The center of a 1.10km diameter spherical pocket of oil is 1.10km beneath the Earth's surface.
Estimate by what percentage directly above the pocket of oil would differ from the expected value of for a uniform Earth? Assume the density of oil is .
Express your answer using two significant figures.
Subtract from the "normal" value of g the acceleration in the opposite direction due to a hypothetical sphere of negative mass equal to the difference between the mass of earth's crust and the same volume full of oil. This reduction in g will be equal to
a = G (Mcrust-Moil)/(0.55*10^3 m)^2
Insert the appropriate M values for a sphere of earth's crust and a sphere of oil.
To estimate the percentage by which the gravitational field directly above the pocket of oil would differ from the expected value for a uniform Earth, we need to compare the gravitational fields of the oil pocket and the Earth.
First, let's calculate the expected value for a uniform Earth. The gravitational field above the Earth's surface can be calculated using the formula:
g = G * M / r^2
Where:
- g is the gravitational field strength
- G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
- M is the mass of the Earth (approximately 5.972 × 10^24 kg)
- r is the distance from the center of the Earth to the point where the field is measured
Since we are interested in the percentage difference, we can take the expected value as a baseline. Therefore, let's calculate the gravitational field directly above the oil pocket and find the percentage difference.
1. Calculating the expected value:
Using the given diameter of the oil pocket (1.10 km), we can find the distance from the center of the Earth to the oil pocket's center:
d_oil = 1.10 km / 2 = 0.55 km = 550 m
Next, we can calculate the gravitational field directly above the oil pocket using the formula:
g_expected = G * M / (r + d_oil)^2
where r is the radius of the Earth (approximately 6,371 km or 6,371,000 m).
Substituting the values, we get:
g_expected = (6.67430 × 10^-11 m^3 kg^-1 s^-2 * 5.972 × 10^24 kg) / (6,371,000 m + 550 m)^2
2. Calculating the gravitational field above the oil pocket:
Given that the oil pocket is beneath the Earth's surface, the distance from the surface to the oil pocket's center is the Earth's radius minus the distance from the surface to the center:
d_diff = 6,371,000 m - 550 m = 6,370,450 m
Using the same formula as before, the gravitational field directly above the oil pocket can be calculated:
g_oil = G * (M_oil / V_oil) / d_diff^2
where M_oil is the mass of the oil pocket and V_oil is the volume of the oil pocket.
We are given the density of oil, so we can calculate its mass using the formula:
M_oil = density_oil * V_oil
Since the oil pocket is spherical, its volume can be calculated with the formula:
V_oil = (4/3) * π * (r_oil^3)
where r_oil is the radius of the oil pocket (half of the given diameter).
Substituting the values for the oil pocket into the formula, we get:
g_oil = G * (density_oil * (4/3) * π * (0.55 km / 2)^3) / (6,370,450 m)^2
3. Calculating the percentage difference:
To find the percentage difference, we will use the formula:
percentage difference = |(g_expected - g_oil) / g_expected| * 100
Substituting the previously calculated values, we get:
percentage difference = |(g_expected - g_oil) / g_expected| * 100
Now, you can substitute the appropriate values into the formulas and calculate the answer.