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23

I feel sure this is a repeat, but couldn't immediately find it. The only things in the night sky we can see (with the naked eye) that are not part of our own Galaxy are (on a good night) the Andromeda galaxy and the Large and small Magellanic clouds. Every individual star brighter than $V=6$ and visible to the naked eye is in the Milky Way. So it would ...


16

It wouldn't be too apparent, but there are a few objects that you can see in good viewing conditions with the naked eye that would disappear. Here they are in order of brightness. I marked the objects in a reddish color. The Large Magellanic Cloud, apparent magnitude 0.9, located in the constellation Dorado. Only visible from the southern hemisphere. ...


11

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 ...


10

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 ...


8

Solar and lunar eclipses occur about equally often - between two and four times per year. However, when you do not intentionally travel around the world chasing solar eclipses, you are more likely to observe more lunar eclipses. The reason is that solar eclipses can only be observed from a comparatively small area while lunar eclipses can be observed from ...


7

It all comes down to the brightness of objects (not their size). For all intents and purposes we can assume that the most distant galaxies and the small, but much closer, objects in the Oort clouds are unresolved point sources.The Oort cloud objects are too faint to see, with JWST, but it should be able to see bright galaxies and quasars even at 13 billion ...


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.


6

It depends a bit on how precise you would want to be. A very good discussion on how to calculate the orbits of solar system objects is given in the book by Jean Meeus, Astronomical Algorithms (1999), which is at an advanced amateur level. At professional level you have the Explanatory Supplement to the Astronomical Almanac by Urban and Siedelmann. For ...


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 $$\cos \theta = \sin a_1 \sin a_2 + \cos a_1 \cos a_2 \cos (A_1-A_2)$$


5

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 ...


5

The distribution of metallicities appear to be more evenly spread out in logspace than in linear space. The reason for this can be ascribed to there being no preferred scale for the abundance of a given element; rather they span several orders of magnitude. The same can be said for instance about the distribution of dust grain sizes, the distribution of the ...


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

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 ...


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

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. ...


4

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 ...


4

The concept of a month derived from the amount of time it takes for the moon to cycle from new moon to new moon (which is roughly 29.5 days). The modern month has experienced changes from this original concept due to trying to fit a standard number of months within a solar year. If you divide the lunar month into quarters, each quarter is approximately 7 ...


4

The distance you are referring to is the "light travel distance". It is the length of time that light has been travelling from the distant galaxy to us. When the light was emitted, our Galaxy was much closer to that distant galaxy. What has happened in the meantime is that the space between the galaxies has expanded. Furthermore we can ask where is that ...


4

There are many resources online, so I'm making this a community wiki answer. Please feel free to add to it! If you want to visualize the stars/planets/etc (as viewed from Earth or another location), you are looking for planetarium software: https://en.wikipedia.org/wiki/Planetarium_software If you want accurate positions for stars/planets/etc, you are ...


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

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 ...


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 ...



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