Environmental and Exploration Geophysics II
Gravity Methods (II)
tom.h. wilson
wilson@geo.wvu.edu
Department of Geology and Geography
West Virginia University
Morgantown, WV
Subsurface properties can be inferred directly from
measured apparent resistivity and conductivity data.
However, much must be done to gravity observations
before it can be related to specific subsurface geologic
problems.
Last week we briefly discussed several factors that
influence the measured or observed value of
gravitational acceleration at given points on the earth’s
surface.
These influences are generally unrelated to subsurface
geology or site conditions. It is critical that these nongeological effects be eliminated prior to the modeling
and geologic interpretation of gravity data.
To conceptualize the dependence of
gravitational acceleration on various
factors, we usually write g as a sum of
different influences or contributions.
These are -
gn the normal gravity of the gravitational
acceleration on the reference ellipsoid
gE the elevation effect
gB the Bouguer plate effect or the contribution to
measured or observed g of the material between sealevel and the elevation of the observation point
gT the effect of terrain on the observed g
gTide and Drift the effects of tide and drift (often
combined)
These different terms can be combined into an
expression which is equivalent to a prediction of what
the acceleration should be at a particular point on
the surface of a homogeneous earth.
Thus when all these factors are
compensated for, or accounted for, the
remaining “anomaly” is associated with
lateral density contrasts within area of the
survey.
The geologist/geophysicist is then left with the
task of interpreting/modeling the anomaly in
terms of geologically reasonable configurations of
subsurface intervals.
That predicted or estimated value of g is often referred to
as the theoretical gravity - gt
gt  g n ( )  g E  g B  gt  gTide  Drift
If the observed values of g behave according to this
ideal model then there is no geology! - i.e. there is no
lateral heterogeneity. The geology would be fairly
uninteresting - a layer cake ...
We’ll come back to this idea later, but for now let’s
develop a little better understanding of the individual
terms in this expression.
Hopefully you’ve had time to read over pages 317-343.
What I’ll be covering should look familiar to you and
should provide an opportunity for you to cover questions
you might have after reading the text.
Consider individual terms in more detail
From our discussion last period you know there are
several reasons why g may differ from one point to
another in the earth’s surface.
The earth is an oblate spheroid, and if we were to
walk from the equator to the poles we would go down
hill over 21 kilometers. We would be 21.4 kilometers
closer to the center of the earth at the poles.
The variation in earth radius is primarily a function of
latitude.
In addition to that we have another latitude
dependant effect - centrifugal acceleration.
Let’s consider the effects of centrifugal acceleration.
The velocity of a point on the earth’s equator as it
rotates about the earth’s axis is ~ 1522f/s.
For relatively small , we can represent S as a
straight line. Thus we have two similar isosceles
triangles and equality of similar ratios.
Use basic arc-length relationship
to radius and subtended angle.
Combine the preceding relationships
to obtain the centrifugal acceleration
What is the centrifugal acceleration at the equator?
for a 150 lb object
Note that as latitude changes, R in the expression does not refer
to the earth’s radius, but to the distance from a point on the
earth’s surface to the earth’s axis of rotation. This distance
decreases with increasing latitude and becomes 0 at the poles.
The combined effects of the earth’s shape and
centrifugal acceleration are represented as a function
of latitude (). The formula below was adopted as a
standard by the International Association of Geodesy in
1967. The formula is referred to as the Geodetic
Reference System formula of 1967 or GRS67
g n  g e 1  A sin 2   B sin 4   cm
s2
g n  978.03185 1  0.005278895sin 2   0.000023463sin 4   cm
s2
The gradient of this effect is
gn
milligal
 1.307 sin 2

mile
This is a useful expression, since you need only go
through the calculation of GRS67 once in a particular
survey area. All other estimates of gn can be made by
adjusting the value according to the above formula.
As Burger notes the accuracy of your survey can be
affected by an imprecise knowledge of one’s actual
latitude. As the above formula reveals an error of 1 mile
in latitude translates into an error of 1.31 milligals at a
latitude of 45o.
How accurately your position must be known is a
function of the change in acceleration you are trying to
detect.
The difference in g from equator to pole is approximately
5186 milligals. The variation in the middle latitudes is
approximately 1.31 milligals per mile (i.e. sin (2) = 1).
gt  g n ( )  g E  g B  gt  gTide  Drift
The next term in our expression of the
theoretical gravity is gE - the elevation
effect.
In our earlier discussion we showed that
dg/dR could be approximated as -2g/R.
Using an average radius for the earth this
turned out to be about 0.3081 milligals/m.
When the variations of g with latitude are considered
in this estimate one finds that
g
 0.3086  0.00023 cos2  0.00000002 z
R
Where z is the elevation above sea-level.
The influence of variations in z is actually quite
small and generally ignored (see next slide).
As the above plot reveals, the variations in dg/dR,
extend from 0.3089 at the equator to 0.3083 at the
poles. In the middle latitude areas such as Morgantown,
the value 0.3086 is often used. Note that the effect of
elevation is ~ 2/100,000th milligal (or 2/100ths of a
microgal) for 1000 meter elevation.
g
 0.3086  0.00023 cos2  0.00000002 z
R
The variation of dg/dR with elevation - as you
can see in the above graph - is quite small.
gt  g n ( )  g E  g B  gt  gTide& Drift
Next we estimate the term gB - the Bouguer plate term.
This term estimates the contribution to the theoretical
gravity of the material between the station elevation and
sea level. We have estimated how much the acceleration will
be reduced by an increase in elevation. We have reduced our
estimate accordingly. But now, we need to increase our
estimate to incorporate the effect of materials beneath us.
First we consider the plate effect from a conceptual point
of view and then we will go through the mathematical
description of this effect.
We don’t have to have any geologic complexity between the
measurement point and sea level. GRS67 makes predictions (gn)
of g on the reference surface (i.e. sea level). If we want to
compare our observations to predictions we have to account
for the fact that at our observation point g will be different
from GRS67 not only because we are at some elevation h above
the reference surface but because there is additional mass
between the observation point and the reference surface.
V
To determine the effect of the plate on g we must sum
together the contributions of volume elements over
the infinite extent of the plate (see pages 327-331).
Acceleration due to mass
of volume element directly
beneath the gravimeter
To do this we evaluate Newton’s universal
law of gravitation in its integral form.
X=0
Assume that our observation is made at an elevation h above
the plate. We first calculate the effect of a rod which extends
from + to -  along the x axis. m is the perpendicular distance
from the observation point to the rod.
Another perspective
dV
g plate   G 2
r
dxdydz
g plate   G
r2
g plate  

x y
dxdydz
 G r 2
z
We’ll solve this equation by doing the integrations one
variable at a time. We start with integration along the x axis.
This turns out to be simpler than it looks because as we
integrate over x, the variables y and z are held constant and
we only have to consider the relationship of x to r.
m
To solve, we express dx as a function of r and
m. We employ arc length relationships to do so.
The interrelationships yield x as a function of r and .
m
Also remember that the gravimeter measures the
vertical component of g or g cos . In this step
multiplication by cos  yields the component of g in the m
direction. We don’t have the vertical component just yet.
We have the integral form of Newton’s law and basic
relationships of dx to r and r to the perpendicular
distance (m) from the point of observation to the rod.
The relationship of r to m is shown above.
Substitute expressions for dx and r into gV
Rearrange terms to isolate
 cosd
Visualize the integral as an area and recall integral
(antiderivative) relationships for trig functions
Evaluate at the limits of the integral
We’ve evaluated the integral for one rod at a distance m from
the point of observation. Now we must sum the influence of
all rods together. We must now integrate in the y direction.
One by one
We now look out in the y direction and previous m becomes
r. Otherwise we proceed as we did for integration over x.
Determine y as a function of r and .
Again - we must
consider only the
vertical component
of g associated
with the mass of
each rod.
We are left with two simple integrals to evaluate.
 d
and
z dz
We end up with a very simple expression that relates the
acceleration due to the plate of material beneath our
observation point directly to its thickness and density.
Remember this is the basis for the
relationship used by Stewart
t  130 g
This expression comes directly from
g plate  2Gt
Stewart has solved it using a
density  = -0.6 gm/cm3. He
has also included the factor
which transforms meters to
feet so that the user can
input t in units of feet. g is in
units of milligals.
Edge effects
Valley
1000’ wide Valley
g=-3.12mg
g=-4.25mg
5000’
5000’
The 3.12 milligal anomaly implies a valley depth of only 406
feet. The 4.25 milligal anomaly implies 550feet bedrock
depth. We have errors of 8.3% and 32% in these two cases.
gt  g n ( )  g E  g B  gt  gTide  Drift
gB may seem like a pretty unrealistic approximation of the
topographic surface. It is. You had to scrape off all mountain
tops above the observation elevation and fill in all the valleys
when you made the plate correction.
See pages 331-336
- we’re not through yet. We now have to reintroduce
the influence of the valleys that were filled and hills that
were scraped away.
We have to compensate for the effect of topography.
But back to the problem with the topographic surface.
What is the effect of the topography on the observed gravity.
We estimate the effect of topography by approximating
topographic features as ring-sectors whose thickness (z) equals
the average elevation of topographic features within them.
You will recall from your reading of Burger (see Eq 6-30) that




g ring  2G  Ro  Ri  Ri2  z 2 2  Ro2  z 2 


1



Ri = inner radius of the ring
Ro = outer radius of the ring
z = thickness of the ring (average
elevation of the topographic features
inside the sector of interest)
gsector  ring
g
number of sectors

gt  g n ( )  g E  g B  gt  gTide  Drift
The topographic effect gt is always negative.
Again, this may seem like a crude approximation
of actual topography. But topographic
compensation is a laborious process and if done in
detail the estimate is fairly accurate. We can
increase the detail of our computation depending
on the accuracy needed in a given application.
Now that digital elevation data are available, we
can actually let the computer do a very detailed
computation using each digital elevation data.
We’ll discuss methods used to compute the
topographic effect more in the next lecture.
The last term we will look at incorporates the
effects of tide and instrument drift.
Tide and instrument drift
We are used to thinking in terms of ocean tides. The
ocean surface rises and falls under the influence of the
combined gravitational attraction of the sun and moon.
The solid earth also deforms under the influence in
response to the differential pull of the sun and the
moon. The change in surface elevation in addition to
their gravitational pull on the gravimeter spring can be
significant and these tidal effects must be
incorporated into our estimate of theoretical gravity.
The gravimeter is just a mechanical system. Its
parts - while simple - change over time. The spring
for example subjected to the constant tug of
gravity will experience permanent changes in
length over time. These changes fall under the
heading of instrument drift.
In general the influence of tide and drift on the
theoretical gravity is estimated by direct and repeated
measurement of gravitational acceleration at the same
place over and over again, throughout the day.
Usually during a survey a base station is reoccupied
every couple hours or so during the day’s survey.
The “drift curve” is constructed from these
measurements and measurements of acceleration made
at other stations are corrected relative to the drift
durve.
Gravity observations
M
i
l
l
i
g
a
l
s
4
3
2
1
8
Base
9
10
11
S1
S2
Base
TIME (am)
Is the acceleration of gravity measured at 9am the same
as that measured at the base station an hour earlier?
Drift Curve
M
i
l
l
i
g
a
l
s
4
+1 mG
Drift
Curve
3
2
-1 mG
1
8
Base
9
10
S1
S2
TIME (am)
11
Base
In this example, the
acceleration at station 1
(S1) is 1 milligal less than
that at the base station not the same.
At station 2, the
acceleration is only 1
milligal greater - not 3
milligals greater.
M
i
l
l
i
g
a
l
s
4
+1 mG
3
Tide and
Drift Curve
2
-1 mG
1
8
Base
9
S1
TIME
10
S2
11
Base
•Read over the remainder of the chapter 6 - pages
344 to 381. We’ll discuss the terrain correction at
greater length next lecture so spend some extra time
looking over the derivation on pages 350-352.
•In the next lecture we will talk about the different
types of gravity anomalies.
•Be prepared to ask question about text discussions of
simple geometrical objects such as the sphere,
horizontal cylinder, horizontal sheet.
•Be prepared to ask questions about the meaning of
the residual gravity.
•Be prepared to ask questions about the half maximum
techniques in the interpretation of gravity data.
•Also be prepared to ask questions about the gravity
lab and Stewart’s paper.
In addition look carefully at problems 1 through 3
at the end of the chapter.
1. In problem 1, you make a measurement of g at an
elevation of 152.7 meters and are asked to
determine the “free-air” and Bouguer corrections.
2. In problem 2 your observation point is located on
the floor of a circular basin 450 meters below the
surrounding escarpment. The surrounding plateau
can be considered to have infinite extent. You are
asked to determine whether the terrain correction
is necessary.
3. In problem 3 you have to correct for tide and
drift.
Problems will be due
in a week and a half.
In the next computer lab be prepared to ask
questions about problems 1 through 3.
Also bring any questions you might have about
Stewart’s paper to Tuesday’s lab and concentrate on
developing an understanding of how the
gravitational acceleration of an infinite plate, whose
thickness is equal to that of the glacial drift, can be
used to estimate the thickness of the drift.
Realize that the use of the infinite plate to
estimate drift thickness assumes that the drift
valleys are much wider than they are thick. If that
is so then the relationship t = 130g provides an
acceptable estimate of drift thickness.


g drift  2Gt


 density contrast
t drift thickness
Assume that the valley width is much greater than
its thickness, then the gravitational acceleration
due to the drift is proportional to its thickness
gt  g n ( )  g E  g B  gt  gTide  Drift
As we conclude the day - do you have any questions
about the model we’ve proposed to explain the
gravitational acceleration at an arbitrary point on the
surface of our theoretical (but geologically unrealistic)
earth?
As geologists we expect there will be
considerable subsurface density contrast
associated, for example, with structure - or
stratigraphy, drift thickness, caves, trenchs …
In preparing our gravity data, we start by
computing the theoretical gravity but usually find
that the theoretical gravity we compute at a given
latitude and elevation does not equal the observed
gravity at that location.
We have an anomaly -
The gravity anomaly  gobs  gt
And therein lies the geology.