No gravity at center of earth= no pressure?

[sophiecentaur]

Are you quite sure?

Absolutely. The Earth can be regarded pretty much as liquid, in the long term (even the lithosphere) and the mutual attraction of all the separate bits creates 'hydrostatic' pressure which increases and increases as you get nearer the centre.

The numbers involved are gzillions!

 

[Studiot]

Well I am not so sure.

Let us examine the situation more closely.

Gravity exerts a body force on matter. It can be exerted on a single point mass. It does not directly exert pressure.

Pressure is not a force it is a surface action. Its action is such that two or more bodies in contact experience a contact force. Pressure is numerically equal to this force divided by the contact area. It requires at least two masses to achieve this action.

So let us look first at the surface of the earth.
There is most definitely a force due to gravity here, but does any mass at the surface of the Earth experience pressure?

Now let us delve into the interior. As I understand it, a mass somewhere in the interior experiences the above mentioned contact force from the material 'above' it as pressure.

Now consider a point mass just to the left of the centre of gravity.
Because it is off centre there is more mass to the right than to the left, yet it experiences a pressure due to the mass above it ie further to the left, not due to the mass to its right.
The further left we go the greater this discrepancy.

I thought that somewhere in this thread a small cavity was mooted at the C of G. A point mass in such a cavity will experience no pressure, since there is no contact with another mass.
It will however experience a balance of holding force as previously discussed.

 

[chingel]

I don't understand, is it that the gravitational waves actually cancel each other out and interfere with each other, or is it that every atom is pulled in all directions with equal force so that the forces cancel each other out?

 

[I like Serena]

So let us look first at the surface of the earth.
There is most definitely a force due to gravity here, but does any mass at the surface of the Earth experience pressure?

Yes, a mass at the surface of the Earth does experience pressure.
It is air pressure which is already quite impressive being the equivalent of a column of 10 meters of water.

As we go down, and add a column of Earth to the column of air, the pressure goes up very quickly.

 

[sophiecentaur]

Well I am not so sure.

Let us examine the situation more closely.

Gravity exerts a body force on matter. It can be exerted on a single point mass. It does not directly exert pressure.

Pressure is not a force it is a surface action. Its action is such that two or more bodies in contact experience a contact force. Pressure is numerically equal to this force divided by the contact area. It requires at least two masses to achieve this action.

So let us look first at the surface of the earth.
There is most definitely a force due to gravity here, but does any mass at the surface of the Earth experience pressure?

Now let us delve into the interior. As I understand it, a mass somewhere in the interior experiences the above mentioned contact force from the material 'above' it as pressure.

Now consider a point mass just to the left of the centre of gravity.
Because it is off centre there is more mass to the right than to the left, yet it experiences a pressure due to the mass above it ie further to the left, not due to the mass to its right.
The further left we go the greater this discrepancy.

I thought that somewhere in this thread a small cavity was mooted at the C of G. A point mass in such a cavity will experience no pressure, since there is no contact with another mass.
It will however experience a balance of holding force as previously discussed.

You're getting carried away with this one. How come the conditions at the centre of a star are sufficient to produce fusion if pressure somehow 'cancels itself out'?
If the substance of an object can flow (which it can, in the case of the Earth) then pressure is transferred. The outer part of the Earth is not like an arch, protecting the inner bits from any pressure that exists. The pressure at the bottom of a core, drilled through to near the centre will be due to the gravitational attraction of all the parts to the centre added up -i.e. very high.

 

[Malibuguy]

To those who say that there is no gravity at the
Center of the earth, consider this :
Then a 1 meter spherical volume at the exact center of the Earth would have no gravity just like a 1 meter sphere in far reaches of outer space.

 

[Doc Al]

To those who say that there is no gravity at the
Center of the earth, consider this :
Then a 1 meter spherical volume at the exact center of the Earth would have no gravity just like a 1 meter sphere in far reaches of outer space.

And so?

 

[Studiot]

Hello ILS,
talking about the atmosphere is a bit of a cop out.
You can simply consider the Earth+atmosphere as a variable density sphere and move my question to its surface.
The question remains the same.

Consider a surface point mass element.
There is oodles of mass under it/adjacent to it.
It experiences the gravitational attraction of all this mass.
Nevertheless all this mass exerts no pressure on it.

Hello SC

Yes I am aware of the limiting process in continuum analysis that defines the pressure at a point and the resolution of the stress tensor into hydrostatic and deviatoric stresses.
So in the limit, if there is material all the way to the C of G, the conditions for stress and therefore pressure exist.

No I did not say anything about arches which sustain significant stress or shielding, which does not.

I think you are both missing the points.

These are that there are substantial differences between force and pressure.

One of these is that you can exert a force, but not a pressure on a single point mass.

(which is why I said there is force, but zero pressure on a point mass in a cavity.)

The other I obviously didn't explain very well and perhaps led to your comment about the arch. Sorry I was trying to do without a diagram.

Anyway to try again with reference to the attached diagram.

Take a solid sphere with centre C. Consider an differential element E as shown.

All the material in the sphere experiences a gravitational attraction, directed towards the centre.

Because of this the the material in column AE presses on E with a force equal given by Newton's law.

The material in BE presses back with an equal and opposite force, again by another of Newton's laws.

If this were not so then the element would be in motion.

Now the important point is that the force exerted is determined solely by the material in column AE, which is smaller than the column BE and therefore exerts a smaller gravitational attraction on E

It is this force whcih , when divided by the cross section of E, yields the pressure on E.

And again this shows that this pressure cannot exist without at least two bodies (in the limit as E becomes infinitesimal) in contact viz AE and BE.

 

[Doc Al]

(which is why I said there is force, but zero pressure on a point mass in a cavity.)

In a hollowed out cavity at the center of the Earth (making the usual assumption of spherical symmetry) a point mass would feel no gravitational force and certainly no pressure (since it's not supporting the weight of the mass of the earth, the cavity is).

 

[Studiot]

In a hollowed out cavity at the center of the Earth (making the usual assumption of spherical symmetry) a point mass would feel no gravitational force and certainly no pressure (since it's not supporting the weight of the mass of the earth, the cavity is).

I think we are converging.

The resultant would certainly be zero, but I still contend there would be attraction from every point mass in the non hollowed out part of the Earth.

 

[Malibuguy]

In this spherical chamber we now place an instrument which can measure gravitational forces. Are we saying that the instrument would read zero gravity?

(Someone was mentioning something about atmosphere. A planet needs to have a certain amount of gravity in order to maintain an atmosphere. Otherwise it literally will float off into space.)

 

[BobG]

If there were a hollow center in the earth, then an object placed in the center would experience the gravity of the surrounding mass of the earth. Gravity would be pulling from all around. Sort of like someone pulling on your head and your feet and pulling on your right arm and your left arm.

Well, this certainly validates DocAl's concern about the semantics being clearer by just saying the strength of the gravity field at the center of the Earth is zero.

The net force of gravity on each molecule of your body would be zero. So it would not be like someone pulling on your head, feet, arms, etc. Every piece of your body is feeling the same net force - zero.

You can't draw a diagram with your body surrounded by lines of force going outward. Instead, your diagram would have a whole bunch of lines all through your body going all kinds of different directions. Except every single of those lines you drew would cancel each other out, essentially giving you a gravity field with a strength of zero.

 

[Studiot]

You can't draw a diagram with your body surrounded by lines of force going outward

You have to be very careful with analogies here.

Which body would you rather inhabit?

One that has zero forces applied?

or

One that has zero net force applied by two wild horses pulling your arms equally in opposite directions?

The 'feeling' effect only applies to point masses. All others would be rent asunder or crushed.

 

[Doc Al]

The resultant would certainly be zero, but I still contend there would be attraction from every point mass in the non hollowed out part of the Earth.

No, Newton showed in his famous shell theorem that the field everywhere within the hollowed out shell is exactly zero. There will be gravity from every piece of mass outside the hollow, but it adds up to zero everywhere within the hollow.

 

[Doc Al]

In this spherical chamber we now place an instrument which can measure gravitational forces. Are we saying that the instrument would read zero gravity?

Yes.

 

[Doc Al]

You have to be very careful with analogies here.

Which body would you rather inhabit?

One that has zero forces applied?

or

One that has zero net force applied by two wild horses pulling your arms equally in opposite directions?

The 'feeling' effect only applies to point masses. All others would be rent asunder or crushed.

A body is made up of point masses. If the gravitational force on every point mass is exactly zero, there is no crushing or tension exerted on the body. Your analogy would work if gravity pulled one way on one arm and another way on the other arm--but it doesn't pull at all on either arm! Or any other part.

 

[Studiot]

A body is made up of point masses. If the gravitational force on every point mass is exactly zero, there is no crushing or tension exerted on the body. Your analogy would work if gravity pulled one way on one arm and another way on the other arm--but it doesn't pull at all on either arm! Or any other part.

But this is plainly impossible.

We have agreed that there is only one point where the gravity resultant is zero.

So your assembly of point masses (even as few as two) is impossible.

 

[Doc Al]

But this is plainly impossible.

We have agreed that there is only one point where the gravity resultant is zero.

No, within a hollowed out cavity at the center of a spherically symmetric mass all points have zero gravity. Any body placed within that cavity would feel zero force from the mass outside.

 

[Studiot]

No, within a hollowed out cavity at the center of a spherically symmetric mass all points have zero gravity. Any body placed within that cavity would feel zero force from the mass outside.

You are quite right there, I was getting confused between solid Earth's and hollow ones.
Thank you for pointing that out.

 

[Malibuguy]

Then how about a region right next to the center? Would the gravity it
Feels be very smaller or larger or the same as on the surface of the earth

 

[Doc Al]

Then how about a region right next to the center? Would the gravity it
Feels be very smaller or larger or the same as on the surface of the earth

If that nearby point is within the hollow cavity, gravity is still zero as discussed. If within the solid earth, then much smaller than the strength of gravity on the Earth's surface.

 

[sophiecentaur]

"The gravity it Feels"?
How is this object going to feel the gravity? It can measure it, though.
1. The object inside a hollow spherical cavity at the dead centre. (Strong enough etc.)
By observing its motion relative to the sides of the spherical hole it's in: it will not observe acceleration in any direction. If it were 'suspended' on a string from a point on the inner surface of the space, the string would be slack.

2. The object is somewhere near the centre of a solid Earth (in a rigid cavity).
It will weigh mgx/r where x is the distance away from the centre and r is the Earth's radius

3. The object is in a thin tube through the centre of an otherwise solid Earth.
It will observe that it exhibits simple harmonic motion - demonstrating that the force towards the centre is proportional to its distance away from the centre.

 

[Malibuguy]

So then gravity gets larger as you go away from the center of the earth? When you hit the surface gravity gets smaller as you go away from surface.

 

[Malibuguy]

This idea that gravity gets smaller when you approach the center of the Earth until it goes to zero makes no sense at all.

Do you weigh more or less when you are on top of mount Everest cOmpared to the bottom of the grand canyon?

 

[sophiecentaur]

This idea that gravity gets smaller when you approach the center of the Earth until it goes to zero makes no sense at all.

Do you weigh more or less when you are on top of mount Everest cOmpared to the bottom of the grand canyon?

This is just not a matter for argument. The sums tell you the whole story. I sometimes wonder why people who don't actually 'know' about stuff like this have such definite opinions.
The top of Mount Everest is not the same as an Earth with that much bigger radius! If you Were on the surface of a bigger planer of the same density then you Would weigh more. If you were on the surface of a smaller planet you would weigh less. When you are down a hole, the whole of the 'shell' above you makes no contribution to your weight at all. Weight is proportional to distance from the centre - until you reach the surface.

Clearly, when you leave the surface, that well known Inverse Square Law kicks in and your weight drops accordingly.

 

[Malibuguy]

It's not just the size it's really the density that determines gravitational force. You could have a large planet made out of a low density material and it's gravity would be much smaller than a very small planet made out if say plutonium a very dense material. The
Smaller planet would have a greater gravitational pull.

But the idea that gravity gets smaller once you move from the surface in either direction (towards space or towards the center) doesn't make sense. Gravity pulls toward the center of a mass. If you are beneath the surface it makes sense for gravity to pull the same or more. Not less

 

[Malibuguy]

Dear Sophiecentaur,

if you weigh yourself on the top of Mount Everest, not leaving the surface of the earth, guess what! you would weigh a little less that at sea level.

 

[Doc Al]

But the idea that gravity gets smaller once you move from the surface in either direction (towards space or towards the center) doesn't make sense.

So if you were to move an object thousands of miles into space, you think the gravitational pull would be the same?

Gravity pulls toward the center of a mass. If you are beneath the surface it makes sense for gravity to pull the same or more. Not less

As far as what happens when you go below the surface, you actually need to do the calculation. 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. (And I hope it's obvious that as you keep going above the surface, the gravitational field gets smaller.)

 

[rcgldr]

A summary of what I've seen posted.

Although gravity is zero at the center of the earth, pressure will be greatest at the center of the Earth unless there is some solid shell surrounding the center of the Earth that is supporting some or all of the weight outside that shell, in which case the pressure inside the shell could be anything.

If there is a shell supporting all the weight outside the shell, and the inside of the shell is a vacuum, then depending on the Earth's mass distribution (such as a uniform sphere, or a sphere made of continuous uniform spherical shells), there would be no gravity at any point within that hollow shell.

If an object were placed inside inside the hollow shell with zero gravity, then the object would create it's own gravitation field. I'm not sure about interaction between the object and the mass outside the shell, other than the object's gravitational field would increase the pressure on the shell.

 

[Malibuguy]

Is the core of the Earth less dense than the outer layers? if the core is less dense then the gravity field would be less. If the core were more dense then the gravity field would be greater. If the same density then the same.

If there was a person who was placed inbetween two large but symmetrically opposing g forces they would not move . But they would experience a greater field of gravitation compared to being weightless in outer space far away form any masses.

Gravity does not cancel itself out to zero.

so if you are in the middle of the Earth you experience a different g field than in the middle of space with no masses around you.