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1

Ironically, it's actually harder to measure the mass of the Milky Way than that of other galaxies. You'd think that with it being RIGHT THERE it would be easy, but alas. Most of the difficulty comes from (1) the galaxy spans a huge part of the sky, so it takes an extremely long time to observe any particular feature in detail across the whole thing (say ...


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A star with a higher metallicity will have a larger interior opacity and a thicker convection zone for a given mass. The thickness of the convection zone is equally important in determining magnetic activity in stars that have an interface between a radiative inner region and convective envelope. The dependence is through the Rossby number - the ratio of ...


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It turns out that it is the distribution of birth stellar masses and most importantly, the lifetimes of stars as a function of mass that are responsible for your result. Let's fix the number of stars at 200 billion. Then let's assume they follow the "Salpeter birth mass function" so that $n(M) \propto M^{-2.3}$ (where $M$ is in solar masses) for $M>0.1$ ...


2

Yes, your qualitative argument is correct and the number of stars brighter than the Sun is almost certainly much smaller than 10 billion. The reason is that the luminosities are hugely variable. Due to the mass-luminosity relation, each doubling of the stellar mass corresponds to increasing the luminosity 10 times. So many if not most of the "stars ...


0

I think that whatever answer is provided, is hypothetical and untestable. So I answer: No, a supernova is too large a phenomenon to affect. Yes, if we can find the appropriate "seed". For example, huge storm clouds can be triggered to rain, by seeding the cloud with silver nitrate crystals, dumping their content to the ground. Do I think humankind can ...


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Any method that turns off the core will work (again, if the star is massive enough), the collapse and bouncing back of the shell will do the rest. For instance, suck up the core to the outside (of course we are no near to have that technology yet). Or alternatively, you can pump in antimatter to the core instead (protected in magnetic vessel/pipe so it ...


22

Why shouldn't the orbits of stars be Keplerian? The answer is simple. Keplerian orbits are predicated on a single central point mass. That assumption fails to some extent even in a solar system. It fails massively in a galaxy. A galaxy is not a point mass.


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Elliptical orbits are direct consequence of orbiting entirely outside a spherically symmetric mass. Even if you assume that a galaxy has a spherically symmetric mass distribution, the amount of mass at a radial distance less than that of the star would be changing (assuming some eccentricity). Once that happens, the orbit is no longer an ellipse.


5

Not Keplerian, because it is not a conic-section. It is not even explained by Newtonian gravity. In contrast, Kepler's laws are explained by newtonian gravity. The lowest orbital-energy from Keplerian orbit is circular. And the orbits of stars are observed to be approximately circular. Hence: $$ \frac{mv^2}{r} = \frac{GMm}{r^2} \quad\Longrightarrow\quad v = ...


1

The other answers cover your second question well enough, but there's some detail still missing on the first one - what happens to the protons and electrons in a star when it collapses into a neutron star. The basic answer is simple: they become those neutrons. The reason that this happens is that, as it turns out, an {electron, proton} pair is sort of ...


3

An ADS search for "star formation" turns up about 142,000 articles with "star formation" in the title or abstract. The first article is a 43 page review paper of Star Formation in Galaxies in the Hubble Sequence, written by Robert Kennicutt, Jr, one of the leaders of the field. He never defines anything else to mean star formation and one of the "key words" ...


2

There are several reasons. One is that when a cloud of gas and dust collapse into a star forming region, it becomes unstable to gravitational fragmentation and usually forms filamentary structures. The gas that lies outside of the densest regions is often not dense enough to be itself then gravitationally unstable. This behaviour is clearly shown in modern ...


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Measuring the star formation history of the Universe is a key component in understandng the evolution of galaxies. It is closely related to the other uestion, recently asked by the same person: Are stars getting more metal-rich, less massive and shorter-lived with cosmic time? Although this question pertains to the Universe as a whole, an understanding ...


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Neutron stars are a special case. When the neutron stars collide a black hole is formed within milliseconds. A gamma ray burst of less than 2 seconds is expected to be observable from a great distant, as well as gravitational waves. About 1% of a solar mass is expected to be ejected and include heavy elements. On June 3rd 2013 the Swift gamma ray ...


1

The light we see from the stars is not "just light". The spectrum contains the absorption and emission lines of many of the elements in the star. So it is not just a black body radiator like most light bulbs - we can see what they are made of.


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I think your question relates to the gravitational lensing. The problem of locating a body (finding its actual position) given only its apparent location (as seen by some one on Earth -by considering their light has travelled in a straight line only) is studied only recently (though the problem existed for almost a century till now). As far as I have ...


3

it can be said that the γ photon loses energy and also its velocity This isn't a good description of what happens. In a material like plasma that contains free electrons photons are absorbed and their energy converted to kinetic energy of the electrons. The electrons then collide with, or otherwise interact with, other electrons and reradiate photons. ...


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Explanation: Photons, quanta of light always travel at the speed of light: $c$, since you are talking about light, classical kinetic energy equation: $E=\frac{mv^2}{2}$ no longer applies to it, instead you have to use this relativistic energy equation: $$ E^2=(m_0c^2)^2+(pc)^2\tag{1} $$ since photons don't have rest mass i.e. their rest mass equals to zero: ...


0

Yes, about 96% of the energy released by fusion in the Sun is due to the proton-proton cycle. The first step of this cycle is p-p fusion. The energy required to bring two protons into contact with each other is about 600 keV (Calculate the potential energy of a system of two protons with a center-to-center separation of about 2.4 fm.) As quick estimate of ...


1

There are lots of questions here, let's take them one at a time. Are star getting less massive as a result of nuclear fusion? Yes. If a star has a luminosity $L$ then this corresponds to a mass change $c^2 dM/dt$. For the Sun $L = 3.83\times10^{26}$ W, so the Sun is getting less massive at a rate of 4 million tonnes per second. However, over the Sun's ...


0

Well, stars do loose mass over the course of their lifetime. And during the course of a star's lifetime it will go through several stages generating heavier materials. However there is a minimum mass for a star. Below this mass the star does not achieve high enough temperaturs to ignite nuclear fusion. Objects below this critical mass are called "brown ...


1

Yes. Instead of allowing the energy to radiate into space, you're containing it and sending it back to the source. Think fire in a room vs. fire outside: outside the heat gets lost to the environment but in a room it stays and warms the room to a much higher temperature than the fire outside warmed the surrounding air.


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The route to the answer is somewhat anti-intuitive. By reflecting some of the Sun's energy back towards the sun at a point you are effectively reducing the flux of energy that can emerge from the photosphere and escape. The global effect of this on the Sun must be similar to that of blocking the flux at the photosphere - in other words, similar to the ...


11

LDC3 and Kitchi addressed your main question, but I'd like to comment on your second paragraph. Isn't this something like a machine is running, I am getting work out of it and supplying back to the machine to 'accelerate' it? I am not expert in physics, but intuitively thinks that this may not be possible. Actually, we do this all the time! Electricity ...


3

There won't be a violation of thermodynamics because you are not creating energy from nothing. The total energy of the system is still conserved, it is just fed back into the system. Here's what will probably happen - The concave mirror will not be perfectly reflecting, so will reflect something like $99\%$ of the incident energy. This energy (although very ...


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If it was possible to reflect the energy back at the sun, yes, the location where the energy strikes will become hotter. If fact, if you could insulate the sun from radiating energy, then the sun would get even hotter.


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I think what you mean is - is it possible for a planetary system to exist such that the planets do not orbit in a single plane, but the planets have a large scatter of inclination angles? Our solar system has a relatively modest range, providing you ignore Pluto, of orbital inclination values (and eccentricities); zero to 7 degrees (Mercury). This is ...


2

"Only about 10 percent of the total baryonic matter is sufficiently condensed by gravity to form stars and galaxies. More than 90 percent was left between the galaxies." http://hubblesite.org/hubble_discoveries/science_year_in_review/pdf/2008/searching_for_baryonic_matter_in_intergalactic_space.pdf 6% of baryonic matter is within stars according to the ...


3

An alternative method to John's answer is to look at the total number of atoms in the observable universe. Thanks to measurements of the cosmic microwave background, we have a fairly precise estimate of this number. Indeed, we know that ordinary matter makes up about 4.9% of the energy content of the universe. In this previous post, I calculated that this ...


1

Nothing stops the core contraction until He ignition takes place! Yes, the core is partly supported by electron degeneracy, this slows the contraction and allows a relatively stable period of shell H-burning, but the central density continues to rise and the core contracts until He is actually ignited. This is especially true where there is a high degree of ...


3

You are confused. When the star expands to become a red giant it is burning hydrogen in a shell around an inert He core. How things proceed from there depend on the mass of the star, but generally speaking, the core contracts and heats up to maintain pressure and hydrostatic equilibrium. The shell moves inwards, heats up and the nuclear burning rate ...


4

The simplest answer is that in order to maintain helium fusion, a certain pressure and temperature are necessary. Therefore, given the fact that helium fusion is occurring in the core, and the mass pushing down on the core is X from dynamic concerns, you therefore must conclude that the temperature in the core is Y, irrespective of the size of the core.


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First, a star does not become a red giant when helium fusion begins, instead it becomes a red giant earlier when an inert degenerate core of helium forms and a shell of hydrogen begins fusion. When shell hydrogen fusion begins, the star expands to be a red giant. The core is degenerate (sustained from collapse by electron degeneracy pressure) and ...



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