Leskov Island (Antarctica)
  Leskov Island Location in Antarctica | |
| Geography | |
|---|---|
| Location | Antarctica |
| Coordinates | 66°36′S 85°10′E / 66.600°S 85.167°ECoordinates: 66°36′S 85°10′E / 66.600°S 85.167°E |
| Area | 49.5 km2 (19.1 sq mi) |
| Length | 9 km (5.6 mi) |
| Width | 5.5 km (3.42 mi) |
| Highest elevation | 185 m (607 ft) |
| Administration | |
| Administered under the Antarctic Treaty System | |
| Demographics | |
| Population | Uninhabited |
Leskov Island is an ice-covered island in the West Ice Shelf of Antarctica, rising to 185 metres (600 ft), 11 kilometres (6 nmi) northwest of Mikhaylov Island. It was discovered by the Soviet expedition of 1956, who named it for Lieutenant A. Leskov of the sloop Vostok on the Bellingshausen expedition of 1819–21.[1]
References
- ↑ "Leskov Island". Geographic Names Information System. United States Geological Survey. Retrieved 2013-06-11.
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