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      Distance from Alexander City to Chateauneuf-les-Martigues

      The shortest distance (air line) between Alexander City and Chateauneuf-les-Martigues is 4,753.68 mi (7,650.30 km).

      How far is Alexander City from Chateauneuf-les-Martigues

      Alexander City is located in Alabama, United States within 32° 55' 22.44" N -86° 3' 51.12" W (32.9229, -85.9358) coordinates. The local time in Alexander City is 17:38 (26.01.2025)

      Chateauneuf-les-Martigues is located in Bouches-du-Rhône, France within 43° 22' 59.16" N 5° 9' 51.12" E (43.3831, 5.1642) coordinates. The local time in Chateauneuf-les-Martigues is 00:38 (27.01.2025)

      The calculated flying distance from Chateauneuf-les-Martigues to Chateauneuf-les-Martigues is 4,753.68 miles which is equal to 7,650.30 km.

      Alexander City, Alabama, United States

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      Chateauneuf-les-Martigues, Bouches-du-Rhône, France

      Related Distances to Chateauneuf-les-Martigues