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Unformatted text preview: (questions adapted from Astronomy Today, page 631, problems 1,2) 600 AU = dist (AU) x 6 o / 57.3 o , therefore dist = 5730 AU 1 pc 1 AU 1'' 50 pc 600 AU ? 3. The figure shows the periodluminosity relation, discovered by Henrietta Leavit, for Cepheid variables. a. What is the approximate luminosity of a Cepheid variable whose period is 5 days. (Solar luminosity = 4 x 10 26 Watts) b. The flux observed for this source is 1015 Watts/m 2 . What is the distance to this object in parsecs? Solution: a. From the graph L ~1000 solar = 4 x 10 29 Watts b. 1015 Watts/m 2 = 4 x 10 29 Watts / 4 x π x distance 2 . Distance = 5.5 x 10 21 m = 180 kpc...
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This note was uploaded on 11/01/2009 for the course PHYS 229 taught by Professor Mehmetyali during the Fall '09 term at Sabancı University.
 Fall '09
 MehmetYali
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