# Gravity / pressure as we move towards centre of Earth

1. Jul 9, 2014

### Gerinski

I may have some confusion with the terms "pressure" and "gravity" here.
If hypothetically we could sink towards the centre of the Earth, say with our feet pointing towards it, as we start to sink intuitively pressure on top of us would gradually increase, as it does when we sink into the ocean dephts. As for the force of gravity, when we are at the surface all of it pulls us towards the center, but it would gradually reduce, and once in the very centre we would not feel any gravity at all, since an equal amount of mass would be pulling us in every direction towards the outside (towards the surface).
Now, I intuitively attribute pressure to the force of gravity, when in the depth of the ocean we have the pressure of the water on top because it is being pulled towards the centre by the Earth's gravity. If the net effect of gravity at the center is zero, it seems that pressure should also be zero, since the stuff just around us is being pulled away towards the outside as much as is being pulled towards us in the opposite direction. There's no reason it should still be pulled towards the center since at the center there is no mass, the mass is now in the sphere surrounding us so the stuff is being pulled outwards equally in every direction.
It seems that as we get deeper and deeper, we would experience an increase of pressure but a decrease in gravity, and at some point in the voyage things would start to reverse and pressure would start decreasing until becoming zero at the center. And yet I guess that the center is the most dense point, so what is wrong with my reasoning?

2. Jul 9, 2014

### A.T.

You must be more precise, and distinguish between the direct effect of gravity on you and the indirect effect on you caused by gravity on the other masses.

3. Jul 9, 2014

### D H

Staff Emeritus
The pressure at some depth is due to the weight of everything above. Pressure increases with depth, with $dP(z)/dz = \rho(z) g(z)$, where $z$ is depth ($z=0$ at the surface), and $P(z)$, $\rho(z)$, and $g(z)$ are the pressure, density, and acceleration due to gravity at that depth.

If the Earth was of a constant density, gravitational force would decrease linearly with increasing depth, reaching zero at the center of the Earth. Pressure never decreases because even though gravitational force is decreasing, the weight of everything above is increasing (albeit slower).

The Earth is not of a constant density. Increasing pressure does three things. It results in differentiation of materials, with the lightest stuff in the crust and the densest stuff (iron and other metals) in the core. Secondly it compresses the rock, making it denser with increasing depth. Finally, it changes the rock chemically. Rock comprising the same basic components will form different chemicals at different depths. This means gravitational force inside the Earth reaches a maximum at the core-mantle boundary. The hydrostatic equilibrium differential equation, $dP(z)/dz = \rho(z) g(z)$, still applies, but the conditions described above make the pressure at the center of the Earth very large.

4. Jul 9, 2014

### Gerinski

Sorry but I do not understand what you mean.

5. Jul 9, 2014

### Gerinski

It seems to me that "pressure at some depth is due to the weight of everything "above", minus the weight of everything in the opposite direction". That's where I get confused, as we get closer to the centre the weight of the mass "above" gets smaller, since it is no more being pulled so strongly towards the centre but it gets increasingly pulled towards the surface until reaching a zero balance at the very centre where it is pulled equally in every direction. So if it does not weigh anymore, why should it cause pressure?

6. Jul 9, 2014

### Staff: Mentor

Direct effect: gravity pulls on you.
Indirect effect: gravity pulls on something else which then pushes or pulls or squeezes you.

At the center of the earth the direct pull of gravity on you is zero. But the rock to the left of you is not at the center of the earth and it's being pulled towards you at the center. So is the rock to your right. Now you're between two masses of rock being pulled together... You feel pressure from them as they crush you to a pulp.

7. Jul 9, 2014

### Staff: Mentor

No, pressure is just due to the weight of every thing above. You may find this easier to see if you think about how water pressure behaves: it's zero at the surface, increases as you go deeper, and reaches a maximum at the bottom. Using your "weight-above minus weight-below" formula it would be negative at the surface (don't confuse pressure with buoyancy!) and would be zero halfway down - and that's not what we observe.

8. Jul 9, 2014

### Gerinski

The rocks to my left and my right are barely not pulled towards me noticeably, since they have equal mass towards the outside as they have towards me.

9. Jul 9, 2014

### Gerinski

Alright but this is because most of the Earth's mass lays in the direction of the bottom of the sea so it pulls the water downwards. When approaching the Earth's centre this is no more the case, water would be pulled outwards at the very centre (equally in every direction so no net pull effect in any direction).

10. Jul 9, 2014

### D H

Staff Emeritus
I suggest you read up on Newton's shell theorem. For a body with a spherically symmetric mass distribution (density is a function of distance from the center), gravitational force depends only on the stuff closer to the center than the point in question. The stuff at greater distances contributes *nothing*.

11. Jul 9, 2014

### A.T.

The mass right next to you is transmitting to you the forces from all the mass above it.

Yeah but pressure is not pull by gravity, it is push by other objects. All the water not in the center is still pulled by gravity, so it pushes on the water in the center.

12. Jul 9, 2014

### olivermsun

As a side note, the earth is actually sufficient non-symmetric such that the gravitational displacements can actually be measured!

But yes, if your concern is being crushed at great depth, then the horizontal gradients are among the least of your problems.

13. Jul 9, 2014

### olivermsun

Okay, so one thing you have to keep in mind is that every "sector" of the earth has to support all the weight above it just to keep it from collapsing inward.

Maybe you can try this thought experiment to convince yourself: suppose you have a skyscraper that weighs pretty much nothing at the bottom floors (it's superstrong material!) but the floors are progressively heavier toward the top. Even though the bottom floors contribute almost nothing to the weight on the base, nevertheless the entire weight of ALL the upper floors is transferred through the structure, and so the base has to support the ENTIRE load.

14. Jul 9, 2014

### Gerinski

So, if I get it right, at the centre of the Earth we will feel weightless but we will feel crushed from everything around us, right?
And on the way towards the centre, we will feel increasingly light but we will feel increasingly crushed by the pressure around us.
In any case it must be true that pressure comes mostly from the stuff closer to the surface which is the one being most strongly pulled towards the centre. The closer to the centre the less the stuff is pulled towards the centre and the more it is pulled outwards towards the surface. Right?

15. Jul 9, 2014

### A.T.

Yes.

16. Jul 9, 2014

### Staff: Mentor

Everything, all 4000 miles of rock to your left, is being pulled towards you by the 4000 miles of rock to your right; and vice versa. There's nothing above those 4000 miles of rock to counter the pull.

17. Nov 7, 2015

### jules K

The relationship between gravity and mass is linear. The relationship between gravity and radius is exponential - Inverse square rule. As radius tends to zero gravity tends to its maximum.

18. Nov 8, 2015

### Staff: Mentor

We can let this old thread rest quietly, I think.

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