A satellite of mass m = 500 kg is in a circular orbit at an altitude of l = 600 km above the Earth’s surface . The satellite returns to Earth as a result of the frictional forces and impacts with a speed of v = 2.00 kms-1. How much energy was absorbed by the atmosphere?
The gravitational potential energy of the satellite at an altitude of 600 km from the Earth's surface is given by the formula PE = m*g*h where g = 9.8 m/s^2 is the acceleration due to the gravitational force of attraction between the satellite and the Earth.
When the satellite lands on the surface it has a velocity of 2 km/s. The kinetic energy of the satellite is (1/2)*m*v^2 = (1/2)*500*2000^2 = 10^9 J. The potential energy of the satellite at 600 km was 500*9.8*600*1000 = 2.94*10^9. As the total energy is conserved the difference between the energy of the satellite in orbit and on the surface has been absorbed by the atmosphere. This is equal to 2.94*10^9 - 10^9 = 1.94*10^9 J
The energy absorbed by the atmosphere is 1.94*10^9 J