Question

A particle near earth's surface traveling faster than about $11km/s$has enough kinetic energy to completely escape from the earth, despite earth's gravitational pull. Molecules in the upper atmosphere that are moving faster than this will therefore escape if they do not suffer any collisions on the way out.

(a) The temperature of earth's upper atmosphere is actually quite high, around $1000K.$Calculate the probability of a nitrogen molecule at this temperature moving faster than $11km/s$, and comment on the result.

(b) Repeat the calculation for a hydrogen molecule $\left(H2\right)$and for a helium atom, and discuss the implications.

(c) Escape speed from the moon's surface is only about $2.4km/s$. Explain why the moon has no atmosphere.

Short answer

(a) The probability of a nitrogen molecule at this temperature moving faster than $11Km/s$is $P\approx 0$thus $N_2$molecule won't be escaping to all.

(b) The calculation for a hydrogen molecule $\left(H_2\right)$and for a Helium $\left(He\right)$atom and the implications islocalid="1650900603347" $P\cong 1.4e-12\approx 0$Thus, all these gas molecules are not able to escape the atmosphere.

(c) All gas molecules escape and there is no atmosphere left on the moon.

Step-by-step solution

Part(a) Step1: Given information

We are given that,

escape velocity $=11km/s$

Temperature,$T=1000k$

Part(a) Step2: simplify

The probability density function for velocity in a gas is given by the maxwell Boltzmann distribution

$f\left(v\right)={\left(\frac{m}{2\mathrm{\pi RT}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$$}\right.2}4{\mathrm{\pi v}}^2{\mathrm{e}}^{\raisebox{1ex}{$-{\mathrm{mv}}^2$}\!\left/ \!\raisebox{-1ex}{$$}\right.}\mathrm{RT}$

therefore probability of velocity being greater than $11km/s$is

$P(v>11km/s)={\int }_{11,000}^{\infty }f\left(v\right)dv$

$P={\int }_{11,000}^{\infty }{\left(\frac{m}{2\mathrm{\pi RT}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}4{\mathrm{\pi v}}^2\mathrm{e}\frac{-{\mathrm{mv}}^2}{\mathrm{RT}}dv$

Replacing $v\leftrightarrow x\&a=\sqrt{\frac{RT}{2m}}$

we get,

$P={\int }_{11,000}^{\infty }\sqrt{\frac{2}{\mathrm{\pi }}}\frac{x^{2}e{\displaystyle \raisebox{1ex}{$-x^2$}\!\left/ \!\raisebox{-1ex}{$2a^2$}\right.}}{a^3}dx$

This, integral is not simple & we will need to use error function.

$P=erfc\left(\frac{x}{\sqrt{2}a}\right)+\sqrt{\frac{2}{\mathrm{\pi }}}\frac{xe{\displaystyle \raisebox{1ex}{$-x^2$}\!\left/ \!\raisebox{-1ex}{$2a^2$}\right.}}{a}$

where, $erfc$is the error function complement

$\left\{Here,erfc\left(x\right)=\frac{1}{\sqrt{\mathrm{\pi }}}{\int }_{-x}^x{e}^{-t^2}dt\right\}$

not possible by hand calculation, so using web tool

$a=\sqrt{\frac{RT}{2m}}$

where $m=$molecular mass of gases$=28amu$

$a=\sqrt{\frac{1.38\times {10}^{-23}\times 1000}{2\times 28\times 1.67\times {10}^{-27}}}$

$a=385.3$

$\frac{x}{\sqrt{2}a}=\frac{11000}{\sqrt{2}\times 385.3}=20.188$

$erfc(20.188)\approx 0.0$

$P=0.0+2e-176$

$P\approx 0$thus, the $N_2$molecule wont be escaping at all.

Part(b) Step1: Given information

We are given that,

For$H_2$,

$m=2v$energy thing else is same.

$m=2\times 1.67\times {10}^{-27}$

Part(b) Step2: simplify

Now we put the values in formula,

$a=\sqrt{\frac{1.38\times {10}^{-23}\times 1000}{2\times 2\times 1.67\times {10}^{-27}}}$

$a=1441.63$

$\frac{x}{\sqrt{2}a}=5.39$

localid="1650109697287" $erfc\left(\frac{x}{\sqrt{2}a}\right)\approx 0$

$P=0+1.4e^{-12}$

$P\cong 1.4e^{-12}\approx 0$

thus, probability is still very small of velocity $>11km/s$.

Since the mass of helium lies in between that of hydrogen and $N_2$,the velocity $>11km/s$is also$\approx 0$.

Thus all these gas molecules are not able to escape the atmosphere.

Part(c) Step1: given information

We are given that,

Since, the escape velocity of moon is $2.4km/s$

Part(c) Step2: Explanation

The molecule of atmospheric gases on surface of moon can move with thermal velocities greater than the velocities of moon,

Thus, all gas molecules escape and there is no atmosphere left on moon.