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Change in gravity due to oil deposit

  1. Sep 22, 2015 #1
    1. The problem statement, all variables and given/known data
    "A mining company finds gravity is 2 parts in 10^7 less over a certain area. Assume there is an oil deposit 3000m below. Take the density of rock to be 3000 kg/m^3, and the density of oil to be 1000 kg/m^3. Estimate the mass of the deposit, assumed spherical."

    2. Relevant equations
    $$F_g = \frac{Gm_1m_2}{r^2}$$

    3. The attempt at a solution
    I remember my professor mentioning in class that the oil deposit's mass can be treated as negative. Thus when the acceleration due to gravity from the oil deposit was subtracted from the total acceleration due to gravity, this would be the same as measured gravity. Thus: $$\frac{Gm_{earth}}{r_{earth}^2}-\frac{Gm_{oil}}{r_{toOil}^2}=\frac{Gm_{earth}}{r_{earth}^2}-\frac{2}{{10}^{7}}$$

    where ##r_{toOil}## is the radius between the surface of the earth and the oil (3000m). This simplifies to: $$\frac{Gm_{oil}}{r_{toOil}^2}=\frac{2}{{10}^{7}}$$

    And when I plugged in the numbers I got the result that the mass of the oil was 2.697e+10 kg, which was wrong. I have no clue where to go from here. Any guidance would be enormously appreciated.
     
  2. jcsd
  3. Sep 23, 2015 #2

    BvU

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    2 parts in 107 is probably relative:$$
    \frac{Gm_{earth}}{r_{earth}^2}-\frac{Gm_{oil}}{r_{toOil}^2}=\frac{Gm_{earth}}{r_{earth}^2}\left ( 1-\frac{2}{{10}^{7}} \right ) $$
     
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