Distance from Jefferson Valley-Yorktown to Alexandroupoli
The shortest distance (air line) between Jefferson Valley-Yorktown and Alexandroupoli is 4,859.99 mi (7,821.40 km).
How far is Jefferson Valley-Yorktown from Alexandroupoli
Jefferson Valley-Yorktown is located in New York, United States within 41° 19' 4.8" N -74° 11' 57.12" W (41.3180, -73.8008) coordinates. The local time in Jefferson Valley-Yorktown is 15:54 (14.03.2025)
Alexandroupoli is located in Evros, Greece within 40° 51' 0" N 25° 52' 0.12" E (40.8500, 25.8667) coordinates. The local time in Alexandroupoli is 22:54 (14.03.2025)
The calculated flying distance from Alexandroupoli to Alexandroupoli is 4,859.99 miles which is equal to 7,821.40 km.
Jefferson Valley-Yorktown, New York, United States