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Chapter 5: Problem 18

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 \(2800 \mathrm{~kg} \mathrm{~m}^{-3}\).

Short Answer

Expert verified
The depth of the volcanic plug is approximately 8 km.

Step by step solution

01

Understanding the Gravity Anomaly Equation

The gravity anomaly (\( \Delta g \) ) due to a vertical cylinder is given by the formula:\[\Delta g = \frac{2\pi G \rho h}{\sqrt{d^2 + h^2}} - \frac{2\pi G \rho_0 h}{\sqrt{d^2 + h^2}}\]where \( G \) is the gravitational constant, \( \rho \) is the density of the plug, \( \rho_0 \) is the density of the surrounding rock, \( h \) is the height (or depth) of the plug, and \( d \) is the radius of the plug.
02

Simplifying the Gravity Anomaly Equation

Since the plug is a cylinder whose top is at the surface, we simplify the equation using the density contrast. The formula can be written as:\[\Delta g = \frac{2\pi G \Delta\rho h}{\sqrt{d^2 + h^2}}\]where \( \Delta\rho = \rho - \rho_0 \) is the density contrast between the plug and the surrounding rock.
03

Calculate Density Contrast and Simplify Further

First, compute the density contrast: \( \Delta\rho = 3000 \mathrm{~kg/m^3} - 2800 \mathrm{~kg/m^3} = 200 \mathrm{~kg/m^3} \).Substitute the values to simplify the formula to:\[0.3 \times 10^{-3} = \frac{2\pi (6.674\times 10^{-11}) \times 200 \times h}{\sqrt{(5000)^2 + h^2}}\]
04

Solve for Height (Depth) h

Rearrange the equation to isolate \( h \): \[0.3 \times 10^{-3} \times \sqrt{(5000)^2 + h^2} = 2\pi (6.674\times 10^{-11}) \times 200 \times h\]Use algebraic manipulation or numerical methods to solve for \( h \). This requires multiple iterations to find a value that satisfies the equation, which can be complex by hand but feasible computationally.

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Key Concepts

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

Density Contrast
The concept of density contrast is crucial in understanding gravity anomalies. It represents the difference in density between two substances. In geological terms, it refers to the difference in density between a geologic structure, like a volcanic plug, and the surrounding material.
In this exercise, the volcanic plug has a density of \(3000 \mathrm{~kg/m^3}\), while the surrounding rocks have a density of \(2800 \mathrm{~kg/m^3}\).
Calculating density contrast is straightforward:
  • You subtract the density of the surrounding material (\(\rho_0\)) from the density of the geologic body (\(\rho\)).
  • In this case, \(\Delta\rho = 3000 \mathrm{~kg/m^3} - 2800 \mathrm{~kg/m^3} = 200 \mathrm{~kg/m^3}\).
This difference is what contributes to the gravity anomaly, influencing the gravitational readings observed at the surface.
Vertical Cylinder Model
When modeling geological structures such as volcanic plugs, the vertical cylinder model is commonly used. It assumes that the shape of the structure can be approximated as a cylinder standing upright from the Earth's surface. This model greatly simplifies the calculations needed to estimate depth from gravity anomalies.
The formula used for this type of model focuses on these key parameters:
  • \( h \): Depth or height of the cylinder.
  • \( d \): Radius of the cylinder's base, derived from half the diameter (for our exercise, \( d = 5000 \) meters).
This constraint makes the model practical for surfaces like volcanic plugs where the diameter and density of the plug are known, allowing for a more manageable computation of depth.
Gravitational Constant
The gravitational constant, denoted as \(G\), is a fundamental component in the formula for calculating gravity anomalies. It is a constant value that is essential in the equations governing gravitational forces.
In the case of gravitational interactions on Earth, it is approximately \(6.674 \times 10^{-11} \mathrm{~m^3~kg^{-1}~s^{-2}}\).
Why is this important?
  • It allows for the calculation of gravitational effects over distances, including the effect of geological bodies on surface gravity.
  • It appears in the gravity anomaly equation, representing the influence of mass (specifically, the density contrast) on gravitational readings.
Understanding \(G\) and its role in these calculations helps you comprehend why gravity readings can change due to subsurface features.
Volcanic Plug
A volcanic plug is a unique geological feature formed when magma hardens within the vent of an active volcano. Over time, as the surrounding rock erodes, the resistant plug becomes exposed.
These structures are significant geologically and can be modeled to understand subsurface properties such as density and depth.
In our exercise, the volcanic plug is assumed to be a vertical cylinder reaching from the surface, representing a solid mass of more dense material compared to its surroundings. This feature's density and shape cause a gravity anomaly detectable by specialized instruments, revealing important clues about its depth and composition.
  • Interpreting gravity anomalies from volcanic plugs helps geoscientists understand volcanic activity and the history of the region.
  • By modeling these plugs as vertical cylinders, the complexity of their gravitational influence is simplified for analysis.

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Most popular questions from this chapter

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 $$ J_{2}=\frac{1}{5}\left(1-\frac{c^{2}}{a^{2}}\right) $$

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?

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

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.
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