What is the definition of a "pole" of a celestial body?

Earth's pole is defined as it's rotational pole. The North and South Poles are the two points on Earth where its axis of rotation intersects its surface.

Apparently, (according to everywhere I've read), the poles of astronomical bodies are determined based on their axis of rotation in relation to the celestial poles of the celestial sphere. But this is an inadequate definition.

What about Pluto? Pluto's axis of rotation does not intersect its surface. Pluto is tidally locked with Charon, and Pluto orbits a barycenter that is located outside of Pluto. And yet Pluto has a North and a South Pole. Why are the North and South Poles of Pluto where they are, as opposed to any other location on the surface of Pluto?

How are these poles defined for Pluto?

What is the definition of a "pole" of a celestial body?


2 Answers 2


You are mixing the rotation of the body around a barycenter with its moons, and the rotation of the body around its own center.

For a bound rotation like Pluto and Charon both have to have the same rotation period - yet both have to be present. Thus Pluto's axis of rotation of course intersects its surface - it rotates around its own axis at exactly the same speed as it orbits around the common barycentre with its moon Charon. Were it different, it would not be a bound rotation.

The North pole usually is defined as the place where the body's rotation axis (about it's own center of mass) with the right-hand-rule intersects the surface (thus the vector of the rotational momentum in positive direction intersects the surface when placed in the centre of rotation).

There is virtually no non-rotating body.

  • 1
    $\begingroup$ For Planets and their Satellites, the IAU defines the North Pole as the one that shares the same Celestial Hemisphere as the Earth's North Pole. So while you can use the Right-Hand-Rule to pinpoint the North Pole for most of the planets, it will give you the wrong official answer for Venus and Uranus. Dwarf planets, minor planets, comets, and their satellites, it's the Right-hand rule, as you say. $\endgroup$
    – notovny
    Dec 5, 2020 at 0:40

I'll just add a supplement to @planetmaker's answer.

As long as a body is distinct and not connected to anything else, it will have a center of mass.

If the body is roughly spherical its center of mass will be near it's middle.

The body's rotational axis by definition passes through its center of mass, and is parallel to it's own angular momentum vector.

If there are two bodies orbiting each other, to a good approximation (but not exactly, see Which mass distributions guarantee two bodies have non-Keplerian orbits? Which non-spherical distributions still allow noncircular Keplerian orbits?) we can consider one's center of mass orbiting around the other's center of mass, and the center of both of their masses to be the pair's barycenter.

Their mutual rotation and orbital angular momentum will be defined by another axis passing through that barycenter, which may be inside one of them (like the Earth-Moon or Sun-Jupiter system) or in space between them like the Pluto-Charon system. It doesn't matter.

One is the rotation of a single body around its own center of mass, the other this the rotation of two centers of mass about their common center of mass. Apples and oranges.

If however a body is crazy-shaped, like a big letter "C" perhaps, that center of mass might be outside the body. That poses a conundrum for placing the "pole" of the body since the axis of rotation will not intersect the body's surface. IN that case the body will simply not have a true pole. But poles are constructs and not fundamental. It still has a center of mass and a rotational axis and angular momentum vector, and those are really what matter.

I don't know where the center of mass of comet 67P/Churyumov–Gerasimenko is exactly, nor where its poles are, so I think that that would be an excellent follow-up question!



  • $\begingroup$ You should have written angular velocity vector rather than angular momentum vector. The angular velocity vector is (somewhat) observable while the angular momentum vector is not observable at all. Angular momentum is inferred from observations that relate to angular velocity combined with estimates of the object's inertia tensor. $\endgroup$ Dec 5, 2020 at 11:50
  • $\begingroup$ @NilayGhosh for questions about known bodies (such as the one I propose as the end) those should be asked as new questions; I'm just talking about definitions here. For other than C shaped bodies, how about O shaped! :-) The [tennis racket theorem]() tells us that bodies tend to rotate about either their axes of largest or smallest moments of inertia, so if the O and C rotate about an axis perpendicular to the page there are no poles and the extremes are at the equator. But if they rotate around a vertical axis they both have poles, and if horizontal, the O has a two opposite poles but C... $\endgroup$
    – uhoh
    Dec 5, 2020 at 22:48
  • $\begingroup$ @DavidHammen don't be so impatient, if we watch the motion of well-characterized bodies evolve over a hundred million years we'll learn about their angular momentum as well :-) $\endgroup$
    – uhoh
    Dec 5, 2020 at 22:49
  • $\begingroup$ @uhoh Thank you. I have asked: What should be the “poles” for irregular shaped bodies? $\endgroup$ Dec 6, 2020 at 9:09
  • $\begingroup$ This should not be the accepted answer. $\endgroup$ Dec 6, 2020 at 9:20

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