A 500 kg satellite Is in a circular orbit at an altitude of

A 500 kg satellite Is in a circular orbit at an altitude of 450 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 2.10 km/s. How much energy was transformed into internal energy by means of air friction?

Solution

mass of satellite , m = 500 Kg

height , h = 450 km

speed, u = 2100 m/s

let the energy transformed into internal energy is W

Using conseravtion of energy

0.5 * m * v^2 - G * M * m/R + Wf = -0.5 * G * M * m/(R + h)

0.5 * 500 * (2100)^2 - 6.673 *10^-11 * 500 * 5.98 *10^24/(6.371 *10^6) + Wf = -0.5 * 6.673 *10^-11 * 500 * 5.98 *10^24/(6.371 *10^6 + 450 *10^3)

solving for Wf

Wf = 1.559 *10^10 J

the increase in internal energy of satellite is 1.559 *10^10 J