91Ó°ÊÓ

Escape speeds of different gas molecules is equal to the rms speeds.

The RMS speed of the molecules in a gas is given by

$V_{rms}=\sqrt{\frac{3RT}{M}}$…… (1)

Where,$\text{T}$is the temperature and$\text{M}$is the molar mass of the gas.

The gravitational potential energy of a particle with mass $m$at Earth's surface is

$U=-\frac{GMm}{r_e}$……. (2)

Also, the gravitational force between the Earth of mass$M$and particle of mass$m$is

$F=\frac{GMm}{r_e^2}$

The force of gravity,$F=mg$

Comparing these two forces,

$\begin{array}{l}⇒g=\frac{GM}{r_e^2}\\ GM=g{r}_e^2\end{array}$

Substituting this value ofGMin equation (2), we get

$U=-mg{r}_e$

The total energy of the particle is kinetic energy plus its potential energy.

$E=\frac{1}{2}m{V}^2+(-mg{r}_e)$

If the particle is just able to travel far away, its kinetic energy must tend towards zero as its distance from Earth becomes large without bound and its potential energy at infinity becomes zero. Therefore, its total energy is equal to zero.

$0=\frac{1}{2}mv{V}^2-mg{r}_e$

$V=\sqrt{2g{r}_e}$…….. (3)

Equating equations (1) and (3),

$\sqrt{3RT/M}=\sqrt{2g{r}_e}$

Squaring both the sides and solving forwe get

$T=2greM/3R$……. (4)

Where$R$is the gas constant and its value is$8.31\frac{\text{J}}{\text{mol.k}}$

Temperature at which RMS speed of${\text{H}}_{\text{2}}$equals the escape speed from Earth:

The molar mass of hydrogen is $2.02×{10}^{−3}\text{kg}/\text{mol},$ so for that gas,

$\begin{array}{c}\text{T}=\frac{2×9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}×(6.37×{10}^6\text{m})×(2.02×{10}^{−3}\frac{\text{kg}}{\text{mol}})}{3(8.31\frac{\text{J}}{\text{mol.K}})}\\ =1×{10}^4\text{K}\end{array}$

Therefore, the temperature at which the RMS speed of$H_2$ (Molecular hydrogen) equals the escape speed from Earth is$1×{10}^4\text{K}$ .

Temperature at which RMS speed of${\text{O}}_{\text{2}}$equals to the escape speed from the Earth:

The molar mass of oxygen is $32×{10}^{−3}\text{kg}/\text{mol},$ so for that gas,

$\begin{array}{c}\text{T}=\frac{2×9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}×(6.37×{10}^6\text{m})×(32×{10}^{−3}\frac{\text{kg}}{\text{mol}})}{3(8.31\frac{\text{J}}{\text{mol.K}})}\\ =1.6×{10}^5\text{K}\end{array}$

Therefore the temperature at which the RMS speed of $O_2$ (Molecular oxygen) equals the escape speed from Earth is $1.6×{10}^5\text{K}$.

Temperature at which RMS speed of${\text{H}}_{\text{2}}$equals to the escape speed from Moon:

$T=\frac{2g_m{r}_{m}M}{3R},$

Here${\text{r}}_m=1.74×{10}^6\text{m}$ is the radius of the Moon and

${\text{g}}_m=0.16\text{g}$ is the acceleration due to gravity at the Moon's surface.

The molar mass of hydrogen is $2.02×{10}^{−3}\text{kg}/\text{mol},$ so for that gas,

$\begin{array}{c}\text{T}=\frac{2×0.16×(9.8\text{m}/{\text{s}}^2)×(1.74×{10}^6\text{m})×(2.02×{10}^{−3}\frac{\text{kg}}{\text{mol}})}{3(8.31\frac{\text{J}}{\text{mol.K}})}\\ =4.4×{10}^2\text{K}\end{array}$

Therefore the temperature at which the RMS speed of $H_2$ (Molecular hydrogen) equals the escape speed from the moon is $4.4×{10}^2\text{K}$.

Temperature at which RMS speed of${\text{O}}_{\text{2}}$equals the escapes speed from Moon:

The molar mass of oxygen is $32×{10}^{−3}\text{kg}/\text{mol},$ so for that gas,

$\begin{array}{c}\text{T}=\frac{2×0.16×(9.8\text{m}/{\text{s}}^2)×(1.74×{10}^6\text{m})×(32×{10}^{−3}\frac{\text{kg}}{\text{mol}})}{3(8.31\frac{\text{J}}{\text{mol.K}})}\\ =7×{10}^3\text{K}\end{array}$

Therefore the temperature at which the RMS speed of${\text{O}}_{\text{2}}$ (Molecular oxygen) equals the escape speed from the moon is $7×{10}^3\text{K}$.

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 Earth’s upper atmosphere is depleted of hydrogen.

On the other hand, very few oxygen atoms escape. So, there should be much oxygen high in Earth’s upper atmosphere.