ch13-p058 - 58. (a) We make use of 3 m2 v 3T = (m1 + m2 ) 2...

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11 12 1 21 1 22 1 3.7 10 m . mm m rr r §· + =− = = × ¨¸ ©¹ Dividing this by 1.5 × 10 11 m (Earth’s orbital radius, r E ) gives r 2 = 2.5 r E . 58. (a) We make use of 33 2 2 () 2 m vT G π = + where m 1 = 0.9 M Sun is the estimated mass of the star. With v = 70 m/s and T = 1500 days (or 1500 × 86400 = 1.3 × 10 8 s), we find 3 23 2 2 Sun 2 1.06 10 kg . (0.9 ) m Mm + Since M Sun 2.0 × 10 30 kg, we find m 2 7.0 × 10 27 kg. Dividing by the mass of Jupiter (see Appendix C), we obtain
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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