Careful measurements of local variations in the acceleration due to gravity can reveal the locations of oil deposits. Assume that the Earth is a uniform sphere of radius \(6370 \mathrm{~km}\) and density \(5500 . \mathrm{kg} / \mathrm{m}^{3},\) except that there is a spherical region of radius \(1.00 \mathrm{~km}\) and density \(900 . \mathrm{kg} / \mathrm{m}^{3}\) whose center is at a depth of \(2.00 \mathrm{~km} .\) Suppose you are standing on the surface of the Earth directly above the anomaly with an instrument capable of measuring the acceleration due to gravity with great precision. What is the ratio of the acceleration due to gravity that you measure compared to what you would have measured had the density been \(5500 . \mathrm{kg} / \mathrm{m}^{3}\) everywhere? (Hint: Think of this as a superposition problem involving two uniform spherical masses, one with a negative density.)

Step by step solution

01

Calculate gravitational force due to entire Earth

Recall Newton's law of gravitation formula: \(F = G\frac{m_1m_2}{r^2}\) Where F is the force of gravity, G is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the two objects, and r is the distance between their centers. For the entire Earth without considering the anomaly, the mass can be calculated using the volume and density: \(V_{Earth} = \frac{4}{3}\pi(6370 \times 10^3)^3 \,\mathrm{m^3}\) Mass \(m_{Earth} = \rho_{Earth} V_{Earth} = 5500 \,\mathrm{kg/m^3} \times V_{Earth}\) The gravitational force due to the entire Earth can be expressed as: \(F_{Earth} = G \frac{m_p m_{Earth}}{(6370 \times 10^3)^2}\) Where \(m_p\) is the mass of the person(we cancel this out later).

02

Calculate gravitational force due to anomaly

First, we calculate the mass of the anomaly due to difference in density: \(V_{anomaly} = \frac{4}{3}\pi(1000)^3 \,\mathrm{m^3}\) Mass \(m_{anomaly} = (\rho_{anomaly} - \rho_{Earth}) \times V_{anomaly} = (900 - 5500) \,\mathrm{kg/m^3} \times V_{anomaly}\) Notice that the mass has a negative value, which represents the "missing" mass due to the anomaly. Now we need to find the distance between the center of the Earth and the center of the anomaly: \(r_{anomaly} = (6370 - 2) \times 10^3 \,\mathrm{m}\) The gravitational force due to the anomaly can be expressed as: \(F_{anomaly} = G\frac{m_p m_{anomaly}}{r_{anomaly}^2}\)

03

Calculate the total gravitational force

Now we can find the total gravitational force which is the sum of the forces due to Earth and the anomaly: \(F_{total} = F_{Earth} + F_{anomaly}\) The ratio between the two forces simplifies and can be calculated as: \(\frac{F_{total}}{F_{Earth}} = 1 + \frac{F_{anomaly}}{F_{Earth}}\) Substitute the expressions for \(F_{anomaly}\) and \(F_{Earth}\) and rearrange to get: \(\frac{F_{total}}{F_{Earth}} = 1 - \frac{(900 - 5500) \times V_{anomaly}}{(6370 - 2)^2} \times \frac{6370^2}{5500 \times V_{Earth}}\)

04

Calculate the ratio

Calculate the ratio by substituting the values of \(V_{anomaly}\) and \(V_{Earth}\): \(\frac{F_{total}}{F_{Earth}} = 1 - \frac{(900 - 5500) \times \frac{4}{3}\pi(1000)^3}{(6370 - 2)^2} \times \frac{6370^2}{5500 \times \frac{4}{3}\pi(6370 \times 10^3)^3}\) The acceleration due to gravity is directly proportional to the force. Thus, the ratio of the accelerations is the same as the ratio of the forces. We can simplify the expression and calculate the final value for the ratio.

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