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Large Scale Gravity and Isostasy. Chapter 9. Isostasy: The Concept of Floating Blocks. Isostasy = the concept that large topographic features effectively float on the asthenosphere Iso = same – stasis = standstill Involves a state of constant pressure at any point Pressure: P = ρ g h

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## Large Scale Gravity and Isostasy

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**Large Scale Gravity and Isostasy**Chapter 9**Isostasy: The Concept of Floating Blocks**• Isostasy = the concept that large topographic features effectively float on the asthenosphere • Iso = same – stasis = standstill • Involves a state of constant pressure at any point • Pressure: P = ρ g h • If g were measured from a balloon at constant elevation you would expect a mass excess (i.e. larger g) above mountains. • We commonly observe considerably less g over large mountains • Consistent with low density roots beneath large mountains**Isostasy: The Initial Discovery**• The hypothesis that large mountains have low density roots was first proposed during topographic surveys of India and the Himalayan Mountains • Questions: • How does this low density root form? • As a mountain range becomes eroded why are there not large negative anomalies due to the low density roots? • Explanation: • Large topographic features effectively ‘float’ on the asthenosphere • Follows Archimedes principle**Archimedes’ Principle**• Supposedly, Archimedes was taking a bath one day and noticed that the tub overflowed when he got in. • He realized that displaced water could be used to • Detect volumes of odd shaped objects • Detect density differences (helped determine that a crown wasn’t 100% gold) Archimedes Principle: • As an object is partially submerged in a fluid, the weight of fluid that is displaced = the weight reduction of the object • When the displaced fluid = the weight of the object • It floats**Isostatic Compensation**• Take the case of a layered block floating in a fluid • Remove two layers and… • The bottom is now less deep below the surface of the liquid • The top is less elevated above the surface of the fluid • Thus, floating blocks will seek isostatic equilibrium and are said to be isostatically compensated • In the case of the continental crust, density differences are small, so any significant topography in isostatic equilibrium must be balanced by a thick crustal root • E.g. the Himalayas are < 9 km above sea level, but have roots > 70 km • So, mountains in isostatic equilibrium are like the tip of an iceberg**Isostasy and Gravity**If blocks are in isostatic equilibrium… • Gravity measurements made at some constant elevation above the blocks would detect (almost) no variation in g • There would be small variations at the edges of blocks due to the nearby topography**Isostatic Calculations**• Take the simple example of two floating blocks A and B • Each has different layers of varying thickness and density • Includes an asthenospheric layer and an air layer • If A and B are in isostatic equilibrium • Their total weights are the same These are referred to as the weight equation**Isostatic Calculations**• If A and B are in isostatic equilibrium • Their total heights are the same • This is referred to as the height equation In practice… • Choose the top to be the highest elevation of rock (or water/ice) • Choose the bottom to be the lowest elevation of lithosphere**Example Isostatic Calculation #1**• Take the case of adding a 2 km thick glacier on top of a continent • The weight of the ice causes the block to sink deeper until isostatic equilibrium is reached • Treat the before and after as two separate blocks that are both in equilibrium and use weight equation (3 x 2.0) + (30 x 2.7) + (70 x 3.1) + (ha x 3.2) = (2 x 0.9) + (3 x 2.0) + (30 x 2.7) + (70 x 3.1) ha = 0.56 km • So what does this mean? • Adding 2 km of ice on top of this continent caused 0.56 km of isostatic subsidence • The elevation of the top of the ice sheet after isostatic equilibrium is reached will be only 1.44 km above the original block (from height equation) • hair + 3 + 30 + 70 + 0.56 = 2 + 3 + 30 +70 • hair = 2 – 0.56 = 1.44 km**Example Isostatic Calculation #2**• Suppose that there was a large 2 km deep lake (in equilibrium) • Gradually, sediments are brought in by rivers that eventually replace the water and fill the lake up to its current water level • How thick would the sediments be? (why not 2 km?) (2 x 1.0) + (3 x 2.0) + (30 x 2.7) + (90 x 3.1) + (ha x 3.2) = (hs x 1.8) + (3 x 2.0) + (30 x 2.7) + (90 x 3.1) Two unknowns! Need another equation. Use the height equation. 2 + 3 + 30 + 90 + ha = hs + 3 + 30 + 90 hs = ha + 2 Substitute hs = ha + 2 into weight equation 2 + ha x 3.2 = hs x 1.8 2+ ha x 3.2 = (ha + 2) x 1.8 3.2 ha + 2 = 1.8 ha + 3.6 3.2 ha = 1.8 ha + 1.6 1.4 ha = 1.6 ha = 1.142857142 km ha = 1.1 km Use Height equation to figure out hs hs = ha + 2 = 3.1 km**Example Isostatic Calculation #2**• So, ha = 1.1 km and hs = 3.1 km • It took 3.1 km of sediment to completely fill up to the previous lake level • This 3.1 km of sediment added weight and caused an isostatic compensation of -1.1 km (subsidence) • Likewise, removing material would cause isostatic uplift • But the overall elevation would decrease even after isostatic adjustment**Airy and Pratt Models of Isostasy**• Two end member models have been proposed to account for isostasy. • Airy Model: All blocks have the same density but different thicknesses • Pratt Model: All blocks float to the same depth but have different densities**Airy Model of Isostasy**• Airy Model: All blocks have the same density but different thicknesses • Thicker blocks have higher elevation and much thicker roots • Higher ground is where the lithosphere is thicker • The weight equation becomes:**Pratt Model of Isostasy**• Pratt Model: All blocks float at the same depth, but have differing density • higher elevations indicate less dense rocks • Higher ground is where the lithosphere is thicker • The weight equation becomes: • The height equation is the same for both Airy and Pratt models**Airy vs. Pratt: Which is Correct?**• The Airy and Pratt are not the only possible models • They are end member models • A combination of density changes and lithosphere thickness may occur But, in general, most data supports… • The Airy model for continental mountain ranges • Continental mountain ranges have thick crustal roots • The Pratt model for mid-ocean ridges • Mid ocean ridges have topography that is supported by density changes • Increased temperature at ridges ---> rocks expand ---> lower density**The Isostatic Anomaly**• Recall that earlier we learned that floating blocks (in equilibrium) should produce no anomaly. • Gravity measurements at B and D will yield the same values if the free-air correction is applied. • When only the latitude and free-air corrections have been made the result is called the free-air anomaly • If the Bouguer correction is made, the extra mass above B is removed and results in a negative Bouguer anomaly**The Isostatic Anomaly**• If a region is not isostatically compensated • The free-air anomaly will be positive • The Bouguer anomaly will be zero • Partially compensated regions are common • free-air anomaly > 0 • Bouguer anomaly < 0

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