Distance from Rogersville to Banovo Polje
The distance between Rogersville, United States and Banovo Polje, Serbia is 8,092 km (5,028 miles) in a straight line. If a road route exists, expect roughly 10115 km (6285 miles) by car — around 126 h 26 min of driving.
| Measure | Value |
|---|---|
| Straight-line distance | 8,092 km / 5,028 mi |
| Road distance (estimate) | ≈ 10115 km / 6285 mi |
| Driving time (estimate) | ≈ 126 h 26 min |
| Flight time (estimate) | ≈ 10 h 1 min |
Road distance and travel times are estimates based on the straight-line distance (road factor 1.25, average speed 80 km/h) — actual routes vary.
| Place | Latitude | Longitude |
|---|---|---|
| Rogersville, United States | 36.407893 | -83.007123 |
| Banovo Polje, Serbia | 44.9092485 | 19.4461467 |
Frequently asked questions
How far is Banovo Polje from Rogersville?
The straight-line (air) distance from Rogersville, United States to Banovo Polje, Serbia is 8,092 km (5,028 miles). By road it is roughly 10115 km (6285 miles), depending on the route.
How long does it take to drive from Rogersville to Banovo Polje?
Approximately 126 h 26 min, assuming an 80 km/h average over an estimated road distance of 10115 km. Actual time depends on the route, traffic and border crossings.
How long is the flight from Rogersville to Banovo Polje?
A direct flight covers 8,092 km in about 10 h 1 min at typical cruise speed, plus airport time. Not all city pairs have direct flights.
How do I calculate distances for a whole list of locations?
Upload a CSV or Excel spreadsheet to the batch distance calculator — it computes the distance for every origin–destination row at once. The first records are free to try.
Related distances
Distance from Banovo Polje to Rogersville (reverse route)
GPS coordinates of Rogersville GPS coordinates of Banovo Polje