Distance from Jefferson City to Lynbrook
Distance from Jefferson City to Lynbrook is 1,602 kilometers. This air travel distance is equal to 995 miles.
The air travel (bird fly) shortest distance between Jefferson City and Lynbrook is 1,602 km= 995 miles.
If you travel with an airplane (which has average speed of 560 miles) from Jefferson City to Lynbrook, It takes 1.78 hours to arrive.
Jefferson City
Jefferson City is located in United States.
| GPS Coordinates (DMS) | 38° 34´ 36.1200'' N 92° 10´ 24.6720'' W |
|---|---|
| Latitude | 38.57670 |
| Longitude | -92.17352 |
| Altitude | 195 m |
| Country | United States |
Jefferson City Distances to Cities
| Jefferson City | Distance |
|---|---|
| Distance from Jefferson City to Sacramento | 2,544 km |
| Distance from Jefferson City to Boston | 1,834 km |
| Distance from Jefferson City to San Francisco | 2,640 km |
| Distance from Jefferson City to Los Angeles | 2,386 km |
| Distance from Jefferson City to Dallas | 767 km |
Lynbrook
Lynbrook is located in United States.
| GPS Coordinates | 40° 39´ 17.3880'' N 73° 40´ 18.4800'' W |
|---|---|
| Latitude | 40.65483 |
| Longitude | -73.67180 |
| Altitude | 8 m |
| Country | United States |