Total rotational angular momentum estimates for Jupiter? I wrote a quickie answer where I estimate the total rotational angular momentum of Jupiter as about 7E+38 kg m^2/s. However it bothers me because one of the sources gets this number assuming uniform density, and the other two cite 6.9E+38 kg m^2/s but it's not clear to me where this comes from.

Are there better estimates that use reasonable models for Jupiter's density distribution?

update: If not answerable due to uncertainties in density distribution and/or gradients in rotation rate (i.e. non rigid-body rotation) which I sort-of remember reading about recently, then an explanation why a better value isn't available would be an acceptable answer.

A better estimate might be to use the moment of inertia of a $n=3/2$ (fully convective) polytrope, which will be a good approximation in an object like Jupiter, even when it approaches electron degeneracy in its interior.
You can look this up and it is given by $I = kMR^2$, with $k=0.205$ (almost exactly a factor of two smaller than a uniform sphere because the mass is concentrated towards the centre).

If we use a mass $M$ of $1.898 \times 10^{27}$ kg, the equatorial radius of Jupiter ($R=71500$ km) and the rotation rate determined from its magnetosphere ($P= 9.93$ hours), then the angular momentum ($2\pi I/P$) is $3.496 \times 10^{38}$ kg m$^2$ s$^{-1}$.

If instead, we use the volumetric average radius of $R=69900$ km, this reduces to $3.34\times 10^{38}$ kg m$^2$ s$^{-1}$.

This simple approach neglects the more complicated equation of state of Jupiter, it's non-sphericity and the probable presence of a solid core. Consideration of these, together with constraints provided by the measured even harmonics of the gravitational field (referred to as $J_2$ and $J_4$, led Helled et al. (2011) to suggest that $k=0.264$, with an uncertainty below 1%. Combined with the average radius, this gives $4.30\times 10^{38}$ kg m$^2$ s$^{-1}$.

More recent work by Ni (2018) has used refinements to the gravitational harmonics from Juno measurements, along with a more sophisticated interior model to suggest $k=0.274$ with 0.5% uncertainty.


Do meteor showers start slowly each night? Does the frequency of the number of visible meteors during a daily shower start off lower at the beginning only to increase as the night goes on or does the frequency remain the same for the entire night?

The number you are able to detect increases as it gets darker.

The number occurring is apt to be greater when the zenith of the sky is closest to pointing in the direction the earth takes orbiting the sun
(like a windshield hitting more bugs than back window when a car is moving forward)

The last factor I can think of at the moment is the density of comet tail material earth happens to be passing through (in its orbit around the sun)
Which could unfortunately be not when your local the zenith is 'forward in our orbit' and when it is dark.


What are the Kinematics / Dynamics behind the motion? https://www.timeanddate.com/eclipse/in/@8133395?iso=20200621
The link I have attached is a predicted model for the eclipse in my area! I am very much confused about the movement of the moon analogous to the video in the link. Community, The movement seems like a parabola curve and no-doubt the model is telling it very well. In other ways, the motion also appears as bouncing of an object! I might have interpreted it in wrong sense or maybe not but that's no matter of concern ;-). A detailed answer would be a piece to take away from this question. Also, I suppose it to be a matter of Orbital Dynamics.
-

Imagine you are watching a car drive down the street from right to left (in front of a building.  It goes from right to left.
Now imagine watching that car with your head on your left shoulder.  The car goes from the bottom of your point of view to the top.
If you move your head as the car is driving past, you will see it come from the right but leave to the top of your point of view.  This doesn't mean that the car changes direction, only your head is turning.
This is relevant because the rotation of the Earth means that over the time of the eclipse, your head turns. If you make a representation that keeps the horizon level and the sun in the centre, the motion of the moon relative to the sun will appear to curve.  This is nothing to do with the actual motion of the moon it is just an effect of the turning of the observer by the spinning of the Earth.


Moon's unusual gravity We know that the Moon's gravity is about one-sixth that of Earth. Then I recently read that the mass of the Moon is about one-eightieth the mass of Earth. Since gravity depends on the mass of the 2 objects, shouldn't the Moon's gravity be one-eightieth the gravity of Earth? Is the Moon extra dense?

As you said, the mass of the Moon is 1.2 percent that of the Earth.
Now, if you mean the gravitational acceleration at the surface, it is calculated like this $G\frac{M}{R^2}$, where $M$ is the mass, and $R$ is the radius of the celestial body. The moon's mass is a hundred times smaller, but the radius is four times smaller, meaning its surface gravity will be $100/16 \approx 6$ times smaller. Considering the factor is mass over radius to the power of two here, density alone does not help you determine the ratio of surface accelerations. On a side note, the moon's density is around 60 per cent that of the Earth.


In what parts of the world is it impossible to view the night sky due to light pollution? Growing up far from any city, I have always been able to view the full night sky. It was just the way how things were. Of course, I knew that light pollution is a thing, and that the sky is a lot less clear if you live in a city.

It just never occurred to how bad it was before talking with people who had experienced it.

Apparently:

Inside a city, it is pretty much impossible to view anything.
It is not a lot better just outside a city, the sky is still mostly empty.
Most of Europe, parts of the US, India, and eastern Asia is so polluted with light that you can not see the Milky way.

Does anyone have a map showing the affected area? How many people have never seen our own galaxy?

Googling for 'light pollution map' gives a pretty good looking result. Those things usually are created using satellite measurments that are susceptible to scattered light (by measuring the polarity of light) in order to separate direct illumination from actual light pollution.

I also suggest you visit a city and then compare the light pollution index on this site to what you know as home, in order to get a feeling for the severity of the numbers.

I cannot answer your question about how many people have never seen our galaxy, as this is a function of travel willingness of each individual living in a city and therefore maybe even impossible to answer for our intelligence agencies.


Does this smartphone photo show Mars just below the Sun? I live in Essex, England (51.7678° N, 0.0878° E). On 25 July 2019 (hottest day ever in the UK, btw), 06:43 BST, I took this smartphone photo of the Sun. Is that white dot just below and left of the Sun the planet Mars? I've checked with my SkySafari app and it looks like it could be Mars, but I'm just surprised that my relatively cheap phone could take a picture of the planet. Although the photo is quite dark, it was a bright day. Thank you.

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

Here is a photo I took with the same artefact. It appears for me as a cyan dot with a magenta outside, but other photos I’ve taken have one with the same colour as the sun.

Notice if you zoom in on yours, there’s a halo around the artefact as well.

Also notice that the artefact is directly below the sun in my image, whereas in yours it’s slightly to the left. However, this is because in your image the sun is slightly to the right, and hitting the lens at an angle.

As uhoh says, you can see that the sun is the same distance from the centre of the image as the lens flare (or very close to):

It should be possible to take a photo with the artefact wherever you want, by re-angling the phone - including inside the sun, where it will be drowned out.

I actually took my phone into an Apple Store to find out what this was, where they explained the lens flair effect. Unfortunately I don’t have a source for this, as it was in person. Hopefully the above is convincing enough.


Could liquid water have existed in open space 15 million years after the Big Bang? Around 15 million years after the Big Bang, the ambient temperatures was about $24^\circ {\rm C}$, which is in a range where water could be liquid. Could liquid blobs of water be existent then?
PS: I am not talking about water on the surface of any solid planet.

Let's interpret your question to be about whether the conditions would permit blobs of water to remain liquid, whether or not water existed yet. And the answer is No, because the pressure was by then far too low. Basically, space was already a vacuum, just not as hard a vacuum as intergalactic space is now.
It is appealing to imagine an era when the universe was simultaneously dense enough and cool enough for liquid water (and thus perhaps humans) to exist. But alas it is not so. At the time of the creation of the cosmic microwave background, around 370 thousand years after the Big Bang, the temperature was around 3,000K, but the pressure was around $10^{-17}$ atmospheres (see the Wikipedia article Chronology of the Universe, and search for "Recombination").


The largest discrepancy in the history of science I watched a cosmology lecture given by Leonard Susskind that convinced me to put my money on dark energy being a pressure caused by energy that is evenly distributed throughout the "vacuum" of space. However, quantum physics predicts that there should be a much larger energy density (i.e. many magnitudes larger). Have there been any more recent findings or theories that could shed light on this discrepancy?

There are roughly two possibilities: either there isn't a large vacuum energy, this would imply that there is something missing from quantum field theory. We don't know what it could be.

Or the zero point energy really is as large as QFT predicts, but there is something else that prevents it from having a large cosmological effect. We have no idea what that could be.

We might get more of an idea if we find out what Dark Matter consists of, or a good quantum theory of gravity we might know more.


How fast can and do gravitational waves travel? I was thrilled by the recent announcement of the detection of gravitational waves. The media has done a great job explaining them and the theory behind them and I get the concept fairly well.

But if it's a warping of spacetime, not a particle phenomenon,  (or is it somehow?) How fast do these waves travel through space? Is it  constant or variable? Slower than the speed of light or even faster? If spacetime has no resistance per se, the what would keep them from traveling infinitely fast?

I'd love to understand it better and hope my question makes sense.

Gravitational waves travel at the same speed as the speed of light in a vacuum, in general relativity. In special relativity, this is the maximum speed for any interaction in nature.

If gravity was faster than the speed of light, then predictions we make according to the bending of light would be incorrect.
If gravity was faster than the speed of light, then the orbits and positions of the planets would be completely different.

Those are my two main points, to help me prove to you that gravity travels at the speed of light.

Also, light is not the only thing in the universe which travels at the speed of light. It is about all massless particles! Photons and gluons are two good examples I can think of of massless particles, I'm not sure about gravitons though - but I'll still include them.


What percentage of a spiral galaxy is the center/bulge? Is there a credible source that can tell me how big the bulge of a spiral galaxy is compared to the rest of the galaxy? Unfortunately, I could not find any.

The term you're looking for is called the bulge-to-disk (size) ratio. Sort of by definition, the answer depends on the morphology of the galaxy, i.e. how "late-type" spiral it is. "Sa" spirals are the ones that resemble ellipticals the most, and hence have large size ratios (of order, but below, unity), whereas bulges in Sc galaxies are (less than) one-tenth the size of the disk.

(EDIT: In fact, the way you phrase your question, you're looking for bulge-to-total, rather than bulge-to-disk, but there's a 1:1 correspondence between. Also, in retrospect I think you're interested in the mass ratio, but I interpreted "how big" as the size ratio. Peter Erwin's answer discusses the masses.)

The ratio is obtained by fitting the luminosity distribution of the bulge and the disk separately. Typically both are fitted as exponentials, with scale lengths $R_\mathrm{b}$ and $R_\mathrm{d}$, respectively, but other forms (e.g. Sérsic profiles) are also used. In this case, the effective radius $R_\mathrm{eff}$ is used, i.e. the radius inside which half the light is emitted. Furthermore, the answer will depend the band in which you observe the galaxy (i.e. IR, optical, UV, …)

The following figure (from Möllenhoff 2004), shows the ratio $R_\mathrm{eff,b}/R_\mathrm{d}$  as a function of "Hubble type" going from 1 (Sa), to 3 (Sb), to 5 (Sc). I've annotated examples of galaxies of Hubble type 1, 3, and 5.

You see that the ratio goes from $\sim0.5\text{–}1$ for early-type spirals, to $\sim0.01\text{–}0.2$ for late-type spirals.

The different symbols correspond to different filters going from the I band (infrared) denoted by squares, to the U band (ultraviolet) denoted by circles; they have been offset slightly along the $x$ axis for visualization purposes — you see that the ratio decreases slightly for shorter wavelengths. In other words, the bulge is less prominent, the bluer the light you consider.


Visitors that became satellites? To anyone's knowledge, has there ever been an object that approached the earth and was captured into an orbit? Is it thought that there might be such objects simply too small to spot or just not discovered yet?

I may not qualify as more knowledgable but here's what I can add to @called2voyage's comment.
In the question Have there been any documented mini-moons since 2006 RH120? I link to  Granvik, Vaubaillon and Jedicke 2012 (The population of natural Earth satellites Icarus, 218 (1) March 2012, pp262-277) available without paywall in ArXiv.

Abstract
We have for the first time calculated the population characteristics of the Earth’s irregular natural satellites (NES) that are temporarily captured from the near-Earth-object (NEO) population. The steady-state NES size-frequency and residence-time distributions were determined under the dynamical influence of all the massive bodies in the solar system (but mainly the Sun, Earth, and Moon) for NEOs of negligible mass. To this end, we compute the NES capture probability from the NEO population as a function of the latter’s heliocentric orbital elements and combine those results with the current best estimates for the NEO size-frequency and orbital distribution. At any given time there should be at least one NES of 1-meter diameter orbiting the Earth. The average temporarily-captured orbiter (TCO; an object that makes at least one revolution around the Earth in a co-rotating coordinate system) completes (2.88 ± 0.82) rev around the Earth during a capture event that lasts (286± 18) d. We find a small preference for capture events starting in either January or July. Our results are consistent with the single known natural TCO, 2006 RH120, a few meter diameter object that was captured for about a year starting in June 2006. We estimate that about 0.1% of all meteors impacting the Earth were TCOs.

The paper describes an exploratory simulation of the probability of objects being captured into at least temporary 3-body orbits in the Earth-Moon system, also taking into account perturbations from the Sun. Usually the orbits last a few months. Longer orbits are less common.
There is certainly the small possibility that an object could spend several years or longer, but the probability of this naturally decreases as the the lifetime increases. You can think of it as the "keyhole" for entering a long term 3-body orbit shrinking as the duration of that orbit increases.
The population of near Earth objects to be captured decreases as the size increases. Thus there could be many temporarily captured objects of micrometeor size, but few large enough to be easily observed. The chaotic nature of these loosely-bound orbits makes them harder to track and analyze; you can't assign Keplerian orbital elements and look for repeat passes.
What is really surprising is the first sentence of the conclusion (Section 4):

We provide the first estimate of the orbit and size-distribution for temporarily captured natural irregular satellites of the Earth. We predict that there is a one-meter-diameter or larger NEO temporarily orbiting the Earth at any given time.

You can also listen to the conference call with Bill Bottke (MP3 plus other documents) about "minimoons" presents an alternative to Asteroid Redirect first got me interested in orbital mechanics.

Here is an example of one theoretical orbit from here or here:


What would happen if a wormhole gets swallowed by a black hole? Assume wormholes exist. One hole allows matter to pop out of another hole somewhere else in the universe.

If one end of the wormhole is swallowed by a black hole, then no information can get from that end to the other hole.

If the wormhole acts as some kind of gateway to another part of the universe, wouldn't that mean that the entire universe would be swallowed by the black hole? ie. the wormhole end is an entry portal to the entire universe, so if that is inside the black hole, wouldn't that mean the entire universe is then in the black hole?

The concept of a wormhole is, of course, highly speculative.  None the less, such a phenomenon could take many forms - most simple something which to outside observers looks like a blackhole on each end of the wormhole.  Alternatively, a traversable wormhole (see for example Morris & Thorne 1988; which has the same basic topology but no horizon), could still behave for the most part* like a normal, massive object.

In either case, if one end were 'swallowed' by a blackhole, it is unclear to me whether it would remain stable (i.e. remain a wormhole).  In the case of a non-traversable wormhole, if it remained stable, it would continue to look like a blackhole on each end --- one end of which now merged with another blackhole, and subsequently more massive.  A traversable wormhole on the other hand could (extremely hypothetically) still remain traversable if there were enough exotic matter, or it could retain the horizon from the blackhole it merged with and be no-longer traversable.

All wormhole metrics that I've seen are more or less symmetric, and thus either have no horizon, or horizons on both ends... I don't think it would be possible to have a horizon on only one side --- allowing something to fall into an apparent blackhole on one side, and come out of a wormhole on the other.  Even so, however, remember that a blackhole isn't a magic vacuum --- and wouldn't 'suck in' the 'entire universe'.

*A traversable wormhole is believed to require some sort of 'exotic matter' in the neck, which may behave somewhat differently...


Is expanding universe adding potential energy? A system with two massive bodies has potential energy proportional to their separation. Since the universe is expanding, is the potential (and total?) energy of such a system slowly increasing? What is the scale of this effect if it exists?

Yes, in an expanding universe there is an increase in potential energy (with same caveats).

The (Friedmann) equations of a homogeneous and isotropic universe with no spatial curvature or cosmological constant can, in fact, be derived very easily from nonrelativistic Newtonian gravity. (This derivation is shown, e.g., in Mukhanov's Physical Foundations of Cosmology.) So it is actually valid to think of such a cosmos in Newtonian terms, using randomly but approximately homogeneously distributed point masses. At early times, the point masses are flying apart at high speed but the average distance between them is small. At late times, they will have slowed down, but the average distance between them will have increased. All that kinetic energy is converted into gravitational potential energy. This also remains true if we take into account Newtonian perturbation theory (the growth of overdense and underdense regions from initial perturbations).

As for the caveats: First, as I mentioned above, when spatial curvature is present, the Newtonian solution is no longer valid. Second, there is a century of literature with a variety of attempts to define a meaningful stress-energy tensor for the gravitational field in general relativity, none completely satisfying. Last but not least, a cosmological constant/dark energy term adds its own interesting twist to the story, as its energy density remains constant in an expanding universe, causing the expansion to accelerate: in this case, both the kinetic and the potential energy increase, thanks to the role played by dark energy's negative pressure.


Can any information leave a black hole? Referring to this question, is there any information that can leave black holes? Are they causing a permanent information loss in our Universe?

Information cannot "leave" the black hole. There is no way (by our current framework of physics) that an entity inside the event horizon can send a signal out.

However, entropy can leave. The black hole has entropy proportional to its surface area, and this roughly corresponds to "the number of ways that black hole could have been created". The surface area decreases due to Hawking radiation, so it can release entropy back to the universe. Of course, the growth rate due to infalling mas is usually many orders of magnitude more than the shrinkage.


Why don't we have in-between planets? The planets in our system are most often grouped into two categories:

Terrestrial:

Mercury
Venus
Earth
Mars

Gas Giants:

Jupiter
Saturn
Uranus
Neptune

Why is it that we don't see "in-between" planets in our system - large, rocky worlds larger than Earth but with thick atmospheres somewhat similar to those of gas giants? Are they possible? Did they not occur by random chance?

Super-Earths and Mini-Neptunes are the "in-between" types of exoplanets you're looking for. A sweeping generalization would put most in the range of $\sim1$-$10M_{\oplus}$ (Earth masses), with some outliers a bit above that. They may have significant quantities of hydrogen and helium in their atmospheres, as well as water, in liquid or vapor form. The latter should also have what are referred to as volatiles (colloquially called "ices") - compounds such as ammonia and water that are present in the atmospheres of proper ice giants like Uranus and Neptune.

One reason we don't have any of these in the Solar System is simply that no worlds with the proper mass formed. I explained here that there is a transition region of sorts, but most super-Earths require minimum masses of $1.5$-$2M_{\oplus}$ to keep their large hydrogen/helium envelopes (which terrestrial planets may accrete early in their lives but later lose). There would have to be a planet just in that mass range and at the right distance from the Sun (see D'Angelo & Bodenheimer (2016) for general constraints on semi-major axes). The environment in the protoplanetary disk may simply not have been suitable to produce the right body at the right time.


Why is the track of the subgiant stage almost horizontal on the HR diagram? It is stated that:

After the main sequence, as fusion weakens or stops in the core, outward radiation weakens. The helium core contracts and heats up. Gravitational energy is converted to thermal energy again!
The star will appear to cool slowly and will undergo a modest increase in luminosity. During this phase, the path the star will follow in the HR diagram is almost horizontal to the right of its position on the Main Sequence. Stars in this phase are usually referred to as subgiants.

But I still can't understand.

I will replace the previous answer to focus on the "subgiant" branch prior to red giant, rather than the pre-main-sequence or the "horizontal branch" of core helium fusion.  Those are other times that the luminosity is constant, but this question is about the subgiant branch, which I missed before.

The reason the luminosity is nearly constant on the subgiant branch is related to the "mass-luminosity relation" of pre- and main-sequence stars.  It is due to radiative diffusion and how it leads to a luminosity that depends only on mass, for a given composition.  If you compare to pre-main-sequence tracks, you should find that the subgiants more or less retrace that prior evolution, just with a somewhat higher luminosity because many of the electrons have been swallowed into neutrons, reducing the opacity and increasing the all-important rate of radiative diffusion.  It's essentially just a helium-dominated mass-luminosity relation, instead of hydrogen-dominated, as the radius rises due to the details of how the interior is evolving.

The reason the luminosity eventually rises steeply on the red giant branch is that as the degenerate core starts to build in mass, it starts to control the temperature of the fusing region, and this significantly changes the internal structure in ways that get into explaining red giants.


Do tides occur on planets like Neptune, having water? I mean, we experience tides here on earth because of combined effect of gravitational field of moon and sun. So I thought that tides could also occur on other planets.

Let us consider Neptune where the gas transits into a slushy ice and water layer. The water-ammonia ocean serves as the planet's mantle, and contains more than ten times the mass of Earth. Also it has 14  moons. What I was thinking was that more moons would mean more tidal effect would be seen even if distance from sun has increased.

But then I thought that the moons of Neptune would all have different orbits and different time periods of rotation. So this will definitely impact the extent of tides.

-To what extent will this affect tides?

And how will tides formed on Neptune be different than that             formed on earth? I mean, how will they vary in properties?

There will be tides, but they will be not very large.  It's pretty easy to get a good estimate of their size.

There are two things that control tides on Earth and both will be present anywhere else: The tidal forces from celestial bodies, and the size of the body of liquid in which the tides are raised along with resonances int he body of liquid.  Taking them in turn.

The tidal force is a bit funny, since the size of the tides raised on a planet by a celestial body is

(1) proportional to the mass of the celestial body raising the tide,

(2) inversely proportion to the cube of the distance away,

(3) inversely proportional to the surface gravity of the planet,

(4) and (roughly) proportional to the diameter of the body of liquid in which the tides are being raised.

That inverse cube of the distance means that the ability of a body to raise a tide drops off very quickly with distance.

Comparing Earth and Neptune: The Sun is 30 times further from Neptune than Earth, so Solar tides will be roughly 27,000 times smaller on Neptune.  They can be completely ignored.

Triton (Neptune's largest moon) is the jackpot, since it's a third as massive as Earth's Moon, and about the same distance away.  Neptune is 17 times more massive than Earth and four times the diameter.  So Triton tides of Neptune will be about fifteen times smaller than lunar tides on Earth.

All the other moons are much smaller or further away or both and will raise completely negligible tides.

So from gravitational effects alone, Triton will raise small tides and nothing else will be measurable without good instruments and a lot of care.

(Note, I'm talking about tides in oceans comparable in size to the planet -- for smaller bodies of liquid, the tides will scale down.)

On Earth, tides vary all over the place, and this is due to the configuration of the body of water and resonance effects.  Basically, if a body of water has a resonance -- a natural period for sloshing back and forth -- that matches the period of the tidal force, the tide can build up like repeated pushes to a kid in a swing.

Additionally, there can be a funnel effect with two arms of land coming together, or shallowing water that can amplify the tides. (E.g., the Bay of Fundy.)  Since as far as we know, Neptune has no land, it's effectively an planetary ocean and will get the full effect of the gravitational tides, but no resonances.

So, bottom line: Small Triton tides and nothing else.


How to calculate a planet's apparent size when the planet is viewed from a moon in orbit around it? Specifically in this case, how large would Jupiter be seen in the Io's sky in comparison to the moon in the Earth's sky? How to calculate this?

Also, if somebody knows any online software which lets you input moons and planets to see how their sky looks like to also fill in the answer, it's appreciated.

By a straightforward bit of trigonometry, if the distance from the observer to the (centre of) the planet is x km, then the radius, r km, of the planet subtends $\arcsin(r/x)$, and so the angular size of the planet is twice this: $2\arcsin(r/x)$

For Io x = 420,000 km, and the radius of Jupiter is r = 70,000 km, so the angular size is
$2\arcsin(1/6)=19^\circ$. For comparison, the moon from the Earth has an angular size of about half a degree.

Stellarium can be set up to view from other worlds of the solar system


Looking for a detailed Milky Way map I am a UI designer working on a sci-fi strategy video game.
I was wondering wether anybody here might be able to help? I am looking for a top down projection of the Milky Way that labels regions of space. This is in order to start building a server map of linked 'matches'.
The space combat sim Elite:Dangerous has a galactic project split into 42 sectors (see attachment) although I know that most of the regions labelled are made up specifically for the game. Obviously I don't want to copy this, so wondered if there was a more scientific version we could base designs on.
Do regional maps of the Milky Way exist in science? Or is it purely a sci-fi thing?

It doesn't exist.
It is actually rather difficult to see the shape of the milky way, because we are inside it. The more distant parts are obscured behind the nearer and very much of it is invisible. (We think that a supernova would happen every 50 years or so in a galaxy the size of the Milky Way, but none have been seen in nearly 400 years, so even massive bright events are hidden).
There are rough maps based on radio imaging of neutral hydrogen, but these are fuzzy and have artefacts, making part of the map invisible and part of it seems to point strongly towards the Sun.  The well-know images of the milky way on Wikipedia are as good an artist's impression as any other.


Where do Neutrinos Go if They Don't Interact? It is said that neutrinos can travel for billons of years at almost the speed of light without interacting with anything. That being the case,what do they do - is it possible they could orbit the universe indefinitely? They certainly can't leave it.

It depends on the ultimate fate of universe, as with everything else in the universe. Fundamentally, neutrinos are not much different from “ordinary matter”. Yes, they do interact only via weak force and gravity, but this only quantitatively changes the frequency of interaction, but does not change the game qualitatively. As long as there’s non-zero chance for them to interact, and as long as there is non-zero density, there is non-zero probability they will eventually interact with another particle (or black hole), and given infinite timeframe, it will happen for sure.

Accelerating expansion of universe does however complicate things a bit, but again, eventually there is bound to be some random fluctuation which will knock the neutrino off from its track.

Eventually neutrinos would meet anti-neutrinos and annihilate each other. Since the chances for this are really low, I would suspect neutrino-antineutrino collisions being the very last blink of this universe - given heat death is the ultimate fate of this existence.


How can an amateur astronomer verify the position of near Earth objects? Sometimes, hobby-astronomers use rather professional means to observe the big voids of space. Every now and then (think in months, not days) even I can locate an NEO (near Earth object).

Now, I'm assuming governments and institutions worldwide track such NEOs, have the best options to observe them, and most probably accumulated more data than we could wish for.

For hobby-astronomers who spot an NEO, it could be of interest to verify the location of the spotted NEO, and maybe even learn more about a specific NEO from additional data like name, type, path, direction, speed, first observed, etc.

Is there any publicly available, NEO-related database out there? Or is there a specific institution a hobby-astronomer can/should turn to to be able to learn more about individual NEOs?

Is there any publicly available, NEO-related database out there? Or is
  there a specific institution a hobby-astronomer can/should turn to to
  be able to learn more about individual NEOs?

Yes, there is NASA's Near Earth Object Program that catalogues all  detected NEO's and had advanced reporting and seearch capabilities (a bit overwhelming number of options actually).

There is also ESA's own (actually DLR's, the German space agency) database under the EARN (European Asteroid Research Node) project within the SSA-Programme (Space Situational Awareness) called The Near-Earth Asteroids Data Base:

  An on-line data-base of Physical and Dynamical properties NEOs, with
  the corresponding bibliographical references is available at the EARN
  site. This data-base, compiled and maintained at DLR by Gerhard Hahn,
  contains information on all known near-Earth objects
  (Atens-Apollo-Amor Asteroids), and is updated on a regular basis. It
  is also linked to, and reachable from the NEO Dynamics Site NEODyS at
  the University of Pisa, which provides extensive information on all
  dynamical aspects of NEOs, including close approach predictions and
  collision probabilities.

Honestly though, I find the NEO Program one from NASA a lot more convenient to use, even if a bit overwhelming. For extensive research though, I'd suggest referring to both of these databases, with the latter probably more conveniently used through the already linked to NEODyS Project.


Calculate object apparent diameter by distance and radius I'm trying to create a dynamic starfield for game using the HYG database. Because there are no data about the apparent size of the stars from earth, I will need to get it myself. I found a formula which calculate the radius by the temperature and the luminosity of the star. It's working, but the problem is that because a lot of stars are very far, I normalize the distance of the stars so they are all at the same distance in the game engine(they form a dome around the scene). I'm looking for a way to set the size of the stars in the sky, even if normalized, using the radius of the stars, and the real distance from earth. I tried: Distance / radius, but the size of the stars make no sense, betelgeuse is 10 times larger than bellatrix, but they should be almost equivalent.

If I understand correctly, you are trying to create a realistic looking view of the stars rendered (in the game) on a dome surrounding the game play area.  So you don't need to fly through the stars or interact with them in a 3D environment, is that right?
If so, it isn't the star size you want to reproduce - it is the brightness.  We can't see the width of a star from Earth and until very recently no telescope could, either.  It's the brightness of stars which makes them seem "bigger" or "smaller" than other stars.
The good news is this makes the calculations much easier for your game.  You want to represent the apparent magnitude as seen from Earth.
Planetarium programs, such as Stellarium or The Sky, achieve nice results with a combination of making the star dot circle radius larger/smaller as well as brighter/dimmer.  The planetarium programs which look the best also change the brightness level of the pixels across the radius of the star circle (the radial profile) which is called the point spread function in astronomy.
The easiest way to get a realistic star is with a simple bell curve shape (radial profile).  Make the width (in pixels) and height (brightness of pixels) of your bell curve variables and play around with them until you get the look you are happy with for your brightest star... those are the trickiest ones to render in a realistic way.  The faint stars are much easier.
When you have found how big the radius can be (and still look nice), assign it to Sirius' (the brightest star in our night sky) apparent magnitude  of ~ -1.5 and go down from there.  In a very dark location, our eyes can see stars down to a magnitude of 5 or 6.  Positive magnitudes are fainter; a magnitude 0 star is 100x brighter than a magnitude 5 star.
In a nutshell, you want to find the best bell curve width (diameter in pixels of your drawn star dot) and height (brightness of pixels at the center of your drawn star dot) to represent a star with an apparent magnitude of -1.5 (Sirius) and the best bell curve width and height to represent stars around magnitude 5 or 6.  You can, of course, include even dimmer stars but then the difference between the bright stars and faint stars becomes less obvious.  You'll have to find a balance you think looks nice.  The more stars you draw, the less of a difference there is between the brightest and faintest. If you increase the diameter in pixels of the bright ones to compensate, at some point they begin to look very fake.
Since you already have the YBC, you already have the stars positions and magnitudes (the YBC magnitude is the apparent magnitude and is listed as "V Mag").  Hope this helps, good luck with your game!
Edit: I was looking at the YBC earlier and had that stuck in my head, but the HYG should also have an entry for "V Mag".


How many wandering black holes are there in the milky way? I have attempted to research this topic due to curiosity, but I have yet to find a credible answer, the best I could find is this article, which has no listed sources, and the author has also done similar "click bait" style headlines with little to no evidence.

Yes as StephanG said there are thought to be 100 million black holes in our galaxy composed of approximately 100 billion stars. Most of these are thought to be stellar mass black holes. Not the super massive black hole assumed to be at the center of our galaxy. Some references are:http://hubblesite.org/explore_astronomy/black_holes/encyc_mod3_q7.html and https://www.sciencenews.org/article/we-share-milky-way-100-million-black-holes.


Theoretically, can a solar system evolve around lagrange points My question is, during the evolution of a solar system is it possible to end up with a massive planet around a star with two small planets around their $L_4$ and $L_5$.
Since we know it is not possible to have any bodies around $L_1, L_2, L_3$ due to their low stability, hence any planet trying to evolve their can be easily nudges out of its position.

Yes, provided the star is big enough, the "Jupiter" is big enough and the planet is small enough.

The details are discussed herein: "Is there a ceiling for stable L4 or L5 masses?"

To summarise: The sun needs to be 25 times more massive than the giant planet, and the small planet needs to have negligible mass.

It is worth noting that the minor body could not be said to be dominating its orbit, and so would be considered a dwarf planet by the IAU.


Could the E.H.T. produce an image of the human artifacts on the moon? Two days ago the New York Times featured the Event Horizon Telescope (E.H.T.), a huge virtual radio telescope emerging from combining data from several dishes scattered around the world. While the E.H.T. is designed and built to observe black holes, specifically Sagittarius A*, the article claimed that the E.H.T. could "spot a doughnut on the moon".

This was probably just meant as an illustration of its enormous resolution, but: Could it? I'm aware that the moon is not a radio source but it was apparently possible to make radar images of its surface using reflected radar radiation from one of the involved dishes. Would something like that be possible with the E.H.T.? Could we "see" the landing sites from earth? Left behind rovers, descent and ascent stages?

This is a really interesting question!

tl;dr: A definite maybe, but you would have to engineer a clever way to focus the transmitted power to a much smaller spot first; possibly several orders of magnitude smaller than what any one dish can do. Phasing several widely-spaced dishes alone would not be enough.

Let's see what can be checked easily.

Resolution:

Using round numbers of $f$ = 300 GHz, $c$ = 3E+08 m/s gives a wavelength $\lambda$ of 1 millimeter. With a baseline $D$ = 10,000 km, the angular resolution $\lambda / D$ is 1E-10. At a lunar distance $L$ of 360,000 km that's a resolution of 36 millimeters. Not only could you spot the donut, but the donut's hole would be resolved (barely)!

Radar at lunar distance:

The second image in your linked article New Radar Images Uncover Remarkable Features below the Surface of the Moon include images of Luther and Aristillus craters which are one and five million centimeters in size, respectively. The images already look noisy at this scale, so going a factor of 1E+05 smaller in resolution is likely to require a stronger transmitted radar signal, as measured in focused power per unit area on the lunar surface.

You might think that that can be fixed by making many of the EHT dishes (distributed across the globe) into transceivers, using very long baseline to focus the radar spot's power, and that could work to a limited extent but the problem is that it would not work the same way a filled aperture works. A sparsely populated arrayed aperture will still put most of the power into a plethora of ugly sidelobes.

There is a name for exactly this problem. I can't remember it now but I'll look for it.

Something similar, but different:

In the question Why was the 100m Green Bank dish needed together with DSN's 70m Goldstone dish to detect Chandrayaan-1 in lunar orbit? I talk about the use of two very large dish antennas to image detect the existence of a small object at the lunar distance. The experiment is described here and here as well.

They used two "tricks" to make this feat possible with only two dishes.

The DSN's Goldstone 70 meter dish transmitted the signal, and the large diameter of the dish made it possible to put most of the transmitted power into a circle which intercepted the spacecraft at its maximum elongation from the Moon (shown below) so that very little of the power would intercept the much larger Moon.
They took advantage of the Doppler shift of the reflected signal caused by the spacecraft's orbital velocity to further separate reflected signals from the Moon from the reflected signals from the spacecraft.

above: "This computer-generated image depicts the Chandrayaan-1's location at time it was detected by the Goldstone Solar System radar on July 2, 2016. The 120-mile (200-kilometer) wide purple circle represents the width of the Goldstone radar beam at lunar distance. The white box in the upper-right corner of the animation depicts the strength of echo. Inside the radar beam (purple circle), the echo from the spacecraft alternated between being very strong and very weak, as the radar beam scattered from the flat metal surfaces." Credit: NASA/JPL-Caltech.  From here

above: "Radar imagery acquired of the Chandrayaan-1 spacecraft as it flew over the moon's south pole on July 3, 2016. The imagery was acquired using NASA's 70-meter (230-foot) antenna at the Goldstone Deep Space Communications Complex in California. This is one of four detections of Chandrayaan-1 from that day." Credit: NASA/JPL-Caltech.  From here


What is a black hole's "shadow" and the best angle to view it to measure the spin? This excellent answer mentions several ways to try to observationally measure the spin of a rotating black hole. The third one is intriguing, but I don't understand how this works:

  The spin of a black hole also affects how it deflects light. Consequently, the pictures of a black hole's shadow such as taken by the event horizon telescope can be used to determine the spin of the black hole (if we happen to view it under the right angle).

Question: What is a black hole's "shadow" and the best angle to view it to measure the BH's spin?

The black hole shadow is basically the image of the event horizon. As you know the event horizon is the geometric locus of the points from which a light ray pointing in the opposite side of the singularity (the center of the black hole) can't no longer escape from it. Any other light ray emmited in any other direction from those points would never reach an external observer since the one pointing directly outside the black hole was the one with the best chances to do so.

So we might think that because of the absence of light coming from these region inside the event horizon an external observer might see a black ball of 1 Schwarzschild radius (which is the physical size of the even horizon), but this is not correct. Even if the event horizon marks the physical boundary between both regions the fact is that due to the extreme light bending the actual image of the event horizon (its "shadow") is a distorted view of this surface. For a classical non-rotating black hole this "shadow" (we call a shadow the image created by the absence of light rays but we can trace it with hypothetical "dark rays" that behave in the same way) looks like a ball with 2.6 Schwarzschild radii in size. Much larger than the actual event horizon! To visualize this I can only point to this fantastic explanation by Derek Muller from whom I took this animation.

As you can see we are throwing light rays from infinity into the black hole (that's why they are all parallel in the beginning). Our light rays fall reach the event horizon even if they were not directly pointed there, because they curve. Since light paths can be reversed and the physics still hold we can talk about light rays coming from different parts of the event horizon and reaching the outside observer in the reversed paths. So as you can see not only light rays emmited just barely outside the event horizon pointing towards us will reach us but also rays coming from "the back" of the event horizon can reach us. And as you can see those rays would create an enlarged image of the event horizon since they seem to come from a region located farther than the event horizon itself. So when you look to a real black hole from outside you will see this "black shadow", which is a map reprojection of the surface of the true event horizon where you can see even the 100% of the surface of it from a single vantage point.

This phenomenon (which is called relativistic light deflection) is also noticiable in other compact objects like Neutron stars. The intense gravitational distortion around them allows for light rays coming from its bright surface to deflect when going outwards and reach your eye even if they where emmited in regions close to "the back of the neutron star". Even if that region of the star shouldn't be accesible to an observer if light rays moved in straight lines (since those parts of the surface lie behind the curve of the star) you still can see them (which is something that can mess with the calculations of their true brightness).

You can make a square grid on top of the surface of the neutron star and see how much of it you can see from far away in this representation:

As you can see we are able to see more than an hemisphere (more than 50% the surface of the star). In fact you can see both polar regions and their surroundings. Well this is the same thing happening in a black hole but in that case the reprojected map is all black (since the event horizon is a uniform featureless surface all around) and you can see the 100% of that surface not only a small extra percent.

Now, this all changes if the black hole is rotating. Why? Because of relativistic frame-dragging. According to Einstein field equations mass-energy does not only curve spacetime but is also able to "twist it" if the object is rotating. We have measured this "twisting" of the surrounding spacetime in our own planet using exquisite instruments onboard the GRACE satellites.

In our case the important thing is that a non-rotating black hole (a Schwarzschild black hole) has a region outside the event horizon where orbiting the black hole in a stable manner is possible, we call it the innermost stable circular orbit (or ISCO). Getting closer to ISCO makes your orbit unstable and you end up falling into the event horizon. But if the black hole is rotating (a Kerr black hole), then ISCO is different if your orbit is prograde (orbit in the same direction as the black hole rotation) or retrograde (goes in the opposite direction around the black hole) because frame-dragging alters the solution. If you orbit prograde the fact that spacetime is been dragged in the same direction allows you to have some push by the black hole and your orbit can be mantained even much closer to the event horizon in a stable manner. On the contrary if you orbit the black hole in the opposite direction you are fighting against the drag of spacetime and thus you will decay more easily, making the ISCO for retrograde orbits a lot higher than the ISCO for prograde orbits.

If you apply this reasoning to photons you can start to notice something interesting. Light coming from far away stars behind the black hole as viewed from an outside observed gets bent in different ways if it is coming from one side or the other due to this frame-dragging effect. If the light ray is coming parallel over the rotating surface of the black hole it is going to be helped by the black hole itself, and some of the angular momentum is going to be transfered to that light ray from the Kerr black hole. If instead the light ray comes anti-parallel to the rotation (which is going to happen in the other side of the black hole), then it might never reach the observer. This reasoning can be applied to the "dark rays" (which do not exist but is a way to trace the shadow which is the absence of light rays) coming from the event horizon and thus the shadow of the black hole is no longer a perfect black disk but an asymetric D-shaped black region, that tells you if the black hole is rotating clockwise or counter-clockwise.

In this animation you can see the appearance of the shadow of the black hole when we increase its rotational speed, as you can see it goes of-center and asymetric as we increase it.

Since frame-dragging goes as the mass rotates you can't notice it so strongly if you watch the black hole from another inclination. In fact the projected rotational speed if you watch a black hole from the poles ($i = 0^\circ$) is zero, and thus the black hole would look just like a non-rotating one. Here you can see the dependance of the shape of the shadow (in red) of a Kerr black hole with a fixed rotational speed when you see it from different inclinations (from the equator to the poles). The event horizon is represented in blue (but remember, you don't see that, you only see the shadow)

As you can see the effect is the same in both cases (changing view angle for a fixed rotational speed vs. changing rotational speed for a fixed inclination), which means that you can't tell the real rotational speed of a black hole just by measuring the shape of the shadow (a disk like shadow could mean a non-rotating black hole or a Kerr black hole as viewed from above for example), but at least it gives you a minumim estimate for the rotation. To study the exact rotational speed we need some independent measurements like for example the inclination of a disk of material around it. In those cases you would have complete information about the angular momentum of the black hole.

Finally here you have a beautifull simulation on what you would see from orbit around a Kerr black hole (I don't know how to embed YouTube videos so...).


What is a Galactic Eclipse? It is said that when our solar system passes through the dark rift of galaxy (what is it exactly?), a phenomenon named 'Galactic Eclipse' happens, when our solar system witnesses 3 days of darkness. Is it possible even in this case that there would be no light in the night sky, as there are many stars in the night sky. (don't consider city lights and other artificial human interference)

Definitely a hoax as other commenters have said.

Dark rifts exist in the galaxy - large clouds of dust, several can be made out by eye, overlapping the milky way, in ideal conditions.  See for example the Great Rift.  However these are so large it would take far longer than three days for a solar system to pass through them.  Probably thousands or millions of years.  So this is definitely incorrect information.


How did the lighter elements end up in the center of the solar system? Solar System Formation The previous generation of stars famously are the origin of all the heavier elements (up until iron?) in the solar system. So a big portion of the solar system mass actually is made up of Carbon, Silicon, Iron and the like because of that. But in the center, and only in the center, there is a star with presumably almost no heavy elements inside. How can that be? Am I wrong about the actual mass concentrations or is there really an imbalance, i.e. is the element distribution really lighter towards the center of the solar system? I would presume that the previous generation of stars just ended in a more or less uniform cloud of debris, from which the solar system formed. But if so, why aren’t there star systems where the star has a very different composition and is kind of a spluttering, dirty fusion machine (metaphorically, I mean)?

The solar system contains very little of elements heavier than Helium - less than 2% by mass.

This is reflected in the chemical abundances measured in the photosphere of the Sun. i.e. The Sun does contain heavier elements.

Your question is the wrong way around; it is not that the heavier elements have not sunk into the middle, it is that the vast majority of hydrogen and helium that was in the same place as the planets when they formed, did not end up as part of the planets. In fact, even this is only partially true. The mass of planetary material in the solar system is also dominated by the hydrogen and helium in the gas giants.

So the conundrum is only why the smaller planets don't have a similar composition to the Sun. The answer to that is temperature and gravity. A small, hot planet just doesn't have the gravity to retain fast moving hydrogen and helium atoms, unless they are trapped in some compound (like water!).

Thus, the small planets close to the Sun are depleted of light elements.


State vectors of "interesting" multiple stars I'd like to show a demonstration of Runge Kutta integration of real systems using examples of interesting multiple stars, where "interesting" means you can see through a small telescope that they are at least double. I've selected three:

Gamma Andromedae (Almach)
Alpha Geminorum (Castor)
Zeta Ursae Majoris (Mizar and Alcor)

How can I find a state vector (set of relative positions and velocities) - or something similar - for these multiple star systems? An open access link (for everyone) would be great, but a journal reference is fine as well if that's what it takes.

Of course the periods are long and there will be uncertainty - I'd just like to use whatever is known and available, for this demonstration.

The USNO Sixth Catalog of Orbits of Visual Binary Stars (ORB6)
has the orbital elements you need.

$\gamma$ And is WDS 02039+4220
$\alpha$ Gem is WDS 07346+3153
$\zeta$ UMa is WDS 13239+5456

Double star expert Bruce MacEvoy explains orbital elements in maybe enough detail to help you work out some state vectors.
My search also turned up a book chapter by Andreas Alzner.


How could one use the transit of Mercury to try and find the Astronomical Unit? I am trying to use the transit of Mercury to estimate the distance between the Earth and the Sun, or 1 AU.

I know that I need to observe mercury from two antipodes, but I do not understand any of the underlying math behind this, or how to derive the astronomical unit from this.

Any help is appreciated!

Thanks to the Mercury transit, you can measure the parallax from the Earth. That happens due to TRACE , which tracks the transit of Mercury along the polar diameter of the Earth.

During that tracking, the transit of Mercury goes like that:

[

Now notice that, if TRACE remained stationary, the transit would be a straight line. So, if you calculate the maximum separation vertically of the center of Mercury and divide by two (how it's another different story), you got the parallax angle $\theta$:

You know $\theta$, you know Earth radius, so you know D, the distance from Earth to Mercury (in km) because

$$\tan\theta = R/D$$

Now, if you wanna know how many km a AU is, just realise that Mercury is 0.56 AU from us, so divide D/0.56 and you got it!

All images credited to NASA/IMAGE and/or Lockheed Martin/TRACE


Is there likely other significant objects we haven’t discovered in the solar system? In the last few weeks, I read a few news that presented new advances in our understanding of the solar system, which implied we don’t really know our neighboorhood that well. I was wondering, how likely is it that we make a significant discovery around us in the next decade? Like discovering a new massive object (moon or even planet etc.), or that a planet area is habitable

I was assuming it’s very unlikely because we have good measures and models (gravity based or other means I guess) but those few papers (sorry I didn’t bookmark them) have me wonder. How much do we really know of our nearest environment, before looking at interstellar stuff?

Moons.  We've explored the planets up to Saturn with probes that remained in orbit for long periods and found moons down to quite small sizes.  While it is certain that there are more small undiscovered moons, these won't be large spherical objects.  Uranus and Neptune have not been visited by orbiting probes, and so there are surely lots of undiscovered moons. Nothing very large, or we could see it from Earth but relatively substantial moons could be orbiting Neptune but be undetected. However there are no planned missions to either Uranus or Neptune in the next 10 years.

Dwarf Planets.  There are probably spherical bodies (and hence "dwarf planets") in the Kuiper belt, in the region of space around Pluto. There could well be further discoveries of dwarf planets here. We are certain there are no more dwarf planets in the asteroid belt. On the other hand we discover new asteroids all the time.

Major Planets.  There is the intriguing suggestion that the orbits of the known Kuiper belt objects are correlated with each other in a way that could be due to a large, Neptune sized planet in orbit in the outer part of the solar system. It is far from certain if such a body exists, but if it does, it could be discovered within a decade.

Habitable zones.  The only Habitable zone is the one which is approximately 150,000,000 km from the sun.  If you go closer, water boils. If you go further it freezes. Liquid water surely exists under the ice of some of the outer solar system moons, but I would not call them "habitable".

Unknown Unknowns.  If you look back 10 years most of the new discoveries could not have been predicted.


How Does IceCube Trace a Neutrino to its Source? I read the following claim in an article (Lisa Grossman, "Origin of Cosmic Oddities Proposed," Science News, February 17, 2018, 8).

  ... IceCube has already traced an especially high-energy neutrino to a single active black hole ... IceCube [Neutrino Observatory in Antarctica] could eventually trace neutrinos back to [active black holes in] galaxy clusters.

Aren't galaxy clusters at least millions of light-years from Earth?

Such neutrinos took millions of years to get to Earth.
Other sources could have been in the path of the neutrinos in such a distant past.
Are neutrinos affected by gravitational lenses as light can be?
Is such tracing possible because neutrinos react so weakly with matter?

You ask a number of questions. I'll try an answer them.

Yes, galaxy clusters (except the one we are part of) are many millions of light years away and so the neutrinos from them (which travel at a speed only a tiny fraction below that of light) have taken millions of years to get here. However, the light from those clusters has also taken millions of years to get here, so if we look in a certain direction and see a galaxy, and we detect a high energy neutrino coming from the same direction, then the galaxy was there when the neutrino was emitted (or passed through it). It may have moved since, but that would affect neither the light nor the neutrino.

Neutrinos are affected by gravity, and so the neutrino images of distant objects can be distorted in the same way as the optical images when there is a massive object (like another galaxy cluster) between it and us. However, if that were happening, we would see the distortion of the optical image.

It is also possible (though extraordinarily unlikely) that a single neutrino might have been deflected by passing close to a very massive object  such as a black hole which is too small to make much of a difference to the optical image of the cluster. The chance of a neutrino from somewhere else being deflected by something we haven't detected, just enough to appear to come from a galaxy we can see, though, is absolutely amazingly minute. A bit like a bullet ricocheting off a falling meteorite in such a way that it seems to have come from that man over there holding a gun, even though it actually came from somewhere else.

Finally, on the question of detection. It is true that neutrinos interact weakly with matter, but these very high energy neutrinos interact less weakly than others. This paper and in particular its figure 1 suggest that neutrinos of the energy reported ($2\times 10^{15}$ eV according to the article you refer to) would interact about 10 million times more strongly than the ones from the Sun or from nuclear reactors. Even so, most neutrinos will go straight through ice cube without interacting with it at all, despite the fact that it consists of a whole cubic kilometer of ice. Of the few that do interact, the interaction will usually "kick loose" a charged particle (often a muon) from the ice, which will be moving at close to the speed of light (in vacuum) itself, and in the same direction that the neutrino was moving. In particular, it will be moving faster than the speed of light in ice. When a charged particle moves faster than the speed of light in whatever material it's passing through it produces a "shock wave" of light, called Cerenkov radiation. Detectors in ice cube pick up that radiation and track the movement of its source, thereby getting a good idea of the direction from which the original neutrino came.


What visual artifacts are expected from the JWST? What visual artifacts should we expect from the images that will be released from the James Webb Space Telescope? Specifically, do we expect 3-pronged diffraction spikes (any preview how they would look?) and do we expect any artifacts due to the honeycomb-like structure of the main mirror?

Are you asking about the PSF (point-spread function)?

There are some simulations at http://www.stsci.edu/jwst/software/webbpsf ; there are some basic images available at that site as well as a downloadable package you can use to compute the PSF for a particular instrument and wavelength.  Since the telescope hasn't been fully assembled yet these are based on simulations.


Is the difference between LIGO & Virgo and their "Advanced" versions really generational, or were these just planned incremental upgrades? I just learned the term "third generation gravitational wave detector" in answers to What would a kHz gravitational wave detector look like? (mountains on millisecond magnetars) How would it differ from LIGO/Virgo?
Wikipedia's Gravitational wave observatory; Interferometers lists LIGO and Virgo as 1st generation, and Advanced LIGO and Advanced Virgo as second generation.
Is the difference between the originals and the "Advanced" versions really generational or were these just planned incremental upgrades that were just waiting for funding once the basic systems had demonstrated they worked as planned?

I think this is largely a matter of what you decide is incremental as opposed to a generational change.
The basic location and overall size and detection principle of the interferometers did not change. An important change to the interferometer configuration was the addition of the signal recycling mirror that allows a certain amount of "tuning" of the frequency response. A homodyne readout scheme was employed instead of heterodyne - and the interferometer is run slightly off the dark fringe as a result. The test mass suspension system changed from being a single to a quadruple pendulum and the suspension wires are now silica rather than metal. The test masses themselves were increased from 11 kg to 40 kg which reduces the low frequency noise significantly.
I believe most of the other changes were essentially upgrades to the materials and technology being used. New mirrors, better lasers, improved vacuum, better seismic isolation etc.
I don't think it is anywhere near the generational change between aLIGO and the proposed Einstein Telescope, which will be in a different location, have a different size, different number of arms and geometry and will be operated cryogenically etc.
However, as pointed out by @mmeent, the improvement in sensitivity is about an order of magnitude in both cases, which is probably what justifies the "next generation" tag for going from LIGO to aLIGO and from aLIGO to the 3G detectors.


Universe is expanding, why doesn't time expand with it? Reading this response When the universe expands does it create new space, matter, or something else? I understand Universe is not expanding like if it is stretching its space, but it is actually adding space that matter can travel through.

So, if the Universe can create more space, and space and time are related, shouldn't that mean Universe is creating also more time? And if it is the case, shouldn't this effect alter results of our measurements about how much time light spends to reach us from distant stars?

We know that photons have no mass, travel at speed of light, and time has no sense for them. But at the same time they are influenced in their trajectories by folds in the space-time.

If gravity deforms space-time, shouldn't an expanding Universe adding new space doing the same?

The short answer is, "gravity".  The time of "co-moving observers" (observers moving with the expansion of the universe") is expanding, but our "proper time" is not, because we're within a local group of galaxies that are bound together gravitationally.  The principles involved are detailed in the Wikipedia article "Co-moving and proper distances", and its links.

However, in any reading about General Relativity--the currently-accepted theory of gravity-- it may be helpful to understand that "observers" are really any sort of hypothesized material or energetic object, however small:  I believe they were supposed to be hypothetical but possible beings, when use of that terminology began--back around the time when GR's predecessor, Special Relativity, was written--but are now involved in so many inaccessible and inhospitable situations that "they" are not necessarily considered to be sentient, or even alive.  It may also be useful to understand that, although gravity is realistically considered to be the curvature of spacetime, "time" itself is best viewed as comparable to the three well-known spatial dimensions, or at least as something comparably definite, like "duration":  The temporal aspect of spacetime actually includes most of its curvature, as described in a website about General Relativity by the University of Pittsburgh's Dr. Norton.

Regarding your secondary question, the delay in the arrival of particles (photons) from distant stars is actually affected by several factors, generally based on inflationary cosmology, that are in a well-known diagram shown in the first Wikipedia item I suggested, but visible in larger versions with explanatory text:  Searches by the names of its originators, "Lineweaver and Davis", will reach references on various levels of sophistication.


What wavelength to best detect the "9th planet"? We know that the reflected sunlight will make detecting the 9th planet very difficult in the visible light. Is there another band that will be more likely to detect it? What is the surface temperature of this object likely to be, and what would that mean about its optimal detection wavelength?

The possible planet 9 is thought to be about 10 Earth masses and is unlikely to be a gas giant (it may be the core of an "interrupted" gas giant). As such, it will not be generating significant luminosity itself and would be rocky, or more likely, icy in character. It would thus only be seen by reflected light.

The considerations for what wavelength to search in balance the sensitivity of the instruments at hand with the likely spectrum of the object. This in turn depends on the solar spectrum and the wavelength dependence of the reflectivity (albedo).

For most icy objects, including Pluto and Trans-Neptunian objects, the reflectance increases to the red and near-infrared, whilst the solar spectrum peaks at shorter wavelengths. This suggests that searches are best carried out with wide field optical instruments in the R or r' bands at around 600 nm.

A further factor in finding a candidate is that you are going to have to cover a large area. This is only feasible at optical and NIR wavelengths unless the object was bright enough in the mid-IR to show up in WISE (which I'm sure is being thoroughly checked). A press release I saw said SUBARU is being used for the search. I would bet they are using the half degree field of Suprime-Cam at optical wavelengths and not pursuing COMICS mid-IR imaging with it 42x32 arcsecond field!

Confirming a candidate should be easy, given the enormous parallax and proper motion expected.


Is the universe moving through infinite space time as it expands? What I mean by this, is imagine a person blowing up a balloon. As the balloon expands, the person blowing it up is also walking. At a much grander scale, apply this concept to the universe. As the universe expands, is it also moving like the balloon through infinite space time?

I think what you're asking is a very fundamental question that doesn't really have a definitive answer. I interpreted your question slightly differently that Sir Cumference, so I'll answer it differently.

For people just learning about the expansion of the Universe, they are often taught about the analogy of the expanding balloon. The concept here is that you have a deflated balloon that you draw two dots on. These dots may represent two galaxies in the Universe defined by the surface of the balloon. The important concept here is that the surface of the balloon represents the entire universe in this scenario. The balloon is then blown up, and you notice that because of the "expansion" of the universe, the dots (i.e., galaxies) are moving farther apart. This is shown in the diagram below.

This concept is usually used to teach an important fact about our universe, namely that there is no center. At no point on the balloon can you saw that you are the center of the expansion because the expansion is happening everywhere in equal magnitude.

What you seem to be doing though, is trying to take this analogy one step further. Even though the balloon universe (ballooniverse?) is defined on the surface of the balloon and is thus two dimensional, there still exists a three dimensional space that the 2D ballooniverse is expanding into. So you question, what does that say about our 3D universe? Are we expanding into some 4D space?

That, in my opinion, is a distillation of the fundamental concepts behind your question. Assuming I'm interpreting your question correctly, then I would have to say the answer is I don't know.

There is a somewhat (unsatisfying) tautological answer in that the Universe is, by definition, everything that exists. Therefore, there can't be a 3D Universe and also a 4D space into which our 3D Universe is expanding into, because that 4D space would necessarily be a part of our Universe, by definition.

A more reasoned approach to this answer is to look at it from a mathematical perspective. If you begin to learn higher math (and by higher I mean upper undergraduate/graduate level math) you'll learn about manifolds. Manifolds are basically mathematical objects that can define things like the surface of a balloon. What you'll find is that the definition for the surface of a balloon, using the concept of manifolds, does not require that the surface exist in 3D space, even if that manifold is expanding. It is perfectly possible to define a 2D manifold that has the same topology (fancy word for structure) as an expanding balloon, without "embedding" it in 3D space. This concept makes your question kind of moot. It allows for an expanding 3D universe, without requiring the existence of some 4D space to expand into. So ultimately, I'm claiming that it is possible for a 3D Universe to exist which both has no center and is expanding, just not into a 4th dimension.


Can Kepler's 3rd Law be applied in higher dimensions to describe orbits (quadric surfaces)? Motivation for asking:

The 2-D orbital path of a celestial body revolving around another celestial body due to the force of gravity can be described by a conic section; this is the curve formed by intersecting a plane with the surface of a cone. This generalized conic section is given by:

$A x^2 + B xy + C y^2 + D x + E y + F = 0$

The ellipse is one example of a conic section (others being parabola and hyperbola); this happens when $B^2 - 4 A C < 0$.

Kepler's 3rd Law can be applied to elliptical orbits; this relates the orbital period $T$ and the semi-major axis $a$ of the orbit such that:

$\frac{a^3}{T^2}$ $=$ $\frac{G (M + m)}{4 \pi^2}$

As I understand it, Kepler's Law holds when the orbital path is 2-D. What happens when the orbital path is 3-D? I am aware that the conic section in 3-D is referred to as a quadric surface, which is given by:

$A x^2 + B y^2 + C z^2 + D xy + E xz + F yz + Gx + Hy + Jz + K = 0$

Say one were to rotate the 2-D orbit by its polar angle in spherical coordinates - that is, the altitude (as opposed to azimuth) - by a non-zero amount such that the orbit was in 3-D (changing z-coordinates as a function of time). Doing this, we can generate an ellipsoid, which is given by: $(\frac{x}{a})^2$ + $(\frac{y}{b})^2$ + $(\frac{z}{c})^2$ = 1.

Question:

Could the logic behind Kepler's 3rd Law be used to relate the orbital period $T$ of a body on an ellipsoidal path to any of the ellipsoids parameters $a, b, c$? What if the path weren't ellipsoidal, but some other quadric? Can the quadric surface always be reduced to a rotate 2-D conic section? Or are ALL orbits considered Keplerian?

On the risk of not fully answering your question:

Trajectories of a system through some solution space, are in physics always solutions of some set of differential equations. However using the right formalism, one can often derive properties of the solution, without actually knowing it.

For the Kepler-problem, there is such a trick.

Let us consider the Lagrangian $L=E_{\rm kin} - E_{\rm pot}$ of the Kepler-problem:
$$L = \frac{1}{2} m (v^2_x + v^2_y + v^2_z) - \frac{\alpha}{\sqrt{x^2+y^2+z^2}},$$ but to simplify the notation we shall abbreviate this to
$$L = \frac{1}{2}mv^2-\frac{\alpha}{r}$$
The fact that this simplifies to a 1-D problem is of importance later.
The equation of motion of the Kepler-problem can be derived from the corresponding Euler-Lagrange equation and results in Newton's law plus $E_{\rm pot}$.

Now we can multiply the Lagrangian by a constant $L'= c L$, and the resulting equation of motion, hence the solutions of the equation of motion, will remain identical. If we choose $c$ carefully, then we get the desired information about the solution of the equation of motion, by only touching $L$. This principle is called the principle of scale-invariance, and it doesn't apply to all Lagrangians or all systems in physics.

Now introduce a different idea, that of rescaling of space and time: Imagine a 'zoom' in space and a 'fast-forward' or 'slow-down' in time, such that $x$ and $t$ are rescaled according to
$x'\rightarrow c_x x$,
$t'\rightarrow c_t t$,
where $c_x$ and $c_t$ are some new constants, that are arbitrary initially. From those rescalings follow now, for the scaling of v that $v' \rightarrow c_x/c_t v $ and the potential, because it only depends on one length, scales as $\alpha /r' \rightarrow \alpha / (c_x r)$.

Let us plug-in all those results from the scalings into $L'$ and relate this to the unprimed, original quantities,
$$L' = \frac{1}{2}mv'^2 - \frac{\alpha}{r'} = \frac{1}{2}m v^2 \cdot \frac{c_x^2}{c_t^2} - \frac{\alpha}{c_x \cdot r}$$.

Now the latter looks a little like a mess. But we want to write this as $L'=cL$, which is only possible, if we can factor out those factors containing all the $c_x$ and $c_t$s. The condition for this is $\frac{c_x^2}{c_t^2} =  c \cdot \frac{1}{c_x}$ or  $$c_x^3/c_t^2 =c$$,
which is one statement of Kepler's third law, namely that the spatial-scale determining and the temporal-scale determining constants have the above relation w.r.t each other. Note that as long as the potential remains a central potential, it doesn't matter how many coordinates we have, i.e. if $r=\sqrt{x^2+y^2+z^2 + a^2+b^2+...}$ this statement retains validity.


Tripod mount missing! I have received a new Celestron Powerseeker 127 (a 5' short Newtonian), however it came without a tripod mount.

Ideas or suggestions on how to procure the right mount our a suitable alternative would be most welcome.

I am looking for sourcing tips rather than advice on how to choose a tripod.

A high end photographic tripod would probably work but it will be difficult to make small tracking adjustments. So a proper astronomical mount would be much better.

An EQ2 mount would just about manage a scope of that size however an EQ3-2 would be a lot more stable and have the advantage that you can either get a goto version initially or upgrade at a later date.

If you are on a tighter budget and/or don't want an equatorial mount then something like a Celestron Heavy Duty Alt/Az Tripod or Skywatcher AZ3 Alt Az Mount would be suitable.

All of the mounts mentioned are available from numerous online retailers but shop around as prices can vary.


How do black holes lose their mass? As I was searching the answer to 'What comes after Black Holes', I found these two threads talking about the same thing

Do black holes die?

Can a black hole reach a limit where it can no longer attract more matter?

The answers here mention that Black holes do die but with time and they "evaporate"

What do you mean by Evaporation of a Black Hole?

How do these black holes lose their mass?

"Black holes evaporate with time" means they lose mass and energy with time. This phenomenon is termed as Hawking radiation (named after Prof. Stephen Hawking who gave the theory) or simply black hole evaporation. According to him when a particle or antiparticle pop into existence at the event horizon instead of annihilating each other as they are supposed to, one of the the two will fall in the black hole and other will escape. Some of the mass inside the black hole is converted into energy which later converts into particles ( at the event horizon in this case)
The black hole loses energy/mass in form of  particle ( mass and energy are equivalent)


Position and velocity vectors of nearby stars? One of my favorite tools for modeling the solar system is the Horizons Ephemeris generator where I can get position and velocity vectors of various bodies. Is there something similar for neighboring stars?

The closest I've come so far is Ashland Studio's 500 nearest stars in the Hipparcos Catalog. It gives distance of each star. Clicking on a star in this list will often give the star's relative velocity wrt our solar system. But it doesn't the direction of it's motion.

For the time being I'd like the model the stars within about 20 lightyears of Sol.

You may use the XHIP star catalog, using the fields X,Y,Z for the heliocentric position and U,V,W for the heliocentric velocity vector.


Why are there no terrestrial planets with a subsurface ocean? Subsurface oceans in satellites are pretty common: Europa, Enceladus, Ganymede, Callisto, maybe Pluto... This is due to tidal heating of their host planet, Jupiter and Saturn, which heats up the inner ice of those satellites. However, planets don't exhibit this inner ice layer, so they don't usually have subsurface oceans (except Pluto or Ceres, if you can call them "planets"). Why is that? Only small bodies like satellites present this inner layer of ice? Is there any Earth-type exoplanets that exhibit this inner layer of ice that could potentially melt down to liquid water? And if there is, why some cold planets have inner ice layers and others don't?

Some hypothesize that the Earth did have a subsurface ocean during the Cryogenian period, which lasted from 720 to 635 million years ago. The Cryogenian saw the two greatest known ice ages in the Earth's history, the Sturtian and Marinoan glaciations. There is some evidence that the Earth was completely covered with ice and snow during those glaciations. (There is also some evidence that it was not.)
Whether even older periods in the Earth's history also succumbed to snowball Earth episodes is even more debatable. The evidence has been wiped out by a billion plus years of tectonic activity. That even older periods of the Earth's history than the Cryogenian did suffer snowball Earth episodes does however make sense.
The Sun is considerably more luminous now than it was shortly after the Earth first formed. Once the Earth cooled from its formation (and that appears to have happened fairly quickly, in a geological sense of "quickly"), that faint young Sun should have resulted in a cool Earth. That the young Earth had periods where it obviously wasn't covered with ice and snow from pole to pole is the faint young Sun paradox. The apparent paradox is almost uniformly explained away via greenhouse gases. But which ones?
That said, once plants started converting carbon dioxide into oxygen, and once the oxygen stopped combining with iron to form most of the world's iron ore deposits, the greenhouse effect that kept the young Earth from freezing over should have dropped significantly. There are some signs that this happened, some that it didn't. Puzzling out what happened well before the Cambrian has always been problematic because rocks that old are hard to find.


How early could life supporting planets been formed? In thinking about exclusion options for where its not worth to look for habitable planets, the past came to my mind.

Right at the beginning of the universe, there was no possibility for (carbon based) life supporting planets to form, simply because there was not enough material to do so.

Population III stars burnt out within millions of years to leave some of these elements, but was it enough to form planets? I have some problem getting things into perspective, considering that

there are so many processes that produce different kinds and amounts of isotopes
these isotopes are spread out in various different ways (and thus also different concentrations from how they were created) throughout the interstellar medium
it takes probably a lot of time to gravitationally condense enough metals into a planet once it was spewed out from e.g. a super nova
the majority of stars are smaller so it would probably take a clustering of heavier stars to provide the conditions earlier in the universe.

I would consider a similar distribution of metals like in earth as sufficient for life, and would concentrate only on those found in DNA (N,H,C,P,O), fats (C,H,O) and proteins (C,O,H,N) as well as most of the essential minerals (Ca,Cl,K,Na,Mg,P,S) and maybe at least a bunch of (possibly essential) trace elements (Co,Fe,I,Cu,Mn,Mo,Se,Zn,As,B,Cr,F,Rb,Te,V,Sn,Ni).

So what is the best approximation that we can currently give on when these elements were abundant enough to condense into earth like rocks with earth like elemental distribution?

You can get relatively high metallicity rather quickly in parts of the early universe -- especially some globular clusters and the centers of massive galaxies -- because star formation rates in those places were very high, which means lots of massive stars and several rounds of "massive stars form, go supernova and seed surrounding gas with metals, new massive stars form out of enriched gas" -- the timescale from formation to supernova can be as short as 2 million years for really massive stars.

So how high a metallicity do you need to form planets, and how early did (parts of) the universe get there?

One the one hand, we know of a star with a metallicity (iron
abundance) only a quarter of the Sun's ([Fe/H] = -0.63) with a
Neptune-mass planet, and Kepler has found roughly Earth-sized planets
around stars almost as metal-poor.
So you don't need very high metallicities to form planets.
On the other hand, there is evidence of gas around quasars enriched
to above solar metallicity by redshifts of 6 to 7 (about 800
million years after the Big Bang). There are also rather old stars in
our galaxy with high metallicities: for example, this paper estimates that
the stars in the globular cluster NGC 6258 are 10-12 billion years
old with near-solar metallicity.

So the short answer is that you could have planets forming within the first 1-2 billion years after the Big Bang, and quite possibly within a few hundred million years after the first stars formed. Because the metals would mostly be from core-collapse supernova (as opposed to the iron-dominated products of Type Ia supernovae), there would be an excess of so-called "alpha elements" -- particularly oxygen, magnesium, neon, silicon, sulfer, argon, and calcium -- relative to iron; I don't think you'd have to worry about missing too many life-critical elements.

These planets would not all be very good places for life, however, because they would be in dense regions with lots of stars and lots of star formation, meaning lots of supernovae going off nearby, and possibly an active quasar as well if we're talking about a galaxy center.


Why don't we see more supernovae in our galaxy? I found this on the Wikipedia page on supernovae:

  The total supernova rate in our galaxy is estimated to be about 4.6 per century, or one every 22 years, although we haven't actually observed one for several centuries.

That sounds like a freak occurrence to me.

It made me curious about the numbers so I refreshed my knowledge on the Poisson distribution. It is not specified how many centuries we have not seen a supernova, but let's assume it's only two.

If on average 9.2 supernovae happen per 2 centuries then the chance of having exactly 9 in any two century period can be calculated with:

$$p(k) = {\lambda^k \times e^{-\lambda} \over k!}$$

Filling in $k=9$,

$$p(9) = {9.2^9 \times e^{-9.2} \over 9!}=0.1315$$

The chance that there is not a single supernova in a two century period is:

$$p(0) = {9.2^0 \times e^{-9.2} \over 0!}=0.00010104$$

That's about a 1 in 10,000 chance. If "several" means 3 centuries, then the chance of not having any supernova is as little as 1 in 100,000.

So from this I can only come to the conclusion that supernovae did in fact happen in the Milky Way in the last two centuries, but that we didn't see any of them.

But now I'm confused. I read about supernovae that have happened in galaxies billions of light-years from here that lit up as the brightest source of light in the sky for days. Surely we would be able to see all of the supernovae that happen in our Milky Way (no more than 100,000 light years away) with the naked eye, let alone with telescopes?

Are some supernovae so much more super than others? Are all of the Milky Way's supernovae visible to the naked eye?

So from this I can only come to the conclusion that supernovae did in
  fact happen in the Milky Way in the last two centuries, but that we
  didn't see any of them.

  But now I'm confused. I read about supernovae that have happened in
  galaxies billions of light-years from here that lit up as the
  brightest source of light in the sky for days. Surely we would be able
  to see all of the supernovae that happen in our Milky Way

One problem with estimating how many supernovae have happened in the Milky Way in the last two centuries, is that the remnant from a supernova is far dimmer than the nova itself and far harder to find. Several supernovae could have happened in the Milky Way in the last 2 centuries and remain undiscovered — and just for clarity, we define the date of a supernova by the date the light from the explosion reaches the Earth, so saying saying a nova "happened" in the last 200 years refers to the date the light from the event reached Earth, not the actual date of the event, which you probably already know, but just to clarify.

So for argument's sake, let's say that a supernova's light reached the Earth about 50 years ago, but it took place on the far side of the galaxy. To find that, we'd have to look for a nebula on the far side of the galaxy and that's a hard thing to see. Similar to looking for the theoretical Planet Nine, finding old supernova remnants on the other side of the Milky Way takes a lot of looking. Even with modern telescopes, it's still a needle in a haystack, and especially if the view is blocked by dust like much of the Milky Way is.

A Milky Way supernova remnant was discovered in 1985, supernova remnant G1.9+0.3. It was thought to have "happened" around 1868, though it went unobserved and probably wasn't visible to the naked eye at the time.   There's probably been several others more recent than that one.  G1.9+0.3 would have been visible if not for interstellar dust. From article above:

  It was a type Ia supernova believed to have exploded about 25,000
  years ago, and the signal began reaching Earth around 1868. The light
  from the supernova would have been visible to 19th century
  astronomers, had it not been obscured by the dense gas and dust of the
  Galactic Center.

I'm not sure the 4.6 supernovae per century from your article is accurate. It might be, but the number I'm used to hearing is about one per century. But regardless of which number is actually correct, it still doesn't imply high mathematical improbability because many Milky Way novae would have gone unnoticed if they were far enough away. In short, I agree with what you said here:

  I can only come to the conclusion that supernovae did in fact happen in the Milky Way in the last two centuries, but that we didn't see any of them.

Here's a related article on Milky Way visibility.

Observation of a distant galaxy nova is possible if we have telescopes looking in that direction. A nova is much more detectable at the time it goes nova.  Much less so, years later.

Today, however, with neutrino detection in 7 locations around the globe, I think it's virtually impossible that we'd miss a supernova in the Milky Way, so we have a good chance of seeing one in our lifetime and it's likely that none have occurred since somewhere around 1980. A supernova was detected in the Andromeda galaxy in 1987 by that method, and our neutrino detection has improved since then to give us early warning and pinpoint location.

As far as visibility, size matters, but what matters more is how close and how much dust is in the way. Most of the recorded supernovae were quite bright and fairly noticeable to someone who was familiar with the stars in the sky (list of known supernovae).

Eight Milky Way novae have been recorded by history and observed by the naked eye in the last 2,000 years, well below the number that should have happened in that time. Five of those eight had brightness greater than $-3$, which is brighter than Jupiter and would have been immediately noticed by anyone familiar with star charting. Two others had magnitudes around zero, which is still brighter than most stars. SN386 was less bright, but is still easily visible and recorded by Chinese astronomers. And finally, Cassiopeia A was quite dim, but it was still observable. Only about 10% (ballpark estimate) of the Milky Way is close enough and unobstructed enough to provide supernovae that would visibly get noticed.  Most would have gone unnoticed until recently.

Hope that's not too wordy, I can try to clean up if needed.


Has the effect of other planets on Earth's orbit been ever measured? Since other planets have their own gravitational pull, I'm wondering if their effect on Earth's orbit has ever been measured? I know that in comparison with the Sun's pull the effect might be rather small, but I'm wondering if it's measurable...

For decades, geologists have understood that climate changes are due in part to gradual shifts in the Earth’s orbit, which are caused primarily by Venus for its proximity and Jupiter for its mass, and repeat regularly every 405,000 years. In 2018 a team of geologists and Earth scientists unearthed the first evidence of these changes – sediments and rock core samples that provide a geological record of how and when these changes took place.

https://www.pnas.org/content/115/24/6153
"Empirical evidence for stability of the 405-kiloyear Jupiter–Venus eccentricity cycle over hundreds of millions of years"

So given the timescales involved, this is as close to measuring this effect as one could expect to get.


If we could travel far enough would we be able to see a constelation in reverse? I am enjoying Orion high in the sky this winter as always. I wondered if I could travel beyond it, say 2730 light years* and turn around, would I see Orion in reverse? If so, what of the neighboring constellations. I know brightnesses would be different but would I recognize the shape?

Happy new year to all.

I chose this number because Alnilam is 1360 light years away so I doubled the distance.

Source: https://www.space.com/3380-constellations.html

If the stars are all similar distance from us, like the Pleiades, then yes, you could find a place from where the stars would appear in a similar pattern and brightness, but the pattern would be a mirror image.

You can't do it when the stars are significantly different distances because they wouldn't look the same.   Most constellations are a range of distances, which wouldn't look like a mirror image from any distance.

Take Canis Major for example.   Sirius is the brightest star in the sky but in large part because it's only 8 light years away.   The 2nd brightest star in that constellation, Epsilon Canis Majoris is about 430 light years away.   At that distance, Sirius might not even be visible to the human eye.  It certainly wouldn't be constellation worthy.


Why planets and stars are always round/oval in size? What kind of stability it provides to each and every celestial body that each one them are round in shape?

The gravitational is radial only. That means if you tried a different shape then you would have regions, like hills, where a change in the angle would experience a much less attraction towards the center. This region would, over time, be attracted to a different angle with a greater force of attraction. Or spherical. This results in a spherical-like shape.