Question

The mass of Jupiter is $318$ times that of earth, and its radius is $11.2$ times the earth's radius. Estimate the escape velocity of a body from Jupiter's surface given the escape velocity from the earth's surface as $11.2km/s$.

Answer 1

Hence, $M_J=318M_e;R_J=11.2R_e;V_e=11.2km/s$
We know, $V_J=\sqrt{\frac{2G{M}_J}{R_J}}$ and $V_e=\sqrt{\frac{2G{M}_e}{R_e}}$
$\therefore \frac{V_J}{V_e}=\sqrt{\frac{M_J}{M_e}\times \frac{R_e}{R_J}}$
$\Rightarrow V_J=V_e\sqrt{\frac{M_J}{M_e}\times \frac{R_e}{R_J}}$
$V_J=11.2{\{\frac{\left(318M_e\right)}{M_e}\times \frac{R_e}{\left(11.2R_e\right)}\}}^{\frac{1}{2}}=11.2{\left(\frac{318}{11.2}\right)}^{\frac{1}{2}}=59.7km/s$.