Escape of atmospheric gases from the Moon
Da Dao-an1,∗ and Yang Ya-tian2 1Lanzhou Institute of Physics, 730 000 Lanzhou, Gansu, China. 2Fujian Normal University, 350 007 Fuzhou, Fujian, China. ∗e-mail: [email protected]
The escape rate of atmospheric molecules on the Moon is calculated. Based on the assumption that the rates of emission and escape of gases attain equilibrium, the ratio of molecular number densities during day and night, n0d/n0n, can be explained. The plausible emission rate of helium and radioactive elements present in the Moon has also been calculated.
1. Introduction equilibrium. The emission rate on the surface of the Moon is estimated to be about 1023 molecules per A thick atmosphere is present on the Earth and second. Venus but it is only a very tenuous atmosphere, 9 −3 with a number density of n0d =3× 10 m during 10 −3 day and n0n =6× 10 m during night, consist- 2. Expression of escaping lifetime ing mainly of helium and argon which exist on of atmosphere the Moon. Whether a thick atmosphere existed on the Moon in the past or not is related to We derive an escape equation of atmosphere the process of formation of the Moon. One view (Konpaneetz 1957) is that the Moon did have a dN thick atmosphere in the beginning but owing to = −λN, (1) its weak gravity, the atmospheric molecules gradu- dt ally escaped, resulting in the very thin atmosphere where N is the total number of molecules in the existing now. To see if a thick atmosphere could atmosphere and λ istheescaperate.Thesolution be retained or not, we assume that the air hav- of (1) satisfying the initial condition of N = N0 ing pressure of 1 atmosphere (as exists on Earth when t =0is now) existed on the Moon initially and calculate −λt the escape life times for various assumed tempe- N = N0e . rature distributions. The results show that if the atmospheric pressure is equal to or less than 1 atm, When t =1/λ the escape time is much less than the age of the Moon (∼ 4.5 × 109 y), except for one special case, −1 N = N0e . implying that any atmosphere (p ≤ 1atm) onthe Moon would almost totally escape. These cal- The escape life time is defined as culations also explain the observed day to night ratio, n0d/n0n ≈ 0.05 based on the assumption 1 that the emission rate of gases (mainly helium) τ = . (2) from the Moon and the escape rate attain dynamic λ
Keywords. Escape rate; lunar atmosphere; radioactivity; helium.
J. Earth Syst. Sci. 114, No. 6, December 2005, pp. 637–644 © Printed in India. 637 638 Da Dao-an and Yang Ya-tian
The expression of λ with uniform temperature T is a constant within the region ri ≤ r r ≥ rb → l(r) ≥ h(r). ture is Teq . The denominator of the last term in (3A) originates from the normalization of number density: l(r) is the mean free path of molecules at r. h(r)is ∞ decided by: n(r)4πr2dr = N. n(r + h(r)) 1 = . 0 n(r) 10 It should be summed over the whole region, taking ≤ ∞ Here n(r) is the molecular number density at TeI = T∞ in the region of rI r The temperature distribution of atmosphere 22 mainly depends on the pressure (or density) of r0 = 1738 km,M=7.36 × 10 kg,Td =400 K, the atmosphere on a planetary body. If the −2 Tn =90K,g=1.62 ms . atmosphere is thick like that on Earth, Venus and some satellites on Saturn, the temperature distri- bution is non-uniform; it varies with height and, In order to compare the retention ability of for simplicity, we can divide the whole atmosphere atmosphere on the Moon with Earth, we assume into several ‘isothermal regions’ by introducing an that the initial atmosphere of the Moon consisted effective temperature Te foreachregiontofitthe of ‘fictitious’ molecules having a mean molecu- measured data of pressure. That means T = Tei lar weight of 28.8, and the initial pressure of Escape of atmospheric gases from the Moon 639 Table 1. Escape rate λ and life time τ for the assumed temperature distribution in the atmosphere on Moon. Assumed r l r h r T r v λτ temperature b ( b) ( b) ( b) 4 −1 distribution (km) (km) (km) (K) (m/s) ηb f (ηb)(s)(y) −7 −21 13 (i) 2.6r0 794 695 240 104.6 3.98 6.11 × 10 2.3 × 10 1.4 × 10 3 3 −15 7 (ii) 8.5r0 4.57 × 10 4.57 × 10 400 135.0 1.70 0.1228 1.2 × 10 2.6 × 10 2 3 3 −15 τ = (iii) 8.5r0 4.57 × 10 4.57 × 10 d: 400 135.0 1.70 0.1228 λd =1.2 × 10 λd + λn =5.2 × 107 −36 −50 1.32r0 69.0 60.8 n: 90 64.0 9.12 7.9 × 10 λn =6.1 × 10 5 4 −15 6 (iv) 47r0 ≥ 2.7 × 10 5.7 × 10 1600 270 0.725 0.789 7.16 × 10 4.4 × 10 the atmosphere was 1 atm. One can calculate the but the difference of λ between day and escape rate by means of the formula (3) and (3A). night is about 35 orders of magnitude, thus Since we do not know the temperature distribu- τ =2/(λd + λn) ≈ 2/λd and the life time in case tion on the ancient Moon, in order to calculate the (iii) is twice the life time of case (ii) owing to escape rate and life time, we assume four simple almost no escape at night time. Taking the age 9 cases of temperature distributions: of the Moon as tM =4.5 × 10 y,itisshownfrom the calculations that 100τ ≤ tM , for all the three × 13 (i) T =240 K, uniform temperature (= mean tem- cases except (i). In case (i), τ =1.4 10 y, much perature of Earth’s atmosphere), greater than tM and only a small fraction of mole- cules would escape. Therefore, the atmosphere on (ii) T = 400 K, uniform temperature (surface tem- the Moon would continue to exist with pressure perature of Moon at day), near 1 atm. But in real case, the higher the height, (iii) Td = 400 K, Tn = 90 K, surface temperature of the higher the temperature would be for the thick Moon during day and night, respectively, atmosphere, until it approaches T∞ which is higher (iv) Te1 = 240 K(r<(3/2)r0), than 1000 K, depending on the solar activity. Dur- ∼ T = 800 K (r =(3/2)r0), ing quiet periods, T∞ = 1000 K and during solar ∼ Te2 = 1600 K(r>(3/2)r0) active periods, T∞ = 1600 K. From the viewpoint of comparing the retention ability of atmosphere similar to the temperature distribution on Earth on the Moon with Earth, the condition of case during the solar active period. We simply (iv) is closer to the situation of Earth during the suppose that the temperature distribution can solar active period. In this case, be divided into two effective isothermal regions: 6 τ =1.8 × 10 y, (1) r0 ≤ r<(3/2)r0, in which the temperature is constant (Te1 = 240 K), like the average tem- 9 3 tM =4.5 × 10 y =2.5 × 10 τ, perature of the atmosphere on Earth below 80 km; (2) the second effective isothermal region −2.5×103 −1.086×103 N = N0e = N0 × 10 . is (3/2)r0 nd 1 ηT denotes ηb(T ),ηd and ηn denote ηb(Td)and = =0.05. n0n 20 ηb(Tn) respectively. ηd =1.86, ηn =3.91. The measured ratio is only 1/4.5ofthetheore- tical value. For the molecular number density of d 9 −3 10 −3 f(η )=0.0744, n0 =3× 10 m − 6 × 10 m as on the Moon, the mean free path of molecules is very large, −6 f(ηn)=1.04 × 10 , l =1× 104 km − 1.9 × 105 km, so the molecules with v ≥ ve on the surface of the Moon can escape. The energy of α particles emitted from tho- 2 2kTd vd = √ =1.45 km/s, rium and uranium series elements is greater than π m 4 Mev, which corresponds to velocities greater × 4 than 1.4 10 km/s, much greater than the escape 2 2kTn velocity (∼ 2.4 km/s) on the Moon. Taking into vn = √ =0.688 km/s. π m account the energy losses while penetrating the rocks or soil of the Moon, the velocity distribution Substituting this into (3B), one obtains: reduces to Maxwellian distribution, then equation (3) or (3A) is valid. The measurements show that × −5 −1 gases on the Moon are mainly composed of helium. λd =5.31 10 s , It is believed that the amount of gas is the result −10 −1 λn =3.52 × 10 s , of competition between the emission and escape processes of gases. Suppose the atmospheric mole- and the equivalent decay constant for uranium and cular number on the Moon is N and the emission thorium series is particle number from lunar surface is Nr,thenwe have ∼ −10 − −9 −1 ≈ −16 − −17 −1 λr (10 10 )y (10 10 )s . dNr = −λrNr, (4A) dt Therefore, λd λn λr. (5) dN − = λrNr λdN (day), (4B) dt It means the escape rate during daytime is much dN greater (about 5 orders of magnitude) compared to = λrNr − λnN (night). (4C) dt theescaperateduringnight,andtheescaperate of helium gas on the surface of the Moon at day or λr is an equivalent decay constant of various night is much greater (about 7 to 12 magnitudes) radioactive series, λd and λn are the escape rates of than the production rate of helium gas due to the gas during day and night respectively, the escape emission of alpha particles by uranium and thorium border is taken as the surface of the Moon, i.e., series nuclides. In this case, rb = r0, thus (3) is simplified as λn ± λr ∼ λn and λd ± λ ∼ λd. (5A) v GmM λ = f(ηb) 2 . (3B) 4 kTr0 Now we proceed to solve the set of equations (4A)–(4C). Substituting Td = 400 K,Tn =90Kintof(ηb) We choose a suitable time t0 as a starting point of day, regarding t0 to t0 +(R/2) as day and t0 +(R/2) 2 −η2 f(ηT )=√ e T ηT +1− erf(ηT ), to t0 + R as night, R being the rotation period of π Moon. As t−t0 → t, assuming the initial condition Escape of atmospheric gases from the Moon 641 N = N0,Nr = Nr0, at t =0, (6) Under the requirement of periodical solution (12) we integrate (4B) the solution of (4A)–(4C) satisfying (6) is: R/2 R/2 R/2 dN −λr t − Nr(t)=Nr0e , (7) dt = λrNr(t)dt λd N(t)dt. dt 0 0 0 λr −λdt N(t)= N0 − Nr0 e The result is λd − λr R R R r λ −λr t R N − N0 = λrNr0 − λdN d , (14) + Nr0e , 0 ≤ t ≤ (day), 2 2 2 λd − λr 2 (8) R/2 ≡ 1 R λr −λ R −λ t− R N d N(t)dt. (14A) r 2 n( 2 ) N(t)= N − Nr0e e R/2 2 λn − λr 0 → λr −λr t R Similarly, we integrate (4C) from R/2 R,and + Nr0e , N (R)=N0. (12) x>1, due to ∆N(R/2) > 0. From (11) and (12) one obtains: The relation between total atmospheric molecular r r0 number N and the molecular number density on λ N 2 N(R)=N0 = −(λ +λ ) R the surface n0 is N =4πr0n0kT/mg. Since the − d n 2 1 e number densities on the lunar surface during day and night are n0d and n0n respectively, we have 1 −λ R −λ R 1 −λ R × e n 2 1−e d 2 + 1−e n 2 . λd λn n0d TnN d λn Tn (13) = = x . (18) n0n TdN n λd Td 642 Da Dao-an and Yang Ya-tian These molecules can be expected not to escape In order to calculate x,webeginwith∆N(R/2). from the Moon. Then the density ratio between N0 has been given in (13) from which one can day and night should obey obtain the expression of N(R/2) by substituting (13) into (10). By subtraction of these two terms n0d (B) Tn we have = . (22) n0n (B) Td R ∆N 2 Here n0(B) denotes the number density of such heavy molecules. Now we proceed to calculate their R = N0 − N percentage abundance in the atmosphere. Suppose 2 the density of helium is n0d(He) and n0n(He) dur- R R ing day and night respectively, from (21), we get −λd −λn 1 1 λrNr 1 − e 2 1 − e 2 − 0 λn λd = R . −(λn+λd) 1 − e 2 Tn (19) n0d(He) = y n0n(He). (23) Td Since λn is much less than λd,weexpand∆N into series of λnR/2 and keeping only the linear term, Suppose the number density of heavier molecules one obtains at nighttime is n0n(B) λd 2 n0n(B)=βn0n(He). (24) x = , λn 1+(λdR/2) / [1 − exp (−λdR/2)] Here β is the ratio of n0n(B)/n0n(He) which for λd λn,λdR ∼ 1. (20) remains to be determined. We denote the number density of heavier molecules at day time as n0d(B), Substituting into (18), we have which can be obtained from (22) n0d Tn Tn = y , n0d(B)= n0n(B). (25) n0n Td Td 2 y = , (21) We determine β as follows 1+(λdR/2) / [1 − exp (−λdR/2)] R is taken as the synodic month, that is R =29.53 n0d = n0d(He) + n0d(B), earth days (26) n0n = n0n(He) + n0n(B), λdR/2=67.8,y=0.0291 n0d (y + β)Tn/Td = , (27) n0d Tn n0n 1+β =0.0291 =0.00654. n0n Td n0d − Tn n y T But the measured value is (n0d/n0n)measure = ∴ 0n d × 9 × 10 β = . (28) (3 10 )/6 10 =0.05. Tn − n0d We find that the calculated value is less than Td n0n the measured ratio by one order of magnitude. The problem originated from too many molecules Substituting n0d/n0n =1/20, Tn/Td =90/400, y = escaping during the lunar day, leading to the very 0.0291 into (28), we have small density ratio between day and night. Thus we have to consider other types of gas molecules, in β =0.248. addition to helium like Ar and Rn which can hardly escape. These molecules have heavier masses and therefore vp values are small, resulting in a large η, This result shows that the number density of and small f(η), finally making λ small. For exam- helium at night is about 75% and other heavier ple, from (3B), molecules constitute ∼ 25%. One can also evaluate the gas emission rate on −13 −1 λd(Ar) = 2.76 × 10 s λn(He). the Moon Escape of atmospheric gases from the Moon 643 1 1 produce elastic or inelastic scattering. A fraction of λrNr0 = (λdN d + λnN n) ≈ λdN d 2 2 the protons scattered in space immediately escapes since the velocity is far more than the escape 1 2 velocity on the surface of the Moon (∼ 2.4km/s). = πr v(Td)f(ηd)n0d(He). (29) 2 0 A fraction of them are also absorbed by the lunar soil or rocks due to multi-inelastic-scattering and From (23)–(26) others lose their energy to form the atmosphere on the Moon with Maxwellian distribution. Sup- Tn n0n pose each solar wind proton captures an elec- n0d(He) = y · . (30) tron to form a hydrogen atom, which subsequently Td 1+β forms hydrogen molecules then, formula (3A) can be used to estimate the escape rate of hydrogen Substituting y =0.0291,Tn/Td =90/400,β= 10 6 molecules. 0.248,n0n =6× 10 ,r0 =1.738 × 10 m, v(Td)= 3 From formula (3B) 1.45 × 10 m/s,f(ηd)=0.0744 and (30) into (28), one obtains 2GM 2kT ≈ 23 −1 ηd(H2)= 1.32, λrNr0 =1.61 × 10 s . r0 m(H2) ≈ This indicates that the amount of helium emitted ηd(H) 0.93, is about 1 g per second and ≈ fH2 (ηd) 0.324, 1 23 −1 fH (ηd) ≈ 0.632, Nr0 ∼ × 1.61 × 10 s . λr −4 −1 λd(H2) ≈ 1.64 × 10 s , −16 −17 −1 Taking λr ∼ (10 − 10 )s −4 −1 λd(H) ≈ 2.26 × 10 s . 39 40 Nr0 ∼ (1.61 × 10 − 1.6 × 10 ). Then, the variation of the total hydrogen number Of course, it does not correspond to the total num- N is ber of thorium and uranium atoms on the Moon, dN = J − λdN. since in the decay processes one thorium or ura- dt nium atom emits 5 to 12 α particles and then decays into a stable element whereas in the above Here J is the flux of incident solar wind and λdN calculation, only one α particle was considered. is the escape rate on the surface of the Moon, Therefore, the total number of thorium or uranium 2 2 atoms should be revised as (1/5) − (1/12)Nr0 ∼ J = πr0vns,N=4πr0kTnd/mg. (1/8)Nr0, corresponding to 6 −3 Here ns =8×10 m is the number density of solar ∼ × × × −27 × 39 ∼ 40 Mr 1.6 240 1.674 10 (10 10 ) wind and nd is the number density of hydrogen. In 1 steady state case, dN/dt =0. × kg ∼ 8.1 × (1013 ∼ 1014)kg. Then, we have 8 J = λdN or nd =(mgv/4kTλd)ns. This amount is equivalent to 1.1 × (10−9 ∼ 10−8) the mass of Moon. It should be pointed out that in the above Then, we have calculations, we did not consider the solar wind ∼ × 9 −3 ∼ × 9 −3 and cosmic ray implanted ions which may escape nd(H2) 5.4 10 m ,nd(H) 2.8 10 m . during day time in the lunar environment. The main ingredients of solar wind are pro- This is an overestimate of the hydrogen number tons. The average velocity of the solar wind is density since a large fraction of the scattered pro- v =∼ 450 km/s and the number density of solar tons are absorbed by the lunar soil and rocks or wind is 8 × 106 m−3. From the mean speed of the escape immediately to space and only a small frac- proton v = 450 km/s, the energy is calculated to tion becomes a part of the lunar atmosphere. Even −3 be about Ep ∼ 2.1 × 10 Mev. This energy is too so the solar wind contributes a significant factor to small to induce nuclear reactions and can only the number density of the lunar atmosphere. 644 Da Dao-an and Yang Ya-tian 5. Conclusions emission rate on the surface of the Moon is about 1.61 × 1023 s−1, equivalent to the emission of 1 g Based on the above calculations we can draw the of helium per second assuming dynamic equili- following conclusions: brium between emission and escape. • • Solar wind contributes a non-negligible fraction Assuming that there existed an atmosphere of in the number density of the lunar atmosphere. 1 atm on the Moon, with temperature below 240 K, the atmosphere would then be retained and a very small fraction of molecules would Acknowledgements escape. • If there existed an atmosphere on the Moon in We are thankful to the reviewers for many precious the beginning and its temperature distribution and helpful comments. was similar to the current temperature distribu- tion on the Earth, then escape time is calculated to be τ =1.8 × 106 y, and the atmosphere would References have completely escaped a long time ago. • Based on the ratio of the measured number Da Dao-an and Yang Ya-tian 2005 Life time of atmospheres on Earth and Venus; Vacuum and Cryogenics 11(2) density between day and night, it is estimated 70–77. that the tenuous atmosphere on the Moon is Konpaneetz A S 1957 Theoretical Physics; Gostechizdat, composed of 75% He and 25% Ar. The gas 2nd Edition.