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In vector notation Newtonian force of gravity is $$\vec F = \frac{GMm\vec r}{r^3},$$ where $r = \sqrt{(x_1 - x'_1)^2 + (x_2 - x'_2)^2 + (x_3 - x'_3)^2}$ and the radial vector $\vec r = \vec x - \vec x'$. we can consider the unit vector $\hat r = \frac{\vec r}{r}$ We can the write the vector notation as $$\vec F = \frac{GMm}{r^2}\hat r = F_g\hat r.$$ ...

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$$F_g=\frac{Gm_1m_2}{|r_{ij}|^3} \vec{r}_{ij}=\frac{Gm_1m_2}{|r_{ij}|^3} |r_{ij}|\hat{r}_{ij}=\frac{Gm_1m_2}{|r_{ij}|^2} \hat{r}_{ij}.$$ It's just one way textbooks write it, and is exactly equivalent to the right-most expression, which is probably the most obvious way to write the gravitational force in vector notation.

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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 ...

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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 ...

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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 ... -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 ... 0 Maybe you're looking for something like this http://arxiv.org/pdf/astro-ph/0001295v1.pdf or https://arxiv.org/pdf/1205.5484v2.pdf You might also search for "Modelling the evolution of solar-mass stars with a range of metallicities using MESA E.F. Jonesa and P.M. Gore " They give different results, they cannot all be right, and might all be wrong. ... 0 Expanding Universe: The idea of the Universe expanding is often described using the analogy of an inflating balloon. It is tempting as a new physics student to imagine the expansion as galaxies whizzing away from each other through some 'medium', however in actuality it is spacetime itself that is expanding. If we glue some pieces of confetti (representing ... 2 Hints: Note that the derivative of the sign function $${\rm sgn}^{\prime}(z)~=~2\delta(z) \tag{A}$$ is twice the Dirac delta distribution. This fact seems to be at the heart of OP's question. Repeated differentiations of the Mestel disk potential $$\Phi~:=~ v_0^2 \ln(r+|z|), \qquad r~:=~\sqrt{R^2+z^2}, \tag{B}$$ leads to $$\frac{\partial \Phi}{\partial ... 0 The images are taken by very very powerful telescopes, like Hubble and Kepler. The images are also taken by space stations and satellites in form of radio waves which are electromagnetic waves. So the images are not damaged since radio waves need no medium to travel/ But the images are taken in black and white, which can later be coloured according to their ... 1 The question operates under a false premise. It is not the case that fitting a blackbody spectrum to the Sun gives you its "surface temperature". The Sun does not have a black body spectrum, although sometimes that approximation is made. A star also does not have a single "surface temperature". What you can do is divide the luminosity of the Sun by 4\pi ... 3 Many spectral lines are very sensitive to the surface gravity of the star - which enables a distinction between dwarfs and giants, because a giant's surface gravity is factors of \sim 100 lower than that of a dwarf of the same temperature. The reason that surface gravity plays a role is via hydrostatic equilibrium; the densities and pressures in a gas ... 5 It depends on what you mean by "real". Let's start with the Andromeda galaxy, which is easily visible with the unaided eye from the Northern hemisphere (and probably from a large part of the Southern hemisphere, too), but it does not look anything like what you are used seeing in photographs. It looks like a small, pale cloud in a dark patch of the sky and ... 0 How do we get images of galaxies? With cammeras and film, with or without a telescope that magnifies what one sees, as here: NGC 4414, a typical spiral galaxy in the constellation Coma Berenices, is about 55,000 light-years in diameter and approximately 60 million light-years away from Earth . You ask: I mean, suppose you have to take a ... 1 There are many mechanisms which can contribute to broadening spectral lines. Usually, one way or another, the atoms have a broad range of random velocities which cause doppler shifts of varying amounts, broadening the line. One the more fundamental cases is simple 'thermal broadening', where the velocity is from thermal motion. The hotter the gas is, the ... 3 When black holes form from the collapse of stars, we think they usually produce bright supernovae explosions which release tremendous amounts of energy. Some models suggest that a non-negligible fraction of stars which produce black holes may not produce normal supernovae, these are often called 'failed supernovae', see for example astrobites: Gone Without ... 1 A black hole m\approx300,000M_{sun} at the center of galaxy NGC 4845 is thought to have a giant gas planet with the mass no larger than that of a brown star in orbit, though it is currently being drained into the black hole. (And by currently I mean what we are seeing right now, which happened a long time ago). The gas giant is thought to have been ... 1 The key word, to answer your question, is wavelength, as in: which wavelengths are you interested in? So far, GRBs have been seen in all wavebands, from TeV down to radio frequencies (of course, not all GRBs have been observed at all frequencies). The waveband in which GRBs release most of their energy is, unsurprisingly, that of \gamma rays, with hard ... 3 The surface temperature of the star has no direct bearing on its metal content. Most stars in the immediate vicinity of the Sun have a very similar metal content. What you are talking about is how this metal content affects the observed spectrum of the star. If the star's photosphere is very hot then the metals become ionised and you don't see the (for ... 1 In a realistic scenario, not very far, maybe just a few miles if you're discussing fragments large enough to leave impact craters. A fair bit further than that if you're talking smaller debris. The distance depending on how high the meteor broke apart and how fast and at what angle it was approaching. If you consider that the dense part of the atmosphere ... 0 The information paradox has no particular connection to electromagnetism. Hawking radiation is not just photons, it's any sort of particle. And Hawking radiation in itself doesn't solve the information paradox - the problem is that Hawking radiation is supposed to be thermal, so quantum information of infalling objects has been irreversibly lost, but that ... 2 First, let me explain the question a little further. The Hulse-Taylor binary is a binary system composed of two neutron stars orbiting each other. Each star is an extended body, and is in the gravitational field of the other, so should experience tidal forces, because one part of the star is closer than another to the opposite star, so the gravitational ... 1 Yes. Black Holes (BH) can grow from accreting anything with energy --- including dark matter (DM). I'm not entirely clear on the second part of your question, but probably the most important thing to keep in mind is that the black hole information paradox is still unresolved. Answering how information is not lost for any type of particle, including DM, ... 3 The "strain" (usually h) is a measure of the 'distortion' (in the metric) from a gravitational wave (GW). The energy carried by GW (its 'luminosity' of sorts) scales like, L_{gw} \propto h^2, and for energy to be conserved as it radiates away, we need L_{gw} \propto r^{-2}, so we find that,$$h \propto r^{-1}$$So your latter estimate is more ... 1 They both refer to the same thing. The habitable zone is also called the Goldilocks zone, a metaphor of the children's fairy tale of Goldilocks and the Three Bears, in which a little girl chooses from sets of three items, ignoring the ones that are too extreme (large or small, hot or cold, etc.), and settling on the one in the middle, which is ... 3 All the neutrino detectors we have or might build will have course angular resolution because they detect the direction of scattered products of neutrino interactions rather than the direction of the neutrinos themselves. Worse, the solar neutrinos are relatively low energy (a few MeV), which means the scattering angles are large. Yes, we could accumulate ... 0 What you are missing is that when charged particles are accelerated away from the magnetic poles of the neutron star, they are tied to helical paths along the field lines. Since the particles are highly relativistic, the consequent curvature and synchrotron radiation is doppler boosted and doppler beamed in the forward direction. The beaming opening angle ... 0 From FURTHER EVIDENCE FOR COLLIMATED PARTICLE BEAMS FROM PULSARS AND PRECESSION (2007): "An overwhelming majority of the observers and theorists interpreting these observations seem to suggest and endorse the following basic picture. The jetlike features nearly along the symmetry axis, bisecting the arcs and the diffuse glow spread about them, are ... 0 Magnetic fields. The beams are generated from charge particles moving in magnetic fields. The pulsar is the result of a large star collapsing and the magnetic field of the original star is compressed, making a very strong field. 4 Data is taken at different frequencies because different frequencies contain different information. The data here covers a factor of 9 in frequency (equivalently wavelength). Compare this to the factor of 2 accessible by human vision. Given two objects that have the same power at one end of the frequency range, they could very well differ at the other end. ... 0 Hopefully an observer can chime in with better advice... but: Generally, people calculate the (average) brightness in rings around the center, then plot that brightness as a function of the projected radius (i.e. angle, e.g. in arcseconds). You can check to make sure your solid angle is correct (for example you say \delta is the complementary angle to ... 0 The scale-factor of the universe changes significantly over that period of time, so you can't calculate distance as simply d = v \cdot t. You have to actually integrate over the expanding spacetime metric, i.e.$$s = \int_{z_1}^{z_2} \frac{c_s}{H(z)} dz$$Where c_s is the speed of sound (and I think that should be c/\sqrt{3} instead of c/2, ... 7 We want the Newtonian limit of the Einstein Field equations for nonzero vacuum energy(=cosmological constant). As \rho_\mathrm{vac}=\Lambda/4\pi G is a mass(=energy) density, Poisson equation is$$ \Delta\Phi=4\pi G\rho(\boldsymbol r)-\Lambda \tag{1} $$If we assume spherical symmetry, and point-like source \rho\sim\delta(\boldsymbol r), the ... 3 You get an extra term that increases with r:$$a = -\frac{G\cdot M}{r^2} + j\cdot r$$with j as the repulsive component. 8 Background Stars are composed of plasmas, which are an ionized gas that exhibit a collective behavior much like a fluid. There are two important aspects of plasmas to keep in mind. The first is that they act like very highly conductive metals in that the electrons can move very freely in order to cancel out any charge imbalance. The consequence is that ... 10 Overall, a star stays more or less neutral. This is true for all stellar objects beside black holes. I am using a simple calculation that can be found in a footnote of https://arxiv.org/abs/1001.3294 on p. 11 chap. 2. Suppose the star has an overall charge of Z times the elementary charge,$Ze\$, and we consider the Coulomb repulsion of a test particle, say ...

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