slide1 n.
Skip this Video
Loading SlideShow in 5 Seconds..
tom.h.wilson tom. wilson@geo.wvu PowerPoint Presentation
Download Presentation
tom.h.wilson tom. wilson@geo.wvu

tom.h.wilson tom. wilson@geo.wvu

224 Vues Download Presentation
Télécharger la présentation

tom.h.wilson tom. wilson@geo.wvu

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Geology 351 - Geomath More about Isostacy Segment II tom.h.wilson tom. Department of Geology and Geography West Virginia University Morgantown, WV

  2. The gravity anomaly map shown here indicates that the mountainous region is associated with an extensive negative gravity anomaly (deep blue colors). This large regional scale gravity anomaly is believed to be associated with thickening of the crust beneath the area. The low density crustal root compensates for the mass of extensive mountain ranges that cover this region. Isostatic equilibrium is achieved through thickening of the low-density mountain root.

  3. Solving isostatic equilibrium problems

  4. On Tuesday, from the foregoing starting point, we derived a couple basic relationships governing the isostatic equilibrium processes. These included: Where m represents the density of the mantle,  = m - c (where c is the density of the crust), and h represents crustal thickening (r + e).

  5. And - or from which we must also have

  6. Review ... In Class Problem: A 500m deep depression on the earth's surface fills with sandstone of density 2.2 gm/cm3. Assume that the empty basin is in isostatic equilibrium and that normal crustal thickness in surrounding areas is 20km. Calculate the thickness of sediment that must be deposited in the basin to completely fill it. (Use crustal and mantle densities of 2.8 and 3.3 gm/cm3, respectively.) Hint: Compute the initial thickness of the crust beneath the empty basin and assume that the crustal thickness beneath the basin does not change.

  7. See Boardwork Recall that on Tuesday we showed that l=16.7km - hence We also showed that r = 20-e-l-b .. thus

  8. After rearrangement or &

  9. Recall that since l = 16.7km and lt does not change as the basin is filled, we now have the depth to the base of the crust in the rifted region (b + l = 18.2km), after isostatic equilibrium has been re-established. The base of the crust now rests 1.8km above the base of the continental crust in the surrounding un-deformed area. Recall, that when the basin was empty (0.5km deep) the crust extended down to 17.2km and r (the antiroot) was 2.8. It took 1.5km of sediment to fill our half-kilometer deep basin!

  10. As sediment is deposited, the basin floor gradually drops to maintain isostatic equilibrium. Does this really happen? Conodant alteration indices from this area of the Appalachians indicate that rocks currently exposed at the surface were once buried beneath 3km of sediment.

  11. Let’s examine the dynamics of this process using EXCEL. Pick up the EXCEL file Isostacy1.xls from my shared directory.

  12. Take Home Problem: A mountain range 4km high is in isostatic equilibrium. (a) During a period of erosion, a 2 km thickness of material is removed from the mountain. When the new isostatic equilibrium is achieved, how high are the mountains? (b) How high would they be if 10 km of material were eroded away? (c) How much material must be eroded to bring the mountains down to sea level? (Use crustal and mantle densities of 2.8 and 3.3 gm/cm3.) There are actually 4 parts to this problem - we must first determine the starting equilibrium conditions before doing solving for (a).

  13. The preceding questions emphasize the dynamic aspects of the problem. A more complete representation of the balance between root and mountain is shown below. Also refer to the EXCEL file on my shared directory.

  14. A few more comments on Isostacy The Pratt Hypothesis

  15. At BC x 42 = 116 C B A The product of density and thickness must remain constant in the Pratt model. At A 2.9 x 40 = 116 C=2.76 At CC x 50 = 116 C=2.32

  16. For Next Time Complete your reading of chapter 3 Think over problem 3.11