No gravity at center of earth= no pressure?

[Markel]

Hey, Just a general question.

So, I've just learned that the force of Net gravity anywhere in a shell is 0. And the Earth can be thought of as a continuation of shells- like an onion or something. And at the center the net gravity is 0.

But I thought the immense force of gravity is what caused nuclear fusion in the center of stars? Does no force of gravity mean no pressure? After all P= F/A. Could someone help me understand this?

Thanks

 

[Doc Al]

Does no force of gravity mean no pressure?

No, far from it. While the gravitational force at the center might be zero, you still have the enormous weight of the layers of Earth above you crushing down.

 

[Markel]

That's what I thought intuitively. Pressure in a liquid is dependent on height. But it's also dependent on g, and if g is 0...

Sorry if my question sounds dumb, but I don't quite understand.
You say you have the "enormous weight of the layers of Earth above you crushing down"
But crushing down with what force?? It can't be the force of gravity because it's 0. at the center.

Thanks for replying so fast

 

[Doc Al]

That's what I thought intuitively. Pressure in a liquid is dependent on height. But it's also dependent on g, and if g is 0...

But the value of g is only zero at the center of the earth. As you move away from the center, the value of g increases until it reaches its maximum value at the Earth's surface.

Each layer of Earth must support the layers above it. Only the very center is 'weightless'.

 

[Markel]

But the value of g is only zero at the center of the earth. As you move away from the center, the value of g increases until it reaches its maximum value at the Earth's surface.

Each layer of Earth must support the layers above it. Only the very center is 'weightless'.

Hmm, that's a good way of looking at it. If the sun is in static equilibrium (at least radialy, not shrinking or growing) then atoms near the center must supply the normal force to counteract that massive column of weight above it.

And perhaps only one lucky atom, (or maybe better electron if it's point like) at the dead center of the sun will feel no pressure at all.

Do I have the right Idea now?

 

[Doc Al]

Hmm, that's a good way of looking at it. If the sun is in static equilibrium (at least radialy, not shrinking or growing) then atoms near the center must supply the normal force to counteract that massive column of weight above it.

Exactly.

And perhaps only one lucky atom, (or maybe better electron if it's point like) at the dead center of the sun will feel no pressure at all.

I'd think that anything at the dead center would feel maximum pressure, all things being equal. (Not sure it's productive to talk about an individual atom or electron, though.)

Do I have the right Idea now?

I think you've got it.

 

[Markel]

I'd think that anything at the dead center would feel maximum pressure, all things being equal.

Yeah I see it now.
Cool, thanks so much for helping me figure that out!

 

[Lsos]

I have a hard time wrapping my head around it, too.

Imagine putting a huge, heavy iron block on top of a table. That table has to support the weight of this block.

What if we made the table out very light material? Let's make it out of unobtanium, which has a weight of 0. Then we have a weightless table, which is awesome. Unfortunately, that iron block on top of it still weighs the same...and the table still has to support it.

Same with that single atom in the very center of the earth. It might not weigh anything, but everything on top of it still does, and everything on top of it still wants to go where that single atom is.

 

[Barwick]

That's what I thought intuitively. Pressure in a liquid is dependent on height. But it's also dependent on g, and if g is 0...

Sorry if my question sounds dumb, but I don't quite understand.
You say you have the "enormous weight of the layers of Earth above you crushing down"
But crushing down with what force?? It can't be the force of gravity because it's 0. at the center.

Thanks for replying so fast

Think of it this way: Imagine the Earth and moon were 1 foot apart from each other, and you were trapped between the two. Same effect as being at the center of the earth, just the "earth and moon" you're trapped between happens to be umpteen billions of tons of Earth on "either side" of you.

 

[brainstorm]

My understanding is that the reason gravity is zero at the center of a sphere is because the mass cancels its own gravity out in the various directions. This would mean that halfway to the center of the Earth, the mass above you would cancel out the gravity of half the mass below you. Therefore, I would think the gravity would keep decreasing until you get to the center. So if net gravity keeps decreasing as you go deeper, why would pressure continue to increase until the center? If anything I would think there would be a hollow void due to the centrifuge effect caused by rotation.

 

[Subductionzon]

Gravity at the center of a sphere is zero, but its pressure is not. Pressure adds from everything above it. Otherwise the pressure of air at sea level would be zero instead of 15 lbs per square inch.

 

[seto6]

no... no force =/= no pressure...

think of going at center of earth.. all the mass on Earth is concentrated at center.. so there will be pressure

 

[cjl]

My understanding is that the reason gravity is zero at the center of a sphere is because the mass cancels its own gravity out in the various directions. This would mean that halfway to the center of the Earth, the mass above you would cancel out the gravity of half the mass below you. Therefore, I would think the gravity would keep decreasing until you get to the center. So if net gravity keeps decreasing as you go deeper, why would pressure continue to increase until the center? If anything I would think there would be a hollow void due to the centrifuge effect caused by rotation.

Nope.

Halfway to the center of the earth, the mass above you cancels out none of the gravity of the mass below you. Instead, it just cancels out the mass that is farther from the center than you are, but on the opposite side of the earth. Gravity pulls downward (albeit with decreasing strength) all the way to the core, and because of this, everything is trying to get to the core. Even though there is no gravity at the core itself, the entire Earth is trying to squish itself towards this point, so the pressure is enormous to counter this force.

 

[Drakkith]

This is the way i understand it. I could be wrong.

If i were 1 mile under the earth, all the mass above me would be exerting a gravitational pull on me just like the rest of the mass below me and to the sides. Now, the gravity of that 1 mile of material would make me "lighter" in weight than i would be standing closer to the center of the earth, because it is no longer pulling me down with the rest of the earth. However, that 1 mile of Earth above me, IS being pulled down by the rest of the earth, and since it is being pulled down it will exert pressure and i would be crushed.

If i were to be at the center of the earth, i would "weigh" nothing, because there is nothing to pull me predominately in one direction like what happens to us here at the surface. The gravity exerted by all the mass surrounding me effectively cancels out each other in sense. But again, all the mass is attracting itself, being pulled inwards and exerting enormous pressure at the center.

 

[Lsos]

A car in a car crusher has a net force on it of 0, because the forces cancel out. It will nevertheless get crushed.

 

[pallidin]

Let there be no mistake: Gravity DOES NOT cease to exist just because you are in a hollow center area of a massive sphere.
Rather, instead of being pulled down as on the Earth's surface, you are being pulled up in all directions.
It is only the "effect" of gravity that is altered in different scenarios. Gravity itself continues to exist.
To my knowledge, "destructive interference" of gravity has not yet been proven.

 

[Doc Al]

Let there be no mistake: Gravity DOES NOT cease to exist just because you are in a hollow center area of a massive sphere.
Rather, instead of being pulled down as on the Earth's surface, you are being pulled up in all directions.
It is only the "effect" of gravity that is altered in different scenarios. Gravity itself continues to exist.
To my knowledge, "destructive interference" of gravity has not yet been proven.

When you are inside a uniform spherical shell of mass the gravitational attraction on you from that shell is exactly zero. The net effect is that you feel no gravitational force at all. There is zero gravitational field (at least from the mass of the shell) at all points within the shell.

 

[brainstorm]

To my knowledge, "destructive interference" of gravity has not yet been proven.

What about rising and falling tides?

 

[Markel]

A car in a car crusher has a net force on it of 0, because the forces cancel out. It will nevertheless get crushed.

Good analogy.


Yeah thanks everyone for your replies. I understand it now and it totally makes sense. I just didn't really have the right way to look at it in the beginning.

 

[Drakkith]

When you are inside a uniform spherical shell of mass the gravitational attraction on you from that shell is exactly zero. The net effect is that you feel no gravitational force at all. There is zero gravitational field (at least from the mass of the shell) at all points within the shell.

Are you saying that there is no gravitational force on you at all, or is it that it all balances out and because of that you feel no force overall?

 

[Doc Al]

Are you saying that there is no gravitational force on you at all, or is it that it all balances out and because of that you feel no force overall?

It depends on what you mean by that second statement. I prefer the first statement: there is no gravitational force on you at all. The second statement could be misinterpreted as implying that there are forces pulling you in various directions that happen to add to zero, thus creating some tension in your body. That's not the case.

The way I like to think of it is like this. The gravitational force on some mass equals the gravitational field strength times the mass. Within that spherical shell the gravitational field is zero everywhere.

 

[Drakkith]

It depends on what you mean by that second statement. I prefer the first statement: there is no gravitational force on you at all. The second statement could be misinterpreted as implying that there are forces pulling you in various directions that happen to add to zero, thus creating some tension in your body. That's not the case.

The way I like to think of it is like this. The gravitational force on some mass equals the gravitational field strength times the mass. Within that spherical shell the gravitational field is zero everywhere.

I don't think that's accurate though. And i thought that my second statement was correct. The gravity would pull you in all directions, but you wouldn't feel any net force in one direction since the pull was equal in every directions.

 

[Doc Al]

I don't think that's accurate though. And i thought that my second statement was correct. The gravity would pull you in all directions, but you wouldn't feel any net force in one direction since the pull was equal in every directions.

We're getting into semantics, perhaps, but I think saying that the force is zero is more accurate than merely saying that the net force is zero. If two people pull your arms in opposite directions, the net force on you is zero but there are surely forces acting on you. Gravity within a shell is not like that.

 

[OmCheeto]

You could reduce it to a one dimensional problem for clarity. Simply take a core* from one side of the Earth to the other.

000|000

The far left '0' is attracted to all of the '0's on the right, so it, along with the far right '0' experiences the most gravitational force.
But because there are no 0's to it's left, it feels no pressure, except perhaps on it's feet.
When you get to the second 0 from the left, it will experience a pressure because it is being crushed between the far left 0 and all the rest.
When you get to the center, the 0 has been crushed flat into a | because of the attraction of the 0's on each side.
But it feels no gravitational force because on each side, there is an equal and opposite 0 that cancels the pull.

Hence, there is no gravity at the center.

One could construct a mathematical model of the above core section to show that the pressure will be highest at the midpoint, while at the same time there is no gravity.

Pressure being defined as the force of the particles to your left and right tending to squash you.



*core as in ice core, not the Earth's core.

 

[pallidin]

Gravity DOES NOT cease to exist with special arrangements of mass.
The EFFECT is altered. Not gravity.

 

[D H]

We're getting into semantics, perhaps, but I think saying that the force is zero is more accurate than merely saying that the net force is zero. If two people pull your arms in opposite directions, the net force on you is zero but there are surely forces acting on you. Gravity within a shell is not like that.

Even though we are playing semantics here, I am siding with Drakkith. The analogy to two people pulling on your arms is not quite apt. That analogy is better suited to describing the tidal forces felt when one is orbiting very close to a neutron star. Inside a spherical shell the net force is zero everywhere, so how can there be any tidal forces?

I am siding with Drakkith for three reasons. First, that the net force anywhere inside a uniform spherical shell is identically zero is exactly how Newton proved his shell theorem.

Second, imagine some shell of matter that is not a uniform spherical shell. The gravitational force on some test particle inside this shell can be computed by integrating the force generated by each infinitesimal point mass dm that comprises the shell. Now imagine some parametric way of distorting this shell into uniform spherical shell. In the limit that the distribution becomes that of a uniform spherical shell the net force will become identically zero everywhere inside the shell.

Thirdly, even though the gravitational force inside a spherical shell is indistinguishable from the gravitational force due to the shell at some great distance away from the shell (they are both zero), the gravitational potential inside a spherical shell differs from that due to the shell at some great distance away from the shell. Setting the potential at infinite distance to zero, the potential inside the shell is constant but it is not zero.

 

[Doc Al]

Even though we are playing semantics here, I am siding with Drakkith. The analogy to two people pulling on your arms is not quite apt.

Of course it's not apt--that was my point! It was meant as a illustration of how 'net force = 0' does not imply no force. (I didn't mean to imply that Drakkith meant it that way, just that there's some ambiguity in the terminology.)

Perhaps I was overreacting to the use of the term 'net force'. To say that the net gravitational force on some object within a uniform mass shell is zero, while true, is not strong enough. Maybe 'the net gravitational force on any infinitesimal element of mass is identically zero' would be more accurate. I guess that's what was meant by 'the net gravitational force is everywhere zero'.

I still prefer something like 'the gravitational field is everywhere zero'.

I am siding with Drakkith for three reasons.

I agree with all of your 'reasons', of course, but I don't see them as relevant to the semantic issue at hand.

Nonetheless, I think I see your point. (And I'm probably arguing against a view that no one is maintaining. :uhh:)

 

[vardhan_harsh]

see..the force of gravity does not completely vanish inside the earth.
take it like this
the particles of dust ball will b atracted to each other if no pressure acts inside them.this will continue until all the partcles come close enough n shrink to point size.
this is bcz gravity is force which goes on increasing if the systemit is acting upon yields to it.
now just apply the above philos to d situation of a star.
if doubtful pls discus.i m also a bit confused.. :)

 

[Drakkith]

see..the force of gravity does not completely vanish inside the earth.
take it like this
the particles of dust ball will b atracted to each other if no pressure acts inside them.this will continue until all the partcles come close enough n shrink to point size.
this is bcz gravity is force which goes on increasing if the systemit is acting upon yields to it.
now just apply the above philos to d situation of a star.
if doubtful pls discus.i m also a bit confused.. :)

Umm, what? Besides the complete lack of effort in grammar and spelling here, this doesn't even make any sense.

 

[sophiecentaur]

Hmm, that's a good way of looking at it. If the sun is in static equilibrium (at least radialy, not shrinking or growing) then atoms near the center must supply the normal force to counteract that massive column of weight above it.

And perhaps only one lucky atom, (or maybe better electron if it's point like) at the dead center of the sun will feel no pressure at all.

Do I have the right Idea now?

This is not right, I'm afraid. The pressure at the centre will not (cannot) suddenly disappear at the centre. The pressure just gets higher and higher as you get nearer the centre. This should not be confused with the (correct) statement that the gravitational force towards the centre goes to zero in the centre. All the stuff above this central position is still transmitting pressure to the centre. The forces will all be inwards on this 'lucky atom'.

 

[Femme_physics]

Pfffff, you guys clearly haven't seen Journey to the Center of the Earth...

 

[sophiecentaur]

I have to admit that I had completely forgotten about the hollow middle.

 

[pete20r2]

IF THE MIDDLE WAS HOLLOW, and vented to the surface;
1. There would be increased pressure, going up in the atmosphere decreases pressure and going down increases, until you get to the center, passing through means your going up again and it decreases.
2. Due the the EVEN DISTRIBUTION OF MASS around you, every single mass (atom, particle, whatever you want to call it), is pulled in every direction evenly. You do not feel the force because it is balanced at every point in your body.

In a vacuum (good start), if there was a hole in the earth, through to the other side and I jumped down it, this is what would happen:
1. I would accelerate down until I crossed half way, this acceleration would be DECREASING as I approached the center and would be 0 as I crossed.
2. I would continue past the center, since I have a lot of speed and there is little acceleration.
3. The acceleration would now build up in the opposite direction to my velocity and I would slow to a stop with my head level to the ground on the other side of the earth, with my feet in the air.
4. I would fall and do it all again.

In air:
Same as before but with losses, I would certainly not make it to the other side, but I would converge on the center. As I get slower and slower every time I pass the center though, my rate of convergence decreases. This is because the acceleration at the center is 0 so there is nothing to keep me there and I would theoretically forever be oscillating about the center.

 

[sophiecentaur]

Perhaps you could explain why pressure necessarily disappears when the gravity does. Take a pressurised spacecraft into deep space. Will it lose it's pressure?
We're talking two entirely different effects.

 

[I like Serena]

IF THE MIDDLE WAS HOLLOW, and vented to the surface;

[...]

In a vacuum (good start), if there was a hole in the earth, through to the other side and I jumped down it, this is what would happen:
1. I would accelerate down until I crossed half way, this acceleration would be DECREASING as I approached the center and would be 0 as I crossed.

I hope your tunnel is in a curve, since the Coriolis force would otherwise be rather nasty! :)