# Tag Info

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Globular clusters occupy an interesting place in the spectrum of composite stellar systems. As you point out, they are highly concentrated populations of stars, and seem to lack any dark matter component, unlike more massive dwarf galaxies. Binary interactions become very important in simulating globular clusters, and interestingly enough (maybe ...

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In a typical position in a globular cluster (maybe halfway between center and edge), there'd be many more bright stars in the sky due to the star density. These would be distributed unevenly in the sky, with more light coming from the center of the globular cluster. Depending on the globular cluster's orbit, we might be able to see the Milky Way face-on. ...

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Well, at least you did some thinking and proposed a couple of personal thoughts instead of just asking for the answer to a homework question. Both observations you make are pertinent. Remember, a celestial sphere doesn't actually exist. It is an imaginary concept, a simplification based upon our perspective of the universe. A planetarium, whether it is an ...

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It's roughly like this: set limit 100 format object form1 "%IDLIST(1) : %OTYPE(S)" query sample (otypes = '*i*') & (bibcode='2009MNRAS.394.1338B') To get the list of the bibcodes choose "Other" / "Catalogue collection" from the menue bar. The screen should look like Then click the icon "Click to display the menu" at the left border: Here you find ...

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Let us assume the data for a globular cluster to be equivalent to that of M13. Given 300,000 stars and a radius of 1 ly, let us assume uniform density. Another assumption is to consider all stars to be Sun-like. The number density can be calculated as $$\rho = \frac{300000*3}{4\pi(1\ ly)^3} \approx 9 \times10^{-44}\ m^{-3}$$ Now, using the formula for ...

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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 ...

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There is kind of an answer over at Math. All you can do in spherical coordinates is to change the position of your "pole", i.e. you have $(1,0,0)$ in your first coordinate system, which is mapped to some $(r', \vartheta', \varphi')$ in the second coordinate system. The two angles represent a rotation, and the $r$ represents a scaling. I think since we ...

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For the second question: If you know the coordinates of the Zenith, your latitude is exactly Zenith's declination. For your longitude you can not rely on the Zenith: the same star will be at the Zenith at the same sidereous time for every place in the same latitude, so you need a clock besides the telescope. cf. ...

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It is quite easy. In fact you do not need a bolometer. You just need to perform Intensity measurements in several parts of the spectrum, and then fit these to a teorethical black body spectrum. Three uses to be enough if it does not happen that you are measuring on a spike or valley in the spectrum caused by an emission or absortion line. The black body ...

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Click on the link "SIMBAD Objects (1224)" shown in this page It seems Simbad doesn't have an specific acronym for the stars studied by Bychkov et al, because I was unable to find such acronym using Simbad's Dictionary of Nomenclature of Celestial Objects. Since I knew that ADS shows a Simbad link for some articles, I searched for articles written by ...

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If it's a binary, it's fairly simple (in comparison with a ternary system), because the stars of the binary orbit about their common barycenter in Kepler ellipses. An orbit simulator, see here; replace the central star by the barycenter of the binaries. Calculation of an Orbit from Three Observations. Kepler problem on Wikipedia. If you don't find a ...

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Of course it is! How do you think sundials work? That said, there are better, more precise ways to keep time. I don't know how to do it for an image; that is really a computational problem rather than an astronomy question. You might be better off asking it at Stack Overflow.

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In other "words", the connection between the time of transit $t_\mathrm{tr}$ of an object, its right ascension $\alpha$ and the geographical longitude of an observer $l$ is the (apparent) siderial time at Greenwich $\theta_0$ (if you know your local siderial time $\theta$, you don't need $\theta_0$ or $l$): t_\mathrm{tr} = \alpha - \theta_0 - l = \alpha - ...

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Let's think it is september 21 midnight AT GREENWHICH (you didn't specify midnight where). Let's go to http://www.csgnetwork.com/siderealjuliantimecalc.html There input september 21, midnight at Greenwich. It says Sidereal time (ST) is 00:00:6.62 or, in hours, 0.001839h This number is the difference between any star RA and its Azimuth at Greenwich. ...

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