# What accounts for the discrepancies in my calculations of year lengths?

A common exercise in many introductory astronomy texts is to use the lengths of various kinds days to calculate the approximate length of the corresponding year.

For example, ratio $k$ of the length of the sidereal day to the mean solar day $D$, can be used to estimate the length of the tropical year, $a_t$ from1

$$a_t \approx \left(k-1\right)^{-1}$$

and, similarly, the ration of the length of stellar day, $k'$, along with the established value for general precession, can be used to estimate the length of the sidereal year2

$$a_s \approx \left(Pk'-1\right)^{-1}$$

where $P \approx 0.999999894$.

But both of the values I arrive at using these calculations differ from established values. The value I get for $a_t$ is off by about a quarter of a second per year,3 while my value for $a_s$ is off by nearly 20 minutes.4

What accounts for these differences? Specifically,

1. Is my reasoning sound, or have a made an error in my thinking or calculations?
2. In particular, do my methods fail to incorporate some phenomena that should be accounted for, or fail to remove some that should be omitted?
3. If there are such omitted or wrongly included phenomena, do they correspond to the differences I'm seeing between my computed values and the established ones; for example, could I use those differences to determine some constant associated with the phenomena?
4. Are these differences just a result of the fact that each of these established values (for $a_t$, $a_s$, $k$, $k'$, and $D$) are determined independently, so that it is too much to expect them to "line up" any better than they do in these results?

Please forgive the extensive footnotes, but I want to be absolutely sure that my reasoning and calculations are, at least, sound.

1: Given the average angular velocity, $\dot{\alpha }$, of the meridian with respect to the ecliptic coordinate system, it must be the case that in a mean solar day, the sun has passed through an angle in that system of$$\theta =\dot{\alpha} D-24^h$$so that it will take $$\frac{24^h}{\theta}=\left( \frac{\dot{\alpha }}{24^h}D-1\right)^{-1}$$days for the sun to complete the circuit of the ecliptic that determines the mean tropical year. And since by definition$$\dot{\alpha } = \frac{24^hk}{D}$$the length of the tropical year is$$\left(k-1\right)^{-1}$$

2: Similarly, from the average angular velocity of the meridian with respect to fixed stars along the ecliptic, $\dot{\sigma }$, we can calculate the angular progress of the sun against fixed stars over the course of a mean solar day$$\phi =\text{ }\dot{\sigma} D- 360{}^{\circ}$$and since this progress exactly corresponds to the angular progress of the earth on its orbit with respect to fixed stars (i.e. excluding precession of the ecliptic) it will take$$\frac{360{}^{\circ}}{\phi }=\left( \frac{ \dot{\sigma }}{360{}^{\circ}}D-1\right)^{-1}$$mean solar days for a complete circuit of that orbit, which is the definition of the sidereal year; where $\dot{\sigma }$ consists of two components: the angular velocity of the earth on its axis which is by definition exactly $360{}^{\circ}$ per stellar day, $D/k'$, and the rate that that axis precesses relative to fixed stars$$\dot{\sigma } = \frac{360{}^{\circ}k'}{D}-\frac{50.28792\text{''}}{a_j}=\frac{360{}^{\circ}k'}{D} P$$where, where $a_j$ is a julian year and thus$$P\approx 0.999999894$$so that it will take$$\left(Pk'-1\right)^{-1}$$days for the sun to complete full circuit against the background stars, which is the definition of the duration of the sidereal year.

3: Using the IERS values for $D$ and $k$ I arrive at a value for $a_t$ which is too small by $0.265284\text{ s}$.

4: Using the IERS values for $D$ and $k'$ and the rate of precession, I arrive at a value for $a_s$ which is too large by $18.7156\text{ min}$, while, confusingly, ignoring the effect of precession (by setting $P=1$) produces a closer value, that comes up short by $1.6943\text{ min}$, which is more than an order of magnitude smaller ($4.1478\text{"}$) than the expected effect of omitting precession over the course of a sidereal year ($50.2888\text{"}$).

Note that I'm pretty sure my discrepancies are not a result of daily variations (due to the obliquity of the ecliptic and the eccentricity of the Earth's orbit) in the values of ecliptic velocities, since these all average out over the course of a year to exactly the values used.
For example, the daily motion of the celestial sphere is around the celestial pole, not the ecliptic pole, so even if the velocity of the meridian around the pole is constant, the velocity with respect to the ecliptic will vary as the meridian sweeps through the ecliptic at varying declinations. Though the average ecliptic velocity over a complete circuit of the ecliptic will be $\dot{\alpha}$, the meridian completes less than a full circuit of the ecliptic in completing a full equatorial circuit (due to precession), and more than a complete circuit in a mean solar day (due to the sun's movement). However both of these omitted and added segments of the ecliptic follow the sun in the course of a year through the full range of declinations, so that over the course of a tropical year the average is $\dot{\alpha}$. So in the calculations for the tropical year, it is the fact that things average out for complete equatorial circuits that lets me use $24^h$ as a constant in the ecliptic system (as in the first equation in note 1), while it is the fact that things average out for mean solar days that let me apply $\dot{\alpha}$ to $D$.

• Excellent work formatting the footnotes :-) – David Z Nov 16 '12 at 0:14
• Forgive me if I just didn't look carefully enough, but it doesn't seem to me that these calculations account for the slowing of the earth's rotation over time due to tidal interaction with the moon, correct? – kleingordon Nov 16 '12 at 9:20
• @kleingordon: I've made no separate allowance for that, so unless the definitions of the "constants" themselves do, that's not accounted for. – orome Nov 16 '12 at 13:58
• Well, I'm not sure this fully answers your question, but you may wish to consult en.wikipedia.org/wiki/%CE%94T and en.wikipedia.org/wiki/Leap_second – kleingordon Nov 16 '12 at 21:02
• @ChrisWhite: I'm not sure how I come up with these. Mostly from just reading a bit too critically through treatments that are more basic than I'm making them. I expect all of my questions are dealt with directly in more advanced texts (which I'm probably not prepared to grasp). The tropical year discrepancy might be a floating point issue; the sidereal year discrepancy seems way too large though. – orome Nov 19 '12 at 5:13

Your first equation is correct: $$a_T = (k-1)^{-1},$$ and this equation is exact. Different authors use different precession values, and this causes the small discrepancy. As for the sidereal year, you shouldn't include the precession, so $$a_S \approx (k'-1)^{-1}.$$ However, this equation is not exact. The reason for this is subtle, and the explanation involves quite a bit of work. The most recent precession values are found in the paper Expressions for IAU 2000 precession quantities (Capitaine et al, 2003). From equation (37) and (39), we have \begin{align} \varepsilon &= 84381''.406 \ +\ \text{H.O.T.},\\ \psi_A &= 5038''.481507\;t \ +\ \text{H.O.T.},\\ \chi_A &= 10''.556403\;t \ +\ \text{H.O.T.}. \end{align} (I've ignored the higher-order terms). The time $t$ is expressed in Julian centuries of 36525 days, counting from Jan 1 2000, 12$^h$ UT1, which has Julian date 2451545.0. In other words, $$t = \frac{\text{JD}-2451545.0}{36525}.$$ $\varepsilon$ is the obliquity, while $\psi_A$ and $\chi_A$ are the precessions. There are indeed two precessions: $\psi_A$ is the luni-solar precession of the equator on the ecliptic, due to the torque from the Sun and the Moon. However, the planets also have an influence: they perturb the orbit of the Earth, which leads to a precession of the ecliptic. The quantity $\chi_A$ expresses this precession with respect to the equator, and is called the planetary precession of the ecliptic on the equator. The situation is sketched in the figure:

The combination of these is the general precession, but crucially, there are to ways to express this precession: the general precession on the ecliptic is $$p_A = \psi_A - \chi_A\cos\varepsilon = 5029''.796195\;t.$$ But we can also define the general precession on the equator $$p'_A = \psi_A\cos\varepsilon - \chi_A = 4612''.160408\;t.$$ Both have their use: $p_A$ for the orbital motion of the Earth on the ecliptic, and $p'_A$ for the daily rotation of the Earth around its axis. The rotation period on Jan 1 2000 is listed in The IAU resolutions on astronomical reference systems (Kaplan, 2005), p. 16: $$P = 86164.0989036903511\;\text{s}.$$ Thus $$k' = 86400/P = 1.00273781191135,$$ consistent with the value given in the Table. The Earth Rotation Angle is $$\theta = 2\pi(0.7790572732640 + 1.00273781191135\;d),$$ where $d$ is the Julian date, counting from Jan 1 2000, 12$^h$ UT1. The relative angle to the Sun is then $$\theta' = \theta - 2\pi d = 2\pi(0.7790572732640 + 0.00273781191135\;d),$$ and is related to the stellar day. The Sidereal Time is the sum of $\theta'$ and the precession on the equator: $$\text{GMST} = t_0 + \theta' + p'_A,$$ where $t_0$ is the GMST on Jan 1 2000, 12$^h$ UT1 (there's also a small nutation term of $-0''.003872$ which I've ignored. See eq (42) in Capitaine et al). Note that we need $p'_A$, not $p_A$. Since $2\pi=360\cdot3600''=24\cdot3600\,\text{s}$, and $t=d/36525$, one obtains \begin{align} \text{GMST} &= t_0 + 129598159''.760640\;t + 4612''.160408\;t\\ &= t_0 + 129602771''.9210482\;t\\ &= t_0 + 8640184^\text{s}.794736\;t, \end{align} which is equivalent to the right ascension of the so-called Fictitious Mean Sun: $$\alpha_\text{FMS} = \alpha_0 + 8640184^\text{s}.794736\;t.$$ The tropical year is the time needed for $\alpha_\text{FMS}$ to increase by $24^\text{h}=86400^\text{s}$, so that \begin{align} a_T = \frac{86400\cdot36525}{8640184.794736} &= 365.242188109\ \text{solar days}\\ &= 366.242188109\ \text{sidereal days}. \end{align} Therefore $$k = \frac{366.242188109}{365.242188109}=1.002737909345,$$ which is very close to the value listed in the Table.

Now let's look at the orbit of the earth, with respect to the (moving) mean equinox. This is called the Dynamical Mean Sun, and by definition its ecliptic longitude varies in the same way as $\alpha_\text{FMS}$: $$\lambda_\text{DMS} = \lambda_0 + 129602771''.9210482\;t.$$ A tropical year is thus also the time needed for $\lambda_\text{DMS}$ to increase by $360^\circ$. On the other hand, the ecliptic longitude with respect to the fixed vernal point of Jan 1 2000 is the difference between $\lambda_\text{DMS}$ and the precession on the ecliptic: \begin{align} \lambda' &= \lambda_\text{DMS} - p_A\\ &= \lambda_0 + 129602771''.9210482\;t - 5029''.796195\;t\\ &= \lambda_0 + 129597742''.1248532\;t \end{align} A sidereal year is then the time needed for $\lambda'$ to increase by $360^\circ$: \begin{align} a_S = \frac{360\cdot3600\cdot36525}{129597742.1248532} &= 365.25636345\ \text{solar days}. \end{align} The difference between these values of $a_T$ and $a_S$ and those listed in the Table is due to the fact that Simon et al., 1994 used different precession values; in particular, he used $p_A=5028''.8200\;t$.

Note however that the formula $$a_S \approx (k'-1)^{-1}$$ is not exact. The reason is that the difference between a stellar day and a sidereal day is related to $p'_A$, while the difference between a tropical year and a sidereal year is related to $p_A$.

The system of recommended constants is inconsistent. The durations of the sidereal and tropical years are based on the IAU 1976 value for the general precession: p1976 = 5029".0966 per Julian century $\approx \frac{2 \pi}{a_t} - \frac{2 \pi}{a_s}$.

The other problem is the conventional duration of the mean solar day is a little shorter than the actual duration. Using units of seconds (instead of days), the actual duration of the mean solar day turns out to be: D = 86400.000553 s.

Using the assumed values for the solar day and precession:

$a_t = \frac{D}{k-1} = \frac{2\pi}{ \frac{2\pi}{a_s} + p_{1976} }$

$a_s = \frac{D}{p k-1} = \frac{2\pi}{ \frac{2\pi}{a_t} - p_{1976} }$

where $p = 1 - \frac{ p_{1976} D }{ 2 \pi k' } = .9999998940488$ (as before)