Distance from Martinsville to Scherpenheuvel
The shortest distance (air line) between Martinsville and Scherpenheuvel is 4,100.96 mi (6,599.87 km).
How far is Martinsville from Scherpenheuvel
Martinsville is located in Virginia, United States within 36° 40' 57.36" N -80° 8' 11.04" W (36.6826, -79.8636) coordinates. The local time in Martinsville is 16:35 (11.02.2025)
Scherpenheuvel is located in Arr. Leuven, Belgium within 51° 0' 37.08" N 4° 58' 22.08" E (51.0103, 4.9728) coordinates. The local time in Scherpenheuvel is 22:35 (11.02.2025)
The calculated flying distance from Scherpenheuvel to Scherpenheuvel is 4,100.96 miles which is equal to 6,599.87 km.
Martinsville, Virginia, United States
Related Distances from Martinsville
Scherpenheuvel, Arr. Leuven, Belgium