Is there a modern accurate formula for the length of the sidereal year?

Question

Newcomb (1898) gives the following formula for calculating the tropical year:

$$365.24219879 - 0.00000614T$$

And the following for calculating the sidereal year:

$$365.25636042 + 0.00000011T$$

Where T is the number of Julian centuries (36525 ephemeris days) since J1900.0.

Since then, more accurate orbital elements for the planets have been calculated, like Simon et al. (1994). Several better formulae for the tropical year have been published since, such as in Borkowski (1991), which gives:

$$365.242189669781 - 0.000006161870T - 0.000000000644T^2$$

With T as Julian centuries since J2000.0. An even more accurate formula can be found in McCarthy & Seidelmann (2018, p. 267), which claims to have in turn calculated it from Laskar (1986):

$$365.2421896698 - 0.00000615359T - 0.000000000729T^2 + 0.000000000264T^3$$

With T as Julian centuries since J2000.0. A slightly different version can be found in Meeus & Savoie (1992):

$$365.242189623 - 0.000061522T - 0.0000000609T^2 + 0.00000026525T^3$$

Where this time, T is Julian millennia (365250 ephemeris days) since J2000.0.

However: I cannot find a single similar modern formula for the sidereal year. It seems almost incomprehensible to me that nobody would have calculated one in 100+ years.

Does such a formula exist in literature? If not, how could it be calculated?

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_{Citations:
Borkowski, K. M. (1991). The tropical year and solar calendar. Journal of the Royal Astronomical Society of Canada, 85, 121.
Laskar, J. (1986). Secular terms of classical planetary theories using the results of general theory. Astronomy and astrophysics, 157, 59-70.
McCarthy, D. D., & Seidelmann, P. K. (2018). Time: from Earth rotation to atomic physics. Cambridge University Press.
Meeus, J., & Savoie, D. (1992). The history of the tropical year. Journal of the British Astronomical Association, 102(1), 40-42.
Newcomb, S. (1898). Astronomical Papers prepared for the use of the American Ephemeris and Nautical Almanac (Vol. 6). Bureau of Navigation, Navy Department.
Simon, J. L., Bretagnon, P., Chapront, J., Chapront-Touze, M., Francou, G., & Laskar, J. (1994). Numerical expressions for precession formulae and mean elements for the Moon and the planets. Astronomy and Astrophysics, 282, 663-683.}

Answer 1

The length of the sidereal year is a function of the tropical year, and does not exist independently. The ratio of the length of the sidereal year to the length of the tropical year is (Barbieri & Bertini 2021, p. 148):

$$1296000/(1296000 - P)$$

Where P is the yearly rate of the procession along the ecliptic in seconds longitude. A formula for the precession constant is given in Liu & Capitaine (2017) as:

$$5028.796891T − 1.1111298T^2 + 0.01695523T^3 − 0.000020031T^4 − 0.000000017T^5$$

Where T is the number of Julian centuries (36525 ephemeris days) since J2000.0. However, we need the rate of change of this formula, which means taking the first derivative. This is:

$$5028.796891 − 2.2222596T + 0.05086569T^2 − 0.000080124T^3 − 0.000000085T^4$$

Since this formula is per century and not per year, the other parts of our first equation must be multiplied by 100 to match. Thus the ratio of the lengths of the sidereal and tropical years is:

$$129600000/(129600000 - 5028.796891 + 2.2222596T - 0.05086569T^2 + 0.000080124T^3 + 0.000000085T^4)$$

Now, let's use a more recent and accurate formula for the length of the tropical year. The one in Meeus & Savoie (1992) is based off VSOP87. One based off VSOP2013 is given in Institut de Mécanique (2021, p. 869-873) as:

$$365.2421904482 – 0.00006116623T – 0.000000065922T^2 + 0.0000002667909T^3 + 0.000000000947396T^4 + 0.000000000106233T^5 + 0.000000000371993T^6$$

But this one has T as Julian millennia since J2000.0. We should convert it to centuries so that it matches our other formula. This is:

$$365.2421904482 – 0.000006116623T – 0.00000000065922T^2 + 0.0000000002667909T^3 + 0.0000000000000947396T^4 + 0.00000000000000106233T^5 + 0.000000000000000371993T^6$$

Thus our formula for the length of the sidereal year is:

$$129600000/(129600000 - 5028.796891 + 2.2222596T - 0.05086569T^2 + 0.000080124T^3 + 0.000000085T^4)(365.2421904482 – 0.000006116623T – 0.00000000065922T^2 + 0.0000000002667909T^3 + 0.0000000000000947396T^4 + 0.00000000000000106233T^5 + 0.000000000000000371993T^6)$$

Which simplifies to:

$$(47335387882.08672 - 792.7143408T - 0.085434912T^2 + 0.03457610064T^3 + 0.00001227825216T^4 + 0.000000137677968T^5 + 0.0000000482102928T^6)/(129600000 - 5028.796891 + 2.2222596T - 0.05086569T^2 + 0.000080124T^3 + 0.000000085T^4)$$

The Maclaurin series of this formula to the 6th degree is approximately:

$$365.2563632882 - 0.000012380178T + 0.00000014270314T^2 + 0.0000000000409688T^3 - 0.0000000000001447616T^4 + 0.00000000000000106231T^5 + 0.000000000000000372007T^6$$

Which, for convenience, we can convert back to a per millennia value rather than per century:

$$365.2563632882 - 0.00012380178T + 0.000014270314T^2 + 0.0000000409688T^3 - 0.000000001447616T^4 + 0.000000000106231T^5 + 0.000000000372007T^6$$

This polynomial approximation is only appropriate for short term calculations, over the next century or so. Here are some years and the amount of time that the approximation is off by from the original formula:

2020 - 0.00000000000000000003 seconds
2050 - 0.000000052 seconds
2100 - 0.000053 seconds
2200 - 0.0133 seconds
2300 - 0.196 seconds
2500 - 14.41 seconds
3000 - 8.04 hours

Adding more terms to the Maclaurin series, while technically making the result more accurate, would significantly increase the amount of false precision.

However, this is all a secondary concern, since both formulas' divergence is much faster from whatever the true value is than from one another. For 2500, they give a length of over 371 days, and for 3000 they give over 738 days, which are obviously wrong results. These formulas for the sidereal year should be good for the short term, but should really only be used until more accurate versions of the procession and tropical year formulas are published in the coming decades.

It feels incredibly good to finally find an answer over a year and a half after I originally asked the question.

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_{Citations:
Barbieri, C., & Bertini, I. (2021). Fundamentals of Astronomy. 2nd ed. CRC Press.
Institut de Mécanique Céleste et de Calcul des Éphémérides. (2021). Introduction aux éphémérides et phénomènes astronomiques.
Liu, J. C., & Capitaine, N. (2017). Evaluation of a possible upgrade of the IAU 2006 precession. Astronomy & Astrophysics, 597, A83.}

Answer 2

A synodic or a day is sunrise to next sunrise and assigned or divided in $24$ hours as an average. While a sidereal day is one complete rotation of earth and equal to $1$ degree on ecliptic.

Now in a day, there is difference of $1$ degree of earth's rotation between sun and $1$ rotation of earth. As duration of a day is $1$ unit then duration of a sidereal day in terms of the day is,

$\text{P}_{sid} = 1 - \frac{1}{360} = \frac{359}{360}$ days per degree

is amount of a sidereal day in terms of a day, that is $24$ hours. So convert it into minutes gives,

$\text{P}_{sid} = \frac{359}{360}\times24\times60 = 1,436$ minutes per sidereal day.

This is about $4$ minutes less than a day that is $1,440$ minutes.

For calculation of a sidereal year, multiply $\text{P}_{sid}$ by $360$. A sidereal day is $1$ degree on ecliptic, so a sidereal year consists of $360$ sidereal day,

Number of days in a sidereal year = $\frac{359}{360}\times360 = $359 days

As there is no fraction of days in a sidereal year and a complete circle align with both its position to stars as a sidereal year and to sun as a tropical year. So there is no difference in tropical and sidereal year, thus there is no backward motion of stars. This eliminates any calculation for precision of equinox.

As calculated above, there is recession of a sidereal day by $4$ minutes to a day which accumulates as a whole day in a year. Now this recession in term of angular measurement is given by difference of angular movement of a day and a sidereal day that is,

$\frac{360}{359} - 1 = \frac{1}{359}$ degree per day $= \frac{3,600}{359}$ arcseconds per day $= 10.0278551532$ arcseconds per day.