Asatellite of mass 6000 kg orbits the earth and has a period of 5700 s. determine the radius of its circular orbit.

1) The gravitational force F between the earth and the satellite is given by:

F = GMm / R²

where
G is the universal gravitational constant { = 6.67 x 10^-11 Nm²/kg² } ;
M is the mass of the Earth { = 5.97 x 10^24 kg } ;
m is the mass of the orbiting satellite ;
R is the distance between their centres of mass.

Force F provides the centripetal force needed to keep the satellite in a circular orbit.

Centripetal force = mω²R
where ω is the angular velocity. So we can write:

GMm / R² = mω²R

{m cancels out. Divide both sides by ω² and multiply by R² }

GM / ω² = R³

We are given that the satellite completes one orbit (= 2π radians) in 5700 seconds. So:

ω = (2π / 5700) = 1.10 x 10^-3 rad/s

R³ = (6.67 x 10^-11 * 5.97 x 10^24 / (1.10 x 10^-3 )² ) = 3.28 x 10^20

R = 6.89 x 10^6 m

2) Gravitational force = GMm / R²

= 6.67 x 10^-11 * 5.97 x 10^24 * 6000 / (6.89 x 10^6)²

= 5.03 x 10^4 newtons

3) Altitude = radius of orbit - radius of Earth
= 6.89 x 10^6 - 6.37 x 10^6

= 5.2 x 10^5 m