The center of a 1.00 km diameter spherical pocket of oil is 1.00 km beneath the Earth's surface[/B]. 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*102 kg/m3
g = G*m/r2
The Attempt at a Solution
Normal gravity = G*m/r2earth
Gravity 0.5 km under the earth = G*m/(rearth-0.5)2
(6378.1/(6378.1-0.5))^2*100 = 100.01
So the gravity increases .01 percent just above the oil.
But the thing concerning me here is that I haven't used the density of oil. Was I supposed to think of it as an extra mass with an effect of G*moil/r2oil? If so, then all I have to is add the magnitude of gravity caused by oil to the normal gravity already being there - 0.5km under the earth, right?