Escape speed:
$M_{\oplus}=M_{Earth}$

$v_{escape}=\left (\frac{2GM}{r}\right)^{1/2}$
$v_{escape}~(km/s) = 11.2~ km/s~\sqrt{\frac{ M~ (M_{\oplus})}{R~ (R_{\oplus})}}$

So at the surface of the Earth the escape speed is 11.2 km/s.

What is the escape speed at 1000 km above the surface?

$v_{escape}~(km/s) = 11.2~ km/s~\sqrt{\frac{ M~ (M_{\oplus})}{R~ (R_{\oplus})}}$

$R= 1~R_{\oplus} + 1000~ km ~\frac {1 R_\oplus}{6.4 \times 10^3 km}=(1 + 0.16)~R_\oplus= 1.16 ~R_\oplus$

$v_{escape}~(km/s) = 11.2~ km/s~\sqrt{\frac{ 1~ (M_{\oplus})}{1.16~ (R_{\oplus})}}=10.4$ km/s

Average molecular speed:

$v_{molecule}~(km/s)= 0.157~\sqrt{\frac{T_{gas}~(K)}{m_{molecule}~ (\mbox{units of m$_{H}$})}}$

What is the average speed of an oxygen molecule in the
Earth's atmosphere?

$T_{atmosphere}=300~K$

$v_{O_2}~(km/s)= 0.157~(km/s)~ \sqrt{\frac{ T_{gas}~ (K)}{m{_{O_2}} (m_H)}} =0.157~(km/s)~\sqrt{\frac{300~(K)}{32~m_{H}}}$

$v_{O_2}~(km/s)= 0.48~km/s$

What is the average speed of a nitrogen molecule in the
earth's atmosphere?

$v_{N_2}~(km/s)= 0.157~(km/s)~ \sqrt{\frac{ T_{gas}~ (K)}{m{_{N_2}} (~m_{H})}} =0.157~(km/s)~\sqrt{\frac{300~(K)}{28~m_{H}}}$

$v_{N_2}~(km/s)= 0.51~km/s$

What is the average speed of an oxygen molecule in the Moon's atmosphere?

$T (\mbox{moon atmosphere})$ = varies from 100-400 K, use 300 K here

$v_{O_2}~(km/s)= 0.157~(km/s)~ \sqrt{\frac{ T_{gas}~ (K)}{m_{O_2} (~m_{H})} } =0.157~(km/s)~\sqrt{\frac{300~(K)}{32~m_{H}}}$

$v_{O_2}~(km/s)= 0.48~km/s$ Same as on Earth.

Compare average molecular speed of oxygen on the moon to the moon's escape speed at the surface.
$M_{moon}=0.012~ M_{\oplus}$ $R_{moon}=0.27~ M_{\oplus}$

$v_{escape}~(km/s) = 11.2~ km/s~\sqrt{\frac{ M~ (M_{\oplus})}{R~ (R_{\oplus})}}= 11.2~km/s \sqrt{\frac{0.012~ M_{\oplus}}{0.27~ M_{\oplus}}}$

$v_{escape} = 2.4$ km/s

Can oxygen escape from moon?

If $(v_{ave}\times 6) > v_{escape}$ then atom or molecule escapes.

$(0.48~ km/s\times 6)= 2.9~ km/s$
$2.9~ km/s > 2.4~ km/s$, so oxygen will escape.