New answers tagged

1

In general, yes you need to know the orbital inclination angle $i$ in order to fully solve the orbit. The radial velocity amplitude $K$ is just modified to $K \sin i$. Combining this with the orbital period and Keplerian orbits gives you the "mass function" $$ \frac{M_1^3 \sin^3 i}{\left(M_1 + M_2\right)^2} = \frac{K_{2}^3 \sin^3 i\ P_{orb}}{2\pi G},$$ ...


2

The apparent line-of-sight velocity (red shift / blue shift) is $v\cos\theta$ where $\theta$ is the angle between the plane of the stars' orbits and the line-of-sight line from the Earth. If the stars eclipse one another at a certain point in their orbit (eclipsing binaries) then we know that the Earth is in their orbital plane, so $\theta=0$ and the ...


0

A 2013 paper by Shtanov and Sahni (already mentioned by Ben Crowell in the comments) says that the modes grow exponentially in conformal coordinates, and Barrow et al overlooked the fact that the conformal time changes very little during and after inflation. A 2014 preprint by Tsagas, one of the authors of the original paper, cites Shtanov and Sahni and ...


0

galaxy come in many different sizes: some of the small-er ones do rotate ["orbit"] around the edge of a large galaxy ... one can also visualize galaxy-clusters, in which the entire cluster rotates .....


1

The key statement is that the $a_{\ell,m}$ are independent Gaussian random variables. For each $\ell$, there are $2\ell+1$ of them. So their sum is, essentially by definition, a chi-squared distribution with $2\ell+1$ degrees of freedom. Now, it is a known fact that the variance of a chi-squared distribution with $k$ degrees of freedom is just $2k$, so ...


0

The total angular momentum of the disk doesn NOT change. There is an outflow of angular momentum due to the shearing of matter rotating at different velocities. Faster inner annuli get a spin down torque from slower outer ones, so they lose angular momentum in favour of the angular of momentum of the outer annuli. I can express this better and answering ...


10

The broadening of emission lines is not due to something that is happening to each individual star, but rather something that affects the whole population. As stars in a galaxy get older, their orbits change relative to the orbits of other stars of the same age. Most relevant for the discussion here, the "velocity dispersion" of a stellar population ...


0

Yes, from this source seems that even very heavy trans-Uranian element like Californium has possibly formed during Crab Nebula supernova explosion, even if this element does not occur naturally in the Earth (with 351 year half life it is too short-lived to be still present). The synthesis of such elements does not create energy for the supernova but probably ...


10

The star that is studied in the paper that you refer to is a very old, very metal-poor "Uranium giant". This is an evolved star with a very deep convective envelope. The Uranium and Thorium that are seen in the atmosphere of the star were not produced in the star. They would have been produced, via the r-process neutron capture mechanism, in the supernova ...


-1

I'm guessing (and I guess breaking the rules, but I'm interested whether my intuition is at all accurate): I've always assumed that although stellar fusion stops at iron, there is plentiful energy from the prior fusion processes available to create the elements heavier than iron, albeit at a net loss of energy. In decreasing and exponentially smaller amounts ...


18

It is a myth that heavier elements than iron are not produced in stars, slow-neutron-capture-process is a nucleosynthesis process that occurs at relatively low neutron density and intermediate temperature conditions in large stars. For details of what elements are produced and about the process itself, see S-process.


19

The heavier-than-iron elements are not formed during stellar fusion, but they are formed during supernovae. Then the oldest stars cannot have these heavier elements, but new generations, formed from 'recycled' material of other stars that went supernova can. See Stellar populations . There are heavier that iron elements on Earth, the Earth was formed from ...


0

Cosmological redshift actually can be explained kinematically and the special relativistic Doppler's formula $$ 1 + z = \gamma \left( 1 - \frac{ \mathbf v \cdot \mathbf w }{\|\mathbf w\|} \right) $$ remains valid in a curved background as long as you realize that you need to determine the 4-velocity $V^\mu = (\gamma, \gamma \mathbf v)$ via parallel transport ...


0

The expansion of the Universe, described by Hubble's constant, is measured in "velocity per distance"-- km/s per Mpc. So to say it's faster than light is not quite correct. Though the idea that somewhere, faraway, space is moving away from us at faster than light is correct, and it introduces the notion of a cosmological horizon. Rather than recite it, I'll ...


-1

The part of the universe which is expanding faster than the speed of light can't be observed.There is a notion of observable universe. There is no law which states that objects cant move faster than the speed of light. It states that information can't travel faster than speed of light. When universe is expanding faster than speed of light it doesn't violate ...


3

The progenitor bias arises in attempts to study early-type (elliptical) galaxies at higher redshift. The desire is to choose a sample of galaxies at high $z$ that are the analogs of the galaxies that evolved to form the low $z$ sample. The bias arises if one chooses a sample of only early-types at high $z$. Because some late-types eventually evolve into ...


5

Whether the dark energy is constant or not will ultimately be determined by experiment. At the moment there is no evidence that the dark energy is changing, but the experimental errors are still quite large so a change is not ruled out. There are lots of papers on this subject, but as yet no firm conclusions. It is important to be clear that dark energy ...


0

As I mentioned in first comment underneath original question, the Hertzsprung-Russell diagram is probably the single best measure of "average" if you have to choose just one measure. But since there seem to be many more additional answers and comments than I'd assumed would be posted, let me add the following remarks that seem to have been overlooked. In ...


0

The shape of the Sun tells you something about it's mean rotation rate. The faster it spins, the more oblate it gets. See Why is the Sun almost perfectly spherical? Of course you are not sensitive to the nuances of latitudinal or radial differential rotation. For the former you really do need to "see" the surface, for the latter you need helioseismology ...


0

Your question is a bit tricky, because "most common" is a very vague statement. So I'll interpret "most common" as "textbook". Opening Binney & Tremaine's Galactic Dynamics (2ed) text, which is as far as I know the standard text for the dynamics of stellar systems, the models described in the relevant section 4.3 are Plummer, isothermal sphere, King, ...


7

The Sun is decidedly NOT an average star, except it is on the hydrogen-burning main sequence, where $\sim 90$% of stars in the local stellar population are found. A much better appreciation of the Sun's "averageness" is gained from looking at a Hertzsprung-Russell diagram (luminosity vs effective temperature or equivalently, absolute magnitude versus colour)...


11

As a followup to @honeste_vivere's note about the H-R diagram, our sun really is living in the middle of average-town: The image, from Wikipedia plots 22,000 stars. When you plot a star's temperature vs brightness, they seem to follow certain patterns. Our star lies right in the middle of the boring main sequence.


12

Why is the Sun called an “average star”? The sun is a yellow dwarf star and dwarf stars are the most common in the universe. Technically, the sun's spectrum peaks in the range of green light, but we see it as effectively white. I am inclined to think if an astronomer or physicist told you the sun was "average" it is because it is part of the main ...


40

Describing the sun as an average star is probably more of a reaction against the idea that there is something unique about it. Obviously there is for us, since it is the star that we happen to be in orbit around, and much closer to than any other star, and hence historically the sun has been considered rather unique. But over the centuries we've discovered ...


1

It is highly unlikely. Neutrinos are known to account for a small part. The problem with neutrinos is that they are low mass and usually highly relativistic. DM needs to be made up of particles or objects that are slow and non relativistic. DM concentrates around galaxies, and tends to stay around, DM needs to be cold to stay around. As for cosmic rays it'...


2

Not directly, because these two quantities are not known to correlate as strongly as other "intermediate" relations. But they do correlate, with bigger disks in bigger haloes; it's fairly intuitive. You could construct a relation from e.g. The Luminosity-Size and Mass-Size Relations of Galaxies out to z ~ 3 (or more specific to B-band, but more simulation ...


2

Well, first of all, that's not the NFW profile, instead you should have: $$\rho(r) = \frac{\rho_0}{\frac{r}{r_s}(1+\frac{r}{r_s})^2}$$ The radius $r_s$ is usually called the scale radius, and is the place where the logarithmic derivative of the density is $-2$. This isn't especially physically meaningful, but is mathematically convenient. The integral is ...


1

On top of the other excellent answers I'd like to point out that the accretion rate of dark matter particles is believed to be much smaller. The reason matter in accretion disk is being accreted rapidly is because they lose energy from electromagnetic radiation. For dark matter particles, in practice the only way it can be accreted is if the particle ...


0

The mass of a star $M$ is given by the integral over its density distribution: $$ M=\int_0^R 4 \pi r^2 \rho (r) \, dr $$ So only because the star is big (large radius $R$) does not necessarily mean that it is heavy. It depends on its density profile. This profile depends on central pressure, equation of state, temperature-/luminosity-profile and more. The ...


-1

A main sequence star burns hydrogen in a shell around the core, and the shell migrates out as it burns hydrogen into helium. Eventually there is not enough material in the outer regions to maintain the pressure necessary to maintain fusion. The core then begins to collapse and helium burning starts, and it begins with a rapid "helium flash." This causes ...


0

This will be a short answer, not going very deep into how stars work. Basically, a star is a ball of gas which is more or less in equilibrium between collapse due to gravitation and expansion due to heat. The radius of the star is determined by this equilibrium. A star which is more massive can have a smaller radius due to a large gravitational pull ...


1

In principle there is no upper limit to the size of objects that can be gravitationally bound. In practice the largest gravitationally bound objects are galaxy superclusters, which have sizes up to around 100 million light-years. The formation of anything bigger than this has been prevented by the expansion of the universe. From your question I'd guess you ...


1

The gravitational force is always attractive (and never repulsive) so the extra mass cannot be expelled by gravity. From a classical newtonian point of view the magnitude of the gravitational force between two masses is given by the following equation: $$ \mathbf{F} = -G \frac{m_1m_2}{\left|\mathbf{r_{12}}\right|^2}\mathbf{\hat{r}_{12}}$$ So you can see ...


-2

The density is not high enough in Jupiter or Saturn to achieve a sustainable environment for the fusion of hydrogen and helium.


0

What you write down is not the pressure of a BEC, but that of a free Bose gas at zero chemical potential (you also missed a factor of m). The pressure of a non-interacting BEC is equal to this result only at $T=T_c$, where the condensate fraction is zero. Below $T_c$, the BEC is a mixture of a superfluid component (with zero pressure), and a normal component ...


4

Good numbers for this have only been coming out for a decade or so, so its a relatively new topic. There does seem to be a strong tendency for dwarf and satellite galaxies to have much lower mass-to-light ratios, and correspondingly smaller baryon-to-DM ratios. See, for example, Stringer+2009, Strigari+2008. These observations are backed up by simulations ...


-1

The pressure of a Bose-Einstein condensate $p = kT\frac{g}{\lambda^3}\zeta(5/2)$ that composes a star, which is an interesting conjecture in a way, is countered by gravity. The hydrostatic condition for a star is $$ \frac{dp}{dr} = \frac{GM(r)\rho(r)}{r^2} $$ The temperature is set by the number of particles, which for a star is considerable $n \sim 10^{60}$,...


-1

This is based on a more explainable or commonly understandable explanation of the theory of relativity. Imagine the space time fabric to be an infinite sized piece of cloth now if a a ball of iron is my sun then it will bend the cloth and attract the objects on the cloth that come in the range of its bending. ...


3

Well yes, maybe, but they are called planets. So fission in stars? No, but maybe in planets. I do not know what the status of this is, but the core of the Earth is heated by weak and maybe strong nuclear processes. The standard model is that weak nuclear decay. The major heat-producing isotopes within Earth are potassium-40, uranium-238, uranium-235, and ...


0

No. Basically because stars, by definition, are fusion reactions. It's feasible for a star to contain Uranium - the Earth, for example, contains Uranium, and could (in fairly extreme circumstances) collect enough matter to become a gas giant, continue to collect matter and then start fusing to become a star. But even then, you'd just have a star containing ...


2

Apart from the heating due to sound absorption, as per the comment by HolgerFiedler, I don't think you will find a mechanism that can radiate due to polarization effects in the medium. Any EM radiation would be at the frequency of your acoustic waves. With the difference between the speed of sound and the speed of light, that would be very-long-wave ...


5

Cosmic Rays are most often high-energy particles, mostly protons and alpha particles accelerated to high velocities by cosmic magnetic fields. They do not show up in the microwave wavelength range that comprise the CMB. As @ACuriousMind says in the comment, there is contamination in the CMB, but this is mainly due to Galactic dust and Bremsstrahlung from ...


0

I'm fairly certain that translational velocity is equal to real velocity. Remember, translational motion is defined as: the motion by which a body shifts from one point in space to another. On the other hand, radial velocity only applies parallel to the line of sight of the observer, ie. the speed at which it moves towards or away from Earth. Radial velocity ...


2

You need to look in to the GZK cutoff. In 1966, Greisen, Kuzmin and Zatsepin calculated that above a threshold of $5\times10^{19} eV$ cosmic ray protons would lose energy to photo-pion production on the cosmic microwave background fairly rapidly. The consequence of this is that cosmic rays above that energy can't travel more than about $50 Mpc$ without ...



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