Explanation:
Since, the escape velocity of earth can be given as
\({v_e} = \sqrt {2gR} = R\sqrt {\frac{8}{3}\pi G\rho } \)
\(\left[ {\because g = \frac{4}{3}\pi G\rho R} \right]\)
\([\rho = {\text{ density }}\,\,{\text{of}}\,\,{\text{ earth }}]\)
\( \Rightarrow {v_e} = R\sqrt {\frac{8}{3}\pi G\rho } \quad \quad \quad (1)\)
As it is given that, the radius and mean density of planet are twice as that of earth. So, escape velocity at planet will be
\({v_p} = 2\,R\sqrt {\frac{8}{3}\,\,\pi \,G\,2p} \quad \quad \quad (2)\)
Divide Eq. (1) by Eq. (2), we get
\(\dfrac{v_{e}}{v_{p}}=\dfrac{R \sqrt{\dfrac{8}{3} \pi G \rho}}{2 R \sqrt{\dfrac{8}{3} \pi G 2 \rho}}\)
\(\Rightarrow \dfrac{v_{e}}{v_{p}}=\dfrac{1}{2 \sqrt{2}}\)