2. Tom Wilson, Department of Geology and Geography Environmental and Exploration Geophysics I tom.h.wilson tom.wilson@mail.wvu.edu Department of Geology and Geography West Virginia University Morgantown, WV Gravity I

3. Questions about the resistivity lab? Tom Wilson, Department of Geology and Geography Some general comments about your starting models and their importance

4. Note Tom Wilson, Department of Geology and Geography Remember there will be no class next Tuesday _ October 14 th due to fall break.

5. Gravitational field methods Tom Wilson, Department of Geology and Geography Passive source & non-invasive LaCoste Romberg GravimeterWorden Gravimeter

6. Tom Wilson, Department of Geology and Geography x spring extension m s spring mass k Young’s modulus g acceleration due to gravity Colorado School of Mines web sites - Mass and spring Pendulum measurement Hooke’s Law

7. The spring inside the gravimeter Tom Wilson, Department of Geology and Geography The spring is designed in such a way that small changes in gravity result in rather large deflections of the movable end of the beam. Early gravimeters read the mechanical movement of the spring. Today’s gravimeters use electrostatic feedback systems that hold the movable end of the beam at a fixed position between the plates of the capacitor. The voltage needed to hold the beam at a fixed position is proportional to the changes in gravity.

8. Tom Wilson, Department of Geology and Geography Newton’s Universal Law of Gravitation Newton.org.uk m1m1 m2m2 r 12 F 12 Force of gravity G Gravitational Constant

9. The gravitational constant Tom Wilson, Department of Geology and Geography The current value of G is 6.67384(80) x 10 -11 m 3 kg -1 s -2 or N-m 2 kg -2. However, The new measurement of G that's published in the journal Nature — 6.67191(99) X 10‾¹¹ m³kg‾¹s‾² — isn't as precise as the best measures, but because it uses single atoms, scientists can be more confident the results aren't skewed by hidden errors,published in the journal Nature So what? Well this would mean, among other things, that the Earth weighs 1.72x10 21 kg more than previously estimated Measured in terms of water, that translates into a cube 1200km on a side – a volume greater than the Earth’s oceans.

10. Tom Wilson, Department of Geology and Geography g E represents the acceleration of gravity at a particular point on the earth’s surface. The variation of g across the earth’s surface provides information about the distribution of density contrasts in the subsurface since m =  V (i.e. density x volume). m s spring mass m E mass of the earth R E radius of the earth Like apparent conductivity and resistivity g, the acceleration of gravity, is a basic physical property we measure, and from which, we infer the distribution of subsurface density contrast.

11. Basic units – the milliGal Tom Wilson, Department of Geology and Geography Most of us are familiar with the units of g as feet/sec 2 or meters/sec 2, etc. From Newton’s law of gravity g also has units of

12. Tom Wilson, Department of Geology and Geography Using the metric system, we usually think of g as being 9.8 meters/sec 2. This is an easy number to recall. If, however, we were on the Martian moon Phobos, g p is only about 0.0056meters/sec 2. [m/sec 2 ] might not be the most useful units to use on Phobos. Some unit names used in detailed gravity applications include 9.8 m/sec 2 980 Gals (or cm/sec 2 ) 980000 milli Gals (i.e. 1000th of a Gal & 10 -5 m/s 2 ) 10 -6 m/sec 2 =the gravity unit (gu) (1/10 th milliGal) We experience similar problems in geological applications, because changes of g associated with subsurface density contrasts can be quite small.

13. Martian moon Phobos Tom Wilson, Department of Geology and Geography orbit: 9378 km from the center of Mars diameter: 22.2 km (27 x 21.6 x 18.8) mass: 1.08e16 kg See nineplanets.org

14. Tom Wilson, Department of Geology and Geography Using the metric system, we usually think of g as being 9.8 meters/sec 2. This is an easy number to recall. If, however, we were on the Martian moon Phobos, g p is only about 0.0056meters/sec 2. [m/sec 2 ] might not be the most useful units to use on Phobos. Some unit names used in detailed gravity applications include 9.8 m/sec 2 980 Gals (or cm/sec 2 ) 980000 milli Gals (i.e. 1000th of a Gal & 10 -5 m/s 2 ) 10 -6 m/sec 2 =the gravity unit (gu) (1/10 th milliGal) We experience similar problems in geological applications, because changes of g associated with subsurface density contrasts can be quite small.

15. Tom Wilson, Department of Geology and Geography If you were to fall from a height of 100 meters on Phobos, you would hit the ground in a.10 seconds b.1 minute c.3 minutes You would hit the ground with a velocity of a.1 m/s b.5 m/s c.30 m/s How long would it take you to accelerate to that velocity on earth? a.10 seconds b.1 second c.1/10 th of a second =189s =1m/s =0.1s The velocity you would reach after jumping off a brick. 27x22x18km

16. Tom Wilson, Department of Geology and Geography 6km If you could jump up about ½ meter on earth you could probably jump up about 1.7 kilometers on Phobos. (It would be pretty hard to take a running jump on Phobos).

17. Tom Wilson, Department of Geology and Geography 6km That would give you a velocity of 4.43 m/s and on Phobos that would keep you off the surface for 26 minutes (13 up and 13 down). With a horizontal component of about 4 meters per second you’d come down on the opposite rim.

18. Putting objects in scale Tom Wilson, Department of Geology and Geography

19. http://htwins.net/scale2/

20. Phobos is about the size of Deimos & Mt. Everest Tom Wilson, Department of Geology and Geography http://htwins.net/scale2/

21. Tom Wilson, Department of Geology and Geography Diameter 6794 km Diameter 12,756 km 78 x 10 6 km

22. Summary relationships Tom Wilson, Department of Geology and Geography 1 milligal = 10 microns/sec 2 1 milligal equals 10 -5 m/sec 2 or conversely 1 m/sec 2 = 10 5 milligals. The gravity on Phobos is 0.0056m/s 2 or 560 milligals. Are such small accelerations worth contemplating? Can they even be measured?

23. Comet Churyumov-Gerasimenko (67P) Tom Wilson, Department of Geology and Geography The acceleration of gravity on P67 is about 3.69x10 -5 m/s2 How many milliGals is that? How long would it take to fall one meter to the surface of this comet? About 4 minutes Rough scale about 5km across. Object density ~0.4 gm/cc.

24. Spring sensitivity Tom Wilson, Department of Geology and Geography Today’s gravimeters measure changes in g in the  Gal (10 -6 cm/s 2 ) range. If spring extension in response to the Earth’s gravitational field is 1 cm, a  Gal increase in acceleration will stretch the spring by 10 -8 m – a length covered by 100 hydrogen atoms lined up side-by-side. The spring response in today’s modern field portable gravimeters is amplified so that detection of these small changes is possible…. for the modest price of $80,000 to $90,000

25. A geologic example Tom Wilson, Department of Geology and Geography Note that the variations in g that we see associated with these large scale structures produce small but detectable anomalies that range in scale from approximately 1 - 5 milliGals. Calculated and observed gravitational accelerations are plotted across a major structure in the Valley and Ridge Province,

26. Planet-scale variations in g Tom Wilson, Department of Geology and Geography We usually think of the acceleration due to gravity as being a constant - 9.8 m/s 2 - but as the forgoing figures suggest, this is not the case. Variations in g can be quite extreme. For example, compare the gravitational acceleration at the poles and equator. The earth is an oblate spheroid - that is, its equatorial radius is greater than its polar radius. R p = 6356.75km R E = 6378.14km 21.4km difference

27. Tom Wilson, Department of Geology and Geography R p = 6356.75km R E = 6378.14km g P =9.83218 m/s 2 g E =9.780319 m/s 2 This is a difference of 5186 milligals. These kinds of differences, which in this case are a function of latitude need to be corrected for – or eliminated Substitute for the different values of R There are great differences in the acceleration due to gravity on the Earth that, in may instances, are unrelated to the details of subsurface geology

28. Mountain belt scale variations in g Tom Wilson, Department of Geology and Geography Density differences arising from isostatic equilibrium processes represent large scale regional changes of g that are often removed before modeling and interpretation. R. J. Lillie, 1999

29. Tom Wilson, Department of Geology and Geography Isostatic compensation and density distributions in the earth’s crust R. J. Lillie, 1999 Generally geological processes produce linear sheet like distributions of materials It’s generally easier to accept this kind of model See also http://www.encyclopedia.com/video/O-qTAp4zhSg- pgr-101-post-glacial-reboundisostasy.aspx

30. Tom Wilson, Department of Geology and Geography Does water flow downhill? A surface along which the distance to the center of the earth decreases is not necessarily down hill.

31. Tom Wilson, Department of Geology and Geography The notion of downhill is associated with a surface along which the gravitational potential decreases

32. Tom Wilson, Department of Geology and Geography The geoid is a surface of constant gravitational potential. The gradient of the potential is perpendicular to the surface. Thus gravitational acceleration is always normal to the equipotential surface.

33. Tom Wilson, Department of Geology and Geography Geoid height anomalies Contours are in meters 160 meters uphill

34. Tom Wilson, Department of Geology and Geography Aside from wind generated surface waves and ocean scale wind generated swells … Is the ocean surface a flat surface?

35. Tom Wilson, Department of Geology and Geography Map of the ocean floor obtained from satellite radar observations of ocean surface topography. SeaSat

36. Tom Wilson, Department of Geology and Geography Detailed map of a triple-junction on the floor of the Indian Ocean derived from ocean surface topography

37. Tom Wilson, Department of Geology and Geography In the environmental applications of gravity methods anomalies smaller than a milligal can be of interest to the geophysicist. A modern gravimeter is capable of measuring gravity to an accuracy of about 100th of a milligal or better. We’ll spend considerable time discussing the applications of gravity data in groundwater exploration. An example of this application is discussed in Stewart’s paper (see web site link) on the use of gravity methods for mapping out buried glacial Valleys in Wisconsin - so read over this paper as soon as you can. Gravity provides interesting views of objects buried deep beneath the surface - out of our reach

38. Break and consider some introductory gravity problems Tom Wilson, Department of Geology and Geography 1. Given that G=6.672 x 10 -11 m 3 kg -1 s -2, that g = 9.8 m/s 2, and that the radius of the earth is 6366km, calculate the mass of the earth. 2. At birth assume that you were delivered by an obstetrician with a mass of 75kg, and that the obstetrician’s center of mass was 0.5 meters from yours. Also assume that at that very point in time, Mars was closest to the earth or about 78 x 10 6 km from your center of mass. The mass of Mars is approximately 6.42 x 10 23 kg. Determine the acceleration due to the gravitational field of the obstetrician and of Mars. Which was greater?

39. Tom Wilson, Department of Geology and Geography 3. A space traveler lands on the surface of a spherically shaped object that produces an acceleration due to gravity of 0.000003086m/s 2. The object has average density of 5500 kg/m 3. What is the radius of this object? How long would it take you to fall 5 meters assuming a constant g of 0.3086 milliGals? Questions next Thursday – Due the following Tuesday.

40. Tom Wilson, Department of Geology and Geography Start doing some background reading for the gravity lab ….

41. Tom Wilson, Department of Geology and Geography Keep reading Chapter 6. Resistivity paper summary due today Resistivity lab due next Thursday, October 16 th Start looking over problems 6.1 through 6.3 (see today’s handout). Handout problems 1-3 (from today) due next Thursday Writing section outline for Essay II is due October 21 st. Remember – no class next Tuesday (fall break)

42. Tom Wilson, Department of Geology and Geography Form Stewart Bedrock models derived from gravity data Residual gravity data The gravity anomalies associated with these glacial valleys have a range of about 4 milliGals. Why residual? The residual eliminates the influence of the deeper strata which dip uniformly across the area. Their configuration is not relevant to the problem at hand. The residual can eliminate geology we aren’t interested in

43. Tom Wilson, Department of Geology and Geography

44. The anomaly shown here is only 1/2 milligal Karst

45. Tom Wilson, Department of Geology and Geography -76 mGals -32 mGals

46. Tom Wilson, Department of Geology and Geography These variations in gravitational acceleration are very small. To give you some additional perspective on the magnitude of these changes, consider the changes in g as a function of r (or R E ) as indicated by Newton’s law of gravity - Recognize that the above equation quantifies the variation in g as a function of r for objects that can effectively be considered as points. For now, let’s take a leap of faith and assume that we can represent the Earth as a point and that the above equation accurately describes the variations in g as a function of distance from the center of the earth, R E.

47. Tom Wilson, Department of Geology and Geography Given this relationship - RERE h What is g at a distance R E +h from the center of the earth? sl=sea level

48. Tom Wilson, Department of Geology and Geography Is there another way to compute the change in g?

49. Tom Wilson, Department of Geology and Geography What is the derivative of g with respect to R?

50.  g/  h - in Morgantown Tom Wilson, Department of Geology and Geography At Morgantown latitudes, the variation of g with elevation is approximately 0.3086 milligals/m or approximately 0.09406 milligals/foot. As you might expect, knowing and correcting for elevation differences between gravity observation points is critical to the interpretation and modeling of gravity data. The anomalies associated with the karst collapse feature were of the order of 1/2 milligal so an error in elevation of 2 meters would yield a difference in g greater than that associated with the density contrasts around the collapsed area.

51. Tom Wilson, Department of Geology and Geography Just as a footnote, Newton had to develop the mathematical methods of calculus to show that spherically symmetrical objects gravitate as though all their mass is concentrated at their center.

52. Tom Wilson, Department of Geology and Geography The acceleration term in Newton’s law of gravitation. tells us we need to consider mass (m) and its distance(s) (r i ) from some observation point. In practice we usually compute the acceleration of some arbitrarily shaped mass by breaking it up into small parts and summing their individual contributions to g.

53. Tom Wilson, Department of Geology and Geography Integral form of Newton’s law of gravitation Line, surface or volume Depending on symmetry dz dy dx dV

54. Tom Wilson, Department of Geology and Geography Consider the following: what is the gravitational attraction of a buried spherically symmetrical object? Let’s work through this on the board

55. Tom Wilson, Department of Geology and Geography What is the vertical component?

56. Tom Wilson, Department of Geology and Geography A symmetrical Earth holds no riddles for the geophysicist.

57. Tom Wilson, Department of Geology and Geography If the earth were this simple our study would be complete.

58. Tom Wilson, Department of Geology and Geography How thick is the landfill? Gravity methods thrive on heterogeneity. In general the objects we are interested in are not so symmetrical and provide us with considerable lateral density contrast and thus gravity anomalies.

59. Tom Wilson, Department of Geology and Geography How does g vary from A to E? We might expect that the average density of materials in the landfill would be less than that of the surrounding bedrock and thus be an area of lower g

60. Tom Wilson, Department of Geology and Geography At present we’ve only accounted for variations in g as a function of elevation or distance from the center of the earth. But obviously we have further to go in terms of conceptualizing and developing the computations needed to understand and evaluate geological problems using measured gravitational fields. Another variable for us to consider is the elevation at which our observations are made.

61. Tom Wilson, Department of Geology and Geography How do we compensate for the influence of matter between the observation point (A) and sea level? How do we compensate for the irregularities in the earth’s surface - its topography? A hill will take us down the gravity ladder, but as we walk uphill, the mass beneath our feet adds to g.

62. Tom Wilson, Department of Geology and Geography What other effects do we need to consider? Latitude effect Centrifugal acceleration 463 meters/sec ~1000 mph

63. Tom Wilson, Department of Geology and Geography Solar and Lunar tides Instrument drift

64. 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 -

65. Tom Wilson, Department of Geology and Geography g n the “normal gravity” or the gravitational acceleration on the reference ellipsoid  g FA the elevation or free air 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. Terms

66. 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.

67. Tom Wilson, Department of Geology and Geography 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 spend more time with these ideas, but in the next couple lectures we will develop a little better understanding of the individual terms in this expression. The Theoretical Gravity

68. Tom Wilson, Department of Geology and Geography We’ll carry on this discussion in greater detail next time. Make sure you continue reading chapter 6 in Burger et al. We’ll go over some of the basic ideas associated with 1) the Bouguer plate correction and 2) the topographic (or terrain) correction. These two effects are approximated using gravitational acceleration produced by a plate of finite thickness but infinite horizontal extent and by individual sectors from a ring of given thickness and width. Read general introduction from pages 349-355 and continue reading about gravity corrections from page 356 through the top of page 373

69. Tom Wilson, Department of Geology and Geography Keep reading Chapter 6. Resistivity paper summary due today Resistivity lab due next Thursday, October 16 th Start looking over problems 6.1 through 6.3 (see today’s handout). Handout problems 1-3 (from today) due next Thursday Writing section outline for Essay II is due October 21 st. Remember – no class next Tuesday (fall break)