Center of gravity an center of mass

Discussion in 'Physics & Math' started by Saint, Oct 19, 2017.

  1. DaveC426913 Valued Senior Member

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    9,005
    You mean Jupiter.
    It's three times more massive and half as far away.
     
    Confused2 likes this.
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  3. Confused2 Registered Senior Member

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    Yes, Jupiter, thank you.
     
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  5. Forceman May the force be with you Registered Senior Member

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    In reply to ~~~~DM the dimension of gravity for the first ground DM is dimensionally wide but not across the dimension of the equatorial belt but first from the mid center regions such as north and south pole DM(2) plus the amount of force land regions can emit to equate the DM and MD from people on land to mass and energy in the center of the earth gm^2 - 1/2k+ v = g = g + m = d = 1/2C^2 + C ; this is not the same as radiation but the center is just that the center to infinity space atmosphere and air resistance so TV can make the sphere of the earth and its spatial organization be such as magnetic currents rather than gravity.
     
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  7. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    Huh?
    What is DM, MD, DM(2) and TV?
    I assume they are not dark matter, doctor, dark matter 2 and television.

    What is the 'dimension of gravity'?

    Wait, as far as I can tell, none of your post makes any sense - could you rephrase and avoid using abbreviations?
     
  8. DaveC426913 Valued Senior Member

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    9,005
    I would say FM is trying to have a little fun at our expense with his mastery of these tools:

    Please Register or Log in to view the hidden image!

     
  9. Q-reeus Valued Senior Member

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    3,044
    Based on above context it seems evident that 'gravitational field' is meant in the Newtonian sense g = -∇φ. Then it's well known but little known that a uniform g can be constructed over an arbitrarily large spatial extent: https://irodovsolutionsmechanics.blogspot.com/2008/09/irodov-problem-1215.html
    Since the potential in such a region does vary linearly in the direction of g, it follows the metric coefficients vary also, but only linearly i.e. there is zero 'tidal gravity'.
    While above is only strictly true in Newtonian weak field limit, even in GR (or similar metric gravity theory), slight modification to matter densities could always be made such as to maintain perfect field uniformity
     
    Last edited: Dec 2, 2017
  10. QuarkHead Remedial Math Student Valued Senior Member

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    One of us is confused
    This is, of course, the definition of the gradient of the scalar or vector field \(\phi\). In either case it is a vector field - strictly a co-vector field or one-form.
    Surely the only situation that the gradient of a field is uniform is when it is uniformly zero.
    What is the "direction" of g?

    You seem to be conflating 3 things here: the gradient, which you defined, the gravitational potential, which you didn't, and the metric tensor field.
     
  11. Confused2 Registered Senior Member

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    496
    It seems to me that there are a lot of hidden assumptions about 'centre of gravity'. When standing still you need to keep your centre of gravity within the area defined by your feet (else you fall over). You could say you need to keep your centre of mass within the same area but then you'd be introducing the problem introduced by Quarkhead
    because without knowing the "direction" of g it would be difficult to stand upright - I think most of us work out the "direction" of g within the first 20 months of life.
     
  12. Q-reeus Valued Senior Member

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    3,044
    You seem to be reacting to my post without having ever actually studied the quite brief but concise article I linked to. Yes one of us is confused.

    (I thought, given the context, it too trivially obvious to need formally defining φ = -Gm/r as the Newtonian gravitational potential exterior to a point mass m. Obviously in the interior case an integration is required but owing to the spherical symmetries as per that article, it works out very simply as shown there (φ itself never needing to explicitly appear). The GR metric coefficients are then defined using φ, e.g. g_00 = 1+2φ/c^2 and as stated earlier, are not constant throughout the hollow region despite g = -∇φ being constant but *non-zero* throughout.)
     
    Last edited: Dec 3, 2017
  13. QuarkHead Remedial Math Student Valued Senior Member

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    Of course I didn't "study" your link. I never click on 3rd party links when posted on forums; when they are merely a personal blog I see even less reason to do so

    Once again, you are confusing the metric tensor field (your first supposed equality) with the gradient (your second).
     
  14. Confused2 Registered Senior Member

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    496
    Constructive input would be appreciated at this point.
     
  15. Q-reeus Valued Senior Member

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    3,044
    Such a position means I cannot help you further.
    Wrong. But I have no interest in bickering, given your stated attitude above.
     
  16. arfa brane call me arf Valued Senior Member

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    Turning to the chapter Gravitational Interaction in my old physics textbook, I find the following:

    ". . . We may then conveniently say that the mass M produces, in the space around it, a physical situation (gotta love that one) which we call a gravitational field, and which is recognized by the force that M exerts on another mass m', brought into that region.
    . . . The gravitational field strength \( \mathcal {G} \) produced by the mass M at a point P is defined as the force exerted on the unit of mass m' placed at P (this is just using a "test mass" to measure the field strength at P, an arbitrary point). Then

    \( \mathcal {G} = \frac {\vec F} {m'} = - \frac {\gamma M} {r^2} \vec u_r \)
    Thus the gravitational field has the direction opposite to that of the unit vector \(\vec u_r \)." Hence \( \mathcal {G} \) is also a vector.
    But we also have:

    \( \vec F = -\nabla E_p \)​

    With the scalar component \( F_s = - \frac {d E_p} {ds} \).

    And since \( \vec F = m'\mathcal {G} \) and gravitational potential (not potential energy) is defined as \( V = \frac {E_p} {M} \) where M is the gravitating body . . . well, I'm sure you can work out that the field is the negative of the gradient of the potential, and is an acceleration field.

    And my work here is done. Except now I'm confused about the difference between field and field strength; my textbook seems to be saying they are the same thing.
     
    Last edited: Dec 5, 2017
  17. arfa brane call me arf Valued Senior Member

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    5,686
    Ah, I see it now (but then I think I'm remembering something too, it has been a while). There is a gravitational field defined by \( \mathcal {G} \) around any mass M, whether or not there is another mass m' in the region.

    So acceleration is only defined (usually called g) if there is an accelerating test mass, but nonetheless the "physical situation" exists and there is a field around M which has an intensity or "strength" defined on it.
    One way to visualise this (as my textbook does) is with radial lines of force (diverging!) and concentric equipotential surfaces (the gradient of the field!) around such a gravitating body. That is, the field is there regardless of whether there are 'real' forces acting on any test (unit) masses.

    Q.E.D.
     
  18. QuarkHead Remedial Math Student Valued Senior Member

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    Yes, exactly. Or, to quote Einstein himself, "space without a metric is unthinkable" (he means spacetime, since in the next sentence he says that the 10 components of the metric tensor are required for the metric field - i.e. a 4- space).

    Go back to Pierre de Laplace. He stated that , given a field and nothing to "disturb" it, i.e. no source, the divergence of the gradient is zero. He wrote \(\nabla^2 \phi =0\) for the field (scalar or vector) \(\phi\) for this situation. Note that the Laplacian is a second-order differential operator with respect to Cartesian coordinates

    In GR, the Laplacian operator is replaced (in part) by the curvature tensor, which is itself a field. But it has the property that it is also a second-order derivative, this time of the metric field (with respect to the spacetime coordinates).

    Note that from the rules of differential calculus that we learn in school, the vanishing of a second-order derivative of the curvature field implies the metric field is constant, and conversely - a constant metric field mandates a zero curvature i.e. quasi-Euclidean (Minkowskian) spacetime.
     
  19. arfa brane call me arf Valued Senior Member

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    5,686
    Is there a difference though, between a center of mass and a center of gravity.

    I'd say yes, because an object can have a center of gravity but no mass at its geometric center.
    For instance a hollow spherical shell of matter with mass uniformly distributed has a gravitational field at all points outside it; this field is identical to that of a particle with the same mass located at the center of the shell. Moreover the field inside a uniform spherical shell is zero and the potential is constant.

    However this extends to a solid sphere in a like way: the external field (and potential) of a solid sphere of mass M is as described previously, but identical to a point mass at the center (i.e. you could replace the solid sphere with a pointlike 'particle' with the same mass and have the same field, sound familiar?). The external field strength varies as \( 1/r^2 \) in either case. Internally the field is proportional to \( 1/r \) for a solid sphere.
     
  20. QuarkHead Remedial Math Student Valued Senior Member

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    Difficult question to answer, because mass is not the only gravitational source.

    This is because GR is a non-linear theory, that is, given a massive body that induces a gravitational field, that field itself is a gravitational source
     
    Last edited: Dec 6, 2017
  21. Q-reeus Valued Senior Member

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    Can't recall how many times I have made it clear that in GR gravity does NOT self-gravitate, but that message just keeps collectively going in one ear and coming out the other. Once again: http://www.physicsforums.com/showthread.php?t=768604
    https://www.physicsforums.com/insights/does-gravity-gravitate/

    Note carefully that the 'yes' bit is basically about still hypothetical graviton-graviton interactions in an as yet unobserved quantum gravity regime not relevant to classical GR at all.
    Field self-interaction i.e non-linearity is often said to be evidence that gravity self-gravitates but it's not the same thing at all.
     
    Last edited: Dec 7, 2017
  22. psikeyhackr Live Long and Suffer Valued Senior Member

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    1,055
    They are different but in the vast majority of cases the difference is not significant. Consider a skyscraper over 1000 ft tall. Say we determine the center of mass is 350ft up. But gravity gets weaker the higher you go though the change is very small. So even though the mass above the center of mass is the same as the mass below it weighs less because the gravitational field is slightly weaker. So the center of gravity is lower than the center of mass by some fraction of an inch.

    This difference would be miniscule for a skyscraper on the Earth's surface because the Earth is so large in comparison. For two asteroids in close proximity the difference could matter in affecting their trajectories after they pass.
     
    Last edited: Dec 9, 2017
  23. DaveC426913 Valued Senior Member

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    I don't see how this defends your initial question, namely "[yes, there's] a difference ... between a center of mass and a center of gravity."

    Even with a hollow shell, the centre of mass is co-located with the centre of gravity. If there's any doubt where its CoM is, set it spinning and see what point it rotates about.
     

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