vₑ and vₚ denotes the escape velocity from the earth and another planet having twice the radius and the same mean density as the earth. Then

The correct option is vₑ = vₚ/2.

To understand the relationship between the escape velocity of the Earth and the other planet, we start with the formula for escape velocity from the surface of a spherical body:
$v=\sqrt{\frac{2GM}{R}}$
where:

• G is the universal gravitational constant,
• M is the mass of the body, and
• R is the radius of the body.

Since the question specifies that both bodies have the same mean density, we can express the mass M of a spherical body of radius R and constant density ρ as:
$M=\rho \cdot V=\rho \cdot \frac{4}{3}\pi R^3$

Substituting this expression for mass into the escape velocity formula, we get:

Simplifying the expression under the square root:
$v=\sqrt{\frac{8\pi G\rho }{3}}\cdot R$

Since the mean density ρ and the gravitational constant G are the same for both the Earth and the other planet, the term under the square root is a constant. Therefore, the escape velocity is directly proportional to the radius of the body:
$v\propto R$

Let vₑ and Rₑ be the escape velocity and radius of the Earth, and vₚ and Rₚ be the escape velocity and radius of the other planet. We can set up the ratio of their escape velocities:
$\frac{v_e}{v_p}=\frac{R_e}{R_p}$

We are given that the radius of the other planet is twice that of the Earth (Rₚ = 2Rₑ). Substituting this value into our ratio:
$\frac{v_e}{v_p}=\frac{R_e}{2R_e}=\frac{1}{2}$

Rearranging the equation yields:
$v_e=\frac{v_p}{2}$