Question

The escape velocity from the earth is $11km{s}^{-1}$. The escape velocity from a planet having twice the radius and same mean density as that of earth is

• A. $5.4km{s}^{-1}$
• B. $11km{s}^{-1}$
• C. $22km{s}^{-1}$
• D. None of the above

Answer 1

The correct option is C $22km{s}^{-1}$

Given : $v_e=11$ $km/s$ $R_p=2R_e$ ${\rho }_p={\rho }_e$

Escape velocity $v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2G\times \frac{4}{3}\pi R^3\rho }{R}}=\sqrt{\frac{8}{3}\pi G{R}^2\rho }$

$⟹$ $v\propto \sqrt{\rho R^2}$

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

Thus we get, $\frac{v_p}{11}=\sqrt{\frac{{\rho }_e(2R_e{)}^2}{{\rho }_e{R}_e^2}}=2$

$⟹$ $v_p=22$ $km/s$