Problem At what altitude is atmospheric pressure 50%50\%50% of that at sea level? And 10%10\%10%? Solution From the barometric formula P(h)=P0 e−αhP(h) = P_0\,e^{-\alpha h}P(h)=P0e−αh with α≈1,13×10−4\alpha \approx 1{,}13\times 10^{-4}α≈1,13×10−4 m−1^{-1}−1 (at T=293T=293T=293 K). At 50%50\%50% (P/P0=0,5P/P_0 = 0{,}5P/P0=0,5): e−αh=0,5 ⟹ h=ln2α=0,69311,13×10−4≈6 135 me^{-\alpha h} = 0{,}5 \implies h = \frac{\ln 2}{\alpha} = \frac{0{,}6931}{1{,}13\times 10^{-4}} \approx \boxed{6\,135\ \text{m}}e−αh=0,5⟹h=αln2=1,13×10−40,6931≈6135 m At 10%10\%10% (P/P0=0,1P/P_0 = 0{,}1P/P0=0,1): e−αh=0,1 ⟹ h=ln10α=2,30261,13×10−4≈20 400 me^{-\alpha h} = 0{,}1 \implies h = \frac{\ln 10}{\alpha} = \frac{2{,}3026}{1{,}13\times 10^{-4}} \approx \boxed{20\,400\ \text{m}}e−αh=0,1⟹h=αln10=1,13×10−42,3026≈20400 m Links Topics: Differential equations Concepts: Decay · Barometric formula · Separable variables Functions: Exponential function Skills: Calculating · Modelling Exercise type: Modelling problem