$P$$, then $$\frac{v_e}{3}\sqrt{x} \text{ ms}^{-1}$$ 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

Escape velocity ($$v$$), is given by the formula: $$v = \sqrt{\frac{2GM}{R}}$$

$$M_e = 9M_p \implies M_p = \frac{M_e}{9}$$

$$R_e = 2R_p \implies R_p = \frac{R_e}{2}$$

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

$$v_p = \sqrt{\frac{2GM_e}{R_e}} \times \sqrt{\frac{2}{9}}$$

$$v_p = v_e \times \frac{\sqrt{2}}{3}$$
$$v_p = \frac{v_e}{3}\sqrt{2} = \frac{v_e}{3}\sqrt{x}$$:

$$x = 2$$