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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 ...


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You can't just tilt the mirror (for the reasons described in JEB's answer). You can have off-axis parabolic mirrors but they are tricky to make. You either have to machine the surface with a diamond tip CNC machine, or saw an off axis region out of a larger mirror At radio frequencies they are a lot more common. Where you can make the shape from metal ...


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Aberrations. A parabolic mirror perfectly focuses rays along the optic-axis, while off axis rays are blurred--tilt the mirror and you get more light, but it's out of focus. I would guess that the non-tilted mirror has 2nd order aberrations, while tilted one is 1st order: it's a lot worse.


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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 ...


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The Oort cloud is a theoretical object. Its existence is more or less accepted on the basis of "comets have to come from somewhere", but no Oort cloud object has ever been observed (except for some comets that pass near us that presumably came from the cloud). It is theorized that it has two components, one disky and one spherical, both centered on the Solar ...


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If the Oort cloud is included to the Solar System and the calculation, it has a potential to influence the barycenter significantly. The distance of the Oort cloud from the Sun is some 100,000 AU (plus minus a factor of two) and its total mass is around 5 times the Earth's mass. If there were a single orbiting object at the location of the Oort cloud, its ...


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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 ...


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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 ...


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No. If you go to a random spot in the visible universe, you will usually be far from any galaxies because the separation between galaxies is large compared to the size of the galaxies themselves. Since distant galaxies are so dim that we can't even see them, you certainly cannot see your reflection by them.


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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 ...


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Yes. This can be done (and is done typically) using systems of pulsars, especially in 'Pulsar Timing Arrays'. See for example https://en.wikibooks.org/wiki/Pulsars_and_neutron_stars/Using_pulsar_timing_to_study_(and_navigate)_the_solar_system. Pulsars (specifically millisecond pulsars, MSP) can be incredibly accurate clocks. Relative motion between the ...


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I'll add a few more options for getting the ages of stars, beyond the HR diagram technique mentioned in Chris White's answer. If you can get a R=50,000 optical spectrum of a star with decent signal to noise ratio will quite easily give you the temperature (to 100K), surface gravity (to 0.1 dex) and metallicity (to 0.05 dex), plus a host of other elemental ...


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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 ...


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The principle way of detecting planets and discovering galaxies are completely different. In general, distant galaxies are faint and their apparent magnitude is high. You need a collect a lot of photons from these objects to obtain an image with desired signal to noise ratio. The total sky area is in the order of 40000 square degrees. This area has been ...


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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 ...


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For a telescope with an objective lens and an eyepiece the magnification $M$ is defines as $$M = \dfrac{\text{angle subtended by object when viewed through telescope}}{\text{angle subtended by object by naked eye}}$$ If you do not have an eyepiece then you form an image of the object in the focal plane of the objective lens and the angle subtended by the ...


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In the case of the second order acceleration, two clouds are approaching, therefore the energy the charged particle gains comes from the energy of the clouds. In the case of the first order acceleration, the charged particle gains energy as it moves repeatedly through the shock front. The region before the shock front (upstream) moves at higher speed than ...


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Note You should clarify your statement from "...a charged particle cannot gain energy from a magnetic field..." to "...a charged particle cannot gain energy from a static magnetic field..." There is nothing wrong with energy transfer from time-varying magnetic fields. Background If the spatial gradient in the magnetic field is slow enough such that the ...



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