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Gravitational force and density.

  1. Jan 7, 2008 #1
    A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center. Is this true and can someone explain this? For example, wouldn't earth as a black hole attract the moon with slightly more force because it's volume is condensed? Doesn't the outer portion of the earth produce a gravitational field line not parallel to the line between the centers of the moon and earth thus canceling the the perpendicular components of the gravitation forces?
     
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  3. Jan 7, 2008 #2

    nicksauce

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    Yes this is true. The reason comes from Gauss's law. The electric form (as I can't recall the gravitational form off the top of my head) is (this will all be essentially the same for the gravitational force, as they both follow 1/r^2 laws)

    [tex]\oint_S{\vec{E}\cdot d \vec{S}} = \frac{Q_{enc}}{\epsilon_0}[/tex]

    Therefore, if we have a (gaussian) surface in which, the field is parallel to the surface normal, we can write:

    [tex]\vec{E} = \frac{Q_{enc}}{A\epsilon_0}[/tex].

    Such a surface is a sphere, since the electric field is radially outward. If we are outside the sphere, say a distance R from the centre, the field will always be

    [tex]\vec{E} = \frac{Q_{enc}}{4 \pi R^2\epsilon_0}[/tex]

    It does not matter whether the charge is uniformly distributed throughout the volume of a sphere, over the surface of a sphere, or is a point charge at the origin, as long as the enclosed charge is the same. (If we are inside a sphere, this does not work, as the enclosed charge varies with R)
     
    Last edited: Jan 7, 2008
  4. Jan 7, 2008 #3

    Shooting Star

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    As long as you are outside the event horizon of a non-rotating BH, you can't say whether it's a BH or just a spherically symmetric distribution of mass. The force is different from the Newtonian case, of course.

    Actually, it does not matter in whatever way the charge or mass is distributed inside as long as only the monopole term survives outside the boundary of the sphere.
     
  5. Jan 7, 2008 #4

    Shooting Star

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    Some mathematics is necessary, which proves that the field of an infinitesimally thin shell of uniform mass density outside the shell is the same as that of a point mass of equal mass situated at the centre. For a spherically symmetric distribution, a sphere can be mentally broken up into infinite such shells and the field of the whole sphere outside then will be like that of a point mass equal to the mass of the sphere sitting at the centre.

    (En passant, Newton had to invent Integral Calculus to prove this.)
     
  6. Jan 8, 2008 #5

    rbj

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    here's how to recall the gravitational form. first recall the two inverse-square forms:

    [tex] F = k \frac{q Q}{r^2} = \frac{1}{4 \pi \epsilon_0} \frac{q Q}{r^2} [/tex]

    [tex] E = \frac{F}{q} = \frac{1}{\epsilon_0} \frac{Q}{4 \pi r^2} [/tex]



    [tex] F = G \frac{m M}{r^2} [/tex]

    [tex] \frac{F}{m} = 4 \pi G \frac{M}{4 \pi r^2} [/tex]


    so you can replace symbols in equivalent roles.
     
  7. Jan 8, 2008 #6

    pam

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    Newton (or Leibnitz) did invent IC, but in his Principia he actually used an ingenious, but complicated, geometric method.
     
    Last edited: Jan 8, 2008
  8. Jan 8, 2008 #7

    Shooting Star

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    Yes, I have seen it. It's almost completely geometrical, and very difficult for us (or at least for me) to understand. His reason for doing so was for the benefit of his contemporaries, since they too wouldn't have understood the newly invented IC, which itself was written very cumbersomely.
     
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