• If the masses are unequal, the lighter body must move faster so that linear momentum remains zero in the barycentric frame.
• Both stars share one angular frequency $\omega$. Period equality follows from shared rotation about the same center.
• The total mass determines the period; the mass ratio determines how the motion is partitioned.
• Changing inclination leaves the orbit geometry alone but changes the projected radial-velocity signal and the amount of line wobble you can observe.

• This model enforces Newton's laws exactly for two point masses in circular orbit.
• It conserves total momentum and uses the barycentric frame (center of mass at rest).
• It does not include eccentricity, tides, relativity, or mass transfer.
• Units: distance in $\mathrm{AU}$, time in $\mathrm{yr}$, and masses in $M_{\odot}$. We use $$G = 4\pi^2\,\frac{\mathrm{AU}^3}{\mathrm{yr}^2\,M_{\odot}}.$$

Start from gravitational force and circular motion in the barycentric frame:

$$F = \frac{G M_1 M_2}{a^2}, \qquad F = M_1\omega^2 a_1.$$

Using $a_1 = a\,M_2/(M_1+M_2)$ gives:

$$\omega^2 = \frac{G(M_1 + M_2)}{a^3}.$$

Then with $\omega = 2\pi/P$ and the teaching normalization for $G$:

$$P^2 = \frac{a^3}{M_1 + M_2}.$$

Once orbital speeds are projected along the line of sight, the measurable quantities are the RV semi-amplitudes:

$$K_1 = \frac{2\pi a_1 \sin i}{P}, \qquad K_2 = \frac{2\pi a_2 \sin i}{P}.$$

For a double-lined binary, the ratio tells you the mass ratio directly:

$$\frac{K_1}{K_2} = \frac{M_2}{M_1} = q.$$

For mass inference, astronomers use the SB1 mass function and the SB2 minimum masses:

$$f(m) = \frac{P K_1^3}{2\pi G}, \qquad M_1\sin^3 i,\; M_2\sin^3 i.$$