Distance from Sokolow Podlaski to Asheville
The shortest distance (air line) between Sokolow Podlaski and Asheville is 4,869.96 mi (7,837.44 km).
How far is Sokolow Podlaski from Asheville
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 21:24 (11.08.2025)
Asheville is located in North Carolina, United States within 35° 34' 14.52" N -83° 26' 46.68" W (35.5707, -82.5537) coordinates. The local time in Asheville is 15:24 (11.08.2025)
The calculated flying distance from Asheville to Asheville is 4,869.96 miles which is equal to 7,837.44 km.
Sokolow Podlaski, Siedlecki, Poland
Related Distances from Sokolow Podlaski
Asheville, North Carolina, United States