Question

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.)

Short answer

Answer: The ratio of the acceleration due to gravity measured above the spherical anomaly to the acceleration due to gravity if the density was uniform throughout is approximately 1 - 0.00437, or 0.99563. This means the acceleration due to gravity above the anomaly is about 99.563% of what it would be if the density was uniform.

Step-by-step solution

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).

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: \(F_{anomaly} = G\frac{m_p m_{anomaly}}{r_{anomaly}^2}\)

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: \(\frac{F_{total}}{F_{Earth}} = 1 - \frac{(900 - 5500) \times V_{anomaly}}{(6370 - 2)^2} \times \frac{6370^2}{5500 \times V_{Earth}}\)

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.

Key concepts

Newton's Law of Gravitation

Newton's Law of Gravitation is a fundamental principle that describes the attractive force between two masses. It is given by the formula \[ F = G \frac{m_1m_2}{r^2} \] where \(F\) is the force of gravity between the masses, \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the two masses, and \(r\) is the distance between the centers of the two masses.
This law is universal, meaning it applies to all objects with mass regardless of their state or location. The essence of this law is that more massive objects exert a stronger gravitational pull. Additionally, the gravitational force decreases with the square of the distance between the objects. This is known as an inverse-square law. In practical terms, if the distance between two objects doubles, the gravitational force between them goes down by a factor of four (since \(2^2 = 4\)).
Using Newton's Law of Gravitation, scientists can calculate the gravitational force between the Earth and another object, such as a person or an instrument detecting gravitational anomalies, enabling us to detect variations in gravity due to different mass distributions.

Gravitational Constant

The gravitational constant, denoted by \(G\), is a key factor in the equation for Newton's Law of Gravitation. It is a fundamental physical constant that quantifies the strength of the gravitational force between two objects. The value of \(G\) is approximately \(6.674 \times 10^{-11} \, \text{m}^3\text{kg}^{-1}\text{s}^{-2}\). This constant is crucial for calculating gravitational forces and is used widely in physics to explore the effects of gravity on various scales, from tiny particles to massive celestial bodies.
Because \(G\) is such a small number, it implies that gravitational forces are extremely weak compared to other fundamental forces, like electromagnetism. This is one reason why everyday objects do not seem to exert gravitational forces on each other that we can easily notice. However, when it comes to massive objects like planets and stars, the force becomes significant and governs the motion of objects in the universe.
In the context of gravitational anomalies, the gravitational constant sits at the heart of calculations, helping to predict deviations from expected gravitational behaviors due to unusual distributions of mass.

Negative Density Mass

In gravitational anomaly problems, the concept of a negative density mass can be tricky but interesting. Here, it's a theoretical construct used to simplify calculations by modeling areas where the mass is less than expected. Basically, it represents the density of a region being less than the surrounding uniform density, appearing as if it were subtracting mass.
The anomaly in the exercise is treated as having 'missing' mass because it has a lower density compared to its surroundings. To calculate its effect on gravitational measurements, we assign this anomaly a mass based on its lesser density, leading to a 'negative' contribution to the gravitational force.
This idea helps in situations following the superposition principle, where the gravitational field from complex mass distributions is computed by adding the field from each part individually if it were isolated, including these 'negative density' components as subtractions from the total.

Superposition Problem

The superposition principle in physics allows the total gravitational field resulting from multiple sources to be considered as the sum of the individual fields. Each source's field can be calculated as if other sources aren't present and then combined to yield the total field.
In the context of gravitational anomaly detection as seen in the exercise, this principle is key. It allows scientists to account for a non-uniform distribution of mass, identifying anomalies by adding the gravitational effects of separate mass components, even those with negative density. This results in more accurate gravitational field measurements.
In simple terms, if there's an area with less mass (an anomaly), its gravitational effect is subtracted. The total gravitational force observed is a mix of the force due to the normal uniform mass of Earth plus the deviation introduced by the anomaly. The superposition principle allows these sources to be treated systematically, yielding precise measurements crucial in fields such as geophysics and oil exploration.
Using superposition, geophysicists can detect underground features by understanding how they might alter the local gravitational field, guiding decisions about resources like oil deposits.