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The Earth’s Shells

Quantitative Concepts and SkillsWeighted averageThe nature of a constraintVolume of spherical shellsConcept that an integral is a sum

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, = 5.5 g/cm3 is a constraint on theories of the internal constitution of the Earth.

Module 1-3A

A. Thicknesses and Densities

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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.

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 produce an aggregate density of the Earth of 5.5 g/cm3.

The Earth’s Shells

Crust

Mantle

Outer core

Inner core

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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.

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

The simplest guess, to set up the spreadsheet, is to make all the shells the same thickness.

Make a spreadsheet that calculates the average density from these guesses for the shell thicknesses and densities.

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Here is one approach: Calculate the aggregate density as the weighted average of the densities using thickness as the weighting factor.

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?

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.

B C D

2 Shell Thickness (km) Density (g/cm3)3 crust 1592.5 2.84 mantle 1592.5 55 outer core 1592.5 76 inner core 1592.5 978 Sum (km) 6370

9 Sumproduct (km-g/cm3) 37901.5

10 weighted average (g/cm3) 5.95

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The weighting factor should be volume, not thickness.

Explanation:

Density is mass over volume,

(1)

The mass of the Earth is the sum of the masses of all of the shells,

(2)

The volume of the Earth is the sum of the volumes of all of the shells,

(3)

Combining the three equations produces the weighted average

(4)

Earth

EarthEarth V

M

shells

iiEarth VM

shells

iEarth VV

shellsi

shellsii

Earth V

V

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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.

shellsi

shellsii

Earth V

V

Explanation: Starting with the basic equation,

you can get to the weighted average using thickness () as the weighting factor

if each of the volumes has the same area (i.e., Ai = A ). Thus

ii

iii

ii

iii

iii

iiii

ii

iii

A

A

A

A

V

V

layersi

katersii

average

This will work for a stack of rectangular layers, but it won’t work for concentric shells. Why?

Layers vs. shells

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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.

m ic ro she llth ic kne ss = d r

r1r2

shellsi

shellsii

Earth V

V To use you need to calculate the volume of spherical shells.

Problem: Find the volume of a spherical shell with inner radius = r1, andouter radius = r2.

Let r be the internal radius of any given microshell.

Then, write down equation for volume of each microshell drrV ii24

And, add them up: 3132

22

3

444

2

1

rrdrrdrrVr

rsmicroshelli

Recall the area of a spherical shell with radius r:24 rA

An integral is a sum.

Thickness, dr, is infinitesimal.

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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 C D E F G

2 depth to r1 thickness volume density

3 base (km) (km) (km) (km3) (g/cm3)4 surface 0 63705 crust 1592.5 1592.5 2.86 mantle 3185 1592.5 57 outer core 4777.5 1592.5 78 inner core 6370 1592.5 99

10 Sum 637011 Sumproduct (km3-g/cm3)

12 weighted average (g/cm3)

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B C D E F G

2 depth to r1 thickness volume density

3 base (km) (km) (km) (km3) (g/cm3)4 surface 0 63705 crust 1592.5 4777.5 1592.5 6.26E+11 2.86 mantle 3185 3185 1592.5 3.21E+11 57 outer core 4777.5 1592.5 1592.5 1.18E+11 78 inner core 6370 0 1592.5 1.69E+10 9910 Sum 6370 1.08E+12

11 Sumproduct (km3-g/cm3) 4.34E+12

12 weighted average (g/cm3) 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.

What a difference.The prior resultwas 5.95 g/cm3.

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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.

What do these values have to be …

for this value to comeout to 5.54 g/cm3?

Trial and Error

B C D E F G

2 depth to r1 thickness volume density

3 base (km) (km) (km) (km3) (g/cm3)4 surface 0 63715 crust 50 2.86 mantle 28917 outer core 51508 inner core 6371 139

10 Sum11 Sumproduct (km3-g/cm3)

12 weighted average (g/cm3)

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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).

Does your answer match this or did you find other values that work? What are some other values that conform to the constraint?

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.

B C D E F G

2 depth to r1 thickness volume density

3 base (km) (km) (km) (km3) (g/cm3)4 surface 0 63715 crust 50 6321 50 2.53E+10 2.86 mantle 2891 3480 2841 8.81E+11 4.67 outer core 5150 1221 2259 1.69E+11 10.58 inner core 6371 0 1221 7.62E+09 139

10 Sum 6371 1.08E+1211 Sumproduct (km3-g/cm3) 6.00E+12

12 weighted average (g/cm3) 5.54

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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 1

and 2, 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 1 with that of 2, 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.”