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Imagine, if you will, a fantasy world so fantastical in regards to physics that one of its telltale signs of disregarding the laws of physics is how atmospheric density is tied to latitude. At zero degrees (shortened and popularized into the "equator"), the atmosphere is twice as dense as Earth's atmosphere, which would make the already hot, wet and humid tropics even hotter, wetter and more humid than on Earth, as thicker atmospheres hold more heat and therefore more moisture. But at 90 degrees on both hemispheres, the density has dropped to 0.24 (or 24% as dense as Earth's), which might make the poles even colder than they already are on Earth. (Though, in order to clarify for the sole purpose of deciding the answer, the ozone layer is one inch thick, eight times thicker than Earth's.)

This scenario is related to my previous question, but this next question is much simpler: On a planet where the reduction in atmosphere detailed above is linearly tied to latitude, which parallel (the term used for a single latitudinal line) would one find an atmospheric density of 1.0 Earths?

For further clarification, this question does not concern climate or wind currents or weather--just atmospheric density.

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2 Answers 2

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You say there is a linear dependency between latitude and atmosphere density, and you give us 2 points on the line. This turns out to be a trivial high school geometry problem.

The slope of the line is $m=\frac{\Delta y}{\Delta x}=\frac{0.24-2}{90-0}=-0.01956$ which gives the following equation: $P(lat)=-0.01956\cdot lat + 2$.

Solving for $P=1$ gives $lat = \frac{1-2}{0.01956}=51.13636$

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In an ideal scenario you're going to lose 1.76/90=~0.02atm per degree of latitude as you head away from the equator. That puts the idealised 1atm line at ~51.14° north and south. Obviously there will be some local variation dependent on topography, weather systems, etc...

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