Barometric pressure obeys ln p/(p_0) = ch, where p_0 is
Barometric pressure obeys $\ln\dfrac{p}{p_{0}} = ch$, where $p_{0}$ is the pressure at sea level and $h$ the altitude. At $h = 5540\ \text{m}$ the pressure is half the normal value. Find the rate of change of pressure with altitude.
Explanation of Method
An exponential law, whose derivative is proportional to itself.
Step 1: Determine the constant
Putting $p = \tfrac12p_{0}$ at $h = 5540$: \[\ln\frac12 = 5540\,c \;\Longrightarrow\; c = -\frac{\ln2}{5540} \approx -1.2512\cdot10^{-4}\ \text{m}^{-1}.\]
The constant is negative, as it must be for a pressure that falls with height.
Step 2: Solve for $p$ and differentiate
\[ p(h) = p_{0}e^{ch} \;\Longrightarrow\; \frac{dp}{dh} = cp_{0}e^{ch} = cp. \]
The rate of change is proportional to the pressure itself — the defining property of exponential decay.
Step 3: Verification
\[\frac{1}{|c|} = \frac{5540}{\ln2} = \frac{5540}{0.693147} \approx 7992.5\ \text{m} \;\;\checkmark\]
\[p(5540) = p_{0}e^{-\ln2} = \frac{p_{0}}{2} \;\;\checkmark; \qquad p(11080) = \frac{p_{0}}{4} \;\;\checkmark\]
\[\text{At sea level, } p_{0} = 101{,}325\ \text{Pa}:\ \frac{dp}{dh} \approx -12.7\ \frac{\text{Pa}}{\text{m}} \approx -0.095\ \frac{\text{mmHg}}{\text{m}} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\frac{dp}{dh} = -\frac{\ln2}{5540}\,p \approx -1.2512\cdot10^{-4}\,p \approx -\frac{p}{7993}\ \text{per metre}\,}\]
Remark
Remark. The quantity $\dfrac{1}{|c|} \approx 7993\ \text{m}$ is the scale height: the rise over which the pressure falls by a factor $e$. The half-value height $5540\ \text{m}$ is $\ln2$ times it. The familiar rule of thumb that pressure drops about $1\ \text{mmHg}$ per $11\ \text{m}$ near the ground is the same statement, since $0.095\times11 \approx 1$.