The center of a 1.00km diameter spherical pocket of oil is 1.00kkm beneath the Earth's surface.
Estimate by what percentage g directly above the pocket of oil would differ from the expected value of for a uniform Earth? Assume the density of oil is 8.0*10^2kg/m^3.
Δg/g = ? %
To estimate the percentage by which g directly above the pocket of oil would differ from the expected value for a uniform Earth, we can use the concept of gravitational acceleration and the principle of superposition.
To start, let's first calculate the expected value of g for a uniform Earth. The gravitational acceleration, g, for a uniform Earth can be determined using the equation:
g = G * M / R^2
Where:
- G is the gravitational constant (approximately 6.67430 x 10^-11 m^3 kg^-1 s^-2)
- M is the mass of the Earth
- R is the radius of the Earth
The mass of the Earth, M, can be calculated using the equation:
M = density * V
Where:
- density is the average density of the Earth (approximately 5,515 kg/m^3)
- V is the volume of the Earth
The volume of the Earth, V, can be calculated using the formula for the volume of a sphere:
V = (4/3) * π * R^3
Now, let's calculate the expected value of g for a uniform Earth:
1. Calculate the mass of the Earth:
M = density * V
M = 5,515 kg/m^3 * (4/3) * π * (6,371 km)^3
M ≈ 1.08 x 10^24 kg
2. Calculate the radius of the Earth:
R = 6,371 km * 1,000 m/km
R ≈ 6.37 x 10^6 m
3. Calculate g for a uniform Earth:
g = G * M / R^2
g = (6.67430 x 10^-11 m^3 kg^-1 s^-2) * (1.08 x 10^24 kg) / (6.37 x 10^6 m)^2
g ≈ 9.81 m/s^2
Now, let's calculate the gravitational acceleration, g', directly above the pocket of oil. We can consider the system to be the combination of the uniform Earth and the oil pocket.
The pocket of oil is spherical with a diameter of 1.00 km, so its radius is 0.50 km. The center of the pocket is 1.00 km beneath the Earth's surface. Therefore, the distance from the center of the Earth to the center of the pocket is:
Distance = R + Depth
Distance = 6,371 km + 1.00 km
Distance = 6.372 km
The mass of the oil pocket, M', can be calculated as follows:
M' = density_oil * V_pocket
M' = (8.0 x 10^2 kg/m^3) * (4/3) * π * (0.50 km)^3
M' ≈ 6.283 x 10^8 kg
Now, let's calculate g' for the combination of the uniform Earth and the oil pocket:
g' = G * (M + M') / Distance^2
g' = (6.67430 x 10^-11 m^3 kg^-1 s^-2) * (1.08 x 10^24 kg + 6.283 x 10^8 kg) / (6.372 km * 1,000 m/km)^2
g' ≈ 9.809 m/s^2
Finally, let's calculate the percentage difference between g' and the expected value of g for a uniform Earth:
Δg/g = (g' - g) / g * 100%
Δg/g = (9.809 m/s^2 - 9.81 m/s^2) / 9.81 m/s^2 * 100%
Δg/g ≈ -0.001%
Therefore, the gravitational acceleration directly above the pocket of oil would differ from the expected value for a uniform Earth by approximately -0.001%.