Question

the ratio of escape velocity at earth v_e to the escape velocity of planet whose radius and mean density are twice than that of earth

Answer 1

$$\begin{gathered} Escapevelocityatearthis: \\ {v}_e=\sqrt{\frac{2GM}{R_e}}=\sqrt{\frac{2G{\rho }_e\times 4\pi {R_e}^3}{3R_e}}=\sqrt{\frac{2G{\rho }_e\times 4\pi {R_e}^2}{3}}-----\left(1\right) \\ Theescapevelocityattheplanetis: \\ {v}_p=\sqrt{\frac{2G{\rho }_p\times 4\pi {R_p}^2}{3}}------\left(2\right) \\ Dividingeqn\left(2\right)byeqn\left(1\right)weget: \\ \frac{v_p}{v_e}=\sqrt{\frac{{\rho }_p{R_p}^2}{{\rho }_e{R_e}^2}}=\sqrt{\frac{2{\rho }_p\times 4{R_p}^2}{{\rho }_p\times {R_p}^2}}=\sqrt{8}=2\sqrt{2} \\ Thereforetheratioof{v}_p:v_e=2\sqrt{2}:1. \end{gathered}$$