Distance from Petersburg to Aleksandrow Lodzki
The shortest distance (air line) between Petersburg and Aleksandrow Lodzki is 4,511.64 mi (7,260.79 km).
How far is Petersburg from Aleksandrow Lodzki
Petersburg is located in Virginia, United States within 37° 12' 15.48" N -78° 36' 31.32" W (37.2043, -77.3913) coordinates. The local time in Petersburg is 05:03 (03.03.2025)
Aleksandrow Lodzki is located in Łódzki, Poland within 51° 49' 9.84" N 19° 18' 14.04" E (51.8194, 19.3039) coordinates. The local time in Aleksandrow Lodzki is 11:03 (03.03.2025)
The calculated flying distance from Aleksandrow Lodzki to Aleksandrow Lodzki is 4,511.64 miles which is equal to 7,260.79 km.
Petersburg, Virginia, United States
Related Distances from Petersburg
Aleksandrow Lodzki, Łódzki, Poland