The center of a 1.40 km diameter spherical pocket of oil is 1.50 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*10^2kg/m^3.

Express your answer using two significant figures.

To estimate the percentage by which the gravitational acceleration (g) directly above the pocket of oil would differ from the expected value of g for a uniform Earth, we need to calculate the change in g caused by the presence of the oil.

The formula for gravitational acceleration is given by:

g = (G * M) / R^2

Where:
G is the gravitational constant (approximately 6.67430 × 10^-11 m^3⋅kg^-1⋅s^-2),
M is the mass of the object,
R is the distance from the center of the object.

In this case, we have a spherical pocket of oil with a radius (r) of 0.7 km (half of the diameter), which is equivalent to 700 m. So, the distance from the center of the oil pocket to the Earth's surface (R') is given by:

R' = 1.5 km + 700 m
= 2.2 km
= 2200 m

We also assume the density of the oil is 8.0 × 10^2 kg/m^3.

To calculate the change in gravitational acceleration caused by the oil, we need to determine the mass of the oil pocket. Using the formula:

Mass = Density * Volume

The volume of the pocket can be calculated using the formula for the volume of a sphere:

Volume = (4/3) * π * r^3

Substituting the values:

Volume = (4/3) * π * (700 m)^3
≈ 1.437 × 10^9 m^3

Mass = Density * Volume
= (8.0 × 10^2 kg/m^3) * (1.437 × 10^9 m^3)
≈ 1.15 × 10^12 kg

Now, we can calculate the change in gravitational acceleration caused by the presence of the oil using the formula:

Δg = (G * Mass) / R'^2

Substituting the values:

Δg = (6.67430 × 10^-11 m^3⋅kg^-1⋅s^-2 * 1.15 × 10^12 kg) / (2200 m)^2
≈ 7.20 × 10^-7 m/s^2

Finally, we can calculate the percentage change as:

Percentage change = (Δg / g_uniform) * 100%

Where g_uniform is the expected value of g for a uniform Earth, which is approximately 9.81 m/s^2.

Substituting the values:

Percentage change = (7.20 × 10^-7 m/s^2 / 9.81 m/s^2) * 100%
≈ 7.3%

Therefore, the estimated percentage by which g directly above the pocket of oil would differ from the expected value of g for a uniform Earth is approximately 7.3%.

To estimate the percentage by which g directly above the pocket of oil differs from the expected value of g for a uniform Earth, we need to calculate the gravitational force exerted by the oil pocket and compare it with the gravitational force exerted by the rest of the Earth.

The formula for calculating the gravitational force is:

F = (G * m1 * m2) / r^2

where F is the gravitational force, G is the gravitational constant (approximately 6.67430 x 10^-11 N m^2/kg^2), m1 and m2 are the masses of the objects, and r is the distance between their centers.

In this case, we are interested in comparing the gravitational force at the Earth's surface directly above the oil pocket (Point A) and the gravitational force for a uniform Earth (Point B). We can assume that Point B is a point directly above the center of the Earth.

1. Calculate the mass of the oil pocket:

The volume of the oil pocket can be calculated using the formula for the volume of a sphere:

V = (4/3) * π * r^3

Given that the diameter of the spherical pocket is 1.40 km, the radius (r) can be calculated as 1.40 km / 2 = 0.70 km = 700 m.

Plugging in this value into the formula, we get:

V = (4/3) * 3.14159 * (700 m)^3 ≈ 1436029870 m^3

The mass (m1) of the oil pocket can be calculated using the formula:

m1 = density * V

Given a density of 8.0 * 10^2 kg/m^3, we can calculate:

m1 = (8.0 * 10^2 kg/m^3) * 1436029870 m^3 ≈ 1.1488 * 10^9 kg

2. Calculate the gravitational force at Point A:

The radius (rA) from the center of the Earth to Point A is the sum of the distance from the surface of the Earth to the center of the oil pocket (1.50 km) and the radius of the Earth:

rA = 1.50 km + 6,371 km = 6,372.5 km = 6,372,500 m

Now, we can calculate the gravitational force (FA) at Point A using the formula:

* m1 * m2) / rA^2

The mass of the Earth (m2) is approximately 5.972 * 10^24 kg.

Plugging in the values, we get:

FA ≈ (6.67430 x 10^-11 N m^2/kg^2) * (1.1488 * 10^9 kg) * (5.972 * 10^24 kg) / (6,372,500 m)^2

3. Calculate the gravitational force at Point B:

The radius (rB) from the center of the Earth to Point B (directly above the center) is the radius of the Earth:

rB = 6,371 km = 6,371,000 m

Now, we can calculate the gravitational force (FB) at Point B using the same formula:

FB = (G * m1 * m2) / rB^2

Plugging in the values, we get:

FB ≈ (6.67430 x 10^-11 N m^2/kg^2) * (1.1488 * 10^9 kg) * (5.972 * 10^24 kg) / (6,371,000 m)^2

4. Calculate the percentage difference:

The percentage difference can be calculated using the formula:

% difference = (FA - FB) / FB * 100

Plugging in the values, we get:

% difference ≈ (FA - FB) / FB * 100

Note that the gravitational acceleration (g) is proportional to the gravitational force (F). Therefore, the percentage difference in g directly above the pocket of oil compared to the expected value of g for a uniform Earth will be the same as the percentage difference in F at Points A and B.

Gravitational field= GMe/re^2-Mass/700^2 + G Massoil/700^2

= 9.8N/kg+ G/700^2 (volumemhole)(densityoil-densityEarth)
= 9.8+G/490000 * 4/3*PI*700^3 (800-6000)
(I estimated Earth density at 6times that of water)
= 9.8+6.67E-11/4900 (1.44E9)(-5200)
=9.8-.12 or 9.7N/kg

check all this.