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Newton demonstrated that the gravitational force at a distance r from the center of a uniform sphere is "as if" all the mass was concentrated at the center. The proof I have seen starts by first integrating over the 2d surface of an infinitely thin shell, and thus finding the gravitational field of such a shell. Outside the shell the above "as if" is demonstrated, and then since it is true for all concentric shells, it is true for the sphere.

But Newton's proof for a shell also proves that the gravitational field INSIDE the shell is exactly zero everywhere internal to the shell. That is, when the gravitational field of all parts of the shell are added up, you get identically zero.

Thus if you are in a deep mine shaft the gravitational field is only dependent on your distance from the center of the earth. That is, that portion of the mass of the earth in the shells above you has no effect on the gravitational field you experience. Further analysis says that if it were possible to drill a hole to the center of the earth, the gravitational field will be directly proportional to the linear (not inverse square) distance from the center of the earth. In particular, the gravitational field at the center of the earth would be zero, since all the shells above it have no net gravitational effect.

If the above argument is valid, why it is commonly said that the center of a sun is very highly compressed by the gravity of the sun's mass, and therefore hot enough to cause nuclear fusion? I have never understood this conclusion. Can someone explain it to me?

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    $\begingroup$ Well that gravitational force pulls the rest of the mass above it inward $\endgroup$ – Triatticus Jun 23 '18 at 17:47
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/184032 $\endgroup$ – Martin C. Jun 23 '18 at 17:48
  • $\begingroup$ I think that you're somehow confusing the idea of net gravitational force on a body with the pressure on the body due to gravitational forces acting on other objects. If you're at the center of the Earth, you won't feel any net gravitational force on yourself, but you will have gravity exerting a net force on all the matter surrounding you that will be trying to pull all that matter towards you. The pressure on you will therefore be enormous. $\endgroup$ – Samuel Weir Jun 23 '18 at 18:35
  • $\begingroup$ I think I got it. Imagine I am lying in a six-foot deep grave with an oxygen tank and no coffin. With open sky above me I feel gravity attracting me to the earth, that is, my weight. It's slightly less than my weight on the surface, since I am six feet closer to the center. As somebody starts piling dirt on top of me, the weight of all that dirt will crush me! It does not help that the six-foot spherical shell above me does not contribute to the force due to the sphere below me. Only the weight of the dirt just poured in affects the crushing force I feel. $\endgroup$ – Allen Simon Jun 23 '18 at 23:28
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We can possibly think about an object in the middle of a celestial body as squeezed between columns of matter pressing from all directions due to the attraction to each other: the sum of all forces is zero, but the squeeze could be significant.

If the celestial body was solid, the pressure could be, at least partially, relieved by stress in the solid matter. For the son, thpugh, which is made out of hot gas, there won't be much of relief.

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  • $\begingroup$ See my comment to Qmechanic $\endgroup$ – Allen Simon Jun 23 '18 at 23:31
  • $\begingroup$ @AllenSimon Yes, you've got it. Of course (if you don't mind), if the grave was filled with big rocks, they could jam and that would reduce the pressure near the bottom of the grave. This would not be the case with liquid or gas. Therefore, someone on the bottom of the ocean or in the middle of the sun, will surely fill tremendous pressure, not mitigated by any structure. $\endgroup$ – V.F. Jun 23 '18 at 23:44
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In particular, the gravitational field at the center of the earth would be zero, since all the shells above it have no net gravitational effect.

If the above argument is valid, why it is commonly said that the center of a sun is very highly compressed by the gravity of the sun's mass ...

The pressure is due to the weight of the material above you. That material experiences a force due to gravity that depends on it's radius from the center (which is not zero). The fact that the force is zero at the center does not matter to the weight of the material outside the center, which is not zero.

Some maths :

The force due to the shell at $r$ is it's mass times the gravity that shell experiences, which is :

$$\delta F = 4\pi r^2 \sigma g(r) \delta r$$

and the gravity for a uniform sphere is :

$$g(r) = G \frac 4 3 \pi r^3 \sigma \frac 1 {r^2} = \frac 4 3 G\pi \sigma r$$

The force due to the weight of material above you at a radius $r$ is given by :

$$F = \int^R_r 4 \pi r^2 \sigma \frac 4 3 G \sigma r dr = \frac 4 3 \pi^2G\sigma^2(R^4-r^4)$$

And the pressure is force over area so :

$$P=\frac 1 3 \pi G \sigma^2\frac{R^4-r^4}{r^2}$$

And you can easily see that this increases as $r$ increases.

It's more complicated for the Sun where you have to take account of the pressure outward from the heat coming from the core, but you can see the idea.

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  • $\begingroup$ Trying to follow your math. Your first equation includes a bunch of factors that look like the total mass of the shell (its area: 4-pi-r-squared) times the shell's thickness (delta-r) times its density (signma) and a mysterious factor g(r) that you expand in your second equation (which I don't understand yet). Still trying to understand the first equation. I think delta-f represents the incremental gravitational force $\endgroup$ – Allen Simon Jun 23 '18 at 22:54
  • $\begingroup$ Previous comment was truncated because it exceeded the maximum character count. Bit that's OK since: With everyone's help, I now get it, as explained in my comment to Qmechanic. $\endgroup$ – Allen Simon Jun 23 '18 at 23:40
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The pressure comes from matter below stopping the matter above from compressing any more. Think about the very center. It's true that there is no net gravity acting there. But just above it, there's mass with a slight gravitational attraction to the center. So it wants to accelerate and occupy the center as well, but the matter already in the center pushes back to stop it from compressing further. This is the pressure that develops. Then there's mass on top of that mass, etc. all the way to the surface of Earth.

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  • $\begingroup$ See my comment to Qmechanic $\endgroup$ – Allen Simon Jun 23 '18 at 23:32

protected by Qmechanic Jun 23 '18 at 19:36

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