Question

7. A 75000 kg space shuttle starts to re-enter the atmosphere at 17500 mph (7600 m/s) at an altitude of 170 km. When it reaches an altitude of 3 km it is traveling at 300 mph (150 m/s). Draw an energy interaction diagram to determine how much heat was dissipated in the atmosphere. Ignore the thermal energy of the heat shield. Part 1: How many energy systems are changing? 0 4 O 2 O 1 O 3

Answer 1

There are 3 energies changing (1) Kinetic energy (KE) (2) Potential energy (PE) (3) Heat (theta) So correct answer is (D) 3 All 4 energy is lost in form of heat so Delta PE_gra + Delta KE = Q Delta PE_gra = mg Delta L = 75000 times 9.8 (170000 - 3000) = 1.22745 times 10^11 J Delta KE = 1/2 m Delta (v^2) = 1/2 times 75000 (7600^2 - 150^2) Delta KE = 37500 (57760000 - 22500) Delta KE = 2.0816 times 10^12 Q = 2.2 times 10^12 J Q = 2.2 times 10^9 KJ (D)