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Unless the observer is at a geographic pole, the celestial equator is not parallel to the horizon but crosses it at an angle equal to the observer's colatitude. A star's path across the sky is parallel to the celestial equator, with maximum altitude midway between the eastern and western horizons, i.e. on the meridian. Rendered by Stellarium


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These are the names of recombination lines. In terms of the Bohr model of the atom: The electron in a Bohr atom can fall from the level ($n+\Delta n$) to $n$, where $\Delta n$ and $n$ are any natural numbers (1, 2, 3, …) by emitting a photon whose energy equals the energy difference $\Delta E$ between the initial and final levels. Such spectral lines are ...


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Zero shadow isn't required. Eratosthenes could have arranged with someone to measure the shadow of a stick, any stick, at local apparent noon, on any given day, in some location known to be due north or south of the location where Eratosthenes would be doing the same thing. The someone would then report to him the length of the shadow and the height of the ...


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It's not necessary that there be zero shadow. The altitude $\alpha$ of an object when it crosses the local meridian on the celestial sphere is $\alpha = \delta + 90^\circ - \lambda$, where $\delta$ is the object's declination and $\lambda$ is the latitude of the observation point. This means that if you carefully measure the altitude of the same object at ...


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As previously mentioned, most nebulous Deep Sky Objects are difficult to view with a traditional eyepiece. Electronic Eyepieces are a recent technological development that replaces the traditional eyepiece with a small camera. The cameras image is viewed on a monitor screen instead of through the eyepiece. These cameras are much more sensitive than the eye. ...


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If an object has a measured brightness, then it's extinction-corrected brightness will always be higher. Hence the extinction-corrected magnitude will always be smaller than the uncorrected (observed) magnitude.


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The Sun will not become a red giant for approximately 7 billion years, so there is little chance that Proxima Cen will be anywhere near it. The Sun goes through TWO red giant phases. In the first, hydrogen shell-burning phase, the Sun will achieve a peak absolute V magnitude of around -3 at the tip of the first ascent red giant branch. It wil have a ...


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The sun, when it becomes a red giant, will be more 100 times more luminous than it is now, but that only gives it an absolute magnitude of about -1, and an apparent magnitude at 4 light-years of about -5: similar to Venus. It is possible for this to cast a very faint shadow, but you need a very dark place to see it. The sun will still be much too small to ...


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Unfortunately, the limit of the observable universe isn't just "photons from the big bang arriving to us now" - so we won't see anything past it because nothing existed before; anything outside that sphere didn't arrive yet. This limit is called the Hubble Sphere, and is only the "first stage" of unobservability - and indeed, as time ...


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The Great red spot looks oval from above but is actually saucer shaped as the storm curves towards the center and the deepest point lies at the center. It is 16350 km wide, so the height is negligible compared to the width. The storm has its roots beneath the surface of the atmosphere. NASA’s Juno spacecraft calculated that the storm penetrate about 300 ...


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I think I get it - you are trying to calculate a luminosity based on some assumption about how much light is reflected/scattered from the Sun towards the Earth - hence the relevance of the flux from the Sun at the comet. I think it's very hard to adopt this approach. There is uncertainty about what is the appropriate geometry to use for the comet and how the ...


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In principle yes, in practice no. The telescope is good enough, but the CCD camera will saturate, preventing a good positional measurement. Salient facts. The parallax to Betelgeuse as seen between Earth and New Horizons will be about 250 mas. The current distance uncertainty is ~20%, so to better that, the position of Betelegeuse measured by New Horizons ...


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Here's a screenshot from Stellarium (it's free and a fantastic tool for observers). I've set the location to Amsterdam and the time to 21:42 (local time) on the 15th of October. At this time Mars was just over 116° from north and 25° above the horizon. It's very bright at the moment. I was observing it a couple of nights ago and watched an airliner pass ...


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Is the Object in this Photo Mars? Almost certainly yes if it didn't move like a plane or hover and make noise like a quadcopter. in-the-sky.org' skymap shows Mars at 117 degrees East and 25 degrees altitude at 21:43 local time in Woerden in central Netherlands yesterday. Clicking on Mars yields more information including a charts showing that it's close to ...


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The Moon is visible only when it is above the horizon and not too near the Sun. Those times depend on its position in the sky relative to the Sun, and the lunar phase is related: 🌒 evening only 🌓 noon to midnight (90° east of Sun) 🌔 afternoon to early morning 🌕 sunset to sunrise (opposite Sun) 🌖 late evening to late morning 🌗 midnight to noon (90° west ...


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Yes, in the time it takes light — or, in this case, gravitational waves (GWs) from the black hole merger event GW190521 — to travel from a source to an observer, the Universe expands, thus increasing the distance further. Various distance terms In the following, "$\mathrm{Glyr}$" means a distance of a billion lightyears, while "$\mathrm{Gyr}$&...


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If you are looking for images of a particular moving object, the Solar System Object Image Search (SSOIS) at the Canadian Astronomy Data Centre (CADC) has indexed the Pan-STARRS collection. You can enter an object name, orbital elements, an arc or an ephemeris and find images of that object. https://www.cadc-ccda.hia-iha.nrc-cnrc.gc.ca/en/ssois/


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There are several different quantities of this sort that you can define, and the definitions are fairly confusing. Hopefully the following diagram will make things clearer. Z <- future infinity / \ / \ / \ D C B A B C D <- now ...


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Like most asteroids and comets, 1I/'Oumuamua's trajectory was determined entirely by measuring its position in optical images over several days. The earliest data came from automated, ground-based surveys such as Pan-STARRS and the Catalina Sky Survey, then targeted follow-ups from various other asteroid observers. The object's evident extrasolar origin ...


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You're completely correct! The farthest we can see (in principle, not in practice) is called the particle horizon. Currently, the distance to the particle horizon is $d_\mathrm{P} \simeq 46\,\mathrm{Glyr}$, but as time goes on, light from more and more distant regions will reach us. If the Universe contains only "regular stuff" such as normal ...


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They look like radio telescopes, because that is what they are. Some radio telescopes have been outfitted to transmit rather than receive, requiring using a different radio telescope to receive the reflected signal. Other radio telescopes have been outfitted to transmit as well as receive in a pulsed manner. These can be used for radar astronomy without the ...


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While this is not exactly how folding works, one-dimensional data encourages trying some simple statistical tallying first. What very easily reveals those periodicities is just tallying up the deltas, with an appropriate fuzziness for what periodicities are relevant: Deltas [0, 9308, 28857, 36709, 28857, 9308, 28858, 36708, 28858, 9308, 28857, 36708, 9309,...


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An alternative prospect might be direct imaging, although the systems that can detect raise the question of whether or not they count as planets/exomoons. As a possible example, see Lazzoni et al. (2020) "The search for disks or planetary objects around directly imaged companions: A candidate around DH Tau B." DH Tauri B is estimated as having 8–22 ...


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