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 :

CONCEPT:

Escape velocity - Escape velocity is defined as the velocity at which an object travels to break free from the earth's gravity or planet's gravity without any development of propulsion. It is written as;

\({V_e} = \sqrt {\frac{{2GM}}{R}} \)

G is the universal gravitational constant.

M is the mass.

R is the radius.

CALCULATION:

According to it, the escape velocity from the Earth's surface is written as;

\({V_e} = \sqrt {\frac{{2GM}}{R}} \) ---(1)

As we know, \(density = \frac {mass}{volume}\)

\(\rho =\frac{M}{V}\\M=\rho V\)

Here, \(\rho\) is the density of the material, M is the mass and V is the volume of the material.

where, \(V = \frac{4}{3}\pi {R^3}\)

Putting these values in equation (1) we get;

\( V_e= \sqrt {\frac{{2Gp\frac{4}{3}\pi {R^3}}}{R}} \)

\( V_e= \sqrt {\frac{{8Gp\pi }}{3}{R^2}} \)

Here, V_e ∝ R (For same density)

and V = R ----(1)

Now, it is given that the other planet has a radius, four times that of Earth. It is written as;

V1 = 4R ----(2)

By dividing equations (1) and (2) we get;

\(\frac{V}{{{V_1}}} = \frac{R}{{4R}}\)

⇒ V1 = 4v

Hence, option 1) is the correct answer.