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6

How opaque is that -- would we be able to see a couple of meters, some kilometers, or nothing at all? The photosphere of our sun is somewhere on the order of 500 km thick. For a quick ballpark, you can imagine an exponential decrease in the transmission of light which about this characteristic thickness. It might be a little less, but it's still ...


2

Well, I can give you a definitive answer to Q1, but my answer to Q2 would only be educated speculation. Perhaps some of the astrophysicists on here can be more help with that one. However, before I tackle Q1, a very important disclaimer: Temperature is a measure of the average kinetic energy of the particles of an object, and cannot be used all by itself ...


4

A brief history of what science thought about the sun can be found here . It is reasonable that once thermodynamics advanced to the point of measuring and calculating energies the discrepancy between heat output of the sun and the age of the earth had to be explained. They tried with gravitation, but until the discovery of nuclear energy and E=m*c^2 it ...


6

I like to explain this using a figure from a talk by Marco Limongi some years ago. Based on a given set of models, the $x$-axis shows the initial mass of the models and the $y$-axis the final mass. The different coloured layers show the composition of the star at the moment of collapse. The mass ejected in the supernova is the difference between the curve ...


65

It has more to do with physiology of the eye rather than the spectrum of light produced by stars. Stars emit light over the full range of visible wavelengths. Hot stars emit more blue/violet light, cool stars emit more red light. The Sun is relatively neutral in that regard, so does not have a strong colouration, but many other stars in the sky have ...


10

Short answer: Many of them are not. Stars can be modeled as black-bodies. That means that the spectrum of light that they emit depends on their temperature, and the color of a star can be described as its Color Temperature. The color temperatures of stars can vary quite greatly, because stars have very different temperatures and emission spectras. One of ...


5

The sun is white. I've seen people say that it just looks yellow because the blue light is being scattered by the atmosphere. Since starlight is also scattered your question is still valid. But based on these pictures, the sun still seems to look pretty white through our atmosphere. My guess is that, since you can safely look directly at a star, you learned ...


2

Our sun is actually white. Sun seen from space. It's just that when the sun rays enter the earth, our atmosphere scatters the white light resulting in different colors. The reason why the Sun appears to be white sometimes is usually because it's directly overhead. Then the rays coming from the sun have to travel the least distance and hence encounter less ...


4

By estimating the distance is the most obvious method, but you are correct, the parallax will be too small to measure. If we can tell what type of star it is (by measuring its spectrum, or using its colour(s)), then we know roughly how intrinsically luminous the star is. The actual brightness then tells us how far away it is. Fortunately, the diameter of ...


3

Lyman Alpha absorption systems The Lyman-$\alpha$ Forest and Gunn-Peterson troughs are two extremes on the scale of absorption features that are left by neutral hydrogen in intergalactic space. When ultraviolet light from a background source, typically a Quasar or a young, strongly star forming galaxy, travels through intergalactic space, it is ...


4

Black body radiation is given by Planck's formula (see link for variables) Here is the measured irradiance of the sun and the attempt to fit it with the black body formula: The effective temperature, or black body temperature, of the Sun (5,777 K) is the temperature a black body of the same size must have to yield the same total emissive power. ...


8

Maybe the simplest way to think about this is that the Sun is in approximate thermal equilibrium and would absorb any photon, of any frequency, that is incident upon it. This is essentially the definition of a BB. There are many radiative processes that can absorb (and hence emit) radiation at all frequencies, not just those corresponding to atomic ...


2

You would be very interested in one of the recent Kepler discoveries - Kepler 444. The star is estimated to be 11.2 billion years old (using asteroseismology) and is surrounded by a number of rocky exoplanets. These planets are all too close to their parent K-dwarf star to be in the habitable zone, but there is no reason there couldn't be planets further ...


4

I think you already know the answer... Pop III stars, by definition, are born from primordial gas that is basically Hydrogen, Helium with trace amounts of deuterium, tritium, lithium and beryllium; they initially contain almost no C, N, or O. Therefore the primary fusion in massive Pop III stars has to be (well, initially the deuterium is burned but this is ...


1

After (re)combination (I never understand why the "re" is used) and the formation of the CMB, the universe was transparent and the only light in it was from the rapidly cooling CMB. The baryonic universe was composed almost entirely of neutral hydrogen and helium. After perhaps 100 million years, the first galaxies and stars (assisted by dark matter) were ...


0

the answer is actually very simple. Electrons, protons and neutrons and other subatomic particles don't exist in the degree of proximity that a collapsing start forced them into. The massive gravitational force overcomes the equilibrium forces that exist in matter in its "normal" state ("normal" here refers to the state of the gas/dust cloud before ...


4

I'm not going to attempt to usurp Chris White's perfectly good answer - but just fill in some detail and answer the edit. For a star like the Sun, the collapse proceeds in 4 basic stages, each takes about 10 times as long as the previous one. Pseudo-spherical collapse of the cloud - not far from a free fall timescale, often quoted as a few $10^4$ years. ...


5

Short answer: gravitational potential energy is converted into heat. Let's look at the Sun as an example. Its mass is $M_\odot = 2.0\times10^{30}\ \mathrm{kg}$ and its radius is $R_\odot = 7.0\times10^8\ \mathrm{m}$. If its density were uniform, its gravitational binding energy would be $$ U_{\odot,\,\text{uniform}} = -\frac{3GM_\odot^2}{5R_\odot} = ...


1

A star is neither "flaming" nor "fire" in the sense that we use those words about things on Earth. It's just a big, hot ball of ionized gas. The only thing that happens "to" it is that it gets hotter and denser. At some point the temperature rises high enough to ionize the gas. Later still fusion becomes possible at non-vanishing rates. The energy for the ...



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