389_Physics ProblemsTechnical Physics

# 389_Physics ProblemsTechnical Physics - Chapter 13 P13.25...

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Chapter 13 391 P13.25 g g MG r a 1 2 2 2 = = + g g y y 1 2 = − g g g y y y = + 1 2 g g g x x 1 2 2 = = cos θ cos θ = + r a r 2 2 1 2 e j g i = 2 2 g x ± e j or g = + 2 2 2 3 2 MGr r a e j toward the center of mass FIG. P13.25 Section 13.6 Gravitational Potential Energy P13.26 (a) U GM m r E = − = − × × + × = × 6 67 10 5 98 10 100 6 37 2 00 10 4 77 10 11 24 6 9 . . . . . N m kg kg kg m J 2 2 e je j b g a f . (b), (c) Planet and satellite exert forces of equal magnitude on each other, directed downward on the satellite and upward on the planet. F GM m r E = = × × × = 2 11 24 2 6 67 10 5 98 10 100 569 . . N m kg kg kg 8.37 10 m N 2 2 6 e je j b g e j P13.27 U G Mm r = − and g GM R E E = 2 so that U GMm R R mgR E E E = − F H G I K J = 1 3 1 2 3 U = × = × 2 3 1 000 9 80 6 37 10 4 17 10 6 10 kg m s m J 2 b g e je j . . . . P13.28 The height attained is not small compared to the radius of the Earth, so U mgy = does not apply; U GM M r = − 1 2 does. From launch to apogee at height h , K
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