Distance from Petersburg to Bad Liebenwerda
The shortest distance (air line) between Petersburg and Bad Liebenwerda is 4,297.20 mi (6,915.68 km).
How far is Petersburg from Bad Liebenwerda
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 13:48 (08.02.2025)
Bad Liebenwerda is located in Elbe-Elster, Germany within 51° 31' 0.12" N 13° 24' 0" E (51.5167, 13.4000) coordinates. The local time in Bad Liebenwerda is 19:48 (08.02.2025)
The calculated flying distance from Bad Liebenwerda to Bad Liebenwerda is 4,297.20 miles which is equal to 6,915.68 km.
Petersburg, Virginia, United States
Related Distances from Petersburg
Bad Liebenwerda, Elbe-Elster, Germany