The escape velocity from the Earth’s surface is v. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is :

The correct option is 4 v.

Let us derive this step-by-step to understand the relation between escape velocity, planet radius, and mass density.

Step 1: Formula for Escape Velocity
The escape velocity ($v$) from the surface of a planet of mass $M$ and radius $R$ is given by the formula:
$v=\sqrt{\frac{2GM}{R}}$
where $G$ is the universal gravitational constant.

Step 2: Expressing Mass in Terms of Density and Radius
Assuming the planet is a uniform sphere, its mass ($M$) can be written in terms of its volume and mass density ($\rho$):
$M=\text{Volume}\times \text{Density}=\frac{4}{3}\pi R^3\rho$
Substituting this expression for $M$ back into the escape velocity formula:
$v=\sqrt{\frac{2G\left(\frac{4}{3}\pi R^3\rho \right)}{R}}$
Simplifying the terms inside the square root:
$v=\sqrt{\frac{8}{3}\pi G\rho R^2}$
Since $R$ is squared, we can take it out of the square root:
$v=R\sqrt{\frac{8}{3}\pi G\rho }$
This shows that for planets with the same mass density ($\rho$), the escape velocity is directly proportional to the radius of the planet:
$v\propto R$

Step 3: Calculating the Escape Velocity for the New Planet
Let $v_e$ and $R_e$ be the escape velocity and radius of the Earth, and $v_p$ and $R_p$ be the escape velocity and radius of the other planet.
We are given that the density is the same, and the radius of the other planet is four times that of the Earth:
$R_p=4R_e$
Using the proportionality relation:
$\frac{v_p}{v_e}=\frac{R_p}{R_e}$
Substitute $R_p=4R_e$ into the equation:
$\frac{v_p}{v}=\frac{4R_e}{R_e}=4$
Therefore, the escape velocity of the planet is:
$v_p=4v$