1. Tom Wilson, Department of Geology and Geography Environmental and Exploration Geophysics II tom.h.wilson tom.wilson@mail.wvu.edu Department of Geology and Geography West Virginia University Morgantown, WV Gravity Methods (III)

2. Tom Wilson, Department of Geology and Geography To conceptualize the dependence of gravitational acceleration on various factors, we usually write g as a sum of different influences or contributions. These are -

3. Tom Wilson, Department of Geology and Geography That predicted or estimated value of g is often referred to as the theoretical gravity - g t g n the normal gravity or the gravitational acceleration on the reference ellipsoid  g T the influence of topographic features on observed g – the terrain effect  g Tide and Drift the influence of tide and drift (often combined)  g B the contribution to measured or observed g of the material between sea-level and the elevation of the observation point - the Bouguer plate effect  g FA the change in gravity with increase or decrease of elevation – the free air effect

4. Tom Wilson, Department of Geology and Geography 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.

5. Tom Wilson, Department of Geology and Geography If the observed values of g behave according to the ideal model then there are no lateral density contrats of unknown origin in the subsurface – so no geology! – That is no interesting geology only a series of homogeneous layers No anomaly – no geology

6. Tom Wilson, Department of Geology and Geography Thus g plate = 4.192 x 10 -7 cm/s 2 (or gals) for a t = 1 cm and  = 1gm/cm 3. This is also 4.192 x 10 -4 mgals since there are 10 3 milliGals per Gal. Also if we want to allow the user to input thickness (t) in meters, we have to introduce a factor of 100 (i.e. our input of 1 meter has to be multiplied by 100) to convert the result to centimeters. This would change the above to 4.192 x 10 -2 or 0.04192.

7. Tom Wilson, Department of Geology and Geography When the factor of 0.04192 is used, thickness can be entered in meters and densities in grams per cubic centimeter, which would be standard units for most of us. Thus - Where  is in gm/cm 3 and t is in meters

8. Tom Wilson, Department of Geology and Geography Stewart uses different conversion factors to convert inputs in different units to obtain 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 centimeters to feet so that the user can input t in units of feet. g is in units of milligals. where t is in feetor

9. Tom Wilson, Department of Geology and Geography Consider this without the vertical exageration of the display 4000 ft 600 ft

10. Tom Wilson, Department of Geology and Geography Thus g plate = 2  G  t = 4.192 x 10 -7 cm/s 2 (or gals) for a t = 1 cm and  = 1gm/cm 3. This is also 4.192 x 10 -4 mgals since there are 1000 milligals per gal. But, if we want to allow the user to input thickness in feet, we have to introduce a factor of 30.48 (i.e. our input of 1 foot has to be multiplied by 30.48cm/foot) to convert the result to centimeters. This would change the above to 0.01277. Note that if we then fix the density  = 0.6 gm/cm 3 then we have g = 0.00767 t in mgals and 1/0.00767 is approximately 130. As noted earlier and

11. Tom Wilson, Department of Geology and Geography Stewart’s formula is developed as follows -

12. Tom Wilson, Department of Geology and Geography The free air and Bouguer plate terms are often combined into a more general elevation term that accounts for both influences on g. The free air effect is often simplified by ignoring the influence of latitude and z When we calculate the theoretical gravity, the free air term is subtracted and the plate term is added. When we are correcting the observed gravity to obtain the anomaly, the free air is added and the Bouguer plate term is subtracted. & the Bouguer effect of

13. Tom Wilson, Department of Geology and Geography The elevation correction is: in milligals or in g.u.  is in units of gm/cm 3 and h is in meters

14. Tom Wilson, Department of Geology and Geography Carry the minus sign through The observation is corrected by subtracting the terms in the theoretical gravity. The elevation correction  g e

15. Tom Wilson, Department of Geology and Geography What is Stewart doing with his formula? What assumptions does he make in using it?

16. Tom Wilson, Department of Geology and Geography

17. The 3.12 milliGal anomaly implies a valley depth of only 406 feet. The 4.25 milliGal anomaly implies 550foot bedrock depth. We have errors of 8.3% and 32% in these two cases. 5000’ g=-4.25mg Edge effects g=-3.12mg 1000’ wide Valley Valley The Glacial Valley

18. Tom Wilson, Department of Geology and Geography If the valley width is much greater than its thickness, then the gravitational acceleration due to the drift is approximated by the infinite plate  density contrast t drift thickness     The Valley and the Plate

19. Tom Wilson, Department of Geology and Geography  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. Topographic correction See figure 6.3

20. Tom Wilson, Department of Geology and Geography Recall the effect of reintroducing topography onto our flat plate

21. Tom Wilson, Department of Geology and Geography 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.

22. Tom Wilson, Department of Geology and Geography 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) For a derivation, see Burger et al. (2006) Also see Keary, Brooks and Hill (p 135) where constants are incorporated to simplify the computation-

23. Tom Wilson, Department of Geology and Geography 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 we use digital elevation data and let the computer do a very detailed computation. But the principle is the same 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.

24. Tom Wilson, Department of Geology and Geography 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 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 Berger (1992)

25. Tom Wilson, Department of Geology and Geography 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. Berger (1992)

26. Tom Wilson, Department of Geology and Geography 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 (a base station) over and over again, throughout the duration of your survey. Usually during a survey a base station is reoccupied every couple hours or so. The “drift curve” is constructed from these measurements and measurements of acceleration made at other stations are corrected relative to the drift curve.

27. Tom Wilson, Department of Geology and Geography 1 2 3 4 Base TIME (am) S1S2 891011 Gravity observations Is the acceleration of gravity measured at 9am the same as that measured at the base station an hour earlier? milliGals

28. Tom Wilson, Department of Geology and Geography 1 2 3 4 MilligalsMilligals Base TIME (am) S1S2 891011 Tide & Drift Curve +1 mG -1 mG Drift Curve

29. Tom Wilson, Department of Geology and Geography 1 2 3 4 MilligalsMilligals Base TIME S1S2 891011 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.

30. Tom Wilson, Department of Geology and Geography 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?

31. Tom Wilson, Department of Geology and Geography 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.

32. Tom Wilson, Department of Geology and Geography An anomaly exists - and therein lies the geology.

33. Tom Wilson, Department of Geology and Geography Items on the list …. Turn in your gravity paper summaries (Thursday, Nov. 5 th ). Keep reading Chapter 6 review remainder of the chapter (past page 378). Any questions about problems 6.1 and 6.2. They are due next Tuesday, Nov. 3rd. Questions about the Gravity Lab?

34. Tom Wilson, Department of Geology and Geography 6.1 If gravity determination is made at an elevation of 152.7 m, what is the value of the free-air correction (assuming sea level as the datum)? What is the Bouguer correction (assuming a 2.5 gm/cm3 reduction density)? 6.2 A gravity station at an elevation of 0 m is located in the center of an erosional basin. The floor of the basin has virtually no relief. The plateau escarpment is located at a distance of 450 m from the gravity station. The surface of the plateau has a relatively constant elevation of 400 m. Will the terrain correction be necessary? Assume a density of 2.5 gm/cm3. Hint: where Ri is the inner radius of the ring R0 is the outer radius of the ring z is the thickness of the ring

35. Tom Wilson, Department of Geology and Geography timedial reading Converted to milliGals relative difference Tide & Drift Drift corrected Base Station0762.7166.279499000 122 774.16 254 759.72 377768.9566.8217550.542256-0.093070.635326 499 771.01 Base Station110761.1866.146542-0.132957-0.132960 See handout and bring questions to class next Tuesday

36. Tom Wilson, Department of Geology and Geography If the valleys do not extend to infinity how will this change the observed gravity?