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Gravitation Force on the Earth's Centre

  1. Aug 20, 2008 #1
    I am having some problem understanding this.. By Newtons Gravitation Law we know:

    F=Gm1m2/r^2

    centre of the particle means r=0, am I right? if so, the equation says that the force will be infinite on the centre of the earth.. but we see that it has 0 force on the centre.. why such thing happened??
     
  2. jcsd
  3. Aug 20, 2008 #2

    cristo

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  4. Aug 20, 2008 #3

    HallsofIvy

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    That is the formula for the gravitational force of a point mass which can also be shown, by integrating over the volume of a sphere, to be the force of a spherical mass on any point outside the sphere.

    Inside the sphere, if, say r< r0, where r0 is the radius of the sphere, it can be shown that the force for the mass outside distance r cancels. (The mass on "this" side pulls upward, on "that" side pulls downward and so cancel.) That is, only the mass for radius < r gives a non-zero total contribution to the force. Since mass (assuming constant density) is proportional to volume which is proportional to r3, the mass of a sphere of radius r inside a uniform sphere of radius r0 of mass M is Mr3/r03.

    Putting that together with GmM/r2, the force on an object at distance r from the center of the sphere is Gm(Mr3/r03)/r2 and so is proportional to r, not 1/r2. Of course, at the center of the earth r= 0 and so the gravitational force is 0. (Again, because of the symmetry of the sphere and the assumed constant density, the gravitational force from any "bit" of matter in the earth is cancelled by the force of a diametrically opposite "bit" of matter.)

    Oh, and that does not mean that the PRESSURE on something at the center of the earth due to the mass above it would be 0. That's a completely different calculation.
     
    Last edited: Aug 21, 2008
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