Compare the velocity of oxygen molecules in planet atmosphere and escape velocity for an imaginary planet with diameter and mass both one half of those of earth. The day's temperature of this planet surface reaches up to 800 K. (Escape velocity from earth's surface = 11.2 km/s, $\mathrm{k}=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$,  mass of oxygen molecule = $5.3\times {10}^{-26} \mathrm{kg})$

As, escape velocity : $v_e=\sqrt{\frac{2GM}{R}}$

so,

$\frac{{\mathrm{v}}_{\mathrm{P}}}{{\mathrm{v}}_{\mathrm{E}}}=\sqrt{\frac{{\mathrm{M}}_{\mathrm{p}}}{{\mathrm{M}}_{\mathrm{E}}}\times \frac{{\mathrm{R}}_{\mathrm{E}}}{{\mathrm{R}}_{\mathrm{P}}}}=\sqrt{\frac{1}{2}\times \frac{2}{1}}=1$

i.e.,

$v_P=v_E=11.2\mathrm{km}/\mathrm{s} ....\left(\mathrm{i}\right)$

Now according to kinetic theory of gases,

$\begin{array}{l}{\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{RT}}{\mathrm{M}}}=\sqrt{\frac{3\left(\mathrm{Nk}\right)\mathrm{T}}{\mathrm{Nm}}}\\ [\text{ as }\mathrm{R}=\mathrm{Nk}\text{ and }\mathrm{M}=\mathrm{Nm}]\\ \text{ i.e., }{\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{kT}}{\mathrm{m}}}=\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 800}{5.3\times {10}^{-26}}}\\ =0.79\mathrm{km}/\mathrm{s}\dots \left(\mathrm{ii}\right)\end{array}$

From Eqns.(i) and (ii) it is clear that velocity of oxygen molecule in the atmosphere of planet (∼0.8 km/s) is much lesser than escape velocity from the planet (= 11.2 km/s); so oxygen molecules cannot escape and so the planet's atmosphere may contain oxygen.