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September 30, 2014

September 30, 2014

Posted by **Charles** on Friday, February 8, 2008 at 5:47pm.

- Physics -
**drwls**, Friday, February 8, 2008 at 6:26pmAdd 2500 m to the radius of the earth to get the distance R from the center. Make sure it is in meters.

Then use the equation GM/R^2 = V^2/R to get the velocity V

G is the universal constant of gravity and M is the mass of the Earth.

You will have some numbers to look up, but they should be easy to find.

- Physics -
**tchrwill**, Friday, February 8, 2008 at 7:44pmThe 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.

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