Question

At what temperature is the root mean square velocity of hydrogen equal to its escape velocity from the surface of the moon? Radius of moon $=1.74\times {10}^{6}m$ and acceleration due to gravity at moon's surface $=1.67m/s^2$.

Answer 1

$V_{rms}=\sqrt{\frac{3kT}{m}}$ and the escape velocity is given by $v_e=\sqrt{2gR}$
If $V_{rms}=V_e$
$\sqrt{\frac{3kT}{m}}=\sqrt{2gR}\Rightarrow T=\frac{2mgR}{3k}$
$=\frac{2}{3}\times \frac{\left(\frac{2\times {10}^{-3}}{6.02\times {10}^{23}}\right)\times 1.67\times 1.74\times {10}^6}{1.38\times {10}^{-23}}=466K$.