The escape velocity from the earth is about 11km /s . The escape velocity from a planet having twice the radius and the same mean density as the earth, is

The correct option is 22 km/s.

To find the escape velocity from the planet, let us express the escape velocity formula in terms of the planet's radius and mean density.

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

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

where $G$ is the universal gravitational constant.

Assuming the planet is a sphere, its mass $M$ can be written in terms of its mean density $\rho$ and radius $R$ as:

$M=\text{Volume}\times \text{density}=\frac{4}{3}\pi R^3\rho$

Substituting this expression for $M$ into the escape velocity formula:

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

$v=R\sqrt{\frac{8\pi G\rho }{3}}$

Since $G$ is a constant and both Earth and the planet have the same mean density $\rho$, the term inside the square root remains constant. Consequently, the escape velocity is directly proportional to the radius of the planet:

$v\propto R$

Let $v_e$ be the escape velocity from the Earth, and $v_p$ be the escape velocity from the planet. Similarly, let $R_e$ be the radius of the Earth, and $R_p$ be the radius of the planet.

We are given:
$v_e=11\text{km/s}$
$R_p=2R_e$

Using the proportionality relation:

$\frac{v_p}{v_e}=\frac{R_p}{R_e}$

$\frac{v_p}{11\text{km/s}}=\frac{2R_e}{R_e}$

$v_p=2\times 11\text{km/s}=22\text{km/s}$

Thus, the escape velocity from the planet is 22 km/s.