Space Science : Planetary Atmospheres Part-6 Early Out-gassing Venus, Earth and Mars Water Loss from Venus Planetary Escape Energy Flux Distribution Jeans Escape Isotopic Fractionation Other 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 the local volatiles are trapped in condensing solid material in the inner solar system and these leak out as the planet cools. Assume H 2 O is the principal species being out-gassed

VENUS, EARTH and MARS Simple Evolutionary Model H 15-16, G+W Model:---- H 2 O is the principal greenhouse gas in early epochs based in solar abundance (what about CH 4, NH 3 ? Abundances O~2xC~5xN; or are they oxidized?) ---- H 2 O comes out of cooling rocks (or pasted on by comets) With increasing time, vapor pressure of H 2 O increases ---- Partial pressure of H 2 O determines T: use radiative transport solution ---- Track T at the surface vs. time as H 2 O out-gasses

Relate Pressure to T g (from radiative transport solution)

Temperature vs. time in an Early Epoch (out-gassing of water from cooling solid) Dashed lines: Surface T: T g = T e [1 + c p H2O ] 1/4 Solid lines: phase diagram p vs. T curve for water Partial pressure of water in a background gas Time --> (H 2 O: present ~1% of 10 6 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% H 2 O in atmosphere ~10 -5 of earth’s oceans

VENUS (Summary)   92 bars, mostly CO 2 Lapse Rate ~ 10 o K / km Horizontally Uniform Atmosphere Rotates ~ 243 days Clouds Rotate ~ 4 days Thin Haze 20 – 50 km Thick Haze (  ~ 20) 50 – 60 km Hazes: H 2 SO 4 and SO 2 Where is the water?

Venus Runaway Greenhouse? H 2 O should be in vapor phase, but observations show it is not. Since H 2 O does not condense and H 2 O is much lighter than N 2 or CO 2 it can reach regions where UV can dissociate H 2 O + h  OH + H +K.E.  O + H 2 +K.E. Both H and H 2 have large scale heights They can diffuse to top where they can escape If the H or H 2 escape what happens to the O ? Oxides surface + oxidizes carbon (CO 2 ) (Disagreement on whether the carbon in the inner regions of the solar nebula was all CH 4 + organics)

Planetary Escape Define 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 / n x  col  col = collision cross section n x = density at exobase  Isotropic scattering col ≈ 1 /(2 1/2 n x  col )] Exobase altitude occurs when scale height ~ mean free path H x ~ col  H x ~ 1 / n x  col since n x H = N x column density N x ≈ 1 /  col

Exobase Probability of collision is small Exobase Exobase: N x  col   x = column of atmosphere  col  ~ molecular size ~ cm 2 Therefore, N x  col -1  atom / cm 2 or, n x  x  col ] -1 Earth: ~1000K, O : g x ~850cm/s 2 ;  x  ~ 1000km n x ~ 10 7 O/cm 3 ; z x ~ 550km

Escape (continued) Escape Energy = E es = ½ m v es 2 = mg x R x g x = acceleration of gravity at exobase R x = distance of exobase from center of planet [surface v es : V 10.4; E 11.2, M 4.8km/s] At Earth T x  1000 K, v=1km/s BUT, for a given T x there is a distribution of v

Escape (continued) Things are often written in terms of the mean speed *

Escape (continued) * * *

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Escape Flux (review) Flux Across Exobase ¼ n x v 8 kT v = ; n x = density  m Probablity of Escape  P es = f + (E) dE mgR f + (E) = energy distribution of flux at exobase

Total Loss by Jeans Escape Flux + x P es x Area x Time or N es =Column Lost = Flux + x P es x Time ( Pressure change = N es m g ) Example: H from the earth Present escape flux ~ 10 7 H/cm 2 /s If constant: t ~ (3x10 7 s/yr) x 4.5 x10 9 yr ~ 1.4 x10 17 s N es ~ 1.4 x H/cm 2  p ~ 2x10 3 dynes/cm 2 ~ 2 x10 -3 bar or Total column at Earth N ~ 2 x10 25 mol.(N 2,O 2 )/ cm 2 (~10 meters frozen) Equivalent water loss: ~0.5 m

Escape (cont.) Problem Approx. Present Values z x : V 200km; E 500km, M 250km, T1500km T x : V 275K, E 1000K, M 300K, T 160K Jupiter: no surface, use T x =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 n A (z h ), n B (z h ): densities at homopause (turbopause) determined by atmospheric concentrations, Ratio:R BA (z h ) = n B (z h )/ n A (z h ) If amount above z h is small, they have same H below z h Then R BA (z h ) = N B (z h )/ N A (z h ) n A (z x ), n B (z x ) densities at exobase Use n A (z x ) = n A (z h ) exp[-  z/H A ] and n B (z x ) = n B (z h ) exp[-  z/H B ]  z = z x -z h Ratio at exobase: R BA (z x ) = n B (z x )/ n A (z h ) R BA (z x ) = R BA (z h ) exp[-  z(1/H B -1/H A )] R BA (z x ) = R BA (z h ) exp[-  z  m g x /kT x ] Therefore, position of exobase above the homopause strongly affects isotope fractionation at z x.

Isotope Fractionation (cont.)

Enrichment of Heavy Isotope Indicates Atmospheric Loss D / H SUN 1.5 X COMETS ~ 3 X EARTH’S WATER ~1.6 X VENUS Atmosphere ~ 2 x D enriched relative to Sun Other species V E M S 36 Ar / 38 Ar Ar / 36 Ar ( 40 K  40 Ar + e + + e - ~ 10 8 years) Xe, N, O and C are also fractionated Need escape processes other than Jeans escape for heavy species

D/H Ratio Earth and Venus t H << t D (your problem)

Hot hydrogen corona at Venus (first detected hot corona) Cold fraction ~275 K Exospheric T Hot fraction ~1020 K Implies nonthermal processes.

ESCAPE PROCESSES 1. Jeans Escape 2.Hydrodynamic Escape (Blow Off) 3.Photo Dissociation 4.Dissociative Recombination 5.Interaction with the Local Plasma 6.T – Tauri Sweeping Very heavy species 2? When molecules are at the exobase 3 and 4 are important Venus H 2 O + h     Present Mars  CO e  C  O e  In the absence of a protecting magnetic field 5 and 6 are important

Photo -dissociation Rates average sun (Heubner)

Fractionation: nonthermal escape processes Diffusive 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 f A is the fraction of an atmospheric species remaining at present Applies to Mars for processes 3, 4, 5 non-fractionation of 18 O / 16 O, 13 C / 12 C means there are large reservoirs: carbonatcs and permafrost? 36 Ar / 35 Ar no need for hydrodynamic episode 14 N / 15 N models give too mach loss (buffered by CO 2 ?)

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 H 2 ) has a mean thermal speed to greater than the escape speed. atmospheric density: n(r)  1/r 2, r is distance from center Light species collide with and drag off heavier species

Isotope Fraction(cont.) Hydrodynamic fractionation is different than other processes Mass difference again affects loss rate: here by diffusion rate at the homopause

#6 Summary Things you should know Greenhouse Model for Terrestrial Planets Venus vs. Earth vs. Mars Water Loss from Venus Planetary Escape Energy Flux Distribution Jeans Escape Isotopic Fractionation Other Escape Processes Hydrodynamic Escape