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

In summary, we are trying to estimate the difference in g at a point directly above a spherical pocket of oil, which is 1.2 km beneath the Earth's surface and has a diameter of 1 km. Using the equations g=GM/r^2 and density=mass/volume, we can calculate the mass of the oil and plug it into the gravity equation to find the difference in g. However, the answer is negligible and can be rounded to 2 significant figures. The use of SI units of distance (meters) is also important in this calculation.
  • #1
Charanjit
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?
 
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  • #2
Are you using the SI units of distance?
 
  • #3
Yes, meters.
 
  • #4
Wow, its a tough one. I got it solved. Thanks anyways.
 

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

1. How is Delta(g)/g calculated for a 1.00 km diameter spherical pocket of oil?

The Delta(g)/g, or the change in gravitational acceleration, can be calculated by using the formula Delta(g)/g = 4/3 x pi x G x (rho_oil - rho_water) x r^3, where G is the gravitational constant, rho_oil and rho_water are the densities of oil and water respectively, and r is the radius of the pocket of oil.

2. What units are typically used for Delta(g)/g?

Delta(g)/g is typically measured in units of meters per second squared (m/s^2) or in parts per million (ppm).

3. How does the depth of the pocket of oil affect the value of Delta(g)/g?

The depth of the pocket of oil does not affect the value of Delta(g)/g, as it is dependent on the radius of the pocket and the difference in densities between the oil and surrounding water.

4. Can Delta(g)/g be used to estimate the size of the pocket of oil?

No, Delta(g)/g only provides information on the change in gravitational acceleration due to the presence of the pocket of oil. Other factors, such as the shape and density of the pocket, must also be considered when estimating the size of the pocket.

5. How accurate is the estimation of Delta(g)/g for a 1.00 km diameter spherical pocket of oil?

The accuracy of the estimation of Delta(g)/g depends on the accuracy of the input parameters, such as the density of the oil and water and the radius of the pocket. With precise measurements, the estimation can be accurate to a few parts per million.

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