<p>Yes, the reference plane for exoplanet <a href="https://en.wikipedia.org/wiki/Orbital_inclination#Exoplanets_and_multiple_star_systems" rel="nofollow noreferrer">orbital inclination</a> is our sky plane.
From the NASA Exoplanet Archive <a href="https://exoplanetarchive.ipac.caltech.edu/docs/parhelp.html#i-l" rel="nofollow noreferrer">documentation</a>:</p>

<blockquote>
  <p>The Observed Inclination is the orbital inclination with respect to the plane of the sky....
  0 degrees correspond to an orbit in the plane of the sky, face on with respect to our line of sight.
  90 degrees corresponds to an orbit that is edge-on with respect to our line of sight, perpendicular to the plane of the sky.</p>
</blockquote>

<p>If you <a href="https://exoplanetarchive.ipac.caltech.edu/cgi-bin/TblView/nph-tblView?app=ExoTbls&amp;config=planets" rel="nofollow noreferrer">query the archive</a> for exoplanets with discovery method "Transit," you get inclinations mostly between 85&deg; and 90&deg;.</p>

<p>If the only available observation is a fluctuation in the star's radial velocity, then the planet's mass <strong><em>m</em></strong> and orbital inclination <strong><em>i</em></strong> are unknown, but a <a href="https://en.wikipedia.org/wiki/Minimum_mass" rel="nofollow noreferrer">minimum mass</a> <strong><em>m</em> sin <em>i</em></strong> can be computed.
The actual mass is greater unless <strong><em>i</em></strong> = 90&deg;.</p>

<p>The orbit diagram from Wikipedia can apply to extrasolar systems with one change: the reference direction is celestial north instead of &#9800;.</p>

CC BY-SA 4.0
2019-09-09T18:35:20.487Z
2019-09-09T18:35:20.487Z
Mike G
<p>This is just a quick question. In this <a href="https://en.wikipedia.org/wiki/Longitude_of_the_periapsis#/media/File:Orbit1.svg" rel="nofollow noreferrer">picture of the orbit of an exoplanet</a>, is the observer's line of sight perpendicular to the plane of reference, which is in the +z direction on the diagram? In the case when the inclination equals to <span class="math-container">$90^{\circ}$</span>, it is ideal for us to measure the radial velocity. Is that right?</p>

CC BY-SA 4.0
2019-09-07T23:00:05.517Z
That is not a picture of the orbit of an exoplanet. That is a diagram showing the meaning of various orbital elements.
James K
CC BY-SA 4.0
2019-09-07T23:04:28.923Z
@James K  But when studying an exoplanet orbiting its host star, the diagram is similar to this. There's not too much difference right?
Consideration
<orbit><astrophysics><exoplanet><binary-star>
By convention, when observing an exoplanet, is the observer's line of sight (on Earth) perpendicular to the plane of sky?
Consideration
# By convention, when observing an exoplanet, is the observer's line of sight (on Earth) perpendicular to the plane of sky?

(Asked by: username_0 on 2019-09-07)

This is just a quick question. In this [picture of the orbit of an exoplanet](https://en.wikipedia.org/wiki/Longitude_of_the_periapsis#/media/File:Orbit1.svg), is the observer's line of sight perpendicular to the plane of reference, which is in the +z direction on the diagram? In the case when the inclination equals to $90^{\circ}$, it is ideal for us to measure the radial velocity. Is that right?

**Tags:** orbit, astrophysics, exoplanet, binary-star
> **Comments:**
> **username_1**: That is not a picture of the orbit of an exoplanet. That is a diagram showing the meaning of various orbital elements.  
> **username_0**: @James K  But when studying an exoplanet orbiting its host star, the diagram is similar to this. There's not too much difference right?  

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## 1 answers

### Answer by username_2 on 2019-09-09. Score: 1 | Accepted answer

Yes, the reference plane for exoplanet [orbital inclination](https://en.wikipedia.org/wiki/Orbital_inclination#Exoplanets_and_multiple_star_systems) is our sky plane.
From the NASA Exoplanet Archive [documentation](https://exoplanetarchive.ipac.caltech.edu/docs/parhelp.html#i-l):

> The Observed Inclination is the orbital inclination with respect to the plane of the sky....
> 0 degrees correspond to an orbit in the plane of the sky, face on with respect to our line of sight.
> 90 degrees corresponds to an orbit that is edge-on with respect to our line of sight, perpendicular to the plane of the sky.

If you [query the archive](https://exoplanetarchive.ipac.caltech.edu/cgi-bin/TblView/nph-tblView?app=ExoTbls&config=planets) for exoplanets with discovery method "Transit," you get inclinations mostly between 85° and 90°.

If the only available observation is a fluctuation in the star's radial velocity, then the planet's mass ***m*** and orbital inclination ***i*** are unknown, but a [minimum mass](https://en.wikipedia.org/wiki/Minimum_mass) ***m* sin *i*** can be computed.
The actual mass is greater unless ***i*** = 90°.

The orbit diagram from Wikipedia can apply to extrasolar systems with one change: the reference direction is celestial north instead of ♈.


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