Distance from Johnston Atoll to Cebu City
Distance from Johnston Atoll to Cebu City is 7,212 kilometers. This air travel distance is equal to 4,481 miles.
The air travel (bird fly) shortest distance between Johnston Atoll and Cebu City is 7,212 km= 4,481 miles.
If you travel with an airplane (which has average speed of 560 miles) from Johnston Atoll to Cebu City, It takes 8 hours to arrive.
Johnston Atoll
Johnston Atoll is located in United States Minor Outlying Islands.
| GPS Coordinates (DMS) | 16° 45´ 24.0120'' N 169° 31´ 59.0160'' W |
|---|---|
| Latitude | 16.75667 |
| Longitude | -169.53306 |
| Altitude | -11 m |
| Country | United States Minor Outlying Islands |
Johnston Atoll Distances to Cities
| Johnston Atoll | Distance |
|---|---|
| Distance from Johnston Atoll to Cebu City | 7,212 km |
Cebu City
Cebu City is located in Philippines.
| GPS Coordinates | 10° 19´ 0.1920'' N 123° 53´ 26.5560'' E |
|---|---|
| Latitude | 10.31672 |
| Longitude | 123.89071 |
| Altitude | 37 m |
| Country | Philippines |