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From an outsider's perspective, how can a black hole grow if nothing ever crosses the event horizon?
Due to time dilation, an outside observer never sees a falling object actually cross the event horizon. I'm not referring to the optical illusion of red-shifted light making objects appear to fade away before crossing the event horizon. I'm referring to the very real effect time dilation has on objects near the event horizon, causing them to slow down to a halt from the frame of reference of an outside observer.
This outside observer can see, however, that the black hole has a non-zero mass and volume. How can this be if from the frame of reference of the outside observer, nothing has ever crossed the event horizon and been integrated into the black hole?
In other words, how can an outside observer say that a black hole has a non-zero mass and volume, but also, nothing has ever crossed its event horizon and contributed to said mass and volume?

in my opinion, the classical definition of the “event horizon” is too strict and imposes a number of extreme restrictions on the capabilities of the observer

Clearly related https://astronomy.stackexchange.com/questions/2441/does-matter-accumulate-just-outside-the-event-horizon-of-a-black-hole

How can a snowball grow if no snow ever enters it?

Good question and I think you could cross-post on the physics site to attract a theoretical answer.

@dodo Is cross-posting a question on multiple Stack Exchange sites permitted if the question is on-topic for each site? Answer: please try not to.

What makes you think the event horizon can't expand?

First off, there's no real time dilation effect. In coordinate systems that cover the event horizon (Kruskal-Szekeres, Eddington-Finkelstein, Gullstrand-Painlevé, Kerr-Schild), objects fall through it without so much as a hiccup. The only real slowing-down effect is from redshift (as seen from outside).
Suppose you watch from the side as someone lobs a basketball into a black hole, against a background of stars. The basketball has mass, so even when it's far from the black hole, it gravitationally lenses the stars behind it. As it approaches the black hole, that lensing will overlap/merge with the lensing of the black hole, so the shape of the black hole will seem to have a slight bulge in the direction of the basketball. As more time passes, the basketball will appear to flatten against the horizon of the black hole (while redshifting into invisibility), and the bulge will spread around so that the combined system appears spherical again, but with a slightly larger mass.
Technically, you never see the object fall through and you never see a "true" black hole, but it makes no practical difference, because the difference between what you see and the ideal limit decreases exponentially with time. The time constant is comparable to the light crossing time of the black hole, which is around 1 day for a huge black hole like M87's, and less than a millisecond for a stellar-mass black hole. So after even one year, never mind billions, you'll have no hope of distinguishing the frozen star from a "true" black hole.

So, purely theoretically, from the perspective of a faraway observer, any black hole that formed a finite time ago is just a bunch of highly redshifted objects spread very close to the event horizon, but not under it. Right?

@Ruslan  That's pretty much the take away I'm getting as well.

@Ruslan: I believe this is one proposed solution to the black hole information paradox but it's not known for certain to be true.

Actually, now I realize that as the conglomerate of the objects gets more mass, its Schwarzschild radius grows, so newer objects will appear on larger distances from the center. Then I suppose the whole system should be filled with matter during its growth, not just have a thin layer above the event horizon.

@Ruslan The apparent location of old objects may appear to move as the horizon grows. I'm not sure. I wish someone would do a numerical simulation of it, but I suppose they have more important things to do.

@benrg I would like to shift to the point of reference of the infalling object. I keep hearing conflicting information about whether or not the universe speeds up behind them; again, not referring to optical illusions, but real effects of time dilation. If it does, do they actually experience crossing the event horizon before the end of time? If it doesn't, wouldn't that conflict with the events experienced by the outside observer?

@AxiomaticNexus: The universe does not speed up behind them. It would if they could hover there.

@AxiomaticNexus See Does someone falling into a black hole see the end of the universe?

@benrg Unless I'm misunderstanding the answers in that question, I think they all tackle the question from the optics angle. I'm not really interested in what the infalling object "sees", I'm actually interested in what really happens to the universe behind them.

@AxiomaticNexus You can calculate what someone sees from a given vantage point, but you can't calculate the nature of reality from a vantage point—there's one shared reality for everyone. Special relativity is badly taught, leading people to think otherwise. See e.g. the end of this answer.

@benrg That's exactly my point, there's one shared reality for everyone; but there's so much disagreement among observers I still can't reconcile. I'll never get GR.

@AxiomaticNexus "What really happens to the universe behind them" can only have one answer, and you already know it: it goes on as normal. I still think your problem is thinking in terms of the SR idea of "observer", which amounts to a kind of private reality even if people wouldn't phrase it that way. It kinda works in SR, but doesn't generalize. A GR observer is just somebody who makes observations. To get GR you have to unlearn the SR observer stuff and understand SR in terms of 4D geometry without coordinates, which does generalize.

Due to time dilation, an outside observer never sees a falling object actually cross the event horizon.

This is correct, but perhaps not quite in the way that you visualize it.
It's trivially true, in a literal sense, that an outside observer can never see anything cross the event horizon, since this would require seeing inside the event horizon, and no light (or other information) can escape from there to the outside.  The most the outside observer could hope to see, even in principle, would be the object hitting the event horizon and disappearing (but they can't even quite see that, for the reasons discussed below).
It's also true that, the closer the falling object gets to the event horizon, the longer it will take for any light it emits to get away from the black hole (and the less of it will make it out at all, and the more redshifted the light that does escape will be).  In fact, as the object approaches the event horizon, the time it takes for light emitted by it to escape the black hole tends to infinity.
However, trying to interpret and visualize this as:

[an] effect time dilation has on objects near the event horizon, causing them to slow down to a halt from the frame of reference of an outside observer

is probably not the best way to think about it.  A better way of looking at what's happening is that time dilation is causing light (and other signals) to take longer to escape the black hole, and thus the outside observer is seeing things that happened a longer time ago in the past.
In other words, the object falling into the black hole isn't slowing down.  As benrg notes, it actually speeds up quickly and falls through the event horizon in finite time.  And in doing so it adds mass to the black hole and causes the event horizon to grow bigger.
All that's slowing down (from the outside observer's perspective) is the light that the falling object emits as it's falling in.  So the outside observer can in fact "see" the black hole form and grow, at least as much as it's meaningful to speak of seeing a black hole at all.  (They obviously cannot receive any light or other information directly from the black hole itself.  All they can observe is the infalling mass and the extreme spacetime curvature and its associated phenomena that are consistent with there being a black hole in the middle.)  But the outside observer is also seeing (very small amounts of very strongly redshifted) light emitted by objects that fell into the black hole a very long time ago, including some emitted by objects that fell in when the black hole (and specifically its event horizon) was smaller than it is now.
Interpreting these observations in a particular way, the outside observer could claim that they're seeing light coming from inside the black hole's current event horizon (as calculated based on indirect observations).  But there's no actual contradiction there, since that light was emitted back when the event horizon didn't extend that far yet!

Of course, in practice, all of this is mostly an exercise in pedantry and deliberately quirky interpretation of observations.  What a real outside observer actually sees, as you correctly note, is objects fading away as they fall into the black hole, and the approximate place where they fade away (accounting for the massive optical distortion around the black hole) being approximately where they can calculate that the event horizon should be.
All the rest is just theory, and not particularly fruitful theory either.  Sure, you can say that in theory you could sit in a spacecraft orbiting a black hole and see light emitted by an object that fell into it a million years ago, because it's theoretically possible for a photon emitted by the object very close to the event horizon to take a million years to reach you.  But in practice you just can't, no matter what kind of incredible sensor technology you may postulate you'd have, because the probability that you'll ever detect a photon emitted during that vanishingly small portion of the object's descent is for all practical purposes zero.

Why we say the objects become "invisible" due to redshift? It's not just a matter of using a different sensor (i.e. IR) instead of our eyes limited to the visible range?

@Mark Redshift shifts the frequencies of individual photons, but it also shifts the rate at which you can detect them. When the detection rate is too low, you can no longer form an image.

So it's similar to the doppler effect in the sense that the passing car's real engine sound (= falling object's real speed) doesn't change but the observer does hear a changing sound (= seeing the object falling slower), and eventually the engine sound just slowly fades away (as does your view on the falling object)?

" But in practice you just can't, no matter what kind of incredible sensor technology you may postulate you'd have, because the probability that you'll ever detect a photon emitted during that vanishingly small portion of the object's descent is for all practical purposes zero." This is possibly still under selling how fast this happens! (For stellar mass black holes), the expected number of photons emitted by the falling object for the rest of eternity drops below zero in a fraction of a second, and continues dropping exponentially.

Objects that approach the event horizon(from an outsider's perspective),will indeed appear to slow down and become increasingly redshifted as they get closer. As a result, it will take an infinite amount of time (as perceived by the outside observer) for an object to cross the event horizon. This is the essence of the "no hair" theorem for black holes, which states that all the information about an object that falls into a black hole is seemingly lost to an outside observer.
However, from the infalling object's perspective itself, it will indeed cross the event horizon in a finite amount of time. For the infalling object, its journey toward the singularity continues, even though it is never visible to the outside observer after crossing the event horizon.
So, while the outside observer sees objects near the event horizon slowing down and never crossing it (due to time dilation and redshift), the objects themselves do cross the event horizon and contribute to the mass and volume of the black hole as they move closer to the central singularity.

Would time speed up behind the infalling object before crossing the event horizon due to time dilation?


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calculating mass of galaxy from luminosity
A very elementary question, just to make sure I got this thing right.
Suppose we are given that the mass to luminosity ratio of a galaxy is $2 \frac{M_{\odot}}{L_{\odot}}$.
Let us say, that the luminosity of the galaxy = $L = 100\,L_{\odot}/pc^2$.
Then the mass density of the galaxy  = $\rho = 100 L_{\odot}/pc^2 \times 2 \frac{M_{\odot}}{L_{\odot}} = 200 M_{\odot}/pc^2$.
Is this the right calculation?

The mass surface density.

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Does the Oort Cloud lose mass with time?
Can blocks of ice escape from the Oort cloud and away from the solar system forever?

Comets constantly gets ejected from Oort cloud. So, you can say it lose mass.

The Oort cloud objects are only very loosely bound to the Sun. That is, their total energy (kinetic plus gravitational potential) is only just negative.
That means you only need a slightly positive perturbation to make those objects "fall out" of the Solar System. Conversely, a small negtive perturbation would cause them to fall inwards - and can give us long period comets.
What could cause these perturbations? The tidal field of other passing stars or structures (spiral arms, molecular clouds) within the Galaxy. Remember that the influence of these will not be felt immediately because the orbital periods of Oort clouds objects are at least thousands, if not millions of years.
As some of the Oort cloud objects leave the Solar System then the cloud as a whole must lose some mass. The Sun will also capture some objects, as it moves through the Galaxy, but this is unlikely to be a symmetric process.

+1, Nice answer!. I have a question: What is the needed escape velocity for an icy comet in Oort's cloud  to escape the Sun's hill sphere?

@Arjun $v_{esc} \simeq (2GM/r)^{0.5}$. So from 10,000 au, $v_{esc} \simeq 0.4$ km/s.

"unlikely to be a symmetric process" I think the sun will lose far more than it gains. This is because the escape velocity in the oort cloud (up to ~1km/s) is much less than the stellar dispersion velocity (very roughly ~20 km/s). This makes it hard to capture objects.

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Can telescopes under $200 help you see saturn clearly by eye without editing the image on a computer?
I had a telescope that only cost 50 dollars  for viewing the moon and it was good for that but the planet's were always badly out of focus. I am thinking of getting a better telescope for up to $200 but reviews on YouTube of these telescopes  usually say they are good but after taking videos and superimposing the best frames on one another.  Can a cheap telescope give clear views of Saturn or Jupiter without the need for making videos or taking photos?

I assume you mean in the US, but perhaps consider writing explicitly which country you're in, as this is an international community.

I can tell that in my country, where everything is more expensive than in the US (except getting sick), you should be able to find a 200 USD telescope that will show you the rings of Saturn, although it will look very small.

Lots depends on what you mean by "clear".  Photo processing can bring out a lot of detail that is difficult or impossible to catch by eye, when the image is constantly shifting/distorting from atmospheric variations.  A $200 telescope will let you see the most prominent bands on Jupiter, for example, and maybe the Great Red Spot on an excellent night, but visually you can't see all the detail you might in well-stacked photos from the same telescope.

A 100mm or wider reflector might serve pretty well. Avoid the ones with short tubes.

@pela I am in the uk.

In the UK we use pounds sterling, not dollars.  I suggest you rephrase your question to use pounds.

Short answer is no. Such low-priced telescopes are toys at best. I don’t mean you can’t enjoy them, but their performance will always be disappointing. It is worth saving up and investing in a better instrument that will cost around 500–1,000 £ at least, if not more.

@James K IS dollars are the internationally accepted standard that most people are familiar with.

@Pierre Paquette IMore expensive telescopes need collimating and I don't think that would be easy to do.

This is the telescope I would get for that amount of money: UK (£195) US ($249). It was within your $200 limit until this year when the price of everything shot up :-(

@Adrdav22: It’s actually not that complicated. There are also refractors and catadioptric telescopes (SCTs, Maksutovs, etc.), that don’t need collimation or can do without…

@Adrdav22 you can't buy telescopes in the UK denominated in dollars.  Exchange rates vary and I, for one, am not familiar with the exchange rate to dollars at the moment.  As you are not able to buy a telescope in dollars, you should edit your question to use your local currency, ie UKP

This completely depends on what your expectations are. Saturn is a easy target, in part because its shape is unique so it's obvious what you're seeing, and it adds a high novelty factor. If you want a dot with rings, you can even use good binoculars. On the other side, you might need to spend $3 billion for a spacecraft to take your pictures.

My simple answer to your question is that you're probably going to have to increase your budget to find something worthwhile.  But there are some options.
For planetary and lunar observing, there's a few key things to keep in mind.  The first is your observing conditions, specifically astronomical seeing.
When we talk about "seeing" we're referring to the steadiness of the air between you and the object you're trying to see. Let's use any fairly bright star as an example.  If you point your scope - regardless of the telescope here - at the star, you'd ideally want to see it as a perfectly steady, sharp point of light that's not moving, flickering, or shifting color.
Regardless of the magnitude of the star (or its actual size and distance), it should appear as a pinpoint of light.  But the atmosphere will alter this appearance.  First, depending on the brightness, the air will scatter some of the light and bloat the star somewhat.  This is why brighter stars appear larger - there's just more light to scatter.
If you're observing a star and its color appears to be shifting -- this is more obvious for stars closer to the horizon -- then there's something in the atmosphere that's causing minor refractions of the light.  In a sense, the air is acting a bit like a prism, and the various wavelengths (i.e. colors) of light are affected slightly differently.  There also may be particulates or gasses in the air that are, in effect, filtering the light a little bit.  As I mentioned, this is more obvious for stars closer to the horizon because when you see them you're looking through a thicker slice of the atmosphere and there's more stuff in the way.
The same thing goes for twinkling.  If the brightness of the star is shifting, this is due to the air and stuff in the air messing with the light reaching your telescope and/or eyes.
Then there's the shifting or dancing of a star.  I remember one incident when I was observing Jupiter (ok, not a star, but the same issues apply).  At low magnification, it was jumping all around the field of view.  This was due to air currents at a higher altitude.
To understand this better, imagine you're standing on the edge of a swimming pool and someone throws a coin in and it settles on the bottom.  If the water is perfectly still, you might be able to see that it's a quarter versus a dime or nickel, or even a half dollar (as uncommon as those are). But what if the water isn't still?  What if someone's swimming around at the other end of the pool and making waves?  You may still be able to see the coin, but not clearly enough to identify it: the image you see is moving around due to the waves and currents in the water.
Air currents have the same effect.  Yes, the air isn't as dense as the water in a pool, but there's a whole lot more of it between you and space.  And while the winds where you are on the surface may be calm, as you go up in altitude, there may be different conditions.  A couple hundred feet above your head there might be a stiff breeze.  A few thousand there might be heavy winds.  Some of those winds may be turbulent, others smoothly flowing.  As the light from whatever star/planet/etc... you're trying to view passes through all of this, the air will refract the light slightly differently - often enough to make for poor seeing conditions.
This is one of the reasons why major observatories are built on top of mountains: it gets them out of the lower, denser, more humid air and away from the worst of the atmospheric conditions in general.
This is regardless of the telescope you're using: it will affect a 50 mm, 500 mm, or 5 m telescope.
Ok, so another consideration is magnification.  This is something that's often misunderstood.  Magnification is important, particularly for planetary observing, but there's things to keep in mind.
For visual observing (i.e. telescope and eyepiece, no camera), you calculate your magnification by dividing the focal length of the telescope by that of the eyepiece used.  If your telescope has a focal length of 1,000 mm (focal length is nearly always expressed in millimeters) and you're using a 25 mm eyepiece, you get a magnification of 40X (1,000 / 25 = 40).  If you switch to a 10 mm eyepiece, you get 100X.  Technically speaking, there is no limit to magnification.  Let's say your shortest eyepiece is 5 mm, with a 1,000 mm telescope that gets you 200X. If you add a 2X Barlow Lens, it effectively doubles the telescope's focal length to 2,000 mm, so you get 400X with your 5 mm eyepiece.  Let's say you add a second Barlow (yes, you can do this, though for several reasons it's probably not worth doing): now you're at 800X.  You can keep doing this as long as you want and keep doubling the magnification.  But at a certain point, all you're doing is magnifying a blur.
This is because while magnification has no limits, detail resolution does.  The amount of detail you can resolve with a given telescope is dependent upon its aperture due to the diffraction of light.  We call this the Diffraction Limit.  The actual calculation for this is specific to a given wavelength of light (e.g. the diffraction limit of a telescope for blue light will be different than that for red light), but we can use a fairly simple formula to get a rough estimate.  This is the formula for Dawes' Limit.  (For those who know about this, yes, it's not really the same, but it's an easy way to get a rough estimate).  To calculate the Dawes' Limit you divide 116 by the aperture in millimeters to get the angular size of the limit in arcseconds.  So for my 80 mm refractors, the Dawes' Limit is 1.45 arcseconds, while my 8 inch SCT gives me 0.57 arcseconds, and my 16 inch Dob gives me 0.29 arcseconds.  This means I should be able to discern details this small through each scope.
So what does that mean for details?  Well, let's take Saturn for example.  The biggest gap in the rings, known as the Cassini Division, is about 4,800 km wide.  So, with my 8 inch telescope, can I see it?  At its closest approach to the Earth, Saturn is about 1.2 billion kilometers away.  At that distance, using a little trigonometry, I can calculate that an angle of 0.57 arcseconds is equal to a linear size of about 3,317 kilometers. Since that's smaller than the Cassini Division, I should be able to see it.  but with my 80 mm refractor, an angle of 1.45 arcseconds gives me a linear size of about 8,439 km, so I shouldn't be able to make it out, or, at least, it will be so blurry as to be difficult to discern.
Again, the Dawes' limit isn't an exact measure to be used here, just a rough guideline (it was actually formulated to determine how far apart two point-sources of light [i.e. stars] have to be separated before you can determine that there's actually two points and not a single one... which is a little bit different, but still depends on angular size/separation).
Since detail is going to be a key concern for planetary observing, aperture is going to be important.  But, again, you have to contend with the atmosphere.
Tele-Vue makes what are frequently considered the best eyepieces on the market (not everyone will agree here, but they're at least near the top of the list).  On their website, they offer a little calculation web app which asks you to enter a couple details about your telescope (any two of these three factors: aperture, focal length, focal ratio) and it will give you a list of performance stats for their various eyepieces.  In the resulting list, you might find some of the magnifications pointing to a footnote that says they recommend up to 60X of magnification per inch of aperture up to a maximum of about 350X regardless of aperture.  This is due to atmospheric effects.  They're essentially saying that 350X is really the best you can hope for with any telescope and they don't recommend going higher.  Since they're essentially telling you to not bother with several of their eyepiece I find this to be good information.  They'd rather you get good performance with what works than waste your money on stuff that won't likely give you good views (though I'm sure they'll happily take your money anyway if you insist on buying them).
To be fair, I've seen great views at higher magnifications a few times.  I was at my club's dark site one night and one of the guys on the field had a 20" Dob and we had great seeing - I don't know what the magnification we viewed Saturn and Mars with, but it was definitely higher than 350X.  But this was an exceptionally good night with a very large telescope, and not something you can expect on a regular basis.
Ok, so what does this mean for you?
First, all other things being equal, you want to get the largest aperture you can afford.  But you should also keep in mind the quality of the optics.  I'd much rather have a Takahashi or Astro-Physics 100 mm apochromatic refractor than an 8 inch Meade or Celestron SCT.  The quality of the optics from a premium brand is going to provide a better view, even if they only have about 1/2 the aperture.
Another important factor is the mount. If you take a smaller aperture telescope on a solid mount and compare it to a larger aperture on less stable mount, the smaller aperture will give a better viewing experience.  This is one of the reasons I STRONGLY recommend against low-price telescopes, especially ones on flimsy tripods.  The Celestron PowerSeeker line comes to mind (some of the worst scopes available  - so bad there's an entire subreddit devoted to how bad they are: www.reddit.com/r/dontbuyapowerseeker).
You can find several 114mm (4.5 inch) scopes on the market these days.  Many of them come on tripod-based mounts.  Others come on what are often referred to as "tabletop-Dobsonian" mounts (these are not actually Dobsonian mounts, just simple alt-az mounts, but I'm splitting hairs here).   An example of this is the Orion StarBlast 4.5.  It's a little above your price point (currently on sale for \$230, normally \$250), but would probably beat out what you currently have.
Orion offers a few smaller, and less expensive, similar options, like the SkyScanner BL102, which is a 102 mm (roughly 4 inch) scope. It's in your price range (Currently \$135), but I'm hesitant to outright recommend it.
Generally speaking, unless you're able to afford a premium scope like those from Astro-Physics or Takahashi, I wouldn't recommend anything smaller than about 5 inches (130 mm).  Below that size, a decent pair of binoculars is usually a better buy.  I generally recommend saving up for something like a 6 or 8 inch Dobsonian (I consider an 8 inch Dob the best beginner scope option, and one that will be useful well into your future as you grow in experience).  But if you're bound and determined to get something now,  that's the direction I'd go.
Of course, this doesn't take into account used equipment.  You can find some really good deals on used scopes if you know where to look. Unfortunately, the best and most reliable sites for used gear typically don't offer much in the way of lower-end equipment (you can often save 30% or so on a mount that otherwise would cost \$2,000, but odds are you won't find something like you're looking for).  But if you're a member of a club, then you might well find something in your local club.
And here is where my final and most important piece of advice comes: find and join a local astronomy club or society.
There's several reasons here.  First, most such clubs put on events known as star parties on a relatively regular basis.  At a star party, you're likely to get a chance to look through a wide range of scopes of all sorts of prices, sizes, and quality levels.  This will help you learn more about what you're really looking for, as well as get a chance to see stuff you might not otherwise see.  Many clubs have their own observatory or dark site which members can learn to use, often with some pretty nice equipment available.  And some clubs have loaner telescopes that members can borrow.  On top of that, members of the clubs I'm in or am familiar with are constantly buying/selling/trading equipment between themselves.  Nearly all of my equipment has been bought used from members of my clubs that were selling stuff off to fund their next upgrade or because they were getting out of the hobby.  You might find someone in a local club selling of an older Dob that's in your price range... or at least a decent upgrade from what you have.  And, of course, with most clubs you have a built-in support network to help you learn more.
So I'd hold the money, join a club, and see what options that brings you.  The membership dues for most clubs in the US is under \$50/yr (mostly well under, I pay about \$55 total for membership in the club local to me in New Hampshire and the club where I used to live in Texas).  There are more benefits still, so I'd definitely look into it.
One final recommendation: if you want to learn more about a variety of telescopes, I strongly recommend the Scope Reviews website and YouTube channel of Ed Ting.  I was aware of Ed's reviews and advise long before moving to NH where I was pleasantly surprised to find out he's a local here and member of the same club I joined.  Ed is well known in the amateur astronomy community and has done reviews on a wide range of telescopes from cheap refractors and Newtonians to high-end premium equipment.  You can take or leave my advice, but you really should listen to Ed, he really knows his stuff.
Good luck and clear skies!

Use \$ to prevent Markdown from interpreting text between dollar signs as a MathJax formula.

Wonderful and wonderfully thorough answer (as always!) +1

Can telescopes under $200 help you see saturn clearly by eye without editing the image on a computer?

Can a cheap telescope give clear views of Saturn or Jupiter without the need for making videos or taking photos?

Hmm... you might get a great telescope under \$200 by finding a good deal on a used telescope that originally cost more, or cost \$200 new thirty years ago :-)
But that's probably not what you are asking.
You haven't explained what "clear image" means to you, but I assume you want to see distinct bands of Jupiter (perhaps even the red spot) and maybe the dark spaces between Saturn and its rings when they are at maximum tilt.
For those -- by eye -- it's Earth's atmosphere that's going to stop you, not the telescope.
Take your low cost telescope to a location and time that has good astronomical seeing which means has the least atmospheric turbulence both

right above you (standing in a still-warm concrete jungle (1, 2) after sunset as the air cools would be bad)
many kilometers higher where astronomical seeing effects are always present.

Usually this means some combination of

going away from cities
up in altitude (mountains)
waiting later into the night

but the planet's were always badly out of focus

Only after working on those points first will you begin to experience what your telescope can do in terms of resolving the planets. If you haven't found a good place in space and time where "seeing" was good, that could certainly explain your feeling that they were "out of focus".
Also, make sure there was not moisture condensation on your optical surfaces and that they were all clean and free of oil from accidental contact with skin.
Finally, if the images were too bright and you eyes were night-adapted, they might appear a bit washed out. You might consider a filter - neutral density to make them darker, and a color filter may even enhance the contrast between bands of clouds on Jupiter.
Only after getting familiar with astronomical seeing and checking for dirt and moisture should you just go out and buy a more expensive telescope thinking it will look better!
My guess is that as long as you want a good visual presentation of the planets without any photography and image sorting/stacking, the internet is where you'll find them, not a telescope eyepiece, unless you happen to have a perfect night with unusually good astronomical seeing!


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How do the orbits of star systems V871 Centauri (septuple), Gamma Cassiopeiae (octuple), and QZ Carinae (nonuple) work?
Several years ago, I asked the question on here, "How do the orbits of Nu Scorpii and AR Cassiopeiae work?" At the time, these two stellar systems were the most populated (I don't know the technical term), meaning these were the only two septuple star systems known at the time, and there were no star systems containing more stars. On that question, I received a fantastic answer that perfectly described the orbits of all 7 of the stars in a way that was really easy to understand.
Then, today, I looked to see if any other multiple star systems have been found that are bigger. According to Wikipedia, there are three new star systems of note: a new septuple named V871 Centauri, an octuple named Gamma Cassiopeiae, and a nonuple named QZ Carinae. None of these seem concrete, however, and none of the sources provided explain the orbits completely.
For V0871 C., the source from wikipedia (Multiply eclipsing candidates from the TESS satellite) says "V0871 Cen is probably a septuple-star system of architecture (Aa-Ab)-B-C-D." In that paper, it describes the probable orbits thusly:

The most inner two pairs are the 2.8 and 2.09 days binaries (we name it Aa-Ab), accompanied with a more distant component B with its only poorly-constrained orbit by Zasche et al. (2009). And much more distant C and D components are probably bound (due to their similar proper motion), but only very weakly.

I take this to mean that there are two separate binary star systems in the center. The first binary pair (Aa) are orbiting each other very closely every 2.8 days, while the second binary pair (Ab) orbit each other even faster at 2.09 days. Apparently, they don't know the rate at which the two binaries orbit each other? Then, a fifth star (B) orbits these four a little further out, but its exact orbit is unknown? Finally, sixth (C) and seventh (D) stars orbit each other distantly while also orbiting Aa-Ab much more distantly, but again, how is unclear? Hopefully, someone can fill in the gaps
Then for Gamma Cassiopeiae, the source given on Wikipedia (Surveying the Bright Stars by Optical Interferometry III: A Magnitude-Limited Multiplicity Survey of Classical Be-Stars), so far as I can tell, does not give any orbital periods but does confirm that it is likely an octuple star system. On page 6 there is a graph whose last column is multiplicity, Gamma Cassiopeiae is listed 8. Further, on page 40, a description of the star system is given. It is incredibly technically dense, but so far as I can tell it is largely talking about stellar magnitudes and angular distances. Maybe the orbital periods can be determined from this information I don't know.
One note on that last source (Surveying the Bright Stars...). Gamma Cassiopeiae is referred to as "Γ Cas". In order to type the Greek letter gamma, press the alt button, and while holding that down, press and release 2, then 2, then 6, then release the alt key. The document doesn't mention it much, but this will help you skim the document for related material.
Lastly, the source given on Wikipedia for QZ Carinae (Towards a consistent model of the hot quadruple system HD 93206 = QZ Carinæ), a description of some of the 4 major, inner orbits of this system. From my understanding, there is a binary pair, Ac1 and Ac2, collectively known as Ac, that have an orbital period of 5.9987 days. Then, there is another binary, Aa1 and Aa2, collectively known as Aa, that have an orbital period of 20.734 days. Then, I believe it is saying that these two binary systems, Ac and Aa, orbit each other at approximately 14,500 days (just shy of 40 years). As for the rest of the stars in the system, it says that "all other known components [were] too faint to affect
the photometric measurements, even if observed with an electronic detector with larger pixel size. The rest of this study will therefore deal with the close quadruple system only." So, no data on the orbital periods of the other 5 stars, which is especially frustrating, because what the heck is going on with Ab Ad? Wouldn't their names suggest they are involved in the orbit with Ac and Aa? Also, on their graph in that paper, on page 2, is component A a different star, or is that the combination of Aa, Ab, Ac, and Ad? Because, if it is its own star, then 8 stars are listed, but Aa and Ac are binaries, so that would make 10 stars. So, I am pretty sure that the A component is made up of the other A's.
So, the question is, how do the orbits of these systems: V0871 Centauri, Gamma Cassiopeiae, and QZ Carinae work? As an example for how to answer this question, I will again suggest this answer to how the orbits of two other septuple star systems works. If the answer is in that type of format, I will easily understand. Now, given that all of these papers were published rather recently (the oldest being 15 September 2021) and they do not seem to have these orbits fully worked out, I imagine it's just the case that humanity does not have the answers to these questions, yet. But, I am confident that we will someday. Still, I had to wait over three years for an answer to my last question, and those systems had been known for a long time, so I will be patient with this answer. Hopefully, this will be popular enough to pop back up in 2030 when we have answers to all these questions.
As always, I appreciate any help. Thanks.


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Anomaly in official video of the today's solar eclipse
I watched an official nasa video of the solar eclipse that happened today.
https://t.me/uniannet/114510 - link to the news channel where the video was posted, not original.
At the fifteenth second of the video, an anomaly is clearly visible, which does not at all resemble a filming defect.
I have no idea how this could be explained.
Maybe someone else has it?
a few screenshots:

That absolutely looks like a digital filming defect. You see them all the time in pretty much any digital media.


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Do the discs of observed solar systems have random orientations?
Is there any bias in the orientation of solar systems?

Good question. I think the best answer is still given here (in which case your question should be closed as a duplicate), but perhaps Rob Jeffries might have some updates (?).

Some additional insight might come from considering the orientations of protoplanetary disks in young (e.g., pre-main-sequence) stars; as far as I know these are random.


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Which place in the world should be best for astronomy with an optical telescope?
Would it be somewhere in a desert a long way from water and mist, or high on a mountain above the lower atmosphere? Where is the best location in theory or in practice?

Why specifically "refractor", given that nearly all large telescopes are reflector telescopes. Moreover what do you mean by "best" - best for what?  Large refractors are nowadays only used in specialist applications, such as solar telescopes.

@James K I have used the word optical instead of refractory now.

Probably at the summit of Mt. Everest, but that's not super practical.

@T.E.D. Almost certainly not. I imagine the seeing would not be good from the top of Mt Everest and nor are there likely to be ~80% clear nights per year as there are at the best sites.

Are there mountain ranges in North Korea or the middle of the Pacific Ocean?

@Valorum Hawai'i.

Could you not first look at a few sites of real observatories and - ruling out politics or economics - try to analyse why they might have been chosen? Look, for instance, at T.E.D.'s Comment, above.

High altitude, so there is less atmosphere to look through.
Dry climate, so there is less water vapour, fewer clouds, less rain.
Low latitude, so more of the sky is visible for at least part of the year.
Remote from non-natural light sources.

And there are lots of non-science considerations: political stability, good transport links, good network connectivity, skilled local labour, no indigenous land rights issues. All these factors have to be balanced against each other.
There isn't a single "best" place, but  Mauna Kea in Hawaii, La Palma in the Canary Islands and Paranal in the Atacama desert in Chile tick a lot of the above boxes.

Surely the steppes can be quite an attractive option if they meet all the requirements?

Well, if the meet all the requirements.

Where are the promising steppes located in this regard?

Also, geologically stable.

There is arguably an additional factor promoting good seeing, which is being on a mountaintop near the ocean, because the inland airflow will tend to be smoother (more laminar) going up and over the mountains than it will be further inland. This is part of why observatories on mountainous islands (Mauna Kea, La Palma) or mountain ranges next to the ocean (Chile) are particularly good.

@gerrit I'm not sure geological stability is actually that important compared to other considerations. Mauna Kea and La Palma are active volcanoes, and the Atacama Desert is right next to a major fault line.

You also often want one place in the Northern Hemisphere and one in the Southern to cover the sky.

The best sites for telescopes are where they currently are and where the even bigger next generation of telescopes are being built. These sites are Mauna Kea in Hawaii, Roque de los Muchachos on La Palma in the Canary Islands, the Chilean Andes (sites near La Serena—Cerro Tololo, Las Campanas—and in the Atacama desert south of Antofagasta—Cerro Paranal), and in the Arizona mountains near Tucson—Kitt Peak and Flagstaff—Lowell Observatory.
For optical astronomy, the main factors that determine the "quality" of the site are the "seeing" (this is the extent to which the atmosphere around and above the observing site blurs the images of stars), the fraction of clear observing nights and the darkness of the night sky when the Moon is below the horizon. All of the sites listed above do pretty well on all these scores.
It's unlikely that there are significantly better sites elsewhere in the world. There has been extensive site testing at various other promising locations, but the ones I've listed generally come out on top.
Of course if you move to longer (infrared) wavelengths then the answer would be different. High altitude is then really important, to reduce the mass of water vapour overhead, and also the atmosphere and telescope are significant thermal sources of background—and so in theory putting a large infrared telescope on a mountain near the South Pole would be a good idea, but very difficult in practice.

Is it possible that the telescope will be placed at some suitable height above the south pole (in the format of an orbital observatory) and maintain a fixation above it?

@dtn There would be no advantage to that and it isn't possible.

interesting, but it turns out that there is a telescope on earth located at the south pole https://en.m.wikipedia.org/wiki/South_Pole_Telescope

@ProfRob If it's not too large, you could attach the telescope to a tethered balloon.  At least airspace is not a problem on the south pole. But indeed, there's no good reason to do so.

@dtn That's not an optical telescope.  Considerations for radiotelescopes are different.

@dtn I have used a (small) infrared telescope sited at the south pole (no longer there), called SPIREX.

@gerrit you're right, I didn't pay attention to that

@ProfRob Have you worked with this telescope remotely or perhaps been able to visit the south pole?

@dtn remotely unfortunately. Before the Spitzer spacecraft it was world-beating at 3-5 microns. Observing trip to La Palma next month though :)

@ProfRob Then good luck on your trip to this beautiful and very picturesque place! :)


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Does the Oort cloud have any large holes in it and if not does this mean nothing large has entered the solar system for a long time?
Would holes in the Oort cloud be a record of what has passed through it?

You should be aware that nobody has ever observed anything in the Oort cloud.  It's existence is infered from long period comets. But anything about the detailed distribution of objects is hypothetical.  Remember also that objects in the cloud are not fixed, they move in orbits around the sun.

So the answers to your questions are "nobody knows" and "probably not"

A buddy asked me for documentaries on either the Oort Cloud or the Kuiper Belt. There's like one and a half of them, all of which are speculation and neither of them were very good.

Recently asked: https://astronomy.stackexchange.com/questions/55039/whats-the-reason-for-the-hollow-region-in-oort-cloud

Note that Oort cloud entities are prone to perturbations and so they can gravitate/accumulate to bodies with bigger mass. Have a read: https://articles.adsabs.harvard.edu//full/1992MNRAS.259...37C/0000037.000.html

@Nilay Gosh That reference is a very interesting read. Stars passing the Oort Cloud reduce the angular momenta of comets so they head in towards the planets. Also the cloud density oscillates and the cloud may gradually disappear as planetisimals get pulled out of it.

"“Space is big. You just won't believe how vastly, hugely, mind-bogglingly big it is..."  Douglas Adams

The Oort Cloud has not been directly observed.
The Oort cloud is still theoretical. We're pretty sure it's there, but we don't have direct evidence. The objects that make it up are too small, too dark, and too far away to observe. We only see comets when they get close enough to the sun for the surface ice to start to burn off and produce a tail of dust behind it. In the "freezer" of the theorized Oort cloud, it would be like trying to see a lump of coal on a moonless night.
The question of whether there are holes in the Oort cloud presupposes we have a good map of the area, which we don't.
If an object made a big hole in the cloud, it would vanish over time.
The Oort cloud (assuming it exists) is made up of many small objects in orbits in a wide range of orientation, distance, and eccentricity. What that means is the orbital period of any given two objects are almost certain to be wildly different, and their paths crisscross throughout the cloud.
If you were to, say, fire a massive laser into the cloud, burning away every object in a huge area (this is not recommended), the hole would vanish over time. It might be possible for a while to detect a "thinness" in one direction over another, but it would probably be impossible to spot after a few thousand years (if you could even see oort objects). Removing those objects would end up just thinning out the oort cloud in general, but not with any directionality.

There would be more high energy cosmic rays coming through holes perhaps.

Although we don't know exactly how many object are in the OOrt cloud, they are certainly pretty spaced out (it's not like a cloud on earth, nor like a starwars asteroid field) There's enough gaps to let light and cosmic rays through.

Long period comets come from an Oort cloud. There is AFAIK no alternative credible model.

@James K We can at least rule out the likelihood that a large star or black hole has passed through the solar system. There are a billion comets estimated and these would have scattered all over the place and perhaps have struck Earth. Of course a star passing more closely to Earth could have disrupted all planet orbits.

@ProfRob Like I said, we're pretty sure it exists, but we haven't mapped it.

@Adrdav22 We are pretty sure that a star passed through the Oort cloud 70,000 years ago.  This is Scholz's star. We know its trajectory and we can see that it came well within the supposed oort cloud.  Gl710 will do so in the not-so-distant future too. We don't know about any "holes"

We're able to detect variations in the CMB that were created when the universe was in its infancy. Once we're able to observe the Oort Cloud, maybe we'll be able to see the remnants of holes.

Doubtful. As I said, the chaotic orbits of oort objects are probably going to mean any disruptions smooth out across the whole cloud in a (astronomically) short time.

Maybe, but not likely.
Let's look at something similar - the Asteroid Belt.
The Asteroid Belt between Mars and Jupiter (not really "between" since some of it extends past Jupiter, and depending on your definition, some is inside Mars' orbit, all the way close to the sun).
The Belt has several gaps and clusters. This is Not even close to scale. Most of those dots are as small as 300 feet.

The Belt's total mass is quite tiny, about the total mass of Pluto. We've tracked many objects there and have a good idea of the total mass.
The Oort cloud's total mass is estimated to be somewhere between the mass of Earth and the mass of Saturn. Saturn isn't really a "large" object astronomically. The Oort cloud is estimated to be 1 to even 3 light-years thick. Scattering the mass of Saturn across even one light year would basically be dust. The distance between objects, even tiny ones, could be multiple AU.
If we go with 3 ly, then the area of the cloud in 2 dimensions is about 27ly squared. It's in 3 dimensions, and lets say it is 1 ly thick. That's 27 cubic light years of volume to fit the planet Saturn.
It is much less dense than the Belt.
As a side note, when NASA sends out a probe past the Belt, they track some of the larger objects and plot a course to not hit them, and ignore everything else since the distance between asteroids is mind-bogglingly-huge.
Thus, since we don't track anything through the Belt this way, it would work even less for the cloud.
But, we have to define "large" in your question. If you mean like a black hole,  or Betelgeuse, then probably not in a billion years. Its gravity would have totally wrecked the whole system, since its size is about the orbit of Jupiter. (Not the size of Jupiter, its diameter is the orbit of Jupiter. It is extra-large.)
Since we are estimating the size of the cloud, and Pluto's orbit is about 5 light-hours vs a few light-years of the cloud, Pluto's orbit gets rounded out.
It is a very interesting theory, and if we could measure the objects in the Cloud, closely, and in real-time, if it exists, then we might could track something planet or star-sized that moved through it.

@Adrdav22 If I'm not mistaking that is the average density in between galaxies, and considering how much empty space there is that is still a mind boggling large amount of matter.

"The Oort cloud is estimated to be 1 to even 3 light-years thick...." Interesting: if Alpha Centauri has its own Oort cloud, the gap between clouds should be of the same order as an average Oort cloud.

@Adrdav22, A=pi r ^2. If r is ~3, then A ~ 27.
Since all these are speculation/hypothesis, it very well could be wrong.

@MikeP is the "three LY thick" counting the radius (i.e. distance from sun or us to the outer "edge") or the diameter?

@Michael I got my calculation wrong. It should have been the mass of Saturn 10^26 kg in 27 light years cubed (10^48 m^3) This is 10^ -22 kgm^3  one billion times larger than my first calculation.


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Have gas or dust or rock accumulated in our solar system on its 20 orbits around the Milky Way?
Also,does the absence of the accumulation of anything large such as rogue planets by the solar system suggest the Milky Way has been orderly and stable for billions of years?

Our solar system has been orderly and stable, our existence proves this.  Dust and gas wouldn't "accumulate" as there is nothing to slow it down and keep it in the solar system. Dust and rocks that enter the solar system from the interstella medium fall towards the sun, speeding up, and then sail away, unless they hit something

@James K  The stability of our solar system over billions of years as it circles the galactic centre suggests that there are no large accumulations of randomly moving cold dark bodies .

You assume that the solar system moves through stuff. Yet everything orbits the galactic center at about the same speed, thus it won't accumulate stuff like you do when driving a Cabrio, but more like you overtaking or being overtaken by other runners at a marathon past half distance: something which happens rarely. And even then James's logic applies

@planetmaker  I was  assuming that there could be some matter moving randomly around in the Galaxy over the last 4 billion years for example stars orbiting the central black hole can be ejected from its vicinity.I think the stability of the solar system shows that most matter moving too fast or too slow to orbit the Galaxy moved to the centre of  or was ejected from the Galaxy early on in its history.

Or that the Galaxy formed initially in a very orderly manner.

How does objects in space accumulate other objects? If we ignore direct, inelastic collisions, the way to capture a passing object is to have an interaction where it loses enough momentum to get gravitationally bound. If it doesn't, it will just sail past. For slow objects tidal interactions might do it, but for fast ones you typically need a third object to pick up the momentum in some kind of three-body interaction.
However, the probability of this being enough goes down rapidly with higher relative velocity. Typical interplanetary velocities are about 10 times the typical interplanetary velocities: it takes a very precise angle and moment for some interloping dust grain or planet to get captured. Old calculations suggest this is very rare (about one comet per 60 million years), but modern multi-planet simulations suggest about 1200 captured (asteroid sized) objects per million years.
Note that since these interactions are time-reversible the solar system also likely loses objects over time.
Another thing to note is that there are many more tiny dust grains than interstellar asteroids or comets, and they in turn outnumber rogue planets. So while we may capture some dust and asteroids, the number of rogue planets captured (or just messing up orbits) is very low even under 20 galactic rotations.

How many rogue black holes or stars are there ?  Would these have left the Galaxy a long time ago?

An article about stars that  escape from their birth place https://phys.org/news/2023-10-astronomers-infant-star.html     Whether or not they can become rogue stars that could affect our solar system isn't discussed.


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If the Earth had another moon would it be better protected from asteroids?
Would the moons get struck by asteroids instead of the Earth or would the moons attract more asteroids and make it more likely the planet gets hit?

It would make hardly any difference. Earth (like all neighborhood-clearing planets) protects itself from asteroids quite well.

I don't quite buy JamesK's claim that "the moon only covers less than 0.001% of the sky, and so leaves us vulnerable to 99.999% of asteroids."
That argument would work if typical asteroids 1) came from random directions, 2) travelled approximately straight through the Earth–Moon system and, if they didn't hit anything, 3) departed never to return.  However, while these might be decent approximations for interstellar objects, none of them are true for typical near-Earth objects (NEOs) that make up the majority of potential impactors.

In fact, a much better approximation would be to assume that an object that passes by the Earth without hitting anything will return to the vicinity of the Earth some number of orbits later, and will keep doing so until it eventually hits either the Earth or the Moon.  Replacing assumption (3) above with this one, while keeping the other two, would imply that the ratio of objects that will eventually hit the Earth vs. the Moon should be approximately proportional to the ratio of the squared radii of the two objects.
Doing the math, this suggests that about 7% of all asteroids with orbits taking them close to the Earth (and the Moon) should eventually hit the Moon, while the remaining 93% should eventually hit the Earth.
Of course, this is still neglecting the gravitational dynamics of the system and the fact that orbits of typical NEOs passing through the Earth–Moon system aren't quite straight lines.  Still, it's at least roughly in line with various claims I found by Googling, such as this passing mention on a NASA web page that "about twenty asteroids strike Earth for every one that strikes the Moon".

Adding a second moon of similar size as the one we already have (and ignoring the unlikely stability of such a system) should approximately double the fraction of asteroids that hit the moon(s), reducing Earth's share of the impacts from 93% down to about 86%.  (In fact it should be a little higher, about (93%)² ≈ 86.5%, to account for the fact that some of the asteroids hitting the second moon would've instead hit the first moon if the second one wasn't there.)
That's way more than a 0.001% reduction in impacts, but still less than a 10% reduction.  So the protective effect of the second moon would be noticeable over geological timescales, but most of the asteroids on course to hit the Earth would still get through.

Wouldn't this all depend on how large the second moon is and how far out it orbits Earth. A large moon a long way from Earth can deflect even large fast moving asteroids enough to stop them hitting Earth. So a large moon could presumably reduce the number of high momentum impacts.

@Adrdav22: If you don't offer a specific size for this second moon (nor any orbital properties), the reasonable assumption is that you're asking about a moon with similar properties. We can't read your mind.

@Adrdav22 Deflection goes both ways: it might be true, but I don't see why gravitational deflection should direct more objects away from Earth that would otherwise hit than direct objects towards Earth that would otherwise miss.

@wizzwizz4  You are right . I suppose asteroids would need to be launched towards Earth with  non random trajectories at non random times for deflection to favour protection.

Neither effect is significant.
Asteroids hit the Earth because they happen to be on a collision course with the Earth.  The Earth's gravity can deflect some that might just miss the Earth onto a collision, but this effect is not great.  So asteroids don't hit the Earth because they have been attracted towards it by the combined gravity of the Earth and moon.
Now it is possible that some asteroids that would have hit the Earth hit the moon instead.  But the moon only covers less than 0.001% of the sky, and so leaves us vulnerable to 99.999% of asteroids.  A second moon might decrease this to 99.998%, but it's not going to be a big change.
So if we had had a second moon, the rate at which asteroids hit us would be almost exactly the same.

this effect is not great Do you know of any calculations that back this up?

So what you're saying is that we need lots and lots of moons...

@Allure You could look at the sizes of the Hill spheres of the earth and moon, which are the regions where one body's gravity dominates. The moon's is only 60,000km in radius, while the earth's is 1,500,000km - whatever effect the moon has on  approaching asteroids, the earth's is much greater over most of the approach. The earth's Hill sphere itself isn't too large at only 0.01 AU, other bodies like the Sun have larger gravitational effects unless you're already pretty darn close to earth.

I don't quite buy this. Your claim that "the moon only covers less than 0.001% of the sky, and so leaves us vulnerable to 99.999% of asteroids" seems to assume that asteroids 1) approach the Earth equally often from all directions, 2) move much faster than the Moon relative to the Earth, and 3) only pass by the Earth (and the Moon) once. While these would be good approximations for interstellar objects, none of them are true for typical NEOs that make up the majority of potential impactors.

…in fact, I seem to vaguely recall reading that the Moon's impact cross-section isn't that much smaller than the Earth's, so that a significant fraction of objects with orbits passing near the Earth in fact eventually end up hitting the Moon rather than the Earth. I fact, just a quick comparison of the physical radii of the Moon and the Earth squared suggests that, ignoring gravitational effects, the Moon should receive about 7.5% as many impacts as the Earth. But the actual figure might be even higher.

I think the question deserves to be entertained even if the magnitude of the effect is insignificant, at least in so far as: does it make it more protected because it's covering more of the sky or does it make it more vulnerable because it's attracting more asteroids.  Or is it literally so insignificant that there's no way to tell?

To deflect asteroids, we'd need a moon about which someone says "That's no moon". But I have a bad feeling about this

If the Earth had another it would have to be a very small moon that would wouldn’t really due any good about taking an asteroid impact it would probably be the same size or smaller than the impacting asteroid. Our planet current moon is actually to big for our planet and has been slowly but surely escaping our planets orbit. If we had another large moon in our orbit asteroid headed toward wouldn’t be our our concern. They gravity if the new moon would pull on our planet along with the old moon as well as them pulling on each other if it doesn’t cause a collision or sling shoots out of orbit into who know where it would most definitely cause gravitational friction that heat our
Planet up and we would lakes and ocean of molten rock. As far as what we have discovered planet wise our planet is special in the fact that we have such a large moon for our planet and whatever event that may have happens for us to have our moon rarely happens but our solar system is the only place it has been observed. Planet similar to Earth any where if they have a moon they are small like mars


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When the sun formed did it have the same chemical elements as the rest of the solar system and does it now?
Did the solar system start out with a uniform distribution of elements or was the distribution uneven?

Does this answer your Q? https://astronomy.stackexchange.com/questions/50258/are-heavy-elements-equally-distributed-throughout-the-solar-system

The Sun was born with abundances that reflect those of the initial gas cloud it formed from. These abundances have since been modified by interior nuclear processes. The rest of the objects in the Solar System have abundance mixtures that are informed by a variety of chemical differentiation processes and in general don't reflect the initial abundances of the protostellar gas cloud. The exception are meteorites, which apart from having little hydrogen, helium or noble gases are otherwise though to be quite representative.
Details
The chemical elements that the Sun formed with are thought to be representative of the gas cloud from which the Sun and Solar System formed.
Those elemental abundances have since been modified by nuclear reactions within the Sun. For example, most (99%) of the initial Li has been burned; the initial deuterium has all gone; some of the initial hydrogen (about 40% of the core hydrogen, which is a bit less than 10% of the total hydrogen the Sun was born with) has been transformed into helium; there have been much smaller changes in the abundances in carbon, nitrogen (increased) and oxygen as a result of the CNO cycle which operates at a low level in the Sun. Note that working out the true average abundances in the Sun is complicated by the fact that there is some chemical differentiation and diffusion that will take place over billions of years (e.g., Asplund et al. 2021).
The raw material for the rest of the Solar System was the same, but it was drastically affected by processes that occured before and during the formation of planetary and other solid bodies.
For example, there is a tendency for solids to form close to the mid plane of the circumsolar disk that forms around the Sun; only solids with low volatility can form close to the Sun and these tend to be rich in heavier elements; more abundant ices, rich in hydrogen can form further out. Much of the hydrogen and helium in the initial gas cloud was not incorporated into solids and is lost when gas is expelled from the early Solar System, though some fraction of it was accreted onto the gas giants. The gas giants Jupiter and Saturn are the closest in terms of the overall abundance mixture to the composition of the Sun and the protosolar gas cloud.
Meteorites are an interesting exception. These were formed very early in the protosolar environment. Aside from having very little hydrogen, helium, noble gases and being somehwat depleted in carbon, nitrogen and oxygen, the abundance mixture of everything else (often expressed as a ratio to silicon) is probably very similar to the protosolar gas cloud and indeed to the present day Sun, aside from the elements discussed above (e.g., Lodders 2003).

When magnesium from Earth ' s crust is low in concentration in the sea then life is less abundant because it is required for enzymes. If we looked at the emission lines of a distant star in a protosolar disc and didn't see magnesium we could conclude that there would be no life in the developing solar system comparable to life on Earth because in keeping with the information you provided we would expect no magnesium orbiting the protosolar sun. Do you think this a reasonable deduction.

Should have said protoplanetary instead of protosolar!

It would be a reasonable deduction that any planetary system would have little magnesium. @Adrdav22

Note though, that if all metallic elements were low it might still be the case that any solids that exist around the star could still have an Earth-like composition. It's the ratio of say Mg/Si or Mg/Fe that is important, not Mg/H. The overall metallicity of the star probably just determines how likely planets are to form at all.

If I was an alien looking at our sun from another solar system then perhaps the 0.07 % of magnesium in our sun would mean to the alien that there is sufficient magnesium for enzymes on a planet orbiting our sun because magnesium is  widely distributed across our solar system.  Using other elements the alien could determine which stars are most likely to have life on them with biochemistry similar to ours.


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Why does the metallicity of stars decrease with increasing galactocentric radius and height above the galactic plane?
Metallicity decreases with the Galactocentric radius and height above the Galactic plane (Bergemann et al. 2014; Duong et al. 2018).
https://www.aanda.org/articles/aa/abs/2014/05/aa23456-14/aa23456-14.html

Did you read the paper and follow up any of the references in it discussing that topic?

The paper does not explain why metallicity decreases. It is a summary of how metallicity is distributed. Thanks to ProfRob I now have a better understanding of why.

There's a combination of things going on.
Stars form in the disk and bulge of the Galaxy where most of the gas gathers. A lot of the chemical enrichment, particularly of "alpha elements" - Mg, Si, S, Ne then happens promptly (within 10-20 million years) from exploding core collapse supernovae. In such short lives, the progenitor massive stars have not moved far from their birth places, so the gas just gets more enriched in situ.
Iron-peak elements are formed mostly in type Ia supernovae. These exploding white dwarfs are older than the progenitors of core collapse supernovae and do migrate from their birth places. Nevertheless there is still some tendency for the enrichment to take place closer to the central bulge and Galactic plane and of course, gas that is above the Galactic plane tends to fall towards the Galactic plane in any case.
Stars that are found above the Galactic plane tend to be older, because the vertical speeds of stars are gradually increased by various tidal interactions during their lives. If they are older stars then they were formed from gas that was less enriched.
Very old, very metal poor stars may have formed before the gas disk formed and these are now more evenly, even spherically, distributed compared with "population I" stars in the disk.
The decrease with radius is again because if there are fewer progenitor stars being born at larger radius (the rate of star birth and the stellar density decreases with increasing radius) then there is less enrichment of the interstellar medium at larger radii.

Our solar system is vertically half way across the Galaxy and horizontally too. Where there is not too much and not too little metal presumably.

Planet sizes are influenced by metallicity, so there should be more  planets closer to the galactic centre that are large (high metallicity) and planets further out should generally be smaller ( low metallicity).


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Could the Methuselah star be evidence that there was a universe before this one?
If the Methuselah star proved to be older than the universe could it be from a previous big bang big crunch cycle in which not everything was crunched? What would be the smallest radius the universe could have  without the temperature getting so high that Methusala would be completely ionized and destroyed by a crunch?

Related: https://astronomy.stackexchange.com/questions/19446/how-can-hd-140283-be-older-than-the-universe?rq=1

I’m voting to close this question because the premise is false, so no reasonable answer is possible.

@James K    Christopher James Huff gave a reasonable answer below that shows  my premise is wrong. Methuselah would be expected to have higher metallicity if I had been correct.

@ProfRob The link you have points out that it is surprising a star like Methusala could form so soon after the Big Bang.Perhaps the explanation relates to why galaxies are too large too soon.

I find this question interesting mainly for the demonstration of a subtract operation exhibiting total loss of precision.

First, if the Big Bang did originate with the collapse of a cyclic universe, the collapse reached densities and temperatures sufficient to not only ionize matter from the previous cycle, but to convert it into a quark-gluon plasma and restart the process of baryogenesis and nucleosynthesis, beginning this cycle with essentially nothing but hydrogen and a small amount of helium. A star would not survive this.
Second, HD 140283 is a metal poor subgiant. This is consistent with formation in the very early universe, before significant metals were formed, and is the opposite of what you'd expect from one of the last stars to form in a previous cycle.


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Why would rocky earth-like planets be rare around white dwarf stars?
I came upon an article here https://phys.org/news/2023-11-white-dwarfs-life-planets.html
The author thinks that rocky earth-like planets could be rare around white dwarfs.What could be the reason for this?

Short answer
Most planets of this type may have formed too close to their parent star (the progenitor of the white dwarf) to have avoided being swallowed when it became a red giant.
Another possibility is that such planets, even if they could survive are either not scattered inwards so that they can be observed as transiting exoplanets or they do not survive that scattering process and get broken up by tidal forces.
Details
Rocky planets are likely to be formed close to their parent stars. It is only in the hot, dense parts of the protoplanetary nebula that sufficient rocky solids can form and aggregate into planets without also accumulating a big gas envelope.
The progenitors of white dwarfs will have gone through a red giant and an asymptotic red giant phase. During these phases their radii would swell to encompass a large fraction of the region where it is most probable that rocky planets would have formed.
If those planets are not engulfed immediately during that process then it is quite likely that they will spiral into the red giant star anyway on a short timescale, because their orbital periods will likely be faster than the rotation periods of the expanded giants. This raises tides in the giant star and extracts angular momentum from the orbit in order to spin the star up. As a result, the orbit decays and the planet will be engulfed.
It is possible (see below) that there are orbital radii from 1 au to a few au where rocky planets could form and might survive the red giant phase. For those planets then to be detected as transiting the remant white dwarf as short-period exoplanets they would need to be scattered inwards somehow. Another class of explanation for the lack of observed rocky transiting planets is that either they do not get scattered inwards efficiently or that they do not survive the scattering process, getting too close to the white dwarf and being broken up by tidal forces.
Further details, comments and context
This question has a number of nuances that my broad-brush answer rather skips over. If interested, then read on...
The context of this question is the recent paper by Kipping 2023. The paper addresses the curious fact that the only confirmed transiting planet around a white dwarf is a giant planet. Kipping points out that if the distribution of stellar mass/radii is similar to that inferred from Kepler/TESS observations of short-period transiting exoplanets around main sequence stars then, that the first discovered transiting planet around a white dwarf is a giant planet would not be expected (with a probability $<0.5$%). Since white dwarfs are small light sources (roughly the size of the Earth) then finding transits from the the more common (in the Kepler/TESS samples) Earth-sized planets should be possible and in fact ought to be much more likely. Hence, Kipping claims that rocky transiting exoplanets around white dwarfs are rare.
i.e. The claim is actually that rocky planets in short period orbits are rare around white dwarfs, not necessarily that rocky planets are rare - though that is clearly one possibility to explain what is going on. Another is to think about how a transiting exoplanet gets to be where it is found. Such objects around a white dwarf cannot have formed in situ and neither can they have migrated there during the lifetime of the progenitor star. They must have scattered inward after the white dwarf had formed. So another possibility is that there is something inherently unlikely about scattering a rocky planet inwards and so even if they are relatively common at large distances from the white dwarf, they may not be found as transiting exoplanets. A variant of this explanation is that the rocky close-in planets invariably end up being broken into pieces by tidal forces and that might account for the observation of a number of debris disks around white dwarfs and the pollution of white dwarf surfaces by heavy elements.
A further nuance to this question (or at least to my answer) is that the likely critical dividing line between the formation of rocky planets and the accumulation of gas to produce giant planets is probably the snowline (or frost line) beyond which more abundant ices could form. This line is at about 2 au for a solar-type star and this coincides (roughly) with the size to which a red giant Sun might swell up to.
However, the progenitors of the white dwarfs that exist today are predominantly stars that are more massive than the Sun, since these are stars that must have had short enough lives to have created white dwarfs within the lifetime of the Milky Way. The position of the snowline scales roughly with stellar mass, so the 1.5-2 solar mass progenitors of most of the white dwarfs we can actually observe today could have had snowlines at 3-4 au, that are beyond their radii as red giants.
Further, the progenitors of these white dwarfs must have lost significant mass during the red giant phase - via a powerful, radiatively driven stellar wind. This results in any planets that are around them moving outwards because of the reduction in gravitational binding energy. For a 1.5-2 solar mass progenitor, which loses 60-70% of its mass before becoming a white dwarf, this would increase orbital radii by factors of 2-3. Thus a planet at say 2 au (well inside the snowline for a $2M_\odot$ star) might end up in an orbit at 6 au and probably far enough away from any red giant to survive engulfment or tidal inspiral (tidal forces scale as $a^{-3}$).
Thus my answer is a simplification. It could be that there is parameter space to form terrestrial planets around 1.5-2 solar mass stars in wide enough orbits that they survive the red giant phase - in which case the lack of any detection of inwardly scattered examples around white dwarfs remains to be explained, which is the conclusion of Kipping's paper.
We actually know very little about the frequency of Earth-sized (and smaller) exoplanets in wide orbits around stars. Transit detection is only sensitive to short-period close-in planets. Doppler surveys are not yet sensitive enough to have detected Earth-mass planets at $>1$ au separations from stars. Microlensing has some sensitivity at orbital radii of a few au, but perhaps not to low enough masses. The plot below shows the inferred mass versus orbital radius diagram for exoplanets, colour-coded by discovery method - red for transits, blue for Doppler detection, green for microlensing and orange for direct imaging. I have marked on a domain where we might want to detect terrestrial planets at say $1-4$ au such that they might be able to form and yet survive the red giant phase. I have also circled in black the information for exoplanets around stars more massive than $1.5M_\odot$.
What you can see is a total absence of detected small exoplanets in this region. The absence might be genuine, in which case the lack of rocky transiting exoplanets around white dwarfs would be explained. But it could still be observational selection, since none of the techniques are really sensitive enough to detect exoplanets in this region of parameter space.

Hm, I wonder, if the mass loss during the supernova will also have an effect, in the other direction. If the star kept those rocky planets during the red giant phase, the sudden loss of mass will lead to destabilized orbits. Could this be enough to eject some of the planets? (maybe only when interacting with a third body)

@Chieron I have aded much more about this. Yes, mass loss could lead to some rocky planets surviving the red giant phase. If they are ejected though they could not be observed as transiting exoplanets! There is no supernova involved in the creation of a white dwarf - the mass loss is just via a stellar wind.


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Why did Theia create only one moon if it struck Earth?
If I dropped a heavy ball into some magma it would splash all over the place.  So why would Theia create only one large drop of liquid (the moon) and if it impacted the Earths mantle (Bits of Theia might be in Earth's Mantle)?
What did it do to the Earth's core?

The impact created lots of splash, but that was a long time ago and all the bits either became part of the Earth, part of the Moon, or left the Earth-Moon system.

Do note that one recent theory is that two moons were created in the impact and those then merged together.  This helps to explain the vast geological differences between the front and back sides of the moon. https://www.space.com/12529-earth-2-moons-collision-moon-formation.html

Most simulations generate multiple moonlets and streams that then fall into Earth, merge into the moon or escape (e.g. like this or this).
The initial orbit tends to be close to the Earth. This means that gravitational interactions will will be strong, making it hard for there to remain multiple moons since they will tend to destabilize each other's orbits. Simulations have shown that Earth at most can keep three moon-sized moons, and that requires careful arrangement so their Hill spheres do not overlap and they do not get into the wrong orbital resonances. The moon's Hill sphere has a radius of 60,000 kilometres, so its early very near-Earth orbit after the impact would have interfered with anything else orbiting nearby. Had it been lighter maybe extra (small) moons might have made it.

The simulations look convincing. But I still think that at a greater distance from Earth there could have been blobs of magma orbiting. I suspect that magnetic and electric fields have not been fully accounted for here. And over time they have repelled some magma into space and the rest onto the Earth and moon.

@DaveTheWave - To send a blob of magma into a high orbit it needs to have a high velocity and then have it changed somehow to circularize it. Otherwise it will just have a very eccentric orbit and likely fall back in. This requires forces stronger than any plausible magnetic effects (and electric charge will be neutral on this scale).

@Jon Custer  Where is the magma that left the Earth moon system.Is it in the asteroid belt? Did the sun and other planets attract it all.


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What happens when two galaxies each with relativistic jets collide?
Would the relativistic jets create new particles where they meet and would all the gas from the two accretion discs be blown away leaving just two black holes?


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What are the effects of mass concentration along a planet's axis of rotation?
Reading about geological history of Mars and coming across the Martian dichotomy, which is about having the two hemispheres' geography differ in elevation by 1 to 3 km.
Does this mean that Mars' center of mass is slightly shifted to the south? and if so, more generally, what is the influence of a concentration of mass aligned with the rotational axis of a planet that is not tidally locked?

Does this mean that Mars' center of mass is slightly shifted to the south?

Yes, by about 3 kilometers, and it is also shifted by about 1.5 km along the y axis in the direction of Tharsis. See Table 4 in Smith et al. (2001), which shows, among other geodetic parameters, the ellipsoid offset of COF (center of figure) from COM (center of mass):

parameter
value

Δx
−233 m

Δy
−1428 m

Δz
−2986 m

What is the influence of a concentration of mass aligned with the rotational axis of a planet?

According to the same source, this morphology have impacted sediment transportation:

The 3-km offset along the z axis between Mars' COM and COF (cf. Table 4) results in a non-zero $C_{1,0}$ term in the spherical expansion of topography (equation (8)). [...] While this zonal elevation change is complicated by both large- and regional-scale topographic features, it is apparent that this fundamental aspect of Martian topography has controlled the surface and subsurface flow of volatiles and sediments throughout much of Martian history. As discussed more below, the global-scale elevation change associated with the $C_{1,0}$ term of the topography field resulted in the northern hemisphere becoming a global-scale sink for transported material.


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If w bosons can create dark matter neutrinos by decay, can they also create dark energy?
Dark energy is often associated with vacuum energy fluctuations. But the experimentally measured Casimir effect suggests that there is almost no vacuum energy.Because most of the vacuum energy is supposed to be electromagnetic (virtual photons) and the weak force , mediated by w bosons is linked to electromagnetism through electoweak theory, has the energy of electromagnetism in the vacuum been converted into weak force energy - did virtual photons become w bosons at the time of the big bang and are  these a bosons now decaying into various other particles to yield dark energy?

The W boson has a lifetime of $3\times 10^{−25}$ seconds. So any big bang W bosons will have long decayed. So they are not a dark energy source. Also, they produce "hot" relativistic decay products, so they are not the source of the "cold" dark matter we see forming halos.

The w boson can only decay quickly if its decay products enter an environment in which their energy is low. The fact that dark matter has high energy (5 times mass of bayons) means that the decay products may not be entering a low energy environment. Although why nuclear decay W bosons seem now to be in a low energy environment and the dark energy e bosons are not would need to be explained.


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If 42000 Starlink satellites interfere with telescope observations from Earth would someone observing from another solar system detect them too?
SpaceX's Starlink constellation could place up to 42000 satellites around the Earth. Would aliens with their own telescopes notice them? They would be about 150 km apart and could reflect light of this wavelength produced by the Sun's corona. Waves bouncing off the ionosphere could perhaps form a detectable diffraction pattern.

The Earth itself would be way easier (but still very difficult) to detect.

It's not clear what you mean by "distant planet"  Do you mean like "Mars" or do you mean a planet around a different star.

@Greg Miller The satellites would be about 10km apart and the Sun's corona produces these wavelengths which can bounce off the satellites.

@James K I have edited the question and it's body to clarify that I mean another solar system and 10 km wavelength light waves.

It would take millions of satellites to cover the Earth with 10km gaps.  I don't really get the connection to ultralong wavelength radio waves, or why you think a space network of satellites would reflect ultralong wavelength radiowaves.

@James K.  I have got the satellite separation wrong.Should be around 150 km.  The satellites I was thinking could diffract waves bouncing off the ionosphere (which 10km waves are known to do) and someone in another solar system might detect the diffraction pattern.Diffraction patterns are strong when slit width is about the same as wavelength. But would the signal be intense enough to detect?

Utterly undetectable. Even assuming millions of satellites, from a nearby solar system the separation of the Earth and the sun would be less than an arcsecond.  One would need a colossal interferometer to even separate the Earth from the sun.  And then the sun would be so much brighter than the reflection from the Earth. Of course you can say "alien technology" and who knows what those colourful characters can build... but no.  Starlink satellites would not be easily detected from an exoplanet

If I hold a dime in front of your eye, you won't be able to see anything but the dime.  The dime is still invisibly small from a mile away, though.

@James K   Alien technology  might involve a gravitational lensing black hole that can make photons converge onto a smaller interferometer!

Yes, that's why this question is unanswerable! To write an answer we would first need to discover aliens.

Not in any way that is special to the Starlink satellites. Many satellites can interfere with earth-based astronomy because they are very close to the earth. A reflection off a satellite that is a few hundred miles away can be brighter than a star that is a few tens of trillions of miles away. But, from a different star, both our sun and the satellites are a few tens of trillions of miles away, and the satellites' reflected light is negligible.
Of course, it is possible to imagine faraway supertelescopes that could detect human artifacts. But such a telescope could also detect the ISS and Iridium flares, among many other things. The Starlink satellites' nuisance effect for earthly astronomers doesn't really make them easier to detect from Alpha Centauri.


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Consistent values for density of galaxies between degree squared and steradian
I have a table of densities of galaxies :
Expected number density of galaxies for photometric survey per unit area and redshift intervals, $\mathrm{d} N / \mathrm{d} \Omega \mathrm{d} z\left[\mathrm{sr}^{-1}\right]$ and the corresponding density of galaxies per $\operatorname{arcmin}^2$ for each redshift

I wonder if the second row values are correct : indeed, I hesitate between both calculus, for example for the bin :

case 1

3 / 11818102.860 * 0.119 = 4219062.72 (rounded to 4219063) in units $\text{d}N/\text{d}\Omega\text{d}z$
OR should I set rather :

case 2

3 / 11818102.860 / 0.119 = 297935366.218 (rounded to 297935366) in units $\text{d}N/\text{d}\Omega\text{d}z$
One of both is wrong since I don't know if the units are $\text{d}N/\text{d}\Omega\text{d}z$ or $\text{d}N/\text{d}\Omega/\text{d}z$.
Could anyone help me what is the convention for the units of the writing $\text{d}N/\text{d}\Omega\text{d}z$ that causes some confusions ( we don't know if we have to multiply or divide by $\Delta z$ ?


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How to plot right ascension with time of the retrograde motion of mars?
Measure the right ascension, declination of mars and sun. plot the variation of right ascension with time.

Would [astronomy.se] be a better home for this question?

No.  There is no question here, just a request.  @123Vega Please clarify what your question is.  Note that we aren't going to do your homework for you.  You need to measure the right ascension of mars and plot the variation with time.  We also expect you to do basic work first, like know how to plot a graph. I've voted to close, as in its current state, it is unsuitable for [Astronomy.se]


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Does dark matter exist because as Einstein said the laws of physics should be the same inside and outside a black hole?
Spacetime is said to be time like inside a black hole ( two events can happen at the same place at different times). Spacetime is called space- like outside a black hole ( two events can happen simultaneously at different places outside the hole.
Albert Einstein didn't think the laws of physics should change like this. I am asking : could dark matter be created by a collapsing star so that neutrons don't speed up too much as they fall into the hole because the energy they should have gained from gravity is used to create dark matter .This could mean that because neutrons haven't sped up that for them gravity doesn't get stronger and spacetime remains space like as Einstein would have wanted. Is there anything obviously wrong with this idea - are there experimental observations that could rule it out?

The simple answer is: no.
The reason is that there is no mechanism preventing matter from speeding when falling into a black hole (it cannot move faster than light according to the local light cones, even though they are weirdly tilted according to an outside observer), and that there is much more dark matter than there has ever been stars or black holes.
Black holes are defined using Einstein's physics, and the reason their interiors work as they do is exactly that general relativity is assumed to be valid inside and outside the event horizon.

If the entire universe collapsed then the gravitational energy could become greater than the mass/ energy of all the stars and black holes. Any dark matter created by my proposed mechanism would be dispersed by inflation to create galaxies.


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How to convert flux to flux density for a synthetic spectrum?
I have several model SEDs generated by the CLOUDY spectral synthesis code. The data is formatted with one wavelength column and one flux column. I wish to perform Hubble synthetic photometry on these spectra using the Python synphot package. Unfortunately, the package requires the flux density units in my spectrum to be in erg/s/cm$^2$/AA, while my synthetic models have flux expressed in erg/s/cm$^2$.
I've tried to convert the flux to flux density by dividing each flux value by its corresponding wavelength, though I don't think this is right, as most sources say that flux density must be obtained by dividing flux over a wavelength/bin range. Theoretically, since I'm trying to calculate flux densities for individual wavelengths, wouldn't the flux density become infinite due to the width being 0?
In total, the question is: How would I convert flux to flux density for tabular spectral data where flux is given for each wavelength?


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Red supergiants with luminosities significantly above the Humphreys-Davidson limit?
The M33 red supergiants (RSDs) LGGS J013312.26+310053.3 and LGGS    J013339.28+303118.8 both have luminosities significantly above the Humphreys-Davidson (HD) limit (logL 5.76 and 5.68 [source], current research puts the RSG luminosity limit at around 5.5). Even the most luminous cool supergiants like Stephenson 2 DFK 49 and HD 269551 A have much lower luminosities (logL is around 5.59 for both [source] [source]). So how are those two stars this luminous? They probably aren't foreground stars.

You should explain the abbreviations you use at least once. RSD and HD are not self-explanatory, and ambiguous, and probably only understood by people who deal with these topics on a daily basis. I did that for you here now. Maybe even the link I used to explain HD already helps as those authors investigate the variation of the HD limit wrt metallicity


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What does the Earth look like when viewed from Saturn?
Are there any data from spacecraft that visited that planet? How large the Earth is when viewed from that planet?

Note that data from spacecraft is unnecessary. We know the characteristics of the Earth and its distance from Saturn, and mathematics can do the rest. We would never have to visit Saturn to know what Earth would look like from it.

Please check the answer of https://astronomy.stackexchange.com/questions/47335/which-planet-viewed-from-which-other-planet-would-appear-the-largest

different but related: How will planets behave in the night sky as seen by Mars colonists?

Perhaps the question is, when viewed from the "surface" of Saturn?

Odin, a very simple way to think about it.  All the planets are "about" the same size. (It's not like there's a planet that's the size of a sun, a galaxy or a pick-up truck!)  If you look up tonight all the planets look "about" the same size to your eyes - basically a dot. The Earth's going to be the same.

Visit a solar system model close to you and see for yourself!  The angular sizes are preserved.

Yes!
It's a tiny blue dot of just about 1.6 arc seconds diameter. Cassini made the most famous image of the "In Saturn's Shadow – The Pale Blue Dot", in remembrance of the first such image from Voyager one: a view of Earth and Moon from Saturn (image: NASA/ESA).

The dark side of Saturn, its bright limb, the main rings, the F ring, and the G and E rings are clearly seen; the limb of Saturn and the F ring are overexposed. The ‘breaks’ in the brightness of Saturn’s limb are due to the shadows of the rings on the globe of Saturn, preventing sunlight from shining through the atmosphere in those regions. The E and G rings have been brightened for better visibility.

Earth, 1.44 billion km away in this image, appears as a blue dot at centre right; the Moon can be seen as a fainter protrusion off its right side. The other bright dots nearby are stars.

JPL shows a few more images, including a zoom image taken of the Moon-Earth-system.
See also the wiki entry on "pale blue dot" as part of the impressive "family portrait" series taken by Voyager 1 in 1990 from a distance of about 40 AU which show most of our solar system planets from 6 billion kilometers distance. Also compare with the successor series taken by the messenger spacecraft 20 years later in 2010 from about the orbit of Mercury.

Not sure whether the OP wanted an angular or pixel size, but could address that part and provide a size of Earth in arcseconds ?

pixel size depends on used camera (and can be checked by following the link to the full-size image), but angular size is fixed for a given distance. I added that. Thanks :)

So this image is fake ?

No, what makes you believe that?

Wikipedia article about this image: https://en.wikipedia.org/wiki/The_Day_the_Earth_Smiled

Just to clearify: Voyagers "pale blue dot" was part of the "family portrait"; it was not taken when Voyager was near Saturn.

@Abigail thanks noted and clarified in the text.

Maybe @An_Elephant was referring to the fact that the image had been intentionally modified, e.g. "The E and G rings have been brightened for better visibility." I wouldn't call that "fake," though.

As a once-upon-a-time photographer I'm used to thinking in terms of "was this taken with a 50mm lens (human eye comparable for 35mm film format)" or a long lens that can change perspective effects. We've all seen the dramatic moon/landscape photos. What focal length lens was used for the Saturn/Earth photo?

Thanks planetmaker for providing this absolutely breathtaking photo! I've long admired the 'Blue Dot' photo that Sagan finally got them to take. I think this Saturn/Earth photo is equally breathtaking.

The specs are readily avaliable: "Wide Angle Camera [WAC](20 cm f/3.5 refractor; 380-1100 nm; 18 filters; 3.5°x3.5°)", e. g. on https://science.nasa.gov/mission/cassini/spacecraft/cassini-orbiter/imaging-science-subsystem/


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How to calculate 2-10keV Flux or Luminosity by Power Law from 0.2-12keV with known $\Gamma = 1.7$
I have $0.2-12$keV flux observations, distance observations and $\Gamma =1.7$. I want to apply power law to calculate 2-10keV flux. How can i do that?