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0

You seem to be neglecting the change in the moment of inertia of the planet in the definition of angular momentum: $$ \mathbf L=I\boldsymbol\omega $$ Since $\boldsymbol\omega$ increases, then $I$ must decrease to satisfy angular momentum conservation. There isn't any need for angular momentum to somehow "reach out" and take it from a nearby planet.


1

You say: But for the observer on the planet, since the total angular momentum of the star about its axis is zero it should remain zero. But the observer on the planet does not occupy an inertial frame. An observer in a rotating frame measures fictitious forces. So there is no reason why angular momentum should be conserved.


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With the given information, several results are possible: All the kinetic energy is converted into heat and the required temperature for "burning" deuterium is not reached (no star). If the temperature is reached, then the energy released by the deuterium burning might be enough to keep the star from contracting any further. The star might expand a little ...


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The answer is that 41% of the stars have masses below 0.25$M_{\odot}$. To check this I integrated the Kroupa initial mass function. This is that $N(m)$ the number of stars per unit mass is proportional to $m^{-1.3}$ for $0.08<m/M_{\odot}<0.5$ and proportional to $m^{-2.3}$ for higher masses. If I integrate this I find that the ratio of stars with ...


6

The stellar mass distribution is the distribution of numbers of stars within a range of masses in a galaxy (or cluster or what have you), not the mass of the stars. So if you looked at the $\sim10^{11}$ stars in the galaxy, you would observe that about $4\times10^{10}$ of them will have a mass less than 0.25 $M_\odot$, and so on with the rest of the masses. ...


3

According to this source, 100% is the number of stars, not the total mass. Same from another source. The reason is that they usually calculate these pies straight from the H-R diagram. The H-R diagram plots individual stars and shows how stellar mass varies along the main sequence. Actually the mass distribution tends to reverse. Even if larger stars are ...


3

You ask "what would we see", meaning you want to know what kind of radiation is given off. (i) sample from the surface looks like the surface of the Sun. i.e. it emits close-to-blackbody radiation at a temperature of 5800K.Except it now can't be a blackbody since you've taken it out of its environment. A black body is both perfect absorber and a perfect ...


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Observations are not made with a single photograph frame, they're made over longer periods of time (for example, a friend takes observations of Herbig stars over 2 minute periods in the J-band with Gemini North). The emissions from stars range from UV to IR (depending on your opinion, it could also range from IR to UV ;) ), these emissions are continuous ...


1

The formula $$ L = \epsilon A \sigma T^4,$$ where $L$ is the luminosity in Watts, can be used for a "grey body" i.e. one that has a constant emissivity with frequency. Here you were told to "assume the Sun to be a perfect blackbody". This means that its emissivity is 1 because a blackbody absorbs everthing incident upon it and because it is in thermal ...


5

I've added this because I don't think the accepted answer is very clear. Estimating the number of stars in the Galaxy relies mostly on two things. We estimate the present day mass function (that is the number of stars that exist per unit mass per unit volume) in the solar neighbourhood. We construct a model for the overall density distribution of the ...


1

If you assume that there are 200 billion stars - that is objects with mass between say $0.075 M_{\odot}$ and $100 M_{\odot}$ you can use this to normalise a mass function - the number of stars per unit mass - and then integrate stellar mass, weighted by this mass function, to estimate the total mass in stars. If you do that then what you find is (1) high ...


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Black holes do explode after their life is over, thus vomiting out all the matter they ingested. Apologies for not providing credible explanation for my claim (in the form of mathematical equations). I read that (that black holes do explode) in Stephen Hawking's book titled Black Holes And Baby Universes. He stated that there is an inverse relation between ...


4

No, throwing matter out isn't contrary to a BH; there is often an accretion disc surrounding the black hole and that is what forms the jet. No, the ejected material cannot condense to form a galaxy. A galaxy requires the material to be gravitationally bound to some central point, a jet moving at $\sim c$ is moving too fast to be gravitationally bound to ...


2

In another closely related question (According to the initial mass function, should there be more brown dwarfs than red dwarfs? ), I showed that the number of brown dwarfs (with $M<0.075M_{\odot}$) is a factor of five smaller than the number of red dwarf stars (stars with $0.075<M/M_{odot}<0.5$), using the widely adopted Chabrier (2005) lognormal ...


3

There are several commonly used analytic approximations for the initial (birth) mass function (IMF) that cover both stars and brown dwarfs. It is not yet absolutely certain which of these is more correct at the low-mass end, whether there is a lower mass cut-off as one approaches planetary masses, or whether the fraction of brown dwarfs (BDs) to stars varies ...


3

This is difficult to answer in an unarguable way because the old bimodal classification of population I and II is more nuanced these days - e.g. thin disk, thick disk, bulge population etc. However, if you define population II as meaning those stars that were born in the first billion years of our Galaxy's evolution, then the following rough calculation ...


3

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 ...


1

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 ...


6

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 ...


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 ...


5

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 ...


0

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.


1

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 ...



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