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FUNDAMENTALS of GEOPHYSICS
3rd edition
SOLUTIONS TO EXERCISES
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William Lowrie and Andreas Fichtner
ETH Zürich
2019
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CONTENTS
Significant digits in a calculation
Exercises to
1. The Solar System
2. Plate Tectonics
3. Gravity and the Figure of the Earth
4. Gravity Surveying
5. Rheology of the Earth
6. Seismology
7. Earthquakes and the Earth’s Internal Structure
8. Geochronology
9. The Earth’s Heat
10. Geoelectricity
11. The Earth’s Magnetic Field
12. Paleomagnetism
NOTE
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Individual sections of these solutions were checked by several helpful colleagues, who
drew attention to errors and inconsistencies. The authors thank (in alphabetic order) Mark
Bukowinski, Chris Finlay, Ann Hirt, Dennis Kent, Luca Lanci, Hansruedi Maurer, Felix
Oberli, Henry Pollack, Jan Van der Kruk, and Tony Watts.
The authors are responsible for any errors that may still be present in the solutions. They
would be grateful for having them drawn to their attention, and especially grateful if the
correct solution – or a better one – is provided!
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SIGNIFICANT DIGITS IN A CALCULATION
The exercises accompanying each chapter in “Fundamentals of Geophysics” are intended
to give the student practice in applying the knowledge that has been acquired from the
chapter. Some of the exercises are extensions of theoretical passages in the text; they involve
deriving relationships, or setting up and solving equations. Others are of a numerical nature,
requiring the evaluation of an answer when certain values are given for parameters in an
equation. In this last type it is important to use the correct number of digits in quoting the
answer, as this reveals much about the student’s understanding.
Any electronic calculator can be programmed to execute calculations with a desired
number of digits after the decimal point, but the result often does not make sense. For
example, the circumference c of a circle of diameter d is given by c = πd. Suppose we
measure the diameter of a circle to be 25.4 mm (one inch) and want to know its
circumference. A calculator, set to deliver 6 places after the decimal point, returns the value
79.796453 mm. The answer implies that the diameter is known to the precision of the final
figure in the result, i.e., one nanometer. This is an absurd result because we have only
measured the diameter with a ruler and estimated it to the nearest one tenth of a millimeter.
Closer inspection of the equation c = πd shows that the number π is irrational and has an
infinite number of digits, whereas the diameter is given with only 3 digits. The appropriate
answer can not be more precise than the poorest measurement, and in this case should also
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have only 3 digits, i.e. the circumference should be given as 79.8 mm. This example
illustrates a general rule for computations: the answer to every numerical exercise must be
quoted to the correct number of significant digits. If intermediate steps are involved in a
computation, their results should be calculated to at least one significant digit more than will
be in the final result. Rounding off to the correct number of significant digits should be done
only at the end.
Which digits are significant?
The number of significant digits in an answer to a calculation depends on the number of
significant digits in each parameter in the given data. Some simple rules determine when
digits are significant.
A non-zero digit is always significant. For example, 25.4 has 3 significant digits, and
79.796453 has 8 significant digits.
Zeros complicate the situation; where they are located in a number influences the number
of significant digits.
(1) Zeros before other digits are not significant; the number 0.00123 has 3 significant digits.
(2) Zeros between other digits are significant; the number 1.023 has 4 significant digits
(3) Zeros placed after other digits are significant if they follow a decimal point; the number
1.2300 has 5 significant digits
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(4) Zeros at the end of a number have unknown significance unless they follow a decimal
point. Thus, the number 12300 could have 3, 4, or 5 significant digits. To avoid this
uncertainty, when trailing zeros are involved, scientific notation should be used, where the
zeros are placed behind a decimal point. For example, the number should be written 1.2300
× 104 if it has 5 significant digits and as 1.230 × 104 if it has only 4.
Significant digits resulting from computations
Addition and subtraction.
The number of significant digits when numbers are added or subtracted is fixed by the
number of decimal places in the numbers. The number of decimal places should equal the
least number of decimal places in any of the numbers being added or subtracted. Thus,
3.75 (2 decimal places) + 7.242 (3 decimal places) = 10.99 (2 decimal places)
[Question: what would the result be if the second number were 7.249?]
Multiplication and division, mathematical functions.
The number of significant digits when numbers are multiplied or divided, or a
mathematical function is used, should equal the least number of significant digits in any of the
numbers involved. A whole number has an unlimited number of significant digits. The
number of significant digits in a computation involving whole numbers is not affected by the
number of digits in the whole number. Thus,
12.7 × 3.14159 / 7.162 should have 3 significant digits; the answer is 5.57.
5.23 sin(62°) should have 3 significant
the
TESTBANdigits;
KSEL
LEanswer
R.COisM4.62.
2 pounds of flour, each weighing 0.453 kg (3 significant digits), together weigh 0.906 kg.
Precision and accuracy
The number of significant digits has to do with the precision of an answer. In scientific
experiments there are two goals, equivalent to hitting a target. One goal is to hit the centre of
the target as closely as possible; this is called the accuracy of the experiment. The second
goal, when several attempts are made, is to repeat the experiment – or in target-shooting to
group the shots – as closely as possible; this is called the precision of the experiment. The
figure above illustrates the difference between these concepts. There are statistical techniques
for measuring the repeatability, which affect the number of significant digits in a result.
Statistical methods are not needed for the exercise sets in “Fundamentals of Geophysics”.
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SOLUTIONS TO EXERCISES
1 The Solar System
1.
Measured from a position on the Earth’s surface at the equator, the angle between the
direction to the Moon and a reference direction (distant star) in the plane of the Moon’s
orbit is 11° 57’ at 8 p.m. one evening and 14° 32’ at 4 a.m. the following morning.
Assuming that the Earth, Moon and reference star are in the same plane, and that the
rotation axis is normal to the plane, estimate the approximate distance between the
centers of the Earth and Moon.
The geometry of the problem is given in the following diagram.
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It is convenient to convert the angular positions of the moon to decimal values:
α1 = 11° 57’ = 11.950° and α2 = 14° 32’ = 14.533°. The angular difference between
these two positions consists of two parts: one part is due to the motion of the moon in its
orbit, and the other to the rotation of the Earth about its own axis. The two observations
of the moon are 8 hours apart, and in this time the moon has moved forward in its orbit.
Given that the lunar orbital period is 27.32 days, in 8 hours (1/3 day) it will have moved
through 1/(81.96) of a cycle, or 360/81.96 = 4.392°. If the moon had not moved in its
orbit, its position at 4 a.m. would have made an angle α0 = (14.533–4.394) = 10.139°
with the reference direction.
The moon is so far from the Earth that, to a first approximation, the triangle with the
moon at an apex may be treated as an isosceles triangle, with the angle at the apex equal
to
α = (α1 – α0) = (11.955–10.139)=1.816°.
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Because A and B are one third of a revolution apart, angle ACM = angle BCM = 60°,
3
a = Rsin 60 = R
and the distance d between M and C is given by
2
d = s + R cos(60) =
d=
a
R
+
tan (α / 2 ) 2
⎞
3R
R R⎛
3
+ = ⎜
+ 1⎟
2 tan (α / 2 ) 2 2 ⎝ tan (α / 2 ) ⎠
Inserting numerical values: α = 1.816° from the observations and R = 6371 km, gives
d=
⎞
6371 ⎛
3
+
1
= 351, 313
2 ⎜⎝ tan ( 0.908° ) ⎟⎠
The approximate distance of the Moon from the Earth is estimated by these
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measurements to be 351,000 km.
[Note: This estimate is about 9% too low. What might be important sources of error?]
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2.
The eccentricity e of the Moon’s orbit is 0.0549 and the mean orbital radius rL is
384,100 km.
(a) Calculate the lengths of the principal axes a and b of the Moon’s orbit.
(b) How far is the center of the Earth from the center of the elliptical orbit?
(c) Calculate the distances of the Moon from the Earth at perigee and apogee.
(a)
The equation in Cartesian coordinates (x, y) of an ellipse with eccentricity e, semi-major
axis a and semi-minor axis b is
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x 2 y2
2
+
=
1
,
where
b
= a 2 (1 − e2 ) .
a2 b2
Let the mean lunar orbital radius be rL = (a + b) / 2 . Inserting b = a 1 − e2 gives
(
2rL = a 1 + 1 − e2
a=
(1 +
2rL
1 − e2
)
)
Inserting the numerical values rL = 384.1 • 103 km and e = 0.0549 gives the semimajor axis
a = 384.4 × 10 3 km
and the semi-minor axis
b = a 1 − e2 = 383.8 × 10 3 km .
(b)
The distance of the focus of an ellipse from its center is, by definition, (ae). Inserting
the appropriate values for the lunar orbit, the Earth is (384,400 • 0.0549) = 21,100 km
from the center of the elliptical orbit.
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(c)
The distance from the center to the nearest point of the orbit (perigee) is P = a (1 – e).
Inserting the values for a and e, the distance of perigee from the center of the Earth is
384,400 • (1 – 0.0549) = 363,300 km.
The distance from the center to the furthest point (apogee) is A = a (1 + e). Inserting
the values for a and e, the distance of apogee from the center of the Earth is
(384,400 • (1 + 0.0549) = 405,500 km.
3.
If the Moon’s disc subtends a maximum angle of 0° 31m 36.8s at the surface of the
Earth, what is the Moon’s radius?
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Let the optical ray to the Moon’s rim be tangential to the Moon at P as in the figure.
Let the angle subtended by the Moon’s diameter be 2φ, so that angle POM equals φ. Let
the distance between the mid-points of Earth and Moon equal the mean radius rL (=
384,100 km) of the Moon’s orbit, and let the Earth’s radius be R (=6,731 km). The
distance from the Earth’s surface to the Moon’s center is (rL – R), so that in the triangle
OMP,
sin(φ ) =
RL
( rL − R )
RL = ( rL − R ) sin(φ )
Now insert the following numerical values:
(rL − R) = 384,100 − 6, 371 = 377, 629 km
2φ = 31m 36.8s = 0.526889°
φ = 0.26344°
The radius of the Moon is found to be RL = 1736 km.
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4.
Bode’s Law (Eq. 1.5) gives the orbital radius of the nth planet from the Sun (counting
the asteroid belt) in astronomical units. It fits the observations well except for Neptune
(n = 9) and Pluto (n = 10). Calculate the orbital radii of Neptune and Pluto predicted by
Bode’s Law, and compare the results with the observed values (Table 1.2). Express the
discrepancies as percentages of the predicted distances.
Bode’s law is: dn = 0.4 + 0.3 × 2 n − 2
The predicted orbital radius for Neptune (n = 9) is: d9 = 0.4 + 0.3 × 2 7 = 38.8 AU
Neptune’s observed orbital radius (30.07 AU) differs from the predicted value by 8.73
AU, or 22.5%.
The predicted orbital radius for Pluto (n = 10) is: d10 = 0.4 + 0.3 × 28 = 77.2 AU
The value in Table 1.2 for Pluto’s orbital radius (38.62 AU) differs from the predicted
value by 38.58 AU, or 50%.
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5.
An ambulance passes a stationary observer at the side of the road at a speed of 60
km/hr. Its dual tone siren emits alternating tones with frequencies of 700 and 1700 Hz.
What are the dual frequencies heard by the observer at 22°C (a) before and (b) after the
ambulance passes? [Assume that the speed of sound in air, c, in m s–1 at temperature T
(°C) is given by c = 331 + 0.607 T ].
Using the given equation, the speed of sound at 22 °C is 344.4 m s–1, neglecting the
effects of humidity and altitude.
As the ambulance approaches the observer, the sound waves it emits are crowded
together, and after it passes the waves are spaced further apart, as in the diagram below.
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If the frequency of a stationary source is ƒ, and the speed of sound c, the wavelength
of the sound wave is c/ƒ. When the source is moving with speed v towards an observer,
the distance between successive waves is shortened, so that the effective wavelength
becomes (c – v)/ƒ. The frequency of the sound heard from the approaching source, ƒ+, is
given by
c
f+ =
f
c−v
After the moving source has passed, the distance between successive waves reaching
the observer is lengthened and the frequency heard from the departing source, ƒ–, is
given by
c
f_ =
f
c+v
In the case of the dual-tone siren of the ambulance, the frequencies heard as it
approaches are 736 Hz and 1786 Hz; these fall to 668 Hz and 1622 hz, respectively,
after it has passed the stationary observer.
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6.
A spacecraft landing on the Moon uses the Doppler effect on radar signals transmitted
at a frequency of 5 GHz to determine the landing speed. The pilot discovers that the
precision of the radar instrument has deteriorated to ±100 Hz. Is this adequate to ensure
a safe landing? [speed of light = 300,000 km s–1]
The propagation of an acoustic wave requires a medium, and its velocity is relative
to the medium. In contrast, an electromagnetic wave does not require a medium and can
travel through a vacuum, with the same speed, c, in all directions. However, if the speed
of the source or detector is much less than the speed of light, classical arguments may
be used.
Let the signal frequency emitted by the spacecraft be f0 . A detector on the Moon,
observing the signal from the spacecraft approaching at speed v, would observe a radar
frequency ƒ+, which, as explained in the previous example, is given by
f+ =
c
f0
c−v
This signal would be reflected from the Moon’s surface at the same frequency, ƒ+.
The reflecting surface is equivalent to a stationary source ‘emitting’ a signal with
frequency ƒ+. The spacecraft is moving towards it, so its pilot would encounter an
TESTBper
ANK
SELLThe
ER.frequency
COM of the returning radar
increased number of wavelengths
second.
signal is
f =
c+v + c+v
f =
f0
c
c−v
Rearranging, and writing ∆ f = f − f0  f0 , we get
v ∆f
=
c 2 f0
Using the values given in the problem, the uncertainty in the speed v of the
spacecraft is ± 3 m s–1 (~11 km hr–1). Would you risk the landing (think of running into
a brick wall at this speed)?
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7.
Explain with the aid of a sketch the relationship between the length of a day and the
length of a year on the planet Mercury (see Section 1.4.2).
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The period of Mercury’s rotation about its own axis (one Mercury day) is 58.646
Earth days, and the period of its orbital rotation about the Sun (one Mercury year) is
87.97 Earth days. The axial rotation speed and orbital rotation speeds are in the ratio
3:2. The different phases of Mercury’s day and year are illustrated above.
(a) Suppose an observer (the black dot) is on Mercury, initially at sunrise.
(b) After one half rotation of Mercury about the Sun the planet has made 3⁄4 turn
about its own axis, and it is midday on the planet.
(c) At the end of one Mercury year, the planet has executed 1 1⁄2 turns (days) and it is
now sunset.
(d) After 1 1⁄2 years and 2 ¼ days it is midnight.
(e) After 2 Mercury years and 3 Mercury days the observer is back at the starting
point, and it is again sunrise on Mercury.
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8.
The rotations of the planet Pluto and its moon Charon about their own axes are
synchronous with the revolution of Charon about Pluto. Show with the aid of simple
sketches that Pluto and Charon always present the same face to each other.
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Figure (a): The rotations of Pluto (P) and Charon (C) about their own axes and of the line that
joins their centers about the barycenter at B are all in retrograde sense.
Figure (b): After the line of centers has rotated through 45°, the planet and moon have each
also rotated through 45° in the same sense. So the facing surfaces (marked with small
semicircles) are always presented toward each other.
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9.
The barycenter of a star and its planet – or of a planet and its moon – is the center of
mass of the pair. Using the mass and radius of primary body and satellite, and the
orbital radius of the satellite, as given in Tables 1.1–1.3, calculate the location of the
barycenter of the following pairs of bodies. In each case, does the barycenter lie inside
or outside the primary body?
(a) Sun and Earth;
(b) Sun and Jupiter;
(c) Earth and Moon;
22
(d) Pluto (mass = 1.27 × 10 kg; radius = 1137 km) and Charon (mass = 1.9 × 1021 kg;
radius = 586 km); the radius of Charon’s orbit is 19,640 km.
Let the mass of the primary body be M and that of its satellite be m, and let the
distance between their centers be r. The center of mass of the pair is located at distance
d from the center of the planet, given by
Md = m(r − d)
m
d=
r
M +m
Inserting the values of these parameters for each primary body and satellite gives:
(a)
Sun and Earth.
24
M = 1.989 × 1030 kg, m = 5.974
TEST×B10
ANKkg,
SErL=L149,600,000
ER.COM km.
Distance from center of the Sun to the barycenter = 449 km. The radius of the Sun is
71,492 km, so the barycenter of the Sun-Earth pair lies inside the Sun.
(b)
Sun and Jupiter.
M = 1.989 × 1030 kg, m = 1.899 × 1027 kg, r = 778,100,000 km.
Distance from center of the Sun to the barycenter = 742,200 km The radius of the Sun is
71,492 km, so the barycenter of the Sun-Jupiter pair lies outside the Sun.
(c)
Earth and Moon.
M = 5.975 × 1024 kg, m = 7.35 × 1022 kg, r = 384,100 km.
Distance from center of the Earth to the barycenter = 4,670 km. The mean radius of the
Earth is 6,371 km, so the barycenter of the Earth-Moon pair lies inside the Earth.
(d)
Pluto and Charon.
M = 1.27 × 1022 kg, m = 1.9 × 1021 kg, r = 19,640 km.
Distance from center of Pluto to the barycenter = 2,600 km. The mean radius of Pluto is
1,137 km, so the barycenter of the Pluto-Charon pair lies outside Pluto.
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10.
A planet with radius R has a mantle with uniform density ρm enclosing a core with
radius rc and uniform density ρc. Show that the mean density of the planet, ρ, is given
by
ρ − ρm ⎛ rc ⎞
=⎜ ⎟
ρc − ρ m ⎝ R ⎠
3
The exercise requires expressing the separate masses of core and mantle as follows:
Radius of planet = R; mean density = ρ; mass of planet, =
4
π R 3ρ
3
4 3
π rc ρc
3
4
4
Volume of mantle = (volume of Earth – volume of core) = = π R 3 − π rc 3
3
3
4
Mass of mantle = = π R 3 − rc 3 ρm
3
Radius of core = rc; density of core = ρc; mass of core, =
(
)
Mass of planet = mass of mantle + mass of core
(
)
4
4
4
π R 3 ρ = π rc 3 ρc + π R 3 − rc 3 ρm
3
3
3
R 3 ρ = rc 3 ρc + R 3 ρm − rc 3 ρm
R 3 ( ρ − ρm ) = rc 3 ( ρc − T
ρmE)STBANKSELLER.COM
Mean density of the planet:
ρ − ρm ⎛ rc ⎞
=⎜ ⎟
ρc − ρ m ⎝ R ⎠
3
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11.
Assuming that the internal structure of the Moon is a set of concentric shells with
different densities (Fig. 1.5) as simplified in the following table, compute the Moon’s
mean density.
Radius [km] Density [kg m–3]
Internal layer
Solid inner core
0-240
8000
Fluid outer core
240-330
5000
Mantle
330-1738
3300
The mass M of a spherical shell of constant density 𝜌 , inner radius r1 and outer
radius r2 is, from the previous exercise:
M=
4
π ( r23 − r13 ) ρ
3
Applying the formula to each concentric shell of the Moon’s structure give the
following masses:
• Solid core, M =
4
π ( 240 3 × 10 9 ) × 8000 = 4.63 × 10 20 kg
3 TESTBANKSELLER.COM
• Fluid core, M =
4
π ( 330 3 − 240 3 ) × 10 9 × 5000 = 4.63 × 10 20 kg
3
• Mantle, M =
4
π (1738 3 − 330 3 ) 3300 = 7.21× 10 22 kg
3
Mass of the Moon = (4.63+4.63+721) x 1020 = 7.30 x 1022 kg
Volume of the Moon =
4
π (1738 3 ) × 10 9 = 2.20 × 1019 m 3
3
Mean density of the Moon = 3320 kg m–3
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12
The moment of inertia I of a spherical shell of density 𝜌 and outer and inner radii R2 and
R1, respectively, about its axis of symmetry is given by the formula:
I=
8π
ρ ( R25 − R15 )
15
(a) Assuming the internal structure of the Moon from the previous exercise, compute
the Moon’s moment of inertia about its rotation axis.
2
(b) The moment of inertia of a sphere can be written I = kMR , where M is its mass and
R its radius. Calculate the value of k for the Moon.
The formula for the moment of inertia gives the following moments of inertia for the
concentric shells of the Moon’s structure:
• Solid core, I =
5
8π
240 × 10 3 ) 8000 = 1.067 × 10 31 kg m 2
(
15
• Fluid core, I =
8π
330 5 − 240 5 ) × 1015 × 5000 = 2.612 × 10 31 kg m 2
(
15
• Mantle, I =
8π
1738 5 − 330 5 ) × 1015 × 3300 = 8.766 × 10 34 kg m 2
(
15
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Moment of inertia of the Moon, I = 8.770 x 1034 kg m2
Mass of the Moon, M = 7.30 x 1022 kg
Radius of the Moon, R = 1.738 x 106 m
Product MR2 = 2.215 x 1035 kg m2
k = I/MR2 = 0.3959
(measured value = 0.3932)
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13
Using the data in Table 1.2, show that the planets conform to Kepler’s 3rd law. What is
the meaning of the constant in your result?
Kepler’s 3rd law for the solar system (Eq. 1.5) is
GS =
4π 2 3
a = Ω2a3
2
T
where G is the gravitational constant, S is the mass of the Sun, T is the period of the orbit, Ω
is the angular velocity of the planet about the Sun, and a is the semi-major axis of the
elliptical orbit. From the data in Tables 1.2 and 1.3 we can construct the following table:
Semi-major axis of
Orbital rate, Ω
Ω2a3
orbit, a (109 m)
(10–9 rad s–1)
(1020 m3 s–2)
Mercury
57.91
827.3
1.329
Venus
108.2
323.9
1.329
Earth
149.6
199.2
1.329
Mars
227.9
105.9
1.327
Jupiter
778.6
Saturn
1434
6.76
1.348
Uranus
2873
2.37
1.332
Neptune
4495
1.209
1.328
Pluto
5906
0.804
1.332
Planet
TESTBAN16.8
KSELLER.COM1.332
The mean value of 1.332 × 1020 m3 s–2 is an estimate of the heliocentric gravitational constant,
GS. (2016 value 1.327 × 1020 m3 s–2)
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SOLUTIONS TO EXERCISES
2 Plate Tectonics
1.
Summarize the geological and geophysical evidence resulting from plate tectonic
activity in the following regions: (a) Iceland, (b) the Aleutian islands, (c) Turkey, (d)
the Andes, (e) the Alps?
Every teacher will have his or her own preferences for handling this question, which
is useful in the context of essay assignments or class discussion. The individual plate
margins are characterized by different types of geophysical evidence, which are
mutually supportive. The student should be able to describe each type of evidence and
explain its origin. The following (incomplete) list may provide points for guiding
discussion.
Region
Iceland
Aleutians
Turkey
Andes
Alps
Type of
location
Spreading
ridge
Subduction
zone
Transform
fault
Subduction
zone
Plate
collision
zone
Geology
&
tectonics
Basaltic
volcanism,
normal
faulting
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Island arc,
oceanic
trench, calcalkaline
volcanism,
thrust
faulting
Strike-slip
faulting
(rightlateral)
Andesitic
volcanism
Mountainbuilding
Seismicity Shallow EQ
Benioff zone
Shallow
EQ
Benioff
zone
Shallow EQ
Gravity
Negative due
to lowdensity
trough
---
Positive
Negative
over edge of due to
continent
subduction
of crust
Magnetics Symmetrical
anomaly
pattern
Truncation of
anomaly
pattern
---
---
---
Heat flow
Low
---
Low
---
Negative due
to hot magma
chamber
High
1
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2.
Using the data in Fig 5.77, compute the approximate spreading rates in the age interval
25-45 Ma at the oceanic ridges in the S. Atlantic, S. Indian, N. Pacific and S. Pacific
oceans.
The vertical axis in Fig. 5.77 shows both ages and distances for sea-floor spreading
at the South Atlantic ridge axis. For the other ridges the distances to particular
anomalies are on the horizontal axis and the corresponding ages on the vertical axis. A
crude estimate of the spreading rate in the 25-45 Ma interval is obtained by
interpolating the distances at these ages directly from the plot. Alternatively, the points
on Fig. 5.77 that lie between the two age-limits can be read off digitally, and a
regression line fitted to each set of points. The two methods give similar results, as
shown in the following table:
Oceanic
Ridge
Interpolated
distance in km
(Age = 25 Ma)
Interpolated
distance in km
(Age = 45 Ma)
Half-spreading
rate (mm/yr)
from
interpolation
Half-spreading
rate (mm/yr)
from
regression
S. Atlantic
ELLER.COM 18.9
472 TESTBANKS851
18.9
N. Pacific
754
1826
53.6
55.3
S. Pacific
619
1013
19.7
19.3
S. Indian
687
1330
32.2
31.7
2
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3.
Three ridges A, B and C meet at a triple junction. Ridge A has a strike of 329°
(N 31° W) and a spreading rate of 7 cm yr–1; ridge B strikes at 233° (S 53° W) and has a
spreading rate of 5 cm yr–1. Determine the strike of ridge C and its spreading rate.
The strike of ridge A is 329°, thus plate 2 separates from plate 1 at an azimuth of
239° with a velocity 1v2 = 7 cm yr–1. Similarly, on ridge B plate 3 has velocity 2v3 = 5
cm yr–1 in a direction 143°. The two velocity vectors meet at an angle of 84°, as in the
velocity triangle. Applying the law of cosines to the side 3v1, we get:
( 3 v1 )
TESTBANKSELLER.COM
2
= 7 + 5 − 2 ⋅ 7 ⋅ 5 cos(84)
2
2
v = 8.17
3 1
Rounding the answer, the velocity of separation on ridge C is 8.2 cm yr–1.
To determine the direction of spreading, apply the law of sines to angle φ in the
triangle:
sin φ sin(84) sin(84)
=
=
5
8.17
3 v1
φ = 37.5° ≈ 38°
This is the angle between the velocity vectors on ridges A and C. The angle between the
strikes of the ridges is therefore (180 − φ ) = 142° ; the strike of ridge A is N31°W, and
so the strike of ridge C is 111° (N111°E).
3
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4.
Three sides of a triangle on the surface of the spherical Earth measure 900 km, 1350
km, and 1450 km, respectively. What are the internal angles of the triangle? If this were
a plane triangle, what would the internal angles be?
Let the sides be a = 900 km, b = 1350 km, and c = 1450 km. In a spherical triangle
these sides must first be converted to the angles they subtend at the center of the Earth.
Using the relationship s = Rφ, where R is the Earth’s radius (6,371 km), the lengths of
the sides are first calculated in radians; by multiplying by (180°/π) the lengths are
converted to degrees. The sides of the spherical triangle become: a = 8.09°, b = 12.14°,
and c = 13.04°.
From the law of cosines for a spherical triangle (Box 2.1, Eq. (4) ), the angle A is
given by:
cos A =
cos a − cosb cos c cos(8.09) − cos(12.14)cos(13.04)
=
= 0.7929
sin b sin c
sin(12.14)sin(13.04)
Entering successively the appropriate sides, this gives for the angles of the spherical
triangle:
A = 37.5°; B = 65.6°; C = 77.7°
TEangles
STBAofNthis
KSE
LLER.triangle
COM is 180.8°.
Note that the sum of the
spherical
In the case of a plane triangle, the ordinary law of cosines (Box 2.1, Eq. 2) gives for
angle A:
cos A =
a 2 − b 2 − c 2 (900)2 − (1350)2 − (1450)2
=
= 0.7957
−2bc
−2(1350)(1450)
In this way the angles of the plane triangle are found to be:
A = 37.3°; B = 65.3°; C = 77.4°
The triangle on the spherical surface is in this case so small that it makes little
difference if it is regarded as spherical or plane.
4
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5.
An aircraft leaves a city at latitude λ1 and longitude φ1 and flies to a second city at
latitude λ2 and longitude φ2. Derive an expression for the great circle distance between
the two cities.
First it is necessary to convert the coordinates of each city to direction cosines (see
Box 1.5). Using subscripts 1 and 2 for the first and second city, respectively, the two
sets of direction cosines are
l1 = cos λ1 cos φ1;
l2 = cos λ2 cos φ2 ;
m1 = cos λ1 sin φ1; n1 = sin λ1
m2 = cos λ2 sin φ2 ; n2 = sin λ2
If ∆ is the angle between these directions, then
cos∆ = l1l2 + m1m2 + n1n2
= cos λ1 cos λ2 cos φ1 cos φ2 + cos λ1 cos λ2 sin φ1 sin φ2 + sin λ1 sin λ2
= cos λ1 cos λ2 ( cos φ1 cos φ2 + sin φ1 sin φ2 ) + sin λ1 sin λ2
cos∆ = cos λ1 cos λ2 cos (φ2 − φ1 ) + sin λ1 sin λ2
This formula gives the angular distance ∆ between the two cities along the great
circle that joins them. The distance in kilometres is equal to (R∆), where R is the Earth’s
radius and the angle ∆ is expressed in radians (π radians = 180°).
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6.
Apply the above formula to compute the great circle distances between the following
pairs of cities:
(a) New York (40°43’ N, 74°01’ W) – Madrid (40°25’ N, 3°43’ W);
(b) Seattle (47°21’ N, 122°12’ W) – Sydney (33°52’ S, 151°13’ E);
(c) Moscow (55°45’ N, 37°35’ E) – Paris (48°52’ N, 2°20’ E);
(d) London (51°30’ N, 0°10’ W) – Tokyo (35°42’ N, 139°46’ E).
Assume a spherical Earth with mean radius 6371 km. The formula gives the following
distances between the pairs of cities:
(a)
New York – Madrid:
λ1 = 40°43’ N =40.72°; φ1 = 74°01’ W = –74.02°;
λ2 = 40°25’ N =40.42°; φ2 = 3°43’ W = –3.72°
cos∆=cos(40.72)cos(40.42)cos(–74.02+3.72)+sin(40.72)sin(40.42) = 0.61749
∆ = 51.87° = 5,767 km
(b)
Seattle – Sydney:
λ1 = 47°21’ N =47.35°; φ1 = 122°12’ W = –122.20°;
λ2 = 33°52’ S =–33.87°; φT2 E
=S151°13’
TBANKES=E151.22°
LLER.COM
cos∆=cos(47.35)cos(–33.87)cos(–122.20–151.22)+sin(47.35)sin(33.87) = –0.37649
∆ = 112.12° = 12,467 km
(c)
Moscow – Paris:
λ1 = 55°45’ N =55.75°; φ1 = 37°35’ E = 37.58°;
λ2 = 48°52’ N =48.87°; φ2 = 2°20’ W = 2.33°
cos∆=cos(55.75)cos(48.87)cos(37.58–2.33)+sin(55.75)sin(48.87) = 0.92491
∆ = 22.35° = 2,485 km
(d)
London – Tokyo:
λ1 = 51°30’ N =51.50°; φ1 = 0°10’ W = –0.17°;
λ2 = 35°42’ N =35.70°; φ2 = 139°46’ E = 139.77°
cos∆=cos(51.50)cos(35.70)cos(139.77+0.17)+sin(51.50)sin(35.70) = 0.06976
∆ = 86.01° = 9,563 km
Is it justifiable to give the answers to this question to the nearest kilometer?
6
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7.
Calculate the heading (azimuth) of the aircraft’s flight path as it leaves the first city in
each pair of cities in the previous exercise.
Let the heading from the first city (coordinates λ1, φ1) to the destination (coordinates
λ2, φ2) be D, as in the diagram. The sides of the spherical triangle that includes angle D
have lengths ∆, (π/2 – λ1), and (π/2 – λ2), respectively. The law of sines for a spherical
triangle (Box 2.1, Eq. 3) defines
anAangle
limited
TESTB
NKSEthat
LLisER
.COMto the range –π/2 to +π/2. In
this problem it must be possible for the range of possible headings to be –π to +π, and
the law of cosines (Box 2.1, Eq. 4) is more suitable. Applying this law to the side
enclosing angle D in the triangle:
⎛π
⎞
⎛π
⎞
⎛π
⎞
cos ⎜ − λ2 ⎟ = cos ⎜ − λ1 ⎟ cos∆+ sin ⎜ − λ1 ⎟ sin∆ cos D
⎝2
⎠
⎝2
⎠
⎝2
⎠
cos D =
sin λ2 − sin λ1 cos∆
cos λ1 sin∆
⎛ sin λ2 − sin λ1 cos∆ ⎞
D = ± cos –1 ⎜
⎟⎠
cos λ1 sin∆
⎝
There is still ambiguity in the answer, because positive D and negative D have the
same cosine. Visual inspection of the geometrical problem determines if the heading to
the destination is eastward or westward.
Using the parameters for the pairs of cities as in the previous question:
(a)
New York (40.74°N 74.02°W) to Madrid (40.42°N 3.72°W), ∆ = 51.87°:
⎛ sin(40.42) − sin(40.74)cos(51.18) ⎞
D = ± cos –1 ⎜
⎟⎠ = ±65.7°
⎝
cos(40.74)sin(51.18)
The directional heading from New York to Madrid is N 65.7° E.
7
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(b)
Seattle (47.35°N 122.20°W) to Sydney (33.87°S 151.22°E) , ∆ = 112.12°:
⎛ sin(33.87) − sin(47.35)cos(112.12) ⎞
D = ± cos –1 ⎜
⎟⎠ = ±116.5°
⎝
cos(47.35)sin(112.12)
The directional heading from Seattle to Sydney is N 116.5° W.
(c)
Moscow (55.75°N 37.58°E) to Paris (48.87°N 2.33°E), ∆ = 22.35°:
⎛ sin(48.87) − sin(55.75)cos(22.35) ⎞
D = ± cos –1 ⎜
⎟⎠ = ±93.0°
⎝
cos(55.75)sin(22.35)
The directional heading from Moscow to Paris is N 93.0° W.
(d)
London (51.50°N 0.17°W) to Tokyo (35.7°N 139.77°E), ∆ = 86.01°:
⎛ sin(35.7) − sin(51.5)cos(86.01) ⎞
D = ± cos –1 ⎜
⎟⎠ = ±31.6
⎝
cos(51.5)sin(86.01)
The directional heading from London to Tokyo is N 31.6° E.
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SOLUTIONS TO EXERCISES
3 Gravity and the Figure of the Earth
1.
Using the data in Table 1.1 calculate the gravitational acceleration on the surface of the
Moon as a percentage of that on the surface of the Earth.
GE
R2
Gm
Gravitational acceleration on the Moon: aM = − 2
RL
Gravitational acceleration on the Earth: aE = −
a
−Gm / RL 2 ⎛ m ⎞ ⎛ R ⎞
Ratio: M =
=⎜ ⎟
aE
−GE / R 2 ⎝ E ⎠ ⎜⎝ RL ⎟⎠
2
Inserting values from Table 1.1 for the mass E (5.974 × 1024 kg) and radius R (6371
km) of the Earth and the mass m (0.0735 × 1024 kg, or 0.0123 E) and radius RL (1738
km) of the Moon gives
2
aM
⎛ 6371 ⎞
= 0.0123 ⎜
= 0.165
⎝ 1738 ⎟⎠
aE
Gravitational acceleration
onTthe
Moon
16.5%
ofOthat
TES
BAN
KSEisLL
ER.C
M on the surface of the Earth.
The mean value of gravity on the Earth is 9.81 m s–2, on the Moon it is 1.62 m s–2.
2.
An Olympic high-jump champion jumps a record height of 2.45 m on the Earth. How
high could this champion jump on the Moon?
The general relationship between initial velocity (u), final velocity (v), constant
acceleration (a) and distance (s) is v 2 = u 2 + 2as .
In the case of a high-jumper, the final velocity is 0, the distance is the height (h)
jumped, and the acceleration is –g, so u 2 = 2gh .
The gravity (gL,) and height jumped (hL) on the Moon are different from gravity (gE)
and height jumped (hE) on Earth, but u is the same, being determined by the highjumper’s ability. Thus,
u 2 = 2gE hE = 2gL hL
hL =
gE
( 9.81) 2.45 = 14.8 m
hE =
( )
gL
(1.62 )
22
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3.
(a) Calculate the escape velocity of an object on the Earth, assuming a mean
gravitational acceleration of 9.81 m s-1 and mean Earth radius of 6371 km.
(b) What is the escape velocity of the same object on the Moon?
The escape velocity v of an object with mass m is reached when its kinetic energy
( /2 mv2) is equal to the work required to move the object from the surface of the planet
to infinity, which in turn is equivalent to its potential energy on the surface of the planet
(mass M, radius RP). This gives the equation
1
1 2
mM
mv = G
2
RP
2GM
v2 =
= 2gP RP
RP
(a)
The Earth’s escape velocity (using gE =9.81 m s–2, RE = 6371 km) is
vE = 2gE RE = 2(9.81)(6.371)10 6 = 11.2 km s-1
This is equivalent to 40,250 km hr–1.
(b)
The lunar escape velocity (using gL =1.62 m s–2, RL = 1738 km) is
vL = 2gL RL = 2(1.62)(1.738) ⋅10 6 = 2.37 km s-1
EST
This is equivalent to 8,550Tkm
hrB–1A
. NKSELLER.COM
4.
The equatorial radius of the Earth is 6378 km and gravity at the equator is 9.780 m s–2.
Compute the ratio m of the centrifugal acceleration at the equator to the gravitational
acceleration at the equator. If the ratio m is written as 1/k, what is the value of k?
Earth's rotation period, T = 24 hr = 86,400 s
Earth's rotation speed,
ω = 2π/T = 7.272 × 10–5 rad s–1
Centrifugal acceleration at the equator:
(
ac = ω 2 r = 7.272 × 10 −5
) ( 6378 × 10 ) = 3.373 × 10
2
3
−2
m s −1
Gravity at the equator = gE = 9.780 m s–1
Gravitational acceleration at the equator: aG = gE – ac
a
ac
Acceleration ratio: m = c =
aG g − ac
m=
0.03373
0.03373
1
=
=
9.780 − 0.03373
9.746
288.9
k = 288.9
23
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5.
Given that the length of a month is 27.32 days, the mean gravity on Earth is 9.81 m s–2
and the Earth’s radius is 6371 km, calculate the radius of the Moon’s orbit.
Moon’s rotation period,
T = 27.32 days = 27.32 × 86,400 s
Moon’s rotation speed,
ω L = 2π / T = 2.662 × 10 −6 rad s –1
Centripetal acceleration at distance rL:
ac = ω L 2 rL
Gravitational acceleration at distance rL:
aG =
Equating these expressions: ω L 2 rL =
2
⎛ R⎞
⎛ r ⎞ GE ⎛ R ⎞
ω L R ⎜ L ⎟ = 2 ⎜ ⎟ = aG ⎜ ⎟
⎝ R ⎠ R ⎝ rL ⎠
⎝ rL ⎠
GE
rL 2
GE
rL 2
2
2
⎛ aG ⎞
⎛ rL ⎞
⎜⎝ ⎟⎠ = ⎜ 2 ⎟
R
⎝ ωL R⎠
3
3
9.81
⎛ rL ⎞
TESTBANKS=E2.173
LLER×.10C5OM
⎜⎝ ⎟⎠ =
2
−6
6
R
2.662 × 10
6.371 × 10
(
)(
)
rL = 3 2.173 × 10 5 ⋅ R = 60.12R = 383, 000 km
Radius of Moon’s orbit = 383,000 km
24
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6.
A communications satellite is to be placed in a geostationary orbit.
(a) What must the period and orientation of the orbit be?
(b) What is the radius of the orbit?
(c) If a radio signal is sent to the satellite from a transmitter at latitude 45°N, what is the
shortest time taken for its reflection to reach the Earth?
(a)
A geostationary orbit is one for which the period of rotation of the satellite about the
Earth is equal to the rotation of the Earth about its own axis. This keeps the satellite
“stationary” above a given location. The orbit must be in the plane of the equator.
(b)
The radius of the satellite’s orbit is found as in the previous exercise.
Satellite rotation period,
T = 1 day = 86,400 s
Rotation speed,
ω s = 2π / T = 7.2722 × 10 −5 rad s –1
Centripetal acceleration at distance rs:
ac = ω s 2 rs
Gravitational acceleration at distance rs:
aG =
ω s 2 rs =
Equating the accelerations:
2
GE
rs 2
GE
rs 2
2
⎛ TRE⎞STBANKSELLER.COM
GE ⎛ R ⎞
rs = 2 ⎜ ⎟ = aG ⎜ ⎟
R ⎝ ωs ⎠
⎝ ωs ⎠
3
2
⎛ 6.371 × 10 6 ⎞
rs = 9.81⎜
= 7.5297 × 10 22
−5 ⎟
⎝ 7.272 × 10 ⎠
3
The radius of the stationary orbit is 42,226 km (≈ 42,200 km).
(c)
The quickest reflection travels along the shortest path to the satellite from the point on
the Earth’s surface at 45°N where the transmitter and receiver are located (point P in the
diagram); the satellite is above the equator (point S in the diagram).
25
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The side d is given by
⎛ 1 ⎞
d 2 = R 2 + rs 2 − 2Rrs cos ( 45° ) = (6371)2 + (42226)2 − 2(6371)(42226) ⎜
⎝ 2 ⎟⎠
Distance from station to satellite, d = 37,989 km
Speed of light, c = 299,792 km s–1
The two-way travel time of the signal is 2(d/c) = 0.253 s
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7.
Calculate the centrifugal acceleration due to the Earth’s rotation of an object at rest on
the Earth’s surface in Paris, assuming a latitude of 48° 52 N. Express the result as a
percentage of the gravitational attraction on the object.
Perpendicular distance of Paris (latitude λ) from rotation axis = R cosλ
= 6371 cos (48.87°)
=4191 km
Earth's rotation speed,
ω = 2π/T = 7.272 × 10–5 rad s–1
Centrifugal acceleration due to Earth’s rotation at 48° 52’N:
(
ac = ω 2 r = 7.272 × 10 −5
) ( 4191 × 10 ) = 2.216 × 10
2
3
−2
m s −1
Gravitational acceleration aG ≈ g ≈ 9.81 m s–1
Fraction of gravitational attraction = (ω2 R cosλ )/aG =(0.02216)/9.81
= 0.226 %
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8.
A solid angle (Ω ) is defined as the quotient of the area (A) of the part of a spherical
surface subtended by the angle, divided by the square of the spherical radius (r): i.e.,
Ω = A/r2 (see Box 5.4). Show with the aid of a diagram that the gravitational
acceleration at any point inside a thin homogeneous spherical shell is zero.
Let P be any point inside the hollow spherical shell. The angle α at point P subtends
an element of the surface with area A1 at a distance r1 from P. If the thickness of the
spherical shell is t and its density ρ, the volume of the surface element is (A1t) and its
mass m1 is (ρ A1t). The angle at P also subtends an area A2 on the opposite surface of the
TESTBANKSELLER.COM
sphere at a distance r2 from P. This element has mass m2 equal t (ρ A2t).
The gravitational acceleration at P due to the surface element A1 is
a1 = −G
m1
ρA t
= −G 21
2
r1
r1
The acceleration at P due to the surface element A2 acts in the opposite direction and is
equal to
a2 = −G
m2
ρA t
= −G 22
2
r2
r2
The net gravitational acceleration at the point P is
⎛A A ⎞
a = ( a1 − a2 ) = −G ρt ⎜ 21 − 22 ⎟ = −G ρt (α − α ) = 0
r2 ⎠
⎝ r1
This relationship holds for any size of the angle α, thus at any position inside the hollow
spherical shell the gravitational acceleration is zero.
28
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9.
Assuming that the gravitational acceleration inside a homogeneous spherical shell is
zero, show that the gravitational acceleration inside a homogenous uniform solid sphere
is proportional to the distance from its center.
Let P be a point inside a solid sphere of uniform density ρ and radius R at distance r
from its center. The gravitational acceleration at the point P has two sources: (1) the
inner part of the solid sphere between P and the center and (2) the outer part between P
and the surface. This outer part may be subdivided into numerous thin concentric
spherical shells. Because P
lies
zero net acceleration
TE
STinside
BANKeach
SELshell,
LERit.experiences
COM
from each one. Thus the gravitational acceleration at P due to the outer part of the
sphere is zero.
The acceleration at P due to the mass m(r) of the inner part of the sphere of radius r is
m(r)
aG = −G 2 = −G
r
(
4
3
π r 3ρ )
r
2
4
= − π G ρr
3
Thus, the gravitational acceleration inside a homogeneous uniform solid sphere is
proportional to the distance from its center.
29
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10.
Show that the gravitational potential UG inside a homogenous uniform solid sphere of
radius R at a distance r from its center is given by
UG = −
2π
G ρ 3R 2 − r 2
3
(
)
In this problem we also have to consider the contributions to the potential from the
part of the sphere inside the radius r (say, U1) and the part between r and R (say, U2).
The potential U1 due to the inner sphere is easy to calculate: the gravitational
acceleration inside the solid sphere at distance r from its center is obtained by
differentiating U1 with respect to r:
a = −G
m(r)
dU
=− 1
2
r
dr
U1 = −G
m(r)
4
= − π G ρr 2
r
3
The gravitational attraction at a point P inside a thin homogeneous shell is zero, as
shown above. Because acceleration is the derivative of potential, the gravitational
potential of the shell must be constant throughout its interior, and this must equal the
potential of its surface (otherwise there would be a discontinuity of the potential whose
TEan
STinfinite
BANKacceleration).
SELLER.CThus
OM the potential at P of a thin
infinite gradient would give
shell of radius x, and thickness dx, where x lies between r and R, is
m(x)
4πρ x 2 dx
dU 2 = −G
= −G
= −4π G ρ xdx
x
x
Integrating over all the thin shells between r and R:
R
U 2 = −4π G ρ ∫
R
r
(
⎡ x2 ⎤
xdx = −4π G ρ ⎢ ⎥
⎣ 2 ⎦r
U 2 = 2π G ρ r 2 − R 2
)
Combining the two contributions, the gravitational potential UG inside the solid
sphere at distance r from its center is
(
4
U G = − π G ρr 2 + 2π G ρ r 2 − R 2
3
UG =
2
π G ρr 2 − 2π G ρ R 2
3
UG =
2
π G ρ r 2 − 3R 2
3
(
)
)
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11.
Sketch the variations of gravitational acceleration and potential inside and outside a
homogeneous solid sphere of radius R.
Radial variation of gravitational acceleration (aG) for a homogeneous solid sphere. The
acceleration is directed inwards toward the center of the Earth.
TESTB
ANKSE(U
LGL)Efor
R.aChomogeneous
OM
Radial variation of gravitational
potential
solid sphere. The
gravitational potential is negative; the minimum value at the center of the Earth is 50%
greater (more negative) than the surface value at r = R.
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12.
A thin borehole is drilled through the center of the Earth, and a ball is dropped into the
borehole. Assume the Earth to be a homogenous solid sphere. Show that the ball will
oscillate back and forth from one side of the Earth to the other. How long does it take to
traverse the Earth and reach the other side?
At any point during its fall through the borehole the gravitational acceleration of the
ball depends only on the mass between it and the center of the Earth (see exercise 9). Its
acceleration aG is proportional to its distance r from the center and is given by
d 2r
4
= − π G ρr
2
dt
3
2
d r 4
+ π G ρr = 0
dt 2 3
d 2r
4
+ n 2 r = 0 where n 2 = π G ρ
2
dt
3
aG =
This is the equation of a simple harmonic motion with frequency n. Its solution is
r = A cos nt + Bsin nt
The ball is dropped from the Earth’s surface, so the boundary condition at t = 0 is r = R.
Thus the equation of motion is
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r = R cos nt
This expression for n can be re-written
3
E 1 g
⎛4
⎞R
n2 = ⎜ π Gρ⎟ 3 = G 2 =
⎝3
⎠R
R R R
The ball will oscillate in the hole from one side of the Earth to the other with period
T=
2π
R
= 2π
n
g
Substituting R = 6371 km and g = 9.81 m s–1, gives
T = 2π
R
6.371 × 10 6
= 2π
= 5063.5 s
g
9.81
The time taken for the ball to traverse the Earth is one half the period, namely 2532 s, or
42 min 12 s.
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13.
The Roche limit is the closest distance an object can approach a planet before being torn
apart by the tidal attraction of the planet. For a rigid spherical moon the Roche limit is
given by Eq. (6) in Box 2.1.
(a) Using the planetary dimensions in Table 1.1, calculate the Roche limit for the Moon
with respect to the Earth. Express the answer as a multiple of the Earth’s radius.
(b) Show that, for a planet whose mean density is less than half that of its rigid moon,
the moon would collide with the planet before being torn apart by its gravity.
(c) Given that the Sun’s mass is 1.989 × 1030 kg and that its radius is 695,500 km,
calculate the Roche limit for the Earth with respect to the Sun.
(d) The mean density of a comet is about 500 kg m–3. What is the Roche limit for
comets that might collide with the Earth?
(e) The mean density of an asteroid is about 2000 kg m-3. If an asteroid on collision
course with the Earth has a velocity of 15 km s–1, how much time will elapse between
the break-up of the asteroid at the Roche limit and the impact of the remnant pieces on
the Earth’s surface, assuming they maintain the same velocity as the asteroid?
1
(a)
⎛ ρ ⎞3
The Roche limit for the solid moon of a planet is given by d R = 1.26R ⎜ E ⎟
⎝ ρM ⎠
TEplanet.
STBAThe
NKS
ELLERof.C
OM
where R is the radius of the
densities
the
Moon and Earth (Table1.1) are,
–3
–3
respectively, ρM = 3347 kg m and ρE = 5515 kg m . Inserting these values into the
equation gives the Roche limit for solid behavior of the Moon
1
⎛ 5515 ⎞ 3
d R = 1.26R ⎜
= 1.49R
⎝ 3347 ⎟⎠
This is equivalent to a distance of 9480 km between the centers of Earth and Moon. If
the Moon behaved as a fluid its break-up would occur further from the Earth at a
distance of
1
⎛ ρ ⎞3
d R == 2.42R ⎜ E ⎟ = 2.86R = 18, 200 km
⎝ ρM ⎠
(b)
If ρ p < 12 ρ M the Roche limit becomes
1
⎛ 1⎞ 3
d R = 1.26R ⎜ ⎟ = 1.00R for the case of a rigid moon. The separation of the centers is
⎝ 2⎠
equal to the radius of the planet. Thus, the moon would collide with the planet before
being disrupted. However, if the moon is fluid, the Roche limit would be at
1
⎛ 1⎞ 3
d R == 2.42R ⎜ ⎟ = 1.92R . If the moon’s radius is appreciably smaller than that of the
⎝ 2⎠
planet, the moon would break up before colliding.
33
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(c)
The volume of the Sun is 43 π (695, 500 × 10 3 )3 = 1.409 × 10 27 m 3
The mean density of the Sun is ρS = (1.989 × 1030)/(1.409 × 1027) = 1411 kg m–3
The Roche limit for the Earth with respect to the Sun is
1
1
⎛ ρ ⎞3
⎛ 1411 ⎞ 3
d R = 1.26RS ⎜ S ⎟ = 1.26RS ⎜
= 0.8RS
⎝ 5515 ⎟⎠
⎝ρ ⎠
E
Thus the Earth would plunge into the Sun before being broken up by tidal forces.
(d)
Mean density of the Earth ρE = 5515 kg m–3
Mean density of a comet ρc = 500 kg m–3
1
1
1
1
⎛ρ ⎞3
⎛ 5515 ⎞ 3
Roche limit for a rigid comet: d R = 1.26R ⎜ E ⎟ = 1.26R ⎜
= 2.8R
⎝ 500 ⎟⎠
⎝ρ ⎠
c
⎛ρ ⎞3
⎛ 5515 ⎞ 3
Roche limit for a fluid comet: d R = 2.42R ⎜ E ⎟ = 2.42R ⎜
= 5.4R
⎝ 500 ⎟⎠
⎝ρ ⎠
c
The comet would disintegrate at a distance between 2.8R (18,000 km, solid body) and
5.4R (34,000 km, fluid body) from the center of the Earth.
(e)
The Roche limit for the (solid) asteroid is
1
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⎛ρ ⎞3
⎛ 5515 ⎞ 3
d R = 1.26R ⎜ E ⎟ = 1.26R ⎜
= 1.77R
⎝ 2000 ⎟⎠
⎝ρ ⎠
A
The closest distance of a point on the Earth to the break-up of the asteroid is 0.77R. If
the comet remnants maintain a speed of 15 km s–1, the minimum time elapsed until they
impact on the Earth is
t = 0.77 ×
6371
= 327 s (i.e. 5 min 27 s).
15
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14.
The mass M and moment of inertia C of a thick shell of uniform density ρ, with internal
radius r and external radius R are given by
M=
(
)
4
πρ R 3 − r 3 ,
3
C=
(
8
πρ R 5 − r 5
15
)
The Earth has an internal structure consisting of concentric spherical shells. A simple
model with uniform density in each shell is given in the following figure:
(a)
(b)
(c)
(d)
(a)
Compute the mass and moment of inertia of each spherical shell.
Compute the total mass and total moment of inertia of the Earth.
If the moment of inertia can be written C = kMR2, where M is Earth’s mass and R its
radius, what is the value of k?
ESifTthe
BAdensity
NKSELwere
LERuniform
.COM throughout the Earth?
What would the value of kTbe
Using the given formulae and dimensions, the masses and moments of inertia of the
spherical shells are
Volume
(× 1018 m3)
Mass
(× 1024 kg)
Moment of inertia
(× 1037 kg m2)
upper mantle
307.0
1.013
2.472
lower mantle
599.2
2.996
4.613
outer core
168.9
1.858
0.936
inner core
0.761
0.0989
0.00589
Shell
(b)
The totals are obtained by summing the individual columns in the above table:
Total mass = 5.966 × 1024 kg ≈ 5.97 × 1024 kg
Total moment of inertia = 8.027 × 1037 kg m2 ≈ 8.03 × 1037 kg m2
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(c)
By definition, k =
C
MR 2
Substituting the values calculated in (b) gives
k=
(
)
8.027 × 10 37
C
=
MR 2
5.966 × 10 24 6.37 × 10 6
(
)(
)
2
k = 0.332
(d)
Let the entire Earth have uniform density ρ. The mass is then given by
M=
4
πρ R 3
3
and the moment of inertia by
C=
8
πρ R 5
15
Dividing the moment of inertia by MR2 gives
⎛ 8
⎞
πρ R 5 ⎟
⎜
⎝ 15
⎠
C
k=
=
2
MR
⎛4
3⎞
2
⎜⎝ πρ R ⎟⎠ R
3
k = 0.4, exactly.
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15.
The gravity potential for a rotating sphere (or spheroid) is given by Eq. (3.56). Show
that this expression does not satisfy the Laplace equation, and give reasons why this is
the case. (Hint: use spherical coordinates as in Box 3.4, Eq. (2), but assume azimuthal
symmetry about the rotation axis, i.e. the potential does not vary with φ).
In Box 3.4, Eq. 2, the variation with φ disappears in the case of azimuthal symmetry
about the rotation axis. The Laplace equation in spherical coordinates then becomes
∇ 2U =
1 ∂ 2 ∂U
1
∂
∂U
r
+ 2
sin θ
=0
2
r ∂r ∂r r sin θ ∂θ
∂θ
The gravity potential Ug for a rotating spheroid (Eq. 3.56) is the sum of the gravitational
1
potential UG and the centrifugal potential U c = − ω 2 r 2 sin 2 θ :
2
U g = UG + U c .
Inserting this expression into the modified Laplace equation gives
1
∂
∂U ⎞ ⎛ 1 ∂ ∂U c
1
∂
∂U c ⎞
⎛ 1 ∂ 2 ∂U G
+ 2
sin θ G ⎟ + ⎜ 2 r 2
+ 2
sin θ
⎜⎝ 2 r
⎟ =0
r ∂r
∂r
r sin θ ∂T
θ ESTB∂AθN⎠KS⎝ErLL∂rER.∂r
∂θ ⎠
COM r sin θ ∂θ
The gravitational potential is known to satisfy the Laplace equation (see §2), so the first
term in brackets is zero. The second term in brackets may be evaluated stepwise:
1 ∂ 2 ∂U c 1 ∂ 2 ∂ ⎛ 1 2 2 2 ⎞
r
= 2 r
⎜ − ω r sin θ ⎟⎠
r 2 ∂r
∂r
r ∂r ∂r ⎝ 2
1 ∂ 3
= −ω 2 sin 2 θ 2
r
r ∂r
= −3ω 2 sin 2 θ
( )
1
∂
∂U
1
∂
∂ ⎛ 1
⎞
sin θ c = 2
sin θ ⎜ − ω 2 r 2 sin 2 θ ⎟
⎠
r sin θ ∂θ
∂θ
r sin θ ∂θ
∂θ ⎝ 2
1 ∂
= −ω 2
sin θ ( sin θ cosθ )
sin θ ∂θ
1 ∂
= −ω 2
sin 2 θ cosθ )
(
sin θ ∂θ
1 ∂
= −ω 2
cosθ − cos 3 θ )
(
sin θ ∂θ
1
=ω2
sin θ − 3cos 2 θ sin θ )
(
sin θ
2
= ω − 3ω 2 cos 2 θ
2
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Combining both parts, the second term in brackets becomes
−3ω 2 sin 2 θ + ω 2 − 3ω 2 cos 2 θ = ω 2 − 3ω 2 ( sin 2 θ + cos 2 θ ) = −2ω 2
Thus, for the gravity potential of a rotating spherioid,
∇ 2U = −2ω 2 ≠ 0
This shows that the gravity potential does not satisfy the Laplace equation.
This is because gravity does not act towards the center of the Earth. It is deflected from the
radial direction by the centrifugal component, which acts perpendicular to the rotation axis.
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SOLUTIONS TO EXERCISES
4 Gravity Surveying
1.
(a) By differentiating the normal gravity formula given by Eq. (3.59) derive an
expression for the change of gravity with latitude. Calculate the gravity change in
milligals per kilometer of northward displacement at latitude 45°.
(b) Repeat the exercise using the International Gravity Formula, Eq. (3.62).
(a)
The normal gravity formula is
gn = ge (1+ β1 sin 2 λ + β 2 sin 2 2 λ )
where ge = 9.780 327 m s–2 = 978,032.7 mgal; β1 = 5.30244 × 10–3; β2 = –5.8 × 10–6.
Hence, geβ1 = 5186 mgal, , geβ2 = –5.67 mgal
Differentiating with respect to λ:
dgn
d
d
= geβ1
sin 2 λ ) + geβ 2
sin 2 2 λ )
(
(
dλ
dλ
dλ
d
sin 2 λ ) = 2sin λ cos λ = sin 2 λ ;
(
dλ
d
sin 2 2 λ ) = 4 sin 2 λ cos 2 λ = 2sin 4 λ
(
dλ
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dgn
= geβ1 sin 2 λ + 2geβ 2 sin 4 λ
dλ
This expression gives the change of gravity in m s–2 with latitude, where the latitude is
in radians. To convert to kilometers of north-south displacement (s) we use the
relationship s =(Rλ), where R is the Earth’s radius.
dgn 1
= ( geβ1 sin 2 λ + 2geβ 2 sin 4 λ )
ds R
Inserting numerical values for the parameters in the equation, with R = 6371 km:
dgn
1
=
{( 5186 ) sin 2λ + 2 ( −5.67 ) sin 4 λ }
ds 6371
= 0.8140sin 2 λ − 0.0018sin 4 λ
This gives the following changes of normal gravity with latitude:
Lat. 30°N, dgN/ds = 0.7034 mgal/km
Lat. 45°N, dgN/ds = 0.8140 mgal/km
Lat. 60°N, dgN/ds = 0.7065 mgal/km
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(b)
The International Gravity Formula (Somigliani formula) is
gn ( λ ) =
gn ( λ ) =
age cos 2 λ + cgp sin 2 λ
a 2 cos 2 λ + c 2 sin 2 λ
)
age + cgp − age sin λ
gn ( λ ) = ge
k=
(
=
age − age sin 2 λ + cgp sin 2 λ
a 2 − a 2 sin 2 λ + c 2 sin 2 λ
2
a 2 − ( a 2 − c 2 ) sin 2 λ
= ge
⎛ cg − age ⎞ 2
1+ ⎜ p
sin λ
⎝ age ⎟⎠
⎛ a2 − c2 ⎞ 2
1− ⎜
sin λ
⎝ a 2 ⎟⎠
(1+ k sin λ )
2
(1)
1− e2 sin 2 λ
cgp − age
= 1.932 × 10 −3 and
age
e2 =
a2 − c2
= 6.694 × 10 −3
a2
To simplify the derivation, let x = sin2𝜆, thus dx = sin2𝜆 d𝜆 and Eq. (1) becomes
gn ( λ ) = ge
(1+ kx )
1− e2 x
The change of normal gravity with latitude is
dgN dgN sin 2 λ dgN ge
d (1+ kx )
=
=
= sin 2 λ
ds
Rd λ
R dx
R
dx 1− e2 x
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⎛
(1+ kx )( 12 )( −e2 ) ⎞
d (1+ kx )
1
2
=
⎜ k 1− e x −
⎟
dx 1− e2 x (1− e2 x ) ⎝
1− e2 x
⎠
⎛ k (1− e2 x ) + 12 e2 (1+ kx ) ⎞
1
=
⎟
(1− e2 x ) ⎜⎝
1− e2 x
⎠
=
k + 12 e2 − 12 ke2 x
(1− e x )
2
dgN ge
= ( k + 12 e2 ) sin 2 λ
ds
R
3/2
⎛
ke2
2 ⎞
⎜⎝ 1− 2k + e2 sin λ ⎟⎠
(1− e
2
sin 2 λ )
3/2
Inserting the appropriate values for the constants the following changes of gravity with
latitude:
Lat. 30°N, dgN/ds = 0.7034 mgal/km
Lat. 45°N, dgN/ds = 0.8140 mgal/km
Lat. 60°N, dgN/ds = 0.7065 mgal/km
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2.
The following gravity measurements were made on a traverse across a rock
formation. Use the combined elevation correction to compute the apparent density of
the rock.
Elevation (m) Gravity (mgal)
100
-39.2
150
-49.5
235
-65.6
300
-78.1
385
-95.0
430
-104.2
The measured gravity anomaly contains (1) a variation due to the elevation above the
reference ellipsoid and the corresponding changing thickness of the “Bouguer plate”
beneath the traverse, and (2) a contribution from deep sources. Ignoring finer details
such as tidal and topographic corrections, the Bouguer anomaly ∆gB is
∆ gB = gm + ∆ ge − gN
(1)
where gm is the measured gravity, given in the table, and gN is the theoretical value on
the reference ellipsoid. ∆ge is the combined elevation correction for elevation h :
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∆ ge = (0.3086 – 0.0419 ρ × 10 -3 )h
(2)
The mean rock density ρ is unknown. Using the Nettleton method (§2.5.5.4) trial values
of density are inserted in Eq. (2) and added to the observed gravity values to give an
elevation-corrected gravity at each measurement station. The optimum density is found
when the corrected anomaly shows the least correlation with the elevation profile.
Elevation
(m)
(1) Anomaly
elevationcorrected
with ρ=2500
(2) Anomaly
elevationcorrected
with ρ=2700
(3) Anomaly
elevationcorrected
with ρ=2500
Free-air
anomaly
(mgal)
100
-18.8
-19.7
-20.5
-8.3
150
-18.9
-20.2
-21.4
-3.2
235
-17.7
-19.7
-21.6
+6.9
300
-16.9
-19.5
-22.0
+14.5
385
-16.5
-19.7
-23.0
+23.8
430
-16.5
-20.1
-23.8
+28.5
The table above gives the gravity anomaly at each station corrected with the
combined elevation correction using densities of 2500, 2700 and 2900 kg m–3.
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A plot of these data (below) shows that using ρ = 2500 kg m–3 the anomaly is
parallel to the elevation profile; this indicates that the chosen density is too low. Using ρ
= 2900 kg m–3 the data are over-corrected; with increasing eleavation the corrected
anomaly becomes ever more negative. With ρ = 2700 kg m–3 there is minimum
correlation with the elevation.
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A better method is to draw a graph in which the gravity anomaly, corrected only for
the free-air effect (last column in the above table), is plotted against the station
elevation.
The points fall on a straight line whose equation may be found using linear regression.
In this case it is
∆ g = 0.1129h − 19.73
The first term corresponds to the Bouguer correction; the second term is the
background anomaly, equal to –19.7 mgal. Thus the coefficient 0.1129 is equal to
0.0419ρ × 10–3. This gives the optimum density, ρ = 2695 kg m–3 ≈ 2700 kg m–3 as in
the first method.
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3.
Show that the “half-width” w of the gravity anomaly over a sphere and the depth z to
the center of the sphere are related by z = 0.652 w.
The equation for the gravity anomaly over the center of a sphere (Eq. 4.25) is
3/2
⎡
⎤
1
⎢
⎥
∆ gz = ∆ g0
⎢ 1 + ( x / z )2 ⎥
TE⎦ STBANKSELLER.COM
⎣
⎛ ∆ ρR3 ⎞
4
where ∆ g0 = π G ⎜ 2 ⎟ is the maximum amplitude of the anomaly over the center
3
⎝ z ⎠
(
)
of the sphere. By definition, the “half-width” w is the width of the anomaly where it has
half its maximum amplitude, i.e. where ∆gz = ½ ∆g0. Because of the symmetry of the
anomaly this occurs at ± x1/2. In this case
⎡
1
⎢
⎢ 1 + ( x / z )2
1/2
⎣
(
)
⎤
⎥
⎥
⎦
3/2
=
1
2
1 + ( x1/2 / z ) = 2 2 / 3
2
Rearranging and solving for x1/2 in terms of z gives
x1/2 = z 2 2 / 3 − 1 = ± 0.7664z
from which the depth in terms of the half-width w is given by
z = 0.652 w
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4.
Assume the "thin-sheet approximation" (Eq. 4.42) for the gravity anomaly over a
vertical fault of density contrast ∆ρ and height h with mid-point at depth z0.
(a) What is the maximum slope of the anomaly and where does it occur?
(b) Determine the relationship between the depth z0 and the horizontal distance w
between the positions where the slope of the anomaly is one half the maximum slope.
(a)
The gravity anomaly of a vertical fault, assuming the thin-sheet approximation, is given
by
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⎡π
⎛ x ⎞⎤
∆ gz = 2G ∆ ρh ⎢ + tan −1 ⎜ ⎟ ⎥
⎝ z0 ⎠ ⎦
⎣2
The slope of the anomaly, m, is found by differentiating with respect to x.
m=
⎞
⎛ x⎞
d
d
h⎛
1
∆ gz ) = 2G ∆ ρh tan −1 ⎜ ⎟ = 2G ∆ ρ ⎜
(
2⎟
dx
dx
z0 ⎝ 1 + ( x / z0 ) ⎠
⎝ z0 ⎠
The maximum slope, m0, is found at x = 0, over the edge of the fault. It is given by
m0 = 2G ∆ ρ
(b)
h
z0
The equation of the slope of the anomaly may be written as
⎛
⎞
1
m = m0 ⎜
2⎟
⎝ 1 + ( x / z0 ) ⎠
There are two places (x1 and x2) where the slope has half its maximum value, as
illustrated in the diagram. Let m = m0/2
⎛
⎞
m0
1
= m0 ⎜
2⎟
2
⎝ 1 + ( x / z0 ) ⎠
1 + ( x / z0 ) = 2
2
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x = ±z0
Thus the two locations where the slope of the anomaly is one half its maximum slope
are at x1 = –z0 and x2 = +z0. The distance between these two positions, w, is therefore
equal to twice the depth to the middle of the fault, z0:
w = 2z0
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5.
Calculate the maximum gravity anomaly at ground level over a buried anticlinal
structure, modeled by a horizontal cylinder with radius 1000 m and density contrast 200
kg m–3, when the depth of the cylinder axis is (a) 1500 m and (b) 5000 m.
According to Eq. (4.36), the maximum gravity anomaly over the axis of a buried
anticline (horizontal cylinder) is given by
⎛ ∆ ρ R2 ⎞
∆ g0 = 2π G ⎜
⎝ z ⎟⎠
where R is the radius of the cylinder, z the depth of its axis, and ∆ρ the density contrast.
(a)
Inserting numerical values: R = 1000 m, z = 1500 m, ∆ ρ = 200 kg m–3,
⎛ 200 ⋅1000 2 ⎞
∆ g0 = 2π G ⎜
= 5.59 × 10 −5 m s –2
⎟
⎝ 1500 ⎠
∆g0 = 5.6 mgal
(b)
Inserting numerical values: R = 1000 m, z = 5000 m, ∆ ρ = 200 kg m–3,
⎛ 200 ⋅1000 2 ⎞
∆ g0 = 2π G ⎜
= 1.68 × 10 −5 m s –2
⎟
⎝ 5000 ⎠
∆g0 = 1.7 mgal
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6. The peak A of a mountain is 1000 meters above the level CD of the surrounding plain, as in
the diagram. The density of the rocks forming the mountain is 2800 kg m–3, that of the
surrounding crust is 3000 kg m–3. Assuming that the mountain and its “root” are
symmetric about A and that the system is in local isostatic equilibrium, calculate the
depth of B below the level CD.
A
C
ρ = 2800
D
ρ = 3000
B
Isostasy is a geological application of the law of Archimedes: “a floating object
receives an upthrust from the supporting fluid equal to the weight of fluid it displaces”.
The diagram shows a cross-section and is two-dimensional; we can compute the forces
ESdiagram.
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acting per meter normal toTthe
ADBC, the mountain and its root, represents the floating object while the “fluid” is
represented by the surrounding mantle.
The volume of displaced fluid is represented by the triangle BCD. If the distance of
B beneath the level CD is d, the volume displaced is ½d(CD). The density of the
displaced material is 3000 kg m–3, so the weight of “fluid” displaced and therefore the
upthrust FU on the block is given by FU = ½d(CD)(3000)g.
This upthrust is counterbalanced by the weight of the mountain with its root.
The volume of ADBC is equal to triangle ACD plus triangle BCD. Point A is 1000
m above CD and point B is distance d below it; the volume of ADBC is equal to
½(d + 1000)(CD). The weight W of the mountain and its root is equal to its volume
times the density 2800 kg m–3: i.e., W = ½(d + 1000)(CD)(2800)g.
Equating the two expressions: FU = W
1
2
d(CD)(3000)g = 12 (d + 1000)(CD)(2800)g
1
2
d(CD)(3000)g = 12 (d + 1000)(CD)(2800)g
d(3000 − 2800) = (1000)(2800)
d = 14000 m (14 km)
9
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7.
A crustal block with mean density 3000 kg m–3 is initially in isostatic equilibrium with
the surrounding rocks whose density is 3200 kg m–3, as in the figure (a). After
subsequent erosion the above-surface topography is as shown in (b). The distance L
remains constant (i.e. there is no erosion at the highest point A) and Airy-type isostatic
equilibrium is maintained. Calculate in terms of L the amount by which the height of A
is changed. Explain why A moves in the sense given by your answer.
First, calculate the height of A above the reference level GG in diagram (a).
Let the width of the base of the block be w and the height of A be h. Applying the
method of the previous exercise,
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wL(3000)g = w(L − h)(3200)g
3200h = (3200 − 3000)L = 200L
h = L / 16 = 0.0625L
Next, calculate the height of A above the reference level HH in diagram (b). In this
case the upper rectangle of the block has been eroded into a triangle of height h which
sits on top of a rectangle of height (L – h). Balancing the weight and upthrust:
w( 12 h + (L − h))(3000)g = w(L − h)(3200)g
3000(L − 12 h) = 3200(L − h)
1700h = 200L
h = (2 / 17)L = 0.1176L
The change in height of the point A above the reference level HH is
1⎞
⎛ 2
∆ h = ⎜ − ⎟ L = 0.055L
⎝ 17 16 ⎠
10
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8.
An idealized mountain-and-root system, as in the figure, is in isostatic equilibrium. The
densities in kg m–3 are as shown. Express the height H of the point A above the
horizontal surface RS in terms of the depth D of the root B below this surface.
A
R
ρ = 2500
ρ=
2000
H
S
H/2
D
ρ = 3000
B
As in the previous exercise, let the width of the base of the block be w. Applying the
same method, the volume of displaced material of density 2500 kg m–3 is w(½H) and
the volume of displaced material of density 3000 kg m–3 is w(D – ½H). The weight of
displaced material, and therefore the upthrust on the block, is
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w( 12 H )2500g + w(D − 12 H )(3000)g
This is counterbalanced by the weight of the block, which is
( 12 w)(D + H )2000g + 2( 12 w)( 12 D)2500g
Equating the weight of the mountain with the weight of material displaced gives
w( 12 H )2500g + w(D − 12 H )(3000)g = ( 12 w)(D + H )2000g + 2( 12 w)( 12 D)2500g
1250H + 3000D − 1500H = 1000D + 1000H + 1250D
H = 0.6D
11
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SOLUTIONS TO EXERCISES
5 Rheology of the Earth
1.
Propose an experiment to measure Poisson's ratio of a rock sample in the laboratory.
Poisson's ratio may most easily be measured via the experiment illustrated in Fig. 5.4. A
force is applied to the rock sample in x-direction. This will lead to a relative
deformation, i.e., strain in x-direction, denoted εxx. The rock sample will also deform in
the perpendicular y-direction, meaning that a strain εyy can be measured. Poisson's ratio
can then be calculated as υ=- εyy/εxx.
2.
A cube of some material is submerged in water where it experiences hydrostatic
pressure of p=10 kPa. The pressure results in the strains εxx = εyy = εz = –0.01 and
εxy = εxz = εyz = 0.
(a) Write down the strain matrix.
ESTBANvolume
KSELchange)
LER.Cthat
OM corresponds to this strain?
(b) What is the dilatation T
(fractional
(c) What is the bulk modulus of the material?
a)
The strain matrix, defined in Eq. (5.14) is given by
⎛ −0.01
0
0 ⎞
⎜
0
−0.01
0 ⎟⎟
⎜
⎜⎝
0
0
−0.01 ⎟⎠
b)
The dilatation is the fractional volume change of the rock sample. Using the diagonal
strains, it may be computed with the help of Eq. (5.8):
θ = ε xx + ε yy + ε zz = −0.03
Thus, as expected, the volume change under hydrostatic compression is negative. The
volume shrinks by 3 %.
c)
The bulk modulus K is the negative ratio of pressure p and dilatation θ . Thus, it
expresses how much the volume of a material changes in response to the application of
an external pressure. In our case, we have
p
10
K =− =−
= 333.3 kPa .
θ
−0.03
1
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3.
Express the Lamé parameter λ in terms of Poisson's ratio and the shear modulus. Can λ
be negative?
From Eq. (19) in Box 5.1 we know that Poisson's ratio can be expressed in terms of λ
and the shear modulus µ as
ν=
λ
.
2(λ + µ)
Solving for λ yields
λ=
2νµ
.
1− 2ν
For the vast majority of Earth materials, ν varies between 0 and 0.5, meaning that λ is
positive. For some engineered materials, ν can be slightly negative, which implies a
slightly negative λ .
4.
Consider a water layer with thickness 1 m, e.g., a shallow river. The flow velocity is 0
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km s-1 at the bottom and 5 km s-1 at the surface. The viscosity of water, at 10°C
temperature, is 1.3 mPa s. What is the shear stress within the water layer?
To solve this problem, we may directly apply Eq. (5.35). The velocity gradient across
the water layer is
dv x
dz
=
5000 m/s
= 5000 s −1
1m
Multiplication by the viscosity yields the shear stress
σ xz = η
dv x
dz
(
) (
)
= 0.0013 × 5000 = 6.5Pa
2
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5.
In a rock deformation laboratory, you expose a sample to a fixed stress σ. As the
experiment proceeds, you increase the temperature (in Kelvin) by 10 %. Knowing that
the activation energy of your sample is Ea=500 kJ mol-1, how do you expect the
deformation rate to increase as a result of heating?
The deformation of solids is a thermally activated process, with a strain rate closely
following Eq. (5.36). Assuming that the shear modulus µ is only weakly affected by
the 10 % temperature change, we find that the ratio of strain rates at different
temperatures T1 and T2 is
(dε /dt )2
(dε /dt )1
=
e
− Ea /kT2
e
− Ea /kT1
≈e
0.909 Ea /kT1
where we have substituted T2 = 1.1 T1. From this we see that the deformation rate
increases, and that the increase itself depends exponentially on the initial temperature
T1. The increase is particularly relevant for low initial temperatures.
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6.
In your laboratory, you also investigate samples of finely-grained sedimentary rock.
Which kind of experiment would you conduct in order to reveal the creep mechanism
by which the samples deform? Argue on the basis of the constitutive equations for
diffusion and dislocation creep.
The experiment must distinguish between diffusion and dislocation creep, the
constitutive equations of which are given in Eqs. (5.36) and (5.37), respectively. A
major difference of the two deformation mechanisms is the dependence of strain rate
dε /dt on the externally applied stress σ . This suggests that one should perform a
series of experiments where the rock sample is subjected to a linearly increasing stress,
while all other conditions such as temperature are kept constant. In the case of
dislocation creep, the strain rate will gradually increase as σ n with n ≥ 3 . In contrast,
for diffusion creep, the strain rate will increase linearly, that is, as σ .
3
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7.
A visco-elastic material is subjected to a constant stress σ, as a consequence of which it
deforms gradually towards a final strain εm. The measured Maxwell time is τ. Derive an
equation for the viscosity of the material in terms of σ, εm and τ.
From the definition of the Maxwell time τ , we directly obtain the viscosity η in terms
of Young's modulus E:
η = τE
Substituting the definition of ε m , we finally find
η=
8.
τσ
.
εm
Consider the deformation of the ocean floor in response to loading in the form of an
island. How is the wavelength of the deformation related to the thickness of the
lithospheric plate? Argue in terms of equations and draw a diagram of the thicknesswavelength relation.
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From Eq. (5.49) we know that the wavelength of the deformation is given by λ = 2πα ,
where α is defined in Eq. (5.50) as
⎡ 4D
⎤
α=⎢
⎥
⎣ (ρm − ρi )g ⎦
1/4
.
For the flexural rigidity D we have from Eq. (5.46)
Eh3
.
D=
12(1− ν2 )
Combining the previous two equations, we finally find
⎡
⎤
4Eh3
λ=⎢
⎥
2
⎢⎣ 12(1− ν )(ρm − ρi )g ⎥⎦
1/4
.
Thus, the wavelength λ is proportional to h3/4 . An increasing thickness of the plate
leads to a slightly sub-linear increase in wavelength of the deformation.
4
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9.
Post-glacial uplift occurs nearly exponentially over time (Eq. (5.55)). Based on the
dating of sedimentary rocks, you know that sample 1, located at a present elevation of
10 m, formed around 1000 years ago. Sample 2, at 5 m elevation, is 500 years old.
(a) What is the corresponding relaxation time of the upper mantle?
(b) The wavelength of the glacial depression is around 1000 km. Provide an estimate of
the upper-mantle viscosity.
a)
Naming the two times t1 and t2, and the corresponding surface depressions w1 and w2,
we obtain two equations from Eq. (5.55):
w1 = w0e
−t1 /τ
, w2 = w0e
−t2 /τ
.
We can eliminate the unknown w0 by dividing one of these equations by the other:
w1
w2
=e
(t2 −t1 )/τ
.
Taking the natural logarithm and solving for the relaxation time τ , we obtain
τ=
(t 2 −t1 )
ln(w1 / w2 )
.
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Inserting the numbers from above, we find τ = 721 years = 22.7 ×109 s .
b)
Rearranging Eq. (5.56), we find an equation for the viscosity η :
η=
1
τρ gλ
4π m
Inserting the relaxation time from part a), an approximate upper-mantle density of
ρm = 4500 kgm-3 , the gravitational acceleration near the surface g = 9.81ms-2 , and the
wavelength λ = 1000×103 m , we finally obtain η = 7.97 ×1019 Pa s .
5
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SOLUTIONS TO EXERCISES
6 Seismology
1.
Calculate the bulk modulus (K), the shear modulus (µ) and Poisson’s ratio (ν) for the
lower crust, upper mantle and lower mantle, respectively, using the values for the Pwave (α) and S-wave (β) velocities, and density (ρ) in the following table.
Region
α [km s–1]
7.4
8.5
Depth [km]
Lower crust
33
Upper mantle
400
Lower mantle
2200
β [km s–1]
4.3
4.8
ρ [kg m–3]
3100
3900
7.0
5300
12.2
The seismic velocities α and β are related to the elastic parameters as follows:
µ
β 2 = , from which µ = ρβ 2
ρ
α2 =
4
µ
3 = K + 4 β 2 , from which K = ρ ⎛ α 2 − 4 β 2 ⎞
⎜⎝
⎟
3 ⎠
ρ
ρ 3
K+
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In terms of the Lamé constants:
2
K =λ+ µ
3
2
λ = K − µ = ρ α 2 − 2β 2
3
(
)
and the Poisson ratio is
λ
ν=
2 (λ + µ )
ν=
ρ (α 2 − 2 β 2 )
((
(α − 2β )
=
) 2 (α − β )
2
)
2 ρ α 2 − 2β 2 + β 2
2
2
2
Inserting the values given in the table, we obtain the following values for the elastic
constants in different depth regions:
Region
Lower crust
Depth
K
µ
10
10
[km] [10 kg m–1 s–2] [10 kg m–1 s–2]
ν
33
9.3
5.7
0.245
400
16
9.4
0.251
Lower mantle 2200
44
26
0.255
Upper mantle
1
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2.
The table below gives the densities and seismic P- and S-wave velocities at various
depths in the Earth.
Depth (km) ρ (1000 kg m–3) α (km s–1) β (km s–1)
100
3.38
8.05
4.45
500
3.85
9.65
5.22
1000
4.58
11.46
6.38
2000
5.12
12.82
6.92
2890
5.56
13.72
7.27
2900
9.90
8.07
0
4000
11.32
9.51
0
5000
12.12
10.30
0
5500
12.92
11.14
3.58
6470
13.09
11.26
3.67
(a) From these quantities calculate the rigidity modulus (µ), bulk modulus (K), and
Poisson’s ratio (ν) at each depth.
(b) Discuss in your own words the information that these data give about the deep
interior of the Earth.
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(a)
The equations for computing the elastic parameters are the same as in the previous
exercise. The values given in the table for various depths in the Earth are converted to
depth-profiles of K, µ and ν, as in the following table.
500
Κ (106)
130
219
µ (106)
67
105
ν
0.280
0.293
1000
353
186
0.275
2000
515
245
0.294
2890
655
294
0.305
2900
645
0
0.5
4000
1,024
0
0.5
5000
1,286
0
0.5
5500
1,383
166
0.442
6470
1,425
176
0.441
Depth (km)
100
(b)
The Earth’s shell-like internal structure of mantle, fluid outer core, and solid inner core
are evident from the velocities and elastic parameters ratios. The inner core is solid, but
the high values of α/β, K/µ and Poisson’s ratio show that it is less rigid than the mantle.
2
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3.
Why might one expect an interface with a small critical angle to be a good reflector of
seismic energy?
A seismic P-wave incident on an interface is partly reflected, partly refracted, and
partly converted into refracted and reflected S-waves as shown in Fig. 6.21. The
refracted and reflected fractions of the energy of an incident P-wave are shown for a
critical angle of 38° in Fig. 6.29. (Note that a small fraction of the incident P-wave
energy that was converted into an S-wave can still be refracted through the interface.)
Rays incident at larger angles than the critical angle are reflected. When the critical
angle is small (i.e. further to the left in Fig. 6.29), more rays are incident at larger than
the critical angle and more of the incident energy is reflected. That is, an interface with
a small critical angle will reflect a large proportion of the incident seismic energy.
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4.
(a)
(b)
(a)
A plane seismic wave, travelling vertically downwards in a rock of density 2200 kg m–3
with seismic velocity 2,000 m s–1, is incident on the horizontal top surface of a rock
layer of density 2400 kg m–3 and seismic velocity 3300 m s–1.
What are the amplitude ratios of the transmitted and reflected waves?
What fraction of the energy of the incident wave is transmitted into the lower medium?
The amplitude ratios for vertical incidence are described by the reflection coefficient R
and the transmission coefficient I:
R=
Z 2 − Z1
;
Z 2 + Z1
T=
2Z1
Z 2 + Z1
where Z1 = ρ1V1 and Z2 = ρ2V2 are seismic impedances, and ρ and V are the density and
seismic velocity, respectively, in each medium
In the upper layer,
Z1 = ρ1V1 = (2200)(2000) = 4.4 × 106 kg m–2 s–1
and in the lower layer
Z2 = ρ2V2 = (2400)(3300) = 7.92 × 106 kg m–2 s–1
Thus
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Z − Z1 7.92 − 4.4
R= 2
=
= 0.29
Z 2 + Z1 7.92 + 4.4
T=
(b)
2Z1
2(4.4)
=
= 0.71
Z 2 + Z1 7.92 + 4.4
The fraction ER of the incident energy reflected at the interface is ER = R2; the fraction
ET of the incident energy transmitted through the interface is equal to (1 – ER).
ET = (1 − ER ) = (1 − R 2 ) = 0.92
92% of the incident energy is transmitted into the lower medium.
4
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5.
A plane seismic wave travels vertically downwards at a velocity of 4800 m s–1 through
a salt layer with density 2100 kg m–3. The wave is incident upon the top surface of a
sandstone layer with density 2400 kg m–3. The phase of the reflected wave is changed
by 180° and the reflected amplitude is 2% of the incident amplitude. What is the seismic
velocity of the sandstone?
The 180° phase change at the surface of the sandstone layer implies that the seismic
impedance Z2 of the sandstone is less than the seismic impedance Z1.of the salt layer.
The reflection coefficient of 2% is thus negative.
R=
Z 2 − Z1
= −0.02
Z 2 + Z1
Z 2 − (2100)(4800)
= −0.02
Z 2 + (2100)(4800)
Solving gives
Z 2 = ρ2V2 =
0.98
(2100)(4800) = 9.685 × 10 6 kg m–2 s–1,
1.02
and by substituting the density ρ2 = 2400 kg m–3 the seismic velocity of the sandstone is
found to be V2 = 4000 m sT–1E. STBANKSELLER.COM
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6.
(b)
(c)
(a) Calculate the minimum arrival times for seismic reflections from each of the
reflecting interfaces in the following section. Consider the base of the lowermost bed to
be also a reflector.
Rock type
Density
(kg m–3)
Thickness
(m)
Formation velocity
(m s–1)
Alluvium
1500
150
600
Shale
2400
450
2700
Sandstone
2300
600
3000
Limestone
2500
900
5400
Salt
2200
300
4500
Dolomite
2700
600
6000
What is the average velocity of the section for a reflection from the base of the
dolomite?
Using the listed densities calculate the reflection coefficient for each interface (except
the base of the dolomite). Which interface gives the strongest reflection and which the
weakest? At which interfaces does a change in phase occur? What does this mean?
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(a)
The formation thicknesses and velocities give the following two-way travel-times in
each formation, and cumulative two-way travel-times for each reflecting interface:
Formation
(b)
Thickness
(m)
Formation
velocity (m s–1)
Formation
Cumulative
two-way
two-way
travel-time (s) travel-time (s)
Alluvium
150
600
0.500
0.500
Shale
450
2700
0.333
0.833
Sandstone
600
3000
0.400
1.233
Limestone
900
5400
0.333
1.567
Salt
300
4500
0.133
1.700
Dolomite
600
6000
0.200
1.900
TOTAL
3,000
1.900
The average velocity for a reflection from the base of the dolomite is the total two-way
distance divided by the total two-way travel-time
V=
s 2(3000)
=
= 3160 m s–1
t
1.900
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(c)
The reflection coefficient for each interface between successive rock types is computed
with the formula for R in exercises 9 and 10:
Rock type
Density
(kg m–3)
Formation
velocity (m s–1)
Reflection
coefficient
Reflection
characteristic
Alluvium
1500
600
0.756
strongest
Shale
2400
2700
0.031
weakest
Sandstone
2300
3000
0.324
---
Limestone
2500
5400
–0.154
phase-change
Salt
2200
4500
0.241
---
Dolomite
2700
6000
---
---
The strongest reflection is at the interface between the alluvium and shale; the
weakest is between the shale and the sandstone. The negative reflection coefficient at
the interface between the limestone and salt indicates a phase change of π radians.
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7.
A reflection seismic record in an area of relatively flat dips gave the following data:
Distance shot-point
to detector (m)
Travel-time (t1),
1st reflection (s)
Travel time (t2),
2nd reflection (s)
30
1.000
1.200
90
1.002
1.201
150
1.003
1.201
210
1.007
1.202
270
1.011
1.203
330
1.017
1.205
390
1.023
1.207
(a)
(b)
(c)
(d)
Plot the t:x curves for these reflections to show the “moveout” effect.
On a different graph, plot the t2:x2 curves (i.e., squared data) for the reflections.
Determine the average vertical velocity from the surface to each reflecting bed.
Use these velocities to compute the depths to the reflecting beds.
(a)
The t:x curves are plotted in the left part of the figure below: their curvature is a result
of the "moveout" effect; TESTBANKSELLER.COM
(b)
The t2:x2 curves are plotted in the right part of the figure: the equations of the best-fit
straight lines are shown on the plot. The theoretical t2:x2 relationship is
2
x2
x2
2
⎛ 2d ⎞
t = ⎜ ⎟ + 2 = (t0 ) + 2 ,
⎝V ⎠
V
V
2
where V is the average vertical velocity (called the stacking velocity) of a reflection and
t0 =
2d
is the two-way vertical "echo" time.
V
8
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(c)
The t2 : x2 plots are straight lines, with typical equation y = y0 + mx, in which m = 1/V2 is
the slope and y0 = (t0)2 = (2d/V)2 is the y-intercept of the line. Using the slopes of the
best-fit lines in the right-hand figure, we get for the stacking velocities, V1 and V2 ,
respectively, of the two reflections:
1st reflection:
1
m1 =
(V1 )
3.064 × 10
1
m2 =
(V2 )
2
−7
= 1807 m s–1
= 1.065 × 10 −7
1
V2 =
(d)
= 3.064 × 10 −7
1
V1 =
2nd reflection:
2
1.065 × 10
−7
= 3064 m s–1
The y-intercept of the line is y0 = (t0)2 = (2d/V)2 in which d is the vertical depth to the
horizontal reflector and V the average (or stacking velocity) of the reflection. Using the
intercepts of the best-fit lines in the right-hand figure, we get:
st
1 reflection:
(t01 )
2
⎛ 2d ⎞
= ⎜ 1 ⎟ = 1.00
⎝ V ⎠
2
1
TESTBAdN1 K=S2EV1LL1.00
ER.=C0.5
OM× 1807 × 1 = 903 m
1
nd
2 reflection:
(t02 )
2
2
⎛ 2d ⎞
= ⎜ 2 ⎟ = 1.44
⎝ V ⎠
2
d2 = 12 V2 1.44 = 0.5 × 3064 × 1.2 = 1839 m
9
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8.
The following table gives two-way travel times of seismic waves reflected from
different reflecting interfaces in a horizontally layered medium.
Geophone to
Shot-point
Distance [m]
(a)
(b)
(c)
(d)
(e)
Travel-time [s] to
First
Reflector
Second
Third
Reflector Reflector
500
0.299
0.364
0.592
1000
0.566
0.517
0.638
1500
0.841
0.701
0.708
2000
1.117
0.897
0.799
2500
1.393
1.099
0.896
Draw a plot of (travel-time)2 against (distance)2.
Determine the vertical two-way travel-time (“echo-time”) and average velocity to each
reflecting interface.
Compute the depth of each reflector and the thickness of each layer.
Compute the true velocity (interval velocity) of each layer.
Verify your results by computing the total vertical travel-time for a wave reflected from
the deepest interface.
TESTBANKSELLER.COM
(a)
Plot of (travel-time)2 against (distance)2
The best-fit straight lines (1), (2) and (3) to the data in the diagram have the following
equations, with x = (distance [km])2 and y = (travel-time [s])2:
(1) y = 0.3086x + 0.01240
(2) y = 0.1792x + 0.0880
(3) y = 0.07567x + 0.3319
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(b)
The equation of each line in the (t2 – x2) diagram has the form
t 2 = t0 2 +
x2
2d
, where t 0 =
is the vertical travel-time ("echo-time") and V is the
2
V
V
average (or stacking) velocity to a reflector. Applied to the equations of the three
straight lines:
(1) average velocity V = 1 / 0.3086 = 1.80 km s–1
vertical two-way travel-time t 0 = 0.0124 = 0.111 s.
(2) average velocity V = 1 / 0.1792 = 2.36 km s–1
vertical two-way travel-time t 0 = 0.0880 = 0.297 s.
(3) average velocity V = 1 / 0.07567 = 3.64 km s–1
vertical two-way travel-time t 0 = 0.3319 = 0.576 s.
(c)
From the vertical two-way travel-time t0 and the average velocity V the depth d to each
reflector is equal to (Vt0)/2
(1) depth d1 = (0.5)(0.111)(1800) = 100 m
thickness of layer (1) is 100 m.
(2) depth d2 = (0.5)(0.297)(2360) = 350 m
thickness of layer (2) is (350 – 100) = 250 m.
(3) depth d3 = (0.5)(0.576)(3640) = 1050 m
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thickness of layer (3) is (1050 – 350) = 700 m.
(d)
The interval velocities of the layers require the layer thicknesses d as well as the
interval travel-time in each layer, which is found from the differences between the
successive "echo-times". Thus,
2di
Vi =
(t 0 )2 − (t 0 )1
(1) the interval velocity of the top layer is the same as the first average velocity:
V1 = 1800 m s–1 = 1.8 km s–1
2(250)
= 2700 m s–1 = 2.7 km s–1
0.297 − 0.111
2(700)
(3) the interval velocity of layer (3) is Vi =
= 5000 m s–1 = 5.0 km s–1
0.576 − 0.297
(2) the interval velocity of layer (2) is Vi =
(e)
The total travel-time for a reflection from the deepest interface is the sum of the twoway travel-times in each layer.
250
700 ⎞
⎛ 100
t = 2⎜
+
+
= 0.576 s, which is equal to the vertical "echo-time"
⎝ 1800 2700 5000 ⎟⎠
from the deepest reflector (solution (3) in part (b) above).
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9.
Assume the horizontally layered structure from the previous problem.
(a) If a seismic ray leaves the surface at an angle of 15° to the vertical, how long does it
take to return to the surface after reflecting from the basement?
(b) At what horizontal distance from the shot-point does this ray reach the surface?
(a)
The path and travel-time of the ray reflected from the basement will mirror the
downward ray. For horizontal layers the angle of refraction at each interface becomes
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the angle of incidence at the next. The angles θ1, θ2, and θ3 at each interface are related
by the law of refraction:
sin θ 2 =
V2
V
V
sin θ1; sin θ 3 = 3 sin θ 2 = 3 sin θ1
V1
V2
V1
Inserting θ1 = 15°, and the interval velocities V1 = 1.8 km s–1, V2 = 2.7 km s–1, V3 = 5.0
km s–1, we get for the angles θ2 and θ3
⎛V
⎞
⎛ 2.7
⎞
θ 2 = sin –1 ⎜ 2 sin θ1 ⎟ = sin –1 ⎜
sin(15°)⎟ = 22.8°
⎝ 1.8
⎠
⎝ V1
⎠
⎛V
⎞
⎛ 5.0
⎞
θ 3 = sin –1 ⎜ 3 sin θ1 ⎟ = sin –1 ⎜
sin(15°)⎟ = 46.0°
⎝ 1.8
⎠
⎝ V1
⎠
The ray traverses each layer along the hypotenuse s of a triangle with the speed V of
the layer, so the travel-time in the layer is given by (s/V). The distance traveled in each
layer is related to the vertical thickness d of the layer and the angle θ of the ray, so the
time spent in each layer is given by:
t=
s 1 d
=
V V cosθ
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This gives interval travel-times for the downward ray in each layer equal to:
t1 =
1 d1
100
=
= 0.0575 s
V1 cosθ1 1800 cos(15.0°)
t2 =
1 d2
250
=
= 0.1005 s
V2 cosθ 2 2700 cos(22.8°)
t3 =
1 d3
700
=
= 0.2014 s
V3 cosθ 3 5000 cos(46.0°)
The total travel-time of the ray reflected from the basement is twice the sum of the
interval times, and is therefore equal to 0.719 s.
(b)
The horizontal distance x traveled by a ray in a layer of thickness d and velocity V is
x = d tan θ
As in the previous case, the upward reflected path is symmetric with the downward
path. The total horizontal distance from the shot point to the point of emergence of the
basement reflection is
x = 2 ( d1 tan θ1 + d2 tan θ 2 + d3 tan θ 3 )
= 2 (100 tan15° + 250 tan 22.8° + 700 tan 46.0° )
TESTdistance
BANKSof
EL1710
LERm.
.COM
The ray emerges at a horizontal
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10.
Assume that the three horizontal homogeneous rock layers in the previous problems
have densities of 1800, 2200, and 2500 kg m–3 respectively. The lowest layer overlies
basement with velocity 5.8 km s–1 and density 2700 kg m–3.
(a) Compute the reflection and transmission coefficients at each interface for a plane Pwave traveling vertically downwards.
(b) Calculate what fraction of the initial energy of the wave is transmitted into the
basement.
(c) Calculate the fraction of the initial energy carried in the reflection that returns to the
surface from the basement.
(a)
Let the layer above an interface have density ρ1 and velocity V1 and let the layer below
the interface have density ρ2 and velocity V2. The seismic impedances of the upper and
lower layer are Z1 = ρ1V1 and Z2 = ρ2V2, respectively. The reflection coefficient R and
transmission coefficient T are defined for the wave traveling vertically downwards as
R=
Z 2 − Z1
;
Z 2 + Z1
T=
2Z1
Z 2 + Z1
Using these formulas with the given information, we get the following values for the
reflection and transmission coefficients at the base of each layer:
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ρ (kg m )
V (km s )
Z (kg m–2 s–1)
× 106
Layer (1)
1800
1.8
3.240
0.294
0.706
Layer (2)
2200
2.7
5.940
0.356
0.644
Layer (3)
2500
5.0
12.50
0.112
0.888
Basement
2700
5.8
15.66
***
***
–3
(b)
–1
Reflection
coefficient
Transmission
coefficient
The fraction ER of the incident energy that is reflected at each interface is equal to R2;
the fraction ET transmitted equals (1 – ER). We can make up another table that shows
these fractions for each individual interface in the exercise:
Downward
traveling wave
Reflection
coefficient R
ER
ET
Layer (1)=>(2)
0.294
0.087
0.913
Layer (2)=>(3)
0.356
0.127
0.873
Layer (3)=>B
0.112
0.0126
0.987
Basement (B)
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The fraction ET1 of the initial energy that is transmitted at the first interface is partly
reflected at the second interface; the fraction that passes the second interface is ET2 of
the incident energy, and is thus equal to (ET1)(ET2) of the initial energy. This process
repeats at each successive interface. The fraction ETB of the initial energy that passes
into the basement is
ETB = ET 1ET 2 ET 3 = ( 0.913) ( 0.873) ( 0.987 ) = 0.788
i.e., 78.8% of the initial energy is transmitted into the basement.
(c)
The energy incident on the basement has passed through two interfaces and is equal to
(0.913)(0.873) = 0.798 of the initial energy. At the basement interface only 0.0126 of
this energy is reflected, equivalent to (0.798)(0.0126) = 0.010 of the initial energy.
During its return to the surface this energy must again be transmitted through (and
partially reflected at) each of the two overlying interfaces. The reflection coefficients at
each interface are the same as for the downward traveling wave, except that their signs
are changed. The fractions of incident energy reflected and transmitted are unaltered.
The fraction of the initial energy that returns to the surface in the reflection from the
basement is therefore equal to
TESTB=A0.0080,
NKSELorLE
R.CofOthe
M initial energy.
(0.798)(0.0126)(0.873)(0.913)
0.8%
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11.
An incident P-wave is converted into refracted and reflected P- and S-waves at an
interface. Calculate all the critical angles in the following 3 cases, where α and β are the
P-wave and S-wave velocities, respectively:
Layer:
Seismic
wave
Case (a)
(km s–1)
Case (b)
(km s–1)
Case (c)
(km s–1)
Above
interface
α
β
3.5
4.0
5.5
2.0
2.3
3.1
Below
interface
α
β
8.5
6.0
7.0
5.0
3.5
4.0
The critical angle of incidence ic at an interface corresponds to the incident angle on
the interface that gives a refracted angle of 90. Rays incident at angles greater than the
critical angle are totally reflected at the interface. The critical angle is derived from
Snell’s law:
V
sin ic = 1
V2
where V1 is the velocity of an incident phase and V2 is the velocity of the refracted
phase.
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At each interface, the incident P-wave and S-wave can each be partitioned into a
refracted P-wave and a refracted S-wave.
Using the values given in the table, we get the following table of critical angles:
Wave
conversion
P- to P-wave
P- to S-wave
S- to S-wave
S- to P-wave
Wave
velocities
(a)
Critical angle
(b)
Critical angle
(c)
Critical angle
α => α
α => β
β => β
β => α
24.3
41.8
51.8
44.4
***
***
23.6
41.1
50.8
13.6
22.5
26.3
NOTE: *** In these cases, the velocity of the refracted phase is less than the velocity
of the incident phase, the refracted ray is deflected toward the normal to the interface,
and there is no critical angle (or head wave).
16
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12.
An incident P-wave is converted into refracted and reflected P- and S-waves at an
interface that is inclined at 20° to the horizontal, as in the figure below. The respective
P- and S-wave velocities are 5 km s–1 and 3 km s–1 above the interface and 7 km s–1
and 4 km s–1 below the interface. If the incident P-wave strikes the interface at an angle
of 40° to the horizontal, calculate the angles to the horizontal made by the reflected and
refracted P- and S-waves.
The first step is to expand the given sketch, as above, by adding the normal to the
interface, the reflected and refracted P- and S-rays, and the corresponding angles iP and
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iS for the reflected rays and rP, and rS for the refracted rays; all angles are measured
from the normal to the interface. From the geometry, the interface is inclined at 20° to
the horizontal, so the normal is inclined at 70° to the horizontal.
Snell’s law gives the relationships between the angles of incidence, reflection and
refraction:
sin iP sin iS sin rP sin rS
=
=
=
=p
α1
β1
α2
β2
The incident P-wave makes an angle of (40° + 20°) to the interface, i.e., 30° to the
normal, and so the angle of incidence iP = 30°.
Inserting the appropriate values we can determine the angles between the reflected
and refracted rays and the normal to the interface.
REFLECTED RAYS
Reflected P-wave:
The angle of reflection is equal to the angle of incidence, so the reflected P-wave is
inclined at 30° to the normal and at 60° to the interface; it is inclined at 80° to the
horizontal.
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Reflected S–wave:
sin iS =
β1
3
sin iP = sin(30°) = 0.3
α1
5
The angle of reflection of the S-wave is sin–1(0.3) = 17.5°. This ray is inclined at 17.5°
to the interface, which itself is inclined at 70° to the horizontal; the ray is inclined at
87.5° to the horizontal.
REFRACTED RAYS
Refracted P–wave:
sin rP =
α2
7
sin iP = sin(30°) = 0.7
α1
5
The angle of refraction of the P-wave is sin–1(0.7) = 44.4°. This ray is inclined at 45.6°
to the interface and at 25.6° to the horizontal.
Refracted S–wave:
sin rS =
β2
4
sin iP = sin(30°) = 0.4
α1
5
The angle of refraction of the S-wave is sin–1(0.4) = 23.6°. Note that this is less than the
angle of incidence of the P-wave.
TESTBThis
ANKray
SEis
LLinclined
ER.COatM66.4° to the interface and at
46.4° to the horizontal.
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13.
(a)
(b)
(c)
(d)
(e)
(f)
A seismic refraction survey gave the following data for the first arrival times at various
distances from the shot-point.
Distance (km)
Time (s)
Distance (km)
Time (s)
3.1
1.912
13.1
6.678
5.0
3.043
14.8
7.060
6.5
3.948
16.4
7.442
8.0
4.921
18.0
7.830
9.9
5.908
19.7
8.212
11.5
6.288
Plot the travel-time curve for the first arrivals.
Calculate the seismic velocities of the layers.
Calculate the critical angle of refraction for the interface.
Calculate the minimum depth to the refracting interface.
Calculate the critical distance for the first arrival of refracted rays.
Calculate the crossover distance beyond which the first arrivals correspond to head
waves.
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(a)
Travel-time plot:
The travel-time plot shows two straight lines with equations
y = 0.6056x ; direct arrivals (line must pass through origin)
y = 0.2348 x + 3.592; double-refracted (head wave) arrivals
(b)
The direct ray has equation t =
x
. The velocity V1 of the upper layer is the inverse of
V1
the slope; it gives V1 = 1.65 km s–1.
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x
+ ti , where ti is the intercept-time, given
V2
2d
t
=
cosic . The velocity V of the
i
by the intersection of the line with the time-axis:
2
V1
The double-refracted ray has equation t =
lower layer is the inverse of the slope of this line; it gives V2 = 4.26 km s–1.
(c)
The critical angle of refraction ic is obtained from Snell’s law:
⎛V ⎞
⎛ 1.65 ⎞
−1
ic = sin −1 ⎜ 1 ⎟ = sin −1 ⎜
⎟⎠ = sin ( 0.3877 ) = 22.8°
⎝
V
4.26
⎝ 2⎠
(d)
The depth to the refracting interface is obtained from the intercept-time ti
d=
(e)
V1ti
(1.65)(3.592)
=
= 3.22 km
2 cosic
2 cos(22.8)
The critical distance xc is the distance where the first head-wave arrival is recorded. The
ray is simultaneously a reflection at the critical angle. The critical distance is given by
xc = 2d tan ic = 2(3.22) tan(22.8) = 2.71 km.
(f)
The crossover distance xcr is where the travel-times of the direct ray and first head-wave
arrival are equal. It is given by
xcr xcr 2d
=
+
cosic , from which
V1 V2 V1
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V2
4.26
xcr = 2d
cosic = 2(3.22)
cos(22.8) = 9.69 km
( 4.26 − 1.65 )
(V2 − V1 )
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14.
A seismic refraction survey is carried out over a layered crust with flat-lying interfaces.
In one case the crust is homogeneous and 30 km thick with a P-wave velocity 6 km s–1
and overlies mantle with P-wave velocity 8 km s–1. In the other case the crust consists
of an upper layer 20 km thick with P-wave velocity 6 km s–1 overlying a lower layer 10
km thick with P-wave velocity 5 km s–1. The upper mantle P-wave velocity is again 8
km s–1. On the same graph, plot the first arrival time curves for the two cases. What is
the effect of the low-velocity layer on the estimation of depth to the top of the mantle?
The geometries of the two structures, and the corresponding travel-time curves are as
follows:
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Near to the shot-point, the first arrival is the direct wave, which travels with the
velocity of the surface layer. At horizontal distances greater than the crossover distance,
the first arrival is the double-refracted head wave in the mantle.
Case (1): single-layer crust.
The critical angle ic at the interface between the crust and mantle is given by
⎛V ⎞
⎛ 6⎞
ic = sin –1 ⎜ c ⎟ = sin –1 ⎜ ⎟ = 48.6°
⎝ 8⎠
⎝ Vm ⎠
The distance xcr to the crossover point is found by equating the travel-times of the direct
wave and the refracted wave:
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xcr xcr 2d
=
+
cosic
Vc Vm Vc
xcr = 2d
Vm
V + Vc
cosic = 2d m
Vm − Vc
(Vm − Vc )
Inserting the given values for the single-layer crust, xcr = 2(30)
8+6
= 159 km
8−6
Beyond the crossover distance of 159 km the first arrival will plot as a straight line with
slope (1/Vm). The intercept of this line on the time-axis is t1, given by
2d
cosic
Vc
2(30)
t1 =
cos(48.6°) = 6.6 s in this exercise.
6
t1 =
Case (2): two-layer crust.
In this case the lower crust is a low-velocity layer, in which the downward ray is
refracted towards the normal to the interface between the upper crust and lower crust.
The ray then impinges more steeply on the mantle-crust interface than in the first case.
The return path to the surface is a mirror-image of the downward path.
The travel-time t of the head wave arrival at horizontal distance x is given by
⎛d
⎞
x
d
t=
+ 2 ⎜ 1 cosi1 + T2 Ecosi
ST2B⎟ ANKSELLER.COM
Vm
V2
⎝ V1
⎠
where i1 is the incident angle at the boundary between the upper and lower crust and i2
is the incident (and critical) angle at the crust-mantle boundary. The head wave arrivals
from the mantle plot as a straight line with slope (1/Vm), as in the single-layer case.
The critical angle i2 at the interface between the lower crust and the mantle is given
by
⎛V ⎞
⎛ 5⎞
i2 = sin –1 ⎜ 2 ⎟ = sin –1 ⎜ ⎟ = 38.7°
⎝ 8⎠
⎝ Vm ⎠
This angle is the same as the angle of refraction i2 at the interface between upper and
lower crust, which is related to the angle of incidence i1 from the upper crust by
sin i1 =
V1
V V
V
sin i2 = 1 2 = 1
V2
V2 Vm Vm
⎛V ⎞
⎛ 6⎞
i1 = sin –1 ⎜ 1 ⎟ = sin –1 ⎜ ⎟ = 48.6°
⎝ 8⎠
⎝ Vm ⎠
Using the layer thicknesses, d1 and d2, we can compute the crossover distance xcr
beyond which the head wave from the mantle is the first arrival at surface
seismometers. This is again obtained by equating the direct and refracted travel-times
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⎛d
⎞
xcr xcr
d
=
+ 2 ⎜ 1 cosi1 + 2 cosi2 ⎟
V1 Vm
V2
⎝ V1
⎠
Inserting the given values in this equation gives a crossover distance of 181 km.
The intercept time for the refracted arrivals is
⎛d
⎞
d
ti = 2 ⎜ 1 cosi1 + 2 cosi2 ⎟ = 7.5 s.
V2
⎝ V1
⎠
The low-velocity layer can not be detected from the refraction profile. Without
knowledge of its existence, the increased intercept-time and crossover distance would
be interpreted to give a crust-mantle boundary that is deeper than it really is.
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15.
The table below gives up-dip and down-dip travel-times of P-wave arrivals for
refraction profiles over an inclined interface. The geophones are laid out in a straight
line passing through the alternate shot-points A and B, which are 2700 m apart on the
profile.
Distance from
shot-point [m]
Travel-time [s]
from A
from B
300
0.139
0.139
600
0.278
0.278
900
0.417
0.417
1200
0.556
0.556
1500
0.695
0.695
1800
0.833
0.833
2100
0.972
0.972
2400
1.085
1.111
2700
1.170
1.170
3000
1.255
1.223
3300
1.339
1.276
3600
1.424
1.329
TESTBANKSELLER.COM
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Plot the travel-time curves for each shot-point.
Calculate the true velocity of the upper layer.
Calculate the apparent velocities of the layer below the refractor.
In which direction does the refracting interface dip?
What is the angle of dip of the interface?
What is the true velocity of the layer below the refractor?
What are the closest distances to the refractor below A and B?
What are the vertical depths to the refractor below A and B?
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(a)
Travel-time curves: The best-fit straight lines to segments of the curves are given for a
coordinate system (x, y) with origin at shot-point A (x = distance from A in kilometers,
y = travel-time in seconds).
(b)
The velocity of the upper layer V1 is obtained by inverting the slope of the direct arrival
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line; i.e., for both shot-points, V1 = (0.463)–1. The upper layer velocity is 2.16 km s–1.
(c)
The apparent velocities of the lower layer, V2A and V2B, are obtained by inverting the
slopes of the refracted arrival lines, for shot-points A and B respectively.
For the profile from shot-point A, V2A = (0.282)–1.= 3.54 km s–1;
for the profile from shot-point B, V2B = (0.181)–1.= 5.53 km s–1.
(d)
The time-intercepts at the shot-points indicate the direction of dip of the interface. The
intercept at A is less than at B, so the refractor is closer to A. The interface dips from A
to B.
(e)
The angle of dip θ is obtained from Eq. (6.84) using the true and apparent velocities:
θ=
(f)
⎫ 1 ⎧ −1 ⎛ 2.16 ⎞
1 ⎧⎪ −1 ⎛ V1 ⎞
−1 ⎛ V1 ⎞ ⎪
−1 ⎛ 2.16 ⎞ ⎫
−
sin
⎨sin ⎜
⎬ = ⎨sin ⎜⎝
⎟⎠ − sin ⎜⎝
⎟ ⎬ = 7.3°
⎟
⎜
⎟
2 ⎩⎪
3.54
5.53 ⎠ ⎭
⎝ V2 A ⎠
⎝ V2 B ⎠ ⎭⎪ 2 ⎩
The true velocity V2 of the lower layer is obtained from Eq. (6.86):
1
1 ⎛ 1
1 ⎞
1
1 ⎞
⎛ 1
=
+
=
+
⎜⎝
⎟ = 0.233
⎜
⎟
V2 2 cosθ ⎝ V2 A V2 B ⎠ 2 cos(7.3) 3.54 5.53 ⎠
The true velocity of the lower layer is 4.28 km s–1.
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(g)
The closest distances to the interface at the shot-points are the perpendicular distances,
dA and dB, which are obtained from the corresponding intercept times, tiA and tiB. First,
it is necessary to calculate the critical angle of refraction, using Eq. (6.85):
ic =
⎫ 1 ⎧ −1 ⎛ 2.16 ⎞
1 ⎧⎪ −1 ⎛ V1 ⎞
−1 ⎛ V1 ⎞ ⎪
−1 ⎛ 2.16 ⎞ ⎫
+
sin
=
sin
+
sin
⎨sin ⎜
⎬
⎨
⎜
⎟
⎜⎝
⎟⎠ ⎬ = 30.3°
⎜⎝ V ⎟⎠
⎝
⎠
2 ⎪⎩
2
3.54
5.53
⎝ V2 A ⎟⎠
⎩
⎭
2B ⎪
⎭
Perpendicular distance at shot-point A, Eq. (6.80):
tiA =
dA
Vt
(2.16)(0.408)
= 0.510 km, or 510 m.
cosic , from which d A = 1 iA =
V1
cosic
cos(30.3)
Perpendicular distance at shot-point B, Eq. (6.82):
The equation requires the time-intercept at B. The equation in the figure gives the yintercept at x = 0, i.e. at shot-point A. Substitution x = 2.7 km gives the intercept at B:
tiB = 0.680 s.
dB =
(h)
V1tiB (2.16)(0.680)
= 0.850 km, or 850 m.
=
cosic
cos(30.3)
The vertical distances DA and DB to the interface at the shot-points are obtained from
the corresponding perpendicular distance, using the computed slope θ of the interface
and the perpendicular distances dA and dB.
TESA:
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Vertical distance at shot-point
DA =
dA
510
=
= 514 m.
cosθ cos(7.3)
Vertical distance at shot-point B:
DB =
dB
850
=
= 857 m.
cosθ cos(7.3)
26
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SOLUTIONS TO EXERCISES
7 Earthquakes and the Earth’s Internal Structure
1.
A strong earthquake off the coast of Japan sets off a tsunami that propagates across the
Pacific Ocean (average depth d = 5 km).
(a) Calculate the velocity of the wave in km hr–1 and the corresponding wavelength,
when the wave has a dominant period of 30 min.
(b) How long does the wave take to reach Hawaii, which is at an angular distance of 54°
from the epicenter?
(a)
The phase velocity and group velocity of a tsunami are equal when the water depth is
much less than the wavelength. In this case, the velocity V of the tsunami is related to
the ocean depth d by the equation
V = gd
The wavelength λ and period T of the wave are related to the velocity by
λ = VT
TEthe
STvelocity
BANKSof
ELthe
LEtsunami
R.COMover the open ocean is
For an ocean depth of 5 km,
V=
( 9.81) ( 5000 ) = 221
m s–1 = 797 km hr–1
The wavelength corresponding to a period of 30 min (1800 s) is
λ = (221)(1800) = 397800 m = 398 km
(b)
The time taken for the tsunami to reach Hawaii is the great circle distance divided by
the velocity of the wave.
The great circle distance is s = R θ where R = 6371 km is the Earth’s radius and θ is
the angular distance in radians, in this case (54/180)π = 0.3π.
s = 6371( 0.3) π = 6005 km
The time for the tsunami to reach Hawaii is therefore
t = 6005 / 797 = 7.5 hr
1
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2.
The dispersion relation between frequency ω and wave number k of seismic water
waves for water depth d is (Box 3.3)
ω 2 = gk tanh(kd)
(a) Modify this expression for wavelengths that are much shorter than the water depth.
(b) Determine the phase velocity of these waves.
(c) Show that the group velocity of the waves is half the phase velocity.
(a)
The wave number is defined as k = 2π/λ, where λ is the wavelength. Thus,
d⎞
⎛
tanh(kd) = tanh ⎜ 2π ⎟
⎝ λ⎠
For λ << d, the argument (kd) becomes very large.
tanh(kd) =
ekd − e− kd 1 − e−2 kd
=
≈ 1 for large values of (kd)
ekd + e− kd 1 + e−2 kd
Thus, for wavelengths much shorter than the water depth, ω 2 = gk .
(b)
The phase velocity is defined as c = ω/k. In the case λ << d,
c=
gk
ω
=
=
k
k
c=
g
λ.
2π
g
k
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The speed depends on wavelength; the waves are dispersive.
(c)
The group velocity is defined as U= ∂ω/∂k.
U=
∂ω
∂
1 g 1
=
gk =
= c
∂k ∂k
2 k 2
Thus for water waves with wavelength much shorter than the water depth (e.g., normal
water waves on the ocean surface), the group velocity is one half the phase velocity.
[NOTE: The velocity of tsunami waves depends only on the water-depth. They are
non-dispersive. Ocean-surface waves are dispersive: their velocity depends on the
wavelength, and the longest wavelengths travel fastest. A group of these waves moves
across the ocean surface at half the speed of the individual waves in the group, with
longer wavelengths from the back of the group overtaking shorter wavelengths. Seismic
surface waves propagate in a similar way (see §3.3.3.3).]
2
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3.
In a two-layer Earth the mantle and core are each homogeneous and the radius of the
core is one-half the radius of the Earth. Derive a formula for the travel-time curve for
the arrival time t of the phase PcP at epicentral distance ∆. Verify the formula for the
maximum possible value of ∆ in this model.
The geometry of the problem is shown in the diagram above. The distance traveled
by the PcP phase is (2s), where
2
⎛ R⎞
⎛ R⎞
⎛ ∆⎞
s = R + ⎜ ⎟ − 2R ⎜ ⎟ cos ⎜ ⎟
⎝ 2⎠
⎝ 2⎠
⎝ 2⎠
2
s=
2
R
⎛ ∆ ⎞ TESTBANKSELLER.COM
5 − 4 cos ⎜ ⎟
⎝ 2⎠
2
The travel-time for the arrival at epicentral distance ∆ is
t=
2s R
⎛ ∆⎞
=
5 − 4 cos ⎜ ⎟
⎝ 2⎠
V V
The maximum travel-time tm occurs when the PcP path is a straight line tangential to
the core at the point P. Triangle OPS0 has sides R and R/2 enclosing angle (∆/2), which
is then equal to 60°, because
⎛ ∆ ⎞ ( OP ) (R / 2) 1
cos ⎜ ⎟ =
=
=
⎝ 2 ⎠ ( OS0 )
R
2
The maximum travel-time is then equal to
tm =
S0 P + PG0
=2
V
(
3R / 2
V
)= R
V
3
This is the same value as obtained from the travel-time equation with ∆ equal to 120°,
the maximum possible epicentral angle for a PcP phase in this problem.
3
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The travel-time equation can be written in terms of tm as
t = tm
5 4
⎛ ∆⎞
− cos ⎜ ⎟
⎝ 2⎠
3 3
The travel-time curve looks like the following diagram:
There are no PcP arrivals at epicentral distances greater than 120°.
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4.
The P-wave from an earthquake arrives at a seismograph station at 10:20 a.m. and the
S-wave arrives at 10:25 a.m. Assuming that the P-wave velocity is 5 km s–1 and that
Poisson’s ratio is 0.25, compute the time at which the earthquake occurred and its
epicentral distance in degrees from the seismograph station.
Given that Poisson’s ratio ν = 0.25 = ¼ , we can write
(α − 2β ) = 1
ν=
2 (α − β ) 4
2 (α − 2 β ) = (α − β ) , or α
2
2
2
2
2
2
2
2
2
= 3β 2 , from which α = 3β
Time of occurrence. Assuming that the velocities α and β are constant for local
earthquakes, the Wadati plot is a straight line with equation:
⎛α
⎞
t s − t p = t p ⎜ − 1⎟ , where ts and tp are the travel-times of the S- and P-waves,
⎝β
⎠
respectively. The difference in travel times is known from the two arrival times: it is 5
minutes (300 s). The ratio α/β is known from the Poisson ratio. Thus, the travel-time of
the P-wave is
ts − t p
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300
tp =
=
= 410 s
(α / β ) − 1
3 −1
The earthquake occurred 410 s, or 6 min 50 s, before the P-wave arrived at 10:20 a.m.
The time of the earthquake was around 10:13:10 a.m. (10 seconds after 10:13 a.m.).
Angular epicentral distance. If the velocity of the P-wave α was constant along its
path, the distance through the mantle to the earthquake focus is 410 × 5 = 2050 km.
If the seismic velocities are constant, the ray is a straight line as in the figure below.
Inserting s = 2050/2 = 1025 km and R = 6371 km, the epicentral angle ∆ is given by
⎛ ∆ ⎞ s 1025
sin ⎜ ⎟ = =
= 0.1608 . This gives an angular epicentral distance ∆ = 18.5°.
⎝ 2 ⎠ R 6371
5
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5.
The following table gives arrival times of P-waves (tp) and S-waves (ts) from a nearby
earthquake:
Recording
Station
Time of day
[hr:min]
tp [s]
ts [s]
A
23:36 +
54.65
57.90
B
23:36 +
57.34
62.15
C
23:37 +
00.49
07.55
D
23:37 +
01.80
10.00
E
23:37 +
01.90
10.10
F
23:37 +
02.25
10.70
G
23:37 +
03.10
12.00
H
23:37 +
03.50
12.80
I
23:37 +
06.08
18.30
J
23:37 +
07.07
19.79
K
23:37 +
08.32
21.40
L
23:37 +
11.12
26.40
M
23:37 +
11.50
26.20
N
23:37 +
17.80
37.70
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(a)
Plot the arrival-time differences (ts – tp) against the arrival times of the P-wave to
(b)
(c)
produce a Wadati-diagram.
Determine the ratio α/β of the seismic velocities.
Determine the time of occurrence (t0) of the earthquake.
(a)
Wadati diagram:
⎛α
⎞
The equation of the Wadati plot is t s − t p = t p ⎜ − 1⎟
⎝β
⎠
6
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(b)
The best-fit straight line to the points on the Wadati diagram in this exercise is
y = 0.737x − 37.0 , where x = tp is the arrival time of the P-wave, and y = (ts – tp) is
the difference in travel times of the P- and S-waves. The coefficient of x gives
⎛α
⎞
⎜⎝ β − 1⎟⎠ = 0.737 from which the velocity ratio is
α
= 1.737
β
(c)
The intercept of the best-fit straight line on the y-axis is –37.0; using the slope of the
line gives the intercept on the x-axis.
0.737x = y + 37.0
x=
y
37.0
+
= 1.36y + 50.5
0.737 0.737
The earthquake occurred at 50.5 seconds after 22:36.
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SOLUTIONS TO EXERCISES
8 Geochronology
1.
How many half-lives must elapse before the activity of a radioactive isotope decreases to
1% of its initial value? How long is this time for 14C, which has a decay rate of 1.21 × 10-4
yr-1?
(1) Radioactive decay is described by the equation
A = A0e − λt
where A0 is the initial activity when the isotope was formed, A is the activity after time t,
and λ is the decay rate. The half-life t1/2 is the time at which A has decayed to one half of
its initial value:
0.5 A0 = A0 e − λ t
t1/ 2 =
ln(2)
λ
When the activity has decayed to 1% of its initial value, A = 0.01 A0. Inserting this in the
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decay equation,
0.01 A0 = A0 e− λt
t 0.01 = −
ln( 0.01) ln(100 )
=
λ
λ
t 0.01 ==
ln(100) ln( 2) ln(100 )
=
t1/ 2 = 6.644 t1 /2
ln(2) λ
ln(2)
Thus 6.64 half-lives must elapse for the activity to decrease to 1% of its initial value.
(2) In the case of 14C, with a decay rate λ = 1.21 × 10-4 yr-1
t1/2 =
ln(2)
0.69315
=
= 5728 yr.
λ
1.21× 10 −4
The time for the activity of 14C to decay to 1% of its initial value is
t 0.01 = 6.644t1/2 ≈ 38,100 yr.
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2.
Radiocarbon dating of a sample of wood from the tomb of an Egyptian pharaoh gave
concentrations of 9.83 × 10–15 mol/g for 14C and 1.202 × 10–2 mol/g for 12C. Assuming that
the initial 14C/12C ratio in the sample corresponded to the long-term atmospheric ratio of
1.20 × 10–12, determine the age of the tomb, the percentage of 14C remaining, and the
original 14C concentration in the wood.
(1) Let the isotopic carbon ratio at present be R = [14C/12C] and let the initial ratio when the
tomb was made be R0 = [14C/12C]0. The decay of radioactive carbon changes the isotopic
ratio with time according to the equation
R = R0 e− λt
Due to radioactive decay of 14C, its proportion has decreased relative to the stable 12C. The
isotopic ratio is now R = (9.83 × 10–15)/(1.202 × 10–2) = 8.178 × 10–13; the initial ratio is
assumed to have been R0 = 1.20 × 10–12.
Setting these values in the equation and using the decay rate λ = 1.21 × 10–4 yr-1 :
⎛ R⎞
ln ⎜ ⎟ = − λt
⎝ R0 ⎠
t=
1 ⎛ R0 ⎞ ln(1.20 × 10 −12 / 8.178 × 10 −13 )
ln ⎜ ⎟ =
TE1.21×
STBA10N−4KSELLER=.3169
COM yr
λ ⎝ R⎠
The age of the tomb is ≈ 3,170 yr.
− λt
(2) The factor e describes the proportion of the radioactive isotope remaining after time t.
After 3,170 yr the product (λt) = 0.38345. The fraction of 14C that remains after 3,170 yr is
therefore e–0.38345 = 0.6815; the percentage remaining is 68.2%.
(3) Applying the decay equation to the radioactive 14C component only, we can write
14
C=
( C)
14
0
e− λt =0.6815 ( 14 C )0
The present measured concentration of 14C in the wood is 9.83 × 10–15, so the initial
concentration was (9.83 × 10–15 /0.6815) = 1.44 × 10–14 mol/g.
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3.
The decay constants of 235U and 238U are λ235 = 9.8485 × 10–10 yr–1 and λ238 = 1.55125
× 10–10 yr–1. Calculate the half-lives of these uranium isotopes.
The half-life t1/2 of a radioactive isotope with decay constant λ is given by
t1/2 =
ln(2)
λ
For the uranium isotope 235U the decay constant is λ235 = 9.8485 × 10–10 yr–1 and the halflife is
235
t1/2
=
ln(2)
ln(2)
=
= 7.038 × 10 8 yr
λ
9.8485 × 10 −10
235
The half-life of 235U is t1/2
= 704 Myr
For the uranium isotope 238U the decay constant is λ238 = 1.55125 × 10–10 yr–1 and the halflife is
238
t1/2
=
ln(2)
ln(2)
=
= 4.468 × 10 9 yr
−10
λ
1.55125 × 10
238
The half-life of 238U is t1/2
= 4.47 Gyr
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4.
Assuming that the isotopes 235U and 238U were created in equal proportions in a common
event, such as a supernova, and given that their abundances are now in the ratio
235
U/238U = 1/137.88, calculate how long ago they were created.
The decay equation for the radioactive uranium isotope 235U is
235
U=
where
235
235
U 0 e− λ235 t ,
U and
235
U 0 are the present and initial concentrations and λ235 is the decay
constant, equal to 9.8485 × 10–10 yr–1.
The decay equation for 238U with decay constant λ238 = 1.55125 × 10–10 yr–1 is
238
U=
238
U 0 e− λ238 t .
Dividing the second equation by the first:
238
U
=
235
U
U 0 e− λ238 t
235
U 0 e− λ235 t
238
initialratio
= 137.88 = e( λ235 −λ238 )t
present ratio
ln(137.88) = ( λ235 − λ238 ) t
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ln (137.88 )
ln (137.88 )
t=
=
= 5.937 × 10 9 yr.
−10
λ
−
λ
9.8485
−
1.55125
×
10
)
( 235 238 ) (
The age of formation of the isotopes is 5.937 Gyr.
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5.
The analysis of strontium and rubidium isotopes in whole rock samples from a granitic
batholith gave the following atomic concentrations in p.p.m.:
87
Sample
87
Sr
Rb
86
Sr
A
2.304
8.831
2.751
B
0.518
29.046
0.420
C
1.619
111.03
1.232
D
1.244
100.60
0.871
(a) Calculate the 87Rb/86Sr and 87Sr/86Sr isotopic ratios for these samples.
(b) Determine the age of the batholith and the initial 87Sr/86Sr ratio.
(a)
The isotopic ratios for the four samples are obtained by simple division of their measured
concentrations:
Sample
A
(b)
87
Rb/86Sr
87
Sr/86Sr
0.8375
TESTBA3.830
NKSELLER.C
OM
B
56.1
1.233
C
68.6
1.314
D
80.9
1.428
Eqs. (4.12) and (4.14) describe the Rb–Sr dating system. A plot with the 87Rb/86Sr ratios on
the abscissa (x) and 87Sr/86Sr on the ordinate (y) axis gives a straight line as in the figure.
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The best-fit straight line in this plot – the Rb-Sr isochron – has the equation
y = 0.00757x + 0.807 .
The age of the batholith is found from the slope m = 0.00757 using Eq. (8.14):
t=
1
1010
ln(1+ m) =
ln(1+ 0.00757) = 531× 10 6 yr
λ
0.1420
Thus the age of the batholith is, to 3 significant digits, 531 Ma.
The initial ratio of 87Sr/86Sr is the intercept on the y-axis where x = 0. From the equation of
the best-fit line the initial ratio is equal to 0.807.
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6.
Argon-argon dating of muscovite in a Late Cretaceous granite gave the following isotope
ratios for the plateau stages during incremental heating:
Maximum Heating
Temperature (°C)
39
Ar/36Ar
40
Ar/36Ar
750
1852
8855
830
1790
8439
895
1439
6867
970
3214
15380
1030
2708
12970
(a) Calculate the 40Ar/39Ar ratios for each incremental heating step.
(b) A calibration constant J = 0.00964 was determined for the monitor mineral. Using the
40
Ar/39Ar ratios from (a) in Eq. (8.18), calculate the apparent ages at each heating step.
(c) Draw an 40Ar/39Ar isochron diagram by plotting each 40Ar/36Ar ratio as ordinate
against the corresponding 39Ar/36Ar ratio as abscissa. Draw a best-fitting line – the
isochron – through the data points, and determine its slope and intercept.
(d) Compute the age of the muscovite from the slope of the isochron. Is the intercept on
the ordinate axis significant?
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(a)
The 40Ar/39Ar ratios, found by simple division, are given in the following table:
Heating
Temperature
(°C)
(b)
39
36
Ar/ Ar
40
36
Ar/ Ar
40
39
Ar/ Ar
Heating
Step Age
(Ma)
750
1852
8855
4.781
81.3
830
1790
8439
4.715
80.2
895
1439
6867
4.772
81.1
970
3214
15380
4.785
81.4
1030
2708
12970
4.790
81.4
Eq. (8.18) describes how the 40Ar/39Ar ratio is used to make an age determination. The age
t is given by
⎛
t = 1.804 × 10 9 ln ⎜ 1 + J
⎝
40
39
Ar ⎞
Ar ⎟⎠
The factor J in this equation is the empirical calibration factor of the equipment,
determined for a control sample of known age. J includes, among other terms, the 39Ar
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production yield from neutron irradiation of 39K; in the present exercise J = 0.00964. This
gives for the individual heating steps in the experiment the age estimates in the right
column of the above table. Their average age is 81.1 Ma and the standard deviation of the
ages is 0.5 Myr.
(c)
The 40Ar–39Ar plot is a straight line, the isochron.
(d)
The equation of the best-fitting straight line is
y = 4.820x − 105
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The slope of the line gives the optimum 40Ar/39Ar ratio for all the data, which can be
inserted in Eq. (8.18) to get the age:
t = 1.804 × 10 9 ln (1 + 0.00964 × 4.820 ) = 8.19 × 10 7 yr
The age of the muscovite by this method is 81.9 Ma.
The non-zero intercept suggests that the measured 40Ar/36Ar ratios are systematically
low by an amount equal to 105 p.p.m. This causes the estimated age at each temperature to
be low. Consequently, the mean of the estimated ages found in part (b), 81.1 Ma, is 0.8
Myr lower than the best-fit result of 81.9 Ma.
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7.
The following isotopic ratios were measured in U-Pb age determinations on three zircon
grains extracted from a granite:
Sample
207
Pb/235U
206
Pb/238U
zircon 1
27.4
0.600
zircon 2
33.3
0.680
zircon 3
37.9
0.740
(a) With the aid of Eqs. (8.19) and (8.20), plot a concordia diagram on graph paper or with
a plotting routine. Enter the measurements from the above table on the graph, and draw the
straight discordia line through the points.
(b) Determine the coordinates of the intersection points of the concordia and discordia
lines.
(c) Using the coordinates of the upper intersection point together with Eq. (8.19) and Eq.
(8.20), calculate the age of formation of the zircons.
(d) Calculate when loss of lead occurred in the zircons.
(a)
206
Pb/238U and 207Pb/235U isotopic ratios calculated from the equations are listed below.
The curved “concordia” line
the
plot; this curve may be plotted
TEis
STshown
BANKonSE
LLfollowing
ER.COM
by hand on graph paper, or the values may be entered in a spreadsheet-style computer
program (e.g., MS Excel), from which a plot of the data can be obtained.
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The x-coordinate is the 207Pb/235U ratio, which at time t is given by
x=
207
Pb
= eλ235 t − 1
U
235
(1)
where λ235 = 9.8485 × 10–10 yr–1 is the decay constant of 235U.
The y-coordinate is the 206Pb/238U ratio at time t, given by
y=
206
Pb
= eλ238 t − 1
U
238
(2)
where λ238 = 1.5513 × 10–10 yr–1 is the decay constant of 238U.
The numbers in italics on the concordia curve are the corresponding ages in Gyr (109 yr).
The data points for the three zircons given in the table are shown as small squares. A bestfit straight line through the three points has equation
y = 0.2348 + 0.01334x
(b)
(3)
The straight line is the discordia line; it intersects the concordia curve at two points. The
coordinates of the intersection points cannot be found analytically but must be obtained
numerically. There are mathematical methods for doing so, but they are beyond the scope
of this book. As a first attempt, however, a graphical solution may be attempted. This gives
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an approximate solution for the coordinates of the upper intersection point as (x = 44, y =
0.8), and the lower intersection point as (x = 4, y = 0.3).
The intersection points must satisfy the discordia line and the concordia curve
simultaneously. Starting from the “rough” values obtained graphically, an iterative method
may be used to find more accurate values that satisfy both equations.
Iterative solution:
Starting with a value x = 43.0 slightly below the expected intersection value, the
corresponding y-value is calculated for the discordia line, Eq. (3); it equals 0.8086.
Inserting the same x-value in Eq. (1) we solve for the time t:
1
1010
t=
ln(1 + x) =
ln(1 + 43) = 3.842 × 10 9 yr.
λ235
9.8485
This value is inserted in Eq. (2) to obtain the y-value of the concordia line at x = 43.
y = eλ238 t − 1 = e(0.15513× 3.8424 ) − 1 = 0.8149 .
The value of x is now incremented by a small amount (0.5) and the procedure is
repeated; further repetition gives a table of results like the following:
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TESTBANKSELLER.COM
x-value
43.0
43.5
44.0
44.5
43.95
discordia y
0.8086
0.8152
0.8219
0.8286
0.8212
concordia t
3.842
3.854
3.865
3.876
3.864
concordia y
0.8149
0.8182
0.8214
0.8245
0.8211
y-difference
0.0064
0.0029
–0.0005
–0.0040
0.0002
The differences in y-values show that the discordia line crosses the concordia curve at
an x-value slightly less than 44.0. Thus, a better starting value for the iteration would be
around x = 43.9. The procedure is now repeated with smaller increments of x to determine
the intersection point more exactly.
Using this iterative approach, the coordinates of the upper intersection of the
discordia and concordia lines are found to be: x = 43.95; y = 0.821.
The same iterative method can be applied to the lower intersection. Its coordinates
are: x = 3.986; y = 0.288.
(c)
The age of formation of the zircons is given by the upper intersection point of the
discordia and concordia lines. As evident from the table for the iterative solution, their
age of formation is 3.86 Ga.
(d)
The departure of the lower intersection point from zero implies loss of lead in an event
that occurred at the time corresponding to the lower intersection point. The xTESTBANKSELLER.COM
coordinate of the point, found by iteration, is x = 3.986. Inserting this value in
Eq. (8.20) gives the age of the event:
t=
1
1010
ln(1+ x) =
ln(1+ 3.986) = 1.631× 10 9 yr.
λ235
9.8485
Rounding the answer to fewer significant digits gives the age as 1.63 Ga.
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Calculation of points on the Concordia curve
The Pb/U isotopic ratios listed below define the concordia diagram for the last 5 Ga. They
were computed with λ238 = 1.55125 × 10-10 yr–1 and λ235 = 9.8485 × 10-10 yr–1 as decay constants
in the following formulae:
206
207
Pb
Pb
λ238 t
=
e
−
1
= eλ235 t − 1
238
235
U
U
Age (Ga)
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
206
Pb/238U
0.0156
0.0315
0.0476
0.0640
0.0806
0.0975
0.1147
0.1321
0.1498
0.1678
0.1861
0.2046
0.2234
0.2426
0.2620
0.2817
0.3018
0.3221
0.3428
0.3638
0.3851
0.4067
0.4287
0.4511
0.4738
207
Pb/235U
Age (Ga)
0.1035
2.6
0.2177
2.7
0.3437
2.8
0.4828
2.9
0.6363
3.0
0.8056
3.1
0.9925
3.2
1.1987
3.3
1.4263
3.4
1.6774
3.5
1.9545
3.6
TE2.2603
STBANKSELLE3.7
R.COM
2.5977
3.8
2.9701
3.9
3.3810
4.0
3.8344
4.1
4.3348
4.2
4.8869
4.3
5.4962
4.4
6.1685
4.5
6.9105
4.6
7.7292
4.7
8.6326
4.8
9.6296
4.9
10.7297
5.0
206
Pb/238U
0.4968
0.5202
0.5440
0.5681
0.5926
0.6175
0.6428
0.6685
0.6946
0.7211
0.7480
0.7753
0.8030
0.8312
0.8599
0.8889
0.9185
0.9485
0.9789
1.0099
1.0413
1.0732
1.1056
1.1385
1.1719
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207
Pb/235U
11.9437
13.2834
14.7617
16.3930
18.1931
20.1795
22.3716
24.7905
27.4597
30.4052
33.6556
37.2424
41.2004
45.5681
50.3878
55.7063
61.5752
68.0517
75.1984
83.0847
91.7873
101.3906
111.9878
123.6818
136.5861
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8. The following isotopic ratios were measured in a K-Ar age determination on an
ignimbrite as part of a combined radiometric-paleomagnetic study of geomagnetic
polarity.
40
Sample
K/36Ar
40
Ar/36Ar
A
4,716,000
822
B
8,069,000
1200
C
12,970,000
1730
D
27,670,000
3280
(a) Plot the isotope ratios, draw the isochron, and compute its slope and intercept.
(b) Calculate the isochron age of the ignimbrite.
(c) Correct the observed 40Ar/36Ar ratios for the initial 40Ar/36Ar concentration, and
compute the individual sample ages.
(d) Calculate the mean age and its standard deviation. Compare the mean age with the
isochron age.
(e) With reference to the radiometric time scale in Fig. 12.29, what magnetic polarity
would you expect the ignimbrite samples to have?
TESTBANKSELLER.COM
(a)
The values listed in the above table give the following isochron plot:
The best-fit straight line has equation
y = 1.0673 × 10 − 4 x + 332.5
1
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(b)
The isochron age is obtained from the slope m of the straight line, using Eq. (8.17):
⎛
t = 1.804 × 10 ln ⎜ 1 + 9.54
⎝
9
Ar ⎞
40
K ⎟⎠
40
In this equation 40Ar refers to the pure in-situ generated component. Both the ordinate
and abscissa in the plot contain the same 36Ar so the ratio 40Ar/40K needed for the age
determination is equal to the slope of the isochron. The age is then
t = 1.804 × 10 9 ln (1 + 9.54 × 1.0673 × 10 − 4 )= 1.83 Ma
(c)
The non-zero intercept on the isochron plot gives an initial 40Ar/36Ar concentration of
332.5. Subtracting this from each of the measurements of 40Ar/36Ar gives the following
table with corrected 40Ar/36Ar ratios:
Ar/40K
( × 10–4)
Age
(Ma)
489.5
1.038
1.785
8,069,000
867.5
1.075
1.849
C
12,970,000
1397.5
1.077
1.853
D
27,670,000
2947.5
1.065
1.832
Sample
observed
40
K/36Ar
corrected
40
Ar/36Ar
A
4,716,000
B
40
TESTBANKSELLER.COM
(d)
The mean of the individual ages is 1.83 Ma and its standard deviation is 0.03 Ma. This
agrees well with the age determined from the isochron.
(e)
The magnetic polarity time scale in Fig. 12.32 consists of a composite scale determined
from radiometric measurements and a scale determined from the spacing of marine
magnetic anomalies. By comparison Fig. 12.29 is based only on radiometric data. The
age 1.83 Ma places the ignimbrite within the Olduvai normal polarity event.
2
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SOLUTIONS TO EXERCISES
9 The Earth’s Heat
1.
List and compare the various factors that may influence the measured temperature
gradient at a depth of 5 m in (a) a deep drillhole in oceanic sediments and (b) a
continental well that encounters the groundwater table at 2 m depth.
(a)
Oceanic sediments are shielded from temperature changes at the ocean surface by the
great depth of water. The effects of seasonal and climatic temperature changes can be
neglected. The ocean bottom is generally flat except near to an oceanic ridge, and so
only in this vicinity are corrections for bottom topography needed. In the undisturbed
sediment column there is little or no circulation of sea water, so thermal conditions are
stable. When oceanic heat flow is measured with a deep-sea corer as in Fig. 4.26, the
temperature is measured with thermistors mounted a few centimeters from the barrel of
the corer. The temperature measurements are made as soon as possible after penetration
of the sediment, before thermal disturbance from the coring can reach the sensors.
When a drillhole is drilled, the thermal stability in the sediments is disturbed by the heat
generated from the friction
ofSthe
TE
TBdrilling
ANKSEprocess
LLERand
.COthis
M must be compensated, for
example by monitoring temperature over a length of time.
(b)
Annual fluctuations of surface temperature have a penetration depth of around 4 m in
continental rocks. At a depth of 5 m in a continental well, these fluctuations can be
ignored. However, longer term temperature changes, such as those associated with
paleoclimatic variation, penetrate much deeper and must be corrected. At 5 m a sensor
is probably in the groundwater, which may be in motion; this can cause time-dependent
(e.g., seasonal) fluctuations in temperature at that depth. If the surface topography
around the well is significant, a correction must be made for its effect on the shallow
subsurface temperature profile. The influences of vegetation, hydrology, latitude and
sun angle, as well as seasonal snow cover must be taken into account where necessary.
Finally, as mentioned in (a) above, if the well was drilled recently, residual heat from
the drilling process may influence the temperature profile.
1
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2.
A shallow circular pond 100 m in diameter freezes solid during a very cold night. The
pond is in a geothermal area in which the temperature at 200 m depth is 40°C higher
than at the surface. The thermal conductivity of the intervening rock is 3.75 W m–1 K–1
and the latent heat of fusion of ice is 334 kJ kg–1. Neglecting other heat sources,
calculate the mass of ice that melts per hour due to the geothermal gradient.
The temperature gradient beneath the pond is (40/200) = 0.2 °C m–1, which is
equivalent to 200 °C km–1. Combining this gradient with the thermal conductivity
k = 3.75 W m–1 K–1 gives the local heat flow q:
q=k
dT
= 3.75 × 0.2 = 0.75 W m–2
dz
The surface area A of the circular pond with diameter d = 100 m is
A=
π 2 π
2
d = (100 ) = 7.854 × 10 3 m2
4
4
The thermal energy Q flowing into this area per hour (3600 s) is
Q = qAt = 0.75 × 7.854 × 10 3 × 3600 = 21, 206 kJ
This energy is used to melt an unknown mass m of ice. The latent heat of fusion L of
STBAof
NKheat
SEenergy
LLER.needed
COM to melt 1 kg of the material.
a material is defined as theTE
amount
For ice, L = 334 kJ kg–1. The mass of ice melted by the thermal energy entering the
pond per hour is
m=
Q 21, 206
=
= 63.5 kg hr–1
L
334
2
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3.
Assuming a constant geothermal gradient of 30 °C per kilometer, estimate what
percentage of the Earth’s volume is hotter than the temperature of molten lava at
atmospheric pressure. Why is the deeper interior of the Earth not entirely molten?
At atmospheric pressure molten lava has a temperature around 1200 °C when it
flows and a temperature around 800 °C when it solidifies. If the geothermal gradient is a
constant 30 °C per km, this temperature is reached in a depth of 40 km. The radius of
the Earth’s interior that is hotter than 1200 °C is therefore (6371 – 40) = 6331 km. The
volume of this spherical region is proportional to the cube of its radius, and so it can be
expressed as a fraction of the Earth’s total volume by the ratio
3
⎛ 6331 ⎞
⎜⎝
⎟ = 0.981
6371 ⎠
More than 98% of Earth’s volume is hotter than molten lava at atmospheric pressure.
Despite high temperatures the Earth’s mantle is not molten because the increase of
pressure with increasing depth causes the melting point to increase faster than the
increase of temperature. Throughout the crust and mantle the temperature is below the
local melting point and so the material remains solid.
TESTthere
BANK
LLERof
.Ccomposition
OM
At the core-mantle boundary
isSaEchange
from silicate mantle to
iron core. The temperature of the outer core is above the local melting point of iron and,
as a result, the outer core is molten. With increasing depth in the core the melting point
(solidus) curve rises more steeply than the actual temperature and eventually crosses it.
This gives rise to an inner iron core that is solid.
3
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4.
The mean global heat flow at the Earth’s surface is 87 mW m–2. Calculate the time in
years needed for the mantle and core to cool by 100 °C, with the following assumptions:
(i) the Earth’s mantle and core cool as a homogeneous unit, (ii) 20% of the observed
heat flow at the Earth’s surface is from the mantle, (iii) the lithospheric thickness is 100
km, (iv) thermal effects from the lithosphere itself may be ignored. Relevant properties
of the mantle and core are: mean density 5650 kg m–3, specific heat 400 J kg–1 °K–1.
The total amount of heat flowing out of the Earth per second (dQ/dt) is equal to the
mean heat flow per square meter (q), multiplied by Earth’s surface area (A):
dQ
= qA
dt
The mean global heat flow is 87 mW m–2.
(
)
2
= 5.101 × 1014 m2.
Earth’s surface area:
A = 4π R 2 = 4π 6.371 × 10 6
Total heat loss per second:
dQ
= 87 × 10 −3 × 5.101 × 1014 = 4.438 × 1013 J s–1
dt
In the exercise, 20% of this heat comes from below the lithosphere and thermal effects
of the lithosphere may be neglected. The mantle heat loss is thus 8.875 × 1012 J s–1.
The mantle and core cool T
asEaSunit,
TBAwith
NKSradius
ELLEequal
R.Cto
OM(6371–100) = 6271 km.
The volume of this cooling sphere is 43 π R 3 = 1.033 × 1021 m3.
Assuming the given mean density of 5650 kg m–3, the mass of mantle and core is
5.836 × 1024 kg.
The specific heat is 400 J kg–1 °C–1. For an object of mass m with specific heat c that
cools through a temperature difference ∆T, the heat loss ∆Q is given by
∆Q = cm ∆T
For a temperature drop of 100 °C, the Earth’s heat loss is
∆Q = 400 × 5.836 × 100 × 10 24 = 2.335 × 10 29 J
If the heat is lost at a constant rate, the length of time required for this temperature
change is
t=
∆Q
2.335 × 10 29
=
= 2.630 × 1016 s = 834 Myr s.
12
(dQ / dt) 8.875 × 10
The time for mantle and core to cool by 100°C is 834 Ma.
4
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5.
A temperature gradient of 35 °C km–1 is measured in the upper few meters of sediments
covering the ocean floor. If the mean thermal conductivity of oceanic sediments is 1.7
W m–1 °C–1, calculate the local heat flow. How far do you think the sampling site is
from the nearest active ridge?
By definition, heat flow = thermal conductivity × temperature gradient
q=k
dT
dz
The temperature gradient is given as 35 °C km–1 = 0.035 °C m–1. Combining this values
with the thermal conductivity k – 1.7 W m–1 gives the local heat flow:
q=k
dT
= 1.7 × 0.035 = 0.0595 W m–2,
dz
i.e., the heat flow is 59.5 mW m–2.
The relationship between heat flow (in mW m–2) and distance from a spreading ridge
depends on the age t of the ocean crust. Two relationships between heat flow and age
are given in Eq. (4.62). Assuming the simplest, which is applicable to ages younger than
55 Ma:
2
TESTBANK⎛ S
EL⎞LER.COM
510
510
q=
mW m–2, from which t = ⎜
Ma. The heat flow of 59.5 mW m–2
⎟
⎝ q ⎠
t
2
⎛ 510 ⎞
⎛ 510 ⎞
corresponds to a crustal age of t = ⎜
=⎜
= 73.5 Ma. This is older than the
⎟
⎝ 59.5 ⎟⎠
⎝ q ⎠
2
range for which the assumed age-heat flow equation is valid.
The alternative equation for ages older than 55 Ma is
q = 48 + 96e−0.0278t mW m–2.
Inserting the calculated heat flow in this equation gives an age
t=
⎛ 96 ⎞
1
1
96 ⎞
⎛
ln ⎜
=
ln ⎜
⎟ = 76.3 Ma.
⎟
0.0278 ⎝ q − 48 ⎠ 0.0278 ⎝ 59.5 − 48 ⎠
The distance of a location with this age from a ridge axis depends on the spreading rate.
Assuming a low half-spreading rate of 1 cm yr–1 (as in the North Atlantic Ocean) the
distance from the ridge is about 760 km. With a fast spreading rate of 6 cm yr–1 (as on
the Pacific-Nazca ridge), the distance from the ridge could be around 4500 km.
5
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6.
What heat flow values would you expect at the locations of the oceanic magnetic
anomalies with numbers C5N, C10N, C21N, C32N, M0? Interpret the ages of the
anomalies from Fig. 12.32 and use the heat-flow model GDH1 (Eq. 9.40) for the
cooling of oceanic lithosphere.
From the polarity time scale in Fig. 12.32 the ages of anomalies C5N, C10N and C21N
are less than 55 Ma. To compute the oceanic heat flow at the locations of these
anomalies the first part of Eq. (4.62) is used:
q=
510
mW m–2.
t
To compute the heat flow at the anomalies older than 55 Ma, the alternative equation
must be used
q = 48 + 96e−0.0278t mW m–2.
Combining the results, the following table of heat flow values is obtained:
Anomaly
number
Age from Fig. 5.78
(Ma)
Heat flow
(mW m–2)
10.5
157
28.5
96
C21N
47
74
C32N
72.5
61
M0R
121
51
C5N
C10N
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7.
Using the relationships in Eq. (9.41), estimate the approximate depths of the ocean at
these locations?
Eq. (9.41) gives two relationships between the ocean depth d (in m) and age t (in Ma) of
the ocean crust:
for t < 20 Ma
d = 2600 + 365 t
d = 5651 − 2473exp ( −0.0278t ) for t ≥ 20 Ma
Of the anomalies in this exercise, only anomaly C5N is younger than 20 Ma. The
first equation is used for C5N, thes second eaquation for the remaining anomalies. The
Calculations give the following age-depth relationships:
Anomaly
number
Age from
Fig. 5.78
(Ma)
Ocean
Depth
(m)
C5N
10.5
3783
C10N
28.5
4531
C21N
47
4981
C32N
72.5
5321
M0
5565
TES121
TBANKSELLE
R.COM
7
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8.
Assuming that the Earth initially had a uniform temperature throughout and has been
cooling by conduction only, use the solution for the 1-dimensional cooling of a semiinfinite half-space (Eq. 9.34 and Box 9.2) to derive Eq. (8.2) for Kelvin’s estimated age
of the Earth.
The 1-dimensional model of cooling represented by Eq. (9.34) gives the temperature
T at depth z and time t after a half-space starts to cool as:
T = Tm erf (η ) , in which erf(η) is the error function (Box 9.3), with argument
η=
z
. Here κ is the thermal diffusivity of the half-space, given by
2 κt
κ=
k
, where k is the thermal conductivity, c is the specific heat at constant
cρ
pressure, and ρ is the density.
In order to compute the heat flow we need to know the temperature gradient.
dT dT dη
=
dz dη dz
Evaluating separately the differentiations on the right side of the equation:
η
dT
d
2TTEdSTB−u
A2NKSE2T
LLER2 .COM
= T0
erf(η ) = 0
e du = 0 e−η , where T0 is the initial temperature;
∫
dη
dη
π dη 0
π
dη d ⎛ z ⎞
1
= ⎜
=
⎟
dz dz ⎝ 2 κ t ⎠ 2 κ t
Combining these expressions gives the temperature gradient at depth z and time t:
2
dT
T0 −η2
T0 − 4zκ t
=
e =
e
dz
πκ t
πκ t
The surface temperature gradient at z = 0 at time t after the start of cooling is:
T0
ρc T0
⎛ dT ⎞
=
⎜⎝
⎟⎠ =
dz z = 0
πk t
πκ t
This is Eq. (8.2) from which Kelvin estimated the age of the Earth.
Kelvin assumed an initial temperature of 7000 °F (4144 K) and a surface temperature
gradient of 0.036 K m–1. Combining these with representative values for surface rocks
(e.g, k = 3.0 W m K–1, ρ = 2700 kg m–3, cp = 800 J kg–1 K–1), we get a cooling time
2
2
⎞
ρc ⎛
T0
2700 × 840 ⎛ 4144 ⎞
15
t=
=
⎜⎝
⎟⎠ = 3.19 × 10 s , which equals 101 Ma.
⎜
⎟
π k ⎝ ( dT dz )0 ⎠
3.0 × π
0.036
8
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9.
The temperature in the near-surface layers of the Earth’s crust varies cyclically with
daily, annual and longer periods. For a surface temperature variation given by
T = T0 cosωt, the temperature variation at depth z and time t is described by:
z⎞
⎛ z⎞
⎛
T (z,t) = T0 exp ⎜ − ⎟ ⋅ cos ⎜ ω t − ⎟
⎝ d⎠
⎝
d⎠
d=
2κ
;
ω
κ=
k
;
ρc p
ω=
2π
τ
where τ is the period of the variation, k is the thermal conductivity, cp is the specific
heat, and ρ is the density. For surface sediments assume k = 2.5 W m K–1, cp = 1000
J kg–1 K–1, and ρ = 2300 kg m–3.
(a) Calculate the phase difference (in days) between the temperature variation at the
surface and at depths of 2 m and 5 m, respectively. Perform the calculations for both the
daily and annual temperature fluctuations.
(b) Assuming that the range in surface temperatures between summer and winter is
40 °C, calculate the depth at which the annual temperature range is 5 °C. How large (in
weeks and days) is the phase difference between the surface temperature and the actual
temperature at this depth?
(a)
The expression for temperature at depth z and time t can be written
TESTBANKSELLER.COM
z⎞
⎛ z⎞
⎛
T (z,t) = T0 exp ⎜ − ⎟ ⋅ cos ⎜ ω t − ⎟ = T1 ⋅ T2 , where
⎝ d⎠
⎝
d⎠
⎛ z⎞
T1 = T0 exp ⎜ − ⎟ describes how the amplitude of a temperature change at the
⎝ d⎠
surface decreases with depth, and
z⎞
z ⎞
⎛
⎛
T2 = cos ⎜ ω t − ⎟ = cos ω ⎜ t −
= cos ω ( t − t 0 ) describes how temperature varies
⎝
⎝ ω d ⎟⎠
d⎠
with time at a given depth z. The function (t – t0) is the phase of this variation and t0 is
the phase difference. t0 is the amount of time by which the fluctuation at depth z is
delayed with respect to the surface variation; it is sometimes referred to as the time-lag.
t0 =
z
τ z
, where τ is the period of the fluctuation.
=
ω d 2π d
Note: If the period is expressed in days, the phase difference will also be in days.
Using the given values for surface sediments:
κ=
k
2.5
=
= 1.087 × 10 −6 m2 s–1
3
ρc p 2300 × 10
9
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The “penetration depth” is d =
2κ
κτ
=
. In order to compute this depth the period
ω
π
of the signal must be expressed in seconds.
Daily temperature variation
For the daily temperature variation t = 86,400 s. This gives a penetration depth of
d=
κτ
86400 × 1.087 × 10 −6
=
= 0.173 m.
π
π
Inserting this in the expression for the phase difference, and expressing the times in
days, we find the phase difference of the daily temperature variation at depth 2 m to be
t0 =
τ z
1 2
=
= 1.8 days
2π d 2π 0.173
Similarly, the phase difference of the daily temperature variation at depth 5 m is
t0 =
1 5
= 4.6 days
2π 0.173
Annual temperature variation
The period of the annual variation is 31,536,000 s and the “penetration depth” is
d=
κτ
=
π
3.1536 × 10 7 × 1.087 × 10 −6
= 3.3 m.
TES
π TBANKSELLER.COM
The phase difference of the annual variation at a depth of 2 m is
t0 =
τ z 365 2
=
= 35 days, or 5 weeks, and
2π d 2π 3.3
the phase difference of the annual variation at depth 5 m is
t0 =
(b)
τ z 365 5
=
= 88 days, or 12 weeks and 4 days.
2π d 2π 3.3
To find the depth at which an annual surface temperature range of 40 °C is reduced to
5°C, we use the expression T1 above, and the annual penetration depth of 3.3 m for d:
⎛ z⎞
T1 = T0 exp ⎜ − ⎟
⎝ d⎠
⎛T ⎞
40
z = d × ln ⎜ 0 ⎟ = 3.3 × ln
= 6.9 m
5
⎝ T1 ⎠
At this depth the phase delay of the annual variation is
t0 =
τ z 365 6.9
=
= 121 days = 17 weeks 2 days
2π d 2π 3.3
10
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10.
The daily average temperature in northern Canada is +10 °C in July and -20 °C in
January. Using the heat conduction equation calculate the thickness of the permafrost
layer (below which the ground is permanently frozen). Relevant physical properties of
the ground are: thermal conductivity k = 3.0 W m K–1, specific heat cp = 840 J kg–1 K–1,
density ρ = 2700 kg m–3.
Taking the average of the January and July temperatures as the mean value for a
year, the mean is found to be –5 °C and the temperature range is ±15 °C about this
value. The permafrost level is the depth below which the temperature is always less than
0 °C so that the ground remains frozen. For this problem we can ignore the geothermal
gradient (about 30 °C km–1) as its effect would not be significant.
The penetration depth for the annual temperature change at the site can be computed
with the given parameters and the equation
d=
κτ
kτ
3.0 × 3.1536 × 10 7
=
=
= 3.644 m
π
πρ c
π × 2700 × 840
The amplitude of the temperature variation at the surface is 15 °C; the mid-value of
the temperature fluctuation is –5 °C. We need to find the depth z at which the amplitude
of the temperature changeTisE5S°C;
TBAthe
NKsummer
SELLEmaximum
R.COM of +5 °C will then equal the
mean temperature of –5 °C and permafrost will (just) melt. Using the surface amplitude
T0 = 15 °C, desired amplitude T1 = 5 °C, penetration depth of 3.64 m, and the equation
from the previous exercise
⎛T ⎞
15
z = d × ln ⎜ 0 ⎟ = 3.644 × ln = 4.0 m.
5
⎝ T1 ⎠
This is the permafrost depth at the particular location with the given ground
parameters; at greater depths the ground will remain frozen. Note that the permafrost
depth depends on the local thermal parameters and will vary with the type of ground
(e.g., soil, sediment, hard rock, etc.)
11
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11.
The half-spreading rate at an oceanic ridge in the middle of a symmetric ocean basin
bounded by subduction zones is 44 mm yr–1. The ridge is 1,000 km long and the
distance from the ridge to each subduction zone is 2,000 km. If the oceanic heat flow
varies with crustal age as in Eq. (9.40), calculate how much heat is lost per year from
the ocean basin.
The ridge is in the middle of the symmetric ocean basin, so the total heat lost is
double that lost from each half of the basin. At a half-spreading rate of 44 mm yr–1, the
age tSZ of oceanic crust at a subduction zone 2,000 km away is
t SZ =
2000 × 10 3
= 45.45 Ma.
0.044
The heat flow q from oceanic crust younger than 55 Ma after a cooling interval t (in
Ma) is given by the first part of Eq. (9.40):
q=
510
0.51
(mWm –2 ) =
(J s –1m –2 )
t
t
The heat flow at the subduction zone is qSZ =
0.51
=
t SZ
0.51
= 7.565 × 10 −2 J s–1 m–2
45.45
Consider the heat loss through a narrow strip of crust parallel to the ridge axis at a
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distance from the ridge corresponding to age t. The area of the strip is dA =(Lv)dt,
where L is the length of the ridge and v the spreading rate. The heat loss dQ is
dQ = qdA = q ( Lv ) dt = ( Lv )
0.51
dt
t
The total amount of heat lost from one half of the basin is found by integrating this
expression with respect to time from the ridge axis to the subduction zone:
Q=
0.51
∫ ( Lv ) t dt = ⎡⎣ 2 ( Lv ) 0.51
t SZ
0
t ⎤⎦
t SZ
0
⎛ 0.51 ⎞
Q = 2 ( Lv ) 0.51 t SZ = 2 ( Lv ) ⎜
⎟ t SZ = 2 ( Lv ) qSZ t SZ
⎝ t SZ ⎠
Here the units of (Lv) are m2 yr–1, qSZ is in W m–2, and the age tSZ is in Ma (=106 yr).
Correcting units, the heat loss per second from each half of the basin is
(
)
(
)
Q = 2 1000 × 10 3 × 44 × 10 −3 ( 0.07565 ) 45.45 × 10 6 = 3.03 × 1011 J
To find the annual heat loss from the ocean basin, this result must be doubled and
multiplied by the number of seconds in a year (3.1536 × 107), which gives
Q = 3.03 × 1011 × 2 × 3.1536 × 10 7 = 1.91× 1019 J yr–1
12
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SOLUTIONS TO EXERCISES
10 Geoelectricity
1.
At the interface between two layers with electrical resistivities ρ1 and ρ2, as in the figure
below, the electrical boundary conditions are: (i) the component of current density Jz
normal to the interface is continuous, and (ii) the component of electric field Ex
tangential to the interface is continuous. A current flow-line makes angles θ1 and θ2
before and after refraction, respectively.
z
J1z = J2z
E1x = E2x
J1
θ1
ρ1
ρ2
θ2
x
J2
Derive the electrical “law of refraction” given by Eq. (4.102):
TESTBANKSELLER.COM
tan θ1 ρ2
=
tan θ 2 ρ1
Ohm’s Law relating current density J, electric field E, and resistivity ρ (Eq. 10.10) is
E = ρJ .
The normal component of current density above the interface is J1z = J1 cosθ1 ; below
the interface it is J 2 z = J 2 cosθ 2 . The tangential component of electric field above the
interface is E1x = E1 sin θ1 ; below the interface it is E2 x = E2 sin θ 2 .
Equating the normal components of current density we get J1 cosθ1 = J 2 cosθ 2
Equating the tangential components of electric field we get E1 sin θ1 = E2 sin θ 2 ,
which can be rewritten using Ohm’s law as
ρ1 J1 sin θ1 = ρ2 J 2 sin θ 2 . Thus,
⎛ J cosθ 2 ⎞
ρ1 ⎜ 2
sin θ1 = ρ2 J 2 sin θ 2
⎝ cosθ1 ⎟⎠
⎛ sin θ1 ⎞
⎛ sin θ 2 ⎞
ρ1 ⎜
= ρ2 ⎜
⎟
⎝ cosθ1 ⎠
⎝ cosθ 2 ⎟⎠
tan θ1 ρ2
=
tan θ 2 ρ1
1
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2.
What is the effective resistivity of a slab of thickness L composed of two half-slabs each
of thickness L/2 and with resistivities (2ρ) and (ρ/2), respectively, as in the diagram?
0
L/2
2ρ
L
ρ/2
The resistance R of a conductor of length l and cross-sectional area A, made of
material with resistivity ρ, is given by
R=ρ
l
A
Consider the flow of electrical current along a path of arbitrary cross-sectional area A
normal to the interfaces of the given slab. The resistance of the left half of the slab is
RL = ( 2 ρ )
(L / 2) ρ L
=
A
A TESTBANKSELLER.COM
The resistance of the right half of the slab is
RR = ( ρ / 2 )
(L / 2) ρ L
=
A
4A
The total resistance of the slab is the sum of these two parts: R = RL + RR
R=
ρL ρL 5 L
+
= ρ
A 4A 4 A
The effective resistivity ρeff is defined by
R = ρeff
L
A
Thus, the effective resistivity of the slab is ρeff =
5
ρ
4
2
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3.
Seawater is contaminating an aquifer that is the source of drinking water for a seaside
town. The following measurements of apparent resistivity (ρa) were made at various
electrode separations (a) with the expanding-spread Wenner method to investigate the
leak.
(m)
ρa
(Ωm)
(m)
ρa
(Ωm)
(m)
ρa
(Ωm)
10
29.0
140
19.8
280
8.7
20
28.9
160
18.0
300
7.8
40
28.5
180
16.3
320
7.1
60
27.1
200
14.5
340
6.7
80
25.3
220
12.9
360
6.5
100
23.5
240
11.3
400
6.4
120
21.7
260
9.9
440
6.4
a
a
a
(a) Estimate the electrical resistivity of each layer.
(b) Divide the apparent resistivity at each position by the resistivity of the upper layer,
then plot the normalized resistivity against electrode separation on a log-log diagram on
the same scale as the model curves in Fig. 10.15.
(c) Match the measured curve
TESwith
TBAthe
NKmodel
SELLcurves
ER.Cand
OM estimate the depth to the
interface.
(a)
The apparent resistivity varies with electrode separation as follows:
The curve is characteristic of a two-layer structure (see Section 10.5.6), in which a good
conductor (the aquifer) lies beneath a more resistive near-surface layer.
3
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At small electrode separations the current lines are confined to shallow depths and
the apparent resistivity is that of the near-surface layer, which is approximately 29 Ω m.
At large electrode separations the current-lines penetrate more deeply, so that most of
the current flows in the conducting aquifer. The apparent resistivity becomes asymptotic
to the resistivity of the aquifer, which, from the values in the table, is 6.4 Ω m.
(b)
Normalizing the apparent resistivities ρa by the resistivity ρm of the top layer gives:
a
ρa/ ρm
(m)
ρa/ ρm
a
(m)
a
ρa/ ρm
(m)
10
1.000
140
0.683
280
0.300
20
0.997
160
0.621
300
0.269
40
0.983
180
0.562
320
0.245
60
0.934
200
0.500
340
0.231
80
0.872
220
0.445
360
0.224
100
0.810
240
0.390
400
0.221
120
0.748
260
0.341
440
0.221
A double-logarithmic plot of these values is shown in the following figure:
TESTBANKSELLER.COM
(c)
Using the estimated resistivities of the upper and bottom layers, the normalized k-factor
used in the interpretation of Wenner type-curves (Fig. 10.15) has the value
k=
ρ2 − ρ1 6.4 − 29
=
= −0.64
ρ2 + ρ1 6.4 + 29
4
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Further interpretation requires use of the type-curves. These are plotted on doublelogarithmic scales. In the figure below the measured data are shown (in blue) against the
relevant part of the type-curves (in red). The ordinate is the observed resistivity
normalized by the resistivity of the upper layer; the abscissa of the type-curves is the
normalized electrode separation (a/d), where d is the unknown depth to the interface
between the layers, and the abscissa of the observed curves is a logarithmic plot of
electrode separation.
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The curves must be shifted horizontally until an optimal visual agreement of the
measured curve with the type curves is obtained. This is found for a k-value in the range
–0.6 ≥ k ≥ –0.7, in broad agreement with the value k = –0.64 found above. The fit is not
perfect: for a < 200 m the measured curve fits better the type-curve k = –0.6, while for
a > 300 m it fits better the type-curve k = –0.7.
At the optimal match of the curves, the ratio (a/d) = 1 agrees with an electrode
separation of a ≈ 95–100 m, so this is also the estimated depth d of the interface.
5
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4.
In the Schlumberger resistivity method the separation of the current electrodes L is
much larger than the separation a of the voltage electrodes. Suppose that the mid-point
of the voltage pair is displaced by a distance x from the mid-point of the current
electrode pair. Show that, for (L – 2x) >> a, the apparent resistivity is given by
2
2
π V ( L − 4x )
ρa =
4 I a ( L2 + 4x 2 )
2
The geometry of the problem is shown in the following figure; the electrode
arrangement is asymmetric:
The general formula for a 4-electrode resistivity investigation is given by Eq. (4.89):
TESTBANKSELLER.COM
⎧
⎫
⎪
⎪
V⎪
1
V
⎪
ρ = 2π ⎨
⎬ = 2π F
I ⎪⎛ 1
I
1 ⎞ ⎛ 1
1 ⎞⎪
−
−⎜
−
⎜
⎟
⎟
⎪⎩ ⎝ rAC rCB ⎠ ⎝ rAD rDB ⎠ ⎪⎭
where F is a geometric factor determined by the electrode array. For the Schlumberger
method shown in the figure the separations of the potential and current electrodes are
L
a
−x−
2
2
L
a
= −x+
2
2
L
a
+x+
2
2
L
a
= +x−
2
2
rAC =
rCB =
rAD
rDB
The geometric factor F is in this case
1 ⎛ 1
1 ⎞ ⎛ 1
1 ⎞
=⎜
−
−⎜
−
⎟
F ⎝ rAC rCB ⎠ ⎝ rAD rDB ⎟⎠
=
1
1
1
1
−
−
+
⎛L−a
⎞ ⎛L+a
⎞ ⎛L+a
⎞ ⎛L−a
⎞
− x⎟ ⎜
+ x⎟ ⎜
− x⎟ ⎜
+ x⎟
⎜⎝
⎠ ⎝ 2
⎠ ⎝ 2
⎠ ⎝ 2
⎠
2
⎛
1
1
1
1
⎞
= 2⎜
+
−
−
⎝ (L − a) − 2x (L − a) + 2x (L + a) + 2x (L + a) − 2x ⎟⎠
6
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Rearranging the order of the terms, this is equivalent to
1
⎛
1
1
1
1
⎞
= 2⎜
−
+
−
⎝ (L − 2x) − a (L − 2x) + a (L + 2x) − a (L + 2x) + a ⎟⎠
F
⎛ (L − 2x) + a − (L − 2x) + a (L + 2x) + a − (L + 2x) + a ⎞
= 2⎜
+
⎟⎠
⎝
(L − 2x)2 − a 2
(L + 2x)2 − a 2
⎛
a
a
⎞
= 4⎜
+
2
2
2
2⎟
⎝ (L − 2x) − a
(L + 2x) − a ⎠
If we now apply the condition that (L – 2x) >> a, the equation for F reduces to
1
⎛
1
1
⎞
= 4a ⎜
+
2
2⎟
⎝ (L − 2x)
F
(L + 2x) ⎠
⎛ (L + 2x)2 + (L − 2x)2 ⎞
= 4a ⎜
⎟⎠
(L2 − 4x 2 )2
⎝
= 8a
(L2 + 4x 2 )
(L2 − 4x 2 )2
The apparent resistivity is therefore given by
−1
V
V⎛
(L2 + 4x 2 ) ⎞
, which, on simplifying, gives
ρ = 2π g(r) = 2π ⎜ 8a 2
I
I ⎝ (L − 4x 2 )2 ⎟⎠
ρ=
π V (L2 − 4x 2 )2 TESTBANKSELLER.COM
4 I a(L2 + 4x 2 )
7
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5.
In the double-dipole resistivity method it is common to keep the separation of the pairs
L an integer multiple n of the distance a between the electrodes in each pair, i.e. L = na.
(a) Rewrite the formula for the apparent resistivity with this assumption.
(b) If L is very large compared to a, modify the formula to show that the apparent
resistivity is proportional to n3.
(a)
The apparent resistivity for the double-dipole method is given by Eq. (4.89):
ρ=π
(
)
2
2
V⎛L L −a ⎞
⎜
⎟
I⎝
a2
⎠
Substituting L = na,
(
)
2 2
2
V ⎛ na n a − a ⎞
V
2
ρ=π ⎜
⎟ = π n n −1 a
2
I⎝
a
I
⎠
(b)
(
)
If L >> a, then n >> 1, and the term (n2 – 1) reduces to n2. The expression for the
apparent resistivity becomes
ρ=π
V 3
na
I
Thus the apparent resistivity
n3R. .COM
TEis
STproportional
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6.
Consider a double-dipole configuration in which the electrode pairs are not collinear but
are broadside to each other (i.e., normal to the line joining them). The electrode
separation is a and the distance between the mid-points of the pairs is L = na. Show
that, for large values of n, the apparent resistivity is given in this case by
ρa = 2π n 3a
V
I
The geometry of this double-dipole configuration is shown in the following figure:
The inter-electrode distances are:
rAC = L
rCB = L2 + a 2
rAD = L2 + a 2
rDB = L
Substituting in the geometric factor F for the 4-electrode resistivity method gives:
TESTBANKSELLER.COM
1 ⎛ 1
1 ⎞ ⎛ 1
1 ⎞
=⎜
−
−
−
F ⎝ rAC rCB ⎟⎠ ⎜⎝ rAD rDB ⎟⎠
=
1
−
L
1
L +a
⎛1
= 2⎜ −
⎝L
2
2
−
1
L2 + a
1
L +a
2
2
+
1
L
⎞
2 ⎟
⎠
⎛ L2 + a 2 − L ⎞
= 2⎜
⎟
⎝ L L2 + a 2 ⎠
If L = na, this simplifies to
1 2 ⎛ n2 + 1 − n ⎞
= ⎜
⎟
F a ⎝ n n2 + 1 ⎠
The apparent resistivity is therefore
ρ = 2π
V
V a ⎛ n n2 + 1 ⎞
F = 2π
I
I 2 ⎜⎝ n 2 + 1 − n ⎟⎠
9
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This can be simplified further to
ρ =π
V ⎛ n n2 + 1 ⎞ ⎛ n2 + 1 + n ⎞
a
I ⎜⎝ n 2 + 1 − n ⎟⎠ ⎜⎝ n 2 + 1 + n ⎟⎠
(
2
2
V n n +1 n +1 + n
=π a
I
( n2 + 1) − n2
=π
V
an n 2 + 1
I
(
n2 + 1 + n
For n >> 1, we can write
ρ = 2π an 3
)
)
n 2 + 1 ≈ n and so
V
I
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7.
Calculate the velocity of an electromagnetic wave in (a) basalt (dielectric constant
κ = 12) and (b) water (κ = 80.4).
The speed v of an electromagnetic wave is obtained from Eq. (10.55):
v2 =
1
, where µ0 is the magnetic field constant and ε is the permittivity of the
µ0ε
medium, which is related to the permittivity ε0 of a vacuum by the dielectric constant κ :
ε0 = κ ε0
Thus v 2 =
1
1
1
1
and c is the speed of light in a vacuum
=
= c 2 , where c 2 =
µ0ε κµ0ε 0 κ
µ0ε 0
(299,792 km s–1).
In applied geophysics the velocities of electromagnetic waves are commonly
given in units of meters/nanosecond (m ns–1) or meters/microsecond (m µs–1); in these
units, c ≈ 0.3 m ns–1 ≈ 300 m µs–1.
(a)
In basalt, with κ = 12,
v=
1
1
c=
c = 0.2887c = 86,540 km s–1 (0.0865 m ns–1)
κ
12
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(b)
In water, with κ = 80.4,
v=
1
c=
κ
1
c = 0.1115c = 33, 400 km s–1 (0.0334 m ns–1)
80.4
11
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8.
Calculate the “skin depths” of penetration in (a) granite (ρ = 5,000 Ω m) and (b) a
pyrrhotite ore-body (ρ = 5 × 10–5 Ω m) for electromagnetic waves in surveys employing
(i) electromagnetic induction (ƒ = 1 kHz) and (ii) ground penetrating radar (ƒ = 100
MHz). Would these methods detect the conducting bodies if they were buried under a
water-saturated soil layer, 3 m thick with resistivity 100 Ω m?
The skin depth for an electromagnetic wave is given by Eq. (4.126):
2
ρ
d=
=
µ0σω
πµ0 f
(a)
For the granite body (ρ = 5,000 Ω m) the skin depth for electromagnetic induction at a
frequency of 1 kHz is:
ρ
5000
d=
=
= 1100 m.
2
πµ0 f
4π × 10 −7 × 10 3
The skin depth for ground penetrating radar at 100 MHz is:
ρ
5000
d=
=
= 3.6 m
2
πµ0 f
4π × 10 −7 × 10 8
(b)
For the pyrrhotite ore-body (ρ = 5 × 10–5 Ω m) the skin depth for electromagnetic
induction at a frequency of 1 kHz is:
d=
ρ
=
πµ0 f
5 × 10 −5
m.LER.COM
TESTBAN=K0.11
SEL
4π 2 × 10 −7 × 10 3
The skin depth for ground penetrating radar at 100 MHz is:
ρ
d=
=
πµ0 f
5 × 10 −5
= 0.36 × 10 −3 m = 0.36 mm
2
−7
8
4π × 10 × 10
Buried conductor under 3 m wet soil
The penetration depth in wet soil (ρ = 100 Ω m) is found similarly.
ρ
100
For the EM induction method d =
=
= 160 m .
2
πµ0 f
4π × 10 −7 × 10 3
For the GPR method d =
ρ
=
πµ0 f
100
= 0.5 m .
4π × 10 −7 × 10 8
2
The skin depth d is where the signal is attenuated to e–1 of its surface value. In fact
the electromagnetic waves penetrate far beyond this depth. At depth 5d the signal still
has 1% (e–5) of its surface amplitude. The weakened signal may penetrate further and
induce a detectable response from greater depths.
The 3 m thick layer of wet soil in this exercise would be penetrated and the buried
conductor surely detected by the 1 kHz EM induction signal. It might also be detected
by the 100 MHz GPR signal, if the transmitter is strong.
12
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SOLUTIONS TO EXERCISES
11 The Earth’s Magnetic Field
1.
The IGRF for 2015 gives the following Gauss coefficients for the geomagnetic dipole:
g10 = –29,442 nT, g11 = –1501 nT, h11 = +4,797 nT.
(a) Show that the Earth’s dipole magnetic moment is 7.724 × 1022 A m2
(b) Calculate the locations of the geomagnetic poles.
(a)
The dipole magnetic moment m in terms of the Gauss coefficients is
2
2
2
4π 3
R ( g10 ) + ( g11 ) + ( h11 ) which on substituting the given values is
µ0
3
4π
2
2
2
m=
6.371× 10 6 ) ( 29442 ) + ( −1501) + ( 4797 )
−7 (
4π × 10
3
4π
m=
6.371× 10 6 ) ( 29868 )
−7 (
4π × 10
m=
m = 7.724 × 1022 A m2
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(b)
Let the latitude of the geomagnetic pole be 𝜆p and its longitude be 𝜙p.
Using the Gauss coefficients, the latitude of the geomagnetic pole is
tan λ p = g10 /
( g ) + (h )
1 2
1
1 2
1
= 29442 /
(1501)2 + ( 4797 )2
= 5.857
λ p = 80.3°N
and the longitude of the pole is
tan φ p = h11 / g11 = −3.196
φ p = −72.6 = 72.6°W = 287.4°E
1
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2.
Assuming an axial dipole with magnetic moment 7.724 × 1022 A m2, calculate the
variation of the total field with latitude in nT/km at 30°N?
The potential of a dipole at distance r from its center and azimuth θ from its axis is
µ m cosθ
W= 0
4π r 2
By differentiation we get the field components
Br = −
∂W
µ 2m cosθ
;
= 0
∂r
4π
r3
Bθ = −
1 ∂W
µ m sin θ
= 0
r ∂θ
4π r 3
The total magnetic field Bt is given by
Bt =
Bt =
( Br )2 + ( Bθ )2
µ0 m
µ m
4 cos 2 θ + sin 2 θ = 0 3 3cos 2 θ + 1
3
4π r
4π r
The magnetic latitude is the complement of the azimuth from the dipole axis ( 𝜆 = 90–𝛳).
Thus the total field at latitude 𝜆 is
Bt = Beq 1+ 3sin 2 λ where Beq =
µ0 m
4π r 3
(1)
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Beq is the magnetic field at the equator (λ = 0).
At the surface of the Earth, r = R = 6371 km. With the given magnetic moment m
and the magnetic permeability µ0 = 4π × 10–7 N A–2, the equatorial field is found to be
Beq =
µ0 m 4π × 10 −7 7.724 × 10 22
=
= 2.9869 × 10 −5 = 29,869 nT .
6 3
4π r 3
4π
( 6.371× 10 )
The change of Bt with latitude λ is found by differentiating Eq. (1):
∂Bt (λ ) Beq 3( 2sin λ cos λ )
3sin 2 λ
=
=
B
eq
∂λ
2
1+ 3sin 2 λ
2 1+ 3sin 2 λ
A change in latitude ∂λ corresponds to a distance along the surface ∂s = R ∂λ; the
change in Bt per northward km at 30°N is found by substitution:
(
)
⎛
⎞
∂Bt (λ ) 1 ∂Bt (λ )
⎛ 29869 ⎞ ⎜ (1 2 ) 3 2 ⎟
=
= 3⎜
= 4.6 nT km–1.
⎝ 6371 ⎟⎠ ⎜ 1+ 3 1 2 2 ⎟
∂s
R ∂λ
( ) ⎠
⎝
2
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3.
Compute the inclination and declination of the magnetic field that would be observed in
Boulder, Colorado (40°N, 105°W) if the Earth’s field corresponded to a perfect
geocentric dipole whose axis penetrates the Earth’s surface at 80°N, 72°W.
Assuming a geocentric dipole field, magnetic inclination I is related to angular
distance θ from the magnetic pole. The angular distance between two locations is found
by using the direction cosines of axes from the center of the Earth through each location
(see Box 2.2). For an axis through the site with latitude λ and longitude φ, the direction
cosines (l, m, n) are
l = cos λ cos φ
m = cos λ sin φ
n = sin λ
Let the direction cosines through Boulder (λ = 40, φ = –105) be (l1, m1, n1) and those
through the magnetic pole (λ = 80, φ = –72) be (l2, m2, n2). Then
l1 = cos(40)cos(–105) = −0.1983
m1 = cos(40)sin(–105) = −0.7399
n1 = sin(40) = +0.6428
l2 = cos(80)cos(–72) = +0.0537
and
m2 = cos(80)sin(–72) = −0.1651
n2 = sin(80) = +0.9848
If the angle between the axes is θ, then
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cosθ = l1l2 + m1m2 + n1n2 = 0.7446 and the angular distance is θ = 41.9°.
This enables us to compute the inclination I of the magnetic field at Boulder.
tan I = 2 cot θ = 2 / tan θ = 2.229 ,
from which the inclination is I = 65.9°.
The declination is equivalent to the (magnetic) heading from Boulder to the pole (see
Exercise 2.7). The pole lies north and east of Boulder. If the great circle distance is θ,
the declination is given by
cos D =
sin λ2 − sin λ1 cosθ
cos λ1 sin θ
Using the parameters from this exercise,
cos D =
sin(80) − sin(40)cos(41.9)
= 0.9899
cos(40)sin(41.9)
The declination is D = 8.1° East.
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4.
Show that, for a small displacement along a meridian of magnetic longitude at magnetic
latitude 45 °N, the change of inclination is exactly 4/5 the change in latitude.
Again we use the dipole equation relating inclination and latitude:
tan I = 2 tan λ
We need further the trigonometric identities
sec 2 ϕ = 1 + tan 2 ϕ
∂
tan ϕ = sec 2 ϕ = 1 + tan 2 ϕ
∂ϕ
Differentiating the dipole equation with respect to λ, we get
∂
∂
tan I = 2 tan λ
∂λ
∂λ
sec 2 I
∂I
= 2 sec 2 λ
∂λ
(1 + tan I ) ∂∂Iλ = 2 (1 + tan λ )
2
(
(
2
)
(
)
)
2
2T1E+Stan
TB2AλNKSELLER.COM
∂I 2 1 + tan λ
=
=
∂λ
1 + tan 2 I
1 + 4 tan 2 λ
) (
At latitude 45 °N, tan λ = 1 and
∂I 2 (1 + 1) 4
=
=
∂λ (1 + 4 ) 5
Thus a small change in latitude ∆λ results in a change of inclination ∆ I =
4
∆λ .
5
4
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5.
The north magnetic pole is at 86.3°N 160.1°W, the south magnetic pole is at 64.3°S
136.6°E.
(a) Why are the poles not antipodal (exactly opposite)?
(b) What is the closest distance between the center of the Earth and the straight line
joining the poles?
(a)
The magnetic poles are the locations where the inclination of the real field is vertical.
For a pure geocentric dipole field the poles are exactly antipodal. For the real field the
poles are not antipodal because of the presence of non-dipole components in the
geomagnetic field.
(b)
Let the distance between the center C of the Earth and the magnetic axis be d, as in the
figure, and let PN and PS be the north and south magnetic poles.
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If (2∆) is the angular distance between the poles, then d = R cos∆, where R is the
Earth’s radius. The angular distance is computed as in earlier exercises by computing
the direction cosines of axes through the two poles:
l1 = cos(86.3)cos(–160.1) = +0.0651
m1 = cos(86.3)sin(–160.1) = −0.0250
and
n1 = sin(86.3) = 0.9976
l2 = cos(−64)cos(137) = −0.3206
m2 = cos(−64)sin(137) = +0.2990
n2 = sin(−64) = −0.8988
cos(2∆) = l1l2 + m1m2 + n1n2 = −0.8832
so the angular distance is 2∆ = 152.5°, from which ∆ = 76.2°.
With R = 6371 km, the distance d of the offset axis from the center of the Earth is
d = R cos∆ = 1,517 km.
NOTE: The straight line joining the poles in this exercise has no geophysical significance.
5
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6.
Measurements of the magnetic field elements at a geomagnetic observatory gave the
following results:
N-component = 27,200 nT; E-component = –1,800 nT; V-component = –40,100 nT.
(a) Is the observatory in the northern or southern hemisphere?
(b) What is the total field intensity at the site?
(c) What are the local values of inclination and declination of the field?
(a)
The negative value of the vertical component shows that the field is directed upwards,
and the magnetic inclination is therefore negative. Because of the difference between
geographic and magnetic latitudes, it is possible for locations just north of the
geographic equator to have small but negative inclinations (see Fig. 5.34a). However, in
this case the vertical component is large so the site is in the southern hemisphere.
(b)
The total intensity T is obtained from
T = N 2 + E2 + V 2
T=
( 27200 )2 + ( −1800 )2 + ( −40100 )2
= 48,488 nT
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NKSELLER.COM
= 48,500 nT to 3 significant
digits
(c)
The declination D is obtained from
tan D =
E −1800
=
= −0.06618
N 27200
which gives a declination D = –3.8° = 3.8 °W.
The inclination is obtained from
sin I =
V −40100
=
= −0.8270
T
48488
which gives an inclination I = –55.8°.
6
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7.
The IGRF for 2015 gives the values of the Gauss coefficients for the dipole and
quadrupole components of the geomagnetic field shown in the following table.
Gauss coefficients in nT
g10 –29,442 g20 –2445
g11
1
1
h
–1,501
g12
+3,013
+4,797
1
2
–2,846
2
2
2
2
g
+1,677
h
–642
h
The mean squared value, Rn, of the intensity of the geomagnetic field at Earth's
surface due to the component of degree n is given by
n
((
Rn = (n + 1) ∑ gnm
m=0
) + (h ) )
2
m 2
n
(a) Calculate the root mean square intensity of the dipole field at the Earth’s surface.
(b) Calculate the corresponding r.m.s.intensity of the quadrupole component and
express it as a percentage of the dipole field intensity.
The Gauss coefficients are explained in section 11.2.4.
(a)
ESgeomagnetic
TBANKSEfield
LLEisR.described
COM by all coefficients with
The dipole component of T
the
n = 1. Using the given formula, the mean squared intensity BD of the dipole field at the
Earth's surface is
((
RD = 2 g10
) + ( g ) + (h ) )
2
1 2
1
1 2
1
Using the values given in the table, the root mean square intensity of the dipole field is
(
BD = RD = 2 ( −29442 ) + ( −1501) + ( 4797 )
(b)
2
2
2
) = 42,240 nT
The quadrupole component is described by all coefficients with n = 2. The r.m.s.
intensity BQ of the quadrupole field is computed in a similar way to the dipole intensity:
((
RQ = 3 g20
) + ( g ) + (h ) + ( g ) + (h ) )
2
1 2
2
(
1 2
2
2 2
2
2 2
2
BQ = RQ = 3 ( −2445 ) + ( 3013) + ( −2846 ) + (1677 ) + ( −642 )
2
2
2
2
2
) = 8,896 nT
The ratio of the r.m.s. quadrupole field to the r.m.s. dipole field at Earth’s surface is
BQ
8896
=
= 0.2106 = 21.1%
BD 42240
7
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8.
The values of the Gauss coefficients in the previous exercise are given for the Earth’s
surface. Re-calculate the root mean square intensities of the dipole and quadrupole field
components at the core-mantle boundary (Earth’s radius = 6371 km, core radius = 3485
km).
The geomagnetic dipole is characterized by the Gauss coefficients n = 1. Eq. (11.36)
shows that the potential of the dipole is proportional to 1/r2, where r is the distance from
the center of the Earth. The magnetic field is determined by differentiating the potential.
Thus the dipole field varies with radial distance as 1/r3. Similarly, the quadrupole
potential (for which n = 2) is proportional to 1/r3 and so the quadrupole field varies with
radial distance as 1/r4.
For example, if the distance r is halved, the dipole field increases by a factor 23 = 8,
whereas the quadrupole field increases by a factor 24 = 16.
Let r = kR. Comparing the dipole field intensity at distance r with the intensity at the
surface R:
3
BD (r) ⎛ R ⎞
1
=⎜ ⎟ = 3
BD (R) ⎝ r ⎠
k
TESfield
TBAintensity
NKSELatLthe
ER.
COdistances
M
Similarly, for the quadrupole
two
r and R:
4
1
⎛ R⎞
=⎜ ⎟ = 4
BQ (R) ⎝ r ⎠
k
BQ (r)
In the exercise, the radius of the core Rc = 3485 km; the radius of the Earth’s surface is
R = 6371 km the distance. At the core-mantle boundary (CMB) r = Rc and
k=
Rc 3485
=
= 0.5470
R 6371
Thus the root mean square dipole field at the CMB has intensity
BD =
42240
= 258,069 nT
( 0.547 )3
The root mean square quadrupole field at the CMB has intensity
BD =
8896
= 99, 362 nT
( 0.547 )4
The ratio of the r.m.s. quadrupole field to the r.m.s. dipole field at the CMB is therefore
BQ
99, 362
=
= 0.385 = 38.5%
BD 258,069
8
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9.
What are the values of n and m in the designation Ynm for the spherical harmonic
functions illustrated in Fig. B11.3.2 in Box 11.3? Sketch how these patterns would
appear on the opposite side of the reference sphere to the one you are looking at?
The order n and degree m of a spherical harmonic function determine the number of
‘latitude’ and ‘longitude’ circles on a sphere that are nodal lines, i.e., where the value of
the spherical harmonic is zero. The geometries of spherical harmonics (Box 11.3) are
determined by m nodal lines of longitude and (n – m) nodal lines of latitude. The
geometric pattern is drawn as a frontal view of a projection of the sphere. Each circle of
longitude continues on the rear of the sphere, with appropriate shading for the zones
between the nodal lines.
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This sectorial harmonic has no nodal lines of latitude. Thus, n – m = 0, and n = m.
The pattern has m = 5 nodal lines of longitude. Thus n = m= 5.
The spherical harmonic function is denoted Y55 .
This tesseral harmonic has one nodal line of latitude. Thus, n – m = 1, and n = m+1.
The pattern has m = 5 nodal lines of longitude. Thus n = (m + 1) = 6.
The spherical harmonic function is denoted Y65 .
9
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10.
In an aeromagnetic survey at a flight altitude of 2,000 m above sea-level, the maximum
total field anomaly over an orebody was 30 nT. In a repeated measurement at 2,500 m
altitude the maximum amplitude of the anomaly was 20 nT. Calculate the depth of the
orebody below sea-level assuming (i) a monopole source and (ii) a dipole source for the
anomaly.
The field of a magnetic monopole varies with distance r from the source as 1/r2
(Eq. (11.2)), while the field of a dipole varies as 1/r3 (Eq. (11.18-19)).
Let the source be at depth d below sea-level and the flight altitude be h; the distance
from the aircraft to the source is r = h + d. Comparing the anomaly signals B1 and B2 at
two altitudes h1 and h2, respectively,
n
⎛ h + d⎞
B1 ⎛ r2 ⎞
=⎜ ⎟ =⎜ 2
B2 ⎝ r1 ⎠
⎝ h1 + d ⎟⎠
n
where n = 2 for a monopole source and n = 3 for a dipole source.
In the present exercise: B1 = 30 nT at h1 = 2000 m and B2 =20 nT at h2= 2500 m.
Depth of orebody for monopole source: n = 2
2
⎛ h2 + d ⎞
B1
⎜⎝ h + d ⎟⎠ = B
1
2
2500 + d
=
2000 + d
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30
= 1.2247
20
2500 + d = 1.2247(2000 + d) = 2449.48 + 1.2247d
d=
50.52
= 225 m
0.2247
Depth of orebody for dipole source: n = 3
3
⎛ h2 + d ⎞
B1
⎜⎝ h + d ⎟⎠ = B
1
2
2500 + d
=
2000 + d
3
30
= 1.1447
20
2500 + d = 1.1447(2000 + d) = 2289.4 + 1.1447d
d=
214.6
= 1455 m
0.1447
10
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11.
The vertical field magnetic anomaly ∆Bz over a vertically magnetized anticline
(represented by a horizontal cylinder) is given by Eq. (11.33). Draw a sketch of the
anomaly on a profile normal to the structure. Observe the horizontal positions where the
anomaly is zero.
(a) Calculate the horizontal positions where the anomaly has extreme values.
(b) Calculate the peak-to-peak values of the anomaly.
Let the depth to the axis of the horizontal cylinder be z and the horizontal position on
the profile be x. Eq. (5.54) gives the following equation for the vertical field anomaly:
(
(
)
)
z2 − x2
1
2
∆ Bz = µ0 R ∆ M z
2
2
z2 + x2
A graph of this anomaly is shown in the figure below, with x normalized by the depth z:
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(a)
The anomaly is zero where ∆Bz = 0, i.e., where the factor
(z − x )
(z + x )
2
2
2
2 2
=0
The zero values occur where x = ± z (and also where x = ± ∞, but this solution is of little
interest here). The depth z of the axis of the anticline can be read directly from a plot of
∆Bz against horizontal position x by observing the zero-crossing positions.
(b)
The extreme values of the anomaly are found where
(
)
2
(
)
∂
∆ Bz = 0 :
∂x
)(
)
z 2 + x 2 (−2x) − 2(2x) z 2 − x 2 z 2 + x 2
∂
1
∆ Bz = µ0 R 2 ∆ M z
=0
4
∂x
2
z2 + x2
(
Simplifying the numerator in this expression,
11
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∂
∆ Bz = 0 where (−2x) 3z 2 − x 2 = 0
∂x
(
)
The extreme values are therefore at x = 0 and x = ± 3z
The maximum value of ∆Bz is at x = 0:
2
1
⎛ R⎞
∆ Bmax = µ0 ∆ M z ⎜ ⎟
⎝ z⎠
2
The minimum values of ∆Bz are at x = ± 3z :
∆ Bmin
(
(
)
)
2
z 2 − 3z 2
1
1 ⎡1
1
⎛ R⎞ ⎤
2
= µ0 R ∆ M z
= − ⎢ µ0 ∆ M z ⎜ ⎟ ⎥ = − ∆ Bmax
2
⎝ z ⎠ ⎥⎦
2
8 ⎢⎣ 2
8
z 2 + 3z 2
The peak-to-peak amplitude of the anomaly is ∆Bpp = ∆Bmax – ∆Bmin:
2
∆ Bp − p
9
9
⎛ R⎞
= µ0 ∆ M z ⎜ ⎟ = ∆ Bmax
⎝ z⎠
16
8
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12.
Assume that the core of an anticline is made of basalt with susceptibility 1.8 × 10–1 (SI),
the host formation is limestone with susceptibility 3.0 × 10–4 (SI), and the rocks are
vertically magnetized.
(a) Compute the induced magnetization contrast when the vertical magnetic field
intensity is 40,000 nT.
(b) If the anticline is modelled by a horizontal cylinder whose radius is one fifth the
depth of its axis, calculate the maximum amplitude of the vertical field anomaly
over the structure.
(a)
Susceptibility k relates the magnetization M and the magnetic field H:
M = kH .
The auxiliary field H is computed from the magnetic field B using B = µ0H
In this exercise the field B is vertical and has intensity 40,000 nT = 4 × 10–5 T, so
4 × 10 −5
H=
= 31.83 A m–1
4π × 10 –7
The median susceptibility of basalt is 1.8 × 10–1 (SI) and that of limestone is 3.0 × 10–4
(SI). The magnetization of the basalt core of the anticline is thus
M 1 = k1 H = 1.8 × 31.83 × 10 −1 = 5.7296 A m–1
while the magnetization of the host limestone is
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M 2 = k2 H = 3.0 × 31.83 × 10 −4 = 0.0095 A m–1
The vertical magnetization contrast is
∆ M z = M 1 − M 2 = 5.7296 − 0.0095 = 5.72 A m–1
(b)
The formula for the maximum amplitude of the vertical magnetic field anomaly over an
anticline, modeled as a horizontal cylinder, was evaluated in the previous exercise:
∆ Bmax
1
⎛ R⎞
= µ0 ∆ M z ⎜ ⎟
⎝ z⎠
2
2
If the depth z is 5 times the cylinder radius R, as given, the maximum amplitude of the
anomaly with the computed magnetization contrast is
2
∆ Bmax
1
⎛ 1⎞
= ( 4π ) ( 5.72 ) ⎜ ⎟ × 10 −7 = 1.44 × 10 −7 T
⎝ 5⎠
2
In more convenient units, ∆Bmax is equal to 144 nT.
13
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13.
Assuming the gravity anomaly over an anticline as given by Eq. (4.35), apply the
Poisson relationship (Box 11.5) to obtain the horizontal field magnetic anomaly ∆Bx of
a vertically magnetized anticline with radius R and magnetization contrast ∆Mz.
Poisson’s relation between the magnetic potential W of a vertically magnetized body
and the vertical gravity anomaly gz of the body is:
W=
µ0 ⎛ ∆ M z ⎞
gz
4π ⎜⎝ G ∆ ρ ⎟⎠
where ∆Mz is the vertical magnetization contrast, ∆ρ the density contrast, and G the
gravitational constant. The gravity anomaly over an anticline modelled by a horizontal
cylinder with radius R is
⎞
⎛ ∆ ρ R2 ⎞ ⎛
1
z ⎞
2⎛
∆ gz = 2π G ⎜
⎜
2 ⎟ = 2π G ∆ ρ R ⎜ 2
⎟
⎝ z + x 2 ⎟⎠
⎝ z ⎠ ⎝ 1 + ( x / z) ⎠
The magnetic potential of a vertically magnetized anticline, using Poisson’s relation is
W=
µ0 ⎛ ∆ M z ⎞
⎛ z ⎞ 1
⎛ z ⎞
2π G ∆ ρ R 2 ⎜ 2
= µ0 ∆ M z R 2 ⎜ 2
2⎟
⎜
⎟
⎝z +x ⎠ 2
⎝ z + x 2 ⎟⎠
4π ⎝ G ∆ ρ ⎠
The horizontal field magnetic anomaly ∆Bx is found by differentiating with respect to x:
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⎛ −z ( 2x ) ⎞
∂ ⎛1
z ⎞⎞
1
2⎛
2
∆ Bx = − ⎜ µ0 ∆ M z R ⎜ 2
= − µ0 ∆ M z R ⎜
⎟
⎝ z + x 2 ⎟⎠ ⎟⎠
∂x ⎝ 2
2
⎜⎝ z 2 + x 2 2 ⎟⎠
(
∆ Bx = µ0 ∆ M z R
where α =
2
(z
zx
2
+ x2
α
⎛ R⎞
= µ0 ∆ M z ⎜ ⎟
⎝ z ⎠ 1+α2
)
2
)
2
(
)
2
x
is the normalized horizontal distance on the profile.
z
The shape of the anomaly is shown in the figure below; the extremes are at α = ±
14
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.
3
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SOLUTIONS TO EXERCISES
12 Paleomagnetism
1.
Assuming that the geomagnetic field corresponds to a geocentric axial dipole, calculate
the latitude of a site where the field inclination is 45°.
For a dipole field the relationship between magnetic inclination I , the angular
distance θ from the magnetic pole, and magnetic latitude λ is given by Eq. (11.19):
tan I = 2 cot θ = 2 tan λ
In the present exercise, assuming the field to be that of a geocentric, axial dipole, the
geographic and magnetic latitudes are equivalent. Inserting I = 45° in the equation:
tan λ =
1
tan I = 0.5
2
because tan(45°) = 1.
Evaluation gives the latitude of the site:
λ = tan −1 (0.5) = 26.6°
2.
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With the same assumption, calculate the inclination of the geocentric axial dipole field
at latitude 45°N.
The same equation is applicable as in exercise 1:
tan I = 2 cot θ = 2 tan λ
In this case we are given that the latitude λ = 45°. Insertion in the equation gives:
tan I = 2 tan ( 45° ) = 2
The inclination is thus
I = tan −1 (2) = 63.4°
1
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3.
The north (N), east (E) and vertical (V) components of a magnetization can be
calculated from its intensity (M), declination (D) and inclination (I) using the following
relationships:
N = M cos I cos D
E = M cos I sin D
V = M sin I
In the progressive thermal demagnetization of a sample of Cretaceous limestone for a
paleomagnetic study the following remanent magnetizations were measured at different
temperatures T:
200
M (10-5 A/m)
5.08
D (°)
60.8
I (°)
60.1
300
3.87
62.1
59.8
400
3.02
61.2
62.1
500
2.10
62.2
61.6
550
1.18
63.1
60.9
T (°C)
(a) Calculate the north (N), east (E) and vertical (V) components of the magnetizations
at each demagnetization stage.
(b) Plot the N-components against the E-components, fit a straight line, and determine
the optimum declination
magnetization
TESfor
TBthe
ANstable
KSEL
LER.COM direction.
(c) Plot the V-components against either the N- or E-components, fit a straight line, and
compute the optimum inclination for the stable magnetization direction.
(d) The straight lines do not pass through the origin of the plot. What does this imply?
(a)
The North (N), East (E) and vertical (V) components of the magnetization (in units of
10–5 A/m), as calculated with the given set of equations, are listed in the following table
T (°C)
200
Magnetization
5.08
D (°)
60.8
I (°)
60.1
East
2.21
North Down
1.24
4.40
300
3.87
62.1
59.8
1.72
0.91
3.34
400
3.02
61.2
62.1
1.24
0.68
2.67
500
2.10
62.2
61.6
0.88
0.47
1.85
550
1.18
63.1
60.9
0.51
0.26
1.03
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(b)
The North component (N) is plotted against
the East component (E) in the upper part of
the figure (solid points). The best-fitting
straight line to these points has the equation:
N = 0.566E − 0.033
dN
= 0.566
dE
The gradient of this line is
The tangent of the declination is
tan D =
dE
1
=
= 1.767
dN 0.566
The declination is therefore
D = 60.5°
(c)
The vertical component (V) is plotted
(downward for positive inclination) against
the East component (E) in the lower part of
the figure (open points). The best-fitting
straight line to these points has the equation:
V = 1.939E + 0.113
TdV
ESTBANKSELLER.COM
The gradient of the line is
dE
= 1.939
This line makes an angle I0 with the East-axis, measured in the V-E plane, as in the
figure. Note that this angle is not the inclination of the magnetic vector, because the
inclination measured in the vertical plane passing through the horizontal component
(see Fig. 11.4). The horizontal component (∆H) is found by combining the north and
east components:
(
)
∆ H 2 = ∆ N 2 + ∆ E 2 = ( 0.566 ) + 1 ∆ E 2 = 1.32 ∆ E 2
2
∆ H = 1.149 ∆ E
tan I =
∆V ⎛ ∆V ⎞ ⎛ ∆ E ⎞ 1.939
=⎜
= 1.687
⎟⎜
⎟=
∆ H ⎝ ∆ E ⎠ ⎝ ∆ H ⎠ 1.149
The inclination is therefore
I = 59.3°
(d)
The fact that the straight lines do not pass through the origin of the plot indicates that
the magnetization contains a component of unknown but different direction, which has
not been removed in the demagnetization procedure.
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4.
In the same paleomagnetic study, the following stable directions of remanent
magnetization, corrected for the dip of bedding in the limestone formation, were
measured in five samples from the same site:
Sample
D (°)
I (°)
SR-04
329.7
40.6
SR-05
336.6
24.7
SR-07
326.2
46.0
SR-10
321.1
40.9
SR-12
322.7
44.9
(a) Calculate the direction cosines (λN, λE, λV) of the north, east and vertical (V)
components of each stable direction, using the relationships
λ N = cos I cos D
λE = cos I sin D
λV = sin I
(b) Add up the values for each direction cosine. Let the sums be X, Y and Z, where
X = ∑ λN ;
Y = ∑ λE ;
Z = ∑ λV
Calculate the vector sum of the directions, R, and the declination Dm and inclination Im
of the mean directionTusing
ESTthe
BArelationships
NKSELLER.COM
R = X2 + Y 2 + Z2 ;
tan Dm = Y / X;
sin I m = Z / R
(c) Using the computed value of R, calculate the precision parameter k of the data and
the 95% confidence error (α95) for the mean direction.
(a)
Applying the given formulas, the direction cosines of each vector are as listed below:
Sample
D (°)
I (°)
SR-04
329.7
40.6
SR-05
336.6
24.7
SR-07
326.2
46.0
SR-10
321.1
40.9
SR-12
322.7
44.9
λN
0.6556
0.8338
0.5772
0.5882
0.5635
λE
-0.3831
-0.3608
-0.3864
-0.4746
-0.4292
λV
0.6508
0.4179
0.7193
0.6547
0.7059
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(b)
Summing the direction cosines in the last three columns of the table gives:
X = ∑ λ N = 3.2183;
Y = ∑ λE = −2.0342;
The vector sum of the directions is R =
Z = ∑ λV = 3.1486
X 2 + Y 2 + Z 2 = 4.94055
The declination of the mean vector is
⎛Y ⎞
⎛ −2.0342 ⎞
Dm = tan −1 ⎜ ⎟ = tan −1 ⎜
= tan −1 ( −0.6321) = 327.7°
⎝ X⎠
⎝ 3.2183 ⎟⎠
The inclination of the mean vector is
⎛ Z⎞
⎛ 3.1486 ⎞
I m = sin −1 ⎜ ⎟ = sin −1 ⎜
= sin −1 ( 0.6373) = 39.6°
⎝ R⎠
⎝ 4.94055 ⎟⎠
(c)
The formula for the paleomagnetic precision parameter k is given in Eq. (12.6):
k=
N −1
4
=
= 67.28
N − R 5 − 4.94055
The precision parameter is used to calculate the angle of 95° confidence of the mean
direction (Eq. (12.7))
α 95 =
140
=
Nk
140
= 7.6°
5 × 67.28
The mean direction estimated in part (b) is based on a very small set of samples. If a
TEmeasured
STBANKthe
SEtrue
LLmean
ER.Cdirection
OM
very large number could be
would lie with 95%
confidence within 7.6° of the estimated mean at Dm = 327.7°, Im = 39.6°.
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5.
The samples analyzed in the previous exercise were gathered at a site in Italy with
latitude 43.4°N, longitude 12.6 °E. Calculate the latitude and longitude of the
paleomagnetic pole position for the Italian limestone.
The latitude and longitude of the paleomagnetic pole location are calculated as
explained in Fig 12.21, using Eqs. (12.8) and (12.9). The latitude λp of the pole is
obtained from
sin λ p = sin λs cos p + cos λs sin p cos D ,
where λs is the latitude of the sampling site, p is the angular distance from there to the
paleomagnetic pole, and D is the declination of the mean direction of the samples
(327.7° in this exercise). The polar distance p is computed from the inclination Im =
39.6° of the mean direction:
⎛
⎞
⎛ 2 ⎞
2
p = tan −1 ⎜
= tan −1 ⎜
= 67.5°
⎟
⎝ tan I m ⎠
⎝ tan ( 39.6 ) ⎟⎠
sin λ p = sin ( 43.4 ) cos ( 67.5 ) + cos ( 43.4 ) sin ( 67.5 ) cos ( 327.7 ) = 0.8301
The latitude of the pole is λp = 56.1 °N.
TESTBANKSELLER.COM
The longitude of the pole must be calculated in two stages. First, the auxiliary angle
β is found from Eq. (12.9):
sin β =
sin p sin D sin ( 67.5 ) sin ( 327.7 )
=
= −0.8856 ,
cos λ p
cos ( 56.1)
from which β = –62.3°. However, sinβ and sin(180 – β) are equivalent. Thus there are
two possibilities for the longitude of the pole
φ p = φs + β ,
for cos p ≥ sin λs sin λ p
φ p = φ s + 180 − β ,
for cos p < sin λs sin λ p
Inserting values from the above calculations:
cos p = cos ( 67.5 ) = 0.3822
sin λs sin λ p = sin ( 43.4 ) sin ( 56.1) = 0.5704
Thus the second formula must be used
φ p = φ s + 180 − β = 12.6 + 180 − (−62.3) = 254.9°
The longitude of the pole is φp = 254.9 °E (105.1 °W).
The paleomagnetic pole is therefore located at 56.1 °N, 105.1 °W, in northern Canada.
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6.
Assume that the location of the Late Cretaceous paleomagnetic pole for the European
plate is at 72 °N, 154 °E and the corresponding pole for the African plate is at 67 °N,
245 °E.
(a) Calculate the expected ‘European’ and ‘African’ directions at the Italian site in the
previous exercise.
(b) Compare the expected directions with the observed directions and explain how the
paleomagnetic results from the Italian limestone should be interpreted.
(a)
To compute the expected directions at the sampling site we use the same procedure as in
exercise 4 of this chapter, with the site location in place of Boulder and each
paleomagnetic pole, in turn, in place of the geomagnetic pole.
First we need to find the angular distance θ from each paleomagnetic pole to the site
at 43.4°N, 12.6 °E.
European pole. The paleomagnetic pole for the European plate is at 72 °N, 154 °E. Let
the direction cosines of an axis through the site be (l0, m0, n0) and those through the
European pole be (l1, m1, n1). Then using
l0 = cos(43.4)cos(12.6) = 0.7091
m0 = cos(43.4)sin(12.6) = 0.1585
and
l1 = cos(72)cos(154) = −0.2777
m1 = cos(72)sin(154) = 0.1355
TESTBANKSELLERn.1 C=Osin(72)
M
n0 = sin(43.4) = 0.6871
= 0.9511
The angular distance of the site from the European pole is given by
cosθ1 = l0l1 + m0 m1 + n0 n1 = 0.47799 and the angular distance is θ1 = 61.4°.
Assuming the dipole relation between magnetic inclination and polar distance
(Eq. (11.19)), the expected inclination for a European pole is obtained from
tan I1 = 2 / tan θ1 = 1.0884 . This gives an expected inclination I1 = 47.4°.
The 'European' declination is equivalent to the (magnetic) heading from the site to the
European pole (see Exercise 5.4). The pole lies north and east of the site. If the great
circle distance is θ, the declination is given by
cos D1 =
sin λ1 − sin λ0 cosθ1 sin ( 72 ) − sin ( 43.4 ) cos ( 61.4 )
=
= 0.9756
cos λ0 sin θ1
cos ( 43.4 ) sin ( 61.4 )
The European declination is D = 12.7° East.
African pole.The paleomagnetic pole for the African plate is at 67 °N, 245 °E. Let the
direction cosines of an axis through the site be (l0, m0, n0), as above, and those through
the African pole be (l2, m2, n2). Then
l2 = cos(67)cos(245) = −0.1651
m2 = cos(67)sin(245) = −0.3541
n2 = sin(67) = 0.9205
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The angular distance of the site from the African pole is given by
cosθ 2 = l0l2 + m0 m2 + n0 n2 = 0.45925 and the angular distance is θ2 = 62.7°.
The dipole relation between magnetic inclination and polar distance is again used:
tan I 2 = 2 / tan θ 2 = 1.0340 . This gives an expected inclination I2 = 46.0°.
If the great circle distance is θ2, the declination at the site due to the African pole is
given by
cos D2 =
sin λ2 − sin λ0 cosθ 2 sin ( 67 ) − sin ( 43.4 ) cos ( 62.7 )
=
= 0.9373
cos λ0 sin θ 2
cos ( 43.4 ) sin ( 62.7 )
The African declination is D2 = –20.4° (339.6°), i.e. 20.4° West.
(b)
The observed and expected directions, rounded to the nearest degree, are summarized in
the following table:
Data type
Declination
Inclination
Italian observed directions
328
40
Predicted African directions
340
46
Predicted European directions
13
47
The observed mean inclination
inNlimestone
TESTBA
KSELLEsamples
R.COMat the Italian paleomagnetic site
(40°) is slightly shallower than the expected inclinations (46-47°). This may imply that
the magnetization vector has experienced some depositional flattening as a result of
compaction. However, the Italian declination (328°) is much closer to the expected
African declination (340°) than it is to the European declination (13°). This suggests
that, when the limestone was deposited, the site lay on the African plate. This and
similar evidence led to the interpretation of the Italian peninsula as a promontory of the
African plate.
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7.
In a paleomagnetic reversal test the normally magnetized samples have a mean direction
DN = 313°, IN = 38°, and the circle of 95% confidence has radius α95 = 8°. The
reversely magnetized samples have DR = 114°, IR = – 33°, and α95 = 7°. Calculate the
angle between the normal and reverse mean directions. Do these directions pass the
reversals test?
The mean direction of the group of normally magnetized samples is a unit vector. Its
direction cosines (lN , mN ,nN ) relative to North, East, and vertically downward axes are
given by
lN = cos I cos D
mN = cos I sin D
nN = sin I
Substituting DN = 313° and IN = 38° in these equations we get the direction cosines
for the normally magnetized samples:
lN = cos 38°cos 313° = 0.5374
mN = cos 38°sin 313° = −0.5763
nN = sin 38° = 0.6157
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while the directions DR= 144° and IR = –33° for the reversely magnetized group have
direction cosines (lR , mR ,nR )
lR = −0.6785
mR = 0.4930
nR = −0.5446
The cosine of the angle ∆ between the normal and reverse mean directions is given by
cos Δ = lN lR + mN mR + nN nR = −0.9841
The angle ∆ is 169.8° so the normal and reverse mean directions are 10.2° from being
exactly opposite. This is greater than the radius of either 95% confidence circle (8° and
7°, respectively). This means that each (inverted) mean direction falls outside the
confidence circle of the other group, so the directions fail the reversals test. This could
be due to an unresolved secondary magnetization component in either group, or to
tectonic disturbance of the sampling sites.
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