How to Find the Density of an Ideal Gas Around a Planet?

[goulio]

Hi,

I need to find the density \rho(r) of an ideal gas at constant temperature T surrouding a planet of mass M and radius R. The gas is attrated by the planet and is also self-attracting. First, I used the hydrostatic equilibirum equation
<br /> \frac{dP}{dr}=-\rho(r)\frac{d\phi}{dr}=-\frac{GM(r)\rho(r)}{r^2}<br />
and the equation of state of the ideal gas
<br /> P(r)=k T \rho(r)\frac{\rho(r)}{m}<br />
where M(r)=M + \frac{4\pi}{3}r^3\pho(r) is the total mass contained in the spherical shell or radius r, m is the mass of the gas molecules, and \phi is the gravitationnal potential, to find a differential equation for \rho(r), but this equation turns out to be nonlinear:
<br /> \frac{k T}{m}\frac{d\rho}{dr}=-\frac{GM\rho(r)}{r^2}-\frac{4\pi G}{3}r \rho(r)^2<br />
And I don't know how to solve it. Then I tried using the poisson equation
<br /> \nabla^2 \phi = 4\pi G \rho(r)<br />
but the equation I ended up with was again nonlinear inlcuding a square of the first derivative.

Any ideas how to solve the preceeding equation or another way to solve the problem?

Thanks

 

[Physics Monkey]

A few quick points. First, the ideal gas law for a uniform system is
P = \frac{N}{V} k T,<br />
so it seems like the appropriate generalization would be to say
P(r) = \frac{\rho(r)}{m} k T,<br />
so I don't know where you got the extra factor of \rho from.

Second, your use of the hydrostatic equilibrium condition is appropriate, but you have the mass wrong. How much mass is contained in a spherical shell of radius r and thickness dr? Hint: it isn't just \rho(r) dr.

Third, I don't understand your expression for the total mass enclosed. I think you have a typo there. The total enclosed mass will actually be an integral of the mass density.

Fourth, when you put all this together, you will need Poisson's equation to complete your set equations i.e. 3 unknowns (P, \rho, \phi) and 3 equations (hydrostatic, ideal gas, Poisson's).

Once you've addressed these points, and if you still can't solve the equation, post again and we can try to move forward from there.

Hope this helps.

 

[goulio]

Ok yeah the equation of the gas is a typo and I got wrong the mass enclosed by spherical shell. You're right it is an integral. Actually it is something like:
<br /> M(r)=\int_0^r du 4\pi u^2 \rho(u)<br />
But I put this in the hydrostatic equation I end up with and integro differential equation, something like
\frac{df}{dx} = -a\frac{f(x)}{x^2}\left (\int_R^r u f(u) du - b \right )
I tried to use poisson without any success.

Thanks for your help, you can't imagine how it is appreciated.

 

[Physics Monkey]

Ok, your mass enclosed formula looks about right, just remember that the gas density drops to zero at r = R, so the mass enclosed should look something like
<br /> M(r) = 4 \pi \int^r_R u^2 \rho(u) \,du + M,
I think you just have a typo (only one power of u) when you used the formula in the hydrostatic equation (also b must be -M).

In obtaining your final equation you have already used Poisson's equation to find \phi in terms of \rho, this is all I was getting at.

Finally, you do end up with an integro-differential equation. To make it into an ordinary differential equation, try isolating the integral and then differentiating both sides with respect to r. Use the fundamental theorem of calculus to evaluate the derivative of the integral. What you should be left with is a second order nonlinear differential equation. If you can get to this point then we can try to make some more progress.

 

[goulio]

In obtaining your final equation you have already used Poisson's equation to find \phi in terms of \rho, this is all I was getting at.

What do you mean by this? I'm not sure to get it.

 

[goulio]

Here is another way I tried to solve the equation:

Ok here is what I have done for the poisson equation. The gas alone has a Maxwell-Boltzmann Distribution, so the gas under the influence of a conservative potential has the following distribution
<br /> n(r,v)=n_0(v)\exp(-\phi(r)/k t)<br />
where n_0 is the MB distribution. If I expand this to the first order in \phi I get
<br /> n(r,v)=n_0(v)(1-\phi(r)/k t)<br />
Then the density as a function of r is just the planet density plus the gas density
<br /> \rho(r) = \frac{3M}{4 \pi R^3} + m n_0(v)(1-\phi(r)/k T)H(r-R)<br />
where m is the mass of the gas particles and H(r-R)[/iex] is the unit step function. <br /> I substitued this in the poisson equation but I can&#039;t solve the resulting equation and Maple won&#039;t solve it<br /> either.<br /> <br /> Any other suggestions.

 

[Physics Monkey]

You used Poisson's equation (perhaps without knowing it) when you calculated the force (per unit mass) as
<br /> - \frac{G M(r)}{r^2}.<br />
This result is easy to obtain from Gauss' Law which is just another way to phrase Poisson's equation.

As for your second attack, you must first integrate over the velocity distribution since you want the unconditional distribution (also its obvious that there shouldn't be any velocity dependence anywhere). The equation you have written down is valid, but it represents a self consistent approach since you don't actually know \phi and it depends on what \rho(r). It is still an integral equation, but if you did things right then you can recover the final equation found in your first attack.

 

[goulio]

Thank you thank you thank you.

Can you tell me how I should differentiate this integral?

Thanks again.

 

[goulio]

Ok forget about that last one, it is getting late here... It's just the function evaluated at r.

 

[Physics Monkey]

The fundamental theorem of calculus says
<br /> \frac{d}{dx} \left( \int^x_a f(t) \, dt\right) = f(x),<br />
pretty much exactly as you would expect. Isolate the integral and then differentiate to obtain a pure (and horribly complicated) differential equation. I happen to know that this equation isn't solvable in general (it comes up, for instance, when trying to describe plasmas). I must go now, but perhaps you can find a solution for the special case, just be careful to get all the terms right. If nothing else, you can numerically integrate the equation and obtain a perfectly good solution.