Distance from Seymour to Mercato San Severino
The shortest distance (air line) between Seymour and Mercato San Severino is 5,005.14 mi (8,055.00 km).
How far is Seymour from Mercato San Severino
Seymour is located in Indiana, United States within 38° 56' 51.36" N -86° 6' 32.04" W (38.9476, -85.8911) coordinates. The local time in Seymour is 12:54 (18.03.2025)
Mercato San Severino is located in Salerno, Italy within 40° 46' 59.88" N 14° 46' 0.12" E (40.7833, 14.7667) coordinates. The local time in Mercato San Severino is 18:54 (18.03.2025)
The calculated flying distance from Mercato San Severino to Mercato San Severino is 5,005.14 miles which is equal to 8,055.00 km.
Seymour, Indiana, United States
Related Distances from Seymour
Mercato San Severino, Salerno, Italy