FUNDAMENTALS of GEOPHYSICS
SOLUTIONS TO EXERCISES
William Lowrie
Zürich
2007
, CONTENTS
TOPIC PAGE
Significant digits in a calculation 1
Exercises to
1. The Earth as a planet 3
2. Gravity, the figure of the Earth, and geodynamics 22
3. Seismology and the internal structure of the Earth 45
4. Earth's age, thermal and electrical properties 77
5. Geomagnetism and paleomagnetism 113
NOTE
Individual sections of these solutions were checked by the following colleagues, who drew
attention to errors and inconsistencies. I wish to 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 responsibility for any errors that may still be present in the solutions is mine. I would
be grateful for having them drawn to my attention, and especially so if the correct solution –
or a better one – is provided!
, SIGNIFICANT DIGITS
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
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
1
,(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..162 should have 3 significant digits; the answer is 5.57.
5.23 sin(62°) should have 3 significant digits; the answer is 4.62.
2 pounds of flour, each weighing 453 gm (3 significant digits), together weigh 906 gm.
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”.
2
, §1 EXERCISES
The Earth as a planet
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.
It is convenient to convert the angular positions of the moon to decimal values:
1 = 11° 57’ = 11.950° and 2 = 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 0 = (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
= (1 – 0) = (11.955–10.139)=1.816°.
3
, 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
a R
d = s + R cos(60) = +
tan ( / 2 ) 2
3R R R 3
d= + = + 1
2 tan ( / 2 ) 2 2 tan ( / 2 )
Inserting numerical values: = 1.816° from the observations and R = 6371 km, gives
6371 3
d= + 1 = 351, 313
2 tan ( 0.908° )
The approximate distance of the Moon from the Earth is estimated by these
measurements to be 351,000 km.
[Note: This estimate is about 9% too low. What might be important sources of error?]
4
,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
x 2 y2
2
+ 2 = 1 , where b 2 = a 2 (1 e2 ) .
a b
Let the mean lunar orbital radius be rL = (a + b) / 2 . Inserting b = a 1 e2 gives
(
2rL = a 1 + 1 e2 )
2rL
a=
(1 + 1 e2 )
Inserting the numerical values rL = 384.1 • 103 km and e = 0.0549 gives the semi-
major 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.
5
, (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?
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 2, so that angle POM equals . 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,
RL
sin( ) =
( rL R )
RL = ( rL R ) sin( )
Now insert the following numerical values:
(rL R) = 384,100 6, 371 = 377, 629km
2 = 31m 36.8s = 0.526889°
= 0.26344°
The radius of the Moon is found to be RL = 1736 km.
6
SOLUTIONS TO EXERCISES
William Lowrie
Zürich
2007
, CONTENTS
TOPIC PAGE
Significant digits in a calculation 1
Exercises to
1. The Earth as a planet 3
2. Gravity, the figure of the Earth, and geodynamics 22
3. Seismology and the internal structure of the Earth 45
4. Earth's age, thermal and electrical properties 77
5. Geomagnetism and paleomagnetism 113
NOTE
Individual sections of these solutions were checked by the following colleagues, who drew
attention to errors and inconsistencies. I wish to 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 responsibility for any errors that may still be present in the solutions is mine. I would
be grateful for having them drawn to my attention, and especially so if the correct solution –
or a better one – is provided!
, SIGNIFICANT DIGITS
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
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
1
,(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..162 should have 3 significant digits; the answer is 5.57.
5.23 sin(62°) should have 3 significant digits; the answer is 4.62.
2 pounds of flour, each weighing 453 gm (3 significant digits), together weigh 906 gm.
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”.
2
, §1 EXERCISES
The Earth as a planet
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.
It is convenient to convert the angular positions of the moon to decimal values:
1 = 11° 57’ = 11.950° and 2 = 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 0 = (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
= (1 – 0) = (11.955–10.139)=1.816°.
3
, 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
a R
d = s + R cos(60) = +
tan ( / 2 ) 2
3R R R 3
d= + = + 1
2 tan ( / 2 ) 2 2 tan ( / 2 )
Inserting numerical values: = 1.816° from the observations and R = 6371 km, gives
6371 3
d= + 1 = 351, 313
2 tan ( 0.908° )
The approximate distance of the Moon from the Earth is estimated by these
measurements to be 351,000 km.
[Note: This estimate is about 9% too low. What might be important sources of error?]
4
,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
x 2 y2
2
+ 2 = 1 , where b 2 = a 2 (1 e2 ) .
a b
Let the mean lunar orbital radius be rL = (a + b) / 2 . Inserting b = a 1 e2 gives
(
2rL = a 1 + 1 e2 )
2rL
a=
(1 + 1 e2 )
Inserting the numerical values rL = 384.1 • 103 km and e = 0.0549 gives the semi-
major 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.
5
, (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?
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 2, so that angle POM equals . 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,
RL
sin( ) =
( rL R )
RL = ( rL R ) sin( )
Now insert the following numerical values:
(rL R) = 384,100 6, 371 = 377, 629km
2 = 31m 36.8s = 0.526889°
= 0.26344°
The radius of the Moon is found to be RL = 1736 km.
6
