## Modelling Atmospheric Pressure at Sea Level

We can model the Earth's atmosphere as decreasing exponentially with increasing height:
$\rho (h)=Ae^{-k h}$
.
At sea level
$h=0, \; \rho (0)=1.29 kg/m^3$
and at
$h=10,000m, \; \rho (10000)=0.43 kg/m^3$
.
Using the first of these,
$1.29=A \times e^{-k(0)}=A \rightarrow A=1.29$

Using the second,
$0.43=1.29 \times e^{-k(10000)}=A \rightarrow k=- \frac{1}{10000} ln(\frac{0.43}{1.29}=1.0986 \times 10^{-4}$

Hence
$\rho (h)=1.29 e^{-1.0986 \times 10^{-4} h}$
.
Above every square metre of the Earth's surface there is
\begin{aligned} M &= \int^{\infty}_0 1.29 e^{-1.0986 \times 10^{-4} h} dh \\ &= 1.29[- \frac{1}{1.0986 \times 10^{-4}} e^{- 1.0986 \times 10^{-4} h} ]^{\infty}_0 \\ &= 1.29(0-- \frac{1}{1.0986 \times 10^{-4}}) \\ &= 1,174 \times 10^4 kg \end{aligned}

The pressure at the Earth's surface according to this model is then
$p=\frac{Mg}{1}=1.15 \times 10^{5} N/m^2$
.