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

AI Thread Summary
The discussion revolves around estimating the percentage difference in gravitational acceleration (g) above a spherical pocket of oil located 1.20 km beneath the Earth's surface. The density of the oil is given as 800 kg/m³, and the calculations involve using the gravitational formula g=GM/r² and the density equation D=m/v. The initial calculations led to confusion regarding the negligible difference in g, prompting further clarification on the use of SI units. Ultimately, the problem was resolved, indicating that the difference in gravitational acceleration is minimal. The thread highlights the complexities involved in gravitational calculations related to subsurface structures.
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?
 
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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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