Question

A planet has radius \( \dfrac{1}{4} \)th of the radius of earth and acceleration due to gravity double than that of the earth. Then the ratio of escape velocity on the surface of the planet to that on the earth’s surface will be

• A. \( \dfrac{1}{\sqrt{2}} \)
• B. \( 2 \)
• C. \( \sqrt{2} \)
• D. \( 2\sqrt{2} \)

Hint:

Escape velocity depends on both gravity and radius of the planet.

Answer 1

The Correct Option is A

Step 1: Escape velocity formula.
Escape velocity is given by:
\[ v_e = \sqrt{2gR} \]
Step 2: Given data.
For earth:
\[ v_e = \sqrt{2 g R} \] For planet:
\[ g_p = 2g, \quad R_p = \frac{R}{4} \]
Step 3: Escape velocity of the planet.
\[ v_p = \sqrt{2 \times 2g \times \frac{R}{4}} \] \[ v_p = \sqrt{\frac{gR}{1}} \]
Step 4: Ratio of escape velocities.
\[ \frac{v_p}{v_e} = \frac{\sqrt{gR}}{\sqrt{2gR}} = \frac{1}{\sqrt{2}} \]
Step 5: Conclusion.
The required ratio is \( \dfrac{1}{\sqrt{2}} \).