New answers tagged

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If the black hole and asteroid have the same mass, the black hole would likely tunnel right through the asteroid and continue on its way. The Schwarzschild radius of an asteroid is well under a millimeter in size, and unless we're talking about a very large asteroid like Ceres the gravitational field of both bodies is very weak. EDIT: of course the exact ...


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From our outside frame of reference, the asteroid would freeze as it arrived at the event horizon. Matter that falls into a black hole experiences large tidal force that disrupt it and spread it out into an accretion disk. The disk gets compressed and heated as it flows into the black hole. I believe the asteroid would be spread out in a belt all around the ...


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The most recent effort to establish current elemental quantities in the sun’s photosphere that Im aware of is this article from 2003: https://iopscience.iop.org/article/10.1086/375492/pdf Table 1 on page 3 shows the estimate for the current sun, and later tables are protosolar. The first number is roughly natural log (For anyone new to that, think of it as ...


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This is not quite the answer to your question, since nobody can reliably put a numerical probability on something that has never been observed to occur in nature. i.e. A close-to single-mass population has never been observed as the outcome of a star formation process and we have never seen another galaxies made exclusively of solar-type stars. This is also ...


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In the solar neighbourhood G-type stars are about $p\approx $ 7%. So were you to generate $N$ stars entirely randomly and independently, the probability of them all being G-type would be $0.07^N$. For $N=10^{11}$ this is $10^{-15490195998.6}\approx 0 $. Changing the exact numbers does not change the conclusion much. Clearly to get a G-type galaxy you need to ...


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It is likely that the Milky Way (or, at least, a cluster of stars that later became the Milky Way) was formed soon after the Big Bang. We know this because we have found several stars in the Milky Way that are estimated to be more than $13$ billion years old. These old stars are called Population II stars and they contain very, very small amounts of elements ...


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There are many, many factors in the formation of stars that determine their type and size. Before stars form, they are called protostars, which are collections of gas that have collapsed from large molecular clouds. The protostar's phase of evolution lasts about 100,000 years where over time, gravitational forces and pressure increase, causing the protostar ...


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The biggest short-term effect on the apparent position of a nearby star relative to background stars is the annual parallax due to the Earth's motion in its orbit. Over longer periods of time the proper motion of the star will have an effect, as will the precession and nutation of the Earth's axis. (Note that this question has been completely changed since I ...


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Your second logic - calculating group velocity via a dispersion relation - is reasonably sound and does produce the correct answer, although I have two minor caveats: We can assume that the plasma frequency $f_p\ll f$ (apologies for using different notation) given the typical properties of the interstellar medium. The electron number density is on the order ...


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I think what you are looking for is the Vis-viva equation. $$v^2 = GM\left(\frac2r-\frac1a\right)$$ $v$ is the relative speed of the two bodies $r$ is the distance between the two bodies $a$ is the length of the semi-major axis ($a > 0$ for ellipses, $a = \infty$ or $1/a = 0$ for parabolas, and $a < 0$ for hyperbolas) $G$ is the gravitational constant $...


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When the sun was forming, the heat from compression would have been the primary source of radiation. It was also the heat that initiated the fusion reactions at the core. Today, very little compression is going on because the radiation from the core supports the weight of the gasses at most levels. What little heat from compression that occurs would result ...


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An interesting question. It's complicated because its hard to say how much infra red radiation "causes" radiation from the sun. Those terms aren't typically what we use. Or, if we do, its to point out that almost 100% of the energy we get from the sun is from infrared radiation off the upper layers. Only the tiniest sliver of the energy of ...


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It depends on what the dust is made of, specifically how many coulombs it has (a surplus or deficit of electrons). As mass can be measured in energy, electronvolts/speed-of-light2, and eV is the energy of 1 electron subject to 1 joule of electric potential difference (or one volt [force exerted over distance behind a coulomb's worth of electrons] with the ...


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A very approximate back of the envelope calculation. The escape velocity for an orbiting body is Sqrt(2) x orbital-velocity. The moon is at radius ~3.84E8 metres so circumference of orbit is ~2.2E9 metres. One orbit takes ~28 days or 2.0E6 seconds, so orbital velocity is ~1E3 m/s. We need to increase that by ~4E2 m/s in order for it to achieve escape ...


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I interpret your question in the following way. You have a theoretical model for the $C_\ell$ which says that the $C_\ell$ are independent of $\ell$. We can express this by saying that \begin{equation} C_\ell = \mathcal{D} \end{equation} where $\mathcal{D}$ does not depend on $\ell$. In this expression, all quantities are theoretical in the sense that they ...


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Yes, inverse beta decay results in the removal of electrons from the degenerate electron gas. At fixed volume, this would lower the electron number density and hence lower the degeneracy pressure. It is possible that the star would then contract, the electron number density would be raised and the increased pressure would still support the star. However, ...


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The decay of the ringdown modes is exponential, with a decay "half-time" determined by the mass and spin of the black hole (and the mode numbers of the mode). So (at least in the context of classical general relativity) this decay takes forever. However, at some point the decaying field will be so weak that we need to start worrying about the ...


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The energy comes from the binary system itself and ultimately from the mass-energy of the binary components. The binding energy of a binary system (the sum of its kinetic and potential energies) is negative. The acceleration of these masses, or more precisely, the accelerating mass quadrupole moment, produces gravitational waves that carry away energy and ...


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any moving object having a mass would create gravitational waves This is not true in general. At the lowest order, the rate of energy lost as gravitational waves is proportional to the third time derivative of the quadrupole moment tensor of the system. For example, a mass moving at constant velocity will not radiate gravitational waves. Neither will a ...


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The objects that are radiating gravitational waves must be losing the energy and angular momentum they had in their orbits. For example, when two black holes spiral in toward each other, they radiate away this energy and angular momentum as their orbit decays and they eventually merge. How does a system, for instance earth-moon orbit, can be stable and not ...


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In addition to the other answers, here is an illustration showing how the collision might look like as seen from earth: (Open image in new tab to enlarge) This series of photo illustrations shows the predicted merger between our Milky Way galaxy and the neighboring Andromeda galaxy. First Row, Left: Present day. First Row, Right: In 2 billion years ...


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The size of a neutron star's magnetosphere (in-so-far as it can be approximated as something spherical) is, to within a small numerical factor, given by the Alfven radius. This is where the magnetic energy density equals the kinetic energy density of the surrounding gas/plasma. Using Gaussian units, the energy density is $B^2/8\pi$, where $B$ is the magnetic ...


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These frequencies $\omega_R$, $\omega_\phi$, $\omega_z$ would be the same for a Keplerian orbit, but could be quite different otherwise. If the orbit is non-Keplerian then you need to define what is meant by "the orbital period". As an example, for the Sun, the three periodicities would be about 250 million years in $\phi$, about 140 million years ...


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Most likely nothing special will be seen from Earth within the lifetime of the Sun. The most recent results (see discussion in Will Milky way and Andromeda collide for sure? ) suggests that the Milky Way and Andromeda will have an initial "glancing blow" with a pericentre (i.e. closest approach) of 75 kpc (220,000 light years) about 5 billion years ...


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The merger would be indirectly noticeable due to a dramatic burst of star formation and supernovas. The gas of the two galaxies will meet at high velocity, clump, and produce new stars. Some will be very heavy and bright, resulting in supernovas and gamma ray bursts: the merged galaxy may become a bit too risky for planet-bound civilizations dependent on an ...


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The Andromeda galaxy is now approaching us at $110$ km/s. Do you notice anything? No. It takes millions of years for the position to change noticeably. It will be the same when the collision occurs. Our galaxy is somewhere around $1000$ light years thick. Since the Andromeda galaxy is traveling at about $10^{-3}$ c, it would take about a million years to ...


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Dark energy does NOT increase the distance between stars or planets. It does NOT affect matter at the molecular or particle level. It does not even act on galaxies. All these objects are bound objects, and thus the tiny dark energy is over-compensated by far. Dark energy is only relevant in essentially empty space in between galaxy clusters, or the universe ...


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