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49

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 ...


29

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 ...


21

Yes, you are right. We don't only see the Sun 8 minutes in the past, we actually see the past of everything in space. We even see our closest companion, the Moon, 1 second in the past. The further an object is from us the longer its light takes to reach us since the speed of light is finite and distance in space are really big.


21

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.


20

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,...


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 ...


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 ...


12

The formula in the wikipedia article explicitly uses integer division with round toward zero. Python's integer division uses round toward negative infinity (i.e., floor). The wikipedia article formula repeatedly uses (m-14)/12. This evaluates to -1 for months 1 and 2 (January and February), zero otherwise. You can use the wikipedia article formula in python3 ...


10

And that is why you don't do the calculations in a frame that is moving at lightspeed. If you have two observers that are moving relative to each other you can use the Lorentz transformation to change between their frames of reference. But if one of the observers is a photon the lorentz transformation becomes singular, because $\gamma$ is infinite. Simply, ...


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 ...


10

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 ...


9

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. ...


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 ...


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

The year is not defined as a multiple of days. The year is the time for the Earth to orbit the sun once. This is not a constant amount of time. So it is not defined in seconds or days or any other length of time. However it only varies a little. The (synodic) day is the amount of time for the Earth to rotate once relative to the Sun. This is also not a ...


7

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 ...


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 $\...


7

You are essentially asking the following: if someone falls from the Earth from some way beyond the event horizon of a black hole, how long after they have left can an observer on Earth still signal to them with a light beam? The answer of course depends on exactly how far the Earth is from the black hole. It is also often forgotten that it is not just light ...


7

The formula is correct. However, the "problem" is with Python1. https://stackoverflow.com/questions/5535206/negative-integer-division-surprising-result. Here is the fix to the formula: jd = (1461 * (y + 4800 + int(float(m-14)/12))) // 4 + (367 * (m - 2 - 12 * (int(float(m-14)/12)))) // 12 - (3 * ((y + 4900 + int(float(m-14)/12)) // 100)) // 4 + d -...


7

Julian dates offer one nice feature: All astronomical observations recorded by humankind have a positive timestamp. A key downside of Julian dates is that current timestamps on computers that use 64 bit IEEE Standard for Floating-Point Arithmetic (IEEE 754) have a resolution of 40 microseconds. Almost all computers use the IEEE 754 floating point standard. ...


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

The modern 7-day week was adopted in the late Roman Empire, replacing an 8-day "nundinal week" that had previously been in use mainly for market intervals. As far as I know, there was no particular reason for the 8-day length other than that it was a convenient size to separate market days. In the early Empire, Christians and Jews both used a 7-day week for ...


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

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

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 ...


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