4 Question: How would we find a closer star than Proxima Centauri? Problem 3. The Parallax of the red giant Betelgeuse is just barely measurable and has a value of about arc seconds. What is its distance? Suppose the measurement is in error by + or – arc seconds. What limits can you set on its distance.Answer: p(Betelgeuse) = arc seconds. Therefore its distance is d = 1/p = 200 pc.The error in this measurement is +/–0.003", so we have to add and subtract this figure from the parallax angle to give the lower and upper limits to the distance estimate.d(pc) = 1/0.008 = 125 pc, the larger angle corresponds to the closer distanced(pc) = 1/0.002 = 500 pc, the smaller angle corresponds to the farther distance
5 Problem 5. The star Rigel radiates most strongly at about 200 nm Problem 5. The star Rigel radiates most strongly at about 200 nm. What is its temperature? How does this compare to the Sun.𝝀m for Rigel = 200 nm. Use Wien’s Law to find its temperature?T = 3 × 106 K nm / 𝝀m = 3 × 106 K nm / 200 nm = 15,000 K.This is much hotter than the surface of the Sun, which is about 6000 K. (2.5 times hotter). (Note that we have approximated the constant here as 3 x 106, it is closer to 2.9 x 106
6 Problem 6. The bright southern star Alpha Centauri radiates most strongly at about 500 nm. What is its temperature? How does this compare to the Sun.. Alpha Centauri has its peak radiation at 500 nm. Applying Wien’s Law,T = 3 × 106 K nm / 500 nm = 6000 K.The temperature of this star is approximately the same as our Sun’s, so Alpha Centauri has a similar spectral type to the Sun (G2).
7 Problem 7. Arcturus is about ½ as hot as the sun but is about 100 times more luminous. What is its radius compared to the Sun.. For Arcturus, T = To/2, where To is the temperature of the Sun. L = 100Lo. To find the star’s radius, use the Stefan-Boltzmann Law, which relates a star’s luminosity, L, to its temperature, T, and radius, R.L = 4𝜫R2 𝝈 T4 or LArc = 4𝜫R2Arc 𝝈 T4Arc and Lo = 4𝜫R2o 𝝈 T4oWe want to find RArc in terms of Ro.100 Lo = LArc = 4𝜫R2Arc 𝝈 T4Arc100 (4𝜫R2o 𝝈T4o) = 4𝜫R2Arc 𝝈 T4Arc100 (R2oT4o) = R2Arc T4Arc = R2Arc (To/2) 4100 (R2oT4o) = R2Arc (T4o) / 24100 (R2o) = R2Arc / 161600 (R2o) = R2ArcThereforeR Arc = (1600) 1/2 Ro = 40 RoArcturus is 40 times wider than the Sun.