I Effect of time on density distribution+shape of uniformly dense sphere

[madchemist]

I agree with Doc. Al that, "For the simplified case of a uniform density spherical planet, the gravitational field varies linearly from 0 at the center to its full value at the surface." But, what is the effect, if any, on the shape and density distribution of such a sphere over time (e.g., billions of Earth years)?

For example, over such time would the sphere begin to sort itself by density (e.g., from least dense at the core to most dense at approx. r/2)? And if so, would the newly distributed density affect the shape of the sphere (e.g., from a solid sphere to a hollow one)?

 

[Baluncore]

For example, over such time would the sphere begin to sort itself by density (e.g., from least dense at the core to most dense at approx. r/2)?

The densest material will move downward and so the sphere should end up with a dense core, a medium density mantle, and a scum of the lowest density material such as water and atmosphere on the surface.

Gravity will always accelerate things toward the centre of the sphere, even if it is hollow. Hydrostatic pressure increases toward the centre, where it reaches a maximum.

What mechanism do you expect could make it hollow with all that mass floating above ?

 

[madchemist]

The densest material will move downward and so the sphere should end up with a dense core, a medium density mantle, and a scum of the lowest density material such as water and atmosphere on the surface.

Gravity will always accelerate things toward the centre of the sphere, even if it is hollow. Hydrostatic pressure increases toward the centre, where it reaches a maximum.

What mechanism do you expect could make it hollow with all that mass floating above ?

Again... "I agree with Doc. Al that, 'For the simplified case of a uniform density spherical planet, the gravitational field varies linearly from 0 at the center to its full value at the surface.'"

Also, the question concerns a sphere, not a spheroid and not a planet.

 

[Baluncore]

Your quote by "Doc. Al" whoever, refers to a "planet".
Is the sphere liquid, or solid ?
Is it static, or spinning ?
Is it in space, or on your table ?

It is true that gravitational acceleration is canceled inside a hollow spherical shell, so there is no gravity at the centre of a sphere, but there is hydrostatic pressure within the shell material that pushes the inner spherical surface toward the centre. A shell will collapse into a sphere that is not hollow. That is a minimum potential energy solution.

Gravity is zero at the centre, then rises to a maximum, which for Earth is at the surface, before it falls again while moving away, due to 1/r². With a low density crust, the gravity can actually be greater below the surface than at the surface. Our atmosphere is an example of that effect.

Again I ask ...

What mechanism do you expect could make it hollow with all that mass floating above ?

 

[madchemist]

Is the sphere liquid, or solid ?
Is it static, or spinning ?
Is it in space, or on your table ?

Solid, compressible (as is everything) and capable of densitiy redistribution within the time stated. The sphere is in empty space with no external forces acting on it.

What mechanism do you expect could make it hollow with all that mass floating above ?

Newton's shell theorum. www.physicsforums.com/threads/no-gravity-at-center-of-earth-no-pressure.428928/

 

[madchemist]

What mechanism do you expect could make it hollow... ?

To be clear, while I have suspicions, i do not actually know the answer. That is the reason why I asked the question. I welcome all thoughts and ideas...

 

[Baluncore]

To be clear, while I have suspicions, i do not actually know the answer. That is the reason why I asked the question.

An outer shell is attracted to all material inside that shell. The weight of the outer spherical shell applies hydrostatic pressure to the outside of the inner sphere. So, hydrostatic pressure accumulates from zero at the surface to a maximum at the centre.

Gravity falls from the surface towards the centre, where it cancels due to symmetry.

The result is a solid sphere, with a dense core, and a low density outer layer.

 

[madchemist]

Having difficulty following you. Coud you please rephrase?

 

[Baluncore]

Which is the first sentence in post #7 that you do not understand.

https://en.wikipedia.org/wiki/Hydrostatics#Pressure_in_fluids_at_rest

 

[madchemist]

The terms, "outer shell," "inside that shell," "weight" (which depends on gravitational field which I think we agree is not constant in our model), and "outside of the inner sphere" are unclear.

By "Gravity falls from the surface towards the centre, where it cancels due to symmetry," do you mean, "the gravitational field varies linearly from 0 at the center to its full value at the surface" or are you trying to say something else?

 

[Baluncore]

The terms, "outer shell," "inside that shell," "weight" (which depends on g which I think we agree is not constant in our model), and "outside of the inner sphere" are unclear.

Separate the complete sphere into a hollow sphere, with a central, inner sphere, completely filling the aforesaid hollow.
"outer shell" + "inside that shell" = Total sphere.
The "outside of the inner sphere" = the surface of the central sphere.
"Weight" is the force that acts on an accelerated mass. F = m⋅a .

By "Gravity falls from the surface towards the centre, where it cancels due to symmetry," do you mean, "the gravitational field varies linearly from 0 at the center to its full value at the surface" or are you trying to say something else?

I certainly did not say "the gravitational field varies linearly".
Read and analyse what I wrote. Do you agree?

Since the gravitational attraction of outer spherical shells cancels everywhere inside those shells, start at the surface and work your way inwards, eliminating the effect of mass further from the centre than your point of analysis. You will then agree that gravity is canceled completely at the centre.

 

[madchemist]

Baluncore: I know you're trying to help but I feel like we're getting lost in the weeds here. Let's try it a different way... Do you agree that "For the simplified case of a uniform density spherical planet, the gravitational field varies linearly from 0 at the center to its full value at the surface"?

 

[Baluncore]

Do you agree that "For the simplified case of a uniform density spherical planet, the gravitational field varies linearly from 0 at the center to its full value at the surface"?

No.
I expect it to increase with radius, but I do not know that it is a linear function.

 

[madchemist]

I understand, "NO," but you lost me on your subsequent qualification of "NO."

 

[Baluncore]

What do you mean by the words "varies linearly" ?

 

[madchemist]

The net gravitational field is zero at the core, but increases as a function of distance from that core. That function does not necessarily give a straight line. What is important is that the field simply increases with distance.

 

[Baluncore]

The net gravitational field is zero at the core, but increases as a function of distance from that core. That function does not necessarily give a straight line.

I interpret "linearly" to be a certain dead straight line.
Since the mass of a sphere is proportional to, r³ ; and gravity is reduced as, 1/r² ; so for an ideal uniform-density sphere, g will therefore be proportional to, r³ / r² = r ; then the line may well be a straight line for r ≥ 0 . That will not be the case for a planet, or for a hollow sphere.

 

[madchemist]

Please Baluncore, do you agree with the original premise of the question (regardless of whether it can be graphed as a straight line or a curve)? If we can't agree on the original premise, then everything that follows lacks value as I'm sure you realize.

 

[Baluncore]

Please Baluncore, do you agree with the original premise of the question (regardless of whether it can be graphed as a straight line or a curve)? If we can't agree on the original premise, then everything that follows lacks value as I'm sure you realize.

For example, over such time would the sphere begin to sort itself by density (e.g., from least dense at the core to most dense at approx. r/2)? And if so, would the newly distributed density affect the shape of the sphere (e.g., from a solid sphere to a hollow one)?

No, that is quite impossible. Maximum density will be at the centre when r=0, NOT at r/2.

The body cannot become hollow, as that ignores hydrostatic pressure. It also ignores the buoyancy of the different density materials. A hollow is the lowest density material, so will rapidly rise to the surface, as it is backfilled by the most dense material.

I don't think you understand hydrostatic pressure, nor gravitational attraction.

 

[madchemist]

To Baluncore: Thank you for your input.

To everyone else: in the interest of truth, please respond ONLY if you first understand and agree with the original premise of this question which has already been hashed out at, inter alia, www.physicsforums.com/threads/no-gravity-at-center-of-earth-no-pressure.428928/

Thank you.

 

[Baluncore]

To everyone else: in the interest of truth, please respond ONLY if you first understand and agree with the original premise of this question which has already been hashed out at, inter alia, www.physicsforums.com/threads/no-gravity-at-center-of-earth-no-pressure.428928/

Thank you.

In that thread, the OP understood gravity is zero at the centre. The OP asked; "Does no force of gravity mean no pressure?".

The participants then discussed how hydrostatic pressure is caused by gravity, and why hydrostatic pressure reaches a maximum at the centre.

 

[madchemist]

To Baluncore: Thank you for your input.

To everyone else: in the interest of truth, please respond ONLY if you first understand and agree with the original premise of this question which has already been hashed out at, inter alia, www.physicsforums.com/threads/no-gravity-at-center-of-earth-no-pressure.428928/

Thank you.

 

[PeroK]

I welcome all thoughts and ideas...

Let's take Newton's second law: $\vec F = m\vec a$ and consider a mass at centre of a planet. The mass is at rest, so there no acceleration and, therefore, no "force".

But, the $\vec F$ in Newton's second law is net external force. You can crush an object by applying a uniform force to all sides. The (centre of mass of the) object does not accelerate in the sense of Newton's second law because there is no net force. But, the object is subject to a crushing pressure.

Let's take an non-gravitational analogy. We have a line of blocks (perhaps 11 blocks) and we apply an inward force to each end block. There are no external forces applied to any other block. But, there is still pressure on the centre block because of the external force on the end blocks.

Now, we apply an additional (external) inward force to the second block from each end. The internal pressure on the middle block increases.

Finally, we imagine an external force on each block except the centre one, decreasing linearly perhaps towards the centre. There is no external force on the middle block, but it is still under pressure according to the sum of the external forces on the other blocks.

The pressure on the centre block is a consequence of Newton's second and third laws. The forces on each block must be balanced (zero acceleration and Newton's second law). And, each block has an equal and opposite normal force with each of its neighbours. If you specify the external forces, you can easily calculate the force on both sides of each block. And, the centre block is under the greatest pressure.

The situation for a large mass under gravity is the spherical equivalent of this.

Note that a perfect large shell of zero thickness would be an unstable equilibrium. But, any deviation from perfection would lead to the shell's collapse under the mutual gravitational attraction of its constituent particles. Likewise, for any material, there would be a maximum size of hollow sphere before the pressure on the inner part of the sphere becomes too great and the sphere collapses under its own gravity.

 

[Doc Al]

Baluncore: I know you're trying to help but I feel like we're getting lost in the weeds here. Let's try it a different way... Do you agree that "For the simplified case of a uniform density spherical planet, the gravitational field varies linearly from 0 at the center to its full value at the surface"?

Yes, for a uniform-density sphere (it doesn't matter if it's a "planet" -- this is an idealization), the gravitational field will vary (linearly) from zero at the center to the maximum value at the surface. So?

Be careful with Newton's Shell theorem. While it's certainly true that a uniform shell of mass will create zero gravity within it, it does not mean that the shell has no forces on it. Especially if it's a shell in that uniform-density sphere.

 

[madchemist]

Thank you Doc Al for weighing in on the premise of the question.

Now (finally) for the actual question... "What is the effect, if any, on the shape and density distribution of such a sphere over time (e.g., billions of Earth years)?" Again, the sphere in our model is solid, compressible (as is everything) and capable of density redistribution within the time stated and it is in empty space with no external forces acting on it.

Be careful with Newton's Shell theorem.

Without getting too sidetracked, just confirming that it does not matter how thick the shell is for Newton's Shell theorem to apply to a point inside that shell?

While it's certainly true that a uniform shell of mass will create zero gravity within it, it does not mean that the shell has no forces on it. Especially if it's a shell in that uniform-density sphere.

No sure I follow. Are you saying the solid sphere in our model can be thought of as an onion with multiple shells? If so, what forces do you expect would act upon them besides gravity?

 

[Doc Al]

Without getting too sidetracked, just confirming that it does not matter how thick the shell is for Newton's Shell theorem to apply to a point inside that shell?

Sure. You can always think of a thick shell as being composed of a bunch of thin shells. And the shell theorem just tells you that the material of the shell exerts no gravitational force on anything within it. (Note that the stuff within the shell exerts gravitational forces on the shell.)

No sure I follow. Are you saying the solid sphere in our model can be thought of as an onion with multiple shells?

Sure. As pointed out, a thin shell by itself would not be stable. So you can think of the solid sphere as composed of many shells.

If so, what forces do you expect would act upon them besides gravity?

There are plenty of internal forces and pressures. Each "layer" is pulled down by the layers beneath it and is squashed between the layers above and below it.

 

[madchemist]

Each "layer" is pulled down by the layers beneath it and is squashed between the layers above and below it.

I agree. That is why I think things will settle at approx. r/2. Specifically, the outer-most layer is being pulled in the direction of the core while the inner-most layer is being pulled in the direction of the outer-most layer. Everything between those layers is compacted with the layer at approx. r/2 feeling the most compaction because it has an approximately equal amount of layers (or matter for those who wish to play semantics) on both sides of it. Yes, I realize that the layer at exactly r/2 would have more matter on one side than the other. And, yes I realize that since r/2 is in 3d space, some adjustment must be made to account for vector forces. That is why I called r/2 an "approximation."

a thin shell by itself would not be stable

I agree probably for the same reasons as I think a solid sphere would also not be stable. I don't want to jump too far ahead, but I think our model sphere is destined to transform repeatedly under the influence of gravity and, after gravity has set the stage, other forces as well.

 

[Doc Al]

I agree. That is why I think things will settle at approx. r/2. Specifically, the outer-most layer is being pulled in the direction of the core while the inner-most layer is being pulled in the direction of the outer-most layer.

Why do you think the inner-most layer is being pulled by the outer-most layer?

 

[Doc Al]

To elaborate on my previous post: Pick any layer. The only force that a layer exerts on the layers beneath it is a downward force (towards the center).

 

[madchemist]

Why do you think the inner-most layer is being pulled by the outer-most layer?

I think the inner-most layer (which is a layer and therefore not the center/core) is being pulled in the direction of the outer-most layer because the force of gravity in that direction is greater (albeit slightly) that the force of gravity in the other (i.e., the other direction is the one from the inner-most layer, through the core, to the other side of the outermost layer). I think the forces of gravity affecting the inner-most layer are un-equal (again, ever so slightly) due to the inner-most layer's position as slightly off-center.

 

[Doc Al]

I think the inner-most layer (which is a layer and therefore not the center/core) is being pulled in the direction of the outer-most layer because the force of gravity in that direction is greater (albeit slightly) that the force of gravity in the other (i.e., the other direction is the one from the inner-most layer, through the core, to the other side of the outermost layer). I think the forces of gravity affecting the inner-most layer are un-equal (again, ever so slightly) due to the inner-most layer's position as slightly off-center.

Sorry, I'm not understanding this. An outer layer exerts no gravitational force on the layers beneath it. The shell theorem, which you referred to before, tells us that.

 

[madchemist]

An outer layer exerts no gravitational force on the layers beneath it. The shell theorem, which you referred to before, tells us that.

I've heard this as well, but have also heard the exact opposite... "Note that the stuff within the shell exerts gravitational forces on the shell" [post #26]. What would happen to a bowling ball whose position is so off-center that it is almost touching the inner-side of the shell?

 

[Doc Al]

I've heard this as well, but have also heard the exact opposite... "Note that the stuff within the shell exerts gravitational forces on the shell" [post #26].

What makes you think that's the opposite? Does the shell exert a gravitational force on something within it? No. Does something within the shell exert a gravitational force on the shell? Sure.

What would happen to a bowling ball whose position is so off-center that it is almost touching the inner-side of the shell?

See above. (The shell will exert no gravitational force on it.)

 

[madchemist]

I think the inner-most layer (which is a layer and therefore not the center/core) is being pulled in the direction of the outer-most layer

To clarify, I think the components of the inner-most layer are being pulled in the direction of the outer-most layer, not the layer as a unit because the layer as a unit is in gravitational equilibrium.

Does the shell exert a gravitational force on something within it? No. Does something within the shell exert a gravitational force on the shell? Sure.

How do we harmonize this with Newton's Third Law?

 

[Baluncore]

To clarify, I think the components of the inner-most layer are being pulled in the direction of the outer-most layer, not the layer as a unit because the layer as a unit is in gravitational equilibrium.

You have that backwards. Think of the core as a single unit, but the shell as many particles.

The spherical core is not going anywhere, because it is being pulled equally in all directions by all the parts of the spherical shell above. That is the shell theorem.

For the analysis of gravitational forces, a sphere can be treated as a point mass, all located at the centre of the sphere.

Every separate particle of the outer shell is being individually attracted towards the centre of the spherical core. So it is the mass of the core, that brings the pressure onto the core itself, by attracting the particles above.

 

[Doc Al]

To clarify, I think the components of the inner-most layer are being pulled in the direction of the outer-most layer, not the layer as a unit because the layer as a unit is in gravitational equilibrium.

Nope. An outer layer exerts no gravitational force on anything within it. That includes every little piece of any inner layer.

How do we harmonize this with Newton's Third Law?

You must be careful when considering composite objects (like shells). The way to understand it is to consider the gravitational forces between each tiny element of mass, in which case Newton's third law applies easily. Then you add up those forces.

Consider a tiny mass element within a shell. Then consider the mass elements making up the shell. You'll find that each piece of the shell exerts a gravitational force on the mass inside -- and, in turn, the mass inside exerts an equal and opposite force on each piece of the shell. But the net force on the mass within the shell is zero -- that's the shell theorem. But there's still a non-zero net force on each element of the shell.

 

[madchemist]

An outer layer exerts no gravitational force on anything within it. That includes every little piece of any inner layer.

I assume you mean an outer layer exerts no gravitational force on anything at any position within it. If that's true, then why would the gravitational field vary linearly from 0 at the center to its full value at the surface (as I thought we already agreed)?

Consider a tiny mass element within a shell. Then consider the mass elements making up the shell. You'll find that each piece of the shell exerts a gravitational force on the mass inside -- and, in turn, the mass inside exerts an equal and opposite force on each piece of the shell. But the net force on the mass within the shell is zero -- that's the shell theorem

I agree, but only when the "tiny mass element within a shell" is at the exact center. I think that any deviation from the center, and the "tiny mass element" will prefer one side of the shell (the closer side) over the other (the farther side). Yes, I've heard over and over again that the position of the "tiny mass element" is irrelevant provided it lies inside the shell, but that doesn't make too much sense to me.

 

[madchemist]

Think of the core as a single unit, but the shell as many particles.

I agree except that I think even if the single unit core is one particle, it is still subject to endless rounds of division, but that's a different story.

 

[Doc Al]

I assume you mean an outer layer exerts no gravitational force on anything at any position within it. If that's true, then why would the gravitational field vary linearly from 0 at the center to its full value at the surface (as I thought we already agreed)?

I don't see your reasoning from one thing to the other. Actually do the math!

The gravitational field strength is proportional to the mass and inversely proportional to the radius squared. Since the mass contained within a radius ($r$) is proportional to $r^3$ (the volume), the gravitational field strength at any point a distance $r$ from the center (of our uniform density solid sphere) is proportional to $\frac{r^3}{r^2} = r$. Thus it varies linearly, from zero to its full value at the surface of the sphere.

I agree, but only when the "tiny mass element within a shell" is at the exact center. I think that any deviation from the center, and the "tiny mass element" will prefer one side of the shell (the closer side) over the other (the farther side). Yes, I've heard over and over again that the position of the "tiny mass element" is irrelevant provided it lies inside the shell, but that doesn't make too much sense to me.

You don't understand the shell theorem. Look it up! (Don't go by intuition, until you've done the math.)

 

[madchemist]

what is the effect, if any, on the shape and density distribution of such a sphere over time (e.g., billions of Earth years)?

In conclusion, the answer to the original question is...?

 

[Baluncore]

Yes, I've heard over and over again that the position of the "tiny mass element" is irrelevant provided it lies inside the shell, but that doesn't make too much sense to me.

Moving away from the centre, towards one side, the steradian on that side will contain less mass, but will be closer. The area of the shell, with its mass, is reduced by a square law, but the gravitational attraction is increased by the inverse square law. Those two effects elegantly cancel, so the attraction per steradian is constant for a particle anywhere inside a shell.

 

[Baluncore]

In conclusion, the answer to the original question is...?

Maybe you should repeat or identify the original question.

 

[madchemist]

Maybe you should repeat or identify the original question.

Now (finally) for the actual question... "What is the effect, if any, on the shape and density distribution of such a sphere over time (e.g., billions of Earth years)?" Again, the sphere in our model is solid, compressible (as is everything) and capable of density redistribution within the time stated and it is in empty space with no external forces acting on it.

See also, Post #'s 1 and 5. Baluncore offered his/her opinion in Post #2. Welcome further opinions.

 

[PeroK]

In conclusion, the answer to the original question is...?

That a) you're wrong; b) you're being wilfully contrary; c) trying to explain Newton's shell theorem to you seems like a waste of time.

Planets and stars are not hollow. The centre of a planet or star is under an enormous pressure. It only take some basic mathematics to confirm why this is so. To imagine otherwise is nonsense.

 

[madchemist]

Doc Al: Thank you for your patience. I think I get it now. Here is what I've learned (I think) so far about our model compressible solid uniform-density sphere in empty space with no external forces acting on it and capable of density redistribution.

1. The net gravitational field varies linearly from 0 at the center to its full value at the surface [post #'s 24, 39].

2. If and only if our model sphere is hollowed-out to form a shell however, then the net gravitational field varies from 0 everywhere inside the hollow of the shell [post #'s 26, 36] to its full value at the surface (the exterior of the shell). It follows then that if a uniform-density spherical mass is placed inside the hollow of such a shell at the center, then the net gravitational field between it and the shell as a unit is 0. The gravitational field between the mass and a particular component of the shell however will vary as a function of distance between them [post #’s 26, 36]. In other words, when the mass is off-center, then some components of the shell will feel a pulling in the direction of the mass with greater force than other components of the shell [post #36]. If the force pulling a particular component of the shell in the direction of the mass is greater than the component’s ability to resist, then the component will separate from the shell and the shell theorem will cease to apply.

 

[madchemist]

Planets and stars are not hollow. The centre of a planet or star is under an enormous pressure. It only take some basic mathematics to confirm why this is so. To imagine otherwise is nonsense.

As pointed out in post #3...

the question concerns a sphere, not a spheroid and not a planet.

 

[jbriggs444]

Does something within the shell exert a gravitational force on the shell? Sure.

It is possible to misunderstand this assertion.

Something within the shell exerts a net gravitational force of zero on the shell-as-a-whole. It exerts a force with a positive downward component on every little bit of the shell-as-individual-pieces. These two statements are not in conflict since "downward" is not a single direction for the shell-as-a-whole.

 

[jbriggs444]

If the force pulling a particular component of the shell in the direction of the mass is greater than the component’s ability to resist, then the component will separate from the shell and the shell theorem will cease to apply.

As I understand it, you are contemplating a small compact mass in the interior of a huge (and hugely massive) uniform spherical shell.

You have already understood that the compact mass is subject to zero net force from the shell. You may have also understood that the shell as a whole is subject to zero net force from the compact mass. The latter would be an immediate consequence of Newton's third law.

You are now considering what happens if the compact mass were to drift off center, close to the shell wall. It would then pull harder on the nearby portions of the shell and less so on the far away portions.

If we imagine the shell as being very thin then it might have very good compressive strength. But it would be weak against buckling because of being large (thus low curvature) and thin. The compact mass could result in local buckling and a resulting collapse of the shell. Or, in your scenario, it could out-gravitate the shell and suck off some material off of the inside surface with a similar catastrophic end result.

I do not consider either scenario likely. We do not need a compact mass on the inside in order for the shell to collapse.

If the shell is not extremely thin then we can consider the effect of the shell's gravity on itself. The outer layers would be pulled inward exactly as if the entire mass of the shell was a point at the center. The middle layers would be pulled inward with roughly half that much force. The innermost layer would not be pulled at all. Roughly speaking, every part of the spherical shell would be subject to an inward force about half as large as if all of the rest of the shell had collapsed to a point in the center. [This holds regardless of how thin the shell becomes except for exactly zero thickness. Exactly zero thickness is not physically achieveable and is an indeterminate case].

This means that the entire shell would have to resist compression. Like a ping pong ball subject to a large and uniform squeezing force. But on a planetary scale, no material is strong enough withstand compression due to self-gravitation strongly enough to avoid collapse. Under long time scales and large forces, all materials are fluid.

That's part of the definition of "planet" (if we were discussing planets).

 

[madchemist]

Under long time scales and large forces, all materials are fluid.

I agree.

If the shell is not extremely thin then we can consider the effect of the shell's gravity on itself. The outer layers would be pulled inward exactly as if the entire mass of the shell was a point at the center. The middle layers would be pulled inward with roughly half that much force. The innermost layer would not be pulled at all.

And the density distribution in such a three-layered shell under a long time scale would be (working from the exterior of the shell to the interior of the shell) dense, very dense and not very dense OR dense, very dense and extremely dense?

 

[jbriggs444]

I agree.And the density distribution in such a three-layered shell under a long time scale would be (working from the exterior of the shell to the interior of the shell) dense, very dense and not very dense OR dense, very dense and extremely dense?

You are expecting the material to become more dense proportional to pressure? I thought that we were talking about a solid, not an ideal gas.

But this is not a simple pressure. It is a large pressure in two horizontal dimensions and a low pressure in the vertical dimension. Together with an undetermined amount of shear stress. You might consider looking at the definition of the Cauchy Stress Tensor.

If the material is able to relax to relieve stress, you end up with a non-hollow sphere in hydrostatic equilibrium.

If the material is not able to relax to relieve stress, the situation is statically indeterminate.

Statically Indeterminate:

You might have an inner shell under pressure like an egg shell, supporting the [uncompressed] layers above with vertical compression. You might have an outer shell under pressure like an egg shell, supporting the [uncompressed] layers below with vertical tension. You might have all layers under uniform horizontal compression with vertical compression holding the top layer up and the bottom layer down.

Also, it is not just three layers. It is a continuum of layers extending from the top of the shell to bottom. I just simplified to three layers to give some intuitive appeal.