Question

Jupiter has a mass 318 times that of the Earth and its radius is 11.2 times the radius of the Earth. Calculate the escape speed of a body from Jupiter’s surface. (Given: escape speed on Earth is 11.2km/s)

Answer 1

Escape velocity, $V= \sqrt{ \frac{2GM}{r} }$

$M_j=318\ M_e\\R_j=11.2\ R_e\\ V_e=11.2\ km/s\\ V_j=\ ?$

$\frac{V_j}{V_e} =\sqrt{ \frac{2GM_j}{R_j}/ \frac{2GM_e}{R_e}}\\ \\ \Rightarrow \frac{V_j}{11.2} =\sqrt{ \frac{2GM_j}{R_j} \times \frac{R_e}{2GM_e}}\\ \\ \Rightarrow\frac{V_j}{11.2} =\sqrt{ \frac{R_e}{R_j} \times \frac{M_j}{M_e}}\\ \\ \Rightarrow V_j=11.2 \times \sqrt{ \frac{1}{11.2} \times 318 }\ km/s \\ \\ \Rightarrow V_j=59.7\ km/s$

So escape velocity from jupiter's surface is 59.7 km/s.