# Density of Atmospheric Gas Using Newtonian Gravity

1. Aug 2, 2010

### ObsessiveMathsFreak

Hello. I've recently been working on the mathematics of atmospheric/stellar density, but I've run into an apparent paradox with the assumptions I have been using. I hope this is the right forum for this question.

At the moment, I'm working on the equations for the atmospheric pressure/density above a planet using Newtonian gravity instead of the more usually seen constant gravity. The paradox appears to be that, under Newtonian gravity, an ideal gas will apparently always escape the planets gravity—or so it seem to me(I am a mathematician not a physicist).

I will now state the problem and the steps and assumptions I have been making.

Firstly; let the mass of the planet be $$M$$, and let its radius be $$R$$. To try and avoid the mass of the atmosphere affecting the problem, I have assumed the planet has no atmosphere, and that a small finite amount of an ideal gas exists only in a thin vertical tube which has one end on the surface of the planet and which goes straight up towards infinity. For simplicity, the entire system is assumed to have the same temperature.

Let $$h$$ be height above the surface within the tube. Assuming that the mass of the gas in the thin tube does not affect gravity overly much, the gravity at height $$h$$ in the tube is simply given by
$$g(h)=g_0\left(\frac{R}{R+h}\right)^2$$
where $$g_0=GM/R^2$$ is the surface gravity of the planet at $$h=0$$

Next, the gas in the tube is assumed to have reached hydrostatic equilibrium in which case its pressure $$P$$ and density $$\rho$$ obey the hydrostatic equation
$$\frac{dP}{dh}=-\rho(h) g(h)$$
To relate pressure and density, I have assumed that in an ideal gas, these are related by the constant $$C=\frac{m}{k_B T}$$, which comes from ideal gas theory, with
$$\rho=C P$$
(I suspect this assumption may be the problem in the argument.)

Then, the density of the gas in the tube obeys the equation
$$\frac{d\rho}{dh}= - C \rho(h) g_0 \left(\frac{R}{R+h}\right)^2$$
This equation is separable and can be integrated directly and eventually gives
$$\rho(h) = A e^{C g_0 R^2 \frac{1}{R+h}}$$
Assuming that the initial density of the gas at $$h=0$$ is $$\rho_0$$, this gives
$$\rho(h) = \rho_0 e^{-C g_0 R} e^{C g_o R^2 \left(\frac{1}{R+h}\right)}$$
And with a little algebra this becomes
$$\rho(h) = \rho_0 exp\left(-C g_o \left(\frac{h}{1+\frac{h}{R}}\right)\right)$$

This equation seems to be in some sense correct as in the limit as $$h/R \rightarrow 0$$, the usual atmospheric pressure equation $$\rho(h)=\rho_0 exp(-C g_0 h)$$ is recovered.

However, the "paradox" is that in the limit as $$h \rightarrow \infty$$, it can be seen that
$$\lim_{h \rightarrow \infty} \rho(h) = \rho_0 exp(-C g_0)$$
Which is a constant value, implying that the volume of gas in the tube is infinite. So the only way the equation can properly hold is for $$\rho_0=0$$, that is, the gas has spread out and "escaped" out to infinity.

This is the essence of the "paradox" I have encountered. I suspect it is a result of mistaken physical assumption on my part, but I do not know which ones. Attempting to enumerate the (known) assumptions in order, they have been

1. The entire system has the same temperature.
2. The gas in the tube does not affect gravity in the problem.
3. The gas has reached hydrostatic equilibrium.
4. The (ideal) gas has a linear relationship with pressure $$\rho=C P$$

Of these assumptions, I would suspect that either 2 or 4 are incorrect or must be modified. I would appreciate any insight that a physicist, cosmologist or other expert in this field could give to this problem. Thanks in advance for your time.

2. Aug 3, 2010

### D H

Staff Emeritus
Assumptions 1 and 4 are very much incorrect. You need to model temperature. Temperature roughly decreases linearly with altitude up to an altitude of 11 km, then is more or less constant up to 20 km, The temperature rises for the next 27 km (altitude=47 km), then falls again, up to about 86 km. Beyond 86 km assumption 3 fails. Things get a lot hairier above 86 km.

3. Aug 3, 2010

### Chalnoth

While DH is correct that these assumptions are wrong, I'm not so sure that this fact is material to the problem at hand.

My first criticism is that when you're talking about an atmosphere, the proper column of air to take is not a column of fixed width, but instead a wedge of fixed angular size. This is actually very easy to do, and makes the math even easier. Basically, when considering your equation for hydrostatic equilibrium, the $\rho(h)$ picks up an extra factor of $(R + h)^2$ due to the increase of volume with radius, so that the equation for hydrostatic equilibrium becomes:

$$\frac{dP}{dh} = \rho(h)g_0\left(\frac{R}{R+h}\right)^2\left(R+h\right)^2 = \rho(h)g_0 R^2$$

From this we find that due to the dimensionality of space, in this idealized case, the usual atmospheric density function holds exactly (the change in falloff of gravity is exactly compensated for by the increased volume of gas at a given height).

4. Aug 3, 2010

### D H

Staff Emeritus
That isn't correct, Chalnoth. Check your units. It isn't even correct dimensionally.

5. Aug 3, 2010

### Chalnoth

Hmm, you're right. I was thinking far too simplistically. Unfortunately the coordinate change from 1-D to 3-D isn't nearly as simple. One way to think of it is the following:

Consider a shell of gas. The pressure on this shell of gas will be equal to the weight of all of the gas above the shell divided by the area. In other words:

$$P = \frac{\int_0^{2\pi}\int_0^{\pi}\int_r^\infty g \rho dV}{4\pi r^2}$$

Now, of course, we want to solve for $\rho(r)$, and to do that we have to take a derivative. But first we can simplify by using spherical symmetry to get rid of the integral in the angular direction (this just picks up a factor of $4\pi$ which can be canceled with the factor on the bottom). I'll also multiply both sides by $r^2$:

$$Pr^2 = \int_r^\infty g\rho r^2 dr$$

The extra factor of $r^2$ comes from $dV = r^2 sin\theta dr d\theta d\phi$, with the integral over $\theta$ and $\phi$ being equal to $4\pi$.

So this is a bit more difficult of a formula to deal with. We can use the fundamental theorem of calculus to take the derivative of both sides. This, however, requires evaluating the argument of the integral at infinity. As long as the density drops off to zero at infinity, however, this just evaluates to zero, and we have:

$$\frac{d}{dr}(Pr^2) = -g(r)\rho(r)r^2$$

So I was right about the right hand side of the equation, but I got the left hand side wrong. If we don't evaluate the derivative on the left hand side, and merely consider the whole thing to be a function of r, this differential equation is no more difficult to solve.

$$\frac{d}{dr}\left(P(R+h)^2\right) = -g_0 \left(\frac{R}{R+h}\right)^2 CP (R+h)^2$$

$$\frac{d\left(P(R+h)^2\right)}{P (R+h)^2} = -g_0 C \left(\frac{R}{R+h}\right)^2$$

Solving this differential equation gives one that looks almost like the equation you had, but with a crucial difference:

$$P(R+h)^2 = P(R)R^2 exp\{-C g_0 \left(\frac{R}{R+h}\right)^2\}$$

So it looks almost like the equation you wrote, but with an extra factor of $(R/(R+h))^2$ in front of the exponent. This has the feature of ensuring that the pressure (and density) go to zero as $h\to \infty$.

6. Aug 3, 2010

### D H

Staff Emeritus
That's not how hydrostatic equilibrium works, Chalnoth. The correct equation is indeed

$$\frac{dP}{dr} = -\rho g$$

It falls right out of the Navier-Stokes equations for a fluid at rest. This equation, not yours, is what stellar physicists use to model the behavior of stars, meteorologists to model the Earth's atmosphere, and hydrologists to model pressure under water.

7. Aug 3, 2010

### Chalnoth

Yeah, you're right, I'm sorry. It's been too long since I've done stellar physics. The gradient in spherical coordinates in the radial direction is indeed just the partial derivative with respect to the radius, so there's no change there. I think my mistake was that I was implicitly assuming that the pressure is a vector field, which would mean it would require a divergence, providing the answer I gave. Obviously this is wrong.

So I guess there is indeed this problem if we take the assumption of an atmosphere that is uniform with respect to height. If I remember correctly, what you get in a real atmosphere with a given thermal distribution is a situation where some percentage of the atoms exist in the high-energy tail of the thermal distribution, meaning some are always escaping the gravity of the planet. But like I said, it's been a while since I've done any stellar physics.

8. Aug 3, 2010

### Ich

should be
$$\lim_{h \rightarrow \infty} \rho(h) = \rho_0 exp(-C g_0 R)$$
This is somwhere in the region << 10^-350.
It's not a paradox: you can't have zero density at the end of a finite potential, no matter what its form is. That means, indeed, that all atmospheres evaporate sooner or later - where "later" can easily be much longer than the age of the universe.
So, as DH said, it's your #3 that is physically unrealistic. #1 and #4 don't change the picture qualitatively.
BTW, even Black Holes evaporate, and globular clusters.

9. Aug 3, 2010

### ObsessiveMathsFreak

So you are saying that in fact atmospheres under Newtonian gravity cannot be stable? That is interesting. Then I would suppose that $$\rho_0=0$$ was the appropriate way out.

I'm not sure what you mean about zero density in a finite potential though. Could you elaborate

P.S.
I did try to consider the hydrostatic equation in spherical coordinates, but as was already mentioned, it comes out to be the same equation as in the 1D case. The derivation is a bit tricky though. You have to consider sperical coordinate segments as before, but you must also take the effect of pressure on the sides for it to work out.

10. Aug 3, 2010

### Ich

Better take $\rho>0$ and ignore the minuscule density that you had far away if there was an equilibrium. $\rho_0=0$ at high altitudes is not a good constraint, as you can't trust the model there. I guess that's the difference between physics and mathematics: even a wrong model is ok as long as you use it in the region where it is approximately right.
Let's simplify that pressure falls exponentially with the potential. That's at least qualitatively ok. So you map a finite potential to a nonzero pressure, which is essentially what you got.

11. Aug 3, 2010

### D H

Staff Emeritus
That doesn't make a lick of sense. A non-zero density at infinity means an infinite mass is required to form a single planet or star.

12. Aug 3, 2010

### Chalnoth

Yes, but that's just a result of the assumption that the system is in hydrostatic equilibrium. What Ich is saying is that assumption is reasonable for the lower part of the atmosphere, but breaks down as you go too high, so that you can use the equation just fine for some tens of kilometers above the Earth's surface.

13. Aug 3, 2010

### D H

Staff Emeritus
The assumption of hydrostatic equilibrium is a very good assumption up to about 86 km (the mesopause). Assuming hydrostatic equilibrium is simply wrong for altitudes above the exobase (250-500 km). The continuum mechanics assumptions implicit in the development of the hydrostatic equilibrium equation simply do not apply when the mean free path is tens of thousands of kilometers or more.

The thermosphere lies between those limits. Modeling the thermosphere, is pretty hairy, is not particularly accurate, and has rather limited predictive power. Good models give two decimal places of accuracy, maybe. A new sunspot or solar flare makes the predicted values for tomorrow become worthless.