Mikhaylov Island
  Mikhaylov Island Location in Antarctica | |
| Geography | |
|---|---|
| Location | Antarctica |
| Coordinates | 66°48′S 85°30′E / 66.800°S 85.500°ECoordinates: 66°48′S 85°30′E / 66.800°S 85.500°E |
| Area | 630 km2 (240 sq mi) |
| Length | 42 km (26.1 mi) |
| Width | 15 km (9.3 mi) |
| Highest elevation | 240.8 m (790 ft) |
| Administration | |
| Administered under the Antarctic Treaty System | |
| Demographics | |
| Population | Uninhabited |
Mikhaylov Island is an ice-covered island in the West Ice Shelf of Antarctica, rising to 240 metres (800 ft), 11 kilometres (6 nmi) southeast of Leskov Island. It was discovered by the Soviet expedition of 1956, who named it for Pavel N. Mikhaylov, artist on the Bellingshausen expedition of 1819–21.[1]
References
- ↑ "Mikhaylov Island". Geographic Names Information System. United States Geological Survey. Retrieved 2013-09-24.
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