Monday

May 30, 2016
Posted by **Macy** on Tuesday, April 6, 2010 at 10:00am.

- Math -
**tchrwill**, Tuesday, April 6, 2010 at 12:56pmFrom your given information,

r = 6,380,000 + 200,000 = 6,580,000 m or

....6,380 + 200 = 6,580 km.

The velocity required to maintain a circular orbit around the Earth may be computed from the following:

Vc = sqrt(µ/r)

where Vc is the circular orbital velocity in feet per second, µ (pronounced meuw as opposed to meow) is the gravitational constant of the earth, ~1.40766x10^16 ft.^3/sec.^2, and r is the distance from the center of the earth to the altitude in question in feet. Using 3963 miles for the radius of the earth, the orbital velocity required for a 250 miles high circular orbit would be Vc = 1.40766x10^16/[(3963+250)x5280] = 1.40766x10^16/22,244,640 = 25,155 fps. (17,147 mph.) Since velocity is inversely proportional to r, the higher you go, the smaller the required orbital velocity.

Insert your data to derive your orbital velocity.