Is there any way to calculate declination of Moon for any day?
I'll subtract inclination of Earth on the ecliptic from the result. I'll get Moon's the shortest angular distance from Sun.

Thanks for all answers!

  • $\begingroup$ It seems doubtful if the subtraction that you mentioned would get you a meaningful result. Could you say more precisely what result you are trying to achieve? That would facilitate helpful answers. (And btw, what makes you think there 'must be' an equation for what you seek? That's by no means a given. A relevant calculation is no doubt possible, but likely to involve many equations, not just 'an equation'.) $\endgroup$
    – terry-s
    Commented Mar 10, 2019 at 10:09

1 Answer 1


The complexity of the formula depends on the precision you need. We can make a crude approximation with two sine waves: one for the Moon's travel eastward around the ecliptic over a tropical month

$$\delta_1 = 23.4^\circ \sin {2 \pi (u - 10.75) \over 27.32158}$$

and one for the Moon's travel north or south of the ecliptic over a draconic month

$$\delta_2 = 5.1^\circ \sin {2 \pi (u - 20.15) \over 27.21222}$$

both modified as the Moon goes faster or slower over an anomalistic month

$$u = t - 0.48 \sin {2 \pi (t - 3.45) \over 27.55455}$$

where t is the number of days since 2000-01-01 12:00 TT (J2000.0). Then

$$\delta = \delta_1 + \delta_2$$

is within 1° of the Moon's geocentric declination most of the time. For a topocentric declination, subtract 1° sin (observer's geographic latitude).

If you need higher precision, see van Flandern and Pulkkinen 1979 or Meeus 1999 for more complicated formulas.

Note: the u and t offsets above are near a 0° longitude crossing, an ascending node, and an apogee in January 2000 but are tweaked to minimize RMS differences from JPL HORIZONS lunar positions for 2000-2018.

  • $\begingroup$ $\delta_2$ is really ecliptic latitude $\beta$. An expression for ecliptic longitude $\lambda$ instead of $\delta_1$ and a transformation of ($\lambda$, $\beta$) to equatorial coordinates would be more accurate, but other sources of error in this approximation are larger. $\endgroup$
    – Mike G
    Commented Mar 19, 2019 at 15:27

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