By Christopher J. Foot

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**Example text**

Molecule centers in this cylinder is n0 ×σc v t (number per unit volume × volume of the cylinder). Then the number of collisions per unit time (the collision frequency) is fcoll = n0 σc v [collision frequency]. 3) We may calculate the average distance between collisions to be the [distance per unit time] = [velocity] × [the time between collisions] (this last factor is the inverse of collision frequency): λ=v× 1 1 = n0 σc v n0 σc [mean free path, approximate form]. 4) The Greek letter λ is used in most texts to denote the mean free path.

A numerical model of the atmosphere has horizontal resolution (grid boxes) 2◦ × 2◦ . What is the area of one of these boxes at the Equator, and at 30◦ N? 81 m s−2 ). 7 Newton’s Law of Gravity says that the force on a particle of mass M is F =− GME M . 1). 8 A particle falls from outer space to the Earth’s surface. Far away its potential energy is zero. 1). What is the velocity of the mass when it strikes the Earth’s surface? (This is exactly the vertical velocity it would have to have if it were to escape the Earth’s gravity ﬁeld after being projected upwards from the surface.

3) We may calculate the average distance between collisions to be the [distance per unit time] = [velocity] × [the time between collisions] (this last factor is the inverse of collision frequency): λ=v× 1 1 = n0 σc v n0 σc [mean free path, approximate form]. 4) The Greek letter λ is used in most texts to denote the mean free path. Actually, it is possible to solve the problem when all the particles are in motion, and the derivation can be found in books on the kinetic theory of gases. 414 . . in the denominator: λ= √ 1 2n0 σc [mean free path, more exact form].