Gravity at the center of the Earth?

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[Jim]: how does the model of inceasing pressure keep the pressure from going to infinity as I said early on?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

The pressure is just a measure of all the weight above that point. How can the weight above a point go to infinity? -|Tom|-

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
If the pressure goes higher because the area of the sphere is less and less it seems to me at the mass center the area gets so small as the radius of the sphere goes to a very small size.
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Not all functions taken to their limits go to either "nothing" (infinitessimly small) or infinity. Neither, too, do all inverse functions whose limits are "nothing" always go to infinity. It happens that the function that describes the pressure at the infinitessimly small point at the center of the sphere goes to a finite value that is the sum of all the "weight" of all the mass "above" it. The arithmetic is worked out in a reference I provided way at the top of this thread.

If the pressure increases with depth and a shell model indicates this is true then when the shells get smaller and smaller won't the area of each smaller sphere or shell be less? And is pressure mass per unit of area? In the SI system newtons/m2=pascal, right? So, as the mass center of the sphere is neared the pressure will get up there and somehow it must be stopped.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[Jim]: If the pressure increases with depth and a shell model indicates this is true ...<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

The "shell model" had some interesting things to say about force and potential, but not about pressure. You can learn nothing about pressure from shells because pressure is omni-directional. (That means it is the same in every direction at any given point. It is <i>not</i> directed downward.)

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>... then when the shells get smaller and smaller won't the area of each smaller sphere or shell be less?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

This is a case in point. The area of the shells gets smaller, so the force of each gets smaller, and goes to zero at the center. But zero force means zero weight, so none of the matter near the center adds much to the overall pressure. Anyplace near the center will give you almost the same pressure reading.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>And is pressure mass per unit of area?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Pressure is force per unit area. (But force is sometimes represented as a weight, making it look mass-like or volume-like, so we can measure it in non-standard units such as pounds per square inch or inches of mercury in a barometer.)

The "unit of area" in the definition is allowed to approach zero, and pressure at any point is the limit of force over area as area goes to zero. (Taking such limits is the subject matter of calculus.)

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>In the SI system newtons/m2=pascal, right? So, as the mass center of the sphere is neared the pressure will get up there and somehow it must be stopped.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

The "square meter" used to measure pressure could be anywhere near the center, with any orientation. But it must be flat, not a portion of a shell. If you want the pressure on a curved surface such as a shell, you need to take the limit as area goes to zero.

To show how arbitrary the "center" is, suppose we scooped out a big mass from one side of the Earth and transported it to the opposite side. Then Earth's center of mass would shift. But nothing would happen to its internal pressures near either old or new center because the weight supported is the same. -|Tom|-

Are you now saying the pressure inside a sphere is about the same throughout the volume? The shells are spheres and since the pressure goes down(I guess) at the center of mass as the force goes to zero, it seems there must be a change in the pressure profile at several points along the radius. Since the pressure stops rising what is the radius at which this happens?

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[Jim]: Are you now saying the pressure inside a sphere is about the same throughout the volume? The shells are spheres and since the pressure goes down(I guess) at the center of mass as the force goes to zero, it seems there must be a change in the pressure profile at several points along the radius. Since the pressure stops rising what is the radius at which this happens?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

No, pressure increases with depth because pressure is equal to the weight of all the layers above, and that weight never decreases with increasing depth.

However, as one approaches the center (the point of maximum pressure), force tends toward zero, which means very little additional weight is being added by the deep layers. So pressure rises rapidly with depth near the surface, but rises more slowly at greater depth, and tends toward a constant as force tends toward zero near the center. -|Tom|-