# The center of a 1.00 km diameter spherical pocket of oil

1. Nov 17, 2014

### hitemup

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*102 kg/m3

2. Relevant equations

g = G*m/r2

3. 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?

Last edited: Nov 17, 2014
2. Nov 17, 2014

### Staff: Mentor

This is both wrong and not relevant.
You cannot use the equation for a point mass if you are inside this mass. And you are supposed to calculate the gravitational force on the surface, not below it.

What is the gravitational force from the oil?
What else changes apart from additional oil?

3. Oct 23, 2016

### NotAPhysWiz

Were you able to figure this out? I have the exact same question right now and I'm stuck.

4. Oct 23, 2016

### Staff: Mentor

Where are you stuck? Do the hints I gave in post #2 (2014...) help?

5. Oct 23, 2016

### Staff: Mentor

What have you tried? Where did you get stuck? You need to show an attempt.