GG 450 Spring, 2008

Methods of Geophysics

Fred Duennebier

- call me Fred -

POST 809

Office: x64779

Cell: (808) EXTINCT

email: fred@soest.hawaii.edu

Teaching Assistant: Adrienne Oakley

phone: 956-4235,

POST 821,

email: aoakley@hawaii.edu

Office Hours: Thursdays: 10:20-11:20

WEB SITE

http://www.soest.hawaii.edu/GG/FACULTY/FRED/gg450/

The site will contain:

News

Important Numbers

Old Quizes and Exams

Course Schedule

Class Notes

Scope: We will study geophysical exploration methods of the sub-surface. Half of the term will be on seismology, half on potential field methods, mainly gravity and magnetics. The emphasis is on methods used in exploration for resources and study of shallow planetary structures.

Labs: There are three major labs (3-week) gravity, magnetics, and seismic refraction. Each of these involves a detailed report that you will present as you would to a customer of a geophysical company.

Expectations of students:

Search the Web and understand the terms and concepts to be covered in the next class.

Get homework and labs in ON TIME.

I expect you to know algebra and basic trig, vectors, integration, and differentiation. We will put it to work in practical situations. We will work with differential equations, but nothing difficult.

A working knowledge of MatLab, Excel, Word, and a drawing program, like Illustrator or Canvas, will be a big help.

Keep in mind that we are all geologists, and our primary goal is understanding the earth and planets and the processes that shape them.

Much of what geophysics provides is constraints on models of the earth’s interior that would be difficult or impossible to obtain using other means. We will look at some of the methods used to obtain this information, their advantages and disadvantages, uses, and limitations.

CLASS PICTURE

GG450 Physics Review

A basic understanding of physics is required for geophysics. And more-than-basic understanding is required as one goes deeper into the various disciplines.

So - let's review some basic physics and math. If this is NEW to you, you may have some trouble BUT - don't be scared!

Ignorance is curable, stupidity isn't. Don't be stupid - Ask questions!!!

FIRST - we need to know where we are.

Locating a point in space:y

x

z

R

u

v

Φ

origin

P

How do we describe the location of the point P?

We need a "number" for each "degree of freedom", or each independent direction that P can move in:

If we assume that P lies in the X - Y plane, then we need two numbers:

We could give just the numbers u and v, (Cartesian coords)

or: length of line from origin:

and angle from x axis in x-y plane:

Cylindrical coords if distance along z is specified as a number with R and Φ .

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R = u2 + v2

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tan Φ = v / u

In many situations, such as on the earth, where we have three dimensions, we use R, and Φ for our three numbers, where R is the distance from the center of the earth, is the latitude (or co-latitude), and Φ is the longitude. These are spherical coordinates.All these are equivalent, use

the one that makes the problem easiest. Coordinate transformation from one to the other is fairly straight forward.

Now we know how to locate a point; what if that point is MOVING.

Moving implies what – in terms of P?

(Let's assume that the origin is NOT moving (Eulerian), although moving coordinate systems (Lagrangian) are also very useful.Let's make the problem easy, and say our point is confined to the x axis. What is the value of y? How many degrees of freedom?

Can you think of a body that moves in only one dimension that you ride every day?

How many numbers do we need to locate P? One, the value of u. Except that now u changes with (or is a FUNCTION of time). So we need some numbers to tell us how u changes with time. u, the dependent variable (because its value depends on time) is called the displacement of whatever is at P, such as an atom, car, swimmer, or whatever. Time is called the independent variable (because it doesn't depend on anything….)What if u does not change with time? Then body at P is at rest, or has a zero velocity. What’s the difference between velocity and speed?

If P is MOVING, then it has VELOCITY. Velocity is a vector that tells us in what direction the point P is moving and how fast:

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velocityx = limΔt⇒ 0

Δu

Δt=

du

dt= ˙ u

Look at some curves of displacement vs. time.

What are the velocities associated with these displacement curves? Take a minute to figure this out.

disp

lace

men

t

Since each curve is a straight line, the slopes, and thus the velocities, of each must be CONSTANT. Since curve A is horizontal, it’s location is constant and its velocity is zero.

Curves B abd C have opposite slopes of d/g and -d/g, thus the velocities in the x direction are:

Can we get the displacement if we know the velocity?

We can actually do this by inspection, but we're looking for a formula that expresses u as a function of time. And a useful formula for a straight line is : , where a is the slope, and b is the y intercept (the value of u when t=0).

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u = at + b

Note that the curve “C” in the displacement figure crosses the u axis at d, and the time axis at g. This means that the point P moved -d units of u in g units of time, so the slope is -d/u, and the intercept is d, so, for curve C above:

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u =−d

gt + d

We can also get this by integrating the velocity:

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u = ˙ u dt∫ = −d

gdt∫ = −

d

gt + u0 = −

d

gt + d

This is the same as getting the AREA under a velocity-time curve. If all we have are the velocities, the displacement, d, where we started is unknown. In this case, d is the “constant of integration” given by the initial conditions. If all we know is the velocity, we cannot get the displacement relative to a fixed origin.

What if velocity changes with time?

This is acceleration. Just as displacement changes with time give velocities, velocity changes with time are accelerations.

acceleration =

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limΔt⇒ 0

Δ ˙ u

Δt=

d ˙ u

dt=

d2u

dt2

Acceleration is the slope of a velocity vs. time curve. Acceleration that is negative means that the velocity is decreasing (but – it could be getting faster in the negative direction!). When you apply the brakes of your car, you are accelerating in a negative sense.

In many important situations, acceleration is a constant. The most obvious situation is when a body is at the surface of the earth, where the acceleration of GRAVITY is (nearly) constant:

Where w is the displacement in the vertical direction, and g is called the acceleration of gravity. On earth, g is about 980 cm/s2. 1 gal is 1 cm/s2, and 1 milligal, is 10-3 gals. The milligal is the common unit of gravity measurement - about 1 millionth of the earth's surface gravity.

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˙ ̇ w = g

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˙ w = ˙ ̇ w dt = gdt = gt + ˙ w 0∫∫

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˙ w 0

Velocity:

where is the initial velocity.

Displacement:

where is the displacement when time=0.

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w = ˙ w dt = gt + ˙ w 0( )∫∫ dt =1

2gt2 + ˙ w 0t + w0

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w0

We can integrate acceleration to obtain the velocity of P under the influence of gravity, and integrate again to get the displacement of P in the vertical direction.

These formulas allow us to solve many important physical problems. One of which is direct measurement of gravity by measuring the time of fall of a mass through a known distance:

A mass is to be dropped 1 meter to measure g to 1 milligal. Neglecting friction, how precisely must the drop time and the drop distance be measured?

g=980 gals, 1 gal = 1 cm/s2, 1 mgal=10-

3gal.

From above, ; and

This is the time needed to fall 1 meter.

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w =1

2gt2

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˙ w 0 = w0 = 0

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w = 0.5∗980∗t2 = 100cm

t = 0.45175395sec

If we add one milligal to g, (g=980.001 cm/s), the time needed to fall is 0.45175372 sec. The difference is 0.23 microseconds. So that, if the distance of fall could be measured exactly, the time precision would have to be this good. How far will the mass fall in the additional 0.23 microsec?

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w1 =12g(t+Δt)2

w1 −w=12g(t2 +2tΔt+Δt2)−

12gt2 ≈gtΔt

w1 −w≈980∗.45175395*.23*10−6 =.0001 cm

This is the distance measurement precision needed.

Ballistics

The equations above allow us to solve some pretty interesting problems:

A stone is shot into the air from the top of a 20 m cliff at an angle of 30° from the horizontal at a horizontal velocity, c, of 20m/s. How high will the stone reach? When will it hit the ground? How far from the ground will it hit?

20 m

u

w

c

This problem is easy if it is divided into two parts, noting that there are no forces in the horizontal direction to change the velocity (except friction, which we will ignore), and the acceleration in the vertical direction is -9.80 m/s2 (gravity). We can separate the problem like this because the horizontal and vertical directions are orthogonal, and what happens in one direction does not affect the other. So, working in the vertical direction first,

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w =1

2gt2 + ˙ w 0t + wo

w0 = 20m

˙ w 0 = c∗sin(30°) = 20*1

2= 10m / s

w = −0.5* 9.80* t2 +10t + 20

from this equation, we can calculate the height, w, of the stone at any time , t. But we really want to know the time when the vertical component of velocity is zero, which is when the stone will be at the peak of its arc, and the time when w=0, when the stone hits the ground.

To get the time at the highest point, we differentiate the above to get the velocity, and then see what the time is when the velocity = 0:

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˙ w = −9.80t +10 = 0

t ˙ w =0 =10

9.80= 1.02sec

If we then plug this time back into the equation for w, we get:

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wmax = −0.5* 9.80* (1.02)2 +10*1.02+ 20

wmax = 25.1m above the cliff.

Plotting the paths of objects like this is easy in MatLab - see the file “Ballistic.m”

What we’ve really plotted here are the highest section of the ORBITS of the objects around the center of the earth.

What are “weight”, “force”, “mass”, and “density” ?

Mass: a measure of the amount of actual material present. Measured in grams or kilograms.

Force: something that causes mass to accelerate. Measured in dynes(gm cm/sec2) or newtons (kg m/sec2)

Weight: The force that gravity exerts on a mass.

Density: mass per unit volume

At the earth's surface, weight = g*mass, where g is the acceleration of gravity (as above), and mass is the property of all materials that gives them substance, In orbit, where gravity is not felt since it is balanced by other forces, a body has no weight, but it still has mass.

We often measure the weight of a body using a spring, and measure Weight=Kw . Where K is a constant of the spring, and w is the length that the spring is extended. This is a very simple form of Hooke's Law - which states that the deformation of a body is directly proportional to the force exerted on that body.

In this case, the force or weight caused a spring to get longer, but forces can also cause acceleration: F=ma, where a is the acceleration of the body. Acceleration does not necessarily change the SPEED of a body, it may only change the body's direction of travel. Example: chalk on string.

DENSITY: mass/unit volume. High density - harder to change velocity

Why are things weightless in orbit?

Can you feel forces?

Can you feel accelerations?

What different types of forces are there? gravity, electromagnetic, nuclear, (strong & weak) - What about mechanical?? Chemical??

What is potential energy? Think of it as a stored force - such as a stretched rubber band, a rock on the edge of a cliff, a moving car, a bomb, or a charged battery. Each of these have potential energy.

How do we measure energy? Potential energy is stored energy, such as a rock on a cliff. Kinetic energy is energy of motion, such as the energy of a meteor impact. Energy comes in many forms, mechanical, chemical, nuclear, electrical, geothermal, etc. Oil contains stored chemical energy - originally from the sun. Geothermal energy contains thermal energy from inside the earth. Ocean waves contain energy from wind.

For example:

the energy of motion:

How much more energy does a car going 80 mph have than a car going 40 mph?

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E =1

2mv 2

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E80

E40

=

1

2m(80)2

12

m(40)2 = (80)2 /(40)2 = 6400 /1600 = 4

Potential energy takes many forms, one of the most common being that afforded by a change in elevation of mass, such as water.

Consider 1000 gallons of water in a reservoir. How much energy is available if the water drops through 100 m?

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P.E . = mgh =1000gallons* 3.785kg /gallon *9.8m /s2 *100meters

P.E . = 3,708,300 Joules

How fast would a 100 gram bullet with this energy be moving?

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E = mv 2 /2

v = 2E /m = 2 * 3708300 /0.1 = 8612 m/s

oil:45 mJ/kg, so this energy is about 80 grams of oil….

LAB TODAY!

MEET HERE AT 1:30

THIS LAB WILL BE DONE INSIDE THE POST BUILDING, AND YOU SHOULD FINISH THE NECESSARY MEASUREMENTS BY ~3 PM.