<p>If you discover an exoplanet via the Doppler (radial velocity) method, then the amplitude of the radial velocity variations depends on the <em>inclination</em>, <span class="math-container">$i,$</span> of the exoplanet's orbital axis with respect to your line of sight. Conventionally, <span class="math-container">$i=90^{\circ}$</span> corresponds to viewing an orbit &quot;edge-on&quot;, which maximises the velocity variations, while a face-on orbit with <span class="math-container">$i=0$</span> would not be detectable.</p>
<p><a href="https://i.sstatic.net/JSWTR.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/JSWTR.png" alt="enter image description here" /></a>
<a href="https://www.researchgate.net/figure/The-inclination-angle-i-is-defined-as-the-angle-between-the-plane-of-the-exoplane-ts_fig1_228392024" rel="nofollow noreferrer">source</a></p>
<p>This means that you cannot directly estimate the exoplanet mass from your radial velocity data, only <span class="math-container">$M \sin i$</span>, as plotted on your histogram. <span class="math-container">$M \sin i$</span> is a <em>lower limit</em> to the mass.</p>
<p>The <span class="math-container">$M_{\rm Jup}$</span> is there to indicate that the axis is labelled in units of Jupiter masses.</p>

CC BY-SA 4.0
2021-11-14T06:43:21.193Z
Why do you say that Msini is a lower limit to the mass? Why not the upper, because there could be other planets contributing to this Doppler effect?
lazearoundallday
CC BY-SA 4.0
2021-11-14T06:45:07.027Z
Wouldn't a face-on viewing mean that I = 90°, as the plane of the orbit would be making a 90° angle with the line of sight?
lazearoundallday
CC BY-SA 4.0
2021-11-14T07:18:17.093Z
$M\sin i$ is a lower limit to $M$ because $\sin i \leq 1$. A face-on orbit has $i=0$; that is how $i$ is defined - the angle between the orbital axis and the line of sight.
ProfRob
CC BY-SA 4.0
2021-10-28T07:33:08.060Z
2021-11-14T09:04:52.733Z
2021-11-14T09:04:52.733Z
ProfRob
<p><a href="https://i.sstatic.net/soMYl.png" rel="noreferrer"><img src="https://i.sstatic.net/soMYl.png" alt="enter image description here" /></a></p>
<p>This is some data presented in a lecture on exoplanets that depicts the distribution of the sizes of super-Earths in comparison to the mass of Jupiter.<br> I would like to know what the argument of the sine function i.e. 'i' implies and why it was multiplied by the mass of Jupiter.</p>

<exoplanet><data-analysis>
What is sin i in this graph and why is it there?
Ambica Govind
# What is sin i in this graph and why is it there?

(Asked by: username_0 on 2021-10-28)

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

This is some data presented in a lecture on exoplanets that depicts the distribution of the sizes of super-Earths in comparison to the mass of Jupiter.  
 I would like to know what the argument of the sine function i.e. 'i' implies and why it was multiplied by the mass of Jupiter.

**Tags:** exoplanet, data-analysis


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## 1 answers

### Answer by username_1 on 2021-10-28. Score: 20 | Accepted answer

If you discover an exoplanet via the Doppler (radial velocity) method, then the amplitude of the radial velocity variations depends on the *inclination*, $i,$ of the exoplanet's orbital axis with respect to your line of sight. Conventionally, $i=90^{\circ}$ corresponds to viewing an orbit "edge-on", which maximises the velocity variations, while a face-on orbit with $i=0$ would not be detectable.

[![enter image description here](https://i.sstatic.net/JSWTR.png)](https://i.sstatic.net/JSWTR.png)
[source](https://www.researchgate.net/figure/The-inclination-angle-i-is-defined-as-the-angle-between-the-plane-of-the-exoplane-ts_fig1_228392024)

This means that you cannot directly estimate the exoplanet mass from your radial velocity data, only $M \sin i$, as plotted on your histogram. $M \sin i$ is a *lower limit* to the mass.

The $M\_{\rm Jup}$ is there to indicate that the axis is labelled in units of Jupiter masses.

> **Comments:**
> **username_2**: Why do you say that Msini is a lower limit to the mass? Why not the upper, because there could be other planets contributing to this Doppler effect?  
> **username_2**: Wouldn't a face-on viewing mean that I = 90°, as the plane of the orbit would be making a 90° angle with the line of sight?  
> **username_1**: $M\sin i$ is a lower limit to $M$ because $\sin i \leq 1$. A face-on orbit has $i=0$; that is how $i$ is defined - the angle between the orbital axis and the line of sight.  

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