Ali Onceloncel@kfupm.edu.sa

Department of Earth SciencesKFUPM

Gravity Methods 3

Fc

Introduction to GeophysicsIn

trod

uct

ion t

o G

eop

hysi

cs-K

FUPM

Previous Lecture

•Gravity Methods 2•Bouguer Correction •Bouguer Gravity Anomaly on Land •Bouguer Gravity Anomaly on Sea

•Mid-Term Exam

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

0

20

40

60

80

100

1 2 3 4 5

MIDTERM

MIDTERM

Gra

de

Students

Grade Distribution in Mid-Term Exam

2043

17

2142

69

2227

36

2262

02

2343

45

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

Mid-Term Exam Solutions

smV

smV

/100002.0

2000

/30004.0

1200

2

1

12

2112

VV

VVhxcr

300010000

10000300021200 1

h

7000

1300021200 1h

186.1

1200h mh 3.4401

h1? km

Bedrock = V2 = xxxx ft/sec

Alluvium = V1 = xxxx ft/sec ? km

Bedrock = V2 = xxxx ft/sec

Alluvium = V1 = xxxx ft/sec

XCR

h1

V 1

V2

1200 3200

0.4

0.6

Meters

Sec

ond

s

V1, V2, h1?

0.3

0.2

0.1

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

Simplified version of Assignment 2

From Kearey, Brooks, and Hill, 2002

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

For example, a double spike function, 2, 0, 1 convolved with an impulse response function 4, 3, 2, 1 .

It was class example of Lecture 9.2

34

pv

sv3

3

/2950

/95.2

mkg

cmg

GPamkgsmv s 04.43)/2950()/3820( 322

GPa

Pamkgsm

vk P

02.79

)10*304.4(3

4)/2950()/6800(

3

4

1032

2

Given that the P-velocity of a rock is 6.8 km/s, the S-velocity is 3.82 km/s, and density 2.95 g/cm3, calculate µ and k?

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

Modified version asked in Quiz 1

?=2.08377.21.8258.4(c)

3.82?=55.332.5533.3(b)

3.272.54.36?=?(a)

V2 (km/sec)R (° )V1 (km/sec)I (°)Case

?=2.08377.21.8258.4(c)

3.82?=55.332.5533.3(b)

3.272.54.36?=?(a)

V2 (km/sec)R (° )V1 (km/sec)I (°)Case

Where possible, supply the missing quantity in each case in the following table, where: i = angle of incidence r = angle of refraction v1 = wave speed in upper layer, v2 = wave speed in lower layer. The wave originates in the upper layer.

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

Modified version asked in Quiz 2

The same question solved in Class

21

11

2,

n

iii

n

iinrms ttvv v1=1500 m s-1

v2=2000 m s-1

v3=2345 m s-1

t1=2.14 s

t2=1.21 s

t3=1.13 s

vrms at the base of layer 3?

1-s m 064.1882

13.121.114.2

)13.12345()21.12000()14.21500( 21

222

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

vrms at the base of layer 1?

1-s m150014.2

)14.21500( 21

2

1-s m 672.1697

21.114.2

)21.12000()14.21500( 21

22

vrms at the base of layer

2?

v1=2.5 km s-1

ρ1 =2.5 g/ cc

v2=1.7km s-1

ρ2 =2.02 g/ cc

v3=2.2 km s-1

ρ3 =2.27 g/ cc

Shale

Sand + gas

Sand + oil

v4=2.3 km s-1

ρ4 =2.28 g/ ccSand + oil

h1=100m

h2=200m

h3=150m

Explosion

v1=2.5 km s-1

ρ1 =2.5 g/ cc

v2=1.7km s-1

ρ2 =2.02 g/ cc

v3=2.2 km s-1

ρ3 =2.27 g/ cc

Shale

Sand + gas

Sand + oil

v4=2.3 km s-1

ρ4 =2.28 g/ ccSand + oil

h1=100m

h2=200m

h3=150m

Explosion

29.0)5.2(5.2)02.2(7.1

)5.2(5.2)02.2(7.1

1122

112212

vv

vvR

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

v1=2.5 km s-1

ρ1 =2.5 g/ cc

v2=1.7km s-1

ρ2 =2.02 g/ cc

v3=2.2 km s-1

ρ3 =2.27 g/ cc

Shale

Sand + gas

Sand + oil

v4=2.3 km s-1

ρ4 =2.28 g/ ccSand + oil

h1=100m

h2=200m

h3=150m

Explosion

v1=2.5 km s-1

ρ1 =2.5 g/ cc

v2=1.7km s-1

ρ2 =2.02 g/ cc

v3=2.2 km s-1

ρ3 =2.27 g/ cc

Shale

Sand + gas

Sand + oil

v4=2.3 km s-1

ρ4 =2.28 g/ ccSand + oil

h1=100m

h2=200m

h3=150m

Explosion

19.0)02.2(7.1)27.2(2.2

)02.2(7.1)27.2(2.2

2233

223323

vv

vvR

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

v1=2.5 km s-1

ρ1 =2.5 g/ cc

v2=1.7km s-1

ρ2 =2.02 g/ cc

v3=2.2 km s-1

ρ3 =2.27 g/ cc

Shale

Sand + gas

Sand + oil

v4=2.3 km s-1

ρ4 =2.28 g/ ccSand + oil

h1=100m

h2=200m

h3=150m

Explosion

v1=2.5 km s-1

ρ1 =2.5 g/ cc

v2=1.7km s-1

ρ2 =2.02 g/ cc

v3=2.2 km s-1

ρ3 =2.27 g/ cc

Shale

Sand + gas

Sand + oil

v4=2.3 km s-1

ρ4 =2.28 g/ ccSand + oil

h1=100m

h2=200m

h3=150m

Explosion

024.0)27.2(2.2)28.2(3.2

)27.2(2.2)28.2(3.2

3344

334434

vv

vvR

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

Recall GradingIn

trod

uct

ion t

o G

eop

hysi

cs-K

FUPM

Final Exam 35%Mid Term Exam 25%Quiz 15%Homework 15%Class Participation 10%

Homework StatusIn

trod

uct

ion t

o G

eop

hysi

cs-K

FUPM

Homework StatusIn

trod

uct

ion t

o G

eop

hysi

cs-K

FUPM

Homework StatusIn

trod

uct

ion t

o G

eop

hysi

cs-K

FUPM

Summary of Equations for Free Air and Bouguer Gravity Anomalies

Standard parameters used to compute gravity anomalies on land and at sea.

•FAC =Free-Air Correction; BC =Bouguer Corrections;

•BCs =Bouguer correction at sea

•ρ =reduction density; h =elevation (m) and hw =water depth (m)In

trod

uct

ion t

o G

eop

hysi

cs-K

FUPM

Bouguer Gravity Anomaly on SeaIn

trod

uct

ion t

o G

eop

hysi

cs-K

FUPM

Δgfa= g – gt + FAC

ΔgB= Δgfa-BC BC= 0.419 ρ h

FAC= h x (0.308 mGal/m)

Terrain CorrectionIn

trod

uct

ion t

o G

eop

hysi

cs-K

FUPM

ΔgBC= ΔgB+TC

terrain correction is no needed in areas of low relief but is required where changes in topography is significant and equation for a complete Bouguer anomaly is:

Terrain correction is needed in areas of high relief in order to account decrease of observed gravity due to

1. mountains above the slab (1),

2. and overcorrection due to valleys (2).

Terrain CorrectionIn

trod

uct

ion t

o G

eop

hysi

cs-K

FUPM

ΔgBC= ΔgB+TC

Terrain Correction

For both situations, the terrain correction is positive, making the complete Bouguer anomaly higher than the simple Bouguer anomaly.

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

Gravity MeasurementIn

trod

uct

ion t

o G

eop

hysi

cs-K

FUPM

True gravitational acceleration (ρ)

Reflecting the difference in gravitational acceleration (Δρ) at one station (ρ1) compared to another (ρ2).

Measurement of relative gravity

A gravimeter measures the length of a spring (L), which is proportional to the gravitational acceleration (g).

+ΔL Increase in gravity-ΔL Decrease in gravity

Map of relative gravity survey. The traverse starts with a measurement at the base station, the each of the 16 stations, followed by re-measurement at the base station.In

trod

uct

ion t

o G

eop

hysi

cs-K

FUPM

Microgravity surveys require that we measure g with

precision of better than 1 part in 40 million

• we measure either absolute gravity or more commonly, relative gravity (change in g between two stations)

Methods:

1. Falling body

2. Pendulums

3. Mass-spring gravimeters

Gravity MeasurementIn

trod

uct

ion t

o G

eop

hysi

cs-K

FUPM

Gravitational acceleration can be measured directly by dropping an object and measuring its acceleration.

Free Fall Methods

• distance a body falls is proportional to the time it has fallen squared

z = ½gt2

g = 2z/t2

ginst = 2 ·50 m/(3.19 s)2

= 9.8 m/s2

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

PendulumsGravity first measured by Pierre Bouguer in 1749

(257 years ago) using Pendulum: period of oscillation T of pendulum is inversely proportional to g

T = 2K/gK = constant related to pendulum design

T

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

Gravity Units

212

E

E

sE

R

mG

m

Fg

Most of us are familiar with the units of g as feet/sec2 or meters/sec2, etc.

From Newton’s law of gravity g also has units of

kilogram

newtonsEg

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

Using the metric system, we usually think of g as being 9.8 meters/sec2.This is an easy number to recall. If, however, we were on the Martian moon Phobos, gp is only about 0.0056meters/sec2 [m/sec2] might not be the most useful units to use on Phobos.

Some unit names you will hear when gravity applications are discussed include:

9.8 m/sec2

980 Gals (or cm/sec2)980000 milligals (i.e. 1000th of a Gal)

We experience similar problems in geological applications, because changes of g associated with subsurface density contrasts can be quite small.

Gravity Units

http://www.esa.int/SPECIALS/Mars_Express/SEM21TVJD1E_0.html

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

1 mGal = 10 microns/sec2

1 milligal equals 10-5 m/sec2 or conversely

1 m/sec2 = 105 milligals.

The gravity on Phobos is 0.0056m/s2 or 560 milligals.

Are such small accelerations worth contemplating?

Can they even be measured?

GalGal = 1 cm/sec2= 0.01 m/sec2

1 mGal = 10-3 Gal

Istituto e Museo di Storia della Scienza

Intr

od

uct

ion t

o G

eop

hysi

cs-K

FUPM

1564-1642