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1) a) Using M1-M2= -2.5 log (L1/L2) Star A: M=4.8 (given) Star B: M1-4.8 = -2.5log(10^3)  M1 = 4.8-7.5 = -2.7 Star C: M1-4.8 = -2.5log(10^-3)  M1 = 4.8+7.5 = 12.3 We can now get the distances using the relationship m-M = 5log (d/10). We change that relationship into the form d=101+(m-M)/5 and compute the distances. Star A: d=101+(12-4.8)/5 =102.44 = 275.4 parsecs Star B: d=101+(14.7)/5 =103.94 = 8709.6 parsecs Star C: d=10 1+(12-12.3)/5 = 100.94 = 8.7parsecs. b) The measured parallax can be computed using the relationship parallax = 1/d parsec Star A: parallax = 1/275.4 = 0.003631” Star B: parallax = 1/8709.6 = 0.0001148” Star C: parallax = 1/8.7 = 0.1149” Distance determination not involving parallax would be required for star B. c) The statements presented in the following answers were deduced using the HR diagram with combination of the stars’ luminosities and absolute magnitudes. Star A, having the mass, luminosity, and absolute magnitude of the sun is very likely to be just like our own sun, a medium-sized star lying along the main sequence. Probably just a couple of billion years old, glowing yellow or orange, with a temperature between 3000K and 6000K, it has not yet exhausted its nuclear fuel and the hydrogen reactions are still going on, just like in our own sun. Star B, with a luminosity of 10^3 and absolute magnitude of -2.7 is very likely to be some type of giant. Having exhausted the supply of hydrogen in its core and converted it to helium, it has resorted to fusing hydrogen in a shell right outside the core. The gravitational forces pull the hydrogen surrounding the core closer towards the centre, increasing the rate by which it is used as fuel, and making the star shine much brighter which would explain the luminosity recorded. Star C, with a very low luminosity and high absolute magnitude, is most likely a white dwarf. A medium-sized star the mass of our sun, which has exhausted the hydrogen fuel in its core, and is not massive enough to fuse hydrogen,