RENEE CONDORI APAZA, JULIOVALDIVIA SILVA, Christopher P. McKay
A principal reason for studying planetary atmospheres isto try to understand the origin and evolution of theearth’s atmosp...
It is also clear now that the earth’s gaseous envelope ischanging and has changed. In fact it is abundantlyclear that the ...
Basic Properties of AtmospheresCompositionSizeEquilibrium TScale HeightAdiabatic Lapse RoleMixing in TroposphereRadiation ...
Type Name Mass Escape p T*(eV/u) (bar) (K)H/He Jupiter 318 18 128Gas Balls Saturn 95 6.5 98Uranus 14.5 2.3 56Neptune 17.0 ...
Type Name Mass Escape p T*(eV/u) (bar) (K)Collisionless Mercury 0.053 0.093Moon 0.012 0.029Other moonsT*: for Jovian they ...
MolecularSunH (H2) 0.86He 0.14O 0.0014C 0.0008Ne 0.0002N 0.0004Jupiter Saturn Uranus NeptuneH2 0.898 0.963 0.825 0.80He 0....
MolecularEarth Venus Mars TitanCO2 0.0031 0.965 0.953N2 0.781 0.035 0.027 0.97O2 0.209 0.00003 0.0013CH4 0.00015 0.03H2O* ...
Pressure is the weight of a column of gas: forceper unit areap = mg N (column density: N)Thickness if frozen: Hsp(bar) Hs(...
p, T, n (density) Equation of StateConservation of SpeciesContinuity Equation: Diffusion and FlowSources / Sinks: Volcanoe...
Equilibrium TemperatureHeat In = Heat OutorSource (Sun) = Sink (IR Radiation to Space)Planetary body with radius a it abso...
Fraction of radiation absorbed in atmosphere vs. wavelengthPrincipal absorbing species indicated
Source=AbsorbArea heat flux amount absorbedpa2 x [F / Rsp2] x [1-A]A = Bond Albedo: total amount reflected(Complicated)Sol...
Bond Albedo, A, isfraction of sunlightreflected to space:Surface, clouds, scattered
Set EqualHeat In = Heat OutTe = [ (F / Rsp2) (1-A) / 4 ]1/4Rsp A Te TsMercury 0.39 0.11 435 440Venus 0.72 0.77 227 750Ear...
Right handaxis meltingpoint
Pressure vs. AltitudeHydrostatic LawForce Up = Force Downp- A=area---------------------------------------------Draw forces...
dp/dz = - gNow Use Ideal Gas Lawp = nkT (k=1.38 x 10-23 J/K) =kT/morp = (R/Mr)T [Gas constant: R=Nak =8.3143 J/(K mole)...
Pressure vs. Altitudep = po exp( - ∫ dz / H)(assuming T constant)p = po exp( - z / H)orDensity vs. Altitude = 0 exp( - z...
Pressure: pp = weight of a column of gas (force per unit area)1bar = 106 dyne/cm2=105 Pascal=0.987atmospheresPascal=N/m2 ;...
PARTIAL PRESSURESLower AtmosphereMixing dominates: use m or MrUpper atmosphereDiffusive separationPartial Pressure (cons...
ShowingRegion wheregasesdiffusivelyseparate
Convection Dominates  Adiabatic Lapse RateIn the troposphereRadiation Dominates  Greenhouse EffectIn the troposphere and...
Shows layeredatmosphereRadiationAbsorptionIndicated
Imagine gas moving up or down adiabatically: noheat in or out of the volumeEnergy = Internal energy + Workdq = cvdT + p dV...
Differentiatep dV + dp V = (R/Mr) dTorcv dT = - (R/Mr) dT + V dp(cv +R/Mr) dT = dp / cp (dT/dz) = (dp/dz) / Apply Hydros...
(dT/dz) = -g / cp = - dHeating at surface + Slow vertical motion.T= [Ts - d z]T falls off linearly with altitudecp (erg/...
cp = Cp / m = cv + (R/Mr)= Cv + kmCvT = heat energy of a moleculeAtom = Cv = (3/2)k ; kinetic energy only3-degrees of free...
But potential energy of internal vibrationsneeded.Cv  3.5 k = 4.8 x 10-16 ergs/K1 mass unit = 1.66x 10-24 gmcv  1.0 x 10...
Now have p(z) with T dependence.Use (dT/dz) = -g / cp and dp/dz = - ρ g and p = nkTdp/p = - mgdz/kT = [m cp/k] dT/T = x dT...
 = T (po/p)1/xAdiabatic  Entropy = ConstantGas can move freely along constant  linesUsing dq = T dS where S is entropyC...
Things you should knowTe and how is it obtainedThe average albedoThe hydrostatic law for an atmosphereThe atmospheric scal...
Planetary Atmospheres I
Planetary Atmospheres I
Planetary Atmospheres I
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Planetary Atmospheres I

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La razón principal para el estudio de atmósferas planetarias es tratar de entender el origen y evolución de la atmósfera de la tierra. Por supuesto, en el intento de comprender el funcionamiento de nuestro sistema solar o incluso la evolución de la Tierra como un organismo, la atmósfera de la tierra es esencialmente irrelevante, ya que su masa es despreciable.

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Planetary Atmospheres I

  1. 1. RENEE CONDORI APAZA, JULIOVALDIVIA SILVA, Christopher P. McKay
  2. 2. A principal reason for studying planetary atmospheres isto try to understand the origin and evolution of theearth’s atmosphere. Of course, in trying to understandthe workings of our solar system or even the evolutionof the earth as a body, the earth’s atmosphere isessentially irrelevant since its mass is negligible. Forthat matter, the mass of the earth is only a smallfraction of the mass of the sun. So we are consideringa thin skin of gravitationally bound gas attached to aspeck of matter in a dynamic and, in thepast, violent, system. Therefore, it is a formidableproblem.However, it is in that thin skin of gas and on that speckof matter that we live, and therefore, it is interesting tous.
  3. 3. It is also clear now that the earth’s gaseous envelope ischanging and has changed. In fact it is abundantlyclear that the present atmosphere barely resembles theoriginal residual gas left when the earth formed.Because of this it is also important to study the otheratmospheres in the solar system, since they are eitherdifferent end states or in different stages of atmosphericevolution. They may all have had roughly similarmaterials as sources, but either these atmospheres areon objects of a very different size or at a very differentdistance from the sun. Since, we can not carry outmany experiments to see how the earth’s atmosphere isevolving, Interpreting the data on otheratmospheres, given to us by Spacecraft and telescopedata, is crucial and is one goal of this theme.
  4. 4. Basic Properties of AtmospheresCompositionSizeEquilibrium TScale HeightAdiabatic Lapse RoleMixing in TroposphereRadiation AbsorptionAbsorption Cross SectionHeating by AbsorptionChapman LayerOzone Production:StratosphereThermospheric StructureIonospheresGreen House EffectAtmospheric EvolutionWater:Venus, Earth, MarsLoss by EscapeIsotope RatiosCO2 cycle:Earth, Venus, MarsAtmospheric CirculationCoriolis EffectLocal CirculationBoundary LayerGlobal CirculationZonal BeltsCloud FormationTopical Problems in PlanetaryAtmospheresOverview of Solar System
  5. 5. Type Name Mass Escape p T*(eV/u) (bar) (K)H/He Jupiter 318 18 128Gas Balls Saturn 95 6.5 98Uranus 14.5 2.3 56Neptune 17.0 2.8 57Terrestrial Venus 0.81 0.56 90 750Earth 1 0.65 1 280Mars 0.11 0.13 8mb 240Titan 0.022 0.051 1.5 94Triton 0.022 0.051 17b 38Escaping Io 0.015 0.034 10nb 130Europa 0.008 0.021 .02nb 120Ganymede 0.024 0.024 .01nb 140Enceladus 0.000013 0.00024 150?Pluto 0.002 0.008 1b 36Comets small ~0
  6. 6. Type Name Mass Escape p T*(eV/u) (bar) (K)Collisionless Mercury 0.053 0.093Moon 0.012 0.029Other moonsT*: for Jovian they are Teq ; for the terrestrialthey are mean surface temperatures; for icysatellites they are the subsolar T1eV = 1.16x104 K1 bar = 105 Pa = 105 N/m2.
  7. 7. MolecularSunH (H2) 0.86He 0.14O 0.0014C 0.0008Ne 0.0002N 0.0004Jupiter Saturn Uranus NeptuneH2 0.898 0.963 0.825 0.80He 0.102 0.0325 0.152 0.19CH4 0.003 0.0045 0.023 0.015NH3 0.0026 0.0001 <10-7 <6x10-7
  8. 8. MolecularEarth Venus Mars TitanCO2 0.0031 0.965 0.953N2 0.781 0.035 0.027 0.97O2 0.209 0.00003 0.0013CH4 0.00015 0.03H2O* 0.01 <0.0002 0.00039Ar 0.009 ~0.0001 0.016 0.01?*Variable
  9. 9. Pressure is the weight of a column of gas: forceper unit areap = mg N (column density: N)Thickness if frozen: Hsp(bar) Hs(m) Ma/Mp(10-5)Mars 0.008 2 0.049Earth 1 10 0.087Titan 1.5 100 6.8Venus 90 1000 9.7How big might Mars atmosphere have been (in bars) basedon its size? How big might the earth’s have been?
  10. 10. p, T, n (density) Equation of StateConservation of SpeciesContinuity Equation: Diffusion and FlowSources / Sinks: VolcanoesEscape (top)Condensation/ Reaction (surface)Chemical Rate EquationsConservation of EnergyHeat Equation: Conduction, Convection, RadiationSources: Sun and InternalSinks: Radiation to Space, Cooling to SurfaceRadiation transportConservation of MomentumPressure BalanceFlowRotating: CoriolisAtomic and Molecular PhysicsSolar Radiation: Absorption and EmissionHeating; Cooling; ChemistrySolar Wind: Aurora
  11. 11. Equilibrium TemperatureHeat In = Heat OutorSource (Sun) = Sink (IR Radiation to Space)Planetary body with radius a it absorbs energy overan area pa2Cooling: IR radiation outIf the planetary body is rapidly rotating or haswindsrapidly transporting energy, it radiates energyfrom all of its area 4pa2
  12. 12. Fraction of radiation absorbed in atmosphere vs. wavelengthPrincipal absorbing species indicated
  13. 13. Source=AbsorbArea heat flux amount absorbedpa2 x [F / Rsp2] x [1-A]A = Bond Albedo: total amount reflected(Complicated)Solar Flux 1AU: F =1370W/m2Rsp= distance from sun to planet in AULoss=Emitted (ideal radiator)Area radiated flux4pa2 x T4 = Stefan-Boltzman Constant= 5.67x10-8 J/(m2 K4 s)Fig. Radiation/ Albedo
  14. 14. Bond Albedo, A, isfraction of sunlightreflected to space:Surface, clouds, scattered
  15. 15. Set EqualHeat In = Heat OutTe = [ (F / Rsp2) (1-A) / 4 ]1/4Rsp A Te TsMercury 0.39 0.11 435 440Venus 0.72 0.77 227 750Earth 1 0.3 256 280Mars 1.52 0.15 216 240Jupiter 5.2 0.58 98 134*If the radiation was slow but evaporation was fast,like in a comet, describe the loss term that would theIR loss.Fig. Sub T
  16. 16. Right handaxis meltingpoint
  17. 17. Pressure vs. AltitudeHydrostatic LawForce Up = Force Downp- A=area---------------------------------------------Draw forces Δz---------------------------------------------p+ mg = (ρ A Δz) gResult:Net Force= 0 = - (Δp A) - (ρ A Δz) gwhere p = p-- - p+
  18. 18. dp/dz = - gNow Use Ideal Gas Lawp = nkT (k=1.38 x 10-23 J/K) =kT/morp = (R/Mr)T [Gas constant: R=Nak =8.3143 J/(K mole)with Mr the mass in grams of a mole]substitute for dp/dz = - p(mg/kT)= -p/HH is an effect height=Gravitational Force/ Thermal EnergySame result for a ballistic atmosphere
  19. 19. Pressure vs. Altitudep = po exp( - ∫ dz / H)(assuming T constant)p = po exp( - z / H)orDensity vs. Altitude = 0 exp( - z / H)Scale Height: HH = kT/mg (or H = RT / Mr g)Mr g(m/s2) Ts(K) H(km)Venus CO2 44 8.88 750 16Earth N2 ,O2 29 9.81 288 8.4Mars CO2 44 3.73 240 12Titan N2 , CH4 28 1.36 95 20Jupiter H2 2 26.2 128 20Note: did not use Te , used Ts for V,E,M
  20. 20. Pressure: pp = weight of a column of gas (force per unit area)1bar = 106 dyne/cm2=105 Pascal=0.987atmospheresPascal=N/m2 ; Torr=atmosphere/760= 1.33mbarsVenus 90 barsTitan 1.5 barsEarth 1 barMars 0.008 barColumn Density: Np = m g NSurface of earth: N  2.5 x 1025 molecules/cm2.What would N be at the surface of Venus?If the atmosphere froze (like on Triton),how deep would it be?n(solid N2)  2.5 x 1022 /cm3N/n = 10m
  21. 21. PARTIAL PRESSURESLower AtmosphereMixing dominates: use m or MrUpper atmosphereDiffusive separationPartial Pressure (const T)p =  pi(z) =  poi exp[ - z/Hi ]Hi = kT/ migFig. Density vs. z
  22. 22. ShowingRegion wheregasesdiffusivelyseparate
  23. 23. Convection Dominates  Adiabatic Lapse RateIn the troposphereRadiation Dominates  Greenhouse EffectIn the troposphere and stratosphereConduction Dominates  Thermal ConductivityIn the thermosphereFig. T vs. z
  24. 24. Shows layeredatmosphereRadiationAbsorptionIndicated
  25. 25. Imagine gas moving up or down adiabatically: noheat in or out of the volumeEnergy = Internal energy + Workdq = cvdT + p dV(energy per mass of a volume of gas V = 1 / )Adiabatic = no heat in or out: dq = 0cv dT = - p dVIdeal gas law [p = nkT = (R/Mr)T ]pV = (R/Mr)T
  26. 26. Differentiatep dV + dp V = (R/Mr) dTorcv dT = - (R/Mr) dT + V dp(cv +R/Mr) dT = dp / cp (dT/dz) = (dp/dz) / Apply Hydrostatic Law(dp/dz) = - g
  27. 27. (dT/dz) = -g / cp = - dHeating at surface + Slow vertical motion.T= [Ts - d z]T falls off linearly with altitudecp (erg/gm/K) d (deg/km)Venus 8.3 x 106 11Earth 1.0 x 107 10Mars 8.3 x 106 4.5Jupiter 1.3 x 108 20
  28. 28. cp = Cp / m = cv + (R/Mr)= Cv + kmCvT = heat energy of a moleculeAtom = Cv = (3/2)k ; kinetic energy only3-degrees of freedom each with k/2N2: One would think that there are6-degrees of freedom: 3 + 3or 3 (CM) + 2 (ROT) + 1 (VIB)Cv = 3k
  29. 29. But potential energy of internal vibrationsneeded.Cv  3.5 k = 4.8 x 10-16 ergs/K1 mass unit = 1.66x 10-24 gmcv  1.0 x 107 (ergs/gm/K)fortuitous as Cp  3.5Define  = Cp/CvUsing the above  - 1 = k/Cvor ( - 1) /  = k/ Cp = k/(mcp)
  30. 30. Now have p(z) with T dependence.Use (dT/dz) = -g / cp and dp/dz = - ρ g and p = nkTdp/p = - mgdz/kT = [m cp/k] dT/T = x dT/Tx = /(-1)=cp/cv1/x = ~0.2 for N2 ; ~0.17 for CO2 ; ~0 for large molecule(~5/3, 7/3, 4/3 for mono, dia and ployatomic gases)Solve and rearrange(p/po) = (T/To)xusing T= [Ts - d z]p(z) = po[1 - z/(xH)]x --> po exp(-z/H) for x small
  31. 31.  = T (po/p)1/xAdiabatic  Entropy = ConstantGas can move freely along constant  linesUsing dq = T dS where S is entropyCan show S = cp ln + const
  32. 32. Things you should knowTe and how is it obtainedThe average albedoThe hydrostatic law for an atmosphereThe atmospheric scale heightThe adiabatic lapse ratePotential Temperature
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