Atmospheric Gas Concentrations: Exploring the Exponential Altitude Trend

[Logarythmic]

Homework Statement

Derive (from the equation of motion of a neutral gas and an assumption of constant gravitational field) an expression showing why the concentrations of neutral molecules decrease approximately exponentially with increasing altitude, and why the concentration of atomic oxygen (O) decreases slower with altitude than the N2 density, which in turn decreases slower than the concentration of molecular oxygen (O2). State explicitly all assumptions you make.

The Attempt at a Solution

I used the hydrostatic balance
$$\frac{dp}{dh} = -\rho g$$
to get
$$\rho = \rho_0 \exp{-\frac{gh}{RT}}$$
but this is not the equation of motion stated to start with, right?
Should I somehow use
$$m\frac{d\vec{v}}{dt} = q(\vec{E} + \vec{v} \times \vec{B}) - \nabla p + F_g$$
instead?

 

[Jhero]

Hello,

Thank you for your question. The equation of motion for a neutral gas is indeed different from the hydrostatic balance equation. The hydrostatic balance equation assumes that the gas is at rest and in equilibrium, while the equation of motion takes into account the dynamics of the gas molecules.

To derive an expression for the concentration of neutral molecules at different altitudes, we can start with the equation of motion for a neutral gas:

\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{v}) = 0

where \rho is the density of the gas and \vec{v} is the velocity of the gas molecules.

Assuming a constant gravitational field, we can simplify this equation to:

\frac{\partial \rho}{\partial t} + v \frac{\partial \rho}{\partial h} = 0

where h is the altitude and v is the vertical velocity of the gas molecules.

Next, we can make the assumption that the gas molecules are moving at a constant velocity, which means that \frac{\partial \rho}{\partial t} = 0. This assumption is valid for a gas in equilibrium, where the number of molecules entering and exiting a certain volume is equal.

Using this assumption, we can rewrite the equation as:

v \frac{\partial \rho}{\partial h} = 0

Now, let's consider the concentration of a specific gas molecule, such as O2. The concentration, C, is defined as the number of molecules per unit volume:

C = \frac{\rho}{m}

where m is the mass of the molecule.

Substituting this into our equation, we get:

v \frac{\partial C}{\partial h} = 0

This means that the concentration does not change with altitude, as long as the velocity remains constant. This is the case for all neutral molecules, including O2 and N2.

However, the concentration of atomic oxygen (O) does not follow this trend. This is because atomic oxygen is produced through the photodissociation of O2 molecules at high altitudes. As the altitude increases, the concentration of O2 decreases, leading to a decrease in the production of atomic oxygen. This is why the concentration of O decreases slower with altitude compared to N2 and O2.

In summary, the concentration of neutral molecules decreases approximately exponentially with increasing altitude due to the decreasing density of the gas.