Post on 01-Sep-2018
Space Science I:Planetary Atmospheres
BookThe New Solar System
Chapters8,9,11,13,15,17,18,20
GoalsTo understand-
The physical and chemical processes thatcontrol our atmosphere
The origins and evolution of atmospheres
OutlineOverview of Solar System* Basic Properties of Atmospheres Composition Size Equilibrium T Scale Height
Mixing vs. Diffusion Adiabatic Lapse Rate Radiation Absorption Absorption Cross Section Heating by Absorption Chapman Layer Ozone Production: Stratosphere Thermospheric Structure Ionospheres Green House EffectAtmospheric Evolution Water: Venus, Earth, Mars Loss by Escape Isotope Ratios CO2 cycle: Earth, Venus, MarsAtmospheric Circulation Coriolis Effect Local Circulation Boundary Layer Global Circulation Zonal Belts Cloud FormationTopical Problems in Planetary Atmospheres
Space Science: AtmospherePart -1
Basic Properties of Atmospheres
Composition: TypesSizeEquilibrium THydrostatic LawScale HeightMixing vs. DiffusionAdiabatic Lapse RateVertical Convection
TYPES OF ATMOSPHERES
Type Name Mass Escape p T*(eV/u) (bar) (K)
H/He Gas Balls Jupiter 318 18 128Saturn 95 6.5 98Uranus* 14.5 2.3 56Neptune* 17.0 2.8 57
Terrestrial-like 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? 17µb 38
Escaping Io 0.015 0.034 10nb 130Europa 0.008 0.021 .02nb 120Ganymede0.024 0.024 .01nb 140Enceladus 0.000013 0.00024 150?Pluto 0.002 0.008 1µb 36Comets small ~0
Collisionless Mercury 0.053 0.093Moon 0.012 0.029Other moons
T*: for gas balls they are Te (to be discussed): for the terrestrial T is mean surface temperature; for icy satellites T is the subsolar T
1eV = 1.602 x10-19 J; for kT= 1eV --> T = 1.16x104 deg K1 bar = 105 Pa = 105 N/m2 = 106 dyne/cm2
COMPOSITION of Atmospheres
SunH (H2) 0.86He 0.14O 0.0014C 0.0008Ne 0.0002N 0.0004
Giant Planets (because of size U,N differ from J,S)Jupiter Saturn Uranus Neptune
H2 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
Terrestrial 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
Giant Planets ComparedNote: Jupiter and Saturn are related to each other
and to observed giant extrasolar planetsUranus and Neptune are transitions from those to
the terrestrial planets
Amount or Size of Atmosphere
Typically Compare pressures: p = Force per unit area or Weight of a column of gas
p = Column density (molecules stacked upper unit area, N)
x weight of each molecule (mg)
mg
p = Σforces/areap = mg N
Amount or Size (cont.)p = m g N
N = column density 1bar = 106 dyne/cm2=105 Pascal=0.987atmosphere Pascal=N/m2 ;
Torr = atmosphere/760 = 1.33mbarsFor Earth: use 1 bar and assume m ~ 29amu
show N ≈ 2.5 x 1025 molecules/cm2
p N(1025/cm2)Venus 90 bars ?Titan 1.5 bars ?Earth 1 bar 2.5Mars 0.008 bar ?
When using column: doesn’t matter how they are stacked
Amount or Size (cont.)p = mg N with N = column density
If Earth’s atmosphere froze (like Pluto’s), how thick would it be?
Hs = Column density/ density of solid = N/ns
Using ns(solid N2 or O2) ≈ 2.5 x 1022mol./cm3
Show ~ 10m for earth’s atmosphere
Atmosphere is very thin:Hs/RE ~ 10m / 6400km = 1.6 x10-6
p(bar) Hs(m) Ma/MpMars 0.008 2 0.049 x10-5
Earth 1 10 0.087 x10-5
Titan 1.5 100 6.8 x10-5
Venus 90 1000 9.7 x10-5
How big might Mars atmosphere have been (in bars) based on its size?How big might the earth’s have been?
Vertical Structure of an Atmosphere What do we need to describe it? Why are we interested?
p, T, ρ Equation of State of a Gas
Conservation of Species Continuity Equation: Diffusion and Flow
Sources / Sinks: Volcanoes Escape (top); Condense/ React (surface)
Chemical Rate Equations
Conservation of Energy Heat Equation: Conduc. Convect., Radiat.
Sources: Sun and Internal Sinks: Radiate to Space, Cool to Surface
Conservation of Momentum Pressure Balance
Flow Rotating: Coriolis
Atomic and Molecular Physics Solar Radiation: Absorb, Emit, Chemistry Solar Wind Ions: Aurora, Stripping
Heat In = Heat Out or
Source (Sunlight) = Sink (IR: Radiate to Space)This gives equilibrium temperature: Te
Consider a planetary body with radius aHeating: Absorbs energy over an area (πa2) Cooling: IR out If the body is rapidly rotating or
has winds rapidly transporting heat then it radiates uniformly from (4πa2)
First simple rule for the atmosphere:Heat Balance (Energy Equation)
Te
Solar Flux In IR Out
Fraction of radiation absorbed inatmosphere vs. wavelength
Principal absorbing species indicated
Wave length
Heat Balance (cont.)
(~250K)
~6000K
Source = Absorb Sunlight Area x heat flux x fraction absorbed πa2 x [F / Rsp
2] x [1-A]
A = Bond Albedo: total amount reflected (Complicated: next fig)
Solar Flux 1AU: F =1370W/m2
Rsp= distance from sun to planet in AU
Loss = Emitted (ideal IR radiator) Area x radiated flux 4πa2 x σTe
4
σ = Stefan-Boltzman Constant = 5.67x10-8 J/(m2 K4 s)
(Note: assumes a perfect emitter; otherwise multiply σTe
4 by the emissivity --here we assume it is 1)
Heat Balance (cont.)
Bond Albedo, A, is fraction of sunlight reflected to space: Surface, clouds, scattered
If cloud cover is increasingdoes Te increase or decrease?
Heat Balance (cont.)
Absorption and Reflectionsof Solar Radiation
Set Equal
Heat In = Heat Outπa2 x [F / Rsp
2] x [1-A] = 4πa2 x σTe4
Te = [ (F / Rsp2) (1-A) / 4σ ]1/4
Rsp A Te Ts
Mercury 0.39 0.12 432 440Venus 0.72 0.75 232 733Earth 1. 0.29 256 288Mars 1.52 0.16 217 220Jupiter 5.20 0.34 110 165*
* T at 1bar level
At what altitude would our atmosphere have T = Te
Another model:If the radiation is slow but evaporation is fast, like in a comet, describe the loss and the Te
Heat Balance (cont.)
Heat Balance (cont.) (not required)
Temperature can be limited bySublimation (cooling by evaporation)
Solar Heat In = Sublimation Rate +
IR Radiation Rate
Right hand axismelting point
Pressure vs. Altitude Hydrostatic Law Force Up = Force Down
p- A = area ---------------------------------------------Slab of atmosphere Δz ---------------------------------------------
p+ mg = (ρ A Δz) g
Balance: up - down = 0(p+ A) - [(p- A) + (ρ A Δz) g] =0
writing Δp = p-- - p+
- (Δp A) = (ρA Δz) g
Simplify Δp /Δz = - ρgor dp/dz = - ρg
Hydrostatic Law for an Atmosphere
Second simple rule: Force Balance (Momentum Equation)
Force Balance (cont.)Hydrostatic Law for an Atmosphere
dp/dz = - ρg
Use Ideal Gas Law(in some forn)p = nkT = ρ kT / m
[k=1.38 x 10-23 J/K ]or
p = (R/Mr)ρT[Gas constant: R=Nak =8.3143 J/(K mole)
with Mr the mass in grams of a moleand Na Avagadro’s number]
pV=nRT, in this case n is the number of molesThese are all equivalent (check that)
Substitute for ρ (= pm/kT) and get
* dp/dz = - p (mg/kT)= - p / H
H = (kT/mg)H is an effective height =Thermal Energy/ Gravitational Force
Solve * and find:p = po exp (-∫dz / H)
Pressure vs. Altitude (general)p = po exp( - ∫ dz / H) ; H= kT/mg
(assume T constant: isothermal + g is const.) p = po exp( - z / H) also Density vs. Altitude ρ = ρ0 exp( - z / H) ; mass density
n = n0 exp( - z / H) ; number density
Scale Height: H
H = kT/mg (or H = RT / Mr g)
Mr g(m/s2) Ts(K) H(km)Venus CO2 44 8.88 733 16Earth N2 ,O2 29 9.81 288 8.4Mars CO2 44 3.73 220 11Titan N2 , CH4 28 1.36 95 21Jupiter H2 2 26.2 110* 17
• Used Te• Note: did not use Te , used Ts for V , E, M and T
amu = 1.66 x10-27 kg
Force balance (cont.)
Force balance (cont.)Actual Data
Earth’s atmospherenearly exponential
pressure p~po exp(-z/H)
n=number density n~noexp(-z/H)
Force Balance (cont.)
Pressure and Density vs. z
Opps! Something happens at ~120km!
H changes drastically: T?,g? or m?
Region showedearlier
Force Balance (cont.)
Pressure and Density vs. zof each species
Shows how gases diffusively separate
Note: the average m in H=kT/mg goes from ~29 to ~16 to ~4
Force Balance (cont.)
Mean molecular mass vs. Altitude the m in the H =kT/mg!
Note: constant m in troposphere: implies it is well mixed
Force Balance (cont.)
Mixed RegionLower Atmosphere
Mixing dominates:use an average molecular mass m or Mr p = po exp(-z/H); H =kT/ m g
Diffusive separation: Unmixed(`light stuff rises to the top’)Upper atmosphere
Partial Pressures(using const T for simplicity)
p = ∑ pi(z) = ∑ poi exp[ - z/Hi ]
Hi = kT/ mig
Force Balance (cont.)
Hydrostatic: spherical atmosphere(a problem)
r is radial distance from center of planetMp = mass of planet
!
Radial Case
dp
dr= "g(r)#
g(r) = G M p
r2
Assume Isothermal
Problem
(a) Show
p(r) = p(r0) exp "G M pm
r
1
KT
r " r0
r0
$
% &
'
( )
(b) Show how this reduces to the flat atmosphere case
discussed when
r - r0 = z with z << r0
Temperature vs. Altitude
Earth’s Atmosphere is Layered
Radiation Absorption Indicated
Layers give the names
See atmosphericstructure of otheratmospheres in book
Diffusive
Mixed
Heats thesurface
288
T(K)200 1000
Temperature vs. Altitude (cont.)
Convection Dominates In troposphereAdiabatic Lapse Rate (+Greenhouse)
Radiation Dominates In stratosphere/ mesosphereOzone +Greenhouse Effect
Conduction Dominates In the thermosphere Thermal Conductivity
This is great for teaching!!
Thermosphere
Temperature vs. Altitude: Troposphere
First Law: Energy Conservation
Vertical convection is required to remove heat from the surfaceTherefore: imagine a volume of gas moving slowly up or down
But pressure also changes as the volume moves up or downTherefore, volume expands against
atmosphere when moving up
What happens to the Tif there is no heat exchanged?
Energy of a volume gas= Internal heat energy of vol. + Work done on vol.
Q = ∫CvdT + ∫pdV Adiabatic: the total energy stays same
dQ = 0Reversible
Troposphere : T(z) (cont.)Adiabatic expansion of a volume of air
T(z)
p(z)
IRcoolingignore !
At edges gases mix-ignore !
T(0)p(0)
As gas volume moves up it expands against atmosphere
∴ does work -- this energymust come from internal heat
warm surface
zSolar heating--ignore!
Troposphere Temperature: T(z) (cont.)
Energy = Internal energy of vol. + Work done on vol.
(Typically use q, energy per unit mass and cv, the heat capacity per unit mass, volume per mass, V = 1 / ρ)
Then the change in energy per mass of a volume of gas isdq = cvdT + p dV
Adiabatic: dq = 0
cv dT = - p dV
Use Ideal gas law [p = nkT = ρ(R/Mr)T ]pV = (R/Mr)T
Differentiate p dV + dp V = (R/Mr) dT or
cv dT = - (R/Mr) dT + V dp(cv +R/Mr) dT = dp / ρcp (dT/dz) = (dp/dz) / ρ
Finally(dT/dz) = (dp/dz) / (ρ cp )
(therefore, gas is not isothermal)
Troposphere: T(z) (cont.)
First Law gave: (dT/dz) = (dp/dz) / (ρ cp)
Use Hydrostatic Law to relate p and ρ (dp/dz) = - ρg
Substitute and find(dT/dz) = -g / cp = - Γd
Γd = Adiabatic Lapse Rate (dry)
Solve: Heating at surface + Slow vertical motion gives
T= [Ts - Γd z]
T falls off linearly with z due to convection!
cp (erg/gm/K) Γd (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
Troposphere: T(z) (cont.)
Adiabatic lapse rate = Γ ~ 6.5o / km!Due to vertical motion of air parcels
Goes to zero at ~45km for slope above Something else happens > ~10km
z less than ~10km --> Troposphere Vertical convection dominates
Why is lapse rate < 10o/km??
T=Ts - Γ z
Troposphere: T(z) (cont.)
Wet Lapse Rate (not required)Heat of Condensation/ Evaporation
!
dq = cvdT + p dV + L dw s
w s = fraction of atmospheric
mass condensed (H2O)
L = latent heat of vaporization
per unit mass
dT
dz= "#w = "
g
cp
1
[1+ (L/cp)(dw s/dT)]
Earth
#d $10o /km
#w $ 5o /km
#ave $ 6.5o /km
Latent heat released as gas cools slows the drop in T
Troposphere: T(z) (cont.)STABILITY of a PARCEL of AIR in a dry atmosphere
Buoyant Force Temperature of a parcel of air, ρab, moving slowly up roughly follows
adiabatic lapse rate.
Therefore, Tab of the parcel (solid line)can differ from ambient T of
the atmosphere
stable
unstable
Γd
T T0
Z Actual Lapse rate γ = dT/z
γ<Γd
γ>Γd
Can itrise? ordoes itfall?
!
"ab ˙ ̇ z = - ( "ab - ")g # z
˙ ̇ z = ( "
"ab - 1)g
= ( Tab
T - 1)g
Tab $ Ts - %dz ; T $ Ts - & z
˙ ̇ z $ -g z
T(%d ' &)
often written ̇ ̇ z $ -B gz (see 2nd prob. set)
Troposphere: T(z) (cont.)Calculate the buoyant force:ρab(z) = density vs. z for which air can move up or down adiabatically
ρ(z) is the actual atmospheric density
pab must adjust to ambient, p, therefore, ρabRTab = ρRT
Gives vertical convection (mixing) in troposphere
Troposphere: T(z) (cont.) Evaluate cp (not required)
cp = Cp / m = cv + (R/Mr) = Cv + k
m Cv depends on number of molecules in the volume
so we can consider heat capacity per molecule CvT = heat energy of a molecule, T in oK Cv = number of degrees of freedom x (k/2) (ways the particle can have energy)
For an atom ; kinetic energy only 3-degrees of freedom each with k/2; Cv = (3/2)k
N2 : One would think that there are 6-degrees of freedom: 3 + 3 or 3 (CM) + 2 (ROT) + 1 (VIB) Cv = 3kBut potential energy of internal vibrations is
an extra dimension in which energy can be storedCv ≈ 3.5 k = 4.8 x 10-16 ergs/K 1 mass unit = 1.66x 10-24 gmcv ≈ 1. x 107 (ergs/gm/K)
Define γ = Cp/Cv = cp/cvUsing the above γ - 1 = k/Cv
or (γ - 1) / γ = k/ Cp = k/(mcp)
Combine p(z) and T(z)
ADIABATIC + HYDROSTATICSince T is not constant and we know T(z)
Use (dT/dz) = -g / cp and dp/dz = - p/H H =kT/mg Details
dp/p = - mg dz/kT = [m cp/k] dT/T
Write p vs. T in troposphere asdp / p = x dT / T
where x = m cp/k
Can write x= γ/(γ-1) = m cp/k =cp/R γ=cp/cv
(γ~5/3 for an atom; ~ 7/3 for a diatomic molecule)
1/x = ~0.2 for N2 ; ~0.17 for CO2 ;1/x ~ 0 for large molecule
Combine p(z) and T(z) (cont.)
Resultdp / p = x dT /T
Solve this to get: (p/po) = (T/To)x
with T= [Ts - Γd z]
Note: p(z) = po[1 - z/(xH)]x --> po exp(-z/H) for x smallx small --> large molecules
This relationship lets us definePOTENTIAL TEMPERATURE
To = T (po/p)1/x
Call θ = T (po/p)1/x = T (po/p)R/cp
Note: only property of the gas need is cp!
Adiabatic Gas can move freely along constant θ lines (or Entropy = Constant)
Using dq = T dS where S is entropy
Can show S = cp lnθ + const
#1 Summary
Things you should understandRough differences between planetsEffective Temperature: TeThe average albedo: AThe hydrostatic law for an atmosphere
dp/dz = - ρgThe atmospheric scale height: H=hT/mgDiffusive separationAdiabatic lapse rate
Γd= -g/cpTroposphere: vertical convectionBuoyant force/ convectionp-T Law in troposphere: (p/po)=(T/To)x
Potential temperature: θ = T(po/p)R/cp