Distance from Aleksandrow Kujawski to Doctor Phillips
The shortest distance (air line) between Aleksandrow Kujawski and Doctor Phillips is 5,070.25 mi (8,159.78 km).
How far is Aleksandrow Kujawski from Doctor Phillips
Aleksandrow Kujawski is located in Włocławski, Poland within 52° 52' 36.12" N 18° 41' 36.96" E (52.8767, 18.6936) coordinates. The local time in Aleksandrow Kujawski is 05:38 (10.12.2025)
Doctor Phillips is located in Florida, United States within 28° 26' 50.64" N -82° 30' 30.96" W (28.4474, -81.4914) coordinates. The local time in Doctor Phillips is 23:38 (09.12.2025)
The calculated flying distance from Doctor Phillips to Doctor Phillips is 5,070.25 miles which is equal to 8,159.78 km.
Aleksandrow Kujawski, Włocławski, Poland
Related Distances from Aleksandrow Kujawski
Doctor Phillips, Florida, United States