Distance from Peekskill to Sentrupert
The shortest distance (air line) between Peekskill and Sentrupert is 4,220.75 mi (6,792.63 km).
How far is Peekskill from Sentrupert
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 08:37 (22.05.2025)
Sentrupert is located in Jugovzhodna Slovenija, Slovenia within 45° 58' 36.84" N 15° 5' 22.92" E (45.9769, 15.0897) coordinates. The local time in Sentrupert is 14:37 (22.05.2025)
The calculated flying distance from Sentrupert to Sentrupert is 4,220.75 miles which is equal to 6,792.63 km.
Peekskill, New York, United States
Related Distances from Peekskill
Sentrupert, Jugovzhodna Slovenija, Slovenia