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Change of g due to oil deposit

  1. Oct 25, 2015 #1
    1. The problem statement, all variables and given/known data

    The center of a 1.00 km diameter spherical pocket of oil is 1.00 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 (800) kg/m^3

    2. Relevant equations:

    Law of Universal Gravitation Fg= G(m1*m2/r^2)
    Density= mass/volume
    Volume of a Sphere: 4/3πr^3
    Gravitational Field g=Gm/r^2
    g on the surface of the earth is 9.80 m/s^2

    3. The attempt at a solution

    The first thing I attempted for this problem was to visualize the given scenario, we happen to have a sphere of oil that is 1.00 km in diameter, the center of this oil deposit is directly 1.00 km beneath the surface of the earth. Which implies that the top of the sphere is only 0.5 km or 500m beneath the surface of the earth.

    The next step I took was to find all the values of each sphere (earth, oil).

    Sphere of Oi:
    Radius: 500m
    Volume: 5.24×10^8m^3
    Density: 800 kg/m^3
    Mass: 4.192*10^11kg

    Earth:
    Radius: 6.380*10^6m
    Volume: 1.087*10^21m^3
    Density: 5540 kg/m^3
    Mass: 5.97*10^24kg

    Having all this information, I do not know what to do next to figure out the g within the top of the sphere and how I find the percentage it would differ. I know that if I calculate the gravitational field for earth i'll get 9.80 m/s^2. If I do the same for the sphere of oil ill get 0.00011184256m/s^2 which is a reasonably small number for a gravitational field. I do not know if I should subtract both gravitational fields and then do a ratio with my result and 9.80m/s^2 to get a percentage. Any help will be appreciated!
     
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  3. Oct 25, 2015 #2

    ogg

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    The Earth, below the bottom of the oil should be considered uniform and a point-like mass at its center.
    So the only 'important' mass is contained in the shell extending from the surface down to 1.5 km, the shell that the sphere of oil is embedded.
    That's about as far as I get...I'd need to integrate the force at the point on the surface directly above the oil over the entire shell...but the problem contains so many symmetries that there's got to be a better far more simple way to solve this. If that shell were uniform then the force would be the same as that mass at the center of the Earth (again, a point-mass). Hmm. so picture the sphere being filled with "Earth" and its contribution would be its mass at its center and the WHOLE Earth's would be g...yeah you DO have enough to do the problem. You subtract out (from g) the sphere of dirt and add back the sphere of oil (I speak of their contribution to the total g force).
     
    Last edited: Oct 25, 2015
  4. Oct 25, 2015 #3
    The top of a 1km diameter sphere is 0.5km beneath the earth surface, since the problem states center of the sphere of oil is 1 km.
     
  5. Oct 25, 2015 #4
    center of the sphere of oil is 1km beneath the surface of earth.
     
  6. Oct 25, 2015 #5
    https://ton.twitter.com/1.1/ton/data/dm/658364940089032707/658364940126846976/-y9GZJfc.jpg [Broken]

    I don't know if I am visualizing this correctly, also how do I find the mass of a spherical shell? Is that what I even have to find for?
     
    Last edited by a moderator: May 7, 2017
  7. Oct 25, 2015 #6

    haruspex

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    You say you calculated the field from the sphere of oil, but is that at the edge of that sphere or at the surface of tne Earth?
    If you were to take away the oil and replace it with the rocks that would otherwise have been there, how much would that increase the field by?

    You do not need to worry about spherical shells.
     
  8. Oct 25, 2015 #7
    3PRpjND.jpg
    I don't know why the image link was broken, but I got it here to work. This is what I am visualizing.
     
  9. Oct 25, 2015 #8

    SteamKing

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    What are the units of G?
     
  10. Oct 25, 2015 #9
    meters per second squared
     
  11. Oct 25, 2015 #10

    SteamKing

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    No, you are thinking of little g, the acceleration due to gravity.

    My question is, what are the units of big-G, the universal gravitational constant, G = 6.67×10-11 {Insert Units Here} ?
     
  12. Oct 25, 2015 #11
    The units for that are G = 6.67×10-11 N*(m^2/kg^2)
     
  13. Oct 25, 2015 #12

    haruspex

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    In your calculation of the field due to the oil sphere, you used r=500m. I ask again, is this supposed to be the field at the edge of the oil sphere or the field at the Earth's surface?
     
  14. Oct 25, 2015 #13
    The field is suppose to be of the edge of the oil sphere, cause I assumed that the distance from the sphere to earth is 500m from the Earth's surface.
    Or should I subtract the 500m from the radius of the earth since the sphere is beneath the surface?
     
  15. Oct 25, 2015 #14

    haruspex

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    But you want to know how the oil sphere affects the total field at the Earth's surface, no? The field at the edge of the sphere is not interesting.
     
  16. Oct 25, 2015 #15
    I thought that when the question asked "Estimate by what percentage g directly above the pocket of oil would differ from the expected value of g for a uniform Earth?" It was asking to find the field directly above the pocket of oil, which is equivalent to 500m below the surface of the earth due to the it's stated fact that the center of the pocket of oil is 1km directly under the surface of the earth.
     
  17. Oct 25, 2015 #16

    haruspex

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    No, I don't think that's the right interpretation. g is the Earth's field at its surface, so the question is asking for how the field at the surface changes. "Directly above" just means it is in a vertical line above the centre of the sphere. To imply touching the oil sphere, it should say something like "immediately above".
     
  18. Oct 25, 2015 #17
    If that is the case then how do I calculate for that?
     
  19. Oct 25, 2015 #18

    haruspex

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    Same formula, different r.
     
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