# In a binary system, why $m_1 a_1 = m_2 a_2$?

In the center of mass coordinate, $$m_1 r_1 = m_2 r_2$$, which is straightforward. Yet in this detailed deviation of radial velocity page 27, it says that the $$r_1$$, which is the magnitude of the vector pointing from the CM to the star, is simply the semi‐major axis of the star’s orbit around the mutual CM, $$a_1$$. With the same reasoning, the $$r_2$$, which is the magnitude of the vector pointing from the CM to the planet, is the semi‐major axis of the planet’s orbit around the mutual CM, $$a_2$$. Therefore $$m_1 a_1 = m_2 a_2$$.

However, I do not see why $$r_1 (r_2)$$ can be identical to $$a_1(a_2)$$ since the $$r's$$ are both changing (in a elliptical orbit for example) while the $$a's$$ are fixed. Or the other way to ask this question is I do not see why $$r_1/a_1 = r_2/a_2$$?

It’s just poorly written. He doesn’t really mean that $$r$$ is $$a$$. He means that $$r$$ is $$a$$ at one point on the orbit, so if $$m_1r_1=m_2r_2$$ for the whole orbit than $$m_1a_1=m_2a_2$$.
• Now I see that $m_1 a_1 = m_2 a_2$ is a special case of $m_1 r_1 = m_2 r_2$. But at the moment when $r_1 = a_1$, how can you know $r_2 = a_2$? – consideration Sep 8 at 0:18
• I think a clearer argument is to say that $2a$ is $r_\text{max}+r_\text{min}$. The two bodies have their min/max $r$ at the same time. So $m_1r_{1,\text{min}}=m_2r_{2,\text{min}}$, etc. That leads to $m_1a_1=m_2a_2$. – G. Smith Sep 9 at 0:09