The center of a 1.00 km diameter spherical pocket of oil

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Homework Help Overview

The problem involves estimating the difference in gravitational acceleration (g) directly above a spherical pocket of oil located 1.00 km beneath the Earth's surface, with a diameter of 1.00 km and a given oil density of 8.0 x 10^2 kg/m³.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of gravitational equations and the relevance of the oil's density in calculating the gravitational force. Some express confusion about whether to consider the oil as an additional mass affecting gravity.

Discussion Status

There are multiple interpretations of how to approach the problem, particularly regarding the gravitational effects of the oil and the correct application of gravitational equations. Some participants have provided hints and questions to guide understanding, but no consensus has been reached.

Contextual Notes

Participants note the importance of considering the gravitational force from the oil and the implications of being inside a mass when calculating gravitational effects.

hitemup
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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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