The moon faces the earth with one fixed side. However there is a small vibration.

So I am not sure the south and north poles of the moon is fixed. Even if they are fixed points on the moon, the pole axis should point at different locations in the sky.

What are the general positions of the two points in equatorial coordinate system of our earth?


3 Answers 3


Here's a NASA movie showing Libration: http://svs.gsfc.nasa.gov/cgi-bin/details.cgi?aid=4000

As given in the details here, the position angle is the angle between the Moon's north and celestial north. This changes because of the inclination of ecliptic to equator + inclination of moon's orbit to ecliptic and plus, there might be some contribution from the inclination of moon's rotation axis to its orbital plane. Note that all these changes are apparent. The pole need NOT physically shift at all for these effects.

Having said that, whether the pole does shift, requires one to correct for all those effects and the remaining effects would be cause by some precession or nutation of the lunar poles, which is likely to be negligible given that it is tidally locked to the earth.

  • $\begingroup$ do you know the exact RA and Dec which the moon points at? $\endgroup$ Commented Jul 16, 2014 at 12:19
  • $\begingroup$ No, not really. There must be some database which has these details, but I'm not sure where you'd find it. EDIT: As Envite pointed out in the comment on the other answer, there seems to be a better explanation with some concrete numbers in that link. $\endgroup$
    – Takku
    Commented Jul 16, 2014 at 20:32
  • $\begingroup$ It is empty here. en.wikipedia.org/wiki/Poles_of_astronomical_bodies $\endgroup$ Commented Sep 9, 2014 at 12:02

The "small vibration" is Libration, and it does not imply that lunar axis changes orientation. It is caused by Moon's movement not being a perfectly circular orbit parallel to Earth's Equator with aligned axis.

Here's a description of the causes of libration from the Royal Observatory (check the linked page for graphics of each source):

Synchronous rotation and libration

The 27.3 days it takes the Moon to complete one orbit around the Earth is about the same as the time taken for it to complete one rotation. This synchronous rotation means that it always shows the same face to the Earth.

However, libration means that it is possible to see around 59% of the surface. It has three components:

i) Diurnal libration

As the Earth rotates the view of the Moon changes. An observer on the surface of the Earth sees slightly around the eastern limb of the Moon at moonrise. At moonset the same observer sees slightly more around the western limb of the Moon.

This arises from the changed position of the observer because of the rotation of the Earth. A parallax effect means that the Moon looks slightly different at moonrise and moonset.

ii) Longitudinal libration

The Moon moves in an elliptical orbit around the Earth. As a result it does not move at a constant speed (this is Kepler's second law)

However the Moon rotates at a constant speed. When it is moving faster in its orbit, this allows observers on Earth to see around the trailing edge of the Moon. When it is moving slowly it is possible to see around the Moon's leading edge.

iii) Libration in latitude

The Moon's axis is tilted by 6.7° to the plane of its orbit around the Earth. This means that sometimes the Moon's North Pole is tilted towards the Earth and sometimes tilted away.

As a result it is sometimes possible to see beyond the lunar North Pole and the South Pole is hidden. Conversely it is sometimes possible to see beyond the South Pole and the North Pole is hidden.

The combination of the three kinds of libration allows observers on Earth to see 59% of the lunar surface. However 41% remained hidden from view until the space age.

So in order to get Moon's poles position you just need to know where the Moon's center is, which is quite standard, and the position of the poles referenced to that center, which is always the same (not in RA/Dec but in true space).

There is a secondary effect of "wobbling" called Nutation, which is caused by the Moon on the Earth, not a movement on Moon itself.

  • $\begingroup$ en.wikipedia.org/wiki/Libration en.wikipedia.org/wiki/Poles_of_astronomical_bodies The main reason should be caused by the change of lunar pole which should be very small. But why is there not position of lunar poles in the second wiki link? $\endgroup$ Commented Jul 16, 2014 at 6:40
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    $\begingroup$ @questionhang en.wikipedia.org/wiki/Orbit_of_the_Moon#Inclination $\endgroup$
    – Envite
    Commented Jul 16, 2014 at 12:37
  • $\begingroup$ Yes. This helps. The rotation axis changes. However I am still not able to find a value which can specify the general position of its axis direction and the circle size draw on the sky. $\endgroup$ Commented Jul 17, 2014 at 1:27
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    $\begingroup$ @questionhang Double check the link I have just provided. The angle between the ecliptic and the lunar equator is always 1.543° and the rotation axis of the Moon precesses with the same rate as its orbital plane, but is 180° out of phase (see Cassini's Laws). This is all you need. You just need to follow the link to Cassini's Laws. Did you really investigate before asking? $\endgroup$
    – Envite
    Commented Jul 17, 2014 at 5:38
  • $\begingroup$ @Envite Thanks, I saw this. I need a RA&Dec and a general region pjojected on the sky. $\endgroup$ Commented Jul 17, 2014 at 6:21

@questionhang, the moon's celestial north pole is always 1.543 degrees from the ecliptic north pole (Earth's orbital pole), which, according to en.wikipedia.org/wiki/Orbital_pole was at RA 18h 0m 0.0s, D +66° 33′ 38.55″ at J2000.

If an error or about 1.5 degrees if small enough for you, then use the ecliptic north pole.

If not, the problem is that the moon's north pole precesses through that whole circle every 18.6 years, so it changes fairly rapidly. If one knew accurately where it was on a given date, one could to calculate where it should be now, but I don't find those numbers either. For my purposes, I am ok using the ecliptic north pole.

  • $\begingroup$ @Envite's post is right. $\endgroup$ Commented Jan 9, 2018 at 11:22

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