Distance from Secaucus to Janow Lubelski
The shortest distance (air line) between Secaucus and Janow Lubelski is 4,364.20 mi (7,023.50 km).
How far is Secaucus from Janow Lubelski
Secaucus is located in New Jersey, United States within 40° 46' 51.6" N -75° 56' 2.76" W (40.7810, -74.0659) coordinates. The local time in Secaucus is 18:11 (12.11.2025)
Janow Lubelski is located in Puławski, Poland within 50° 43' 0.12" N 22° 25' 0.12" E (50.7167, 22.4167) coordinates. The local time in Janow Lubelski is 00:11 (13.11.2025)
The calculated flying distance from Janow Lubelski to Janow Lubelski is 4,364.20 miles which is equal to 7,023.50 km.
Secaucus, New Jersey, United States
Related Distances from Secaucus
Janow Lubelski, Puławski, Poland