Distance from Gainesville to Bad Sassendorf
The shortest distance (air line) between Gainesville and Bad Sassendorf is 4,661.87 mi (7,502.55 km).
How far is Gainesville from Bad Sassendorf
Gainesville is located in Florida, United States within 29° 40' 49.44" N -83° 39' 14.76" W (29.6804, -82.3459) coordinates. The local time in Gainesville is 10:39 (13.08.2025)
Bad Sassendorf is located in Soest, Germany within 51° 34' 59.16" N 8° 10' 0.12" E (51.5831, 8.1667) coordinates. The local time in Bad Sassendorf is 16:39 (13.08.2025)
The calculated flying distance from Bad Sassendorf to Bad Sassendorf is 4,661.87 miles which is equal to 7,502.55 km.
Gainesville, Florida, United States
Related Distances from Gainesville
Bad Sassendorf, Soest, Germany