This is just a quick question. In this picture of the orbit of an exoplanet, 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?

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    $\begingroup$ That is not a picture of the orbit of an exoplanet. That is a diagram showing the meaning of various orbital elements. $\endgroup$
    – James K
    Sep 7, 2019 at 23:00
  • $\begingroup$ @James K But when studying an exoplanet orbiting its host star, the diagram is similar to this. There's not too much difference right? $\endgroup$ Sep 7, 2019 at 23:04

1 Answer 1


Yes, the reference plane for exoplanet orbital inclination is our sky plane. From the NASA Exoplanet Archive documentation:

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 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 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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