Question

At what temperature does the rms speed of (a)$H_2$ ( molecular hydrogen and ( b) $O_2$ ( molecular oxygen) equal the escape speed from Earth? At what temperature does the rms speed of (c) $H_2$ and (d) $O_2$ equal the escape speed from the Moon ( where the gravitational acceleration at the surface has magnitude $0.16g$? Considering the answers to parts (a) and (b), should there be much (e) hydrogen and (f) oxygen high in Earth's upper atmosphere, where the temperature is about $1000K$?

Answer 1

The rms speed of molecules in a gas is given by $v_{rms}=\sqrt{3RT/M}$, where $T$ is the temperature and $M$ is the molar mass of the gas. The speed required for escape from Earth's gravitational pull is $v=\sqrt{2g{r}_e}$, where $g$ is the acceleration due to gravity at Earth's surface and $r_e(=6.37\times {10}^{6}m)$ is the radius of Earth. To derive this expression, take the zero of gravitational potential energy to be at infinity. Then, the gravitational potential energy of a particle with mass $m$ at Earth's surface is
$U=-GMm/r_e^2=-mg{r}_e$,
where $g=GM/r_e^2$ was used. If $v$ is the speed of the particle, then its original energy is $E=-mg{r}_e+\frac{1}{2}m{v}^2$. If the particle is just able to travel far away, its kinetic energy must tend toward zero as its distance from Earth becomes large without bound. This means $E=0$ and $v=\sqrt{2g{r}_e}$. We equate the expressions for the speeds to obtain $\sqrt{3RT/M}=\sqrt{2g{r}_e}$.
The solution for $T$ is $T=2g{r}_{e}M/3R$.
(a)The molar mass of hydrogen is $2.02\times {10}^{-3}kg/mol$, so for that gas
$T=\frac{2\left(9.8m/s^2\right)(6.37\times {10}^{6}m)(2.02\times {10}^{-3}kg/mol)}{3(8.31J/mol.K)}=1.0\times {10}^{4}K$.
The solution for $T$ is $T=2g{r}_{e}M/3R$.
(b) The molar mass of oxygen is $32.0\times {10}^{-3}kg/mol$, so for that gas
$T=\frac{2\left(9.8m/s^2\right)(6.37\times {10}^{6}m)(3.20\times {10}^{-3}kg/mol)}{3(8.31J/mol.K)}=1.6\times {10}^{5}K$.
(c) Now, $T=2g_m{r}_{m}M/3R$, where $r_m=1.74\times {10}^{6}m$ is the radius of the Moon and $g_m=0.16g$ is the acceleration due to gravity at the Moon's surface. For hydrogen, the temperature is
$T=\frac{2\left(0.16\right)\left(9.8m/s^2\right)(1.74\times {10}^{6}m)(2.02\times {10}^{-3}kg/mol)}{3(8.31J/mol.K)}=4.4\times {10}^{2}K$.
(d) For oxygen, the temperature is
$T=\frac{2\left(0.16\right)\left(9.8m/s^2\right)(1.74\times {10}^{6}m)(32.0\times {10}^{-3}kg/mol)}{3(8.31J/mol.K)}=7.0\times {10}^{3}K$.
(e) The temperature high in Earth's atmosphere is great enough for a significant number of hydrogen atoms in the tail of the Maxwellian distribution to escape. As a result, the atmosphere is depleted of hydrogen.
(f) On the other hand, very few oxygen atoms escape. So there should be much oxygen high in Earth's upper atmosphere.