A 600 kg satellite is in a circular orbit at an altitude of

A 600 kg satellite is in a circular orbit at an altitude of 550 km above the Earth\'s surface. Because of air friction, the satellite eventually falls to the Earth\'s surface, where it hits the ground with a speed of 1.50 km/s. How much energy was transformed into internal energy by means of air friction?

Solution

let PE be the gravitation potential energy of a satellite at a distance of 550 km from the surface of the earth

                   at a height , h the total energy of the satellite is in the form of potential energy

                      PE = m g h

         = 600 kg * 9.8 m/sec^2 * 550 * 10^3 m

                           = 3.234 * 10 ^ 9 Joules

by the time the satellite hit the ground the some of the energy was transformed into internal energy by means of air friction, and the rest of the potentential energy converted into kinetic energy KE

Frictinal energy = PE- KE

                          KE = m * v ^2 / 2 = 600 kg * (1.5 * 10^3 ) ^2 / 2

                                                        KE = 6.75 x 10^ 8 Joules

                Frictinal energy = PE-KE = 32.34 * 10 ^ 8 Joules - 6.75 x 10^ 8 Joules

                                                              = 25.59 * 10^ 8 Joules is the energy which was transformed into internal energy by means of air friction.