What is the difference between solar degrees and Calendar days. Is there any formula to convert one to another.

for example Starting from 1/1/2000

1/1/2000 + 300 calendar days=some date (easy)

1/1/2000 + 300 solar degrees=?

  • $\begingroup$ By "solar degrees" do you mean the Right Ascension (RA) of the Sun ? $\endgroup$ – astrosnapper Oct 20 '19 at 19:51
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    $\begingroup$ Some tips to write a better question. Don't say "I need to know". You want to know (for some reason that may be important to you). Do explain the context: Why do you want to know this, if it is a "homework" task please say so. Do describe what you have already done to answer this question. You might have a text book or done a google search. Do check your grammar, spelling and punctuation before posting. You don't need to add "Thanks" or "Please help" those are implict. Have a look at the help center to see more of how to ask. $\endgroup$ – James K Oct 20 '19 at 20:53
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    $\begingroup$ @astrosnapper Degrees of ecliptic longitude is another option. TheExpat Wanderer: In either case, if you just want a rough number use degrees = days × 360 / 365.25. If you need more accuracy you have to account for the variation in the Earth's orbital speed. It moves fastest at perihelion (early January, and slowest at aphelion (early July). $\endgroup$ – PM 2Ring Oct 20 '19 at 20:57
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    $\begingroup$ This actually seems related to an astrology based system for timing stock market prices. I'm not sure that there is much actual astronomy here. Please edit to explain if I'm wrong. One solar degree is the time taken for the earth to move one degree in its orbit of the sun. An acurate calculation of the time of a solar degree would need an orbital model for the Earth. $\endgroup$ – James K Oct 20 '19 at 21:03
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    $\begingroup$ My formula is accurate to a few degrees. For more accuracy, see the Wikipedia article on the Sun's position. You want the Sun's ecliptic longitude. $\endgroup$ – PM 2Ring Oct 21 '19 at 12:15

Warning: If you try to time the stock market using the position of the sun, you will lose money as often as you win it. If there was a system that worked, everyone would be using it (and then it wouldn't work any more)

You want to calculate the ecliptic longitude of the sun. This roughly increases by about one degree per day, but that is neither exact nor constant. It varies due to the elliptical orbit of the Earth.

There is a formula. If $n$ is the number of days since midday UTC Jan 1 2000.

Then the longitude of the sun in degrees is

$$ 280.46^\circ + 0.9856474^\circ n \\ + 1.915^\circ \sin(357.528^\circ+0.9856003^\circ n) \\ + 0.02^\circ\sin(355.056^\circ +1.9712006^\circ n) $$

Applying that formula you will find that the sun will be at 300 degrees after 306.07785 days on November 7th 2000 at 13:52

The formula (derived from Wikipedia)is empirically derived to fit the observed orbit of the Earth. We approximate the position of the sun as an initial angle (280.46) plus a linear increase (0.9856474n). If the Earth's orbit was perfectly circular, that would be the complete formula. But as it is elliptical there is periodic variation. That is approximated by a combination of two sinusoidal terms. These can be understood as the first two terms in a Fourier series that could describe the actual variation. You can see that the second term has a very small coefficient (0.02) and double the frequency of the first. In principle, further terms could be added, but this formula is now already accurate to a small fraction of a degree.

The formula is good until about 2050.

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    $\begingroup$ What is the source of that formula? $\endgroup$ – Mike G Oct 22 '19 at 1:39
  • $\begingroup$ thanks James K for a valuable addition - could you explain what are these numbers. $\endgroup$ – TheExpat Wanderer Oct 23 '19 at 16:57

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