How does gravity really work I am only 12 years old and I'm constantly wondering and trying understand how gravity really works. On YouTube everyone always talks about objects warping space time around themselves and uses the analogy of a trampoline. I still don't understand gravity because if space were like a trampoline, then earth would be spiraling in towards the sun along with all the other planets, right? So could someone explain to me how gravity really works without the trampoline analogy?

First of all: "How gravity really works" is a deep question, and any serious scientist would quickly concede that all we have is an incomplete working model. You certainly have heard about  General Relativity; the first image on the page is your trampoline.

Our working model, General Relativity, is working because it explains a lot of observations very nicely. (Careful, here is another deep question lingering: "Explains" means that we can predict some observations from other observations with the model of gravity we have in our  mind. It does not necessarily mean that we understand the "real nature" of the underlying issues.) But we are very confident that the model is working over a wide range of observations. One of the last "first-time" observations which followed the predictions and thus gave us more confidence in the model was the two black holes colliding lately. Lately? Well, billions of years ago. We just learned about it lately.  Here is a link to a New York Times article with an impressive video. (I think one can still read a limited number of Times articles for free, so try it out.)

Our model of gravity is incomplete because it doesn't connect well to the model of nature we have for other things (elementary particles, quantum physics). For a while (like 70 years or so) it didn't connect at all; Einstein himself completely failed to connect the dots, which was probably not encouraging since he had received the Nobel Price for laying one of the foundations of quantum physics and was the obvious authority about gravity. If he couldn't do it, who could?

If I'm not mistaken, the physicists today are making progress, slowly. This connection between quantum physics and gravity is one of the main unresolved problems in modern physics.

Last, let me address your concern about the planets spiraling into the sun. This idea probably comes from actual balls on an actual trampoline spiraling in, I suppose. You probably know that the balls lose speed due to friction, much the same way you slow down on your bike when you stop pedaling. Some of the kinetic energy is transformed into heat.

And you know what? You are right. Given enough time, the planets would eventually fall into the sun. Low-flying satellites fall back to earth after a few years, because there are still traces of atmosphere slowing them down out there. The reason is that there is "friction" in the wider sense involved in all large-scale processes in the universe. That is actually one of the fundamental physical principles making up the world we know. It's just that the near-vacuum between the planets doesn't provide that much friction, and the planets are fairly massive bodies with an enormous mass and kinetic energy. It will take a long long time for them to lose enough energy that they'll be so close as to touch the sun. (Perhaps too long to happen at all.) In fact, over human life times the planets, moons and stuff are almost perfect examples for movement without friction. But in the astronomical time scale -- billions of years --, there certainly is friction. For example, the moon is showing us always the same side because friction slowed its rotation so that the rotation is now "locked" with its orbit.

Bottom line: The idea that gravity bends space and time "explains" all large-scale observations so far; the "trampoline" is a good model for a 2-dimensional "space", i.e. a surface, if you ignore friction.


Event Horizon Telescope Data Cleanup I was wondering how the researchers were able to minimize or neutralize confirmation bias regarding the data collected?

As I understand it, the algorithm pruned much of the data, which would make sense, since it would likely be useless noise. See for example the video Event Horizon Telescope Press Conference. First Image Of a Black Hole. at about 38:40.

  Five petabytes is a lot of data… The image you saw of course isn’t 5 petabytes in size, it’s a few hundred kilobytes, so our data analysis has to collapse this five petabytes of data into an imaget that’s more than a billion times smaller.

But apart from that, could certain features be misidentified as artifacts, and be thrown out in favour of preselecting what researchers expected to see?

The only reason I ask is that the image looks almost identical to simulated predictions from two years ago - which may raise a slight cause for alarm.

Thank you.

To follow up (based on answers section below):

In the first stage process - "common features" - are they like existing stamps of astronomical features (and image comparatives) that have been already observed in our universe (in both the visible and invisible spectrum and used for comparison)? And a second question - if we have never observed the earmarks of a black-hole before, how are we to use existing stamps of knowledge of astronomical features that we have seen to assess and validate an as of yet unknown (which would seem to directly impact what artifacts are preserved and what artifacts are thrown out and discarded)?

This process is described in great detail in this paper. The abstract says:

  To assess the reliability of these results, we implemented a two-stage
  imaging procedure. In the first stage, four teams, each blind to the
  others' work, produced images of M87 using both an established method
  (CLEAN) and a newer technique (regularized maximum likelihood). This
  stage allowed us to avoid shared human bias and to assess common
  features among independent reconstructions.

They then generated a large number of "plausible" images, reconstructed the radio signals they would have given and ran them through the image analysis pipeline to see if it showed any biases.

  In the second stage, we reconstructed synthetic data from a large survey of imaging parameters and then compared the results with the corresponding ground truth images. This stage allowed us to select parameters objectively to use when reconstructing images of M87. Across all tests in both stages, the ring diameter and asymmetry remained stable, insensitive to the choice of imaging technique.

Incidentally, it's not really appropriate to think of the rejected data as image data in the sense you would get from a camera. It is (almost) recordings of the exact electromagnetic waveforms received by the telescopes. The image arises from subtle correlations between these datasets. So most of what is thrown away (apart from noise) is information about the exact phase of the radio signal on a nanosecond-by-nanosecond basis, which is not really interesting.


Does Doppler shift affect apparent pulsar frequency? We know that redshift and blueshift is the result of the frequency of light waves reflected or emitted by an object lengthening or shortening due to the relative velocity of the observer moving away or toward the object. This is the Doppler effect.

However, it was stated in response to a question I asked on a sister site that the Doppler effect would also result in the apparent frequency of a pulsar's pulses being longer or shorter due to the relative velocity of the star. Given that the speed of light is constant regardless of the frame of reference (a concept I have a really hard time understanding by the way), is this correct?

My thinking is that it is incorrect. Using the analogy of a launcher lobbing medicine balls at me at a fixed rate, I can "see" the rate of launch by the frequency of when I get hit. However, if the launcher is moving away from me, the velocity of the balls lobbed at me will be slower (relative to my frame of reference), and thus will hit me with less frequency. However, if the launcher were to adjust its firing speed (i.e. speed of the ball relative to itself; frequency remains constant in this analogy) so that the balls always hit me at the same speed (thereby simulating the concept of the always-fixed speed of light), the time between them would not change.

So, given the always-constant speed of light, is it correct that a pulsar's pulse rate would appear to be different at different relative velocities?

Yes, it is correct.

Consider the classic example of a (say) $1000$ Hz siren on an ambulance approaching you.  Once a single wave is produced, the source moves towards you before the next wave is produced.  The wavelength of the sound waves in the air is reduced, so the frequency you hear is increased.  Simple, basic Doppler shift...

But consider the case where the siren is being switched on and off with a five-second overall period.  So you have a $1$ kilohertz signal modulated at $0.2$ Hertz.

But the same argument applies to this modulation frequency.  A chain of $1$ kilohertz waves is generated for $2.5$ seconds and sent on its way.  The ambulance approaches you for the next silent $2.5$ seconds, then starts the next chain on its way, closer to the last chain because of the distance travelled in the silent phase.  The same Doppler shift factor applies to the carrier frequency and the modulation frequency.

The same physics almost derailed the Cassini-Huygens probe. The extreme velocity changes of the probe caused a Doppler shift in both the transmission frequency and the data rate carried by this transmission. The first was handled;  the second was not!

See http://www.thespacereview.com/article/306/1

or

http://spectrum.ieee.org/aerospace/space-flight/titan-calling


Delete key does not always work in IRAF's xgterm on Mac Sometimes it does not let me delete text in an xgterm terminal and outputs ^H instead.

Example:

Run IRAF:

xgterm -e cl

It opens the xgterm, there I type:

splot black_hole.fits[0]
Image line/aperture to plot (0:) (12): 192

Then I realise I made a mistake and press the Delete key on Mac, and it writes the ^H text instead:

Image line/aperture to plot (0:) (12): 192^H^H

This is very annoying. Is there a way to fix it?

Note that delete key works in xgterm initially, but breaks and writes ^H in the case above.

I've installed IRAF 2.16 through astroconda iraf-all package on Mac.

Update

The only solution I found was to use PyRAF instead. PyRAF uses the normal Mac's terminal, which does not have those Delete key issues. I can still run the same IRAF commands in PyRAF terminal, so it works fine for me.

This sounds like a problem with your terminal settings.  Quoting the Setting up IRAF section this page, "We assume the user has defined this correctly when issuing the MKIRAF and no longer key off the unix TERM to set a default."  The page goes on to say that one should add stty xgterm the to your login.cl file (directly above the final keep line might be a good place).  The page linked above goes into more detail.


Is there a more modern/updated paper than Hurley's 2000 SSE? I've implemented equations from Hurley's 2000 paper on Single Stellar Evolution (SSE) to model the stages of stellar evolution without having to use the full SSE code (or MESA for that matter).

Is there a more updated version of such a paper, where numerical formulae are given to model stages of stellar evolution? Has the knowledge base of the field of stellar evolution changed enough in the last 18 years to warrant an updated numerical formula recipe?

Here's the arxiv of the paper.

My guess would be no because, as of this writing, from the paper link you provided you can see there are 949 citations to that paper with a good chunk of them in this year. You could always contact the author(s) to discuss it. Perhaps your effort could lead to an update.


What does half-lit means? Aristarchus estimated the relative distance of the Sun and Moon by
  observing the angle between the Sun and the Moon (α in the diagram)
  when the Moon is exactly half lit. Angle β must be 90° for the Moon to
  be half lit. By observing the angle α, he could then set the scale of
  the triangle and thus the relative lengths of the sides. (Sizes and
  distances are not to scale.)

Does half-lit mean exactly 45 degrees?

Half lit means the Moon is at a quarter phase, either First Quarter or Third Quarter.

The original question asks "Does half-lit mean exactly 45 degrees?".  I assume this is asking if the angle $\alpha$ is 45 degrees. The answer is no. The angle $\alpha$ is found from trigonometry to be $cos(\alpha)=EM/ES$ where EM is the distance from the Earth to the Moon and ES is the distance from the Earth to the Sun. EM/ES is approximately 1/390 (=238000 miles/93 million miles), so $\alpha$ works out to be 89.85 degrees.

Aristarchus was trying to do the reverse calculation to find the distances. If the angle $\alpha$ could be measured accurately, the relative distances EM/ES could be calculated using the same formula. The problem is the difference between 89.85 degrees (the Sun is 390 times farther than the Moon) and 90 degrees (the Sun is infinitely farther than the Moon) is a very small angle and difficult to measure exactly. If the Sun were a lot closer, then $\alpha$ would be a smaller value, and the error in the measured angle of $\alpha$ would be restrict the value of EM/ES to a more meaningful range than "infinity".


Can things move faster than light inside the event horizon of a black hole? Black Holes are regions of space where things get weird [Citation Needed]. Past the event horizon of a black hole, any moving particle instantaneously experiences a gravitational acceleration towards the black hole that will cancel out it's current velocity, even light. That means that the gravity well of the black hole must be able to accelerate from -C¹ to 0 instantly². Given that fact, we can assume the gravitational acceleration of black holes is C/instant³. Given this, it stands to reason that in successive instants, the particle will be moving at speeds greater than C, because it is experiencing greater gravitational forces and continuous gravitational acceleration.

Does this actually make sense? Is there something I'm missing here? By this logic, it seems like anything inside of the event horizon of a black hole could and should move faster than C due to gravitational acceleration.

Edit: I showed this question to a friend and he questioned if the hypothetical particles that were radiating from the singularity (The photon traveling exactly away from the black hole) might be hawking radiation; that is, the gravitation acceleration of a black hole is only strong enough to curve the path of light around a non-zero radius (thus not actually stopping it, but altering it's course), and not powerful enough to decelerate light. Is this actually what hawking radiation is, or is he as confused as I am?

¹ Where movement towards the singularity would be considered a positive value, movement away from the singularity is a negative value, that is, anything moving at the speed of light away from the singularity would be moving with a velocity of -C relative to the singularity.

² If it couldn't accelerate from -C to 0 instantly, any photon traveling exactly away from the black hole would be able to escape the event horizon.

³ An instant is an arbitrary amount of time, it could be a fraction of a second, a second, a minute....

I think your initial question is a good one, but the text gets a bit more jumbled and covers a few different points.

  Can things move faster than light inside the event horizon of a black
  hole?

Nice question.

  Black Holes are regions of space where things get weird.

I'm 100% OK with this statement.  I think it's a true enough summary and I'm sure I've heard physicists say this too.   Even if "Weird" isn't a clearly defined scientific term, I'm 100% fine with this (even without a citation).

  Past the event horizon of a black hole, any moving particle
  instantaneously experiences a gravitational acceleration towards the
  black hole that will cancel out it's current velocity, even light.
  That means that the gravity well of the black hole must be able to
  accelerate from -C* to 0 instantly✝.

Are you quoting somebody here?   Anyway, this isn't quite true.   Black holes don't accelerate things from -c (which I'm guessing would be a light beam trying to fly away from the singularity but inside the event horizon), to 0 "instantaneously".

Perhaps a better way to look at it is to consider curvature of space, and inside a black hole, space curves so much that all directions point to the singularity.   It's the "all roads go to Rome" scenario, even if you do a complete 180, you're still on a road that leads to the singularity.

I understand the temptation to look at that as deceleration, but I think that's a bad way to think about it.  Light doesn't decelerate, it follows the curvature of space.

  Given that fact, we can assume the gravitational acceleration of black
  holes is C/instant**. Given this, it stands to reason that in
  successive instants, the particle will be moving at speeds greater
  than C, because it is experiencing greater gravitational forces and
  continuous gravitational acceleration.
  Does this actually make sense? Is there something I'm missing here? By
  this logic, it seems like anything inside of the event horizon of a
  black hole could and should move faster than C due to gravitational
  acceleration.

outside of a black hole, continuous acceleration would never lead to a speed greater than C.  You can accelerate for billions and trillions of years and all you'd do is just add more 9s to the right of the decimal point.

You seem to be assuming that inside a black hole this can happen, but I'm not sure why you'd assume that.

"continuous gravitational acceleration" - no matter how strong, is no guarantee for faster than light travel.   That's logically inconsistent with the laws of relativity.

  Edit: I showed this question to a friend and he questioned if the
  hypothetical particles that were radiating from the singularity (The
  photon traveling exactly away from the black hole) might be hawking
  radiation; that is, the gravitation acceleration of a black hole is
  only strong enough to curve the path of light around a non-zero radius
  (thus not actually stopping it, but altering it's course), and not
  powerful enough to decelerate light. Is this actually what hawking
  radiation is, or is he as confused as I am?

I think, a more correct way to look at hawking radiation is to see it as something that forms just outside of the black hole, a particle/anti particle pair and one escapes and the other falls inside, and that's probably not 100% correct either, but the singularity itself doesn't send out particles.   Hawking radiation has to do with quantum properties of space.  It's not a property of black holes.  The black hole just happens to be unique in that it can capture one half of a virtual particle pair and the other half can escape.

This also is a pretty different topic than your original question.

  *Where movement towards the singularity would be considered a positive value, movement away from the singularity is a negative value, that
  is, anything moving at the speed of light away from the singularity
  would be moving with a velocity of -C relative to the singularity.
  ✝If it couldn't accelerate from -C to 0 instantly, any photon
  traveling exactly away from the black hole would be able to escape the
  event horizon.

  **An instant is an arbitrary amount of time, it could be a fraction of a second, a second, a minute....

I think it's a good idea to differentiate mass-less pure energy particles and particles with mass.   You seem to be saying that a ray of light can be traveling away from a black hole at the speed of light, get caught in the gravity, slow down and then fall back into the black hole like a ball that's tossed straight up into the air from the surface of the Earth.   That's probably not what happens.   The ray of light follows the path of space time ahead of it, which happens to be curved so much that it points into the black hole, even if, in the classical sense, the light begins by pointing away.   All space curves into the singularity once you're inside the event horizon, so there is no "away from" anymore.   At least, that's how I think it works.


Is this star group a constellation recognized by the IAU? Below is an arrangement of stars I've found in a TV series. As an amateur astronomer I cannot identify if this is a constellation, after having looked through what is recognized by the IAU. Is this just a depiction of a random group of stars?

The IAU recognizes 88 constellations. The charts for each constellation can be found at the bottom of this page from the IAU website. You can go through them yourself, but I haven't found any that really resembles your image. So it probably isn't a constellation recognized by the IAU.

But it could still be part of the sky, without being a constellation. So I also ran the image through astrometry.net, but it didn't find what field the image could belong to.

This suggests that it is just a random group of stars, rather than an actual image of the sky.


which or what kind of star has a very stable luminosity? I mean those kind of stars should be very stable and barely have activity.

Their light curves should be very very flat.

Our sun, as a G type star, are not so stable. At lest there are lots of floating solar spots.

Hipparchos numbers of the 26 most stable stars known:
2021, 2854, 5542, 16611, 19747, 24927, 32537, 38414, 42913, 45556, 50191, 57363, 71053, 73555, 74666, 76440, 74946, 90139, 94648, 96052, 10239, 102488, 104732, 111169, 116631, 118322.

Compare Hpmax and Hpmin: They differ 0.01. Standard error e_Hpmag is 0.0003 mag.
There are 681 stars in the Hipparchos catalog with an amplitude of 0.01 mag.

More detail in this paper, including a link to the appropriate VizieR query.


Explosions of black holes I was bopping around YouTube and observed this enjoyably produced video. In it, when describing the behavior of a black hole with the mass of a US nickel, the narrator says, "Its 5 grams of mass will be converted to 450 terajoules of energy, which will lead to an explosion roughly three times bigger than the bombs dropped on Hiroshima and Nagasaki combined."

Of all the fun things illustrated there, that was the one whose pretense I didn't understand. Do black holes explode after they've radiated away all of their mass? Or would the "explosion" just come from the rapid pace at which the black hole would consume nearby matter?

The Googling I've done thus far has provided no firm answer. The closest I've come is from the Wikipedia page on Hawking Radiation, which states, "For a black hole of one solar mass, we get an evaporation time of 2.098 × 10^67 years—much longer than the current age of the universe at 13.799 ± 0.021 x 10^9 years. But for a black hole of 10^11 kg, the evaporation time is 2.667 billion years. This is why some astronomers are searching for signs of exploding primordial black holes."

Some other websites refer to the last "explosion" of the Milky Way's supermassive black hole being some 2 million years ago, but is that the same mechanic mentioned on the Wikipedia page? Or the YouTube video, for that matter?

Thanks in advance. =)

What this video is talking about is Hawking Radiation, as you've linked. Hawking Radiation is a proposed hypothetical (by no means verified or proven) way for a black hole to radiate its energy into space. The basic idea is that a black hole is nothing but mass/energy compressed to an infinitesimal point, which is radiating its energy into space over time. For large black holes (such as solar mass or bigger), this radiation process is tiny and the time taken to leak all the black hole's energy into space (and thus for the black hole to "evaporate") is exceedingly long. For tiny black holes however, the time to radiate all the black hole's energy is exceedingly short.

You can calculate how long it will take for a black hole of mass $m$ to evaporate (and release all its mass/energy) with the equation

$$t_{\text{ev}} = \frac{5120\pi G^2 m^3}{\hbar c^4} = (8.41\times 10^{-17}\:\mathrm{s}\:\mathrm{kg}^{-3})\:m^3$$

For $m = 5\:\mathrm{g}=0.005\:\mathrm{kg}$, you get $t_{\text{ev}} \simeq 4\times10^{-19}\:\mathrm{s}$. Now that means in this tiny amount of time, the black hole will radiate all of its mass/energy away and completely evaporate. But the output of all the energy in 5 g of mass is a huge output. Putting out 450 Terajoules of energy in $10^{-19}\:\mathrm{s}$ is basically just an explosion. You can determine the total energy output from the famous equation

$$E=mc^2$$

Just plug in $m = 0.005\:\mathrm{kg}$ and $c = 3\times 10^8\:\mathrm{m/s}$ and you'll get $E=4.5\times 10^{14}\:\mathrm{J} = 450\:\mathrm{Terajoules}$.

So in short, hypothetical calculations (not even theory at this point) suggest that a tiny black hole with the mass of a nickel would immediately explode out in a huge amount of energy. Whether such a black hole can form, or if such an evaporation would/could occur is still highly debated and ultimately unknown at this point.


Conflicting information about the star Delta Pavonis? Delta Pavonis has about the same mass as the sun so I would think that the evolutionary path would be about the same (the mass is given as O.991 M(Sun)/ not sure how we get this so accurately). It has been estimated to be about 6.6 to 6.9 Ga with a luminosity of 1.22 L(Sun). This all makes sense if you put it on the main sequence where we will be in a couple of billion years (after T(max) in about a billion years). The problem is that the surface temperature is only 5550K (spectral) and for this reason is thought to be in subgiant phase which isn't supposed to happen to us for about another 5.5 billion years. Delta Pav also has a very high metallicity [Fe/H] of +0.33 which also screams "subgiant". What am I missing?? (data from Wikipedia sources)

To answer a question, we do need the source of information (wikipedia just isn't good enough).

According to the SIMBAD database, delta Pavonis is a G8 subgiant, with a temperature of 5512K and a $\log g$ of 4.23 (in cgs units) (Gray et al. 2006), which looks correct for a subgiant classification. The metallicity is given as $+0.13$ in the same source.

There are a large number of other spectroscopic determinations of the star's parameters listed in the SIMBAD measurements. All seem to agree on the subgiant nature, but for example Bensby et al. (2014) makes the star a little warmer (5635 K) and more metal rich (+0.37).

So what appears to be at odds with this is the age/mass combination that you quote.

The wikipedia entry and information you quote appears to be taking information from Takeda et al. (2007). They estimate the age and mass of the star from the temperature and gravity using a Bayesian fitting method and the YREC stellar evolution models.

If this is really important to you, then you need to read the Takeda et al. paper. They show that the posterior probability distribution of age and mass can have multiple peaks. In the case of delta Pavonis it appears that wikipedia has got things wrong (what a surprise). The age that is quoted is actually for a secondary peak in the posterior probability distribution. The secondary peak in the mass distribution is for $M= 1.101M_{\odot}$. I think that age, mass pair makes much more sense.

They do not provide detailed information/discussion on individual cases. Like you, I cannot see how they arrive at 1 solar mass for a star of any age, unless they have picked up a pre main sequence solution in the HR diagram - for which the age corresponding to 1 solar mass would be very small. However, they claim that their models start at the ZAMS. The fact that they list no primary peak in the age posterior pdf suggest to me that the pdf look like a declining function from zero (the ZAMS), with a secondary peak at the 7Gyr (quoted by wikipedia). This then suggests that this secondary solution is also not a brilliant fit and that the fitting process would really have liked to make this younger than ZAMS - i.e. a PMS star (which it isn't). In turn then, this may suggests that the gravity used by Takeda et al. might be too low.

Takeda et al. used spectroscopically derived parameters of temperature 5590 K, $\log g= 4.31$ and $[M/H]=+0.26$ from Valenti & Fischer (2005). These authors also did their own isochrone fits to these parameters finding $M=1.045 \pm 0.058 M_{\odot}$ and an age of 5.3-6.8 Gyr. They say that the mass at the best-fitting isochronal age is $1.07M_{\odot}$.

I think part of what is puzzling you is that you have to remember that having a metal content that is twice the Sun's does affect the evolutionary tracks considerably. It is not that the evolutionary timescales are massively different, but the effective temperature of the star is changed considerably for a given mass and age.


Aren't there more naked-eye-visible stars in the Milky Way plane? Most stars which are visible to the naked eye are within 1,000 light years. The Sun is inside the Orion arm which has a diameter of about 3,500 light years. Thus, all stars (with very few exceptions) we can see unaided are inside the Orion arm and should be equally distributed in all directions. The structure of the galaxy is too large to affect the distribution of visible stars.

Still, we do see the Milky Way disk. What is it we see? The dim light of stars we can't make out individually? Or star light reflected by gas clouds? And is the density of stars to the unaided eye really not higher in the plane of the Milky Way? (Since I'm living in a light polluted city I can't easily check it out myself)

You can tell a lot about Galactic structure by just looking. The ~5000 stars that can be seen with the naked eye have a roughly "lognormal" distribution of distance. I show plots below which were generated from the most recent version of the Hipparcos parallax catalogue. Fig.1 shows results for all stars with $5.5<V<6.5$  (i.e. very faint naked eye stars). The Hipparcos catalogue is almost complete for these stars, though some stars are so far away that the distance is highly uncertain - nevertheless, the general picture should be ok.

The median distance is about 440 light years. But the premise of your question is I think, that you dispute that this is far enough for the non-spherical distribution of stars in our Galaxy to become apparent. The answer is actually that it is, but only just. The Sun is very close to the plane of the disk of our Galaxy. The scale height of stars above this disk varies depending on stellar age and mass. Very approximately, the exponential scale height is 300-500 light years for most of the stars in our Galaxy.

This is just small enough that if we look at the demographics of stars in two Galactic latitude regions we do see a difference. Fig. 2 below shows stars with $5.5<V<6.5$ at low Galactic latitudes (within 15 degrees of the plane in green) and more than 45 degrees from the plane (in blue). There is a clear and significant difference. More stars can be seen at greater distances towards the plane, betraying the non spherically-symmetric nature of stars in our Galaxy.

In other words the structure of the Galaxy is not too small to affect the distribution of naked eye stars.

When we look at the Milky Way we are seeing millions of unresolved stars that are in general more distant still. The effects of the disk-like nature of the Galaxy become more apparent as one goes to greater distances. In particular, at high Galactic latitudes one simply runs out of stars and hence there is no uniformly illuminated sky - leading to Olber's paradox, as correctly stated in another answer. But at low latitudes, there are sufficient stars that in general, within the resolution of our eyes there are many stars whose light sums to provide a visual stimulus. This picture is interrupted by dust. Dust in the Galaxy is even more concentrated towards the plane than the stars. It is for this reason that beyond a few thousand light years, dust plays a major role in shaping the Milky Way structures that we can see, effectively blocking light at very low Galactic latitudes.

Fig 1: Probability distribution of distance for naked eye stars in the Hipparcos catalogue

Fig 2: Probability distribution of distance for naked eye stars divided into low and high Galactic latitude regions (i.e. towards and away from the Galactic plane).

IMPORTANT EDIT:

After the answer was accepted I did a bit of double checking. I split the Hipparcos sample into bright ($V<6$, 5000 stars) and very bright ($V<3$, 173 stars) and looked at the distribution of stars per unit area on the sky as a function of Galactic latitude (Fig. 3). The results are shown below and more obviously make the point. It turns out that the asymmetry between in an out of the plane is clearly seen, even in the brighter sample. The reason is that the very bright stars are not all that much closer than the bright stars. Perhaps a median of 200 light years vs 350 light years. Thus the scale height of the Galactic disc must be small enough that the difference from spherical symmetry is already appearent at these kinds of distances.

Fig. 3: Normalised plot of number of stars per unit area versus Galactic latitude for a sample of bright and very bright stars. Note the concentration towards the Galactic plane.

FURTHER EDIT: You can even see from the plot above that there is also a slight concentration towards negative latitudes; the peak density is around -5 to -10 degrees in both cases. This is likely because the Sun is above the Galactic plane at present (though I also wonder if dust plays a role). The Sun is currently 60 light years above the plane and heading upwards (see How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it? ). The combination of the results I have shown may be sufficient to conclude that the Galactic disc has a "characteristic thickness* not more than a few hundred light years, but more than 60 light years!


On gravitational wave radiation and arrangement of galaxies post- big bang Orbits of planets and stars decay due to gravitational wave radiation. An elliptical orbit would become more circular with time. This is especially observed well in binary systems.

Taking the example of binaries, we can compute the rate of change of eccentricity of the orbit and the rate of change of the semi-major axis as a function of time. If we consider a highly elliptic orbit of a binary, with eccentricity, $\epsilon\rightarrow{1}$, we can prove that with time the orbit would eventually become a circular one, as follows:

  The power emitted by gravitational waves is given by:
  $$P_{GW}= \frac{c^5}{G}\left(\frac{GM}{c^5l}\right)^5$$
  Very compact binaries will lose energy rapidly by GW radiation.
  If we assume that the two bodies making up the binary lie in the $x-y$ plane and their orbits are circular ($\epsilon=0$), then only non-vanishing components of quadrupole tensors are:
  $$\boxed{Q_{xx}=-Q_{yy}=\frac{1}{2}(\mu)a^{2}\cos2\Omega t}$$ and
  $$\boxed{Q_{xy}=Q_{ya}=\frac{1}{2}(\mu)a^{2}\sin2\Omega t}$$
  Where $\Omega$ is the orbital velocity, $\mu=\dfrac{m_{1}m_{2}}{m}$ is the reduced mass and where $m=m_{1}+m_{2}$
  The luminosity of the system can be deduced as:
  $$L^{GM}=\frac{32}{5}\frac{G}{c^5}\mu^2 a^4\Omega^6=\frac{32}{5}\frac{G^4}{c^5}\frac{M^3 \mu^2}{a^5}$$
  The latter part is obtained from Kepler third law: $\Omega^2=\frac{GM}{a^3}$
  As the gravitating system loses energy by emitting radiation, the distance between the 2 bodies shrinks at a rate:
  $$\boxed{\frac{da}{dt}=\frac{64}{5}\frac{G^{3}M\mu}{c^5 a^3}}$$
  The binary would hence colase at a time :
  $$\tau=\frac{5}{256}\frac{c^5}{G^3}\frac{a_{0}^4}{\mu M^4}$$

  By applying a similar treatment to elliptical orbits , the following can be computed:
  $$\left<\frac{da}{dt}\right>=-\frac{64}{5}\frac{G^{3}m_{1}m_{2}(m_{1}+m_{2})}{c^{5}a^{4}(1-e^{2})^{\frac{7}{2}}}\left(1+\frac{73e^2}{24}+\frac{37e^4}{96}\right)$$

  $$\left<\frac{de}{dt}\right>=-\frac{304}{15}\frac{G^{3}m_{1}m_{2}(m_{1}+m_{2})}{c^{5}a^{4}(1-e^{2})^{\frac{5}{2}}}\left(1+\frac{121e^2}{304}\right)$$

  Solving this system of ODE's ultimately results in the equation:

  $$T(a_{0},e_{0})=\frac{12(c_{0}^4)}{19\gamma}\int_{0}^{e_0}{\frac{e^{29/19}[1+(121/304)e^2]^{1181/2299}}{(1-e^2)^{3/2}}}de\tag1$$
  Where $$\gamma=\frac{64G^3}{5c^5}m_{1}m_{2}(m_{1}+m_{2})$$
  Upon solving this one can find the time taken for the orbit to decay into a circle from an ellipse due to gravitational wave radiation.

A similar treatment can be done for the orbital decay of the planets of our solar system. Now my question is if we apply time reversal symmetry, the orbits become more and more elliptical and eventually tend to become a straight line(an elliptical orbit with $\epsilon\rightarrow{1}$) with a chunk of mass(moving in the straight line) from which all the planets and the Sun  of our solar system arose. Now the Milky Way "orbits" the Local Group, which in turn "orbits" the Virgo Supercluster.
Beyond that, the expansion of the universe starts to dominate over gravitation.

So reversing time to a space after the Big Bang, there must have been many galaxies and solar systems that were forming; so according to the above result the mass chunks from which galaxies, stars and planets arose must have had a highly elliptic orbits. Thus, the spread of matter in space must have been along approximate lines(which were orbits for the massive mass chunk a that formed them). This moves towards a linear arrangement of galaxies, contrary to what is observed. Can anyone please resolve my doubt. Any help is appreciated.

When you go back in time, take collisions into account. A very highly elliptical orbit primarily leads to chunks colliding at the very first periapsis.

Even if they are not chunks but very dense and small objects (point-like), if you extrapolate back in time so far as to the straight line "orbit", what does it mean? Extreme case of objects with zero relative initial velocity going against each other on a collision course - ups, this "orbit" does not have a chance to "decay due to gravitational wave radiation". Two smaller chunks collide and form a larger chunk, which has now a non-zero velocity (probably, if the chunks weren't exactly equal momentum). The chunk won't enter a straight line "orbit" around (towards) anything, but either an elliptical or hyperbolic.

Your extrapolation does suggest that initial "chunks" hadn't ideally equal momenta (masses*velocities); that's all.


What is the temperature inside a Black Hole? Is it insanely hot? Cause of so much gravity?
Is it close to absolute zero? Cause the matter is so closely packed, there is hardly any space for particles to move?
Is it room temperature? Cause Cooper didn't die in Interstellar when he went into a Black Hole?

The temperature of a black hole as seen from the outside is the temperature of the Hawking radiation, which @jyoti has discussed.

The temperature you would experience as you fell freely through the event horizon, is a different question. I think the answer depends on what else is falling into the black hole at the same time. From its own viewpoint, everything that falls into a black hole reaches the singularity in a fixed maximum time (a few days for the very largest supermassive black holes) so you could only exchange energy with things that fell in around the same time as you did. If there was very little such stuff, then I don't think you'd notice anything out of the ordinary -- you'd be exposed to the cold of space, just as you were before (assuming the black hole is big enough that you haven't been heated up the tides of the black hole distorting you. On the other hand if lots of stuff was falling into the black hole with you then the whole lot would get churned up, and you'd be falling in as part of a mass of very hot plasma. Eventually, at least in a non-spinning black hole, you'd approach the singularity and be ripped apart by tides.


How distorted will our galaxy be if we are viewing it from several thousand light years away? I saw this answer and read this sentence regarding the distortion of our galaxy when viewed above the galactic plane:

Once you got to a point where the entire [sic] galaxy was within your vision, the entire galaxy would look distorted, since our galaxy is 100,000 light years [sic] across, the light from the edges of the Galaxy would be thousands of years old, and would not match up with how the spiral should appear.

I am thinking that this could potentially be true, but I am not sure why or how distorted it would be. Does this fact hold, and for what distances will it no longer be distinguishable?
Note: I have thought that it will be well-blended because the distance does not spike, it increases gradually, but I need verification.

The light travel time of 100,000 years is quite small compared to the time it takes the Milky Way's spiral arms to complete an appreciable fraction of one rotation. The arms have a pattern angular speed of $\Omega_{\text{p}}=28.2\pm2.1\text{ km}\text{ s}^{-1}\text{ kpc}^{-1}$ (Dias et al. 2019), so they should complete one full rotation on the order of $\tau=2\pi/\Omega_{\text{p}}\approx218\text{ Myr}$.
An observer a distance $h$ above the center of the galaxy would observe light coming from a radius $R$ to have to travel an additional time
$$\Delta t=\frac{1}{c}(\sqrt{h^2+R^2}-h)=\frac{R}{c}\left(\sqrt{\left(\frac{h}{R}\right)^2+1}-\frac{h}{R}\right)$$
which is a decreasing function of $h/R$. Its maximum is at $h=0$, where it would be on the order of 100,000 years for stars on the outer edge of the disk. Clearly, $\Delta t\ll\tau$, so there would not be an appreciable distortion, per se, because of the additional distance. While it's true that the objects you'd see would indeed have moved some amount by the time the light reached you, it wouldn't be significant.


What type of planetary-mass object would Planet Nine be? Given that Planet Nine is hypothesized to be 10-20 times the mass of the Earth orbiting the sun some 200-2500 AU out, what kind of planetary-mass object would it be?

Even given that it has apparently "influenced" several Trans-Neptunian objects, since its orbit is so large and so slow (approximately 25,000 years) it would take an awful long time to clear out its own orbit (especially given that we do not know how long it has been in that orbit), so it seems like it would not qualify as an actual planet under the rules that re-classified Pluto.

Since it orbits our sun, it wouldn't be an exoplanet nor a rogue planet.  And it sure seems too big to be called a dwarf planet.

A captured planet seems closest in other attributes, but it probably was not either a capture nor a re-capture, but more likely a partial ejection.

So what classification would it fall into?

It is a planet - either a gas giant core, a mini-Neptune, or a super-Earth.

First off, Mike Brown has stated outright

  “It is a planet—there’s virtually no doubt,” he said. “What we now call planets are objects that can gravitationally dominate their neighborhood. Pluto is a slave to the gravitational influence of Neptune. By area, Planet Nine dominates more of the solar system than any other known planet—it’s only because of this that we can infer its existence. And because of this we’re pretty sure it’s not a small object: it’s at least ten times more massive than Earth and five thousand times more massive than Pluto. In many ways, you could argue that this is more of a planet than anything else in the solar system.”

I wrote an answer to How was the hypothetical ninth planet kicked so far out of the Solar System? that discussed the possibility that the ninth planet is the hypothetical 5th gas giant proposed by Nesvorný (2011). The details can be found in that answer, but basically, Brown & Batygin's reference to previous papers by Morbidelli et al. suggest that they think that this is what remains of an ice giant, like Uranus and Neptune, that formed near Saturn and Jupiter but was ejected in a series of gravitational interactions that led it to the Kuiper Belt and beyond.

That said, Batygin's estimates place the ejection of this planet way before previous models of a 5th gas giant predict, so either we accept models with a  6th gas giant (also an ice giant) and say that both of these were ejected independently of one another - not impossible - or we go for some other option. The bottom line is that this is a planet, if we go with the above assumptions.

Two other guesses have been made based on the planet's mass, estimated by Brown & Batygin at about 10 Earth masses. One is that it is a very massive super-Earth, hanging around the proposed upper mass limit for those objects. This implies that the planet is rocky, but does not imply anything else. The other guess is that it is a mini-Neptune, with a predominantly gaseous composition but at about the same mass. The super-Earth hypothesis arose in part because of claims a long time ago about the ejection of a 5th terrestrial planet.

I've used the term "planet" throughout this entire answer, and that's because this object may be one - and most likely was in the past. Let's look at the IAU's three major criteria to be a planet:

It orbits the Sun. This is definitely true, if it exists. Also note that if this is a captured rogue planet, as some have proposed, then we can still call it a "planet", because it now orbits the Sun.
It is in hydrostatic equilibrium. This is most likely true, and is true if any of the three models I suggested above are correct.
It clears its neighborhood. This is a bit iffy. It clearly influences objects around it, but we don't know much about those influences besides observations of the six Trans-Neptunian Objects (TNOs) Batygin & Brown studied. That said, an object with a mass of 10 Earth masses will most likely clear the neighborhood around its orbit if it has a stable orbit, which should be the case here. See the quote at the start of this answer for more information.


Figure out the location of celestial objects relative to the Milky Way in x, y, z co-ordinates I am trying to place various celestial objects the correct distance from the Milky Way, using x, y, z co-ordinates.
So assuming the centre-of-mass of the Milky Way is at 0, 0, 0 - how can I figure out where to place a galaxy or star given it's celestial coordinates?
I'm just an amateur, and pushing 60 years old, so please excuse me if I have something wrong here...

The spherical $(r, l, b)$ can be converted to
rectangular
$(U, V, W)$ as:
$$\begin{align}
U &= r \cos b \cos l \\
V &= r \cos b \sin l \\
W &= r \sin b
\end{align}$$
Using $(r, l, b) = (770 \pm 40 ~\text{kpc}, 121.2^\circ, -21.6^\circ)$,
$$\begin{align}
U &= -371 \mp 19 ~\text{kpc} \\
V &= ~~~ 612 \pm 32 ~\text{kpc} \\
W &= -283 \mp 15 ~\text{kpc}
\end{align}$$
That's relative to the Sun.
For a position relative to the center of our galaxy,
subtract $8 \pm 0.5 ~\text{kpc}$ from $U$.


what would the mass and density of Sirius A be if the mass of Sirius B is roughly the same as the Earth? I am writing a blog post and while I am interested in astronomy I havent done any physics like this since high school. My question is: what would the mass and density of Sirius A be if the mass of Sirius B is roughly the same as the Earth? If Sirius B is actually only as heavy as the Earth then Sirius A would need to be far less dense. How dense would it need to be to be the same size it currently is and what would its mass be?
The orbit of Sirius A and Sirius B has been observed. But the true mass and density of Sirius A and Sirius B cannot be known with certainty like with the orbit. So, it is possible that Sirius B actually has the mass and density roughly equal to the Earth. The light we see when we look at Sirius B is some kind of alien technology they are using to block out cosmic radiation.
If this fictional scenario were the case then what would the mass and density of Sirius A need to be in order for the orbits to be the same given that Sirius B has the mass and density roughly equal to the Earth?
Additionally, how dense would Sirius A be compared to Betelgeuse in this scenario?

The orbital motion reveals that Sirius A, the brighter star, has 2.02
± 0.03 solar masses and Sirius B, the white dwarf, has 1.00 ± 0.02
solar masses.
Betelgeuse has a radius 1,100 times that of the Sun but its mass is
only about 8 – 20 times the Sun. This means the density of Betelgeuse
is much, much lower than the Sun. The average density of Betelgeuse is
about a million times less dense than Earth’s atmosphere at sea level.
That’s about the same as a vacuum found in an insulating Thermos
bottle.

The masses of the two objects are completely determined by a careful measurement of their relative positions on the sky over decades, combined with an accurately known distance.
There is no degeneracy in this solution. For a given orbital period and separation of the two objects, the masses are entirely determined. If Sirius B were not a massive white dwarf then the orbit would be different.  Specifically, if you made the total system mass smaller, but with the same orbital period, then the size of the orbit would be smaller. If you tried to compensate by adding mass to Sirius A, then it would almost coincide with the centre of mass of the system and would show almost no orbital motion.
The density is then found by combining these masses with radii estimated from the measured luminosities and temperatures of the components and is therefore also fixed (to about $\pm 10$% because of the observational uncertainties in the mass and radius determinations).


How do we know what the biggest star is? Kurzgesagt claims that the largest (observable) star in the universe would be Stephenson 2-18 which is in line with Wikipedia:

It is among the largest known stars, if not the largest, and one of the most luminous red supergiants, with an estimated radius around 2,150 times that of the Sun, which corresponds to a volume nearly 10 billion times bigger than the Sun.

Furthermore, Wikipedia says

The open cluster Stephenson 2 was discovered by American astronomer Charles Bruce Stephenson in 1990 in the data obtained by a deep infrared survey.

I stumbled upon that statement for two reasons:

The survey was made in 2010, so if we again would make a survey with newer, higher resolution data (which I assume we might have in the mean time), would we find other objects even bigger?
How are we sure that St2-18 was indeed the largest object within the given survey? Technically speaking, one would have to calculate at least the distance or lumnosity for each object of the survey?

In a nutshell: How sure are we (in percent) that St2-18 is right now, in 2020, still the largest observed star?
Related

What is the largest hydrogen-burning star?
Is there a theoretical maximum size limit for a star?

As @uhoh stated, there is a crucial distinction between what is in principle "observable" and what we have to-date "observed."
Also, you have to specify whether you're talking about biggest in radius or biggest in mass. For stars with very large radii, like the St2-18 system you've referenced, they often can't even constrain the mass! And when they are able to, they end up having pretty small masses, i.e. UY Scuti. Stars, especially high-mass stars, generally increase in radius as they age, but they lose mass as they age. So it's not surprising to find red supergiants with very large radii (because they're at a late stage in their stellar evolution) and with small masses. But the theory behind supergiant stars is very complicated and uncertain still, making mass measurements very very difficult (And thus are not usually attempted) since mass measurements require some kind of binary interaction or a well-understood model of the stellar atmosphere (which is lacking for supergiants as I understand).
Regarding what's "observable," keep in mind that this is a fast-changing landscape: for example, gravitational wave detections of intermediate-mass black holes suggest that they possibly form from core-collapse of super-massive stars (among other possibilities). And the link to the question "what is the size limit theoretically of a star" that you provided has many answers which explain how uncertain such a theoretical limit is for mass and for radius.
Regarding "observed" systems, you must take into consideration the error bars on mass measurements of stars. The largest observed star with a known mass that has respectable error bars is the Wolf-Rayet star R136a1. See here for a comprehensive list.
With that said, you asked,

How sure are we (in percent) that St2-18 is right now, in 2020, still the largest observed star?

Of course there is always the possibility that with a new survey with better resolution, larger field of view, etc... even larger stars may be in the universe. It's a zoo! We were surprised to find stars as big as St2-18, so why would the universe stop there, eh? There COULD BE SOME reason in principle that disallows larger stars, but the universe keeps providing bigger and bigger stars for us to ponder, so there's no reason a priori to assume so. Regarding the survey you're asking about in particular, the wiki article has a nice discussion explaining the great uncertainty in determining the radius of St2-18, i.e. do we rightly consider it as part of the cluster or not in order to make assumptions about its distance, etc...:

A calculation for finding the bolometric luminosity by fitting the Spectral Energy Distribution (SED) gives the star a luminosity of nearly 440,000 L☉, with an effective temperature of 3,200 K, which corresponds to a very large radius of 2,150 R☉ (1.50×109 km; 10.0 au; 930,000,000 mi),[a] which would be considerably larger and more luminous than theoretical models of the largest, and most luminous red supergiants possible (roughly 1,500 R☉ and 320,000 L☉ respectively).[11][6] An alternate but older calculation from 2010, still assuming membership of the Stephenson 2 cluster at 5.5 kpc but based on 12 and 25 μm fluxes, gives a much lower and relatively modest luminosity of 90,000 L☉.[7] A newer calculation, based on SED integration and assuming a distance of 5.8 kpc, gives a bolometric luminosity of 630,000 L☉ although the authors are doubtful that the star is actually a member of the cluster and at that distance.

Lastly, you asked

How are we sure that St2-18 was indeed the largest object within the given survey? Technically speaking, one would have to calculate at least the distance or lumnosity for each object of the survey?

I quickly looked at the survey report paper, and it seems that this is actually unclear. Keep in mind that they were surveying a few different clusters, the Stephenson#2 being only one of them. The tables at the end of their paper make that clear. However, they only list the technical properties, and do not even attempt to constrain the radii for all of these stars. So I'd take it with a grain of salt.


How do the orbits of Nu Scorpii and AR Cassiopeiae work? I apologize for the way this question is worded but I don't think I know the proper verbiage. Basically, I want to know how the stars orbit one another in the two septuple star systems. For example, are there two massive stars in the middle that have the five remaining stars orbiting similar to the planets of our system; does one of the orbiting stars have one or more smaller star-"moons;" maybe there are stars trapped in one of the Lagrange Points. I hope that clears up the question.

I am aware that the orbits of these systems may not be known/understood. If that is the case, then I would appreciate the orbital relationships between the stars in the highest-multiple star system. If that is confusing, I mean, if we know the orbits for the stars in a sextuple system; if not then a quintuple... etc.

Finally, as an added bonus, I would appreciate being told the correct verbiage. Thank you, much.

I checked the Multiple Star Catalog for data on these systems. It has hierarchical information on Nu Scorpii and AR Cassiopeiae, which I also used for this answer. I'll try to explain the orbital structure of both systems.
Nu Scorpii
Think of Nu Scorpii as two subsets of stars, Nu Scorpii AB and Nu Scorpii CD, orbiting each other with a period of 105,000 years. AB contains four stars:

Nu Scorpii Aa, Ab and Ac: Aa and Ab are a binary system (period: 5.5 days) orbited by Ac (period: 8.3 years).
Nu Scorpii B: This star orbits Nu Scorpii A - the three stars above - with a period of 675 years.

Nu Scorpii CD contains three stars:

Nu Scorpii C: C is a single star.
Nu Scorpii Da and Db: Da and Db are a binary system (period: unknown), which together orbit C over 3,100 years.

AR Cassiopeiae
AR Cassiopeiae is more complicated; it contains three main subsets:

AR Cassiopeiae AB: The contains three stars: Aa and Ab, a binary orbiting each other every 6 days, and B, orbiting A every 545 years.
AR Cassiopeiae CD: These pair of binary stars orbit each other once every 2,200 years, and orbit component AB in 438,000 years.
AR Cassiopeiae FG: This is another binary, with a period of 58,000 years. It orbits AB every 381,000 years.

Multiple star systems are usually nested like this - binaries and triple systems inside more binaries. I would image it's rare to have any other setup, like four stars orbiting together, because such a system would be highly unstable. You really need to nest the stars, with large separations between components (i.e. the distance between Nu Scorpii AB and CD must be much larger than the distance between Nu Scorpii A and Nu Scorpii B).


Obtain IR diffuse data from WISE I need some help obtaining IR background diffuse data from the WISE observation. There is already some work done in this regard; see: http://wise.skymaps.info/ . These are .fits files. How do I find out the photon irradiance for a range (say glat b/w 0 and 30) from these files? Any help is appreciated. Also, how do I correlate this with UV flux for the same range using python?

Assuming you have downloaded their software from the link you provided and have astropy installed, you should be able to do:

from wssa_utils import wssa_getval
from astropy import units as u
from astropy.coordinates import SkyCoord

# Declare your wanted point in Galactic co-ordinates
c = SkyCoord(l=0, b=30, unit=(u.deg,u.deg), frame='galactic')
ra = c.icrs.ra
dec = c.icrs.dec
vals = wssa_getval(ra, dec)

The results look like they will be in MegaJanskys per steradian, which would be u.MJy/u.sr in astropy units.
For UV data, I'm not sure what is best, maybe GALEX ?


Good eyepieces for Saturn and Jupiter I have a celestron powerseeker 114eq telescope. Aperture : 114mm Focal Length: 900mm
I have 20mm and 4mm eyepieces and a 3x Barlow. I have seen Saturn's rings using 4mm eyepiece but it's blurry and with 3x it becomes impossible to see any detail. With 20mm with 3x barlow they are too small. I found a kit on amazon with 6mm and 12.5mm kellener eyepieces with a moon filter. I wanted to view a little larger than 20mm with 3x barlow and with details like color contrasts. Could anyone tell whether the kit that I found will be useful or anything else to suggest to solve my problem?

From my limited experience, being able to see the rings on Saturn is more or less the extent of how powerful eyepieces can get for a home telescope. To get better and larger images would require a better and probably larger telescope.


face-on galaxy and edge-on galaxy I already googled a face-on galaxy and an edge-on galaxy, but I couldn't find any proper information. Would you explain them to me?

The terms refer to the viewing angle, i.e. from which direction do we observe a (disk) galaxy. If we happen to be located roughly in its plane of rotation, we see it from the "edge", whereas if we are more or less above or below its plane, we see its "face".

To make a more quantitative statement, the orientation is described by the inclination angle $i$, which is the angle away from "face-on". That is, the inclination of a galaxy that is observed face-on is $i=0$, and the inclination of a galaxy observed edge-on is $i = 90º$. For randomly oriented galaxies, the mean inclination angle will be around $i = 60º$.

Wrt. your second question, you're right that the text is not correct. The three small constellations lie north of hydra. "North" means "toward higher values of declination". Perhaps it's more easily seen in this image (from Wikipedia):


If we stood on the surface of a Jovian moon, Saturn would appear as a dot. Why doesn't Saturn appear like this in a telescope? Images from the conjunction like these show Saturn just a bit smaller than Jupiter. However, if you were in the vicinity of Jupiter, Saturn would still appear as a dot to the naked eye, wouldn't it? If so, why is Saturn recognizable as a planet through a telescope in which Jupiter is also seen as a planet? As if they were actually very close to each other, at about the same distance from the Sun. I mean, if you recognize Jupiter in a telescope, shouldn't Saturn behind it look like a dot, unless you have a much, much larger zoom in which Jupiter would appear too close to match into the whole image while being too unsharp anyway?

Telescopes magnify, they don't bring you closer.
So if from Earth Jupiter has an apparent radius of 0.01 degrees (measured as an angle because it is the apparent size)
And if Saturn has an apparent angle of 0.005 degrees, then if you magnify 100x then Jupiter will have an apparent size of 1 degree, and Saturn would have a size of 0.5 degrees.  Magnification just increases the angular size in proportion
But if You go to Jupiter you have travelled less than half the distance to Saturn.
So saturn is still small. Travelling closer does not make everything increase in size in proportion.
You can see this simply:  Stand where you can see something 20 metres of so away, and where you can see into the hills (etc) in the background.  Walk towards the thing.  Note that the thing appears to get bigger as you approach, but the distant hills don't change size.  Travelling closer does not magnify everything in proportion.
Telescopes don't "bring things closer" they "magnify".


Why are event horizons described in Schwarzschild radii rather than diameters, and either way can they change equator to pole? I never heard of Schwarzschild diameters, only of Schwarzschild radii. Why is it like this? Wouldn't it be better to portray the (possible) size of a black hole in a diameter?
Either way, can we use the terms radius or diameter term precisely? Is an event horizon always perfectly spherical or can it have something like an equatorial bulge even though it's a black hole's immaterial event horizon.

We use the Schwarzschild radius $r_s$ (rather than a diameter) because it's convenient. We're want to describe what happens in the vicinity of a black hole, so it's natural to talk about the distance from the black hole.
For example, there's the photon sphere at $3r_s/2$ (for a Schwarzschild BH), and the innermost stable circular orbit or ISCO, the smallest circular orbit in which a test particle can stably orbit a massive object, at $3r_s$

The ISCO plays an important role in black hole accretion disks since it marks the inner edge of the disk. [...] Circular orbits are still possible between the ISCO and the photon sphere, but they are unstable.
For a massless test particle like a photon, the only possible circular orbit is exactly at the photon sphere, and is unstable. Inside the photon sphere, no circular orbits exist. [...] The case for rotating black holes is somewhat more complicated.

Another phenomenon which involves the distance from a BH is gravitational lensing. The amount of deflection of a light beam passing near a BH (or indeed any source of gravity) depends on the distance $r$ of the beam's path from the centre of mass. The deflection angle is given by $\theta=2r_s/r$. I guess you could use diameter in that formula, but it's more natural to think in terms of radius. It's a bit like shooting at a target (with light beams). When you shoot at a circular target, you're concerned with how far your shots are from the centre of the target.

As I said in the comments, our usual notions of geometry are not very useful in the highly curved spacetime near a BH. And if you want to get close to a BH, you need to be in an orbit travelling near the speed of light, and / or your spacecraft needs to be capable of the extremely high acceleration required to hover near a BH (and you & your ship need impossible strength to cope with the insanely huge g force resulting from such acceleration). Travelling at high speed adds additional relativistic effects on top of those GR effects due to the BH's gravity.
To describe the locations and times near the black hole, you need to choose a coordinate system. Schwarzschild coordinates are often used to describe a non-spinning black hole, but they have a coordinate singularity at the EH (event horizon), so they're annoying to work with when you want to talk about objects crossing the EH. It's a lot like how the latitude & longitude on Earth break down near the poles. You can't travel further south than the South Pole at 90°S, and the pole itself doesn't have a well-defined longitude. If I tell you that I'm 1 km north of the South Pole, I could be anywhere on a circle of 1 km radius.
Fortunately (as I mentioned in the comments), GR is very flexible regarding coordinates, and there are quite a few standard coordinate systems used in various circumstances, you can see a list of some of them on Wikipedia's page Category:Coordinate charts in general relativity. However, that flexibility makes things complicated, and difficult to describe properly without using advanced mathematics. So popular treatments of GR gloss over those difficulties (or avoid them completely), and that has led to a popular understanding of GR elements that is somewhat distorted.


What is the completeness of an observation and how do I calculate it? Many publications write about "completeness". I have a vague idea that it has to do with the ratio of how many sources from an observation are detected and how many sources are actually in that field. However, can someone please explain to me:
a) What is the definition of completeness?
b) How do you calculate the completeness of an observation?
As an example: Smolčić et al. 2008 are writing about completeness. Please have a look at A New Method to Separate Star-forming from AGN Galaxies at Intermediate Redshift: The Submillijansky Radio Population in the VLA-COSMOS Survey

a) What is the definition of completeness?

Completeness is the number of objects in a data set that are detected over the number that exist.  In astronomy, completeness is often estimated for a particular apparent magnitude or flux density.  As an example, for sources that are as bright as the Sun (-27 magnitude), we have a completeness of 1.  That is, we’ve discovered all sources in the celestial sphere with the Sun’s magnitude.  As we get down to very dim objects like tiny pieces of space junk in Earth orbits, our completeness ratio drops dramatically.

b) How do you calculate the completeness of an observation?

On a ROC curve, completeness is the True Positive Rate or the y-axis if all the objects have been observed (thanks @ProfRob!). If only a fraction of the objects have been observed, completeness is that fraction times the true positive rate. Note that a classifier only yields better completeness at the cost of higher false positive rates.  Here is an excellent example of ROC curves for different algorithms which differentiate quasars from stars:

The left panel shows data used in color-based photometric
classification of stars and quasars. Stars are indicated by gray
points, while quasars are indicated by black points. The right panel
shows ROC curves for quasar identification based on u - g , g - r , r - i
, and i - z colors.

If you don’t know the true positive rate, you may be able to run a simulation in which you inject objects into your data set and determine what portion of these are found by your algorithm.  This will give you an estimate of your completeness.  Matlab has a specific function for ROC curve Monte Carlo simulation, for example.


What is the apparent magnitude limit for the naked eye? If you only read the question, you will answer: ''That depends on the light pollution''. Yes, it does, but if you are in the darkest night sky (Bortle class 1), what magnitude would have the faintest star? In the Wikipedia article about the magnitude, it's written that +6.0 would be the typical limit, but in the article about the Bortle scale, I saw the fainter magnitude +8.0. So, what is the real limit?

If the question has been already asked, please give me the link.

The apparent magnitude classification was done arbitrarily by Ptolemy. His thought was to set the 20 brightest stars to the first position, the less bright stars to the second position and so on up to the faintest stars which were given the sixth position. After the use of Pogson's law we were able to give to stars not only natural numbers, but also numbers with decimal digits. When we started to use the first telescopes, we realised that the numbers which are greater than 6 have meaning. So the natural limit of naked eye is the apparent magnitude of +6. Here you can find a nice work from the italian Light Pollution Science and Technology Institute which shows as the typical limit of naked eye compared to the light pollution of each area.


Calculating orbits using observational data How did astronomers in the 18th and 19th centuries used to calculate a comet's or planet's orbit using observational data, given that this data is relative to a non static reference point (i.e. the Earth)?

For instance, here one can find the measurements of Mars' declination by Tycho Brahe throughout two decades. Not only it seems very difficult to me to deduce the right curve from the original observations points, but also the peaks of the curve (which resembles a sinusoid) do not seem perfectly equally distant from each other.

After researching a bit more, I think I can answer my own question (or at least share what I have found). When Kepler derived his laws using Tycho Brahe's observational data of Mars, he followed the procedure below:

first, the orbital period of Mars must be known. This value was already known in the time of Kepler (aprox. 687 days) $^1$, and can be derived as follows: first, one takes notes of the synodic period of Mars (which "[...] is the time that elapses between two successive conjunctions with the Sun–Earth line in the same linear order."$^2$). This can be done by calculating the elapsed time between two consecutive observations in which Mars is in the same apparent position in the sky. Then to calculate the orbital period of Mars, one can use:$\dfrac{1}{T_{M}} = \dfrac{1}{T_{E}} - \dfrac{1}{T_{M\mathrm{syn}}}$,$^3$ where $T_{M}$ is the orbital period of Mars, $T_{E}$ is the orbital period of Earth and $T_{M\mathrm{syn}}$ is the synodical period or Mars.
knowing the orbital period, Kepler then selected some of Brahe's observations that are exactly 687 days apart from each other (that is, one Martian year apart). This means that regardless of Mars relative position in the sky, the planet will be in the same position in relation to the Sun:

using triangulation, Kepler then calculated the distance of Mars to the Sun in terms of the distance of Earth to the Sun (1 AU). He did this for several positions of Mars.$^1$
while any three points define a circle, four or more do not necessarily do so. Kepler noticed that the collected data of distances between Mars and the Sun could be fit into an ellipse with the Sun in one of the focuses.


Star System Formation questions I'm making a game in which I'd like to randomly generate star systems. Which is cool.

The trouble I'm having is working out exactly what limits I need to place on the generation. So far, the wikipedia hasn't been all that helpful. So some questions:

Rocky planets form within the frostline of a protostar. What's the radius of this line? What I'm sort of asking is that given a particular mass and density, what radius from that star is the line.
If the jovian planets form outside of this line; why have so many been found within this line in other star systems from our own? Is this only selection bias or is it more likely that our models are incorrect at this point?
How does the creation of the gas giants differ from the creation of the ice giants? Is it simply like a second "frost line" where there wasn't enough gaseous material to form a large jovian planet but plenty of ices to form ice giants? If so, how is the creation of the ice giants different to Dwarf planets?

I will edit this with the appropriate equations and timeframes in a few hours when I get back to my office. For now I will address the foundations of your questions with some basics.

1: The radius of the frost line is determined by the star itself (edit: More precisely the nebula that the star forms from.), more in-depth answer will come with edit.

2: The fact that most of the data for these systems is from methods that favour large planets close to the parent star, means that our selection isn't biased, but the method instrument combination is. Another possibility is that migration and loss of angular momentum could be the culprits in this scenario. It is also worth noting, that in the next decade there should be significant data to state whether our solar system is actually common, an anomaly or just younger than many of the other solar systems observed thus far.

3: Larger gas giants "Jovian" planets like all orbiting objects migrate inwards as they loose angular momentum. During the oligarchic growth phase of a soon to be large planet the forming core accrets a lot of material while clearing "all" of the material within it's orbit and associated jeans radius, as the core migrates inwards the supply of material is once again increased.

In short that is why there are "Gas Giants" within the frost line, things migrate inward as they loose angular momentum, and "Ice Giants" form beyond the line because they took too long to migrate inward and gather enough material to move to the next level. Also, if the protoplanet takes too long to accret a ~10-15 earth-mass core, all of the accretable material will be ejected from the system by the star before any significant atmosphere can be obtained.

The dwarf planet part, has to do with the amount of material and how long the protoplanet took to accret it. In that respect, the ice giants and the dwarf planets share a common shortcoming.

Update:

I find myself overwhelmed at work and unable to fully explain the entireties of the process at this time, please forgive me.

However, I do offer a link to an upper division astronomy class' notes that gives a very good explanation of what I believe will help you with your question.


If an exoplanet transit we are seeing is 13000 light years away, are we seeing a 13000-year-old orbit? If a star is 13000 light years away, doesn't that mean we are seeing 13000-year-old light?  If it does, then does that mean when we discover a planet with dimming star light, we are seeing a planet that is 13000 years old?  Are we seeing a 13k-year-old orbit?

In one word, yes. Anything and everything we see, we see the way it was a certain time ago—about 1.3 seconds for the Moon, about 13,000 years for your hypothetical planet. Like @Richyt pointed out, there is no way for information to travel faster than the speed of light.
Because any and all celestial objects gravitationally affect any and all others, orbits are not frozen in time and do “evolve.” If this planet is the only “major” thing gravitating around its star, its orbit will change very slowly, but there is still a change, which may or may not be noticeable in 13,000 years, depending on many factors such as the proximity of the planet to its star (the closer it is, the stronger relativistic orbital precession would be—e.g. it is 43″/century for Mercury, 8.6″/century for Venus, and 3.8″/century for the Earth [source for these latter two: https://arxiv.org/pdf/0802.0176.pdf).


If Earth is tilted, why is Polaris always above the same spot? Why is Polaris, the North Star, always above (or near) the North Pole? If Earth is tilted, Polaris' path should be in winter 23 degrees away from its path in summer, or not?

You are correct that the axis of the Earth's rotation is tilted with respect to the plane of its orbit by 23 degrees. But it is incorrect that the direction that the axis points changes by a large amount (it should be 2*23 degrees) over a 6 month time span.

Your assumption: If axis it pointed at Polaris at (1), then it should be pointed at a different star at (2).

The axis remains pointed in the same direction throughout the entire year because the laws of physics are that the axis of a spinning object remains pointed in the same direction unless a torque acts on the body to change its orientation. There are torques acting on the Earth (namely the Sun and Moon), but it takes 13000 years to change the direction from "one way to the opposite way", not 6 months.

Correct: The orientation of the Earth's spin axis remains pointed at the same star throughout the year (ignoring the slow, 26000 year precession cycle).


Why can you see the space station on some days but not on others? In my area, from December 17th - 26th, you can see the space station every day, twice

But from December 27th - January 6th, you can only see it twice

My question is, why does it appear for almost the whole month of December, and then not appear for a few days? Is it based on my location? I live in Alberta, Canada if that helps.

Two basic conditions need to be met for anyone on the globe to observe the International Space Station (ISS) from any location:

ISS needs to pass roughly overhead of your location, and
it needs to do that during night so it's visible to the naked eye

Now, obviously there are other requirements, like e.g. weather (if it's overcast we won't be able to see it through the clouds), but those screens you attach don't account for that, since it would be nigh impossible to predict weather so far in advance.

Let's discuss the two basic conditions mentioned a bit more. ISS has an orbital inclination of 51.65°. Why is that is explained in this answer on Space.SE, but to quickly recap, this orbital inclination was chosen because it's near ideal for launches from the Baikonur Cosmodrome, where most launches towards the ISS are made from, it is still achievable from the US launch facilities and most US built launchers, and it allows the ISS to go over majority of the populated areas on Earth. So it was decided both for economic reasons, as well as to allow ISS to do as much science as possible.

But what this inclination means is that the ISS will make a ground track (path along the surface of the Earth directly below it) that moves towards the West with each new orbit, even though the ISS itself moves West to East (or saying it otherwise, it's in a prograde Low-Earth orbit). Its ground track could look something like this:

                  Ground track of the International Space Station with an area where it is observable from the Earth's surface                  shown in the white circle (Image source: ISS Tracker)

ISS orbital period, the time it takes it to circumnavigate the globe and complete one orbit, is roughly 93 minutes. As you can see in the screen grab from the ISS Tracker above, its ground track moves westward on each new orbit, which is due to Earth's rotation on its axis, so this shift towards the West will be equal to how much the Earth rotates in the meantime, which in those 93 minutes comes out at roughly 22,5° or a good 2,505 km (1,556 mi) of the Earth's total equatorial circumference.

But ISS's orbit isn't Sun-synchronous, meaning it won't be over any given Earth latitude at the same local mean solar time, so when it will be observable depends also on its orbital position relative to the position of the Sun. We require that it is dark, remember? ISS isn't bright enough (doesn't reflect enough light with its truss and solar panels) during the day to be brighter than the day skies are. So we have another parameter, which is local time. As you probably anticipated by now, the length of the day isn't really equal at any day of the year, and moving further away from the equator only increases this difference of the length of the day in winter, or the length of it during summer. This is another range or values that your software you're attaching print screens from has to account for.

So, when your software collects data from you regarding your geolocation and current date and time (might do that automatically if you're using it on a device that has GPS enabled), it will establish the times (timetable) of when the ISS ground track is over or around your area, and if that happens during night. The ISS doesn't have to be exactly overhead, at zenith, so it will also calculate angle to the zenith for any area that the ISS goes over, at which you would still be able to see it.

Understanding this, even if we don't look at your timetable, we could establish that if you're at latitude to the equator that enables you to see the ISS, you'd be able to do that roughly a few times in a row separated by around 93 minutes, then wait for the ISS ground track to come back towards your longitude (16 orbits or about one day and an hour), and do that during night when you'd be able to observe it. Nights are long now during December at Northern latitudes, so you see a long row of couples of such chances during each night. As long as the weather holds, of course.


Stellar core mass as a function of metallicity Using Hurley's 2000 paper on Single Stellar Evolution, I have graphed the core mass of a star at three stages as a function of metallicity, Z, for a chosen ZAMS mass: at the beginning of the HG (MS), at the end of the Hertzprung-Gap (HG), and at the Base of the Asymptotic Giant Branch (BAGB).

All three functions have a cusp at Z = 0.0005 which I 1) do not understand physically and 2) wonder if it is incorrect? Is this expected? I would have expected that the core mass is monotonic in metallicity, Z. The cusp is present for any ZAMS mass chosen.

EDIT: The equations I use are...

Core mass at beginning of HG:
$Mc_{MS}$ is the product of equation 29 and the core mass at the end of the HG described below ($\rho$ is always $< 1$).

Core mass at end of HG:
$Mc_{HG}$ is, using equation 28 over the relevant mass range the HG ends at Helium Ignition, given by equation 44, where I solved for the time, $t$, in equation 35 and substituted that into equation 34 to find that $M_{c} = \Big(\frac{L}{D}\Big)^{1/5}$, and following the procedure given in the paragraph below equation 65 (where it references equation 44). I used equation 49b for the Luminosity here.

Core mass at BAGB:
$Mc_{BAGB}$ is equation 66.

I'm only interested in stars above $10 \odot{M}$, so I do not consider the possibility of a Blue Loop phase for simplicity.

I feel silly, feel free to laugh at this, but I found my mistake. In the Hurley paper, he uses "log" for the logarithm of base 10. I had been using the natural logarithm. With the fix, the graph looks more realistic:


Relationship between Dark Energy and Dark Matter I have encountered two seemingly contradictory theories on this(I don't know which one of them is correct, or where I am wrong):-

From Wikipedia:

  Measurements of cosmic microwave background (CMB) anisotropies indicate that the universe is close to flat. For the shape of the universe to be flat, the mass/energy density of the universe must be equal to the critical density. The total amount of matter in the universe (including baryons and dark matter), as measured from the CMB spectrum, accounts for only about 30% of the critical density. This implies the existence of an additional form of energy to account for the remaining 70%."

The above statement, if correct, I suppose indicates that there is a critical density for which mass/energy has to account for. This means that if suppose there was no Dark Matter, our calculations/assumptions regarding the amount of Dark Energy would have been much higher(because then Dark Energy would have to account for a greater share of critical density).
The universe is accelerating in expanding. Dark Energy is supposed to cause this acceleration in expansion. While ordinary matter plus Dark Matter, is supposed to reduce this acceleration/expansion through gravity. So, according to this theory, if there would have been no Dark Matter, then there would be less Dark Energy required to account for the observed acceleration in expansion, so our calculations/assumptions regarding the amount of Dark Energy would have been lower.

First of all, if there were no dark matter (DM), you wouldn't ask this question, since structures — including galaxies, stars, planets, and you — wouldn't have had the time to form in the early Universe before it had expanded too much for gravitational collapse to occur. But let's use magic and make the galaxies anyway:

The CMB (specifically the power spectrum of the CMB) shows that the total density $\rho_\mathrm{tot}$ of mass/energy in the Universe is extremely close to the critical density $\rho_\mathrm{c}$. That is,
$$
\frac{\rho_\mathrm{tot}}{\rho_\mathrm{c}} \equiv \Omega_\mathrm{tot} \simeq 1.
$$
The "$\Omega$" is a common way to express densities; as a ratio to the critical density. The CMB also gives some constraints on the total amount of mass (DM + "normal" matter, i.e. baryons), but it is better at constraining the ratio of baryons-to-DM. Together with the matter density $\Omega_\mathrm{M}=\Omega_\mathrm{b}+\Omega_\mathrm{DM}$ obtained from observations of supernovae and, in particular, baryonic acoustic oscillations, we then obtain the amount of DM. This fraction is roughly $\Omega_\mathrm{DM}=0.26$ (Planck Collaboration et al. 2016).
If there were no DM, then $\Omega_\mathrm{tot}$ just wouldn't be $1$, but rather $\Omega_\mathrm{tot} - \Omega_\mathrm{DM} \simeq 0.74$.
Similarly, if there were no DM, there would be less matter to counteract the expansion of the Universe. That means that we wouldn't observe the same relation between the brightness and the distance of supernovae, from which we infer the presence of dark energy (DE). Instead, the brightnesses would be somewhat lower, because the supernovae would be farther away due to the faster expansion.

In other words, you are right that if there were no DM, and if we observed the same thing as we do, then there would be a contradiction. But if there were no DM, we  wouldn't see what we see. Therein lies the resolution.


Why do we continue to find a discrepancy in the Hubble Constant? After taking a look at some recent papers (examples below) that attempt to derive the value of the Hubble Constant, they all conclude with dramatically different values (speaking in context). Why is this? Each of the techniques seems sufficiently rigorous and equally well rooted in understood physics. Is this simply some form of error or is it due to some yet-to-be-discovered physics?

Papers:

A gravitational-wave standard siren measurement of the Hubble constant, H0 = 70
Planck 2015 Results, H0 = 67.8
New Parallaxes of Galactic Cepheids from Spatially Scanning theHubbleSpaceTelescope: Implications for the Hubble Constant, H0 = 73.48

I do understand that this is a hot topic right now in cosmology, however I just don't yet understand why we're having trouble with getting our measured values of the Hubble Constant to converge.

To make a long story short, the measurements from Planck and the Hubble Space Telescope disagree, and the reason behind this isn't known.

First, let's look at the values with the uncertainties. We then have three different results that are, perhaps, not as inconsistent as they originally seemed:

$70.0^{+12.0}_{-8.0}\text{ km s}^{-1}\text{ Mpc}^{-1}$ from LIGO.
$67.8\pm0.9\text{ km s}^{-1}\text{ Mpc}^{-1}$ from Planck.
$73.48\pm1.66\text{ km s}^{-1}\text{ Mpc}^{-1}$ from Hubble Space Telescope (HST) observations.

The LIGO measurement is consistent with the others, as the authors happily note in their conclusions. However, there is a discrepancy between the results from Planck and the results from the HST. This isn't a new problem; the Hubble group previously determined a value for the Hubble constant of $73.24\pm1.24\text{ km s}^{-1}\text{ Mpc}^{-1}$ in 2016. The uncertainties in the measurements describe one standard deviation; it doesn't mean that all values beyond the interval of the measurement plus or minus one standard deviation are impossible.

Now, the difference between Planck results and the 2016 HST data is $3.4\sigma$, which is a lot. The group came up with several explanations for the discrepancy:

Systematic errors in the data collection, including in the use of the Cepheid variables used for the measurements.
Small-scale variations in the Hubble constant, which appear unlikely to have propagated through the data.
Systematic errors in the Planck results.
Corrections needed to the $\Lambda$CDM model of cosmology, used by Planck.

In the 2018 paper, the authors note that the conflict with the Planck data is even stronger - $3.7\sigma$ (and we're now getting results that say it could be $4.4\sigma$; I should check those out). They haven't been able to point a figure at any single problem that could be behind the discrepancy; they do optimistically say that it could be the sign of new physics.


Does this galaxy resemble the Milky Way, and is it relatable as the "template" for the Milky Way? Taking a look at this photo of comet 2I/Borisov and a background galaxy (2MASX J10500165-0152029):

And compare it to this rendition of the Milky Way:

The two galaxies slightly resemble each other. Is there a relationship between these two? Was this galaxy used in defining the Milky Way's shape?

In the case of the image of the Milky Way Galaxy that is provided, no.
Wikimedia Commons lists the Milky Way Galaxy image as "Artist's conception of the Milky Way galaxy" and provides the following credit statement, labeling creation date as 25 June 2009:

Nick Risinger - Own work Adapted from the following NASA images:
236084main MilkyWay-full-annotated.jpg Messier51 sRGB.jpg

Given that about 70% of observed galaxies are spiral galaxies, it seems unlikely that 2MASX J10500165-0152029 in particular was specifically chosen to use as a model for formulating the general image of the Milky Way.


What makes some meteor showers continue for days, while the "Unicorn shower" can be shorter than one hour? The Washington Post's One hundred meteors in 15 minutes? ‘Short-lived outburst’ of shooting stars forecast for Thursday night. says:

A “short-lived outburst” is forecast for Thursday evening around 11:50 p.m. Eastern.

and later:

The prediction of an outburst comes from NASA research scientist Peter Jenniskens and Finnish Fireball Network’s Esko Lyytinen and was published in MeteorNews. The pair specialize in sniffing out meteor outbursts and storms, expertly calculating the orbits of various celestial bodies that give rise to meteor showers. Scientists still haven’t pinned down the object depositing the debris anticipated to trigger this display.
However, outbursts in 1925, 1935, 1985 and 1995 offered enough information for the team to produce a model. It indicates the close passage of a dense pocket of spaceborne debris. That narrow but potent debris stream is likely to manifest in a 15-to-40-minute-long barrage of meteors around 11:50 p.m. Eastern on Thursday, according to the researchers.
What will it look like? Jenniskens and Lyytinen indicate events in 1985 and 1995 produced 700 and 400 meteors per hour and wrote that this event could emit anywhere from around 100 per hour to even more than 1,000 per hour, the latter considered a meteor “storm.”
The outburst is calculated to last a fraction of an hour, so those numbers can be sliced in half. And because the radiant point of the meteors will be comparatively low in the sky, we can trim off a bit more.

The linked Meteor News article LIKELY ALPHA MONOCEROTIDS (AMO#246) OUTBURST ON THE MORNING OF NOVEMBER 22, 2019 gives orbital elements derived by different groups of astronomers that can be used to predict the time of maximum.
But what is it that makes this debris trail so narrow that the shower is predicted to last less than one hour while other showers can sometimes extend for several days?

It sounds like this effect:

  ... in addition to the usual shower on August 12th, there might be an extra surge of meteors on August 11th caused by a filament of dust newly drifting across Earth's orbit.
  The filament, like all the rest of the dust in the Perseid cloud, comes from Comet Swift-Tuttle. The difference is, the filament is relatively young. It boiled off the comet during the Civil War, in 1862. Other dust in the cloud is older (perhaps thousands of years old), more dispersed, and responsible for the month-long shower that peaks on August 12th. The filament will eventually disperse, too, but for now it retains some of its original ribbon-shape.

The 2004 Perseid Meteor Shower NASA


When we say the sun takes 230M years to orbit the galaxy, what is this in relation to? When we say the sun takes 230M years to orbit the galaxy, what is this in relation to? We measure the earth's rotation relative to the distant stars. What is the reference for our Sun's motion around the galaxy?

The Sun's velocity is known (with uncertainties) with respect to the Galactic centre (GC), as is the Sun's distance from the Galactic centre. These measurements have a long tortuous history, which I am not going to attempt to summarise here. However, the motion of the Sun with respect to the GC is established by measuring the proper motion of the Sgr A* source with respect to the assumed zero proper motions of distant radio quasars (see for example Backer & Sramek 1999; Reid & Brundthaler 2004).

The astrometry frame of reference now is known as the International Celestial Reference Frame and is defined by hundreds of compact, extragalactic radio sources.

If you then assume the Sun's orbit is circular and that Sgr A* is at the GC, then the job is done - the relative velocity divided by the distance to the GC gives the angular velocity $\Omega$, and the orbital period is $2\pi/\Omega$ . If you have a model for the Galactic potential then you can do a bit better.


Calculating the age of the universe Is the calculated age of the universe that of the visible universe or the entire universe? I dont know how the age is calculated but if it is believed that all we see visibly IS the entire universe then that means it is finite. If it is not finite then the light from the really far astronomcical objects has not reached us yet so there is a possibility it is infinite.
Can the method used to calculate the universe age since the Big Bang guarantee that it has an age or is it possible that it existed infinitely?

It is a common point of confusion, but an infinite universe can still have been at a unique singularity in space-time at a finite time in the past.

This singularity is not a "point in space", since every point in the universe participated in the big-bang. It is merely a time at which the scale factor, which multiplies the separation between all spatial points in the universe,  was zero.

So in cosmology governed by the Friedmann equation, the age of the universe thus calculated is the same whether you are talking about the "visible" universe or an infinite universe. The light travel time really doesn't come into this, because different parts of even a very large universe can be causally separated yet still have originated at the same singularity.


Could iron rain upset tidally locked WASP76b? The Jupiter mass planet WASP76b orbits its 1.5*Sunmass star at 0.03AU and is tidally locked. The day side heats to 2400C which vaporizes metals, including iron observed in its atmosphere. The balmy night side is only 1500C causing ferocious winds which carry the iron vapor from hot to cool side. There it condenses, falling as rain.

Could this redistribution of mass change how the planet faces its star, and what could make that happen? For example, if the winds blew only one direction (I don't know why this would occur) might that act as a small spin that tidal locking forces might try to neutralize?

This is a refugee question from Space Exploration.

The important thing to bear in mind is that WASP-76b is a gas giant (at the time of writing, the NASA Exoplanet Archive gives values of 0.92 Jupiter masses and 1.83 Jupiter radii, so this planet is much less dense than Jupiter is): there's no solid surface for the iron to build up on. Instead what would happen is that the iron would rain down until it reached depths at which the temperatures are sufficiently high for it to vaporise again. This would then allow it to be carried back to the dayside as iron vapour. Once it is back on the day side of the planet, the increased temperature would allow it to rise back up to higher altitudes, where the winds would then carry it back to the night side in an "iron cycle".

Superrotation is predicted to be a general feature of hot Jupiter atmospheres, though the detection of westward displacements of hotspots on planets like CoRoT-2b suggests that things may be more complex: see for example Debras et al. (2019) "Acceleration of superrotation in simulated hot Jupiter atmospheres" for some discussion of how this might occur.

There's also the question of the so-called "thermal tide", where the atmospheric heating affects the rotation. This effect is thought to be responsible for Venus avoiding tidal lock. On a gas giant there isn't a solid surface for the atmosphere to couple to, but in some cases this effect may contribute to at least the outer radiative zone of the planet being in non-synchronous rotation, see Auclair-Desrotour & Leconte (2018) "Semidiurnal thermal tides in asynchronously rotating hot Jupiters" for details.