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I've long been taught that all matter having mass, possesses attractive forces somewhat akin to gravity. As such, imagine we can 'teleport' a gravitonic detection device that can accurately measure the gravitational forces regardless of strength present at each location, it can detect all gravitational forces ordered by direction of pull within a given radius of up to let say [through a range of settings/switches of 0.001 nanometer to 100,000 km. We teleport these detectors equi-distantly apart in the earth, from troposphere to molten mantle to core's center of mass. measure and plot all the forces detected. Supposedly, there is no gravity at earths center of mass...that means gravity can be concentrated and nullified. Logic tends to indicate 'no gravity=no weight. Therefore why does all reference material say massive objects have tremendously large core pressures and squeeze atoms to extremely dense masses. If gravity is nullified, what is source of this crushing, without usurping all gravitational common sense?

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    $\begingroup$ Could you try to state your question a little bit clearer? It seems to me you are asking why there is a high pressure at the core of, say, the earth although the gravitational potential vanishes at the center? But then again there seems to be no connection to the title you chose for your question? $\endgroup$ – André Apr 5 '14 at 23:16
  • $\begingroup$ I think what user43994 is getting at is that while there is no gravity at the center of the Earth, there is extremely high pressure at the center of the Earth, and he's confused as to why no gravity doesn't imply no pressure. $\endgroup$ – DumpsterDoofus Apr 5 '14 at 23:37
  • $\begingroup$ Related: physics.stackexchange.com/q/2481/2451 and links therein. $\endgroup$ – Qmechanic Apr 6 '14 at 0:05
  • $\begingroup$ It isn't that there is no gravity, in fact there is more than when you are on the surface but it is all pulling outwards and in all directions and with the same magnitude an any direction and the force sums to zero. Any tiny deviation from the exact center should show on a gradiometer. This assumes a perfect sphere with perfect spherical distribution of mass. Hmmm. Can you have local minimums in the real Earth? Geophysics theorists must have worked out the implications. Curious people would like to know. $\endgroup$ – C. Towne Springer Apr 6 '14 at 3:16
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As best I can tell, your question can be paraphrased as:

If there's zero gravity at the center of the Earth, why is pressure not also zero? Pressure is caused by weight, so if you're weightless at the center, shouldn't it also be pressureless?

Let's calculate the pressure.

Model the Earth as an incompressible material with density $\rho$ and radial pressure $p(r)$ for $0\leq r\leq R$, where $R$ is the radius of the Earth. The force per unit volume due to pressure is given by $$F_p=-\frac{\partial}{\partial r}p(r).$$

Meanwhile, the force per unit volume due to gravity at a depth $r$ is given by Newton's law: $$F_g=\frac{G\rho\left(\frac{4}{3}\pi r^3\rho\right)}{r^2}=\frac{4}{3} \pi G \rho ^2 r.$$ Setting $F_p=F_g$ and integrating to find $p(r)$ along with the condition $p(R)=0$ yields $$p(r)=\frac{2}{3} \pi G \rho ^2 \left(R^2-r^2\right).$$ Note that the pressure is actually highest at the core, and that as you travel towards the surface, it decreases quadratically to zero.

In short: just because you're weightless at the core doesn't negate the fact that you've still got mass all around you that has weight, and is crushing down on you.

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There will be no gravitational force at the center of the Earth because of Gauss' law for gravity: $\oint_{\partial V} \vec{g}\cdot d\vec{A} = -4\pi GM$ where $M$ is the mass enclosed in the volume $V$. The case of interest here is that of a sphere, so the left-hand side of this integral will evaluate to $4\pi r^2 g(r)$ where $r$ is the distance from the center of the sphere. You can rearrange this to give $g(r) = -GM\hat{r}/r^2$ but the point of all this is that when we get to the center of the Earth, which we can approximate as a spherical mass distribution, the sphere we're integrating over shrinks to enclose no mass, so there is no $\textit{gravitational}$ force acting there. However, if you go just a little bit outside of that center point, there will be gravitational force pulling you towards the center. Consider a spherical shell somewhere within the Earth of radius $r$. It's being pulled towards the center of the Earth by gravity, but it's not moving. So there must be another force opposing it, which we call the normal force. Now by Newton's third law, there must be a reactionary force to the Normal force pushing down on the layer just below the one we're considering. When you add up over all of the spherical shells that make up the Earth, this amounts to an enormous pressure being exerted at the center. We can calculate what this force should be by integrating up the weight of each little piece of the Earth.

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