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9

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 ...


9

Number of stars in the observable universe There are about 100 to 200 billion galaxies in the observable universe. Galaxies range in size from a few billion to hundreds of trillions stars. Using 100 billion galaxies and 1 trillion stars in a galaxy yields 1023 stars in the observable universe as a rough order of magnitude estimate. (After getting this ...


6

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. ...


6

"The worldwide community of professional astronomers is only about 10,000; most are located in the us (with about 1,000 in the UK and 250 in Australia)." From So You Want to Be an Astronomer by Duncan Forbes. Another source stated that the number of professional astronomers is about the size of a small town.


5

Throughout man's history, we have been learning things about the universe we live in. Probably one of the most interesting discoveries that let us know how amazingly large the universe really is was not really very long ago, only in the 1920s when Edwin Hubble demonstrated that the Milky Way was not in fact the total sum of the universe, but just a galaxy, ...


5

You use Spherical Trigonometry Given $A_1$ and $A_2$ are the respective azimuthal coordinates of the two objects, and $a_1$, $a_2$ their respective altitudes, the angular seperation $\theta$ is given by $$\cos \theta = \sin a_1 \sin a_2 + \cos a_1 \cos a_2 \cos (A_1-A_2)$$


4

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 ...


4

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 ...


4

In fact, some people had looked at it more seriously recently and conducted a computer simulation to visualize the night sky as seen from within a globular cluster. The article has appeared recently in Astronomy journal. This is just one example of a typical image inside a globular cluster: Some more discussion can be found here: ...


4

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 ...


4

You can't, not with that information. However if you also have the date and time, you'll be good to go: there are plenty of sites in the web with this information, such as: RA and DEC to ALT and AZ


4

You're asking a very big question - basically, "what is the history of astronomy?" It's a very long story. Our knowledge about the size of the Universe grew slowly, bit by bit. The ancient Greeks knew that Earth is round, like a sphere, and measured its size with surprisingly good precision. http://en.wikipedia.org/wiki/Eratosthenes In the 1600s, with the ...


4

Yours is a very, very interesting work. If you publish it it may happen that 3D diagrams will be used in the future. Please publish it. But this does not mean that there is a flaw in the H-R diagram. It is correct as it is: it displays Luminosity Vs. Color and is perfect at that. And then it happens that some extra relations appear: stars are not randomly ...


4

All matter radiates (except if it's at absolute zero temperature), regardless of its composition (you got that of Mercury badly wrong). The most important form of radiation is the black-body radiation which only depends on the temperature of the material, but line emission and absorption may also be important (but depends on the composition and ionisation ...


4

I personally found Stephen Hawking's "A Breif History of Time" to be very interesting and informative. It is a little bit advanced, but is put forward in plain English. Here is a link to it on Amazon, and you can "look inside" to see part of the book and see what you think. ...


3

I feel this question is so broad, it might well be the topic of an interesting book on the history of astronomy, should someone be inclined to write it. :) Anyway, I think a few points could be made briefly. 1. Collecting data In astronomy, that means observing the cosmos. That means using an instrument of some kind, typically a telescope, and gathering ...


3

The synodic month is the "average period of the Moon's revolution with respect to the line joining the Sun and Earth". However, the Earth also moves in its orbit around the Sun during this month. From our vantage point, the Sun has appeared to move in the sky with respect to the background stars, in the same direction as the Moon moves in the sky with ...


3

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 ...


3

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 ...


3

I know that quoting Wikipedia is frowned upon here, but as there has been no other answer posted, this is what the Wikipedia article on Kepler has to say about the matter: He then set about calculating the entire orbit of Mars, using the geometrical rate law and assuming an egg-shaped ovoid orbit. After approximately 40 failed attempts, in early 1605 he ...


3

If they mention times of night then you have to figure out to what extent the time zone (BST/GMT vs the US time zones) makes any difference, but it shouldn't make more than an hour difference either way for an object that is moving sidereally with the rest of the night sky. The main difference would be that the USA is at lower latitudes than the UK. ...


2

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 - ...


2

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. ...


2

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. ...


2

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 ...


2

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 ...


2

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 ...


2

The Hertzsprung-Russell diagram is an observational tool. The axes are things that can be observed, or at least estimated reasonably well, for most stars (well, those with known distance anyway). We cannot in general measure the masses of stars - only those in some binary systems. The schematic that you show, with an arrow indicating that mass increases ...


2

RA and DEC are the coordinates on sky. RA (Right Ascension) is the angular position measured eastward in a full circle along the celestial equator. Dec (Declination) is the angular position measured from the celestial equator in either the north (positive) or south (negative) direction. $+90\,^{\circ}$ is the north celestial pole, $-90\,^{\circ}$ is the ...


2

If you browse further down the page, there are details for each of the top table data. So RA and Dec give you coordinates (Right Ascension and Declination, like longitude and latitude). There is a designation J2000.0 which tells you the zero point to which the coordinates relate (remember, the Earth wobbles, so any coordinate system tied to Earth changes ...



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