Geomath

Tom Wilson, Department of Geology and Geography

[email protected].

eduDept. Geology and

GeographyWest Virginia University

Start reviewing materials for the final!… current to-do list


1. Problem 9.7 is due today

2. Hand in the gravity computation

3. I will give you till next Tuesday to finish up problems 9.9 and 9.10

4. Start reviewing class materials. Next week is a final review week

No class this Thursday

Take advantage of this day off and …


Turn in any late assignments by Friday afternoon, April 26th.

Put all late assignments in my mailbox (mailroom, 3rd floor Brooks)

In this simple example, we didn’t have to employ calculus


2

GMg

r This formula can be evaluated – as is – for

points and equidimensionally shaped objectsg

gv

Sulfide deposit

r z

x

Sulfide minerals


L. Morgan, 2010, Geophysical characteristics of volcanogenic massive sulfide deposits; USGS report 210-5070-C, 19p.

Excess density or high horizontal density contrast produces observable changes in the g



In this location the gravity anomaly was less distinctive than other geophysical features



In this area, a distinctive gravity anomaly is associated with a sulfide deposit


M. Thomas, 1997, Gravity prospecting for massive sulfide deposits in the Bathurst mining camp, New Brunswick Canada: Proceedings of Exploration 97, Fourth Decenial International Conference on Mineral Exploration, p837-840.

However, we had to modify it algebraically since we wanted to solve for gv and express as a function of x & z (rather than r)


•Find r as a function of x and z

• Rewrite g in terms of x and z

• Express gv in terms of g and

• Replace cos with its spatial equivalent to get gv as a function of x & z

• Simplify by factoring z out of the denominator

We looked at the geometry and did this in a series of steps

2

GMg

r

In the form below it is easy to compute gv at arbitrary x along the surface.


32 2

22 1

v

GMg

xz

z

Given that G=6.6732 x 10-11nt-m2/kg2, x=1km, z=1.7km, Rdeposit=0.5km and =2gm/cm3, you

would find that

Spatial variation in the gravity anomaly over the sulphide deposit


Note than anomaly is symmetrical across the sulphide accumulation

One of the first problems you did in the class was a units conversion problem for acceleration


gv at 1 km is 0.0000155 m/sec2

Remember what a Gal is?

How about a milliGal?

Make the units conversion from m/sec2 to milliGals

The same governing equation but with more complex geometry could be used to calculate gcore and gmantle


11,000 kg/m3

Approximate the average densities

4,500 kg/m32

GMg

r

Here, the gravitational field is associated with shells of differing density and the problem is a

little more complex.

Another slant on the text problem

In the book problem we are just trying to estimate the mass of the earth and simplify the problem by

assuming the earth can be represented by two regions:


11,000 kg/m3

Approximate the average densities

4,500 kg/m3

1) an inner core of average density, i, and 2) an outer shell (mantle and crust) represented by

another average density, o.

What do we get when we integrate the surface area over r?


11,000 kg/m3

We can simplify the problem and still obtain a useful result. Approximate the average

densities

4,500 kg/m3

2

1

4N

i ii

M r r

2

04

RM r dr

1

Surface area of shell density of shell N

i

M r

Area of sphere = volume of shellr

Actually a pretty good approximation


11,000 kg/m3

We can simplify the problem and still obtain a useful result. Approximate the average

densities

4,500 kg/m3

3480 63712 2

0 34804 .11000 4 .4500r dr r dr

2

04

RM r dr

34

3r C

3480 63713 3

1 20 3480

4 4

3 3r r

3480 63713 3

20 3480

44000 18000

3 3r r

The result – 6.02 x 1024kg is close to the generally accepted

value of 5.97 x 1024kg.

We could then pose the question: what is the acceleration of gravity due to the core at the Earth’s surface?


3480 2

04 .11000r dr

34803

10

4

3r

34803

0

44000

3coreM r

Mass of core ~ 1.94 x 1024kg.

Considering only the core, we find it’s mass is

1.94186x1024 kg (about 1/3rd the total mass of the

earth.

With an outer radius of ~6371km (6,371,000m)


2

GMg

r

The core is about 2900km beneath your feet,

We have to keep units consistent and use

G=6.6732x10-11 m3/(kg-sec2)

M=1.94186x1024 kg

And r=6,371,000 m

The contribution to the total acceleration of ~9.8 m/s2 due to the core is 3.29m/s2.

In general we express the acceleration of gravity produced by an object of arbitrary shape as


2 2

GM dMg G

r r

We usually look for some symmetry to help simplify our problem.

Let’s take a look at the acceleration produced by a very long horizontal cylinder


This could be a cave passage or tunnel. Point of observation

m

rr+dr

dx

2 2 2

dM dV dxdydzG G G

r r r

In this example, we can let the cross sectional area = dydz


Point of observation

m

rr+dr

dx

2 2

dxdydz AdxG G

r r

Again, we are interested in the vertical component of g, so

2cos

AdxG

r

Zoom in on the little element dx


rr+dr

Area = R2

R

cos

rddx

Substitute for dx, simplify and also note that r=m/cos



m

rr+dr

dx

2

cos

cos

Ard AdG G

r r

cosv

Adg G

m

Note that the only variable left is and the limits of integration would be from -/2 to /2



m

r=m/cosr+dr

dx=rd/cos

cosv

Adg G

m

/2

/2cosv

Ag G d

m

This is an integral you should be able to evaluate


/2

/2cosv

Ag G d

m

What do you get?

Assume that you run a gravity survey across a roughly cylindrically shaped cave passage


Hint: replace m with r to develop this relationship

ggv

Cave Passage

r z

x

/2

/2cosv

Ag G d

m

Cylinder goes in and out of the slide

Developing g as a function of x and z


Hint2: Once again – take the vertical component!

ggv

Cave Passage

r z

x

/2

/2cosv

Ag G d

m


Li

Lf

, the elongationi

Le

L

f

i

L

L

dL

L ln f

i

L

LL

ln( ) ln( )

ln

f i

f

i

L L

L

L

ln( )S

The total natural strain, , is the sum of an infinite number of infinitely small extensions

In our example, this gives us the

definite integral

Where S is the Stretch

1 , the stretchS e


1 1f i f

i i i

L L LLe S

L L L

ln( ) ln(1 )S e

2 3 4

...2 3 4

e e ee

f

i

L

L

dL

L

Strain (or elongation) (e), stretch (S) and total natural strain ()

Elongation

Total natural strain

expressed as a series expansion of ln(1+e)

The six term approximation is accurate out to 5 decimal places!

Comparison of finite elongation vs. total natural strain



Volume of the earth – an oblate spheroid

22 221ep

zr rr

In this equation r varies from re, at the equator, to r=0 at the poles. z represents distance along the earth’s rotation axis and varies from –rp to rp.

The equatorial radius is given as 6378km and the polar radius, as

6457km.

Problem 9.10


0 exp( / )t t x X

In this problem, we return to the thickness/distance relationship for the

bottomset bed.

Problems 9.9 and 9.10 will be due next Tuesday

Don’t forget to hand in the answer to the gravity problem!


What was the gravitational acceleration produced by the sulfide deposit?

Start reviewing materials for the final!… current to-do list


1. Problem 9.7 is due today

2. Hand in the gravity computation

3. I will give you till next Tuesday to finish up problems 9.9 and 9.10

4. Start reviewing class materials. Next week is a final review week

No class this Thursday

Geomath

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