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      Distance from Lexington Park to San Gregorio di Catania

      The shortest distance (air line) between Lexington Park and San Gregorio di Catania is 4,758.22 mi (7,657.62 km).

      How far is Lexington Park from San Gregorio di Catania

      Lexington Park is located in Maryland, United States within 38° 15' 10.08" N -77° 33' 27.36" W (38.2528, -76.4424) coordinates. The local time in Lexington Park is 15:17 (05.03.2025)

      San Gregorio di Catania is located in Catania, Italy within 37° 34' 0.12" N 15° 7' 0.12" E (37.5667, 15.1167) coordinates. The local time in San Gregorio di Catania is 21:17 (05.03.2025)

      The calculated flying distance from San Gregorio di Catania to San Gregorio di Catania is 4,758.22 miles which is equal to 7,657.62 km.

      Lexington Park, Maryland, United States

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      San Gregorio di Catania, Catania, Italy

      Related Distances to San Gregorio di Catania