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?
Best Answer
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 $a = a_1+a_2$.