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Gravity Ratio

  • Thread starter Charanjit
  • Start date
  • #1
48
0
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?
 

Answers and Replies

  • #2
69
0
Are you using the SI units of distance?
 
  • #3
48
0
Yes, meters.
 
  • #4
48
0
Wow, its a tough one. I got it solved. Thanks anyways.
 

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