Distance from Akhtubinsk to Jeffersonville
Distance from Akhtubinsk to Jeffersonville is 9,307 kilometers. This air travel distance is equal to 5,783 miles.
The air travel (bird fly) shortest distance between Akhtubinsk and Jeffersonville is 9,307 km or 5,783 miles.
If you travel with an airplane (which has average speed of 560 miles) from Akhtubinsk to Jeffersonville, It takes 10.33 hours to arrive.
Akhtubinsk
Akhtubinsk is located in Russia.
| GPS Coordinates (DMS) | 48° 16´ 46.3800'' N 46° 10´ 19.8120'' E |
|---|---|
| Latitude | 48.27955 |
| Longitude | 46.17217 |
| Altitude | 3 m |
| Country | Russia |
Akhtubinsk Distances to Cities
| Akhtubinsk | Distance |
|---|---|
| Distance from Akhtubinsk to South Bend | 9,001 km |
| Distance from Akhtubinsk to Panama City | 10,085 km |
| Distance from Akhtubinsk to Jeffersonville | 9,307 km |
| Distance from Akhtubinsk to Roanoke | 9,136 km |
| Distance from Akhtubinsk to Winona | 8,979 km |
Jeffersonville
Jeffersonville is located in United States.
| GPS Coordinates | 38° 16´ 39.2520'' N 85° 44´ 13.8480'' W |
|---|---|
| Latitude | 38.27757 |
| Longitude | -85.73718 |
| Altitude | 138 m |
| Country | United States |