A spy satellite was launched into a circular orbit with a height of 599 km above the surface of the Earth. Determine its orbital speed and period. (The mass of the Earth is 5.97 1024 kg, and the radius of the Earth is 6370 km.)

Answer 1

we know that centripetal acceleration is a=v^2/r.
If we set this equal to g (g= GM/r^2), then we find that v=sqrt[GM/R].

Then we just plug in our variables. *Note: remember what units to use: meters (m), kilograms (kg). This will produce an answer in m/s.

So:
v= sqrt(6.67e-11 x 5.98e24 kg)/(599e3 m + 6.38e6 m)
= 7559.9 m/s