The escape velocity from the earth is about 11km /s. The escape velocity from a planet having twice the radius and the same mean density as the earth, is :
Answer & explanation
Correct answer: option 2
\(\text{Escape velocity : } v_e = \sqrt{\frac{2Gm}{R}}\)
\(v_e = \sqrt{\frac{8}{3}\pi \rho G R^2}\)
\(v_e \propto R \) if :
\(\rho = constant\).
Since, the planet has double the radius in comparison to earth therefore, the escape velocity becomes twice i.e. 22 km/s.