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
\[\begin{equation} \begin{aligned} M &= \int^{\infty}_0 1.29 e^{-1.0986 \times 10^{-4} h} dh \\ &= 1.29[- \frac{1}{1.0986 \times 10^{-4}} e^{- 1.0986 \times 10^{-4} h} ]^{\infty}_0 \\ &= 1.29(0-- \frac{1}{1.0986 \times 10^{-4}}) \\ &= 1,174 \times 10^4 kg \end{aligned} \end{equation}\]

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

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