# Deriving minimum escape height and velocity

1. Sep 14, 2011

### guitarstorm

1. The problem statement, all variables and given/known data

At which minimum height atoms or molecules are able to leave Earth’s
atmosphere to space? To obtain an approximate estimate use the assumption that molecules can leave if the mean free path, λ, is larger than the scale height H. Write λ and H as a function of height and obtain the minimum “escape” height.

Obtain the minimum escape velocity and compare it to the molecular
velocity of the H-atom and the N2 molecule for conditions of the above
derived escape height.

Use σ=3 10-20 m2, the temperature profile of the US standard atmosphere,
and assume T=1000 K for heights > 100 km. (The real escape height is
about 400 km for the H atom because of the necessary kinetic energy.)

2. Relevant equations

Two equations I started with:

Mean free path: $\lambda=\frac{1}{\sqrt{2}\sigma n_{v}}$

Scale height: $H=\frac{RT}{g_{0}}$

3. The attempt at a solution

I'm assuming I have to set the two equal to each other, so looking through my textbook I found other equations to plug in to try to get similar terms in both equations. I've only attempted the escape height part first, since I can't do the rest until I get that...

$n_{v}=\frac{p}{kt}=\frac{pN_{a}}{R^{*}T}$

So then the mean free path becomes:

$\lambda=\frac{R^{*}T}{\sqrt{2}\sigma pN_{a}}$

But then we need to find the height (z), so I figured I'd try to plug in $p=\rho gz$.

And then I set $\lambda =H$ as:

$\frac{R^{*}T}{\sqrt{2}\sigma \rho gzN_{a}}=\frac{RT}{g_{0}}$

I solved for z and got $z=\frac{1}{\sqrt{2}\sigma \rho N_{a}}$, which doesn't really help me...

When I plug in the numbers, I get a very low number as well (around 10^-5), which is obviously not right. I'll try to play around with formulas some more but right now I feel overwhelmed by this problem!

2. Sep 14, 2011

Are pressure, density, and temperature a function of the height, z? If they are, then the expression that you used to get a z dependence in your equation isn't valid, since $$p=\rho g z$$ assumes that density is constant. You may want to take a look at the temperature profile of the US standard atmosphere.

Last edited: Sep 14, 2011
3. Sep 15, 2011

### guitarstorm

Well, I did a lot of work on this question today, including going to my professor's office hours, and I've gotten through most of it but still have a few questions.

I ended up having to use the equation $p=p_{0}e^{-\frac{h}{h_{0}}}$ to relate the pressure and height. Assume $p_{0}=100,000 Pa$, the surface pressure, and $h_{0}=7 km=7000 m$, the scale height at the surface.

When rearranging and plugging in all the values, I get the minimum escape height to be 160 km, which seems reasonable.

The minimum escape velocity, I assume is the Earth's escape velocity of about 11,179 m/s... I went through a little derivation I found on Wikipedia for that.

The final part is where i ran into trouble, where it says to relate the molecular velocity of hydrogen and nitrogen to the escape velocity:

I used the relation $KE=\frac{1}{2}mv^{2}=\frac{3}{2}kT$, but when plugging in the values I got a velocity of magnitude 10^9 m/s for H and N2... This can't be right can it?

Then I used another formula I found online, $v_{rms}^{2}=\frac{3RT}{m}$, and solving for v I got values which seem to make more sense... 4994 m/s for H and 944 m/s for N2.

So basically I'm wondering if the 2nd method I used is correct and why the other one doesn't work.

Thanks!

4. Sep 16, 2011

$KE=\frac{1}{2}mv^{2}=\frac{3}{2}kT$ and $v_{rms}^{2}=\frac{3RT}{m}$ are the same equation. The difference is in the units for m. In the first case, [kT] has units of Joules, so you need a mass in kg. In the second case, since you've used R instead of k ([itex]R=N_A k[\itex] where [itex]N_A[\itex] is Avagadro's constant), [RT] has units of Joules/mole. You need to use the molar mass, which has units of kg/mole. When you get weird answers, double checking that you have the right units can be useful.

5. Sep 16, 2011

### guitarstorm

I guess I could only use the 2nd equation then, because I have the molar masses...

Is there a way I could have determined the number of moles of each gas at that height in the atmosphere?

6. Sep 19, 2011