Is there an easy way to tell at what angle the solar equator is wrt the horizon at any given time of the day? I would think that it rotates 180 degrees from sunrise to sunset, but not in a linear fashion. But solar noon should always be 0 degrees wrt the horizon right?


3 Answers 3


Depends on the precision you need.. The sun's plane of rotation coincides with the plane of rotation of the solar system.
And for the latter, there is already a name in astronomy, the Ecliptic. Depending on your latitude on the planet $\delta$, the ecliptic plane cuts through your local horizon in east and west at an angle of $\alpha = 90°-\delta - 23.5°$.
This is the maximum height of the sun at midday and the sun's height varies thus between 0 and $\alpha$.

I think this would then be the angle you're looking for.

Another recent answer pointed out the angle between solar axis of rotation and the Ecliptic is $7.25°$. Translating that into the angle unter which one sees the solar equator, it is still $\alpha$, up to a small correction factor of $- \frac{R_{\odot}}{1 AU} \sin(\delta + 23.5°)$ which can be maximally $-0.05°$.

  • $\begingroup$ How does it evolve over time though? This is just sunrise? $\endgroup$
    – Coolcrab
    Apr 17, 2018 at 10:45
  • $\begingroup$ @Coolcrab: Maybe If you made a sketch of the geometrical situation, that would help you? The angle is nearly the same as to the center of the sun, which is why I wrote 'depends on the precision". And the latter angle is known. $\endgroup$ Apr 17, 2018 at 12:16

The angle between the observer's zenith, any given celestial object, and the north celestial pole is the object's parallactic angle. This angle $q$ depends on the object's hour angle $H$ and declination $\delta$, and the observer's geographic latitude $\varphi$: $$ \tan q = {{\sin H} \over {\tan \varphi \cos \delta - \sin \delta \cos H}} $$ At local apparent solar noon, $H_\odot$ = 0 and $q_\odot$ = 0 as you expect.

The Sun's declination is roughly $$ \delta_\odot \approx \varepsilon \sin \left({2\pi {D - 79 \over 365.25}}\right)$$ where $D$ is the day of the year and $\varepsilon$ is the 23.44$^\circ$ tilt of the Earth's axis.

The position angle $P$ of the Sun's north pole relative to the north celestial pole depends on both $\varepsilon$ and the Sun's 7.25$^\circ$ axial tilt relative to the ecliptic plane. These two components combine somewhat out of phase, roughly $$ P \approx -26.3^\circ \sin \left(2\pi {{D - 5} \over 365.25}\right) $$

Meeus's Astronomical Algorithms gives more precise formulas for the above in terms of the Sun's ecliptic longitude $\lambda_\odot$, taking the Earth's orbital eccentricity and other factors into account.

The angle of the Sun's equator with respect to the observer's horizon is the same as the angle of the Sun's north pole with respect to the observer's zenith, $P - q_\odot$.


I think that if you want to determine the angle between the solar equator and the horizon on Earth at a specific location, there are three components to take into account:

1) the angle between the local zenith direction at the specific location and the geocentric polar axis (angle q in the drawing below). It depends on your latitude, it would be 0 degree at the Geographic Pole and 90 degree at the equator.

2) the angle between the geocentric North Pole and the solar rotation North Pole (angle P) measured eastward from geocentric north. It can be between -26.31 and +26.31.

3) the apparent tilt of the sun's axis of rotation with respect to the ecliptic (angle B0) which can be between -7.25 and +7.25. This angle is also the heliographic latitude of the central point of the solar disk.

The angles P and B0 can be found in an ephemeris table for a given date and time.

Here an illustration: enter image description here

  • 1
    $\begingroup$ 1) also depends on solar hour angle. 2) looks like a combination of Earth equator obliquity to ecliptic and 3). $\endgroup$
    – Mike G
    Apr 20, 2018 at 12:32

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