15

A system projected into the sky Pretend all the stars are painted on the inside of a large ball, and you are at the center of the ball. The imagined ball is called the Celestial Sphere. If the Earth wasn't blocking the lower half of your view, and the Sun wasn't making the blue sky so bright, you would be able to look at the stars in any direction. First, ...


14

The equatorial coordinate system is very similar to the system used on a globe or on maps. To specify a point on a sphere or a globe you need only two numbers. These are the longitudes and latitudes. On a globe of the earth you have longitudes from -180º to +180º and latitudes from -90º to +90º. The south pole lies at -90º latitude, the north pole at +90º ...


12

The easiest way to think of equatorial coordinates is by extending lines of latitude/longitude (the geographic coordinate system we usually learn in school) out into space. The earth's equator becomes the celestial equator, and the north and south poles become the celestial north and south poles, respectively. By doing this you have a fairly simple way to ...


7

Assuming you mean the angle between the meridian line through A and the great circle that goes through points A and B, then it goes something like this. Define vectors from the origin to A and B assuming they lie on a unit sphere, such that $x_A= \cos \delta_A \cos \phi_A$, $ y_A = \cos \delta_A \sin \phi_A$ and $z_A= \sin \delta_A$, and similar for B. Here ...


5

The answers to this question are very clear and much better than I could have written. I'm not sure what you mean by "the stellar time at Greenwich at 0h UTC is 22h20min", but I'm assuming that you mean that that is the amount of time since the culmination of Aries over the Prime Meridian (Greenwich). That being the case, the Right Ascension (RA) of your ...


5

For computer software, the easiest way to take a sphere (and/or hemisphere) and flatten it into a flat shape (usually a rectangle) is the equi-rectangular projection (also known as the plate carrée), because it has the simplest formula relating pixels and coordinates: $$x = w \times \frac\lambda{360} + \frac w2$$ $$y = -h \times \frac\phi{180} + \frac h2$$ ...


4

I'm not sure it qualifies as "simple", but, using http://idlastro.gsfc.nasa.gov/ftp/pro/astro/hadec2altaz.pro (and some additional calculations/simplifications): $ \begin{array}{|c|c|c|c|} \hline \text{Event} & \text{Time} & \phi & Z \\ \hline \text{Any} & \text{t} & \tan ^{-1}(\cos (\lambda ) \sin (\delta )-\cos ...


4

The position angle P of a body ($\alpha_1, \delta_1$) with respect to another body ($\alpha_2, \delta_2$) can be calculated from $$tan(P)={sin(\Delta\alpha)\over cos(\delta_2)tan(\delta_1)-sin(\delta_2)cos(\Delta\alpha)}$$ where $\Delta\alpha = \alpha_1-\alpha_2$. If the denominator is negative, the position angle lies in the range of 90 to 270 degrees. ...


4

You are correct that topocentric coordinates are for the position of "close" objects, corrected for observing from the Earth's surface instead of the theoretical center. The topocentric and geocentric coordinates would use the same system (J2000, true, etc.) There is no convention to use one in favor of another (but there may be preferences in some cases).


3

Constant for every star with different declination's. Though you would have to assume the the declination is such that it can have an azimuth of 90 each, not all stars will. Depending on your latitude some stars will never get to an azimuth of 90, they will either be always be further north or south.


3

DecRA is the decimal right ascension RAh, RAm, RAs are the hms form $${\rm DecRA} = {\rm RAh}\times 15.0 + {\rm RAm}/4.0 + {\rm RAs}/240.0$$ $$ {\rm RAh} = {\rm INT}({\rm DecRA}/15.0) $$ $${\rm RAm} = {\rm INT}(({\rm DecRA}-{\rm RAh}\times 15.0)\times 4.0)$$ $${\rm RAs} = ({\rm DecRA}-{\rm Rah}\times 15.0 - {\rm RAm}/4.0)\times 240.0$$ where INT is the ...


3

The number 18.697374558 was the value of GMST (in hours) at noon UT on 1 January, 2000. The number 24.06570982441908 reflects the fact that it takes slightly less than 24 hours (23.934469591898 hours to be precise) for a star to the appear to rotate by 360 degrees. This is a consequence of the fact that the Earth orbits the Sun. Our 24 hour day is based on ...


3

The values are more steady for objects close to Ecliptic poles, but in genenral, they change by a few arcseconds every year. Yes, we lack an absolutely fixed frame of reference.


3

0 RA is the direction from the center of the Earth to the vernal equinox. You've already imagined the plane in space created by projecting the equator into space. Now consider the plane formed by the mean orbit of the Earth; this is the ecliptic. The two planes intersect in a line. The ray from the center of the Earth toward where the Sun is in March is the ...


3

Short answer, no, they don't. Longer answer, it's complicated. There are, in essence, five different measures for the centre of a galaxy cluster, based on different physical properties of the cluster, and occasionally not in agreement with each other. BCG, the brightest cluster galaxy. It should sink to the bottom of the gravitational well, but that's only ...


3

The calculations are not trivial, but they are encapsulated in software libraries such as pyephem, which has examples of finding rise, set and transit If you want to understand how these are calculated, you can readnthe source, which is based on the xephem application. One detail which complicates the calculation is the refraction caused by the atmosphere, ...


3

The complexity of the formula depends on the precision you need. We can make a crude approximation with two sine waves: one for the Moon's travel eastward around the ecliptic over a tropical month $$\delta_1 = 23.4^\circ \sin {2 \pi (u - 10.75) \over 27.32158}$$ and one for the Moon's travel north or south of the ecliptic over a draconic month $$\delta_2 =...


3

The stars should fit $r = \sqrt{x^2 + y^2 + z^2}$ but seem to be plotted in the $x = r$ plane instead. Conventionally the x axis is at (α=0h, δ=0°), the y axis is at (α=6h, δ=0°), and the z axis is at δ=+90°. Using right triangles instead of isosceles triangles, Then $h = r \cos \delta$, and $$\begin{align} x &...


2

The declination of the point overhead (zenith) is the same as the observer's latitude. The RA of the point transiting zenith at any given time is the equivalent of the local sidereal time. (Alternatively, the local sidereal time is RA of the observer's meridian.)


2

After quite an extensive search of dictionaries and books on astronomical nomenclature I came across this article that has a comprehensive review of celestial coordinate systems. And upon reading page 84 (8/14) I came across the answer to your question. It would appear that the reason declination is used, as opposed to inclination, is that declination was ...


2

You are forgetting parallax. Something which is very distant and lies on the celestial equator will have a declination of 0°, but if it is nearby then its declination will only be 0° to an observer located along Earth's equator. From your location in Zagreb, something like the ISS, even if it is crossing Earth's equator at a longitude of ~16° will appear ...


2

If you want a website rather than a software package like James suggests, I recommend use the INGT object visibility tool which has all 3 inputs you request and produces an elevation plot with rise and set times, maximum altitude etc. An example is shown below:


2

The Sun at solar noon (meaning that it is on the meridian) will be in one of these directions: Due south if the declination of the Sun is less that your latitude. (For Brasilia, that occurs from Nov 6 to Feb 5, approximately) Directly overhead (at the zenith) if the declination of the Sun is the same as your latitude. (For Brasilia, that occurs Nov 5 and ...


2

I think I mis-read your question before. Let me rephrase it to make sure I understand. In step D (moving the scope from Sirius to M42), that process took 5 minutes to complete. Or maybe it took 30 seconds to complete, but you then took a 4.5 minute coffee break :-). So during that 5 minutes of time since you calibrated the setting circle, M42 has drifted ...


2

It's not hours, minutes, and seconds, its degrees, arcminutes and arcseconds. There are 360 degrees in a circle, 60 arcminutes in a degree and 60 arcseconds in an arcminute. Since the planets lie close to the ecliptic, and the ecliptic tilted at 23 degrees relative to the equator, the declension of the planets rarely exceeds +-23 degrees, giving an ...


2

The formula you gave is to find the hour angle of the star while setting, not the setting time itself. Suppose the hour angle of star is $HA\star$, and $RA=\alpha$,then the Local Sidereel Time is given by $LST=HA\star+\alpha$. At the time of setting, $HA\star=10^h1^m30.2^s$. Thus, $LST=13^h16^m30.2^s$at the time of setting. Now, $RA\odot=6^h$ on $\text{June}...


2

There are no standards for representing right ascension and declination apart from the abbreviations RA and Dec, and the symbols $\alpha$ and $\delta$. The use of decimal degrees, dms, °'", hours and decimals, hms etc. depend on the [author's] style and intent. For example, determining transit times from local sidereal time and right ascension is easier ...


2

Given a date and time, the position of the Moon can be calculated to provide the declination and right ascension. The sub-point of the Moon (the point on the Earth at which the Moon is at the zenith) is as follows: latitude = declination of the Moon longitude can be found by calculating the local mean sidereal time (LMST) that equals the Moon's right ...


2

The ecliptic is the plane of the Earth's orbit around the Sun. From the Earth's point of view, the Sun appears to migrate along the ecliptic, eastward one cycle per year. The ecliptic and equator intersect at the equinoxes. The Sun appears to cross the equator northward at the vernal equinox (♈) around March 20 and southward at the autumnal equinox (&#...


2

With stars extending to ±90° on the horizontal axis and rather sparse near those limits, I suspect that RA and Dec are swapped. Try this instead: plt.scatter(df['rarad'], df['decrad'], s=1, color='red') Also if you want the plot to resemble the night sky, filter the star list by apparent magnitude, and reverse the horizontal axis so it increases ...


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