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The light that we see coming from the Sun is mainly due to black body radiation at its surface. The spectrum of black body radiation is statistical in origin, and as long as there are enough processes contributing to it the black body spectrum is independant of the microscopic details and depends only on the temperature. There is a discussion of this in the ...


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Most of the brightest stars are spectral types B,A,F main sequence stars (50%), but there are also a bunch of O-type main sequence and giant stars (5%) and another big clump of red giants (about 35%) and a few percent are supergiants. There are no white dwarfs, and definitely no neutron stars! I have attached an image which I created by selecting 4992 ...


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Here is an example for the Sun. The figure below plots a (reliable) estimate for the interior density profile of the Sun, $\rho(r)$. So for a given radius $a$, the mass interior to that radius is given by $$ M(a) = \int^{a}_{0} 4\pi r^2 \rho(r)\ dr $$ And of course the gravitational field strength assuming spherical symmetry will be $$g(a) = - G ...


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The rate of formation is much higher in the presence of dust. There needs to be a mechanism for the energy of formation of the hydrogen molecule to be dissipated. Dissipating energy via a photon involves a forbidden transition. Instead, the energy can be transferred to the vibrational lattice of a dust particle. See The Interstellar Abundance of the ...


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Given the huge reservoir of angular momentum stars are born with it is more a question of how is it that stars can get rid of most of this angular momentum to become the relatively slowly rotating objects we see? One of the best and most comprehensive pictures of stellar rotation that we have comes from the periodic modulation of a star's light caused by ...


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$ L = \int \int {\bf F} \cdot d{\bf s}$ is where you should start, where $F$ is the flux in units of Watts/m$^2$. Blackbody flux is given by $\sigma T^4$ and hence an isotropic flux integrated over a sphere $$ L \int^{2\pi}_{0} \int^{\pi}_{0} \sigma T^4 r^2 \sin \theta\, d\theta\,d\phi = 4\pi r^2 \sigma T^4$$


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Is it just after they have finished core H burning and the core contracts creating high temperatures which result in core He burning...? It is after the core finishes H burning, but He burning is not required. Hydrogen shell burning is sufficient to make it a red giant. Helium burning would make it a Horizontal Branch Star. See good explanation here: ...


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The answer depends upon the mass of the star. For stars of less than 2 solar masses, electron degeneracy pressure stops the collapse. For more massive stars, helium fusion begins which stops collapse, without a degenerate state being reached.


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I think, you mix two or even three kinds of redshift effects. First, the classical Doppler shift due to which light emitted from a moving object is shifted to the red or blue. This is dependent on relative movement of sender and receiver only. For transverse motion we still get a redshift (see Wikipedia URL below), but it is very small for low speeds. ...


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Carbon has to be produced by the triple-alpha process because there is no stable nucleus with 8 or 5 nucleons. The probability of this is very low, because it requires three different particles to be in the same place at the same time. You'll note that the Wikipedia article says: One consequence of this is that no significant amount of carbon was ...


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inner radius of habitable zone (AU) $= \sqrt{\frac{L}{1.1}}$ outer radius of habitable zone (AU) $= \sqrt{\frac{L}{0.53}}$ where $L$ is absolute luminosity of the star. See Calculating the Habitable Zone for more information.


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There is a reasonable chance that yes, planets can form before the star "ignites" (which I take to mean the fusion of hydrogen into helium, not the very brief phase of deuterium burning which certainly will take place before planets can form). Planets form in a disk of circumstellar material around their parent protostars. The "core-accretion" model of ...


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In a neutron star there are mostly "free" neutrons and the question then is why they don't all beta decay into electrons and protons? Well, some of them do, but the point is that when the electron (or proton, there are equal numbers of each) numbers build up then they become degenerate (meaning no more than two electrons can occupy the same energy state and ...


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Indeed, what we infer about stars from the light we see at the Earth is "old news". However for almost all practical purposes in stellar astrophysics this doesn't matter. The phases of a star's life last millions if not billions of years and most of the individual stars that are studied are within say 30 thousand light years of the Earth. The example you ...


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I am not exactly sure which low energy cutoff you refer to; however, there is a low-energy cutoff for photons that I am aware of. Photons with energies on the order of $H_0\sim10^{-33}\text{eV}$ would be super-horizon modes. That is, their wavelengths would be on the order of the Hubble radius, $H_0^{-1}=14.6~Gly$. Larger than this would mean that the ...



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