 # (a) Pressure decreases as one ascends the atmosphere. `
Question:

(a) Pressure decreases as one ascends the atmosphere. If the density of air is ρ, what is the change in pressure dp over a differential height dh?

(b) Considering the pressure p to be proportional to the density, find the pressure p at a height h if the pressure on the surface of the earth is ρo.

(c) If po = 1.03 × 105 N/m2, ρo = 1.29 kg/m3 and g is 9.8 m/s2 at what height will the pressure drop to (1/10) the value at the surface of the earth?

(d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.

Solution:

(a) As we move above, the pressure decreases because the thickness of the gas above us decreases. Let A be the cross-section of the horizontal layer of the air and height be dh.

Then P is the pressure on the top layer and p + dp is the pressure at the bottom.

Since the layer of the air is in equilibrium, the net upward force is balanced

That is net upward force = net downward force

Which is given as:

dp = -ρgdh

(b) Let ρo be the density of air on the surface of the earth such that pressure is proportional to the density.

$\rho \propto p$

$\frac{\text { pressure at some height }(p)}{\text { pressure at the surface of earth }\left(p_{0}\right)}=\frac{\rho}{\rho_{0}}$

Solving the above, we get pressure as:

$\left.p=p_{0} e^{(}-\frac{\rho_{0} g h}{p_{0}}\right)$

(c) We know that

$\left.p=p_{0} e^{(}-\frac{\rho_{0} g h}{p_{0}}\right)$

By substituting the values, we can calculate h = 18.43 km

(d) For the relation

$p \propto \rho$ to be valid, the temperature in the air should be uniform which is known as isothermal situation and this possible only near the surface of the earth and not at greater heights.