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1

For stars (which have huge amount of mass and density), gravity is taken to be responsible for the heat increase. because heat and volume (thus density) thus gravitation of a (massive) star, are related. This is exactly one of the factors that make nuclear fusion (in stars) possible. The two effects thermodynamics (and kinetic energy) and gravity are ...


2

Consider a satellite in orbit about the Earth and moving at some velocity $v$. The orbital velocity is related to the distance from the centre of the Earth, $r$, by: $$ v = \sqrt{\frac{GM}{r}} $$ If we take energy away from the satellite then it descends into a lower orbit, so $r$ decreases and therefore it's orbital velocity $v$ increases. Likewise if we ...


5

So a while ago I did a little project where I grabbed a "standard solar model" from this paper, which gives me some information that's useful for actually making an estimate. (Unsurprisingly the link given to download the data has changed in the last ten years; I haven't sleuthed to see whether the data is still publicly available.) Only about 1.5% of the ...


0

At the end of your question you ask if there are other ways to express the temperature of the Sun in terms of energy. This is probably not exactly what you're looking for, but from Wien's law $\lambda_{\rm max}T=b$ ($\lambda_{\rm max}$ is the peak wavelength of the Sun's $\sim$blackbody spectrum, $b$ is Wien's displacement constant) and ...


1

Fun, So you are asking about the thermal energy content of the sun? If we assume that all the hydrogen is dissociated. (single atoms) Then each atom has three degrees of freedom and carriers 3/2 kT of energy. So count up the number of atoms at each temperature.... That will work until the atoms ionize. Then there will be equal energy in all the ...


0

The sun is more than a cloud of hot gas that is radiating energy. The sun also has energy stored in hydrogen "fuel" that will be "burned" through nuclear fusion into helium, releasing a lot of energy. The sun is $2 \times 10^{30}$ kg, and about 70% hydrogen, so around $1.4 \times 10^{30}$ kg of hydrogen or $8 \times 10^{56}$ protons. Let's estimate that ...


1

I like Ben's answer, but here is my take on this. Degeneracy pressure is not due to a fundamental force; in fact in the simplest model, it occurs in ideal gases of non-interacting fermions. The simple quantum mechanics of particles in an infinite potential well (i.e. trapped in a volume) tells us that only certain quantised wavefunctions are possible. Each ...


2

The force we associate with the Pauli exclusion principle is not a fundamental force, connected with the four fundamental interaction, but rather an entropic force, a consequence of the limitations the principle puts on allowed wave functions. A more general statement of the principle states the the total wavefunction of a system of two (or more ) identical ...


12

Estimating the mass of a "single star" can be a very difficult task, though perhaps your question is too pessimistic. There are a number of suggested relationships linking the mass of a star to its luminosity. These can of course come from stellar evolution models, but they can also be empirically calibrated using stars in resolved binary systems of known ...


7

A binary star system, which is quite common, will allow us to determine mass with great accuracy, using Kepler's Third Law of Planetary Motion which is as follows: $$ \frac{T^2}{r^3} = \frac{4 \pi^2}{GM_{sum}} $$ Using the orbital period, $T$, we can determine the acceleration and affect that stars have on one another to determine mass. Once we have the ...


7

How does the Pauli Exclusion Principle actually create a force? The Pauli exclusion principle doesn't really say that two fermions can't be in the same place. It's both stronger and weaker than that. It says that they can't be in the same state, i.e., if they're standing waves, two of them can't have the same standing wave pattern. But for bulk matter, ...


2

Your first paragraph is not quite right. Gas pressure does not "stop" upon formation of an iron core, it is merely that the star cannot generate further heat from nuclear reactions and becomes unstable to collapse. i.e. The star does collapse! Perhaps what you mean is what halts the collapse (sometimes) before the star disappears inside its own event horizon ...


0

The Chandrasekhar mass is not the dividing line between those stellar remnants that will become black holes and those that will become something else. A compact, cold white dwarf (i.e. one supported by electron degeneracy pressure) may become unstable and collapse at close to the value of 1.44$M_{\odot}$ described by Kyle and derived using simple Newtonian ...


26

For visible stars, the answer is no. In Newtonian physics, a star that would pull something travelling at light speed back to itself, i.e. a star for which the escape velocity were $c$, was called a dark star and seems to have been first postulated by the Rev. John Mitchell in a paper to the Royal Society in London in 1783. The great Simon Pierre de Laplace ...


16

In astronomy parlance, the Sun has a "metal"$^{1}$ mass fraction of about $0.02$. A solar mass is $\sim2\times10^{30}\;\rm{kg}$, so the sun contains about $4\times10^{28}\;{\rm kg}$ of "metals". That's about $20$ times the mass of Jupiter. A lot of that metal mass will be ${\rm C}$ and ${\rm O}$ and other elements a chemist would call non-metals, but I think ...


5

"Storing" something implies the purpose is to put is somewhere safe so that it can eventually be retrieved. Heavy metals should eventually sink to the center of a star, but how are you going to retrieve it? Even in a science fiction context, it's hard to imagine a plausible means or retreiving a large pile of heavy metal from the center of a star. ...


7

Our primary method of understanding stars is using optical telescopes. We can directly measure the temperature of the "surface" of a star by fitting its spectrum to a blackbody. However, this gives us the temperature only at the "surface" or the optical depth of a visible photon. We know that stars are extremely hot through the hydrodynamic/thermodynamic ...


0

r is the distance between the two masses v is the relative velocity a is the relative semimajor axis* * Two bodies orbiting each other trace out two separate ellipses in an inertial frame. The smaller body traces a larger ellipse, and vice versa. The relative semimajor axis (a) is equal to the sum of the semimajor axes of these two ellipses. The relative ...


-2

I learned that the electrons which are separated from their nuclei are being annihilated by the positrons. These positrons are the product of the weak interaction that emits photons from the sun. So, the sun is producing photons by destroying the electrons (and of course, by converting protons into neutrons. Source : ...


8

The electrons are still inside the stars. A stellar plasma is electrically (almost) neutral, the electrons in a plasma are simply not bound to individual nuclei. If we could take some plasma out of a star and we would let it cool down to room temperature, most of the matter in that gas (at least from main sequence stars) would be ordinary neutral hydrogen ...


2

According to Wikipedia The Chandrasekhar limit is the maximum mass of a stable white dwarf star. The limit was first published by Wilhelm Anderson and E. C. Stoner, and was named after Subrahmanyan Chandrasekhar, the Indian-American astrophysicist who improved upon the accuracy of the calculation in 1930, at the age of 19. White dwarfs with masses greater ...


0

Above the Chandrashekhar limit you would obtain a neutron star: https://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit Also see: https://en.wikipedia.org/wiki/Chandrasekhar_limit


4

Have a look at this article. It gives the number as $10^{24}$ rather than $10^{23}$, but it's such a vague estimate that a factor of ten is within the expected error. The number is the number of stars in the observable universe i.e. within 13.7 billion light years of Earth at the time the light we see today was emitted. Note that visible means visible to a ...


1

Here are the systems I found: 6: ADS 9731 Beta Tucanae Gamma Velorum Kappa Tauri Mu Sagittarii 7: AR Cassiopeiae Nu Scorpii ... no physical multiple stars of greater multiplicity yet found.


4

The following passage is extracted from Stephen Hawking's book "A Brief History of Time": In fact, various contemporaries of Newton had raised the problem, and the Olbers article was not even the first to contain plausible arguments against it. It was, however, the first to be widely noted. The difficulty is that in an infinite static universe ...


7

Who is interested can find detailed information at wiki, or here The problem is known (as you added in your edit) as Olbers' paradox, and was posed already in the mid 1500's, by Johannes Kepler in 1610 and even later by Edmond Halley in the eighteen century, and curiously, even the novelist an poet Edgar Allen Poe anticipated possible explanations as to why ...


13

I'm going to respond to (v1) of the question, which asks why the night sky is dark (black and unlit) compared to the day sky even though there are many light sources at night. The updated question references Olber's paradox, which has been answered many times before. Like most things we see in everyday life, there are a number of reasons contributing to ...


0

Even without setting $\sin \theta=1$ you have a cubic equation for $m_2$. Such things can be solved, like the quadratic formula but messier. Various computer algebra packages incorporate the solution. For graphing, you can just pick a bunch of $m_2$ points, solve for $m_1$, and plot the points.



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