Explaining Dark Matter and Dark Energy to layman With my little knowledge, I know this:

Dark Matter

The center of a galaxy controls/attracts its objects (stars, planets, comets etc.) towards itself because of gravity. But the mass of the center of this galaxy appears to be less than what it should take to attract all objects of the galaxy. Hence Dark Matter is there, which makes this attraction or controlling possible.

Dark Energy

This is the force which is pulling galaxies away from each other, and hence making the universe expand. If we don't have this hypothetical force, we cannot explain why distances between galaxies are increasing.

Question

Please let me know how far I am from the actual meaning/definition of these terms? And how can I explain these to a layman, who has no or very little knowledge about astronomy?

Thanks.

P.S.: I have read this question here, which is excellent but I could not get an idea of how to explain to a layman.

Dark Matter
Your understanding of dark matter isn't bad, but here's a few clarifying details.
Orbits: The speed of an object's orbit is related to 2 things: the radius of its orbit and the mass inside of it. In the solar system, over 99% of the mass is concentrated at the centre, so radius is the dominant effect on orbital speed. As we look at planets further away from the sun, their orbital speed decreases. In the galaxy, it's a similar story, but with increasing orbital radius, you're also getting more and more mass inside of your orbit since the galaxy is full of other stars. Even so, we would expect orbital speeds to drop as you look at objects toward the edge of the galaxy, since the stars get more diffuse as you move out and not enough mass is inside the orbit to compensate for the increasing radius. Instead, we see that after a certain point, the orbital speeds stay the same, suggesting that there is more mass there than we can account for with stars and other visible objects.

Here is an example of a so-called rotation curve of the spiral galaxy NGC 3198:

Without dark matter, we would expect the speed of objects as a function of radius (from the center of mass) to follow the curve labeled 'disk'. However, what we see appears to be the sum of the two contributions (the disk and the 'dark matter halo' which surrounds this disk), which is overplotted with actual data.

We also have evidence of this extra mass from gravitational lenses. As light passes by a large object like a galaxy cluster, its path can get bent by that object's gravity. We can see this effect when a galaxy cluster is in front of an even more distant galaxy, the light from the background object gets magnified and distorted. We can calculate the path the light must have taken to appear as we see it, and the mass needed to bend it that way can't be accounted for with just the stars and gas we see in the lensing cluster. This again suggests that there is extra mass that we can't see.

Dark Energy
You're not too far off with this one, with one major distinction: Dark energy is the 'force' that's causing the universe to accelerate its expansion. We expect the universe to be expanding from the Big Bang cosmology model, but we'd expect it to be slowing down as the mutual gravity of everything in the universe acted upon each other. Dark energy appears to be a "pressure" on every part of space to expand, but it's very weak and has been dominated by gravity and other factors throughout most of the universe's history. It's only in the past few billion years it's started to take over.

It works something like this. Imagine you have a 1-D region of spacetime 1 unit long (just for convenience's sake). Here it is: |--------|
Now this region has this dark energy 'expansion pressure' acting on it. Let's say it's 0.07 units per year per unit. This means that every year, this region of spacetime gets bigger by 7%. In 10 years, the length will be double:
|--------|--------|
Now the thing is, each of these regions has the same expansion pressure as the original one! So both will double in 10 years, and then those will double, and so on. So what happens is you get a small expansion locally, but the further away something gets the faster it's accelerating away from you. The real effect of dark energy is orders of magnitude smaller, something like 67.15 ± 1.2 (km/s)/Mpc (Wikipedia), but it still means that any galaxies more than 4.5 Gigaparsecs away from us are currently being carried away from us faster than the speed of light. (We can still see them right now because the light we see was emitted long before the expansion rate got that high.) The expansion adds up over the huge distances we see in the universe.

Dark energy doesn't affect things like planets, solar systems, and galaxies because the effect is so weak on small scales that gravity more than counteracts it. It's effect can mostly be seen over the vast empty stretches of space between galaxy clusters.

Hope that helps!


Is it possible to find what crater the Mars meteorites come from? A few dozen meteorites have been determined to come from Mars. These meteorites likely formed from three separate events.
Each one of these events must have created a big enough crater that rocks would escape Mars' gravitational field. This restricts the number of craters to look at.
The composition of the meteorites can also give clues. For instance, the basalts on the floor of Gusev crater are somewhat more mafic than the meteorites, so they don't come from Gusev.
Using crater size, rock composition,... would it be possible to determine which craters the Mars meteorites come from?

Scientists think they have - the Mojave crater. Also see  https://science.sciencemag.org/content/343/6177/1343


How did Biden become VP? (naming of 2012 VP113) The Carnegie Science article Solar System's Edge Redefined describes the discovery of the object 2012 VP113.
Wikipedia's article 2012 VP113 says:

Nickname:
2012 VP113 was abbreviated "VP" and nicknamed "Biden" by the discovery team, after Joe Biden, who at the time of discovery, was Vice President (VP) of the United States.11
11 Dwarf planet stretches Solar System's edge

and the referenced Nature News item says:

The newfound object's official name is 2012 VP113, but the discovery team calls it VP for short, or just 'Biden' — after US Vice-President Joe Biden. In several years time, after observations have pinned down its orbit, the scientists will submit a name for consideration by the International Astronomical Union (IAU), the organization in charge of celestial nomenclature.

Question: I understand that VP is short for Vice President, but where do those two letter codes come from? Is the VP in the object's name short for something literal (did the team choose these two letters?), or just a random computer-generated code, or something else?

This has to do with how minor planets obtain provisional names (see also Wikipedia). The year is divided into 24 half months, with a particular letter associated with each one. Each of these half-months has a number of cycles of length 25 depending on how many minor planets are discovered; a minor planet is assigned a letter corresponding to its order in the cycle, and the number of the cycle it is in. This means that the provisional name for a minor planet is, in general,

year+half-month+position in cycle+cycle number

2012 VP113 was discovered in 2012, in the first half-month of November (corresponding to the letter V) and was the 15th minor planet discovered in the 113th cycle that half-month.


Are stars NML Cygni, UY Scuti, VY Canis Majoris and VV Cephei near the ends of their lives? Are the four stars mentioned in the title stars that are going to go supernova, or are they stars that are in mid-life, like our Sun, but just happen to be so much larger than our Sun?

NML Cygni and VY Canis Majoris
Zhang et al. (2012) considered several values for $T_{\text{eff}}$ (effective temperature) and $L$ (luminosity) and found that NML Cygni matches up with the evolutionary track of a star of mass $\sim25M_\odot$, near the HR-diagram location of a similar star, VY Canis Majoris (denoted as VY CMa). Using these data points, the model estimates that NML Cygni is approximately 8 million years old and in the post-main sequence phase of its life.
However, NML Cygni is thought to be related to the Cygnus OB2 association, which has an age of 2-3 million years. This is strange because all the stars in the association should be nearly the same age  (see Knodlseder (2008)). That age may be questionable, however, as evidence for older stars nearby has been presented (see Wright et al. (2010)).
The exact location of VY Canis Majoris on the evolutionary tracks is not certain. Massey et al. (2006) created models that placed it on a lower-mass track to help it avoid the "forbidden zone" beyond the Hayashi limit.
VV Cephei
Based on standard parameters, stellar wind measurements and Geneva grids, Bennett (2010) suggested that the red supergiant component of VV Cephei could be near the end of its life, assuming a mass of $\sim20$-$25M_\odot$.
UY Scuti
Wikipedia claims

UY Scuti should fuse lithium, carbon, oxygen, neon, and silicon in its core within the next million years. After this, its core will begin to produce iron, disrupting the balance of gravity and radiation in its core and resulting in a core collapse supernova.

As far as I know, that claim is made only through rough estimates of evolutionary tracks, not precise calculations. We do know that UY Scuti is certainly in the later period of its hydrogen-burning phase, if it is indeed at that point in its life. Arroyo-Torres et al. (2013) believe observations place it near evolutionary tracks of stars originating with masses of $\sim25$-$40M_{\odot}$. Therefore, it may have already lost a substantial fraction of its original mass - not surprising for a red hypergiant.


What does "the depth of the 13.0 Å feature differs by 7 sigma from zero" mean? In a paper I've read, I have seen the following:

In view of the noise levels, it is difficult to perform such a search effectively using statistical significance criteria only. We will therefore appeal to spectroscopic consistency arguments when assessing apparent absorption features in the spectrum. As a guide, however, we estimated the significance of the features by fitting each with a simple gaussian profile, optimizing the continuum, and evaluating their significance in terms of the amplitude of the gaussian. In the early-phase spectrum, the depth of the 13.0 Å  feature differs by 7 sigma from zero, whereas those of the 25.3, 26.3 and 26.9 Å features differ by 3–4 sigma.

What does "the depth of the 13.0 Å  feature differs by 7 sigma from zero" mean?
How these  significance level calculate? The line seems weak and could not fit with normal fitting method. So use chi-squared method seems do not work.
And another way to calculate such weak line significance level is to use equivalent width divide the uncertainty, but to the so narrow absorption line, how to calculate or get the uncertainty of the equivalent width?

In this case itseems to mean that the depth of the line is 7 times its error bar below the continuum level.

Impossible to answer. You say it can't be done, but the authors say that they fitted a Gaussian.

You either use a rough estimate (attributable to Cayrel de Strobel 1988) of
$$\Delta {\rm EW} \sim 1.5\frac{\sqrt{RP}}{{\rm SNR}},$$
where $R$ is the resolution element (the FWHM) of your instrumental wavelength response, $P$ is the pixel size (both in the same wavelength units) and SNR is the signal-to-noise ratio per pixel. Or you use an integration method and combine the SNR of each pixel used to find the uncertainty in the integral. Or you fit a Gaussian but using the area under the Gaussian as one of the fitting parameters.


If I had a horseshoe orbit of two moons... Consider an Earth-like planet, with two moons. I guess the two moons would have to be smaller than our own, for the system to be stable. Is that stable? I did some research, and apparently Janus and Epimetheus (Saturn moons that are in a horseshoe orbit) are like that, but I don't know how their cycles work, or how they look from Saturn.

Hence,:

Would a moon have to be smaller that the other one, or would both have to be the same size?
Would both have the same cycle?
How could nights be seen from the Earth?

I'm looking for that sort of answer.
Thanks in advance! :)

It's not the size of the Moons so much as the proximity to the sun that makes a 2 moon system (and especially a 2 moon horse-shoe orbit) unstable.   Size matters too, but the proximity to the sun matters more.    The sun's tidal force perturbs the Moon's orbit around the earth quite a bit.   If Earth was much further from the sun, it could have several moons, not just one and some a fair bit more distant and the big moon's orbit would be nearly circular due the inverse tidal forces, which tend to circularize an orbit.

Making the moon's smaller helps some, but what you really need, if you want that orbit to be relatively long-lasting, is to push the Earth away from the sun at least a few more astronomical units.

If we ignore the solar-tidal effects, then Earth could have 2 moons in a horse-shoe orbit to each other.    As to what it would look like, the first thing to consider is that everything in the sky rises in the east and sets in the west and that's because the Earth rotates.    If you watch each of the two horseshoe moons individually, they would move much like our moon, rising in the east, moving slowly against the fixed stars, and setting in the west.

The second thing to consider is that the Moons, the planets and the sun move against the fixed background stars, but this movement is slow unless the Moon is very close to Earth.  Our Moon, moves a full 360 degrees across the fixed stars once every 27 days (sidereal orbit).

The motion of the two moons individually would be very similar to our moon now in terms of new, waxing, full, waning, and repeat.

What the horseshoe orbit would do is you'd see one moon slowly catching up to the other, perhaps touching it from the point of view of earth, but never passing it.  it would get close, then it would reverse direction relative to the other moon and in a couple/few weeks, it would approach from the other direction, slow down and reverse.    Each moon individual would behave much like our moon but relative to each other there would be this weird game of attraction-repulsion between them.

I believe (but don't quote me on this), from Earth you'd never see the Moons' cross, but you'd perhaps see them touch.   The nature of the horeshoe orbit prevents crossing and passing, you'd just see them move closer, then move further away from each other.   If you had a really good eye, you might be able to notice the smaller moon change size as it moves closer and further from the Earth as the moon's get close, but that difference would be subtle.

This is the best diagram I've seen for what happens as the moons get close.  Unfortunately it's not in English, but it's not hard to follow:

https://upload.wikimedia.org/wikipedia/commons/9/99/JanusEpimetheus.gif

The horseshoe back and forth, assuming some degree of stability, it should have a regular period, just as the full-moons would have a period.   The optimal time for moon watching would be when those 2 periods coincide and you have a double full moon.   The horseshoe period and the full moon period (currently 29 days - sidereal orbit) should have nothing to do with each other but they could in theory be coincidentally in some kind of resonance.

As pointed out in the other thread, even if the Earth was far enough away from the sun to have two moons in this pattern, it's probably not truly long-term stable.   It might last for a few hundred orbits, or a few thousand.   It's not what is generally considered a stable orbit.

  Would both have the same cycle?

This is a bit of a trick question.   On average, yes.  The orbital periods would need to be the same for the horseshoe to work when averaged out, but individual orbits would need to vary by a little bit, so the individual orbits wouldn't have the same cycle.

Corrections are welcome if I got any of that not quite right.


Collimating with a home-made collimation cap I've made a home-made collimation cap and am trying to use the procedure linked to in another answer.

However, I've run into two issues right off the bat.  In this picture from the linked page, it shows a "primary mirror donut". All I see is a black circle with nothing in between.  I'm using a piece of bright white paper that I printed the alignment patterns on to attach to the eyepiece and I thought this would be bright enough to see, but apparently not.

The second issue is that this method seems to rely on there being four "spider vanes" which are holding the secondary mirror in place.  However, my telescope only has one, making it seemingly impossible to line anything up.

Here is what it looks like through the cap:

It looks like the dark circle in the middle it fairly well aligned, but I can't tell for sure if it needs any adjustment.  And I'm not sure how to deal with the fact that there is nothing inside the circle to align with.  I have tried taking an LED flashlight and shining it on the top of the paper while i look through the pinhole and if I focus the LED circle on the edge of the paper at 90 degree intervals it appears to be aligned, but I don't know if this is sufficient.

Incidentally, the halo around the outside seems to be the edge of part of the inside of the side tube, and it does not move when I adjust the collimation screws.

The Gary Seronik website that you cite is a good one - it's not leading you astray!

There are a couple of important things to look at:

Look straight down the main tube of your telescope at the primary (largest) mirror. Is there a small paper ring or another mark dead in the centre of that mirror? If so, good. If not, read through this webpage http://garyseronik.com/centre-dotting-your-scopes-primary-mirror/
What are the lighting conditions when you're performing this? The photo you included is helpful but seems to imply that you might be pointing the scope at a bright light. This will wash out all contrast and make the procedure difficult. I collimate my scope just using ambient light during the daytime

Don't worry about only having one support rather than four. The key will be to identify your centre dot (you can just see it on Seronik's webpage, first picture ) and to see how it is centred when you look through the collimation cap onto the secondary.

Maybe read this article: http://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/

Or search for Newtonian Collimation on Youtube?


Is dark matter influenced by gravity of baryonic matter? As far as I know galaxies and clusters form along so-called "galactic filaments" which are influenced by a dark matter web that organises them in to this position in the universe. So it looks like dark matter attracts baryonic matter by a kind of gravity. But is it also true the other way around, and that 'our' mass has gravitational forces on dark matter?
And if not why does it only go one way?

In a study scientists have observed the four colliding galaxies of cluster Abell 3827 and discovered that one of the envelopes of dark matter in the galaxy behind it encloses. The backlog in the system stands at 5,000 light years (50,000 million million kilometers - the NASA space probe Voyager would do 90 million years to cover that distance.
Expected to create some lag between a galaxy and its associated dark matter when dark matter affects itself, however small, by other forces than the gravity.

See also https://www.youtube.com/watch?v=qtfTUPhXutQ and
http://spacetelescope.org/news/heic1506/

Dark matter attracts baryons not only by a "kind of gravity", but simply by gravity. Everything that has mass (even if it's only relativistic mass like light) attracts everything else by gravity, and thus baryons also attract dark matter.

The reason dark matter "organizes" galaxies is just that there is much more of it, roughly 7 times more, to be specific. So, on large scales, the dynamics of galaxies are dominated by dark matter. But because baryons can dissipate energy, it may collapse in clumps — such as molecular clouds — which are hence dominated by baryons.


How can the radius of a star be measured using parallax? The radius of Betelgeuse was measured using parallax. From the Wikipedia article:

  the estimated parallax from the Hipparcos mission was 7.63±1.64 mas, yielding an estimated radius for Betelgeuse of 3.6 AU.

I understand how a distance can be determined using parallax, but I don't understand how the radius can.

What is the math behind deriving a radius from a parallax measurement?

Betelgeuse is large and close enough that it's angular diameter can be measured directly (via optical interferometry etc...)

When you have the angular diameter, knowing the distance lets you calculate the radius with simple trigonometry.

The fact that Betelgeuse has such a large angular diameter has actually made parallax measurements more difficult, because you are measuring the relative movement of a disc that's not insignificant in angular diameter compared to its angular parallax movement, instead of a single point.  So the distance to Betelgeuse has been more uncertain than might normally be expected from a relatively nearby star.


Why does the Simbad page "A.A. Michelson's Jovian Galilean-satellite interferometer" show data for Betelgeuse? When searching for things related to How did Michelson measure the diameters of jupiter's moons using optical interferometry? I came across the ui.adsabs.harvard.edu entry A. A. Michelson's Jovian Galilean-Satellite Interferometer at Lick Observatory in 1891 which links to this Simbad page.
But I don't understand what I'm looking at. The title of the Simbad page is

A.A. Michelson's Jovian Galilean-satellite interferometer at Lick Observatory in 1891.

but the date directly below it seems to be Betelgeuse.
I don't use Simbad so I'm not sure exactly how to understand the juxtaposition of the title's
Jovian Galilean-satellite interferometer" with data about Betelgeuse. Is it possible to explain the page's purpose?

The SIMBAD link might be there just because Osterbrock's 2004 AAS presentation about the interferometer mentioned an observation of Betelgeuse.
This would be consistent with the policy stated in Wenger et al. 2000:

No  assessment  is  made  of  the  relevance  of  the  citation in terms of astronomical contents: the paper can be entirely devoted to the object, or simply give a side mention of it - in both cases this gives a reference in SIMBAD.
[...]
SIMBAD approach favours exhaustivity, at the cost of increased information noise.

Other articles about SIMBAD bibliography, from Laloë et al. 1993
to Delacour et al. 2018,
show an evolution of their process from mostly manual to largely automated, but knowledgeable humans are still in the loop.
Sometimes an instrument or project intended primarily to investigate one subject also makes important contributions when reused on another subject.
For example, the Dark Energy Survey of distant galaxies has also contributed to the discovery of several trans-Neptunian objects.
However, that is not exactly what happened here.
Michelson 1920 discusses another interferometer spanning the objective of the 100-inch telescope at Mt. Wilson, noting that atmospheric seeing conditions affected the results less than expected.
Betelgeuse required a longer baseline, so Michelson and Pease 1921 used four 6-inch mirrors mounted on a 20-foot steel beam.


High quality, detailed images of planets on web I have been looking for images of the planets and have found pretty decent ones of Earth and Venus from NASA.

That detail is opposed to this one of Uranus:

And these of Mars, (and this and this)

I am disappointed in the quality of the public domain images from NASA I have found for these planets:

Mars
Jupiter
Saturn
Uranus
Neptune
Pluto
Earth's Moon

I am happy with the ones found for:

Mercury
Venus
Earth

They seem to be high quality in terms of detail and pixel size (in the MBs).

Wondering if one could explain how I go about finding these public domain planet images from NASA (or elsewhere, but I imagine they are all from the Hubble telescope, I don't know too much) that are high quality in terms of detail and pixels/size. If nothing else it would be helpful to simply link to them if there is no pattern to where they are located.

If there is a pattern to where the high quality images are located, I would be interested too to find ones of comets/asteroids/stars. I have found many high quality images of galaxies which is neat, but not of the planets yet.

The best NASA images of planets are not from the Hubble Space Telescope; they are from the spacecraft that flew by or went into orbit around them. For Uranus and Neptune, this means Voyager 2; for Pluto, this means New Horizons. For Jupiter, this means primarily Voyagers 1 and 2, and especially Galileo and Juno; for Saturn, Voyagers 1 and 2 and (especially) Cassini.

If you google for "[planet-name] planet gallery", you should get some pretty good images that way; also try adding the name of a relevant spacecraft (if you're stuck, check the Wikipedia page for each planet to learn which spacecraft visited them). A few NASA web pages:

https://photojournal.jpl.nasa.gov/mission/Viking
https://solarsystem.nasa.gov/planets/saturn/galleries
https://solarsystem.nasa.gov/planets/neptune/galleries
https://solarsystem.nasa.gov/planets/dwarf-planets/pluto/galleries

and so forth. (Note that Uranus, even when seen by Voyager 2, is pretty featureless and boring, at least at visible wavelengths.)

You might also try searching at the Astronomy Picture of the Day site. (Not all of those are necessarily public domain, thoug.)


Why do scientists think life requires water? As far as I know, scientists seeking extraterrestial life are simply looking for plantets containing water in liquid state. To be honest, I find this weird.

It is true that life - in the form we know on Earth - cannot exists without water. But how can one proof that such life isn't but a special case out of a bunch of other different types of life?

Actually, I think we can come out with at least one example of a "organism-like thingy" that would not require liquid water. Well, we could build a simple robot, whose "natural ecosystem" would be a junkyard (for mechanical, electric and electronic garbage). That robot would would fuel itself with solar energy, and reproduce by seeking out appropiate parts in the junkyard and building a copy of itself from them. Of course, that's just a theoretical example, I know it has many faoults, but I still hope the idea is clear - it should be indeed possible for life to exists without water.

I have also heard (sorry, I can't remember the source) that theoretically, there could be an organism whose metabolism would rely on sulphur.

So, unless scientists can proove that no life can exists without water (any life; not just Earth-like life), why do they keep looking just for water?

Unless, of course, we define "life" as "any organism functioning like Earth organisms; therefore, metabolising organic structures with enzymes, built of protein, requiring water, etc"

BTW. This question is a little bit different than Why is liquid water considered a requirement for life? . The answers there were pointing out that water is a good solvent. And I'm (also) asking why is any similar solvent considered needed. See my example above.

When you lose your keys on a dark night you look where the probability of finding them is highest (which is a combination of being likely to see them with their probability of being there), which is not necessarily where they are most likely to be. In the case of the search for extra-terrestrial life we have a fairly good idea what the signature of carbon based life using water as its primary solvent (conventional life) looks like. We also know that such life is possible. Other unconventional forms of life could exist but we have a very poor idea of what signature to look for, so we devote most of our effort to looking for the signature we will recognise.

We do not need to prove that unconventional life is impossible in order to spend most effort on searching for the conventional. It is sufficient to suspect that the probability of finding such life is higher than unconventional just because the probability of detection is so much higher. If we find nothing looking for conventional life, then a larger proportion of the effort will be devoted to searching for the unconventional.


How can professional astronomers keep amateurs engaged given the ever-increasing power of modern telescopes? There're many amateur astronomers out there who build their own telescopes, write their own data-gathering programs, and scan the night sky tirelessly. Sometimes they even happen upon new discoveries simply because they had their telescopes pointed in the right direction at the right time.

However, the telescopes used by professional astronomers get more and more sophisticated all the time. This means that the window for amateur astronomers to contribute gets smaller and smaller all the time. For example, the recently-discovered Comet Borisov was discovered by an amateur, but the number of comets discovered by amateurs has been constantly decreasing. Which is obviously rather depressing. One day someone might write a poignant documentary which goes "I thought I was contributing, but then the professionals built [this telescope] and I've become redundant ..."

These citizen astronomers are obviously great PR for astronomy, but given the above, how can professional astronomers keep them engaged?

The role of amateurs in making discoveries will likely change. Telescopes like the LSST will make it even more unlikely that an amateur with a smaller telescope will detect an object first. But the LSST will produce a vast amount of data that will need to be processed, and amateurs may still play a role in this, as citizen scientists do with projects like Folding At Home.

Professional astronomers will continue the same outreach they currently do. Amateur astronomy is still rewarding without the vain hope of making new discoveries. It's enjoyable for the same reason people still visit scenic wilderness instead of just studying maps and photos of it on their computers.


Gravitational wave distortion Gravitational waves detected are from merging of black holes few millions of light years away. While waves have reached us after travelling so far, they must have encountered many black-holes and have reached us after refracting/gravitational lens effects. Then how do one decide the correct direction of those merging black holes?

Gravitational waves are affected by gravitational lensing in the same way as light. Most lensing events are not caused by black holes, but by more common objects with similar mass.

There largest deflection examples of the light from distant galaxies, quasars etc being diverted by foreground sources (clusters of galaxies) is about an arcminute (e.g. Zitrin et al. 2014). Current gravitational wave detectors are not capable of locating sources even to a few degrees. So source location isn't an issue at present.

Source strength might be. A lensed source can be magnified or even split into multiple images that arrive at different times. This will lead to a luminosity distance underestimate from a gravitational wave source and an overestimate of its mass.

For those that are interested, Oguri et al. (2018) have done an extensive set of simulations, showing the effects of gravitational lensing on estimates of mass and redshift distributions for black hole binary gravitational wave sources.

The conclusion is that for detectors like LIGO there should be a (very) small tail of highly magnified events with underestimated redshifts and overestimated masses. These are predominantly at (chirp) masses $>40 M_{\odot}$ (which is also roughly the minimum mass of the primary component in a binary) and more massive than any of the LIGO events seen so far.

The "optical depth" to lensing as a function of source redshift is shown in Fig. 3 of that paper. What this shows is that the chance of any individual source being strongly ($>$ factor of 10 or split into multiple images) lensed is less than one in a million at typical LIGO source distances of a billion light years (redshifts of less than 0.1). But this probability can increase by several orders of magnitude at the high redshifts that might be probed by detectors in the future.


Explanation of resonance stability and instability I'm having some trouble grasping the idea of planetary resonances due to what seems to me like an ambiguity. It is known that the asteroid main belt is sculpted by the Kirkwood gaps, corresponding to orbits that are in mean motion resonance with the Jupiter. As such, the asteroids reaching such orbits get ejected (unstable) from such resonance orbits. Nevertheless, objects like Pluto are dynamically stable due to their resonance with Neptune. Mercury's spin-orbital resonance makes its elliptical orbit stable around the Sun.

Any ideas how to explain the two-fold nature of the outcomes of planetary resonances ? Any references to some literature / mathematical proof would be highly appreciated.

Thank you in advance,

Mercury and Spin-orbital resonance is pretty straight forward.  Planets and Moons are gravitationally lumpy and large bodies are somewhat fluid, even rocky bodies.   Both aspects are prone to tidal forces and that can lead to spin-orbital resonance if the tidal forces are strong enough.   Mercury's somewhat high eccentric orbit balances out with a 3:2 spin-orbital resonance.

Resonance in 3 body orbital system is much more complicated and the math is above my pay-grade.  A proper answer to this question would have lots of math where orbital stability can be measured over long time periods and thousands of orbits.  Something precision is probably lost in an intuitive explanation.

The standard, 2 paragraph answer on Kirkwood gaps is that, if the Kirkwood object is in resonance with Jupiter, then Jupiter gives it a tug whenever it passes closest to the planet and those tugs over time destabilize the 2:1, 5:2, 7:3, 3:1 and less populated 4:1 orbital resonances.

See this summary.

  The Kirkwood gaps are regions in the main asteroid belt that have been
  cleared of asteroids by the perturbing effects of Jupiter, named for
  Daniel Kirkwood who discovered them. The Kirkwood gaps are due to
  resonances with Jupiter's orbital period. For example, an asteroid
  with a semimajor axis of 3.3 AU makes two circuits around the Sun in
  the time it takes Jupiter to make one and is thus said to be in a 2:1
  resonance orbit with Jupiter. Once every two orbits, Jupiter and such
  an asteroid would be in the same relative positions, so that the
  asteroid would experience a force in a fixed direction. Repeated
  applications of this force would eventually change the semimajor axes
  of asteroids in such orbits, creating gaps at that distance. Gaps
  occur at 4:1, 7:2, 3:1, 5:2, 7:3, and 2:1 resonances

This is only a partial answer to your question, because 3:2 (Hildas) and the much less populated 4:3 (Thules) collect around the resonance, they don't move away from it, and as you noted, Pluto and other Resonant Trans Neptunian Objects (TNOs) or Plutionos collect around the resonance as well.   They also get tiny tugs from the planet each orbit, but those tugs tend to keep them around the resonance rather than pulled away from it.

As a perhaps obvious sidebar, Resonant objects, while they orbit the sun, the need to be "herded" so to speak, by a single planet.   That's why there are relatively few resonant objects between the gas giants, and the vast, vast majority of them are between Mars and Jupiter, because Mars is small enough to not have much gravitational influence, or outside Neptune, because there are no large planetary bodies in the vicinity outside Neptune.   That's why Kirkwoods and Hildas and Resonant TNOs dominate the known orbital resonance objects.

Cis-Jovian, for inside Jupiter's orbit or Trans-Neptunian, for outside Neptune's.   In time, we might discover Planet Nine and it might have a collection of resonant Trans and Cis-Ninian objects, but that remains to be seen, and I only mention it because I wanted to be say the word Cis-Ninian.  Obviously that won't stick if/when the planet is found and given an official name, but for now, there are theoretical resonant Trans and Cis-Ninian objects.   In fact, it's that theoretical resonance that lead to the hypothesis that Planet Nine was out there and gave a clue where to look.

So . . . meat of the answer:

Jupiter's 3/2 Hilda's, for the most part, have their aphelions passing close to one of Jupiter's Lagrange points, L3, L4 or L5.  That's why the Hilda's form a kind of triangle.   They don't orbit in a triangle, they orbit in slightly perturbed elipses, but together, they form a triangle because there are 3 main categories of Hildas, those with aphelions in Jupiter's L3, L4 and L5.

This is explained in more detail here, under Dynamics

  The Hildas taken together constitute a dynamic triangular figure with
  slightly convex sides and trimmed apices in the triangular libration
  points of Jupiter—the "Hildas Triangle".2 The "asteroidal stream"
  within the sides of the triangle is about 1 AU wide, and in the apexes
  this value is 20-40% greater. Figure 1 shows the positions of the
  Hildas (black) against a background of all known asteroids (gray) up
  to Jupiter's orbit at January 1, 2005.7

  Each of the Hilda objects moves along its own elliptic orbit. However,
  at any moment the Hildas together constitute this triangular
  configuration, and all the orbits together form a predictable ring.
  Figure 2 illustrates this with the Hildas positions (black) against a
  background of their orbits (gray). For the majority of these
  asteroids, their position in orbit may be arbitrary, except for the
  external parts of the apexes (the objects near aphelion) and the
  middles of the sides (the objects near perihelion). The Hildas
  Triangle has proven to be dynamically stable over a long time span.

This resonance where the Aphelion is about 60 degrees or 180 degrees away from Jupiter keeps the Hilda objects from getting too close to Jupiter, which assists their long term stability.  Too close for a resonant object is generally bad.

Pluto, may have a similar resonant pattern with Neptune (note, I need to check my sources, I misread the linked answer) - will try to follow up.
 some details in the accepted answer to this question.

The Kirkwood objects are mostly too close to the sun to be gravitationally tied to one of the 3 Trojan points.   Passing close to Jupiter's orbit would make a 2:1 resonance impossible.   As a result, there isn't a stable area for them to settle in, and they are pushed by Jupiter away from resonance, unlike the 3:2 resonant Hilda's which orbit through one of the Trojan regions.  That's the best I can work this out at least.

I'm not sure I explained that as well as I should have, but that's the gist of it, as I understand it.  The requirements for resonance are very specific and the Kirkwood objects don't have a stable zone to gather around, so the periodic tugs of Jupiter sheppard them between the resonances rather than hold them to the resonances.


Converting Light curve to Luminosity So I'm currently working around a specific variable star, α Ori and I want to determine it's luminosity...I do only have the light curve (driven from V-optical photometric band - Johnson V from AAVSO)... Is it allowed to utilize it to compute for Luminosity L? The distance modulus doesn't seem to fit since it is for 'visual' magnitude (Using our eyes) so I guess it requires a different approach?
I tried to check in SDSS tho for the other photometric system...But since it avoids the galactic plane and α Ori is "bright" enough, I'm sure the detectors will be saturated so I opted it.
Since I'm new to this kind of stuff, I'm looking forward for your help/s, suggestions and others. Also, I may have missed out something so I apologize.
Best regards,
CGHA

To calculate a "luminosity light curve" from a time series of V-band photoetry, you need two things.

You need to know the distance. The distance to Betelgeuse is uncertain and that means the absolute value of the luminosity you get will also be shifted systematically up or down by whatever distance you adopt.

The V-band only contains a small fraction of the luminous flux from Betelgeuse (most is in the red and near infrared). To convert a V-band magnetude to a "bolometric magnitude" requires you to add a bolometric correction, which is dependent on the spectrum of the star. In principle, if the star is varying, in the V-band, that might be because the bolometric correction is varying (i.e. the luminosity is constant, but the distribution of flux with wavelength is changing). Or it could be that the bolometric correction is constant and the total luminosity is varying. You can't tell from the V-band data alone.

In summary - you can't do it.
The best you can do is assume a constant bolometric correction and turn your V-band light curve into a bolometric magnitude light curve and hence into a luminosity light curve. But then, since the absolute luminosity is uncertain because of the distance uncertainty, you may as well just work with the V-band flux.
To do any better requires nearly-simultaneous photometry in more bands, especially in the red and near-infrared.


How can I download all of this JPL MRO imaging data at once? There are several files. I think that they all needs to downloaded to have the complete dataset. Is there a particularly good way to download them all in one shot?

https://pds-imaging.jpl.nasa.gov/volumes/mro/release50.html

The page shows over 100 individual icons, one for each volume.

You can do it, I will leave you to decide whether you should... As an experiment, I downloaded the mrox_3374 subdirectory of files and the result came out to 3.7 GB. Assuming the other 99 "volumes"/directories are the same, this multiplies out to 370 GB.

I downloaded these data using wget which is a standard utility on Unix/Linux and Mac computers. It is also available for Windows systems; this Wget for Windows 10 tutorial should be of some help in getting it installed and going.

Once this is installed you can do:

wget -np -nH -c --cut-dirs 4 -r https://pds-imaging.jpl.nasa.gov/data/mro/mars_reconnaissance_orbiter/ctx/mrox_3374/

The key flags are -r which means recurse into directories, -np which means don't ascend to the parent directory (to stop it sucking down the entire website), -c which will continue the transfer if it fails and you restart the wget and the --cut-dirs 4 which cuts 4 levels of the directories off the recursive download (the data/mro/mars_reconnaissance_orbiter/ctx parts)

If you really, really want to suck down the whole thing you could do:

wget -np -nH -c --cut-dirs 3 -r https://pds-imaging.jpl.nasa.gov/data/mro/mars_reconnaissance_orbiter/ctx/

and that should grab everything.


Can there be a three-moon system where only two are visible most of the time? I am trying to build a moon system for a sci-fi work in collaboration with my boyfriend. He's the main writer and wants there to be three moons circling the planet in a way that most of the time, only two moons are visible and seeing all three or none of the moons is equally rare. I don't have much training in astrophysics (only a university level intro class to it), so my main consideration was that, in the case of Earth's moon, it is only fully/almost non-visible about 10% of its rotation (2-3 days of 28). What else should I consider? If it's possible, how long would each of their orbits be?

For reference, the planet would be orbiting around a Sun-like star and be about 1.13AU away from it (still within the habitable zone, but with a slightly longer year). Each of the three moons rotates on itself as fast as it rotates around Earth, just like Moon does.

Start with some basic calculations.  At any time, you can see a hemisphere, which is (gosh! :-) ) half of all space.  This means that no matter what you do the average presence (lit or not) at any time is  1.5 moons.  Since there's only one "sun," all moons present in the observable hemisphere will have an equal amount of the observable face illuminated.

Now, if we decide to have two of the moons orbiting each other in a binary fashion (and their net center is orbiting the planet), then 50%  of the time  these two moons will be in your visible hemisphere, and 50% of that time so will the third.  Conversely, 50% of the time these two moons will not be in your visible hemisphere, and 50% of that time neither will  the third.

All of the above assumes orbits similar to our Moon, i.e. roughly equatorial, around our own planet's rotational axis.  What if these 3 moons somehow manage to be circumpolar?  In this case they'll be permanently "quarter moons," visible whenever in our hemisphere of view.  Now make the third moon circumpolar with such huge ellipticity that it's not visible to the naked eye for most of the orbit.

So the only thing not possible in either  of the above scenarios is managing to make the first two moons visible for "most" but not "all" of the time.   Perhaps someone else can come up with a more esoteric solution.

Note: all this assumes the moons are small enough that their interactions do not lead to short-term chaotic behavior.


Why Is The Expansion of The Universe Allowed To Move Faster Than Light? Why is the expansion of the universe allowed to faster than light if everything that moves within a vacuum can't reach the speed limit of light?

I've read that astronomers looked at the distance between galaxies but I'm still not understanding what's going on. If objects in the universe can't go faster than the speed of light in a vacuum, how are galaxies (which objects within the universe) moving away from each other faster than light?

You are not allowed to move faster than c in sense of classical motion that is velocity can't be greater than  c but ,universe is expanding not because objects are moving away but new space time is being created between them which increases distance between them and that can happen such that expansion is faster than c , there is no violation of theory of relativity.
I hope that's clear


Tables of aphelion/perihelion dates for other planets? It's relatively easy to find tables for Earth aphelion and perihelion dates and times, such as this one:

http://www.astropixels.com/ephemeris/perap2001.html

But what about the other planets in the solar system? Are there tables or lists of aphelion and perihelion dates for them too? I checked Wikipedia, but it only lists perihelion distances and not dates, so it is insufficient.

Update:

This is the error I get on the SPICE web interface.

(too long for a comment, this answers why barycenter data is available for a much longer time span than planet center data)

When NASA computes planetary positions, they treat each planet and its moons as a single point mass (the barycenter of the planet and its moons). They compute these positions +-15000 years from now.

For most planet systems, the barycenter is very close to the planet center (Mars' planet barycenter is only a few inches from Mars' system barycenter because Phobos and Deimos have such little mass-- for the moonless Mercury and Venus, the system barycenters and the planet barycenters are the same).

They then compute the locations of the moons and the planet's center itself. These calculations are done for a much shorter period of time.

Therefore, you can compute the location of the barycenter of Jupiter and its moons for +-15000 years, but the actual barycenter of Jupiter itself for a much shorter period of time, even though Jupiter is massive enough that its own barycenter is very close to the barycenter of Jupiter and its moons.


Ratios of magnitudes instead of differences The color index in the photometric system is defined by the differences in magnitudes between two wavelength filters, which gives the ratio of intensities. For example, to determine the B-V index, you'd need the magnitudes in the B band (which is centered on $\lambda = 445$nm) and in the V band, centered on $\lambda = 551$nm.

But what about other operations? For instance, would there be any significance in taking, say, the ratio $B/V$ or $B\times V$?

Consider for example these bright stars (V and B-V values from Yale Bright Star Catalog).
Near V=0, B/V is ill-behaved, and B*V is near 0.

Name       B     V     B-V    B/V   B*V
Arcturus  1.19 -0.04  +1.23 -29.7  -0.05
Vega      0.03  0.03   0.00   1.00  0.00
Capella   0.88  0.08  +0.80  11.0   0.07
Rigel     0.09  0.12  -0.03   0.75  0.01

Now imagine that each of those stars were 10 times as far away.
Neglecting interstellar extinction, this would decrease the apparent brightness by a factor of 100 and increase the apparent magnitude by 5.0.
As B and V increase together, B/V approaches 1 and B*V approaches V^2, telling us nothing meaningful about a given star.

Name       B     V     B-V   B/V    B*V
Arcturus  6.19  4.96  +1.23  1.25  30.7
Vega      5.03  5.03   0.00  1.00  25.3
Capella   5.88  5.08  +0.80  1.16  29.9
Rigel     5.09  5.12  -0.03  0.99  26.1

B-V, on the other hand, has stood the test of time as a distance-independent metric of a star's color and surface temperature.


Would time go by infinitely fast when crossing the event horizon of a black hole? If you were to fall into a black hole, my understanding is that from your reference point, time would speed up (looking out to the rest of the universe), approaching infinity when approaching the event horizon.  If this is correct, would you see the whole universe's future "life" flash before your eyes as you fall in, assuming you could somehow withstand the tremendous forces, and assuming black holes don't evaporate?  If it is correct that black holes evaporate due to Hawking radiation, would you be "transported" forward in time to where the black hole fully evaporates?

This is considering the "alternate" frame of reference from my question: Does matter accumulate just outside the event horizon of a black hole?  In that question, I thought about what happens to matter falling into a black hole from the perspective of someone observing from the outside (e.g. as seen from earth).  Here I am considering the perspective of the thing falling into a black hole.

This also takes into account ideas discussed in: Why does time get slow near a black hole?

Note: This answer to another question provides some insight here too (refer to the last part of the answer): https://astronomy.stackexchange.com/a/3713/1386

(I will assume a Schwarzschild black hole for simplicity, but much of the following is morally the same for other black holes.)

  If you were to fall into a black hole, my understanding is that from your reference point, time would speed up (looking out to the rest of the universe), approaching infinity when approaching the event horizon.

In Schwarzschild coordinates,
$$\mathrm{d}\tau^2 = \left(1-\frac{2m}{r}\right)\mathrm{d}t^2 - \left(1-\frac{2m}{r}\right)^{-1}\mathrm{d}r^2 - r^2\,\mathrm{d}\Omega^2\text{,}$$
the gravitational redshift $\sqrt{1-\frac{2m}{r}}$ describes the time dilation of a stationary observer at a given Schwarzschild radial coordinate $r$, compared to a stationary observer at infinity. You can check this easily: plug in $\mathrm{d}r = \mathrm{d}\Omega = 0$, the condition that neither the radial nor the angular coordinates are changing (i.e. stationary observer), and solve for $\mathrm{d}\tau/\mathrm{d}t$.

The conclusion is that if you have the rocket power to hover arbitrarily close to the horizon, you will be able to see arbitrarily far into the history of the universe over your lifetime. However, that doesn't actually cover what happens to an observer that crosses the horizon. In that case, $\mathrm{d}r\not=0$, and the coefficient of $\mathrm{d}r^2$ above becomes undefined at the horizon: as in the other question, the Schwarzschild coordinate chart simply fails to cover the horizon and so is ill-suited for talking about situations cross the horizon.

But that's a fault of the coordinate chart, not of spacetime. There are other coordinate charts that are better adapted to questions like that. For example, the two Eddington-Finkelstein charts are best suited for incoming and outgoing light rays, respectively, and the Gullstrand-Painlevé chart is adapted to a freely falling observer starting from rest at infinity.

  If this is correct, would you see the whole universe's future "life" flash before your eyes as you fall in, assuming you could somehow withstand the tremendous forces, and assuming black holes don't evaporate?

No. I think this is best seen from the Penrose diagram of Schwarzschild spacetime:

Light rays run diagonally. In blue is an example infalling trajectory, not necessarily freely falling. Note the two events where it crosses the horizon and where it reaches the singularity. Shown in red are inward light rays that intersect those events. Thus, the events that the infalling observer can see of the external universe consist of the region between those light rays and the horizon. The events occurring after that won't be seen because the the observer will have already reached the singularity by then.

Now suppose the observer tries a different trajectory after crossing the horizon, accelerating outward as much as possible in order to see more of the future history of the external universe. This will only work up to a point: the best the observer can do is hug the outgoing light ray (diagonally from lower-left to upper-right) as much as possible... but since the observer is not actually allowed to go at the speed of light, seeing all of the future of history will be impossible. The best the observer can do is to meet the singularity a bit more on the right of the diagram.

By the way, since the light ray worldlines have zero proper time, trying to do that will actually shorten the the observer's lifespan. If you're in a Schwarzschild black hole, you would live longer if you don't struggle to get out.

The above is for an eternal, non-evaporating black hole, as that's what you're asking about here. (The 'antihorizon' is there because the full Schwarzschild spacetime is actually an eternal black hole and its mirror image, a white hole in a mirror 'anti-verse', which not shown on this diagram. That's unphysical, but not relevant to the situation we're considering here.)

  If it is correct that black holes evaporate due to Hawking radiation, would you be "transported" forward in time to where the black hole fully evaporates?

An evaporating black hole is morally the same as above: only an ideal light ray can reach the point when the black hole fully evaporates; everyone else gets the singularity. (Since this ideal light ray right along the horizon would be infinitely redshifted, arguably not even that.) You can repeat the above reasoning on its Penrose diagram yourself:

Addendum:

  I have thought a bit about this, and does this solution take into account the relativistic time effects near the horizon of the black hole (e.g. is my understanding correct that the observer would observe time in the universe passing approaching infinitely fast when approaching the event horizon)?

How much time dilation happens depends entirely on what coordinates we're talking about (more generally, which frame field). What a given observer will actually see, however, is completely independent of choice of coordinates. In particular, Penrose diagrams illustrate the light cone structure of the given spacetime, and what an observer can in principle see depends entirely on what light rays intersect the observer's wordline. So yes, it's taken into account by default.

If you're actually falling in it, no, your understanding is mistaken, for reasons explained above. For additional motivation, flip the question around: what does the very distant stationary observer see of the infalling object? On the above Penrose diagram, outwardly directed light rays are diagonal, from lower-left to upper-right. Draw some some outward light rays from the blue infalling worldline. You will see that no matter how far into the far future (up on the diagram) you pick an event outside the black hole to be, you can connect that event with an outward light ray originating from the blue infalling worldline before it crosses the horizon. The conclusion would be that an observer that stays outside the black hole would be able to see the infalling object arbitrarily far into the future. No matter how much time passes for someone who stays out of the black hole, the image of the infalling object would still be visible as it was before it crossed the horizon. (In principle at least; in practice it will get too faint to see after a while.)

Thus, the usual result of "infinite gravitational time dilation makes the image of the infalling object hover forever near the horizon" is also straightforwardly deducible from the diagram, and so is completely consistent with the infalling object being able to see a finite part into the future of the external universe. Perhaps it is best to emphasize that the situation is not actually symmetric: what the external observer sees of the infalling object is not some straightforward flip-around of what the infalling object sees of the external universe. The black hole itself breaks that symmetry.


What do astronomers do different now that they know about fusion of the Sun? I would like to know how knowledge of the fusion in the Sun has changed how astronomers explore the universe. What do astronomers do differently now that we know that the Sun’s use of proton-to-proton chain to create energy? Please include sources.

How has knowledge of fusion of the Sun changed astronomy?

TL;DR: It changed everything.
Up until the 1930s, physicists and astronomers argued for a rather young universe while geologists and biologists argued for a markedly older Earth. The problem was that physicists and astronomers couldn't image any known processes that would possibly allow the Earth to be as old as geologists and biologists claimed. Note well: geologists and biologists of that time were not arguing for an Earth that is 4.55 billion years old. They were wanting an Earth that was at least several hundred million years old.
Hans Bethe's 1938 paper, along with a few papers that immediately preceded Bethe's paper, paved the way for addressing that discrepancy. It wasn't until the 1950s when Clair Cameron Patterson showed that the Earth was much older than geologists and biologists thought. The 1950s also were when the apparently ridiculous claims of an expanding universe finally become accepted.


Why is blue light extincted more strongly than red? We know that blue light suffers extinction more strongly than red (examples include ISM, atmosphere etc.), but why? Is it a property of the shorter wavelength itself? Can someone provide a conceptual explanation as to why this occurs? TIA

There are two effects causing this:

The relevant quantity for determining whether or not a photon interacts with a particle is the ratio
$$
x \equiv \frac{2\pi r}{\lambda},
$$
where $r$ is a size of the scattering agent.

Rayleigh scattering

When $x\ll 1$, we're in the Rayleigh regime where the wavelength is so long compared to the particle that the electric field of the photon causes charges within the particle to oscillate with the same frequency, effectively resulting in dipole scattering. This has a very strong wavelength dependence, with the probability of interaction begin proportional to $\lambda^{-4}$.

Mie scattering

When $x \sim 1$, we're in the Mie regime, where the probability of interaction is largest. So if you had an ensemble of randomly sized particles with an equal amount of any size, and you shone a beam of light consisting of photons with the same distribution of wavelengths, you would scatter all photons with equal probability (unless of course your distribution extended to very small values of $x$).

However, as with many other things in the Universe — e.g. galaxies, stars, cities, lakes, and wealths — dust grains are not equally distributed in size, but has a distribution which declines as a power law from small to large size. In the case of dust grains, this power is very steep, with a power law slope of roughly $-3.5$; that is, the density $n$ of dust grains as a function of size drops as
$$
n(r) \propto r^{-3.5}.
$$

Hence, there are many more small dust grains available, casing the blue photons to be scattered much more often.

Larger particles / smaller wavelengths

When $x>1$, but less than $\simeq1.8$, the probability of scattering declines, meaning that in this narrow window, red light is scattered more easily than blue. Coincidentally, these values correspond to optical light and the dust on Mars, leading to blue sunsets, as discussed in this answer.

When $x\gg 1$ — i.e. when the particles become very large compared to the wavelength — we have "geometric scattering", all light is scattered equally. In astronomy, we call this "gray dust", and would be caused e.g. by smalles pebbles.

Scattering vs. absorption

The albedo $A$ of the dust grains — i.e. the probability that an interaction result in a scatter rather than an absorption — is also wavelength dependent, but not as strong. Typical values are of the order of $0.5$, but decreases toward zero for long wavelengths (e.g. Pei 1992)


What is radiation pressure and how does it prevent a star from forming? This is a follow up to: Is there a theoretical maximum size limit for a star?

The answer there talks about the radiation pressure preventing a star from forming.

What reaction is causing this radiation pressure and why does this potentially prevent a star from forming?

Radiation pressure is nothing but electromagnetic interaction.

Imagine a hydrogen atom hit by a stream of photons coming from the same direction. Although the atom as a whole is neutral, the electron and the proton are physically displaced, forming a dipole, i.e. a positive-negative charge couple. Some of the photons therefore scatter against the dipole transferring to it some momentum. So the atom start moving in the same direction as the photons. If the photons are in the ultraviolet, the electron can be exited to higher orbitals and possibly stripped from the atom. In this case the scattering is even more efficient.

Now imagine a star surrounded by a layer of hydrogen. Gravity attracts the layer towards the star. The photons emitted by the star try to push the hydrogen atoms away from it, through the electromagnetic force.

Very massive stars are very luminous and hot, which means that they emit a lot of ultraviolet photons. When the pressure transferred from the photons to the layer is larger than the gravitational attraction, then the layer begins expanding, effectively stopping growth of the star.

In the figure posted by the OP there is also dust. I don't know the details of photons-dust-gas interactions (we need a stellar atmosphere expert, I guess), but the basic principle is nonetheless the same.


Why can we trust Hubble Time if the rate of expansion is not constant? The age of the Universe can be estimated from taking the inverse of the Hubble constant: $t_\text{universe} = 1/H_0 =d/v.$

It seems to me this method assumes that any given galaxy has been receding at a constant apparent velocity for the lifetime of the Universe. However, by including data from larger & larger distances, it can be seen that Hubble's law doesn't hold. The data suggest that the rate of expansion was actually slower in the past.

Why is it appropriate to use only nearby galaxies to estimate the age of the Universe using the Hubble time when the rate of expansion isn't constant?

$H_0^{-1}$ is only a rough estimate for the age of the universe and you have correctly identified the reasons why not.

A correct age estimation relies on knowing $H_0$ and the densities of matter and dark energy, so that the past expansion history of the universe can be correctly modelled. Even this relies on an assumption about how dark energy behaves.

A more interesting question is why a value of $H_0 \simeq 70$ km s$^{-1}$/ Mpc gives an $H_{0}^{-1}$ of 14 billion years, which is within a few per cent of the current best estimate for the age of the universe of 13.8 billion years. In the $\Lambda$CDM cosmology model, the reason for this cosmic coincidence is that the universal expansion underwent a period of decelaraton up until about 4 billion years ago, when it started to accelerate again (the red curve on the plot below).

As a result, a tangent to the red curve, showing the size of the universe versus time, at the present epoch almost goes to ${\rm size}=0$ about 14 billion years ago. If we were to go backwards or forwards in time by 5 billion years, the agreement between $H_0^{-1}$ and the age of the universe would not be so good. At earlier times $H_0^{-1}$ would have overestimated the age of the universe, whereas at later times $H_0^{-1}$ will underestimate the age of the universe.


How to model and subtract nebular emission lines from stellar spectrum I have optical stellar spectra from a star forming region with spatially varying nebular background.

I want to subtract the nebular lines from the stellar spectra (which are varying in intensity from spectrum to spectrum because of the varying background).  How can I model and subtract individual nebular lines from a stellar spectrum?  Are there any known tools for this?  Any help or suggestions would be highly appreciated.

In general, you can't.  If obtaining spectra in regions where there is expected to be a spatially varying background then you either need to do long-slit spectroscopy so that you have a good measurement of the ISM contribution either side of your source, or you do integral field spectroscopy with the same idea.

The problem is that the line strengths for the emission lines you see from the ISM are integrated along the line of sight and are very dependent on the physical conditions of the ISM. This means that you cannot really assume any particular ratio of line strengths to assist in any subtraction.

In some circumstances you may be fortunate that there is a relative velocity between your stellar source and the background - in which case, separation of the nebular and stellar components may be possible. Or it may be that the stellar features are broadened by rotation (if you have the resolving power), whereas the nebular contribution is usually very narrow/unresolved.

It is difficult to provide more details without a lot more information about the observations, the target and the background.


Why don't the inner moons of Jupiter have tidally-induced volcanism? Volcanism on Io is caused by the fact that it is tidally heated. There are four moons that are closer to Jupiter than Io with higher eccentricities, yet they don't seem to have any volcanism at their surface. Why isn't the tidal heating enough to generate volcanism on any other moons of Jupiter?

It’s because they are much smaller than Io.  Tidal forces are differential forces, that is, they result from the difference in gravitational pull on one side of a body compared to the other.  When an object is small, the difference in distance to the two sides of it is necessarily small as well.
According to Wikipedia, Amalthea, the largest of those four innermost moons has a long axis that is only 250 km, and the others are smaller yet. The strength of the tidal heating scales as the body’s radius to the 5th power, quite a strong dependence.   The small mass plays a role, too, but the dependence on radius is stronger.
For this same reason, small objects on Earth don’t feel tidal stretching from our Moon, yet the Earth as a whole does.
On a related note, the inner three of those four satellites are inside Jupiter’s Roche limit, the orbital distance at which a purely self-gravitating body should be pulled apart by tidal forces.  But they survive because most small bodies aren’t held together primarily by gravity (just as your own body isn’t) - the internal electromagnetic forces between atoms (which manifests macroscopically as the rigidity of the rock) is the main source of the bodies’ structural strength.


How to see Comet 46P/Wirtanen? I've got a Celestron Astromaster 114 telescope. I live in Los Angeles and would like to get a look at 46P/Wirtanen. Can anyone tell me when this comet is going to brightest and how best to locate it in the sky? It's my (perhaps incorrect) understanding that it is not especially bright. Any tips/hints/information would be much appreciated.

Source

If you have, or can borrow binoculars, that is probably a much better way to try to find the comet than your telescope. If it's set up like the one in the image in your question (that I've added) you don't have a computer read-out for RA/Dec or any way to position your mount accurately so that the polar axis points at polaris. The bright skies in LA will make doing so manually a bit of a challenge, and lack of previous experience doing so will make it even more of one.

Go to in-the-sky.org and then to Planetarium view. Go to the bottom right and type 46P into the Find box and it will find your object's name and then display it for you. Turn on the Alt/Az grid instead of the RA/Dec grid and you can see that 46P is about 60 degrees up at 8 PM your time. Lay on your back and take some time to look around with your binoculars.

It should set your location to Los Angeles automatically, but double check that.

You can see that there are some very bright stars lined up to make your viewing easier. First is Betelgeuse the bright corner star in Orion (do not say Betelgeuse three times in a row!) then Aldebaran the eye of the bull, then the comet.

At about 4th magnitude you should be able to see it as a green fuzzy star, unlike anything else. You can also zoom in using in-the-sky.org to see the local stars. The comet is going to be moving relative to the stars, so don't make zoomed-in print-outs and try to use them six hours or a day later.

Set up anything you have around the spot where you'll be laying down, to block lights from neighbors or the street, turn your own lights off. You are going to be looking East-Southeast about 2/3 of the way up to the zenith in the evening, and 46P will remain high in the sky most of the night.

Have fun!

Oh, the telescope! Once you've gotten good at finding 46P with binoculars and have some feel for the other stars nearby, then try the absolute lowest magnification you have on your telescope and just look around slowly. Loosen up your mount so you can move the tube by hand, and scan around until you see the blue fuzzy blob again.


Where did the grand design spiral galaxy designation come from? What is the history behind the name that we gave to that class of galaxies?

The term can be traced back to the theoretical work of C.C.Lin in explaining the spirals as due to density waves in galactic discs.

A paper from 1970 discusses the "Existence of 'Grand Design'". Note the use of "scare quotes", suggesting that the term was new and would not be widely recognised by his readers.

Chia-Chiao Lin was an important Chinese/American Mathematician. It is reasonable to believe that the term was coined by him or one of the mathematicians that he worked with at MIT. And was in use since about 1964, when the first of a series of papers "On the structure of Spiral Galaxies" was published.


Does the Hubble constant depend on redshift? I know there are lot of questions on the Hubble constant already, but I am curious to know if it changes with redshift? If at current redshift, $z=0$, we know its value to be 0.7, will it be different at higher redshift ($z=0.1$)? If so, is there any relationship with redshift?

Yes, definitely.

The Hubble constant describes the expansion rate of the Universe, and the expansion may, in turn, may be decelerated by "regular" matter/energy, and accelerated by dark energy.

It's more or less the norm to use the term Hubble constant $H_0$ for the value today, and Hubble parameter $H(t)$ or $H(a)$ for the value at a time $t$ or, equivalently, a scale factor $a = 1/(1+z)$, where $z$ is the redshift.

The value is given by the Friedmann equation:
$$
\frac{H^2(a)}{H_0^2} =
\frac{\Omega_\mathrm{r}}{a^4} +
\frac{\Omega_\mathrm{M}}{a^3} +
\frac{\Omega_k}{a^2} +
      \Omega_\Lambda,
$$
where
$\{
\Omega_\mathrm{r},
\Omega_\mathrm{M},
\Omega_k,
\Omega_\Lambda
\} \simeq
\{
10^{-3},0.3,0,0.7
\}
$
are the fractional energy densities in radiation, matter, curvature, and dark energy, respectively.

For instance, you can solve the above equation at $z=0.1$ and find that the expansion rate was 5% higher than today.

Since everything but dark energy dilutes with increasing $a$, $H(a)$ will asymptotically converge to a value $H_0\sqrt{\Omega_\Lambda} \simeq 56\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$.

The figure below shows the evolution of the Hubble parameter with time:

As noted by KenG, the fact that $H$ decreases with time may seem at odds with the accelerated expansion of the Universe. But $H$ describes how fast a point in space at a given distance recedes. Later, that point will be farther away, and so will recede faster. From the definition of the Hubble parameter, $H\equiv\dot{a}/a$, multiplying by the scale factor shows the acceleration $da/dt$:


What differentiates a "group of rogue stars" from a galaxy? According to Wikipedia, a rogue star or intergalactic star is

  not gravitationally bound to a galaxy.

But according to a citation on the same Wikipedia page, the first intergalactic stars were discovered in the Virgo cluster of galaxies.

  These stars form a massive group  aproximately 300,000 light years away from the nearest galaxy. (Emphasis mine)

Is there any clear, stated definition which differentiates galaxies from "massive groups of rogue stars"?

It is just a poor choice of words. The comment refers to a discovery paper by Ferguson et al. (1998), where they used a very deep image in the Virgo cluster to establish that there were an excess of distant stars compared with a control field outside of the cluster.

This excess of about 630 stars were all in one tiny HST field of view, so form a "group" in some sense, but there is no suggestion that these stars are not found throughout the Virgo cluster. Indeed subsequent work has found evidence for intracluster stars in other areas of the cluster (e.g. Durrell et al. 2002).


Newtonian reflector collimation question After an initial collimation, I took my scope (150mm/750mm) out for a star test. While I could see concentric circles when the bright star is at the centre of the video, when the star is on the side of the view, I get an image typical of an uncollimated scope. Please refer the images below. The left image is how it appears when the bright star is at the centre. The right image is how it appears when the star is on the upper portion of the view.

Does my scope require further collimation, or is it fine the way it is now?

Your scope is f/5. There will always be some coma aberration.
Always, the image quality is worse as you get closer to the edge of the field of view (FOV).
Personally, given that the fact that you will want to observe your targets at the center of the FOV, this collimation would be acceptable.

Happy photon hunting!


In what ways does the mass and energy of the milky way curve space time? Space is curved by mass (or more precisely by mass and momentum as described by the stress-energy tensor). The milky way system has a lot of mass and a lot of energy. In what ways is space-time curved within it?

In short, not by much.

Out by us the Earth the gravitational acceleration towards the centre of the galaxy is $v^2/r = \frac{(2\times2\times 10^{20}\times \pi\  \mathrm{metres} / 230000000\  \mathrm{years})^2 }{ 2\times 10^{20} \mathrm{metres}}$, which is of the order of $10^{-10} ms^{-2}$ : in other words, it is very small. This is because spacetime is not highly curved except close to massive bodies.

This spacetime is approximately flat. It can be well approximated by Minkowski space. If you need to include in your model the distribution of mass in the galaxy, then Newtonian mechanics is sufficient, for example https://physics.stackexchange.com/questions/62637/the-potential-and-the-intensity-of-the-gravitational-field-in-the-axis-of-a-circ

In the centre of the galaxy is a black hole, and it is in the neighbourhood of the black hole that GR really comes into play. Near the black hole, the Kerr solution of GR is a good model (the Kerr solution describes a symmetric uncharged spinning mass)

Now on the massive scale, a galaxy, or a cluster of galaxies can have a significant effect on light, given enough time, and enough distance. Light, travelling for many millions of years, past a galaxy, can have its path significantly bent by gravity. (The Schwartzchild solution is a first approximation to the shape of spacetime around a galaxy, though for a better approximation, numerical relativity can be used) This causes gravitational lensing. For light traveling inside a galaxy this is not a significant factor, There isn't time in a few years for light to be bent significantly by the general gravitational field of the galaxy. However light from a star passes close to another star, then there can be measurable bending.

The measurement of the bending of light of stars behind the sun during a solar eclipse was one of the first experimental confirmations of General Relativity (1919) https://www.wired.com/2009/05/dayintech_0529/ Note that even light that passes right over the Sun's surface would only be deflected by less that 2 arcseconds.


what is a MIA satellite I came across this sentence in a magazine:

  We sit outside the Lorne Community Hall, Victoria, watching the MIA satellite overhead.

I wonder if the author is talking about an artificial satellite. Can one recognize an artificial satellite with naked eyes?

Thanks for response.

As @usethedeathstar said, MIA satellite most likely means 'Missing In Action'. The following lyrics from the song 'MIA satellite' seems to solidify this:

  MIA Satellite-
  You've got no path in sight.
  You leave me hanging in the dead of night-
  In the dead of night.

Regarding the ability to identify artificial satellites, it is possible to identify some satellites. Some identification guides:

Iridium

Iridium satellites are usually faster than most satellites, but their tell-tale sign is that they go from dim to bright to dim as they rotate. This bright flash is known as an Iridium Flare.

ISS

This moves East to West and is brighter than the average satellite.


Using SDSS data / finding a catalog I am looking for a catalog of low redshift galaxies (preferably SDSS) with information on line fluxes and stellar mass. My Google searches keep pointing me to work by the Portsmouth group, so I downloaded some of the fits files listed at the bottom of the page, but there are no line fluxes in those files, even though they seem to suggest that the line fluxes are being made available... I was just wondering if what I'm looking for is publicly available and if I'm looking in the right place? Their page links to a table that lists all the key words I would like to use, but I'm not sure how to obtain the data.
In one of the files I downloaded (portsmouth_emlinekin_full-26.fits) there's a column called BPT, which lists the BPT classification (Seyfert, LINER, Star forming...), so I was expecting to also find the "building blocks" ([OIII], [NII], [SII], [OII], Halpha and Hbeta line fluxes) of the BPT diagram. Is it usually necessary to contact the authors for stuff like this?
Would be very grateful to learn about using publicly available survey data from a more experienced scientist :)

From what I can see, you already found what you were looking for. The table you linked is the description of the data available in the Catalog Archive Server (CAS) database. All you have to do now is use those table names (it seems that masses are on different tables, here and here) and then search whatever you want. If you do not have constraints, you could just search by sky regions, but I would recommend you to use the SQL queries as you can manually select columns an even join tables. Just to give you a quick example:
SELECT TOP 10
   p.specObjID, p.ra, p.dec, p.z, p.bpt, p.Flux_NeV_3425
FROM emissionLinesPort as p
WHERE
   p.z BETWEEN 0 AND 2

You will get a table with the first 10 results that have a redshift between 0 an 2. Columns will be ID, right ascension, declination, redshift, classification and flux of [NeV] 3425 (everything described in the table you linked). Note that for masses you would have to change the table by editing emissionLinesPort and using stellarMassStarFormingPort or stellarMassPassivePort instead (as well as change the columns accordingly).


How are space telescopes stabilised to a perfect standstill? I assume that for space telescopes to get good images, they must be put in perfect standstill, as even the slightest deviation in the viewing angle is amplified by the distance of observed objects, results in observing a different area of space. How is this feat accomplished? How can you stabilize an instrument in space, so that it can take images of galaxies gazillion light years away? And how do you know you have succeeded, apart from getting quality imagery?

Pointing

In some ways it's easier to keep the direction that a telescope is pointing stable in orbit in space than it is on Earth.

There's no wind Why is the opening in the Anglo-Australian Telescope's dome so small?
You aren't on a spinning planet requiring you to constantly move the telescope with electric motors and gears
there's no sagging due to gravity

There are tiny torques due to differential solar photon pressure on different parts of a space telescope, but these are small, very reproducible and can be learned, predicted, and compensated.

The James Webb Space Telescope has a giant sunshield that keeps the infrared telescope cold, and this can produce substantial torque. So it has a second reflector used to passively null out most of the torque. You can read more about it in answers to How will JWST manage solar pressure effects to maintain attitude and station keep it's unstable orbit?

In low Earth orbit (LEO) telescopes like Hubble will also experience tiny torques due to tidal forces; for example, one end of Hubble is in a higher or lower orbit than the other end. Without damping, this could cause it to slowly oscillate see this answer. Again, this will be small, very reproducible and can be learned, predicted, and compensated.

If you can somehow stop something in space from rotating, it will simply stay stopped and point in a stable direction, at least for a short time. If you can predict the torques due to solar pressure and gravitational tidal forces, you can apply tiny controlling torques to minimize those effects and it will stay pointed longer.

How?

Sensitive gyroscopes to measure tiny rates of rotation (inertial stabilization)
Watch the stars at the edge of the telescope's field of view, using them as guide stars

Both are shown in the image below, taken from Hubble Space Telescope Pointing Control System.

The objects labeled "Gyros" are gyroscopes used to measure tiny rates of rotation of the telescope and send those measurement to the attitude control system for the telescope. Systems can use real spinning wheels, or fiber optic rings with light going in both directions, or other techniques, but the result is the same.

Currently Hubble is working without all three gyroscopic sensors on-line because it has redundant systems: You can read more about that in Hubble Space Telescope Reduced-gyro control law design, implementation, and on-orbit performance

The objects labeled Fine Guidance Sensors use CCDs at the edge of the Hubble's field of view to track stars:

  The FGSs use starlight captured by the telescope’s mirrors to find and maintain a lock on guide stars to ensure that the spacecraft’s attitude does not change. One FGS can also be used as a scientific instrument to determine a star’s position with high accuracy. The level of stability and precision that the FGSs provide gives Hubble the ability to remain pointed at a target with no more than 0.007 arcsecond of deviation over extended periods of time. This is the same as holding a laser beam focused on Franklin D. Roosevelt’s head on a dime roughly 200 miles away — which is about the distance from Washington, D.C., to the Empire State Building in New York City — for 24 hours.

Feed this and other bits of information from other star cameras into the computer, combine it with the predicted torques from tidal forces and photon pressure, and the computer decides on a tiny correction torque necessary to keep it pointed stably.

That message then goes to the objects labeled as Reaction Wheels. These are four heavy, spinning wheels whose angular momentum is balanced such that it normally applies no torque to the telescope.

When the computer senses that a little torque is needed to keep the telescope pointed in the right direction, or to move it to a new field, motors on the spinning reaction wheels will speed them up or slow them down a little bit, and that will apply a compensating torque to the telescope.

Vibrations

Vibrations and oscillations of a space telescope's structure can also be a problem. These are addressed through very careful engineering and special materials. Structures are design to minimize vibration and oscillation by damping, isolation of moving parts (pumps, solar panels, etc.) and by use of light weight and stiff materials like silicon carbide to move resonant frequencies as high as possible where they are most easily damped and isolated.

Silicon carbide is a very popular material in newer space telescopes and is found in the optical system of too man of them for me to remember, but here is mention of it in GAIA's optical bench answer and in New Horizion's LORRI telescope answer.

above: Gaya's Silicon Carbide Optical Bench, with the two objective mirrors of it's twin telescopes pointing 106.5° apart. From Spaceflight 101, image credit: ESA/Astrium.

above: The LORRI telescope is described for example in the ArXiv preprint Long-Range Reconnaissance Imager on New Horizons

below: From Hubble Space Telescope Pointing Control System


How bright could Gliese 710 appear in 1.35 million years? I read this article on Gliese 710 that stated it would come within about 1.1 ly (+/- 0.577 ly) of earth in 1.35 million years. Wikipedia states it would have an apparent magnitude of 0.5, and the original paper abstract states it would have a -2.7 apparent magnitude.

From reading the list of brightest stars, apparent magnitude of 0.5 would make it about the 10th brightest star, and -2.7 would make it the brightest (besides the sun).

I'm curious about the discrepancy between these sources.

How bright would Gliese 710 appear if it were at the near end (0.523 ly) and far end (1.677 ly) of the expected closest approach to us?

WolframAlpha says (to the query "apparent magnitude of star with absolute magnitude = 8.23 and distance = 1.1 ly") that the apparent magnitude of Gliese 170 would be about 0.87 on average. I took the value of the absolute magnitude (8.23) from Wikipedia. You can plug in different values for the distance in the formula and see what comes out. The formula used by WolframAlpha is

$$D_l = 10^{{m-M \over 5} +1} pc$$

where $D_l$ = luminosity distance, $m$ = apparent magnitude, $M$ = absolute magnitude, $pc$ = parsec.


Why does it take so long to transmit an image from New Horizons to Earth? I just got the news that the New Horizons space probe has passed by some remote planet on the edge of the solar system.

I was surprised that the guy from NASA says that it might take 24 months from us to get the photo of that planet.

The solar system is not that big, right? It is slow because the signal transmission is slow, right? But why is the transmission so slow?

New Horizons has just passed the Kuiper Belt Object (KBO) 2014 MU69 also known as Ultima Thule. KBOs form a belt of asteroids (the Kuiper Belt) from Neptune's orbit outwards and of which Pluto is the largest member of the Belt. During the encounter with Ultima Thule, all of the 7 instruments on New Horizons were gathering data (although not all at the same time) and the total data collected is expected to be about 50 gigabits of data (compared to 55 gigabits of data taken during the Pluto encounter in 2015).

Since New Horizons is about another billion miles further out than Pluto was and 3 more years have elapsed, there is less power for the (tiny) transmitter and the signals are much weaker. The bit rate is about 1000 bits per second and so the 50 gigabits to transmit this will take 50e9 bits / 1000 bits per second = 50,000,000 seconds or about 579 days. Converting (roughly) to months by dividing by 365.25 and multiplying by 12 shows that it will indeed take about 19-20 months to transmit everything back. The first image at about 300 meters per pixel resolution and so about 100 pixels across the 30 km KBO, should be received on 2019 Jan 1. A second higher resolution image with about 300 pixels across the KBO is expected to be downloaded by 2019 Jan 2. There will be a press conference on 2019 Jan 2 when these images are due to be released and shown. (more details on what to expect when at Emily Lakdawalla's Planetary Society blog entry)

After the initial data download, they expect to perform some analysis to see which images have the best data with 2014 MU69 in the frame. Given the uncertainty in the position of 2014 MU69 and the high speed of the encounter, they had to shoot strips of images and not all will contain the target. These data will be prioritized in the downlink so they arrive on the ground first and can be analyzed first.

As mentioned by @luis-g there is also the Solar conjunction which will cause a 5 day period (according to PI Alan Stern in the 2019 Jan 3 press briefing) when reception of the data won't be possible. We would expect this to re-occur in January 2020 but these approx. 10 days don't make a large difference to the time taken which is dominated by the weakness of the received signal after the 15W transmission travels the ~4 billion miles and falls off due to the inverse square law, the corresponding low bitrate allowed by the need to have the transmitted data decodeable and the amount of data to transfer.


How harmful is Saturn's magnetic field for human life? AFAIK, Jupiter has the strongest magnetic field in the solar system, and it's so powerful that I've been told humans only could survive the radiation generated by it in its mooon Callisto, with proper protection.

But how about Saturn? Saturn has the second strongest magnetic field in the solar system. Which moons of Saturn could men survive from the radiation generated by the planet's magnetic field in?

Wikipedia says that "The saturnian radiation belts are generally much weaker than those of Jupiter" mostly due to the rings and other orbiting bodies absorbing the ions, I think. I have't yet found anything quantitative though


How can the Event Horizon Telescope image Sgr A* when it's not visible from all sites at one time? I went to https://eventhorizontelescope.org/array and read about the ten sites listed as part of the EHT, I have a mashed-up screen shot of them below.

I made a little script also shown below with approximate coordinates, and saw that Sgr A* is at a declination of about -29.5 degrees (South).  I spun it around the earth, calculated the dot product between each of the sites' position vectors and all normals pointing to Sgr A* as it rotates around the Earth.

I then plotted the elevation angle above the horizon for all ten sites. I was quite surprised!

Sgr A* is never above the horizon for the Greenland telescope!
Sgr A* is never simultaneously visible even to all of the remaining nine telescopes

Have I made a mistake? If not, how do these ten sites combine their data to image Sgr A*?

class Site(object):
    def __init__(self, name, lat, lon, alt):
        self.name = str(name)
        self.lat  = rads * float(lat)
        self.lon  = rads * float(lon)
        self.alt  = km   * float(alt)
        (clat, slat), (clon, slon) = [[f(x) for f in (np.cos, np.sin)] for x in (self.lat, self.lon)]
        self.norm = np.array([clon*clat, slon*clat, slat])

data = (('Northern Extended Millimeter Array', (44.634, 5.908, 2550.)),
        ('IRAM 30 meter telescope', (37.066, -3.393, 2850.)),
        ('The Greenland Telescope now near Thule Air Base', (76.531, -68.703, 10.)),
        ('Combined Array for Research in Millimeter-wave Astronomy CARMA', (37.280, -118.142, 2196.)),
        ('Kitt Peak National Observatory 12 meter Submillimeter Telescope (SMT)', (1.9583, -111.5967,  2096.)),
        ('Mt. Graham International Observatory 12 meter ALMA prototye', (32.701, -109.892, 3191.)),
        ('The Large Millimeter Telescope Alfonso Serrano', (18.985, -97.315, 4600.)),
        ('ALMA', (-22.971, -67.703, 4800.)),
        ('Caltech Submillimeter Observatory', (19.823, -155.476, 4140.)),
        ('South Pole Telescope', (-90.0, 0.0, 2800.)))

# https://eventhorizontelescope.org/array
# https://astronomy.stackexchange.com/questions/26413/math-behind-a-uv-plot-in-interferometry

datadict = dict(data)

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Topos, Loader, EarthSatellite
from mpl_toolkits.mplot3d import Axes3D

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads = 180/pi, pi/180
km         = 0.001

sites = [Site(a, *b) for a, b in datadict.items()]

Dec = rads * -29.5  # SgrA_star
cDec, sDec = [f(Dec) for f in (np.cos, np.sin)]
th         = twopi * np.linspace(0, 1, 100001)[:-1]
cth, sth   = [f(th) for f in (np.cos, np.sin)]
zth, oth   = np.zeros_like(th), np.ones_like(th)

SgrA_star = np.stack([cDec*cth, cDec*sth, sDec*oth], axis=1)

for site in sites:
    site.elev = np.arcsin(np.dot(SgrA_star, site.norm))

if True:
    for site in sites:
        plt.plot(degs*site.elev)
    plt.plot(zth, '-k', linewidth=2)
    plt.ylim(-90, 90)
    plt.ylabel('elevation (deg)', fontsize=16)
    plt.show()

I don't think it is necessary for all the telescopes to view the target simultaneously if you can make the assumption that the source you are observing is only varying on timescales longer than it takes the source to be viewable by all the telescopes. Sure, the maximum baseline you can get will be defined by the most distant pair of simultaneously observing telescopes, but that doesn't stop you adding together images taken at other times.

This is very probably the reason why M87 is in the fact the first image presented by the EHT. The timescale for significant variability around this (big) black hole will be of order a few time $2GM/c^2$, which is a few days for the M87 black hole. Imaging Sgr A* is going to be more difficult (or at least more blurry) because the variability timescale is minutes.

It looks like (from the data reduction paper) that the first steps involve treating every baseline in a pair-wise way (as I suggested above) and then combining them across the network, with whatever baselines were observing at the time, to give data at the surprisingly (to me) high time-resolution of 10s.


Why not use only one support vane for the secondary mirror, to avoid multiple diffraction spikes? Each vane holding the secondary mirror in front of the primary causes a diffraction spike. Couldn't the secondary mirror be held with only one vane?

If stability is the issue, couldn't it be substituted be steering the vane to hold the secondary in place with precision, similar to active optics, regardless of winds? And why does the James Webb Space Telescope use three vanes when it will operate in undisturbed microgravity and shadow?

(Maybe the star spikes are desired 'cause they make stars look nice?)

I can't speak for the JWST, but I do work at a company that designs telescopes for spacecraft and I can tell you that a lot of the structure of the telescope is to protect the instruments during the initial ascent to space where the launch vibrations and g-loading will be the worst. Plus mirrors for telescopes that big have really precise alignment tolerances and if the secondary mirror isn't very rigidly attached to the spacecraft even simple microstrains from heat loads or spacecraft pointing can move the mirror out of alignment. Even the fact that the telescope is going from a 1g environment to a 0g environment can move the mirror because it was aligned on Earth when the beam was slightly bent and then it goes back to being straight in microgravity and the mirror is misaligned! It stinks that there is stuff in the way of some beautiful science that you're trying to take a picture of, but it's the only way to make sure the telescope survives up there!