In an isothermal atmospheric layer of constant temperature $T = 216.65\text{ K}$, the acceleration due to gravity is $g = 9.80665\text{ m/s}^2$ and gas constant for air is $R = 287.05\text{ J/(kg}\cdot\text{K)}$. The geopotential height difference $\Delta h$ (in $\text{meters}$) required for the ambient pressure to drop to half of its initial value ($p / p_0 = 0.50$) is closest to:

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### 1. Strategy & Theoretical Principles In an isothermal atmospheric layer, the hydrostatic equation $dp = -\rho g dh$ combined with the ideal gas law $p = \rho R T$ yields: $$\frac{dp}{p} = -\frac{g}{R T} dh$$ Integrating between limits gives the barometric formula: $$\frac{p}{p_0} = \exp\left(-\frac{g \Delta h}{R T}\right)$$ ### 2. Analytical Formulation & Calculation Taking the natural logarithm: $$\ln\left(\frac{p}{p_0}\right) = -\frac{g \Delta h}{R T} \implies \Delta h = \frac{R T}{g} \ln\left(\frac{p_0}{p}\right)$$ Given: - $T = 216.65\text{ K}$ - $R = 287.05\text{ J/(kg}\cdot\text{K)}$ - $g = 9.80665\text{ m/s}^2$ - $p_0 / p = 2.0$ 1. Scale height of the isothermal layer: $$H = \frac{R T}{g} = \frac{287.05 \times 216.65}{9.80665} = \frac{62{,}189.38}{9.80665} \approx 6{,}341.55\text{ m}$$ 2. Height difference for half-pressure: $$\Delta h = H \ln(2) = 6341.55 \times 0.693147 \approx 4{,}395.6\text{ m} \approx 4{,}392\text{ to }4{,}396\text{ m}$$ ### 3. Distractor Forensics - Candidate reporting scale height $H = 6,342\text{ m}$ without multiplying by $\ln(2)$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Distinguish pressure scale height ($e$-folding height $H$) from half-pressure height ($H \ln 2$). - **Standard Reference:** Anderson, J.D., *Introduction to Flight*, Chapter 3.

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