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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 $r_1/a_1 = r_2/a_2$?

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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 $m_1a_1=m_2a_2$.

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  • $\begingroup$ 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 $r_2 = a_2$? $\endgroup$
    – Consideration
    Sep 8, 2019 at 0:18
  • $\begingroup$ 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 $m_1a_1=m_2a_2$. $\endgroup$
    – G. Smith
    Sep 9, 2019 at 0:09

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