# Tag Info

35

This can get a bit confusing, because "arcminute" and "minute" are both sometimes used in celestial coordinate systems but mean two different things. An arcminute is 1/60th of a degree, and an arcsecond is 1/60th of an arcminute. That's simple enough, and when talking about small angular distances, it's often much handier to refer to ...

15

Your trigonometry book isn't wrong: both "minute" and "arcminute" can refer to $\frac1{60}$ of a degree. It's certainly a very good idea to use the term "arcminute" when referring to $\frac1{60}$ of a degree, but it's not essential if there's no ambiguity, eg, in a static geometry problem where there's no mention of time. The ...

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 ...

7

The area of a triangle enclosed by 3 stars on the celestial sphere, in square degrees, is given by: $$A = \frac{180}{\pi}\times E$$ Where E is the spherical excess and is equal to the sum of all angles of the triangle minus 180°. Problem is, we don't know the angles of the Summer Triangle and instead we have to use an alternate equation for the spherical ...

6

Fortunately things aren't as entangled as you thought. The J2000 epoch does not depend on the equinox; it is simply 12:00 TT on January 1, 2000. The time of vernal equinox is unrelated to sunrise anywhere; it is just when the Sun appears to cross the celestial equator northward as seen from Earth. The vernal equinox point ♈ is located at the ...

6

Question 1: How do sidereal time and RA correlate together? Think of the sky as a globe. The constellations and lines of right ascension (RA) are painted on the globe. Because of the Earth's rotation, it appears that the sky globe rotates east to west once every 23 hours 56 minutes 4 seconds. The meridian is the line of RA that goes from due south to the ...

5

The sun's position is irrelevant to when a star will be in its meridian; the Sun's coordinates in the RA/dec grid does change over the year. A star's position is fixed in that system and it will be always at the exact same sidereal time that it culminates (as that's the definition of sidereal time). The difference of the sidereal day to the mean solar day is ...

5

The north celestial pole doesn't have a right ascension. It's like the north terrestrial pole not having a longitude. Asking for it creates an exception. In any case, the RA of a point, other than the poles, isn't dependent on time or the observer's location. Your simulator has an error in its error handling.

4

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 ...

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

The Vernal Equinox is defined by the point where the sun's path across the sky, the Ecliptic, crosses the Celestial Equator. It is always going to be on the Celestial Equator. Given the various ways that objects in orbit over Earth are perturbed, any set of keplerian orbital parameters that describe the orbit of any object over the Earth (including the Moon!...

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).

4

Most of your assumptions are correct. However, right ascension and declination are an equatorial system, while (heliocentric or geocentric) longitude and latitude are an ecliptic system. The reference planes do not coincide, even though the main reference point, the vernal equinox, does—this is because that point is where the ecliptic crosses the equator ...

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 &...

3

There are several meanings to "vernal equinox", and I think you have not picked the right sense here. The vernal equinox is here taken to be a point on the celestial sphere. It is the point where the projection of the Earth's equator to the celestial sphere and the projection of the Earth's orbit intersect. (There are of course two points, the other is the ...

3

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 ...

3

It's a bit late but hopefully it will help others in the future. The calculations can be found on this site, article from Keith Burnett (keith@xylem.demon.co.uk) The first problem is that you use J2000 as time reference but it's not what the site takes as time reference! "Get the days to Dec 31st 0h 2000 - note, this is NOT same as J2000" So you ...

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

An Earth-centered observer looking at that point on Earth’s surface would also be looking toward the point that is the zenith for that location on Earth. Thus, you just need to find the local sidereal time (LST) for that Earth location. That gives you the right ascension on the meridian for that place and time. The meridian passes through the zenith, so ...

3

Wikipedia has formulas for the area of a celestial triangle in steradians (which they call the "excess" and write $E$) in terms of the angles, the side lengths, and side-angle-side. The angle formula is the simplest, but calculating the angles is tricky. It may be easier to calculate the side lengths and use the side-side-side formula. You can ...

3

Note: I'm accepting this as the answer because the main cause of the problem was that I was counting the days wrong. But there is another more important, more fundamental problem which was correctly pointed out by planetmaker and PM2Ring so make sure you read their answers/comments as well. I figured what I've been doing wrong. I was counting the days until ...

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

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

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

The issue appears to be in your understanding of the separation function. You state that it calculates the separation (angle) between those 2 planets with respect to their "x" projection in a plane. however, the documentation states The separation() function computes the angle in degrees between two bodies, as measured by their right ascension and ...

1

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 ...

1

At noon on 1 Jan 2000 (UT) the sun's RA & dec were (J2000): 18hrs 45min 9.36s -23deg, 2 m, 8.2s This is per Guide9 software.

1

Zodiac signs are equal 30° divisions of the ecliptic. In the Western tropical zodiac, these are linked to the equinoxes and solstices, and named after the nearest constellations at the time Ptolemy wrote about them. The equinoxes precess westward 1.4° per century, so the signs currently lie 25° to 30° west of their namesake constellations. ...

1

Will you know the date and location from where the image is taken? If you are sucessfully using the big dipper already you probably using this info. There is a "degeneracy" between date and time. The sky rotates once a siderial day (23hrs 56 min) and once a siderial year (which is 20min longer than tropical year: sun to same position). So the sky tonight at ...

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