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Astronomy of Stars, Galaxies and the Universe - Solved Homework 3 | PHYS 101, Assignments of Physics

Material Type: Assignment; Professor: Piner; Class: Astronomy of Stars, Galaxies and the Universe; Subject: Physics; University: Whittier College; Term: Spring 2009;

Typology: Assignments

Pre 2010

Uploaded on 08/17/2009

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PHYS 101
Spring 2009
HW 3 Solutions
Chapter 8
1) The apparent brightness of a star depends on its distance and its luminosity.
10) These temperatures refer to the surface temperature of the star (technically,
the temperature of the photosphere). This is because these temperatures are
determined by studying the absorption lines in the spectrum of the star, and
these absorption lines are formed in the star’s outer layers close to the surface.
16) Appendix 10 lists information on the 34 nearest stars to the Sun. The
temperature is determined by the spectral type, which is G2 for the Sun. Looking
at the spectral type column in Appendix 10, and remembering the ordering of the
spectral types by temperature (Table 8.1), we see that most of these stars are
cooler than the Sun (most are type M). The stars that emit more energy than the
Sun would have a luminosity bigger than one solar luminosity. There are three
stars in this table that emit more energy than the Sun (Alpha Centauri A, Sirius
A, and Procyon A).
26) There is a difference of 20 magnitudes between the galaxy and the star (from
6 to 26). The galaxy is therefore 2.51220 = 108 times fainter than the star (see p.
184).
27) The Sun has an apparent magnitude of -26.2 (p. 184), while Alpha Centauri
has an apparent magnitude of 0 (the problem has a typo that was corrected in
class). This corresponds to a difference of 26.2 magnitudes, which by the
definition of apparent magnitude is a factor of 2.51226.2 = 3.0x1010 in brightness.
The Sun is therefore 3.0x1010 times brighter than Alpha Centauri, even though
the problem says to assume that they have the same luminosity. The difference in
brightness must then be entirely due to the different distances. Since we know
brightness is proportional to the inverse square of the distance (inverse-square
law from Chapter 4), we can then set up the following ratio:
Bsun/BAC = (DAC/Dsun)2, where Bsun and Dsun are the brightness and distance of the
Sun, and BAC and DAC are the brightness and distance of Alpha Centauri. We just
computed the brightness ratio above, so (DAC/Dsun)2 = 3.0x1010. Taking the square
root of both sides, DAC/Dsun = 1.7x105. So DAC = 1.7x105 Dsun, and since Dsun = 1
AU, DAC = 1.7x105 AU, or 170,000 Astronomical Units. This is about 60% of the
actual distance to Alpha Centauri measured by the parallax technique (so is just a
rough estimate).

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PHYS 101

Spring 2009 HW 3 Solutions Chapter 8

  1. The apparent brightness of a star depends on its distance and its luminosity.
  2. These temperatures refer to the surface temperature of the star (technically, the temperature of the photosphere). This is because these temperatures are determined by studying the absorption lines in the spectrum of the star, and these absorption lines are formed in the star’s outer layers close to the surface.
  3. Appendix 10 lists information on the 34 nearest stars to the Sun. The temperature is determined by the spectral type, which is G2 for the Sun. Looking at the spectral type column in Appendix 10, and remembering the ordering of the spectral types by temperature (Table 8.1), we see that most of these stars are cooler than the Sun (most are type M). The stars that emit more energy than the Sun would have a luminosity bigger than one solar luminosity. There are three stars in this table that emit more energy than the Sun (Alpha Centauri A, Sirius A, and Procyon A).
  4. There is a difference of 20 magnitudes between the galaxy and the star (from 6 to 26). The galaxy is therefore 2.512^20 = 108 times fainter than the star (see p. 184).
  5. The Sun has an apparent magnitude of - 26.2 (p. 184), while Alpha Centauri has an apparent magnitude of 0 (the problem has a typo that was corrected in class). This corresponds to a difference of 26.2 magnitudes, which by the definition of apparent magnitude is a factor of 2.51226.2^ = 3.0x10^10 in brightness. The Sun is therefore 3.0x10^10 times brighter than Alpha Centauri, even though the problem says to assume that they have the same luminosity. The difference in brightness must then be entirely due to the different distances. Since we know brightness is proportional to the inverse square of the distance (inverse-square law from Chapter 4), we can then set up the following ratio: Bsun/BAC = (DAC/Dsun)^2 , where Bsun and Dsun are the brightness and distance of the Sun, and BAC and DAC are the brightness and distance of Alpha Centauri. We just computed the brightness ratio above, so (DAC/Dsun)^2 = 3.0x10^10. Taking the square root of both sides, DAC/Dsun = 1.7x10^5. So DAC = 1.7x10^5 Dsun, and since Dsun = 1 AU, DAC = 1.7x10^5 AU, or 170,000 Astronomical Units. This is about 60% of the actual distance to Alpha Centauri measured by the parallax technique (so is just a rough estimate).