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

[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.

 

[madchemist]

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.

We're talking about a solid of initial uniform density but not necessarily the same type of matter. The matter per unit volume could be a mixture of iron, nickel, uranium or whatever. It starts out with a uniform density regardless of how long it stays that way. The question is in what arrangement does it end?

 

[jbriggs444]

We're talking about a solid of initial uniform density but not necessarily the same type of matter. The matter per unit volume could be a mixture of iron, nickel, uranium or whatever. It starts out with a uniform density regardless of how long it stays that way. The question is in what arrangement does it end?

Both theoretically and experimentally: a non-hollow sphere at hydrostatic equilibrium. If it is fluid enough to convect, it is fluid enough not to be stable as a hollow shell.

 

[madchemist]

Both theoretically and experimentally: a sphere at hydrostatic equilibrium.

I'm talking about before that while the shell is still a shell and the hollow is still a hollow.

 

[jbriggs444]

I'm talking about before that while the shell is still a shell and the hollow is still a hollow.

If it is fluid enough to convect, it is fluid enough not to be stable as a shell.

 

[madchemist]

If it is fluid enough to convect, it is fluid enough not to be stable as a shell.

In other words, at the moment before the shell collapses, would you expect to find heavy tightly-compacted stuff comprising the inner side of the shell and lighter less-compacted stuff comprising the outer side?

 

[jbriggs444]

In other words, at the moment before the shell collapses, would you expect to find heavy tightly-compacted stuff comprising the inner side of the shell and lighter less-compacted stuff comprising the outer side?

I would expect that the shell will collapse before any significant stratification has occurred and before any pressure gradient sufficient to produce a density gradient has occurred. In addition, I would never expect a shell to form in the first place.

 

[madchemist]

I would expect that the shell will collapse before any significant stratification has occurred and before any pressure gradient sufficient to produce a density gradient has occurred.

Even if the volume of the hollow of the shell is infinitesimal?

 

[Baluncore]

Even if the volume of the hollow of the shell is infinitesimal?

An infinitesimal hollow will be filled instantly as a result of the hydrostatic pressure.

 

[Baluncore]

Consider a sphere of water, the size of a planet. You sit in your boat on the surface, and lower a pressure gauge over the side. The measured hydrostatic pressure rises as you lower the gauge deeper.

I can understand how the pressure can stabilise to a maximum approaching the centre, but how might the pressure fall with increasing depth? Any reduction in pressure with depth would result in an instant implosion toward the centre that would correct the imbalance.

The water molecule at the centre would be attracting every other molecule towards it, resulting in a maximum hydrostatic pressure. At the same time, the molecule in the centre is being attracted equally in all directions at the same time, so the net gravitational force is totally canceled on the molecule at the centre.


Systems tend to move towards a point of minimum energy. For a viscous sphere subjected to self gravity, a central cavity cannot be part of a minimum energy solution. A minimum energy solution will have the highest density, and the highest pressure, at the centre.

No physicist is going to trash the fundamentals of their science to agree with an unrealistic hollow sphere model. Your belief system appears to be incompatible with the science of gravity and hydrostatics. You must either study and accept the physics, or walk away from science.