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Ch.2 Isostacy Ch.2 Isostacy 2. Introduction A 1730s French expedition led by Pierre Bouguer to Peru revealed a rather interesting finding – that while a plumb line was deflected from the vertical direction by the mass of the Andes mountains, the degree of deflection was much less than initially estimated. In the 19th century, Sir George Everest found a similar reduced deflection near the Himalayas. Imagine if one made a map of gravity variations near these mountains, one might have expected huge variations arising from these huge masses. However in reality one would see much smaller variations than first expected. What can account for this phenomenon? In 1855 J.H. Pratt and G. Airy independently proposed two hypotheses to explain this phenomenon. These are known as Pratt’s hypothesis and Airy’s hypothesis. No surprise there! Figure 13 (Fowler Fig5.6) illustrates the two hypotheses. (Refer to Fowler Section 5.2.2. for a quantitative discussion.) Figure 13: Airy’s and Pratt’s Hypotheses For both the Andes and the Himalayas, it appears that a reduction in mass must exist either inside or underneath these mountains in order to explain the lower- than-expected gravitational attraction near them. If one calculates the amount of mass reduction (or mass deficit) required, one finds a mass comparable to the entire mountain! This calls to mind Archimedes Principle of hydrostatic equilibrium, which says that a floating body displaces its own weight of fluid in which it is immersed. But wait - who is talking about the Andes or the Himalayas floating on a fluid? 1 Isostacy (or isostasy) is a principle that is capable of accounting for observations that large-scale topographic features on Earth have little effect on large scale, regional and global, gravity field patterns. 2.1 Geoid We can represent the gravity field by introducing a surface that is everywhere perpendicular to it. For example, if we connected all the continental regions of the ocean by canals then the resulting water surface of the sea and connected canal would represent an equipotential surface, very close to what is termed the geoid. The geoid is such a surface, and is shown in Figure 14 (Stacey Fig9.1), where the landmasses are also depicted. Figure 14: Geoid It is clear that there is little relation between the anomalies of the geoid and excess mass of the continents. In order to appreciate the expectation of continents affecting the gravity field, consider the following idealization: if Earth were a uniform, non-rotating sphere, having constant density, objects near Earth's surface would fall directly towards its center with the same acceleration everywhere on the globe's surface. It would also be true that a pendulum anywhere on the surface would point to the center when at rest and have the same period, when put into motion, no matter where it was placed. A pendulum at rest points in the direction of local gravity. For our spherically symmetric Earth, the combined effect of all the mass elements of Earth is such that the gravity vector points to the centre. Now suppose a huge (mountain sized) mass is suddenly placed nearby the pendulum. One expect a certain degree of deflection, as per Figure 15 (to be filled in by hand): 2 Figure 15: Deflection due to “uncompensated” mass One would expect the pendulum to be deflected towards the mass due to the attraction between the mass and the pendulum. Indeed this would happen in our thought experiment. Now consider the real Earth and imagine taking a pendulum near the Himalayas to check for a deflection. In fact when this was done, very much less deflection toward the mountain was seen than was expected. To understand this reconsider our thought experiment above and take the mountain away. Now instead of placing a mountain nearby, make a huge hole in the ground. What would our pendulum do? To answer this consider two regions of equal mass, equidistant on opposite sides of the pendulum in the undisturbed Earth as illustrated below: Figure 16 (to be filled in): Effect of mass excess/deficit Each of the masses attracts the pendulum by the same force so that the pendulum points directly to Earth centre. If we remove one of the masses, its attractive force disappears and the pendulum will then be deflected away from the void left by the mass removed as shown above. We can think of the void as a mass deficiency and the mountain previously added as a mass excess. Thus the deflection of the pendulum toward the excess mass above Earth could be balanced by a mass deficiency below Earth. The mass deficiency would not have to be a void but simply a reduced density as shown in the cartoon below: 3 Figure 17 (to be filled in): Effect of “compensated” mass The mountain shown here has what is known as a root of lower density than the rock in which the mountain “floats”. 2.2 Isostatic Compensation We now consider the elementary physical model of isostatic compensation, which is based on the principle of hydrostatic equilibrium and states simply that hydrostatic pressure is constant at some level in the mantle and this level is called the depth of compensation. It is clear that if this were not the case flow would occur until pressure became the same at this level. Before continuing with this principle, it is useful to review an elementary observation from hydrostatics given by L. Prandtl and illustrated below: Figure 18 (to be filled in): Hydrostatic Equilibrium In this picture there are different volumes (and masses) of fluid all connected at the base by a common reservoir. What is obvious from this figure is that the height of the fluid column is the same regardless of the shape and mass of the fluid above the reservoir. Put another way, the hydrostatic pressure or force per unit area of the fluid depends only on the height of the fluid column above (not on its volume or mass). 4 We can apply this principle to Earth's crust floating on the mantle as illustrated in the figure below: Figure 19 (to be filled in): Isostatic Compensation involving two layers We can find the depth of the root, r, from the height, h, above the surrounding crust of thickness tc under the assumption of hydrostatic equilibrium. We call the density of the crust ρc and the density of the mantle ρm. We calculate the mass of each of the two columns 1 and 2 with cross sectional area A, and then equate the pressures at the base of the columns. Mass of column 1 = Mass of Air + mass of Crust + mass of Mantle: Force at base of column 1 is its weight is Pressure at base of column 1 is Similarly for column 2 Equating these two pressures gives or Since air density << mantle density, then or where the LHS is the mass deficit below and the RHS is the mass excess above the surface. We can express the depth, r , of the root in terms of the height h above the surface as 5 We can find the relationship between the height and root by using appropriate values for the densities. We take: which yields so that for each km of height there is a 5.6 km root! If we define the excess* crustal thickness associated with a mountain of height h as E=r+h, then (Note *: “excess” here refers to the thickness above the depth of compensation.) This means that a 3km mountain is associated with an excess crustal thickness of about 20km. Just as excess mass in a mountain requires a root, a mass deficiency at the surface, due to a basin filled with water, for example, requires excess mass at its root. This is shown in the sketch below: Figure 20 (to be filled in): Isostatic Compensation involving a basin Here the depth of the basin is d, filled with material of density ρf. Equalizing the pressures at depth gives: or 6 Typical values of ρf the fill density are given in the table below: The amount of crust that needs to be removed to make a basin of depth, d, is It is useful to make up a table of relationships between r, d, and E for the three different fill materials air, water and sediments. The interpretation of values in this table is as follows. Top 2 rows: Basin depth d represents only 15% of the total crustal material excavated to produce the basin if air filled, 22% of the excavation if water filled and 63% if filled with sediments. The root r makes up the difference. Lower two rows: For a given (observed) depth of basin, d, these numbers give r and E. Of particular interest is the amount of crustal material that needs to be excavated for a basin of depth d. For a sediment filled basin, E=1.6d; which means for a 10km deep sediment filled basin, 16km of crust must be removed. This is about 50% of total crustal thickness. If instead of sediments the basin is water filled the situation is more extreme, since E=4.6d so that for d=10km we get E=46km, which is greater than mean crustal thickness. Let's consider a shallow water filled basin that is d=100m deep. The table shows that E=460m. This means that a 460m thinning of the crust will be compensated in such a way that the resulting basin is only 100m deep. In other words the base of the crust (the Moho) below a basin must rise as a result of isostatic compensation.
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