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-1

I think any rods-and-cones explanation, such as the Purkinje effect, for the perceived colour of the Moon itself is nonsense, simply because the light reflected from the (full, overhead) Moon is easily bright enough to be perceived by the cones in our retina. As you say, if the contrary were true, it would also prevent us from seeing the Moon as orange when ...


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First of all, a celestial body could appear blue for a simple reason: it may be blue, indeed. This is the case of Neptune. Its being blue simply means that the probability that a photon is reflected off Neptune is higher if the frequency (color) of the photon is basically blue than when it is green, yellow, or red. However, the Moon isn't actually blue. The ...


8

The image below represents the Sun's density gradient, which shows how the density changes with the radius. The ground we stand on should have a density between 2 to 3 $g/cm^{3}$. That should put you just above the water point on the vertical axis. The corresponding radius is then about 0.45 of the solar radius. Note that the vertical axis is in a ...


2

As @KyleKanos point out, "the answer is a google search away", but it's not quite as simple as he suggests. The mean density doesn't answer your question (the mean density turns out to be about 1.4 times the density of water by the way). An ill-defined idea in your question is the "surface" of the sun. Where is the "surface" of the sun given that none ...


-1

So starlight propagates spherically and each human eyeball creates localized photons just at the intersection of wavefront and retina. No matter where you are in relation to the star some part of this wavefront will reveal the photon stream. Some kind of sensor that could image the path of all the photons/wave functions as they were emitted would reveal a ...


0

Molecular clouds are the most common places where stars are born, so it's best to discuss them. Richard Larson (who has studied molecular clouds in depth) discusses two processes in an overview of these structures in The Evolution of Molecular Clouds: Random collisions between particles in the interstellar medium (ISM) grow over time until eventually ...


1

You can check The Best Cheap Budget Telescopes Under $200; this presents the latest list of all affordable telescopes that are competent for beginners. Also you can check The cheap telescopes of 2014. Remember always that though magnification is good, but you must need to have a good apparture in your device & for that I prefer Newtonian telescopes or in ...


-1

To make fusion occur, large amount of activation energy is required. Once running, it will keep going. The reason why the surface of a star of a large mass is a suitable place for fusion to occur is its large gravitational force on small atoms like hydrogen. This strong gravitational force makes for a strong pressure on the gas, and according to the ideal ...


0

The Coulomb barrier for protons is on the order of $10^6$ eV. Treating the Sun's core as a gas at 15.7 million kelvins, the mean kinetic energy of the protons is $2\times10^3$ eV. A quick partition function estimate of the probability of protons having enough kinetic energy to overcome the Coulomb barrier yields a probability of $10^{-257}$. Considering the ...


4

Both From Wikipedia Luminance Versus Luminosity In astronomy, luminosity is the total amount of energy emitted by a star, galaxy, or other astronomical object per unit time. It is related to the brightness, which is the luminosity of an object in a given spectral region. Now your question: Let's assume we have two stars that have the same surface ...


2

Think of 2 hydrogen atoms or, protons more accurately since at those temperature the atoms don't have electrons, it's more of a soup. So, 2 protons, both positively charged so they repel each other, crash into each other pretty rarely, cause it's still a lot of empty space, but they do make contact every so often. The energy required to get 2 protons or ...


0

As I see it, and, correct me if I'm wrong, but there is a way to do it. The problem with using pure Uranium or any other readily fissionable element is that, as one element decays, that releases 2 neutrons which can speed up the decay of nearby elements. If you had a ball of U235 or U238, say, the size of a planet or even a small moon, you'd have a ...


1

In principle yes, though it would be a highly contrived situation and not one likely to arise naturally. A star works because hydrogen to helium fusion is energetically favourable. But the process has a huge activation energy so you need an environment as hot and dense as a star's core to provide that activation energy. Likewise, for any sufficiently heavy ...


1

Answer from astrophysicist. Point 1. The problem with eternal shining is that star loses energy with photons (and some material) and it'll need income of energy from somewhere. There are brown dwarfs, black dwarfs, black holes etc., which are just remaining of stars and don't 'shine' (or, in case of white dwarfs, fade out to the point we cannot detect ...


0

Its just a consequence of what concentrates the fuel/contents of a star together in the first place, I think? Stars usually form by the collapse of gas clouds onto themselves. These collapses generate high temperature and density and are ideal places for nuclei to 'bump' into each other and also, the high amount of energy nuclei can possess in such a ...


4

(By stars I'm assuming you're implying stars like the Sun, which are a majority of the stars we see. @Dirk Bruere's answer about Black Dwarves is correct. ) No, I don't think they can. The primary process that 'fuels' stars is nuclear fusion. In the process of nuclear fusion, lighter elements fuse together, releasing a tremendous amount of energy (because ...


4

The word thing you are looking for is "Black Dwarf" stars. Which are White Dwarf stars which have cooled to match the temperature of the cosmic background. Since this is likely to take more than the current age of the universe, there aren't any. These will exist forever, unless hypothetical proton decay finishes them off or a hypothetical Big Rip due to Dark ...


2

If you have $L$ and you have $T$, then nothing more complicated than Stefan's law is required. If $T$ is the effective temperature of the star then this gives an exact answer. $$ R = \left(\frac{L}{4\pi \sigma_B T^4}\right)^{1/2}$$, where $\sigma_B = 5.67\times 10^{-8}$ in SI units. If on the other hand you are trying to solve the structure from first ...


0

A proto-stellar disk is composed of gas and dust from a giant molecular dust cloud that collapses into itself because of gravitational attraction. It's an early stage in the formation of a star. The collapse may be triggered by shock waves passing through the molecular cloud which disrupt the balance between inward gravitation and outward gas pressure that ...


0

A circumstellar disk is a set of objects that orbit about a star-like object and that collectively look disk-like in some way. A protostellar disk is thus a kind of circumstellar disk, the asteroid belt and the Kuiper belt about our Sun are also circumstellar disks.



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