# Kepler's 3rd law applied to binary systems: How can the two orbits have different semi-major axes?

I suddenly came to the realization that I don't understand something about Kepler's law when applied to binary systems, because I encountered an apparent paradox. There must be an error somewhere in my reasoning, but I can't figure out what.

Consider a binary system of stars with masses $m_1$ and $m_2$. Both stars will be in an elliptic orbit around their common center of mass. If, say, star 1 moves on an ellipse with semi-major axis $a$, then the period of the orbit of star 1 should be given by Kepler's third law, \begin{align} T_1^2 = \frac{4\pi a_1^3}{G(m_1 + m_2)}. \end{align} The same holds for star 2: \begin{align} T_2^2 = \frac{4\pi a_2^3}{G(m_1 + m_2)}. \end{align} But both stars must have the same orbital period (because otherwise the center of mass cannot be at rest), which seems to imply that the semi-major axes of the two stars should also be equal, looking at the above formulas. However, if we consider the Sun-Earth system, for instance (neglecting the other planets, etc) we see that this is clearly not the case.

Where is the error in the above reasoning?

• I have to look into it, but my hunch is they have a different reduced mass. Edit: the jacobi coordinates scale with the mass. Jan 28, 2018 at 18:20
• Possible duplicate of Kepler's third law for binary systems Jan 29, 2018 at 9:14

Kepler's Third law takes a slightly different form when you consider motion around the center of mass. The equations of motion are \begin{align} m_1\ddot{\boldsymbol{r}}_1 &= - \frac{Gm_1m_2}{|\boldsymbol{r}_1 - \boldsymbol{r}_2|^3}\left(\boldsymbol{r}_1 - \boldsymbol{r}_2\right),\\ m_2\ddot{\boldsymbol{r}}_2 &= \frac{Gm_1m_2}{|\boldsymbol{r}_1 - \boldsymbol{r}_2|^3}\left(\boldsymbol{r}_1 - \boldsymbol{r}_2\right).\\ \end{align} If you want to describe the relative motion of one celestial body with respect to the other one, you can combine these equations to obtain $$\ddot{\boldsymbol{r}}_1 - \ddot{\boldsymbol{r}}_2 = - \frac{G(m_1+m_2)}{|\boldsymbol{r}_1 - \boldsymbol{r}_2|^3}\left(\boldsymbol{r}_1 - \boldsymbol{r}_2\right),$$ or in short $$\ddot{\boldsymbol{r}} = - \frac{\mu}{r^3}\boldsymbol{r},$$ where $\boldsymbol{r} = \boldsymbol{r}_1 - \boldsymbol{r}_2$ and $\mu =G(m_1+m_2)$. This is the familiar Kepler problem, with the corresponding 3rd Kepler law $$T^2 = \frac{4\pi^2}{\mu}a^3.$$ On the other hand, if you wish to describe the motion of both celestial bodies with respect to the center of mass, you need to separate $\boldsymbol{r}_1$ and $\boldsymbol{r}_2$ in the equations of motion. You can do this by using the fact that the position of the center of mass remains constant $$m_1\boldsymbol{r}_1 + m_2\boldsymbol{r}_2 = \boldsymbol{0},\tag{1}$$ so that $$\boldsymbol{r}_1 - \boldsymbol{r}_2 = \frac{m_1+m_2}{m_2}\boldsymbol{r}_1 = -\frac{m_1+m_2}{m_1}\boldsymbol{r}_2.$$ Therefore, \begin{align} m_1\ddot{\boldsymbol{r}}_1 &= -Gm_1m_2\left(\frac{m_2^3}{(m_1+m_2)^3r^3_1}\right)\left(\frac{m_1+m_2}{m_2}\boldsymbol{r}_1\right),\\ m_2\ddot{\boldsymbol{r}}_2 &= Gm_1m_2\left(\frac{m_1^3}{(m_1+m_2)^3r^3_2}\right)\left(-\frac{m_1+m_2}{m_1}\boldsymbol{r}_2\right), \end{align} or \begin{align} \ddot{\boldsymbol{r}}_1 = -\frac{\mu_1}{r^3_1}\boldsymbol{r}_1,\qquad\text{and}\qquad \ddot{\boldsymbol{r}}_2 = -\frac{\mu_2}{r^3_2}\boldsymbol{r}_2, \end{align} with $$\mu_1 = \frac{Gm_2^3}{(m_1+m_2)^2},\qquad\text{and}\qquad \mu_2 = \frac{Gm_1^3}{(m_1+m_2)^2}.$$ So once again we have two Kepler problems, but this time the 3rd laws take the form $$T^2 = \frac{4\pi^2}{\mu_1}a_1^3,\qquad\text{and}\qquad T^2 = \frac{4\pi^2}{\mu_2}a_2^3.$$ Note that this implies $\mu_2a_1^3 = \mu_1a_2^3$, which simplifies to $m_1a_1 = m_2a_2$, consistent with Eq. (1). Also, $\mu_1 a = \mu a_1$ and $\mu_2 a = \mu a_2$ lead to $m_2 a = (m_1+m_2)a_1$ and $m_1 a = (m_1+m_2)a_2$, so that indeed $a = a_1+a_2$.
If, say, star 1 moves on an ellipse with semi-major axis $a$, then the period of the orbit of star 1 should be given by Kepler's third law, \begin{align} {T_1}^2 = \frac{4\pi {a_1}^3}{G(m_1 + m_2)}. \end{align} The same holds for star 2: \begin{align} {T_2}^2 = \frac{4\pi {a_2}^3}{G(m_1 + m_2)}. \end{align}
Minor issue: You are missing a factor of $\pi$: You should have $T^2 = (2\pi)^2 \frac {a^3}{G(m_1+m_2)}$.
You simply have your definitions wrong in your application of Kepler's third law. In the case where the masses of both bodies are considered, then $a$ is not the semi-major axis of each orbit, it is the sum of the semi-major axes of the two bodies (and is therefore the same in both equations). $$a = a_1 + a_2$$ $$m_1 a_1 = m_2 a_2$$ $$T^2 =\frac{4\pi^2}{G(m_1+m_2)} a^3$$