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It is known from hydrostatics that for a fluid in equilibrium in a gravitational field,

$$\frac{dP}{dz} = -ρg$$

Let us from now on suppose the atmosphere is isothermal and has temperature $T$.

We can apply the ideal gas law to find

$$ρ = \frac{P\,m}{kT}$$

Where $m$ is the average mass of an individual particle of the gas and $k$ is Boltzmann's constant.

If we let $ g =\large \frac{GM}{z^2}$, we find:

$$\frac{dP}{dz} = -\frac{PmGM}{kTz^2}$$

This may become:

$$\frac{\mathrm{d}P}{P} = -\frac{GMm}{kT} \, \frac{\mathrm{d}z}{z^2}$$

If we define $P_0$ as the pressure on $z = R$ (surface of the planet), we may integrate to find:

$$\ln{\left(\frac{P}{P_0}\right)} = \frac{GMm}{kT}\left[\frac{1}{z} - \frac{1}{R}\right]$$

From this we find:

$$P = P_0 \exp{\left(\frac{GMm}{kT} \left[\frac{1}{z} - \frac{1}{R}\right]\right)}$$

Now this is fine, after all the pressure does indeed decrease as $z$ increases.

My problem comes when we try to find the pressure as $z$ goes to infinity, that is, infinitely far away from the planet's surface. Intuition tells me that the pressure should go to zero, but instead what we find if we calculate the limit is:

$$P_{\text{f}} = P_0 \exp{\left(-\frac{GMm}{RkT}\right)}$$

It's a counter-intuitive result, since it implies that this atmosphere is infinitely massive. In the "end" of the gravitational field, we find an infinitely large gas of pressure $P_{\text{f}}$ and temperature $T$.

This result is also interesting if we compare it to Halley's Law for isothermal atmospheres in constant gravitational fields:

$$P = P_0 \exp{\left[-\lambda z\right]}$$

Where $\lambda$ is a constant. In this case, the limit as z goes to infinity is:

$$Pf = 0$$

The results are incompatible, but this is expected due to approximating the gravitational field as constant.

Anyway, I believe I have committed no mistake, and this result seems to imply that there cannot actually be an isothermal atmosphere. This is observed on real-life atmospheres, since they are never isothermal.

One interpretation for the result is: the fact that the average kinetic energy must remain the same means there are always many particles on each 'layer' $z$ with velocity larger than the escape velocity. This could be verified using the Maxwell-Boltzmann distribution of velocities. Since many particles have enough velocity to escape, they will, and thus the infinite gas cloud at infinity is formed.

Is this explanation correct? Does it make sense? I have a feeling it is off due to many particles also coming back to the surface of the planet, falling from infinity. If this explanation is not correct, what would the correct physical interpretation be?

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  • $\begingroup$ I suspect that you mix up the two $z$'s. The $z$ in hydrostatic pressure is not the same as that in gravitational force. $\endgroup$ – K_inverse Oct 22 '18 at 0:43
  • $\begingroup$ They actually are the same. I set them to be the same. $\endgroup$ – João Vítor G. Lima Oct 22 '18 at 0:53
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    $\begingroup$ Have you actually calculated what that final pressure would be? My estimate is that it would be on the order of about $e^{-1000}$ bars. How many molecules would be in a cubic meter of ideal gas at that pressure and at, say, 300K? What would be the mean free path of a molecule? This basically means that the ideal gas law is not valid at such a low pressure. $\endgroup$ – Chet Miller Oct 22 '18 at 1:15
  • $\begingroup$ Nice post. If I understand correctly what @ChesterMiller says, everything is working fine. Pressure is very low, even using IG Law which isn't valid in those conditions. On the other hand, when you say M needs to be $\infty$, what do you mean? As $M$ is the mass of the earth... $\endgroup$ – user153036 Oct 22 '18 at 3:32
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    $\begingroup$ This result seems correct. In case people would like a reference, it is derived more directly from the Boltzmann distribution by Berberon-Santos et al, Amer J Phys, 65, 404 (1997) as an example. They say it is "well known" that no equilibrium distribution is possible. I have to admit, I hadn't come across it before. They also mention the argument about a certain fraction of molecules having more than the escape velocity, referencing RP Wayne's book Chemistry of Atmospheres. $\endgroup$ – user197851 Oct 22 '18 at 10:33

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