Question

The escape velocity from the earth is about $11{kms}^{-1}$. The escape velocity from a planet having twice the radius and the same mean density as, the earth is

• A. $22{kms}^{-1}$
• B. $11{kms}^{-1}$
• C. $5.5{kms}^{-1}$
• D. $15.5{kms}^{-1}$

Answer 1

The correct option is A $22{kms}^{-1}$

The value of escape velocity from a planet $=\sqrt{\frac{GM}{R}}$ $=\sqrt{\frac{G\rho \frac{4}{3}\pi R^3}{R}}$ $\propto \sqrt{\rho R^2}$

Hence $\frac{v_p}{v_e}=\sqrt{\frac{{\rho }_p{R}_p^2}{{\rho }_e{R}_e^2}}$ $=\frac{2R_e}{R_e}=2$

$⟹v_p=22km/s$