Two stars orbit their common center of mass

AI Thread Summary
In the discussion, participants explore why both masses (3M and M) are combined in Kepler's third law, leading to the equation T^2 = 4π^2R^3/G(3M + M). The key point is that the orbital period depends on both masses due to the gravitational interaction between them. The two-body problem is analyzed by separating the motion into the center of mass and a Kepler problem, which uses the reduced mass concept. The gravitational potential energy is expressed in terms of the total mass, reinforcing that Kepler's laws apply to the combined mass of the system. Understanding this relationship is crucial for accurately determining the orbital period of the star with mass 3M.
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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
 
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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