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 (IV)

2. What physical characteristics of the observation point and it’s surroundings influence the observed acceleration of gravity? Tom Wilson, Department of Geology and Geography

3. 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. What is this gravity anomaly associated with? Tom Wilson, Department of Geology and Geography

6. What is this much smaller anomaly produced by? Tom Wilson, Department of Geology and Geography

7. If the observed values of g behave according to our 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 But – in this case ->

8. We can estimate the geological properties of the subsurface from the anomaly Tom Wilson, Department of Geology and Geography Because we know something about the local geology

9. Tom Wilson, Department of Geology and Geography Thus g plate = 4.192 x 10 -7  t Gals, where t is in centimeters and  in gm/cm 3. This is also 4.192 x 10 -4 mGals since there are 10 3 milliGals per Gal. Influence of materials between the datum and your gravimeter

10. Tom Wilson, Department of Geology and Geography Also if we want to allow the user to input thickness (t) in meters, we have to introduce a factor of 100 into the above relationship. When we enter a value of 1 for t we want the computer or calculator to realize that that is a meter and convert that to 100cm. That way all input values of t in meters are converted to the required units of centimeters. This would change 4.192 x 10 -4  t mGals (in the previous slide) to 4.192 x 10 -2  t or 0.04192  t mGals.

11. Bottom line – revised equation Tom Wilson, Department of Geology and Geography Where  is in gm/cm 3 and t is in meters These equations can be as confusing as they are helpful. If you don’t mix units, you’ll get the wrong answer!

12. What did Stewart do?? Tom Wilson, Department of Geology and Geography 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 He’s made a critical assumption. What is it?

13. Tom Wilson, Department of Geology and Geography Consider this without the vertical exaggeration of the display 10,000 ft 557 ft Vertical exageration is still about 5:1

14. Tom Wilson, Department of Geology and Geography Thus g plate = 2  G  t = 4.192 x 10 -7  t in units of cm/s 2 or Gals. This is also 4.192 x 10 -4  t mGals as mentioned earlier As noted earlier and

15. Tom Wilson, Department of Geology and Geography 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. and  and t are in units of gm/cm 3 and cm respectively, This would change the above to 0.01277  t milliGals, where  and t have units of gm/cm 3 and ft respectively. Note that if we then fix the density  = 0.6 gm/cm 3 then we have g = 0.00767 t in mgals and thus t =(1/0.00767)g or approximately t=130g. But we’re lazy, we’d like to input our thicknesses in feet, not meters

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

17. 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. & the Bouguer effect of When we are correcting the observed gravity to obtain the anomaly, the free air is added and the Bouguer plate term is subtracted.

18. 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 Correction refers to the changes we introduce into the observed gravity to get the anomaly

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

20. Tom Wilson, Department of Geology and Geography What are Stewart’s assumptions?

21. Tom Wilson, Department of Geology and Geography

22. 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 32% and 8.3% in these two cases. 5000’ g=-4.25mg Edge effects g=-3.12mg 1000’ wide Valley Valley The Glacial Valley Depth=600’ Predicted anomaly = 4.62 mG

23. 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

24. 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

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

26. 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.

27. 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) Using the units simplification discussed earlier

28. Tom Wilson, Department of Geology and Geography The topographic effect  g t is always negative, the correction positive. 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.

29. 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)

30. 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)

31. 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.

32. 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

33. 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

34. 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.

35. 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?

36. 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.

37. Tom Wilson, Department of Geology and Geography An anomaly exists - and therein lies the geology. Geology inferred from modeling the anomaly

38. Tom Wilson, Department of Geology and Geography Items on the list …. Turn in your gravity paper summaries (Tuesday, Nov. 15 th ). Any questions about problems 6.1 – 6.3. They will be also be dueTuesday, Nov. 15 th. Gravity Lab will be due on Nov. 17 th. We will clear up any questions you have about the gravity lab in the next couple lectures. Review the remainder of the chapter 6 (past page 378).

39. 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

40. 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 This Thursday

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