Astronomy - Determine the sum of the stellar masses

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Homework Statement


A double star located at a distance of 10 light years from us. The maximum angle between the stars, as seen from Earth, is 2 arcseconds. (1 arcsecond = 1 / 3600 degrees), we can assume that the stellar orbit is circular and that this angle gives us the real distance between the stars. We observe that the rotation period of the stars is 4 years. What is the sum of stellar masses?

Homework Equations


Newton's law of universal gravitation:
[tex]F_{G}=G\frac{m_{1}m_{2}}{r^{2}}[/tex]
Centripetal force:
[tex]F_c=\frac{mv^2}{r}[/tex]
Kepler's third law:
[tex]P^{2} \propto a^{3}[/tex]

The Attempt at a Solution


The stellar orbit is circular, so the gravitational force should be equal to the centripetalforce, but how do I proceed?
 

Answers and Replies

  • #2
gneill
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You can work out the actual distance between the two bodies by simple geometry. You should also realize that the both bodies will orbit a common center of mass, so you can work out the individual radii of their orbits (symbolically) as fractions of that distance which depend upon their relative masses. Their individual velocities can similarly be determined (symbolically) from the resulting circumferences of their orbits and the given period of the orbit.

These, along with your gravitational force and centripetal force expressions, should allow you to find an expression for the sum of the two masses.
 
  • #3
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I rewrite the centripetal force as:
[tex] F_{c}=m\cdot \omega^{2}\cdot r[/tex]

So now I can express the equality between the gravitational force and centripetal force for respective star to be
[tex] M\omega^{2}R=G\frac{Mm}{(R+r)^{2}}[/tex]
[tex] m\omega^{2}r=G\frac{Mm}{(R+r)^{2}}[/tex]

I simply the expression and then and I add the left-hand side of the equations together and the same procedure to the right-hand side and I find this expression for the sum of the masses:
[tex] M+m=\frac{\omega^{2}\cdot (R+r)^{3}}{G}[/tex]

The angular velocity can be expressed as
[tex] \omega=\frac{2\pi}{T}[/tex]

I finally found this expression for the sum of the masses:
[tex] M+m=\frac{(\frac{2\pi}{T})^{2}\cdot (R+r)^{3}}{G}[/tex]
But I’m not sure how I’ll proceed and calculate the distance (R+r), should I relate the distance from Earth and the maximum angle between the stars?
 
  • #4
gneill
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But I’m not sure how I’ll proceed and calculate the distance (R+r), should I relate the distance from Earth and the maximum angle between the stars?
Yes. Basic trig.
 

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