Why are Saturn bands much fainter than Jupiter's? The atmospheric composition of both planets are very similar. Then why are Jupiter's bands more visible (North and South equatorial belt, etc.):

Whilst Saturn's are not:

(except for the storm of course)

I'll give this one a shot. Correction is welcome.

Upper atmosphere temperature.

It's not just elements that give a planet color, but the temperature of elements. When we examine what a planet looks like, we're basically talking about reflected sunlight from the planet's surface or atmosphere. With Earth, its atmosphere is transparent enough that its surface can be seen from space. However, it is the only planet in the Solar System where that is the case. The other planets have thick atmospheres, so all we see is basically just high altitude clouds.

Also, as I understand it, the pictures of planets are often enhanced to make any distinctions more apparent. So if you actually flew above Jupiter, its lines might not be quite so distinct as the pretty pictures you'll usually find on the web. Regardless, that doesn't change your question; its lines are still far, far more distinct than Saturn's are and Galileo was able to see its Red Spot, so I think it's safe to say its lines are pronounced, even if the pictures we get are doctored a bit.

So anyway, Jupiter is at the right distance from the Sun to undergo atmospheric transition from ice to gas. As Wikipedia puts it,

  The upper ammonia clouds visible at Jupiter's surface are organized in
  a dozen zonal bands parallel to the equator and are bounded by
  powerful zonal atmospheric flows (winds) known as jets. The bands
  alternate in color: the dark bands are called belts, while light ones
  are called zones. Zones, which are colder than belts, correspond to
  upwellings, while belts mark descending air. The zones' lighter
  color is believed to result from ammonia ice; what gives the belts
  their darker colors is not known with certainty

Ammonia ice, like all ice, is very reflective, so the colder bands with ice are lighter. Ammonia gas is transparent, but anyone who has ever flown over the ocean knows that if you have enough of a transparent thing (water), it has a distinct color. While the quote from Wikipedia above says the reason for the darker colors is uncertain, less ice in its uppermost atmosphere means less reflected light and darker color.

Saturn, Neptune and Uranus are far enough away from the Sun where they always have ice in their upper most atmospheres, so they have less variation in color.  Earth's clouds are also primarily ice (not water vapor), so that is basically the answer. Jupiter is the right distance from the Sun for its upper atmosphere to transition and have distinct bands, some with ice, some without.

"But why are the bands in straight lines?"

This is due to the Coriolis effect.  Viewed from above, the Coriolis effect creates bands that line up with the equator.  Jupiter and Saturn both rotate quite quickly (9.5 and 10.8 hours respectively), so both have strong Coriolis effects.

Jupiter's atmosphere isn't just moving along those visible lines; it's circulating up from the warmer lower parts of the atmosphere to higher parts by convection. This is because there's a lot of heat to be transferred. Jupiter (along with Saturn, Uranus and Neptune) are all radiating more heat into space than they are receiving from the Sun, so while the heat from the Sun plays a role in not freezing Jupiter's upper atmosphere over warmer bands, it is the Coriolis effect that creates the bands.

Saturn has bands too (as you know). They are not as visible because both the warm bands and cold bands of Saturn are icy. See articles here and here and article with false color picture here, crediting Saturn's colors to ammonia ice.

So, for bands to form, all you need is relatively fast rotation. But for highly visible bands, the gas giant planet needs to be the right distance from the Sun and/or have the right amount of internal heat. It's all about the temperature.

I feel that I have to punctuate this with a "probably", because I can't swear that Saturn has ammonia ice all around its upper atmosphere, but I'm pretty sure that is the primary difference between Saturn's more uniform look (Uranus and Neptune too). It is primarily the ammonia that gives these planets their color, even though both Saturn and Jupiter are 99% hydrogen and Helium. Straight gas molecules (O2, N2, H2) and noble gases tend to have very little interaction with visible light.

Here's a fun article on gas giant planet's color. The science of what gives an atmosphere a specific color, however, is quite complex and above my pay-grade. Also, I also found this article interesting, though I can't swear to the accuracy of it.  Saturn radiates a surprising amount of heat given that it's 1/3rd the mass of Jupiter.

Lastly, this one may provide a more accurate but less sexy picture of Saturn than we're used to seeing.

Final note, on Jupiter's dark spots and dark bands. Wikipedia said the reason for the darkness is unknown (as quoted above). While I think the lack of atmospheric ammonia ice is very much a part of the answer, here's an alternate explanation for Jupiter's dark red spot and (perhaps) to some extent, its dark bands. Jupiter gets, on average, about 3.4 times the solar radiation per square meter than Saturn gets. To add to that, it might have more extensive atmospheric mixing going on than Saturn and more organic compounds in its upper atmosphere, which, combined with UV rays, could well play a role in Jupiter's darker lines too.


Was the discovery of six exoplanets around one star as "easy" as counting six peaks in the FT? The phys.org article Scientists make huge dataset of nearby stars available to public describes the release of a publicly accessible database of Echelle radial velocity measurements; The LCES HIRES/KECK Precision Radial Velocity Exoplanet Survey. See also Keck's HIRES home page.

  For two decades, these scientists have pointed HIRES at more than 1,600 "neighborhood" stars, all within a relatively close 100 parsecs, or 325 light years, from Earth. The instrument has recorded almost 61,000 observations, each lasting anywhere from 30 seconds to 20 minutes, depending on how precise the measurements needed to be. With all these data compiled, any given star in the dataset can have several days', years', ore even more than a decade's worth of observations.

This part caught my interest especially:

  "We recently discovered a six-planet system orbiting a star, which is a big number," Burt says. "We don't often detect systems with more than three to four planets, but we could successfully map out all six in this system because we had over 18 years of data on the host star." (emphasis added)

For very simple cases of one, or maybe two planets with minimal interplanetary gravitational interaction, a Fourier transform of a nice, long, continuous radial velocity measurement would show two main peaks, and possibly other artifacts. If the stellar motion induced by each planet were similar magnitude, the analysis might be fairly simple.

But for the six planet case mentioned in the quote (I don't know which one it is), and patchy time coverage (it's a survey) how was this analysis done? Peaks alone? Or just throw it to a supercomputer simulation of every possible combination and let simulated annealing run for a month?

Or was there some 'detective work' involved as well - assumptions, limitations of the fitting space, or even inclusion of other data from outside the study?

I suspect that the record holder (as of 14/2/2017) is HD 10180 which has at least 7 planets and possible evidence for as many as 9.

Lovis et al. (2011) announced the initial discovery based on 190 radial velocity measurements taken over 6 years. The precision of the measurements was 0.3-0.9 m/s.

Section 4 of that paper describes how they go about finding the planets in the data. It is a hybrid of fourier and fitting methods. As each peak in the periodogram is found it is added to a model which is then iterated towards a best solution.

Successive planets reduce the rms scatter around the predicted radial velocity curve. Ultimately a judgement has to be made whether the improvement in the fitting statistic justifies the addition of another planet (and more free parameters) to the model. The final rms after adding 7 planets was just over 1 m/s, which is worse than the expected precision, but which Lovis et al. attribute to (probable) radial velocity jitter due to stellar activity. The orbital model is then refined to include the effects of planet-planet interactions and tidal forces.

The radial velocity amplitudes attributable to each planet range from 0.8-4.5 m/s. The most marginal detection has the smallest amplitude, but the shortest period (more cycles and hence easier to detect smaller amplitudes).

A later paper by Tuomi (2012) used a more conventional Monte Carlo Markov Chain in a Bayesian framework to fit a model of $n$ non-interacting planets to the radial velocity data. Again, there is lots of discussion (see section 3.3 and 3.4) of exactly how many planets are required to fit the data. Tuomi claims there is strong evidence for an 8th and 9th planet in their analysis.

There are a number of important assumptions made in these types of analysis. The main one is that you have to assume some sort of model for the background noise and it is often assumed to be Gaussian and non-periodic.


How bright is the Sun in S-band? As far as I can tell when the Voyagers listen to Earth they have the Sun in the same beam. I never did calculate how bright a blackbody Sun would be because I have no idea if that's representative of the Sun's output in S-band

How well can Voyager 1 separate Earth signals from Solar noise these days? shows that the Voyagers see the Sun and Earth separated by 0.2 to 0.5 degrees each year.

I also don't know how wide of a slice in frequency the front end is exposed to, or if the Sun would be strong enough to saturate it.Frequency responses of Voyager's S-band high gain antenna's feed-horn and receiver front end & IF? remains unanswered.
This is related Why is the operating temperature for the Voyagers' receiver noise calculation about 1550K? and answers there are informative.
I assume the designers were optimistic that Voyager might survive this long and designed the system to be able to pick out transmissions from Earth from the Sun's output, but I'm not quite able to put the whole puzzle together yet.
So I'd like to ask:
Question: How bright is the Sun in S-band? Is there a measured value for power per unit frequency? How close is it to what a blackbody would produce at say 5800 K? At this frequency is there a big difference between when the sun is quiet and when it is active?
See also Do stars have "radio photospheres"? Are they different from their optical photospheres?
note: The linked questions and answers have a lot of Voyager-specific information which is relevant to Voyager's operation in S-band, but I have nothing on the Sun's behavior at 2.1 GHz Actually the first link gives the downlink frequency as

2113.312500 MHz for Voyager 2

Well, I did some digging and found a helpful chart here. The image itself is located at this link.
Wikipedia defines the S band as the section of the electromagnetic spectrum from 2 to 4 GHz. To interpret the respective values the graph displays at these frequencies, we need to convert these two figures to wavelengths:
$\lambda_1 = \dfrac{c}{\nu} = \dfrac{300\:000\:000\:m/s}{2\:000\:000\:000\:Hz} = 0.15\:m$
$\lambda_2 = \dfrac{c}{\nu} = \dfrac{300\:000\:000\:m/s}{4\:000\:000\:000\:Hz} = 0.075\:m$
And, we might as well find the wavelength at 3 GHz, since it's at the middle of the S band and might help us (hint: since the graph is on a log scale, and this calculation comes out to a nice power of 10, it definitely will help us):
$\lambda_m = \dfrac{c}{\nu} = \dfrac{300\:000\:000\:m/s}{3\:000\:000\:000\:Hz} = 0.1\:m$
So, we can take a look at the graph at 0.1 m. The units seem to be in $W\: m^{-2} \: (c/s)^{-1}$, and I'm assuming that $(c/s)$ is just $Hz$, which would make these units have the same units as Janskys, though, as Connor Garcia mentions in his comment, Janskys are $10^{-26}\: \rm{W\: m^{-2}\: Hz^{-1}}$. So, the quiet sun in the S band lies around $10^{6}\: \rm{Jy}$, and the "disturbed" or "active" sun lies around $10^{7}\: \rm{Jy}$.
And as for the second part of your question, the graph also shows some blackbody curves. You asked for 5800 K, but the closest one on the graph is at 6000 K - hopefully that's okay. As expected, the Sun seems to closely match that line for a good portion of the wavelengths on the graph, but seems to deviate in the S band, which is interesting, to say the least. So, from the blackbody curves in the graph, the sun is brighter than a typical blackbody at ~6000 K in the S band, both in its quiet and disturbed states. This difference is around one order of magnitude in the quiet state, and around two orders of magnitude in the disturbed state.
And lastly, you asked about the difference between the "brightness" of the sun in its active and disturbed states in the S band. From visual inspection of the graph, it seems like the difference in the S band is about one or two orders of magnitude.


Do the terrestrial planets form later than gas giants in our solar system? Sorry I forgot where this statements come from, but I also remember the reason behind it is due to young Jupiter moves inwards and destroys the original super earth in inner solar system, and the current terrestrial planets are formed after that, is that true?

The current ideas are that both terrestrial planets and giant planets start their formation in a similar manner. Dust settles towards the mid-plane of a predominantly gaseous disk, starts to stick together and eventually small (km-sized) planetesimals are formed. This process may be quicker in the outer parts of the solar systems where the gas is colder and the condensed material (ices) is probably "stickier". The planetesimals then interact with each other gravitationally and can grow by merger. The rate of growth is controlled by their spatial densities and their relative velocities, but is thought to occur quite quickly in both the inner and outer parts of the solar system ($\sim$ 1-3 million years, e.g. Righter & O'Brien 2011).

Thereafter, the inner and outer parts of the solar system differ. The gas is cold enough in the outer parts of the solar system to accrete onto rocky cores and build giant planets on timescales of another few million years.
In the inner solar system the gas is too hot to be accreted and instread the next tens of millions of years are characterised by high velocity collisions between planetary embryos and planetesimals. This indeed may be sculpted and influenced by the early migration inwards of Jupiter to about 1.5 au, followed by migration outwards - the so-called "Grand Tack model" (Raymond & Morbidelli 2014).

Overall it probably takes the inner solar system and terrestrial planets of order 100 million years to settle down into its final configuration. The collision that formed the moon may have been several tens of millions of years after the formation of the Sun and certainly long after the giant planets formed.


Seeing a galaxy (quasar) greater than 46.6 billion light years away Assuming the Big Bang was 13.8 Billion Years ago: Is it possible to observe a Galaxy's redshift showing distances slightly greater than that, like 48 Billion Light Years away? I understand that in a static universe, light traveling in a straight line would only be visible at less than 46.6 Bly. But, can Hubble's constant distort this number? Also, do we assume some error due to the light taking a non-linear path because of lensing? If so, is it possible to have an error of several hundred million years?

The edge of the observable universe is actually 46.6 billion light years away, despite the Big Bang being only 13.8 Billion years ago. This is because the light which we are now receiving as the furthest visible stuff had to travel through ever expanding space in between, being redshifted down into what we call the Cosmic Microwave Background Radiation (CMBR). There is a little bit further than that which we are technically receiving, but it has been redshifted infinitely.

To see anything further away than 46.6 Bly, it would have had to existed literally before time itself, or travelled faster than the speed of light. Two highly improbable things


mm-wavelengths Moon map, scientific case I would like to test a telescope with a Moon observation.
It is a low spectrometer-imager at 110-300 GHz with a resolution of 4 arcmin@150 GHz.

Is there any scientific target that concerns the Moon with such an observation?

Millimeter and sub-mm observations (110-300 GHz = 2.7-1.0 mm) are sensitive to the thermal emission and provide a brightness temperature of airless bodies like the Moon or asteroids. The radio emission is from just below the surface/skin of the body (down to $\sim10\lambda$ so ~1 cm in this case) and so is also sensitive to the thermal inertia and conductivity of the surface and regolith. A paper discussing sub-mm observations of the Moon (which is freely available) is Pardo et al. 2005.

A major issue with observation in the mm/sub-mm is atmospheric absorption due to precipitable water vapour (PWV) as seen in the plot below which shows the atmospheric transmission in the mm/sub-mm at the summit of Mauna Kea (~4100 meters):

Note that the worst case shown (in red) with 5mm PWV is already a very dry atmosphere that is rarely found even at other high mountain sites; sea level sites could easily have 50+mm of PWV. This could make these type of measurements very challenging.


What elements and compounds would be present in a red gas giant? What would give a gas giant a red color (dark or light)?

It's a more complicated question than it appears.   Also, a couple points are in order.

First, you're thinking about gas giant color is wrong.  It's not the ingredients as much as the temperature.   The ingredients don't change all that much.  They change some, but gas giant color has more to do with temperature.

Two, color is a very narrow perspective.  We see, only a narrow band of the spectrum.

NASA brings a variety of cameras on their missions, Infra Red, Radio, UV as well as visible light.   See here.

Also, many photos that we see of planets are doctored to make the variations stand out more visibly, and I think it's great that NASA does this.

And three, the planet's star can affect it's color.   Color is different under a blue star compared to a red dwarf, so there's 2 colors to consider, color under a white light and color under the planet's star.

All that said, color's still interesting and fun and it tells us something.   Using visible light only, many gases are largely transparent, but that doesn't mean they're colorless.  Any gas will reflect and diffract some sunlight and gas giants have atmospheres thousands of miles thick.  There's no such thing as a transparent gas giant planet, with an earth like atmosphere that you can see through.

Hydrogen and Helium are both highly transparent to visible light but bluish when thick enough.  (Neptune gets it's blue color from a combination of it's hydrogen and helium and trace amounts of Methane ice).

Upper-atmosphere temperature is also very important.    Saturn, Uranus and Neptune are cold enough that mostly their color is defined by ice, which tends to be more reflective, giving them lighter and generally considered "prettier" colors.   Uranus and Neptune's methane ice give them their blue/green and more pure blue colors.    Saturn is a little too warm for Methane ice, but it has ammonia ice giving it a lighter color and other gases in it's upper atmosphere giving it some darker yellow and orange colors.

Jupiter's bands can be color-differentiated by where there is abundant ice high in it's atmosphere and where there is more gas.   The darker bands are warmer, generally circulating upwards from deeper in the planet, like Hadley cells on Earth, driven in part by Jupiter's fast rotation and strong Coriolis effect and significant internal heat.   It's lighter bands are colder, where ice reflection is the primary factor in it's color.   Jupiter also has some complex hydrocarbons in it's upper atmosphere due to chemistry, intense lightning and atmospheric mixing.

So there's quite a few difficulties in answering this because a gas giant planet's color depends on many factors, How well it's upper atmosphere circulates, it's temperature and what ices, if any, can form in it's upper atmosphere.   Any trace elements that might carry color, even a 1 part in tens of thousands, can be enough for a dominant color if the other gases are mostly transparent.   Take Earth for example - clouds are a very small part of the atmosphere by mass but they change the color of the sky (viewed from space) significantly.  This makes predicting a gas giant planet's color very tricky and borderline guess-work, because one trace gas can be enough to change it.

Gas giants are mostly hydrogen and helium in their upper atmospheres, but, as noted above, it's often the other gases, ices or trace gases, not the hydrogen and helium that will give them their color.

I always thought Jupiter's "great red spot" was misnamed.  It looks more like an orange-brown to me.    But if that counts as red, then red might be relatively common.

It might be UV rays and a trace gas, ammonium hydro-sulfide, that gives Jupiter's spot it's color.  See article here.   I gather there's still some debate on that question.

Here's a question with a nice answer on the colors of the planets.   Venus is unusually bright of the reflective SO2 high in it's atmosphere.


On the day of total solar eclipse why was early sunrise observed in India? I am a middle school teacher from India(UTC+5:30). On August 21, 2017 morning at (17.385044, 78.486671) we observed an unusually early sunrise. It felt like its 7'O clock in the morning at 5:45 AM. And we had an unusually bright day. I'm sure it has to do something with the total solar eclipse but I don't exactly know what it is.

My students are persistently asking me the reason for this early sunrise but I don't have enough knowledge of astronomy to answer them. Can someone kindly answer why we observed this phenomenon?

If your sunrise had been earlier than expected, it would have been the same around the world, and would have been noticed very widely.

Here is a chart showing times of sunrise/set at your location.


how do we know the universe is STILL expanding? According to Hubble more distant galaxies are moving away faster but since the light took billions of years to reach us we should say they WERE moving away faster billions of years ago. How do we know that those galaxies where light is just reaching us know are STILL moving away at all.... let alone faster

We don't know. This is a depressing thing that makes me, at least, feel so small. The only thing we might assume, that if they once were accelerating that fast, they probably will do it still (even faster), but of course we can't be sure if anything have happened.

It's like this with all the long distances in the universe. How do we know if Proxima Centauri (4.2 light years away) have died/exploded? We wait, 4.2 years.


Difference in creation of solar flares and CME? What is the difference in the creation process of solar flares and CME (coronal mass ejections)? How do they relate to sunspots and coronal holes?

Here there are some good explanations about the flares' and CME's differences. But my question focuses on how each of them is created, and what are the respective conditions at the corona.

Edit:
In an effort to make the questions more clear, I could add that as far as I can tell both effects result from magnetic reconnection, but in one case we have only energetic particles emitted while in the other we have whole chunks of plasma. Do we know which is the determining factor about which of the two will happen?

Also, I understand that flares are mostly related to sunspots while ejections are to coronal holes. Is this correct? If yes then it sounds like a pretty big difference.

(Please be reminded that specifying that the answer is not known yet is an answer.)

Edit #2: Please see this comment for further clarification of what the question is about.

The point of my comment was that the connection between CMEs and flares would make for a good 20 page research paper in a college course.  The connections are still only understood at a kind of cartoon level, after decades of concerted research.  But I can tell you what the basic cartoon is, and you can think of that as a kind of operating hypothesis that helps organize many of the observations, but is still a long way from a predictive model.  The basic idea is that the energy for flares and for CMEs both seem to come from magnetic free energy stored in the corona due to the magnetic and turbulent nature of the convection zone.  The magnetic field is capable of trapping pockets of hot gas that has been energized by this magnetic activity, but then when the field lines get significantly twisted up they reach a point where they can release that energy and relax to a smoother configuration.  This relaxation does two things-- it releases energy, and it reconfigures the field such that it can release these pockets of trapped plasma.  The former is the solar flare, the latter is the CME, and that's why the two often come together.

However, there seems to be a full range of weak to strong flares, and weak ones would not be associated with CMEs.  On top of all that, there may be other processes that heat the corona, and other processes that lead to the solar wind, so one big question is if the heating of the corona is mostly from lots and lots of small flares, and the solar wind is lots and lots of small CMEs, or if the general heating of the corona and general driving of the wind are separate processes from flares and CMEs.

Also, you mention closed and open field regions, so these questions can be asked separately in those two domains.  But CMEs would normally come from closed field regions that open up when the field reconfiguration occurs, I believe open field regions tend to be rather sedate and not associated with either strong flares or CMEs, but rather a more steady wind.  In the closed field regions, CMEs are often associated with features called  "helmet streamers," which are regions of trapped plasma that are kind of combed out at their highest point by the solar wind, creating an interface between closed and open fields that is a good setup for getting field reconfiguration that could launch out some plasma.


What is the actual visual characteristics of a black hole? Is there a safe distance in which a black hole can be observed with the human eye? What is a safe distance?

I have seen many computer simulated and artist renditions, and I am not sure if they depicted the true colors and shape of BH.

Below I have a few of examples of what may a BH may look like rendered.

How light from the accretion disk maybe refracted by gravitational lensing.
http://www.physics.utah.edu/~bromley/blackhole/index.html

https://www.bbc.com/news/science-environment-38937141

https://www.nasa.gov/connect/chat/black_hole_chat.html

Should these models be mixed to make a more complete rendition of a BH?

This simulation (video link) has been made to show what a black hole might look like seen from different angles.
From APOD :

If the black hole was surrounded by a swirling disk of glowing and accreting gas, then the great gravity of the black hole would deflect light emitted by the disk to make it look very unusual. [...]  Surrounding the central black hole is a thin circular image of the orbiting disk that marks the position of the photon sphere -- inside of which lies the black hole's event horizon. Toward the left, parts of the large main image of the disk appear brighter as they move toward you.


Can a solar eclipse be directly overhead? I have trouble picturing eclipses and whether the moon can be directly overhead.

They could be directly overhead if you live in the tropics.

The tropics are the regions near the equator, between 23 degrees North and 23 degrees south. In the tropics, the sun and the moon can pass directly overhead. If both pass overhead at the same time there will be a solar eclipse directly overhead.

The eclipse of July 11, 1991 was visible directly overhead, (to within a degree) and in totality, for observers in the Nayarit province in Mexico


Sagittarius star cloud? I was using SkySafari with my phone, but the compass didn't quite work.
So as I knew where the cardinal points were, I think I have shot the Sagittarius star cloud, at least a part of it.

Can you guys tell me if I am wrong?

It was the 5th of July at 00:10 local time. I was at N 45.61°, E 7.35°.

The image scale is a bit small, but I can see the constellation of Lyra just above center. The bright star Deneb (in the constellation of Cygnus) is left of center, and the star Altair (in the constellation of Aquila) is at the bottom (just right of center). Thus, your photo does not include Sagittarius. Sagittarius is off to the bottom right. The Scutum star cloud is the bright patch toward the right corner.

It would be helpful if you include the time of the photograph and location (latitude and longitude).


Formation and capture of Mars' moons, Phobos and Deimos Mars's moons have the appearance of captured asteroids.  How could this capture have taken place.  An asteroid in solar orbit would need to loose some momentum to be captured.  The gas Giants can capture new moons through 3 body interactions between the asteroid and their existing moons,  but Mars doesn't have this issue. Airbreaking would lead to impact, not capture.

An answer here suggests that the moons could have formed recently (millions of years, not billions) after an impact ejected a substantial amount of material in Martian orbit. What is the evidence for this?  Is there an impact crater of the right size, and age? Were both moons formed together?

You're right, Mars' moons likely did not form with from the same dust mass as it; their spectra, albedo, and density are suspiciously similar to those of C- or D-type asteroids. Thus, Mars presumably captured Phobos and Deimos from the asteroid belt. However, the origin of the Martian moons is quite controversial.

Burns (1992) notes that capture requires the loss of energy (and momentum, as you pointed out). So obviously, something would need to dissipate that energy. The immediate candidate is aerobraking, but we run into a problem: Mars' atmosphere is too thin to capture a Phobos-sized object through aerobraking. In fact, even if aerobraking could capture Phobos and Deimos, the densities of both moons indicate that they may not survive structurally.

Several hypotheses have been proposed to answer this. Landis (2001) suggests tidal forces are to credit:

  The existence of asteroid moons provides a mechanism for the capture of the Martian moons (and the small moons of the outer planets). When a binary asteroid makes a close approach to a planet, tidal forces can strip the moon from the asteroid. Depending on the phasing, the asteroid can then be captured. Clearly, the same process can be used to explain the origin of any of the small moons in the solar system.

According to Geoffrey Landis, Deimos and Phobos could have each been moons of parent asteroids, and they were separated due to tidal forces:

This would explain how Phobos and Deimos could survive as moons. I could mention the other hypotheses, but Landis' would answer your question quite well.

Right now, there isn't any clear consensus as to how Phobos and Deimos became moons. More research is still necessary, and we may need to modify our models for the evolution of Mars and the asteroid belt.


Do planetary surface temperatures change in unison in a solar system? Are there any known correlations between the changes in planetary surface temperatures in a solar system?

If so, do the farthest planets have smaller albeit correlated changes?

The simple answer to your question is yes. Taking a simplified equation from Carroll & Ostlie, An Introduction to Modern Astrophysics Second Edition, the temperature of a planet can be estimated as:
$$
T_{p} = T_{\odot}(1-a)^{\frac{1}{4}}\sqrt{\frac{R_\odot}{2D}}
$$
Where $T_p$ is the predicted temperature of a planet in a circular orbit of radius $D$ with an albedo of $a$ around a star with a temperature of $T_\odot$ and a radius of $R_\odot$. If the energy output of the star were to increase, raising $T_\odot$, then there would be a corresponding increase in the temperature of all planets orbiting said star.

In practice there are factors which can make this correlation difficult to measure. The albedo  of a planet during the course of a day can vary greatly and the distance of a planet from the host star changes throughout the year. This equation also assumes that the planet is a perfect black body which most are not which can also change a planets temperature and obscure any changes caused by the host star.


What are some examples of routine human activity times for certain cultures/peopes that are based on the position of the sun? What are some examples of timekeeping that are based on the height/angle/position of the sun. Two examples that come to mind are the times of the Islamic prayers times and the golden hour in photography.

We are now dependent for timekeeping on clocks, most clocks are synced (directly or indirectly) to atomic clocks that keep time without direct reference to the sun.

However, the principal definition of clock time, UTC, is coordinated to the position of the sun, through the insertion of leap seconds every 18 months or so. So anything that happens at a particular time is coordinated to the position of the sun, though not directly.


Opportunities to see a near earth object via unaided eyes or binoculars? I would like to see a near earth object proper with my eyes or binoculars someday.
Has this been possible recently? Are there any upcoming potential opportunities to do so in say the next 20 or 50 years? I have a shot at the first one, the second one is for the benefit of others :-)

Your best viewing opportunity will be 99942 Apophis which will pass Earth on April 13, 2029, where it will pass about 19,600 miles from Earth; look out for that. According to Wikipedia:

On that date, it will become as bright as magnitude 3.1 (visible to the naked eye from rural as well as darker suburban areas, visible with binoculars from most locations)

Other ones include:
(137108) 1999 AN10, passing Earth at a distance of 240,000 miles on August 7, 2027, peaking at apparent magnitude 7.3, visible with binoculars, but questionably visible in urban areas.
(153814) 2001 WN5, approaching at 155,000 miles on June 26, 2028, at magnitude 6.7.
I hope this helps.


Is the Celestron astromaster 130 EQ Model able to see nebulas and galaxies? Is the Celestron astromaster 130 EQ Model able to see nebulas and galaxies?
If this telescope is not able to see nebulae, please suggest me the best one telescopes for nebulas.

The question isn't whether you can see deep-sky objects but how many.
Given a reasonably dark observing site, the Orion nebula and the Andromeda galaxy are visible in 7x50 binoculars.
A 130mm objective should be enough to see most of the nebulae and galaxies in the Messier catalogue.
Once you get the hang of it, you may be able to see a few NGC objects too, but a bigger scope will show more of them.


Is there a fundamental difference between a small galaxy, and a large star cluster? For instance, If a large globular cluster were somehow ejected from its galaxy, and was observed by astronomers, would they call it a galaxy, or would there be characteristics of the star cluster that make it obviously not a galaxy by definition? And why are the Magellanic Clouds dwarf galaxies, and not large star clusters? Could we remove stars until the Magellanic clouds became star clusters, or is there something intrinsically galaxy like about these objects?

There are a couple of important distinctions between the two types of objects. Galaxies are objects which range in mass from about $10^{9}-10^{12} M_{\odot}$, and contain 'halos' of dark matter which represent the majority of the mass of the object. Now, there are things called dwarf galaxies (which are less massive than regular galaxies; I'd imagine that there really could be a bit of a blurred distinction between these two kinds of objects), however dwarf galaxies still seem to have dark matter components to them.

Globular clusters are tightly bound collections of about $10^{6}$ stars, and do not have significant (if any) dark matter contained within or around them. So for these reasons in practice it is not hard to confuse the two.

Some References:

1) Evidence that dwarf galaxies are dominated by a dark-matter component.

2) Dark matter present in globular clusters are either unlikely, or incredibly highly concentrated and approximately the same mass as the stellar component.


Simulation and its necessity in astronomical research I was wondering if mainly everything in Astronomical and Astrophysical was dependent on Simulations then what are the educational requirements are necessary to qualify as the Simulation Expert.

What softwares are needed globally? Are all simulations run on a Supercomputer? Do astrophysicist require coding experience? And, is Knowledge of Programming a must?

I agree with @uhoh that you don't have to an expert, but above-average knowledge of coding is definitely useful, bordering on "a must". Not for writing huge programs with 1000s of lines, but for writing smaller pieces of code that help you in everyday tasks.

As uhoh says, you can very well find your place in a group where other people are in charge of numerical modeling, and focus instead on, for instance,
interpretation of observational results,
physical modeling,
or even the technical aspects of telescope operation.

Software

You ask about what software is used: Some of the more popular if you're working with observational data are IRAF for reducing and analyzing the data and SAOImage/ds9 for visualizing. A long list of astronomical software can be found here.

On the other hand, if you're modeling galaxies or the interstellar medium you may want to familiarize yourself with stellar population synthesis codes such as STARBURST99, semi-analytical galaxy formation models such as GALFORM, or radiative transfer codes such as Cloudy.

Coding

For coding, if you are constructing large codes that require many hours of computation — possibly on a supercomputer — you will probably want your code to be fast. Then languages like FORTRAN and C are often used. For programs that don't necessarily need to be fast, comprehensive languages such as Python are very popular. For smaller tasks, shell-scripting can be very useful.

Even if you don't run simulations, you will most likely eventually need to automatize your work. For instance, rather than reducing 100 images one by one, you create a pipeline that does all the biasing, flat-fielding, cosmic-ray-removing, etc. in one go. And rather than going through a catalogue of a million galaxies one by one to look for those that match your preferred criteria, you create filtering software that finds them for you.

Supercomputing

Not all simulations are run a supercomputer. Huge cosmological simulations, galaxy formation simulations, etc. usually are, because you can usually start a simulation and then let it run for three months without interfering. But if your simulations are smaller, and possibly require that you check the process all the time, it's often sufficient/easier to run it on your local computer.


How near to Earth do comets pass? Siding Spring now passes about 10 Earth diameters away from Mars.

What are the estimates of how often comets fly by Earth at different distances? Are the distances to some famous historical comets known?

The answer is that they don't come very close. As Wikipedia notes,

  It is rare for a comet to pass within 0.1 AU (15,000,000 km; 9,300,000 mi) of Earth.

Even better, though, is an actual list of some of the closest approaches of comets to Earth. The closest one listed, Comet Lexell in 1770, came 0.0151 AU away from Earth. The list only shows 20 comets have come with 0.1 AU of Earth. So Earth has not had a close encounter with a comet (like Mars is having) in a long time, if ever.

This site says that C/1491 came within an incredible 0.0094 AU in 1491, but I'll take the IAU site over it. And it does admit that that comet has an uncertain orbit, so that figure could be wrong.

But when comets come even a tiny bit near to Earth, there are some articles written about them. See here and here.

As for the distances to famous comets:

Closest Ever

Halley's - 0.0334 AU
Great Comet of 1760 - 0.0682 AU
Tempel-Tuttle - 0.0229 AU

Also, some near-Earth asteroids used to be comets. Extinct comets have lost a lot (if not all) of their ice and so may be classified as asteroids. This blurs the lines between the two groups. Some extinct comets have become near-Earth asteroids. This explores the issue pretty well. So if you include extinct comets as "comets," you may have comets that have come even closer to Earth.


Does anyone know why three of Jupiter's largest moons orbit in 1:2:4 resonance? Three of the first four moons ever discovered outside of our own planetary sphere of gravitational influence orbit in very close to perfect resonance. Europa's orbit is almost exactly twice as long as Io's, and Ganymede's is almost exactly twice as long as Europa's. Perhaps it is just coincidence. If you look at enough random things, you start to see patterns that aren't necessarily there. But it would be a pretty big coincidence. When I see something like this in nature, I have to wonder if there is an underlying cause.

The ratios aren't quite exactly 2, so if there is a gravitational influence or something that pulls them into this resonance, the influence must not be overpowering enough to force exact resonance. Or maybe it is, given enough time, but "recent" collisions or the pull of other satellites in the system may have driven them a little bit off of their resonance. Has anyone studied this and found a probable reason for this phenomenon?

When the Galilean moons formed, they weren't in resonance with each other. All of them were in slightly smaller orbits than they are now. Over time after their formation, Io's orbit slowly moved outward due to tides from Jupiter. This is the same effect that is causing our moon to slowly move away from the Earth (at about the same rate your fingernails grow). It goes like this. The Moon's gravity causes tides to form in Earth's oceans. This bulge of water gets carried forward with Earth's rotation, because the Earth is spinning faster than the Moon is orbiting. Because the moon is still attracting the bulge, it causes a drag on the Earth, slowing its spin. At the same time, the bulge is attracting the moon, causing it to go faster in its orbit. As the moon speeds up, its orbit gets bigger. So essentially, the Earth's spin energy is getting transferred into the Moon's orbital energy. The same thing happens with Jupiter and Io, with a bulge in Jupiter's atmosphere causing Io's orbit to get bigger.
As Io's orbit expanded, its 'year' got longer, until it approached a 2/1 resonance with Europa. Once they reached resonance, they got 'locked in', their mutual gravity acting on each other reinforced it. Io was still raising tides on Jupiter, though, and its orbit was still trying to expand. As Io's orbit kept expanding, it gave a gravitational kick to Callisto on each pass, expanding both of their orbits until Callisto reached a 2/1 resonance with Ganymede. This is where the inner 3 Galilean moons got their resonance. The orbits are still expanding, but much more slowly because with each one you add on, it gets harder to transfer the energy. Given enough time, all 4 Galilean moons would probably reach resonance, although the sun will die before that happens.
I may have some of the minor details wrong here, but this is the story as I understand it.
EDIT:  Seven years later, When will Callisto be in orbital resonance with the rest of Jupiter's big moons? was asked.  The newest answer for when the first four moons will be in orbital resonance is much sooner than the Sun will die.  Specifically, about 1.5 billion years from now.


VLT interferometry vs E-ELT? If the VLT (very large telescope) can use interferomerty to act as a 200 meter mirror, then why are they trying to build the 39 meter E-ELT? What advantage does the E-ELT have, if the VLT can act as a 200 meter mirror?

Ok, not sure where you get 200m from, the maximum baseline for the VLT interferometer is 130m. Nevertheless, your question still stands.

There are a number of reasons. Firstly, it is important to understand that with relatively limited number of baselines (4 VLTs = 6 baselines) one does not get actual images out of the interferometer, but rather "visibilities" that enable one to work out whether sources are double, extended or whatever. Second, the interferometry only works at infrared wavelengths. This is a technical issue that might be solved in the next decade or so, but a general rule of thumb is that the longer the baseline, the longer the wavelengths you will be limited to use. Thirdly, the light gathering power is limited to the light that can be gathered by the constituent telescopes. I think this also limits the kinds of instruments you can use since the required delay lines and optical components of the system make everything inefficient. Fourth, there are limitations of the abilities of optical/IR interferometers caused by turbulence in the atmosphere. This limits them to very small fields of view and the performance does not reach the ideal case. It also means that you are limited to bright objects because individual exposure times need to be very short.

The E-ELT is designed to work at optical and infrared wavelengths and will gather enormous amounts of light. Not everything in astronony needs spatial resolution; sometimes what you need is just a bigger light bucket. Examples include: deep imaging of the universe, gathering spectra from faint objects; accumulating enough photons to do transmission spectroscopy using exoplanet transits.

It's horses for courses really. At optical and IR wavelengths the future will likely be big telescopes in space and even bigger interferometers in space.


Why is there no color shift on the photo of the M87 black hole? Last year, the first photo of a black hole in Messier 87 was published:

(Source: EHT)
It is quite obvious that about the lower half of the accretion disk is brighter.
This question (or rather, the asnwers) explain that it is caused by Doppler beaming. Since Doppler beaming partly constitutes of Doppler effect, I would expect to see a blueshift on the brighter areas and a redshift on the fainter areas; similarly to what the original rendering of the black hole from the movie "Interstellar" looked like:

(Source)
So why is this not the case for the photo of the M87 black hole?

The picture isn't a "colour" picture - it is monochrome. i.e. It is obtained at a single microwave wavelength of 1.3 mm, and so not at any wavelength you could see (Akiyama et al. 2019).
There isn't therefore any spectral information that would reveal the expected Doppler effect. Any difference of colour in the "false-colour picture" is purely related to the intensity of the emission, not its wavelength.
If one were able to obtain coverage at multiple wavelengths then you might expect a "bluer" (i.e. shorter average wavelength) colour to be associated with the brightest regions.


Evidence that galaxies are made of billions of stars? What is the evidence that galaxies are made of billions of stars?

Even faint galaxies?

Can spectral analysis distinguish galaxies from individual stars?

The individual stars can be resolved in galaxies close to us...including our own, the Milky Way Galaxy.
More distant galaxies are morphological similar to those that can be resolved into stars, so a natural induction would be that they are similar in nature to those we can resolve into stars.
Galactic spectra are consistent with them being composed of stars in the sort of relative motion we expect.
Supernovae are observed in distant galaxies, and these are similar to those seen close up which we know are stellar in origin.
... etc


Is Earth's climate significantly different when the one solstice occurs at aphelion rather than the other? I mean different in some kind of significantlly non-symmetrical way, such as it being much colder during one period rather than the other? Let's assume other factors, such as tilt due to nutation and variation in the Sun, are the same.

Seasons on the Earth are driven primarily by axial tilt. I'm writing this answer in late August, when the Earth is a bit past aphelion. That event currently occurs in early July, a couple of weeks after the June solstice. Yet it's summer in the northern hemisphere. Precession have a secondary effect on the seasons. The current situation (perihelion close to the December solstice, aphelion close to the July solstice) was reversed ~12000 years ago when perihelion coincided with the June solstice. This made Northern Hemisphere summers hot but short, winters cold and long.

The brutally hot Northern hemisphere summers back then resulted in the end of the last glaciation (colloquially, ice age). Glaciations start when summertime temperatures at far northern latitudes (65° north is the standard proxy) are so mild that snow from the previous winter doesn't completely melt. Glaciations end when summertime temperatures in the Northern Hemisphere become hot enough to melt the ice that has accumulated over forty thousand to a hundred thousand years.

The reason we aren't we seeing a glaciation start now (conditions are right from the perspective of precession) is that the Earth's eccentricity is rather low. Precession, eccentricity, and other factors have to line up to trigger a glaciation.


What zodiac constellation will the Vernal Equinox be directed towards in ~6500? My question is basically asking how the vernal equinox changes with time, and how we can track where it will be at certain points in time.

Due to Earth's axial precession, the vernal equinox gradually travels westward, completing a cycle in roughly 26,000 years, i.e. about 1º every 72 years. Currently, the vernal equinox crosses the ecliptic in Pisces, near its border with Aquarius. In the year 6500, i.e. in 4500 years, it will have moved a further
$$
4500\,\mathrm{yr} \,\, \times \,\, \frac{360º}{26\,000\,\mathrm{yr}} \simeq 62º,
$$
that is, to the point that currently has a right ascension of roughly 20 hours. From the map below (from Wikipedia), you can see that it will just have moved out of Capricornus and into Sagittarius.

As @barrycarter made me aware, my first drawing was wrong. Today, the vernal equinox in this map is a vertical line in the right-most edge of the map, going from Polaris in Ursa Minor and through Cepheus, Cassiopeia, etc., down to the Octans. However, due to the precession this map will become distorted, and Polaris will no longer be the Northern Star; instead the axis of Earth will travel roughly along the green line, in the year 6500 lying somewhere in the middle of Cepheus. The Vernal Equinox will lie roughly along the red line, going from Cepheus through Cygnus, Vulpecula, Aquila to Sagittarius.

I drew this line using Stellarium, setting the date to 6500-01-01, switching on equatorial grid of that date and today, and using the orthographic view. My line on the southern sky may be a somewhat imprecise, but on the northern it should be fine.


Is it realistic to build a telescope powerful enough to take a photo of TRAPPIST-1's planets? Assuming modern and near future technology, could we build a telescope powerful enough to actually see TRAPPIST-1's planets?

It's not out of the question, but it would be very difficult to use direct imaging for any of the exoplanets around TRAPPIST-1:

The exoplanets are small, all not much bigger than Earth. Ideally, you want a target the size of Jupiter or bigger. Likewise, they are very low-mass, and are thus small. Once more, this is a problem. Direct imaging has only had much success with large planets, of many Jupiter masses.
The exoplanets are close to the star, with the furthest one out - admittedly with a poorly-constrained orbit - having a semi-major axis of 0.063 AU (and a year of 20 days), give or take. That is tiny. Mercury's semi-major axis is more than six times that. If the exoplanets are close to the star, it makes it much harder to distinguish them from the star's glare. Granted, there are methods that do this very well, but they're not meant for planets this close.

To see what I'm talking about, look at some examples of directly-imaged exoplanets. Almost all in that list have many times Jupiter's mass and at least its radius. Their semi-major axes are many, many - sometimes thousands - times that of Earth.

I can't speculate as to what technology we'll have decades from now, but if you want to find out more about the exoplanets around TRAPPIST-1, direct imaging is not the way to go, and I suspect it will not be a very popular choice among astronomers.


What does the ']' in the spectral line "CIII] 1909 Å" mean? The above emission line, as I understand, is a useful probe of early star-forming galaxies.

However, I do not understand what the ']' to the right of the 'CIII' means.

I could not find any online reference that can help me decipher this and other spectral line names (for example what does the double lambda in 'O III] λλ1661' mean. Any good reference is appreciated.

Under conditions of very high temperature and very low pressure forbidden conditions can occur, they shift the color of the excitation radiation from the ultraviolet into the visible spectrum giving nebula their beautiful colors. These are represented using brackets.

See Wikipedia's Selection Rules, section Angular Momentum:

  "Semi-forbidden transitions (resulting in so-called intercombination lines) are electric dipole (E1) transitions for which the selection rule that the spin does not change is violated. This is a result of the failure of LS coupling.".

That notation would read:

Element.
Ionization state.
Forbiddenness - superallowed, allowed, first forbidden, second forbidden, third forbidden, fourth forbidden, etc. Think of "forbidden" as meaning "likelihood". Represented with one or two brackets.

See note below: "In many-electrons systems transitions that violate rule 4 change the total spin (semi-forbidden transitions) and are called intercombination lines (e.g. CIII]). Transitions which violate the propensity rule 5 and/or 6 are strictly forbidden, and are labelled by two square brackets (e.g. [CIII]). Table 1 gives examples of the 3 types of transitions.

  Do not confuse the use of two brackets around an element with their usage around abbreviations for massive hot blue emission stars. With massive Wolf-Rayet stars, their spectra groups are enclosed in square brackets: e.g. [WC]. Most of these show [WC] spectra, some [WO], and very rarely [WN].

The "double Lambda", in the example you provided, is a 'half doublet'. A doublet for that element would be OIII]λλ1661,1666 another example would be SiIII]λλ1882,1892. One paper where this is discussed is: "CIII] Emission in Star-forming Galaxies at z ~ 1" (Mar 4 2017), by Xinnan Du, Alice E. Shapley, Crystal L. Martin, and Alison L. Coil. A doublet is two distinct spectral lines formed independently but which are so close that previously inaccurate instruments and poorer understanding of these mechanisms led many to believe a single line at a particular location. See: "Performance Analysis of Standard Fourier-Transform Spectrometers" 2007 by Douglas Cohen.
The Å ($\unicode{x212B}$$\unicode{x212B}$ or $\unicode{x00C5}$$\unicode{x00C5}$, a Latin capital letter "A" with a ring above it) is a Swedish letter, the symbol for the ångström, a unit of length equal to 10$^{−10}$ m (one ten-billionth of a metre) or 0.1 nanometre.

The Wikipedia webpage: "Spectroscopic notation" should be helpful:

  Ionization states

  "Spectroscopists customarily refer to the spectrum arising from a given ionization state of a given element by the element's symbol followed by a Roman numeral. The numeral I is used for spectral lines associated with the neutral element, II for those from the first ionization state, III for those from the second ionization state, and so on.10 For example, 'He I' denotes lines of neutral helium, and 'C IV' denotes lines arising from the third ionization state, C3+, of carbon.

See also: Wikipedia's Ionization energy, section Values and Trends.
More about doublets:

"The Grism Lens-Amplified Survey from Space (GLASS). XI. Detection of C IV in Multiple Images of the z = 6.11 Lyα Emitter behind RXC J2248.7-4431" (2017), by Schmidt, KB; Huang, KH; Treu, T; Hoag, A; Bradač, M; Henry, AL; Jones, TA, et al.

"The UV Properties of the Narrow Line Quasar I Zwicky 1" (Jun 26 1997), by Ari Laor (Technion), Buell T. Jannuzi, Richard F. Green, Todd A. Boroson (NOAO).

Note:

From "Spectroscopy - IPAG Grenoble" (.PDF), page 2:

  "2.3 Selection rules for Hydrogenic atoms

  In general, there is always a non-zero probability for any transition between two states to occur. However, in some instances, the probability is exceedingly small. Under some approximations (e.g. L-S coupling, dipole approximation, ...) the matrix element might be strictly zero. However, higher-order terms generally lead to non-zero, yet small, probability. The cases in which probabilities vanish, under given approximations, are called selection rules.

  The dipole transitions described above are called dipolar electric transitions. The detailed computation of the key term $\scriptsize \left| \overrightarrow \epsilon \cdot \overrightarrow \mu_{if} \right|^2$ allows to determine the cases for which a
  transition is allowed:

  $\scriptsize{\overrightarrow \mu}\!_{i f}$ does not vanish: dipolar electric transition between states $i$ and $f$ is allowed;
  $\scriptsize{\overrightarrow \mu}\!_{i f}$ vanishes but there exists non-zero higher-order terms in the development of $e^{i \overrightarrow k \cdot \overrightarrow r}\!\!$: dipolar electric transition is said to be semi-forbidden, but e.g. electric quadrupolar transition are allowed, though with much smaller rates;
  $\scriptsize{\overrightarrow \mu}\!_{i f}$ and all higher-order terms of the development also vanish: the transition between $i$ and $f$ is forbidden. Other higher-order terms may still allow the emission/absorption of photons.

  The selection rules that apply to one-electron systems are:

  energy level: ∆n any
  orbital angular momentum: ∆l = ±1
  parity must change between i and f
  magnetic quantum number: ∆m = 0,±1
  spin does not change: ∆s = 0 (always true for the H atom)
  total angular momentum: ∆j = 0,±1

  In a multi-electron atom, these rules apply to the jumping electron. These rules completely determine the spectra of one-electron atoms, such as HI and HeII, and also the alkali metals. Diagrams which show the allowed transitions are called Grotrian diagrams.

  2.4 Selection rules for many-electrons systems

  We have thus far discussed the interaction of single-electron atoms with a radiation field. The results can be generalized to many-electron systems, by considering the total electric dipole moment, $\scriptsize\overrightarrow D = \sum{_i} −e\overrightarrow {r_i}$. For multi-electron atoms, the selection rules for dipolar electric transitions are:

  total angular momentum: ∆J = 0,±1 but J = 0 → 0 is strictly forbidden
  magnetic quantum number: ∆M$_J$ = 0,±1
  parity must change between i and f
  spin does not change: ∆S = 0
  one electron jumps, ∆n any, ∆l = ±1
  orbital angular momentum: ∆L = 0,±1, 0 − 0 forbidden

  The first three rules are rigorous and must be satisfied by all dipolar electric transitions. The last three are not necessarily satisfied in complex atoms, and are called propensity rules. Transitions which fulfill rules 1–3 are allowed. Transitions that violate 4 change the total spin and are called intercombination lines (e.g.CIII]). Transitions which violate the propensity rule 5 and/or 6 are strictly forbidden, and are labelled by square brackets (e.g. [CIII]). Table 1 gives examples of the 3 types of transitions.

  $$\begin{array}{ccccccrrr} \\
\text{Species} & f \qquad \leftarrow \qquad i & λ (\unicode{x00C5}) & A_{ul}(s^{−1}) & \Delta J & \text{Parity}^\dagger & \Delta S & \Delta l & \Delta L \\
\hline
\,\text{NII} & 2{p ^{2}} \; {^{3}\!P} ^e_0 \leftarrow 2p3s \; {^{3}\!D} ^o_1
& 1084.0 & 2.18×10^8 & -1 & o \rightarrow e & 0 & -1 & -1 \\
\;\text{CIII]} & 2{s ^{2}} \; {^{1}\!S} ^e_0 \leftarrow 2s2p \; {^{3}\!P} ^o_1
& 1908.7 & 114 & +1 & o \rightarrow e & -1 & -1 & -1 \\
\text{[CIII]} & 2{s ^{2}} \; {^{1}\!S} ^e_0 \leftarrow 2s2p \; {^{3}\!P} ^o_2
& 1906.7 & 0.0052 & +2 & o \rightarrow e & -1 & -1 & -1 \\
\end{array}
$$
  $$\text{Table 1: Examples of allowed, semi-forbidden, and forbidden transitions."}$$


Could it have been possible for the Moon to be split without any observable effect on Earth? Before asking my question, I'd like to clarify why I'm asking about something so absurd. A supersitious claim has been bought to me that in the past, the moon was split and seperated as a sign. My attempts at dissuading them by pointing out that half of the planet failed to observe it has been met with the reply that this is because it took place for a short period of time. After which ofcourse the moon was returned to it's original form, leaving no evidence :)
I would like to know if there would've been any effect observable on Earth resulting from such a massive body being split in two. Perhaps something to do with the effect of the sudden change in the Moon's gravity on earth? Or perhaps some other factor that would make such an act impossible?

No the energy required to do this would cause the moon to melt.
However the story of the spitting of the moon supposes a supernatural cause, and one which is not bound by the physical laws.
God could split the moon and put it back together by his word. And God could arrange so that only some people saw it and others looked away.  When you bring God into your explanations the answer is always "yes he can".  That's what being omnipotent means.


Moon and Sun UV Light How much ultraviolet light comes from the moon at night and from the Sun during the day? I am interested in using a UV camera on a robot to see the UV from the sun/moon to do robot localization.

"A little" (compared to how much visible light there is)

(from a question on physics se

The sun produces relatively little UV radiation, most of its light is visible, or infrared, and anything shorter than about 290nm is completely blocked by the atmosphere (remember to thank the atmosphere today)

UVa (in the range 400-320) and UVb (320-290nm) do get through. The sun is so very bright, that it is by far the brightest natural source of UV on the Earth.

The moon is about 400,000 times dimmer than the sun. Ultraviolet reflected off the moon could reach the Earth. It would be very dim.

At night UV camera systems use active illumination rather than depending on the moon.


Matter and Antimatter Interaction in the early Universe Well it is said that during the Big Bang, things were created in pairs: one matter and one anti-matter. I think its a part of the Big-Bang Theory.

It is also said that matter + anti-matter - Energy

Then shouldn't it be so that there would have been no matter or antimatter in existence now since all of them should have reacted to form pure Energy. I mean how were we created then?

And also where are all the antimatter gone now? Are they still reacting with us?

Thanks

Matter and antimatter aren't perfectly symmetric, when following the "forward" direction of time. This is commonly called charge-parity violation.
It's assumed, that this asymmetry caused an excess of matter in the early stages of the Universe after the Big Bang, annihilating almost all antimatter, and leaving behind matter.


Question from a non-astronomer /non-mathematician regarding rotation speeds I have been watching too much TV related to black holes.  Most black holes are depicted with a disk of matter that spirals into the black hole. Does the mass and diameter of the disk have a bearing on the rotational speed, maximum rotational speed of a black hole? Also, do black holes have gravity? It looks like they must in order to collect matter. Does the gravity of a black hole create gravitational waves much as the electromagnetic waves created around the earth?  Do these waves have any influence on the speed of the black hole's rotation. What about if the rotation if the black hole is not symmetric?

This is a bunch of questions, many of which have answers already, but they are linked.

  Most black holes are depicted with a disk of matter that spirals into the black hole. Does the mass and diameter of the disk have a bearing on the rotational speed, maximum rotational speed of a black hole?

It's much more the other way round. The size of the accretion disk is determined mainly by the mass of the black hole, and by the supply of new material passing close enough to the black hole to possibly get trapped in the disk. The rotation of the black hole only plays a role in the innermost part of the disk, where the strange effects of General Relativity cause the material of the disk to be dragged around.

  Also, do black holes have gravity?

Yes. Indeed they don't really have much else.

  Does the gravity of a black hole create gravitational waves much as the electromagnetic waves created around the earth?  Do these waves have any influence on the speed of the black hole's rotation. What about if the rotation if the black hole is not symmetric?

A black hole is only asymmetric for very brief periods after large objects fall into it (for example when two black holes merge). In this case it produces very large amounts of gravitational wave energy, rapidly shedding the asymmetry. This process is called "ringdown". A symmetrical spinning object does not radiate gravitational waves.

A very good source for seeing these kinds of things are the videos produced by the SXS consortium with the caveat that they are visualisations of aspects of what is happening, not images. There are far too many dimensions of data to show the whole story.


What percent of planets are in the position that they could be viewed edge-on from Earth? (and thus able to undergo transits) Star number 12644769 from the Kepler Input Catalog was identified as
  an eclipsing binary with a 41-day period, from the detection of its
  mutual eclipses (9). Eclipses occur because the orbital plane of the
  stars is oriented nearly edge- on as viewed from Earth. During primary
  eclipses, the larger star, denoted “A,” is partially eclipsed by the
  smaller star “B,” and the system flux de- clines by about 13%

From http://www.sciencemag.org/content/333/6049/1602

Here's the thing though: out of all possible edge-on configurations, there are far more configurations where a planet can never be in the position to be edge-on, rather than configurations where a planet could potentially be in the position to be edge on. (I'd guess that it happens in fewer than one in several hundred cases)

So why are we able to observe so many transits?

Because there are so many planets out there!

There just happens to be an entire web page dedicated to calculating that answer.

  Transits can only be detected if the planetary orbit is near the line-of-sight (LOS) between the observer and the star. This requires that the planet's orbital pole be within an angle of $d_*/a$ (part 1 of the figure below) measured from the center of the star and perpendicular to the LOS, where $d_{*}$ is the stellar diameter (= 0.0093 AU for the Sun) and $a$ is the planet's orbital radius.

  This is possible for all $2\pi$ angles about the LOS, i.e., for a total of $4\pi d_*/2a$ steradians of pole positions on the celestial sphere (part 2 of figure).

  Thus the geometric probability for seeing a transit for any random planetary orbit is simply $d_*/2a$ (part 3 of figure) (Borucki and Summers, 1984, Koch and Borucki, 1996).

  For the Earth and Venus this is 0.47% and 0.65% respectively (see above Table). Because grazing transits are not easily detected, those with a duration less than half of a central transit are ignored. Since a chord equal to half the diameter is at a distance of 0.866 of the radius from the center of a circle, the usable transits account for 86.6% of the total. If other planetary systems are similar to our solar system in that they also contain two Earth-size planets in inner orbits, and since the orbits are not co-planar to within $2d_*/D$, the probabilities can be added. Thus, approximately $0.011 \times 0.866$ $= 1\%$ of the solar-like stars with planets should show Earth-size transits.

That's pretty freaking amazing! Kepler has been up there for a short while, and has a possible list of nearly 2000 planets just looking at about 150,000 stars for only a couple of years! So if only 1% statistically transit, that would mean that just randomly 1500 systems would have the correct orientation (given the results to date, that makes sense). And given that about 7500 stars were eliminated from consideration due to being variable of one sort or another... I think it would be pretty safe to say that pretty much every star out there has at least some sort of planetary body around it.


Does Eddington's variable polytropic index better fit data from the Standard Solar Model? I came across a 1938 paper by Arthur Eddington1 in which he discusses stellar polytropic models where the polytropic index $n$ varies inside the star, decreasing from a value $n_2$ at the surface to a value $n_1$ at the center. He specifically mentions that the case of $n_1=1.5$ and $n_2=3.5$ should represent "ordinary gaseous stars" quite well.

In his conclusion, he states

  We have seen that actual stars are likely to be represented by a model with $n$ continually increasing outwards.

78 years later, we have better data about the structure of the Sun, thanks in part to helioseismology. So I have two questions:

Do researchers use Eddington's variable index treatment in theoretical models?
Does Eddington's model better fit current data from the Sun than the standard polytropic model with $n$ constant throughout the star?

I ask partly because I know that the standard polytropic model (with $n=3$) begins to fail in the outer regions of a star, thanks in part to convection, as shown below, and I wondered if even a simple two-phase model could suitably describe a convective envelope.

Graphs from this page.

1 Eddington mentions earlier work by Milne (1932), who uses a slightly more complex analysis. I haven't yet read Milne's paper, so I'm not familiar with its conclusions and use.

The definition of a polytrope is one for which the pressure and density are related by
$$ P = K \rho^{(n+1)/n}$$
The pressure is thus assumed to be independent of temperature and this simplifies the equations of stellar structure so that certain analytic solutions and approximations are possible.

It is however a  big assumption and is generally only accurate for stars/planets supported by degeneracy pressure, where $1.5<n<3$ depending on whether the degeneracy is non-relativistic or relativistic. The other instance is where the energy transport is by adiabatic convection, which gives an independent relationship between $P$ and $T$ and means that for a completely ionized gas that $P \propto \rho^{5/3}$ and $n=1.5$.

A further polytropic approximation can also be made where the energy transport is predominantly radiative and where it is assumed that the radiation pressure is a fixed proportion of the total pressure. In these circumstances $P \propto \rho^{4/3}$ and $n \simeq 3$.

I found a bit of a primer here, but most stellar astrophysics textbooks will go through this.

The Eddington standard model assumes a radiative ($n=3$) polytrope. The vast majority of the Sun, by mass, is indeed radiative and the polytropic approximation works reasonably well. However the outer parts of the Sun (beyond about 70% of the radius) are convective and so a single polytropic model does not fare so well. So what you can do is have a "mixed polytropic model", where you divide the star up into polytropic shells with different polytropic indices, each forming the boundary conditions for the ones above and below.
A double polytropic model for the Sun, with two zones that have $n_1\simeq 3$ in the interior and $n_2\simeq 1.5$ in the outer convection zone should be much more accurate.

But modern astrophysical calculations do not rely on the analytic approximations of polytropes and the Lane Emden equation. They numerically solve the coupled differential equations of stellar structure, subject to the boundary conditions imposed by an atmosphere on the top of the star.

EDIT: Unashamedly tempted by the bounty, I did a little bit more research. I came across some sort of high-level set of lecture notes by
Robert French of Swinburne University. He discusses modelling the Sun with composite polytropes at some length. On p.8 he describes how the matching criteria at the boundaries between polytropic zones is actually quite tricky and you have to let the matching radius and central density float in order to get a smooth join.

The conclusion he arrives at is that a combination of a $n_1=3$ and $n_2=1.5$ actually appears to do a worse job of fitting the entire Sun than modelling with a single average polytrope (although with $n$ varied depending on which physical quantity you want on the y-axis - e.g. $\bar{n}=3.11$ for density vs radius). If on the other hand you allow the polytropic indices and the interface radius be free parameters in order to get a good match to the standard solar model, you can get a really good fit with $n_1=3.94$ in the radiative zone, $n_2=1.58$ in the convection zone and an interface at 68% of the solar radius. (Plot reproduced below - compare these with the plots in your question!).


Can a probe orbit Pluto given Charon's gravity? Is it possible for a man-made probe to orbit Pluto given its closeness to Charon? The gravitational field interaction of Pluto and Charon is likely to make simple orbit structure difficult for a man-made probe. Thanks for your insight.

Avoiding hard math, which I'm not very good at, the simple answer is yes, provided there's enough distance.   Pluto/Charon have 4 moons orbiting them a bit further away, all in relatively stable orbits.   Source

Here's distances to scale.  - same source.

Because the ratio of gravitational field strength to size is exponential (Power of 1.5), the apparent closeness of Pluto and Charon to the 4 small moons looks unusual, but it's perfectly acceptable for objects of that size.

Lagrange point stability requires a mass ratio of about 26 to 1.  (.0385 to 1 per source).  The Earth has satellites in unstable Lagrange points, so it's certainly doable, it's just not technically "stable".

The hardest part about getting a satellite to orbit Pluto is that Pluto's sufficiently small that any ship that approaches it would need to slow down significantly on it's own to get captured into a Pluto Orbit.   That's why New Horizon was a flyby, not an orbit.


Scholarly work on the gravitational disruption of a rocky planet by a close pass with another planet? note: While the apparently abandoned question Effects of a rogue planet passing between the Earth and the Moon? (inspired by Thundarr the Barbarian) asks for speculation and was answered with it, this question asks for something different. Solar system evolution will likely involve both accumulation of matter into planets and destruction of formed planets by various mechanisms.
I'm interested in finding out if there is scholarly work on the *gravitational disruption of rocky, Earth-like planets due to close passes by, but not collisions with other planets.

An earth-sized rogue planet passing at high speed on a hyperbolic orbit around the sun will generally not have a huge effect on the solar system as a whole.  But if it did happen to pass close to earth (say ~1 million miles), it could seriously disrupt the Earth's orbit around the Sun and the Moon's orbit around the Earth.  Most perturbations would lead to much more elliptical orbits.  These would likely lead to an increase in seasonal extremes and perhaps to larger ocean tides.  These changes would then be relatively permanent.  There are also questions about the transient effects during the 'near miss'.  Would tidal forces from the passing rogue planet induce seismic events on the Earth?
This maybe isn't quite the right forum, but have there been any scientific studies along these lines? It also sounds like a good topic for a science-based novel.  Presumably mankind would have years or decades of warning for such an event and would be able to make reasonable predictions about how the environment on Earth will change.  The long-term survival of mankind or life in general in the new environment would be a whole other question.
I do see this similar question and that answer does discuss some important points.
[Edit] But I was hoping that someone might have investigated this scenario in more detail, perhaps with a specific example. I'm wondering if one can do a back-of-the-envelope calculation of how much delta-V the Earth would get from an Earth-size intruder whizzing past at a distance of, say, 1 million km at a speed of 100 km/s. That might allow the range of perturbed orbits and range of insolation and temperatures, etc. to be calculated.

Scenario: This looks specifically at a near-miss scenario with an Earth-like intruder passing by at a distance of 1,000,000 km at a speed of 100 km/s.  This is about three times the distance to the Moon and at about three times Earth’s orbital velocity. The 100 km/s seems a reasonable number given the relative velocities of our stellar neighbors, the solar infall/escape velocity, and the likely difference in directions of the Earth and intruder velocities.
First Sighting: It's a pure guess, but maybe the intruder will be first recognized when it’s brightness (apparent magnitude) becomes similar to that of Pluto.  The difference in absolute magnitude between the Earth and Pluto is (-3.99)-(-0.70) = -3.29.  This corresponds to a difference in brightness of 100^(3.29/5) = 20.7.  At that distance from the Earth/Sun, the brightness will go as the 4th power of distance, so we should first spot the intruder when it is 20.7^(1/4) = 2.13 times further away than Pluto or about 12.6 billion km.  Just approximating with a constant 100 km/s, this means we will get about 4 years of advance warning.
Night Sky: Assuming the intruder approaches from roughly the opposite direction of the Sun, it will be visible in the night sky with a magnitude -3.99+(1/5)log100(1/s^2/(1+s)^2 where s is the distance from Earth in AU.  At 0.6 AU (90,000,000 km) it will be roughly as bright as Venus and clearly visible in the night sky.  This will occur about 10 days before closest approach.
At its closest approach (1 million km), the intruder will appear about 40% larger diameter than the Moon and will be about 7 times brighter, both because of the larger apparent area (0.73/0.52)^2 and higher albedo (0.43/0.12).  The intruder will take 5.6 hours to travel the 2 million kilometers  across 90 degrees of the night sky while its apparent area goes from 50% to 100% and back to 50% of maximum.  A sight worth watching!
Orbital Disturbance (Earth):  Assuming the deflections of the Earth and the intruder are small, the delta_V can be calculated as (1/2)GMe/d^2 x (2d/v), where Me is the mass of the Earth or the intruder, d is the distance of closest approach and v is the velocity.  The first factor corresponds to the maximum acceleration of the Earth and the second factor to the effective time during which the acceleration is applied.  The first factor is 0.000199 m/s^2 and the second factor is 20,000 seconds giving a delta_V of 3.98 m/s. This is 0.00013 of the Earth’s orbital speed of 30 km/s.  Generally this would change the ellipticity of the orbit slightly and could also change the mean radius if the delta-V were mainly along the direction of the orbit.  The orbital energy could change by +/-0.00026 and the mean radius would change the same fraction.
Corresponding to a change in the Earth’s mean orbital radius, the insolation would change by double that fraction or +/-0.00052.  The temperature based on black body radiation would change by the fourth root of the insolation, i.e. +/-0.00013 or +/-0.013% or +/-0.04 degrees C.   These changes seem quite minor.
Orbital Disturbance (Moon):  The Moon would be more severely affected because its mass is small compared to the intruder, so the acceleration would be roughly twice as large.  Also the Moon could be on the ‘wrong’ side where it would be closer to the intruder by a factor of 2/3.  So the acceleration and the delta_V could be higher by a factor of 2/(2/3)^2 = 4.5.  The resulting delta_V of 18 m/s on the Moon’s orbital velocity of about 1000 m/s is more significant.  This could result in a 3.6% change in orbital energy and in radius and a change of 5.4% in period, changing the 28 day period by more  than a day.  Since tidal effects go as the cube of distance, presumably the lunar tidal ranges would increase (or decrease) by about 11%.
Transient tidal effect: Tidal forces are proportional to mass/distance^3.  Ignoring the constant factor, the peak tidal force from the Moon : Sun : Intruder are in the ratio 7.35E22/(3.84E5)^3 : 1.989E30/(1.48E8)^3 : 5.97E24/(1E6)^3 or 1.30 : 0.61 : 5.97.  So the maximum tidal force as the intruder swings by will be about three times more that the combined lunar and solar tidal force.  Presumably, during the several hours of closest approach, this will raise ocean tides about three times larger than the familiar large spring tides.  Coastal areas would experiences tides tens of feet higher than normal much as in a storm surge or tsunami. I suppose it is also possible that some latent earthquakes could also be released by such large tidal forces
(The last time I did a back-of-the-envelope calculation like this, I was wrong by a factor of 10^6.  But it certainly elicited some pretty quick responses!)


What is the meaning of grain opacity and why does it affect the formation time of gas giants? While doing research for my presentation on the formation of gas giants, more specifically the "core-accretion model", I have been stumbling across the term "grain opacity" and don't quite understand it's meaning. From what I've already read, it's a quantity that heavily influences the time it takes for giant gas planets to form and that a lower grain opactiy leads to shorter formation times. But why does the opacity of dust grains in the protoplanetary disk have such an effect? Is it because of the protoplanet cooling off faster?

The cooling of a gas giants outer envelope (which in turn cools the convective interior) is dominated by radiative transport.
The radiative flux at the photospheric boundary is $$F_{rad}\propto- \frac{ T^3}{\rho \kappa_R}\nabla T,$$ where $\kappa_R$ is the Rosseland-mean opacity (for more, see e.g. these open lecture notes, by K. Dullemond, in 5.5.3). This is what is often called 'the' opacity in the context of gas giant formation.
The opacity quantifies how opaque a medium is to the transport of photons. High opacity means low transport of photons, hence low cooling.
Now depending on the exact density and temperature conditions of a gas layer, it can be that either atomic, molecular or dust-grain contributions dominate the opacity (see a more detailed account for the collapse of protostars, a related problem, in Vaytet et al., (2013), particularly their fig. 2). As the outer layers of gas giants, which set $F_{rad}$, can be quite cold (T<2000K) it is the strong continuum opacity of dust grains that sets the cooling rate, therefore the contraction rate of the envelope, and hence the mass accretion into the inner layers of the proto-gas giant. Note, that when talking about $F_{rad}$ it is equivalent to talk about the gas giants cooling luminosity $L$, as they are trivially related via $L=4\pi r^2_{\tau=1}F_{rad, \tau=1}$, i.e. the radiative flux at the  protogiants photosphere.
The actual self-consistent dust-dependent cooling problem was adressed in Movshovitz et al. (2010), while you are probably aware of the seminal work by Pollack et al., (1996).
For 3-D radiative hydrodynamic simulations of this process see e.g. SPH-based in Ayliffe & Bate (2009), long SPH-based in D'Angelo and Bodenheimer (2013) and grid-based simulations in Schulik et al., (2019).


Solar luminosity proportional to $G^7$? A book I recently purchased describes Dirac's proposed variation of Newton's G, and why that specific proposal wouldn't work because the earth would have been too hot for the known history of life. A footnote explains that the luminosity of the sun is proportional to the seventh power of G, which seems incredible to me. I know that this has to be a simplification, since stellar dynamics are annoyingly complex, but if we put that aside, is this essentially correct, or is it a typo? What would be the theoretical justification for this in broad terms? (eg, "Gravitational pressure is proportional to this exponent of G, but since that also shrinks the size of the star, which increases the pressure, it's actually this exponent of G, and the rate of fusion is proportional to the gravitational pressure squared, so you end up with 7").

I think the argument is the following. The central temperature can be estimated from a form of the virial theorem. At least dimensionally speaking,  total thermal energy $MkT/\mu$ (where $\mu$ is the average mass per particle) is proportional to gravitational potential energy $GM^2/R$. So $T \propto GM/R$.

But for a given mass, standard polytropic theory for a star, says that
$$M \propto \rho_c^{(3-n)/2n} G^{-3/2}$$
$$R  \propto \rho_c^{(1-n)/2n} G^{-1/2}$$
 where $n=3$ is an approximate polytropic index for the Sun and $\rho_c$ is the central density. See for example Does Eddington's variable polytropic index better fit data from the Standard Solar Model?

Putting this together we have
$$ T \propto GMR^{-1} \propto G^2 \rho_c^{-1/3}$$

The power generation in the solar core from the pp cycle goes approximately as $L \propto \rho_c^2 T^4$ and using the proportionality for $T$ derived above, we have $L \propto G^8 \rho_c^{2/3}$.

Finally, we note that for a $n=3$ polytrope, that $\rho _c$ and $M$ are independent, but that $\rho_c^{2/3} \propto G^{-1} R^{-2}$. So for a fixed solar radius we have $L \propto G^7$.


Is scattering efficiency of dust related to albedo? I know that the scattering cross-sections of dust in comets depend on the scattering efficiency. A higher scattering efficiency leads to an increase in brightness. Therefore, does it mean that this will also increase the albedo or these quantities are not related at all?

There will be a relationship but not a simple one. If we model a uniform slab of material by a complex dielectric constant $n + ik$ then we can calculate three things; reflection, absorption and transmission. If we model a dust particle as a uniform sphere of the same, then the far field electric amplitude scattered from an incident plane wave on it will depend on the complex dielectric constant. The reason we can't just use reflection, absorption and transmission is that those are the amplitudes squared, and so phase information is lost, so you can't easily work your way backwards.
There is also the angular distribution to worry about, which will be different for each size and also for each complex index of refraction. So two identically shaped and sized dust particles; one from a high bulk albedo material and one from a low bulk albedo material could end up with a variety of different relative scattering efficiencies.
A cloud of either one might attenuate light from a star, depending on the details, and one might scatter light more strongly at some specific angle than the other.
There might be some rules of thumbs for certain kinds of dust particles made of materials most likely to be observed and discussed by astronomers, but I don't think there are any first-principles-based generalizations that always apply.


How to install Distro Astro operating system? I've recently started a project on Galaxy spectroscopy for which I had to use IRAF software. During the search for IRAF software, I came across a platform which said that Distro Astro is a very good OS Amateur astronomer should check out. I have searched the internet for the procedure and am an amateur in OS installation and other stuff.

Does anybody know the detailed procedure of installing the software?

Distro Astra is a Linux distribution for Astronomy.

Unless you are very familiar with Linux distributions be vary careful about installing it you main computer. If you have an old computer not being used I would suggest you first installed Distro Astra on that computer.

If it is installed as the only operating system for the computer it will wipe whatever was on the disk prior to installing Distro Astra and you will lose everything that was on that disk.

Being an operating system, once installed, it will make permanent changes to the hard disk drive. It may be possible to have a dual boot system that gives the user the option to use the computer via Distro Astra, or the existing (most likely) Microsoft based operating system originally installed, when the computer is booted.

If you do not have an old computer keep your existing hard disk drive in storage but install a new hard drive for Distro Astra. If the installation of Distro Astra becomes totally messed up you will still have the original hard drive to reinsert and your computer will be as it was prior to dabbling with Distro Astra.

If you have no knowledge of Linux operating systems, find a friend to help you or be prepared to learn, make mistakes and sometimes be very frustrated.

Installing Linux on older computers that utilize a BIOS is easier than installing on newer computer that use a UEFI.


How do tip tilt mirrors correct distortion in adaptive optics? I'm currently trying to learn about how adaptive optics correct the blurriness caused by atmospherical seeing effects. This is my current understanding of how adaptive optics works (trying to offer some context, I have a very limited understanding at the moment).

Wavefronts from celestial bodies become distorted as they pass through the Earth's atmosphere due to variations in refraction indices as they pass through different density air, temperature variations and so on. This distortion as far as I know, causes the wavefront to look squiggly instead of being straight parallel lines. Once the light reaches the telescope, it is reflected off a tip tilt mirror, then reflected off a deformable mirror, then a beam splitter splits the light rays and one half is passed through a computer sampling device which detects distortions in the wavefront. This is done up to a 1000 times per second, and somehow the computer causes actuators fitted at the back of the deformable mirror to move such that the reflected light off the deformable mirror is now undistorted (i.e. the wavefronts are a straight line, not squiggly). Thus adaptive optics fixes the distortion issue.

I have a few questions arising from my current understanding:

What exactly does the tip tilt mirror do to help fix the distortion problem? Wiki simply says that it fixes the problem of aberration but doesn't offer a specific explanation of how.
Why are straight wavefronts ideal when viewing celestial bodies? As in, how does the squiggliness of the wavefronts actually make the images look blurry?

I believe I am asking too many different types of questions in one post, so I'll make a separate post for this last question, but for anyone reading this:

I keep reading that increasing the size of the primary mirror will improve resolution. I have seen justification of this using Dawe's limit such as

  By considering the formula R= 2.1*10^5*wavelength/ D, where R is the minimum angular separation required by two point sources for them to be resolved, so as primary mirror diameter increases, D increases, so R decreases meaning that resolution of the telescope is improved.

I don't really see how they can apply the R=... formula to primary mirrors because I have only seen it derived in the context of light diffracting when it passes through a small slit, and not mirrors (which don't allow light to pass through, but rather reflects it). Can someone explain to me why we can apply this formula to mirrors as well?

Not an expert, but offer one solution:

Question 1

The adaptive optics correction is accomplished by a tip-tilt mirror and a deformable mirror. Usually the atmospherical wavefront distortion (or say, the phase screen) looks like this:

Note its ramp-like shape. In other word, the 2nd and 3rd Zernike terms are large.

The deformable mirror has limited correction range: it deals well with the higher-order Zernike terms, but not the 2nd and 3rd Zernike terms. This is why the tip-tilt mirror is here: it introduces a general ramp correction to suppress the 2nd and 3rd Zernike terms, then the afterwards wavefront contains only the higher order Zernike terms, and hence can be corrected well by the deformable mirror.

Question 2

You need a little Fourier optics knowledge to fully understand. Briefly speaking, The image $I$ (this is also called the point-spread-function, the linear system response of a Fourier optical system) captured on the sensor follows:
$$I(\frac{x}{\lambda f}) = |\mathcal{F}\{p(y)*e^{j (2\pi/\lambda) w(y)}\}|^2,$$
where $x$ and $y$ are the coordinates on the sensor and the aperture, $\lambda$ the wavelength, $f$ the lens focal length, $\mathcal{F}$ denotes Fourier transform, $p$ and $w$ are the aperture shape and the distortion wavefront, respectively.

For simplity let $p(y) = 1$ for every $y$ (i.e. infinity aperture size). If $w(y) = 0$, then $I(x)$ is a perfect Dirac function. This simply says your star has no blurring. To the contrary, for example $w(x) = \alpha x^2$ for some $\alpha$, then it results in a defocused blurring point.

The same formula above also helps answering your additional question. Increasing the aperture size (the main mirror will finally be mapped to the aperture) is equivalent to, in the Fourier domain, decreasing the blurring size.


What keeps the planets of the solar system moving in their fixed orbits? Models of the solar system show each planet moving around on its own orbit.

Is is really because of the Sun that every planets remains in its own fixed orbit?

It is ,of course, gravity which keeps the planets circling the sun. They are actually falling toward the sun, but as the sun is a sphere, its surface keeps curving away, so they never get any closer to it. Because they are in a state of free fall, they are in effect weightless with respect to the sun, just as the space station is weightless with respect to the Earth, and as they use no energy in falling they can do it for ever. If you are wondering why you don't feel weightless, that's because of the gravitational field of the Earth. If the space station was a massive body like the moon.the astronauts wouldn't be weightless,but although the space station has a gravitational field of its own it is so trivial that it has no measurable effect on the astronauts.