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4

You are looking for the initial mass function (IMF). This tells the probability of finding a star of mass $m$ (in solar mass units). The prototypical IMF is the Salpeter IMF, $$ \phi(m)\sim m^{-2.35} $$ This gives a decent and quick approximation, though a multi-power-law fit seems to be better; thus we have the Kroupa 2001 model: $$ \phi(m)\sim ...


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If we start with the equation you quote: $$ \rho_c = \frac{3H^2}{8\pi G} $$ and rewrite it as: $$ \rho_c = \frac{3}{8\pi G} H^2 $$ then it's the same as your equation: $$ \rho_c = E H^2 $$ because all you've done is to replace the constant factor of $3/8\pi G$ with the symbol $E$. This is of couse a perfectly reasonable thing to do, but it isn't new in ...


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The ejecta of a supernova does indeed move at a fraction of the speed of light (somewhere around the 10% mark). However, it does not remain at this speed forever. As the supernova ejecta expands outwards, it creates a shell of material that is actually gathering up particles in the ambient medium (typical interstellar densities are around 1 particle per ...


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In short, yes it completely makes sense to keep searching for Dark Matter using Earth-based direct-detection equipment. Even in the case that there is no significant amount of dark matter in the Solar vicinity, that is the only area we are currently able to search using direct-detection. So it makes sense that if we search for dark matter (and we should ...


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I have taken the revised Hipparcos parallax catalogue, produced by van Leeuwen (2007, Astronomy & Astrophysics, 474, 653) and taken a subset of stars with Hpmag <6 (i.e. roughly the naked eye limit) and accepted only those objects with a parallax/error in parallax > 2.5. Anything with a larger fractional error in parallax really can't be trusted. If ...


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This is actually more subtle than you would think it would be. First, remember that a fourier transform is defined, for some time-dependent signal $F(t)$, as${}^{1}$: $$F(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} dt\, e^{i\omega\,t} F(t)$$ Well, this is great in special relativity, but in general relativity, what time do we actually use? ...


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A telescope with two convex (converging) lenses is a Keplerian telescope. The lens with the longer focal length is the objective, and the shorter focal length lens is the eyepiece. Since it is explicitly stated that the lenses are thin, you can use the thin lens equations: $$ \frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f} $$ where $d_i$ is the distance to the ...


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The radial velocity $V_r$ is the velocity resolved along the dashed line (the line of sight from the Sun to the star) with respect to the Sun. If $V_c$ is the velocity of the star, then the component along the dashed line is $V_c \cos (\alpha)$. We then have to subtract the velocity of the Sun resolved in the same direction. This is $V_{c,0} \sin (l)$. Hence ...


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This is a very late response, but there is no accepted answer as of yet, and none of the answer quite hit the mark. Regarding the magical collision hypothesis, that smacks of being rather non-scientific. Scientists as well as Missourians are wont to say, "Show me!" Other than the fact that Venus's rotation is anomalous, what, exactly, is the evidence for ...


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The cosmic microwave background gives us a convenient rest frame in that it represents the average distribution of matter in the universe. So if the CMB looks isotropic, i.e. there is no Doppler shift in different directions then this is a plausible definition of stationary. A velocity relative to the CMB would then indicate a peculiar velocity. The Milky ...


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NASA still images; audio files; video; and computer files used in the rendition of 3-dimensional models, such as texture maps and polygon data in any format, generally are not copyrighted. You may use NASA imagery, video, audio, and data files used for the rendition of 3-dimensional models for educational or informational purposes, including photo ...


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The basis for a stellar engine (according to the Wikipedia link) is that the radiation pressure from the star is used to create thrust, so the question reduces to "Do brown dwarfs emit enough radiation"? Stars like the Sun do emit radiation, and lots of it - in the Sun's case, $3.846 \times 10^{26}$ watts. That's a lot of power! Brown dwarfs can't undergo ...



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