[SOLVED] Binary star system and Apparent magnatudesby AstroPhys Tags: apparent, binary, magnatudes, solved, star 

#1
Mar3008, 12:21 PM

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I was asked this questing in my tetbook and can't figure out how to do it.
Suppose there is a binary system made up of two identical stars, each with an apparent magnitude of +4.1. However, from our 40 cm telescope they appear as one star, i.e., they are unresolved. What would be the apparent magnitude of this ``star?" I figure that the two stars apparent magnatudes must add together to get this magnatude somehow but I am not quite sure how, because 100 magnitude 6 stars = a magnitude 1 star. is there some simple formula to figure this out. 



#2
Mar3008, 12:25 PM

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There should be a formula for relating N stars of different magnitudes to one effective magnitude.




#3
Mar3008, 12:30 PM

P: 4

Yes, you would think so, just that i can't find any thus far.




#4
Mar3008, 12:42 PM

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[SOLVED] Binary star system and Apparent magnatudes
If you can relate intensity to magnitude, that's a start. The two stars in a binary system are at the same distance. Do you have a formula for intensity and magnitude, or size and intensity with magnitude?
If two stars have the same magnitude, then together they would have twice the intensity as either one. If you had one star that doubled in intensity, what would happen to the magnitude? Also, there should be a formula relating apparent and absolute magnitude. 



#5
Mar3008, 12:51 PM

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There are a couple of ways to do this (I say a couple they are the same way effectively). One could remember that a difference in magnitude of 5 is the same as a factor of 100 in brightness. So you can work from there to find what difference in magnitude gives double the brightness. Surely you have been given some sort of formula in your notes relating to apparent magnitude?




#6
Mar3008, 12:53 PM

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#7
Mar3008, 12:56 PM

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by intensity do you mean brightness or luminosity??
mM = 5 logd 5 m2m1 = 2.5 log (b1/b2) 



#8
Mar3008, 12:58 PM

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#9
Mar3008, 12:58 PM

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It isn't magnitudes that add, but rather luminosities (the power of the star) or intensities (the power/detector area at the observation point) that will add.
You need to convert magnitude at least into relative intensity first. You have the absolute magnitude of a star given by M = 2.5 log (L / L_sun) + 4.8 , where M is the absolute magnitude of the star, L is its luminosity and L_sun is the Sun's luminosity (the +4.8 is approximately the Sun's absolute magnitude, but the exact value isn't important at the moment, since it's going to drop out). The difference between the absolute magnitude of two stars is then deltaM = M1  M2 = [ 2.5 log (L1 / L_sun) + 4.8 ]  [ 2.5 log (L2 / L_sun) + 4.8 ] = 2.5 log (L1 / L_sun) + 2.5 log (L2 / L_sun) = 2.5 log (L1 / L2) = 2.5 log (L2 / L1) . We can rearrange this to get L2 / L1 = 10^(deltaM / 2.5) . If the two stars you are comparing are at the same distance, then the difference in their absolute magnitudes is exactly the same as the difference in apparent magnitudes, so we can write L2 / L1 = 10^(deltam / 2.5) . [Note: if you use the same instrumental setup to observe both stars, the ratio of luminosities will be the same as the ratio of intensities, so we also have I2 / I1 = 10^(deltam / 2.5) .] For convenience, we can pick as a reference value the luminosity or intensity of a zeromagnitude star, making our equation Io / I1 = 10^( {m1  0} / 2.5 ) or I1 / Io = 10^( m1 / 2.5 ) . One of the stars in the binary system has m = 4.1 , so I = 10^( 4.1 / 2.5 ) · Io . So what is the intensity of two such stars together? Take this result and put it into the intensity equation to solve for the apparent magnitude of the system. This really ought to be in your text someplace, but maybe the course hasn't reached it yet... 



#10
Mar3008, 01:01 PM

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brightness as it relates to magnitude. Brightness is the luminosity we 'see', i.e. it's visual perception.




#11
Mar3008, 01:29 PM

P: 4

Wow, thank you very much for all your help.



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