PC Chapter 16

PC Chapter 16 - Serway/Moses/Moyer Modern Physics, 3rd...

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Serway/Moses/Moyer Modern Physics, 3 rd edition Chapter 16 Answers and Solutions 3. (a) We can estimate the distance with Hubble’s law: R = v H = (0.55)(3.00 × 10 8 m / s) 23 × 10 - 3 m / (s lightyear) = 7.2 × 10 9 lightyears (b) t = R v = 7.2 × 10 9 ly 0.55 c = 7.2 × 10 9 c (1 y) 0.55 c 13 billion years 4. (a) Let H ( t ) = b , a constant. From equation 16.14 Ý a ( t ) a ( t ) = b , which may be integrated to give da a = bdt + C , where C is another constant. Integrating gives ln a = bt + C or a = Ae bt where A = e C (b) Equation 16.21 for an Einstein-de Sitter universe is a = Ct 2/3 Thus, H = Ý a a = 2 3 t - 1 5. Using Equation 16.22 gives a ( t e ) = a ( t 0 ) 6 so the size is 1/6 the present linear size.
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Using Z = 5 in Equation 16.23 yields t e = 1/6 (29 3 /2 t 0 0.07 t 0 or about 7% of the present age. 6. Answers: (a) 590.2 nm, (b) 599.1 nm, (c) 688.6 nm Solutions: v = HR , H = (2.3 × 10 - 2 m / s) / ly (a) v (2 × 10 6 ly) = 4.6 × 10 4 m/s λ =λ 1 + v / c 1 - v / c = (590 nm)(1.00031) = 590.2 nm (b) v (2 × 10 8 = 4.6 × 10 6 λ = (590 nm) 1 + 0.01533 1 - 0.01533 = 599.1 nm (c) v (2 × 10 9 = 4.6 × 10 7 (590 nm) 1 + 0.1533 1 - 0.1533 = 688.6 nm 7. Answers: (a) 0.383 c (b) 5.0 × 10 9 ly Solutions: (a) Solving Equation 16.1 for v gives v = λ 2 2 λ 2 2 c = 650 2 - 434 2 650 2 + 434 2 c = 0.383 c (b) v = HR R = v / H = (0.383)(3 × 10 8 m / s) / [(2.3 × 10 - 2 m / s) / ly] = 5.0 × 10 9 ly. 8. Solution: In our frame of reference, Hubble’s law is exemplified by v 1 = HR 1 and v 2 = HR 2 . From these we may form the equations - v 1 =- HR 1 and v 2 - v 1 = H ( R 2 - R 1 ) . These equations express Hubble’s law as seen by the observer in the first galaxy cluster, as she looks at us to find - v 1 = H ( - R 1 )
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PC Chapter 16 - Serway/Moses/Moyer Modern Physics, 3rd...

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