# Derivation of exponential density function for air

1. Nov 19, 2007

### tony_cruz

1. The problem statement, all variables and given/known data
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

2. Relevant 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)
3. 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.

2. Nov 19, 2007

### Astronuc

Staff Emeritus
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.

3. Nov 19, 2007

### tony_cruz

Cheers, I got the rest myself. Thanks.