Gravity at the center of the earth

[gre]

What would the gravity be like in the center of the earth? Would it make objects stretch out in all directions?

Edited May 13, 2009 by gre

[UC]

the field would be effectively equivalent in all directions, and the net field would be 0.

[throng]

the field would be effectively equivalent in all directions, and the net field would be 0.

 

If a cavity existed at the centre of earth which way would be up?

 

If you dropped a ball within that cavity which direction would it go?

[swansont]

 

If you dropped a ball within that cavity which direction would it go?

 

If it had zero initial speed, it wouldn't go anywhere. No force, no acceleration.

[Sisyphus]

And every way would be "up," in the same way that every horizontal direction at the north pole is south.

[gre]

Since the force of gravity is tugging on the object in all directions, would an object be larger, than on earth's surface?

[Mr Skeptic]

Since the force of gravity is tugging on the object in all directions, would an object be larger, than on earth's surface?

 

Nope, the forces cancel. It's not tugging the edges of the object, it's tugging every sub-particle equally in all directions, for no net effect. I think that if you use General Relativity there may be a tiny effect, due to the gravitational potential (not force) at the center.

[Sisyphus]

Furthermore, if it was big enough that the extremities were far enough away from the center to make a difference, it still wouldn't be pulled apart, it would be compressed together. If you're a mile away from the center, the gravity from all the mass except a one mile radius sphere cancels out. And that one mile radius sphere still pulls you towards the center (weakly), just as much as if you were standing on a planet that was two miles wide.

[gre]

It's not tugging the edges of the object, it's tugging every sub-particle equally in all directions

Gotcha, so what effect does this have on sub atomic particles exactly? If it's strong enough, does it make them increase in size a little, weaken bonds, etc?

[Sisyphus]

Gotcha, so what effect does this have on sub atomic particles exactly? If it's strong enough, does it make them increase in size a little, weaken bonds, etc?

 

No. Why would it? Being "pulled apart" implies that part of it is being pulled one way, while another part is being pulled another way. This is not the case with gravity. It just cancels out.

[proton]

No. Why would it? Being "pulled apart" implies that part of it is being pulled one way, while another part is being pulled another way. This is not the case with gravity. It just cancels out.

 

The gravitational field is zero at r = 0 but the tidal force tensor is non-zero. In this instance there is a difference between no gravitational field and the field vanishing at the point r = 0. In GR lingo one would say that the gravitational field (Christofel symbols) vanishes at r = 0 but that the spacetime curvature does not vanish there.

[Baby Astronaut]

Being "pulled apart" implies that part of it is being pulled one way, while another part is being pulled another way. This is not the case with gravity. It just cancels out.

Would you mind giving a source for that, which I can look into further?

[swansont]

The gravitational field is zero at r = 0 but the tidal force tensor is non-zero. In this instance there is a difference between no gravitational field and the field vanishing at the point r = 0. In GR lingo one would say that the gravitational field (Christofel symbols) vanishes at r = 0 but that the spacetime curvature does not vanish there.

 

For a solid mass or for one with an empty cavity?

[proton]

For a solid mass or for one with an empty cavity?

In Newtonian gravity Possion's equation is the trace of the tidal force tensor set proportional to the mass density. Therefore a zero tidal force tensor gives zero mass density. If there is a cavity centered at the point then the tidal force tensor may or may not be it zero there. In the present case of, say, a spherical cavity at the center of the earth then it will be zero. Otherwise it will depend on the shape of the cavity. If the center of a spherical cavity is not at r = 0 and the earth is a sphere of uniform mass density then the tidal forces in the cavity will be zero since the field is uniform in such a cavity. A square cavity will give different results of course. Edited May 15, 2009 by proton

[Sisyphus]

Would you mind giving a source for that, which I can look into further?

 

For which part?

[Baby Astronaut]

For which part?

 

The bolded

Being "pulled apart" implies that part of it is being pulled one way, while another part is being pulled another way. This is not the case with gravity. It just cancels out.

[Sisyphus]

Ah, well, that's really just saying that gravity (or any force, actually), can only have one direction at a time. It's a vector, and two vectors add together to make one, third vector: the net force. So, the gravity experienced at any particular point at any particular time is the net gravity, which basically just means everything that isn't canceled out by gravity in the opposite direction. At the exact center of a solid sphere, or inside a spherical hollow at the center of a spherical shell, all of the gravitational force from the sphere/shell gets canceled out. For every point pulling you one way, there's a symmetrical point pulling you the other way with the same force, and you feel nothing.

[darshanchande]

I guess at the center of the earth gravitational force is at it's best.

[Kyrisch]

I guess at the center of the earth gravitational force is at it's best.

 

That was enlightening...

[froarty]

Ah, well, that's really just saying that gravity (or any force, actually), can only have one direction at a time. It's a vector, and two vectors add together to make one, third vector: the net force. So, the gravity experienced at any particular point at any particular time is the net gravity, which basically just means everything that isn't canceled out by gravity in the opposite direction. At the exact center of a solid sphere, or inside a spherical hollow at the center of a spherical shell, all of the gravitational force from the sphere/shell gets canceled out. For every point pulling you one way, there's a symmetrical point pulling you the other way with the same force, and you feel nothing.

 

I agree with Sisyphus regarding cancellation but no one has mentioned time dilation vs the surface, the increased vectors represent an increased dilation regardless of cancellation so a person inhabiting the center of a planet would age more slowly than on the surface.

Edited October 5, 2009 by froarty
no instead of no one

[Bob_for_short]

What would the gravity be like in the center of the earth?...

 

It is a gravity force from, say, Sun.

[froarty]

It is a gravity force from, say, Sun.

 

I don't follow you but ok lets say a dead non rotating sun in deep space with an observer on the surface and an observer in a central cavity - the gravity cancells for the observer in the center but he has far more gravitatinal vectors regardless of cancellation than the observer on the surface. This should equate to being in a deeper gravitational well and therefore in a slower clock rate compared to the observer on the surface. I am taking the position that dilation is only related to the quanity of vectors and not orientation.

Regards

Fran