An aircraft of gross weight $W = 80{,}000\text{ N}$ and wing planform area $S = 25\text{ m}^2$ has a parabolic drag polar $C_D = 0.020 + 0.045 C_L^2$. In steady, level flight in standard sea-level air ($\rho = 1.225\text{ kg/m}^3$), the airspeed $V_{md}$ (in $\text{m/s}$) for **minimum drag (minimum thrust required)** is closest to:

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### 1. Strategy & Theoretical Principles For a parabolic drag polar $C_D = C_{D,0} + K C_L^2$: Minimum thrust required occurs at maximum lift-to-drag ratio $(L/D)_{\max}$, where parasite drag equals induced drag ($C_{D,0} = K C_L^2$): $$C_{L,md} = \sqrt{\frac{C_{D,0}}{K}}$$ ### 2. Analytical Formulation Given $C_{D,0} = 0.020$ and $K = 0.045$: $$C_{L,md} = \sqrt{\frac{0.020}{0.045}} = \sqrt{\frac{4}{9}} = \frac{2}{3} \approx 0.6667$$ ### 3. Step-by-Step Calculation for Airspeed In steady level flight, $L = W = \frac{1}{2} \rho V^2 S C_L$: $$V_{md} = \sqrt{\frac{2 W}{\rho S C_{L,md}}} = \sqrt{\frac{2 \times 80{,}000}{1.225 \times 25 \times (2/3)}}$$ Denominator: $$1.225 \times 25 \times \frac{2}{3} = 30.625 \times \frac{2}{3} = 20.41667\text{ kg/m}$$ Numerator: $$2 \times 80{,}000 = 160{,}000\text{ N}$$ Evaluating $V_{md}$: $$V_{md}^2 = \frac{160{,}000}{20.41667} \approx 7{,}836.73\text{ m}^2/\text{s}^2$$ $$V_{md} = \sqrt{7{,}836.73} \approx 88.53\text{ to }89.2\text{ m/s}$$ With exact fraction $160000 / (61.25 / 3) = 480000 / 61.25 = 7836.734 \implies 88.53\text{ m/s}$. Let's set correctAnswer to $88.5$ (tolerance $86.5-90.5$). ### 4. Distractor Forensics - Candidate evaluating minimum power required speed gets $V_{mp} = V_{md} / (3)^{1/4} = 88.53 / 1.316 = 67.3\text{ m/s}$. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Distinguish speed for minimum thrust required ($C_L = \sqrt{C_{D0}/K}$) from speed for minimum power required ($C_L = \sqrt{3 C_{D0}/K}$). - **Standard Reference:** Anderson, J.D., *Introduction to Flight*, Chapter 6.

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