Gravitational acceleration on the surface of a planet is $(√\frac{6}{11}) g,$ where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\frac{2}{3}$ times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km/s, the escape speed on the surface of the planet in km/s will be

Step 1: Identify the given parameters.

Let the subscripts $e$ and $p$ denote the Earth and the planet, respectively.

$$g_p = \frac{\sqrt{6}}{11} g_e$$ $$\rho_p = \frac{2}{3} \rho_e$$ $$v_e = 11 \text{ km s}^{-1}$$

Step 2: Relate escape velocity to surface gravity and radius.

The escape velocity $v$ from the surface of a spherical body of mass $M$ and radius $R$ is given by:

$$v = \sqrt{\frac{2GM}{R}}$$

The acceleration due to gravity $g$ at the surface is:

$$g = \frac{GM}{R^2}$$

Substitute $GM = gR^2$ into the escape velocity formula.

$$v = \sqrt{2gR}$$

Step 3: Express radius in terms of gravity and density.

The mass of a uniform spherical body of density $\rho$ is:

$$M = \frac{4}{3}\pi R^3 \rho$$

Substitute this mass into the expression for $g$.

$$g = \frac{G}{R^2} \left(\frac{4}{3}\pi R^3 \rho\right)$$ $$g = \frac{4}{3}\pi G \rho R$$

Solve for the radius $R$.

$$R = \frac{3g}{4\pi G \rho}$$

Step 4: Determine the proportionality relation for escape velocity.

Substitute the expression for $R$ back into the escape velocity formula.

$$v = \sqrt{2g \left(\frac{3g}{4\pi G \rho}\right)}$$ $$v = g \sqrt{\frac{3}{2\pi G \rho}}$$

This yields the proportionality relation:

$$v \propto \frac{g}{\sqrt{\rho}}$$

Step 5: Calculate the escape velocity on the planet.

Set up the ratio of the escape velocity of the planet to that of the Earth.

$$\frac{v_p}{v_e} = \frac{g_p / g_e}{\sqrt{\rho_p / \rho_e}}$$

Substitute the given values into the ratio.

$$\frac{v_p}{v_e} = \frac{\frac{\sqrt{6}}{11}}{\sqrt{\frac{2}{3}}}$$ $$\frac{v_p}{v_e} = \frac{\sqrt{6}}{11} \cdot \sqrt{\frac{3}{2}}$$ $$\frac{v_p}{v_e} = \frac{3}{11}$$

Solve for $v_p$.

$$v_p = v_e \cdot \frac{3}{11}$$ $$v_p = 11 \cdot \frac{3}{11}$$ $$v_p = 3 \text{ km s}^{-1}$$

Answer: The escape speed on the surface of the planet is $3\text{ km s}^{-1}$.