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Simple question regarding gravity

  1. Jul 20, 2011 #1
    We know g is inversely propotional to d2 but at the center d will be zero and we know that g= MG\d2 and if d2 at center will be zero then anything by zero will be infinite so at earth core g should be infinite which is not the case?????
     
  2. jcsd
  3. Jul 20, 2011 #2
    M's inside sphere of radius d
     
  4. Jul 20, 2011 #3
    d is the distance from the mass. At the center the mass is distributed all around you at some distance. d is not zero at the center.
     
  5. Jul 20, 2011 #4
    But suppose we move a little further from the center will the g be same 10m\s2 and if not why
     
  6. Jul 20, 2011 #5
    It would be a summation of the mass and distance for each piece of mass around you.

    When you say gravity is 0 at the center what your really doing is a quick summation in a symmetrical situation.
     
  7. Jul 20, 2011 #6
    When you're inside the Earth, the gravitational acceleration is given by:

    [tex]g = -kr[/tex]

    where:
    [itex]k = \frac{4}{3} \pi G \rho[/itex]
    [itex]G[/itex] is the Gravitational constant
    [itex]\rho[/itex] is the average density of the planet
    [itex]r[/itex] is the displacement from the center
     
  8. Jul 20, 2011 #7
    see, you know what happens when one of the objects its center with the other.. the attraction of gravity is same in all the direction. [that really depends on the geometry of the first object] now going by the equation, mathematically u can't place two things on the same place and call them two different objects. that means, the m1 and m2 in the equations now vanishes. m2 = 0 while mass of m1 = will actually become [m1 = m1+ m2] so now the equation becomes 0/0. indeterminate. the doesn't have sufficient info to handle the situation. this is what came to mind firstly, correct me if i am wrong.
     
  9. Jul 21, 2011 #8
    The Earth is not a point mass. Outside the Earth, you can treat it as if all the mass were located at the center of the Earth. But this isn't valid in the Earth. Actually, you need to sum the contributions to the gravitational field of each atom of the Earth. At the center of the Earth, the gravity from atoms in all directions cancel out, leaving 0 gravity in the Newtonian sense.

    In fact, the gravity from a spherically symmetric source at some arbitrary radius R is the same as the gravity from a point source at the center of the sphere with mass equal to the total mass that is contained in the sphere of radius R. Any mass outside R doesn't contribute (it cancels out).
     
  10. Jul 21, 2011 #9

    Drakkith

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    Staff: Mentor

    They key difference here is the NET force from the earths gravity compared to the force itself. The net force here on the surface exists because almost all of the mass is UNDER us. Once you are inside the earth the mass is spread all around you. You still experience the force of gravity about as much as you did before, perhaps a little more now that you are in the densest part of the earth, but it pulles about equally in all directions on every point in your body so you would experience not NET force.
     
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