Weight of mass on surface > center of Earth

In summary, according to the ether theory, mass on Earth would weigh less than mass at the center of the Earth because of the curvature of space. However, this has not been confirmed by experiment. Instead, experiments have shown that the mass on the surface of the Earth weighs more than at the center. This is explained by the theory by saying that the center of a uniform spherical mass in empty space lies exactly on the surface or horizon.
  • #1
tdunc
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Logic tells us that a mass on the surface of the Earth weighs more than at the center(or very deep underground). My question is has this been confirmed by experiment? Wouldn't it be truly bizarre if the mass weighed the same?

In any case, I can explain why mass would weigh less according to my ether theory (I think), however I'd like to hear why it would weigh less according to GR (fabric of space), of which I have no answer. I mean we are talking about gravity. Quite frankly GR makes no sense to me period I'll say that right now for the record.
 
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  • #2
Gravitational force is given by GMm/r2 where G is the "universal gravitational constant", M is the mass of the earth, m is the mass of the object, and r is the distance from the center of the earth. It is pretty easy to show that, because of the "1/r2", that the total mass from a sphere outside that radius is 0: opposite parts cancel. So the "net" force is only due to the mass between the object and the center of the earth. Assuming a uniform Earth (reasonably accurate), the mass "below" the object is 4/3 times density times r3 so we have
Gm(4/3 density r3)/r2= (4/3) Gm density r. That clearly decreases as r decreases and is 0 exactly at the center of the earth.

No one has ever measured anything at the center of the Earth but there have been experiments on object a couple of miles down which support the calculation.

" Quite frankly GR makes no sense to me period I'll say that right now for the record."

In other words "I don't understand it, therefore it is wrong."
 
  • #3
Relating that to GR, we can say that the curvature of space is greater at the surface.
 
  • #4
Upon further thought

Assume for the moment that the Earth is a perfect sphere with uniform matter. You would feel no gravity at the very center. The same principle can be applied regardless of the circumference of sphere, so you would feel no gravity at the center of the Sun also. This is explained perfectly by GR believe it or not! There is no curvature of space at the center! The center of a uniform spherical mass in emtpy space lies exactly on the surface or horizen of that space!

Against all my better judgement that a bending of space is illogical, it works in this example! And not only does it work, I cannot give a proper explanation of this fact according to my model!

So now if anyone cares to comment, I would like to hear your explanation of such according to your alternate theory of gravity.
 
  • #5
Actually I don't know that that statement is accurate. How can we say that the curvature of space is zero at the center? A black hole would not agree with that too well for one. Nor would even a significant mass such as our Sun.

So if no curvature of space equals no gravity felt, then you must conclude that any curvature of space equals gravity felt. Yet that is not the case if the space at center of the Sun for example is curved.

Does that make sense?

The only conclusion you can reach is that at the center you are being pulled equally in all directions, therefore no gravity felt even though you are lying on curved space. With that understanding, we still don't prove a curvature of space is the cause of gravity or even that a curvature exsists.
 
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  • #6
tdunc said:
Logic tells us that a mass on the surface of the Earth weighs more than at the center(or very deep underground). My question is has this been confirmed by experiment? Wouldn't it be truly bizarre if the mass weighed the same?

In any case, I can explain why mass would weigh less according to my ether theory (I think), however I'd like to hear why it would weigh less according to GR (fabric of space), of which I have no answer. I mean we are talking about gravity. Quite frankly GR makes no sense to me period I'll say that right now for the record.
Who's logic? It seems to me that you're referring to Newton's theory; HallsofIvy's response is within this framework. While differences between GR and Newtonian theories of gravity are observable here on the surface of the Earth, they're pretty small.

IIRC, quite a number of pretty sensitive measurements of gravity were taken a few years ago, when there was a lot of interest in possible deviations from the inverse square relationship. These included measurements down deep mines. No deviations from inverse square found (to the limits of experimental error).
 
  • #7
tdunc said:
Actually I don't know that that statement is accurate. How can we say that the curvature of space is zero at the center? A black hole would not agree with that too well for one. Nor would even a significant mass such as our Sun.

So if no curvature of space equals no gravity felt, then you must conclude that any curvature of space equals gravity felt. Yet that is not the case if the space at center of the Sun for example is curved.

Does that make sense?

The only conclusion you can reach is that at the center you are being pulled equally in all directions, therefore no gravity felt even though you are lying on curved space. With that understanding, we still don't prove a curvature of space is the cause of gravity or even that a curvature exsists.

The spacetime geometry for the interior of the Earth does not match that of a black hole. No curvature of space also is not equivalent to no gravity felt. Actually, one never feels the force of gravity. You feel the normal force of the floor up on your feet. And your definition of proof is useless. The spacetime geometry for the interior of the Earth is given by the line element
[tex]ds^{2} = (1 + \frac{GM_{earth}r^{2}}{R^{3}c^{2}} - \frac{3GM_{earth}}{Rc^{2}})dct^{2} - \frac{dr^{2}}{1 - \frac{2GM_{earth}r^{2}}{R^{3}c^{2}}} - r^{2}d\theta ^{2} - r^{2}sin^{2}\theta d\phi ^{2}[/tex]
The r-r spatial componant of the metric does reduce to the value of spherical coordinates in an otherwise flat spacetime at r = 0, but the time component does not. Even so the affine connections do vanish at r = 0 meaning that no gravitational acceleration is observed to occur there.
 
  • #8
"The spacetime geometry for the interior of the Earth does not match that of a black hole."

I don't immediatly know what your trying to get at, if you want to elaborate and make a point I'd be willing to listen.

"No curvature of space also is not equivalent to no gravity felt."

Would you like to give an example?

"Actually, one never feels the force of gravity. You feel the normal force of the floor up on your feet."

That statement reads as if it came straight out of a textbook and in no relation or thought to the given concepts we are discussing. I would just rather not comment on that. Actually, take a satellite for example, are you telling me that it does not feel the force of gravity?
 
  • #9
tdunc said:
"The spacetime geometry for the interior of the Earth does not match that of a black hole."

I don't immediatly know what your trying to get at, if you want to elaborate and make a point I'd be willing to listen.

I made a point.

"No curvature of space also is not equivalent to no gravity felt."

Would you like to give an example?

I did.

"Actually, one never feels the force of gravity. You feel the normal force of the floor up on your feet."

That statement reads as if it came straight out of a textbook and in no relation or thought to the given concepts we are discussing.

It isn't my fault that you missed the point.

I would just rather not comment on that.

Then why did you?

Actually, take a satellite for example, are you telling me that it does not feel the force of gravity?

Yes.
 

1. What is the weight of an object at the surface of the Earth?

The weight of an object at the surface of the Earth is dependent on both the mass of the object and the strength of the Earth's gravitational pull. On average, the weight of an object at the surface of the Earth is approximately 9.8 newtons per kilogram (9.8 N/kg). This is often referred to as the acceleration due to gravity.

2. Does the weight of an object change as it moves closer to the center of the Earth?

Yes, the weight of an object changes as it moves closer to the center of the Earth. This is due to the inverse square law of gravity, which states that the force between two objects is inversely proportional to the square of the distance between them. As an object moves closer to the center of the Earth, the distance between the object and the Earth's center decreases, resulting in an increase in gravitational force and therefore an increase in weight.

3. How does the weight of an object on the surface of the Earth compare to its weight at the center of the Earth?

The weight of an object on the surface of the Earth is greater than its weight at the center of the Earth. This is because, at the surface, the object is further away from the Earth's center and therefore experiences a weaker gravitational pull. As the object moves closer to the center, the gravitational pull increases, resulting in a greater weight.

4. Is the weight of an object on the surface of the Earth affected by its size or shape?

Yes, the weight of an object on the surface of the Earth is affected by its size or shape. This is because the mass of an object is a factor in determining its weight, and a larger or differently shaped object will have a different mass. However, as long as the mass of the object remains the same, its weight will not change regardless of size or shape.

5. How does the weight of an object on the surface of the Earth differ from its weight on other planets?

The weight of an object on the surface of the Earth differs from its weight on other planets due to variations in gravitational pull. Each planet has its own unique mass and radius, which affects the strength of its gravitational pull. For example, an object would weigh less on the surface of the Moon due to its smaller mass and weaker gravitational pull compared to Earth. On the other hand, an object would weigh more on a planet with a greater mass and stronger gravitational pull, such as Jupiter.

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