<p>Gauss's divergence theorem applied to the gravitational field $\vec{g}$ is that
$$ \oint \vec{g} \cdot d\vec{A} = \int \nabla \cdot \vec{g}\ dV,$$
where the left hand side is the flux of gravitational field into/out of a closed surface and the right hand side is the integral of the divergence of that field over the volume enclose by the surface.</p>

<p>The fundamental definition of how mass produces gravitational field is that
$$ \nabla \cdot \vec{g} = -4\pi G \rho,$$
where $\rho$ is the mass density.</p>

<p>If you are at a point <em>outside</em> the mass then the right hand side of the divergence theorem  becomes a <em>constant</em>.
$$ \oint \vec{g} \cdot d\vec{A} = -4\pi G \int \rho\ dV = -4\pi GM.$$
Therefore the left hand side is also constant when calculated over any surface that encloses <em>all</em> the mass.</p>

<p>If you allow the mass to contract <em>in a spherically symmetric fashion</em>, then as the gravitational field on the left hand side is then always in the radial direction then $\vec{g}$ must stay the same at any position in space <em>outside</em> the mass distribution.</p>

CC BY-SA 3.0
2017-02-15T13:47:53.040Z
Do you mean at any fixed (given) position outside, g will stay the same no matter how compressed it is. g outside of any spherical distribution can be calculated by replacing it with a point mass at the center. Right?
uhoh
CC BY-SA 3.0
2017-02-15T13:50:38.563Z
@uhoh Yes, that is the shell theorem.
ProfRob
CC BY-SA 3.0
2017-02-15T13:52:27.033Z
OK so the gravity *on the surface* of the body will of course definitely get stronger as it compresses. Great. I thought there was something strange about the linked comment in my question. I could not understand how the divergence theorem applied there. Thanks!
uhoh
CC BY-SA 3.0
2017-02-15T13:27:08.713Z
2017-02-15T13:27:08.713Z
ProfRob
<p>Below <a href="https://astronomy.stackexchange.com/q/20058/7982">this question</a> there is a <a href="https://astronomy.stackexchange.com/q/20058/#comment33190_20058">comment</a> that caught my interest:</p>

<blockquote>
  <p>First, there's a misconception. Compressing a body does not increase the force of gravity on outer-lying masses. That's Gauss's famous divergence theorem.</p>
</blockquote>

<p>I understand that one can use the divergence theorem to address (derive) the <a href="https://en.wikipedia.org/wiki/Shell_theorem" rel="nofollow noreferrer">shell theorem</a> - one consequence of quick is that a particle inside a spherically symmetric shell experiences no net gravitational force from the shell, but this seems different as it refers <em>changes in average density</em> of a body and gravitational forces on <em>outer-lying masses</em>.</p>

<p>So I'd like to understand better: <em>What does Gauss' divergence theorem say about compression of a body under self-gravitation?</em></p>

<p>For reference, <a href="https://astronomy.stackexchange.com/a/8032/7982">this answer</a> covers the <a href="https://astronomy.stackexchange.com/a/8032/7982">shell theorem</a> nicely.</p>

<gravity><planetary-formation>
What does Gauss' divergence theorem say about compression of a body under self-gravitation?
uhoh
# What does Gauss' divergence theorem say about compression of a body under self-gravitation?

(Asked by: username_0 on 2017-02-15)

Below [this question](https://astronomy.stackexchange.com/q/20058/7982) there is a [comment](https://astronomy.stackexchange.com/q/20058/#comment33190_20058) that caught my interest:

> First, there's a misconception. Compressing a body does not increase the force of gravity on outer-lying masses. That's Gauss's famous divergence theorem.

I understand that one can use the divergence theorem to address (derive) the [shell theorem](https://en.wikipedia.org/wiki/Shell_theorem) - one consequence of quick is that a particle inside a spherically symmetric shell experiences no net gravitational force from the shell, but this seems different as it refers *changes in average density* of a body and gravitational forces on *outer-lying masses*.

So I'd like to understand better: *What does Gauss' divergence theorem say about compression of a body under self-gravitation?*

For reference, [this answer](https://astronomy.stackexchange.com/a/8032/7982) covers the [shell theorem](https://astronomy.stackexchange.com/a/8032/7982) nicely.

**Tags:** gravity, planetary-formation


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## 1 answers

### Answer by username_1 on 2017-02-15. Score: 5 | Accepted answer

Gauss's divergence theorem applied to the gravitational field $\vec{g}$ is that
$$ \oint \vec{g} \cdot d\vec{A} = \int \nabla \cdot \vec{g}\ dV,$$
where the left hand side is the flux of gravitational field into/out of a closed surface and the right hand side is the integral of the divergence of that field over the volume enclose by the surface.

The fundamental definition of how mass produces gravitational field is that
$$ \nabla \cdot \vec{g} = -4\pi G \rho,$$
where $\rho$ is the mass density.

If you are at a point *outside* the mass then the right hand side of the divergence theorem becomes a *constant*.
$$ \oint \vec{g} \cdot d\vec{A} = -4\pi G \int \rho\ dV = -4\pi GM.$$
Therefore the left hand side is also constant when calculated over any surface that encloses *all* the mass.

If you allow the mass to contract *in a spherically symmetric fashion*, then as the gravitational field on the left hand side is then always in the radial direction then $\vec{g}$ must stay the same at any position in space *outside* the mass distribution.

> **Comments:**
> **username_0**: Do you mean at any fixed (given) position outside, g will stay the same no matter how compressed it is. g outside of any spherical distribution can be calculated by replacing it with a point mass at the center. Right?  
> **username_1**: @username_0 Yes, that is the shell theorem.  
> **username_0**: OK so the gravity *on the surface* of the body will of course definitely get stronger as it compresses. Great. I thought there was something strange about the linked comment in my question. I could not understand how the divergence theorem applied there. Thanks!  

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