# Mass of secondary normalized to 1 au

What does Mass of secondary normalized to 1 au in this catalogue https://vizier.cfa.harvard.edu/viz-bin/VizieR?-source=J/A+A/623/A72 mean, please? I tried to multiply that by Jupyter mass and the resulting masses are extremely low. Thank you

I read the section from the paper Kervella, 2019. Do I understand that correctly, that $$M_2(\mathrm{Msun}) = \frac{\mathrm{M2n}(\mathrm{Mjup})}{\mathrm{OrbG2(au)}^{1/2} } \cdot \frac{1.89813 \cdot 10^{27}}{1.9891 \cdot 10^{30}}$$, where $$\frac{1.89813 \cdot 10^{27}}{1.9891 \cdot 10^{30}}$$ is the conversion between solar and jupiter masses?

• please read the related paper, section 3.5. Companion mass Commented Jul 9, 2022 at 13:45
• I read it. I edited my question. Commented Jul 13, 2022 at 12:04
• Thank you. To help others understand the question, could you add a link to the paper and explain where you get those numbers from? Commented Jul 13, 2022 at 14:06
• The numbers are just the conversion between solar and Jupiter masses because I need the result in solar mass. Commented Jul 19, 2022 at 13:36
• instead of adding text as screenshot you might want to actually quote that paper (it works better with screen readers, thus accessibility) and - more important - really show in exemplum what calculation does not work (or why you think it does give incorrect results, quoting both, what you get and what you expect and why. The formulas are numbered, so you can walk us through your calculation and thoughts step by step Commented Jul 19, 2022 at 15:02

I can only guess the definition of the quantities that you use in the first equation (since you don't define them), but if the units are the ones indicated inside the parentheses, then your equation has some dimensional issue, because you have a mass on the left and a mass divided by a length$$^{1/2}$$ on the right.

If $$m_2$$ is the mass of the secondary star, and $$m_2^\text{norm}$$ is the normalized mass defined in the paper, then you have

$$m_2 = m_2^\text{norm} \sqrt{r},$$

where $$r$$ is the orbital radius. Since the authors express $$m_2^\text{norm}$$ in units of $$M_J \text{AU}^{-1/2}$$ and you need $$m_2$$ in units of $$M_\odot$$, one could also write

$$m_2 = \left({m_2^\text{norm} \over M_J \text{AU}^{-1/2}}\right) \left({r \over \text{AU}}\right)^{1/2} \left({M_J \over M_\odot}\right) M_\odot$$

This convenient notation is unit-independent, and basically means that if you plug in the values of $$m_2^\text{norm}$$ and $$r$$ in units of $$M_J \text{AU}^{-1/2}$$ and $$\text{AU}$$ respectively, and then multiply by the adimensional factor $${M_J \over M_\odot}$$, you get the value of $$m_2$$ in units of $$M_\odot$$.