I Gravity between extended bodies

[FlamesOfIgnorance]

Hello!
Can you tell me how to calculate the gravitational force between two extended inhomogeneous bodies? (In the general case, we don't know anything about the shapes, we only know the density of masses)

 

[vanhees71]

If we know the mass density of the bodies we also know about their shapes.

 

[Ibix]

Write down the gravitational acceleration felt at position (X,Y,Z) due to the elementary mass at (x,y,z). Then, since Newtonian gravity is linear you can just integrate over x, y and z.

 

[FlamesOfIgnorance]

If we know the mass density of the bodies we also know about their shapes.

Yes that is right. Then I have to modify my statement. Their shapes are totally free. Gravitation is the only force.

 

[FlamesOfIgnorance]

Write down the gravitational acceleration felt at position (X,Y,Z) due to the elementary mass at (x,y,z). Then, since Newtonian gravity is linear you can just integrate over x, y and z.

Sorry for double posting I'm a beginner. I don't really understand what you wrote. At (x,y,z) the mass may be 0, because the system could be continous. Would you clarify a bit for me?
I am not an expert in Physics in fact I have just read about the Newtonian axioms for about 10 years and I would like to generalize them to continuous bodies.
edit: I mean not that I want to generalize them but I am interested if there are any generalizations. Because I have only read about forces between a continuous body and a point of nonzero mass and zero volume. But what if everything is continuous?

 

[Ibix]

The mass in a small cuboidal volume with sides of length $\delta x$, $\delta y$, and $\delta z$ containing the point (x,y,z) is approximately $\rho (x,y,z)\delta x\delta y\delta z$. That's not quite right because $\rho$ isn't constant, but if you let the volume get very, very small the error becomes very, very small. And the mass at the point (x,y,z) becomes $\rho (x,y,z)dxdydz$. Now you can write down the acceleration from that tiny mass and integrate to get the total.

You may wish to look up the shell theorem, which simplifies things when the mass is spherically symmetric. And at distances much larger than the size of the object the answer will be pretty close to the same as a point mass at the centre of mass.

 

[FlamesOfIgnorance]

Thank you for your answer!
And we get the acceleration of that cube by integrating by (X,Y,Z) because the other body is continuous aswell. So we get an acceleration at every point of the body right? (Not quite an acceleration but the lim of accelerations)
The method is the same I guess when there is only one body and we want the accelerations at every point of the body.

 

[vanhees71]

The gravitational potential of mass distribution 1 is
$$\phi(\vec{x})=-\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \frac{\gamma \rho_1(\vec{x}')}{|\vec{x}-\vec{x}'|},$$
and thus the total force acting on mass distribution 2
$$\vec{F}=-\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \vec{\nabla} \phi(\vec{x}) \rho_2(\vec{x}),$$
i.e.,
$$\vec{F}=-\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\gamma \rho_1(\vec{x}') \rho_2(\vec{x}) (\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3}.$$

 

[FlamesOfIgnorance]

Well, thank you Sir!

 

[Ibix]

And we get the acceleration of that cube by integrating by (X,Y,Z) because the other body is continuous aswell. So we get an acceleration at every point of the body right?

You are assuming your body is perfectly rigid, which may or may not be totally plausible. But basically, yes.

The method is the same I guess when there is only one body and we want the accelerations at every point of the body.

Yes. But anything which has any significant gravitational effect on itself (like, say, the Earth) is going to collapse into a sphere under its own weight unless it's made of magically strong stuff.

 

[FlamesOfIgnorance]

I didn't want to assume that any of the bodies is rigid. Nobody assumed that right?

 

[Ibix]

In that case prepare yourself for scary maths. You need a model of how your material reacts to stress and how the density changes as it does

You'll get (I guess) nine simultaneous differential equations to solve for each point in your body. I'm afraid it's beyond my experience. Make sure you've got a big computer, though.

 

[FlamesOfIgnorance]

I don't want to solve the problem just to formalize it. You say that the solution provided by vanhees71 is not the general case?

 

[Ibix]

Not if the body isn't rigid. To handle that you need to add a time dependence to $\rho$ and then provide some kind of model of the material's response to forces. I strongly suspect anything non-trivial is going to be a numerical-solution-only job, and quite possibly beyond the reach of home computing.

There might be specific situations that are more tractable, and it's always possible that I'm over-estimating the complexity - as I say it's beyond my experience.

If you treat the gravitational field as uniform and have just one non-rigid body, mechanical engineers do that all the time. The engineering forums might offer some insight.

 

[vanhees71]

Of course, there is a time dependence of $\rho_1$ and $\rho_2$ since the bodies will move under the mutual gravitational interaction. What I've given is just the formula for the force at one given point in time.

Of course, if the bodies are not rigid you have to solve the corresponding fluid-dynamical equations, which is much more efford.