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I understand that the right ascension of the mean sun changes (at least over a specified period) by a constant rate, but where is it zero? I had naively assumed that it would be zero at the most recent vernal equinox, but when I try to calculate the equation of time using this assumption and true sun positions, all my values are about 7.5 minutes larger than they should be.

When (at what date and UT time) is the right ascension of the mean sun 0? And why?

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I have erased my answer for the moment, because I am having some doubts. I will undo the erasing when I am sure about the question. –  Eduardo Guerras Valera Dec 15 '12 at 1:12
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The EOT changes sign in the September Equinox, so there must be another explanation. It is amazing how many websites are supposed to explain the EOT calculation but leave the important point of your question unclear (proving that they blindly copy from each other, both the theory and the algorithms without really understanding what they are doing). I am visiting the library of my old faculty next Monday, to clarify the question in a real book. I had always understood like you, that the Mean Sun would be synchronized with the Apparent Sun on the Vernal Equinox, but it seems it is not so. –  Eduardo Guerras Valera Dec 15 '12 at 1:59
    
@EduardoGuerras: Yes, I think that's a good point. All these sources seem to borrow blindly from each other, leaving an amateur like me lost as to where the real foundation is. –  raxacoricofallapatorius Dec 15 '12 at 5:14
    
Right ascension of the mean sun is 0 only when it reaches vernal equinox. If we keep pondering over the same thing, we are never reaching the error, since that assumption is probably correct. It would be good if you tell us what calculations you took. The problem does come and has come probably from the assumptions you made for eccentricity of Earth's orbit. Last thing. Did you calculate the true sun's position too? OR you calcuated mean sun's position and measured true Sun's position. –  Cheeku Mar 15 '13 at 0:00
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2 Answers

No, the right ascension of the mean Sun is NOT zero at the vernal equinox. It is in fact nearly identical to the ecliptic longitude of the mean Sun (the difference is due to UT vs ephemeris time), and this is defined such that it coincides with the ecliptic longitude of the apparent Sun when the Earth is at perihelion. So that should be the starting time to calculate the equation of (ephemeris) time.

Using an ephemerides site like this one, one can check that the Earth was at perihelion on Jan 2 2013, at 4:37 UT, and will return to perihelion on Jan 4 2014, at 11:58 UT. The difference between these is called an anomalistic year. The same site shows that on Jan 2 2013, at 4:37 UT, the Sun had an apparent right ascension

$$ \alpha_p = 18^\text{h}51^\text{m}56^\text{s}, $$

and the corresponding ecliptic longitude is

$$ \lambda_p = \tan^{-1}(\tan\alpha_p/\cos\varepsilon) = 18^\text{h}47^\text{m}46^\text{s}, $$ where $\varepsilon=23^\circ 26' 21.4''$ is the obliquity of the ecliptic. The equation of ephemeris time is then $$ \Delta t = M + \lambda_p - \alpha, $$ where $M = 2\pi(t-t_p)/t_Y$ is the mean anomaly, $t_p$ is the moment of perihelion and $t_Y$ is the length of the anomalistic year. The quantity $M + \lambda_p$ is the ecliptic longitude of the mean Sun. The apparent right ascension $\alpha$ can be calculated from $$ \begin{align} M &= E - e\sin E,\\ v &= 2\tan^{-1}\left[\sqrt{\frac{1+e}{1-e}}\tan\frac{E}{2}\right],\\ \lambda &= v + \lambda_p,\\ \alpha &= \tan^{-1}(\tan\lambda\cos\varepsilon), \end{align} $$ with $E,e,v,\lambda$ the eccentric anomaly, orbital eccentricity, true anomaly and apparent ecliptic longitude (see also wiki). This should give you an approximation of the equation of time, accurate to a few seconds. If you want a really detailed calculation, have a look at this paper.

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Thanks, this is promising and may explain what I'm seeing in my related question, though I'm not clear right now (it's been a while since I asked) how to modify my "naive" EOT derivation to accommodate this information to confirm that it solves my problem there (help with that would be an answer). –  raxacoricofallapatorius May 14 '13 at 13:38
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Try the same formula, but replace $t_0$ with $t_p$, and set $\alpha_M(t_p)=\alpha_p$. For 2012, aphelion was on Jan 5, at 31:13 UT, and $\alpha_p=19^\text{h}1^\text{m}3^\text{s}$. –  Pulsar May 14 '13 at 17:48
    
Of course, I meant perihelion in my previous comment. –  Pulsar May 14 '13 at 19:22
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Im not entirely convinced that light-time is the culprit here. The Earths orbit is not circular, its elliptical, as the Earths orbital eccentricity is non-zero. Because of this, the Earths distance to the sun throughout the course of the year is not constant, and therefore the time delay due to lights finite speed will not be a constant either. If you have computed this value of 7.46988 to a high enough accuracy, there should be some variation throughout the course of the domain, a year. According to you, there is not.

Might I suggest a possible explanation? Its trivial. And if Im right, its a naive mistake. Are you taking into account your longitudinal offset from Greenwich (or from your time-zones defining longitudinal meridian)?

You said Right Ascension, so your location on the Earth should not be a variable, as Right Ascension is measured with respect to the stars. If you truly are measuring RA then Im at a loss for help and you can disregard my previous paragraph.

As I understand it, any body that passes through the Vernal equinox has a RA of zero. However, there is a distinction between mean and apparent vernal equinox. There are precessional and nutational factors involved.

Furthermore, one thing I am VERY fuzzy on, is whether or not we use the true vernal equinox at all, or if RA is based on the original vernal equinox defined thousands of years ago, known today as the First Point of Aries. They are well out of synch today.

These previous two paragraphs, if not taken into account, could cause significant errors that appear constant in the short term.

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I've split off the aspects of this question dealing with my EOT derivation into a separate one. See my comment on Eduardo's answer. –  raxacoricofallapatorius Dec 14 '12 at 21:13
    
Well, as far as I know RA is calculated using the first point of aries as reference, and not present day vernal equinox. I'm still trying to figure out where 7.5 minutes comes from though. –  Kitchi Dec 31 '12 at 21:12
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