The gravitational attraction between the Earth and the satellite provides the centripetal force that keeps the satellite in circular motion:
[tex]m \frac{v^2}{r}= G \frac{Mm}{r^2} [/tex]
where
m is the satellite mass
v is its speed
r is its distance from the Earth's center
G is the gravitational constant
M is the Earth's mass
Re-arranging the formula, we get
[tex]v= \sqrt{ \frac{GM}{r} } [/tex]
The satellite orbits at a distance equal to one Earth's radius (R) above the surface. This means that its distance from the Earth's center is twice the Earth radius:
[tex]r=2R=2 \cdot 6400 km = 12800 km = 1.28 \cdot 10^4 m[/tex]
Therefore, its velocity is
[tex]v= \sqrt{ \frac{GM}{r} }= \sqrt{ \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.98 \cdot 10^{24} kg)}{1.28 \cdot 10^4 m} }=1.76 \cdot 10^5 m/s [/tex]
