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7

There are a number of different frames of references. For the velocities of celestial objects we use: (i) The geocentric frame: This is a velocity measured with respect to the Earth's centre. Obviously this is quite useful for artificial satellites, but also for things like meteors. (ii) The heliocentric frame: this is the velocity as seen from the centre ...


17

There are two separate questions there. The easiest one to answer is how we measure the vleocity of the Earth, Milky Way etc, because we measure it relative to the cosmic microwave background (or CMB). If you measure the CMB in all directions and find it's the same in all directions then you are stationary in comoving coordinates. However if you find the ...


1

Speed is a distance (separation between two well defined points in space) traveled over a time. The speed of Earth you quote is the orbital velocity. We know how far away the Sun is and we know the shape of the orbit, so we know how far the Earth travels relative to the Sun (distance) per year (time). Likewise the speed of the Milky Way is also given ...


2

I think the following image, which comes from Tomczak et al. (2014) and the so-called ZFOURGE/CANDELS galaxy survey should do the trick. It shows how the galaxy stellar mass function (i.e. the number of galaxies per unit mass per cubic megaparsec that have a certain stellar mass) evolves as a function of redshift. As you might imagine this is not just a ...


1

Its complicated. There wasn't just one Hubble Ultra Deep Field (UDF) observation, but several, taken at different times, with different instruments, but pointed in (almost) the same direction. The first image you refer to was taken with the WFC Infra red camera in 2009, in near infrared bands (1-1.6 microns). The area covered by this camera is $2.4$ ...


4

Your method is correct. When the angle is "small" so that we can ignore curvature, then the rectangular solid angle is just the product of the two side angles. (This doesn't hold for "large" angles). I cannot find a perfect explanation, but one source of this confusion may be that the UDF was imaged by two separate instruments, one optical, one ...


3

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 ...


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 ...


-2

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 general the planes of solar systems are not aligned with the plane of the Galaxy, but are oriented in all different directions. The size of a solar system is so much smaller than the size of the Galaxy, that the Galaxy's structure has no impact on the orientation of a solar system. What determines their orientations is the direction of the angular ...


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 ...


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 ...


4

It is difficult to estimate the masses of either galaxies or clusters of galaxies, and it depends on what source you consult. You can get different values from different sources, for instance in this reference: the local group is estimated to have $5.27\times 10^{12} M_\odot$ (which does include dark matter). I didn't have access to the wikipedia source.


4

I think the following image sums up why your model, at least for our galaxy, is wrong rather nicely: These are the orbits for 6 stars in the inner region of the galaxy. The orbital period for S2, for instance, is 15 years for an orbit that is roughly twice the size of Sedna's orbit--which takes it 12 thousand years to complete its orbit. Using Kepler's ...


8

There are many problems with this line of reasoning. The most common galaxy types are elliptical galaxies and spiral galaxies, and there might be a parallel with star systems, where the most common types are systems with a single star, and binary systems with two stars in the middle. There is simply no justification for this. The dynamics of stellar ...


-1

The stars in the galactic disk rotates with almost the same orbital velocity 200-230 km/s around the galactic center. Unlike a star system where the planets follow Kepler's third law. To explain the almost constant orbital speed of the stars in the galaxy, we have calculated dark matter, which is distributed in such a way, that it gives stars almost the same ...


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.


11

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 = ...


84

This whole question is a mistaken premise. There are spherical (or at least nearly spherical) galaxies! They fall into two basic categories - those elliptical galaxies that are pseudo-spherical in shape and the much smaller, so-called "dwarf spheroidal galaxies" that are found associated with our own Galaxy and other large galaxies in the "Local Group". Of ...


6

You mentioned elliptical galaxies, which the other answers haven't touched upon. Contrary to your statement about the galaxies being 2D, elliptical galaxies are "3 dimensional" in the sense that the stars are not confined to one plane; You could think of them as being "egg shaped". So why are elliptical galaxies not confined to a plane? Mostly because they ...


16

Actually, there are parts of a galaxy that extend beyond the galactic plane: Galactic halo: This is actually the primary part of a galaxy that is not in the main galactic disk. It's made up of multiple sections, and is composed or an array of objects. Dark matter halo: This is a section of the galaxy's dark matter that exists in a semi-spherical shape. ...


14

All matter in the galaxy has to rotate (not necessarily in the same direction) so that a centrifugal force acts. Without the centrifugal force, all matter contained in the galaxy will collapse into the center of the galaxy due to gravitation. The rotation happens about an axis, a line about which all matter revolves in the galaxy. Now, the manner in which ...


9

It is due to the combined effect of rotation and "dissipation". A rotating cloud of gas consists of particles which interact strongly with each other (colliding physically) on relatively short timescales can radiate away some of their energy and momentum by emitting photons. For both of these reasons, a dense cloud of rotating gas will collapse to form a ...


0

I am no specialist in gravity or cosmology. Though, I know (without details) that A. Peres proved that the light velocity wasn't the same all along the history of the universe. The reference is Int. J. Mod. Phys. D, 12, 1751 (2003). DOI: 10.1142/S0218271803004043 International Journal of Modern Physics D (Gravitation; Astrophysics and Cosmology) Volume ...



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