Question

The ratio of escape velocity at earth $v_e$ to the escape velocity at a planet $v_p$ whose radius and mean density are twice as that of earth is :

• A. $1 : 2 \sqrt{2}$
• B. $1 : 4$
• C. $1 : \sqrt{2}$
• D. $ 1 : 2 $

Answer 1

The Correct Option is A

Escape Velocity and Density Relation 

The escape velocity \( V_e \) from a spherical body is given by the formula:

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

Where:

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

 

Substituting the expression for mass \( M \) in terms of density \( \rho \) (where \( M = \frac{4}{3} \pi R^3 \rho \)), we get:

\(V_e = \sqrt{\frac{2G}{R} \left(\frac{4}{3} \pi R^3 \rho \right)}\)

This simplifies to:

\(V_e \propto R \sqrt{\rho}\)

Conclusion:

Therefore, the ratio of escape velocities for two bodies with different radii and densities is:

\(1 : 2 \sqrt{2}\)