Is it a coincidence that Earth's axial tilt and maximum declination are both approximately 23 degrees?

Mechanically speaking, do planets physically cross the Solar equator, or is declination due to the axial tilt angle? The later is difficult to visualize given variety of planetary axial tilts i.e Mercury 0, Jupiter 4, Earth and Saturn 23 degrees approximately.

I'll try and give a concise example. At JPL's Horizons you can download data of RA and DEC for all planets and objects placing either Earth or the Sun at the center. This will calculate the Declination of a planet geocentrically or heliocentrically.

The calculations suggest all planets will obtain a maximum declination North and South between 23° and 28° (i.e Mercury reaches 28° when viewed from the Earth). Earth's axial tilt (obliquity) at 23.45° matches its maximum declination N/S.

However, Mercury's Axial tilt is 0° and Jupiter 3.13° so if declination is the angle between the ecliptic plane and the celestial equator how do Mercury and Jupiter obtain maximum declination of 23° and 28° N/S?

I'll appreciate any help clarifying this.

  • 2
    $\begingroup$ Maximum declination, axial tilt, and obliquity are synonyms. $\endgroup$ Jul 14 '19 at 19:25
  • $\begingroup$ What do you mean by the "Earth's maximum declination"? Declination is determined by the Earth, so the other planets and Sun have nothing to do with the "definition" of declination. $\endgroup$
    – JohnHoltz
    Jul 15 '19 at 1:43
  • 1
    $\begingroup$ As @david-hammen notes: if you mean the Sun's maximum declination, this is the same thing as the Earth's axial tilt with respect to the Earth-Sun plane, so it's not a coincidence these numbers are equal. $\endgroup$
    – user21
    Jul 15 '19 at 4:12

A planet's apparent position in the sky is determined by orbital geometry alone. The tilt of its axis only affects its orientation as we see it in a telescope.

It's easier to think of the solar system in terms of ecliptic coordinates. The Sun's geocentric latitude is always near 0°, as is the Earth's heliocentric latitude. A planet's orbital inclination is measured relative to the ecliptic, and its maximum excursion in latitude is related to that inclination. Each planet crosses the ecliptic northward at its ascending node and southward at its descending node.

For example, Saturn's orbit has a semimajor axis of 9.6 au and is inclined 2.5° to the ecliptic. Its heliocentric latitude runs that far north or south of the ecliptic over its 29.5 year period. Earth's orbit brings us 1 au closer at opposition, so Saturn's geocentric ecliptic latitude can reach ±2.8°.

Mercury's orbit has a semimajor axis of 0.4 au and is inclined 7°, so its heliocentric ecliptic latitude range is ±7°. At inferior conjunction ~0.6 au away, its geocentric ecliptic latitude range is about ±5°.

The transformation from ecliptic to equatorial coordinates is a 3D rotation by the 23.4° obliquity, not a simple sum. Also a planet's orbit and the Earth's axis may be tilted in different directions. Saturn's ascending node is 114° longitude east of the J2000 vernal equinox; its orbital inclination and our axial tilt are that far out of phase. A JPL HORIZONS ephemeris for 29 years of Saturn oppositions and conjunctions gives geocentric declinations in a range of ±22.6°.

Saturn's 26.7° axial tilt relative to its orbital plane does not affect its declination at all. The effect we do see is that at different points in its orbit, its rings appear open to the north, open the south, or edge-on and slanted east or west, as shown here.

Since you mentioned the solar equator, it's inclined 7.2° to the ecliptic. The planets do cross it, but their declinations are not related to it either.

  • $\begingroup$ Much obliged Mike G. Just shows the complexity of Solar system motion while moving forward at 20kilometers per.sec. Planets moving more helically than elliptically. $\endgroup$
    – Dave Hook
    Jul 21 '19 at 14:39
  • $\begingroup$ @DaveHook It's complex enough if, as this answer does, you ignore the Sun's motion around the galaxy. $\endgroup$
    – Mike G
    Jul 21 '19 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.