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Space Science : Planetary Atmospheres Part-6
Early Out-gassingVenus, Earth and MarsWater Loss from VenusPlanetary EscapeEnergy Flux Distribution Jeans EscapeIsotopic FractionationOther Escape Processes
How do we account for the principal differences
in the Terrestrial Planets?Simple Model (still 1D)
Assume giant planets condensing from the solar nebula maintained most of the local solar abundance of elements, because of their size and distance from the sun
Assume a large fraction of thelocal volatiles are trapped in condensing solid material in the inner solar system and these leak out as the planet cools.
Assume H2O is the principal species being out-gassed
VENUS, EARTH and MARS Simple Evolutionary Model H 15-16, G+W 132-136
Model:---- H2O is the principal greenhouse gas in early epochs based in solar abundance
(what about CH4, NH3? AbundancesO~2xC~5xN; or are they oxidized?)
---- H2O comes out of cooling rocks (or pasted on by comets).
---- With increasing time, vapor pressure of H2O increases
---- Partial pressure of H2O determines T: use radiative
transport solution
---- Track T at the surface vs.time as H2O out-gasses
Relate Pressure to Tg
(from radiative transport solution)
€
Tg ≈ Te [ 1 + τ g* / 2 ]1/4
Partial Pressure of H2O
pH2O = mg NH2O
This gives an optical thickness
τ g ≈ σ abs NH2O
τ g * ≈ c pH2O ; c ~ 5 mg / 3 σ abs
Using our solution for the surface temperature
Tg ≈ Te [ 1 + c pH2O ]1/4
As vapor pressure of H2O increases :
For a given Te,
the surface temperature,Tg, increases.
Temperature vs. time in an Early Epoch(out-gassing of water from cooling solid)
Dashed lines: Surface T: Tg = Te [1 + c pH2O]1/4
Solid lines: phase diagram p vs. T curve for waterPartial pressure of water in a background gas
Time -->
(H2O: present ~1% of 106 dynes)
Same albedo: 0.3
Result of Simple Model
Mars The Ice Planet Water primarily in ice caps and regolith as a permafrost
Earth The Water Planet
Venus The Water Vapor Planet Run away Greenhouse Effect?
Caveats: 1. Water implies clouds
Clouds play a complex role reflect radiation from sun but also trap IR radiation from below
2. Where is the water now ?~ 0.01% H2O in atmosphere~10-5 of earth’s oceans
VENUS (Summary)
92 bars, mostly CO2 Lapse Rate ~ 10o K / km
Horizontally Uniform Atmosphere Rotates ~ 243 days
Clouds Rotate ~ 4 days
Thin Haze 20 – 50 km Thick Haze ( ~ 20) 50 – 60 km Hazes: H2 SO4 and SO2
Where is the water?
VenusRunaway Greenhouse?
H2O should be in vapor phase, but observations show it is not.
Since H2O does not condense and H2O is much lighter than N2 or CO2
it can reach regions where UV can dissociate H2O + h OH + H +K.E. O + H2 +K.E.
Both H and H2 have large scale heightsThey can diffuse to top where they can escape
If the H or H2 escape what happens to the O ? Oxides surface + oxidizes carbon (CO2)
(Disagreement on whether the carbon in the inner regions of the solar nebula was all CH4 + organics)
Planetary EscapeDefine Exobase : Top of Atmosphere If an atom or molecule has an energy sufficient to escape + is moving upward it will have a high probability to escape
Top: Probability of escape is high if collision probability is small; Since this is diffusive region mean free path for a collision
col ≈ 1 / nx col
col = collision cross section nx = density at exobaseIsotropic scattering
col ≈ 1 /(21/2 nx col )]
Exobase altitude occurs whenscale height ~ mean free path
Hx ~ col Hx ~ 1 / nx col
since nxH = Nx column density Nx ≈ 1 / col
Exobase
Probability ofcollision is small
Exobase
Exobase: Nx col x = column of atmospherecol ~ molecular size ~10-15 cm2
Therefore,Nx col
-1 1015 atom / cm2
or,nx x col]-1
Earth: ~1000K, O : gx ~850cm/s2;x ~ 1000km
nx ~ 107 O/cm3 ; zx ~ 550km
Escape (continued)
Escape Energy = Ees = ½ m ves2
= mgxRx
gx = acceleration of gravity at exobase Rx = distance of exobase from center of planet
[surface ves: V 10.4; E 11.2, M 4.8km/s] At Earth Tx 1000 K, v=1km/sBUT, for a given Tx there is a distribution of v
€
f(v v ) =
1
[2π kT/m]3/2exp −
mv2
2kT
⎡
⎣ ⎢
⎤
⎦ ⎥
= f(vx ) f(vy ) f(vz) ; f(r v ) d3v = 1∫∫∫
f(vz) = 1
[2π KT/m]1/2exp −
mvz2
2kT
⎡
⎣ ⎢
⎤
⎦ ⎥ ; f(vz)
−∞
+∞
∫ dvz = 1
Flux of molecules across exobase
in the + z direction
Φes = nx vz f(v v ) d3v ; vz > 0, v > ves∫∫∫
Escape (continued)Things are often written in terms of the mean speed
€
v ≡ v f(v v ) d3v∫∫∫
Problem : Verify that the mean speed is
v = 8 K T
π m
Flux across a surface vz > 0, (all v)
Φ+ = nx vz f(v v ) d3v∫∫∫
= nx vz
[2 π kT/m]1/20
∞
∫ exp −mvz
2
2KT
⎡
⎣ ⎢
⎤
⎦ ⎥dvz
= nx (kT/m)/[2 π kT/m]1/2 = nx
kT
2πm
⎡ ⎣ ⎢
⎤ ⎦ ⎥
1/ 2
Φ+ = 1
4 nx v ; 1/4 due to isotropic assumption
*
Escape (continued)
€
Mean energy of molecules in a gas
E = E f(v v ) d3v ; E =
1
2∫∫∫ m v2
Problem : verify that
E = 32 kT
For particles crossing a surface
E + = E (n vz) f(v ) d3v∫∫∫[ ]1
4nv
⎡ ⎣ ⎢
⎤ ⎦ ⎥
Problem : Using E = 12 mv2; dΩ = dcosθ dϕ
Verify that you can rewrite
E + = E0
∞
∫0
1
∫0
2π
∫ cosθ
πf+(E) dE dcosθ dϕ
where f+(E) = E
(kT)2exp[−E/kT]
This is the energy destribution for a Boltzmann Flux,
Problem : Verify that
E + = 2 k T
Note : E + > E (for molecules in a gas)
*
*
*
Escape (continued)
€
Use E > E es
Energy distribution for escape flux
f+(E) = E
(kT)2exp[−E/kT]
Solid angle distribution
fΩ+( θ , ϕ ) = cosθ
π
f+(E) dE0
∞
∫ = 1 ; fΩ+( θ , ϕ ) ∫ dΩ = 1
Problem :
Use the expressions above to rewrite
Φes = (nx vz) f(v v ) d3v∫∫∫
Then show = 1
4 (nx v ) Pes
with
Pes = f(E) dEE es
∞
∫
Jeans (thermal) Escape : using f(E) above
Pes = 1 + Ees
kT
⎡ ⎣ ⎢
⎤ ⎦ ⎥ e
-E eskT
dePL λ esc ≡ E eskT
or using, E es = mgxRx ,
λ esc ≡ mgx RxkT = Rx /Hx [or Hx = Rx (E es/kT)]
*
Escape Flux (review)
Flux Across Exobase
¼ nx v
8 kT
v = ; nx = density m
Probablity of Escape Pes = f+(E) dE mgR
f+(E) = energy distribution of flux at exobase
Total Loss by Jeans Escape
Flux+ x Pes x Area x Time or
Nes =Column Lost = Flux+ x Pes x Time ( Pressure change = Nes m g )Example: H from the earth
Present escape flux ~ 107H/cm2/sIf constant: t ~ (3x107 s/yr) x 4.5 x109yr ~ 1.4 x1017 sNes ~ 1.4 x 1024 H/cm2
p ~ 2x103 dynes/cm2 ~ 2 x10-3bar or
Total column at EarthN ~ 2 x1025 mol.(N2,O2)/ cm2
(~10 meters frozen)Equivalent water loss: ~0.5 m
Escape (cont.)Problem
€
Escape Energy
Venus 0.56 eV / u
Earth 0.65 eV / u
Mars 0.13 eV / u
Jupiter 18 eV / u
Titan 0.036 (0.024*) eV/u *(at exobase)
Assume Tx = 1000 K (not true on present Venus)
Use a time t = 4.5 byr ≈ 4.5 × 109 × 3 × 107 sec
Assume Nx = 1015 atoms / cm2
Calculate the net column loss, Nes,
of H, H2(or D), N from
each planet assuming H = k Tx / m gx
and the exobase is either all H, H2, or N.
Approx. Present Valueszx: V 200km; E 500km, M 250km, T1500kmTx: V 275K, E 1000K, M 300K, T 160KJupiter: no surface, use Tx =1000K
*
Isotope Fractionation(preferential loss of lighter species)
Relative loss rate is determined by relative escape probabilities and relative exobase densities: hence, on differences in diffusive separation
A = lighter; B = heavier
nA(zh) , nB(zh): densities at homopause (turbopause) determined by atmospheric concentrations,
Ratio: RBA(zh) = nB(zh)/ nA(zh)
If amount above zhis small, they have same H below zh
Then RBA(zh) = NB(zh)/ NA(zh)
nA(zx), nB(zx) densities at exobase
Use nA(zx) = nA(zh) exp[-z/HA]
and nB(zx) = nB(zh) exp[-z/HB]
z = zx -zh
Ratio at exobase: RBA(zx) = nB(zx)/ nA(zh)
RBA(zx) = RBA(zh) exp[-z(1/HB-1/HA)]
RBA(zx) = RBA(zh) exp[-z m gx/kTx]
Therefore, position of exobase above the homopause strongly affects isotope fractionation at zx.
Isotope Fractionation (cont.)
€
Loss rate of a column of atmosphere
dNA
dt= − Φes
A
For Jeans (thermal) Escape :
ΦesA = −
nA vA
4Pes
A
with PesA = 1 +
E es
kT
⎡ ⎣ ⎢
⎤ ⎦ ⎥e
-EeskT
nA(zx ) ≈ nA(zh )exp(−Δz /HA)
Assume the total column density is dominated
by column below the homopause
Then nA(zx ) ≈ (NA/H)exp(−Δz /HA)
Can now write :
dNA
dt= − Φes
A ≈ − (NA/H)exp(−Δz /HA) vA
4Pes
A
dNA
dt ≈ − NA/tA or NA(t) ≈ NA
o exp(-t/tA) !!
(tA)−1 ≈ [(vA/4)Pes
A]
H exp(−Δz /HA)
or the fraction of species A still in the atmosphere is
fA(t) ≈ NA /NAo = exp(−t/tA)
Depends only on mass of A and the temperature at the exobase
Isotope Fractionation (cont.)
€
The ratio of total atmospheric concentrations vs. t is
rAB(t) =NA(t)
NB(t)
⎡
⎣ ⎢
⎤
⎦ ⎥ /
NAo
NBo
⎡
⎣ ⎢
⎤
⎦ ⎥ ≈ exp[−t (
1
tA
−1
tB
)]
≈ exp[−t
tA
(1−tA
tB
)] = [exp(−t
tA
)](1−
t A
t B
)
Writing mB
mA
=1+ δ
Then assuming δ << 1 (not true for D/H)
one can write
rAB(t) ≈ [fA (t)]x with x = (1- tA/t B)
and fA(t) the fraction of A remaining at time t.
Keeping the largest terms only and assuming
and assuming E esA /kT >> 1
x ≈ 1- (1 +δ/2) [exp(-Δz /HA - E esA /kT)]δ
δ →0 ⏐ → ⏐ ⏐ δ [Δz /HA +E esA/kT]
For 36Ar vs. 38Ar this is accurate -
-for D vs. H (δ =1) calculate more carefully
Isotope Fractionation (cont.)
€
Using the time dependence of the ratio
rAB(t) ≈ fA (t)x with x = (1 - tA/tB)
and with rAthe rate of loss for the principal isotope
One can , in principle, calculate the
isotopic fractionation in an atmosphere
Note : vice versa
- - - knowing the present ratios, the δ, and the initial solar abundances
one can extract the atmospheric loss of A, rA (t), (to be discussed)
All of this depends on there being no limiting flux at the homopause
However, often there is (e.g., for D/H ratio at earth : evaporation and
condensation limit the escape flux).
At the homopause : if A is a light, trace species
with a diffusion coefficient DA
and H is the scale height of the principal species and HAe
the equilbrium scale height of trace species A, then we showed
the limiting flux from our notes on diffusion is
ΦD A ≈ DAnA[1/H -1/HAe] ≈ (nA/n)(v /σ col) /H
From Chamberlain, many people (e.g. dePL) write
(DA n) ≡ b = (v /σ col);
they call b the collision parameter which is primarily a funtion of T.
Enrichment of Heavy Isotope Indicates Atmospheric Loss
D / H SUN 1.5 X 10-5
COMETS ~ 3 X 10-4
EARTH’S WATER ~1.6 X 10-4
VENUS Atmosphere ~ 2 x 10-2
D enriched relative to SunOther species V E M S36Ar / 38Ar 5.1 5.3 4 5.540Ar / 36Ar 1 296 3000 ---( 40K 40Ar + e+ + e- ~ 108 years)
Xe , N, O and C are also fractionatedNeed escape processes other than Jeans escape for heavy species
D/H Ratio Earth and Venus tH << tD (your problem)
€
rHD( )now=
NH
ND
⎛
⎝ ⎜
⎞
⎠ ⎟now
NH
ND
⎛
⎝ ⎜
⎞
⎠ ⎟solar
≈ fH(t)
Earth : D / H 1.6 × 10-4
rHD = 1.5 × 10-5 / 1.6 × 10-4 ≈ 0.09
fH(t) = 0.09
That is, only ~ 10% of original water released is still in the
atmosphere and oceans (assuming well mixed);
Disagrees with present loss rate must have been an earlier
epoch of rapid loss
Venus : D / H 2 × 10-2
fH(t) ≈ 0.0008 !
or − -assuming the same original water budget as the Earth
≈ 0.008 of Earth's present water budget
present value of loss rate (~ 107 /s) too small
Need wet hot early atmosphere with H2O well mixed
(to hot to condense ar all altitudes inspite of lapse rate)
For both V and E :
did hotter (EUV) early Sun and/or T - Tauri caused blow - off
Hot hydrogen corona at Venus (first detected hot corona)
Cold fraction ~275 KExospheric T
Hot fraction ~1020 K Implies nonthermal processes.
ESCAPE PROCESSES
1. Jeans Escape2. Hydrodynamic Escape (Blow Off)3. Photo Dissociation 4. Dissociative Recombination 5. Interaction with the Local Plasma6. T – Tauri Sweeping
Very heavy species 2?
When molecules are at the exobase3 and 4 are importantVenus
H2O + h Present Mars 2. CO2
+ + e C O2
+ + e
In the absence of a protecting magnetic field 5 and 6 are important
Photo -dissociation Rates average sun (Heubner)
€
Energy Rate
(eV ) (10−7 /s)
H2 + hν → H(1s) + H(1s) 8.22 0.78
→ H(1s) + H(1s,2p) 0.43 0.58
N2 + hν → N + N 3.38 1.11
O2 + hν → O(3P) + O(3P) 5.12 1.80
→ O(3P) + O(1D) 1.44 52.30
→ O(1S) + O(1S) 0.74 0.66
H2O + hν → H + OH 3.73 139.
→ H2 + O(1D) 3.89 10.4
→ H + H + O 0.7 13.3
Note : Excess Energy shared for
A 2 → A + A
H2O → H + OH E H = E mOH
mH2O
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
= (17/18) E
→ H2 + O E H2= (16/18) E
Therefore, light species can be energetic
Fractionation: nonthermal escape processesDiffusive separation gives an exobase ratio between a lighter (A) and heavier (B) species
If the loss mechanism is not strongly affected by then our earlier formula can be written
That is, only their relative abundance at the exobase is important: Rayleigh fractionation law
roughly applies for energetic ejection processes.Again fA is the fraction of an atmospheric speciesremaining at present
Applies to Mars for processes 3, 4, 5 non-fractionation of 18O / 16O, 13C / 12C means there are large reservoirs:
carbonatcs and permafrost? 36Ar / 35Ar no need for hydrodynamic episode 14N /15N models give too mach loss (buffered by CO2?)
€
RAB
exo = exp −Δzx
1
HA
−1
HB
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
= [exp(Δzx HA)]δ ; δ =mB - mA
mA
€
rAB = fA(R AB
exo −1) or fraction remaining : fA = rAB1/(R AB
exo -1)
Hydrodynamic Escape (Hunter, Pepin, Walker, Icarus 69, 532, 1987)Hot atmosphere (early sun, or a large planet close to its star).
Light species ( typically H or H2)has a mean thermal speed to greater than the escape speed.atmospheric density: n(r) 1/r2 ,
r is distance from centerLight species collide with and drag off
heavier species ∝
€
ΦAes = escape flux for outflowing light species
Upper limit for escape flux of heavier species : ΦBes
ΦBes ≈ (XB/XA) ΦA
es; mB >> mA
XB ( XA) are fractional concentrations
Diffusion limited at homopause : diffusive flux of B = nB D 1
HB
−1
H
⎡
⎣ ⎢
⎤
⎦ ⎥
ΦBes ≈
XB
XA
ΦAes − nB D
1
HB
−1
HA
⎡
⎣ ⎢
⎤
⎦ ⎥
[full drag] [relative diffusion]
≈ nB
nA
ΦAes mC − mB
mC − mA
⎡
⎣ ⎢
⎤
⎦ ⎥ ; mC = mA +
kT ΦA
D nA g
m C is called the "cross over" mass
i.e., if mB is too heavy (> mC) there is no drag effect :
That is, it can not be supplied fast enough at homopause
Isotope Fraction(cont.) Hydrodynamic fractionation is different than other processes
Mass difference again affects loss rate: here by diffusion rate at the homopause€
NB
NB0 =
NA
NA0
⎡
⎣ ⎢
⎤
⎦ ⎥
ξ
; ξ = mC - mB
mC - mA
A = very light; B = heavy
For two different heavy species : 36Ar, 38Ar
B = lighter , B' = heavier ; δ =mB' - mB
mB
rBB' = NA NAo
[ ]ξ −ξ '
= NA NAo
[ ]−δq
q = nAD HA[ ] /ΦA = diffusive flux
escape flux
€
Solar EUV Flux vs. Time :
FEUV(t) = FEUVo ( 4.5 × 109 yrs / t)5/6
with t in years and FEUVo the present value
Heating leading to blow - off possibly explains some
Xe isotope differences from carbonaceous chondrites
#6 Summary Things you should know
Greenhouse Model for Terrestrial Planets
Venus vs. Earth vs. MarsWater Loss from VenusPlanetary EscapeEnergy Flux Distribution Jeans EscapeIsotopic FractionationOther Escape ProcessesHydrodynamic Escape