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Gravitational Influence of Stars in a Galaxy

  1. Aug 9, 2015 #1
    Can anyone tell me where I can find an explanation that only the gravity of the stars inside the orbit of your star is important in a galaxy? Someone told me this and the integration of gravitational effects I did seems to disagree so I wanted to look at what has already been done.
     
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  3. Aug 9, 2015 #2

    phinds

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    That is simply one example of the fact that for ANY spherical body with uniform distribution of matter, the matter outside (not inside as you stated) the sphere defined with you as a radius from the center is irrelevant to the gravitational attraction you feel from the body.

    Have you studied calculus?
     
  4. Aug 9, 2015 #3
    Assuming a spherical homogenous earth. If I drill down 1/2 radius (3,189KM),then the radius from the center is 1/2 what it was, and the mass with this radius (4/3 pi r^3) is 1/8 what it was at the surface and I should feel 1/8g pull towards the center?
     
  5. Aug 9, 2015 #4

    phinds

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    Yep, and when you get to the center, you'll be weightless (and not just because you'll have burned to a cinder :smile:)
     
  6. Aug 9, 2015 #5
    This does not seem correct. Do you have a reference book that goes through the math involved? Or do you know the math involved?
     
  7. Aug 9, 2015 #6

    phinds

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    No, I learned this (with the math) almost 50 years ago but these days I can hardly add 2+2. I'm sure one of our members can give you a good reference.
     
  8. Aug 10, 2015 #7

    Bandersnatch

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    That is not correct!

    You'll feel 1/2g at half the radius. While there's less mass pulling you in, you are also 2 times closer to the centre of the field, which makes the acceleration 4 times as strong (the inverse square law in the law of gravity).
    In a uniform, ideally symmetric sphere, the strength of gravitational acceleration a test particle experiences goes down linearly with radius as you descend beneath the surface.

    The maths here rely on the 'shell theorem', in two ways (Wikipedia and hyperphysics cover it well):
    - so that all the mass below a certain radius can be treated as concentrated in the centre of the sphere of that radius
    - so that all the mass above a certain radius can be disregarded as contributing nothing to the acceleration of a test particle below

    The applicability of the shell theorem to galaxies is valid only as an approximation (e.g. when you can treat all mass of a galaxy as concentrated in the bulge; often used as the first approximation in introductory texts), since it does not hold for disc-shaped distributions of mass.
    You can find more details about the mathematics of gravity in a disc here:
    http://applet-magic.com/gravdisk2.htm

    I should add that there is some overlap between the applicability and non-applicability:
    - in the bulge the shell theorem can be said to hold
    - in the disc it doesn't hold
    - the spherical distribution of dark matter again lends itself to
     
    Last edited: Aug 10, 2015
  9. Aug 10, 2015 #8

    phinds

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    Bandersnatch, you are saying that you are NOT weightless at the center of the Earth?
     
  10. Aug 10, 2015 #9

    Bandersnatch

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    No, I'm saying at 1/2 the radius the pull is not 1/8g. I was objecting to your confirming the OP's error, not to the latter part of your post.
     
  11. Aug 10, 2015 #10

    phinds

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    Fair enough. It's just that your quote made it seem as though you were specifically saying weightless at center is incorrect.
     
  12. Aug 10, 2015 #11

    D H

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    The shell theorem applies to spherical bodies with any spherical mass distribution (density is a function of radial distance only), not just a uniform mass distribution.

    It does not apply to two dimensional disks in three dimensional space.

    One of the early successes of general relativity was fully explaining the precession of Mercury. Newtonian mechanics explains over 90% of Mercury's observed precession rate. One way of modeling that non-relativistic precession is to replace Jupiter, Venus, and the Earth with rings of material of the same masses as those planets, and located at the same distances from the Sun as are the planets. There would be no effect on Mercury from those rings if the shell theorem truly did apply. It doesn't apply; there is a small perturbative effect.

    Nonetheless, the effect is small, and acts primarily to cause an elliptical orbit to precess. There is very little effect on circular orbits. That the effects are small is what lets one ignore the stars that orbit in the galactic disk outside of some star.
     
  13. Aug 10, 2015 #12

    phinds

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    Just to be sure I'm getting this correctly, you are saying that, for example, a body that has spheres of varying density would not have uniform density at the scale of the whole body but the shell theorem would still give the same result. That makes sense, I just hadn't thought about it that way.
     
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