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387_Physics ProblemsTechnical Physics

387_Physics ProblemsTechnical Physics - Chapter 13 P13.19...

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Chapter 13 389 P13.19 GM R d R d T J J J + = + d i d i 2 2 2 4 π GM T R d d d J J 2 2 3 11 27 2 2 7 3 7 4 6 67 10 1 90 10 9 84 3 600 4 6 99 10 8 92 10 89 200 = + × × × = × + = × = π π d i e je j b g e j . . . . . N m kg kg m km above the planet 2 2 P13.20 The gravitational force on a small parcel of material at the star’s equator supplies the necessary centripetal force: GM m R mv R mR s s s s 2 2 2 = = ω so ω = = × × × GM R s s 3 11 30 3 3 6 67 10 2 1 99 10 10 0 10 . . . N m kg kg m 2 2 e j e j e j ω = × 1 63 10 4 . rad s *P13.21 The speed of a planet in a circular orbit is given by F ma = : GM m r mv r sun 2 2 = v GM r = sun . For Mercury the speed is v M = × × × = × 6 67 10 1 99 10 5 79 10 4 79 10 11 30 10 4 . . . . e je j e j m s m s 2 2 and for Pluto, v P = × × × = × 6 67 10 1 99 10 5 91 10 4 74 10 11 30 12 3 . . . . e je j e j m s m s 2 2 . With greater speed, Mercury will eventually move farther from the Sun than Pluto. With original
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