That actually depends on latitude. The closer you are to the poles, the darker the sky to the unaided eye at 10km.
Why is that?
My best guess would be because the blue gets scattered out as it has to travel through “more” atmosphere due to the angle of light traveling to reach the poles.
The atmosphere is thicker at the equator. Not only does the rotation of Earth drive more atmosphere to the equator, but that atmosphere is warmer than the atmosphere at the poles. As a result, you’re above a greater percentage of the local atmosphere at high latitude locations, as compared to low latitude locations, at a given height above sea level.
Well that makes complete sense. Thanks for the explanation.
Just to clarify, that’s because centrifugal forces are larger at the equator than at the poles?
True, good point.
I forgot about the bulging effect near the equator.
Does anyone actually know how big the difference in atmosphere is?
I do know that the oceans bulge about 30 km, or 20 miles.
Yes, exactly.
It’s the same force that keeps water on the bottom of a bucket if you spin it fast enough.
I wrote a MATLAB sim a few weeks ago to calculate centripetal/coriolis forces as a function of position on Earth. I’m not sure how I would translate that into macroscale estimations of atmospheric densities though(or if you’d need to! Turns out I don’t.).
Here’s the equation for the density of the atmosphere as a function of height:
You’d probably want to vary g with height since we’re trying to be accurate here. I’d integrate over the height of the atmosphere up to the troposphere(since that’s the constraint on this equation) at the equator and then compare that with a seperate g at the poles. That could probably give you an idea for the difference in densities.
Edit: It says that g is actually just the surface acceleration of gravity, so you could probably just use the two different g’s. One for the pole and one for the equator. Depends how accurate you want to be I guess. I’ll work this out later when I have a bit of time to play in matlab.
Is the performance of the Helion sitting in data files somewhere? Can somebody go in and knock the thruster power down to about 1.5g? That would vaguely match the Saturn V launch profile.
No use of warp allowed.
Midway through writing the script I found the answer.
So it looks like the atmopshere would be about 13km thicker at the equator than at the poles.
Running the simulation up to 100km anyways because why not.
Almost. The force that keeps water in a revolving bucket is a centripetal force. The centripetal force that keeps the Earth’s atmosphere, regardless of latitude, is gravity. The centrifugal “force” is a byproduct of the fact that things move more quickly at greater radii in solid body rotation.
Maybe? As aa result of Earth’s rotation the atmosphere moves more quickly at the equator, and a larger centripetal force is required to keep it constrained. The atmosphere stretches at the equator, finding a new equilibrium. This, however, results in a lower atmospheric pressure at the equator, , as a fluid, the atmosphere needs to balance. It does this, of course, by increasing the amount of air at the equator.
Can this difference be also a cause of how the light is reflected/refracted according to Earth’s tilt or it has no impact on it ?