question_answer 74)

                  (a) How many astronomical units (A.U.) make 1 parsec?                 (b)  Consider a sunlike star at a distance of 2 parsec. When it is seen through a telescope with 100 magnification. What should be the angular size of the star? Sun appears to be \[{{(1/2)}^{o}}\] from the earth. Due to atmospheric fluctuations, eye can?t resolve   objects smaller than 1 arc minute.                 (c) Mars has approximately half of the earth?s diameter.                  When it is closest to the earth it is a about 1/2 A.U. from the earth. Calculate what size it will appear when seen through the same telescope.                 (Comment: This is to illustrate why a telescope can magnify plants but not star.)                

Answer:

                  (a) 1 parsec \[=\frac{1\,AU}{1''}=\frac{1\,AU}{(1/{{3600}^{o}})\,(\pi /{{180}^{o}})\,\text{rad}}\]                 \[=\frac{3600\times 180}{\pi }AU=206369\,AU\approx 2\times {{10}^{5}}\,AU\]                 Here, symbol \[('')\] stands for arc second.                 (b) At 1 A.U. , diameter of the sun \[={{\left( \frac{1}{2} \right)}^{o}}\]                 \[\therefore \] At 1 parsec \[(\,\tilde{\ }\,2\,\times \,{{10}^{5}}AU)\],                 Diameter of the sun \[=\frac{1}{2\times \,2\,\times \,{{10}^{5}}}\]                 \[=\frac{1}{4}\,\times \,{{10}^{-5}}\] degree \[=\,\,\frac{60}{4}\,\,\times \,\,{{10}^{-5}}\]                 \[=\,15\,\times \,{{11}^{-5}}\] are minute                 At 2 parse, diameter of the sunlike star \[=\,30\,\times \,{{10}^{-5}}\,\]are minute                          Diameter through the telescope                 \[=30\,\times \,{{10}^{-5}}\,\times \,100=\,30\,\times \,{{10}^{-3}}\]                 \[=\text{ }3\times \,{{10}^{-2}}\] are minute                 Since eye can?t resolve objects smaller than 1 arc minute, so the telescope can not used to magnify the sunlike star.