Do the axes of rotation of most stars in the Milky Way align reasonably closely with the axis of galactic rotation? The axis of rotation of the Solar System makes a large angle of about 60 degrees relative to the axis of rotation of the Milky Way.  That seems unusual - for example, most of the bodies within the  Solar Sytem are better behaved than that.
Do most stars or planetary systems in the Milky Way rotate in close accord with the galactic rotation?  Or is there a large scatter, so that, in fact, our Sun is not atypical?

There is very likely to be a random scatter.
Unlike planets orbiting the Sun in the Solar System, most of the stars in the Galaxy did not form at the same time as the Galaxy itself. There is therefore no strong reason to suspect that the angular momentum vectors would be aligned for similar reasons. On the other hand, the Galactic gravitational potential does depart from spherical symmetry in its inner regions, because the visible matter, which becomes dominant in the inner regions, is concentrated into a disc - so presumably this, or perhaps the tidal forces exerted by this on molecular clouds, could imprint some angular momentum preference.
The evidence is sketchy but suggests random orientations, at least in the solar neighbourhood. I refer you to Detection of exo-planets , where I discuss this in the context of detecting transiting exoplanets.
In a series of papers, myself and colleagues have investigated the distribution of spin-axes in clusters of stars. The idea here, which is not far-fetched, is that large clouds from which clusters form will have some angular momentum. The question is how much of that angular momentum is inherited by the stars it forms, or to what degree can turbulence in the collapsing gas essentially randomise the spin vectors of collapsing fragments. Our technique was to combine rotation periods (latterly from Kepler observations) with careful measurements of projected equatorial velocities ($v \sin i$, where $i$ is the spin inclination to the line of sight) to get projected radii ($R \sin i$) and then to model the distribution of $R \sin i$ with various assumptions about the spin-axis distribution. In all three of the clusters we have studied (Pleiades, Alpha Per, Praesepe), the distribution was consistent with a random distribution, with quite strong limits on the amount of alignment that was possible (Jackson & Jeffries 2011; Jackson, Deliyannis & Jeffries 2018; Jackson et al. 2019). The technique has been replicated in a fourth cluster, NGC 2516, by Healy & McCullough (2020), with the same conclusion.
Other authors have claimed alignments in some cases. Notably, using Kepler asteroseismology of red giants in two clusters in the Kepler main field, Corsaro et al. (2017) claimed a quite tight alignment of spin axes, pointing almost towards us in each case. Since the Kepler field is not far from the Galactic plane and these were distant clusters, then the spin axes would be almost in the Galactic plane (a bit like Uranus and the Sun). However, the likelihood of finding such a result if individual clusters had random average angular momentum vectors raised question marks - the probability of seeing that vector pointing towards you is very low. Work by Kamiaka et al. (2018) shows that the asteroseismological estimates may be systematically biased towards low inclinations.
A further piece of evidence for some alignment was in the orientations of bi-polar planetary nebulae towards the Galactic bulge. Rees & Zijlstra (2013) found a non-random distribution that suggested that the orbital angular momenta of binary systems, responsible for the bipolar shape of the nebulae, were oriented aligned with the Galactic plane (again, like Uranus around the Sun). The result is highly statistically significant but as far as I know has not been followed up despite its obvious implications for estimations of transit yields from exoplanetary surveys.
I think the biggest argument that there is no significant effect for average stars in the field of the Galaxy, is that the exoplanet people working on the TESS survey (which covers the whole sky), would have found a drastic spatial dependence on their yield of transiting planets as a function of Galactic latitude. The majority of transiting planets (or at least hot Jupiters) have orbital axes coincident with the spin axis of the star (like planets in the Solar System). If these orbital axes were aligned with Galactic north (or any other direction) it would mean you would see far fewer transiting planets when looking towards those directions. I have heard no reports of such a spatial dependence.


Apollo 11 Under Different conditions Howdy, this might be an odd question, but I'm using whatever information I gather here to help write an alternate universe version of the 1950s through 1980s. Due to some convoluted plot contrivance involving several delays to the moon landing, the very last pass behind the moon the Apollo crew takes place during a lunar eclipse. Now, I unfortunately don't have access to any simulation software, so I'm not sure what that would look like to them! How much of the lunar surface would the crew be able to see in passes that occur beforehand, and what does the phenomenon look like from space? Do we know what it might look like?
Time is not of the essence, but I still need to know these details before writing this particular section of the story.

Lunar eclipses occur when the moon is full.  When the moon is full, the far side of the moon is unlit. With neither the Sun nor the Earth in the sky, the only ambient light would come from stars, and this is not bright enough to see anything by.  The moon would look completely black. The fact that a lunar eclipse is happening is largely irrelevant. The far side would be unlit during any full moon.

The near side of the moon would be lit by light refracted around the Earth. This would be bright enough to see the moon surface. It may be bright enough to read by. The exact brightness would depend on lots of unpredictable factors such as the amount of volcanic dust in the stratosphere. It would certainly be red just like we can see the moon during a lunar eclipse, so would the astronauts orbiting it.

In previous passes the Earth would at most partially cover the moon and the nearside surface would be well lit. The far side would be unlit and dark.

In fact the landing happened when the moon was just less than half full and so much of the far side would have been well lit.


Do extraterrestrial landscapes or planets count as nature? If I am talking about the surface of Mars, is this still nature, comparable to uninhabited places on earth, or does nature exclusively describe not man-made things on earth.

I am asking this question in Astronomy, because I would like to know what astronomers or other people dealing with extraterrestrial science on a regular basis think about it.

ARTFL Nature:

  The existing system of things; the world of matter, or of matter and mind; the creation; the universe.

That definition includes the surface of Mars, and the core of Betelgeuse.


what does it mean of different eyepieces? what is the difference between 4mm and 24mm and 30mm eyepieces?
could I buy an eyepiece from another brand?
I have the Gskyer telescope.could I buy for Celestron eyepieces?

thanks.

The 4, 24 and 30mm eyepieces give different magnification (equal to the the focal length of the telescope in mm, divided by the number for the eyepiece).
You can use eyepieces from other brands, but make sure they have the same size barrel (the bit that slides into the telescope).  I don't know the Gskyer brand, but the eyepiece size will be either .965 inch or 1.25 inch.  Provided you get the same size, any other brand will work.  It's probably best to get familiar with what is provided by Gskyer before buying anything else.

You will likely find that using lower magnification at first gives the best outcome.


Hypothetically, if the sun stays the same size forever, when will the Earth fall into it? Hypothetically speaking (once again), if the Sun stays the same size forever (i.e. it somehow maintains a stable hydrogen supply and the helium product just disappears), how long would it take before the decay of the Earth's orbit sends it plummeting into the Sun and disintegrating?

As has already been said, the major sources of change to Earth's orbit are interactions with other planets and passing stars.

We're ruling out mass loss of the Sun, so the next consideration is probably tidal interactions between the Earth and Sun. This paper suggests that the Earth is receding from the Sun by about 15cm/yr for this reason. This is 150 km per million years, so over a trillion years or so, would move Earth quite a bit further from the Sun. The effect would reduce as the Earth got further away from the Sun, but I think it would always be slowly receding. I haven't done the calculation, but I'm not sure it would ever reach tidal lock -- the angular momentum of the Sun is too high.

Independently of this, though, orbital energy and angular momentum would slowly be lost by gravitational radiation. This wikipedia page gives a formula

$$t={\frac {5}{256}}\,{\frac {c^{5}}{G^{3}}}\,{\frac {r^{4}}{(m_{1}m_{2})(m_{1}+m_{2})}}.$$

for the time to collision ignoring tidal effects.

We can plug in known values for the parameters $r = 1.5 \times 10^{11} m$, $c = 3\times 10^8 m/s$, $G = 6.7 \times 10^{-11} m^3 kg^{-1} s^{-2}$, $m_1 = 6\times 10^{24} kg$, $m_2 = 2 \times 10^{30} kg$ and get  about $3.3\times 10^{30} s$ as the time to collision. I suspect that's the time until the centres of the Earth and Sun coincide, so the actual collision would be a bit earlier, but probably not enough to notice at this precision. So that's about 10 million million times the lifetime of the universe so far.

I can't find enough information to work out how these two effects would interact. Clearly in the first $10^{12}$ years tidal evolution would be more important, but that conserves angular momentum and slows as the Suns rotation slows, whereas gravitational radiation actually carries angular momentum away, so might dominate in the long term. On the other hand the radiation drops as a high power of the separation, so we might have the Earth moving ever more slowly away from the Sun, while radiating gravitational waves of ever weaker strength, and never reaching tidal lock or starting to move back in.


Help plotting a trip to the Moon! Take me to the moon!

I know this is literally rocket science but, I would like any sources or formulas that would help me plot a trip to the Moon. I'm not sending anything, nor do I ever plan to, I just want to do the next best thing by plotting the trip. So if anyone could give me any info as to how I can plot the trip, that'd be great! :)

P.S. I'm perfectly aware that the moon isn't always the same distance away, that it's like hoping onto a taxi in New York City, and how it's got a lot of work (I wouldn't mind doing it really), I'd just like to plot the trip. And please don't hate on me for asking, let's just be civil adults here. :)

This website has a mathematical in depth analysis of the Apollo 11 translunar trajectory. It looks to be a reasonably reliable resource but I've only had a quick skim through so you should check on the sources cited.
There should be enough information there to make a computer program to calculate the orbits.


Calculating the distance of objects using the Gaia DR1 The Gaia DR1 was released middle of September with ~1.1 billion objects. The schema contains right ascension, declination and parallax, which could be used to calculate Cartesian co-ordinates (xyz).

The overwhelming majority (>99.998%) of objects do not contain an entry for parallax. I suspect this needs to be computed using other columns (e.g. phot_g_mean_mag?).

How do you compute the parallax for an object when its not included?

The data model is documented here .

The early 1990s Hipparcos mission
yielded parallaxes for 118000 stars (Hipparcos catalog) and positions without parallax for another 2.4 million (Tycho-2 catalog).
The Tycho-Gaia astrometric solution (TGAS) combines those data with preliminary Gaia observations to get 2 million parallaxes, "only" 17 times as many as Hipparcos, with better precision.
If this is a small fraction of the total number of star positions, it's not a deficiency; it's just easier to get a position than a parallax.
Lindegren et al. 2016 explain what they could or could not do with the first 14 months of Gaia observations.
Future Gaia data releases should have still more parallaxes.


Is Io's orbit or rotation affected by its volcanism? The rotations of comets and asteroids are affected by outgassing volatilities. Io is very volcanically active. Does this give Io a varying orbit and a slow rotation? Tidal forces should take longer than the effect of a large eruption. Tidal locking means that any volcano alway pushes Io in the same direction (radially, tangential or inclined) accumulating the same kind of orbital change during all of its its active lifetime.

Is Io's orbit or rotation affected by its volcanism?
The rotations of comets and asteroids are affected by outgassing volatilities. Io is very volcanically active. Does this give Io a varying orbit and a slow rotation?

Jupiter's moon Io has a lot of volcanoes.
A similar answer to your question is provided at Quora's question: "Could the eruptions of all the volcanoes on Earth simultaneously knock it off its rotation or change its orbit?".
There are two differences: Eruptions on Io can reach escape velocity (2.558 km/s), and matter is subject to gravity as it is propelled away from the moon allowing some to escape and some to return; so it is a source of wobbling and rotational speed changes.
"Jupiter's rocky moon Io is the most volcanically active world in the solar system, with hundreds of volcanoes, some erupting lava fountains dozens of miles (or kilometers) high. Io’s remarkable activity is the result of a tug-of-war between Jupiter's powerful gravity and smaller but precisely timed pulls from two neighboring moons that orbit farther from Jupiter – Europa and Ganymede.
...
Orbit and Rotation
Although Io always points the same side toward Jupiter in its orbit around the giant planet, the large moons Europa and Ganymede perturb Io's orbit into an irregularly elliptical one. Thus, in its widely varying distances from Jupiter, Io is subjected to tremendous tidal forces.
These forces cause Io's surface to bulge up and down (or in and out) by as much as 330 feet (100 meters). Compare these tides on Io's solid surface to the tides on Earth's oceans. On Earth, in the place where tides are highest, the difference between low and high tides is only 60 feet (18 meters), and this is for water, not solid ground.
Io's orbit, keeping it at more or less a cozy 262,000 miles (422,000 kilometers) from Jupiter, cuts across the planet's powerful magnetic lines of force, thus turning Io into a electric generator. Io can develop 400,000 volts across itself and create an electric current of 3 million amperes. This current takes the path of least resistance along Jupiter's magnetic field lines to the planet's surface, creating lightning in Jupiter's upper atmosphere.
Surface
The tidal forces generate a tremendous amount of heat within Io, keeping much of its subsurface crust in liquid form seeking any available escape route to the surface to relieve the pressure. Thus, the surface of Io is constantly renewing itself, filling in any impact craters with molten lava lakes and spreading smooth new floodplains of liquid rock. ...".
The two-body problem explains how multiple interacting objects affect the orbit of each other. While the orbital eccentricity of Io is quite low (0.0041) the orbits of three of the Galilean moons (Ganymede, Europa and Io, excluding Callisto) have a 1:2:4 resonance. The "pushing" will act in periodic repetition to have a cumulative effect on the motion.

Orbital resonances greatly enhance the mutual gravitational influence of the bodies, i.e. their ability to alter or constrain each other's orbits. In most cases, this results in an unstable interaction, in which the bodies exchange momentum and shift orbits until the resonance no longer exists. Under some circumstances, a resonant system can be stable and self-correcting, so that the bodies remain in resonance.
The small effect of the volcanoes is corrected by the much greater effect of resonance. The website space.com estimates Io's age at about 4.5 billion years old, the same age as Jupiter. If the orbit was unstable and changing it may well have been destroyed long ago.


Age of Jezero crater on Mars How long ago the Jezero crater was formed on Mars?
And is it a meteorite impact  crater?

Jezero crater is roughly 50 km across, and it is emplaced within the wall of Isidis crater ("basin").  Isidis itself has seen a lot of erosion, but it has been dated by lots of different people through the use of superposed crater counts (craters on top of it, with the number-age relationship tied to the Moon) to somewhere around 3.85–4.05 billion years old, including by me (Robbins et al. 2013).  Jezero cannot be older than Isidis, so it is capped at 3.85–4.05 billion years.
Jezero must have been formed when there was liquid water on the surface of Mars in enough quantity to form a delta deposit.  That would mean that it should have formed by the Late Hesperian chronostratigraphic epoch, which dates, in various chronologies, to about 3.00–3.65 billion years ago.  So, Jezero formed at least 3.00* billion years ago, but no later than 4.05* billion years ago.
Beyond that, no one that I know of has done superposed crater counts on Jezero's rim to try to model its age.  Jezero is highly degraded, and it has only one larger crater emplaced on its remaining rim, making any such crater counts highly suspect.  This is why the work I linked to above, I only tried to date craters >150 km, and even that is somewhat suspect.  Jezero certainly has deposits in the floor (such as the delta), so any dating of the floor from craters would only be a minimum age because it can't be younger than its floor deposits.  Therefore, even if someone does try to do superposed crater counts to get an age for the crater, I don't think they will be successful in getting a very good one.  For example, this paper found that the age of the floor deposits is 2.6±0.5 billion years, so again Jezero is at least 2.6 billion years old, but that's a minimum, and the floor deposits must be younger than the crater.
*The Mars crater chonology has an uncertainty of a factor of 2-ish, so even though there are two decimal points on my numbers, treat that with a grain of salt.
And yes, an impactor formed the crater, either an asteroid or a comet.


Claim that 30-m class telescopes will have resolution far superior to Hubble: true? This article makes the claim that the Giant Magellan Telescope (GMT, number 4 in the list) will have resolution 10 times better than that of Hubble, while the Thirty Meter Telescope (TMT, number 3 in the list) will have resolution 12 times better than that of Hubble.  These are claims I find impossible to believe.

A simple application of, for instance, the Raleigh Criterion ($\theta = 1.22~\lambda/D$), where $\theta$ is the angular resolution of the telescope, $\lambda$ is the wavelength of the light in question, and $D$ is the diameter of the telescope, would show that a comparison between angular resolutions is as simple as comparing the diameters of the telescope apertures.  Hubble's mirror is 2.4 m; comparing with the 24.5 m GMT and the 30 m TMT we see that the GMT will have 10.2 times the angular resolution of Hubble, while the TMT will have 12.5 times the angular resolution of Hubble.  I think a calculation similar to the one I have described is how the article linked above came up with the numbers they did about the angular resolution of these telescopes compared to Hubble.

However, the Raleigh criterion only applies to telescopes working at their diffraction limit.  Space telescopes (if they're designed well and built correctly) can work close to the diffraction limit (maybe even at the diffraction limit).  Ground-based telescopes, however, are limited in angular resolution by the atmosphere, which at best will limit resolution to about an arc-second at the best sites on Earth. Thus the GMT and TMT by themselves will not have better image resolution than Hubble.

My question, then, is whether this article is correct (possibly because of one of the reasons I list below) or whether it seems this article naively applied the Raleigh Criterion for angular resolution with no thought about how the atmosphere will affect the resolving capabilities of these large ground-based telescopes.

Possible reasons the article may still be correct:

Adaptive optics, a developing technology which can allow telescopes to correct for distortions to the image produced by the atmosphere.  Perhaps GMT and TMT will have very fancy adaptive optics systems.
Another technique, such as speckle imaging or lucky imaging.
Article could be referring to spectral resolution, rather than angular resolution (or image resolution).  However, obtaining good spectral resolution is as much the job of the spectral instrument as it is of the telescope so I don't consider "spectral resolution" to be an inherent property of a telescope.

Adaptive optics, as it says on the GMT website:

  One of the most sophisticated engineering aspects of the telescope is
  what is known as “adaptive optics.” The telescope’s secondary mirrors
  are actually flexible. Under each secondary mirror surface, there are
  hundreds of actuators that will constantly adjust the mirrors to
  counteract atmospheric turbulence. These actuators, controlled by
  advanced computers, will transform twinkling stars into clear steady
  points of light. It is in this way that the GMT will offer images that
  are 10 times sharper than the Hubble Space Telescope.

and on the TMT web site:

  In addition to providing nine times the collecting area of the current
  largest optical/infrared telescopes (the 10-meter Keck Telescopes),
  TMT will be used with adaptive optics systems to allow
  diffraction-limited performance, i.e., the best that the optics of the
  system can theoretically provide. This will provide unparalleled
  high-sensitivity spatial resolution more than 12 times sharper than
  what is achieved by the Hubble Space Telescope. For many applications,
  diffraction-limited observations give gains in sensitivity that scale
  like the diameter of the mirror to the fourth power, so this increase
  in size has major implications.


Why do astronauts move so slowly in zero gravity? When astronauts are floating about in a spaceship or space station, they nearly always move very slowly. After doing a bit of research I can't see why being in zero gravity would restrict movement to such a degree. It's almost as if there's resistance to their movement, like they're moving through water.

It's more for safety than anything else. Space is a very dangerous place for so many reasons. And making mistakes can very easily cause death.

Being weightless does not mean you lose mass, so momentum is just as difficult as ever. But whereas on the ground you can easily use friction to stop, in space if you try to stop against the floor you will just move off it. You can only stop by holding something, or pressing against something close to perpendicular to your movement.

As an example, imagine you jumped with all your force from one wall in the ISS. You will notice as you approach the other end that you are travelling at speed, head first, with no safe way to stop. Even reaching out to a handhold on a side wall will whip you round and into that wall, possibly injuring yourself or damaging instruments on the wall.

Look at any video from the ISS to see how carefully they move.

Similarly, outside the ISS, you want to do everything slowly so you don't damage your suit, miss a handhold or otherwise cause death.


Is there a mechanism that makes small moons more rounded than comets? 67P/Churyumov–Gerasimenko has a highly irregular dumbbell shape. But the sample of comet shapes observed is very small, so I wonder if irregular shape is the norm for comets and for small moons. Many known moons are no larger than this ~4 km diameter comet. The smallest moons imaged are modestly irregular, basically just elongated, at least they don't have a waist like 67/P. The most elongated moon is perhaps ~135/60 km Prometheus.

Is there reason to believe that moons which are too small for hydrostatic equilibrium get more rounded than comets of similar mass and composition? Moons differ from comets in several ways. AFAIK: Moons in general have a very different gravitational environment, a much more stable distance to the Sun, experience more frequent impacts, another formation history if not captured. Composition and density depends on the formation distance from the Sun, although even ~300 km Hyperion has a similar density as 67/P.

Should we expect the small moons in general, and the moons of Pluto especially, to be rounded or dumbbell shaped? If 67P is a merged binary, isn't it more likely for two objects to merge if they are moons instead of comets, since neighboring moons have low relative speed? ~100-200 km Janus and Epimetheus look as if they could merge fairly calmly. Does the lack of (observed) dumbbell shaped moons tell us something about 67P, that its shape is a rare outlier for example?

I can only speculate but here goes nothing.

67P/Churyumov–Gerasimenko apparently came from the Kuiper Belt. There are a lot of KBO (Kuiper Belt Objects) in that region, so it is not unreasonable to expect some of these objects to have crashed into other objects of about the same mass, resulting in irregular shapes.

Small moons (or other moons in the Solar System) have been isolated from other objects of the same size. This allows gravity work work undisturbed for a longer period of time than those in the Kuiper Belt, molding the object into more or less a spherical shape.

That is my best stab at the question, though I can't imagine some other cause that would result in such a drastic irregular shape in something as large as a comet (relatively speaking).

I just found something interesting. So the smallest moon in the solar system is Aegaeon. It happens to be a couple of orders of magnitude smaller than the comet mentioned above. In addition, this moon appears to have an oblate spheroidal shape to it. It may just be the case that we have only seen spherical moons, but there are many more moons out in the cosmos that have odd shapes to them.


Has anyone filmed a total lunar eclipse from the Moon's perspective? With all of the footage we have of total lunar eclipses, has NASA, ESA, or any other space program ever taken a photo or video of it from the Moon itself?

What can you say but "Wow..."

http://www.space.com/32245-total-solar-eclipse-video-dscovr.html

"The moon's dark shadow travels across Earth's face in spectacular new views of Tuesday's (March 8) [2016] total solar eclipse captured by a satellite in deep space."

I'm sorry my answer is off by about 700,000 miles compared to what you asked :)

I'm fairly sure there did not happen to be, a time when NASA had cameras on the moon, when, there was a lunar eclipse.

Note that cameras today (such as this one on DSCOVR) are unbelievably better than cameras from Apollo era, so this would blow away any merely 700,000 miles closer imagery captured during the Apollo era.


Full Moon Position of Sun I am trying to figure out the position of sun with respect to moon on a full moon day.

Here's an image from certain wiki which I do not find convincing.
1

If the sun rays point as shown in the image and the earth stands between the moon and sun on Full moon's day, I dont see any reason why the moon should be shining in its full glory.

Similarly when the moon is directly in between earth and sun, it should have maximum brightness but the image shows it as the new moon.

In fact, I cannot deduce how at all can moon receive light across the entire sphere at any given angle from the sun.

Putting it specifically, what would the position of sun w.r.t to earth and moon on a full moon day ?

The orbits are not all in exactly the same plane, that is why we do not have eclipses every month. With the orbits at a slight skew it is possible to have the geometry very near to that in your diagram without the Earth blocking the Sun.

Lunar eclipses, the Moon passing through the Earth's shadow which is what your post is about, do occur relatively frequently compared to Solar eclipses because the Earth is bigger than the Moon.

At New Moon the lit side of the Moon is facing away from the Earth and so we do not see it.

You will find a fairly good explanation, and a better diagram, of all this here


Acquirable Raw Data in Amateur Astrophotography What raw data can I possibly acquire from an 8" Classical Dobsonian Telescope, and a DSLR? Could anything eye-opening to amateur astronomers be computed or calculated first-hand with such equipment? I'm sure scientists must've considered this equipment "advanced technology" at some point in history not too far back...Could I rediscover or calculate some Laws (like Kepler's laws) or some other things amateur astronomers would be amazed to calculate themselves (like the distance to a planet) using this equipment?

First off, pairing a classic dob with a DSLR is a bit like a shotgun marriage. A dobsonian is fundamentally a visual telescope. Most manufacturers don't even consider the possibility that these instruments could be used for data collection via a sensor. There are 2 issues here:

1. The dobsonian is not tracking

The sky is moving, the dob stays still. You have to push the dob to keep up with the sky. Any long-exposure photo would be smeared. To remedy this, you'll need an equatorial platform, which will move the dob in sync with the sky.

Please note that only the best platforms allow reasonably long exposure times. Then the results can be fairly good.

2. There isn't enough back-focus

The best photos are taken when you remove the lens from the camera, plug it into the telescope directly, and allow the primary mirror to focus the image directly on the sensor. This is called prime focus photography. But most dobs can't reach the sensor within the camera, because their prime focus doesn't stick out far enough. There are several remedies for this, like using a barlow, moving the primary mirror up in its cell, etc.

The bottom line is that it takes some effort to make a dob and a DSLR play nice together. Is it doable? Yes. Is it simple and immediate? No. So the literal answer to your question is that there isn't much you can do with just a dob and a DSLR.

You can take photos of the Moon and the Sun, because the short exposure there does not require tracking, but that's pretty much it. Here is an image of the Moon I took with a home-made 6" dob (with home-made optics) and a mirrorless camera (prime focus, about 1/320 sec exposure):

Makes a cute little desktop background, I guess, but it's definitely not research-grade.

Now add a tracking platform and things become more interesting, and the possibilities open up quite a lot.

In a more general sense:

There are telescopes that are specifically made for astrophotography. They have lots of back-focus, they are short and lightweight and therefore can easily be installed on tracking mounts. More importantly, there are tracking mounts made specifically for imaging - very precise, delicate mechanisms that follow the sky motion with great accuracy. In fact, the mount is more important than the scope.

A typical example would be a C8 telescope installed on a CGEM mount, or anything equivalent. Barring that, a dob with lots of back-focus sitting on a very smooth tracking platform (probably not as accurate as a GEM, but good enough for many purposes).

Make sure you don't exceed the load capacity of the mount. If the mount claims it can carry X amount of weight, it's best if the telescope weight doesn't exceed 1/2 of that amount. Close to the weight load limit, all mounts become imprecise. The exceptions are high end (the most expensive) mounts which cost many thousands of dollars and usually honor their promises in terms of load capacity 100%.

Once you have: a tracking mount, a good camera, and a telescope (listed here from most important to least important), you can start imaging various portions of the sky for research. There are 2 main classes of objects that you could image:

1. Solar system objects

They're called "solar system objects" but the class includes anything that's pretty bright, not very big, and it's high resolution. Tracking is important but not that crucial.

You need a sensitive, high speed camera that can take thousands of images quickly (a movie, basically). These are called planetary cameras. They generally have small sensors, are high sensitivity, and can operate at high frame rates (hundreds of frames per second).

As a cheap alternative in the beginning you could use a webcam, there are tutorials on the Internet about that. A DSLR in video mode in prime focus might work, but it's going to do a lot of pixel binning, so resolution would be greatly reduced unless you use a very powerful barlow (or a stack of barlows).

You'll load all those images in a software that will perform "stacking" to reduce them all to one single, much clearer image.

The scope needs to operate at a long focal length, f/20 being typical, so a barlow is usually required. The bigger the aperture, the better.

2. Deep space objects (DSO)

These are anything that's pretty faint and fuzzy, like galaxies, but some comets are also DSO-like in their appearance. You need to take extremely long exposures; usually a dozen or a few dozen images, each one between 30 sec and 20 min of exposure, sometimes even longer. Extremely precise tracking is paramount, so you need the best tracking mount you could buy. Autoguiding is also needed to correct tracking errors.

The scope needs to operate at short focal ratios, f/4 is pretty good, but as low as f/2 is also used; focal reducers (opposite of barlows) are used with some telescopes, like this or like this. Aperture doesn't mean much; small refractors are used with good results.

The camera needs to be very low noise; DSO cameras use active cooling that lowers their temperature 20 ... 40 C below ambient. Typically they have large sensors.

DSLRs can also provide decent results, but their noise is typically higher than dedicated cameras, so you need to work harder for the same results.

Specific software is used for processing, stacking, noise reduction, etc.

So what can you do with such a setup?

Comet- or asteroid-hunting works pretty well. Terry Lovejoy has discovered several comets recently using equipment and techniques as described above. Here's Terry talking about his work.

Tracking variable stars is also open to amateurs. This could also be done visually, without any camera, just a dob, meticulous note-taking, and lots of patience.

With a bit of luck, you could also be the person who discovers a new supernova in a nearby galaxy. You don't need professional instruments, you just need to happen to point the scope in the right direction at the right time and be the first to report it. This also could be done purely visually, no camera, just a dob.


Earth's perihelion vs winter solstice I found that Earth's perihelion occurs roughly two weeks after winter solstice similar thing happens for summer solstice and aphelion. I understand what those terms mean but I don't understand the reason for the two-week gap. I actually thought they occur on the same day.

What am I not understanding here? Can anyone explain the reason for the two weeks?

My understanding is that the earth's aphelion and perihelion occur on the major axis of earth's elliptical orbit. Why are they not the extremities of analemma which correspond to solstices? Where is my thinking wrong?

There is no connection between the date of the solstice and the perihelion. It is merely a minor coincidence that perihelion occurs close to the solstice.

The relationship isn't fixed. Precession in the Earth's orbit (caused by gravitational perturbations of Jupiter and other planets) will change the relative time of perihelion and solstice over a period of thousands of years.


What is the cause of all of these sharp, concentric rings around bright stars in this HST image? ESA's HST page heic1819 — Photo Release; Hubble reveals cosmic Bat Shadow in the Serpent’s Tail is of course beautiful and stunning, but my eyes are drawn to the diffraction artifacts of the bright stars.

I'm assuming the crosses are due to four vanes supporting the secondary mirror, but are the tight concentric rings due to Hubble's large aperture, or it's smaller secondary mirror blocking that aperture, or something else, perhaps image processing?

Without a scale for reference, it's hard to get an angular frequency and compare to an Airy-like diffraction pattern to get a diameter, and that's where it gets more puzzling, because you need a narrow wavelength range to get a coherent oscillation for so many cycles (I think I can see perhaps 15 or more sharp, distinct rings), and to zeroth order stars are mostly black-body.

Is this a bit of a puzzle, or am I missing something obvious (e.g. filters)? Or both?

Here's a cropped, monochromed, ROI:

Further stretched in contrast and size:

The diffraction pattern at the focal plane created by a circular aperture is called an Airy Disk or Airy Pattern.  Both the outer opening and the inner hole plus secondary contribute to the exact function.  This is usually not easily observed with ground based telescopes because the seeing fluctuations due to air turbulence smears it out.  These images use filters that are a few 100 Angstroms wide, so although not monochromatic, they are narrow enough to clearly see the pattern.  The spectral shape of the star's light within the filter makes the effective width of frequencies more narrow.


Is there an equivalent of the red shift effect for cosmic rays? I had read somewhere that light from very distant sources can be measured to be increasingly red shifted the further away the object is (due to cosmic inflation?).
Suppose you had an object emitting cosmic rays or neutrinos other physical matter (not photons), is there an equivalent effect for these objects? ex: a red-shifted helium nuceli in a cosmic ray or red-shifted neutrinos coming from a neutrino source?

Yes, there is a "retardation of the co-moving velocity" of particles.  It is important to take it into account to understand the time history of peculiar velocities of galaxies and for determining the energy distribution of cosmic rays coming from other galaxies.
Peculiar velocities, the difference between an object's velocity and the local rest velocity with respect to the microwave background, decay as 1/a(t), if there are no forces.  The global scale factor a(t) is the relative size of the universe compared to today.  So there is a "redshift" that applies to particles similar to that of the frequency of light.
The ratio of "observed" velocity to "emitted" velocity is $\frac{v_o}{v_e} = \frac{1}{1 + z}$. (Peebles, Principles of Physical Cosmology). The $z = \frac{1}{a_e}-1$ here is different from the $z$ of the galaxy because the time of the particle's emission is different from the emission time of the light that we see now.


Why do Uranus' and Pluto's moons orbit the equator? What makes a moons' inclination tilt with the rotational axis of their planet? Uranus and Pluto have their axes of rotation almost 90 degrees towards the ecliptic. But why do their moons tilt the same way?

Does it mean that their tilts were caused by passing near some external mass that affected the planet and its moons equally, rather than some planet specific process such as a comet impact or tectonics?

Mars' rotational axis tilts over time, but its moons today are spot on over its temporary equator. Earth's moon is aligned with the ecliptic and doesn't seem to care about our planetary tilt, although (or because of?) oceanic masses tugging on its tides. Exoplanets have been found that don't orbit in alignment with their star's rotation, and why would they do so billions of years after formation?

I have managed to find two hypotheses for Uranus's tilt that explain why its moons are also orbiting on a tilted plane.

The multiple impacts hypotheses lies on the fact that, if Uranus had become tilted by the force of a single large impact (as was commonly believed), then the moons should have stayed on their original plane. A corrected version of this hypothesis said that, if the accretion disc from which Uranus was forming was still around, an impact would have destroyed it and recreated it on the same place as the new equator. The moons would then have accreted from the disc on that plane. This, however, would have produced moons with retrograde motion (which is not the case for most). So the latest idea is that Uranus suffered multiple impacts, which, according to the simulations, works to produce the planetary system as we see it today. (There's another problem with this hypothesis.)

The collision-free theory (or collissionless scenario as described in the paper) proposes that Uranus could have become tilted progressively as it formed without the need to postulate an impact (or series of impacts), if it had an additional satellite and started with a large inclination (which is allowed by current models); the extra moon could have been ejected later. The simulations worked for an initial inclination of more 17 degrees and a moon of 0.01 Uranian mass and at 50 Uranian radii.


Can I have a career in observational astronomy if I study astrophysics or does it have to be particularly astronomy? So I'm in year 11 and have an aspiration to be an observational astronomer but I'm worried that if I study astrophysics I won't be able to be an astronomer in particular. It probably sounds like a really dumb question. I am open to doing any job in astronomy/astrophysics but I just need to know. Thanks :)

I don't think there really is a difference between "astronomer" and "astrophysicist". Both are physicists using their knowledge to learn about the Universe. Their is perhaps a tendency to call yourself an astronomer if you're doing observational stuff, and an astrophysicist if you're more theoretical/numerical. If people ask me what I do, I use "astronomer" if I want to sound romantic, and "astrophysicist" if I want to sound clever. :)

So, you are interested in doing observations, rather than doing simulations or pure theory. But in all cases, you start by gaining basic knowledge of both observational astronomy, "normal" physics, theory, and coding. Depending on which country you study in, at some point you will probably have the opportunity to have more influence on the courses you take, so that you could skip courses of, say, particle physics and general relativity, in favor of data analysis and, even better, observing summer schools (which I would recommend even for students planning to do theory, because it's a fantastic experience). Finally, your Master's thesis (to a small degree) and eventually your Ph.D. (to a large degree) will have an impact on the path you choose for your career. But many astro*s have been known to switch.


Ionization Fraction of Helium in alternate, He dominated early universe I'm asked to verify an expression for the ionization fraction of helium in a universe in which helium dominates baryonic matter.

I'm given that the ionization fraction $X = \frac{n_{He^+}}{n_{He^+}+n_{He}}$ with a given ionization energy $Q_{He}$, and number densities by the formula $n_i=g_i(\frac{m_ik_BT}{2\pi \hbar})^{3/2}\exp(-\frac{m_ic^2}{k_BT})$. Also the baryon-photon ratio in this case is $\eta=\frac{4(n_{He^+}+n_{He})}{n_\gamma}$.

I'm trying to show that by putting all this together I get something with the form:

$\ln(\frac{1-X}{X^2})=A+\ln(\eta)-B\,\ln(\frac{Q_{He}}{k_BT})+\frac{Q_{He}}{k_BT}$. where $A$ and $B$ are constants.

When I plug everything in, I got $\frac{1-X}{X^2}=\frac{n_\gamma}{8}\eta \, \exp(-\frac{Q_{He}}{K_BT})$, which gives:

$\ln(\frac{1-X}{X^2})=\ln(\frac{n_\gamma}{8})+\ln(\eta)+\frac{Q_{He}}{k_BT}$

Any ideas of where I might have gone wrong?

My answer is incomplete (but still possibly helpful, only posting because I don't have enough "points" to comment).

I did a bit of the math involved but did not substitute terms with n's and eta's. However, from the result at the bottom of that page, it looks to me like a Binomial Expansion (truncated Taylor Expansion). The link shows the highest order for your terms (which is okay since higher order terms can usually be regarded as negligible) with the same +/- pattern you expect in your answer.


In how far is Hipparcos data still being used? I remember being excited when ESA launched Hipparcos in 1989 and was in operation until 15 August, 1993. Later, Gaia came into operation in 2013.
Reading What is the overlap between the Gaia and the Hipparcos catalog? I started wondering whether any of the Hipparcos data is still being used and for what. Could somebody please guide me with a concise explanation?
Reference

https://www.cosmos.esa.int/web/hipparcos/catalogues
https://www.cosmos.esa.int/web/gaia/data

The Hipparcos results can be used as an earlier epoch to improve the astrometry produced by Gaia. This was the linchpin of data release 1 (DR1), but less important in DR2 and DR3, because the astrometry of Gaia alone is so much better in general.
An exception to this is bright stars where much data is still missing in DR3, but can be found in the Hipparcos catalogue.
Apart from precision-based concerns, the earlier Hipparcos data, combined with Gaia, can be used to look for changes in proper motion: i.e. accelerations (see Brandt 2018).


Subterranean Oceans On Other Planets/Planetoids: How Do Astronomers Deduce This Recently I have been looking into planetoids in our asteroid belt and I have found one that caught my interest, Ceres. One of the main points that was said about it was that it had a subterranean ocean. But I'm puzzled as to how astronomers can come to this conclusion. Any explanations would be much appreciated.

How they have concluded that there could b.e a subterranean ocean on Ceres is by Spectralscopy.
Spectral Signature can be summarised to like this:

Different elements emit different emission spectra when they are
excited because each type of element has a unique energy shell or
energy level system. Each element has a different set of emission
colors because they have different energy level spacings. We will see
the emission spectra or pattern of wavelengths (atomic spectra)
emitted by six different elements in this lab. We will then identify
an unknown element by comparing the color of the unknown with the
flame color of our knowns.

And another:

When something is hot enough to glow (like a star), it gives you information about what it is made of, because different substances give off a different spectrum of light when they vaporize. Each substance produces a unique spectrum, almost like a fingerprint.

So how the scientists would have concluded that there was water because of the water vapours in the atmosphere. It gives off a certain wavelength that could be compared the closest element that gives off the same wavelength which is water.
Then the scientists would have used Galileo's magnetometer(an instrument which measures the strength and direction of magnetic fields) to conclude that there could be an ocean on the dwarf planet. The strength and response of the induced field would tell the scientists a rough estimate of the dwarf planet's sub surface which in return they could deduce the fact the there is a large amount of water present which equals an ocean.


Gravitational Properties of Asteroids How large would an asteroid have to be in order to hold a person so that the person could not escape?

Of course you would need to specify who the person is - an Olympic athlete? Let us assume so and then you can scale downwards accordingly.

So an Olympic high jumper can jump hard enough to raise their centre of gravity about 2m off the ground.

Let us assume this is a ballistic problem. The athlete actually gives themselves sufficient upward speed to get their centre of gravity from about 1m to 2m in the Earth's gravitational field. Using the usual equations for uniform acceleration, the required initial velocity is $v \simeq \sqrt{2gh} = 4.4$ m/s.

Now let us assume that the athlete could deliver something similar on an asteroid. This is doubtful, because getting a good run up, while wearing a spacesuit, is probably not going to happen. But if it were possible, then we just equate the jump velocity to the escape speed
$$ v = \sqrt{\frac{2GM}{R}},$$
where $M$ and $R$ are the mass and the radius of the (assumed spherical) asteroid.

Thus we do not have separate constraints on the mass and radius of the asteroid, only on their ratio. If
$$ \frac{M}{R} > \frac{v^2}{2G},$$
then even an Olympic athlete couldn't jump into space.

To get something more definite we would need to assume a density, $\rho$, for the asteroid. This depends on what type of asteroid you are talking about, but could be between 1500 and 5000 kg/m$^3$ (Carry 2012).

If we assume (spherical asteroid again) that $M = 4\pi R^3 \rho/3$ and substituting this for the mass, we get a constraint on the asteroid radius such that someone is trapped if:
$$ R > 4.2 \left(\frac{v}{4.4\ {\rm m/s}}\right) \left(\frac{\rho}{2000\ {\rm kg/m}^3}\right)^{-1/2}\ {\rm km}$$

You can mess about with this and assume a different take-off speed (you might want to argue, as MarkP does, that you could work up a larger horizontal speed by just running fast - but I doubt that in a "milligravity environment" - the surface gravity is of order 0.002 m/s$^2$) or different asteroid densities to modify the answer (trans-Neptunian objects or comets have densities lower than 1000 kg/m$^3$). Or you could come up with an equivalent constraint on the mass by substituting for $R$ in terms of the mass and density instead.


How do we determine a redshift error? I am struggling to calculate the redshift error of a galaxy. I was able to determine the redshifts of the absorption and emission lines from the galaxy's spectrum and then took the average of the redshift of the lines to determine the (overall) redshift of the galaxy. I know "error" is associated with the true and measured values but I don't know the true value of the redshift of the galaxy. Is there an explicit formula or something that I can use to calculate the redshift error of the galaxy without reference to the true redshift of the galaxy? All I know are the emitted and observed wavelengths of the emission and absorption lines of the galaxy spectrum.

You're confusing error and uncertainty.

In school labs or other situations where the true value of a quantity is known, the difference between a measured value and that true value is sometimes referred to as the "error" in the measurement. A lower error is taken as an indication that the results are successful.

When making observations, the true value of a quantity is seldom known. Indeed, that's usually the purpose of making observations: not as a pedagogical tool but as a way to determine something about the world. In this case, we're concerned with the uncertainty in the measurement - that is, essentially how confident we are in the value we determine. More quantitatively, we might give something like a $3\sigma$ uncertainty that explains how likely it is that the true value falls within 3 standard deviations of the measurement. That's a way of showing how robust the results are.

It's also helpful for comparing sets of measurements - if one group says that (and I'm making up numbers here) that their measurements have determined the Hubble constant to be $H_0=67.0\pm1.2\text{ km}\text{ s}^{-1}\text{ Mpc}^{-1}$ and another group says that their methods have found it to instead be $H_0=74.00\pm0.75\text{ km}\text{ s}^{-1}\text{ Mpc}^{-1}$, then we can see from the uncertainties that the two results are in fairly strong tension with one another.

If you're determining something like redshift, you presumably have some raw data measuring other quantities. Perhaps you've identified H$\alpha$ emission from a source and found it to be at $\lambda=8540\pm30$ angstroms, while the rest wavelength is $\lambda_0=6562.8$ angstroms. Then you can use standard uncertainty propagation methods to determine the uncertainty in the redshift of the source. In this case, the redshift is
$$z=\frac{\lambda}{\lambda_0}-1$$
The uncertainty in $z$ is then
$$\sigma_z=\sqrt{\left(\frac{\partial z}{\partial\lambda}\right)^2\sigma_{\lambda}^2+\left(\frac{\partial z}{\partial\lambda_0}\right)^2\sigma_{\lambda_0}^2}$$
where I've included $\sigma_{\lambda_0}$, though realistically it should be much smaller than $\sigma_{\lambda}$. Keep in mind also that $\sigma_{\lambda}$ has itself presumably been determined through error propagation from the raw data collected.

(Side note: As pela pointed out, this assumes that the redshift is entirely due to the motion of the object you're looking at, and that's not always the case. I originally picked Lyman $\alpha$ has my line, which turns out to not be particularly useful. Additionally, the toy redshift ended up being quite low and therefore even more sensitive to other not-recessional contributions, like stellar winds. H$\alpha$ is, I think, a better example, especially with this higher redshift.)


Do planetary radio frequency change upon alignments? If Neptune would align Uranus, would the radio frequency perceived on Earth increase?

Are the radio antennas on NM capable of getting the 2 different radio frequencies feom both planets or would they get a bigger range of frequencies?

If you're talking about the electromagnetic radiation (in the radio spectrum) produced by the planets themselves, then the answer is no.

BTW, not all planets generate a lot of radio EM. Jupiter does generate some amount, but the other planets are quite a bit more quiet. Still, a very small amount is probably generated by most planets out there.

In any case, this is a strictly local phenomenon. Local conditions within and on that planet predominate in driving the production of radio EM.

By comparison, distances between planets are HUGE. Whether they are "aligned" or not is of no importance to what happens on the surface. Anyway, they appeared "aligned" only to us, on Earth, and even then the "alignment" is always imperfect (they are never exactly on a line).

Shortly: no.


What existed before the big bang What existed before the big bang?

What all thing are there inside big bang object ?.

We don't have theories that can describe the big bang (or whatever happened) properly so we can't say anything definite about the big bang itself, let alone if/what was before it.  We would need (at a minimum) something like a quantum field theory for gravity and we don't have one yet.  We've no reason to think that's all we'd need.

It's not even possible (as I understand it) to define "before" the big bang.  Time, in essence, started at the big bang.  It has no common sense meaning beyond that.


If fusion only happens near the center of the sun, what's with all the "fire" on the surface? I understand that it isn't "fire" in the combustion of oxygen sense, but I can't find an answer online about what we're witnessing happening at the visible surface of the sun if all actual fusion explosive stuff is occurring near the center.

Would love a primer or a link to an explanation.

99% of nuclear fusion of hydrogen nuclei into heavy nuclei happens at the core of the Sun. This is because only at the core is the temperature high enough to overcome the coulomb barrier caused by the electrostatic repulsion between the nuclei. The solar core produces a lot of energy $(3.828×10^{26} W)$ by fusion and since all that energy has to go somewhere, it eventually reaches the surface making the surface of the Sun hot (6000K).

Most objects like a red hot iron or tungsten filament of a bulb glow when heated. An idealized version of this is called black body radiation. The colour at which the body glows or the frequency of the electromagnetic radiation depends on the temperature. Since the surface of the Sun is hot, it glows. This is not combustion but natural heat glow. Moreover glowing hot plasma rises up in columns while cooler denser gases settle at the surface of the Sun at speeds upto 7km/s in a process called solar granulation causing the Sun to look like a fierce fire ball.


How to find the distance between two quasars I want to calculate the distance between two quasars of which I know the angular position and the red shift. Let $Q_1=(\alpha_1,\delta_1, z_1)$ and $Q_2=(\alpha 2,\delta 2, z_2)$ and suppose $z_2 > z_1$.

I know how to find the angular separation $\theta$ between them by means of the angular coordinates. But how to find the comoving distance (at epoch $z=0$) from them ? I know how to find the comoving distance from the Earth of the two quasars, can we find the distance between them using these two distances?

A related question is to find the redshift of $Q_2$ as seen by $Q_1$ at the epoch when $Q_1$ received the light emitted from $Q_2$ .

There is some standard method to solve this problem ?

Edit: I found you a very relevant paper!

http://arxiv.org/pdf/astro-ph/0007341v1.pdf seems to be solving exactly your problems for sources with arbitrary angular separation $\alpha$. Equations (12) and (14) will give you the comoving distance $\chi_2'$ and redshift $z_2'$ between them at the epoch when $Q_1$ received light from $Q_2$. You will have to solve these equations numerically to find the solutions, which will be more complicated depending on the cosmological model you use.

Have a read of these notes http://arxiv.org/abs/astro-ph/9905116. They are my go-to when trying to work out cosmological distances.


When are the symbols ⊕ and ♁ used to designate Earth? In astronomy I usually see Earth designated with a circled plus or cross. NASA says this is the Earth's astronomical symbol. Wikipedia tends to use it, for example designating the Earth–Moon distance by $\Delta_{\oplus L}$. But Wikipedia also says that the globus cruciger ♁ is used to symbolize Earth.

Does the latter find any usage in astronomy?

(The actual Unicode code point for the circled plus Earth symbol is U+1F728 but this does not render for me, so I used the graphically similar U+2295 instead.)

Yes, I don't know how common the globus cruciger is, but it is used at least here for the electron neutrino pp flux at Earth: NEUTRINO 2006: Proceedings of the XXIIth International Conference on Neutrino Physics and Astrophysics.


How to calculate the orbital period and the density of a planetary body? I'm a student studying in 9th grade, I am making this fictional habitable planet named "Darwin B" for a planet making competition. It orbits a sun-like star at a distance of 1.15 AU or 172 million kilometres in a nearly circular orbit. Its rotational period is 19 hours, 38 minutes. Its mass is $6.15×10^{24}kg$ and its radius is about 6,743 kilometres. I have to calculate its orbital period and density but I'm weak in maths and don't know how to. Please help.

The formula for orbital period is given on Wikipedia:
$$T=2\pi \sqrt\frac{a^3}\mu$$
where:

$T$ is the orbital period in seconds
$a$ is the orbit's semi-major axis in meters
$\mu = GM$ is the standard gravitational parameter

$G$ is the gravitational constant
$M$ is the mass of the more massive body in kilograms

So $T = 2 \pi \sqrt \frac { (172 \cdot 10^9) ^ 3 } { 6.674 \cdot 10^{-11} \cdot 1.9884 \cdot 10^{30} }$. Can you take it from here?
As for density, the volume of a sphere is given by the formula $V = \frac43\pi r^3$; density ($\rho = \frac MV$, with $M$ the mass) is usually given in grams per cubic centimeter, so it makes sense to convert to those units. That gives the following calculation:
$$\frac { 6.15 \cdot 10^{27} } { \frac43\pi (6.743 \cdot 10^8)^3 }$$


Any studies of the detection of dark matter in or near the solar system? Has there been any theoretical or observational study of the possibility of detecting "local dark matter", either within the solar system or near the Sun in interstellar space?

If so, what methods could possibly be used to detect it?

We don't know what dark matter is, and we have never directly detected the exotic dark matter that is believed to make up much of the mass of the universe.

We expect dark matter to be in the form of a particle that interacts only with gravity and the weak nuclear interaction. Such a particle would be invisible (as light wouldn't interact with it) and it could pass through normal matter.

Since such dark matter doesn't interact much, except gravitationally, it would be spread out in a large elliptical cloud around the galaxy. There would be dark matter everywhere in the galaxy, including in the solar system, but there are no clumps of dark matter. There are no dark matter planets or stars.

In fact, we would expect about 1 proton mass of dark matter in every 3 cubic cm. That is a very small amount of mass. In the solar system, there is a billion-billion times more normal matter than exotic dark matter. And as I've noted, nobody has ever directly observed a dark matter particle, we don't know what they are.


Is there an online planetarium where the observer is on another celestial body? I know about in-the-sky.org and its online planetarium feature, but that's from the Earth's perspective. Is there a planetarium (preferably online but if not, that's fine) where the observer is on another celestial body, like the Moon or Mars?

Yes, the free software Stellarium (also Wikipedia) can do that. It has a list of celestial bodies and you select which one you want to be on. When you open Stellarium go to Location Menu (or the button F6 for Windows OS) and that is where you set your viewing point.

There might be additional options in Where can I find/visualize planets/stars/moons/etc positions?


What range of exit pupils work for observing the full moon? I'm observing the full moon, from a major city, with heavy light pollution and dust.

The objective size is fixed, for this comparison, something between 4 and 6 inches. The question is about the tradeoff between magnification and dimness of the image. I'd like to know what range of exit pupils works best. Again, the goal is to observe the full moon.

To put it differently:

What's the smallest exit pupil at which the moon gets so dim that you can't take advantage of the increased magnification? ("Empty magnification").
What's the largest exit pupil at which the moon is bright enough, so a larger exit pupil doesn't help?

The only reference I could find is this, which claims that 1-2mm works best for the moon. Is that correct?

Light pollution does not matter for the Moon. Even transparency doesn't matter that much. What does matter is seeing, a.k.a. air turbulence.

It is very rare that an exit pupil smaller than 0.5 mm is useful for anything - perhaps for some tight double stars, but that's about it. So take that as a hard lower limit.

In terms of a "soft" limit, it depends. If seeing is excellent, if the optics are very good, if the instrument is in perfect collimation and at perfect thermal equilibrium, then you can push e.p. below 1 mm, especially for a small aperture like the range you mention.

If seeing is less than perfect, if optics are so-so, if the instrument is miscollimated or too hot, then the minimum usable e.p. goes up. Each one of these factors influences the observation, so it's hard to pin a number.

Bottom line is, use what works for you.

  claims that 1-2mm works best for the moon. Is that correct?

2 mm seems too big for a small 4...6" aperture when observing the Moon. It might be okay if you want to capture the whole disk, but it seems large if you're interested in small details.

Again, it's all relative.

I have a 6" reflector with a self-made primary mirror of excellent quality. I always take care to collimate the scope, and to keep it at thermal equilibrium. I live in California, so seeing is often good. It's not uncommon for me to use less than 1 mm exit pupil when observing the Moon. I use 0.84 mm as a starting point, sometimes pushing it down as low as 0.6 mm.

If I use an e.p. of 1.1 mm for the Moon that means seeing is pretty bad - but in those cases I'd rather do something else, instead of being frustrated by bad seeing. But at that e.p. in a wide eyepiece (82 degrees apparent field) you see the whole Moon at once, which is always nice when you do a public demo (sidewalk astronomy).

For much larger apertures, the ideal exit pupils tend to be somewhat larger even when observing the Moon.

To predict seeing, go on the Clear Sky Chart site, find a location near your home, and look at the third row in the chart (labeled Seeing); when it's dark blue, seeing is good.

For the largest practical e.p., don't worry about it. The Moon is so bright, losing light is not something you should pay any attention to at all.


How precise are the observational measurements for the speed of gravity? General Relativity says that gravity moves at the speed of c.  How precise are the measurements?

The accuracy is extremely good thanks to LIGO and a binary neutron star merger.  It's accurate to about 3 parts in 1 quadrillion!
See the paper Gravitational Waves and Gamma-rays from a Binary Neutron Star Merger: GW170817 and GRB 170817A
The paper constrains "the difference between the speed of gravity and the speed of light to be between $−3×10^{-15}$ and $+7×10^{−16}$ times the speed of light."
Older response before the LIGO observation:
One of the more recent of gravity was first measured in 2003 using Jupiter and a quasar.  The measurements were for the purpose of proving a finite speed and weren't very precise.  See the paper for more.
More recently in 2013 a Chinese group built a model using Earth's tides that helped them narrow it down.

... [T]he speeds of gravity are from 0.93 to 1.05 times the speed of light with a relative error of about 5%. This provides first set of strong evidences to show that the speed of gravity is the same as the speed of light.

This is so far the most accurate measurement I've seen.  See the paper for more.
In the near future, LIGO may be able to provide more accurate measurements by comparing the distance among detectors and the delay of observation.
Also related, see the question "How is the most accurate value of  G  measured?"


what determines the density of atmospheres What factors determine the atmospheric density on various planets and moons? Some larger bodies have less dense atmospheres, while some smaller ones have a more dense atmosphere.

Most simply, the density is determined by the number of molecules, the surface area of the planet, the temperature, and the gravity.  But it sounds like what you really care about would be called the "column density", which depends only on the number of molecules and the surface area, and you are wondering why there is not some direct correlation between the two.

I would say that the reason you don't see a clear correlation is that atmospheric gas generally comes through the surface of the planet, so all else being equal, you'd expect all planets to have the same column density.  But all else is not equal, because you have issues of how volcanic is the interior (so how much gas gets liberated), and how much escape into space happens.  The amount of volcanism is higher for bigger planets (they take longer to cool), and bigger planets also have stronger gravity and cling better to the gas, so one might generally expect larger planets to have higher column density (and indeed the Earth and Venus have higher column densities).  But a higher temperature will lead to escape, and a lower temperature will lead to freezing out many of the species, so temperature is a key factor as well.  Magnetic fields can also play a role, by protecting the atmosphere from being stripped by the stellar wind, and it's still not real clear what planets will have strong fields (Earth) and what planets won't (Venus, whose interior is almost the same as the Earth but doesn't spin much).  Finally, you can have even more complicated effects, such as the way Earth's oceans have dissolved and locked up carbon dioxide.  So when there are a lot of potential factors, you can expect a lot of variation.


How does a telescope measure parallax angle? I know how stars' distances are measured by using parallax, but I want to know how the actual telescope measures the parallax angle. Any helpful links/ explanations are needed.

No, the telescope doesn't measure the parallax. A sextant or any other angle measuring device fit on the telescope does.

And, we don't(can't) directly measure the parallax angle. Instead, we just track the position of the star/object throughout the year. A little bit of spherical astronomy math shows us that the path of a star in the celestial sphere defined by a fixed reference through the year is an ellipse around it's mean postion, called the parallactic ellipse

Here is a link which does the rough math involved.

The semi-major axis of this ellipse is equal to the parllax angle P, while the semi-minor axis is equal to P*sin(b), where b is the stars ecliptic latitude.

EXTRA NOTE: These parallactic ellipses are often rotated and parameters modified as the abberational ellipse is superposed on it. In practice, things get nigh complicated.


What measures are currently in place to detect asteroids on possible collision courses with earth? I know some measures must exist, but I have no idea what the programs are, what organizations are involved, or how effective they are.

This site is an asteroid tracking program started up by NASA. It seems to be fairly effective at detecting asteroids. Read up on this article for further information about NASA's plan.


What does BJD-2450000 mean? In different papers I've encountered time indications in BJD-2450000. I know that BJD stands for Barycentric Julian Date, but I don't understand the meaning of the suffix -2450000.

This is an example of what I'm talking about (table is taken from this paper).
I'd like to know how I can convert this in JD or just "normal" BJD if that makes sense?

Julian dates offer one nice feature: All astronomical observations recorded by humankind have a positive timestamp. A key downside of Julian dates is that current timestamps on computers that use 64 bit IEEE Standard for Floating-Point Arithmetic (IEEE 754) have a resolution of 40 microseconds. Almost all computers use the IEEE 754 floating point standard. This lack of precision can be a problem with some modern astronomical observations.
One way around this limitation is to use a pair double precision numbers to represent time, the sum of which logically represent the Julian date. This is the approach used by the Standards of Fundamental Astronomy (SOFA). Another approach is to use an offset.
One example is Modified Julian Date, subtracts 2400000.5 from the Julian date. This slices off the leading 24 which is always present on any recent Julian date (JD 2400000.5 was midnight on 17 November 1858, and JD 2500000.0 will occur at noon on 31 August 2132). The additional half of a day in the MJD offset effectively sets the start of a day at midnight rather than at noon.
Subtracting 2450000.0 is equivalent the slicing off the leading 245 from recent Julian dates. This simple slicing technique is valid from noon 9 October 1995 (inclusive) to noon 24 February 2023 (exclusive). The precision using IEEE 754 doubles at the latter date will be two picoseconds, which is more than precise enough for timestamping any astronomical observation for the past two decades and the next decade.


Name for 1-e and 1+e terms? In several equations of orbital elements (such as the determination of true anomaly from mean anomaly), the terms 1-e and 1+e appear. These are the ratios of the orbital periapsis and apoapsis to the semi-major axis, but do the ratios themselves have names?

As the commenter states, $e$ is indeed called the orbital eccentricity. If you add a radial scale length (e.g., semi-major axis) to both of those values the $1-e$ describes the periapsis (closest approach of orbit) and the $1+e$ apoapsis (furthest) of an elliptical orbit. They don't have a specific special name, as they are dimensionless measures, but can be quite useful in determining the orbit of planets and other Keplerian systems.

Periapsis:
$$
r_{p}=a(1-e)
$$
Apoapsis:
$$
r_{a}=a(1+e)
$$

Perhaps the could be named the maximum radial eccentricity and minimum radial eccentricity, if you needed to describe these parameters in a report or class homework etc.


Why are planets spherical? I am a complete novice when it comes to astronomy, but someone asked me the above question casually. It seems as though planets are all roughly spherical, whereas other celestial bodies aren't necessarily. Is there are particular reason for this?

That's mainly because of the gravity, and of size. Small objects, e.g. a stone, create only a very weak gravitation. The stone is much harder than necessary to withstand this force caused by its own gravity.

Earth is much larger. The gravity at the surface causes the weight we feel. A huge mountain on Earth would be flattened under its own weight, because the matter it's made of isn't hard enough to withstand this weight.
By hills filling up valleys the planet as a whole becomes spherical.

A fast-rotating planet gets a slightly flattened ellipsoid shape.


Classification and discovery of stars How do astronomers find the difference between a 'Cephid Variable Star' and distinct normal stars with opaque objects blocking it's light, like stars with planets revolving around them while both create a dip in the brightness-time graph at regular intervals of time, when this is how new stars are discovered?

Cepheid variable stars and exoplanets transiting stars have very different light curves (the relation between brightness and time).

Exoplanet light curve from NASA:

Cepheid variable light curve from astronomynotes.com:

Also, as Rory Allsop points out, the scale of the change in brightness is very different. Cepheid variables can be seen at great distances, which allows us to use them to measure distances by their regular period-brighness relation, whereas exoplanets can only be found more locally by this method.


Do falling bodies ever slow down? I've had some discussions recently about black holes, and the issue of infalling bodies taking forever to reach the event horizon. That's essentially what Einstein said in his 1939 paper on a stationary system with spherical symmetry consisting of many gravitating masses. He said "it is easy to show that both light rays and material particles take an infinitely long time (measured in “coordinate time”) in order to reach the point r = μ/2 when originating from a point r > μ/2”.

Now, I'm a big fan of Einstein. But there seems to be two issues with this:

One is that Einstein concluded that black holes cannot form, but we
have good evidence that there are black holes out there. The obvious
example is Sagittarius A*. That's where's there’s something
with a mass 4.28 million times the Sun, with a diameter of less than
44 million kilometres, and we can't see it. Surely it's just got to be a black hole.
The other issue is that falling bodies don't slow down. Imagine you
drop a body at elevation A and it falls down to elevation B. The
"force" of gravity relates to the first derivative of gravitational
potential. Hence the bigger the difference in gravitational time
dilation between elevations A and B, the faster the body falls past
B. Then if you drop a body at elevation B it falls down to elevation
C. Again the bigger the difference in gravitational time dilation
between elevations B and C, the faster the body falls past C. In a
typical gravitational field the force of gravity at B is greater than
at A. Hence as the body descends, the acceleration increases as well as the falling speed.

Imagine a gedanken spaceship from which we've suspended a cable. We have clocks at different elevations, so we can measure the gravitational time dilation at each elevation. We can also release test bodies at each elevation and record the  clock readings as they fall past other elevations:

At the end of the experiment we can reel in the cable and upload the recorded measurements to ascertain how our test bodies behaved. My understanding is that we will always find that time dilation always increases as we descend, that the falling body always accelerate downwards, and that both the acceleration and the falling speed always increases as the body descends. Is this correct? Or do falling bodies somehow stop accelerating? And do falling bodies ever slow down?

According to the standard interpretation of General Relativity (e.g. as presented in "Exploring black holes" by Taylor & Wheeler chapter 3, or "Black holes, white dwarfs & neutron stars" by Shapiro & Teukolsky, pp 343-345) then yes they do. But it depends on the frame of reference of the observer - there is no absolute answer.

According to an observer far from the black hole, the rate of change of radial coordinate with time (for an object that started falling radially inwards far from a non-rotating black hole) is given by
$$\frac{dr}{dt} = -\left(1 - \frac{r_s}{r}\right)\left(\frac{r_s}{r}\right)^{1/2}$$
where $r_s$ is the Schwarzschild radius and $r$ and $t$ are Schwarzschild coordinates.

If we call this the infalling speed as measured by a distant observer then we can see by differentiation that it goes through a maximum at $r=3r_s$ and that $dr/dt \rightarrow 0$ as $r \rightarrow r_s$.

However, an observer accompanying the falling particle would totally disagree. To them, their velocity is given by $dr/d\tau$, the rate of change of $r$ with respect to the time on their clock.
$$\frac{dr}{d\tau} = -\left( \frac{r_s}{r}\right)^{1/2}$$
which continues to increase up to and below the event horizon.

The latter appears to admit the possibility of faster than light travel, but no more so than me (correctly) saying that if you travel at close to the speed of light you can get to a star 10 light years away in much less than 10 years (as measured on your clock).

Finally we could have the point of view of stationary "shell" observers at fixed radial distances (outside the event horizon, because no stationary observer is possible below the event horizon). They would measure the speed of objects falling past them to be
$$\frac{dr_{\rm shell}}{d\tau_{\rm shell}} = -\left(\frac{r_s}{r}\right)^{1/2}.$$
This means that the reports of stationary observers (which is the gist of your question I think) at increasingly lower heights, is indeed that the velocity of the falling object is increasing as it falls.

There is no paradox to these apparently contradictory points of view. Measurements of non-local events and phenomena are not bound to agree in General Relativity, where there isn't even agreement on what is meant by "now" or whose coordinate system in what frame of reference should be used in any particular circumstance.


Can we track matter through time by looking at different depths in space? If we look back far enough we can see all the origins of the universe, so is it possible, even if not feasible, that we could trace the history of some matter as it moves through space-time? I want to understand how looking at different depths in space and time are correlated in regards to the matter being observed.

For example would it be possible to look deep into a certain part of space and time to find some galaxy that contributed to the matter that makes up the Milky Way today? Then somehow follow it through space-time by looking at different depths and locations in space, and see how it came to be part of the Milky Way?

Would it be possible to look deep into a certain part of space and
  time to find some galaxy that contributed to the matter that makes up
  the Milky Way today?

No, that's not possible. If we could do that, it'd mean that the matter traveled from there to here faster than its light got here, and matter can't travel faster through space than light does.

All we can do is look at similar galaxies to the Milky Way at earlier times. And because of the expansion of space those galaxies are now even further away from us than they were when they emitted the light that we're seeing now.

Galaxies develop (mostly) in isolation from one another, apart from the occasional merger or collision between neighbouring galaxies. Intergalactic distances are fairly huge, so it takes vast amounts of time for matter to travel from one galaxy to another, and matter is mostly bound by gravity to the galaxy it's in. Galactic escape speeds are pretty high, although the occasional star does get flung out of the galaxy by cataclysmic events like supernova explosions. But even then, such rogue bodies mostly end up in intergalactic space. The odds of them ending up in a different galaxy are pretty slim.


Why do astronomers use supernova to measure distance in space? Why can't the distances between our Sun and the celestial objects be measured directly and instead had to rely on looking around for supernova? Is these events frequent, evenly distributed and lasting enough for astronomers to measure any place in our observable universe? Is this method of measurement more preferable than others currently?does this method had any limitations such as an absent of supernova nearby?

Type Ia supernovae are not common; they are quite rare events, happening maybe once per 100-200 years in a big galaxy. Nevertheless they have two properties that make them fantastically useful for distance measurement.

They are (very close to) "standard candles". The physics of the supernova detonation, thought to be when a white dwarf accretes matter and exceeds the Chandrasekhar limit, is very "standardised". The bomb goes off in exactly the same way with the same amount of identical explosive. That means to a good approximation, measuring the apparent brightness of a type Ia supernova and comparing it with nearby examples means that the distances to these events can be accurately estimated.
They are really luminous and last 2-3 weeks. This means that they can be seen at enormous distances, they almost outshine the galaxies that they are in, and they last long enough for astronomers to discover them in automated surveys and still have time to follow them up and measure their light curves and estimate their peak brightness.

The limitations are that you can't choose which galaxies you measure the distance to. You have to wait for a supernova to go off and that might take 100-200 years or more for a particular galaxy. There are also continuing debates about just how standard a candle these objects are. It is possible that more distant galaxies that we see in the very early universe made stars with a different composition (far less elements heavier than helium) and that this alters things a bit. Other problems are associated with "de-reddening" the supernova light curves to account for the possibility of obscuring dust in the host galaxy. It is also possible that there is more than one way to produce a type Ia supernova (accretion of additional mass or merger of two white dwarfs).

I am puzzled by your question "Why can't the distances between our Sun and the celestial objects be measured directly?". Galaxies are much too far away to have a trigonometric parallax distance. Methods like using Cepheids or RR Lyrae variables don't work for very distant galaxies because we can't resolve the individual stars. You could estimate a distance using a measurement of redshift and a cosmological model for the expansion of the Universe, but really the point of using these supernovae was and is to test and improve the cosmological models.


How do accretion disks around black holes evaporate via winds? I'm curious as to how accretion disks around black holes evaporate. I've heard winds play a crucial role in this. But what winds specifically and how do they form? Is it a stellar wind of sorts? If so how effective would it be?
I've also heard radiation pressure causes these winds, but that shouldn't work unless the black hole is still in an X-ray binary. Any help is appreciated.

Not an expert, but here are my 2 cents until there will be a better answer:
Winds are not directly responsibly for the evaporation. Rather can they consist of evaporated material.
The cause for the evaporation of gas out of the accretion disk is primarily heating from the much hotter corona. Or as Liu & Taam put it:

The ions in the corona are directly heated by viscous
dissipation, partially transferring their energy to the electrons by
means of Coulomb collisions. The energy gained by electrons
cannot be effectively radiated away and is conducted to the
lower, cooler, and denser coronal layers by electron–electron
collisions. In the transition layer, the conductive heat flux
is radiated away through bremsstrahlung only if the number
density in this layer reaches a critical value. If the density
is too low to efficiently radiate the energy (lower than the
critical value), a certain amount of lower, cooler gas is heated
up, whereby the bremsstrahlung cooling rate is raised, until
an energy equilibrium is established between the conduction,
radiation, and heating of the cool gas. The transfer of gas
from the disk to the corona, to establish the equilibrium, is
called evaporation.

So, all you need for evaporation is a hot corona above the accretion disk and more heat transfer as you can radiate away. Then gas will "boil off" and the corona will "dig into"   the disk until a critical density and equilibrium is reached.
For all this you do not need any winds. A wind will originate from the disk or the corona when gas is further accelerated outwards. This acceleration can be primarily radiation or line-driven when the intense radiation from the central regions illuminates the outer parts of the disk (see here). This also works for AGN. See e.g. these simulations.
Alternatively, these winds can also be (partly) magnetically or thermally driven. Which of these mechanisms dominates depends on the type of source (e.g., AGN/XRB), state and position on the disk. (Someone else might be able to give details here.)
And because images are always nice, here is a graphic how one can imagine the geometry for an AGN (from this paper):


Does Sun have a reflection on Earth? The new Google Maps1 presents an actual view of Earth, with the current position of Sun illuminating half part of Earth in real time. It is quite an exquisite view.

My question is based on the following image:

As you can see, Sun's reflection from southern Atlantic Ocean looks very charming. But does it really happen like that? If we travel far2 from Earth, can we actually see Sun's reflection or it is just something Google added for aesthetics?

1: Yes, I don't like most of its new features too, especially when they broke several of the old ones. But that's a separate discussion.

2: Although, the camera's supposed position would be close to geostationary orbit, I guess the reflection can also been seen as close as ISS or Hubble.

It's called sunglint. This can be problematic for Earth-observing satellites in low Earth orbit. Such satellites typically don't take a "picture". They instead continuously scan the Earth a line at a time. This means the sunglint moves with the satellite.

You can see this effect yourself while flying in an airplane. Little ribbons of rivers and lakes can appear to be on fire. Here's a nice short YouTube video that shows this perfectly.


Stopping down a telescope to reduce brightness by 5 magnitudes Suppose we have a variable star with a change in magnitude between maximum and minimum of 5 mag.

How much would you need to stop down a telescope (block some of the area) so that the star at maximum brightness appears to be the same brightness as when it is at minimum magnitude with no stop.

Treating as a school question

Each change of 1 magnitude changes brightness by how much? So by what factor will a change of 5 magnitudes give? (5 magnitudes is a very convenient number for this question)

You will need to reduce the area of the telescope by the same factor.


Did Earth "tilt" seasonally before it was accepted that the world was not flat? Before it was widely accepted that the Earth was round, how did astronomers describe the yearly movement of the sun to south in the Northern Winter that produced the seasons? Did the Earth wobble or did the Sun just not have a perfectly circular trajectory around the Earth? Did this perspective change when we realized the Earth was a globe? What about before Heliocentrism was the norm?

This is not really a physics or astronomy question, but I reckon that when you model the Earth as flat, the stars and the Sun are merely minor objects and their motion is governed by heavenly spirits rather than regular orbits. So, rather then the flat Earth tilting, the Sun's trajectory changed with the seasons.

Btw, what I find more concerning is how the (infinitely extended) flat Earth can be reconciled with the Sun rising in the East after setting in the West: how did it get there?

Btw2, I don't think that elites in classic times (Greek/Egyptian/Middle East Cultures ~1000Bc and more recent) seriously considered the Earth to be flat. All the constructions of epicyles etc for explaining the data assume a spherical Earth in the centre of the cosmos. Flat is much more arcane than Geocentric.


Can you test solar eclipse glasses with a remote control? By putting solar eclipse glasses directly between a remote control IR emitter and device receiver, could some unsafe glasses be detected? My guess is most fake glasses are just blocking visible light at best, while safe ones should block the IR signal as well. It would not guarantee they are 100% safe, but it might be able to detect bad ones.

I know to check for the ISO logo, cert number, and buy from a reputable vendor recommenced by the American Astronomical Society (which I have). It would still be nice to double check pairs of glasses before giving them out to friends and family.

I was considering a similar line of thinking.  However, there were a few hangups which I believe prevent yee-average-human from testing the glasses.

Many sites suggest a transmittance of 0.0032 (welding shade 12) to be the minimum to be safe (and typically recommend 13 or 14, which has even lower transmittance).  That means you're only seeing 0.3% of the light get through the goggles.  Your test would need to have enough sensitivity to pick up that tiny amount of IR light, and to not only receive a signal but be able to meaningfully measure its amplitude. You would also need to ensure that UV light is blocked as well... and all the other IR bands (I am not confident as to how wide-band protection one actually needs).

Even then, you'd have to deal with the issue of having built your own contraption to test a device which must protect your eyes from a permanent damage that you literally cannot see.  I'd want to be quite the confident electrical engineer and systems test expert to make sure that the tests I do actually test what needs to be tested.

Note: I have a related question asking how one can track down the ISO certifications and see if your particular vendor has legitimate glasses or not.  I run into the same issue: how can you trust that this device performs as indicated, because your eyesight depends on it.


Jupiter vs brown dwarf star What would happen to Jupiter if a brown dwarf star collided with it, and vice versa what would happen to the brown dwarf star?

Brown dwarfs are heavier than Jupiter (jupiter has mass 0.000949 solar masses), although about the same size. Their compositions are fairly similar, with brown dwarfs depleted in deuterium if they are heavy enough (it is burned over a few million years after formation).

Adams & Laughlin modelled the collisions of brown dwarfs, mainly to estimate whether two dwarfs could merge to make a star heavier than 0.075 solar masses that would begin hydrogen fusion properly. They found that this could happen, and not too much mass appears lost as they splatter into each other if it is a somewhat glancing collision. Obviously, a near glancing hit might just scatter some of their atmospheres but not stop them.

Jupiter would likely behave the same. The result would be a heavier brown dwarf. It would be heated significantly by the energy of the merger, which in turn would make it expand to a larger size until it cooled down. Jupiter would also contribute deuterium so if the dwarf was heavy enough it would further heat up for a while. Given Jupiter's minimal mass compared to the 0.075 limit it is not likely it would be enough to turn the new object into a red dwarf.


Why is the outline of a crescent moon brighter? Some quick googling has given me the answer to whether seeing part of the moon that receives no sunlight is an optical illusion or not: it isn't. Earthshine lights the moon enough for the human eye to notice.

However, it seems clear to me that the outline of the unlit part of the crescent moon is brighter than the surface. I was riding my bike home yesterday on a clear winter night, and became confused on why that could be. What mechanism is responsible for making the edge brighter than the dark part of the moon? Or is it my brain trying to make a circle?

Here's an example of what I mean.

In the image you are now refering to the only significant brightness difference is between the mare and the "highlands".

The retina enhances edges. So you'll see edges enhanced, even if they aren't.

By image processing edges can also be enhanced to make contours clearer. This is not the case for the referred image.

To better assess, which part of the perception is real, and which is illusion, here the illusion part:


Do we get any benefit from being in a galaxy? If our solar system were to somehow be created outside of a galaxy as a single star in the vast nothingness between galaxies, would life on Earth change any?

Does being a member of a galaxy actually provide some kind of benefit to our solar system or the creation/sustaining of life?

I've had my eye on this question for some time, but I haven't had the time to answer it. Sorry for not getting to it sooner.

As zibadawa timmy said, you'd be hard-pressed to find the materials necessary for a star system if you're outside a galaxy. For one thing, there aren't any stellar nurseries in the intergalactic void (well, none that we know of. But I'd guess it's highly unlikely that they exist plentifully). So you have no hydrogen or helium, or dust, or other elements - in fact, you have nothing that you could use. That's a wee bit of a problem.

There is a way you could get around this, though - if your star system is in a globular cluster. They're groups of stars that orbit a galaxy. We don't know just how they got out there. They could be from dwarf galaxies, which then lost them, or they could have been partially chucked out of the larger galaxy but stayed in an orbit. The trouble is, as Wikipedia says,

  However, they are not thought to be favorable locations for the survival of planetary systems. Planetary orbits are dynamically unstable within the cores of dense clusters because of the perturbations of passing stars. A planet orbiting at 1 astronomical unit around a star that is within the core of a dense cluster such as 47 Tucanae would only survive on the order of $10^8$ years.

Yep. Not a great place to live.

But I take it you didn't intend for our solar system to be in a globular cluster, so I'll go back into intergalactic space. It's not actually a vacuum - not that most parts of outer space are. There's a lot of plasma and hot gas there (the intergalactic medium), sometimes getting to a toasty $10^5$ or $10^7 \text { K}$. Could it harbor a stellar system? That depends on just what material the star brings with it. If planets haven't formed around it yet, and the star has a protplanetary disk, then perhaps that disk could dominate over the intergalactic medium. Remove the ICM and maybe planetary formation could proceed normally.

What happens then? Well, there would be some interesting effects. For example, the Oort Cloud might not have become spherical, because it was partly deformed by the galactic tide. Here, tidal forces probably wouldn't be as strong (although this depends in part on just how far away we are from the galaxy). We probably wouldn't have a lot of negative effects, because we don't have a lot of extreme benefits here in the Milky Way. Our galaxy does not appear to have an AGN, there aren't any monstrous black holes nearby, and we aren't undergoing any catastrophic galactic collisions. And the Milky Way-Andromeda collision won't be too bad.

We probably wouldn't be in too bad of a position if we were in intergalactic space. There's not a lot of benefits from being in a galaxy - well, no significant ones, at least. Well, besides star formation depositing all the materials we need for the solar system to form. That's a biggie.


Why do small mirror imperfections matter with modern computers Modern telescopes go to great lengths to have perfectly shaped parabolic mirrors. My question is, why go to the trouble of having a perfect mirror? Why not take a mirror roughly the right shape, and then correct for the distortion using computers?

correct for the distortion

An imperfect mirror does not produce a distorted image - it produces a blurry image. With light-field sensors and phase imaging, one could possibly correct for the blur, but it is much more challenging problem than normal lens distortion correction.
Distortion refers to a systematic change in how shapes are projected in an image. It results from a lens or mirror with good, accurate geometry that just does not produce a rectilinear projection.
Random imperfections in a mirror do not cause distortion. Every point in the surface of a mirror contributes to every pixel in the result image. If a single part of the mirror is at slightly wrong angle, it does not cause a distortion in one point of the image. Instead, it projects the same image at a slightly different alignment on the same sensor. (1)
In the case of a starfield, this would cause ghost images of very dim stars to appear next to the real stars. Repeat this for a thousand imperfections, and the result is just blurry dots. Deconvolution is a process that can be used to remove blurriness, but noise and other uncertainties limit its effectiveness.
(1) This may be a bit unintuitive if you think about funhouse mirrors where the image is distorted. Those work differently because they act the part of a planar mirror, where indeed each part of image is reflected by a single part of a mirror. But planar mirrors cannot form an image by themselves, instead the lens in your eye is the critical component of the image accuracy.