Question

A 475 kg satellite is in a circular orbit at an altitude of 525 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 1.50 km/s. How much energy was transformed into internal energy by means of air friction?

Answer 1

475 kg. Solution Given that mass of satellite is altitude is 525 km. e final velocity is a 1500 m/s. Agi distance of satellite is r= radius of earth + altitude r= 6371 x 1034 525x103 mtr. r = 689 6x10mtr. and for orbit, centripetal acceleration = gravitational acceleration pas Grim Timis mass of 7 satellite Earth in then K.E. of satellite KIE = t mul KE = GMM 2 orbit ram is mass of 1 Satellite s value from eph ] 27 and energy is Gravitational potential P.E = - GMm.

Total Energy is TE = KET PE = GMM + (-) GMm TE= - GMM - ③ and KE when satellite touch earth's surface KE = I m u 2 = I x 475 % (15002 KE = 5:34 3 8 108 T. and (P. E) on earth - 63 1103 Now, from Energy Conservation. (KE+ P.E) in orbit = (KE+P-E) on earth friction. 8 mm ) + LGMM + Wt - 5.343 x 10 - GMM 27 5.343 x 108 WE= -6-6748101x5097281049 X475 2x 68968 103 + 6.674 X 10" x 5.972 X 1024 x 475 6371 X 103 WF = -13-720109 - 0.5343x109 +29.71 x 103 | WF = 15:45*103 T the Internal Energy transtered by is 15.458109 J friction

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