Estimating Delta(g)/g for a 1.00 km Diameter Spherical Pocket of Oil

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Homework Help Overview

The problem involves estimating the percentage difference in gravitational acceleration (g) directly above a spherical pocket of oil located beneath the Earth's surface. The context includes the dimensions of the pocket and its density, with the goal of comparing the calculated g to the expected value for a uniform Earth.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the mass of the oil pocket using its density and volume, then apply this to the gravitational equation. There are questions about the use of SI units and the interpretation of the results.

Discussion Status

The discussion includes attempts to clarify the calculations and units used. Some participants express uncertainty about the approach, while one participant claims to have solved the problem, indicating a potential shift in focus or understanding.

Contextual Notes

There is mention of significant figures in the context of the expected answer, and the original poster expresses feeling lost in the process of reaching a conclusion.

Charanjit
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1. Homework Statement :
The center of a 1.00 km diameter spherical pocket of oil is 1.20 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*10^2 (kg/m^3).

Delta(g)/g=



2. Homework Equations
g=GM/r2

D=m/v

3. The Attempt at a Solution :

Well I calculated that the pocket of oil is 0.7km beneath the earth. And using density=mass/volume to get the M and plugged it into the gravity equation. And subtracted it from 9.8 and then divided by 9.8. The answer is neglegable since they want me to answer using 2 sigfigs. So I am kind of lost, what do I need to do?
 
Physics news on Phys.org
Are you using the SI units of distance?
 
Yes, meters.
 
Wow, its a tough one. I got it solved. Thanks anyways.
 

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