The center of a 1.00 km diameter spherical pocket of oil

In summary, the conversation is about estimating the percentage difference in gravitational force above a spherical pocket of oil that is 1.00 km beneath the Earth's surface. The conversation discusses using the equation g = G*m/r^2 to calculate the expected value of g for a uniform Earth and how to factor in the density of oil (8.0*10^2 kg/m^3) to the calculation. The conversation ends with a request for the person to show their attempt and where they got stuck.
  • #1
hitemup
81
2

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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  • #2
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?
 
  • #3
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.
 
  • #4
Where are you stuck? Do the hints I gave in post #2 (2014...) help?
 
  • #5
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.
 

1. What is the center of a 1.00 km diameter spherical pocket of oil?

The center of a 1.00 km diameter spherical pocket of oil is the point at which the oil is equidistant from all points on the surface of the sphere. It is the exact midpoint of the sphere.

2. How is the center of a spherical pocket of oil determined?

The center of a spherical pocket of oil can be determined using mathematical equations and measurements of the diameter and volume of the oil pocket. It can also be located using advanced imaging technology such as CT scans or MRI.

3. Is the center of a 1.00 km diameter spherical pocket of oil fixed or can it change?

The center of a 1.00 km diameter spherical pocket of oil is fixed, as long as the diameter and volume of the pocket remain constant. However, if there are changes in the diameter or volume of the pocket, the center may also shift.

4. Why is it important to know the center of a spherical pocket of oil?

Knowing the center of a spherical pocket of oil is important for accurately calculating its volume and for determining the best methods for extracting the oil. It is also important for understanding the structure and behavior of the pocket.

5. Can the center of a 1.00 km diameter spherical pocket of oil be located without physical measurements?

Yes, the center of a 1.00 km diameter spherical pocket of oil can be located using mathematical equations and advanced imaging technology, without needing to physically measure the diameter and volume of the pocket.

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