5. a 500 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 2.00 km/s. how much energy was transformed into internal energy by means of air friction?

the energy transformed into internal energy is E fr = 1.57*10¹⁰ J

Explanation:

From energy conservation , the internal energy gained by friction should be equal to the loss of total energy of the satellite

Since the total energy of the satellite can be decomposed into kinetic and potential energy

E fr = ΔE = E₂-E₁

E₁=  K₁+V₁ = 1/2*m*v₁² + m*g*h₁

E₂ = K₂+V₂=  1/2*m*v₂² + m*g*h₂

first, we can choose our reference state so that h₂=0 and h₁=h=500 km

second , we can calculate the approximate the inicial velocity as the velocity required for a stable circular orbit

g = v₁²/(h+R) → v₁² = g*(h+R)

as the velocity diminishes, h diminishes, falling into the earth

assuming the radius of the Earth as R= 6371 km , then

v₁² = g*(h+R) = 9.8 m/s² * (500 km+ 6371 km) *1000 m/km = 6.73 * 10⁷ (m/s)²

replacing values

E₁ = 1/2*m*v₁² + m*g*h₁ = 1/2* 500kg *6.73 * 10⁷ (m/s)² + 500kg* 9.8m/s² * 500 km = 1.67*10¹⁰ J

E₂= 1/2*m*v₂² + m*g*h₂ =  1/2* 500kg *(2000 m/s)² + 0 = 1*10⁹ J

therefore

E fr = ΔE = E₂-E₁ = 1.67*10¹⁰ J - 1*10⁹ J = 1.57*10¹⁰ J