  Leskov Island Location in Antarctica | |
| Geography | |
|---|---|
| Location | Antarctica |
| Coordinates | 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 First Russian Antarctic Expedition of 1819–21, and named after Lieutenant Leskov, one of the members of that expedition.[1]
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References
This article incorporates public domain material from "Leskov Island". Geographic Names Information System. United States Geological Survey.
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This page was last edited on 8 October 2021, at 16:19