Two stars orbit their common center of mass

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SUMMARY

The discussion focuses on the application of Kepler's Third Law to a binary star system where two stars with masses 3M and M orbit their common center of mass. The equation T² = 4π²R³/G(3M + M) is derived, emphasizing that the total mass (4M) is crucial for determining the orbital period. The two-body problem is addressed, highlighting the significance of the reduced mass in gravitational interactions and the derivation of the Kepler potential based on the combined masses of the stars.

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  • Knowledge of gravitational potential energy and Newton's law of gravitation
  • Concept of reduced mass in orbital mechanics
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mancity
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Homework Statement
Two stars orbit their common center of mass as shown in the diagram below. The masses of the two stars are 3M and M. The distance between the stars is d.
Determine the period of orbit for the star of mass 3M.
Relevant Equations
T^2=4pi^2R^3/GM
Why do we add the two masses (3M+M=4M) and use that for M in the equation of kepler's 3rd law?
Namely why is it T^2=4pi^2R^3/G(3M+M)
 
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mancity said:
Homework Statement: Two stars orbit their common center of mass as shown in the diagram below. The masses of the two stars are 3M and M. The distance between the stars is d.
Determine the period of orbit for the star of mass 3M.
Relevant Equations: T^2=4pi^2R^3/GM

Why do we add the two masses (3M+M=4M) and use that for M in the equation of kepler's 3rd law?
Namely why is it T^2=4pi^2R^3/G(3M+M)
The period must depend on both masses. After an extensive Internet search I found this:

https://imagine.gsfc.nasa.gov/features/yba/CygX1_mass/binary/equation_derive.html
 
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Stated somewhat differently: The two-body problem separates into the linear motion of the center of mass and a Kepler problem for the separation vector. The Kepler problem has a mass that is the reduced mass ##\mu = m_1 m_2/(m_1 + m_2)## of the system and the Kepler potential is the usual Newtonian gravitational potential based on the two masses ##m_1## and ##m_2##. Because of this, the potential per reduced mass is given by
$$
- G\frac{m_1 + m_2}{r}.
$$
This Kepler problem is what Kepler's laws apply to and therefore the mass in the 3rd law is ##m_1 + m_2##.
 
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