Q: If Earth has mass nine times and radius twice that of the planet mars, calculate the velocity required by a rocket to pull out of the gravitational force of Mars. Take escape speed on surface of Earth to be 11.2 Km/s

Sol: Here, M_{e} = 9M_{m} , and R_{e} = 2R_{m}

v_{e}(escape speed on surface of Earth) = 11.2km/s

Let V_{m} be the speed required to pull out of the gravitational force of mars.

We know that $\large v_e = \sqrt{\frac{2 G M_e}{R_e}}$

$\large v_m = \sqrt{\frac{2 G M_m}{R_m}}$

On dividing ,

$\large \frac{v_m}{v_e} = \sqrt{\frac{M_m}{M_e}\times \frac{R_e}{R_m}}$

$\large \frac{v_m}{v_e} = \sqrt{\frac{1}{9} \times 2} = \frac{\sqrt{2}}{3}$

$\large v_m = \frac{\sqrt{2}}{3}\times v_e = \frac{\sqrt{2}}{3}\times 11.2 $

= 5.3 km/s