The Fine-Tuning of Earth's Atmosphere

December 13, 2017

Earth is truly a magnificent place. Humans are likewise a magnificient species of hominid. The fact that we are so diverse, and yet so unified; so strong, and yet so fragile. The way the universe operates and the way the laws of nature are structured, when one ponders on these things, it should make you pause on the edge of anxiety that you are alive to experience life the way it is. Here's an example from the science of thermodynamics.

 

The density of atmospheric air decreases with increasing altitude. At sea level we are ~6,377 km from the center of the Earth, and thus the air density is 1.225 kg/m^3. However, if we travel to the top of Mount Everest, we would be ~6,387 km from the center of the Earth, and thus the air density would be 0.4135 kg/m^3. Now here's the fascinatingly creepy part. Using this data, how much does Earth's atmosphere weigh? Indulge with me into some math. Volume of Earth (assuming a perfect sphere):

 

dV = 4π(r+z)^2 dz

 

Where r = radius and z = atmosphere thickness. Therefore, m = ∫ρdV where we integrate to a thickness of ~25 km, where m = mass of atmosphere and ρ = density of air. Therefore, the density polynomial from:

 

z = 0 km | ρ = 1.225 kg/m^3

z = 25 km | ρ = 0.04008 kg/m^3

 

Becomes:

 

ρ(z) = (1.20252 - 0.10167z + 0.00223747z^2) kg/m^3

 

By substituting ρ(z) and dV into the integral mass equation along with all the parameters, and solving, we get the mass of the atmosphere: 5.09234 x 10^18 kg. That means the total mean mass of the atmosphere is 5 million trillion kg. Therefore, every square meter is ~10,000 kg (hence 101 kPa at sea level). Next time you breathe, be thankful that the air pressure is the way it is.

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