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problem12_63

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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12.63: One can solve this problem using energy conservation, units of J/kg for energy, and basic concepts of orbits. , or , 2 2 1 2 r GM a GM v U K E - = - + = where U K E and , are the energies per unit mass, v is the circular orbital velocity of 1655 m/s at the lunicentric distance of m. 10 79 . 1 6 × The total energy at this distance is Kg. J 10 37 . 1 6 × - When the velocity of the spacecraft is reduced by 20 m/s, the total energy becomes , m) 10 (1.79 kg) 10 35 . 7 ( ) kg / m N 10 673 . 6 ( s) / m 20 s / m 1655 ( 2 1 6 22 2 2 11 2 × × × - - = - E or kg. J 10 40 . 1 6 × - = E Since , 2 a GM E - = we can solve for m, 10 748 . 1 , 6 × = a a the semi –major axis of the new elliptical orbit. The old distance of
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Unformatted text preview: m 10 79 . 1 6 × is now the apolune distance, and the perilune can be found from m. 10 706 . 1 , 6 2 × = = + p r r r a p a Obviously this is less than the radius of the moon, so the spacecraft crashes! At the surface, kg. J 10 818 . 2 or, 6 m ×-=-= U U R GM Since the total energy at the surface is kg, J 10 40 . 1 6 ×-the kinetic energy at the surface is h. km 6057 s m 10 682 . 1 or kg, J 10 415 . 1 So, kg. J 10 415 . 1 3 6 2 2 1 6 = × = × = × v v...
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