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Physics · PYQ NEET UG 2016(1)
हिन्दीThe 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
Correct answer · Option 1
\( 1:2\sqrt{2} \)
Explanation
Escape velocity \( v = R\sqrt{\frac{8}{3}\pi G \rho} \). Since \( R_p = 2R_e \) and \( \rho_p = 2\rho_e \), the ratio \( v_e/v_p = \frac{R_e\sqrt{\rho_e}}{R_p\sqrt{\rho_p}} = \frac{1}{2 \times \sqrt{2}} = \frac{1}{2\sqrt{2}} \).
