Distance from Peekskill to Drobeta-Turnu Severin
The shortest distance (air line) between Peekskill and Drobeta-Turnu Severin is 4,579.00 mi (7,369.19 km).
How far is Peekskill from Drobeta-Turnu Severin
Peekskill is located in New York, United States within 41° 17' 18.24" N -74° 4' 38.28" W (41.2884, -73.9227) coordinates. The local time in Peekskill is 02:27 (20.05.2025)
Drobeta-Turnu Severin is located in Mehedinţi, Romania within 44° 37' 59.88" N 22° 38' 60" E (44.6333, 22.6500) coordinates. The local time in Drobeta-Turnu Severin is 09:27 (20.05.2025)
The calculated flying distance from Drobeta-Turnu Severin to Drobeta-Turnu Severin is 4,579.00 miles which is equal to 7,369.19 km.
Peekskill, New York, United States
Related Distances from Peekskill
Drobeta-Turnu Severin, Mehedinţi, Romania