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The Sun does NOT revolve around an even bigger star. If it did, you'd see two stars, assuming the larger one didn't outshine the Sun. In that case, you'd only see one. Also, the earth would be a lot hotter, getting heat from two stars. We'd also be in danger of solar wind, more solar wind than we'd usually get from just the Sun. So, because we don't see two ...


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Does the Sun turn around a big star? No. Such a star, if it existed, would easily be the brightest star in the sky. You would have been taught about it early on in school if it existed. But it doesn't. For a while it was conjectured that the Sun had a small companion star to explain a perceived periodicity in mass extinction events. This too has been ...


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The Sun is not within the gravitational sphere of influence of any other star. The centre of mass of the solar system (which is very close to the Sun) instead orbits in the general Galactic gravitational potential. Because this has a roughly cylindrical symmetry (the Galaxy is basically a disk with a bulge in the middle), this means that it executes a ...


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In addition to the answers by Florin and Rob, I'd like to point out that the gravitational energy in the end dominates over all other sources, like fusion, at least for massive stars. When a massive star eventually collapses to a neutron star, the gravitational pressure is so strong that protons and electrons are pressed into neutrons and all heavy nuclei ...


3

Gravitational contraction will always release gravitational potential energy. In most systems, where this happens slowly, you can apply the virial theorem to say that half the released energy is radiated and half is used to heat the contracting gas. The real question is whether the release of gravitational PE is significant compared with other sources. In ...


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It is correct. Contracting stars heat up due to their own contraction. This justifies the initial temperature raise in a star, before fusion begins in earnest. Kelvin was not 100% wrong, it's just that fusion, over the life time of the star, makes much, much more energy than contraction. The collapse heats up the star at the very beginning, then fusion ...


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I just found the answer to my question at this place: http://astronomynow.com/2015/02/18/suns-close-encounter-with-scholzsstar/ Here is the answer: Currently, Scholz’s Star is a small, dim red dwarf in the constellation of Monoceros, about 20 light-years away. However, at the closest point in its flyby of the solar system, Scholz’s Star would have ...


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It's not completely clear what you are asking, but if this is a multi-choice quiz, then the only option that could be correct is (a). (b) Is not correct, because a white dwarf that just passes the Chandrasekhar mass is comfortably below the maximum mass that is supportable by a neutron star. So neutronisation followed by neutron degeneracy pressure and the ...


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Unless the stars comes so close that they actually collide, two stars will not be able to catch each other gravitationally. The reason is energy conservation: As they approach each other, their potential energy is converted into kinetic energy, increasing their velocities. When they are closest, their velocities are at their highest, but since there's ...


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This is the classic misconception about the expansion of the Universe: there is no center of the Universe. The whole universe expands as one, not from a single point nowhere. Knowing this it makes no sense speaking of the direction of an object due to the Universe. You could compare the directions of two objects in it though. Therefore star's direction ...


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I am not from astro background. but from what I understood about the diagram, why not try the third axis to be time?. hope u he some data for the time-age for stars in the main sequences and the age for few stars in the giants /WD region. try plotting the available star-age as third axes and try to see what data u get for the other star. I think u will get ...


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HDE226868's answer is perfectly correct. Just to add the following important information though. The s-process is limited to elements lighter than lead and produces little peaks of abundance at certain elements along the s-process path - eg Ba, Sr, Eu, Y. These are often called s-process elements. All the elements heavier than lead are produced by explosive ...


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The processes at work are the r-process and the s-process. The r-process occurs in supernovae. Under the extreme conditions leading up to it, heavy elements are initially blocked from undergoing beta decay to other elements. If a heavy atom captures a nearby neutron, it can go to a heavier isotope or element. This happens multiple times to an atom, ...


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Stars are far apart from other celestial bodies because of mass, temperature, elements. The way in which a star burns, henceforth its mass, depending on its given place in the cosmos. A large star burns hot and dies fast, while a small or dwarf star such as our own star (the sun), burns at a more consecutive speed, and burns longer, now to ask a question ...


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Planets, if any in that star system, would be visible to the naked eye, the way you can see Mercury, Venus, Mars, Jupiter, Saturn and Uranus in our system. On a virtually empty sky, they would draw that much more attention. Of course, if you had a Moon (or several moons), that would be visible too. Very rarely and briefly, asteroids passing by very close ...


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The answer is given here. One minute of time corresponds to 15 arcminutes (written as 15'). This is because in 24 h the Earth revolves 360º, so $$\textrm{angle per time} = \frac{360º}{24 \textrm{ h}} =\frac{21,600'}{1440 \textrm{ min}} = 15'/\textrm{min}.$$ If you turn this fraction upside down, you see that 1' corresponds to 1/15 min, or 4 seconds. That ...


5

Since Andromeda is already visible to the naked eye, to a civilization located at half the distance from the Milky Way, Andromeda would be still be visible. Its total brightness would be four times higher, but since its area would grow by the same factor, its surface brightness would stay constant. The Milky is less bright by a factor of ~2.5, but also ...


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You can see the Andromeda galaxy with a naked eye, even with some level of light pollution. So if the star you are asking about would be even closer to Andromeda, you would see at least that galaxy. There are some other galaxies that can be seen from Earth with a naked eye or with binoculars, so sky of such a lonely planet would still have some night lights ...


3

But as far as I know entropy is the amount of disorder. Entropy is a measure of the number of possible microscopic states consistent with an observed macroscopic state1, $S = k_\text{B}\ln N$. Fundamentally it has nothing to do with disorder, although as an analogy it sometimes works. For example, in simple situations like an $n$ point-particle gas in ...


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This is basically a trigonometry question. I wasn't sure from the question if the apparent location from your friend's position was six degrees higher or lower. I chose higher for the example, but you could change it. You have two triangles with the same altitude, but different bases and angles. $$\text{altitude} = x \tan(81^{\circ}) = (x + 20km) \tan ...



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