The question:

At what temperature is the RMS speed of Hydrogen molecules equal to the escape speed from the earth's surface? Values of radius of earth($r$) and gas constant $R$ has been supplied only.

We know that the escape velocity of a body on earth is given by $(2rg)^{1/2}$. Putting in the values of $r$ and $g$ we get $11.2\ \text{km/s}$ as the answer.

We also know that for a mole of ideal gas $PV=RT=\dfrac13M[c_{rms}]^2$,where $M$ is molecular weight of the gas.

So substituting for $T$ we get $T=\dfrac{M[c_{rms}]^2}{3R}$ I have put $11.2\ \text{km/s}$ in place of $c_{rms}$ and $M=2$ and put the value of $R$ and I have got $904.8\ \text{K}$ as my answer, which is equal to $631.8\ ^o\text{C}$.

For moon the answer came as $98\ ^o\text{C}$.

My question is that moon certainly has a temperature was lower than $98\ ^o\text{C}$ so then shouldn't it have an atmosphere of hydrogen? But the moon has no atmosphere!!

I cannot understand where i have gone wrong.

  • 1
    $\begingroup$ I made a terrible mistake! The temperature for rms velocity to be equal to escape velocity for moon comes out to be -84.15 degree centigrade. $\endgroup$ May 20, 2013 at 19:04
  • $\begingroup$ I think you're right that something is wrong. I've done this calculation several times before and have always obtained on the order of 5,000 Kelvin, which is close to the temperature of the sun, which allows the qualitative point. Google "(1 amu)*(11.2 km/s)^2/(3*(boltzmann constant))", this is the calc that gets me 5,000. The equation is $ \tfrac{1}{2}m \overline{v^2} = \tfrac{3}{2} k T $, which I think is the same between us. $\endgroup$ May 20, 2013 at 19:06
  • $\begingroup$ But the temperature on the moon is variable...I am very confused! Does this mean that there was an atmosphere on it which gradually disappeared after the molecules attained escape velocity? when sunlight falls on moon its temperature can go very high, as high as 123 degree centigrade. $\endgroup$ May 20, 2013 at 19:07

1 Answer 1


I believe your mistake is with units, and it is the following:

$$T=\dfrac{M[c_{rms}]^2}{3R} = \dfrac{\left( 1 \text{amu} \right) [11.2 \frac{km}{s}]^2}{3 \left( 8.3144621 \frac{\mathrm{J}}{\mathrm{\text{mol} K}} \right)} $$

This doesn't even cancel out because you're left with a $\text{mol}$ unit. Add avogadro's number.

$$T = \dfrac{\left( 1 \text{amu} \right) [11.2 \frac{km}{s}]^2}{3 \left( 8.3144621 \frac{\mathrm{J}}{\mathrm{\text{mol} K}} \right)} \left( 6.022 \times 10^{23} \frac{1}{\text{mol}} \right) = 5,028 K $$

This is the case for Earth. For the Moon:

$$T = \dfrac{\left( 1 \text{amu} \right) [2.4 \frac{km}{s}]^2}{3 \left( 8.3144621 \frac{\mathrm{J}}{\mathrm{\text{mol} K}} \right)} \left( 6.022 \times 10^{23} \frac{1}{\text{mol}} \right) = 230 K $$

This is negative in Celsius units. This is not a problem. It is merely saying that even freezing temperatures are enough for a lone Hydrogen atom to escape the gravity of the moon with. All we required was that this number be less than the temperature of the sun, which it is. Any surface that faces away from the sun will again see those photons within a month, unless it's in a crater on one of the poles, which we know can have ice near the surface. So that makes sense.

  • $\begingroup$ Why require the result to be less than the Sun's temperature, as opposed to $200\ \mathrm{K}$ (temperature of a rapidly spinning blackbody sphere at $1\ \mathrm{AU}$) or better yet $400\ \mathrm{K}$ (temperature of a constantly-illuminated surface facing the Sun at $1\ \mathrm{AU}$)? Basically, can we assume protons thermalize between photon absorptions? $\endgroup$
    – user10851
    May 21, 2013 at 2:17
  • $\begingroup$ @ChrisWhite There's a lot of room for discussion, I focused on a free proton on the surface of the moon absorbing the sun's photon. The 400 K scenario would require that the photons are absorbed by other material, which then slowly transfer energy to the proton (or H atom). Even then it would appear that the Hydrogen escapes. Both approaches strongly suggest a surface devoid of free Hydrogen atoms, consistent with our expectation. $\endgroup$ May 21, 2013 at 12:44

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