planetary atmospheres i
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RENEE CONDORI APAZA, JULIO
VALDIVIA SILVA, Christopher P. McKay
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.
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.
Basic Properties of AtmospheresCompositionSizeEquilibrium TScale HeightAdiabatic Lapse RoleMixing in Troposphere
Radiation AbsorptionAbsorption Cross SectionHeating by AbsorptionChapman LayerOzone Production:
StratosphereThermospheric StructureIonospheresGreen House Effect
Atmospheric Evolution
Water:
Venus, Earth, Mars
Loss by Escape
Isotope Ratios
CO2 cycle:
Earth, Venus, Mars
Atmospheric Circulation
Coriolis Effect
Local Circulation
Boundary Layer
Global Circulation
Zonal Belts
Cloud Formation
Topical Problems in Planetary
Atmospheres
Overview of Solar System
Type Name Mass Escape p T*(eV/u) (bar) (K)
H/He Jupiter 318 18 128 Gas Balls Saturn 95 6.5 98
Uranus 14.5 2.3 56Neptune 17.0 2.8 57
Terrestrial 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 17b 38
Escaping 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 1b 36Comets small ~0
Type Name Mass Escape p T*(eV/u) (bar) (K)
Collisionless Mercury 0.053 0.093Moon 0.012 0.029Other moons
T*: for Jovian they are Teq ; for the terrestrialthey are mean surface temperatures; for icysatellites they are the subsolar T
1eV = 1.16x104 K1 bar = 105 Pa = 105 N/m2.
Molecular
SunH (H2) 0.86He 0.14O 0.0014C 0.0008Ne 0.0002N 0.0004
Jupiter 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
Molecular
Earth 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
Pressure is the weight of a column of gas: forceper unit area
p = mg N (column density: N)
Thickness if frozen: Hs
p(bar) Hs(m) Ma/Mp
(10-5)Mars 0.008 2 0.049Earth 1 10 0.087Titan 1.5 100 6.8Venus 90 1000 9.7
How big might Mars atmosphere have been (in bars) based on its size? How big might the earth’s have been?
p, T, n (density) Equation of State
Conservation of SpeciesContinuity Equation: Diffusion and FlowSources / Sinks: Volcanoes
Escape (top)Condensation/ Reaction (surface)
Chemical Rate EquationsConservation of Energy
Heat Equation: Conduction, Convection, Radiation Sources: Sun and InternalSinks: Radiation to Space, Cooling to SurfaceRadiation transport
Conservation of MomentumPressure Balance FlowRotating: Coriolis
Atomic and Molecular Physics Solar Radiation: Absorption and Emission
Heating; Cooling; ChemistrySolar Wind: Aurora
Equilibrium Temperature
Heat In = Heat Outor
Source (Sun) = Sink (IR Radiation to Space)
Planetary body with radius a it absorbs energy over an area pa2
Cooling: IR radiation outIf the planetary body is rapidly rotating or haswindsrapidly transporting energy, it radiates energy
from all of its area 4pa2
Fraction of radiation absorbed in atmosphere vs. wavelength
Principal absorbing species indicated
Source=Absorb
Area heat flux amount absorbedpa2 x [F / Rsp
2] x [1-A]
A = Bond Albedo: total amount reflected (Complicated)
Solar Flux 1AU: F =1370W/m2
Rsp= distance from sun to planet in AU
Loss=Emitted (ideal radiator)Area radiated flux4pa2 x T4
= Stefan-Boltzman Constant= 5.67x10-8 J/(m2 K4 s)
Fig. Radiation/ Albedo
Bond Albedo, A, isfraction of sunlightreflected to space:Surface, clouds, scattered
Set Equal
Heat In = Heat Out
Te = [ (F / Rsp2) (1-A) / 4 ]1/4
Rsp A Te Ts
Mercury 0.39 0.11 435 440
Venus 0.72 0.77 227 750
Earth 1 0.3 256 280
Mars 1.52 0.15 216 240
Jupiter 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
Right hand axis melting point
Pressure vs. Altitude
Hydrostatic Law
Force Up = Force Down
p- A=area---------------------------------------------
Draw forces Δz---------------------------------------------
p+ mg = (ρ A Δz) gResult:
Net Force= 0 = - (Δp A) - (ρ A Δz) g
where p = p-- - p+
dp/dz = - g
Now Use Ideal Gas Lawp = nkT (k=1.38 x 10-23 J/K) =kT/m
or
p = (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/H
H is an effect height= Gravitational Force/ Thermal Energy
Same result for a ballistic atmosphere
Pressure vs. Altitudep = po exp( - ∫ dz / H)
(assuming T constant)
p = po exp( - z / H)
or
Density vs. Altitude
= 0 exp( - z / H)
Scale Height: H
H = kT/mg (or H = RT / Mr g)
Mr g(m/s2) Ts(K) H(km)
Venus CO2 44 8.88 750 16
Earth N2 ,O2 29 9.81 288 8.4
Mars CO2 44 3.73 240 12
Titan N2 , CH4 28 1.36 95 20
Jupiter H2 2 26.2 128 20
Note: did not use Te , used Ts for V,E,M
Pressure: p
p = weight of a column of gas (force per unit area)
1bar = 106 dyne/cm2=105 Pascal=0.987atmospheres
Pascal=N/m2 ; Torr=atmosphere/760= 1.33mbars
Venus 90 barsTitan 1.5 barsEarth 1 barMars 0.008 bar
Column Density: N
p = m g N
Surface 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 /cm3
N/n = 10m
PARTIAL PRESSURES
Lower Atmosphere
Mixing dominates: use m or Mr
Upper atmosphere
Diffusive separation
Partial Pressure (const T)
p = pi(z) = poi exp[ - z/Hi ]
Hi = kT/ mig
Fig. Density vs. z
ShowingRegion wheregasesdiffusivelyseparate
Convection Dominates Adiabatic Lapse RateIn the troposphere
Radiation Dominates Greenhouse Effect In the troposphere and stratosphere
Conduction Dominates Thermal ConductivityIn the thermosphere
Fig. T vs. z
Shows layered atmosphereRadiation Absorption Indicated
Imagine gas moving up or down adiabatically: no heat in or out of the volume
Energy = Internal energy + Work
dq = cvdT + p dV(energy per mass of a volume of gas V = 1 / )
Adiabatic = no heat in or out: dq = 0cv dT = - p dV
Ideal gas law [p = nkT = (R/Mr)T ]
pV = (R/Mr)T
Differentiatep dV + dp V = (R/Mr) dT
or
cv dT = - (R/Mr) dT + V dp
(cv +R/Mr) dT = dp /
cp (dT/dz) = (dp/dz) /
Apply Hydrostatic Law(dp/dz) = - g
(dT/dz) = -g / cp = - d
Heating at surface + Slow vertical motion.
T= [Ts - d z]
T falls off linearly with altitude
cp (erg/gm/K) d (deg/km)
Venus 8.3 x 106 11
Earth 1.0 x 107 10
Mars 8.3 x 106 4.5
Jupiter 1.3 x 108 20
cp = Cp / m = cv + (R/Mr)
= Cv + k
m
CvT = heat energy of a molecule
Atom = Cv = (3/2)k ; kinetic energy only
3-degrees of freedom each with k/2
N2: One would think that there are 6-degrees of freedom: 3 + 3or 3 (CM) + 2 (ROT) + 1 (VIB)Cv = 3k
But potential energy of internal vibrations needed.
Cv 3.5 k = 4.8 x 10-16 ergs/K
1 mass unit = 1.66x 10-24 gm
cv 1.0 x 107 (ergs/gm/K)
fortuitous as Cp 3.5
Define = Cp/Cv
Using the above - 1 = k/Cv
or ( - 1) / = k/ Cp = k/(mcp)
Now have p(z) with T dependence.
Use (dT/dz) = -g / cp and dp/dz = - ρ g and p = nkT
dp/p = - mgdz/kT = [m cp/k] dT/T = x dT/T
x = /(-1)
=cp/cv
1/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)
x
using T= [Ts - d z]
p(z) = po[1 - z/(xH)]x --> po exp(-z/H) for x small
= T (po/p)1/x
Adiabatic Entropy = Constant
Gas can move freely along constant lines
Using dq = T dS where S is entropy
Can show S = cp ln + const
Things you should know
Te and how is it obtained
The average albedo
The hydrostatic law for an atmosphere
The atmospheric scale height
The adiabatic lapse rate
Potential Temperature
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