Question

The escape velocity of the earth is $11km{s}^{-1}$. The escape velocity from a planet having twice the mass and twice the radius will be :

• A. $(11\times \sqrt{12})km{s}^{-1}$
• B. $11km{s}^{-1}$
• C. $\frac{11}{2}km{s}^{-1}$
• D. $(11\times \sqrt{2})km{s}^{-1}$

Answer 1

The correct option is B $11km{s}^{-1}$

The escape velocity from the earth is $v_e=\sqrt{\frac{2G{M}_e}{R_e}}$

For planet, $M_p=2M_e$ and $R_p=2R_e$

Thus, escape velocity from planet is $v_p=\sqrt{\frac{2G{M}_p}{R_p}}=\sqrt{\frac{2G\left(2M_e\right)}{2R_e}}=v_e=11km/s$