Distance from Janow Lubelski to Jacksonville
The shortest distance (air line) between Janow Lubelski and Jacksonville is 5,186.02 mi (8,346.09 km).
How far is Janow Lubelski from Jacksonville
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 13:54 (14.12.2025)
Jacksonville is located in Florida, United States within 30° 19' 55.92" N -82° 19' 30.36" W (30.3322, -81.6749) coordinates. The local time in Jacksonville is 07:54 (14.12.2025)
The calculated flying distance from Jacksonville to Jacksonville is 5,186.02 miles which is equal to 8,346.09 km.
Janow Lubelski, Puławski, Poland
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Jacksonville, Florida, United States