Distance between Lubusz and Evansville

The distance between Lubusz (Poland) and Evansville (Indiana, USA) is ca. 10.980,8 km (air line).

Lubusz

Koordinaten: 2°11′43″N 15°20′51″E
Latidute: 2.19527778
Longitude: 15.3475

Evansville

Latidute: 37.977223
Longitude: -87.550551

Distance between Lubusz and other cities and places worldwide

Distance between Lubusz and other major cities