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Weight of mass on surface > center of Earth

  1. Apr 26, 2004 #1
    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 truely 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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  3. Apr 26, 2004 #2

    HallsofIvy

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    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."
     
  4. Apr 26, 2004 #3
    Relating that to GR, we can say that the curvature of space is greater at the surface.
     
  5. Apr 26, 2004 #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 explaination of this fact according to my model!

    So now if anyone cares to comment, I would like to hear your explaination of such according to your alternate theory of gravity.
     
  6. Apr 26, 2004 #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 blackhole 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.
     
    Last edited: Apr 26, 2004
  7. Apr 26, 2004 #6

    Nereid

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    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).
     
  8. Apr 26, 2004 #7

    DW

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    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.
     
  9. Apr 27, 2004 #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?
     
  10. Apr 27, 2004 #9

    DW

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    I made a point.

    I did.

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

    Then why did you?

    Yes.
     
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