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# Unassigned%20Problems%203 - Chapter 19 Star Formation...

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Chapter 19: Star Formation 1 Chapter 18 Problems 2. The distance to Alpha Centauri is 1.33 pc × 3.1 × 10 16 m/pc = 4.1 × 10 16 m. The volume of space is 4.1 × 10 16 m 3 . Multiplying this by the mass density from the previous question gives a total mass of 7.0 × 10 -8 kg. This is equal to 0.070 mg. 4. Assume the shape is a cylinder of length L and cross-sectional area π r 2 , the product of which is the volume. From the scale of the diagram, the diameter is about 10 -7 m and a length of about three times this. Using this to calculate the volume and multiplying by the density will give the mass. 3 × 10 -7 × π × (5 × 10 -8 ) 2 × 3000 = 7.1 × 10 -18 kg 5. Every 5 pc it gets fainter by a factor of 2. For example, in 10 pc it gets 2 × 2 = 4 times fainter. It gets reduced by 2 a total of 60/5 = 12 times as it passes through the cloud. This is 2 × 2 × 2 × ..... , or more easily 2 12 = 4096. The light is seen 4096 times fainter than it was before it entered the cloud. On the magnitude scale, the light is reduced in intensity by 2.5Log(4096) = 9 magnitudes. 7. M = -6 and m = 14, 14 - (-6) = 5Log(d) - 5, d = 10 5 pc. Solving for m , from the known distance of 5,000 pc, gives m - (-6) = 5Log(5000) - 5, m = 7.5 It has been dimmed by 6.5 magnitudes over 5 kpc or 1.3 mag/kpc. 9. M = -5 and m = 10. 10 - (-5) = 5Log( d ) - 5 + ( d /1000) × 2. Reduce this equation to 4 = Log( d ) + d /2500. You can not solve directly for d , so try various values for d , solve the right hand side and if the result is greater than 4, choose a smaller d ; if smaller than 4, choose a larger

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Unassigned%20Problems%203 - Chapter 19 Star Formation...

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