From 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.
A satellite is launched into orbit 200 kilometers above the Earth.The orbital velocity of a satellite is given by the formula v=√GmE/r, where v is the velocity in meters per second, G is a given constant, mE is the mass of Earth, and r is the radius of the satellite"s orbit. The radius of Earth is 6,380,000 meters. What is the radius of the satellite's orbit in meters?
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