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

In summary, the gas is attracted to and self-attracts to the planet, but is not Uniform so different equations are needed to solve for the gas density.
  • #1
goulio
15
0
Hi,

I need to find the density [itex]\rho(r)[/itex] of an ideal gas at constant temperature [itex]T[/itex] surrouding a planet of mass [itex]M[/itex] and radius [itex]R[/itex]. The gas is attrated by the planet and is also self-attracting. First, I used the hydrostatic equilibirum equation
[tex]
\frac{dP}{dr}=-\rho(r)\frac{d\phi}{dr}=-\frac{GM(r)\rho(r)}{r^2}
[/tex]
and the equation of state of the ideal gas
[tex]
P(r)=k T \rho(r)\frac{\rho(r)}{m}
[/tex]
where [itex]M(r)=M + \frac{4\pi}{3}r^3\pho(r)[/itex] is the total mass contained in the spherical shell or radius [itex]r[/itex], [itex]m[/itex] is the mass of the gas molecules, and [itex]\phi[/itex] is the gravitationnal potential, to find a differential equation for [itex]\rho(r)[/itex], but this equation turns out to be nonlinear:
[tex]
\frac{k T}{m}\frac{d\rho}{dr}=-\frac{GM\rho(r)}{r^2}-\frac{4\pi G}{3}r \rho(r)^2
[/tex]
And I don't know how to solve it. Then I tried using the poisson equation
[tex]
\nabla^2 \phi = 4\pi G \rho(r)
[/tex]
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
 
Last edited:
Physics news on Phys.org
  • #2
A few quick points. First, the ideal gas law for a uniform system is
[tex] P = \frac{N}{V} k T,
[/tex]
so it seems like the appropriate generalization would be to say
[tex] P(r) = \frac{\rho(r)}{m} k T,
[/tex]
so I don't know where you got the extra factor of [tex] \rho [/tex] 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 [tex] r [/tex] and thickness [tex] dr [/tex]? Hint: it isn't just [tex] \rho(r) dr [/tex].

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, [tex] \rho [/tex], [tex] \phi [/tex]) 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.
 
Last edited:
  • #3
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:
[tex]
M(r)=\int_0^r du 4\pi u^2 \rho(u)
[/tex]
But I put this in the hydrostatic equation I end up with and integro differential equation, something like
[tex]\frac{df}{dx} = -a\frac{f(x)}{x^2}\left (\int_R^r u f(u) du - b \right )[/tex]
I tried to use poisson without any success.

Thanks for your help, you can't imagine how it is appreciated.
 
  • #4
Ok, your mass enclosed formula looks about right, just remember that the gas density drops to zero at [tex] r = R [/tex], so the mass enclosed should look something like
[tex]
M(r) = 4 \pi \int^r_R u^2 \rho(u) \,du + M, [/tex]
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 [tex] \phi [/tex] in terms of [tex] \rho [/tex], 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 [tex] r [/tex]. 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.
 
  • #5
Physics Monkey said:
In obtaining your final equation you have already used Poisson's equation to find [tex] \phi [/tex] in terms of [tex] \rho [/tex], this is all I was getting at.

What do you mean by this? I'm not sure to get it.
 
  • #6
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
[tex]
n(r,v)=n_0(v)\exp(-\phi(r)/k t)
[/tex]
where [itex]n_0[/itex] is the MB distribution. If I expand this to the first order in [itex]\phi[/itex] I get
[tex]
n(r,v)=n_0(v)(1-\phi(r)/k t)
[/tex]
Then the density as a function of [itex]r[/itex] is just the planet density plus the gas density
[tex]
\rho(r) = \frac{3M}{4 \pi R^3} + m n_0(v)(1-\phi(r)/k T)H(r-R)
[/tex]
where [itex]m[/itex] is the mass of the gas particles and [itex]H(r-R)[/iex] is the unit step function.
I substitued this in the poisson equation but I can't solve the resulting equation and Maple won't solve it
either.

Any other suggestions.
 
  • #7
You used Poisson's equation (perhaps without knowing it) when you calculated the force (per unit mass) as
[tex]
- \frac{G M(r)}{r^2}.
[/tex]
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 [tex] \phi [/tex] and it depends on what [tex] \rho(r) [/tex]. It is still an integral equation, but if you did things right then you can recover the final equation found in your first attack.
 
Last edited:
  • #8
Thank you thank you thank you.

Can you tell me how I should differentiate this integral?

Thanks again.
 
  • #9
Ok forget about that last one, it is getting late here... It's just the function evaluated at [itex]r[/itex].
 
  • #10
The fundamental theorem of calculus says
[tex]
\frac{d}{dx} \left( \int^x_a f(t) \, dt\right) = f(x),
[/tex]
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.
 
Last edited:

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

What is an ideal gas?

An ideal gas is a theoretical gas that follows the ideal gas law, which describes the relationship between pressure, volume, and temperature. It is made up of point particles that have no volume and do not interact with each other.

How does a planet become surrounded by an ideal gas?

A planet can become surrounded by an ideal gas through a process called outgassing. This occurs when gas trapped within a planet's surface is released due to geological activity or when a planet's atmosphere is formed during its formation.

What are the properties of an ideal gas?

An ideal gas is characterized by the following properties: it has no volume, the particles do not interact with each other, and it follows the ideal gas law (PV = nRT). It also has a constant temperature, pressure, and density throughout the system.

Can any planet be surrounded by an ideal gas?

No, not all planets can be surrounded by an ideal gas. Planets that have a solid or liquid surface, such as Earth, cannot be surrounded by an ideal gas because the gas particles would interact with the surface and not follow the ideal gas law.

What are the applications of studying a planet surrounded by an ideal gas?

Studying a planet surrounded by an ideal gas can help us understand the formation and evolution of planets, as well as their atmospheres. It also has practical applications in fields such as meteorology and space exploration, where understanding the behavior of gases is crucial.

Back
Top