Astr 2100 Descriptive Astronomy
Fall 2017 - Assignment 11 - Answers

Assignment

Find the distances to the following
    1. Star A, with a parallax angle of 0.1 arc seconds d = 1/0.1 = 10 pc
    2. Star B, with a parallax angle of 0.012 arc seconds d = 1/0.012 = 83.3 pc
    3. The parallax angle for the star Arcturus
      1. The parallax angle is 0.0883 arc seconds
      2. d = 1/0.0883 = 12 pc
    4. Star C with a surface temperature of 4,500 K and a relative magnitude of 9
      1. From the HR diagram, for a temperature of 4,500 K the absoluet magnitude is about 5.2
      2. m - M = 5 log(d) - 5
      3. 9 - 5.2 = 5 log(d) - 5
      4. log(d) = 1.76
      5. d = 101.76 = 58 pc
    5. Star D with a surface temperature of 20,000 K and a relative magnitude of 7.6
      1. From the HR diagram, for a temperature of 4,500 K the absoluet magnitude is about -25
      2. m - M = 5 log(d) - 5
      3. 9 - (-2.5) = 5 log(d) - 5
      4. log(d) = 3.02
      5. d = 103.01 = 1050 pc
    6. Star E with a light curve whose maximum is at 450 nm and a relative magnitude of 5.9
      1. Using Wien's Law T = 2,898,000 / 450 = 6,440 K
      2. From the HR diagram, for a temperature of 6,440 K the absoluet magnitude is about 3
      3. m - M = 5 log(d) - 5
      4. 5.9 - 3 = 5 log(d) - 5
      5. log(d) = 1.58
      6. d = 101.58 = 38 pc
    7. Star F with a light curve whose maximum is at 800 nm and a relative magnitude of 8
      1. Using Wien's Law T = 2,898,000 / 800 = 3,625 K
      2. From the HR diagram, for a temperature of 3,625 K the absoluet magnitude is about 7
      3. m - M = 5 log(d) - 5
      4. 8 - 7 = 5 log(d) - 5
      5. log(d) = 1.2
      6. d = 101.2 = 16 pc
    8. A Cepheid variable G whose period is 8 days and whose relative magnitude is 11
      1. With a period of 8 days the absolute magnitude is about -2.6
      2. m - M = 5 log(d) - 5
      3. 11 - (-2.6) = 5 log(d) - 5
      4. log(d) = 3.72
      5. d = 103.72 = 5,250 pc
    9. A Cepheid variable H whose period is 40 days and whose relative magnitude is 15
      1. With a period of 40 days the absolute magnitude is about -5.1
      2. m - M = 5 log(d) - 5
      3. 15 - (-5.1) = 5 log(d) - 5
      4. log(d) = 5.02
      5. d = 105.02 = 105,000 pc
    10. A Type I supernova with a relative magnitude of 8.9 (Hint, first search for the absolute magnitude of a Type I supernova)
      1. All supernovae have an absolute magnitude of -19.3
      2. m - M = 5 log(d) - 5
      3. 8.9 - (-19.3) = 5 log(d) - 5
      4. log(d) = 6.64
      5. d = 106.64 = 4.4 x 106 pc = 4.4 Mpc

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