Question

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 :

• A. v
• B. 2 v
• C. 3 v
• D. 4 v

Answer 1

The Correct Option is D

To solve the given problem, we need to calculate the escape velocity from the surface of another planet given specific conditions and compare it with the escape velocity from Earth.

Escape velocity is the minimum velocity required for an object to break free from the gravitational attraction of a planet without any additional propulsion. The formula for escape velocity \(v_e\) from the surface of a planet is given by:

\(v_e = \sqrt{\frac{2 G M}{R}}\)

Where:

• \(G\) is the gravitational constant.
• \(M\) is the mass of the planet.
• \(R\) is the radius of the planet.

For Earth, let's denote the escape velocity as \(v\), radius as \(R_E\), and mass as \(M_E\).

Given, for the second planet, the radius is four times that of Earth, i.e., \(R_P = 4R_E\). The mass density (mass per unit volume) is the same as Earth. The mass of a planet can be written in terms of its density \(\rho\) and volume \(V = \frac{4}{3} \pi R^3\):

\(M_P = \rho \times V_P = \rho \times \frac{4}{3} \pi (4R_E)^3\)

For the Earth, \(M_E = \rho \times \frac{4}{3} \pi R_E^3\).

Since the density is the same, substitute to find \(M_P\):

\(M_P = \frac{4}{3} \pi (4R_E)^3 \times \rho = 64 \times \frac{4}{3} \pi R_E^3 \times \rho = 64 M_E\)

Now, we substitute in the escape velocity formula for the second planet:

\(v_{e,P} = \sqrt{\frac{2 G M_P}{R_P}} = \sqrt{\frac{2 G \times 64 M_E}{4 R_E}}\)

\(v_{e,P} = \sqrt{16 \frac{2 G M_E}{R_E}}\)

\(v_{e,P} = 4 \sqrt{\frac{2 G M_E}{R_E}} = 4v\)

Thus, the escape velocity from the surface of the other planet is \(4v\).

Therefore, the correct answer is 4 v.