JOHN J. ALEXANDER
University of Cincinnati Cincinnati. OH 45221
An Attractive Problem on Atmospheric Escape Michael P. S. Collins1 Bayero University Kano, Nigeria Introducing "fresh air" into problems on the gas laws and elementary kinetic theory is not easy. The question below which is essentially a simple problem on "escape velocity" (with a sneaky aside on 0 2 dissociation) is representative of a wealth of material that can he exploited from applying the eas laws to atmosuheric and meteoroloeical uhenomena. The only prerequisite for the student is a familiarity with the concept of escape velocity and its relationship to gravitational attraction. This can come from the briefest of mentions in a lecture and course work review urohlems. Some useful references are:
molecules thermally oi. photochemically dissociate in the region of the exosphere? (2) Spitzer has proposed that there is a relationship hetween planetary half-lives, t t n , of exospheric constituents and the ratio of their root mean square velocity, @ , a n d u,. For an exospheric temperature of 1500 K he proposes
- .
v
(1) Smith. E. P. and Jacobs, K. C., "lntroduetory Astronomy and Astrophysics," W. B. Saunders. Philsdelphia, 1973. (2) Cole, F. W., "Fundamental Astronomy," John Wiley. New York, 1974.
(a)
R ~ w ~ ~ (~ ,~J .dA i.
JS"l vS
tqt2EaRh years
0.4-0.5 0.3-0.4 0.2-0.3
10' 10'"
0.1-0.2
1060
Assuming T.
= 1500 K
Show that by crediting a past temperature of 1500 K to the Martin atmosphere a possible explanation of the present lack of oxygen and abundance of nitrogen on that planet is provided.
(?)'12 = (3RTIM)1/2 or = (GMM.,.I~)~/~ i ~ ~ ) , - ~ h ~ ~ i ~ ~ ~ f t h ~ ~ ~ ~ ~ ~ t ~ ~ ~ h ~ ~ ~ , " ~ ~ ~ d ~ ~ i ~ ~ ~ ~ . ~ ~ ~ G is the constant of Universal gravitation = 6.67 X 10-'1 S.I. units
197C (4) Mc1ntosh.D. H. andThom,A. S.,"EssenLisls ofMeterolq?.," WykehsmPuhiiesfions.
London, 1969.
Question
The outermost fringe of a planetary atmosphere is termed the exosphere. For the earth, the exosphere is considered to extend 700 km uuwards. In this reeion. - . the densitv of air is so low that it is onG possible to talk of temperatureUasa means of defining the mean kinetic energy of molecules present; an accepted temperature of the earth's exosphere, T,,is 1500 K. Individual air molecules uerform lone elliutical orbits between collisions. However, if acollision is violent enough to impart a velocity greater than the escape velocity, u,, then upward moving molecules with this velocity will he projected out of the atmosphere. Because of the relative lack of shielding in the exosphere, the molecules in this region are susceptible to a considerahle flux of ultraviolet radiation from the sun, the earth's exosphere a t 2000 A (1P\ = 10-'0 m). (1) Given the following dissociation energies: Hz: 435 k J mole-'; Nz: 945 kJ mole-'; 0 2 : 498 kJ mole-'. Will any of these
'
While this article was in press, the Editor was informed of Dr. Collin's death. All correspondence regarding this article may be addressed t o Professor B. J. Salter-Duke,Head, Department of Chemistry, Bayero University, P.M.B. 3011, Kano, Nigeria.
284 / Journal of Chemical Education
M is the molar mass of the relevant gas R is the ideal gas constant = 8.314 J deg-I mole-' MM... is the mass of Mars = 6.5 X 102Qg r is the radial distance to the exosphere = 3.5 X %T.I
Comment on the possibility of any primordial methane and ammonia remaining presently in the Martian atmosphere. (Approximate age of Universe is said to he 101° Earth years.) Planck's constant = 6.62 X 10F4J sec Avogadro's number = 6.023 X 1OZ3 mole-' Velocity of light in vacua = 2.998 X 108m sec-' Acceptable Solution: (1) Thermal energy of the systems is of the order of 1X RT - 2 X RT per mole 1 X 8.31 X 1500 - 2 X 8.31 X 1500 J mole-'
-
= 12.5 - 24.9 kJ mole-'
Clearly there is not enough energy thermally to dissociate any of Nz, HI and 02. Photons of light ofwavelength X, 2000 A, correspond to quantaof energy E. E=
he
his Planck's constant; e is the velocity of light in vacuo
E=
6.62
X
lo-"
J sec X 2.998 X m 2.000 X
lo8 m sec-' = 9.92 X 10-I9 3 molecule-'
9.92 x 10-'9 X 6.023 X Hence 0~and
loz3= 598 kJ mole-'
Hz can be phornrh~mi~~lly disso~.iatedi n the upper
atmusphere; mdted O7 is prncr~cnllyall diisociarcd in this regmn. 6.67 X lo-'' X 6.5 X (2) Escape velocity of Mars = 3.5 X 106
(3
8.314 X 1500)lm 28 X = 1.16 X 103 m sec-I (3 X 8.314 X 1500)lm Velocity of dioxygen molecules = 32 X = 1.08 X lo3 m seer1
Velocity of dinitrogen molecules =
X
From the above table we get for ratios of (?)"%. molecule
(3"*lvs
predicted time of half-life
N2
0.33 0.31
1010 years 10'O years
02
This clearly indicates that slower moving diaxygen molecules should have a comparable half-lifeto dinitrogen. However, this does not seem to he the case empirically. An explanation comes forward if Oz is considered to be dissociated in the Martian exosphere-as on Earth. X 8.314 X 1500)'/1 velocity of oxygen atoms = 16 X
(3
The ratio ( ~ ) ' " l u . here = 0.44 which indicates a half-life for oxygen of -10" "ears..much smaller than the Universal ace. Hence the Martian ntmmphcre isdepleted in O1hy this t w - d a y ? mechanism. CH4 rmdnrmns~16 X 10- kg)nnd I\'HI
