Distance from Janow Lubelski to Oxon Hill
The shortest distance (air line) between Janow Lubelski and Oxon Hill is 4,570.41 mi (7,355.36 km).
How far is Janow Lubelski from Oxon Hill
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 21:25 (19.12.2025)
Oxon Hill is located in Maryland, United States within 38° 47' 18.24" N -77° 1' 38.28" W (38.7884, -76.9727) coordinates. The local time in Oxon Hill is 15:25 (19.12.2025)
The calculated flying distance from Oxon Hill to Oxon Hill is 4,570.41 miles which is equal to 7,355.36 km.
Janow Lubelski, Puławski, Poland
Related Distances from Janow Lubelski
Oxon Hill, Maryland, United States