Distance between Sevastopol and Lenins'ke
Distance between Sevastopol and Lenins'ke is 201.35 km. This distance is equal to 125.11 miles, and 108.65 nautical miles.
The distance line on map shows distance from Sevastopol to Lenins'ke between two cities.
If you travel with an airplane (which has average speed of 560 miles per hour) between Sevastopol to Lenins'ke,
It takes 0.22 hours to arrive.
| Sevastopol | 44.61665 | 33.525367 |
|---|---|---|
| Lenins'ke | 45.2521204 | 35.9208413 |
| Distance | 201.35 km | 125.11 miles |
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