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The scape velocity is:

$v_{\text{esc}}=\sqrt{\frac{2GM}{R}},$

The Rms speed is:

$v_{\text{rms}}=\sqrt{\frac{3k_{B}T}{m}},$

(a)

By setting $E_{\text{mechanical}}=0$, an expression for the escape velocity $v_{\text{esc}}$in the Earth is:

$v_{\text{esc}}=\sqrt{\frac{2GM}{R}},$

where M , is the Earth's mass, G is the gravitational constant, and R is the Earth's radius.

To find its numerical value, we plug in$M=5.98脳{10}^{24}\text{kg},G=6.67脳{10}^{-11}{\text{Nm}}^2/{\text{kg}}^2$ , and$R=6.37脳$${10}^6\text{m}$ we get:

$\begin{array}{rcl}{v}_{\text{esc}}& =& \sqrt{\frac{2\left(6.67脳{10}^{-11}\right)\left(5.98脳{10}^{24}\right)}{6.37脳{10}^6}}\\ & =& 1.12脳{10}^4\text{m}/\text{s}.\\ & & \end{array}$

The rms speed of a gas molecule is

$v_{\text{rms}}=\sqrt{\frac{3k_{B}T}{m}},$

where R is Boltzmann's constant, T is temperature, and m is the mass of the molecule. If we consider atomic Hydrogen with $m=1.66脳{10}^{-27}\text{kg}$, its rms speed at $T=1000\text{K}$is:

$\begin{array}{rcl}{v}_{\text{rms},\text{H}}& =& \sqrt{\frac{3\left(1.38脳{10}^{-23}\right)\left(1000\right)}{1.66脳{10}^{-27}}}\\ & =& 4993.97\text{m}/\text{s}\\ & & \end{array}$

As for a Nitrogen molecule with$m=4.65脳{10}^{-26}\text{kg}$ , its rms speed at that same temperature is

$\begin{array}{rcl}{v}_{\text{rms},{\text{N}}_{\text{2}}}& =& \sqrt{\frac{3\left(1.38脳{10}^{-23}\right)\left(1000\right)}{4.65脳{10}^{-26}}}\\ & =& 943.57\text{m}/\text{s}.\\ & & \end{array}$

Consider atomic hydrogen.

Taking the ratio between the escape speed in the Earth and its rms speed at$\text{T}=1000\text{K}$ we get:

$\begin{array}{rcl}\frac{v_{\text{esc}}}{v_{\text{rms,H}}}& =& \frac{1.12脳{10}^4}{4993.97}\\ & =& 2.24\\ & & \end{array}$

Now, to determine whether Earth should be able to hold on to Hydrogen, the fraction of molecules faster than $2v_{\text{rms}}$is $~{10}^{-2}$. Since the escape speed is twice as large as the rms speed of hydrogen, and since we know that a fraction of${10}^{-2}$ particles could only have speeds larger than twice the rms speed, then:

No. of$H\text{w}/$ speed larger than$2v_{\text{rms}}={10}^{-2}\left({10}^{48}\right)=1脳{10}^{46}$

Since there are$1脳{10}^{46}$ atoms of H with speed larger than twice the rms speed, the Earth may not be able to hold hydrogen very well in the atmosphere.

Now consider Nitrogen molecule.

Taking the ratio between the escape speed in the Earth and its rms speed at$\text{T}=1000\text{K}$ we obtain:

$\begin{array}{rcl}\frac{v_{\text{esc}}}{v_{\text{rmsN}}\text{2}}& =& \frac{1.12脳{10}^4}{943.57}\\ & =& 11.87.\\ & & \end{array}$

Now, to determine whether Earth should be able to hold on to Nitrogen, the fraction of molecules faster than $10v_{\text{rms}}$is$~{10}^{-64}$ . Since the escape speed is a little more than 10 times as large as the rms speed of nitrogen, and since we know that a fraction of ${10}^{-64}$particles could only have speeds larger than ten times the rms speed, then:

Number of/speed larger than$10v_{\text{rms}}$

$\begin{array}{rcl}& & {\text{10v}}_{\text{rms}}{\text{= 10}}^{\text{- 64}}\left({\text{10}}^{\text{48}}\right)\\ & & \text{= 1}脳{\text{10}}^{\text{- 16}}\\ & 鈮�& \text{0}\\ & & \end{array}$

Almost zero nitrogen molecules in the atmosphere could have speeds of about times rms speed of gas molecules. Hence, Earth can effectively hold Nitrogen molecules.

(b)

In the case of moon mass is $M=0.0123$, $M_{earth}=7.36脳{10}^{22}kg$and radius is$R=0.26R_{\text{earth}}=1.66脳{10}^6\text{m}$ . Calculating its escape speed, we get:

$\begin{array}{rcl}{v}_{\text{ese}}& =& \sqrt{\frac{2\left(6.67脳{10}^{-11}\right)\left(7.36脳{10}^{22}\right)}{1.66脳{10}^6}}\\ & =& 2434.78m/s.\\ & & \end{array}$

To see whether the moon can hold hydrogen atoms, let us first calculate the rms speed of hydrogen at $T=370\text{K}$:

$\begin{array}{rcl}{v}_{\text{rms, H}}& =& \sqrt{\frac{3\left(1.38脳{10}^{-23}\right)\left(370\right)}{1.66脳{10}^{-27}}}\\ & =& 3037.71\text{m}/\text{s}\\ & & \end{array}$

For nitrogen gas at that same temperature, its rms speed is:

$\begin{array}{rcl}{v}_{\text{rms},{\text{N}}_{\text{2}}}& =& \sqrt{\frac{3\left(1.38脳{10}^{-23}\right)\left(370\right)}{4.65脳{10}^{-26}}}\\ & =& 573.95\text{m}/\text{s}.\\ & & \end{array}$

As we can see, the rms speed of hydrogen at the surface temperature of the moon is much larger than the escape speed in the moon. Hence, hydrogen can easily escape the moon's atmosphere. As for nitrogen, the ratio of the escape speed to its rms speed is:

$\begin{array}{rcl}\frac{v_{\text{esc}}}{v_{{\text{rmsN}}_{\text{2}}}2}& =& \frac{2434.78}{573.95}\\ & =& 4.24\\ & & \end{array}$

Now since the fraction of molecules moving faster than the rms speed exponentially decreases with speed, it is clear that although the moon could hold a certain amount of Nitrogen molecules, afraction of it may still escape its atmosphere since the fraction of molecules moving faster than$6v_{\text{rms}}$ is $~{10}^{-23}$.