387_Physics ProblemsTechnical Physics

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

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Chapter 13 389 P13.19 GM Rd T J J J + = + di 2 2 2 4 π GM T R d d d JJ 22 3 11 27 2 27 3 7 4 6 67 10 1 90 10 9 84 3 600 4 6 99 10 892 10 89200 =+ ×⋅ × × = × + = ej e j bg .. . . . Nm kg kg m km above the planet 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 × × GM R s s 3 11 30 3 3 667 10 2199 10 10 0 10 . m e j 163 10 4 . r a d s *P13.21 The speed of a planet in a circular orbit is given by Fm a = : GM m r mv r sun 2 2 = v GM r = sun . For Mercury the speed is v M = ×× × 6 67 10 1 99 10 579 10 479 10 11 30 10 4 . . e j m s ms 2 2 and for Pluto, v P = × 6 67 10 1 99 10 591 10 474 10 11 30 12 3 . . e j m s 2 2 . With greater speed, Mercury will eventually move farther from the Sun than Pluto. With original
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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