Distance from Lexington Park to San Severino Marche
The shortest distance (air line) between Lexington Park and San Severino Marche is 4,468.17 mi (7,190.82 km).
How far is Lexington Park from San Severino Marche
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 14:53 (05.03.2025)
San Severino Marche is located in Macerata, Italy within 43° 13' 44.04" N 13° 10' 37.56" E (43.2289, 13.1771) coordinates. The local time in San Severino Marche is 20:53 (05.03.2025)
The calculated flying distance from San Severino Marche to San Severino Marche is 4,468.17 miles which is equal to 7,190.82 km.
Lexington Park, Maryland, United States
Related Distances from Lexington Park
San Severino Marche, Macerata, Italy