Solution:
Escape velocity is given by \(v_e = R\sqrt{\frac{8}{3}\pi G\rho}\). Thus, \(v_e \propto R\sqrt{\rho}\). The ratio is \(frac{v_e}{v_p} = \frac{R_e}{R_p}\sqrt{\frac{\rho_e}{\rho_p}} = \frac{1}{2}\sqrt{\frac{1}{2}} = \frac{1}{2\sqrt{2}}\).
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
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