Why did the distribution of asteroids discovered in 2010 have a radial modulation? This answer links to one of Scott Manley's excellent asteroid videos Asteroid Discovery - 1970-2015 - 8K resolution. The animation highlights the positions of the meteors at the moment of their discovery, and by watching one can see the technology improve and notice patterns as instruments are pointed in different directions to avoid the light from the Sun and (at least sometimes) the Moon. (there is music, adjust volume accordingly)

There are often fan-shaped patterns showing the directions that more sensitive telescopes with modest fields of view are pointed.

However, I noticed that only during the year 2010, roughly between asteroid numbers 500,000 and 520,000 there are radial striations at certain distances from the Sun. I don't see this happen at any other time during the video.

Is this just a rendering artifact, or is it real? If real, what would cause the periodic radial modulation of sensitivity, and only in 2010?

note 1: YouTube allows for playback rates between 25% and 200%, and variable video resolutions. I found 25% and 1080p optimal for my current internet connection and screen.

note 2: For those with GIFs disabled, one image is a GIF.

note 3: 2nd image contains several cropped screenshots highlighting the "periodic radial modulation of sensitivity" in asteroid detection during 2010, for clarification purposes.

I'm pretty sure that the radial pattern found in the data is a result of WISE's approximately 90 minute sampling cadence (dictated by the satellite's orbit), astrometric precision (about 0.2 arcseconds in the stacked images around launch, see Wright et al. 2010), and the number of free parameters in fitting the asteroid orbits based on that data. See, in the actual images the asteroids appear as points of light that shift appreciably between frames. IIRC, they expected 7 to 12 observations per asteroid. So you have 10-ish observations spanning 15 hours or so to fix the asteroid's orbital parameters around the sun. As you can imagine, there will be more parameters than can be perfectly fit with a single pass in this data set alone.

At a guess: it's related to quantization in uncertainty estimates and how that feeds forward into the orbit fitting algorithm.

I don't know the details behind the striping, but I'd bet it's related to the numerical precision used in the early processing of the data. They've either refined the orbits since using observations from passes separated by ~6 months, or modified how they're handling the numerical precision in their astrometric measurements since. More likely the former, but I'm sure if you asked Amy Mainzer (PI of NEOWISE, and head of the asteroid hunting part of the mission), Roc Cutri (head of the database creation and data processing part of the team), or any of the people on Mainzer's team, they could tell you more.

Relevant background: I was Ned Wright's grad student (original PI of WISE), and he had me design and test an asteroid hunting algorithm in the run-up to launch (we ultimately didn't use it - it scaled like $N^2$, IIRC, and existing efforts in the literature scaled like $\log(N)$ or $N\log(N)$). I ended up working with the extragalctic part of the team, hence my uncertainty on the precise details of what the solar system team did. I think I asked them about the striping during a presentation, but I don't recall the answer, so I'm pretty sure the answer was mundane and unavoidable.


Life on other planets If all living things like animals require oxygen to live, how can people think that there could be life on other planets such as mars for example ? Living things require oxygen and carbon dioxide, and these planets do not have plants and such to provide this. So why do scientists try to still look for life on these planets ?

TL;DR: Not all life requires atmospheric oxygen or carbon dioxide. There are many organisms here on Earth that use chemosynthesis to process chemicals like methane or hydrogen sulphide for energy.

The search for life on Mars has concentrated on finding evidence of autotrophic microorganisms – organisms that convert simple chemicals in the environment into complex organic compounds. Most autotrophs use water as the reducing agent, but some can use other chemicals like hydrogen sulphide.

The energy source for these autotrophs can either be sunlight, in which case we call them phototrophs (which includes algae and green plants here on Earth), or obtained through the oxidation of electron donors in their environment, in which case they're called chemotrophs.

There's a wide variety of chemotrophs here on Earth, typically living in completely dark and often hostile environments such as next to deep-sea volcanic vents. For example, iron-oxidizing bacteria even colonise new lava beds in the deep ocean, and in effect use the oxidation of available ferrous iron as their respiratory process.

There's no reason to think that autotrophs couldn't find niche environments on Mars in which to exist, but the biggest hurdles are the significantly lower temperature and pressure, and the significantly greater radiation. The latter can be overcome by colonising sub-surface environments, which eliminates the phototrophs from contention. While UV radiation penetrates soils only in the millimetre range, one study found that viable dormant cells would have to be at least 7.5m below the planet's surface to survive the cumulative damage from cosmic radiation.


How does the dense atmosphere of Venus affect the planet's rotation? Reading through questions about my favourite planet, Venus, I came across this answer to the question What is the current accepted theory as to why Venus has a slow retrograde rotation?, specifically where the answer suggests one of the possible reasons for the slow retrograde motion as being

  The spin slowed to a standstill and then reversed, caused by the sun's gravity, the dense atmosphere and friction between core and mantle

Based on a fairly recent study published in the Nature article "The four final rotation states of Venus (Alexandre C. M. Correia & Jacques Laskar, 2001).

Given the relatively high density of the Venusian atmosphere, this leads to the question:

How does the dense atmosphere of Venus affect the planet's rotation?

How does the dense atmosphere of Venus affect the planet's rotation?

I don't know all the details, but I can give a partial answer, with links to further reading if interested.

Tidal locking, and orbits slowing down due to a tidal bulge and tidal forces and/or a denser or greater gravitational gradient on one side of the tidally locked object are all pretty straight forward, and I don't think that part needs to be explained too much.  The tidal bulge has mass and that mass creates drag on the rotation (or speeds it up if the bulge is behind the object causing the tides).

And it's worth adding that rocky worlds are gravitationally lumpy, where the heavier side (gravitational gradient), tends to, over time and in tidal locking, face the object it orbits.   The near side of the Moon, for example, is slightly denser on average than the far side.  See here for a better explanation and here for gravitational map of the moon (the Earth side is on the left in the picture below).

Venus' atmosphere, particularly it's upper atmosphere, behaves very differently.   There's still an atmospheric tidal bulge, but because of the heat from the sun, the hot gas on the sun side of Venus upper atmosphere has lower density, so the tidal bulge may have less mass, not more mass, even with greater size.   Atmosphere doesn't behave like a typical tidal bulge orbiting close to a star because the hot atmosphere is less dense and the night-side of Venus' upper atmosphere is more dense.  The (by mass) true tidal bulge is on the far side of Venus, away from the sun, not the near side.

Another effect is that the heat from the sun causes something like a jet-stream on Venus, a very powerful jet stream.   The wind speed on Venus' upper atmosphere is faster than any hurricane on Earth and the upper atmosphere tends to rotate in the same direction around Venus.   This solar driven rotation (may be) a factor in Venus' surface having a slow retrograde rotation where, without an atmosphere it might be tidally locked.

It's difficult to imagine an upper atmosphere wind could "spin" a planet, and it would only do so very very slowly, (Venus' atmosphere is about 1/10,000th it's total mass, not too far off from the ratio of Earth's oceans to Earth's total mass) and the tidal slowing of a planet is very gradual, at Venus distance taking perhaps 100 million years to slow a rocky body of that size to tidal locking.     So the rotation may be a balance between Venus' solar driven jet stream and it's gradual tidal locking.

At least, that's what I came up with doing some reading on this.    From Wikpedia:

  The rotation period of Venus may represent an equilibrium state
  between tidal locking to the Sun's gravitation, which tends to slow
  rotation, and an atmospheric tide created by solar heating of the
  thick Venusian atmosphere

and link to the source, which is in French but English translation is available.   (see footnote 100 and 101)

On Venus "jet-stream" like upper atmosphere, note that while the jet-stream on Earth is driven by surface temperature variation perhaps more than solar and by Earth's Coriolis effect, and it varies by seasons.   Venus doesn't have surface temperature variation to speak of and essentially no Coriolis effect due to it's slow rotation, so it's jet-stream tends to be much more along latitude lines and much less curved than Earth's.    That's not to say that there's no variation.   Venus' wind seems to operate on a 5 day cycle and it appears to have grown stronger in the years since we've been observing Venus.  Neither of those is well understood.

Article here

  At the very top of the cloud layers on Venus, wind speeds reach 355
  km/hour (or 100 meters/second). This is the same the jet stream here
  on Earth. As you descend through the cloud layers, though, the wind
  speeds pick up. In the middle layer, the winds can reach speeds of
  more than 700 km/hour. That’s faster than the fastest tornado speed
  ever recorded on Earth.

  But then as you descend further down through the clouds, the
  thickening atmosphere slows the winds down, so that they act more like
  currents in the ocean than winds in the atmosphere. Down at the
  surface, the winds only move at a few km/hour. That’s not much, but
  the thick atmosphere can still kick up dust and push around small
  rocks.

and

  The winds on Venus travel in a westerly direction, the same backwards
  direction that Venus rotates. Seen from above, Venus rotates in a
  clockwise direction. This is backwards from the other 7 planets, which
  rotate counter-clockwise.

another article here.

and, related, quote from Wikipedia and source article.

  Venus's rotation has slowed down by 6.5 min per Venusian sidereal day
  in the 16 years between the Magellan spacecraft and Venus Express
  visits

So, short summary, Venus atmosphere may play a key role in it's reverse rotation, or, maybe it's rotation was reversed by an impact a long time ago and it's still slowing down from that impact.   The impact theory might make the most sense, but impacts large enough to spin a planet are quite rare so, it shouldn't be assumed.   I don't believe it's known 100% whether Venus' atmosphere caused it's backwards rotation, but it seems reasonable and possible.   My personal favorite idea, though it may not be likely, is the gyroscope flipping hypothesis.  (Posting video cause it's really cool, but it has nothing to do with your question).

https://www.youtube.com/watch?v=1n-HMSCDYtM


Calculate the absolute magnitude for a multi-star system We know that the absolute magnitude of a single star can be calculated with the formulae:
$$M_{\rm bol} = 4.75 -2.5 \log_{10} \left(\frac{L_*}{L_{\odot}}\right)\ ,$$
$$M = -2.5  \log_{10}(L_* / L_0),$$
where $L_*$ is the luminosity of the given star, and $L_0$ being the luminosity of a magnitude 0 star at a distance of 10 pc ($79L_\odot)$.
However, this only works for a single star, not multiple star systems. Is the formula for the absolute magnitude of a multiple star system defined as $L_*=L_1+L_2+...+L_n$ (the sum of all luminosities in the system), where $L_n$ is the luminosity of a star in the system, or is it something else?

Yes, you are correct. The luminosities can be added. Luminosity is the amount of electromagnetic energy emitted per unit of time (measured in $ \textrm{J} \cdot \textrm{s}^{-1} $ or $\textrm{W}$). So if you have a multiple star system, the total amount of energy emitted (i.e. the total luminosity) is simply the sum of the energy emitted by each of its components.
This does not take into account any absorption of the radiation of one of the components by one of the others.
In the book by Jean Meeus [1], the formula for adding stellar magnitudes can be found in Chapter 56:
$$ m = -2.5 \log_{10} \sum_{i=1}^{n} 10^{-0.4 m_i} $$
This formula is for the apparent magnitude, but would also apply for the absolute magnitude, which, after all, is only the apparent magnitude at $10\,\textrm{pc}$ distance.
$$ M = -2.5 \log_{10} \sum_{i=1}^{n} 10^{-0.4 M_i} $$
[1] Jean Meeus, 1998, Chapter 56: Stellar Magnitudes, from Astronomical Algorithms, Willmann-Bell Inc. Second Edition.


From which exoplanets is our sun the brightest star on the night sky? The brightest star on the night sky is Sirius, here in the Earth. But are there any exoplanets, where the brightest star on the night sky is our sun?

I wrote https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-13115.m to solve this with the full results at: https://github.com/barrycarter/bcapps/blob/master/ASTRO/brightest-stars-from-other-stars.txt.bz2

As viewed from RigelKentaurusB, the Sun has magnitude 0.48,
and is the 10th brightest star in the sky. This is the only
star system with known exoplanets where the Sun even makes
the top 10.

From EpsilonEridani, the sun shines at magnitude 2.37,
making it the 79th brightest star in their sky. Things go
rapidly downhill from there:

79 EpsilonEridani Sun 2.36646
207 HIP113020 Sun 3.18865
208 HIP85523 Sun 3.11216
247 HIP106440 Sun 3.29606
396 HIP15510 Sun 3.74083
423 HIP74995 Sun 3.81444
685 Fomalhaut Sun 4.25746
878 HIP64924 Sun 4.48202
949 HIP109388 Sun 4.54456
972 HIP99825 Sun 4.55679

This is based on Mathematica's list of 552 exoplanets
orbiting 464 stars, and a total list of 88,637 stars.

If the naked eye visibility limit is magnitude 5.5, the Sun
would only be visible from 28 of these stars:

10 RigelKentaurusB Sun 0.476137
79 EpsilonEridani Sun 2.36646
208 HIP85523 Sun 3.11216
207 HIP113020 Sun 3.18865
247 HIP106440 Sun 3.29606
396 HIP15510 Sun 3.74083
423 HIP74995 Sun 3.81444
685 Fomalhaut Sun 4.25746
878 HIP64924 Sun 4.48202
949 HIP109388 Sun 4.54456
972 HIP99825 Sun 4.55679
1086 HIP57443 Sun 4.65677
1126 HIP21932 Sun 4.69871
1381 HIP57087 Sun 4.87838
1426 HIP83043 Sun 4.89689
1409 Pollux Sun 4.90049
1617 HIP10138 Sun 5.01833
1649 HIP57050 Sun 5.04143
1676 HIP3093 Sun 5.05658
1716 HIP48331 Sun 5.06528
1831 Gl317 Sun 5.13093
2104 HIP22627 Sun 5.24572
2264 Rho1Cancri Sun 5.31849
2263 HIP40693 Sun 5.32723
2397 HIP27887 Sun 5.36943
2355 HIP80337 Sun 5.37669
2382 GJ1214 Sun 5.3879
2689 UpsilonAndromedae Sun 5.47503

If the limit is magnitude 6.5, 33 more star systems can see the Sun:

2838 Alrai Sun 5.52682
2926 HIP53721 Sun 5.571
3399 HIP79431 Sun 5.69393
3637 MuArae Sun 5.74863
3679 HIP113357 Sun 5.76061
3783 TauBootis Sun 5.7949
3941 HIP98767 Sun 5.83457
4071 HIP85647 Sun 5.86725
4318 HIP71395 Sun 5.92513
4580 HIP6379 Sun 5.95739
4783 IotaHorologii Sun 6.01138
4835 HIP7978 Sun 6.02527
4867 RhoCoronaeBorealis Sun 6.03887
5016 HIP1292 Sun 6.0583
4949 NN3634 Sun 6.08757
4993 HIP55848 Sun 6.10352
5252 HIP49699 Sun 6.11214
5300 HIP83389 Sun 6.11214
5031 HIP65721 Sun 6.11803
5343 HIP79248 Sun 6.12236
5385 PiMensae Sun 6.12986
5453 EpsilonReticulii Sun 6.13303
6158 HIP98505 Sun 6.251
6351 HIP113421 Sun 6.30305
6436 Nu2CanisMajoris Sun 6.31593
6343 HIP99711 Sun 6.32197
6737 Hamal Sun 6.35637
6516 HIP64457 Sun 6.37975
6867 HIP58451 Sun 6.4205
7614 HIP25110 Sun 6.43775
7525 HIP109378 Sun 6.46941
7538 HIP54906 Sun 6.47034
7957 HIP96901 Sun 6.48193

OLD ANSWER FOR REFERENCE (note that Sirius does not have known exoplanets, so the Sirius solution I propose below wouldn't work):

As others have noted, since Sirius is 25 times more luminous than our Sun (though still not luminous as Canopus or Rigel), so it would be the brightest star for most nearby star systems, and all the exoplanets we've found are fairly closeby.

The only possible exception would be an exoplanet orbiting Sirius itself, since Sirius would be considered their sun, and not a star in the night sky.

Unfortunately, Sirius' smaller companion, Sirius B, would be much brighter than our Sun there. I'm also pretty sure Sirius B would be in the night sky (not just in the daytime sky), so it would count.

I'd like to poke around a bit more before declaring this answer complete. I think I can show that any point where the Sun is brighter than Sirius must be less than 4.3 light years (and probably much less) away from our solar system, so, unless we discover an star system MUCH closer than Proxima Centauri, there are no exoplanets where our Sun is the brightest star.


What is the maximum transmission distance of the radio signal in the outer space which could still be understood? I’ll put this question in a form of a physics problem:
Transmitter at a point A sends a signal to a receiver at a point B. What is the longest distance in light years from points A to B for which receiver can still understand the signal?

Given:

Signal type: electromagnetic wave.
Signal frequency: frequency for which signal fades the least - probably radio.
Signal form:

Variant 1: beam of highest concentration possible.
Variant 2: full sphere.

Transmitter power: power of the most powerful transmitter known to be built in the Earth.
Receiver: sensitivity of the most sensitive receiver known to be built in the Earth.
Transmission medium: interstellar space.
Understanding of a signal:

Variant A: Can detect that the signal is not a noise.
Variant B: Can decode the information in the signal.

It cannot be said correctly, since we humans have hardly traveled to the moon and sent space probes to explore other planets in our solar system. So, theoretically anything might be possible. I'm trying to be a bit practical here. The only man made object that has gone really far is Voyager 1, which is at a distance of 18.7 billion kilometers (125.3 AU) from the sun. Although launched in 1977, it is the only live transmitter and receiver which is that far.

The radio communication system of Voyager 1 was designed to be used up to and beyond the limits of the Solar System. The communication system includes a 3*.7 meters (12 ft) diameter parabolic dish high-gain antenna* to send and receive radio waves via the three Deep Space Network stations on the Earth. Voyager 1 normally transmits data to Earth over Deep Space Network Channel 18, using a frequency of either 2296.481481 MHz or 8420.432097 MHz, while signals from Earth to Voyager are broadcast at 2114.676697 MHz.
As of 2013, signals from Voyager 1 take over 17 hours to reach Earth.

I agree that there are powerful transmitters in the world than what is present in the Voyager 1, but, most of them still remain untested. So, we can be exact with the measurements.


When will all other galaxies become not visible from Earth/Milky way? I know that because of the universe expansion, at some point we will not see any other galaxies from Earth, and we will be able to see only our own galaxy.
Does anyone have an origin or time estimation when will it happen?

That will never happen. We will always be able to see the galaxies in the Local Group. We won't get separated from them by the expansion of space because the group is bound together gravitationally. But the rest of the galaxies will be beyond the cosmic light horizon in 150 billion years or so.
However, because of the mutual gravitational attraction it's likely that the Local Group will merge into a single giant galaxy. It appears that the Milky Way & the Andromeda galaxy are already on a collision course, and will merge in around 4.5 billion years, about half a billion to a billion years before the Sun starts its red giant phase. The full merger of the group will take a long time, Wikipedia says the median time is around 450 billion years.
Due to the Sun's gradually increasing luminosity, the Earth will be uninhabitable in a billion years or so, and it's likely to get swallowed by the red giant Sun at some stage. But I guess humans may survive elsewhere in the Solar System (or outside it).
For further details, please see Wikipedia's Timeline of the far future.


What happens when an Ultra Massive Black Hole cannibalize another? Say hypothetically an UMBH with 50 billion solar masses (supposedly upper limit) is beginning to merge with another (Super)MBH similar to Sagittarius A* that sits at the center of the Milky Way Galaxy, what would happen next in this clash of the titans?

From this article:

  “Bigger black hole masses are in principle possible – for example, a
  hole near the maximum mass could merge with another black hole, and
  the result would be bigger still. But no light would be produced in
  this merger, and the bigger merged black hole could not have a disc of
  gas that would make light.”

The theoretical size limit has to do with the ability to form an accretion disk, not the hole itself.   While it's probably an extremely rare occurrence due to space expansion, if a 50 billion solar mass black hole was to merge with another 50 billion solar mass black hole, you'd have a 100 billion solar mass black hole.    There might be some interesting spiraling in towards each other and some gravitational waves made in the process.   It would be cool to study, but in such a scenario, there's no force that would prevent the two black holes from merging into a bigger black hole.


How far away are objects whose light will never reach us, because of the expansion of the universe? I thought I had read this question on Stack Exchange before, but now I cannot find it...   In fact, I thought I had posted this question before, somewhere, on Stack Exchange...
I believe the answer was about 62 billion light years...
In other words, if an object is currently 62 billion light years away, or more, the light it is emitting right now will never reach Earth...
Is there a term for this 'line' or 'sphere'?

There are several different quantities of this sort that you can define, and the definitions are fairly confusing. Hopefully the following diagram will make things clearer.
                   Z                      <- future infinity
                  / \
                 /   \
                /     \
      D   C    B   A   B    C   D         <- now
      .   .   /   / \   \   .   .
      .   .  /   /   \   \  .   .
      .   . /   /     \   \ .   .
      .   ./   /       \   \.   .
      .   /   /         \   \   .
      .  /.  /           \  .\  .
      . / . /             \ . \ .
      ./  ./               \.  \.
~~~~~~d~~~c~~~~~~~~a~~~~~~~~c~~~d~~~~~~   <- last scattering (universe opaque below this)

The horizontal axis on this diagram is "comoving distance", with respect to which objects that move with the Hubble flow are at rest. The vertical axis is "conformal time", with respect to which light travels along diagonal lines of a constant slope (in combination with the comoving distance).
A is our current location; Z is the ultimate location in the far future of the same matter, supposing it doesn't deviate much from the Hubble flow.
Light emitted in the current cosmological era from closer than B will reach us at some point in the future. Light emitted in the current era from farther than B will never reach us (if we don't deviate too far from the Hubble flow). The distance from A to B is around 16 Gly (billion light years). This seems like the closest match to the quantity you were asking about in your question. I'm not sure that it has a name.
Light emitted at the last scattering time from c is just reaching us now. This is the cosmic microwave background radiation. The distance in the present era from us to the presumed current location of the matter than emitted that light (if it hasn't deviated too much from the Hubble flow) is the distance from A to C. That distance is around 46 Gly. This is what's normally called the radius of the observable universe. Note that the distance from a to c is around 1100 times smaller, or 42 Mly.
Light emitted at the last scattering time from closer than d will eventually reach us. Light emitted at the last scattering time from farther than d never will (if we don't deviate from the Hubble flow). The present-day distance from A to D, which is the extrapolated current location of that matter supposing it doesn't deviate from the Hubble flow, is around 62 Gly. Again, the distance from a to d is around 1100 times smaller. (And the distance at future infinity is ∞ times larger.)
So your statement that "if an object is currently 62 billion light years away, or more, the light it is emitting right now will never reach Earth" is incorrect, but it would be correct if you replaced 62 by 16, or if you replaced "is emitting right now" by "emitted at the last scattering time, supposing it didn't deviate too much from the Hubble flow". Also, you should probably replace "Earth" by "whatever matter is left at Earth's location in the distant future, where 'location' is defined by the Hubble flow".


Water Ice on Meteorites I was reading an article by David O'Brien et al. where it stated

  "in meteorites only the water bound to the silicates can be found, all the water ice having been lost, whereas on asteroids water ice itself has been detected"

  https://www.sciencedirect.com/science/article/pii/S0019103514002620

Now this was published in 2014, so I'm wondering what is the validity of this statement today. Have we, since the publication of this article, discovered water ice on meteorites?

It looks to me that you probably have some misunderstanding of the term meteorite. A meteorite is a remnant body which has reached the surface of a major or minor body (a planet, a moon, an asteroid, etc.). The meteorite's parent body (a meteoroid) would definitely lose its entire water ice (if it had any at all) no later than during its passage through the Earth's atmosphere. That's why no water ice was or will ever be found in meteorites.

So, unless you compare common meteorite impacts with hazardous asteroid impacts, these two body types are incomparable as water ice carriers. And thus, the statement by O'Brien et al still stands firm and will do so ever.


Free neutrons and stellar nucleosynthesis Both the R-process and the S-process of stellar nucleosynthesis rely on absorbing 1 (S) or more (R) neutrons. Where do they come from? I thought there are no free neutrons in stars. (Also the half-life of a free neutron is only 15 minutes, so the source has to be inside the star)

I have yet to find good information regarding the s-process, so I'll just talk about the r-process here.

The key difference between the two is the conditions under which they take place. The r-process takes place during supernovae, specifically those due to gravitational collapse. In a collapsing star, many of the protons and electrons are squeezed tightly together - so tightly, in fact, that a process called neutronization begins. The protons and electrons actually combine to form neutrons (and neutrinos). This creates a large neutron flux, and so there are now lots of neutrons available for the r-process.

None of this lasts incredibly long, though, because the "explosion" doesn't last long. It soon propels much of the outer material of the star out into space, and the remains become either a neutron star or a black hole. That said, an interesting follow-up might address the relationship between the r-process, neutronization and the formation of neutrons stars in more detail.

Sources:

https://en.wikipedia.org/wiki/R-process

http://www.ucolick.org/~bolte/AY4_00/week8/SNeII.html


How did TESS scientists choose which stars to observe at a "2 minute cadence" in each field? The BBC news podcast Science in Action episode New Planet Hunting Mission covers the upcoming TESS mission between 01:00 and 07:15 with Sara Seager, TESS deputy director of science, Massachusetts Institute of Technology, Cambridge, Massachusetts.

After about 05:00 Dr. Seager mentions that the TESS spacecraft will measure a special set of order 20,000 stars in each camera's 24x24 degree field at a two minute cadence.

How were the stars in each field selected for TESS?

The excellent YouTube video Transiting Exoplanet Survey Satellite (TESS) nicely explains how TESS operates and will slowly tile the celestial sphere.

below: From see TESS Operations. Click for full size.

A combination of algorithm and by committee.

The majority of targets are bright, low-mass dwarf stars. The primary goal of TESS is to discover planets by transit, and these targets offer the best chance of that. The way this is done is described by Stassun et al. (2017). This paper summarises how an input catalogue is constructed (from many data sources) and used to estimate a luminosity, distance, temperature, mass and radius for each star. The catalogue is substantially complete for stars capable of being observed by TESS, since it is based on the 2MASS all-sky catalogue.

The candidate target list is designed to meet the priorities of the NASA TESS mission - to find small (rocky) planets in $<10$ day orbits using the transit technique. A subset of targets (near the ecliptic poles) are observed for longer, and there, there is a possibility of finding small planets in longer orbits - up to 120 days. The steps that are taken to prioritise this list are (i) remove giants -- too big for appreciable transits to be detected; (ii) avoid crowded sources -- the TESS point spread function is quite large; (iii) the prioritisation of small stars (which give a large transit signal) and of bright stars (which yield more signal-to-noise in the light curves). The final priorities are artificially boosted for those stars near the ecliptic poles  (the continuous viewing zone) and reduced for stars near the galactic plane, where the de-reddening calculations are more uncertain.

However, space in the target list was also reserved for targets selected by the "community".
There were a number of working groups setup (by science area - I was in the open star clusters group).

There was then an open call for proposals last September to bid for which stars to include in the 2-minute cadence list. The bids were mainly from the working groups. A committee (of astronomers) then met to decide which bids/targets were meritorious.


Why the chosen orientation for the M87 black hole image? By now we are all familiar with the official image of the black hole in M87:

Now, in the universe, as far as I know, there is no up and down, nor left and right. All is relative. Hence my question: why the particular chosen angle of the image released? Why not instead release the image as, for example, below? Which is the point of reference chosen in the original image? Is it of any particular significance?

North is to the top and East is to the left. I believe it is a standard image in celestial coordinates.

It is important to standardise so that the image can be compared with large scale features like the M87 jet.


Can the electromagnetic counterpart of a gravitational be seen by 'small' telescopes? As follow-up question for my previous question I'm thinking about the search for the electromagnetic (EM) counterpart of a gravitational wave (GW). Is it possible to search for those events with 'normal' telescope, the ones (professional) amateur astronomers have? What I mean with a (professional) amateur astronomer is: a telescope with an aperture of 10" or more (remote) and the ability to do long exposures.
Is that equipment capable of detecting the EM counterpart of a gravitational wave (in theory)? Thanks in advance!
EDIT: I'm not sure if there is an EM counterpart in the visible light at all. Is this the case? Can such an event, that triggers a detectable GW, been seen in the visible light (by extension: with 'amateur' telescopes as stated above)?

Unlikely.
The Neutron star merger GW170817 had an optical counterpart (SSS 17a) The optical magnitude peaked at about +18, this is beyond the abilities of most amateur set ups.
Black hole mergers may be intrinsically more energetic, but less bright in the electromagnetic spectrum, as the black holes don't have a surface to interact electromagnetically.
However each gravitational wave detection is examined to see if there is a corresponding gamma ray burst, or optical transient.  Combining results of gravitational, neutrino and optical observations has been called "multi messenger astronomy".


Is Earth's true anomaly roughly 1 degree currently? I wrote software which uses GMT real time to calculate Earth's mean, eccentric, and true anomaly. I encountered a bug where after reaching 360 degrees, instead of flipping back to zero, it subtracts from 360.

So I checked Wolfram Alpha with search for Earth's true anomaly, just to see if we had passed through 0 degrees triggering the bug. As of a few days ago Wolfram and I were in agreement, approaching 360 degrees. But now Wolfram reads 179 degrees approx. I remember passing 180 in July or August.

So it appears Wolfram is also experiencing an error. To double check, true anomaly is the angle between Earth and perihelion side of major axis. We know Earth reaches its closest point to the Sun in the northern hemisphere winter, January roughly. And aphelion in summer months.

Additionally the j2000 epoch I uses 358 degrees roughly, putting the perihelion or closest approach right around early january roughly.

Thus I can only conclude Wolfram is in error as well?

I've found an almanac for Earth perihelion up to the year 2100. 2019 perihelion occurred at roughly 5:00 January 3, 2019. Thus the angle now of true anomaly is some extent past 0 degrees. And Wolfram Alpha is wrong by negative 180 degrees. By Wikipedia and other standard definitions of mean anomaly which have persisted since Kepler's time.


Are there more images with different filters available for the April 6 Hubble image of Jupiter? This recent HST image of Jupiter near opposition is described here and here and was made under HST proposal 14756. It was taken with WFC3/UVIS. It is made from three images using narrow-band filters at 395, 502, and 631 nanometers.

I am wondering if there were additional images recorded with filters at other wavelengths. If so, how could I go about finding them?

As those observations are from only a few days ago, I think the data will still be embargoed, usually one year, until it's publicly available.

To confirm this, I've looked at the Hubble Legacy archive, where Hubble data is made publicly available. When you look there, the last public data for Jupiter is from February 2016.
Although I couldn't use the retrieved data from 2016, as it's not the FITS format, but it downloaded something.

So either you'll have to wait it seems, or you contact the project PI and ask nicely ;)

In the HST proposal page from the MAST archive (that links to the legacy archive) you see the photos they've taken listed, but the data is not available, I guess due to embargo.

Edit:
However I'm not sure if the legacy archive has all the data, as there's one dataset from 2011 with which I've worked that should be there, and I can't find it. Weird.


Altitude of Midnight Satellite? The other night, I observed what I believe must have been a satellite, however it was at 1:30am so I was not sure that this would be possible. Being in the Southern Hemisphere we are just past Summer Solstice, so theoretically it could have been reflecting the sun's light.

There were no flashing lights, just a star-like point of light. As it slowly traveled in a NE direction across the sky, it began to gradually fade as typical of a satellite passing into Earth's shadow.

Based on my observation, I'd like to get an understanding of the approximate minimum altitude of the object that I observed, allowing for it to be out of the shadow of the Earth sunlight. I'm not a mathematician, but using Space Engine with the correct time & location, I had to position the camera to some 7000km in altitude before the sun was visible over the horizon. The object was N to NE from where I observed it, whist the midnight sun would have been shining away at the South Pole.

Date/Time: 27-Dec-2017, ~01:30 local time (GMT + 8)
Location of observation: Perth, Western Australia (~32S, 116E)
Position in sky : ~ 80 to 60 degrees elevation, moving north due northeast
Duration of observation : ~2 to 3 minutes

I'll venture a guess that the 7000 kilometers (your software's camera position) was with respect to the center of the Earth. Using 6371 km as a value for the Earth's radius, that would make the altitude only 629 km, which it turns out is quite a popular altitude for satellites in LEO.

For the simplifying case of midnight, one can draw a simple diagram. I get

$$h = R_{\textrm{Earth}} \left(\frac{1}{\cos(35°)} - 1\right)$$

which is about 1400 km at midnight, and the height will be lower before and after.

I'll wait to hear about the camera "height" and try to find another tool for you to use. Do you ever use Python programming language?

Red dots are Perth at local noon and midnight.


Is there a formula to calculate any star mass from Luminosity, Radius, and/or Temperature (K)? I know there is the Mass-Luminosity relationship, but I am wondering if there is a more accurate formula I can use based on the data points I have generated:

Absolute Magnitude (based on class and type of star in relation to H.R. diagram)
Temperature (same as above; in Kelvins)
Luminosity (calculated from Absolute Magnitude)
Radius (calculated from Temperature and Luminosity)

Basically the last 2 pieces I need to calculate are Gravity and Mass which is a pain because the one formula I have requires one of them to find the other. So far everything is calculated from another piece and so all of my values are in line with each other.
I could use the Mass-Luminosity relationship if I must, but are there any other formulas? I couldn't find anything that didn't require gravity. I doubt it, but there isn't by chance something to calculate gravity from my data points is there?

I think the issue you are going to have here is that the position of a star in the HR diagram - which amounts to saying what the (2 dimensional) spectral type is - does not just depend on its mass.
The Vogt-Russell theorem says that the position of a star on the HR diagram depends on its mass, its composition and crucially, how its chemical elements are distributed in the interior. Thus we know for example that we could find a 2 solar mass star on the main sequence with a spectral type of say A2V but it is also possible for a K0III giant to be a star of 2 solar masses.
So there really isn't any shortcut or quick formula that is going to give you a route to finding the mass based on a spectral type (or equivalently, based on a luminosity and effective temperature).
If you know the star is on the main sequence then you can use an approximate luminosity-mass relation (although you should be aware that stars change their luminosity during their main sequence lifetime and so the luminosity of a main sequence star depends on its mass, age and its initial chemical composition).
If your star is a giant or at some intermediate evolutionary stage (subgiant, or some short-lived supergiant phase) then it is really difficult to estimate the mass of such an object without also knowing its age. For example if you know a giant star is a billion years old, then there is only a small range of masses which will be in the relatively short-lived giant phase at 1 billion years old.


Is an ice planet made only by Ice? Is an ice planet made only by ice?
What causes the water to conglomerate and form ice in a planet?

Even if the planet is not made entirely by ice but has an high % of ice (e.g. 50% ice), how can it be explained?

I heard about ice planets with surface temperature way above the melting point.
Can it be explained with surface pressure?
Edit: This is probably wrong but feel free to point it out in a full answer

Thanks.

There are several compounds referred to as 'ices' in the solar system, these are mostly hydrogen-based compounds such as water, ammonia and methane (and others) accreted around what is believed to be a rocky core, these compounds are important, as they comprise a great proportion of the two 'Ice Giants' observed in our Solar System (Uranus and Neptune), as seen in the diagram below, compared to the Gas Giants - Jupiter and Saturn:

Image Source: Wikipedia

This answer will just mainly focus on the Ice Giants. as discussion of these answer both your questions regarding the ice planets and what is sometimes described as ices existing at high temperatures.

In terms of planets with proportionally large amounts of ice, as mentioned in the comments, there is a dividing line when the solar system formed known as the Frost Line, illustrated in the diagram below:

Image Source

Inside the Frost Line, the hydrogen compounds remain vapourised, beyond the line, hydrogen compound ices can condense and solidify, as seen for Uranus and Neptune (and an array of dwarf planets, moons and other celestial objects), these 'ices' comprise the mantle of both planets. An important distinction about what these "ices" are is explained in the paper "The Interior Structure, Composition, and Evolution of Giant Planets" (Fortney and Nettelmann, 2009) as being:

  The label ice refers to a mixture of $H_2O$, $CH_4$, and $NH_3$ that are supposed to have been in an ice phase during protoplanetary core formation.

The compounds in the mantle in the Ice Giants are under immense pressure (in excess of 10 GPa), far in excess of those existing on Earth. Greater pressures change the properties of the hydrogen-based ices - so it is one of the reasons that it is sometimes referred to as 'hot ice', according to the AstroTech Gallery page 'Ice Giant', describe a model of the Ice Giant planets as being:

  Much of this mass (perhaps about 65%) is made up of hot ices, which despite being very hot are probably
  kept at least partially solid by the enormous pressures within these massive planets.

The effect of pressure on water (as an example) can be seen in the example phase diagram below (of course there would be differences depending on the actual nature of the mixtures and compounds present):

Image Source: Duke University

However, according to the article The gas (and ice) giant Neptune, the present icy (or 'hot ice') mantle of Neptune is described as:

  The mantle is equivalent to 10 – 15 Earth masses and is rich in water, ammonia and methane. This mixture is referred to as icy even though it is a hot, dense fluid, and is sometimes called a "water-ammonia ocean".

A similar situation is hypothesised to exist for Uranus.

It is known that ice exist on Mercury, Earth, Moon and Mars, but these comprise a small proportion compared to the planet itself, so not included in this answer.


What is meant by "revolution of Sun's apogee"? While reading Translation of the Surya Siddhanta and of the Siddhanta Siromani(1861) - Bapu Deva Shastri, I stuck in the follwing.
Chapter I, verse 41 & 42

In a KALPA (4.32 billion years), there are 387 revolutions of the Sun's Apogee (about the earth), 204 of Mars' apogee, 368 of Mercury's apogee, 900 of Jupiter's apogee, 535 of Venus' apogee and 39 of Saturn's apogee.

Now my question is what is meant by 387 revolution of Sun's apogee? It will be helpful if someone can explain this.
This is the same question I have asked here, but as not getting proper explanation I am posting here too.

The apogee is the direction of the sun when it is at greatest distance from the Earth.  We would normally call this the aphelion because we use a heliocentric model. The sanskrit doesn't have the precise terms for apogee and aphelion, it uses a word meaning "greatest height".  The translator has converted the notion of "greatest height of the sun above the Earth" as "apogee"
This direction changes (relative to the stars) mostly as a result of perturbations of Venus and Jupiter (though the cause could not have been known to ancient authors without a model of gravity).  Currently, aphelion (or apogee from the perspective of the sun) occurs on about July 4th, but this changes slowly.
It takes about 112000 years for the aphelion to revolve by 360 degrees. The calculation you quote gives about 10000000 years for a revolution. What are we to make of these numbers?
Measurement of the rotation of the aphelion is a hard thing to observe.  It is harder than the observation of the precession of the equinoxes, which was well established by the time that the Surya Siddhanta was composed. And the value quoted here is incorrect by a factor of about  100. Apsidal precession could not have been observed at this level, and without a theory of gravity and calculus, it could not have been calculated. It is therefore reasonable to suppose that the value here is found by analogy with the apsidal revolutions of the moon (which were observed in India) although exactly how this analogy was constructed is not explicit in the text.
We should also note that the model of planetary motion was not base on ellipses, but on circles with the Earth placed off-centre and the sun moving around it. In reading ancient accounts of science, we need to put ourselves into an ancient mode of thought, even if the translation uses modern terminology.


In the end what is the ultimate matter/element in the universe, due to fusion process in stars? We know that, in stars, hydrogen is used by the stars and due to the fusion in two hydrogen elements we get helium -> Carbon -> Oxygen ->Other elements -> Iron -> then Super/Hypernova.

If everywhere hydrogen is getting converted to other element then will we(universe) have hydrogen to create/retain stars ?

Or in the end what is the ultimate matter in the universe ?

If we don't wait for too long, where we don't know for sure, whether protons or atomic nuclei stay stable forever, there won't be just one single element.

The most abundant element in the universe will probably stay hydrogen, since it will stay in the intergalactic medium, and thin out by accelerating cosmic expansion.

The second-most abundant element will probably stay helium, mostly for the same reason.

The most frequent stars are probably red dwarfs.
They'll burn hydrogen and helium to oxygen and carbon, and end up as white dwarfs.

Only a small fraction of stars is large enough to burn oxygen and carbon to silicon and iron.

Very heavy elements, like uranium and beyond, as formed in supernovae, will decay to lead, helium (alpha particles), some hydrogen (via neutrons decaying to hydrogen), and a couple of less abundant elements formed by spontaneous fission. Other stable (isotopes of) elements heavier than iron are also formed in, or a consequence of, supernova explosions.

Part of the dead stars will collide, and merge to heavier stars, which will end as intergalactic dust (mixture of elements, e.g. iron and silicon), including planets, after supernova explosions, and as Black Holes. Others will be scattered out of reach of the supermassive black hole in the center of the galaxy (gravitational relaxation).


Is Titan Better or Mars? Isn't Titan better to colonize than Mars? It has a thick atmosphere which is made up of nitrogen like our planet's. It doesn't have a magnetic field, but it does get a type of protection from Saturn. Maybe the only downside I see is that it is too far away. While Mars has neither a thick atmosphere nor a magnetic field, it is bombarded with solar radiation every day.

We have no practical means to travel to either planet and set up even a base, let alone a colony.  As such neither is better.

Also note that we have no reason to do it and we would need an extremely good reason because it would almost certainly involved the expenditure of vast resources to even set up a base, let alone a viable colony.

In a practical sense neither have any real human habitability options other that entirely sealed bases.

However current knowledge suggests that the geology on Titan would be really challenging.  We'd be operating a base at significantly higher temperatures than the local norms and as such we'd always be subject to problems in a atmosphere with active weather.  The only attraction on Titan is studying the locale, but we'd have done that in enormous depth before we could even consider a colony, so it's a mute point.

Mars is closer, less hostile and more familiar.  We'd go there before Titan for the simple reasons it's easier, cheaper and more likely to succeed.  But again, why ?

But we'd be unlikely to do either before we'd establish a viable self-sustaining Moon base of at least "colony" status.  Again that's not even remotely possible and would not be possible without a unified global effort.  Well this particular globe can't even agree the wording of one sentence in a document about it's own report on climate change.  They can't even agree about one lousy word !  So no one is going to even start talking about the possibility of setting up a committee to discuss the likelihood of setting  up another committee to start planing for a colonization effort research program.

In short, even if we solved the technical problems to make it worth deciding which place to colonize, we have no political chance in Hell of actually doing it.  Humanity will die squabbling in a committee room.  No colonists required until we replace humans with something that works.  Skynet anyone ? :-)


Fate of helium moving toward a singularity On the molecular level, if an atom, consider helium, were to fall into a black hole, what would be the ultimate fate of its electrons, protons and neutrons?

I'll take the reference of an observer falling with the helium atom. At the Event horizon, the tidal force across a helium atom is still minute, $10^{-29}N$ much less than the Coulomb forces between the electrons and nucleus, still less than the strong forces inside the nucleus. So the atom will fall through the event horizon without damage. (The region around the black hole can be a pretty violent place, with lots of hard X rays from the superheated matter, which would likely have ionised the atom, but I'll suppose it somehow manages to avoid this)

As it falls past the event horizon it is doomed. Think of the singularity less of a place in the middle of the black hole, and more as a point in time. It is inevitable our He atom will reach that point. As it approaches it the tidal forces will grow larger, eventually sufficient pull the atom apart. To describe what happens now would require a good description of quantum gravity, which we don't have.

The is no physics that describes what happens "after" the singularity. Think of the singularity as a point in time, but for the atom it is also the end of time for the atom. It is possible that something beyond our current understanding prevents a singularity from forming, but that is speculation.

This is explained diagramatically in a Penrose diagram


How do we measure distance in space? I know that you can calculate the distance to another star for example, by using the phenomenon redshift to determine how much it has redshifted because of the expansion of the universe. But how do you know what the original frequency/wavelength of the light was? (And if all of my above statements are wrong, how do you measure the distance?)

Red shift is used for measuring the distance to very distant stars (galaxies mostly, in fact). The secret is to use spectral lines. Specific elements when very hot emit light at very specific colours and you can spot the pattern of those colours when the whole pattern has been shifted towards the red and see how far it has shifted.

For closer stars, there are other techniques, using types of star whose brightness can be worked out in some way, or, for the closest, by observing how the position of the star compared to more distant ones changes as the Earth moves from one side of the Sun to the other.


Does the sun cross other spiral arms in its movement around the galaxy's center? Today, Ansa.it released an article that states:

  [...]. In questo suo peregrinare galattico, il Sole ha attraversato anche
  i due bracci della Via Lattea Perseo e Centauro. "Sono zone di alta
  densità stellare, in corrispondenza delle quali il Sole e le stelle
  intorno rallentano e possono anche fermarsi, [...]

(my) translation to English:

  In this galactic wandering, the sun traversed the two spiral arms of the Milky Way, Perseus and Centaurus.
  These are regions of high density of stars, in correspondence of which
  the Sun and other surrounding stars slow down and can even stop, [...]

The article is complemented with a suggestive video-clip that depicts the Sun orbiting around the galaxy's center and crossing various spiral arms.

Given my lack of an education on the topic, I cannot reconcile the existence of an entity called "spiral arm" with the notion that a star can freely cross several "spiral arms" in its movement around the galaxy's center.

My intuition is that if every star within the Milky Way could freely cross the "spiral arms" several times during their life-span, then there would be no such thing as a "spiral arm" at all (because this is supposed to be a mass grouping wherein all matter moves --please forgive me the abuse of the word-- "together").

Where am I wrong?
Is the above description of the movement of the Sun accurate?
In case of an affirmative answer, is it a very special case or is it a defining characteristic of the entire Orion's arm?
In the latter case, can other spiral arms cross each other?

What is a spiral arm?

The reason that the Sun, in principle (but see below), may cross spiral arms is that galactic spiral arms are not rigid entities consisting of some particular stars; rather they are "waves" with a temporary increase in density. An often-used analogy is the pile-up of cars behind a slow-moving truck: At all times, all cars are moving forward, but for a while, a car behind the truck will be moving slow, until it overtakes and speeds up.

Similarly, stars may overtake, or be overtaken by, the spiral arms. Inside a certain distance from the center of the galaxy called the corotation radius ($R_\mathrm{c}$) stars move faster than the arms, while outside they move slower.

Since the stars and the interstellar gas follow the rotation of the arms for a while once they're inside, the density of the arms is higher than outside, but only by a factor of a few (e.g.
Rix & Rieke 1993).

When interstellar gas falls into the potential well it is compressed, triggering star formation. Since the most luminous stars burn their fuel fast, they will mostly have died once they leave an arm. Hence, what we see as spiral arms is not so much the extra stars, but mostly due to the light from the youngest stars which are still inside the arms. Since most luminous also means hottest, their light peaks in the bluish region — hence spiral arms appear blue.

Origin of the spiral structure

At least the most prominent spiral arms (especially grand designs) are thought to be created by these long-lived, quasi-stationary density waves (Lin & Shu 1964). The reason that the density waves exist in the first place is not well-understood, I think, but may have to do with anisotropic gravitational potentials and/or tidal forces from nearby galaxies (e.g. Semczuk et al. 2017).

But in fact even small perturbations may spawn gravitational instabilities that propagate as density waves. In computer simulations of galaxy formation, even numerical instabilities may cause this, so the fact that your simulated galaxy has spiral arms doesn't necessarily mean that you got your physics right.

When the luminous and hence massive stars die, they explode as supernovae. The feedback from this process, as well as that exerted by the radiation pressure before they die, may help maintaining the density waves, at least in flocculent galaxies (Mueller & Arnett 1972). Perhaps this so-called Stochastic Self-Propagating Star Formation may also initiate the density waves (see discussion in Aschwanden et al. 2018).

The rotation speed of the material in the galactic disk is roughly constant with distance from the center (this is mainly due to the dark matter halo hosting a galaxy). Hence, stars close to the center complete a revolution faster than those farther away. In contrast, the spiral pattern rotate more like a rigid disk such that, in an intertial frame, the pattern can be described by a constant angular speed $\Omega_\mathrm{p}$ throughout the disk. However, note that spiral arms are transient phenomena; they appear and disappear with lifetimes of the order of (a few) Myr (e.g. Grand et al. 2012; 2014). Sometimes you also see multiple spiral patterns propagating with different velocities.

The Sun in the Milky Way

Note: This section first contained errors based on dubious values for the angular speed of the spiral arms, as pointed out by @PeterErwin and @eagle275.

In the case of our Sun, we happen to be located very near the corotational radius $R_\mathrm{c}$; we sit at a distance of $R_0 = 8.32\,\mathrm{kpc}$ from the center of the galaxy (Gillessen et al. 2017), while
$R_\mathrm{c} = 8.51\,\mathrm{kpc}$ (Dias et al. 2019).

Using Gaia data, Dias et al. (2019) find a pattern angular speed $\Omega_\mathrm{p} = 28.2 \pm 2.1\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{kpc}^{-1}$. At the location of the Sun ($R_0$) this implies a pattern speed of $\simeq235\,\mathrm{km}\,\mathrm{s}^{-1}$. This is only a little bit slower than (and in fact statistically consistent with) the Sun's velocity of $239\pm5\,\mathrm{km}\,\mathrm{s}^{-1}$ (Planck Collaboration et al. 2018). If the spiral arms were "permanent"*, the timescale for the Sun crossing an arm would hence be gigayears.

However, because as described above they're quite transient, we might once in a while overtake a spiral arm. I haven't been able to find firm evidence for whether we have or haven't crossed any arms; Gies & Helsel (2005) argue that we have crossed an arm four times within the last 500 Myr, but base this on matching glaciation epochs with passages through spiral arms (and admit that this requires a lower but still acceptable pattern speed).

The article you link to

I now wrote to Jesse Christiansen (who the linked article quotes) and asked her if she knows whether or not we are moving in and out of spiral arms; she replied within roughly 8 seconds, tagging Karen Masters who chimed in even faster — they both agree that this is an ongoing debate with no conclusive evidence.

Anyway, the article seems to have misunderstood the tweet from Jesse Christiansen. In her animation she shows the journey of the Sun, but shows the galaxy itself as being static, which she did on purpose to keep it simple. Hence, you see the Sun traversing the arms unnaturally fast.


What is the farthest point of light that is emitted by a torch? If I focus my torch into the night sky,what would be the farthest point the light can reach? Recently I looked at the night sky and noticed a light spark moving around the clouds for about 15 min.After observation I jumped into a conclusion that its probably because of the light house near by or any powerful light emitter.

If you point a flashlight up, then the light will reach as far as the other ends of the universe, if one has a sensitive enough measuring device, so long as it is not absorbed. Of course, said device might require a lot of work, and a huge telescope. Lights can easily be seen by satellites, but it would take a huge telescope to see a single light from, say, the Moon, and even larger the further you go.

In the case of your moving light, I would suspect a laser of some kind. Lasers are focused enough where one could easily see a reflection from the clouds, and pretty focused as well. It could be a moving light source, such as a plane or balloon of some kind as well.


V838 Monocerotis "light-echo" images morphed into nice video, but why so few original images? This NYTimes video shows what I think is a series of real Hubble images which have been "morphed" artistically to appear to be continuously expanding. (also see The V838 Monocerotis Star Still Has Astronomers’ Heads Exploding) If I understand correctly it is the light-echo, not actual material that is doing the expanding at this rate. They also give four discrete images available in 2006 (shown below) and you can find a link to a video there which also seems to be made of just four images.

This is (very roughly speaking) tomography of the cloud - the parabolic "shell of reflection" expands and each image highlights a view of the material in that shell. That seems to be what the diagram illustrates.

my question is: why weren't there more images taken of this - technically speaking - awesome event, enough to make an animation without the artistic "morphing"?

note: I'm not asking "why do you think there weren't...". I'm looking for the actual reason. Are these "found images" - nobody thought of doing it and later just pulled these from archives, or where these thought to be sufficient? Are there more images available?

Here are some links to video I have found

YouTube:
"Hubble: Timelapse of V838 Monocerotis (2002-2006)"

https://www.youtube.com/watch?v=U1fvMSs9cps

...which can also be seen at spacetelescope.org

http://www.spacetelescope.org/videos/heic0617a/

... and this Wikipedia article links to this video

https://upload.wikimedia.org/wikipedia/commons/1/1b/Evolution_of_the_light_echo_around_V838_Monocerotis_%28Heic0617a%29.ogv

and here are some "stills"

Images from NYTimes

To get time on the Hubble Space Telescope is a difficult, peer-reviewed process. The time is usually about ten times oversubscribed. The scientific justification to get more images must be more than "it would make a nice movie".

The submission of proposals for HST time is a confidential process; so I don't know whether the principal investigators for these images asked for more at different epochs and were turned down, or whether they thought this many images were sufficient for the science they wanted to accomplish.

There is no technical reason why they couldn't have obtained more images. In fact more images were obtained (see this link to the Hubble archive) at later epochs between 2005 and 2012. A problem I guess with incorporating these images into the movie sequence were that the light echo gets fainter, more diffuse and starts to exceed the size of the detector, which would necessitate more than one pointing. They were also taken with a different camera, which might raise aesthetic compatibility problems. I show below a fairly crude colour composite I extracted from the archive for an observation in 2007.


How to separate binary and optical stellar pairs? I have list of stellar pairs from The Washington Double Star Catalog (WDS), i.e. I have next parameters:

angular separations
position angles
stellar magnitudes of both components

Additionally I have GAIA DR2 parallaxes (and other parameters) from SIMBAD database. How to distinguish between binary and optical stellar pairs?

If they are both in the Gaia DR2 catalogue, then a physical binary should have components with similar parallax and proper motion.


Why nuclear fusion is a controlled process in stars? Nuclear fusion is the main way for stars to generate energy. But we know that nuclear fission and fusion are not controlled, then it causes chain reaction and sudden explosion. But why in the case of stars it does not happen? I mean, that nuclear fusion happen in stars but it take few billion years to complete the amount of hydrogen. Is there any controlling mechanism? If no, what happens here?

Because in most stars, the pressure where nuclear burning takes place is proportional to temperature. In order for the nuclear reaction rate to increase, the temperature and hence pressure must increase. This would cause the gas to push out the layers above it, then to expand and cool.

Conversely, if the nuclear reaction rate fell, so would the temperature and pressure. The smaller pressure would allow the outer layers to move inwards, compressing the gas, heating it up and speeding up the nuclear reactions.

In this way the nuclear reaction rate (for hydrogen burning) can be kept almost constant. The small changes (increase) that occur during a star's main sequence life are driven by the increasing mean atomic mass as hydrogen turns to helium. This means you need a gradually higher temperature to maintain the same pressure.

Where the temperature and pressure are decoupled - for instance in the degenerate helium core of a low(ish) mass star near the red giant tip, where electron degeneracy pressure is independent of temperature, then runaway nuclear reactions can be initiated - the "helium flash".


I need help getting mars to look like a circle I am having a lot of trouble getting a sharp image in my telescope. It is a skywatcher with a focal length of 1200mm and a diameter of 254mm. I have a laser colimater that I got for Christmas, so I have been using that but when I try to look at mars I see three mars in an overlapping triangle. I am not sure what to do. I have made sure that the laser is straight by 1. spinning it in the eyepiece holder and the laser stayed in the same spot 2. zooming in and out and again the laser stayed in the same place. the laser seems to be centered so I put it in the middle of the doughnut on the mirror and the laser bounces back into the middle of the colimating lens, but when I look at mars there is still three and there is no detail its just a red fuzz ball. I also checked the collimation of the telescope by zooming all the  way out to see where the shadow was and it seemed to be right in the middle.hopefully this makes sense, but let me know if I should say something in a different way. thank you In advance.

I have made sure that the laser is straight by 1. spinning it in the
  eyepiece holder and the laser stayed in the same spot 2. zooming in
  and out and again the laser stayed in the same place. the laser seems
  to be centered so I put it in the middle of the doughnut on the mirror
  and the laser bounces back into the middle of the colimating lens

Very good. You're doing all the right things. Kudos for taking care of your telescope.

Now, about Mars looking like a fuzzball.

The angular size of Mars right now is 6.5 arcsec. The maximum theoretical resolving power of your scope is 0.4 arcsec. In ideal conditions (assuming everything was perfect, which it never is), you'd get something like 16 "pixels" diameter for the image of Mars in your scope. That, I'm sure you realize, is a very low resolution. No wonder it looks like a fuzzball.

Mars is a tricky target. The best times to observe it is around oppositions, which happen once every two years (plus a few months). Mars is nice and big a few weeks before and after opposition, and otherwise it's pretty small. The last opposition was in July 2018, when it measured 24 arcsec. The next one is in October 2020, and it will measure 22 arcsec.

http://www.nakedeyeplanets.com/mars-oppositions.htm

Stick to a few weeks before and after opposition. In the rest of the time, Mars is just too small.

Another source of problems is seeing - or atmospheric turbulence. When seeing is bad, all the high resolution targets (like all planets and the Moon) look bad. Nothing you can do about it. When seeing is good, your scope can work closer to its theoretical resolving power.

https://en.wikipedia.org/wiki/Astronomical_seeing

To predict seeing, go on the Clear Dark Sky site, choose a location closest to where you live, and check the forecast they give you for that location.

https://www.cleardarksky.com/csk/

The row called Seeing is a seeing forecast. When it's dark blue, seeing is good. When it's white or light blue, seeing is bad. No point in looking at Mars when seeing is bad.

Finally, make sure collimation is actually good. You seem to be doing all the good stuff with the laser collimator. That's great. Now, when seeing is good, point the scope at the North Star, defocus a bit (move the focuser in or out of perfect focus a tiny bit) and look at the star image. It looks like a shooting target, a bunch of concentric rings. If those rings are perfectly round and concentric, your collimation is probably good.

http://www.astrophoto.fr/collim.html

This is called a star test, and it's the most accurate test of collimation. It's also the most difficult. So use your laser collimator on a daily basis, which is much easier, and maybe do the star test once or twice, just to keep your laser honest. Once you're convinced your laser technique works well, there's no need for the star test.

Oh, one more thing. Make sure your scope is at thermal equilibrium with the environment. If you're storing your scope indoors, it's probably way hotter than the cold night air. That will create internal turbulence within the scope. It's basically like bad seeing, except it's all inside your scope.

https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/

https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/

No need to install a fan on the primary (although that would definitely work if you're willing to make the extra effort). Just take the scope outside 1 hour before you begin using it. It will "breathe out" all the heat and it will be close to equilibrium.

Good luck and clear sky!


If oxygen were pumped into space where would it go? If an astronaut went into space with a sealed container of oxygen and opened it up in space and let all the oxygen go, what would happen to it? would it all just dissipate into some other particle or matter or would it just stay there until something consumed it

Gases tries to occupy all the volume of the container in which they are encased in equilibrated pressure. This happens because the gas molecules are always colliding and kicking each other in a 3D high velocity brownian motion. The result is that the gas molecules occupies all the container's volume homogeneously.

So, if the sealed container is opened up in space, the oxygen would quickly leak out, because the molecules near the vacuum edge would be instantaneosuly kicked out by the gas innermost molecules. This would happen rather quickly and in a chained manner, but is dependent on the size of the open hole. Macroscopically, the result would be that all the gas would be running out in a form of a wind/jet out of the open hole into outer space. This would only stop when all the molecules becomes too far apart from each other to have any further intermolecular interaction.

This means that the oxygen (or any other gas) would be dissipated into outer space. The gas molecules would initially be spread out by the intermolecular collisions and then would go outwardly due to the momentum acquired from each individual gas molecule.

When freed in outer space, each molecule would eventually reach something else to interact, so they could be attracted and retained by some gravity well (most likely from a planet or moon), or interact with some magnetosphere (if near a planet) or interact with some other molecule wandering around, or with a cosmic ray or be catched by the solar wind particles (or any of the equivalent if this happens near some other star).

If this happens in intergalactic space where there is no body with appreciable gravity or magnetic field nearby, no stellar wind and the cosmic rays are very sparse and rare and the very few that exists are randomly directed, then the molecules would just spread out in virtually linear trajectories and travel the intergalactic space lonely, quietly and virtually undisturbed for some billions or trillions of years. In their journey, each molecule would only very occasionally interact with some other molecule (likely to be hydrogen) or some other lost particle wandering around.


What determines the configuration of orbits in a binary system? There are essentially two configurations of orbits a system with two bodies can have:

A configuration where the paths do not cross:

A configuation where the paths cross:

Note: Size of the bodies relative to one another is unimportant here.

What determines which of the two configurations a binary system will develop into?

Following the convention $\mathbf{r} = \mathbf{r}_2-\mathbf{r}_1$, $M = m_1+m_2$, in the center-of-mass frame we have, by definition,
$$\begin{eqnarray*}\mathbf{r}_1 = -\frac{m_2}{M}\mathbf{r}\text{,}\quad&\mathbf{r}_2 = \frac{m_1}{M}\mathbf{r}\text{.}\end{eqnarray*}$$
Hence, $\ddot{\mathbf{r}} = -GM\hat{\mathbf{r}}/r^2$ implies that the individual orbits are similar conic sections in this frame, and moreover in the bound case they are ellipses that share a common focus at the center of mass, with all three of the distinct foci being collinear and with the center between them. Although there is a degenerate case of a circular orbit.

Therefore, there's a geometrically simple condition for the intersecting configuration: there is intersection if, and only if, the distance to apoapsis of the larger mass (smaller orbit) is greater than or equal to the distance to periapsis of the smaller mass (larger orbit). Since a general ellipse in polar coordinates about a focus can be described by
$$r = \frac{p}{1+e\cos(\phi-\phi_0)}\text{,}$$
where $e$ is the eccentricity, and the individual orbits are proportional, we have intersection if, and only if,
$$\frac{1-e}{1+e}\leq \frac{m_1}{m_2}\leq\frac{1+e}{1-e}\text{,}$$
as the apsides occur when the cosine term is $\pm 1$.


When the Moon formed and early on, did it orbit over the Earth's equator? Or was it always at about 5.14 degrees inclination or has the inclination changed over time?

See diagram.

James K's answer to this question got me thinking about this, and I don't mean to call him out, but this part of his answer doesn't seem quite right when he wrote:

  Our moon is unique in being close to the plane of the ecliptic, and
  not in the plane of the equator, which suggests its formation was not
  like that of other moons in the solar system.

For starters, I imagine the physics of the initial orbit of an impact moon would be somewhat complicated, but it seems likely, given an angled impact like the one that's believed to have formed our moon (and hitting at an angle is statistically more common than a direct impact anyway),

So, in this scenario, the planet gets significant angular momentum and this angular momentum should dictate the planet's new equator and axial tilt relative to it's orbit.  I would think the moon should form roughly along that same equatorial plane, but I'm just guessing.   Perhaps it could form several degrees off - not sure.

The 2nd point, is, a rapidly rotating planet with a large equatorial bulge and a close moon, if the Moon formed off the equator, would it's orbit migrate over the equator where the gravitation was greatest and would that happen relatively quickly or not at all?

That's basically the question.   Was the Moon always at roughly 6 degrees off the Earth's equator or has it only moved off an orbit over Earth's equator more recently, perhaps due to the gravitational effect of the Sun?

Or are there other factors.   Mars' axis is thought to have changed rather significantly due to Jupiter's gravitational effects perhaps 100,000 years agao and it's Moons orbit over Mars' equator which suggests that Mars' equatorial bulge dragged the Moons with it? - or is my thinking way off on that?

My thinking is that a planet's equatorial bulge would drive moons towards a 0 degree inclination around it's equatorial bulge and our Moon is different because of proximity to the sun which also has a strong gravitational effect.   The 5.14 degrees of inclination is a balance between the Earth's equator and the solar gravity.

Yes, the Moon was likely formed in an orbit over the equator. I'm going to recycle my answer to a different question about the Moon's orbit as most of it is relevant to this one as well.
The main concept here is the Laplace plane, which is the plane about which a satellite's orbit precesses. In general, the inclination of a satellite with respect to this plane gets damped to small values. Close to the planet, the main influence on the Laplace plane is the equatorial bulge of the planet, further away the Sun (and to a lesser extent, the other planets in the Solar System) becomes a more significant influence.
The transition between satellite orbits aligned with the planet's equator and satellite orbits aligned with the planet's orbit occurs around a distance called the Laplace radius $r_{\rm L}$, from Nesvorný et al. (2014), which can be thought of as the scale of a planet's regular satellite system. This distance is given as:
$$r_\mathrm{L}^5 = J_2' R_\mathrm{p}^2 a_\mathrm{p}^3 \left(1-e_\mathrm{p}^2 \right)^{3/2} \frac{M_\mathrm{p}}{M_\odot}$$
where $R_{\rm p}$ is the planetary radius, $a_{\rm p}$ is the planet's semimajor axis, $e_{\rm p}$ is the planet's orbital eccentricity, $M_{\rm p}$ is the mass of the planet and $M_\odot$ is the mass of the Sun.
$J_2'$ is the quadrupole coefficient of the planet $J_2$, adjusted for the presence of $n$ equatorial inner satellites:
$$J_2' R_\mathrm{p}^2 = J_2 R_\mathrm{p}^2 + \frac{1}{2}\sum_{i=1}^n a_i^2 \frac{m_i}{M_\mathrm{p}}$$
In the case of the Moon, there are no inner satellites so $J_2' = J_2$.
The angle between the planet's spin axis and the Laplace plane $\phi$ is given by:
$$\tan 2\phi = \frac{\sin 2\theta}{\cos 2\theta + 2r_\mathrm{L}^5/a^5}$$
Where $\theta$ is the planet's obliquity and $a$ is the semimajor axis of the satellite's orbit. For $a \ll r_{\rm L}$, the denominator becomes very large, resulting in $\phi \approx 0$. For $a \gg r_{\rm L}$, the $2r_\mathrm{L}^5/a^5$ term becomes very small, and the equation becomes $\tan 2\phi \approx \sin 2\theta / \cos 2\theta$, giving $\phi \approx \theta$, which is the behaviour described earlier.
Using the $J_2$ value of Earth of 1.08263×10-3 from the NASA Earth Fact Sheet, the Laplace radius works out to 53600 km, or 8.4 Earth radii. Note that this may not necessarily have been the exact value billions of years ago, due to differences in the Earth's shape caused by rotational spindown, deformations due to ice ages, etc.
The Moon likely formed closer to the Earth than this distance, so would have been aligned with the equator. As it migrated outwards across the transition region, the orbit would have undergone rapid changes in orientation, which would also have an effect on the Earth's obliquity, potentially allowing the Earth to transition from a high-obliquity state which might explain the Faint Young Sun paradox (Jenkins, 2000) to the less-severe obliquity we observe today.
Sarah Stewart-Mukhopadhyay created a video of a simulation of the migration of the Moon across the transition region, with the Earth initially in a high-obliquity state.


Are wormholes even possible? While this question could fall under Space Exploration or Physics (I don't have the "best" reputation on THAT forum), I feel that it is very much related to how the universe itself functions, especially regarding Einstein. While looking to see if anyone has confirmed our universe having a 4th spatial dimension (or some non temporal or spatial dimension) due to the fact that space curves, for me curvature requires a higher dimension (paper bending or a stick bending, always +1 to its dimension), I was met with the explanation that this "curvature" was intrinsic and not extrinsic (requiring another dimension). That was fine and all until I realized that wormholes would be exclusively extrinsic, how could you make a wormhole if bending space is intrinsic? Obviously you can't. But Einstein  put a lot of time into something that apparently wouldn't work, so personally I am siding with Einstein here. But that still leaves the question; Are wormholes possible thus making our universe a 3+1+1 dimensional universe (I wouldn't group it up with the 3 because that would say that matter could move like that, it can't) or are wormholes impossible and so our universe remains 3+1 dimensional?

The idea of wormholes is completely abstract.  We have no evidence they actually exist and they are a result of following the strict mathematics of general relativity to and beyond it's intended limits.

When you push a theory past it's "design limits" you can't expect the results to be valid.  Sometimes they are, sometimes they aren't.

  for me curvature requires a higher dimension (paper bending or a stick bending, always +1 to its dimension)

The problem here is that you're thinking in a common, everyday sense about something that fundamentally is not understandable using common everyday sense.  One of the major problems people face learning even special relativity is that common sense ideas (like simultaneity) go out the window, and this is even more difficult in general relativity.

To understand how space-time is distorted in such extreme ways requires an understanding of the mathematics.  The simple explanations that involve things like rubber-sheet analogies just don't give you the right idea at all.

A typical tool to explore space-time and how it distorts is a Carter-Penrose diagram.  Another one is the Kruskal-Szekeres coordinates.

So really the problem here is you trying to use your common everyday sense in a situation that it does not work in.  The reason physicists do all that complex mathematics in the first place is because that is what works.


Can a single location on earth witness solar eclipse totality 2 times during a single eclipse event? The question came to my mind when I saw a visualisation of the path of the moon (relative to the sun - the sun is kept centered in the graphic, and only the moon moves) as its seen from a particular location during a solar eclipse.

This reddit post has the visualisation - https://www.reddit.com/r/delhi/comments/h0u878/in_10_days_there_will_be_a_partial_solar_eclipse/

In the visualisation, the moon appears to move towards the sun, and then retraces its path (almost).

If the moon crosses the sun, and then makes its way back, it is possible for the moon to cause 2 totality events at the same location on the earth.

Is this actually possible? Why or why not?

Has it even happened in the past? When and in which location?

When and where will it happen again?

No.

The video has been made so that the sun remains at the centre, but so that "up" is still "up relative to the horizon". This means that the shape of the crescent of the moon is how it will appear in the sky.  The turning of the Earth makes the direction of the moon relative to the sun appear to change.


Flatness of celestial systems like galaxies and Solar System If we look around the cosmos, we will see most of the systems are just flat disks. Either it is solar system, galaxies, or the rings of Saturn. What is the matter, does it depend on how they were formed, or anything else?

Flatness is caused by conservation of angular momentum. As solar system/galaxy collapses under its own gravity, its velocity increases to conserve the angular momentum. This high velocity leads to formation of disc shaped structure. The similar manner in which piece of dough becomes flat when it is rotated.


Small changes to Earth's orbit caused by a large impact on the Moon Is there any scenario where a large impactor striking the far side of the Moon would perturb the Earth's orbit enough to reduce the distance from the Earth to the Sun and still remain within the habitable zone?

... and still remain within the habitable zone?

Everything perturbs all the time.  Every change in the distribution of mass in the solar system (or in the universe) perturbs the orbit of the Earth... so the literal answer is "yes".
But suppose someone asked instead:

Can an 'invisible' impact perturb the orbit enough to have a measurable effect on the the climate

The answer to that question would be "no"  The Earth's orbit already takes it closer and further from the sun each year, reaching its closest in January. To move the Earth enough that it has a climatic effect would require a huge force on the Earth-moon system, and an impact on the far side of the moon that is large enough to cause such a perturbation would cause other effects (like melting the moon, disrupting it completely, blasting enough of the moon into space and having it fall to Earth....) These would affect the climate much more.
An undetected asteroid impact on the far side of the moon might change the orbit of the Earth-moon system around the sun by millimeters.
It is just really really hard to change the orbit of a planet.


Changing Appearance of the Moon During a Single Night I have noticed that the inclination of the moon at a given location varies throughout the night. From the UK, the terminator in a half-moon is clearly sloping from top left to bottom right early in the evening, but later on, as the moon is setting, it is clearly sloping from top right to bottom left. I know that the shape of the moon changes from night to night, but I just can't see why the inclination would change during a single night. I know that the appearance of the moon is due to a combination of the tilt of the earth's axis and the latitude of my position (the moon slopes oppositely in the southern hemisphere), but I just can't visualise in three dimensions exactly what is going on. Please can you help me to do this.

If you take a line from the middle of the terminator, but at 90 degrees to the termination, it more or less points towards the sun.  Now consider the moon at first quarter: as the moon rises, the sun is due south, so imagine a line from the moon to the sun, and the terminator will be at 90 degrees to this, so pointing from bottom right to top left.  At mid-day, the sun is in the west, and the terminator will be sloping bottom left to top right.


Knowing the RA/DEC of a star how do I locate a star from ground? Standing in Melbourne at 4 am in the morning on 14th April 2019,  Sirus RA = 06h 45m 08.9s, Dec = -16° 42' 58"..How do I Locate this. Standing facing east I turn myself towards south approx 90 Deg and then look up about 16 Deg, I do not find any star.

Calculating local sidereal time exactly is not trivial as shown by the calculation for local sidereal time given at Approximate Sidereal Time:

  The Greenwich mean sidereal time in hours is: $$GMST = 6.697374558 + 0.06570982441908 D_0 + 1.00273790935 H + 0.000026 T^2$$
  where $T = D/36525$ is the number of centuries since the year 2000.

Fortunately thereare a number of online calculators that will do that for you.

You can, however ballpark it knowing that at the equinox in March (~21st, depending on the year, because you also need to take leap years into account), local sidereal time is 12 hours different than local time (without daylight saving). i.e. 6 pm (18:00) is 06:00 LST. This means that RA 06h00m is overhead at 18:00 local time.

Then add 1 hour for each $\frac{1}{2}$ month or 2 hours for each month since then. So, mid-April is about 1 month after the March equinox which means that RA 08h00m is overhead at 18:00 local time, and mid-May will be 2 months so RA 10h00m is overhead at 18:00.

From there, it is just like a clock - midnight is 6 hours later than 18:00 so you add 6 hours to the local sidereal time calculated earlier meaning that in mid-April RA 14h00m is overhead at midnight.

What this tells us is that around the March equinox (~21st), Sirius (RA 06h45m) is overhead at about 18:45, which we can round off to around 19:00 (7 pm) given the roughness of the calculation so far. We can guess then that it will set then around 6 hours later, at about 1 am.

In mid-April, RA 21h00m will be overhead at around 19:00 (7 pm), which means Sirius will have been overhead about 2 hours earlier - around 17:00 (5 pm), which means it will set around 11 pm, and in mid-May will set around 9 pm.

Melbourne is located at ~145° E, ~38° S. From this you can work out the relative angle from the declination (Sirius dec. is -16°43'). Melbourne is at 38° south, so declination -38 is directly overhead. Sirius is at dec. -16°43' (so about -17°) so it will be about 20° north of directly overhead.

So, you would face towards north and look upwards almost to directly overhead. Knowing the LST, you know the RA that is directly overhead and can count towards the west (or east, depending on the month), and knowing that in mid-April Sirius was overhead at around 5 pm and will set around 11 pm, you can find it.

NB. James K's answer is really worth reading.


Why is right ascension measured on a 24 hour scale rather than a 23 hours and 56 minutes scale? From several texts I have read, I have learned that right ascension is measured on a scale of 24 hours. I understand that Earth rotates almost exactly 360 degrees relative to distant objects, which is what RA is used for. The 24 hour system should only apply to the 361 degree rotation period (solar day). Why then do we split the sidereal day of the earth into 24 hours, when if it were split into 23 hours and 56 minutes we could know where an object would be x hours away after x hours?

More (possibly) relevant information:
I understand the Hour Angle HR is the measure of how far Aries is from the local meridian (RA of current meridian of longitude). This way you could calculate how many hours away the object is, but once again, this would not be exactly correct as 4 minutes has to be subtracted from the 24 hour scale, multiplying the whole time to something like 99.7% (14.44 degrees per hour), correct?

Thanks, this is really confusing me and I would appreciate some help. I did see a similar question but that had nothing to do with the sidereal day v. solar day issue I'm having.

Right ascension is a historical oddity.

To specify a point in the sky you need a coordinate system, the one which we have come to use has it's origin at the Point of Aries on the Equator and the Ecliptic (a reasonable choice), It uses the Equator for one axis, and the meridian through the point of Aries for the other, again these are convenient and reasonable.

The coordinate system also requires a unit to be chosen for measuring angular distance. For measuring declension, we use degrees. The more mathematically pure may prefer radians, but degrees are a commonly used unit for measuring angles.

For measuring the angular distance around the equator we have come to use "hours", by dividing the full turn into 24 parts. This is a historical oddity. It has a few conveniences: The sidereal day is only slightly shorter than 24 hours, so a star with RA of 1hr will be due south about 1 hour after a star with a RA of 0hr (not exactly but close enough for a rule of thumb). A telescope with an equatorial mount that is set to make a full turn in 24 hours will track stars well enough for a human observer.

Dividing the equator into (23+56/60) parts would be inconvenient since its not a whole number. For these reasons Flamsteed used Hours for his catalogue in 1712, and we have followed his tradition.

So the use of "hours" may be roughly related to the apparent motion of the stars, but it is really just a way of specifying an angle, and using 24 is easier than (23+56/60)


When do Mercury/Venus reach greatest elevation at sunset/twilight for a given location? On what day does Mercury reach its greatest elevation (in degrees from
the horizon) at sunset a given location?

The obvious answer is the day of Mercury's greatest elongation from
the Sun, but, since the ecliptic is slanted with respect to the
horizon, I'm not convinced this is correct.

In other words, on the day after greatest elongation, Mercury's total
angular distance from the Sun will be smaller, but it's vertical
distance in elevation (for a given location) at sunset might be
higher.

Same question for Venus, and for when the sun is 6 degrees below the
horizon (ie, civil twilight), and for sunrise/dawn.

I'm guessing the date might vary based on position (mostly latitude) since the ecliptic's slant varies at different locations.

I googled and found nothing. My (preliminary and possibly wrong)
expierments with stellarium show that Mercury's elevation at sunset IS
higher 1-3 days after its greatest elongation, but by less than 1/2
degree.

So, it's possible that the date of greatest elongation is a close
enough approximation.

You can take a look at this table - it's in Dutch, but the dates and numbers in the table shouldn't be too hard to understand.  The first date is that of the greatest elongation (GE), the second (halfway the columns) that of the best visibility (BV).  You should ignore the lines with red dates - they're unfavourable appearances, and the code may find a date far from that of the GE.  The table shows that the difference between the moment of GE and that of the BV is typically a few days, although it can be about a week. (You can click on the first date for a page which includes a sky map of the appearance at constant Sun altitude (-6° I believe)).

Venus is a different story, because it is usually visible for months.  As you may know, in the northern hemisphere morning appearences of the inner planets are unfavourable in spring and favourable in autumn, and for evening appearances this is the other way around.  The table shows that Venus had its GE (west/morning) on 8 June 2001, but was best visible on 22 August.  In the mean time, the planet had moved somewhat closer to the Sun, but the effect of moving from spring towards autumn was clearly stronger (click the date link for the weird map that goes with this).  Hence, for Venus the difference can be several months.

Of course, these are just examples for one location (52°N).  I think the process may be too complex to be described by a single, simple equation.


Are Sun-Earth L1 & L2 points generally considered outside the Earth's Hill Sphere? This image from Wikipedia of the SE Lagrange points and the Hill Spheres suggests that the SEL points are outside the Earth's Hill Sphere. (The Hill spheres are the circular regions surrounding the two large masses.)  Is this correct?

But I thought I've read that both the Hill Sphere and L1 & 2 are 1.5 million kilometers from the Earth.

I'm going to answer this in a definition-based manner.

Here is the definition of a Hill Sphere according to Wikipedia:

  The Hill sphere or Roche sphere of an astronomical body is the region
  in which it dominates the attraction of satellites.

Here is the definition of a Lagrange point according to Wikipedia:

  In celestial mechanics, the Lagrangian points are the points near
  two large bodies in orbit where a smaller object will maintain its
  position relative to the large orbiting bodies.

So, if you think about it - L1 and L2 are on the edges of Earth's Hill Sphere - technically, neither inside nor outside, because neither of the two objects dominate gravitational attraction. This is stated on the Wikipedia page for the Hill sphere - and userTLK alluded to this in his comment as well:

  In the example to the right, Earth's Hill sphere extends between the
  Lagrangian points L1 and L2, which lie along the line of centers of
  the two bodies.

The Lagrange points are where the Sun's and Earth's gravitational influences have no net effect on an object. So, if an object at L1 moved ever so slightly towards the Sun, then the Sun would dominate the attraction of that object, meaning it would be inside the Hill sphere of the Sun. The same goes for the Earth, just in the opposite direction. (The conditions for L2 are different, but detailing them likely isn't relevant to the answer.)


Nebula and its colors A Nebula has bright awesome colors that include red, blue, green, orange, white, etc?

They're result of excitation of electrons.

Is there a clear explanation as to which color is attributed to an atom for Color properties in Cosmos!?

Yes, nebulae can often have very distinct colours. What produces those colours can depend on what elements are in the nebula material and what the temperature and density are.

Generally speaking, green colours in a nebula are due to forbidden transitions in ionised Oxygen, though can feature the hydrogen $\beta$ Balmer line. Red colours can be due to Hydrogen $\alpha$, but there are also lines due to ionised sulphur and nitrogen that often make contributions.

If you take a spectrum of a nebula it usually completely obvious what elements are responsible, because the spectra consist of very sharp emission lines at the wavelengths that correspond to the well-known electronic transitions in particular species. An example is shown below, which is the spectrum of a typical planetary nebula.
Here, the labels mean the following: OIII means oxygen with its outer two electrons removed; NII is nitrogen with its outer electron removed etc., H$\alpha$, $\beta$, $\gamma$ are the $n=3 \rightarrow 2$, $n=4 \rightarrow 2$ and $n=5 \rightarrow 2$ transition of the hydrogen Balmer series. The wavelength scale is in angstroms where 10 angstroms is 1 nm. So 5000 angstroms is the green part of the spectrum, 6500 angstroms is red and 4000 angstroms is blue.


Arago Spot in the Shadows of Celestial Bodies Recently, I've watched this video by Veritasium describing Poisson's Spot, or the Arago Spot. It is explained in the video that near circular (or spherical) objects can produce this optical effect, given that there aren't many obstructions about the circumference of the object to diffuse and scatter the light and the light producing the shadow is more or less in phase.

My question is, Are Arago Spots possible/Have Arago Spots been observed in the solar shadows of any celestial bodies?

The Arago spot phenomenon requires the light source to be point-like and very bright. The sun is not a point-like light source, as it has an apparent diameter of about 0.5 degrees

The only body large enough to cast a full solar shadow is the moon.  Arago's spot has not been observed during a total solar eclipse.