Distance from Lexington Park to Sint-Joost-ten-Node
The shortest distance (air line) between Lexington Park and Sint-Joost-ten-Node is 3,869.27 mi (6,226.99 km).
How far is Lexington Park from Sint-Joost-ten-Node
Lexington Park is located in Maryland, United States within 38° 15' 10.08" N -77° 33' 27.36" W (38.2528, -76.4424) coordinates. The local time in Lexington Park is 00:55 (24.01.2025)
Sint-Joost-ten-Node is located in Arr. de Bruxelles-Capitale/Arr. Brussel-Hoofdstad, Belgium within 50° 51' 0" N 4° 22' 59.88" E (50.8500, 4.3833) coordinates. The local time in Sint-Joost-ten-Node is 06:55 (24.01.2025)
The calculated flying distance from Sint-Joost-ten-Node to Sint-Joost-ten-Node is 3,869.27 miles which is equal to 6,226.99 km.
Lexington Park, Maryland, United States
Related Distances from Lexington Park
Sint-Joost-ten-Node, Arr. de Bruxelles-Capitale/Arr. Brussel-Hoofdstad, Belgium