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?

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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**

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