2.15: S17. Parallax, Cepheid Variables, Supernovae, and Distance Measurement - SOLUTIONS
Exercise 17.1.1
- Answer
-
It's about 6 cm between one pupil and the other. I got about 3 thumb widths for the shift from viewing with one eye to the other. This is six degrees. So the distance is about 6cm / sin(six degrees) = 6cm/(6*3.14/180) = 60 cm where in the first equality we used the small angle approximation after converting from degrees to radians. (Note the possibility here for making some mistakes with factors of two. The angle we measure here is 2p, as p is defined below, but the distance we are using is twice the baseline, as the baseline distance is defined below. These factors of 2, at least in the small-angle approximation, cancel out.)
Exercise 17.2.1
- Answer
-
The observed shift is actually 2p, with p as indicated in the diagram. Relative to the background stars the cluster of stars is shifting by 0.03 arc seconds. To understand this, you have to realize the background stars are much, much further away than the Earth-Sun separation -- much further than it looks in the diagram. Because of this large distance, the line of sight from the December position to the blue background star is parallel to the line of sight from the June position to the blue background star. The angle at December formed by red background star -- December -- blue background star you cna then see is 2p. This is the observed shift. So p = 0.015 arc seconds. Looking at the geometry in the figure we have \(d = 1 AU / \sin(p) = 1 AU/ (\pi/(180*60*60)*0.015) = 1.4 \times 10^7 \)AU.
Exercise 17.2.2
- Answer
-
Distance to KNOX0325 is \(2.1 \times 10^{20}\) cm = 220 light years.
Exercise 17.3.1
- Answer
-
d/parsec = 1 arscec/p ==> d = (1/.015) parsec = 66 parsec.
Exercise 17.4.1
- Answer
-
Answer: d/parsec = arcsec/p ==> d = 1,000 parsecs.
Exercise 17.5.1
- Answer
-
\(d = 10^{(26-11+5)/5}\) parsec = \(10^4\) parsec.
Exercise 17.6.1
- Answer
-
We know the distance to Galaxy A is \(10^4\) parsecs. So we can use \(d/parsec = 10^{(m-M+5)/5}\) to solve for M. Or, we could just say that since d is the same for the Cepheid and the supernovae, the m-M values have to be the same. For the Cepheid, m-M = 15, so for the supernova m-M=15. This implies that M=m-15 =-19.3.
Exercise 17.6.2
- Answer
-
\(d=10^{(15.7+19.3+5)/5}{\rm parsec} = 10^8\)parsec = 100 Megaparsec.