Question

A 400 kg satellite is in a circular orbit at an altitude of 500 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 1.60 km/s. How much energy was transformed into internal energy by means of air friction

Answer 1

E = 1.45 x 10⁹ J = 1.45 GJ

Step-by-step explanation:

According to the law of conservation of energy:

Potential Energy Lost by Satellite = Kinetic Energy + Internal Energy

`mgh = (1)/(2) mv^2 + E\\\\E = mgh - (1)/(2) mv^2`

where,

E = Internal Energy = ?

m = mass = 400 kg

g = acceleration due to gravity = 9.81 m/s²

h = height = 500 km = 500000 m

v = speed on ground = 1.6 km/s = 1600 m/s

Therefore,

`E = (400\ kg)(9.81\ m/s^2)(500000\ m)-(1)/(2) (400\ kg)(1600\ m/s)^2\\E = 1.962\ x\ 10^9\ J - 0.512\ x\ 10^9\ J`

E = 1.45 x 10⁹ J = 1.45 GJ