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Gravitational force and distance

  1. Jun 25, 2007 #1
    If F=GmM / d^2 Lets assume we're using it to measure the force between me and the earth, as i go further away the force on my decreases, what would happen if i could go right to the centre of the earth? would i be crushed under the force or what?
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  3. Jun 25, 2007 #2


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    Well for starters that equation is invalid for distances d>r; where r is the radius i.e. beneath the surface of the earth. Beneath the surface, you'll have to apply Gauss' law. Mayb that should get you thinking.

    Are you familiar with Gauss's law? Use geometry to get a feel as to what the resultant force is on a particle in a hollow sphere then apply that thinking to the earth.

    edit: I forgot to say that r is the radius of the earth
    Last edited: Jun 25, 2007
  4. Jun 25, 2007 #3
    Im not talking about a hollow sphere, because it has to have mass to have a gravitational pull, but if i assume there was a well that went to the earths core, i jumped in and made it to the bottom, would i be crushed under the force when i reach the middle?

    Also i dont see how Gauss's law can help me with this problem, although he states the equation i mentioned before and that gravity acts as if the mass is concentrated at the centre, im trying to work out what would happen at the centre if i could get there.
    Last edited: Jun 25, 2007
  5. Jun 25, 2007 #4


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    As ranger indicated the acceleration of gravity depends on the mass enclosed in a sphere. As one goes to the center of the earth, the enclosed mass decreases.

    On the other hand, the earth's mass pulls on itself, and the pressure at the center of earth is on the order of 350 GPa. (1 atm is ~0.1 MPa, or ~0.0001 GPa).

    If one could maintain a column of air to the center of the earth, the pressure would be less since the density of the air is less than rock.
  6. Jun 25, 2007 #5


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    There is a Gauss's Law equivalent for gravitational field.


    You have to recalculate the radial dependence for the inside solution. The 1/r^2 dependence no longer work in this region (as anyone who has done electrostatic problems can tell you).

    If you assume a sphere of uniform density, you'll see that the gravitational field drops with r, not 1/r^2 for the inside solution. So the gravitational force goes to zero as r -> 0.

  7. Jun 25, 2007 #6
    that's pretty cute. isn't there a dichotomy between newtonian gravity and relativistic gravity at the center of the earth?
  8. Jun 25, 2007 #7
    The OP was asking about gravitational force and referred to the Newtonian formula.
  9. Jun 25, 2007 #8
    my question was an aside.
  10. Jun 25, 2007 #9


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    To calculate the force a person feels when below the surface of the Earth, you divide the Earth into two pieces:
    - the solid sphere that is below him
    - the hollow spherical shell that is above him

    The inner sphere below him acts as a point mass.

    The hollow shell above him cancels out its gravitational forces, leaving net zero because the forces felt anywhere inside any hollow sphere are net zero.

    Using these, it is trivial to calculate the net forces felt at the centre of the Earth ( i.e. where the inner sphere is of radius zero - hint hint.)
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