Modeling the atmosphere is a vague statement. What we mean is, we will determine the planet's density profile and its temperature profile. These profiles are functions that track these quantities depending on the distance from the surface.
We've previously determined that the surface temperature of our atmosphere lies around 271K (-3.5°C). This was decided based on a lot of assumptions and simplifications (see the post for more about that), and may not be entirely accurate, but it is the best estimate we have and so we will use this as the surface temperature, \(T_0\). We begin by assuming that the temperature of the atmosphere is adiabatic, meaning that it changes with no heat change (no heat lost, nor gained) in the process, up til the height where the temperature is half of what it is at the surface, \(T=T_0/2\). After this, we say that the temperature is isothermal, meaning we assume it is approximately the same at every point.
Next, we approximate the atmosphere as an ideal gas, meaning that we can use the ideal gas equation, where pressure can be expressed as \(P=nkT\), where \(n\) is the number of molecules, \(k\) is the Boltzmann constant, and \(T\) is the temperature. We can write this equation as a function of the distance, \(r\), from the surface of the planet:
\(P(r)=n(r)kT(r)\)
If we now write \(n(r)=\frac{\rho(r)}{\mu m_H}\), where \(\rho(r)\) is the density as a function of distance (we can write the density as a function of the distance when we assume that the atmosphere is spherically symmetric), \(\mu\) is the average molecular weight, and \(m_H\) is the mass of a hydrogen atom, we can again rewrite the ideal gas equation as:
\(P(r)=\frac{\rho(r)kT(r)}{\mu m_H}\)
This is the first of two expressions we need!
In order to find the second expression we will assume that the atmosphere is in something called hydrostatic equilibrium. Exactly what this means is still disputed, but we will have it mean that the pressure outwards is the same as the force of gravity inwards, so that the gas does not undergo much change.
We can find the equation for hydrostatic equilibrium from Newtons second law:
\(-F_g-F_P(r+dr)+F_P(r)=dm\frac{d^2r}{dt^2}\)
where \(F_g\) is the force of gravity, \(F_P\) is the force from the pressure, and \(r\) is the distance.
Study the picture to the left and see if you understand how this formula is set up. We will not explain exactly how to go from here to the formula for hydrostatic equilibrium, but you're encouraged to either try yourself, or sign up for the class for the full explanation. Or just search for it, it's not secret. Anyway, through some math magic, we end up with
\(\frac{dP}{dr}=-\rho(r)g(r)\)
where we see that the density, \(\rho(r)\), multiplied by the gravitational acceleration, \(g(r)\), both as a function of distance, is equal to the derivative of pressure, P, with regard to the distance.
We now have two expressions:
\(P(r)=\frac{\rho(r)kT(r)}{\mu m_H}\) and \(\frac{dP}{dr}=-\rho(r)g(r)\)
We already know that \(T=T_0\) when the atmosphere is isothermal, and while it is adiabatic (up the point where \(T=T_0/2\)) we can describe it as \(T=CP^{\frac{\gamma-1}{1}}\), where \(\gamma=1.4\). This leaves us with two unknown variables, \(P(r)\) and \(\rho(r)\).
We will find an expression for the pressure of a height, \(r\), by utelizing the Euler-Chromer method. We have used this, and other integration methods, previously. The Euler-Chromer method is a good choice here because of it's simplicity.
Our initial conditions are given as \(T_0=271\) K, \(\rho_0 = 2.209\) kg/m3 (Earth's is \(1.2\) kg/m3). We can then find the furface pressure by using the ideal gas equation \(P_0=\frac{k}{\mu m_H}\rho_0T_0\) kPa. This gives \(C=P_0^{\gamma-1}T_0^\gamma\).
This is the algorithm, where \(i=1,2,3,....N\) . We run the simulation \(N\) times, until we have a good picture of our atmosphere.
\(G(i) = \frac{MG}{(r+i)^2}\) The algorithm finds the gravitational acceleration.
\(\frac{dP}{dr}=-\rho(i)g(i)\) Finds the derivative of the surface pressure.
\(P(i+dr)=P(i)+\frac{dP}{dr}\) Finds the pressure at a height dr.
\(T(i+dr)=CP(i+dr)^{\frac{\gamma-1}{\gamma}}\) Inserts this pressure into adiabatic expression for T.
\(\rho(i+dr)=\frac{k}{\mu m_H}P(i+dr)T(i+dr)\) Finds the density from the ideal gas equation.
This is valid for as long as \(T>T_0/2\), which is our condition for an adiabatic atmosphere. We assume that the mass is uniform, as the mass of the atmosphere is negligible compared to the mass of the planet. Once the atmosphere turns isothermal, the temperature is constant and we set \(T=T_{limit}\), where \(T_{limit}\) is the temperature at \(T=T_0/2\), and keep this value constant. We give new start values for \(P=P_{limit}\) and \(\rho=\rho_{limit}\), and run the same algorithm as above.
We want a model of our atmosphere, and so we can stop when the atmosphere stops. But where exactly is this? Most experts say that space starts at the point where orbital dynamic forces become more important than aerodynamic forces, others say where the atmosphere alone is not enough to support a flying vessel at suborbital speeds. We decide to model until the atmosphere is 1/100 of the surface pressure.
We have the height above the surface along the x-axis, and how the temperature, density, and pressure behaves as we move upwards into the atmosphere along the y-axis. The y-axis only goes from 0 to 1, which is because we have normalized the values. We do this by dividing all of the values by the initial value, resulting in a plot that shows how the values change in relation to the initial state, rather than the actual values. The black, dotted line show the transition between adiabatic and isothermal state of the atmosphere. As expected, the temperature is constant after this point, which in turn affects the density and pressure. Most notably the density, where we can see that the line becomes a bit wonky after we enter the isothermal state.
With this model of the atmosphere, we can finally land! But first, we decide on a proper place to land. Stay put!