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

Aform of energy that is reflected or emitted from objects in the form of electrical and magnetic waves that can travel through space. there are many forms of  electromagnetic  energy including gamma rays, x rays, ultraviolet radiation, visible light, infrared radiation, microwaves and radio waves.