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48

The answer depends on your definition of a day. If you define a day as we usually define it in the Earth (time between the Sun is at noon or average time between sunrises, a 24 hours day in Earth), the length of a day in the Moon is the synodic period of the Moon and it takes 29.530589 days (29 d 12 h 44 min 2.9 s). Anyway, if you define a day as rotation ...


27

You make a great point. The reason behind the discrepancy between the dates is due to a complicated history behind it. The calendar is based on the calendar created by ancient Romans, which is based on one Moon cycle. One lunar cycle is 29.53 days. www.universetoday.com/20620/lunar-year/ which does not evenly divide into the 365.25 days of ...


20

Defining that a day is a rotation on the axis, 1 moon day is approximately equal to 27.3 Earth days. This link on moon rotation may interest you.


18

Leap years exist for two reasons: There are not an integer number of days in a year. People perceive a need to keep the seasons where they are on the calendar. Given the above, there is no way to avoid leap years, or something similar. Defining the calendar year as being a fixed number of days (e.g., 365 days) would result in the seasons shifting by one ...


18

The synodic period of the moon is $29.53$ days, a little shorter than a calendar month, which is on average about $30.4$ days. This is slightly longer than its orbital period, but corresponds to the periodic visual appearance of the moon as viewed from Earth. I mention this to make it clear that we should be forgiving of a little imprecision. Conventionally,...


15

$t$ signifies time; see the Wikipedia article for spacetime, and then the subsection for 4-vectors. The basics are pretty natural to understand. Suppose something happens, an event, like an apple falling off of a tree. In order to tell someone else about it you need the three space coordinates $x, y, z$ and the time coordinate $t$. Without all four, you ...


14

For locating objects in the sky, the horizontal and equatorial coordinate systems are commonly used. These systems describe the position of some object in the sky very well, but do not explain the position of the object in space (if you know the distance you know "where" the object is, but this is relative to the equatorial/horizontal plane and is tricky to ...


10

This Wikipedia article states (correctly) that "The Julian Day Number (JDN) is the integer assigned to a whole solar day in the Julian day count starting from noon Greenwich Mean Time, with Julian day number 0 assigned to the day starting at noon on January 1, 4713 BC, proleptic Julian calendar (November 24, 4714 BC, in the proleptic Gregorian ...


9

Everything is in motion in our Galaxy. The Sun has executed some 20 laps of the Galaxy since it was born and may have migrated inwards or outwards to some extent. The Sun's location has nothing to do with the Gould belt or vice versa. The Gould belt stars formed just 30 million years or so ago. The position of the Sun relative to the Gould belt is a ...


8

LST = 100.46 + 0.985647 * d + long + 15*UT They don't explain what the two constants are (100.46 and 0.985647), could anyone explain what those constants are and how they were calculated in the first place please? There are three constants there, 100.46, 0.985647, and 15. The value of 100.46 degrees is the value needed to make the expression yield ...


7

Another valid definition is the time between earthrises instead of sunrises, since the moon is in orbit around earth. In that case, there is no direct concept of a day, because the moon is tidally locked to earth.


7

So I think it should be JD2451558. An extra 13 days. What's going on here? The switch from the Julian calendar to the Gregorian calendar involved two changes. One change was the frequency at which leap years occur. The Julian calendar had a leap year every four years, making a Julian year 365.25 days long on average. The Gregorian calendar has 97 leap years ...


7

Yes, for stars in our galaxy, if a star is 736 light years away, the light took 736 years to reach here. For very very distant objects, you need to account for the expansion of space. The light might have been travelling for 10 billion years, but in that time space has expanded and so the distance that you would find if you froze time and measured would be ...


7

In general, the closer you are to the primary, the shorter the orbital period, except of course that you can't orbit inside the primary. So, taking the primary to be a sphere of radius $R$ and density $\rho$ we find that the mass is $$\frac{4}{3}\pi R^3\rho$$ so that the acceleration due to gravity at the surface is $$G\frac{4}{3}\pi R\rho$$ For a surface-...


7

Since the Julian day is defined as a continuous count of seqential days starting at Jan 1 4013 BC, and there is no limit to the number of integers available for the count, there is no specified maximum. If you are asking when the Julian Day Number will no longer be useful, possible answers are when there is no one left to count it, or when it is replaced ...


7

I'd like to pick up the point you ask about in your edit. Suppose we have a homogeneous cube and we want to construct some axes so we can locate points in the cube. The obvious choice is to put the origin at the centre and use axes that are like this: Suppose we have two points $A$ and $B$ then we can locate them by their position vectors $\mathbf a$ and $\...


6

Since the rate of lengthening of the day on Mars is three orders of magnitude less than that of the Earth for a first go at calculating how long it will be before the day lengths are equal we can ignore the change in day length on Mars. The difference in day lengths is $\approx 40$ minutes. The Earths day lengthens by $\approx 1.7$ ms/100 years. Therefore ...


6

Analemma is a diagram showing the deviation of the Sun from its mean motion in the sky, as viewed from a fixed location on the Earth. Note that it says nothing about time- it can be any fixed time. Analemmas created at different times of the day have slightly different shapes. For example, see the solar analemmas taken at the same place at different times. ...


6

That should be in western China, since all of China uses Beijing's time zone. Reference: https://en.wikipedia.org/wiki/Time_in_China --- Edit below --- In response to the comment by @adrianmcmenamin: I'll leave this as guesswork since I simply don't know the peculiarities of every timezone there is. Here's a back-of-the-envelope calculation for local noon ...


6

Why all the stunts with leap days and leap seconds, etc... when we now have atomic clocks? Science exists to serve mankind, not to rule it. Calendars were one of the first concepts developed by humankind, arguably predating writing. Timekeeping is even more important to modern society than it was to the ancients. You're not going to get rid of calendars. ...


6

Calculation of the subsolar point is relatively straightforward. The actual point moves at nearly 500m/s in a circle, as the Earth spins. The linked website does the calculation in a few lines of python using a free library, The value is probably correct to within a few 10s of kilometres. But you are looking for a different order of accuracy: If you (for ...


5

To make a clock you need something that changes predictably with time, and stars are nearly constant, making them poor clocks. The stars do move across the sky each night, but since the morning stars of winter become the evening stars of spring, and the sky looks different in different locations you need to know the date of the image, the location of the ...


5

Time on Mars is measured in sols. Venus could be a little trickier, depending on your application. I'm not aware of any timekeeping system for any other planets. There hasn't been a need for one yet. So, again depending on your application, you could come up with various different possible schemes. Speculation (based on the OP's input regarding ...


5

To do so, I'm following what's described in this document and implementing C++ code. That is a very old document you are using. It's using the 1982 IAU precession model. There have been multiple updates since then. The latest set of changes are working toward eliminating the concept of Greenwich Sidereal Time. (In software terms, GMST and GAST are ...


5

It's historical. TT is a successor to The International Astronomy Union's ephemeris time, and the offset between TT and TAI is a result of getting TT to match ET when they switched. Excerpts from https://en.wikipedia.org/wiki/Terrestrial_Time: Terrestrial Time (TT) is a modern astronomical time standard defined by the International Astronomical Union, ...


5

Leap seconds keep the clock in sync with solar time of day. Leap years keep the calendar in sync with the seasons. The two (leap seconds and leap years) have nothing to do with one another. We have leap seconds because a day is now a tiny bit longer than 86400 seconds. We have leap years because a year (tropical year) is about 365.24219 days long. If we ...


5

As a simplification, think of the length of the year as where the Earth is in its orbit. After 1 year of 365 days, the Earth has not returned to the same spot in its orbit. If we did not have leap years, then the date when seasons occur would change. Leap years keep the seasons in synch with the date. After 365+365+365+366 days (that is, 3 regular years and ...


5

As the solar time Wikipedia page illustrates, we can split a solar day into a constant 23h56m to complete a rotation relative to the stars (a sidereal day) and a variable ~4m to compensate for one day's orbital motion around the Sun. The Sun's apparent motion around the equator determines how much more than 360$^\circ$ the Earth must rotate from one solar ...


5

The rotation of the Earth, UT1 and the Earth Rotation Angle are strictly measured agaist extragalactic light sources. Indeed, nowadays, UT1 is reckoned by using the definition you mentioned, a direct linear relationship with the Earth Rotation Angle between the intermediate celestial and terrestrial frames instead of the actual ficticious mean Sun, because ...


5

This question asks in effect: Is there a discrepancy, perhaps progressive, between the official time-measure UT1, and some other credible measure of mean solar time? This is harder to answer than it might first seem. UT1 is the current official representative of mean solar time: it is doubtful that any independent measure of mean solar time (other than ...


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