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Physics Problem 6.20

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The center of a 1.40 km diameter spherical pocket of oil is 1.50 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*10^2kg/m^3.

Express your answer using two significant figures.

  • Physics Problem 6.20 - ,

    Gravitational field= GMe/re^2-Mass/700^2 + G Massoil/700^2

    = 9.8N/kg+ G/700^2 (volumemhole)(densityoil-densityEarth)
    = 9.8+G/490000 * 4/3*PI*700^3 (800-6000)
    (I estimated Earth density at 6times that of water)
    = 9.8+6.67E-11/4900 (1.44E9)(-5200)
    =9.8-.12 or 9.7N/kg

    check all this.

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