Distance from Aleksandrow Lodzki to Petersburg
The shortest distance (air line) between Aleksandrow Lodzki and Petersburg is 4,511.64 mi (7,260.79 km).
How far is Aleksandrow Lodzki from Petersburg
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 02:59 (09.12.2025)
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 20:59 (08.12.2025)
The calculated flying distance from Petersburg to Petersburg is 4,511.64 miles which is equal to 7,260.79 km.
Aleksandrow Lodzki, Łódzki, Poland
Related Distances from Aleksandrow Lodzki
Petersburg, Virginia, United States