by mariuslvasile » Sat Jan 24, 2026 6:30 pm
A simple example to illustrate how this works is by using a binary star system where both stars have equal mass. The stars will not orbit around each other, but around their common center of mass, which is situated at half the distance between their centers of mass. But there is nothing special or supranatural going on in that point, no singularity, no unicorns, and no voodoo magic either.
GOOGLE wrote:Yes, in a binary system with two stars of equal mass, both stars will orbit around a common center of mass (barycenter) located exactly at the midpoint, or one-half of the distance, between their centers. Due to their equal gravitational pull, they move in identical elliptical (or circular) orbits on opposite sides of this central point.
Key details regarding this system:
Center of Mass: For equal masses (m1=m2), the center of mass is equidistant from both stars (d1=d2=dtotal/2).
Orbital Motion: Both stars move in their orbits with the same angular velocity, ensuring they remain diametrically opposite each other at all times.
Orbit Shape: They orbit their common barycenter, which is a key characteristic of binary systems.
COPILOT wrote:In a binary system with two stars of equal mass, both stars orbit around their common center of mass, which is located exactly midway between their centers.
Center of Mass in Equal-Mass Binary Systems
For a system of two bodies with equal mass m1=m2, the center of mass (COM) is calculated as:
RCOM=(m1r1+m2r2)/(m1+m2)
If the stars are separated by a distance d, and their masses are equal, the COM lies exactly at the midpoint, i.e., at a distance d/2 from each star.
Orbital Motion
Both stars orbit the common COM, not each other.
Since they have identical masses, their orbits are mirror images, and they each follow a circular or elliptical path around the COM, with the COM remaining stationary at the midpoint (in an inertial frame where no external forces act).
The orbits are synchronized in period but opposite in direction, maintaining the COM stationary.
Key Implications
The statement in the query is correct: the point around which both stars orbit is situated half the distance between their centers, i.e., the center of mass.
If one star were more massive than the other, the COM would shift toward the more massive star, and the stars’ orbits would have radii inversely proportional to their masses.
Summary
In an equal-mass binary star system:
Each star traces an orbit around the midpoint of their separation.
The orbital radius for each is exactly half the separation.
The COM remains stationary (neglecting external forces), and the system exhibits perfect symmetry.
This principle is fundamental in stellar dynamics and observational astronomy, such as when determining binary star masses from orbital periods and separations
And the same principle applies to star systems made of more than 2 stars, like ternary star systems:
COPILOT wrote:In a ternary system with three stars of equal mass, the calculation of the center of mass (COM) generalizes naturally from the binary case while preserving the concept of a mass-weighted average position. Consider three stars, each of mass mm, located at positions r1, r2, and r3 in a three-dimensional space. The center of mass is given by the vector formula:
RCOM=mr1+mr2+mr3/m+m+m=(r1+r2+r3)/3
Properties for Equal-Mass Stars:
Equidistant Balance: If the stars are positioned symmetrically (e.g., at the vertices of an equilateral triangle), the COM lies at the geometric center of the triangle. More generally, the COM is simply the arithmetic mean of the three positions.
Vector Distances from COM: Define vectors from the COM to each star:
ri COM=ri−RCOM,i=1,2,3
These vectors satisfy the zero vector sum condition due to equal
masses:
r1 COM+r2 COM+r3 COM=0
This is the ternary analog of the binary relation r1+r2=0 (with appropriate vectors) for the equal-mass two-body system.
Orbital Considerations: If the stars orbit each other under mutual gravity, their motion can be described in the COM frame such that vectors from the COM trace paths determined by their initial positions and velocities, preserving the COM at rest in an inertial frame.
Summary:
For three stars of equal mass, the center of mass is the arithmetic mean of their positions:
RCOM=(r1+r2+r3)/3
The relative positions of the stars with respect to the COM satisfy a vector sum of zero:
∑i=13(ri−RCOM)=0.i=1∑3(ri−RCOM)=0.
This ensures the COM remains stationary in an inertial frame, while each star orbits or moves relative to the COM depending on the system's dynamics.
For symmetric layouts (equilateral triangle, linear arrangement with equal spacing), the COM coincides with the geometric center of the configuration.
This vector-based approach generalizes straightforwardly to N-body systems as well: the COM is always the mass-weighted average of positions, and for equal masses, it reduces to the simple arithmetic mean.
ucf.edu
18.2 Measuring Stellar Masses – Astronomy
A simple example to illustrate how this works is by using a binary star system where both stars have equal mass. The stars will not orbit around each other, but around their common center of mass, which is situated at half the distance between their centers of mass. But there is nothing special or supranatural going on in that point, no singularity, no unicorns, and no voodoo magic either.
[quote=GOOGLE]Yes, in a binary system with two stars of equal mass, both stars will orbit around a common center of mass (barycenter) located exactly at the midpoint, or one-half of the distance, between their centers. Due to their equal gravitational pull, they move in identical elliptical (or circular) orbits on opposite sides of this central point.
Key details regarding this system:
Center of Mass: For equal masses (m1=m2), the center of mass is equidistant from both stars (d1=d2=dtotal/2).
Orbital Motion: Both stars move in their orbits with the same angular velocity, ensuring they remain diametrically opposite each other at all times.
Orbit Shape: They orbit their common barycenter, which is a key characteristic of binary systems. [/quote]
[quote=COPILOT]In a binary system with two stars of equal mass, both stars orbit around their common center of mass, which is located exactly midway between their centers.
Center of Mass in Equal-Mass Binary Systems
For a system of two bodies with equal mass m1=m2, the center of mass (COM) is calculated as:
RCOM=(m1r1+m2r2)/(m1+m2)
If the stars are separated by a distance d, and their masses are equal, the COM lies exactly at the midpoint, i.e., at a distance d/2 from each star.
Orbital Motion
Both stars orbit the common COM, not each other.
Since they have identical masses, their orbits are mirror images, and they each follow a circular or elliptical path around the COM, with the COM remaining stationary at the midpoint (in an inertial frame where no external forces act).
The orbits are synchronized in period but opposite in direction, maintaining the COM stationary.
Key Implications
The statement in the query is correct: the point around which both stars orbit is situated half the distance between their centers, i.e., the center of mass.
If one star were more massive than the other, the COM would shift toward the more massive star, and the stars’ orbits would have radii inversely proportional to their masses.
Summary
In an equal-mass binary star system:
Each star traces an orbit around the midpoint of their separation.
The orbital radius for each is exactly half the separation.
The COM remains stationary (neglecting external forces), and the system exhibits perfect symmetry.
This principle is fundamental in stellar dynamics and observational astronomy, such as when determining binary star masses from orbital periods and separations[/quote]
And the same principle applies to star systems made of more than 2 stars, like ternary star systems:
[quote=COPILOT]In a ternary system with three stars of equal mass, the calculation of the center of mass (COM) generalizes naturally from the binary case while preserving the concept of a mass-weighted average position. Consider three stars, each of mass mm, located at positions r1, r2, and r3 in a three-dimensional space. The center of mass is given by the vector formula:
RCOM=mr1+mr2+mr3/m+m+m=(r1+r2+r3)/3
Properties for Equal-Mass Stars:
Equidistant Balance: If the stars are positioned symmetrically (e.g., at the vertices of an equilateral triangle), the COM lies at the geometric center of the triangle. More generally, the COM is simply the arithmetic mean of the three positions.
Vector Distances from COM: Define vectors from the COM to each star:
ri COM=ri−RCOM,i=1,2,3
These vectors satisfy the zero vector sum condition due to equal
masses:
r1 COM+r2 COM+r3 COM=0
This is the ternary analog of the binary relation r1+r2=0 (with appropriate vectors) for the equal-mass two-body system.
Orbital Considerations: If the stars orbit each other under mutual gravity, their motion can be described in the COM frame such that vectors from the COM trace paths determined by their initial positions and velocities, preserving the COM at rest in an inertial frame.
Summary:
For three stars of equal mass, the center of mass is the arithmetic mean of their positions:
RCOM=(r1+r2+r3)/3
The relative positions of the stars with respect to the COM satisfy a vector sum of zero:
∑i=13(ri−RCOM)=0.i=1∑3(ri−RCOM)=0.
This ensures the COM remains stationary in an inertial frame, while each star orbits or moves relative to the COM depending on the system's dynamics.
For symmetric layouts (equilateral triangle, linear arrangement with equal spacing), the COM coincides with the geometric center of the configuration.
This vector-based approach generalizes straightforwardly to N-body systems as well: the COM is always the mass-weighted average of positions, and for equal masses, it reduces to the simple arithmetic mean.
ucf.edu
18.2 Measuring Stellar Masses – Astronomy[/quote]