Mean free path
Suppose the air is filled with balloons and you fire a laser beam in a random direction.
n = Density of balloons in units of balloons/meter^3
R = Radius of balloon in meters (all balloons have the same radius)
A = Cross-sectional area of the balloon
L = Characteristic distance the laser beam has to travel before hitting a balloon.
= The "mean free path"
L = 1 / (n A)
Suppose we replace the balloons with gas molecules.
The mean free path of a gas molecule is the characteristic distance it has to travel
before hitting another gas molecule.
What is the characteristic cross-sectional area of a nitrogen molecule? What
is the density of gas molecules in the Earth's atmosphere and in Mercury's
atmosphere? (in molecules per meter^3). What is the mean free path for a gas
molecule on the Earth and Mercury?
In space, the chief radiation hazard is from high-energy cosmic ray protons.
Protons are stopped if they collide with a nucleus. For nuclei, protons and
neutrons are packed uniformly in a nucleus.
Most atoms have similar size. In the following calculation we will assume that
all atoms have the same size as iron. The size of the nucleus, however, depends on
the number of protons + neutrons.
|Size of atoms
Z = Number of protons in the nucleus
W = Number of protons + neutrons in the nucleus
R = Radius of the nucleus
= 1.25 W^(1/3) * 10^-15 meters
n = Density of iron atoms in metallic iron
= 8.49e28 atoms/meter^3
= Density of nuclei in metallic iron
A = Cross section of the nucleus
= Pi R^2
The mean free path of a high-energy proton passing through matter is
L = 1 / (n A)
= 2.40 W^(-2/3)
For iron, W = 56 and L = .16 meters
The larger the nucleus, the smaller the mean free path. Tungsten is good for
radiation shielding because it has a heavy nucleus and because it is cheap.
Suppose you construct a wall of shielding with thickness L. The "column density"
is mass of the wall per square meter.
M = Mass of atom
= 1.66e-27 W
D = Density of the shielding material
= n M
= 140.9 W
Q = Column density of shielding wall in kg/meter^2
= L D
= 338 W^(1/3)
The lighter the nucleus, the less mass required for the wall.
To block cosmic rays you need a wall with a thickness of 4000 kg/meter^2.
For iron, this corresponds to a thickness of about 3 mean free paths.
The sun has a speed of 20 km/s with respect to the local stars.
Suppose we're wondering if a passing star will disrupt the orbit of the Earth.
n = Density of stars in the solar neighborhood
= .004 stars/lightyear^3
R = Minimum distance a passing star can get to the sun and not disrupt the
orbit of the Earth
~ 10 AU
~ 1.5e12 meters
A = Effective cross section of the solar system for an encounter with a passing star
= Pi R^2
L = Mean free path of the sun for a gravitational encounter with a passing star
S = Diameter of the Milky Way
= 100000 light years
What is L/S?