How high would the atmosphere extend if it were of uniform density throughout, equal to half the present density at sea level?
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This is Giancoli Answers with Mr. Dychko. Atmospheric pressure of this column of air is gonna equal the air's density times g times Δh. We can use this formula only when you have a fluid of uniform density and we are told to let's assume that air fits that description which really isn't true because air is highly compressible so its density is changing as you go lower and lower but we are told to approximate that it's true by saying uniform density is equal to half the density at the sea level down here. So how high would this column of air be? Now there's no atmospheric pressure applied to the top of it, I mean this is the atmosphere so out here there's just outer space. So the total pressure at the bottom then is just this and we can solve this for Δh by multiplying both sides by 2 and dividing by g times density of air at sea level and we have Δh is 2 times 1.103 times 10 to the 5 newtons per square meter divided by 1.29 kilograms per cubic meter times 9.8 newtons per kilogram which gives about 16 kilometers or 1.60 times to the 4 meters.
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