Distance from Sokolow Podlaski to Albuquerque
The shortest distance (air line) between Sokolow Podlaski and Albuquerque is 5,653.53 mi (9,098.48 km).
How far is Sokolow Podlaski from Albuquerque
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 16:19 (11.08.2025)
Albuquerque is located in New Mexico, United States within 35° 6' 19.44" N -107° 21' 12.6" W (35.1054, -106.6465) coordinates. The local time in Albuquerque is 08:19 (11.08.2025)
The calculated flying distance from Albuquerque to Albuquerque is 5,653.53 miles which is equal to 9,098.48 km.
Sokolow Podlaski, Siedlecki, Poland
Related Distances from Sokolow Podlaski
Albuquerque, New Mexico, United States