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## Homework Statement

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*10**

^{2}kg/m^{3}## Homework Equations

g = G*m/r

^{2}

## The Attempt at a Solution

[/B]

Normal gravity = G*m/r

^{2}

_{earth}

Gravity 0.5 km under the earth = G*m/(r

_{earth}-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*m

_{oil}/r

^{2}

_{oil}? 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?

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