Derivation of exponential density function for air

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SUMMARY

The discussion focuses on deriving the exponential density function for an isothermal ideal gas in the atmosphere, expressed as ρ = ρ0 e^(-z/h), where h is the scale height of 8.5 km. The derivation begins with the hydrostatic equilibrium equation dP/dz = -gρ, linking pressure, density, and gravitational acceleration. Participants explore the relationship between pressure and density, ultimately leading to the conclusion that the density decreases exponentially with height.

PREREQUISITES
  • Understanding of hydrostatic equilibrium and the equation dP/dz = -gρ
  • Familiarity with the ideal gas law and isothermal processes
  • Knowledge of exponential functions and their properties
  • Basic concepts of atmospheric science, particularly scale height
NEXT STEPS
  • Study the derivation of the ideal gas law and its applications in atmospheric physics
  • Learn about the implications of scale height in different atmospheric conditions
  • Explore the relationship between pressure, density, and temperature in isothermal processes
  • Investigate the effects of gravitational acceleration on atmospheric density profiles
USEFUL FOR

Students of atmospheric science, physicists, and engineers interested in understanding the behavior of gases in the atmosphere and the mathematical modeling of density variations with height.

tony_cruz
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Homework Statement


If atmosphere can be treated as an isothermal ideal gas of constant mean molecular mass m, show that density drops exponentially with height - ρ= [ρ0]e^-z/h - where h is a constant

Homework Equations


ρ= [ρ0]exp^-z/h (derivation of ...)

ρ=density
ρ0=initial density at sea level
z = height
h = scale height (found in later question to be 8.5km)

The previous question was also a derivation -->
dP/dz = -gρ which I managed. May or not be a starting point to this question

(P = pressure, z = distance, g= acc due to grav, ρ = density)

The Attempt at a Solution



I've read my lecture notes about 100x but can't even begin to see where this derivation can come from. A previous derivation was the equation
dP/dz = -gρ
(P = pressure, z = distance, g= acc due to grav, ρ = density)


____
Sorry ese's, I posted this in the wrong section to start with.
 
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If one has P(z) = ρ(z) g (h-z), the since h >> z, P(z) ~ ρ(z) g (h) => dP(z) = dρ(z) gh

and using P(z) = ρg (h-z) then dP(z) = ρg -dz, but this assumes that ρ(z) is more or less constant.

one wants to end up with

dρ(z) gh = ρ(z) g -dz, which yields,

dρ(z)/ρ(z) = -dz/h

At the moment, I forget the argument for this approach. Hopefully, ones notes addresses the key parts.
 
Cheers, I got the rest myself. Thanks.
 

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