Finding the Masses of a binary star using the distance, angle and orbital period

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SUMMARY

The discussion focuses on calculating the masses of two stars in a binary system with an orbital period of 30 years and a distance of 20 parsecs. The angular radius of the orbit is given as 1", which converts to approximately 0.000290888209 radians. The calculation using the formula \(\frac{a^3}{p^2}=2M\) leads to an incorrect mass estimation of 960,000 solar masses for each star. The error stems from a misinterpretation of the angular measurement, where the contributor mistakenly equated 1" to 1', leading to inflated mass results.

PREREQUISITES
  • Understanding of binary star systems and orbital mechanics
  • Familiarity with angular measurements in radians
  • Knowledge of the mass-luminosity relationship in astrophysics
  • Proficiency in using Kepler's laws for celestial mechanics
NEXT STEPS
  • Review the derivation of Kepler's Third Law for binary star systems
  • Learn about the conversion between angular measurements and their implications in astrophysics
  • Investigate the methods for calculating stellar masses using observational data
  • Explore the implications of mass estimation errors in astrophysical research
USEFUL FOR

Astronomy students, astrophysicists, and anyone interested in binary star dynamics and mass calculations will benefit from this discussion.

PaulWright
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Homework Statement



Two stars are in a circular visual binary system. The orbital
period of the binary is 30 years. The distance to the binary is 20
parsecs. The angular radius of the orbit of each star is 1". What
are the masses of the two stars?

Homework Equations



I am assuming that the two stars are of the same mass.
\frac{a^3}{p^2}=2M

The Attempt at a Solution



as the angle is 1" this is 0.000290888209 radians

using the small angle approx opp/adj should = 0.000290888209 radians
we have adj, which is 20pc, therefore the opp (the radius of the orbit) should be 5.81776418*10^-3pc, which is 1200AU, which is therefore the distance between the two stars in AU, which is required for the equation.

therefore we get \frac{1200^3}{30^2}=M_1+M_2 which is 1920000 Solar masses, therefore each mass is 960000 Solar mass.

This seems way too big, and I would like someone to show me where I have gone wrong.

Cheers,
Paul
 
Last edited:
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1'' = 1/3600 degrees = 4.848 * 10^(-6) radians.
 
willem2 said:
1'' = 1/3600 degrees = 4.848 * 10^(-6) radians.

Jees, I never saw that I assumed 1"=1'

Cheers
 

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