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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
Explanation
$v_e \propto R\sqrt{\rho}$, so doubling both gives $2\sqrt{2}$ times.
Detailed Solution
Escape velocity $V = \sqrt{\frac{2GM}{R}}$, with $M = \rho\times\frac{4}{3}\pi R^3$
For the planet: $R_p = 2R_e$, $\rho_p = 2\rho_e$, so $M_p = 2\rho_e\times\frac{4}{3}\pi(2R_e)^3 = 16M_e$
$V_p = \sqrt{\frac{2G(16M_e)}{2R_e}} = \sqrt{\frac{2GM_e}{R_e}\times8} = 2\sqrt{2}V_e$
$V_e : V_p = 1 : 2\sqrt{2}$
For the planet: $R_p = 2R_e$, $\rho_p = 2\rho_e$, so $M_p = 2\rho_e\times\frac{4}{3}\pi(2R_e)^3 = 16M_e$
$V_p = \sqrt{\frac{2G(16M_e)}{2R_e}} = \sqrt{\frac{2GM_e}{R_e}\times8} = 2\sqrt{2}V_e$
$V_e : V_p = 1 : 2\sqrt{2}$