The gravitational attraction of the Earth provides the centripetal force that keeps the satellite in circular motion around it:
[tex]m \frac{v^2}{r}=G \frac{Mm}{r^2} [/tex]
where the term on the left is the centripetal force, while the term on the right is the gravitational force, and
m is the mass of the satellite
v is its orbital speed
r is the distance of the satellite from the center of Earth
G is the gravitational constant
M is the Earth's mass
Simplifying, we get:
[tex]v= \sqrt{ \frac{GM}{r} } [/tex]
However, we must be careful: r is the distance of the satellite from the Earth's center, so we must add the Earth's radius (6370 km) to the distance of the satellite from the surface (830 km):
[tex]r=6370 km+830 km=7200 km=7.2 \cdot 10^6 m[/tex]
The Earth's mass is [tex]M=5.97 \cdot 10^{24} kg[/tex], so the orbital speed of the satellite is
[tex]v= \sqrt{ \frac{GM}{r} } = \sqrt{ \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(5.97 \cdot 10^{24} kg)}{(7.2 \cdot 10^6 m)} }=7437 m/s = 7.44 km/s [/tex]
