Unassigned%20Problems%206

Unassigned%20Problems%206 - Chapter 27: The Early Universe...

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Chapter 27: The Early Universe 1 Chapter 26 Problems 2. The distance is 1,000 Mpc. Interpolate between the values of 873 Mpc for a redshift of 0.200 and 1080 Mpc for 0.250. This will be only a rough estimate because this type of interpolation requires a linear relationship, which redshift and distance do not have. The result is 0.231. 3. The volume of space out to a distance of one billion parsecs is () 3 92 7 3 4 1 10 pc 4 10 pc 3 π ×= × . There are (10 6 pc) 3 = 10 18 pc 3 in one cubic megaparsec. Dividing this into the previous volume gives 4 × 10 9 . If the galactic density is 0.1 galaxy per cubic megaparsec, then this number would be multiplied by 0.1 to give 400 million galaxies in this volume. 4. 1 Mpc 3 is a volume equal to (3.1 × 10 22 m) 3 = 3.0 × 10 67 m 3 . The 5 billion stars will each have a volume of 6.0 × 10 57 m 3 . A solar radius is 7 × 10 8 m, a solar volume is 1.4 × 10 27 m 3 . How many stars would fit into the volume around one star? 6.0 × 10 57 m 3 / 1.4 × 10 27 m 3 = 4.2 × 10 30 stars. The volume that would hold that many stars is simply 4.2 × 10 30 stars × 6.0 × 10 57 m 3 = 2.5 × 10 88 m 3 . The radius of this volume is 1.8 × 10 29 m = 5.9 × 10 12 pc = 5.9 × 10 6 Mpc. 5. The diagonal of any one of the 10 by 10 Mpc squares is 200 = 10 2. The right triangle to be solved has sides of 10 and 10 2. The hypotenuse is the distance from one corner to the other corner. Distance = 10 2 +(10 2) 2 Distance = 17 Mpc The recessional velocity at this distance is 17 × 70 = 1200 km/s. 10. From Table 25.1, take the “radius” of the universe to be 8430 Mpc. 1 Mpc = 3.1 × 10 22 m, so this distance is equal to 2.6 × 10 26 m. Assuming a spherical volume to the universe, 4/3 π r 3 gives 7.5 × 10 79 m 3 .
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Unassigned%20Problems%206 - Chapter 27: The Early Universe...

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