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

In summary, the conversation discusses the effect of time on the shape and density distribution of a uniform density spherical planet. The experts agree that over time, the densest material will move downward, resulting in a dense core, a medium density mantle, and a low density outer layer. The question is posed of what mechanism could make the planet hollow with all the mass floating above. The experts suggest using Newton's shell theorem and considering the hydrostatic pressure and gravity within the planet. The conversation concludes with a further discussion on the terms and concepts used in the conversation.
  • #1
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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)?
 
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  • #2
madchemist said:
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 ?
 
  • #3
Baluncore said:
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.
 
  • #4
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 ...
Baluncore said:
What mechanism do you expect could make it hollow with all that mass floating above ?
 
  • #5
Baluncore said:
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.

Baluncore said:
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/
 
  • #6
Baluncore said:
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...
 
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  • #7
madchemist said:
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.
 
  • #8
Having difficulty following you. Coud you please rephrase?
 
  • #10
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?
 
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  • #11
madchemist said:
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 .
madchemist said:
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.
 
  • #12
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"?
 
  • #13
madchemist said:
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.
 
  • #14
I understand, "NO," but you lost me on your subsequent qualification of "NO."
 
  • #15
What do you mean by the words "varies linearly" ?
 
  • #16
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.
 
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  • #17
madchemist said:
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, ; 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.
 
  • #18
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.
 
  • #19
madchemist said:
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.

madchemist said:
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.
 
  • #21
madchemist said:
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.
 
  • #23
madchemist said:
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.
 
  • #24
madchemist said:
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.
 
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  • #25
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.

Doc Al said:
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?

Doc Al said:
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?
 
  • #26
madchemist said:
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.)
madchemist said:
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.
madchemist said:
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.
 
  • #27
Doc Al said:
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."

Doc Al said:
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.
 
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  • #28
madchemist said:
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?
 
  • #29
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).
 
  • #30
Doc Al said:
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.
 
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  • #31
madchemist said:
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.
 
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  • #32
Doc Al said:
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?
 
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  • #33
madchemist said:
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.
madchemist said:
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.)
 
  • #34
madchemist said:
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.

Doc Al said:
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?
 
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  • #35
madchemist said:
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.
 

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