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I understand that inclination is the measure of how tilted an orbit is. But is an inclination of 0 degrees like this

enter image description here

or like this?

enter image description here

I know there are many formulas relating inclination, including calculating the true mass, but I am unsure as to which reference frame inclination is measured from. I understand that if one is 0 degrees, the other must be 90 degrees.

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    $\begingroup$ See also en.wikipedia.org/wiki/Orbital_inclination $\endgroup$
    – jcaron
    Commented May 21 at 12:02
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    $\begingroup$ Note that on your figures, the point of view is different, but the orbit is actually (nearly) the same. $\endgroup$
    – jcaron
    Commented May 21 at 12:06
  • $\begingroup$ Also see: astronomy.stackexchange.com/questions/35471/… $\endgroup$ Commented May 22 at 3:00
  • $\begingroup$ Orbital Inclination can be respect to many reference planes - ecliptic(orbital plane around Earth), Sun's equator or invariable plane. See the link from jcaron. $\endgroup$ Commented May 22 at 12:46

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For the solar system the reference plane is usually the plane of the ecliptic, so the Earth has a zero inclination. Sometimes the "invariant plane" defined by the angular momentum of the solar system as a whole is used. As its name suggests, the invariant plane doesn't change (but the ecliptic plane can be perturbed by the gravity of other planets). More rarely the spin axis of the sun can also be used.

For exoplanets the vector from the star to the sun defines 90 degrees. So a zero degree inclination means that the planet orbits its star face on to the Earth, whereas a 90 degree inclination means the orbit is "edge on". Nearly all planets found by the transit method have an inclination of about 90 degrees. Since the inclination is defined relative to the sun, it isn't related to the angle between the spin of the star or the angular momentum of the system.

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    $\begingroup$ Note that this definition is the opposite of the galaxy inclination. $\endgroup$
    – pela
    Commented May 20 at 10:27
  • $\begingroup$ For completeness you may want to add the case of satellites of the planets (natural and artificial), where the reference plane is the equatorial plane, which is the plane perpendicular to the axis of rotation of the central body, so 0 degrees = orbit in the equatorial plane and 90 degrees = polar orbit. $\endgroup$
    – jcaron
    Commented May 21 at 12:05
  • $\begingroup$ @pela for galaxies face-on is 0 degrees and edge-on is 90 degrees, which seems to be the same as in this answer. $\endgroup$
    – Kyle
    Commented May 21 at 16:04
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    $\begingroup$ @pela Ah I see what you mean. I never really thought of those as comparable inclinations because the Solar System one is not really about viewing angle but angular offset from a reference plane... I'm struggling to articulate it but it somehow feels qualitatively different. $\endgroup$
    – Kyle
    Commented May 22 at 7:25
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    $\begingroup$ @pela Inclination is the angle between one plane and a reference plane. For Solar System orbits, it's the angle between the orbital plane and (usually) the plane of the ecliptic. For external systems (exoplanetary orbits, rotating stars, galaxies), it's the angle between the orbital or equatorial plane and the sky plane (which is perpendicular to the line of sight). $\endgroup$ Commented May 22 at 8:17

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