Distance from Sterling to Aleksandrow Kujawski
The shortest distance (air line) between Sterling and Aleksandrow Kujawski is 4,361.67 mi (7,019.42 km).
How far is Sterling from Aleksandrow Kujawski
Sterling is located in Virginia, United States within 39° 0' 18.72" N -78° 35' 42" W (39.0052, -77.4050) coordinates. The local time in Sterling is 07:44 (17.11.2025)
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 13:44 (17.11.2025)
The calculated flying distance from Aleksandrow Kujawski to Aleksandrow Kujawski is 4,361.67 miles which is equal to 7,019.42 km.
Sterling, Virginia, United States
Related Distances from Sterling
Aleksandrow Kujawski, Włocławski, Poland