Binary system of stars (##\alpha## - centauri)

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##\alpha##-Centauri is a binary star system with a separation of 8.0'' and 9.7'' from their center of mass and is located 1.31 parsecs from Earth, with a revolution period of 80.1 years. To calculate the masses of the stars, the equation ##\omega^2 = \frac{G(M_1+M_2)}{a^3}## is used, along with ##\omega = \frac{2\pi}{T}##, allowing for the determination of the combined mass. A separate equation is needed to find individual masses, and the center of mass definition is suggested as a potential solution. The discussion highlights the importance of converting arcseconds to radians for accurate distance calculations. The user has derived reasonable results but is encouraged to present their calculations for verification.
Aleolomorfo
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Homework Statement


##\alpha##-Centuary is in a binary visual system with another star. Their separation, from their CM, is 8.0'' and 9.7''. The distance from the Earth is 1.31pc. Their revolution period around the CM is 80.1 years. I have to find masses and luminosities for each star.

Homework Equations


Third Newton's law: ##\omega^2 = \frac{G(M_1+M_2)}{a^3}##

The Attempt at a Solution


Immagine.jpg
[/B]
With the help of this picture I can calculate ##a=1.7''\times1.31pc##. Then I can solve this system of equations:
$$\omega^2 = \frac{G(M_1+M_2)}{a^3}$$
$$\omega = \frac{2\pi}{T}$$
In this way I find only ##M_1+M_2##. I need another equation but I do not find it.
For the luminosity I can use the scaling relation that ##L\propto M^4##. However, firstly I need to find ##M_1## and ##M_2## separately.
 

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Sorry but I have noticed that I had switched the separations from the CM. I put 9,7'' on the shorter segment and viceversa, but it is not a problem since I have used the sum.
 
Maybe I have an idea: I can use the definition of CM.
$$M_1\times r_1=M_2\times r_2$$ with ##r1=1,31pc\times 9.7''## and ##r_2=1.31pc\times 8''##
I think it is ok, isn't it?
 
You can't just multiply arcseconds by distance to get the angular distance; you have to convert to radians first.
 
tms said:
You can't just multiply arcseconds by distance to get the angular distance; you have to convert to radians first.
Yes, I have implied it, I should have written explicitly, sorry. I have done it and I have found reasonable results, so I think it is ok.
 
Aleolomorfo said:
Yes, I have implied it, I should have written explicitly, sorry. I have done it and I have found reasonable results, so I think it is ok.
You should show your work and the values that you obtained.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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