Environmental and Exploration Geophysics II tom.h. wilson Department of Geology and Geography West Virginia University Morgantown, WV Gravity Methods (II)

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 non- geological 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 -

g n the normal gravity of the gravitational acceleration on the reference ellipsoid  g E the elevation effect  g B the Bouguer plate effect or the contribution to measured or observed g of the material between sea- level and the elevation of the observation point  g T the effect of terrain on the observed g  g Tide 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 - g t 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 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.

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. Consider individual terms in more detail

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 The formula is referred to as the Geodetic Reference System formula of 1967 or GRS67

The gradient of this effect is 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 g n 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 45 o. 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).

The next term in our expression of the theoretical gravity is  g E - 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 milligals/m.

When the variations of g with latitude are considered in this estimate one finds that 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 at the equator to at the poles. In the middle latitude areas such as Morgantown, the value is often used. Note that the effect of elevation is ~ 2/100,000th milligal (or 2/100ths of a microgal) for 1000 meter elevation.

The variation of dg/dR with elevation - as you can see in the above graph - is quite small.

Next we estimate the term  g B - 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 (g n) 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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 ).

Acceleration due to mass of volume element directly beneath the gravimeter Vertical component from mass of distant volume element

To do this we evaluate Newton’s universal law of gravitation in its integral form.

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. X=0

Another perspective

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 g V

Rearrange terms to isolate

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. and

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 This expression comes directly from Stewart has solved it using a density  = -0.6 gm/cm 3. 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.

5000’ g=-4.25mg 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. Edge effects g=-3.12mg 1000’ wide Valley Valley

 g B 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

- 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.

What is the effect of the topography on the observed gravity. But back to the problem with the topographic surface.

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.

R i = inner radius of the ring R o = outer radius of the ring z = thickness of the ring (average elevation of the topographic features inside the sector of interest) You will recall from your reading of Burger (see Eq 6-30) that

The topographic effect  g t 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.

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. Tide and instrument drift

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.

MilligalsMilligals Base TIME (am) S1S Gravity observations Is the acceleration of gravity measured at 9am the same as that measured at the base station an hour earlier?

MilligalsMilligals Base TIME (am) S1S Drift Curve +1 mG -1 mG Drift Curve

MilligalsMilligals Base TIME S1S Tide and Drift Curve +1 mG -1 mG 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.

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 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 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.

Assume that the valley width is much greater than its thickness, then the gravitational acceleration due to the drift is proportional to its thickness  density contrast t drift thickness    

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 - And therein lies the geology.