# Understanding Astronomical Days & Measuring a Year's Length

• I
• fizzy
In summary, the different meanings for the word "day" in astronomy are explained. The days are human noon to noon days, so presumably these equate to mean solar days of 86400 seconds (by definition). However, there again there may be qualifications due to tidal drag etc. It is explained that astronomers in modern times use JD as defined by UT1 (or UTC) (which are, by definition solar days), but when doing calculations, a constant day of 86400 SI seconds (in Geocentric Coordinate Time TGC) is more typical. There is a Supplementary question asking for the best average value for Saros centred on J2000. The length of a Saros cycle as a function of time in units
fizzy
I found an excellent article on the different meanings there are for the word day in astronomy. Very will written:

https://www.physicsforums.com/insights/measuring-how-many-days-are-in-a-year/

Now often astronomers work in Julian days to have a consistent time scale without skips and bound.s. Various parameters are often presented as polynomials using JD , J year of even J millennia.

The days are human noon to noon days, so presumably these equate to mean solar days of 86400 s ( by definition ) but there again there may be qualifications due to tidal drag etc.

http://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html

Code:
                                    J            E           ratio

Sidereal orbit period (days)      4,332.589     365.256      11.862

Tropical orbit period (days)    4,330.595     365.242      11.857

(E=365.256 thus must be msd.)
pJ=4332.589 msd.

What I would like to know is how to compare orbital period of Jupiter , polynomial functions of say lunar precession period in JD and eclipse series Saros periods.

Supplementary question: what is the best average value for Saros centred on J2000 ? I need a value compatible with the previous, not a derivative approximation in year,months,days and hours, or N synodic months etc.

I looked at data from NASA and averaged over series of similar eclipses but always came up with results a little shorter than values found stated elsewhere for Saros.
https://eclipse.gsfc.nasa.gov/SEsaros/SEsaros135.htmlthanks to anyone who can clarify this infernal mix of different "days".

fizzy said:
The days are human noon to noon days, so presumably these equate to mean solar days of 86400 s ( by definition ) but there again there may be qualifications due to tidal drag etc.
I think that astronomers in modern times use JD as defined by UT1 (or UTC) (which are, by definition solar days (for UT1 it is not a mean)). However, when doing calculations, a constant day of 86400 SI seconds (in Geocentric Coordinate Time TGC) is more typical (since a uniform timescale is assumed in analytical and numerical calculations). For short time periods and purposes that do not require great precision there is no significant difference between the two (currently the difference accumulates to 0.9s every few years see UT1-UTC on the International Earth Rotation Service website).

fizzy said:
What I would like to know is how to compare orbital period of Jupiter , polynomial functions of say lunar precession period in JD and eclipse series Saros periods.
I think the best way is to convert everything into JD (or possibly Julian centuries) and then compare (here, since long time periods are involved, a JD should be taken to be 86400 SI seconds). Then convert back to local time at the end of your calculation (if necessary). (I hope this answered the question you were asking here, it was not entirely clear to me what the question was precisely.)

fizzy said:
Supplementary question: what is the best average value for Saros centred on J2000 ? I need a value compatible with the previous, not a derivative approximation in year,months,days and hours, or N synodic months etc.

I looked at data from NASA and averaged over series of similar eclipses but always came up with results a little shorter than values found stated elsewhere for Saros.
For Saros cycles, I found a paper Five Millennium Catalog of Solar Eclipses, which describes the methodology and statistics of the Saros data that you linked to. Table 5-12 (pg. 61 of the paper, pg. 69 of the pdf) gives the length of a Saros cycle as a function of time in units of draconic and anomalistic months (a Saros cycle is defined to be 223 synodic months exactly). This information pertains to length of time that a Saros series lasts. If, instead, you just want the length of a Saros cycle in days, that is given at the beginning of section 5.3 (pg. 48/56) as approximately 6585.3223 days (in the year 2000).

Note that in the paper Terrestrial Dynamical Time (TDT or TD in the paper, or TT in most other places) is used (see section 1.2.4 pg. 2/10 for first mention, section 2.3 pg. 9/17 for a description).

Thanks very much Iso. That is a great help. I will have more questions once I've gone through that in more detail, but just wanted to drop you a courtesy thank you.

## 1. What is the difference between a sidereal day and a solar day?

A sidereal day is the time it takes for a planet to rotate once on its axis, with respect to the stars. This is approximately 23 hours and 56 minutes. On the other hand, a solar day is the time it takes for a planet to rotate once on its axis, with respect to the Sun. This is approximately 24 hours.

## 2. How is a year's length measured in astronomy?

A year's length is measured by the time it takes for a planet to orbit around its star. This is called the orbital period and it varies depending on the distance of the planet from its star. For example, Earth's orbital period around the Sun is approximately 365.25 days.

## 3. Why is a leap year necessary in our calendar?

A leap year is necessary to keep our calendar aligned with the Earth's orbital period around the Sun. Since Earth's orbital period is not exactly 365 days, adding an extra day every four years helps to keep our calendar in sync with the seasons.

## 4. Can a planet have more than one moon?

Yes, a planet can have more than one moon. For example, Jupiter has 79 known moons, while Earth has only one. The number of moons a planet has depends on factors such as the planet's size and distance from its star.

## 5. How do scientists measure the length of a day on other planets?

Scientists use a variety of methods to measure the length of a day on other planets. One method is by using spacecrafts equipped with instruments that can track the planet's rotation. Another method is by observing the planet's surface features, such as volcanoes or mountains, and measuring the time it takes for them to rotate into view. Additionally, scientists can also use radio waves to measure the rotational period of a planet.