# Poisson Problem - Gravitation

1. Feb 11, 2005

### secret2

If I have an infinite slab of incompressible self-gravitating fluid of density rho within the region |z|<a, and I am asked to find the potential both inside and outside the slab, where should I start?

2. Feb 11, 2005

### dextercioby

With writing the (differential) equations which account for the dynamics...??

Daniel.

3. Feb 11, 2005

### clive

Do you mean by "potential" the gravitational potential and by "rho" the mass density?

4. Feb 11, 2005

### dextercioby

Sure he does.

Daniel.

5. Feb 11, 2005

### clive

A very fast solution to this problem can be obtained using a Gauss-like law for the gravitational field (it can be demostrated by direct integration of the Poisson equation and using the divergence theorem). The "gravitational flux" through a closed surface must equal the total mass inside the surface times gravitational constant.
$$\oint_S \vec{\Gamma}\cdot d\vec{S}=-\gamma \int \rho dv$$
Then, if you know $$\Gamma(z)$$, the potential is just
$$V=-\int \Gamma dz$$
(for the integration constant you can impose V(0)=0)
You can choose a cylindrical gaussian surface with its axis parallel to Oz and play with this theorem. For this cylinder, the total flux is $$2\Gamma S$$ (S is basis area)

I think $$\Gamma$$ will vary linearly from z=0 to z=a and would be uniform for z>a. So the potential will be quadratic and linear respectively.....but you must verify that.....

Last edited: Feb 11, 2005
6. Feb 11, 2005

### dextercioby

Your analysis would be okay,if the "infinite slab of incompressible self-gravitating fluid of density rho" would not mean what i think it does:namely a fluid to which u have to apply not only the Poisson equation (for a gravitostatic field),but also Euler's equations and the continuity of mass (for an incompressible fluid).You'd have then 5 equations with 5 unknowns:the gravity potential,the velocity field and the density field...
It would be really nasty,indeed.

Daniel.

7. Feb 14, 2005

### secret2

Thanks Clive. But wouldn't using Gauss Law introduce a factor of 4*Pi? The answers have no 4*Pi in it. And do I have to consider also the pressure and the boundary condition?

By the way, here's the answers provided:
Code (Text):
V = (1/2)G rho (z)^2             |z|<a
= G rho a (|z| - (1/2)a)     |z|>a