What Is the Combined Apparent Magnitude of Two Close Stars?

[sanitykey]

Homework Statement

Two stars have apparent magnitudes of V = 5.1 and V = 4.6 but are too close together to be resolved with the naked eye and appear to be a single object. What is its apparent magnitude?

Homework Equations

I don't know what the relevant equations are if there are any.

The Attempt at a Solution

Apparent magnitude = the sum of the two magnitudes = 9.7 but i know that's probably wrong, any help would be appreciated thanks!

Sorry if this is in the wrong place by the way.

 

[Dick]

Magnitudes don't add. Luminosities add. What's the relation between magnitude and luminosity?

 

[sanitykey]

m = -2.5*log(L/4*pi*d^2) + c

So L = 10^((m-c)/-2.5) * (4*pi*d^2)

So errm i think i was given the c for the visual in another question i think it's -21.58...

So L1+L2 = 10^((5.1+21.58)/-2.5)*(4*pi*d^2)+10^((4.6+21.58)/-2.5)*(4*pi*d^2) = 2.13*10^-11*(4*pi*d^2) + 3.37*10^-11*(4*pi*d^2)

= 5.50*10^-11*(4*pi*d^2)

m = -2.5*log(L/4*pi*d^2)+ c
m = -2.5*log(5.50*10^-11*(4*pi*d^2)/4*pi*d^2) -21.58
m = -2.5*log(5.50*10^-11) -21.58
m = 4.07

Is that right? I just thought woohoo i have a number which is different from the others but i don't know if it's right?

 

[Dick]

It's right. You've got a lot of extra factors you don't really need in this particular problem, but I wouldn't worry about it.

 

[sanitykey]

Thanks for the help :)

 

[tony_cruz]

I have two questions that are very very similar to the one above but I'm pretty sure I have no value for the constant.

Question 1, parts a & b

A binary star has a total apparent magnitude of 15.00. One component star is twice as bright as the other.

(a) Show that the apparent magnitude of the brighter star is 15.44. [2]

This one I can do, but don't understand why it works.

Ie - m + 2.5log(F) = mbright --> F= 1.5
15 + 2.5log(1.5) = 15.44

So why does that work?



(b) The fainter star has an absoloute magnetude of 4.50. How far away is the binary system in Kpc?

Then this next question I'm sure I could do if it wasn't for the confusion with part a but I have a mental block because of that. I know the distance equation D = 10 ^ (m-M)/5 x 10, but I need the apparent magnitude of the fainter star to work that out and I can't do it. I don't think you need to find the total apparent magnitude to find the answer but I could be wrong.



EDIT - Going on a complete guess is the apparent mag of the faint star 16.19 (using 15 + 2.5log(3) ) giving a distance of 2.18Kpc? Even if this is right, it's revision for an exam so I'd like to know how to do it rather than fluking my way there.

 

[Dick]

You are getting the right answers but you don't know why? For problems like this you can simplify the luminosity magnitude relation to just M=(-2.5)*log(L). The real formula has an additive constant and the L is divided by some stuff, but that all cancels out. You have a bright star of luminosity 2L and a dimmer one at L. So the total magnitude of the system is 15=(-2.5)log(3L). The magnitude of the brighter is M2=(-2.5)log(2L). Take the difference of the two. M2-15=(-2.5)*(log(2)-log(3))=0.44 (use stuff like log(3L)=log(3)+log(L)). Similarly the magnitude of the dimmer is M1=(-2.5)log(L). Take the difference again, M1-15=(-2.5)*(-log(3))=1.19.

 

[tony_cruz]

Thanks, that was really helpful. Knowing my luck it probably won't come up on the exam now I understand it :D.