<p>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<span class="math-container">$^{1/2}$</span> on the right.</p>
<p>If <span class="math-container">$m_2$</span> is the mass of the secondary star, and <span class="math-container">$m_2^\text{norm}$</span> is the normalized mass defined in the paper, then you have</p>
<p><span class="math-container">$$m_2 = m_2^\text{norm} \sqrt{r},$$</span></p>
<p>where <span class="math-container">$r$</span> is the orbital radius. Since the authors express <span class="math-container">$m_2^\text{norm}$</span> in units of <span class="math-container">$M_J \text{AU}^{-1/2}$</span> and you need <span class="math-container">$m_2$</span> in units of <span class="math-container">$M_\odot$</span>, one could also write</p>
<p><span class="math-container">$$m_2 = \left({m_2^\text{norm} \over M_J \text{AU}^{-1/2}}\right) \left({r \over \text{AU}}\right)^{1/2} \left({M_J \over M_\odot}\right) M_\odot$$</span></p>
<p>This convenient notation is unit-independent, and basically means that if you plug in the values of <span class="math-container">$m_2^\text{norm}$</span> and <span class="math-container">$r$</span> in units of <span class="math-container">$M_J \text{AU}^{-1/2}$</span> and <span class="math-container">$\text{AU}$</span> respectively, and then multiply by the adimensional factor <span class="math-container">${M_J \over M_\odot}$</span>, you get the value of <span class="math-container">$m_2$</span> in units of <span class="math-container">$M_\odot$</span>.</p>

CC BY-SA 4.0
2022-07-20T06:46:27.853Z
2022-07-20T06:46:27.853Z
Prallax
<p>What does Mass of secondary normalized to 1 au in this catalogue <a href="https://vizier.cfa.harvard.edu/viz-bin/VizieR?-source=J/A+A/623/A72" rel="nofollow noreferrer">https://vizier.cfa.harvard.edu/viz-bin/VizieR?-source=J/A+A/623/A72</a> mean, please? I tried to multiply that by Jupyter mass and the resulting masses are extremely low. Thank you</p>
<p><strong>After advice:</strong></p>
<p>I read the section from the <a href="https://ui.adsabs.harvard.edu/abs/2019A%26A...623A..72K/abstract" rel="nofollow noreferrer">paper Kervella, 2019</a>. Do I understand that correctly, that <span class="math-container">$M_2(\mathrm{Msun}) = \frac{\mathrm{M2n}(\mathrm{Mjup})}{\mathrm{OrbG2(au)}^{1/2} } \cdot \frac{1.89813 \cdot 10^{27}}{1.9891 \cdot 10^{30}}$</span>,
where <span class="math-container">$\frac{1.89813 \cdot 10^{27}}{1.9891 \cdot 10^{30}}$</span> is the conversion between solar and jupiter masses?</p>
<p><a href="https://i.sstatic.net/xYFTI.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/xYFTI.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2022-07-09T13:45:24.943Z
please read the related paper, section 3.5. Companion mass
Prallax
CC BY-SA 4.0
2022-07-13T12:04:49.767Z
I read it. I edited my question.
Anna-Kat
CC BY-SA 4.0
2022-07-13T14:06:00.557Z
Thank you. To help others understand the question, could you add a link to the paper and explain where you get those numbers from?
Prallax
CC BY-SA 4.0
2022-07-19T13:36:36.157Z
The numbers are just the conversion between solar and Jupiter masses because I need the result in solar mass.
Anna-Kat
CC BY-SA 4.0
2022-07-19T15:02:45.097Z
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
planetmaker
CC BY-SA 4.0
2022-07-19T17:54:50.420Z
I quoted that and the screenshot is in addition. I am not saying that the calculation does not work I am just asking for checking. The first fraction I read out from the paper and the second fraction is the conversion from M_jup to M_sun.
Anna-Kat
CC BY-SA 4.0
2022-07-20T13:52:37.473Z
You show no calculation which one can check.
planetmaker
<binary-star><star-catalogues><keplers-laws>
Mass of secondary normalized to 1 au
Anna-Kat
# Mass of secondary normalized to 1 au

(Asked by: username_0 on 2022-07-09)

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

**After advice:**

I read the section from the [paper Kervella, 2019](https://ui.adsabs.harvard.edu/abs/2019A%26A...623A..72K/abstract). 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?

[![enter image description here](https://i.sstatic.net/xYFTI.png)](https://i.sstatic.net/xYFTI.png)

**Tags:** binary-star, star-catalogues, keplers-laws

> **Comments:**
> **username_1**: please read the related paper, section 3.5. Companion mass  
> **username_0**: I read it. I edited my question.  
> **username_1**: Thank you. To help others understand the question, could you add a link to the paper and explain where you get those numbers from?  
> **username_0**: The numbers are just the conversion between solar and Jupiter masses because I need the result in solar mass.  
> **username_2**: 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  
> **username_0**: I quoted that and the screenshot is in addition. I am not saying that the calculation does not work I am just asking for checking. The first fraction I read out from the paper and the second fraction is the conversion from M_jup to M_sun.  
> **username_2**: You show no calculation which one can check.  

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## 1 answers

### Answer by username_1 on 2022-07-20. Score: 2

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 $M\_\odot$.


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