Distance from Sokolow Podlaski to Jacksonville
The shortest distance (air line) between Sokolow Podlaski and Jacksonville is 5,122.18 mi (8,243.35 km).
How far is Sokolow Podlaski from Jacksonville
Sokolow Podlaski is located in Siedlecki, Poland within 52° 23' 60" N 22° 15' 0" E (52.4000, 22.2500) coordinates. The local time in Sokolow Podlaski is 17:55 (18.08.2025)
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 11:55 (18.08.2025)
The calculated flying distance from Jacksonville to Jacksonville is 5,122.18 miles which is equal to 8,243.35 km.
Sokolow Podlaski, Siedlecki, Poland
Related Distances from Sokolow Podlaski
Jacksonville, Florida, United States