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