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Problem with Gravitation inside the Earth

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data
    How deep would a mine shaft have to be for the gravitational acceleration at the bottom to be reduced by a factor of 5 from its value on the Earth's surface?


    2. Relevant equations

    g=r(GME/RE3)
    G = 6.67 * 10-11
    ME = 5.97 * 1024 kg
    RE = 6.38 * 106 m

    3. The attempt at a solution
    I set g/5 equal to the equation g=r(GME/RE3). I got it wrong. I also tried dividing the result by 5, and it still didn't work. What else am I supposed to do?
     
  2. jcsd
  3. Nov 19, 2011 #2

    Doc Al

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    Staff: Mentor

    That should work. First simplify the equation to show g(r) in terms of g at the earth's surface. (You shouldn't have to use G or ME.)
     
  4. Nov 19, 2011 #3
    You have to take into account that the mass of the Earth below you is less. You'll probably want to assume that the Earth is of uniform density to to that.

    Do you see how?
     
  5. Nov 19, 2011 #4
    Good point, Fewmet. However, how would I calculate the mass below if I don't know the depth?
     
  6. Nov 19, 2011 #5

    Doc Al

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    The equation he has assumes just that. It assumes the Earth is of uniform density.
     
  7. Nov 19, 2011 #6
    hmm... I hastily read g=r(GME/RE3) as g=GME/RE2. My mistake.

    I don't immediately see where g=r(GME/RE3) comes from. Is r depth or the radius of the Earth beneath you when you are at depth?

    I assume the derivation involves g=GME/RE2. I was thinking that assumes a point mass rather than a uniform density of the planet. (Sorry if that seems like quibbling.)

    xnitexlitex:
    My reply was in the context of approaching the problem as though you were standing on a smaller planet when down in the mine shaft. You are closer to the center of the Earth and are being attracted by a lesser mass. If you are at depth d, your are at radius RE-d, the mass is ρ(4/3)∏(RE-d)3.

    Alternatively, you could subtract the mass of the outer shell (whose thickness is d).

    While you don't know d, it is the only unknown in g=GME/RE2.
     
  8. Nov 19, 2011 #7
    Looking again at g=r(GME/RE3), doesn't that tell you that acceleration is g/5 at r/5?

    As I said, I do not know what r represents, so I do not know how helpful that observation is.
     
  9. Nov 20, 2011 #8

    Doc Al

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    r is the distance from the center of the Earth.

    The final answer involves g, but that answer is derived assuming a uniform density.

    Let r = the distance from the Earth's center and then derive the effective g(r). You'll get the equation this thread started with.
     
  10. Nov 20, 2011 #9
    Cool. I've got the derivation now. Thanks for taking time to sketch that out.

    I'd long known that the g increased linearly with distance from the center of the Earth, but only as a qualitative piece of information. I hope my clumsy approach hasn't robbed xnitexlitex of the joy of discovering that.
     
  11. Nov 20, 2011 #10
    I'm glad that you figured out the derivation, but it still isn't helping.
     
  12. Nov 20, 2011 #11
    You could go with Doc Al's original suggestion and eliminate G and ME in
    g=r(GME/RE3)
    using
    g=GME/RE2

    If that doesn't make clear how to proceed, go back to what I said in the seventh post in this strand (which probably gives it away too easily). If not, post again to let us know.
     
  13. Nov 20, 2011 #12

    sophiecentaur

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    Why not just work out the mass of Earth below you when you at r from the centre (using volume and density) and then work out the gravitational force from that mass at distance r ( using G and mass).
    It all falls out very nicely and gives a laughably simple answer.
     
  14. Nov 20, 2011 #13

    Doc Al

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    Show exactly what you plugged in where. Even if you didn't simplify the expression as I suggested, it should still work just fine. Realize that the formula gives you r, the distance from the Earth's center; you then have to use that to calculate the depth below the surface.
     
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