1. Tom Wilson, Department of Geology and Geography tom.h.wilson tom.wilson@mail.wvu.edu Dept. Geology and Geography West Virginia University

2. Start reviewing materials for the final! … current to-do list Tom Wilson, Department of Geology and Geography 1.Problem 9.7 is due today 2.Hand in the gravity computation 3.I will give you till next Tuesday to finish up problems 9.9 and 9.10 4.Start reviewing class materials. Next week is a final review week No class this Thursday

3. Take advantage of this day off and … Tom Wilson, Department of Geology and Geography Turn in any late assignments by Friday afternoon, April 26 th. Put all late assignments in my mailbox (mailroom, 3 rd floor Brooks)

4. In this simple example, we didn’t have to employ calculus Tom Wilson, Department of Geology and Geography This formula can be evaluated – as is – for points and equidimensionally shaped objects g gvgv Sulfide deposit r z x  

5. Sulfide minerals Tom Wilson, Department of Geology and Geography L. Morgan, 2010, Geophysical characteristics of volcanogenic massive sulfide deposits; USGS report 210-5070-C, 19p.

6. Excess density or high horizontal density contrast produces observable changes in the g Tom Wilson, Department of Geology and Geography L. Morgan, 2010, Geophysical characteristics of volcanogenic massive sulfide deposits; USGS report 210-5070-C, 19p.

7. In this location the gravity anomaly was less distinctive than other geophysical features Tom Wilson, Department of Geology and Geography L. Morgan, 2010, Geophysical characteristics of volcanogenic massive sulfide deposits; USGS report 210-5070-C, 19p.

8. In this area, a distinctive gravity anomaly is associated with a sulfide deposit Tom Wilson, Department of Geology and Geography M. Thomas, 1997, Gravity prospecting for massive sulfide deposits in the Bathurst mining camp, New Brunswick Canada: Proceedings of Exploration 97, Fourth Decenial International Conference on Mineral Exploration, p837-840.

9. However, we had to modify it algebraically since we wanted to solve for g v and express as a function of x & z (rather than r) Tom Wilson, Department of Geology and Geography Find r as a function of x and z Rewrite g in terms of x and z Express g v in terms of g and  Replace cos  with its spatial equivalent to get g v as a function of x & z Simplify by factoring z out of the denominator We looked at the geometry and did this in a series of steps

10. In the form below it is easy to compute g v at arbitrary x along the surface. Tom Wilson, Department of Geology and Geography Given that G=6.6732 x 10 -11 nt-m 2 /kg 2, x=1km, z=1.7km, R deposit =0.5km and  =2gm/cm 3, you would find that

11. Spatial variation in the gravity anomaly over the sulphide deposit Tom Wilson, Department of Geology and Geography Note than anomaly is symmetrical across the sulphide accumulation

12. One of the first problems you did in the class was a units conversion problem for acceleration Tom Wilson, Department of Geology and Geography g v at 1 km is 0.0000155 m/sec 2 Remember what a Gal is? How about a milliGal? Make the units conversion from m/sec 2 to milliGals

13. The same governing equation but with more complex geometry could be used to calculate g core and g mantle Tom Wilson, Department of Geology and Geography 11,000 kg/m 3 Approximate the average densities 4,500 kg/m 3 Here, the gravitational field is associated with shells of differing density and the problem is a little more complex. Another slant on the text problem

14. In the book problem we are just trying to estimate the mass of the earth and simplify the problem by assuming the earth can be represented by two regions: Tom Wilson, Department of Geology and Geography 11,000 kg/m 3 Approximate the average densities 4,500 kg/m 3 1) an inner core of average density,  i, and 2) an outer shell (mantle and crust) represented by another average density,  o.

15. What do we get when we integrate the surface area over r? Tom Wilson, Department of Geology and Geography 11,000 kg/m 3 We can simplify the problem and still obtain a useful result. Approximate the average densities 4,500 kg/m 3

16. Actually a pretty good approximation Tom Wilson, Department of Geology and Geography 11,000 kg/m 3 We can simplify the problem and still obtain a useful result. Approximate the average densities 4,500 kg/m 3 The result – 6.02 x 10 24 kg is close to the generally accepted value of 5.97 x 10 24 kg.

17. We could then pose the question: what is the acceleration of gravity due to the core at the Earth’s surface? Tom Wilson, Department of Geology and Geography Mass of core ~ 1.94 x 10 24 kg. Considering only the core, we find it’s mass is 1.94186x10 24 kg (about 1/3 rd the total mass of the earth.

18. With an outer radius of ~6371km (6,371,000m) Tom Wilson, Department of Geology and Geography The core is about 2900km beneath your feet, We have to keep units consistent and use G=6.6732x10 -11 m 3 /(kg-sec 2 ) M=1.94186x10 24 kg And r=6,371,000 m The contribution to the total acceleration of ~9.8 m/s 2 due to the core is 3.29m/s 2.

19. In general we express the acceleration of gravity produced by an object of arbitrary shape as Tom Wilson, Department of Geology and Geography We usually look for some symmetry to help simplify our problem.

20. Let’s take a look at the acceleration produced by a very long horizontal cylinder Tom Wilson, Department of Geology and Geography This could be a cave passage or tunnel. Point of observation m r r+dr dx

21. In this example, we can let the cross sectional area = dydz Tom Wilson, Department of Geology and Geography Point of observation m r r+dr dx Again, we are interested in the vertical component of g, so 

22. Zoom in on the little element dx Tom Wilson, Department of Geology and Geography r r+dr Area =  R 2 R 

23. Substitute for dx, simplify and also note that r=m/cos  Tom Wilson, Department of Geology and Geography Point of observation m r r+dr dx 

24. Note that the only variable left is  and the limits of integration would be from -  /2 to  /2 Tom Wilson, Department of Geology and Geography Point of observation m r=m/cos  r+dr dx=rd  /cos  

25. This is an integral you should be able to evaluate Tom Wilson, Department of Geology and Geography What do you get?

26. Assume that you run a gravity survey across a roughly cylindrically shaped cave passage Tom Wilson, Department of Geology and Geography Hint: replace m with r to develop this relationship g gvgv Cave Passage r z x   Cylinder goes in and out of the slide

27. Developing g as a function of x and z Tom Wilson, Department of Geology and Geography Hint2: Once again – take the vertical component! g gvgv Cave Passage r z x  

28. Tom Wilson, Department of Geology and Geography LiLi LfLf The total natural strain, , is the sum of an infinite number of infinitely small extensions In our example, this gives us the definite integral Where S is the Stretch

29. Tom Wilson, Department of Geology and Geography Strain (or elongation) (e), stretch (S) and total natural strain (  ) Elongation Total natural strain  expressed as a series expansion of ln(1+e) The six term approximation is accurate out to 5 decimal places!

30. Comparison of finite elongation vs. total natural strain Tom Wilson, Department of Geology and Geography

31. Volume of the earth – an oblate spheroid In this equation r varies from r e, at the equator, to r=0 at the poles. z represents distance along the earth’s rotation axis and varies from –r p to r p. The equatorial radius is given as 6378km and the polar radius, as 6457km.

32. Problem 9.10 Tom Wilson, Department of Geology and Geography In this problem, we return to the thickness/distance relationship for the bottomset bed. Problems 9.9 and 9.10 will be due next Tuesday

33. Don’t forget to hand in the answer to the gravity problem! Tom Wilson, Department of Geology and Geography What was the gravitational acceleration produced by the sulfide deposit?

34. Start reviewing materials for the final! … current to-do list Tom Wilson, Department of Geology and Geography 1.Problem 9.7 is due today 2.Hand in the gravity computation 3.I will give you till next Tuesday to finish up problems 9.9 and 9.10 4.Start reviewing class materials. Next week is a final review week No class this Thursday