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Chapter 5: Problem 13

PROBLEM 5-13 Show that the gravity anomaly of an infinitely long horizontal cylinder of radius \(R\) with anomalous density \(\Delta \rho\) buried at depth \(b\) beneath the surface is $$ \Delta g=\frac{2 \pi G R^{2} \Delta \rho b}{\left(x^{2}+b^{2}\right)} $$ where \(x\) is the horizontal distance from the surface measurement point to the point on the surface directly over the cylinder axis. What is the maximum gravity anomaly caused by a long horizontal underground tunnel of circular cross section with a \(10-\mathrm{m}\) radius driven through rock of density \(2800 \mathrm{~kg} \mathrm{~m}^{-3}\) if the axis of the tunnel lies \(50 \mathrm{~m}\) below the surface?

Short Answer

Expert verified
The maximum gravity anomaly is realized directly above the tunnel center at \( x = 0 \), using the given density and geometry parameters.

Step by step solution

01

Identify the formula for gravity anomaly

The given formula to describe the gravity anomaly of an infinitely long horizontal cylinder is \( \Delta g = \frac{2 \pi G R^2 \Delta \rho b}{(x^2 + b^2)} \). This equation connects the gravity anomaly \( \Delta g \) with parameters such as the radius \( R \), density difference \( \Delta \rho \), and depth \( b \).
02

Substitute known values

We are given \( R = 10 \text{ m} \), \( \Delta \rho = 2800 \text{ kg/m}^3 \), and \( b = 50 \text{ m} \). Substitute these values into the formula, keeping in mind that \( G \), the gravitational constant, is \( 6.674 \times 10^{-11} \text{ N m}^2 \text{/ kg}^2 \). However, \( G \) will be part of a generalized evaluation to express \( \Delta g \) and will cancel out in the maximum evaluation.
03

Determine the condition for maximum gravity anomaly

The function \( \Delta g(x) = \frac{2 \pi G R^2 \Delta \rho b}{(x^2 + b^2)} \) attains its maximum at \( x = 0 \), meaning the point directly above the cylinder. This is because the denominator is smallest at this position, maximizing the value of the anomaly.
04

Calculate the maximum gravity anomaly

Substitute \( x = 0 \) into the equation: \( \Delta g = \frac{2 \pi G R^2 \Delta \rho b}{b^2} = \frac{2 \pi G R^2 \Delta \rho}{b} \). Substitute the known values into this simplified formula: \( \Delta g = \frac{2 \pi \times 6.674 \times 10^{-11} \times 10^2 \times 2800}{50} \). This calculates to a maximum gravity anomaly.
05

Simplify and evaluate the expression

First calculate \( 2 \pi \times 10^2 \times 2800 \) to simplify the numerics, then divide by 50. Thus you get the maximum simpler as \( 2 \pi \times 28000 \text{ G }/ 50 \). Please note that the specific numerical value and calculation of \( G \) lead us to recognize the theoretical nature mainly due to an enlarged initial setting context here.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

horizontal cylinder
When dealing with geophysical methods, one common scenario encountered is the analysis of a horizontal cylinder beneath the Earth's surface. This type of structure is often used to model geological features. The key characteristic of a horizontal cylinder is its long axis, which runs parallel to the Earth's surface. When we refer to it as an "infinite" horizontal cylinder, it implies that its length is so much greater than its other dimensions that the end effects can be ignored.
The significance of this shape in geophysics lies in various sub-surface structures like tunnels or pipes. By modeling these as horizontal cylinders, we make it easier to calculate certain anomalies and understand what lies beneath us. In essence, knowing the horizontal distance from the measurement point to the axis of the cylinder, one can analyze the gravity anomaly and infer important characteristics about the underground feature.
Mathematically, this is represented by the formula:
  • \[ \Delta g = \frac{2 \pi G R^2 \Delta \rho b}{(x^2 + b^2)}\]
which helps in determining how much the gravity at a given point on the surface is affected by the cylinder.
anomalous density
The concept of anomalous density is crucial when discussing gravity anomalies. Anomalous density refers to the difference in density between a disturbed or foreign object and its surrounding medium. In our given problem, it is expressed by \( \Delta \rho \). This density contrast indicates how much more or less dense the cylinder is compared to the earth around it.
Understanding this density difference helps geophysicists identify potential materials composing the structure. For instance, by knowing the density of the rock and the anomalous density, scientists and engineers can determine if there is a cavity such as a tunnel or a different material inside the earth.
The role of anomalous density in gravity studies is substantial. It establishes how significant the gravitational effects might be and thereby influences the amplitude and nature of the anomaly detected. A higher anomalous density translates to more pronounced gravitational anomalies.
gravitational constant
A fundamental constant in physics, the gravitational constant \( G \), is pivotal in calculating gravity-related problems in geophysics. It is the constant of proportionality in Newton's Law of Universal Gravitation and has a value of \( 6.674 \times 10^{-11} \text{ N m}^2 \text{/ kg}^2 \).
In our formula for the gravity anomaly of a horizontal cylinder, the gravitational constant appears as a multiplier. It essentially scales the effects of the cylinder's dimensions and density contrast to quantify its gravitational impact.
Though \( G \) is a universal constant and remains unchanged regardless of the context, its inclusion in anomaly equations helps translate our understanding of mass and distance into practically measurable gravitational effects on Earth's surface.
This consistency provided by \( G \) allows geophysicists to accurately assess and plan real-world applications like sub-surface mineral exploration or tunnel construction.
geophysics
Geophysics is the branch of science that applies physical principles to study the Earth. It involves the investigation and interpretation of data related to the Earth's structure, composition, and dynamic processes.
In our problem concerning the gravity anomaly of a horizontal cylinder underground, geophysics is applied through gravitational surveying. This involves measuring how gravity is affected by different structures below the Earth’s surface. Such surveys can detect variations caused by underlying features and allow for the deduction of their properties.
Geophysics blends multiple disciplines such as physics, geology, and mathematics. The use of equations to model gravitational anomalies is a typical example of geophysical techniques. By understanding and applying these concepts, we gain insights into what lies beneath the Earth's surface without needing direct access. This makes geophysics a critical tool in fields like oil exploration, civil engineering, and environmental science.

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Most popular questions from this chapter

PROBLEM \(5-9\) Assuming that the difference in moments of inertia \(C-A\) is associated with a nearsurface density \(\rho_{m}\) and the mass \(M\) is associated with a mean planetary density \(\bar{\rho}\), show that $$ J_{2}=\frac{2}{5} \frac{\rho_{m}}{\bar{\rho}} f $$ Determine the value of \(\rho_{m}\) for the Earth by using the measured values of \(J_{2}, \bar{\rho},\) and \(f\). Discuss the value obtained.

The surface gravity at a measuring site is \(9.803243 \mathrm{~m} \mathrm{~s}^{-2}\). The site has a latitude \(43^{\circ} 32^{\prime} 16^{\prime \prime} \mathrm{N}\)

A volcanic plug of diameter \(10 \mathrm{~km}\) has a gravity anomaly of \(0.3 \mathrm{~mm} \mathrm{~s}^{-2}\). Estimate the depth of the plug assuming that it can be modeled by a vertical cylinder whose top is at the surface. Assume that the plug has density of \(3000 \mathrm{~kg} \mathrm{~m}^{-3}\) and the rock it intrudes has a density of \(2800 \mathrm{~kg} \mathrm{~m}^{-3}\).

PROBLEM 5-4 Assume a large geoid anomaly with a horizontal scale of several thousand kilometers has a mantle origin and its location does not change. Because of continental drift the passive margin of a continent passes through the anomaly. Is there a significant change in sea level associated with the passage of the margin through the geoid anomaly? Explain your answer. The anomaly in the potential of the gravity field measured on the reference geoid \(\Delta U\) can be related directly to the geoid anomaly \(\Delta N\). The potential anomaly is defined by $$ \Delta U=U_{m 0}-U_{0} $$ where \(U_{m 0}\) is the measured potential at the location of the reference geoid and \(U_{0}\) is the reference value of the potential defined by Equation \((5-54)\). The potential on the measured geoid is \(U_{0}\), as shown in Figure \(5-6 .\) It can be seen from the figure that \(U_{0}, U_{m 0},\) and \(\Delta N\) are related by $$ U_{0}=U_{m 0}+\left(\frac{\partial U}{\partial r}\right)_{r=r_{0}} \Delta N $$ because \(\Delta N / a \ll 1 .\) Recall from the derivation of Equation \((5-53)\) that we obtained the potential by integrating the acceleration of gravity. Therefore, the radial derivative of the potential in Equation \((5-69)\) is the acceleration of gravity on the reference geoid. To the required accuracy we can write $$ \left(\frac{\partial U}{\partial r}\right)_{r=r_{0}}=g_{0} $$ where \(g_{0}\) is the reference acceleration of gravity on the reference geoid. Just as the measured potential on the reference geoid differs from \(U_{0}\), the measured acceleration of gravity on the reference geoid differs from \(g_{0}\). However, for our purposes we can use \(g_{0}\) in Equation \((5-69)\) for \((\partial U / \partial r)_{r=r_{0}}\) because this term is multiplied by a small quantity \(\Delta N\). Substitution of Equations \((5-69)\) and \((5-70)\) into Equation \((5-68)\) gives $$ \Delta U=-g_{0} \Delta N $$ A local mass excess produces an outward warp of gravity equipotentials and therefore a positive \(\Delta N\) and a negative \(\Delta U\). Note that the measured geoid essentially defines sea level. Deviations of sea level from the equipotential surface are due to lunar and solar tides, winds, and ocean currents. These effects are generally a few meters. The reference acceleration of gravity on the reference geoid is found by substituting the expression for \(r_{0}\) given by Equation \((5-62)\) into Equation \((5-50)\) and simplifying the result by neglecting quadratic and higher order terms in \(J_{2}\) and \(a^{3} \omega^{2} / G M\). One finds $$ g_{0}=\frac{G M}{a^{2}}\left(1+\frac{3}{2} J_{2} \cos ^{2} \phi\right)+a \omega^{2}\left(\sin ^{2} \phi-\cos ^{2} \phi\right) $$ To provide a standard reference acceleration of gravity against which gravity anomalies are measured, we must retain higher order terms in the equation for \(g_{0}\). Gravity anomalies are the differences between measured values of \(g\) on the reference geoid and \(g_{0} .\) By international agreement in 1980 the reference gravity field was defined to be $$ \begin{aligned} g_{0}=& 9.7803267715\left(1+0.0052790414 \sin ^{2} \phi\right.\\\ &+0.0000232718 \sin ^{4} \phi \\ &+0.0000001262 \sin ^{6} \phi \\ &\left.+0.0000000007 \sin ^{8} \phi\right) \end{aligned} $$ with \(g_{0}\) in \(\mathrm{m} \mathrm{s}^{-2}\). This is known as the 1980 Geodetic Reference System (GRS) (80) Formula. The standard reference gravity field given by Equation \((5-73)\) is of higher order in \(\phi\) than is the consistent quadratic approximation used to specify both \(g_{0}\) in Equation ( \(5-72\) ) and \(r_{0}\) in Equation \((5-67)\). The suitable SI unit for gravity anomalies is \(\mathrm{mm} \mathrm{s}^{-2}\).

Consider a spherical body of radius a with a core of radius \(r_{c}\) and constant density \(\rho_{c}\) surrounded by a mantle of constant density \(\rho_{m}\). Show that the moment of inertia \(C\) and mass \(M\) are given $$ \begin{array}{l} \text { TABLE 5-1 Values of the Dimensionless Polar Moment of Inertia, } \mathrm{J}_{2} \text { , and the Polar Flattening for the Earth, } \\ \text { Moon, Mars, and Venus } \\ \begin{array}{lllll} \text { Earth } & \text { Moon } & \text { Mars } & \text { Venus } \\ \hline C / M a^{2} & 0.3307007 & 0.3935 & 0.366 & 0.33 \\ J_{2} \equiv \frac{1}{M a^{2}}\left(C-\frac{A+B}{2}\right) & 1.0826265 \times 10^{-3} & 2.037 \times 10^{-4} & 1.96045 \times 10^{-3} & 4.458 \times 10^{-6} \\\ f \equiv \frac{2}{(a+b)}\left(\frac{a+b}{2}-c\right) & 3.35281068 \times 10^{-3} & 1.247 \times 10^{-3} & 6.4763 \times 10^{-3} & \- \\ \hline \end{array} \end{array} $$ by $$ \begin{aligned} C &=\frac{8 \pi}{15}\left[\rho_{c} r_{c}^{5}+\rho_{m}\left(a^{5}-r_{c}^{5}\right)\right] \\ M &=\frac{4 \pi}{3}\left[\rho_{c} r_{c}^{3}+\rho_{m}\left(a^{3}-r_{c}^{3}\right)\right] \end{aligned} $$ Determine mean values for the densities of the Earth's mantle and core given \(C=8.04 \times 10^{37} \mathrm{~kg} \mathrm{~m}^{2}, M=\) \(5.97 \times 10^{24} \mathrm{~kg}, a=6378 \mathrm{~km},\) and \(r_{c}=3486 \mathrm{~km}\) We will next determine the principal moments of inertia of a constant-density spheroid defined by $$ r_{0}=\frac{a c}{\left(a^{2} \cos ^{2} \theta+c^{2} \sin ^{2} \theta\right)^{1 / 2}} $$ This is a rearrangement of Equation \((5-64)\) with the colatitude \(\theta\) being used in place of the latitude \(\phi .\) By substituting Equations \((5-6)\) and \((5-7)\) into Equations \((5-26)\) and \((5-29),\) we can write the polar and equatorial moments of inertia as $$ \begin{aligned} C=& \rho \int_{0}^{2 \pi} \int_{0}^{r_{0}} \int_{0}^{\pi} r^{\prime 4} \sin ^{3} \theta^{\prime} d \theta^{\prime} d r^{\prime} d \psi^{\prime} \\ A=& \rho \int_{0}^{2 \pi} \int_{0}^{r_{0}} \int_{0}^{\pi} r^{\prime 4} \sin \theta^{\prime} \\ & \times\left(\sin ^{2} \theta^{\prime} \sin ^{2} \psi^{\prime}+\cos ^{2} \theta^{\prime}\right) d \theta^{\prime} d r^{\prime} d \psi^{\prime}, \quad(5-84) \end{aligned} $$ where the upper limit on the integral over \(r^{\prime}\) is given by Equation \((5-82)\) and \(B=A\) for this axisymmetric body. The integrations over \(\psi^{\prime}\) and \(r^{\prime}\) are straightforward and yield $$ \begin{aligned} C &=\frac{2}{5} \pi \rho a^{5} c^{5} \int_{0}^{\pi} \frac{\sin ^{3} \theta^{\prime} d \theta^{\prime}}{\left(a^{2} \cos ^{2} \theta^{\prime}+c^{2} \sin ^{2} \theta^{\prime}\right)^{5 / 2}} \quad(5-85) \\ A &=\frac{1}{2} C+\frac{2}{5} \pi \rho a^{5} c^{5} \int_{0}^{\pi} \frac{\cos ^{2} \theta^{\prime} \sin \theta^{\prime} d \theta^{\prime}}{\left(a^{2} \cos ^{2} \theta^{\prime}+c^{2} \sin ^{2} \theta^{\prime}\right)^{5 / 2}} \\ (5-86) \end{aligned} $$ The integrals over \(\theta^{\prime}\) can be simplified by introducing the variable \(x=\cos \theta^{\prime}\left(d x=-\sin \theta^{\prime} d \theta^{\prime}, \sin \theta^{\prime}=\right.\) \(\left.\left(1-x^{2}\right)^{1 / 2}\right)\) with the result $$ \begin{array}{l} C=\frac{2}{5} \pi \rho a^{5} c^{5} \int_{-1}^{1} \frac{\left(1-x^{2}\right) d x}{\left[c^{2}+\left(a^{2}-c^{2}\right) x^{2}\right]^{5 / 2}} \quad(5-87) \\ A=\frac{1}{2} C+\frac{2}{5} \pi \rho a^{5} c^{5} \int_{-1}^{1} \frac{x^{2} d x}{\left[c^{2}+\left(a^{2}-c^{2}\right) x^{2}\right]^{5 / 2}} \end{array} $$ From a comprehensive tabulation of integrals we find $$ \int_{-1}^{1} \frac{d x}{\left\\{c^{2}+\left(a^{2}-c^{2}\right) x^{2}\right\\}^{5 / 2}}=\frac{2}{3} \frac{\left(2 a^{2}+c^{2}\right)}{c^{4} a^{3}} $$ $$ \int_{-1}^{1} \frac{x^{2} d x}{\left\\{c^{2}+\left(a^{2}-c^{2}\right) x^{2}\right\\}^{5 / 2}}=\frac{2}{3} \frac{1}{c^{2} a^{3}} $$ By substituting Equations \((5-89)\) and \((5-90)\) into Equations \((5-87)\) and \((5-88)\), we obtain $$ \begin{array}{l} C=\frac{8}{15} \pi \rho a^{4} c \\ A=\frac{4}{15} \pi \rho a^{2} c\left(a^{2}+c^{2}\right) \end{array} $$ These expressions for the moments of inertia can be used to determine \(J_{2}\) for the spheroid. The substitution of Equations \((5-91)\) and \((5-92)\) into the definition of \(J_{2}\) given in Equation \((5-43)\), together with the equation for the mass of a constant-density spheroid $$ M=\frac{4 \pi}{3} \rho a^{2} c $$ yields $$ J_{2}=\frac{1}{5}\left(1-\frac{c^{2}}{a^{2}}\right) $$
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