Question

The ratio of the escape velocities from the surface of two planets having densities and radii in the ratio 2 : 1 and 1 : 2 respectively is

• A. 1 : 1
• B. 1 : 2
• C. 1 : \(\sqrt{2}\)
• D. 1 : \(\sqrt{3}\)
• E. 1 : 4

Hint:

Always simplify \(\frac{\sqrt{x}}{x}\) as \(\frac{1}{\sqrt{x}}\). It helps match the options in competitive exams.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Escape velocity (\(v_e\)) is the minimum speed needed for an object to escape from the gravitational influence of a celestial body.

Step 2: Key Formula or Approach:
The standard formula for escape velocity is \(v_e = \sqrt{\frac{2GM}{R}}\).
Expressing Mass (\(M\)) in terms of density (\(\rho\)): \(M = \rho \times \frac{4}{3}\pi R^3\).
Substituting this gives: \(v_e = \sqrt{\frac{2G(\rho \frac{4}{3}\pi R^3)}{R}} = \sqrt{\frac{8}{3}\pi G \rho R^2}\).
Thus, \(v_e \propto R \sqrt{\rho}\).

Step 3: Detailed Explanation:
Given:
Ratio of densities \(\rho_1 / \rho_2 = 2 / 1\).
Ratio of radii \(R_1 / R_2 = 1 / 2\).
Using the proportionality:
\[ \frac{v_{e1}}{v_{e2}} = \frac{R_1}{R_2} \times \sqrt{\frac{\rho_1}{\rho_2}} \]
\[ \frac{v_{e1}}{v_{e2}} = \frac{1}{2} \times \sqrt{\frac{2}{1}} = \frac{\sqrt{2}}{2} \]
Simplifying the ratio:
\[ \frac{v_{e1}}{v_{e2}} = \frac{1}{\sqrt{2}} \]

Step 4: Final Answer:
The ratio is 1 : \(\sqrt{2}\).