Geoscience Reference
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which gives the mass-dependent temperature T e of escape:
2
3
R E Mg
R
T e =
(12.20)
We can now conclude that all the gas of molecular weight
M on a planet of radius
R E are lost when the atmospheric temperature exceeds T e . Overall such a simple argu-
ment explains why there is no atmospheric gas on the Moon, very little on Mars, whereas
the larger planets Venus and the Earth enjoy a thick atmosphere, not to mention Jupiter
and Saturn. Although this calculation gives us a good first-order indication, the lack of
helium in the terrestrial atmosphere and the lack of some gases on Venus and Mars, most
notably N 2 , indicate that the situation is more complicated. For example, the Earth has
an acceleration of gravity of 9.81 m s 2 and a radius of 6.371
<
10 6 m. Helium-3 ( M
×
=
10 3 kg) should be lost at temperatures in excess of (2/3) (6.371
10 6 )(3
10 3 ) 9.8 1/
3
×
×
×
8.31
15 000 K. A first obvious reason is that our ca lculation uses the mean velocity u
of the gases: some molecules must dash faster than u , while some others must be much
slower and remain in the atmosphere. We therefore need a more sophisticated description
of atmospheric velocities.
Let us stay for simplicity with our simple monatomic gas (e.g. Ar). If the atmosphere is
thick enough, molecules, will collide with each other and exchange their energy, a process
referred to as thermalization. An atom of molecular mass M traveling in the direction x
with the velocity
=
v x transports a kinetic energy 2 M
x . The proportion f
v
(v x )
d
v x of atoms
with velocity between
v x and
v x +
d
v x will be given by the Boltzmann distribution:
C exp
2
x
2 RT
M
v
f
(v x )
d
v x =
(12.21)
where C is a constant that will be determined later. We can write the same equation for the
other two dimensions and multiply the three equations to obtain the distributi on of velocity
in any direction, but are mostly interested in the distribution of speeds u
=
v
x + v
y + v
z
(the modulus of the velocity), which, with a bit of mathematics, is given by:
M
2
3 / 2
u 2 exp
Mu 2
2 RT
f
(
u
)
d u
=
4
π
(12.22)
π
RT
This is the famous Maxwell-Boltzmann velocity distribution ( Fig. 12.19 ) giving the prob-
ability that an atom has a speed between u and u
d u and which shows that, to a good
approximation, RT is the variance of the velocity distribution. The ro ot -mea n-square
speed, which is the parameter we can use to describe the kinetic energy, is u
+
= 3 RT
M
We should now be happy with the solution and contend that the fast tail of the Maxwell-
Boltzmann distribution above the escape velocity should leave the planet. It shows that
temperature is rather the equivalent of the variance of the speed than of its mean (molecules
traveling together are cold regardless of their speed). Collision after collision, the slower
molecules would progressively thermalize again and produce more of the fast molecules.
Sooner or later, most of the light gases should therefore be gone. One more time, it does
not quite work that way! The reason is that it is not enough that a molecule travels faster
than the escape velocity, the way out must be clear of collisions. The important concept
is therefore that of the mean free path between collisions: if a molecule traveling faster
/
 
 
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