51

There are two reasons that often — but not always — light from galaxies millions and even billions of lightyears away make it through the Universe and down to us: Particle number and particle size First, the intergalactic medium (IGM) is extremely dilute. The number density of particles out there is of the order $n\sim10^{-7}\,\mathrm{cm}^{-3}$, or roughly ...


44

That is a camera artefact caused by the bright sunlight reflecting within the lens on your phone. It’s more pronounced than on a large camera because of the small lens size. This is a secondary image of the sun, as the brightness of the source allows for the reflection to be still intense enough to be detected. Here is a photo I took with the same artefact....


31

M33 does not appear to contain a supermassive black hole: in fact there's no evidence that it contains a central black hole at all. The upper limit on the mass of a central black hole based on the dynamics of the core region is a few thousand solar masses. Merritt et al. (2001) "No Supermassive Black Hole in M33?" derive an upper limit of 3000 solar masses ...


30

I wholeheartedly recommend astrometry.net for this sort of thing. It Just Works(tm); running on your image produced this output with absolutely no hints or guidance from me: For avoidance of doubt, I have no association with astrometry.net.


30

This NASA page says this photo was taken on April 28 2006. Using Celestia, I managed to find the picture from Cassini that best lines up with the photo. It doesn't match up precisely, but that's to be expected as the calculated orbital elements of all these moons (and cassini) in the software won't necessarily match up to reality precisely. Below is the ...


24

As Rob Jeffries says, the universe is mostly empty space. A photon can easily travel thousands of light years without interacting with anything. Most of the interaction would occur when photons entered the earth's atmosphere. The Hubble avoids this. These photos were most likely from combining several viewing sessions giving basically an extended time period ...


23

There's a misconception in your question I don't think the other answers have addressed. If light emitted from the galaxy travels in all directions, then how is it that we can still map out the entire galaxy Light is emitted from the galaxy in all directions. Only a tiny, tiny fraction of it is directed to Earth, and of that, an even tinier fraction ...


23

The orbital elements are on wikipedia: $$e=0.884\ a=0.125'',\ i=134^\circ,\, \Omega=228^\circ$$ (At an assumed distance of 8kpc, $0.125'' = 1000au$) It is the inclination that means that the black hole is not at the focus of the projected ellipse. Imagine a circular orbit, with a central body. If viewed from a distant but high inclination, the orbit is ...


22

We are situated inside the galactic disk. The first image below, showing our approximate location, is a fabrication rather than a photograph. Such a photo would require a space probe to be tens of thousands of light years above the plane of the disk, which is patently infeasible with current technology. We don't even know exactly how many spiral arms our ...


18

A contrast stretch reveals stars down to magnitude 4 or 5. The stars you asked about are Deneb (center) and Vega (bottom). The constellation in the center and below is Cygnus; we also see Cepheus at upper right and part of Draco at lower right. The stars appear as blobs 10-12 arcminutes wide due to several factors: Rotation around the north celestial pole, ...


17

Is that right? Yes. Is the fuzzy one an extended object? That would certainly be my guess (probably a distant galaxy). What causes so many isolated pixels to be so much brighter than the background? Is this just the tail of a statistical distribution of shot noise, or are there other mechanisms that can produce single pixel noise many ...


16

We can calculate the distance we need to be from an object in order for it to be any arbitrary angular size using the formula $D=\frac{d}{2\tan(\frac12\delta)}$. In this case, $D$ is the distance from the object, $d$ is its diameter, and $\delta$ is the angular size. The Milky Way's diameter is 100,000 light years. If we want the Milky Way to appear as ...


16

Astrometry.net has identified your star field as being part of the Andromeda constellation. The diffuse object in the centre of your image is the Andromeda Galaxy (Messier 31). The bright star to the left of it is Mirach (β Andromedae). Given a reasonably dark sky and averted vision, it is possible to see the core of M31 with the naked eye. Since M31 is ...


16

Congratulations on your purchase. The first pictures dont' show anything much. Just a out-of-focus blur. The last one shows Jupiter and three of its moons. I've overlaid the image onto a simulated image from stellarium (at about 10pm BST, the moons move pretty quick so you need an exact time!): You can clearly see which moon is which, and why Europa is ...


15

A self-sufficient orbiting telescope is basically Hubble mkII and would never get off the ground, literally and metaphorically Hubble was expensive because it was state-of-the-art, requiring development of many new systems. The systems it needed to function as a standalone satellite (compared to being attached to the ISS) were cheap by comparison (reaction ...


15

Stellarium shows Mars close to the Sun and just above the horizon at that date and time. Unfortunately, Mars is only 2° above the horizon (and also on the other side of the Sun to the Earth), so it is very unlikely to be visible against the Sun's glare and through atmospheric haze (atmospheric effects are disabled in the Stellarium image below). Still, it's ...


14

Sorry if this logic seems a bit circular, but we can get unobscured pictures of galaxies because they are unobscured. As has been mentioned - space is really, really big and really, really empty. This is hard for us to contemplate, because there's so much stuff right next to us - but this is actually a really unusual condition. The next star to the Sun is ...


14

The paper in question is now on the arXiv here: "Detection of the Schwarzschild precession in the orbit of the star S2 near the Galactic centre massive black hole". This gives the following orbital elements (Table E.1): a = 125.058 mas e = 0.884649 i = 134.567° ω = 66.263° Ω = 228.171° The orbital elements $i$, $\omega$ and $\Omega$ are essentially ...


13

The JPL Solar System Simulator doesn't show Epimetheus but does show Titan behind the Encke gap at 2006-04-28 08:12 UTC. The simulated surface texture is probably composed of VIMS images in infrared wavelengths where Titan's atmosphere is relatively transparent. On the real Titan, haze scatters visible light so strongly that the surface is indistinct and ...


13

Galaxies are far away but very big. Planets are nearer but very small. Even for planets that are not completely swamped by the much brighter light of their host star, their angular size is much, much smaller than even quite distant galaxies and certainly smaller than the best achievable angular resolution by any telescopes on Earth or in space. A quick ...


13

I expect that all the itelescope.net instruments work at visible wavelengths. Therefore you have no chance at all to image the stars around Sgr A*, since it is behind about 25-30 magnitudes of optical extinction, caused by dust between us and the Galactic centre. The published images you have seen were taken by large telescopes working with adaptive optics ...


12

Let's pretend for the moment that this is possible (it's generally not, as I'll explain below) and see how we would go about it. In principle, this is basic trigonometry: you have a measured angle ($\alpha$, the diameter of the star), you know the distance to the star ($D$, from the six-month-separated parallax measurements), so to determine the linear ...


11

Your guess was correct. It is the Andromeda Galaxy, M31. Here is a map of the part of the sky near zenith at the place and time you provided: Sky map for Taganrog, Russia on 11/23/2013 5:00:00 PM UTC. Even the rotation is small. The sky map is rotated approximately 30° counter clockwise relative to the photo. You were approximately facing south when taking ...


11

There is no straight yes or no answer. Some objects do have significant radiation in visible part of the spectrum which would make them colorful if we would be close enough while observing them (our eye needs large number of photons to distinguish color so that means, astronomy wise, that object needs to be either very bright or very close or both). On ...


11

One thing to keep in mind is that the Kepler instrument is not a telescope like Hubble. It is a photometer and though it uses CCDs to look at the sky, it doesn't return a picture in the usual sense. The way it operates is that you only look at the pixels around the object you're interested in because otherwise you'd never get all those pixels transmitted ...


11

The reason that moon image looks wrong is because it is wrong. It is not a real image of the moon $-$ at least the terminator is not real. The original article you cite has a link just below their image indicating the source of their image of the moon. That source is the night sky planner, hosted by JPL. You'll find the same image on that website, albeit ...


11

A quick check shows that the spot is diametrically opposite the Sun, which although not at all conclusive, adds credence to it likely being lens flare. import numpy as np import matplotlib.pyplot as plt img = plt.imread('sun flare.png')[:, :, :3].copy() s0, s1 = img.shape[:2] X, Y = np.meshgrid(np.arange(s1), np.arange(s0)) x0, y0 = 163, 154 x1, y1 = ...


10

This has to do with the angular resolution of the Hubble telescope and the ratio between the distance of an object in space and its size in space. The galaxies that the Hubble telescope can see are bigger in size than they are far in light years away compared to pluto from earth. Take the galaxy NGC 5584 for example: It spans 50,000 light-years and it's 72 ...


10

Take another photograph of the same field that is less exposed. (Doesn't matter if you do this by shortening the exposure time, decreasing the ISO, etc.) This will give you an image with far fewer stars so you can easily pattern match your images.


10

They are called diffraction spikes, and they're artifacts from a supporting structure inside a reflector-type telescope.


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