The center of a 1.00 km diameter spherical pocket of oil

AI Thread Summary
The discussion focuses on calculating the gravitational difference above a spherical pocket of oil located 1 km beneath the Earth's surface. The initial calculation suggests a minimal increase of 0.01 percent in gravity due to the pocket, but there is confusion regarding the use of the oil's density in the calculations. Participants emphasize that the gravitational effect should be calculated at the surface rather than below the oil. Additionally, there is a need to clarify how to incorporate the oil's mass into the gravitational equation correctly. The conversation highlights the importance of understanding gravitational forces in relation to both the Earth's mass and the additional mass of the oil.
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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*102 kg/m3

Homework Equations



g = G*m/r2

The Attempt at a Solution


[/B]
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?
 
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hitemup said:
Gravity 0.5 km under the earth
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?
 
hitemup said:

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*102 kg/m3

Homework Equations



g = G*m/r2

The Attempt at a Solution


[/B]
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?

Were you able to figure this out? I have the exact same question right now and I'm stuck.
 
Where are you stuck? Do the hints I gave in post #2 (2014...) help?
 
NotAPhysWiz said:
Were you able to figure this out? I have the exact same question right now and I'm stuck.
What have you tried? Where did you get stuck? You need to show an attempt.
 
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