Basic concepts

The problem of permanently removing a molecule from a planet's atmosphere is much the same as the problem of sending a rocket from Earth to Mars: one must impart enough velocity to the object, and in the right direction, to allow the object to overcome the potential energy at the bottom of the gravitational well, and still have enough kinetic energy left over to allow the object to continue moving away. This leads to the central concept of escape velocity, which is the minimum velocity an object needs in order to escape to infinity, provided no drag forces intervene. The escape velocity is obtained by equating initial kinetic energy to the gravitational potential energy. Let m be the mass of the object, v its speed and r its initial distance from the center of the planet. Let MP be the mass of the planet and G be the universal gravitational constant. Then, equating kinetic to potential energy we find that 2 mv2 = GMP m/r, so v = ■\J2GMP/r = \/2gr where g is the acceleration of gravity at distance r from the planet's center. In many situations of interest, the altitude from which the molecules escape is sufficiently close to the ground that g is only slightly less than the surface gravity. Important exceptions include small bodies with a massive atmosphere, such as Titan, or such as the Moon would be if it were given a massive atmosphere. In this section, we'll use g to represent the actual radially varying acceleration g(r), and will use the symbol gs when we need to refer specifically to the surface gravity. The acceleration at distance r is then g(r) = gs(rs/r)2, where rs is the radius of the planet's surface.

In order to allow a molecule to escape, enough energy must be delivered to the molecule to accelerate it to the escape velocity. The study of atmospheric escape amounts to the study of the various ways in which the necessary energy can be imparted. Since the kinetic energy of a molecule with mass m is 2mv2, light molecules like H2 will escape more easily than heavier molecules like N2, given an equal delivery of energy. Dissociation of a molecule like CO2 or H2 into lighter individual components also aids escape. Using the formula for escape velocity, we can define the escape energy of a molecule with mass m as mgr. For escape of N2 from altitudes not too far from the Earth's surface, this energy is 2.9 • 10-18 J, or 2.9 attoJ 2. For H2 the escape energy is only 0.2 attoJ.

In the end, like so many things in planetary climate atmospheric escape is all about energy. There are four principle sources of energy that could potentially feed atmospheric escape:

2 One frequently sees the unit electron volt used for measuring small quantities of energy in a context like this. 1 ev = 0.16 attoJ

• The general thermal energy of the atmospheric gas, which ultimately comes either from absorbed solar radiation or from heat leaking out of the interior of the planet.

• Direct absorption of solar energy in the outer portion of the planet's atmosphere, which may energize particles to escape velocity either directly or through indirect pathways,or which may manifest itself in a hydrodynamic escaping current of gas. The solar radiation responsibe for these mechanisms is typically in the extreme ultraviolet (EUV) portion of the solar spectrum because there is so little mass in the regions involved that absorptionis weak, whence there is a premium on absorbing individual photons which have a great deal of energy.

• Collisions with the energetic particles (usually protons) of the stellar wind which streams outward from the atmosphere of the planet's star.

• Kinetic energy imparted to the atmosphere by the impact of large objects.

For a portion of the atmosphere where collisions are frequent enought to maintain thermo-dynamic equilibrium, each degree of freedom gets an energy of 1 kT on average. Some molecules will have more energy than this, and some will have less, but if the mean energy is coniderably less than the escape energy, only a very small proportion of molecules will have sufficient energy to escape. For T = 300K, the typical energy is only 0.002 attoJ. At this temperature only a small fraction of H2 molecules would have sufficient energy to escape from Earth, and the escape rate becomes only moderately higher if the temperature of the escaping gas is raised to 1000K. Heavier molecules like N2 could hardly escape from a planet with Earth's or Venus' gravity at all, if the only source of energy were thermal motions. Even on a light body like Titan, the escape energy for N2 is still 0.16 attoJ based on surface gravity, so escape will not be easy. Escape due to the energy associated with the thermal motions of particles in thermodynamic equilibrium is called thermal escape, or sometimes Rayleigh-Jeans escape. The ratio of escape energy to kT, namely Xc = mgr/kT, is an important parameter in the theory of thermal escape. In this formula, r is the radius of the atmospheric shell from which particles escape, g is the acceleration of gravity at that radius, and T is the typical temperature there.

For a particle to escape, it is not enought that it have reached escape velocity. It must also have a reasonable chance of escaping the gravitational well of the planet without suffering many collisions, since each collision will divert the particle from its outward path and rob it of some of its velocity. If the particle undergoes many collisions, it will undergo a random walk, leading to slow diffusion of the substance rather than rapid outward streaming. The portion of the atmosphere where particle collisions are so infrequent that a particle with sufficient energy has a good chance of escaping without collision is called the exosphere. In order to define the exosophere quantitatively, we first introduce the notion of mean free path, which is the mean distance traveled by a molecule between collisions. The mean free path depends on the total number density of particles, n, and the effective cross section area of the molecules, x. For simplicity, we'll assume for the moment that all the molecules in the gas are identical. Two molecules are considered to collide if their centers approach within a distance of 2^Jx/n. To estimate the mean free path, construct an imaginary cylinder with axis aligned with the direction of travel of the particle we are tracking, and having a radius \Jx/n. A collision is inevitable if this cylinder contains a particle from the rest of the gas, which becomes likely when nV = 1, where V is the volume of the cylinder. Writing V = ¿na2 = ¿x we find that the mean free path is « 1/n\. This estimate would be precise if the particles being collided with were stationary. In reality, the distance moved by the test particle before collision is affected by the fact that some of the other particles are moving toward the test particle, while others are moving away from it; a more precise calculation making use of the actual distribution




H -H2




H -air




H -CO2
















H2 O-air
















Table 8.1: Effective collision radius for various binary collisions, computed from diffusion data. The radius is given for three different temperatures, in units of picometers. (lpicom = 10-12m). The collision cross section for collision radius a is x = na2.

Table 8.1: Effective collision radius for various binary collisions, computed from diffusion data. The radius is given for three different temperatures, in units of picometers. (lpicom = 10-12m). The collision cross section for collision radius a is x = na2.

of particle velocities in thermodynamic equilibrium yields the slightly modified result

when all the particles have the same mass. When the background particles are very massive compared to the particle we are tracking, then they can be regarded as essentially stationary and the a/2 factor in the denominator can be dropped; for most of the uses to which we will put the mean free path, this effect is of little importance.

The effective particle collision cross-section depends on the pair of molecules which are colliding, and is also a weak function of the energy of the collision, since molecules are not hard spheres but rather can be penetrated when the collision energy is large enough. Still, the effective collision radius does not differ greatly between one molecule and another. The collision radius can be inferred from measured diffusion rates, and a few typical values are given in Table 8.1. For the most part, the effective collision radius is between 100 and 200 picometers. An important exception is atomic hydrogen (H), which has an effective radius on the order of a mere 10 picometers. Curiously, the effective binary collision radius remains this small even when H is colliding with a much larger molecule. Evidently, the H atoms are like little bullets, which punch right through the outer electron clouds of the bigger molecules. This effect gives atomic hydrogen an anomalously large mean free path for a given density, which has a number of important consequences. Data for atomic oxygen (O) is hard to come by, but neon is believed to be an analog and should have a similar collision radius. Air at the Earth's surface with a temperature of 300K has a particle density of 2.4 • 1025m-3. Based on a collision radius of 125 picom the mean free path is about 0.6 ym. Adopting a scale height of 8km, the particle density at an altitude of 100km falls to 9.0 • 1019m-3 and the mean free path increases to 16 cm - about the width of a hand.

The mean free path increases exponentially with altitude, because the particle density in a gravitationally bound atmosphere decreases exponentially with altitude, at a rate given by the density scale height (RT/g for the isothermal case). The exosphere is said to begin where the mean free path becomes sufficiently large that further collisions before escape are unlikely; this critical altitude is called the exobase. Commonly, it is defined as the altitude where the mean free path becomes equal to the scale height, since the exponential increase of mean free path with height means that if a particle doesn't collide with another within the first scale height, it is basically home free and unlikely to find another to collide with. When the exobase is not too far above the ground, in comparison to the planet's radius, g can be approximated by the surface gravity and one can immediately estimate the exobase altitude in terms of the temperature of the exosphere and the scale height for the dominant constituent of the exosphere. In applying this procedure, one must keep in mind that the temperature of the exosphere could be quite different from the temperature of the lower atmosphere, and the composition could differ greatly from the bulk composition of the atmosphere. We will have more to say about both these aspects of the exosphere a bit later. For the most part, one is interested in the exobase particle density, since that is what will determine the flux of particles to space. When one at least knows that the exobase is low enough that g « gs, the exobase density is immediately given by the requirement that I = H, which implies that the exobase particle density is n = gs = mgs (8 11)

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