Distance from Petersburg to Schaesberg
The shortest distance (air line) between Petersburg and Schaesberg is 4,021.45 mi (6,471.89 km).
How far is Petersburg from Schaesberg
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 08:17 (01.03.2025)
Schaesberg is located in Zuid-Limburg, Netherlands within 50° 53' 60" N 6° 1' 0.12" E (50.9000, 6.0167) coordinates. The local time in Schaesberg is 14:17 (01.03.2025)
The calculated flying distance from Schaesberg to Schaesberg is 4,021.45 miles which is equal to 6,471.89 km.
Petersburg, Virginia, United States
Related Distances from Petersburg
Schaesberg, Zuid-Limburg, Netherlands