Calculate Escape Speed on Surface of Planet Using Gravity and Density

To find the escape speed on the surface of the planet, we use the relations for gravitational acceleration, density, and escape velocity.

The gravitational acceleration $g$ on the surface of a spherical body of mass $M$, radius $R$, and uniform density $\rho$ is given by:
$g = \frac{GM}{R^2} = \frac{G \left(\frac{4}{3}\pi R^3 \rho\right)}{R^2} = \frac{4}{3}\pi G \rho R$

From this, the radius $R$ can be expressed in terms of $g$ and $\rho$:
$R = \frac{3g}{4\pi G \rho}$

The ratio of the radius of the planet $R_p$ to the radius of the Earth $R_e$ is:
$\frac{R_p}{R_e} = \left(\frac{g_p}{g_e}\right) \left(\frac{\rho_e}{\rho_p}\right)$

Given that:
$\frac{g_p}{g_e} = \frac{\sqrt{6}}{11} \quad \text{and} \quad \frac{\rho_p}{\rho_e} = \frac{2}{3} \implies \frac{\rho_e}{\rho_p} = \frac{3}{2}$

Substituting these values into the ratio of the radii:
$\frac{R_p}{R_e} = \left(\frac{\sqrt{6}}{11}\right) \left(\frac{3}{2}\right) = \frac{3\sqrt{6}}{22}$

The escape speed $v_{\text{esc}}$ from the surface of a celestial body is given by:
$v_{\text{esc}} = \sqrt{2gR}$

Taking the ratio of the escape speed on the planet $v_p$ to that on the Earth $v_e$:
$\frac{v_p}{v_e} = \sqrt{\frac{g_p}{g_e} \times \frac{R_p}{R_e}} = \sqrt{\left(\frac{\sqrt{6}}{11}\right) \left(\frac{3\sqrt{6}}{22}\right)} = \sqrt{\frac{3 \times 6}{11 \times 22}} = \sqrt{\frac{18}{242}} = \sqrt{\frac{9}{121}} = \frac{3}{11}$

Given that the escape speed on the surface of the Earth is $v_e = 11\text{ kms}^{-1}$:
$v_p = v_e \times \frac{3}{11} = 11 \times \frac{3}{11} = 3\text{ kms}^{-1}$