Distance from Petersburg to Sendenhorst
The shortest distance (air line) between Petersburg and Sendenhorst is 4,069.87 mi (6,549.82 km).
How far is Petersburg from Sendenhorst
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 18:59 (13.02.2025)
Sendenhorst is located in Warendorf, Germany within 51° 50' 38.04" N 7° 49' 40.08" E (51.8439, 7.8278) coordinates. The local time in Sendenhorst is 00:59 (14.02.2025)
The calculated flying distance from Sendenhorst to Sendenhorst is 4,069.87 miles which is equal to 6,549.82 km.
Petersburg, Virginia, United States
Related Distances from Petersburg
Sendenhorst, Warendorf, Germany