Q: A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that escape velocity from the earth is 11 km/s, the escape velocity from the surface of the planet is

Sol: Given M_{p} = 10M_{e} ; R_{p} = R_{e}/10

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

$\large v_p = \sqrt{\frac{2 G M_p}{R_p}}$

$\large v_p = \sqrt{\frac{2 G (10M_e)}{R_e/10}}$

$\large v_p = 10 (\sqrt{\frac{2 G M_e}{R_e}})$

$\large v_p = 10 v_e = 10 \times 11 $

= 110 km/s