Distance from Petersburg to Arandelovac
The shortest distance (air line) between Petersburg and Arandelovac is 4,830.19 mi (7,773.44 km).
How far is Petersburg from Arandelovac
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 19:25 (06.03.2025)
Arandelovac is located in Šumadijska oblast, Serbia within 44° 18' 15.12" N 20° 33' 21.96" E (44.3042, 20.5561) coordinates. The local time in Arandelovac is 01:25 (07.03.2025)
The calculated flying distance from Arandelovac to Arandelovac is 4,830.19 miles which is equal to 7,773.44 km.
Petersburg, Virginia, United States
Related Distances from Petersburg
Arandelovac, Šumadijska oblast, Serbia