# Tag Info

4

It's almost 100% stars. In good conditions, you can see perhaps 2000 stars. (There are about 6000 naked-eye visible stars; of these, 3000 are above the horizon at any time, and about 1000 are hidden because they're too close to the horizon and blocked by the atmosphere.) The number of non-star objects you can see without assistance is tiny in comparison: ...

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This is only a rough estimate. But the number of supernovae in the Milky Way is approximately 2-5 / 100 years (here and here). There is approximately $3 \times 10^{11}$ stars in the Milky Way. Provided the supernovae rate did not change much through the history (which is a big if that probably doesn't work), the total number of supernovae would be ...

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Galaxy Map has detailed information on 5000+ stars. You can download the exel spreadsheet here Don't know if it helps. Good luck ;)

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The classification of stars using spectral class is a very useful classification when considering the properties of (the atmosphere of) a star at that moment. If you consider the different stages in human development (embryo, fetus, infant, toddler, etc...), for instance, here one person also continuously changes its class. So it is not ...

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Why isn't this so? If the mass of the star is about 1/4 the mass of our sun, then the core's (where fusion occurs) temperature and pressure will never be enough to fuse anything other than hydrogen into helium. The death of this star may be something like a fire dying. The flames disappear for a while, then you may get one or 2 popping back into ...

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The temperature in the core of the star may be sufficient to fuse hydrogen into helium but the temperature near the surface will not be high enough. During the main sequence the stellar core will be transformed into a helium core but the hydrogen in the outer layers will not have undergone fusion and will remain hydrogen.

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It would be good if you referenced your sources, because you may be misunderstanding them. We'd be able to see what they actually say, and help you understand them. Nucleosynthesis of iron does not use more energy than it produces. It is, however often referred to as the heaviest element created in fusion that results in more energy produced than consumed. ...

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In addition to Count Iblis wonderful answer, consider that supernovae in other galaxies can sometimes be seen with low-powered binoculars. I seem to remember mention of a naked-eye extra-galactic supernova, but I cannot find the reference.

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The reason that this brown dwarf is so cold compared to other stars is that it was never on the main sequence. More specifically, it never reached the hydrogen-1-burning phase of its life. Most stars (including our Sun) fuse elements together (beginning with hydrogen-1), which generates heat. Brown dwarfs do not do this, and thus are much colder than other ...

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Wikipedia reports the age of HD 140283 as the figure you cited, 14.46±0.8 billion years. Now look at the reference it refers to. This is the paper by Bond et al that presents measurements of some characteristics of the star and the subsequent calculations, as well as a lengthy section on where the error came from (800 million years is, after all, quite a lot ...

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If you'd somehow be able to freeze a star and stop nucleosynthesis (fusion) then it would not be a star anymore. By definition, a star needs a temperature high enough to support fusion. This is why brown dwarfs are not true stars. There are of course stellar remnants (white dwarfs, neutron stars) that have actually crystallised (and are therefore in a solid ...

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The Sun will never run out of Hydrogen. This is a common misconception. At this moment the Sun is fusing Hydrogen into Helium. This fused Helium remains at the core until it will reach a critical mass. At this point the core will begin to collapse. This collapse increases the temperature and pressure around the core where Hydrogen is being fused causing the ...

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I don't know if this is what most astronomers use, but it is certainly a method that could be used. It also makes for some interesting photos. Gravitational lensing was one of the great predictions of Einstein's theory of general relativity, and was actually one of the first pieces of evidence for it - see the solar eclipse of 1919. Gravitational lensing is ...

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Iron spheres There is a hypothesis that after about $10^{1500}$ years (many star generations) iron stars (iron sphere star remnants) will form. This would require all the elements lighter than Fe to fuse into Fe and all the elements heavier than Fe to decay into Fe. Also proton is required not to decay for this hypothesis to work. Mass limit Only stars ...

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they use Newton's law of universal gravitation, example gravity interaction between moon and earth , jupiter Jupiter and its moons, or sun and planets, you can find here , play with m1 and m2, G and r , to calculate , the key is r and G , get m1 for earth and m2 for any body in free fall , after get distance between earth and sun, and get m for the sun , if ...

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The short answer is no. We don't even think the census of stars within 30pc is complete, let along have all the other information you want. For all bright stars (say V<10) and close, there are decent parallaxes and proper motions in the Hipparcos and Tycho catalogues, so you can compile tangential velocities for these. A lot of these will also have ...

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There are many different ways to get spatial information about the surface of a star besides direct imaging. Direct imaging is difficult because the angular resolution available goes as $\lambda/D$. For a 8-m telescope and light at 500 nm, one can resolve $6\times10^{-8}$ radians (assuming the blurring of the atmosphere can be overcome by adaptive optics or ...

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Yes, Edwin Hubble did that for the first time in 1919. Before that time, it was thought that the galaxies we can observe were just nearby gas nebulae located inside our Milky Way. But Hubble was able to resolve the nearby galaxies like the Andromeda nebula into individual stars. By measuring the brightness of so-called Cepheid variable stars, he was able to ...

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