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. 4 v
• B. v
• C. 2 v
• D. 3 v

Answer 1

The Correct Option is A

To solve this problem, we need to understand the concept of escape velocity and how it changes with variations in planetary characteristics.

The escape velocity \(v_e\) from the surface of a planet can be determined by the formula:

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

Here:

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

Given in the problem, the new planet has a radius four times that of Earth (i.e., \(R' = 4R_{E}\)) and the same mass density as Earth. To find out the new mass \(M'\) based on density concepts:

\(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\)

The volume \(V\) of a sphere is given by:

\(V = \frac{4}{3} \pi R^3\)

If the density remains constant and the radius increases to four times:

\(\text{New Mass }, M' = \text{Density} \times \text{New Volume} = \text{Density} \times \left(\frac{4}{3} \pi (4R_{E})^3\right)\)

Therefore, \(M' = 64 \times \text{Mass of Earth }, M_{E}\), because \((4R_{E})^3 = 64R_{E}^3\)

Substitute \(M'\) and \(R'\) into the escape velocity formula:

\(v_e' = \sqrt{\frac{2G(64M_E)}{4R_E}}\)

Simplifying:

\(v_e' = \sqrt{\frac{128GM_E}{4R_E}} = \sqrt{32} \, \times \sqrt{\frac{2GM_E}{R_E}} = 4 \, \times v\)

This means the escape velocity from the surface of the new planet is four times that of Earth.

Therefore, the correct answer is 4 v.