Space Science: Atmospheres Part- 7b. Venus, Earth and Mars Where is the H 2 O on Venus? Planetary Escape Isotope Fractionation Hydrodynamic Escape

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1 Space Science: Atmospheres Part- 7b Venus, Earth and Mars Where is the H 2 O on Venus? Planetary Escape Isotope Fractionation Hydrodynamic Escape

2 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? Where is the water? Since H 2 O does not condense and is lighter than N 2 or CO 2 it can reach regions where UV can dissociate H 2 O + hν OH + H +K.E. O + H 2 +K.E. Large scale heights

3 Vertical Structure Exosphere??

4 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 of escaping Probability of escape is high if collision probability is small; mean free path for a collision λ col 1 / n x σ col σ col = collision cross section n x = density at exobase Exobase altitude occurs when scale height ~ mean free path H x ~ λ col H x ~ 1 / n x σ col since n x H = N x column density N x σ col ~ 1 N x 1 / σ col

5 Exobase Probability of collision is small Exobase Exobase: N x σ col 1 Ν x = column of atmosphere σ col ~ molecular size ~10-15 cm 2 Therefore, N x σ col atom / cm 2 or, n x [Η x σ col ] -1 Earth: ~1000K, O : g x ~850cm/s 2 ; Η x ~ 1000km n x ~ 10 7 O/cm 3 ; z x ~ 550km

6 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 4.8 km/s] At Earth T x 1000 K, v=1km/s!! (no escape???) BUT, for a given T x there is a distribution of v

7 For a given T x there is a distribution of v f( v ) = 1 $ mv2' exp # 3/2 & ) [2" kt/m] % 2kT( = f(v x ) f(v y ) f(v z ) ; f( v r ) d 3 v = 1 *** Can Focus on the z component only! 1 f(v z ) = [2" KT/m] exp $ # mv 2' z 1/2 & ) ; % 2kT( ++ * f(v z ) dv z = 1 #+ Need : Flux of molecules across exobase in the + z direction, es = [n x v z ] f( v ) d 3 v ; * * * v z > 0, v > v es

8 * Escape (continued) Things are often written in terms of the mean speed v " # # # v f( v ) d 3 v Problem : Verify that the mean speed is v = 8 K T $ m Flux across a surface v z > 0, (all v) % + = n x v z f( v ) d 3 v # # # & v ( = n z x # exp ' mv 2 + z * - dv 0 [2 $ kt/m] 1/2 z ) 2KT, ( kt + = n x (kt/m)/[2 $ kt/m] 1/2 = n x ) * 2$m, - 1/ 2 % + = 1 4 n x v ; 1/4 due to isotropic assumption This is in the absence of gravity

9 Escape Flux (cont.) Flux Across Exobase ¼ n x v 8 kt v = ; n x = density π m Probablity of Escape P es = f + (E) de mgr f + (E) = energy distribution of flux at exobase

10 Escape (continued) For Escape Use : E > E es where E es is the escape energy The escape flux is " es = E> E es (n x v z ) f( v ) d 3 v # # # Rewrite as the flux across the exobase times a probability of escape " es = 1 4 (n xv ) P es Obtain P es problem Jeans (thermal) Escape $ P es = 1 + E ' es % & kt( ) e-e es kt E es kt : compares thermal energy to escape energy OR : using, E es = mg x R x, E es kt * mg x R x kt = R x /H x In this form: Compares Radius at the top of the atmosphere to the scale height!

11 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 Δp ~ 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!

12 * Escape (cont.) Problem: Jeans escape Venus Earth Mars Jupiter Escape Energy 0.56 ev / u 0.65 ev / u 0.13 ev / u 18 ev / u Titan (0.024*) ev/u *(at exobase) Assume T x = 1000 K (not true on present Venus) Use a time t = 4.5 byr " 4.5 # 10 9 # 3 # 10 7 sec Assume N x = atoms / cm 2 Calculate the net column loss, N es, of H, H 2 (or D), N from each planet assuming H = k T x / m g x and the exobase is either all H, H 2, or N. 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

13 Isotope Fractionation (preferential loss of lighter species) Relative loss rate is determined by 1. Masses of the escaping species therefore, (P es ) D << (P es ) H!! 2. But they must be present at the exobase! hence, depend on differences in diffusive separation!! Compare loss of H to D: H = lighter; D = heavier to learn about Loss of water

14 Isotope Fractionation (preferential loss of lighter species) Relative loss rate depend on differences in diffusive separation!! Compare loss of H to D: H = lighter; D = heavier N H (z h ), N D (z h ): col densitiesbelow homopause (turbopause) determined by atmospheric concentrations, same H--mixed At homopause Ratio: R D/H (z h ) = N D (z h )/ N H (z h ) = n D (z h )/ n H (z h ) At exobase n H (z x ), n D (z x ) densities at exobase: Δz = z x -z h Different H =kt/mg n H (z x ) = n A (z h ) exp[-δz/h H ]; n D (z x ) = n B (z h ) exp[-δz/h D ] Ratio: R D/H (z x )= R D/H (z h ) exp[-δz(1/h B -1/H A )] = R D/H (z h ) exp[-δz Δm g x /kt x ] Rel. concentration. Height of exobase, Δz, + mass dif., Δm.

15 Isotope Fractionation (cont.) Loss rate of a column of atmosphere dn A A = " # es dt For Jeans (thermal) Escape : with P es A A # es = " n A va 4 P A es $ = 1 + E ' es % & kt( ) e-e es kt Calculate The fraction of species A still in the atmosphere is o f A (t) * N A /N A = exp("t/t A ) (t A ) "1 * [(v A/4)P A es ] exp("+z /H A ) H Depends only on mass of A and the temperature at the exobase

16 Isotope Fractionation (cont.) Two species A and B (e.g., H and D) The ratio of total atmospheric concentrations vs. t is " r AB (t) = N (t) % A $ # N B (t)& ' / " N o % $ A o ' ( exp[)t ( 1 ) 1 )] # N B & t A t B One can re write r AB (t) ( [f A (t)] x with x = (1- t A /t B ) and f A (t) the fraction of A remaining at time t.

17 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 Sun So not only Venus --but earth has also lost water!! (unless you think comet pasting on of water is the starting point!) Note: this does NOT depend on how much water we started with!!

18 D/H Ratio Earth and Venus t H << t D (your problem) ( r HD ) now = " $ # N H N D % ' & now " $ # N H N D % ' & solar ( [f H (t)] x Earth : D / H 1.6 ) 10-4 r HD = 1.5 ) 10-5 / 1.6 ) 10-4 ( 0.09 x ~ 1; f H (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 f H (t) ( ! or * -assuming the same original water budget as the Earth ( of Earth's present water budget present value of loss rate (~ 10 7 /s) too small Need wet hot early atmosphere with H 2 O 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

19 Enrichment of Heavy Isotope Indicates Atmospheric Loss 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

20 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 + hν ΟΗ + Η + ΚΕ Ο + Η 2 + ΚΕ Present Mars CO e CΟ + Ο +ΚΕ O e Ο + Ο + ΚΕ In the absence of a protecting magnetic field 5 and 6 are important

21 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 the mass difference then: Only their relative abundance at the exobase is important: Rayleigh fractionation law! Applies to Mars for processes 3, 4, 5 Result (Ar and N arew fractionated) 36 Ar / 35 Ar no need for hydrodynamic episode 14 N / 15 N models give too mach loss (buffered by CO 2?) BUT! non-fractionation of 18 O / 16 O, 13 C / 12 C means there are large reservoirs!! carbonatcs and permafrost?

22 #7b 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