Astr 3000 Contemporary Astronomy
Fall 2017 - Assignment 12 - Answers

Assignment

  1. Stars A and B are equally bright, but star B is 6 times further away
      1. Set PA = PB = 1
      2. dA = 1
      3. dB = 6dA = 6
      4. then IA = PA / dA² = 1
      5. and  IB = PB / dB² = 1/36
    1. Which one is dimmer from our point of view? Since IB < IA star B is dimmer
    2. By how much? by a factor of 36
  2. Star C has only half the brightness of star D, and is 3 times further away
      1. Set PD = 1
      2. PC = ½PD = ½
      3. dD = 1
      4. dC = 3dD = 3
      5. then ID = PD / dD² = 1
      6. and  IC = PC / dC² = ½ / 3² = 1/18
    1. Which one is dimmer from our point of view? Since IC < ID star C is dimmer
    2. By how much? by a factor of 18
  3. Star E has five times the brightness of star F, and is 3 times further away
      1. Set PF = 1
      2. PE = 5PF = 5
      3. dF = 1
      4. dE = 3dF = 3
      5. then IF = PF / dF² = 1
      6. and  IE = 5 / 3² = 5/9
    1. Which one is dimmer from our point of view? Since IE < IF star E is dimmer
    2. By how much? by a factor of 9/5 = 1.8
  4. Star G is six times the radius of star H, and has a surface temperature twice as high
      1. Set rH = 1
      2. rG = 6rH = 6
      3. TH = 1
      4. TG = 2TH = 2
      5. then PH = rH² TH4 = 1
      6. and PG = rG² TG4 = 6² 24 = 576
    1. Which one produces the most amount of light? Since PG > PH star G has the higher luminosity (power)
    2. By what factor? by a factor of 576
  5. Star K is six times the radius of star L, but has a surface temperature only half that of star L
      1. Set rL = 1
      2. rK = 6rL = 6
      3. TL = 1
      4. TK =  ½TL = ½
      5. then PL = rL² TL4 = 1
      6. and PK = rK² TK4 = 6² ½4 = 2.25
    1. Which one produces the most amount of light? Since PK > PL star K has the higher luminosity (power)
    2. By what factor? by a factor of 2.25
  6. Compared to star M, star N is three times bigger and eight times further away. The surface temperature of start M is 4000 K, whereas the surface temperature of start N is 5000 K
      1. Set rM = 1
      2. rN = 3 rM = 3
      3. TM = 1 (in units of 4000 K)
      4. TN = 5000 K = 5/4 TM
      5. dM = 1
      6. dN = 8dM = 8
      7. then IM = rM² TM4 / dM² = 1
      8. and  IN = rN² TN4 / dN² =  3² (5/4)4 / 8² = 0.34
    1. Which one is dimmer from our point of view? Since IN < IM star N is dimmer
    2. By how much? by a factor of 0.34

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