Question

A 450 kg satellite is in a circular orbit at an altitude of 400 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 2.40 km/s. How much energy was transformed into internal energy by means of air friction?

Answer 1

mass of satelite , m = 450 KG

height of satelite , h = 400 KM , velocity with which it hits ground , v = 2.4 km/s

as it was in orbit so it posses both energies potential energy and kinetic energy

using

GMm/R = muR

this gives  

1/2mu? = 1/2mGM/R

energy transformed =

1/2mv? + mgh - 1/2mv>

=  14069268 * 10^6 KJ (approximately )