If earth has a mass nine times and radius twice to the of a planet P . Then v e 3 x m s - 1 will be the…

The formula to calculate the escape velocity$\left(v_e\right)$ for the Earth is given by

$v_e= \sqrt{\frac{2G{M}_e}{R_e}}.......................\left(1\right)$

where, $G$ is the Universal Gravitational constant, $M_e$ is the mass and $R_e$ is the radius of Earth.

The formula for the escape velocity $\left(v_{\mathrm{esc}}\right)$ of any planet is given by

$v_{\mathrm{esc}}= \sqrt{\frac{2GM}{R}}........................\left(2\right)$

where, $M$ is the mass and $R$ is the radius of the planet.

Given that the mass of the planet is $9$ times the mass of the Earth and the radius of the planet is $2$ times the radius of the Earth.

Substitute $\raisebox{1ex}{$M_e$}\!\left/ \!\raisebox{-1ex}{$9$}\right.$ for $M$ and $\raisebox{1ex}{$R_e$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ for $R$ into equation (2) and simplify to obtain the escape velocity for the planet in terms of that in Earth.

$\begin{array}{rcl}{v}_{\mathrm{esc}}& =& \sqrt{\frac{2G{\displaystyle \frac{M_e}{9}}}{{\displaystyle \frac{R_e}{2}}}}\\ & =& \sqrt{\frac{2}{9}\frac{2G{M}_e}{R_e}}\\ & =& \frac{\sqrt{2}}{3}{v}_e.....................\left(3\right)\end{array}$

Comparing equation (3) with the given expression, it can be written that

$x= 2$.