A 550 kg satellite is in a circular orbit at an altitude of 450 km above the Ear
ID: 1510901 • Letter: A
Question
Explanation / Answer
1) let PE be the gravitation potential energy of a satellite at a distance of 450 km from the surface of the earth
at a height , h the total energy of the satellite is in the form of potential energy
PE = m g h
= 550 kg * 9.8 m/sec^2 * 450 * 10^3 m
= 2.426 * 10 ^ 9 Joules
by the time the satellite hit the ground the some of the energy was transformed into internal energy by means of air friction, and the rest of the potentential energy converted into kinetic energy KE
Frictinal energy = PE- KE
KE = m * v ^2 / 2 = 550 kg * (1.7 * 10^3 ) ^2 / 2
KE = 7.95 x 10^ 8 Joules
Frictinal energy = PE-KE = 24.26 * 10 ^ 8 Joules – 7.95 x 10^ 8 Joules
= 16.31 * 10^ 8 Joules is the energy which was transformed into internal energy by means of air friction.
2) For this orbit, the total energy of the satellite–Earth system which is given as :
Etotal = K.Esat + P.Eearth
Etotal = (1/2) ms v2 + mE g h
Etotal = (0.5) (2.6 kg) (8300 m/s)2 + (5.98 x 1024 kg) (9.8 m/s2) [(9200000 m) + (6371000 m)]
Etotal = (89557000 J) + (912522884 x 1024 J)
Etotal = 1.002 x 1033 J
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.