No gravity at center of earth= no pressure?

[pete20r2]

I was thinking about coriolis as I was writing that, but thought it irrelevant to this discussion, thanks for making people aware of the facts, though.

 

[I like Serena]

I was thinking about coriolis as I was writing that, but thought it irrelevant to this discussion, thanks for making people aware of the facts, though.

Note also that you would not fall through the center of the earth, but more in an elliptic orbit with the geometric center coinciding with the center of the earth.

 

[sophiecentaur]

Take the North South route and avoid coriolis. But still get crushed and roasted to death before you've gone any significant distance down there.

Bear in mind that air pressure doubles every 5.5km downwards, so the air you went through would soon no longer be a gas on the way down.

 

[Drakkith]

Ok folks! Threads over! Nothing to see here! *fireworks are going off from a burning fireworks stand behind me* Nothing to see here! Move along! *Rabid elephant ninjas on fire are running rampant through the town behind me* Move along! Nothing to see here!

 

[pete20r2]

It's a bit like that.. HEY EVERYONE LOOK OVER THERE IT'S A HIGGS BOSON HERE TO RUIN EVERYTHING

 

[sophiecentaur]

Note also that you would not fall through the center of the earth, but more in an elliptic orbit with the geometric center coinciding with the center of the earth.

Elliptic? Are you sure? The attraction law is not inverse square and elliptical orbits are generated by inverse square. It may not 'not' be but I'm not sure. in the detail. If you were a whizz at analysis, you could, no doubt derive the actual curve and show that it is (or not) an ellipse.

 

[Studiot]

This whole discussion so much hypothetical dreaming anyway because it assumes the Earth is an isolated body.

Of course, this is not so. Indeed our tides depend upon the fact that the Earth - Moon system revolves around a common centre of gravity that is not at the centre of the Earth.

Any object (taking the Earth as sphere for simplicity) placed at the centre of mass this sphere will not experience zero gravity for this reason.

 

[sophiecentaur]

However, the time constants involved in any other interaction are much greater - by factors of 28, at least. So we are allowed to go along with it, I think.

ALso, you could do the same experiment with a pebble and an asteroid and still get the same result (assuming the density were similar).

 

[I like Serena]

Elliptic? Are you sure? The attraction law is not inverse square and elliptical orbits are generated by inverse square. It may not 'not' be but I'm not sure. in the detail. If you were a whizz at analysis, you could, no doubt derive the actual curve and show that it is (or not) an ellipse.

Yes, I am!
The attraction law is linear instead of inverse square.
Still an ellipse, but the center of attraction is not in one of the focus points, but in the geometric center. Cool huh!

(Of course that would be if the Earth were an isolated body, which it isn't, as Studiot pointed out. )As for the proof, isn't it enough that a spring or a pendulum makes a harmonic motion?

 

[cjl]

Yes, I am!
The attraction law is linear instead of inverse square.
Still an ellipse, but the center of attraction is not in one of the focus points, but in the geometric center. Cool huh!

Really? An inverse attraction law (1/r) creates an elliptical orbit with the body at the geometric center of the ellipse? I would be surprised if it were the case, but I haven't worked out the math...

As for the proof, isn't it enough that a spring or a pendulum makes a harmonic motion?

I don't see how that is sufficient. It does show that the object will oscillate harmonically given a straight path through the center, but it doesn't really say anything about 2-D orbits...

 

[I like Serena]

Really? An inverse attraction law (1/r) creates an elliptical orbit with the body at the geometric center of the ellipse? I would be surprised if it were the case, but I haven't worked out the math...

Not inverse - linear (~ r).

I don't see how that is sufficient. It does show that the object will oscillate harmonically given a straight path through the center, but it doesn't really say anything about 2-D orbits...

Parametric equation of an ellipse is:
x = a cos phi
y = b sin phi

(Need I say more?)

 

[cjl]

Not inverse - linear (~ r)

Ahh, good point. I wasn't thinking there.

Parametric equation of an ellipse is:
x = a cos phi
y = b sin phi

Yes, but I'm still not sure that is sufficient...

(I might try to work this out on paper in a bit, just to convince myself)

 

[I like Serena]

Yes, but I'm still not sure that is sufficient...

(I might try to work this out on paper in a bit, just to convince myself)

I you go from north to south as sophiecentaur suggested, you'll get a nice sin wave (if the Earth would be an isolated body ).

Now let's add a deviation in a perpendicular direction.
Hey, that will make a sin wave a well, which is independent and possibly shifted in phase and with a different amplitude...

 

[cjl]

I you go from north to south as sophie_centaur suggested, you'll get a nice sin wave (if the Earth would be an isolated body ).

Now let's add a deviation in a perpendicular direction.
Hey, that will make a sin wave a well, which is independent and possibly shifted in phase and with a different amplitude...

That does seem to make sense. That's an interesting result...

 

[I like Serena]

That does seem to make sense. That's an interesting result...

Cool huh?

 

[sophiecentaur]

OMG, it's a lissajou figure! Durrrrrrr, how dumb of me!
As I like Serena says, it's got to be an ellipse around the centre - not a focus. The timing round the ellipse is different from the timing round a planetary orbit, though- it's symmetrical for a start. No Keppler Law at work there.

 

[Malibuguy]

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.

 

[cragar]

If an object was placed in the middle of the Earth in a hollow cavity, it would not experience any forces they would cancel. Gauss's law.

 

[sophiecentaur]

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.

You are just making that up out of your imagination, I'm afraid. In a spherical hole in the centre of a sphere with a spherically symmetrical mass distribution there would be no forces at all on your body except those of mutual attraction between the various parts of your own body. You can't contradict Mr Gauss' law.

 

[Malibuguy]

Let's say we have several large masses very close together. An object placed in the center would experience the gravitational forces of all of the masses. The object might not have a net movement if the the gravitational forces were balanced but it would feel the pull from each mass

 

[Malibuguy]

What if we divided the Earth into 4 parts of a sphere. Now we placed them very close together. In an instant the four parts would come together again due to gravity. But before the instant they did come
Together-an object placed in the center would feel the gravitational pull of the four large masses. Gravity doesn't cancel itself out.

 

[Malibuguy]

The gravity effect would be pulling inward on all directions. large planets are spherical In shape in part due togravity pulling together at the center! Otherwise spinning planets would fall apart due to centrifugal forces

 

[Malibuguy]

So there is a lot of pressure at the center of the earth

 

[Doc Al]

Let's say we have several large masses very close together. An object placed in the center would experience the gravitational forces of all of the masses. The object might not have a net movement if the the gravitational forces were balanced but it would feel the pull from each mass

Assuming the masses were symmetrically placed about the center, the gravitational field at the center would be zero. So the object wouldn't 'feel' anything. For a spherically symmetric shell of mass one can show that the field is zero everywhere inside the shell.

What if we divided the Earth into 4 parts of a sphere. Now we placed them very close together. In an instant the four parts would come together again due to gravity. But before the instant they did come
Together-an object placed in the center would feel the gravitational pull of the four large masses. Gravity doesn't cancel itself out.

Sorry, that's wrong. Sure, the pieces would attract each other, but they would exert no net gravitational pull on a object placed at the center.

The gravity effect would be pulling inward on all directions. large planets are spherical In shape in part due togravity pulling together at the center! Otherwise spinning planets would fall apart due to centrifugal forces

Again, the pull of various parts of a planet on each other is not in question.

So there is a lot of pressure at the center of the earth

That's true.

 

[Studiot]

Originally Posted by Malibuguy
So there is a lot of pressure at the center of the Earth

That's true.

Are you quite sure?

 

[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.