Module 1-3A
The Earth’s Shells
A. Thicknesses and Densities
Finding the density of the Earth was an
immense achievement. It was the first
quantitative indication that the interior
of the Earth is different from the rocks
on the surface.
Any theory of the interior of the
Earth must be consistent with the fact
that its aggregate density is 5.5 g/cm3.
In other words, r = 5.5 g/cm3 is a
constraint on theories of the internal
constitution of the Earth.
Quantitative Concepts and Skills
Weighted average
The nature of a constraint
Volume of spherical shells
Concept that an integral is a sum
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The Earth’s Shells
You have probably learned in
Introduction to Geology that the
Earth consists of concentric
shells: an outermost crust, a
thick shell called the mantle, and
an interior core. You probably
also learned that the outer core
is liquid and the inner core is
solid.
Crust
Mantle
Outer core
Inner core
Knowledge that the Earth is composed of these shells comes from the
interpretation of travel times of seismic waves. Earthquakes occur near the
surface of the Earth (up to depths of ~700 km), and so seismic waves can
travel from one side of the Earth to the other, where their arrivals are recorded
by seismographs. The travel times give an indication of the density of the
material that the seismic waves pass through.
This module explores the combination of densities and shell thicknesses that
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produce an aggregate density of the Earth of 5.5 g/cm3.
PREVIEW
The core mathematics of this module is the weighted average. We want the
overall average density of the Earth. We cannot calculate the overall average
density as the simple average of the shell densities. The mantle is much larger
than the crust, for example, and so the density of the mantle contributes much
more to the density of the earth than the density of the crust does.
Going into the calculation, we do not know either the thickness of the four shells
or their densities. To start, Slides 3 and 4 use a weighted average where the
densities are mere guesses, and the thicknesses are equal (1/4 of the radius of
the Earth). The calculation in Slide 4 uses thickness as the weighting variable in
the weighted average, and Slides 5 and 6 discuss why that is an incorrect
approach.
Slides 7 and 8 gets us on the right track: the weighting parameter must be the
volume of the shells, and slide 10 performs the calculation for the same
combination of thicknesses and densities as in Slide 4. The result is not 5.5
g/cm3, however, and so the rest of the module does follow-up calculations
adjusting the shell thicknesses and densities until the bottom line of the
spreadsheet produces the correct 5.5 g/cm3 for the overall average.
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PROBLEM
Given that the density of the earth is 5.5 g / cm3, what
are the densities of the crust, mantle, inner core, and
outer Core?
Don’t look this up in a book (not yet anyway). For now,
just make some guesses. Remember that the radius of
the Earth is 6,370 km, so the shell thicknesses have to
sum to that value.
The simplest guess, to set
up the spreadsheet, is to
make all the shells the
same thickness.
Here are some guesses
Shell
Density (g/cm3)
Thickness (km)
Crust
2.8
1592.5
Mantle
5.0
1592.5
Outer Core
7.0
1592.5
Inner Core
9.0
1592.5
Make a spreadsheet
that calculates the
average density
from these guesses
for the shell
thicknesses and
densities.
4
Here is one approach: Calculate the aggregate density as the
weighted average of the densities using thickness as the
weighting factor.
B
C
D
Shell
Thickness (km) Density (g/cm 3)
crust
1592.5
2.8
mantle
1592.5
5
outer core
1592.5
7
inner core
1592.5
9
2
3
4
5
6
7
8 Sum (km)
6370
9 Sumproduct (km-g/cm3)
10 weighted average (g/cm 3)
37901.5
5.95
This spreadsheet uses the
SUMPRODUCT function
(Cell D9). Alternatively, you
can add Column E for the
product of Columns C and
D in Rows 3-6, and then
sum all those products.
The result would be the
same as in Cell D9.
Note, the result (5.95 g/cm3) does not conform to the
constraint (5.5 g/cm3).
More importantly, the approach is not correct.
Why is this an incorrect approach?
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The weighting factor should be volume, not thickness.
Explanation:
Density is mass over volume,
M
r Earth  Earth
VEarth
(1)
The mass of the Earth is the sum of the masses of all of the shells,
M Earth 
Vi r i
shells
(2)
The volume of the Earth is the sum of the volumes of all of the shells,
VEarth 
Vi
(3)
shells
Combining the three equations produces the weighted average
r Earth 
Vi ri
shells
Vi
(4)
shells
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The calculation that used shell thickness as the weighting factor is incorrect
because it assumes that the areas of all of the shells are the same.
Explanation: Starting with the basic equation,
r Earth 
Vi ri
shells
Vi
shells
you can get to the weighted average using thickness (b ) as the weighting factor
r average 
 bi ri
katers
 bi
layers
if each of the volumes has the same area (i.e., Ai = A ). Thus
Vi ri  b i Ai ri
i
Vi
i

i
 b i Ai
i
This will work for a stack of
rectangular layers, but it
won’t work for concentric
shells. Why?

A b i r i
i
A b i
i

 bi ri
i
 bi
i
Layers vs. shells
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To use
r Earth 
Vi ri
shells
Vi
you need to calculate the volume of spherical shells.
shells
Problem: Find the volume of a spherical shell with inner radius = r1, and
outer radius = r2.
m icroshell
thickness = dr
r1
r2
Strategy: Divide the shell into a bazillion
microshells each with incredibly small
thickness dr. Find the volume of each of
the microshells and add them up.
Let r be the internal radius of any given microshell.
Recall the area of a spherical shell with radius r:
Vi  4ri2 dr
Then, write down equation for volume of each microshell
r2

4
3
3
And, add them up: V   4ri dr   4r dr   r2  r1
3
microshells
r
2
2
1
A  4r 2

An integral is a sum.
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Thickness, dr, is infinitesimal.
Now that we have the formula for the volume of a spherical
shell, we can lay out a spreadsheet that correctly calculates
the average density (using the same guessed values for the
thickness and density of the shells).
B
2
3
4
5
6
7
8
9
10
11
12
C
depth to
base (km)
surface
0
crust
1592.5
mantle
3185
outer core
4777.5
inner core
6370
D
r1
E
thickness
(km)
6370
(km)
Sum
Sumproduct (km3-g/cm3)
weighted average (g/cm 3)
1592.5
1592.5
1592.5
1592.5
F
volume
(km3)
G
density
(g/cm3)
2.8
5
7
9
6370
9
B
2
3
4
5
6
7
8
9
10
11
12
C
D
r1
depth to
base (km)
surface
0
crust
1592.5
mantle
3185
outer core
4777.5
inner core
6370
(km)
6370
4777.5
3185
1592.5
0
Sum
3
3
Sumproduct (km -g/cm )
weighted av erage (g/cm 3)
E
F
G
thickness
volume
density
(km)
(km )
(g/cm )
1592.5
1592.5
1592.5
1592.5
6.26E+11
3.21E+11
1.18E+11
1.69E+10
2.8
5
7
9
6370
1.08E+12
3
3
What a difference.
The prior result
was 5.95 g/cm3.
4.34E+12
4.01
Note, the result (4.01 g/cm3) still does not conform to the
constraint (5.5 g/cm3), but at least the result is correct for the
assumptions.
What we need now are better values for the depth to the shell
boundaries and the density of the shells.
Time to consult a geology textbook.
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Here are values from Fowler, C.M.R., 1990, The Solid Earth: An Introduction
to Global Geophysics, Cambridge University Press, p. 112.
Shell
Density (g/cm3)
Depth to base (km)
Crust
2.6-2.9
50
Mantle
3.38-5.56
2891
Outer Core
9.90-12.16
5150
Inner Core
12.76-13.08
6371
Redo the spreadsheet using these textbook values. Assume the density of the
crust and inner core are 2.8 and 13 g/cm3, respectively.
B
2
3
4
5
6
7
8
9
10
11
12
C
depth to
base (km)
surface
0
crust
50
mantle
2891
outer core
5150
inner core
6371
D
r1
E
thickness
(km)
6371
(km)
Sum
Sumproduct (km3-g/cm3)
weighted average (g/cm 3)
F
volume
(km3)
G
density
(g/cm3)
What do these values
have to be …
2.8
13
for this value to come
out to 5.54 g/cm3?
Trial and Error
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Using trial and error, you probably
had several wrong answers before
you came up with one that conforms
to the constraint (Cell G12).
B
2
3
4
5
6
7
8
9
10
11
C
D
r1
depth to
base (km)
surf ace
0
crust
50
mantle
2891
outer core
5150
inner core
6371
(km)
6371
6321
3480
1221
0
Sum
3
3
Sumproduct (km -g/cm )
3
12 weighted average (g/cm )
Does your answer match
this or did you find other
values that work? What are
some other values that
conform to the constraint?
E
F
G
thickness
v olume
density
(km)
(km )
(g/cm )
50
2841
2259
1221
2.53E+10
8.81E+11
1.69E+11
7.62E+09
2.8
4.6
10.5
13
6371
1.08E+12
3
3
6.00E+12
5.54
Columns B, C, and G comprise a model for the density structure of the Earth.
The constraint allowed you to eliminate many possible models. Still, there are
many other possibilities that have not been eliminated.
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End of Module
Assignments
1.
Answer the question in Slide 7. Expand your answer to include an explanation of why
the result in Slide 10 (4.01 g/cm3) ) is smaller than the result in Slide 5 (5.95 g/cm3)
2.
According to Dante’s Inferno, Hell is at the center of the Earth. Assume that Hell has a
radius of 1000 km, and that the rest of the Earth is the density of normal rocks (say 2.8
g/cm3). What would the overall density of the Earth be? What would g be?
3.
Suppose that a planet consists of two shells of equal thickness. Derive a formula for the
density of the planet as a function of three variables: R, the radius of the planet, and r1
and r2, the densities of the two shells. Test your formula by using a modification of the
spreadsheet in Slide 12. Does the order of the densities matter? In other words, if you
exchange the value of r1 with that of r2, does the overall average change? Explain.
4.
Find three possible answers for the densities of the mantle and outer core that result in
the correct overall average (Slide 12).
5.
Slide 12 is an example of forward modeling: guessing values of the parameters that
produce a known end result. Write a paragraph on forward modeling, and include the
words “constraint” and “unique solution.”
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