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University of Santo Tomas This is pick out answer NOT an essay.................1. Alfred Wegener included all of the following lines of evidence as supporting his "continental drift" hypothesis except one. Which was NOT part of Wegener's evidence?A. The geometrical fit between South America and Africa B. Late Paleozoic glacial features C. Similar fossils on widely-separated continents D. Stripes of reverse magnetism on either side of oceanic rides 2.During an earthquake near Port Royal, Jamaica, the water-saturated sand on which the city was built vigorously shook. As a result, the sand particles lost contact with one another, giving the mixture the consistency of a thick milk shake. Anything supported by the ground, such as buildings and people, either floated or sank. One eyewitness stated: "Whole streets with inhabitants were swallowed up. . . . Some were swallowed quite down, and cast up again by great quantities of water; others went down and were never more seen." This is a description of __________. A. liquefaction B. tsunami C. landslide D. subsidence 3.The distance between a seismological recording station and the earthquake source is determined from the __________. A. earthquake magnitude B. intensity of the earthquake C. length of the seismic record D. arrival times of P and S waves4. You are dropped off by helicopter at a site on the land. You are told that it is a plate boundary. Almost immediately, you feel a weak earthquake. You look around and note the absence of volcanoes, but you notice an offset stream channel, and find a vertical fault with two kinds of rock on either side of it. What sort of plate boundary are you most likely on? A. Divergent B. Convergent C. Transform D. Plate interior (hot spot)5. Which of the following energy sources is thought to drive the lateral motions of Earth's lithospheric plates? A. Gravitational attractive forces of the Sun and Moon B. Electrical and magnetic fields localized in the inner core C. The upward movement of hot buoyant mantle and the downward pull of cold oceanic lithosphere D. Swirling movements of the molten iron particles in the outer core The motion of tectonic plates is driven by convection in the mantle. In simple terms, convection is the idea
that dense, cold things sink, and buoyant, warm things rise. In the earth the cold sinking things are slabs (subducting plates) and the warm things are plumes, or just rising material from deeper in the mantle. There are three main forces that determine the rate at which tectonic plates move as part of the mantle convection system: This force balance is given by: \[F_{ridge-push}+F_{slab-pull}-F_{viscous-drag}=0\] Because the viscous drag depends explicitly on the plate velocity, it is possible to predict the speed plate motion when the forces are balanced (under certain simplifying assumptions). Recall from Chapter 1 that stress is related to force as \(\sigma=\frac{F}{A}\) where \(A = L W\). Rearranging, \(F=\sigma A\). When considering the force balance on the a plate it is useful to using a simplified geometry as shown below and consider the force per unit length (or ridge or subduction zone) \[F= \sigma A/W = \sigma L \label{force}\] As seen in the figure below, \(W\) is the length of
the plate along the ridge and subduction zone. By considering the force per unit length (in the W direction), we can then consider the forces working along a profiles from the ridge to the subduction zone and ignore any variation in the ridge-parallel direction. One of the mechanisms for plate motion that is a result of convection is viscous drag. This concept relates to ideas we discussed earlier, including the
Couette Flow and viscous stress. The drag on the base of the oceanic lithosphere can both drive or resist plate motion, depending on the relative motion between the plate and the underlying mantle. For this discussion, we will consider that the plate is being pulled or pushed by slab-pull and ridge push and drags the asthenosphere with it. Therefore viscsous drag oppose the motion of the plate (or slab) and acts to slow it down. From the figure above, we can derive an equation to describe plate motion due to viscous drag. \[V=V_o(1-\frac{y}{h})\] \[\dot{\varepsilon}=\frac{1}{2}\frac{dV}{dy}=-\frac{V_o}{2h}\] \[\begin{align*}\sigma &=2\eta\dot{\varepsilon}
\\[4pt] &=2\eta(\frac{-V_o}{2h}) \\[4pt] &=\frac{-V_o\eta}{h} \end{align*}\] The negative sign indicates an opposition to plate motion, i.e., the viscous drag. We now have an equation that describes the magnitude of drag: \[|\sigma|=\frac{V_o\eta}{h}\] placing this into our previous force equation (Equation \ref{force}): \[F=\sigma L = \frac{V_o\eta}{h} L\] Let's do an example calculating viscous drag under a plate. Example Viscous Drag
Under the Plate Assume that: From the strain-rate, we can get the stress as shown above, and then multiply
by the length of the plate Using these numbers, we can calculate the force on the descending plate, \(F=1.59 \times 10^{11}\) N/m. The above example is for the viscous drag just under the plate, but now we will also consider a case where there are stresses acting on both sides of the subducting slab. As the slab sinks into the mantle, the viscous mantle on both sides resists the sinking motion of the slab. For this simple example, \[\sigma=\frac{V_o\eta_{av}}{h}\] \[F=\sigma
L_{slab}\] \[F_{vis.}=2F=2\sigma L_{slab}=\frac{2V_o\eta_{av}}{h}L_{slab}\] Example Viscous Drag on Two Sides of a Plate As we did in the previous example, consider viscous drag on a slab, but now on both sides of the plate. We are given that In this case, \(F = 78 \times 10^{13}\) N/m. If we change \(\eta\) to \(10^{21}\), then equals \(F=1.78x10^{10}\) N/m. Figure \(\PageIndex{5}\): Force on a Descending Plate On the sides of the slab, the above figures demonstrate the forces that are present. We can also see that the slab is sinking vertically, which is not often the case in the real world. From the right figure, \(\eta\approx 10^{21}\).
This is actually too high because of non linear viscosity, but we are ignoring other things so it works for this example. Notice that the stress is proportional to the velocity \(\sigma \alpha V\) so as the velocity increases so does the stress opposing the sinking. However, the biggest changes come from the viscosity because it can changes by x10-100 in the upper mantle, thus this change has the biggest effect on the viscous drag. The viscous drag is also
proportional to the length of the slab, so the resistance also increases as the slab gets longer (but so does the slab pull force). The main force on a subducting plate is the slab-pull force. This is the force due to the density contrast
between a slab and its surroundings, ie., the mass anomaly of the slab in the mantle. We can approximate this force by thinking of the sinking slab a a vertical rectangle (like the viscous drag example) with a width, \(w\), hanging down a distance, \(d\) from the Earth's surface with a temperature \(\Delta T\) different than its surroundings. The density as a function of temperature is given by \[\rho (T)=\rho_m(1+\alpha (T_m - T))\] where \(\rho_m \) is the reference mantle
density at \(T = T_m\) and cold material has a higher density, and \(\alpha\) is the thermal expansion coefficient and is ~\(2 \times 10^{-5}\) 1/K in the sinking slab. For the slab, we can rewrite this as: \[\rho (T)=\rho_m(1+\alpha \Delta T)\] where \(\Delta T=T_m-T_{slab}\) The density anomaly is then \[\Delta \rho= \rho_m\alpha\Delta T\] This is the extra density of the slab relative to the mantle. To calculate the force per unit area, we need to determine
the mass anomaly per unit area, \(\Delta M = \Delta \rho A_{slab} \), where A_{slab} is the cross sectional area of the slab along the profile (width across times length into the mantle). The temperature in the slab is related to the temperature of the subducting plate. Recall from the chapter on thermal diffusion that the temperature of a tectonic plate is given by the half-space cooling model. Once the plate starts sinking, the cold slab warms up, and the surrounding mantle cools down.
However, the total temperature anomaly (temperature difference times volumes) stays the same. (Remember that there is conservation of energy, so the total energy of the cold slab is conserved). So, if we integrate the temperature profile solution from the half-space cooling model over the thickness of this plate, this will also be the thermal anomaly of the the slab. Finally, because the density anomaly is just \[\Delta \rho=\rho_m\alpha\Delta T\], we can get the total density anomaly by just
multiplying by \(\rho_m \alpha\). So, the trick to figuring out the slab pull force is the first determine the density anomaly of the subducting plate just before it
enters the trench. Then imagine taking a length of plate equal to the length of the slab and rotating it en masse in to the mantle. The density anomaly of the sinking slab will be the same. To get the density anomaly, we integrate the temperature equation for the plate from the half-space cooling model. \[T(z, t)=(T_s-T_m)erfc(\frac{z}{2\sqrt{\kappa t}})+T_m\] Here \(z\) is the depth through the subducting plate, before it sinks, or the width through the slab while it is sinking. \[\Delta T=T_m - T(z, t)\] \[\Delta T=-(T_s-T_m)erfc(\frac{z}{2\sqrt{\kappa t}})\] \[\Delta T=(T_s-T_m)erfc(\frac{z}{2\sqrt{\kappa t}})\] Then, multiplying the \(\rho_m \alpha\), \[\Delta \rho=\rho_m\alpha\Delta T\] \[\Delta \rho=\rho_m\alpha\Delta T(z)\] Next we integrate over the width (depth) to the get the mass per unit length (length parallel to the trench) in a single profile, (we will need to sum this over the length of the slab): \[\Delta m=\int_{0}^{z\rightarrow\infty}{\Delta \rho(z)dz}\] \[\Delta m=\rho_m\alpha(T_m-T_s)\int_{0}^{\infty}erfc(\frac{z}{2\sqrt{\kappa t}})dz\] Now we use variable substitution so we can solve the integral Let \(q=\frac{z}{2\sqrt{\kappa t}}\) where \(z\rightarrow\infty\) and \(q\rightarrow\infty\). \[\frac{dq}{dz}=\frac{1}{2\sqrt{\kappa t}}\] \[dz=2\sqrt{\kappa t}dq\] Now substitute into the integral, \[\Delta m=\rho_m\alpha(T_m-T_s)2\sqrt{\kappa t}\int_{0}^{\infty}erfc(q)dq\] The complementary error function integral solution can be found in any integral table (or search online) and equals \(\sqrt{\frac{1}{\pi}}\) \[\Delta m=\rho_m\alpha(T_m-T_s)2\sqrt{\frac{\kappa t}{\pi}}\] The above equation gives the mass anomaly (relative to the surrounding mantle) per unit length. Finally, we can determine the total mass anomaly by multiplying by the length of the slab: \[\Delta M=\Delta m L_{slab}\] The slab pull force is then given by, \[F_{slab-pull}=\Delta m L_{slab} g\] and subsituting for \(\Delta m\), \[F_{sp}=2 \rho_m\alpha (T_m-T_s) g L_{slab}\sqrt{\frac{\kappa t}{\pi} } \] Example Slab-Pull Force We are given that
We know that \(F_{sp}=\Delta m L_{slab} g\) Thus, the force per unit length of the slab-pull is Fsp=2.88x1013 N/m You might notice that this force is 100 - 1000x more than the viscous drag force. Ridge-Push ForceLet's cover a final force a subducting plate would experience, the ridge-push force. This force results from the elevation of oceanic ridges above the seafloor. This difference in height leads to pressure that 'pushes' the plate away from the ridge. Ridge push is the simplest force in some ways, as all its components can be easily examined. Simply put, to determine ridge-push forces, we need to do two things. First, look at the isostatic balance and the depth of oceans w(t). Second, we need to look at the force balance on the plate. \[T(t)=(T_s-T_m)erfc(\frac{z}{2\sqrt{\kappa t}})+T_m\] \[\rho(T)=\rho_m(1+\alpha\Delta T)\] \[\Delta T=T_m-T(t)\] T(t) is <Tm+\(\Delta T\) because of more density. Balance pressures: P1=P2 \[drp_wg+\int_{dr}^{dr+w}\rho (T)gdz+\int_{dr+w}^{z_c}\rho (T)gdz=(dr+w)\rho_wg + \int_{dr+w}^{z_c}\rho (T)gdz\] \[\int_{dr}^{dr+w}\rho_mdz+\int_{dr+w}^{z_c}\rho_mdz=wp_w+\int_{dr+w}^{z_c}\rho (z)dz\] \[w\rho_m-wp_w=\int_{dr+w}^{z_c}\rho (z)-\rho_mdz\] \[\rho (z)=\rho (T)\rightarrow T(z)\] Like we did above, we will use the complementary error function and find its solution in a integral table. \[w(\rho_m-\rho_w)=\rho_m\alpha(T_m-T_s)\int_{0}^{\infty}erfc(\frac{z'}{2\sqrt{\kappa t}})dz'\] The complementary error function integral solution is \(2\sqrt{\frac{\kappa t}{\pi}}\) \[w=\frac{2\rho_m\alpha(T_m-T_o)}{\rho_m-\rho_w}\sqrt{\frac{\kappa t}{\pi}}\] An important thing to note is that the depth of the ocean basins deepen at ~\(\sqrt{t}\) (the square root of age) ⇒balanced pressures (isostasy) Ridge-push is very similar to the balanced pressures idea, but uses balanced forces instead. ⇒\(\Delta P\)⇒drives flow across A. No net force⇒no \(\Delta P\). \[F\backsim\Delta P A\] This is the force per unit width. Example Ridge-Push The forces in this example should balance. After we write the integrals for the various forces, we should have an expression where FRP=F1-F2-F3. \[F_1=\int_{0}^{z_c}\rho_mgzdz=F_s\] Here, \(\rho_mgz\) represents the pressure, dz is XA⇒Area⇒wXL\(\frac{area}{w}\)=L We can do a simple integral in depth, this is the same pressure from the mantle below a plate. Rewriting F1 0→zc, 0→w + w-zL \[F_1=\int_{0}^{w}\rho_mgzdz+\int_{w}^{w+z_c}\rho_mgzdz+\int_{0}^{2L}\rho_mg\bar{z}d\bar{z}\] Where we let \(\bar{z}=z-w\) \[F_2=\int_o^w\rho_wgz=F_4\] \(\rho_wgz\) is P2. Here, we do not need to integrate over shape of w. \[F_3=\int_0^{z_L}P_Ld\bar{z}\] \[P_L=\rho_w g w+\int_0^{\bar{z}} \rho_L (T)g d\bar{z}\] Here, \(\rho_w g w\) is the pressure due to the water and \(\rho_L\) is the temperature-dependent density of the lithosphere. Remember our force balance equation: \[ F_{RP}=F_1-F_2-F_3\] Plugging everything in and skipping some of the basic algebra we get: \[F_{RP}=g[\frac{1}{2}(\rho_m-\rho_w)w^2+\kappa t\rho_m\alpha(T_m+T_s)]\] We get the term w2 because PL~w and we integrated PL. From isostasy, we can plug in \[w(t)=\frac{2\rho_m\alpha(T_m-T_s)}{(\rho_m-\rho_w)}\sqrt{\frac{\kappa t}{\pi}}\] \[F_{RP}=g \rho_m\alpha(T_m-T_s) \kappa t (1+\frac{2p_m\alpha(T_m-T_s)}{\pi(\rho_m-\rho_w)})\] Notes on Ridge-Push A few things to note: isostasy→controlling w(t) The pressure across the lithosphere→ridge-push force FRP\(\alpha\)t age of the oldest part of plate Now that we have our general solution for the ridge-push force, if we are given actual values, we can solve it. If g=9.81, k=1e-6\(\frac{m^2}{s}\), \(\rho_m\)=3300\(\frac{ky}{m^3}\), Tm-Ts=1400 K, \(\alpha\)=2x10-5\(\frac{1}{K}\), \(\rho_m\)=1000\(\frac{kg}{m^3}\), we can finally find the force \[F_{RP}=3.24x10^{12}\frac{N}{m}\] What do you call the process involved in the tectonic movements where the weight of subducting plate pull the trailing slab at the subduction zone *?The process of a tectonic plate descending into the mantle is termed subduction. Slab pull occurs when an oceanic plate subducts into the underlying mantle.
What do you call the process involved that tectonic movement where the weight of the uplifted ridge pushes the oceanic crust toward the trench of the subduction zone?Ridge push (also known as gravitational sliding) or sliding plate force is a proposed driving force for plate motion in plate tectonics that occurs at mid-ocean ridges as the result of the rigid lithosphere sliding down the hot, raised asthenosphere below mid-ocean ridges.
What do you call the process where the weight of the subducting slab drags the rest of the plate down into the mantle?Ridge-push refers to gravity causing lithosphere to slide downhill away from the elevated mid-ocean ridges. Slab-pull refers to the weight of subducting slabs dragging the rest of the plate down into the mantle.
What pulls the trailing slab of plate into the subduction zone during slab pull?"slab pull" As lithospheric plates move away from midocean ridges they cool and become denser. They eventually become more dense than the underlying hot mantle. After subducted, cool, dense lithosphere sinks into the mantle under its own weight. This helps to pull the rest of the plate down with it.
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What do you call the process involved in the tectonic movements where the weight of subducting plate pull the trailing slab at the subduction zone?
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