1. **Problem Statement:** Calculate the approximate geodetic distance between two points G1 and G2 given in geodetic coordinates (latitude, longitude, height) on an ellipsoid. 2. **Given Data:** - G1: $20^\circ 12' 32''$ N, $38^\circ 6' 26''$ E, height = 102 m - G2: $23^\circ 49' 54''$ N, $39^\circ 54' 7''$ E, height = 83 m - Ellipsoid parameters: - Semi-major axis $a = 6377932.253$ m - First eccentricity squared $e^2 = 0.00668467$ 3. **Step 1: Convert coordinates from degrees, minutes, seconds to decimal degrees.** - For latitude and longitude: $$\text{Decimal degrees} = \text{degrees} + \frac{\text{minutes}}{60} + \frac{\text{seconds}}{3600}$$ - G1 latitude: $20 + \frac{12}{60} + \frac{32}{3600} = 20 + 0.2 + 0.008889 = 20.208889^\circ$ - G1 longitude: $38 + \frac{6}{60} + \frac{26}{3600} = 38 + 0.1 + 0.007222 = 38.107222^\circ$ - G2 latitude: $23 + \frac{49}{60} + \frac{54}{3600} = 23 + 0.816667 + 0.015 = 23.831667^\circ$ - G2 longitude: $39 + \frac{54}{60} + \frac{7}{3600} = 39 + 0.9 + 0.001944 = 39.901944^\circ$ 4. **Step 2: Convert decimal degrees to radians** (since trigonometric functions use radians): $$\text{radians} = \text{degrees} \times \frac{\pi}{180}$$ - $\phi_1 = 20.208889 \times \frac{\pi}{180} = 0.3527$ rad - $\lambda_1 = 38.107222 \times \frac{\pi}{180} = 0.6651$ rad - $\phi_2 = 23.831667 \times \frac{\pi}{180} = 0.4159$ rad - $\lambda_2 = 39.901944 \times \frac{\pi}{180} = 0.6969$ rad 5. **Step 3: Calculate the prime vertical radius of curvature $N$ at each latitude:** $$N = \frac{a}{\sqrt{1 - e^2 \sin^2 \phi}}$$ - Calculate $N_1$ at $\phi_1$: $$N_1 = \frac{6377932.253}{\sqrt{1 - 0.00668467 \sin^2(0.3527)}}$$ - Calculate $N_2$ at $\phi_2$: $$N_2 = \frac{6377932.253}{\sqrt{1 - 0.00668467 \sin^2(0.4159)}}$$ 6. **Step 4: Convert geodetic coordinates to ECEF (Earth-Centered, Earth-Fixed) Cartesian coordinates:** $$X = (N + h) \cos \phi \cos \lambda$$ $$Y = (N + h) \cos \phi \sin \lambda$$ $$Z = \left(N(1 - e^2) + h\right) \sin \phi$$ where $h$ is the height above the ellipsoid. - Compute $X_1, Y_1, Z_1$ for G1 - Compute $X_2, Y_2, Z_2$ for G2 7. **Step 5: Calculate the Euclidean distance between the two points in ECEF coordinates:** $$d = \sqrt{(X_2 - X_1)^2 + (Y_2 - Y_1)^2 + (Z_2 - Z_1)^2}$$ 8. **Step 6: Perform calculations:** - $\sin(0.3527) \approx 0.3451$, $\sin^2(0.3527) = 0.1191$ - $N_1 = \frac{6377932.253}{\sqrt{1 - 0.00668467 \times 0.1191}} = \frac{6377932.253}{\sqrt{1 - 0.000796}} = \frac{6377932.253}{0.999602} = 6380500.5$ m - $\sin(0.4159) \approx 0.4039$, $\sin^2(0.4159) = 0.1631$ - $N_2 = \frac{6377932.253}{\sqrt{1 - 0.00668467 \times 0.1631}} = \frac{6377932.253}{\sqrt{1 - 0.00109}} = \frac{6377932.253}{0.999455} = 6384600.7$ m - Calculate $X_1, Y_1, Z_1$: $$X_1 = (6380500.5 + 102) \cos(0.3527) \cos(0.6651) = 6380602.5 \times 0.9381 \times 0.7873 = 4719273.3$$ $$Y_1 = 6380602.5 \times 0.9381 \times \sin(0.6651) = 6380602.5 \times 0.9381 \times 0.6166 = 3689273.7$$ $$Z_1 = \left(6380500.5 \times (1 - 0.00668467) + 102\right) \sin(0.3527) = (6380500.5 \times 0.9933153 + 102) \times 0.3451 = 6338250.3 \times 0.3451 = 2186757.3$$ - Calculate $X_2, Y_2, Z_2$: $$X_2 = (6384600.7 + 83) \cos(0.4159) \cos(0.6969) = 6384683.7 \times 0.9143 \times 0.7663 = 4479277.1$$ $$Y_2 = 6384683.7 \times 0.9143 \times \sin(0.6969) = 6384683.7 \times 0.9143 \times 0.6425 = 3759277.5$$ $$Z_2 = \left(6384600.7 \times 0.9933153 + 83\right) \sin(0.4159) = (6342300.1 + 83) \times 0.4039 = 6342383.1 \times 0.4039 = 2560757.4$$ - Calculate distance $d$: $$d = \sqrt{(4479277.1 - 4719273.3)^2 + (3759277.5 - 3689273.7)^2 + (2560757.4 - 2186757.3)^2}$$ $$= \sqrt{(-240996.2)^2 + (70003.8)^2 + (374000.1)^2}$$ $$= \sqrt{5.81 \times 10^{10} + 4.9 \times 10^{9} + 1.40 \times 10^{11}}$$ $$= \sqrt{2.29 \times 10^{11}} = 478591.5 \text{ m}$$ 9. **Step 7: Final answer:** The approximate geodetic distance between G1 and G2 is **478591.500 meters** or **478.592 kilometers** (rounded to 3 decimal places).