Question

If the radius of a planet is three times the radius of the earth. Both have same mass-densities. \( v_P \) and \( v_E \) are the escape velocities of the planet and the earth respectively, then

• A. \( v_P = 1.5v_E \)
• B. \( v_P = 3v_E \)
• C. \( v_E = 2v_P \)
• D. \( v_P = 2v_E \)

Hint:

For same density planets: \( v \propto R \). Bigger planet $\longrightarrow$ higher escape velocity.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Escape velocity depends on the mass and radius of a planet. If the density is constant, we can express the mass in terms of radius and density, revealing how escape velocity scales with the size of the planet.
Step 2: Key Formula or Approach:
1. Escape velocity: \( v_e = \sqrt{\frac{2GM}{R}} \).
2. Mass in terms of density: \( M = \rho \times V = \rho \times \frac{4}{3}\pi R^3 \).
Step 3: Detailed Explanation:
Substitute the mass expression into the escape velocity formula:
\[ v_e = \sqrt{\frac{2G(\rho \cdot \frac{4}{3}\pi R^3)}{R}} = \sqrt{\frac{8}{3}\pi G \rho R^2} = R \sqrt{\frac{8\pi G \rho}{3}} \]
Since both planets have the same density \( \rho \), everything under the square root is a constant. Thus:
\[ v_e \propto R \]
Taking the ratio for the planet (\( P \)) and Earth (\( E \)):
\[ \frac{v_P}{v_E} = \frac{R_P}{R_E} \]
Given \( R_P = 3 R_E \):
\[ \frac{v_P}{v_E} = \frac{3 R_E}{R_E} = 3 \]
\[ v_P = 3 v_E \]
Step 4: Final Answer:
The escape velocity of the planet is \( 3 v_E \).