Distance from Aleksandrovskoye to Rapid City
Distance from Aleksandrovskoye to Rapid City is 9,618 kilometers. This air travel distance is equal to 5,976 miles.
The air travel (bird fly) shortest distance between Aleksandrovskoye and Rapid City is 9,618 km= 5,976 miles.
If you travel with an airplane (which has average speed of 560 miles) from Aleksandrovskoye to Rapid City, It takes 10.67 hours to arrive.
Aleksandrovskoye
Aleksandrovskoye is located in Russia.
| GPS Coordinates (DMS) | 44° 42´ 51.0120'' N 43° 0´ 2.9880'' E |
|---|---|
| Latitude | 44.71417 |
| Longitude | 43.00083 |
| Altitude | 295 m |
| Country | Russia |
Aleksandrovskoye Distances to Cities
| Aleksandrovskoye | Distance |
|---|---|
| Distance from Aleksandrovskoye to Chesterfield | 9,668 km |
| Distance from Aleksandrovskoye to Cerritos | 11,096 km |
| Distance from Aleksandrovskoye to Bloomfield | 8,656 km |
| Distance from Aleksandrovskoye to Hanford | 10,861 km |
| Distance from Aleksandrovskoye to Rapid City | 9,618 km |
Rapid City
Rapid City is located in United States.
| GPS Coordinates | 44° 4´ 49.9440'' N 103° 13´ 51.6360'' W |
|---|---|
| Latitude | 44.08054 |
| Longitude | -103.23101 |
| Altitude | 990 m |
| Country | United States |