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Berkeley ASTRON C162 - Planetary Astrophysics – Problem Set 8

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Astro 162 – Planetary Astrophysics – Problem Set 8Due Thursday March 31.Readings: Skim all of Chapter 8. Pay particular attention to section 8.3 on “TheDynamic Earth and Drifting Continents,” and the last part of section 8.5 on “InternalHeating from Core Separation.”Problem 1. Dry vs. Wet Adiabatic Lapse RatesHere we derive and numerically evaluate the adiabatic temperature gradient for dryand moist atmospheres. Remember that these adiabatic gradients DO NOT EQUALthe actual gradients. The adiabatic gradients are reference quantities for any planetaryatmosphere, to be compared against the actual gradient to d ecide whether an atmosphereis convective or not. The adiabatic gradients are derived und er the assumption thateach displaced fluid parcel exchanges zero heat with its su rroundings, contrary to whathappens in actuality.(a) Use the adiabatic relation f or (a dry, non-condensing) gasP ∝ ργ, (1)the cond ition of hydrostatic equilibrium,1ρdPdz= − g (2)and the ideal gas law,P =ρkTµ(3)to solve for the s o-called “dry adiabatic lapse rate,” dT/dz, in terms of g, k, µ, and γ.(b) Evaluate the dry lapse rate for conditions in the E arth ’s troposhere. Exp ress in [K/ km ], and make sure you get the s ign right.(c) If, as the temperature decreases with increasing height above the atmosphere, sometrace species (e.g., water) of the atmosph ere begins to condense out and make clouds,latent heat is released by the condensing vapor. This latent heat increases the temper-ature of the atmosphere. Thus, the so-called “wet adiabatic lapse rate” is smaller thanthe “dry” rate; the temperature of the moist atmosphere still decreases with height, but1less steeply than for a dry atmosphere because of the latent heat that is r eleased bycondensing vapor.The first law of thermodynamics can accommodate this extra release of latent heat:dU = −P dV + dQ + Lvapdm (4)where Lvap[erg/gram] is the latent heat of vaporization (condensation) of some vapor,and dm is the d ifferential amount of vapor mass that condenses. Here U, P , and V are theinternal energy, pressure, and volume of our proverbial test parcel of gas. By definition,for adiabatic processes, dQ = 0 (the parcel exchanges no heat with its environm ent).Use the first law as written above, the condition of hydrostatic equilibrium, the idealgas law, and the lecture regarding specific heats for gas to derive the wet adiabatic lapserate,dTdz=−gkµγγ−1− LvapdwdT(5)where w is the mass of condensed vapor per unit mass of (total) atmospheric gas. DONOT use the dry adiabatic relation given in (a), P ∝ ργ, because it does not ap plyunder these moist conditions. DO u se γ ≡ CP/CV; this is merely a definition for thesymbol γ as used in the equation above for the wet adiabatic lapse rate.Remember that we are treating the condensable vapor as a trace (minor) constituent ofthe total atmosphere. This is a fine approximation f or many situations: for Jupiter, thecondensable vapors are ammonia, water, and ammonium hydrosulfide (vs. molecularhydrogen and helium for th e bulk of the atmosphere); for the Earth, the condensableis water (vs. nitrogen, oxygen, and carbon dioxide); for Venus, the condensable ishydrosulfuric acid (vs. nitrogen and carbon dioxide). The quantity dw/dT measuresthe amount of vapor mass that condenses per unit m ass of atmospheric gas, per degreeKelvin change. If dT < 0, then dw > 0; hence dw/dT < 0.(d) ESTIMATE the wet adiabatic lapse rate for conditions in the Earth’s troposphere.Express in [K / km].Assume that the air is completely saturated with water vapor at every height. Use theClausius-Clapeyron formula for the saturation vapor pressure of water,Psat,water= CLexp[−Ls/(RgasT )] (6)where CL= 3 × 107bar, Rgas= 8.3 × 107erg K−1mole−1, and Ls= 5.1 × 1011erg2mole−1. As T decreases, Psat,vapordecreases; in other words, as it gets colder, the aircan’t hold as much water vapor; the water vapor is forced to condens e out as droplets(fog). Use this information and Psat,waterto ES TIMATE dw/dT .Check that you have the right magnitude for the wet lapse rate relative to the dry lapserate. One should be larger than the other.(e) Do you expect moist regions in the atmosphere where vap or is condensing to bemore likely to be convective than dry regions in the atmosphere? Which environmentdo hang-gliders prefer? In the more vigorously convective envir on ment, what is the extrasource of energy?Problem 2. The Lid Makes All the DifferenceThis problem asks you to estimate the thicknesses of the hard, conductive lithospheresfor Venus, Earth, and Mars. In so doing, you will have taken a small step towardsunderstanding why Earth exhibits plate tectonics while Venus and Mars do not.Take the main internal source of energy of every planet to be from radioactive U, Th,and K (not26Al; that fuel source ran out a long time ago). Assume each planet is, onaverage, made of chondr itic material which to day emits 5 × 10−8erg s−1g−1. So giventhe total mass of each planet, you know the internal luminosity of each planet.This radiogenic energy is carried condu ctively through the hard outer lithosphere of eachplanet. Recall the equation for the condu ctive heat flux:F = −KC∇T (7)where KCis the thermal conductivity (proportional to, but having wildly differentunits from, the thermal diffusivity), and T is the temperature. Flux F has units oferg s−1cm−2.At the top of the lithosphere, the temperature is just the surface temperature. Youknow what that is. (And don’t forget that Venus is plenty hotter than you would guessfrom the blackbody formula.)At the bottom of the lithosphere (top of the convective mantle), the temperature is∼1200 K, the temperature at which rock starts to become plastic (that is, flowing; theviscosity of rock is exponentially sensitive to temperature).Use all the f acts above to estimate the thicknesses of the lithospheres for Venus, Earth,and Mars.Geologists tells us that “thin” lithospheres are not good for plate tectonics because they3are too light and therefore do not subduct (sink) to greater depths (like trying to drown arubber duck; it just won’t go down.) Geologists also tell us that “thick” lithospheres arenot good for plate tectonics because upwelling m agma cannot penetrate to the surface.Can you begin to understand why the Earth exhibits tectonic motion but the otherplanets do not?Problem 3. Sea Floor SpreadingThe mid-Atlantic ridge rep resents an


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