Okay, so in another life I was a meteorology major. Had to quit two years in because I love math, but it’s an unrequited love. I failed calc II twice.
But I do remember some things, and want to use what I remember to see what a typical sounding of Yih’s atmosphere looks like. This is an important question that the research monks have to solve. Before they can get to space they have to get in the air.
Yih has the following relevant stats (relative to Earth)
mass: 0.90
radius: 1.01
surface gravity: 0.88
Or in absolute terms:
Mass: 5.37×10^24 kilograms Radius: 6.442×10^6 meters Area: 5.215×10^14 square meters Gravity: 8.63 meters per second squared
Finding the surface pressure is relatively simple. Let’s start by assuming the atmosphere of Yih has the same total mass as Earth’s atmosphere, 5.1441×10^18 kilograms (side note, WolframAlpha is your friend when doing stuff like this). Since pressure is force divided by area, and since force is mass times acceleration (of gravity in this case), we can get the surface pressure by multiplying the mass of the atmosphere by surface gravity and dividing by total surface area of the planet. So we get the following pressure:
85100 pascals (0.8399 atmospheres)
That’s about the same as the station pressure in Denver.
Now that we have a surface pressure, we can try to find out how it changes with height. For reference the pressure at a typical jet cruising altitude is 200 hectopascals.
Per Wikipedia, the change in pressure with height, geopotential height, is the negative of density times gravity.
dP/dh = -rho*g
And the ideal gas law is
P = rhoRT, where R is a constant specific to dry air.
We can rearrange the ideal gas law to solve for density, and substitute it for density in the equation for geopotential height to arrive at
dP/dh = -Pg/(RT)
And that’s kinda where I hit a wall. I’m pretty sure this is a differential equation, and here be dragons as far as my brain is concerned.
I’m not into meteorology, so I’ll just assume the previous steps are correct and jump to the differential equation.
This is a 1st order DE, meaning that it only contains the first derivative of the function, which is the easiest case. If I’m doing it correctly, the solution is in the form P = Ae^(-hg/(R*T)), where e is Euler’s number, and A is a constant that you have to determine from something other than the equation (because any constant would work as far as the math is concerned). In this case, you should can set A equal to the pressure at the surface, so that the solution gives you that same value for h=0, and an exponentially decreasing pressure as the height increases.
Interesting. If I plug all that into Desmos and twiddle with the height and temperature, it says we hit 200 hPa at about 426 meters above sea level. This assumes the air column is A) totally dry and B) the same temperature throughout. This is not the case, so I may have to go with my other option, taking a standard atmosphere and shifting it down so that it’s 85100 Pa at the surface. I don’t know how accurate that is, though.
