Distance from Essex to Aleksandrow Kujawski
The shortest distance (air line) between Essex and Aleksandrow Kujawski is 4,311.79 mi (6,939.16 km).
How far is Essex from Aleksandrow Kujawski
Essex is located in Maryland, United States within 39° 18' 7.56" N -77° 33' 18.36" W (39.3021, -76.4449) coordinates. The local time in Essex is 21:14 (12.03.2025)
Aleksandrow Kujawski is located in Włocławski, Poland within 52° 52' 36.12" N 18° 41' 36.96" E (52.8767, 18.6936) coordinates. The local time in Aleksandrow Kujawski is 03:14 (13.03.2025)
The calculated flying distance from Aleksandrow Kujawski to Aleksandrow Kujawski is 4,311.79 miles which is equal to 6,939.16 km.
Essex, Maryland, United States
Related Distances from Essex
Aleksandrow Kujawski, Włocławski, Poland