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Gravity Ratio

  1. Apr 4, 2010 #1
    1. The problem statement, all variables and given/known data:
    The center of a 1.00 km diameter spherical pocket of oil is 1.20 km beneath the Earth's surface. Estimate by what percentage g directly above the pocket of oil would differ from the expected value of g for a uniform Earth? Assume the density of oil is 8.0*10^2 (kg/m^3).

    Delta(g)/g=



    2. Relevant equations
    g=GM/r2

    D=m/v

    3. The attempt at a solution:

    Well I calculated that the pocket of oil is 0.7km beneath the earth. And using density=mass/volume to get the M and plugged it into the gravity equation. And subtracted it from 9.8 and then divided by 9.8. The answer is neglegable since they want me to answer using 2 sigfigs. So I am kind of lost, what do I need to do?
     
  2. jcsd
  3. Apr 4, 2010 #2
    Are you using the SI units of distance?
     
  4. Apr 4, 2010 #3
    Yes, meters.
     
  5. Apr 4, 2010 #4
    Wow, its a tough one. I got it solved. Thanks anyways.
     
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