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EART 160: Planetary Science 06 February 2008
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Last Time Planetary Surfaces –Summary Planetary Interiors –Terrestrial Planets and Icy Satellites –Structure and Composition: What all is inside? –Exploration Geophysics: How can we tell?
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Today Homework 3 graded Projects – Have you got a topic yet? Midterm Friday! – details Paper Discussion: Stevenson (2001) –Mars Magnetic Field Planetary Interiors –Pressure and Temperature –Heat Sources and Cooling Mechanisms –Rheology
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Mean: 35 St. Dev.:8
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Homework Issues Please talk to me if you have difficulties –Before class is not usually a good time –No more Monday due dates Units: –Pressure:1 Pa = 1N m -2 = 1 kg m -1 s -2 –Energy:1 J = 1 kg m 2 s -2 –Power:1 W = 1 J s -1 Stress –Tectonic stress is not the Lithostatic Pressure –Normal stress is the Pressure + normal component of Tectonic –Shear stress is the tangential component of Tectonic
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Midterm Exam Closed-book –I will provide a formula sheet –You may provide an 8.5” × 11” sheet of paper with whatever you want on it; hand it in with your test. –Formulae won’t help you if you don’t understand them! Several short-answer questions, descriptive 3 quantitative problems, pick 2 to answer –Similar to Homework, but less involved –Show your work! Review Session? What say ye?
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Exam Topics Orbital Mechanics –Kepler’s Laws, Newton’s Laws –Conservation of Energy, Momentum, Angular Momentum –Escape Velocity Solar System Formation –Composition of the Solar Nebula –Jeans Collapse –Accretion and Runaway Growth –Frost Line Meteorites and Asteroids –Chondrites: Remnants from Early Solar System –Role of collisions –Radiometric Dating
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Impacts Crater size depends on impactor size, impact velocity, surface gravity Crater morphology changes with increasing size Simple vs. complex crater vs. impact basin Depth:diameter ratio Crater size-frequency distribution can be used to date planetary surfaces Energetics, Global effects due to impacts Atmospheres and geological processes can affect size-frequency distributions
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Volcanism Solidus & liquidus Magmatism when solidus crosses adiabat –Higher temperatures, reduced pressure or lowered solidus Volcanism when buoyant magma erupts Conductive cooling time t = d 2 / Magma composition controls style of volcanism Flow controlled by viscosity –Viscous materials = d dt
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Tectonics Planetary cooling leads to compression Hooke’s law and Young’s modulus –Elastic materials = E Contraction and cooling Byerlee’s law Styles of tectonicsm: compression, extension, shear
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Gradation Erosion on planets with atmospheres –Aeolian, Fluvial, Glacial Mass Wasting, Sputtering everywhere. Valley networks, gullies and outflow channels
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Planets are like Ogres Compositional Layers 1.Core: Metal 2.Mantle: Dense silicate rock (peridotite) 3.Crust: thin silicate rock (basalt) 4.Ocean: liquid layer 5.Atmosphere: gas layer Mechanical Layers 1.Inner Core: solid metal 2.Outer Core: liquid metal 3.Lower Mantle: High viscosity silicate 4.Aesthenosphere: ductile upper mantle 5.Lithosphere: Brittle uppermost mantle and crust ON an icy satellite, the ocean will be beneath the icy mantle. Other ice phases are denser than water. May have ice – ocean -- ice
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Mercury Venus Earth Moon Mars Ganymede Io Actual Planetary Interiors Only Earth has an layered core The Moon has a TINY core (why?) Icy satellites may have liquid oceans beneath the ice shell High-Pressure Ices beneath that. Interior of Europa -- NASA
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Stevenson et al., 2001 Nature
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Pressures inside planets Hydrostatic assumption (planet has no strength) For a planet of constant density (is this reasonable?) So the central pressure of a planet increases as the square of its radius Moon: R=1800km, P=7.2 GPa Mars: R=3400km, P=26 GPa
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Pressures inside planets The pressure inside a planet controls how materials behave E.g. porosity gets removed by material compacting and flowing, at pressures ~ few MPa The pressure required to cause a material’s density to change significantly depends on the bulk modulus of that material The bulk modulus K controls the change in density (or volume) due to a change in pressure Typical bulk modulus for silicates is ~100 GPa Pressure near base of mantle on Earth is ~100 GPa So change in density from surface to base of mantle should be roughly a factor of 2 (ignoring phase changes)
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Real planets Notice the increase in mantle density with depth – is it a smooth curve? How does gravity vary within the planet?
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Phase Transitions Under pressure, minerals transform to different crystal structure How do we detect this? Transition zone can sore a LOT of water! How do the depths change on other planets?
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Temperature Planets generally start out hot (see below) But their surfaces (in the absence of an atmosphere) tend to cool very rapidly So a temperature gradient exists between the planet’s interior and surface We can get some information on this gradient by measuring the elastic thickness, T e The temperature gradient means that the planet will tend to cool down with time
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Heat Sources Accretion and Differentiation –U = E acc –E acc = m C p T –C p : specific heat Radioactive Decay –E = H m –H ~ 5x10 -12 W kg -1 –K, U, Th today –Al, Fe early on Tidal Heating in some satellites
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Specific Heat Capacity C p The specific heat capacity C p tells us how much energy needs to be added/subtracted to 1 kg of material to make its temperature increase/decrease by 1K Energy = mass x specific heat capacity x temp. change Units: J kg -1 K -1 Typical values: rock 1200 J kg -1 K -1, ice 4200 J kg -1 K -1 E.g. if the temperature gradient near the Earth’s surface is 25 K/km, how fast is the Earth cooling down on average? (about 170 K/Gyr) Why is this estimate a bit too large? –Atmosphere insulates
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Energy of Accretion Let’s assume that a planet is built up like an onion, one shell at a time. How much energy is involved in putting the planet together? early later In which situation is more energy delivered? Total accretional energy = If all this energy goes into heat*, what is the resulting temperature change? Earth M=6x10 24 kg R=6400km so T=30,000K Mars M=6x10 23 kg R=3400km so T=6,000K What do we conclude from this exercise? * Is this a reasonable assumption? If accretion occurs by lots of small impacts, a lot of the energy may be lost to space If accretion occurs by a few big impacts, all the energy will be deposited in the planet’s interior So the rate and style of accretion (big vs. small impacts) is important, as well as how big the planet ends up
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Cooling a planet Large silicate planets (Earth, Venus) probably started out molten – magma ocean Magma ocean may have been helped by thick early atmosphere (high surface temperatures) Once atmosphere dissipated, surface will have cooled rapidly and formed a solid crust over molten interior If solid crust floats (e.g. plagioclase on the Moon) then it will insulate the interior, which will cool slowly (~ Myrs) If the crust sinks, then cooling is rapid (~ kyrs) What happens once the magma ocean has solidified?
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Cooling Radiation –Photon carries energy out into space –Works if opacity is low –Unimportant in interior, only works at surface Conduction –Heat transferred through matter –Heat moves from hot to cold –Slow; dominates in lithosphere and boundary layers Convection –Hot, buoyant material carried upward, Cold, dense material sinks –Fast! Limited by viscosity of material Running down the stairs with buckets of ice is an effective way of getting heat upstairs.-- Juri Toomre
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Conduction - Fourier’s Law Heat flow F T 1 >T 0 T1T1 T0T0 F d Heat flows from hot to cold (thermodynamics) and is proportional to the temperature gradient Here k is the thermal conductivity (W m -1 K -1 ) and units of F are W m -2 (heat flux is power per unit area) Typical values for k are 2-4 Wm -1 K -1 (rock, ice) and 30- 60 Wm -1 K -1 (metal) Solar heat flux at 1 A.U. is 1300 W m -2 Mean subsurface heat flux on Earth is 80 mW m -2 What controls the surface temperature of most planetary bodies?
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Diffusion Equation Here is the thermal diffusivity (=k/ C p ) and has units of m 2 s -1 Typical values for rock/ice 10 -6 m 2 s -1 F1F1 F2F2 zz We can use Fourier’s law and the definition of C p to find how temperature changes with time: In steady-state, the heat produced inside the planet exactly balances the heat loss from cooling. In this situation, the temperature is constant with time
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Diffusion length scale How long does it take a change in temperature to propagate a given distance? This is perhaps the single most important equation in the entire course: Another way of deducing this equation is just by inspection of the diffusion equation Examples: –1. How long does it take to boil an egg? d~0.02m, =10 -6 m 2 s -1 so t~6 minutes –2. How long does it take for the molten Moon to cool? d~1800 km, k=10 -6 m 2 s -1 so t~100 Gyr. What might be wrong with this answer?
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Internal Heating Assume we have internal heating H (in Wkg -1 ) From the definition of C p we have Ht= TC p So we need an extra term in the heat flow equation: This is the one-dimensional, Cartesian thermal diffusion equation assuming no motion In steady state, the LHS is zero and then we just have heat production being balanced by heat conduction The general solution to this steady-state problem is:
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Example Let’s take a spherical, conductive planet in steady state In spherical coordinates, the diffusion equation is: The solution to this equation is So the central temperature is T s +( HR 2 /6k) E.g. Earth R=6400 km, =5500 kg m -3, k=3 Wm -1 K -1, H=6x10 -12 W kg -1 gives a central temp. of ~75,000K! What is wrong with this approach? Here Ts is the surface temperature, R is the planetary radius, is the density
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Convection Convective behaviour is governed by the Rayleigh number Ra Higher Ra means more vigorous convection, higher heat flux, thinner stagnant lid As the mantle cools, increases, Ra decreases, rate of cooling decreases -> self-regulating system Image courtesy Walter Kiefer, Ra=3.7x10 6, Mars Stagnant lid (cold, rigid) Plume (upwelling, hot) Sinking blob (cold)
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Viscosity Ra controls vigor of convection. Depends inversely on viscosity, . Viscosity depends on Temperature T, Pressure P, Stress , Grain Size d. A – pre-exponential constantE – Activation Energy V – Activation VolumeR – Gas Constant n – Stress Exponentm – Grain-size exponent Viscosity relates stress and strain rate
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Viscoelasticity A Maxwellian material has a viscous term and an elastic term. If h is high, we get an elastic behavior. If h is low, we get a viscous behavior. Depends also on the rate of stress. Materials are elastic on a short timescale, viscous on a long one. There are other types of viscoelasticity, but Maxwell is the simplest
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Elastic Flexure The near-surface, cold parts of a planet (the lithosphere) behaves elastically This lithosphere can support loads (e.g. volcanoes) We can use observations of how the lithosphere deforms under these loads to assess how thick it is The thickness of the lithosphere tells us about how rapidly temperature increases with depth i.e. it helps us to deduce the thermal structure of the planet The deformation of the elastic lithosphere under loads is called flexure EART163: Planetary Surfaces
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Flexural Stresses In general, a load will be supported by a combination of elastic stresses and buoyancy forces (due to the different density of crust and mantle) The elastic stresses will be both compressional and extensional (see diagram) Note that in this example the elastic portion includes both crust and mantle Elastic plate Crust Mantle load
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Flexural Parameter Consider a load acting on an elastic plate: TeTe mm load ww The plate has a particular elastic thickness T e If the load is narrow, then the width of deformation is controlled by the properties of the plate The width of deformation is called the flexural parameter and is given by E is Young’s modulus, g is gravity and n is Poisson’s ratio (~0.3)
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If the applied load is much wider than , then the load cannot be supported elastically and must be supported by buoyancy (isostasy) If the applied load is much narrower than , then the width of deformation is given by If we can measure a flexural wavelength, that allows us to infer and thus T e directly. Inferring T e (elastic thickness) is useful because T e is controlled by a planet’s temperature structure
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Example This is an example of a profile across a rift on Ganymede An eyeball estimate of would be about 10 km For ice, we take E=10 GPa, =900 kg m -3 (there is no overlying ocean), g=1.3 ms -2 Distance, km 10 km If =10 km then T e =1.5 km A numerical solution gives T e =1.4 km – pretty good! So we can determine T e remotely This is useful because T e is ultimately controlled by the temperature structure of the subsurface
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T e and temperature structure Cold materials behave elastically Warm materials flow in a viscous fashion This means there is a characteristic temperature (roughly 70% of the melting temperature) which defines the base of the elastic layer 110 K 270 K elastic viscous 190 K E.g. for ice the base of the elastic layer is at about 190 K The measured elastic layer thickness is 1.4 km (from previous slide) So the thermal gradient is 60 K/km This tells us that the (conductive) ice shell thickness is 2.7 km (!) Depth 1.4 km Temperature
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T e in the solar system Remote sensing observations give us T e T e depends on the composition of the material (e.g. ice, rock) and the temperature structure If we can measure T e, we can determine the temperature structure (or heat flux) Typical (approx.) values for solar system objects: BodyT e (km)dT/dz (K/km) BodyTeTe dT/dz (K/km) Earth (cont.) 3015Venus (450 o C) 3015 Mars (recent) 1005Moon (ancient) 1530 Europa240Ganymed e 240
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Next Time Paper Discussion – Stevenson (2001) Planetary Interiors –Cooling Mechanisms –Rheology: How does the material deform? –Magnetism
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