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 |
| Highest elevation | 240 m (790 ft) |
| Administration | |
|
None | |
| Demographics | |
| Population | Uninhabited |
| Additional information | |
| Administered under the Antarctic Treaty System | |
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.
This article incorporates public domain material from the United States Geological Survey document "Mikhaylov Island" (content from the Geographic Names Information System).
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