Distance from Yakutsk to San Benito
Distance from Yakutsk to San Benito is 9,363 kilometers. This air travel distance is equal to 5,818 miles.
The air travel (bird fly) shortest distance between Yakutsk and San Benito is 9,363 km= 5,818 miles.
If you travel with an airplane (which has average speed of 560 miles) from Yakutsk to San Benito, It takes 10.39 hours to arrive.
Yakutsk
Yakutsk is located in Russia.
| GPS Coordinates (DMS) | 62° 2´ 2.0040'' N 129° 43´ 59.0160'' E |
|---|---|
| Latitude | 62.03389 |
| Longitude | 129.73306 |
| Altitude | 126 m |
| Country | Russia |
Yakutsk Distances to Cities
| Yakutsk | Distance |
|---|---|
| Distance from Yakutsk to Alliance | 8,261 km |
| Distance from Yakutsk to Torrington | 8,312 km |
| Distance from Yakutsk to Faribault | 7,607 km |
| Distance from Yakutsk to Tempe | 8,082 km |
| Distance from Yakutsk to San Benito | 9,363 km |
San Benito
San Benito is located in United States.
| GPS Coordinates | 26° 7´ 57.2880'' N 97° 37´ 51.9600'' W |
|---|---|
| Latitude | 26.13258 |
| Longitude | -97.63110 |
| Altitude | 16 m |
| Country | United States |