Barometric height distribution formula problem

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The discussion focuses on calculating the effective thickness of Earth's atmosphere using the barometric height distribution formula, assuming a mixture of 79% nitrogen and 21% oxygen at an average temperature of 10°C. The key challenge is defining thickness in terms of density and correctly integrating the mass density from sea level to infinity. The initial integration attempt led to an incorrect result of -97975m, while the expected value is approximately 9km. The correct approach involves integrating the density as a function of height and dividing by sea-level density to find the average thickness. This method emphasizes the importance of accurately representing density in the calculations.
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Homework Statement


Consider the earth’s atmosphere assuming it is a mixture containing 79% N2 and 21% O2 gas. Furthermore, assume that the atmosphere is at an average constant temperature of 10o Celsius and that the acceleration due to gravity is g = 9.81 ms-2.

Using the barometric height distribution formula integrate over the atmosphere’s mass density
(from sea-level [height ‘0’] to very great heights [‘infinity’]) and thereby determine the earth’s atmosphere’s effective thickness [in terms of the density at sea-level].

Homework Equations


pV=NkBT


The Attempt at a Solution


how can a thickness be defined in terms of a density? and does mass density just mean density? I've integrated the barometric height formula between 0 and infinity and got to:

lnp(h) = -mg/kbT . h

h = -ln(p) kbT/mg = -97975m

which can't be right having researched it its around 9km
 
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show your working

the integral will probably be over density as a function of height
I = \int_0^{infty} dh. \rho(h)

each part of the integral effectively adding up density times height, to get the "average" thickness based on sea level density, just divide by sea level density

h_{average} = \frac{\int_0^{infty} dh. \rho(h)}{\rho_{sea level}}
 

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