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<ri+1, of the atmosphere is given as i =1, 2,...,q,q+1,...,I as rI+1 →∞. With this approximation, we obtain the follo- wing general expression for λ: v GmM GmM 1 − 1 λ = f (ηb) 2 exp . 4 kTr kT rb r0 b v GmM λ = f (ηb) 2 (3) 4 kTrb Here G is the gravitational constant, m is the mole- q−1 GmM 1 1 exp − i=0 − cular mass of gas, M is the mass of the planet, kTei ri ri+1 v = 8kT/πm is the mean velocity of gas mole- × exp − GmM 1 − 1 cules, k is the Boltzmann constant and r0 is the kTeq rq rb × effective radius of the planet. The function f (ηb) I 1 − exp − GmM 1 − 1 is the percentage of number density with speed i=0 kTei ri ri+1 v greater than the escape velocity. rb is radius of ‘escape border’ through which the molecules with rq ≤ rb <rq+1. (3A) velocity v greater than escape velocity ve (v ≥ ve) can escape to space without collisions and is deter- Here T is the real temperature at rb.The mined by the following condition: escape border is within the isothermal region rq ≤ rb <rq+1, in which the effective tempera- 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<rI+1(= )and apointr. The escape velocity ve = (2GM/r). summing up to I. Function f(η) is the percentage of number density The following assumptions have been used to with speed v greater than the escape velocity at the derive the above formula point r over the total molecular number density at • that point: the temperature distribution does not change with time • the escaping process of gas is a quasi-static n(v ≥ ve) 2 2 f(η) ≡ = √ ηe−η +1− erf(η) process n(v ≥ o) π • the molecular number density distribution of gas is given by a convergent formula, discussed else- ∼ 2 −η2 1 where (Da Dao-an and Yang Ya-tian 2005). = √ e η + π 2η 3. Escape rate of atmosphere on Moon η ≡ ve(r)/vp, vp ≡ (2kT/m). erf(x) is the error function. The approximate value of f(η)islessthan 0.1% if η ≥ 3. We can see that η is a function of The effective radius r0,massM, temperature during day and night: Td and Tn, acceleration of T and ve at r,andηb denotes the value of η at gravity g of Moon are taken as follows: r = rb. 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 .