Question

For a satellite to be in a circular orbit 830 km above the surface of the earth, what orbital speed must it be given?

Answer 1

The gravitational attraction of the Earth provides the centripetal force that keeps the satellite in circular motion around it:

`m (v^2)/(r)=G (Mm)/(r^2)`
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:

`v= \sqrt{ (GM)/(r) }`
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):

`r=6370 km+830 km=7200 km=7.2 \cdot 10^6 m`
The Earth's mass is
`M=5.97 \cdot 10^(24) kg`, so the orbital speed of the satellite is

`v= \sqrt{ (GM)/(r) } = \sqrt{ ((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`