Could the word "multiple stars" include binary stars? We are translating some illustrated book for kids about astronomy, and then have a question as in the title.

Referring to Wikipedia etc., It appears that "binary stars" represents the star systems of two stars, whereas "multiple stars" does those of three or more stars.

Thus, our question is whether it is an abuse to use "multiple stars" for binary stars.

Thank you in advance.

I'm not aware of any official (i.e., IAU) definition of a multiple star system.  However, as someone who used to do research in that field, I would interpret multiple star system as encompassing systems with two or more stars.  So a binary system, a triple system, and a quadruple system are all multiple star systems.


A question about star names... E.G. CW Tauri, DZ Bootis, IQ Lupi, EU Eridani Take the following examples of star names/identifications:-

Regulus is the common name for a star.

HIP49669 is the Hipparcos ID for the star

HD87901 is the Henry Daper ID for the star

32 Leonis is the Flamsteed ID for the star

Alpha Leonis is the Bayer ID for the star

And now for a probably unrelated stars at random.

Why and how does stars like 'CW Tauri', 'DZ Bootis', 'IQ Lupi', 'EU Eridani' get their name, is there a catalogue for them?  Who assigns their name? Do they individually mean something or randomly chosen letters?  I'm not just after those stars but all stars with similar style naming.

Thanks in advance

The Bayer catalogue uses Greek letters, then lower case Latin letters and if needed, upper case letters. But it never gets beyond "Q"

In 1855, the German astronomer Argelander proposed naming a particular variable star "R" (as using upper case letters at the end of the alphabet would avoid clashing with the Bayer catalogue.

In 1867 it was agreed by the German astronomical society to name variable stars R,S,T... Z, then RR, RS, RT up to ZZ, followed by the constellation name.

This system provided enough names until 1907, when they reached ZZ Cygni.  It was decided to start again with AA, up to AZ, Then BB to BZ etc.  That system provides 344 names and worked only until 1929, and after that, when more variables than this were found in a single constellation, they switched to a 'Vxyz " where "xyz" is a number starting with 345 e.g. V345 Cyg.

So the "2-letter designation" is particular to variable stars http://cdsarc.u-strasbg.fr/afoev/var/edenom.htx and is used in the General Catalogue of Variable stars, but the scheme doesn't originate in the at Catalogue.


how did James Bradley discover light aberration? I do not understand his timing method. James
The magnitude of the aberration could be v/c?
The speed of earth is about 30km/s.  That means the deviation from the real position of a star is about 30/300000 rad =180*3600/3.14159/10000 ~20arcsecond.
That is a horrible value.
Could the effect of parallex, proper motion and light aberration of a star be separated clearly?
－－－－－－－－－－－－－－－－－－－－－
Sure. Aberration is always the same. Parallax depends on the distance, and the phase is different.
Why is their phase different? They all have a one year period.
Generally we should not worry about the light aberration, right? Even space telescopes have reference stars. The position of the reference stars are fixed. So although there are detectable wobble(20") in one year , because we use reference stars, this kind of effect is removed every day immediately. Every night, when the night assistant lets a telescope point at some direction, he has to use reference stars to calibrate that night's pointing.
Am I right?

That is a horrible value.

That value for aberration of starlight is "horrible" in two ways: One, in that's is very small. And two, in that it's large enough to interfere with parallax measurements.

In the late 1600s, scientists had already adopted the heliocentric model, and so were looking to measure the distance to the stars via parallax. Robert Hooke was among the first to attempt it, and probably the first to claim a measurement of the parallax. His observations were not very convincing at that time, but he probably did observe some effects from aberration of starlight (though they didn't know such a thing existed back then).

In the early 1700s, Bradley attempted to make a more precise measurement of parallax, and did obtain significant data, but the variations in the star positions were out of phase with the model. Using the speed of light estimate by Ole Roemer from Hooke's time, he correctly proposed that this was not parallax, but a speed-of-light phenomenon, now known as the aberration of starlight.

More info here:

http://arxiv.org/pdf/1208.2061.pdf

It took another century, until early 1800, for Bessel to measure the first parallax.

Basically, aberration of starlight produces the same variation in the positions of all stars, a periodic elliptic "wobble" that is the same for all stellar objects. Parallax produces a similar kind of wobble, but the amplitude depends on the distance to the star, and the phase is different from the aberration's.

There are also other effects, such as nutation, which introduce further perturbations.

http://cseligman.com/text/history/bradley.htm

Aberration per se is easy to explain via the old analogy to running through the rain. When you're standing, rain falls vertically. When you're running, rain appears to come from a different direction. Same with starlight. Like in this diagram (by Professor Courtney Seligman, linked above):

Now, small as that value is, it's still 20 arcsec. The resolving power of early telescopes (and modern amateur telescopes) is 1 arcsec or better - which is more than enough. What you need is a very solid mount with good setting circles. Lacking that, you could build a telescope with a fixed position, very rigid, forever pointing at the same place in the sky - that's what Hooke did - and then you're only limited by resolving power.

In any case, the precision required was quite doable with 1700s technology. An amateur could do it today, at very low cost - what you need is persistence to follow through with a long term project.

  Could the effect of parallex, proper motion and light aberration of a
  star be separated clearly?

Sure. Aberration is always the same. Parallax depends on the distance, and the phase is different.

EDIT:

  Why is their phase different? They all have a one year period.

Go to the site by Prof. Seligman, link above, and look at the two diagrams showing the aberration and the parallax. In point B (in April), the star is in the middle of the ellipse due to parallax, but it's at the right-hand tip of the ellipse due to aberration.

Broadly speaking, parallax is a position-based effect, whereas aberration is a speed-based effect. It makes sense that they are out of phase by 90 degrees (remember trigonometry?)

Proper motion appears as a long-term drift of stellar positions that remains even after you subtract aberration, parallax, nutation, etc. There's a back-and-forth-back-and-forth from aberration and parallax, whereas proper motion keeps going in one direction only.

You could measure proper motion with relatively short term observations, but it only works for stars that are nearby and are moving fast. If you're willing to wait decades, proper motion is much more easy to measure.

EDIT:

  Generally we should not worry about the light aberration, right?

Yeah, sort of. You know, at any given moment, the size and the direction of it. So you could simply subtract it from the measured position of the star you're watching.

OTOH, it was an additional source of error, until speed of light and the Earth's orbital parameters were very precisely determined. Perhaps it still contributes to the noise added to the data, to some extent - I'm not entirely sure here.


What would you call death by magnetar? The magnetic field of a magnetar is so strong that it would stretch people's atoms apart. This, would obviously, result in death. But how spaghettification refers to death by extreme tides, and for the most part, refers to death by black hole, what field-specific term would best refer to death by extreme magnetic fields, eg. from a magnetar?

Nobody has ever needed such a word, and it doesn't exist.
"Spaghetification" is, of course, a joke word, but each time you make a new joke based off an old one, the less funny it becomes.
If you just want a (poor) joke:  Teslascind

In order to capture @pela's ephemeral comment in a more durable post:

Magnetaaarrgh!


Which stars, identifiable with the naked eye, have exoplanets? If I look at the night sky, which easily identifiable stars have known exoplanets, and what are the names of those planets?

With the recent discovery of so many exoplanets, it is a nice idea to look at the stars with this new knowledge. They will no longer seem like a bunch of anonymous white dots.

Stars with an apparent magnitude of 7.0 or greater are typically not visible to the naked eye. So if you take a list of exoplanetary host stars, such as this, and you sort by apparent magnitude, in ascending order, then you will see a list of identifiable stars known to have exoplanets, until you reach an apparent magnitude of 7.0 on the list.

Finding the star can be trickier, but Wikipedia will usually show you what constellation you can find the star in. With a little work, and maybe the help of some other posts on this site, you could probably identify which stars are visible in your area and time.

Most exoplanets have not been given formal names themselves, but instead a lowercase letter is appended to the end of the name or designation of the star, for example "Beta Geminorum b (abbreviated β Geminorum b, β Gem b) and HD 62509 b" for Pollux b, the first planet discovered around Pollux. Incidentally in 2014, the International Astronomical Union did release a process for giving proper names to exoplanets, and Pollux b was given the name Thestias. If you look up other exoplanets on Wikipedia, you can find out whether they have been given proper names as well.


How do we know distant stars still exist The Pleiades are 450 light years away. Thus when we see the twinkle in the night sky the light that hits our retina emanates from the year 1569 or so. On that proviso; how can we be sure that the star is still alive given the fact that we are seeing light from hundreds of years ago.

We don't know from sensory evidence.  As far as we can tell, no information ever moves faster than the speed of light, so we are always at least 450 years out of date for anything happening in the Pleiades.

The only evidence we have is inferential: As far as we can tell (and we've looked real hard at this) the world works the same both here and at the Pleiades.  And as far as we can tell (and, once again, we've looked real hard at this) stars like those in the Pleiades don't just disappear or go out in only 450 years -- it takes millions of years.

So, based on what we observe of them as of 450 years ago, and based on what we know of how stars work, we conclude that they are pretty much the same "today" -- though, really, what we're concluding is that when we observe them 450 years from now, they'll still look pretty much like they do today.

This is all inferential, but I'd bet just about anything on it.  (If only I were here in 450 years to collect!)


Find the height of the shooting star from the Earth's ground One night, you and your friend were talking to each other. Your friend far about 20 km. from you. You told your friend that you see the shooting star pass through the sky at altitude of 75 degrees. Your friend also see this shooting star, but has difference apparent location by 6 degrees. Find the location of this shooting star, what is it hight from earth ground.

I think 6 degrees is the different of azimuth angle. But as I live in different lattitude, it may be effect the distance between friend and me. Or the answer is just 75degrees turn to the height from the earth ground. But how? Which and how does the correct way to solve this question?

This is basically a trigonometry question.  I wasn't sure from the question if the apparent location from your friend's position was six degrees higher or lower.  I chose higher for the example, but you could change it.

You have two triangles with the same altitude, but different bases and angles.

$$\text{altitude} = x \tan(81^{\circ}) = (x + 20km) \tan (75^{\circ})$$  Do you see how to calculate it from there?


How are cardinal points on horizon defined? How are the cardinal points (North-East-South-West) defined? On equatorial regions it is somewhat obvious. But what about the higher latitudes? If the pole star is up above the sky instead of horizon, how do we define North? What is the case on the geographic poles?

Geographic Meridian
Every point on Earth is on some meridian:

A (geographical) meridian (or line of longitude) is the half of an imaginary great circle on the Earth's surface, terminated by the North Pole and the South Pole, connecting points of equal longitude, as measured in angular degrees east or west of the Prime Meridian.

From where you are standing, North is the direction on the horizon along this meridian or "line" of longitude that extends to the Earth's geographic North pole, and South is the direction on the horizon 180 degrees opposite, towards the South pole.
East and west are the perpendicular directions, which are also found extending lines of latitude from where you are standing towards the horizon.
Locally you can think of the region under your feet as flat, and so North, South, East and West all form right angles even far from the equator.
Close to the poles, none of that changes mathematically, and exactly on the poles those terms suddenly no longer have any meaning.
Celestial Meridian
meridian:

In astronomy, the meridian is the great circle passing through the celestial poles, as well as the zenith and nadir of an observer's location.
Consequently, it contains also the north and south points on the horizon, and it is perpendicular to the celestial equator and horizon. A celestial meridian is coplanar with the analogous terrestrial meridian projected onto the celestial sphere. Hence, the number of celestial meridians is also infinite.

Source


The CNO cycle in first generation of stars I'm an undergrad interested in astophysics and this question struck my mind:

As the CNO cycle requires carbon to start, and carbon is produced by thermonuclear fusion in stars, is it fair to think that the hotter stars in the first generation didn't have a CNO cycle?
Or should I instead think that the first generation of stars had an initial metalicity and therefore the CNO cycle was possible?

First generation stars didn't do the CNO cycle initally.

It is estimated, that after the Big Bang three quarters of the matter is hydogren, one quarter helium, and trace amounts of heavier elements.  This is then also the composition of (as of now unobserved) first generation stars, having zero metallicity.  Even though those stars probably were very massive, they could not run the CNO cycle initially.

Luckily, the CNO cycle is not the only nuclear fusion process in stars.  In the proton-proton chain reaction, hydrogen is fused into helium.  The star is in equilibrium until the hydrogen in the core is "used up".  Then it may collapse and heat until the triple-alpha process begins to produce carbon.  From there on there are more and more processes producing many different elements.  Due to convection and shell-burning, the CNO cycle might have been running in those first generation stars at later stages.

Interestingly, the most massive of the first generation (population III) stars may have ended up as black holes, gobbling up all of the produced metals (elements heavier than helium).  Many others may have exploded in pair-instability supernovae ejecting all material into interstellar space, leaving no compact remnant behind.

A Heger and S. E. Woosley: The Nucleosynthesis Signature of Population III


Why do some solar eclipses' umbra cross arctic/antarctic regions? The ecliptic (Earth's orbital plane) is inclined 23.4 degrees. This is the same as Earth's axial tilt. The Moon's orbit is inclined 5.145 degrees to the ecliptic.

Therefore, the maximum latitude where a solar eclipse produces an umbra should be 28.545 N and 28.545 S.

But this is clearly not the case. Wikipedia has a huge list of solar eclipses with charts. Here's one that goes very near the south pole. How is this possible?

Edit: Actually I missed something. Since the Moon and Sun must be in the same place, and the Sun can never exceed 23.4 degrees declination, solar eclipses must be bound by 23.4 N/S. I should not have added the two angles. Thanks to user:berrycenter for pointing it out.

I wanted to ask this question even though I realized the correct geometry in the middle of typing it. So I will answer this myself.

First of all, the sub-lunar point never exceeds 28.545 N/S. If you're standing at the sub-lunar point, the Moon will be directly overhead. The Moon never wanders outside this latitude bracket.

But that does not mean the Moon's shadow can't wander outside it. Here's one way it could happen.

Edit: this is my attempt to draw something approximately to scale. I think I got the Earth/Moon sizes about right, but the Moon-Earth distance should around 4x greater. And the solar rays are of course not parallel but I don't think it's possible to draw or perceive tiny angles for this scale.

The green line is the ecliptic. The Sun and Moon are supposed to be aligned along a horizontal line, but for some reason I drew the solar rays half and half beside the green line which is deceptive. I'll try to edit the pic when I can.

In technical parlance, the Moon could be slightly above or slightly below its Node. If the Moon was exactly at its node, then the umbra would absolutely be centered somewhere between 28.5 N/S (actually 23.4 N/S because the Sun never goes beyond that). But as you can see, the farther it gets from its node, the steeper the shadow impacts Earth and the farther towards a pole it gets. Too far from the node, and the shadow will not intercept Earth at all and we won't have an eclipse.

If you draw a line from the center of the Moon to the center of the Earth, you can see that the sub-lunar point is definitely within 28.5 degrees N/S. But the shadow does not necessarily follow that imaginary line. The shadow always follows the line from the Sun to the Moon.

You could even imagine a very extreme case where the shadow falls on "the other side" of the pole. I think that may be the case here since that eclipse looks really short.


How do we measure the brightness of the stars? How do scientists measure the brightness of so distant stars?

The term brightness, or apparent brightness, is used to describe how bright a star appears to us from Earth. The term luminosity is used to describe how bright the star is physically, also called intrinsic brightness.

As light spreads from a star to the viewer, it diverges inversely proportional to the square of its distance (d) from the viewer.  $\begin{equation}B \propto \frac{1}{d^2}
\end{equation}$.

The relationship between brightness (B) and luminosity (L) is described by $\begin{equation}B = \frac{L}{4\pi d^2}
\end{equation}$ (1). So we have three variables, brightness, luminosity and distance.

So out of those three, you'd need two to acquire the third. The distance is generally the hardest to acquire. The gold standard in measuring the distance to a star is stellar parallax. If you can't do parallax, then distance-independent properties of stars sometimes serve as clues to distance.

Measuring brightness is generally the easy one. CCD's are often used to measure apparent brightnesses. Brightness can be expressed in stellar magnitudes or absolute fluxes.

So to sum up, we calculate or measure distance, we measure apparent brightness and finally calculate the luminosity value from (1).


Law of Conservation of Information? I read about Hawking and Leonard Susskind and their debates about whether information is destroyed or not if it falls into a black hole.  It was resolved in favor of information always being conserved, preserved, whatever (i.e. not destroyed).

I see how this could be consistent with determinism.  However, how is this in keeping with quantum physics, when there is so much randomness.  It seems that at any moment, all matter, energy, even every atom in our bodies, is indeterminate and "scrambled" somewhat.  Even if you could pinpoint or describe every atom and its position perfectly in your body at one moment, by the next moment there is uncertainty.  So how is this compatible with conservation of information?

The real reason of conservation of information in quantum mechanics is that the von Neumann entropy $S(\rho) = -Tr(\rho \log_2 \rho)$ of an isolated quantum system with density operator $\rho$ does not change in time. This happens because $S(\rho)$ only depends on the spectrum of $\rho$, and the unitarity of time evolution preserves the spectrum.

If you start out with a perfectly certain state it will evolve unitarily, and remain perfectly certain - but realistically, there are going to be subspaces you did not measure and their uncertainty are going to "leak out" and get mixed with the state. This is basically the no hiding theorem which states that when your system interacts with the outside world and decoheres the information about the original state ends up on the outside and cannot be recovered by operating on just the internal state (but it is indeed conserved for the whole system).

Note that none of this depends on uncertainty per se nor the issue of what measurement does, but rather that the deterministic effects of the Schrödinger equation/unitary evolution makes things dependent on degrees of freedom you have little control over.


How can clouds form in Jupiter's atmosphere of Hydrogen and Helium? Here is a graphic of cloud layers of Jupiter (source: Wikipedia):

There are three distinct cloud layers of ammonia, ammonium hydrosulfide, and water. The temperature and pressure conditions seem to be surprisingly earth-like; temperatures between 200 and 300 K, pressures about 1 to 10 atm, gravity around 1.3g.
Clouds (of water) form on earth because they solar energy causes them to evaporate from a solid surface, rise a few km, then condense to form water droplets (or solid crystal snow). But Jupiter does not have a solid surface, or nearly as much solar energy as the Earth.
All three of the cloud forming compounds should be liquid at the conditions of their cloud layer. Given the density of those liquids (between 0.7 and 1.2 g/cm$^3$) and the density of the bulk of the hydrogen and helium atmosphere, how do the clouds not fall as precipitation into Jupiter's interior and never re-emerge?

First, it's a great question.  Mostly the answer is straight forward, so I can answer it, but it's still a great question.

and I'll add a similar, but slightly more detailed picture to the one you posted.

Source

You're right that there is a clear difference between Earth's surface where liquid water can exist, evaporate, make clouds, rain and repeat.   Earth's water cycle in theory, could go on indefinitely so long as Earth's atmosphere and the solar input are maintained (and lost hydrogen is replaced), but it's a circular system that only needs solar input.

Jupiter is different because over time, the heavier gases in Jupiter probably will sink deeper towards the center and Jupiter's cloud forming gases should decrease given enough time.   Some of Jupiter's "rain" probably does fall too deep in its swirling mix of gases, and leaves Jupiter's cloud cycle permanently, similar to water seeping underground and leaving Earth's water-cloud cycle.    So, in 100 billion or a trillion years or so, Jupiter could lose its clouds and cloud forming gases in its upper atmosphere for the reasons you suspect.

The reason this hasn't happened yet is simply mixing.  While density of gas tends towards layers of increasing density, the internal heat inside Jupiter also wants to even out, so there's enormous convection going on pretty much all the way through the planet.   This keeps some heavier gases in Jupiter's upper atmosphere.   Jupiter's much too turbulent to have just hydrogen and helium in its upper atmosphere.

So, once we begin with the observation that Jupiter's upper atmosphere is (about) 90% hydrogen, 9% helium, 1% other gases and the mixing maintains the 1% of other, after that it's just cloud physics.

Clouds look like puffy collections of water vapor (tiny droplets of ice or water, as water vapor is actually transparent).  They look like objects with shapes, but that's not entirely accurate.   If you are up close to a cloud (flying in a plane for example), the clear edges disappear.  A cloud isn't an object so much, it's a visible phase change.

Atmosphere on Earth is about 78% nitrogen, 21% oxygen, 0.9% argon and (usually not listed because it's so variable), about 0.4% water vapor on average, as high as 1% with high temperature and high humidity and close to 0% in cold temperatures or dry-deserts.   When you take warm surface air that's 0.6-0.8% water vapor, and that air rises (as hot air does), it's the phase change that creates the clouds.   The cloud forms in the hot rising air as it cools.  There's some electrostatic attraction, but mostly it's just a block of similar air undergoing cooling and the cloud looks like it has solid edges, but it doesn't.

The same exact thing happens on Jupiter, different gases phase change at different temperatures/pressures, but the process is the same.  And, just as on Earth, once the droplets or "icelets" form, they are denser and they begin to fall, but falling droplets are very small so they fall very slowly and for the most part, they are falling through rising atmosphere.  Also, as they are a phase change, new cloud is being formed and old cloud is being disbursed or returned to gas all the time, kind of like sea ice.   The clouds have the appearance of semi-permanence, but clouds are dynamic.

If my explanation doesn't work for you, here's an explanation on clouds and how they really aren't bound together even though they look that way.

But that's the gist of it, mixing keeps the upper atmosphere of Jupiter from being pure hydrogen and helium (or pure hydrogen), and after that, the cloud formation is pretty much the same as on Earth, just without a surface.   Some of the heavier gases probably do get lost from the cycle, but the loss is slow enough that Jupiter still has some heavy clouds forming gases in its upper atmosphere and will for probably billions of years to come.

The greater density variation between H/He and other gases likely plays a role in how the clouds behave, as the density variation is greater, but the wind speeds are also higher on Jupiter.   All that's really needed is mixing.  After that, with gases that can become liquid or ice under temperature/pressure variations, the phase changes create the clouds.

It's also possible that Jupiter's cloud forming gases, from time to time, are replenished by asteroid and comet impacts.   Shoemaker-Levy 9 was about 5 km in diameter and a fair percentage of that was probably ammonia and water ice.   That's a lot of cloud forming gas added to Jupiter's upper atmosphere.   Jupiter's faint ring system, which might have been much larger millions of years ago, but since rained onto Jupiter, and eruptions from Io might also play roles in keeping Jupiter's upper atmosphere rich enough in cloud making elements like water and ammonia.


Is it plausible that Mars could have been one of Jupiter's natural satellites at one time? Would more questions regarding the planet be answered if this was the case?

I'm going to say no, for a few reasons, but if anyone wants to give a more detailed answer, feel free.

1) Planets and moons that form inside the frost-line have much less water and other ices than those that form outside.   Mars likely once had water and oceans but if it had formed outside the frost-line it should have a lot more water and a lower density.   The frost line moves away from the sun as the sun grows more luminous and planets move, so there is a degree of imprecision with this argument, but I believe this on it's own is a pretty good reason why it's not likely.

2) A moon that escapes Jupiter's orbit should enter a near Jupiter orbit initially, and at some point later, it could receive a gravity assist and be propelled either in towards the sun or away, but it's unlikely that Mars could have obtained it's relatively circular orbit had it originated as one of Jupiter's moons, unless Jupiter was much closer to the sun when Mars escaped.

3) it's difficult to see how the asteroid belt could have stayed around if Mars originated outside the asteroid belt and moved inside it.   Mars passing through the asteroid belt would have caused significant disruptions to it.


What determines which frequency range a radio telescope has? My simple understanding is that a parabolic dish focuses radio waves to a receiver. It is obvious that any parabola focuses visible light and heat, so I assume that it focuses all radio wavelengths too. So evidently the design of the receiver determines which wavelengths it is sensitive to.

How is such a receiver designed in radio telescopes?
What properties of the receiver determine the wavelengths it is sensitive to?
Is the sensitivity range electronically adjustable?
Why aren't different receivers used so that the same parabola, for example the huge Arecibo dish, could be used for all wavelengths? (Or is it?)
What importance does the size and material of the parabola have for the sensitivity to different radio wavelengths?

A parabola does indeed focus over a broad frequency range. The lower limit is determined by the dish diameter, the upper by the construction (mesh size, parabolic accuracy etc.).

The collector placed at the focal point may be a simple dipole or other fixed frequency antenna, or more commonly, a waveguide that leads the collected signals to a low noise amplifier and receiver. The physical dimensions of the antenna or waveguide set the operating frequency range of the antenna system. More than one range may be tuned by means of multiple switched collectors.

The receiver is fundamentally no different to any other receiver; a tuning element selects the signals of interest which are then detected and amplified.


I want a galaxy count database, also how do I use SDSS skyserver to fetch galaxies? I'm new to all this and trying to implement galaxy counting. For the initial stage I need raw image data, that is images with galaxies and stars along with a count of how many galaxies there are in the image.

I came across this which seems to have a way to label galaxies

http://skyserver.sdss.org/dr13/en/tools/chart/navi.aspx?ra=179.68929&dec=-0.45438&opt=

but I cannot figure it out at all, I used this page

http://skyserver.sdss.org/dr13/en/tools/chart/chartinfo.aspx

to try to get it to label only galaxies are seen but I got nothing (put G in opt section, did not make a difference).

Does anyone know how to operate the above or where I could get images with lots of galaxies (any number would do) in them along with a count of how many galaxies there are in the image?

I found https://data.galaxyzoo.org/ but traversing it seems impossible

You might find the NED database and Astrometry.net useful. NED is NASA's extragalactic database. You can search by a position in the sky with a radius and it will give you a list of known objects in that field.

Astrometry.net lets you upload images and will resolve what you're looking at, also giving you a list of known objects in your image.

I like to use CDS for a quick look up on objects. This will let you browse images from different surveys easily and links you to other sources on that object as well as publications about your object.


What would Earth be like without the Moon? If somehow our moon was destroyed (for example by an asteroid) what would happen to Earth?

Would the Earth be affected by the absence of the moon? Would humans be able to live normal lives without the moon?

Destroying the moon is not easy. It's pretty big. Has quite a lot of gravity and is sitting in the Earth's gravity well. It gets hit by a lot of asteroids and doesn't get destroyed.

The moon has a binding energy of about $10^{29}$ joules. That is a lot of energy. Even if you got the largest asteroid (ceres) to collide with the moon you would not get close to destroying it.

You would need something like the full power of the sun for about 15 minutes.

If something massive enough to disrupt the moon were to hit it, the energy of that explosion, and the massive amount of moon rocks that would subsequently hit the Earth would probably be enough to melt the surface of the Earth. It would be the end of life on Earth.

There is another way to get rid of the moon. Get a wizard to magic it away. The effect of not having a moon would then be pretty minor in the short term (no tides) and fairly significant in the long term. Human life could continue as normal for a while.

The moon stabilizes the rotational axis of the Earth, without the moon the angle would vary more (over thousands of years) this might mean more extreme ice ages and perhaps ultimately a snowball Earth scenario.


Why do Black Holes in the middle of galaxies not suck up the whole galaxy? As stated in several sources, it's supposed that in every galaxy there is a black hole in the middle.

My question is, why do these black holes in the middle of galaxies not suck up all the surrounding matter in the galaxy?

You shouldn't think of black holes as "sucking things in". Black holes interact with matter through gravity, just the same as any other object. Think of our Solar System. All the planets orbit around the sun because it has a lot of mass. Since the planets have some lateral motion (they're not moving directly towards or away from the sun), they circle around it. This is known as conservation of angular momentum.

When talking about gravity, all that matters is the mass of the objects involved. It doesn't really matter what kind of object it is*. If you were to replace the sun with a black hole that had the same mass as our sun, the planets would continue on the exact same orbits as before.

Now, the black holes at the centers of most spiral galaxies do accumulate mass. Some of these black holes have accretion disks around them. These are swirling disks of gas and dust that is slowly falling into the black hole. These gas and dust particles lose their angular momentum through interactions with gas and dust nearby and by radiating energy as heat. Some of these black holes have very large accretion disks, and can generate huge amounts of electromagnetic radiation. These are known as active galactic nuclei.

So, long story short, black holes don't "suck". They just interact with things gravitationally. Stars, gas, and other matter in the galaxy has angular momentum, so it stays in orbit around the center of the galaxy. It doesn't just fall straight in. This is the same reason the Earth orbits around the Sun.

*Disclaimer: When you talk about things like tidal forces, you do need to take into account the size of the objects. But for orbital mechanics, we don't need to worry about it because the distances between the objects are generally much larger than the objects themselves.


What is left of a universe where no matter and energy exist and is there a term for this? In the final energy state of the universe, there exists a hypothesis that all matter will have fallen into black holes and neutron stars. If, through Hawking radiation, after all matter is consumed and black holes eventually do dissipate, what is left of our Universe?

I may be describing Heat Death, but I'm not 100% sure. However, due to the Law of Conservation of Energy, I suspect that energy will be all that is left, only in equilibrium meaning no ability for energy to form matter.

In the event I'm wrong and there exists no matter and no energy, would the term Universe still be applicable here?

However, due to the Law of Conservation of Energy, I suspect that energy will be all that is left, only in equilibrium meaning no ability for energy to form matter.

The law of conservation of energy states that energy is conserved in a closed system. This means the Universe, which is likely an open system, does not necessarily have to obey that conservation law. Not to mention, scientific laws are not absolute, despite their names; they are entirely empirical, and only apply as far as we've observed. As Edwin Hubble put it, "Beyond the limit of the observations, the form of [a] function, or relation, is speculative. For this reason, [a] law is empirical and it must remain empirical until it is explained by an accepted theory."

Anyway, let's get on with the answer. You've just described the Black Hole Era in the heat death scenario. If Hawking radiation occurs, then the black hole will continue to emit particles. During the last stages of its evaporation, a black hole will emit not only massless particles, but also heavier particles, such as electrons, positrons, protons, and antiprotons.

Once all black holes have evaporated, the Universe will be nearly empty, sans some particles that will almost never interract. Eventually, even the protons and antiprotons emitted by the black holes will decay into photons and leptons. Thus begins the Dark Era, when the Universe will be dominated by dark matter, electrons and positrons.

Electrons and positrons will rarely come together and form positronium atoms (in which the electron and positron are electromagnetically bound). Positronium decays very within nanoseconds, since the two particles annihilate. Thus, over time, the energy levels in the Universe will become lower and lower.

What happens next is a matter of speculation. According to Carroll & Chen (2004), quantum tunnelling and quantum fluctuations could hypothetically create another Big Bang, approximately $10^{10^{10^{56}}}$ years from now.

Adams & Laughlin (1997) speculates that quantum effects will dominate once all macroscopic matter is gone. Essentially, the laws of "macro-physics" will break down, while quantum mechanical phenomena prevail. As a result, the smallest perturbations would make the biggest difference in this era, so there is no telling what may happen to space or time.

Furthermore, they speculate that if the Universe is a false vacuum, the vacuum may decay into a lower-energy state, creating a new "child universe":

  It is also possible for the universe to spontaneously create “child
  universes” through a quantum tunneling process roughly analogous to
  that considered above (e.g., Sato et al., 1982; Hawking, 1987; Blau,
  Guendelman, & Guth, 1987). In this situation, a bubble of false vacuum
  energy nucleates in an otherwise empty space-time. If this bubble is
  sufficiently large, it will grow exponentially and will eventually
  become causally disconnected from the original space-time. In this
  sense, the newly created bubble becomes a separate “child universe”.

So yes, something as small as quantum tunnelling will potentially have an enormous effect on the future of the Universe.

In the end, we really have no idea what will happen after enough time, so there's no definitive answer to your question in the long run.


Tilting cycle vs orbital cycle of earth Common sense tells us a year is conveniently divided exactly into four seasons. I'm wondering whether that means the earth's tilting cycle is equal to its orbital cycle. If so, there must be a reason. Do we have a proof?

There is no tilting cycle.

The Earth continues to tilt the same way all year round.  The axis points towards the pole star. But as the Earth goes round the sun, in December the northern hemisphere points away from the sun. In June the Northern hemisphere points towards the sun, as the Ascii art below shows (not to scale)

N                    N
 \                    \
  O         *          O
   \                    \
    S                    S
Earth                 Earth
in Dec     Sun       in Jun

So no special explanation is required. The Earth points the same way all year, it just moves relative to the sun.

The tilt of the Earth relative to the stars doesn't change. So the tilt of the Earth relative to the sun is entirely due to the Earth's orbit.  This proves that the cycle of tilting relative to the sun will take exactly one orbit to complete. It is no coincidence; the cycle of seasons will take exactly one year.

(There is variation in the direction of tilt, called precession, but this is a much slower cycle, lasting 25000 years)


Putting mass-luminosity relation and Hertzsprung-Russel diagrams together leads us to a mass-age relation; so how do stars lose their mass over time? I think that the title is completely clear, but here's an expansion:

I was just reading about Mass-luminosity relation that says massive stars are more luminous than tiny ones. Well, let's talk about main-sequence stars for now. This relation becomes interesting when it's mixed with the Hertzsprung-Russel diagram that says young stars are more luminous than older ones (Originally says that hotter stars are more luminous and we know that hotter stars are younger).

So mixing these two relations produces some kind of Mass-age relation which says young stars are more massive, that is a star loses its mass over time, right? (Please tell me if I'm going wrong)

If this Mass-age relation is right, how do stars lose this extra mass during their evolution? And where does this stray matter go?

There is an alternative interpretation: Massive stars burn their fuel much faster than tiny ones. Hence massive stars are short-lived in comparison to tiny stars.

The lifetime of a star is proportional to about the inverse of the cube of its mass.
The Sun's lifetime is about $10^{10}$ years, hence for stars with 10 solar masses the lifetime is about $10^7$ years, and for stars with 0.1 solar masses it's about $10^{13}$ years.

But stars also loose mass, e.g. by stellar wind and coronal mass ejections. Heavy stars may loose mass (parts of their envelope) e.g. by pulsation or radiation pressure during their red giant or their Wolf-Rayet phases, before they explode as a supernova.


Why are three year old eclipse glasses not recommended? NASA's eclipse safety page says that eclipse glasses should:

  Not be used if they are older than three years, or have scratched or wrinkled lenses

Why shouldn't three year old glasses be used? Is it because the coating somehow degrades with time? Were standards changed three years ago such that the older glasses can no longer be recommended? Is there just a stronger concern for scratches or damage over the years? Will eclipse glasses (or other solar filters) purchased today be unsafe three years from now?

According to the American Astronomical Society (AAS), the warning that eclipse glasses should not be used if they are more than 3 years old is outdated (provided that they are compliant with the ISO 12312-2 standard).

A page on their web site says the following:

  Note: If your eclipse glasses or viewers are compliant with the ISO 12312-2 safety standard, you may look at the uneclipsed or partially eclipsed Sun through them for as long as you wish. Furthermore, if the filters aren't scratched, punctured, or torn, you may reuse them indefinitely. Some glasses/viewers are printed with warnings stating that you shouldn't look through them for more than 3 minutes at a time and that you should discard them if they are more than 3 years old. Such warnings are outdated and do not apply to eclipse viewers compliant with the ISO 12312-2 standard adopted in 2015.


Which things can LIGO see that LISA can't, and vice-versa? CNET's Astronomers discover two ferociously fast stars locked in a death spiral quotes Kevin Burdge, lead author on the new paper in Nature  General relativistic orbital decay in a seven-minute-orbital-period eclipsing binary system (no ArXiv?) and a Ph.D. candidate in physics at Caltech as saying:

  LIGO can see things that LISA can't, and LISA can see some things that LIGO can't, and there are a handful of things that might switch from being visible in LISA to LIGO over time"

Is there a simple way to understand which "things" fall into which categories?

LIGO is a ground-based gravitational wave observatory and is used together with Virgo, whereas LISA stands for Laser Interferometer Space Antenna and will be a GW sensing inteferometer based on a trio of satellites in an Earth-trailing heliocentric orbit.

GIF Source

  LISA spacecrafts orbitography and interferometer -yearly-periodic revolution in heliocentric orbit.

Gravitational wave detectors have a frequency range that they are sensitive to.

In the case of LIGO it is about 10Hz to 1kHz. The lower limit is imposed by seismic noise, the upper limit by "shot noise" (basically not having enough photons to sample the interferometer path difference at high frequencies).

LISA is in space and doesn't have the problem with seismic noise. However, there is still an upper limit determined by shot noise. As I understand it, LISA will not/can not use the same sort of resonant cavity arrangement that LIGO uses to boost the effective laser power and hence number of photons in the apparatus. Thus the upper frequency limit for LISA is more like 1Hz.

So now to your question. What LIGO can see that LISA can not, are gravitational waves with frequencies of 10-1000 Hz. The astronomical phenomena that lead to such waves are the merger of compact binary systems with stellar masses, rapidly rotating, asymmetric pulsars and perhaps supernova explosions.

Lower frequency GWs can only be seen with LISA. This would include stellar binary systems with orbital periods longer than about 10 seconds, merging supermassive black holes and maybe GWs from the big bang.

What could move from being observable in LISA to being observable in LIGO? In principle, the merger of any binary system results in a gradually increasing frequency and amplitude. Stellar mass, neutron star or black hole binaries can get to frequencies around 1kHz before they merge, but sweep through lower frequencies before that, but with much lower amplitudes. Perhaps the best bet would be the merger of intermediate mass black holes with masses of $10^3$ to $10^4$ solar masses, which would have merger frequencies of tens of Hz, but a significant amplitude at much lower frequencies prior to merger.

I suppose the implication of the quote you gave is that this newly discovered inspiralling white dwarf binary might be another possibility. I don't think that is correct. It has a 7 minute orbital period, which puts it in LISAs frequency domain (the GWs emitted have twice the orbital frequency), but because white dwarfs are physically larger than neutron stars, the peak frequency at merger (some time in the future) will only be a few Hz and not visible to LIGO.


Are there currently observed systems of close (narrow) binary brown dwarfs? All over the Universe, there are (gravitationally bound) binary star systems. Considering the huge number of brown dwarfs, are there currently known double systems of gravitationally bound binary brown dwarfs? Given their comparatively low gravitation, how long would they need to melt together to a larger object?
What happens if binary brown dwarfs collide? Is this anyway a matter of any interest?

You've got a lot of questions here, so I'll try to answer them one by one.

  Are there currently known double systems of gravitationally bound binary brown dwarfs?

Certainly there are. I don't study low-mass stars so I'm not up on the literature enough to know any good examples off the top of my head. Kozaky suggested the Luhman 16 system in his comment and a quick google search also brought up the triple star system Epsilon Indi with a central K type star being orbited (at ~1460 AU) by a binary brown dwarf system. Perhaps some other more knowledgeable person can suggest more systems of interest.

  Given their comparatively low gravitation, how long would they need to melt together to a larger object?

That depends on a variety of factors, including their respective masses and initial distance apart. As two objects orbit one another, they continually release gravitational waves due to their interacting gravitational fields. These waves carry away energy and to conserve energy, the stars must lose angular momentum, thus getting closer. The time necessary for two stars of mass $m_1$ and $m_2$, with an initial radius $r$ to fall into each other is given by

$$t = \frac{5}{256} \frac{c^5}{G^3} \frac{r^4}{(m_1m_2)(m_1+m_2)}$$

where $c$ is the speed of light and $G$ is the gravitational constant. We can apply some basic numbers and calculate an order of magnitude time. If I assume $r = 1\:AU=1.5\times10^{11}\:m$ and $m_1 = m_2 = 2\times10^{28}\:kg$ (10 times the mass of Jupiter) then I find the timescale for collision into one object is

$$t = 5\times10^{30}\:s \approx 100\:billion\:trillion\:years$$

This is obviously way longer than the age of the Universe and effectively means that two Brown Dwarfs will never collapse. This fact could have been seen from the equation. Notice that the mass of the objects is in the denominator. That means that with lower mass means longer timescales and the fact that the distance is to the fourth power means the distance is a much stronger controlling factor here. You need huge masses to offset that $r^4$ for you to get a collision on reasonable timescales. Even doing a best case scenario of $r = 0.1\:AU$ and $m_1=m_2=10^9\:kg$ still gives me $\sim100\:thousand\:trillion\:years$.

  What happens if binary brown dwarfs collide?

Not much? I don't know if anyone has ever really tried to work out the theoretical work for such a problem. Given the answer above, the only chance two brown dwarfs are going to collide is by a freak chance that they're on a collision course in space and not in a slowly degrading orbit, but the chances of that are pretty minuscule as well.

One situation that may be of related interest is to consider regions of star formation. While it is somewhat contested, some people believe stars can form "bottom up". That means smaller objects form first, then those coalesce into larger objects and on upwards until you end up with big, bright stars. In this scheme, you would first form many many brown dwarfs in a tight area and these would dynamically interact and collide with each other to form larger stars. This isn't the same scenario of two lone binary brown dwarfs colliding, but its the best you're going to get if you want to see colliding brown dwarfs.

In this scenario, I think that since brown dwarfs aren't fusing and their main energy output comes from core collapse (by some Kelvin-Helmhotz mechanism), it isn't going to be a hugely explosive result (comparatively) to smash two together. They will of course burst out with energy in the process, but it won't be as dramatic as two fully-fledged stars merging. And that's probably a good thing. If you had a star forming region with many brown dwarfs that were slowly coalescing into larger stars, you wouldn't want each merger to blow out huge energies as it would prevent further stars from coalescing.

  Is this anyway a matter of any interest?

From an academic standpoint, I'm sure it is. As I said, I don't study this topic, but understanding the affects of brown dwarf collisions would be very useful for fleshing out star forming theories. I'm sure one could think of numerous other academic benefits for studying such a topic.

From a practical standpoint, I can't say understanding brown dwarf collisions immediately benefits humanity. But as with any topic in science, you never know how knowledge gained today may help in the future.


How frequently do Jupiter & Regulus have triple conjunctions? I am wanting to know if the triple conjunction of Jupiter with Regulus in:
Sep 10, 3 BC
Feb 7, 3 BC &
May 8, 2 BC
with Jupiter later having a conjunction with Venus in Jun 17, 2 BC has happened since, and if so, how often does this triple conjunction happens?

With any given star, Jupiter has 3 or 4 triple conjunctions per century, with either 12 or 71 years in between.
These occur with Regulus in 1873, 1885, 1956, 1968, 2039, 2051, 2063, 2134, and 2146.

Jupiter takes 11.9 years to orbit the Sun, advancing at a fairly steady ~30° per year.
From Earth, Jupiter appears to move ~40° forward and ~10° backward in a 13-month cycle.
The retrograde loop happens when Earth and Jupiter are on the same side of the Sun.

We see a triple conjunction with a star when the retrograde loop encompasses it.
Around the February 2051 opposition, Jupiter has conjunctions with Regulus in October, February, and June:

If Jupiter passes a star while Earth is on the other side of the Sun, we see only a single conjunction, as in August 2003:

Jupiter-Venus conjunctions occur at intervals of 10 or 14.5 months.
The Jupiter-Venus conjunctions of October 1884 and July 2146 occur reasonably close to a Jupiter-Regulus conjunction in a triplet listed above and far enough from the Sun to observe easily.


Would we detect a spacecraft before it landed on Earth? We have satellites and telescopes looking outward and inward to the earth. Would any of these detect an extraterrestrial spacecraft before it landed, or in other words, would we know a spacecraft was approaching before it did (assuming its going at some reasonably detectable speed in our local area)?
Let's assume it is Saturn V like wrt detectability.

Some thoughts about Earth-based detection systems:

We are looking at asteroids using radar: For instance the former Arecibo telescope could actively send out radio waves and calculate object position and size. However, we might be able to detect an incoming spaceship using this technology, assuming that the object approaches Earth within the radio beam, if we we are currently studying another object. In other words: Detecting an incoming object would be purely chance.

An all sky camera network for meteorite detection is another outward-looking system worth mentioning, but that would be very very briefly before landing of a hypothetical extraterrestrial space ship. I am sure you are not after this option since we would be seeing the object already with naked eyes if it would not be incoming at a very remote location. However, this option is to my knowledge working 365/24/7 and covers large portions of the sky, I believe.

Edit suggested by uhoh: There are also initiatives like NASA's Space Fence or ESA's Space Surveillance and Tracking (SST) segment monitoring space debris within the geospace which could possibly detect other objects as well, see also https://space.stackexchange.com/q/45144/12102

Concerning satellite-based observations:
If the objects emits significant radiation e.g. in parts of the spectrum of communication satellites, we might detect anomalities or malfunctions in one or more satellire and by chance figure out the extraterrestrial source of the distortion. This would of course only hold for slow approaching objects.


Sun's position on summer solstice Let's say I am in Brasilia (latitude = 15.78°S). This is obviously south of the equator but north of the tropic of capricorn. So on the summer solstice I know that the Sun's declination is +23.26°. Does this mean that the sun will appear directly north at solar noon? Thanks.

The Sun at solar noon (meaning that it is on the meridian) will be in one of these directions:

Due south if the declination of the Sun is less that your latitude. (For Brasilia, that occurs from Nov 6 to Feb 5, approximately)
Directly overhead (at the zenith) if the declination of the Sun is the same as your latitude. (For Brasilia, that occurs Nov 5 and Feb 6, approximately)
Due north if the declination of the Sun is more than your latitude. (For Brasilia, that occurs from Feb 7 to Nov 4, approximately)


How long does it take for an asteroid to get to Earth? How long does it take an asteroid to get from the asteroid belt, to Earth? I know the asteroids don't go directly from the asteroid belt to Earth, but I know it's not quick, either. The time probably varies, but what's the average time?

The full orbital period of an object orbiting the sun is $$T=2\pi \sqrt{\frac{a^3}{GM_\odot}},$$ where $a$ is the semi-major axis of the orbit and $M_\odot$ the mass of the sun.

If the asteroid has apoapsis in the asteroid belt ($r_{max}= 3$ AU) and periapsis near Earth ($r_{min}=1$ AU) then the semimajor axis will be $a=(r_{max}+r_{min})/2=2$ AU and going in from the belt takes half a period, so the answer is $\pi\sqrt{(2 AU)^3/GM_\odot}=1.4140$ years. An asteroid coming from further out, say 4 AU, would take 1.9762 years. From 2 AU, 0.9184 years.

Of course, were one to just give an asteroid a big push it could move nearly arbitrarily fast (up to the lightspeed limit if we want to be absurd) along an Earth-intersecting orbit, but I suspect the question was about normal asteroids.

However, it is worth noting that long-periodic comets can move fast (tens of km/s). Such a comet (or an interstellar asteroid) would have a velocity on the order of $\sqrt{2GM_\odot/(1AU)}\approx 42.1285$ km/s if it came in from the outer edge of the solar system. That would cross the 2 AU from the asteroid belt to Earth in about 82 days.


How long between the moment a meteor is first visible and the moment a meteorite hits the ground? A meteor is usually only visible for a couple seconds at most, but I would guess that it usually stops being visible high above the ground.
Meteors often don't make it to the ground as they completely burn up before. But among those that do reach the ground, how long does it take for a meteoroid to go through the Earth's atmosphere and hit the ground? How many seconds between the moment it is first visible and the moment it hits the ground?

I'm a researcher at Curtin University working on the Desert Fireball Network. The DFN is the largest fireball observation network in the world, and our primary goal is to recover meteorites with orbital information attached.
The traditional rule of thumb for meteorite-dropping events is a final luminous height below 35 km and a final luminous velocity below 10 km s-1
So when I refer to the luminous part of the trajectory, I am referring to the portion in which ablation is occurring and optical light is emitted. As you said, this portion typically lasts a few seconds depending on the initial size, strength, speed, and slope of the meteoroid. After this, if the object still has mass left, the speed continuously decreases. We call this portion of the trajectory the "dark-flight" (as opposed to the "bright-flight" which is the luminous part) because we can longer observe the meteoroid.
During the dark flight, the rock goes from <10 km s-1 to tens to a few hundred m s-1 when it finally impacts the ground as a meteorite. This part of the trajectory can be greatly affected by wind as seen below from the dark flight modeling done for the recovered Dingle Dell meteorite:

So, in total, from first becoming a meteor to impacting the surface, you should expect about tens of seconds to minutes to pass. Of course, this is dependent on things I have mentioned like mass, speed, and slope notably.


Abbreviations for days of the week Is there a set of standard abbreviations for the days of the week for use in astronomy? I couldn't find any reference to it in the The IAU Style Manual (1989).

Wikipedia uses: mon, tue, wed, thu, sat, sun.

ISO 8601 defines an exchange standard for dates based on weeks and days, which looks like this: 2017-W01-3 or this 2017W011 (Year, ISO week number, number of day of week starting with Monday). Not so surprisingly it does not standardize names of days as far as I can tell.

So you could claim to use ISO-8601 conventions and simply use 1-7 (starting with Monday).

A comprehensive listing of abbreviations of day names for European countries is given by the European Medicines Agency for blistered drugs for the curious.


Has star KIC 8462852's 3 year long Kepler-fading continued after the primary Kepler mission ended? BACKGROUND

May 11, 2013 the Kepler space telescope ended its primary mission. Data captured until then showed that the now famous "Tabby's star" or "Boyajian's star" (KIC 8462852) had been fading throughout the Kepler mission by about 3% cumulatively. (This independently of other reports of brief very deep dimmings and restorations during the mission.)

QUESTION

Today when I post this question, about as much time has passed by since the primary Kepler mission ended, as that mission lasted. So this makes me wonder if other telescopes have observed, or should've been sensitive enough to have observed, a continuation of such a fading trend for that star? Another 3%.

DISCUSSION

If not, what could the explanation be for why this 3+ year fading trend was suddenly interrupted when the primary Kepler telescope mission was terminated? And how long would it take to determine from ground based observatories whether this particular dimming trend continues or not?

Fig. 5 from the paper by Montet and Simon 2016.

Well "if no known astrophysical model can explain it" then nobody told Wright & Sigurdsson (2016) who, cognisant of Montet & Simon's results, explore a number of astrophysical models. They conclude by saying that the most "plausible model" is that of small scale intervening material between us and the star that may be responsible for the short-term dips and the secular dimming. Lacki (2016) appears to agree.

Wright & Sigurdsson also consider and largely rule out or class as unlikely a number of other possibilities, including artificial structures.

The Gaia satellite has been observing this star (and a billion others) since December 2013, but so far the photometry time series has not been released. This should become available towards the end of next year and will probably exceed the accuracy of the Kepler data. In the meantime, it has been claimed by Hippke et al. (2016) that the star was relatively constant between 1934 and 1995 in contradiction to previous claims of a progressive dimming over the last century.


Which telescope's model is being shown to Albert Einstein in this file footage? What event might this be? In the Economist video Why does time pass? there is some file footage of Albert Einstein attending some event where a model of a large telescope is on display, and being described to Prof. Einstein.

Which telescopes model is being shown to Prof. Einstein in this file footage? Is it possible to tell what event this might be, or who some of the individuals might be (any famous Astronomers)?

I'm pretty certain it is the 200inch Hale telescope at Palomar Observatory. There is an image galley of the 200 inch telescope. According to Richard Preston's book First Light: The Search For The Edge Of The Universe, Einstein attended the completion of the telescope truss, yoke and horseshoe bearing at Westinghouse Electric and Manufacturing Company in South Philadelphia (page 41) so it may have been this event that the footage is from.

Searching for "celluloid model of Hale telescope" in the Hagley Digital Archive will yield information such as

  This celluloid mount was part of a 1/32 scale model of Hale Telescope that was constructed by Westinghouse Electric Corporation engineers to test design utility and points of stress.

and images such as these:


What is the role of the random catalogue in calculating the two-point correlation function? To study clustering of galaxies and clusters, one makes use of
the two-point correlation function. However, we can see that all
the estimators of $\xi(r)$ use a random catalogue.

What is the random catalogue used for?

The point of the two-point correlation function (pun not intended) is to describe how clustered the galaxies in the universe are. Astronomers want to know if they're all bunched up in tight bundles with huge voids, or maybe they're more or less uniformly distributed.
The two-point correlation function is a way of measuring the distribution of these galaxies compared to a random distribution. In other words, if your distribution of galaxies is not uniformly random (which it isn't for our universe - our universe has a distinct non-random structure of clusters and voids) then your correlation function will indicate the degree to which it isn't random. Let's take a look at the last equation listed in your link.

The most commonly-used estimator is from Landy & Szalay (1993)
$$\xi(r) = \frac{1}{RR(r)}\Big[DD(r)\Big(\frac{n_R}{n_D}\Big)^2 - 2DR(r)\Big(\frac{n_R}{n_D}\Big)+RR(r)\Big]$$

The symbols in this equation are defined as follows:

$\xi(r)$ - The two-point correlation function.
$r$ - The separation between two galaxies.
$DD$,$DR$,$RR$ - The counds of pairs of galaxies as a function of separation in the data catalogue, between the data and random catalogue, and in the random catalogue, respectively.
$n_D$, $n_R$ - The mean number density of galaxies in the data and random catalogues, respectively.

If your data catalogue happens to be completely random, then $DD=DR=RR$ and you find that $\xi(r)=0$ (assuming also that $n_D=n_R$), showing your data catalogue is indeed indistinguishable from a random distribution. If, however, your data catalogue is not defined with completely random galaxies, you'll find $\xi(r)\ne0$, and the degree to which it is non-zero indicates how non-random, or organized your system is.


Would a Einstein–Rosen bridge change size and/or position in an expanding Universe? I appreciate that a Einstein–Rosen bridge has never been observed but since there are some valid theories in General Relativity I thought I'd ask this.

Would the start and end point of a Einstein–Rosen bridge change in an expanding Universe or would the whole thing stretch but retain it's start and end position?

The latter is closest to the truth, although I wouldn't use the phrasing "stretch". The "mouths" of the wormhole are (more or less) fixed in comoving coordinates (i.e. the coordinate system that expands with the Universe, and in which galaxies lie approximately still). But the bridge is sort of outside our three-dimensional space, and doesn't necessarily follow the same expansion.

If the mouths retained a fixed physical distance, they would accelerate in comoving coordinates beyond bound, eventually moving through space at superluminal velocities, which is forbidden. For example, consider the Milky Way and the galaxy GN-z11 which today lies at a distance of 32 billion lightyears from us. Roughly 13.3 billion years ago, the size of the Universe was 1/10 of today's value; that is $a=0.1$, and the distance to GN-z11 was only 3.2 billion lightyears. If at that time you created a wormhole$^\dagger$ bridging MW and GN-z11, and if the distance between the mouths were fixed in physical coordinates, then today, at least one of the mouths would have moved so far outside its host galaxy that it must have traveled superluminally.

$^\dagger$Creating a wormhole 13.3 billion years ago is left as an exercise for the reader.


What factors determines the size of a galaxy? Recently i saw a post comparing sizes of different galaxies with the Milky Way. The largest one being the Ic 1011.What makes this galaxy so large? What factors decide the size of a galaxy?

Galaxies reside in dark matter halos, and the size of the galaxy is more or less proportional to the size if its halo. The distribution of the sizes of halos is given by the halo mass function (HMF), which is a declining function with mass — that is, the smaller the mass, the larger the number. At low masses, the HMF falls off as a power law, while at higher masses there is an exponential cut-off.

Hence, a galaxy such as IC 1011 is much rarer than a galaxy like M87, which in turn is rarer than Andromeda and the Milky Way.

An analogy is the size distribution of towns and cities. Capitals are rare, towns are common, and villages are even more common.

The reason the galaxy masses follow this distribution is first given by how dark matter in the early Universe fragments into clumps, and start collapsing and thus detach themselves from the Hubble flow. The first analytical attempt to solve the problem of structure formation assumed spherical collapse, and was presented by Press & Schechter (1974). Later, assuming ellipsoidal collapse, Sheth & Tormen (1999) gave an improved solution. More realistic forms cannot be calculated analytically, but must be estimated using large-scale $N$-body simulations (e.g. Tinker et al. 2008, Klypin et al. 2011).

As the Universe evolves, galaxies merge to form larger galaxies. Thus, the distribution depends on the time one considers. The figure below (from Klypin et al. 2011) shows how the HMF evolves from early times ($z=10$) to now ($z=0$). You'll see that the amplitude grows with time, because more galaxies are formed. From $z=2.5$ to $z=0$ you can also see that the number density of small halos actually decrease a little, because small galaxies are "eaten" by larger ones.

Halo mass function, i.e. number density of dark matter halos as a function of mass. Points are from an $N$-body simulation, while curves show the Sheth-Tormen approximation, which provides a very accurate fit at $z = 0$, but overpredicts the number of halos at higher redshifts.


How would the temperature on earth be if the sun was slightly more or less powerful? The temperature on earth is around 14C celsius.
It is around 1370 W/m^2 from our sun. The earth has an albedo of around 0.3
On Tatooine (from Star Wars) i have been able to calculate it goes from 1066 W/m^2 to 2100 W/m^2 (depending if the stars eclipse each other or not). Tatooine is a desert planet with a albedo of around 0.4. How can i make a simplified function to calculate what the temperature ranges from on Tatooine?
I did read on effective temperature on wikipedia, but that is higher than the level in my physics class so i didnt understand.
In the movies they say the temperatures get quite cold at night and warm during the day, so i would expect it to go something like -10C to 40C.
We can just assume the atmosphere is the same as on earth since it is breathable for humans.
Considering as the albedo increases, the temperature should decrease with an increasing albedo.
Therefore my first attempt was
T=k(1-a)*P, where T is temp, k is a constant, a is albedo and P is w/m^2
Since the atmosphere is the same, k should be the same?
Therefore
287=k(1-0.3)*1370, solving for k gives us
k=0.3
therefore temperature on Tatooine should follow the function
f(x)=T=0.3*(1-0.4)*x-273 (to get it measured in celsius)
Where 1066<x<2100.
f(1066)=-81 and f(2100)=99
the temperature are not this extreme doe, so this wont work.
I am aware this is not an accurate science, but im taking my first physics course in high school now, so I dont understand the "proper" way of doing this. Is it completely incorrect to assume temperature is proportional to (1-a) and how much power the sun gives us, when its the same atmosphere (ish)?
If anyone wants how i found out the power the twin suns gives Tatooine i can share it here.

Don't have high enough reputation to comment, but someone should mention the Stefan-Boltzmann Law: https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law
When a planet (or anything) gets warmer, the amount of radiation it emits increases with the 4th power of temperature (measured on an absolute scale like Kelvins).
A planet's temperature reaches equilibrium when it receives about as much radiation as it emits back into space.
This should give you a better approximation than the linear model.


Is the shortest day duration constant? Is the exact duration of the shortest day of a year (at the winter solstice in the northern hemisphere) always the same, for given latitude and longitude, or does it fluctuate? If it does, what is the range?

By looking at the link provided by James K, I was able to write a Python 3 script that calculates the duration of the longest night for a predefined location (Boston, MA) for 160 years. The script uses the PyEphem module. The longest night becomes shorter by ~72 msec per year, and its length also oscillates with the period of ~19 years.

Here's the script that I used:

import ephem
import matplotlib.pyplot as plt
import matplotlib
import numpy as np

location = ephem.Observer()
location.lat, location.lon = '42.3601', '-71.0589'

years = range(1900, 2060)
longest_nights = []
for year in years:
    location.date = ephem.next_winter_solstice('{}-12-12'.format(year))
    sunset = location.next_setting(ephem.Sun())
    sunrise = location.next_rising(ephem.Sun())
    if sunrise < sunset: # Look at the night before!
        location.date -= 1
        sunset = location.next_setting(ephem.Sun())
    longest_nights.append(sunrise - sunset)

longest_nights = np.array(longest_nights)
longest_nights = (longest_nights - longest_nights.mean()) * 24 * 3600 # In seconds

# Find the slope
xp = np.poly1d(np.polyfit(years, longest_nights, 1))
np_years = np.array(years).reshape(len(years), 1)

# Plot
matplotlib.style.use("ggplot")
plt.plot(years, longest_nights, "-o")
plt.ylabel("Deviation from the mean, sec")
plt.xlabel("Year")
plt.title("Duration of the longest night in a year in Boston")
plt.plot(years, xp(years), "-")
plt.arrow(2017, plt.ylim()[0], 0, plt.ylim()[1] - plt.ylim()[0], color='k')
#plt.savefig("solstice.png")
plt.show()


How strong are the tides raised by Io on Jupiter relative to the ones raised by the Moon on Earth? There seems to be 2 ways of calculating tidal forces that appear contradictory. Either:

By calculating the difference of Io's gravitational acceleration on a point on Jupiter's near side and the gravitational acceleration felt by a point on Jupiter's far side, using the equation:

$$\frac{GM_{io}}{(d-r_{jupiter})^{2}}-\frac{GM_{io}}{(d+r_{jupiter})^{2}}$$

For the Io-Jupiter system, this is $24.0807\times10^{-6}$ Newtons, while for the Moon-Earth system it is $2.2009 \times 10^{-6} $ Newtons.

So if you were to calculate tidal forces that way, the tides raised on Jupiter by Io would be $10.94$ times as strong as the tides raised by the Moon on Earth

However there is also another way I found, which is:

By using the derivative of Newton's law of universal gravitation, using the equation:

$$\frac{-2G\times M_{jupiter}\times{m_{io}}}{d^{3}}$$

For Io, this is $3.015 \times 10^{14}$ Newtons per meter, whereas for the Moon it is $1.031 \times 10^{12}$ Newtons per meter.

So if you were to calculate tidal forces that way, the tides raised on Jupiter by Io would be $292.5$ times as strong as the tides raised by the Moon on Earth

So, which one is it? Are the tides raised on Jupiter by Io 10.94 or 292.5 times as strong as the ones raised by the Moon on Earth? My intuition leads me to believe the first answer is correct since it takes the primary's radius into account but I just want a concise answer.

Both expressions are incorrect. The first should be
$$\frac{GM_{\text{moon}}}{(R_{\text{moon}}-r_{\text{planet}})^2} - \frac{GM_{\text{moon}}}{{R_{\text{moon}}}^2}\tag{1b}$$
or
$$\frac{GM_{\text{moon}}}{{R_{\text{moon}}}^2} - \frac{GM_{\text{moon}}}{(R_{\text{moon}}+r_{\text{planet}})^2}\tag{1a}$$
where $R_{\text{moon}}$ is the distance between the center of the planet and the center of the moon. Equation (1a) pertains to the point on the surface of the planet closest to the moon while (1b) pertains to the point on the surface of the planet furthest from the moon.

Note that the results have units of acceleration rather than force. The tidal force on a small mass on the surface of the planet is the product of the tidal acceleration and the mass of that small mass.

Since $R_{\text{moon}} \gg r_\text{planet}$ in the case of Io and Jupiter, and also in the case of the Moon and the Earth, the two expressions are nearly equal to one another, and both in turn are nearly equal to
$$2\frac{GM_{\text{moon}} r_{\text{planet}}}{{R_{\text{moon}}}^3}\tag{2}$$
This is the correct version of your second expression.

On plugging in the numbers one finds that the tidal acceleration exerted by the Moon at the surface of the Earth is 1.068 μm/s^2 while the tidal acceleration exerted by Io at the surface of Jupiter is 9.014 μm/s^2 -- over 8 times the tidal acceleration caused by the Moon at the surface of the Earth. This does not quite show the extent of the tides raised by Io on Jupiter (or by the Moon on the Earth).

It is important to note that equations (1) and (2) pertain only to two special points, the submoon point and its antipode. The tidal acceleration at the point where the moon is on the horizon is half the value of equation (2) and is directed inward. What would be nice would be a potential model whose gradient results equation (2) and also in the inward acceleration at the horizon. This potential function is, to first order,
$$V = \frac{GM_{\text{moon}} {r_{\text{planet}}}^2}{{R_{\text{moon}}}^3}\frac{3\cos\phi^2-1}{2}$$
where $\phi$ is the angle between the line segment between the center of the planet and the center of the moon and the line segment between the center of the planet and the point in question on the surface of the planet.

If the planet consisted of a fictitious frictionless fluid, the fluid would form a prolate spheroid whose height at some angle $phi$ differs from what would be the planet's moonless spherical shape by $V(\phi) / g_\text{planet}$. The height of the tidal bulge is the difference between the maximum at $\phi=0$ and the minimum at $\phi=\pi/2$, or
$$h = \frac32 \frac{M_{\text{moon}}}{M_{\text{planet}}} \frac{{r_{\text{planet}}}^4}{{R_{\text{moon}}}^3}$$

On plugging in the numbers one finds that the tidal bulge created by the Moon on a fictitious frictionless fluid Earth would be 0.5327 meters while the tidal bulge created by Io on a fictitious frictionless fluid Jupiter would be 22.52 meters -- over 42 times the size of the tides raised by the Moon on the Earth.

While neither Jupiter nor the Earth is composed of such a fictitious fluid, the tidal bulge height nonetheless does provide a nice picture of how large the tides that result from a moon can be.


Exoplanet dip in transit light curve when the planet passes behind the star In the animation below, I don't understand why the brightness slightly decreases when the planet is behind the star.

Where does this effect come from?

Just before the planet goes behind the star, we see the light directly from the star as well as the light reflected from the planet's surface.  When the planet is behind the star, we no longer see the reflected light.
Note that the top of the line also curves to reflect how much of the planet's surface is illuminated from the star.  The most illumination is when it is almost "full" just before it passes behind the star.  The least illumination is when it is "new" just before it passes in front of the star.
Edit (to address Fraxinus's comment below): Here is an example from Zhang et. al: Phase Curves of WASP-33b and HD 149026b and a New Correlation between Phase Curve Offset and Irradiation Temperature.

The above graphic shows processed data from the Spitzer space telescope.  From the paper:

Observations were timed to begin before a secondary eclipse and end
after the following secondary eclipse,

These light curves help aid in estimates of planetary albedo, day/night temperature, and atmospheric composition.


Where should I be looking to see Neowise in southern Sweden? I'm in the south of Sweden and just spent over an hour trying to find Neowise, but didn't see it. I live in a rural area, there are no lights nearby and there is a small hill with clear visibility between north and northeast. I went out at 00:45 (GMT+2) and came back in at 02:00, not having seen anything resembling a comet, not even with binoculars.
News outlets around here are saying everything from "Look towards the Big Dipper and you'll see it" to "Between north and north-east", or "To the west in the evening and east in the morning". Many say that it would be somewhere "down to the left of Capella". Frankly it's quite frustrating to not be able to find neither Neowise nor a surefire description of how to locate it.
This photo is taken to the north-east around 01:45 (GMT+2). Where in relation to the stars on the photo would Neowise be? The brightest one is Capella according to Sky Maps.

The comet's position changes from night to night.
It was in Auriga earlier this month but is now in Ursa Major.
Here is a finder chart generated by Stellarium, with cyan dots indicating the position of C/2020 F3 at 01:10 GMT+2 from July 22 to 31.


Why can't the surfaces of stars be observed? If I'm correctly informed, only three stars: Sol, Betelgeuse and Altair have surfaces which have been resolved by telescopes. All other stars are only point sources of light, even in the greatest of telescopes. Is it just the huge distance to the stars which reduce them to dots? Or is there some other optical explanation? I mean, the Sun is about as large as the Moon on the sky, but I cannot see any details on its surface because it blinds me. However, the next nearest star is invisible to the naked eye.

Some exoplanets have been directly imaged by removing the light from their star. It seems funny that we cannot detect any features on the photosphere of such a star, such as spots or flares or its shape. Plenty of photons are available.

Actually, there are several stars which have been directly imaged, maybe my sopposition is quite wrong:
https://en.wikipedia.org/wiki/List_of_stars_with_resolved_images

There are many different ways to get spatial information about the surface of a star besides direct imaging.

Direct imaging is difficult because the angular resolution available goes as $\lambda/D$.
For a 8-m telescope and light at 500 nm, one can resolve $6\times10^{-8}$ radians (assuming the blurring of the atmosphere can be overcome by adaptive optics or similar).

The nearest stars are a couple of parsecs away, so the smallest spatial scales that could be resolved are $\sim 2\times 3.1\times10^{16} \times 6\times10^{-8} =  3.7\times10^{9}$ m, or about 500 solar radii. Hence no surface features or even a disk could be resolved.

Of course you could use interferometric techniques to effectively increase the size of $D$ and measurements of the angular radii are now possible for many nearby stars or giant stars at larger distances.

Surface imaging is harder. Indirect techniques are much more common. These include doppler imaging and eclipse mapping. The former uses the fact that there is a relationship between the position of a bright/dark feature on a rotating star and the doppler shift of the light from that feature. By observing a time-series of spectra, the lumps and bumps in spectral lines can be inverted to produce a "doppler map" of the surface. The technique is usually limited to stars that are considerably more rapidly rotating than the Sun. There is a lot of ambiguity in the image reconstruction process - many surfaces could lead to the same observable sigature and clever statistical techniques (and even philosphies) have to be deployed to choose between them. Many stars have published "doppler maps" of their surfaces. Here is a typical example of such a study and below I show an example of a "doppler image" for the star II Peg (a K-type subgiant), from Gu et al. (2003), showing dark spotted regions. A typical resolution for such an image is about 10 degrees on the star.

Eclipse mapping, for which I can't easily locate a good link, uses the fact that a star/accretion disk is orbited by another star or planet that periodically eclipses it. What happens to the light from the system in and out of eclipse can be used to probe the surface of the eclipsed object. There are of course limitations to the spatial resolution that can be obtained, depending on the size of the eclipsing object, how long the eclipse takes and how wide the orbit is. But useful constraints can be made on the structure of accretion disks, sizes of starspots etc, although "maps" are not usually produced. A recent example using the transits of a planet to probe starspot structure is Roettenbacher et al. 2013.

Another possibility is rotational modulation. Features on the surface that rotate around are self-eclipsed by the star and produce modulation of the observed light. This can be used to try and estimate e.g. the size and location of starspots. Again there are many degeneracies and ambiguities, but this has become a growth industry since the delivery of thousands of extremely high quality light curves from the Kepler satellite.


Could something temporarily break the Moon's tidal lock? Would it be possible for something to cause the Moon to start rotating at a different rate, breaking its tidal lock with Earth, without ripping it apart?

If so, would the Moon's tidal lock eventually stabilize?

Yes, theoretically it is possible to impact the Moon with enough moment and low enough energy to have it change its rotation periode, which actually is the same as its orbital periode (this is the tidal lock). Best candidate will be a high-mass, low-speed asteriod impacting almost tangentially and on the rotation plane.

Is this really possible? I would bet that it will never happen.

Supossing that it happens, will Moon get tidal locked again? Yes. As long as there are water on Earth suffering tides from Moon, these same water will attract one side of the Moon differentially from the other, so Moon will get locked again with the same face towards us.


Centre of the singularity that is the big bang If it's true that the deeper we look into space the further back in time the light is coming from then does that mean that just behind the cosmic background radiation the entire sky is filled with an unviewable image of the singularity that was the big bang and can that mean that we are sat at the centre of the Big Bang?

Since the Universe evolved from something like a singularity, where all of space was contracted to a point, yes, we are at the center of the Big Bang. So is every other point in the universe.

The CMB represents the light from a universe-filling plasma that lasted for roughly 378,000 years. It is a feature of the observable universe, rather than something that is actually there in the present universe; in particular, we can never reach it, as it's the same distance away from every point in the universe. As such, there's not much sense in talking about what's behind it, much like talking about what's behind a shadow.


What is the overlap between the Gaia and the Hipparcos catalog? Is there any kind of concordance available between the sources in the Hipparcos catalog and the Gaia DR2 catalog? I.e., mapping the Gaia DR2 source_id to the Hipparcos HIP number?

As part of the Gaia DR2, cross-matches to several catalogues including HIP2 were included. The main reference for this is Marrese et al. (2018) "Gaia Data Release 2. Cross-match with external catalogues: algorithms and results". The resultant matches are included in the tables gaiadr2.hipparcos2_best_neighbour and gaiadr2.hipparcos2_neighbourhood available via the query facilities in the Gaia Archive.

Unfortunately the match is not particularly complete and it is not entirely clear why. From the paper:

  While we expect to find Gaia counterparts for most of the Hipparcos2 sources, with the exception of the brightest ones, the
  cross-match results include only about two-thirds of them. This
  means that according to the adopted cross-match algorithm, only
  about two-thirds of the Hipparcos2 objects have a Gaia counterpart that is compatible within the position errors (i.e. have at least
  one good neighbour). Hence the Hipparcos2 cross-match results
  clearly show an issue that needs to be investigated.

As a result, they also provide a table of cross-matches based on a 1-arcsecond cone search, available on the Gaia DR2 known issues page. This cross-match is a lot more complete, but there are still a few objects that are present in both catalogues that are missing. Several of these are nevertheless present in SIMBAD, e.g. 85 Pegasi = HIP 171 = Gaia DR2 2853900872207150720.


How to calculate the ground track of the Moon's position on the Earth's surface? I would like to project the Moon's position on to the surface of the Earth, starting with its position expressed as Right Ascension in degrees.

I know that the declination of the Moon corresponds to the terrestrial latitude but how can I "convert" the right ascension of the moon to find my longitude?

I'm not an astronomer, so I'm hoping for a fairly simple answer.

EDIT

My friend used this formula a long time ago:

For the time at which I want to calculate the longitude,

time_hrs = hours + minutes / 60

then using right ascension of Sun,

delta_x = time_hrs * 15 + right_ascension_sun - 180;

The coordinates of the Moon's ground track at this time are then:

longitude = Right_ascension_moon - delta_x;
latitude  = Declination_of_moon

but i don't know if is correct, or how precise it will be.

Given a date and time, the position of the Moon can be calculated to provide the declination and right ascension. The sub-point of the Moon (the point on the Earth at which the Moon is at the zenith) is as follows:

latitude = declination of the Moon
longitude can be found by calculating the local mean sidereal time (LMST) that equals the Moon's right ascension. (LMST dependents on the date, time, and longitude.) Everything is known except for the longitude.

Accurate Method (using GMST)

Calculate the Local Mean Sidereal Time (LMST) from $LMST=GMST+long_{east}$, where GMST is the Greenwich Mean Sidereal Time and $long_{east}$ is the longitude with positive values in the eastern hemisphere. From the post Local Sidereal Time,
$\mathit{GMST}_{\text{deg}} = 100.4606184 + 0.9856473662862 D + 15 H
$ where D is the number of days (including the fraction of days) from J2000 (Jan 1, 2000 at 12 h UT = Julian Day 2,451,545.0) and H is the Universal Time (UT) in hours.

For example, at H=17 hour UT on Nov 1, 2000, I calculate the following values:
$$
RA_{Moon}=18h\;49m\;35s = 282.400° \\
Julian Day = 2,451,850.208 \\
\therefore D=2,451,850.208 - 2,451,545=305.209 \\
\therefore GMST = 296.288° \\
$$
Then
$$
RA_{Moon}=LMST=GMST+long_{East} \\
282.400=296.288+long_{East} \\
\therefore long_{East}=-13.888°
$$

Approximate Method (using Sun's position)

Your method of using the Sun's right ascension assumes that the Sun is over 0° longitude at 12 h UT. Written as a formula and allowing for different times and longitudes, it would be $LMST=RA_{sun}+15(Time \;in \;UT - 12h)+long_{east}$. Setting this LMST to the right ascension of the Moon gives the following:

$$
RA_{Moon}=LMST=RA_{sun}+15(Time \;in \;UT - 12h)+long_{east} \\
\therefore long_{East}=RA_{moon}-RA_{sun}-15(Time \;in \;UT - 12h)
$$

The reason this is an approximation is because of the following:

The Sun is not at 0° longitude at 12:00:00 UT everyday because of the Equation of Time. The Sun can be off by up to 16 minutes of time (which is equivalent to 4° of longitude).
The term 15(Time in UT - 12h) should be 15*1.002738*(Time in UT - 12 h), but this difference is small compared to the first approximation.

Continuing the example of Nov 1, 2000, I calculate the following:
$$
RA_{sun}=14h\;28m\;46s = 217.19° \\
long_{East}=RA_{moon}-RA_{sun}-15(Time \;in \;UT - 12h) \\
long_{East}=282.400-217.19-15(17-12) \\
\therefore long_{East}=-9.79°
$$

This is different by 4° because I chose Nov 1 since it is close to the date of the largest Equation of Time.

Likewise, if you wanted to calculate the sub-point of the Sun, it would be better to use the GMST and the Sun's right ascension by solving for the longitude from $RA_{sun}=LMST=GMST+long_{East}$.


How many Kuiper Belt objects have moons? How do we know this? On 2012 in the New Horizons' The PI's Perspective Alan Stern wrote The Kuiper Belt at 20: Paradigm Changes in Our Knowledge of the Solar System (more also archived) which includes:

Most of the known KBOs are just 100 to 300 kilometers across, about one-tenth of Pluto’s diameter. But some are smaller than 100 kilometers across, and some are larger than 300 kilometers across. In fact, there is great diversity among KBOs:

Some are red and some are gray;
The surfaces of some are covered in water ice, but others (like Pluto) have exotic volatile ices like methane and nitrogen;
Many have moons, though none with more known moons than Pluto;
Some are highly reflective (like Pluto), others have much darker surfaces;
Some have much lower densities than Pluto, meaning they are primarily made of ice. Pluto’s density is so high that we know its interior is about 70% rock in its interior; a few known KBOs are more dense than Pluto, and even rockier!

Question: How many Kuiper Belt objects have moons? How do we know this? What observational techniques have revealed moons of KBOs?

"Moons" is a misnomer here, the majority of Kuiper Belt Objects (KBOs) with companions are binaries.
Now there are several populations in the Kuiper Belt, characterized by their eccentricities and inclinations. The cold classicals are hereby the most important ones, having mostly circular, low inclination orbits and a high binarity fraction of about 30% (Noll et al. 2008).
Hot classicals, with eccentric and inclined orbits possess a binarity fraction of about or less than 1 in 40.
The other important KBO populations, i.e. the scattered disc, the resonant ones and the centaurs possess negligible binarity. In terms of numbers, the classicals make up about 2/3 of the entire KBO population by numbers, and the cold classicals are about 50-60% of the total classical population.
So in total you'd get a fraction of at least $f_{\rm Moon}\approx 0.66*0.5*0.3\approx10\%$ of KBOs having 'moons'.
In order to find and characterize binary orbits, strange light curves pose a first hint, but only high spatial resolution actually helps to discern the two binary components. As most KBOs are low-mass objects, their binary orbital times are long. Prolonged, 5-15 year snapshot observations with the HST allowed recently to characterize the spin direction, for example (Grundy et al. 2019).


What will be the temperature on Earth when Sun finishes its main sequence? We know that presently Sun is 4.5 billion years into its main sequence. It has another 5 billion years before it enters the Red Giant phase. We also know that Sun's luminosity increases by 10% every billion years during the main sequence. I am interested in finding the temperature rise as we approach the end of main sequence. I got two different values for temperature on Earth.

Wikipedia entry says that temperature on Earth would be 422 k in 2.8 billion years. However, if we use the formula for effective temperature as discussed in this answer
https://earthscience.stackexchange.com/a/4274/15299 and L = 1.8 then, the temperature on Earth would be 330K. Also in this book, the author does the same calculations on page 255.

The difference is that your analysis is assuming that the albedo stays fixed, so the surface temperature simply scales like luminosity to the 1/4 power.  The Wiki entry is including feedback from the greenhouse effect, which will tend to further increase the surface temperature.  Note that an analysis that just looks at solar irradiation would get way too low a surface temperature for Venus, for example.  I can't speak to the accuracy of the feedback included-- as I understand it, that is far from a simple effect to include.


My first pictures of Jupiter Well, I'll start off by saying I am certain I am doing it wrong! Now that I got that out of the way here is my image:

I have a feeling I can't see the planet because I need a filter to block out the light surrounding Jupiter, however I honestly didn't expect to have such a profound atmosphere-like disc around Jupiter.

My fiance thought it was my own eye being projected, but it was a DSLR doing the actual looking.

I am using the Celestron NexStar 127 SLT which comes stock with a 25mm and 9mm eyepiece.

Update:
Okay! I focused and was able to see Jupiter, including some aspects of its atmosphere. However, I was not able to take a descent picture of it. Now I'm thinking the need for filters and proper DSLR usage methods will come into play.

Your problem is that Jupiter is completely out of focus.  You need to turn the focus knob so that the bright ring gets smaller and smaller, and then you will find that the bit in the middle is bright, and is Jupiter.  Try turning the focus knob one way, and if the central dark circle gets bigger, then turn the other way, slowly.

What your pictures are showing is a magnified image of the central reflector that is located on the front corrector plate of your scope - that's the dark central circle - surrounded by the unfocussed light of Jupiter.

You could start by practising during the day, focus on distant objects (trees etc) and without the camera. Then try at night, again without a camera. Once you can use the telescope to see, try taking a photograph, again you can practise during the day. The moon might then be a good target before trying to get images of planets.

Good luck


Do the magnetic fields of stars and/or other planets reverse? I've heard of the magnetic field of the earth reversing poles over the course of thousands of years but do stars and other planets (specifically gas giants) also experience this? If so how does it differ and if not why?

Sure. It even happens with the Sun, as part of the solar cycle.

Earth's magnetic field is described by something called the dynamo theory, which states that the movement of fluids in the core generates the magnetic field. A modification of this is used to model the magnetic fields of stars. Given that the cause of geomagnetic reversals is related to the movement of currents deep on Earth's core, it makes sense that the same would be the case in other bodies.