Eventual outcome of tidal acceleration and deceleration So, I know the Moon experiences tidal acceleration from the Earth. And, from what I've read, if not for the fact that the sun would boil away the oceans and engulf both of them first, about 50 billion years from now the Moon would be geosynchronous.

I was curious whether tidal acceleration could ever cause a satellite to escape, and short answer seems to be, yes, it's possible. Is that accurate? And if so, which factors add to it?

A faster rotating planet also results in tides farther ahead, and has more energy to lose before being totally locked. Very closely related, a farther satellite orbits more slowly, resulting in tides farther ahead, and requires less energy to escape. This one seems obvious: the faster the planet is still rotating while the satellite is farther away, the more likely it escapes.
A more fluid planet would experience stronger tides, which would again slow it down faster, but definitely accelerate the satellite faster. A more viscous fluid than water would experience weaker tides, but I think they'd be farther ahead. I think more fluid is definitely more effective, but it seems less obvious than the first point.
A larger planet would have its orbit slowed less by tidal acceleration, but have a stronger pull to escape. A larger satellite would cause stronger tides and more tidal force, but would require more force to accelerate, and would slow the planet down more quickly. I really can't tell which has the stronger effect...

So that's question 1:
Is it really possible for tidal acceleration to throw a satellite into space before the planet is tidally locked, and if so, does either objects' size impact it, or just fluidity and relative periods?

Then, I wondered about tidal deceleration, and crashing into the planet. It seems obvious that any satellite orbiting retrograde would never stop decelerating, so all would eventually get torn apart and crash onto their planet. For non-retrogade satellites, an article I read implied that all of them will have the same fate... Which implies that the planet's rotation will always speed up more slowly than the satellite's falling orbit. Is that true? If not, then it seems there must be cases where the planet's rotation catches up, and they again end up tidally locked.

So that's question 2:
Will all non-retrogade tidally decelerating satellites eventually crash into their planet, or is it possible for the planet to catch up given the right starting conditions?

Edit: Haven't really found any more information on this. Is there someone familiar with relevant equations, who knows the answer?

So that's question 1: Is it really possible for tidal acceleration to
  throw a satellite into space before the planet is tidally locked, and
  if so, does either objects' size impact it, or just fluidity and
  relative periods?

During the locking process, the angular momentum of one is fed into the other until the periods match.  The final destination depends on the initial momentum of both objects and the moment of inertia.  As an example, if you make the moon less massive, or the earth more massive or with a faster spin, then the final configuration of the system would have a larger orbit.

You could certainly create a system where the final theoretical configuration has a distance to the satellite that is beyond the primary's Hill sphere.  Such a system would be expected to eventually separate.

Fluidity might change the rate at which the final destination is reached, but not its configuration.

  Will all non-retrogade tidally decelerating satellites eventually
  crash into their planet, or is it possible for the planet to catch up
  given the right starting conditions?

Each object is transferring the same amount of angular momentum.  But the response of each object depends (at least initially) on the relative magnitudes of their moments of inertia.  For a system where the satellite is (relatively) small, or where the difference in rotation rates is large, impact is inevitable.  The primary cannot speed up fast enough to keep up with the increasing speed of the descending partner.

But when the difference in periods is small, and the primary is small enough/far enough, then the objects would lock before impact.


Delay between neutrino pulse and visible flash of supernova If you were orbiting a star undergoing a core-collapse supernova, what would be the delay between the neutrino pulse from the collapse and the first visible effects on the surface? Basically, I'm trying to get a scientifically-plausible gauge of the warning time given by said neutrino pulse.

The neutrino pulse associated with the core-collapse supernova SN1987A in the Large Magellanic Cloud, were estimated to have arrived at the Earth about 3 hours before the rise of the supernova light curve (e.g. the review by Beall 2006).

If neutrinos are massless or have small masses (which they do), then this is your answer.

If they have masses (and they do), then the answer could depend on the neutrino mass and how far away you are from the supernova, since particles with mass will travel slower than the speed of light. Therefore the "delayed", but faster, optical emission will catch-up with the neutrinos.

The neutrino masses are probably of the order of 0.1 eV, whilst the energies of neutrinos from core-collapse supernovae are around 30 MeV in energy. i.e. The neutrinos travel with a Lorentz factor $\gamma = (1-v^2/c^2)^{-1/2} \sim 3\times 10^{7}$. Thus
$$\frac{1}{1 - v^2/c^2} \sim 9\times 10^{14}$$
and so $v/c \sim 1 - 1/1.8\times 10^{15}$. That is, the neutrinos probably travel at about 1 part in $10^{15}$ slower than the speed of light.

At the distance of the Sun, this amounts to a delay of $5 \times 10^{-13}$ s, at the distance of the Andromeda galaxy about  0.006s, but for a galaxy at a distance of a Gpc, the delay would be about 100 s.

So actually, unless you are detecting the supernova on the other side of the observable universe, it makes no odds and certainly makes no difference at the distance of the Large Magellanic Cloud. So I say your answer is about 3 hours.


If the fifth gas giant in the early Solar System was completely ejected, where would it be now? I've read about the possible 5th gas giant in the Solar System, and about its ejection about ~100 million years after the formation of the Solar System. However, I have not seen anything about its potential whereabouts. Assume this gas giant (let's call it G5) was completely ejected from the solar system. Would G5 still remain in the vicinity (100 ly or less) or would it be very far away from the Sun right now?

It could be anywhere in the Galaxy or even (unlikely) have left the Galaxy.
The kinematics of freely orbiting objects in the Galaxy are heated - that is, the velocity dispersion of objects increases with time.
For objects of age of 4 billion years we might expect a dispersion of about 10 -15 km/s in each velocity coordinate. That's about 10-15 pc/Myr. So in 4000 Myr, an object could have scattered to most parts of the Galaxy, or at least to a reasonably wide and thick annulus around the Galactic centre.
It is for similar reasons that, even though the Sun was probably born in a cluster of $\sim 10^4$ stars, none of those siblings have been firmly identified.


Can light be trapped in orbit? Could rays of light be ever trapped in a constant orbit around a black hole if approached from a certain angle ? Like light hitting a glass or liquid with critical angle.

In principle yes, but in practice no.

At the so-called photon sphere, gravity is exactly so strong that a photon on a tangential trajectory would stay in orbit. For a non-rotating black hole of mass $M$, the radius of the photon sphere is $3/2$ times the radius of the event horizon — i.e. the "surface" of the black hole — which itself is given by $r_s \equiv 2GM/c^2$, where $G$ and $c$ are the gravitational constant and the speed of light, respectively.

However, the orbits are unstable; any perturbation will cause a photon on the photon sphere to either escape or plunge into the black hole.


Identifying Celestial Bodies vs Drones A neighbor called me to ask about drones flying around. I managed to take several pictures and they don't look like any star, planet or other celestial body I have ever seen, but they seem too steady to be drones.

I managed to get one decent picture:

Any advice for figuring out whether that is a real celestial object or man-made?

Any advice for figuring out whether that is a real celestial object or man-made?

Yes - in almost all cases, you can readily identify a celestial object by going out the next clear night, and looking for it in the same position.  (Very few celestial objects move far over a few days.)

In response to comments about geostationary satellites:
I doubt it is a geostationary satellite, answers to this question indicates these are usually too faint to be seen without special equipment. Also they don't move, whereas nearby stars will move a little over the space of half an hour or more.


Are the sub-spectral types (1,...9) based on temperature or spectral lines? The spectral type of stars has an OBAFGKM class, along with a number, from 0 to 9. According to Wikipedia, (sample shown below), the lower the number, the hotter and more massive the star, but absorption lines are not included. Then looking at the second image, we can see that different spectral types have different absorption lines. So I am wondering, are the sub-spectral types (numbers) based off of temperature and/or mass or absorption/spectral lines?

The spectral classes (O, B, A, F, G, K, M) and their 10 subtypes (0 to 9) were initially meant only as differentiators of spectral type. Annie Jump Cannon was the creator of this system. Through her work for/with Edward Pickering, she ended up classifying nearly a third of a million stars over a few decades. She (and many others) did not realize that this was actually a temperature scale - the classifications were based on the spectral lines that Cannon observed and sorted into categories. It wasn't until Cecilia Payne's work, and, most notably, her doctoral thesis, that anyone realized that the OBAFGKM system (and the sub-classes it came with) was actually a temperature scale.
Payne's work showed that the difference in spectral type was not due to differing amounts of elements, but mainly due to the temperatures of stars. This also led to the realization that stars were made of far more helium and hydrogen than other elements. However, the professor who reviewed her thesis, Henry Russell, dismissed her findings as impossible, since at the time, the implications went against much of contemporary knowledge. After a few years, Russell realized that Payne was right, and praised her work, crediting her with the discovery.
(Side note: there's an amazing episode of the TV series Cosmos - the new one with Neil DeGrasse Tyson - called "Sisters of the Sun" that dives into this story quite well. If you have the time, I'd definitely recommend it - it's a great watch.)


What is meant by "Radial Direction" of a galaxy? There is a scientific journal article having a line:

  "In order to study the star formation scenario in the radial direction
  of the LMC...."

What is meant by 'radial direction' of a galaxy ?

'Radial direction' typically means from the center, moving outwards. Typically, these kinds of studies will break a galaxy up into several annuli, or rings, and examine the star formation activity within each ring. (Imagine drawing a series of concentric circles, each one a little bigger than the last.) Then, if the star formation rate or star formation history changes as a function of radius, those changes can be more easily quantified.

It is often difficult to do this well because galaxies are not perfectly symmetric. The LMC, for instance, is a dwarf irregular galaxy, meaning its structure is not nearly as orderly as some other galaxies (e.g. The Milky Way, which is a spiral). The LMC does contain more stars near its center, and it even has a bar, but it is difficult to discern much more structure than that. To define the center of such a galaxy, the total light profile is measured and the centroid calculated.

Since we are rarely fortunate enough to be looking at another galaxy perfectly face-on, we also have to consider the effects of inclination on our observations. (If you were to hold up a plate and angle it a little bit, it would begin to look like an oval, but that doesn't mean it isn't still actually round.) If they have taken inclination into account, they should reference the "deprojected radius." That means that they know that the inclination to our line of sight makes some parts of the galaxy look closer to the center than they actually are, and they corrected for that effect.


What is 'surface acceleration' of a planet or other body? As opposed to its 'g' (gravity)? The Mars Fact Sheet at NASA (their NSSDC or NSSDCA section, whatever that means) lists a value for 'surface acceleration' just below one for 'surface gravity'.  They also have the values for Earth, for comparison.  The values for surface acceleration are very similar to those for gravity, just slightly smaller...

The technical content of NASA's web pages have sadly declined over the years. The planet fact sheets are exemplary of that decline. The December 8, 2002 version of that page provides only one value instead of two, titled "surface gravity". The values are 3.69 m/s2 for Mars and 9.78 m/s2 for Earth.
The gravitational acceleration at the surface of the Earth (or Mars) is not constant. It varies with latitude and altitude. The value of 9.78 m/s2 for the Earth represents the gravitational acceleration, including centrifugal acceleration, at Earth sea level at the Earth's equator. The gravitational acceleration at the Earth's North Pole is a bit greater than this, about 9.834 m/s2. There is no centrifugal acceleration at the Earth's poles, and the Earth's poles are a bit closer to the center of the Earth than is the equator thanks to the Earth's rotation. The same concepts apply to Mars.
The value of 9.80 m/s2 added to the cited page a bit later is arguably erroneous; it should be 9.81 m/s2 (or even better, 9.80665 m/s2) rather than 9.80 m/s2. As I've said elsewhere regarding Wikipedia, "that's Wikipedia for you." Unfortunately, the same applies to NASA web pages, in spades.


Corresponding values for the Sun's ultra violet color maps? In my laboratory, we have an antenna to get data from the GOES-16 satellite and I want to reproduce NASA's nice visualization of ultraviolet images from the sensor SUVI.

I can read the data without any problem and thanks to the project sunpy I got the color tables corresponding to each band. But I have not found a reference for the exact correspondence between values and color indexes, so my version is not remotely as nice.

Where can I find a reference to know how to use this color map with my own software?  I will appreciate any hint.

It’s hard to tell for sure without seeing the data and playing around, but looking at your image, I would suggest trying to subtract off a constant background level to get the sky as close to black as possible.  You could sample the edges of the image to find the average background level.  Then subtract that value from the whole image array and try plotting it.  Even simpler would be just to subtract the minimum value, but that’s prone to being skewed by just a single pixel.
If you try it, post an image back here and let’s see how it looks.


Is it possible to see Saturn from Titan's surface, day and night? Titan is the only moon in our solar system with a thick atmosphere, similar to what our Earth's atmosphere was before life "appeared". It also orbits Saturn close enough to actually see it beautifully in the sky... or can it?

Titan has a very thick and hazy atmosphere. Tholin seems to be the main factor of this "haziness". Also we can't see Titan's surface from space without infrared instruments. So in visible light, Titan looks like an almost uniform ball of beige, yellowish, golden brown because of its atmosphere, hiding the surface below.

So this should probably be the same case the other way around, right? Can we see Saturn from the surface of Titan, or does the haziness hide the parent planet from view during day time and night time (since I believe the atmosphere's opacity changes in day time and night time)?

"since I believe the atmosphere's opacity changes in day time and night time" maybe to a very small factor, somewhere in the upper atmosphere, in the UV-wavelengths where plasma opacities play a role.
But not in the optical, where the bulk density of the atmosphere plays a role, wich at 1.6 bar doesn't change much between day and night.

Your assertion "So in visible light, Titan looks like an almost uniform ball of beige, yellowish, golden brown." is incorrect, where do you take that from? The optical depth integral is symmetrical both ways (ignoring scattering), so if we by far cannot see the surface, then one will certainly not be able to see Saturn from the surface.
There is also only one existing photo taken on Titans surface, so not sure where you get this idea from.

Apart from the other statement "Titan is the only moon in our solar system with a thick atmosphere, similar to what our Earth's atmosphere was before life 'appeared'", which is a questionable statement at best.


Why not take a picture of a closer black hole? There are closer galaxies than Messier 87 for sure, even ours! It sparked my curiosity that they went with one 53 million light years away. Is there a reason for this?

I was surprised too when I first heard they were trying to image M87's black hole.

The short answer is because it's really, really big. It is 1500 times bigger (diameter) than our Sagittarius A*, and 2100 times farther away. This makes its apparent size about 70% of that of Sgr A*, which they are also attempting to image.

A cursory search of wikipedia's List of Largest black holes shows that there's no other black holes with a combination of size and closeness greater than these two.

A couple of other candidates are not too far off. Andromeda's black hole is 50x the size of ours, and at 100x the distance, it would appear half the size of Sgr A*. The Sombrero galaxy is 380 times farther way than Sgr A*, and has a black hole estimated to be 1 billion solar masses, which is 232 times Sr A*, resulting in an angular diameter about 60% of Sgr A*.

There appear to be many other considerations to which black holes were chosen, as explained in this similar question. At a guess these would include how obscured each black hole is with foreground dust/stars etc, how active (and therefore bright) the nuclei are, and their inclination w.r.t earth affecting which observatories could observe them at which times.

Edit: I've found another plausible candidate. NGC_1600 is 200 M light years away with a central black hole estimated to be 17 billion solar masses heavy. This would put it at about 40% the apparent diameter of Sgr A*.

Comparison of the apparent size of the largest nearby black holes

And of course obligatory XKCD to remind us how small these objects really appear.


For how many bodies can there be a stable orbit with no very heavy central body? The question is a bit vague, but let me explain:

Take for example 2 bodies of the same mass. They can orbit around their centre of mass/gravity. Is something like this possible with multiple bodies of varying mass and distance from the centre of gravity without a relatively extremely massive body there?
i.e. would it be possible to have a planetary system that orbited around the centre of mass without there being a massive star (roughly) in the middle?

I know that the Sun also has an orbit around the centre of gravity of the solar system, but that orbit is very small compared to its size, which puts it roughly in the centre.

It is not unusual to have pairs of pairs or even more complex stellar systems.  Two stars make a stable pair.  A pair of these make a stable system and a pair of these would also make a stable system.  It is, however, necessary for stability for each pair to be separated by roughly 10 or more times the separation of the previous level.  This could continue ad infinitum were it not for the tidal field of neighboring systems or the Galactic potential limiting the overall size.


Does the Sun orbit the Milky way in a (kind of) flat or inclined orbit or more of a sine wave? I know it can't be a perfect ellipse or smooth orbit due to all the matter in every direction, and our Sun could be affected by a gravity assist of sorts if it passes close enough to another massive star. I also know that galaxy orbits are different than Solar System orbits because most of the mass is in the dark matter halo and as a result, the outer stars orbit faster than in a standard orbit, so there's a number of variables.

But, if we ignore gravity assists, is the sun mostly in a standard, or lightly inclined orbit around the center of the Milky way or more sinusoidal?

Reason I'm asking - see (attached and in my opinion, bad) article, and picture included.

I've seen pictures like this before that suggests that the Solar System moves in a kind of sine wave motion. I've seen estimates between 26 and 32 million years with the solar-system, rising above then below the plane of the Milky way with consistency (and perhaps, mass extinctions tied to this movement).

I can kind of see that happening, as the sun rises above it might have more gravitational pull below it drawing it back down and you kind of have a horizontal velocity to displacement cycle creating a sine wave in the orbit. That makes logical sense, but I'm wondering if that 26-32 million year sine wave type of movement is generally accepted or more somebody's imagination. Is anything definitive known about stars' orbits within galaxies?

It is correct that the sun bobs "up and down" (relative to the plane of the galaxy as it orbits the galaxy, and it takes about 64 million years  to complete a full cycle, so it does pass through the plane of the galaxy every 32 million years or so (there are quite large error bars on those numbers.)

This idea is uncontroversial, but it gets caught up in other theories that are less well established (for example the idea that there are dense clumps of dark matter in the galactic plane, that can have disrupt the Oort cloud, or otherwise cause mass extinction) or are simply nonsense (the idea that the solar system is a "vortex").

Phil Plait rubbishes this second idea on his blog, and includes some further discussion of the "bobbing" path of the sun. He notes that the disk is about 1000 ly thick, and we bob up and down by about 200 light years every 64 million years.

Lisa Randall's is a serious scientist, who would have no truck with "solar system vortices", and the ideas in her book are reasonable, but not uncontroversial. We have no evidence of dense clumps of dark matter, nor do we know exactly where the Earth was 65 million years ago. We last passed through the galactic plane about 3 million years ago, without serious consequences.


What happens with the black hole's mass when a black hole disappears? If a black hole isn't "fed" by attracted matter, it disappears someday. But how is that even possible? What happens with the mass at its singularity? It can't just disappear or am I wrong with something? Where does the mass disappear to and how is it possible?

The mass (equivalently energy) is radiated away as Hawking radiation.


How to convert sidereal time to local time? I came across something that won't let me go, but no matter how much I calculate (which I do not like), it doesn't seem to work.
Also: I've used so much apps and links already to convert, but it doesn't work in that direction. You see, local time converting to sidereal time = not that hard.

I need to know the local 'normal' time of my local sidereal time.
The sidereal time is exactly 13.30h, from that I would like to know the normal local time.

The time zone = Brussels (CET)
Olson time zone Europe/Brussels
Time difference: same as Amsterdam and Europe UTC+1
no summer time at the moment (and what if that was the case, what would the local time be then?)
Latitude: 50,929315°
Longitude: 5,337367°
Address: Grote Markt 18, 3500 Hasselt Belgium.

The reason I need to know this has to do with a sort of awareness I found in ancient writings and books. It has extra power than to meditate, and I would like very much to try this out.

But of course, I need to know the local time then.

Convert your current local date and time "t" to UTC if necessary.
Use an app to calculate the local sidereal time (LST) for the current date and time t.
Calculate the following equation:

$$local\;time\;for\;13.30\;LST = t+\frac {13.30-LST\;at\;current\;date\;and\;time\;t}{1.0027379}$$

For example, at 3:00 local time (2:00 UTC) on Feb 6, 2019, the LST=11:25:02 (11.4172 hours) at 5 deg 20' 15" east longitude (5.3375 deg). The above formula gives the time for 13.30 hours LST as
$$=3:00+\frac{13.30-11.4172}{1.0027379}$$
which equals 4.8777 hours local time (4:52:40). Plugging that time (converted to UTC if necessary) back into an app confirms the LST is 13:18:00 (13.30 hours).

After you know one date and time, you can calculate the next date and time approximately by adding 23:56:04 to your answer. For a date more than a few weeks in the future, I suggest you repeat the calculation for a the new date.

I was using this online site, but any site should give a similar answer: Calendar Date And Time To Julian Day And Sidereal Times


Ejected planets during the early stages of the formation of the Solar System This is similar to a previously asked question, but I am just asking about theory rather than observational evidence. Assuming there were a much larger number of protoplanets in the early Solar System, what percentage of these would theoretically have been ejected (orbital velocity increased beyond escape velocity due to interaction) and what percentage would have been just moved out to a more distant orbit (say, in the Kuiper Belt or the Oort Cloud)?
In other words, should we expect to find rocky worlds in the far outer Solar System that were formed inside the current radius of the asteroid belt?

We can identify two distinct realms of planetesimal formation - the inner Solar System and the outer Solar System. The initial group of small bodies1 in the inner section of the protoplanetary disk was rather quickly accreted into planets via various interactions; some bodies grew bigger and thus dominated the disk around them. In the outer parts of the disk, many small bodies were accreted into the cores of the giant planets. However, not all were accreted, and there may in fact have been a second generation of planetesimals that formed via gravitational collapse due to disk instabilities.
We can compute timescales for this "cleanup" phase, the period during which small bodies were either accreted onto protoplanets or ejected to larger orbits or out of the Solar System entirely. In the inner section of the disk, among the terrestrial planets, it was on the order of $\sim10^8$ years, while in the outer section, it was on the order of $\sim10^9$ years. However, the giant planets may actually have finished accreting matter before the terrestrial planets.
This means that the small bodies that were ejected or boosted to larger orbits were largely from the outer Solar System, terrestrial planetary migration effects aside2. Various N-body simulations agree that a total of $\sim300M_{\oplus}$ of small bodies was ejected from the outer disk, due to interactions with Jupiter and the formation and evolution of Uranus and Neptune. We know that the Oort Cloud's mass is on the order of $\sim1$ to $\sim10M_{\oplus}$, and the Kuiper Belt and scattered disk have masses an order of magnitude or two lower. Therefore, only a small fraction - certainly less than 10% - of the small bodies could have been ejected into the Oort Cloud.
If you want to visualize some of this, check out this video from a simulation of the Nice Model3. It shows the orbits of the four giant planets (Jupiter in red, Saturn in orange, Uranus in purple, and Neptune in blue). It shows Neptune and Uranus swap orbits, and also shows the dispersal of a large group of small bodies outside the original orbit of Uranus (which began further out than Neptune). The video shows the simulation over the course of a bit more than one billion years, and you can see a drastic clearing out of the extreme outer disk over this timescale.
The above results came before the development of the Nice Model and its variants, as well as the possible discovery of Planet Nine and the discoveries of many small objects in the outer Solar System, beyond Neptune. They should therefore not be taken as totally accurate descriptions of the formation of the Solar System - the video in particular is likely incorrect - and I'm guessing I'll get some flak for using them as the basis of this answer.
Obviously, many N-body simulations of the Solar System have been done, and over the last decade, a substantial number use some variant of the Nice Model. Some use planetesimal disks - excluding the masses of the eight planets - on the order of $\sim10$ to $\sim10^2M_{\oplus}$; Nesvorny & Morbidelli (2012) (two of the architects of the model) found a sweet spot, so to speak, of $20$ to $50M_{\oplus}$, depending on the exact result. As I understand it, masses of the surplus disk around these values fit the Nice Model well, and so perhaps we should drop the total mass of the ejected small bodies by a factor or about ten.
Footnotes
1 I used this term quite a lot here; it refers to bodies with masses in the planetesimal regime and lower. The reason for this is that a substantial amount of the ejection came after the inner planets had formed, and the remaining bodies were small.
2 This should not be taken to mean that no or very little matter from the inner Solar System was ejected, especially if we assume some movement among the orbits of the inner planets (see Meech et al. (2016)).
3 Originally brought to my attention through a question on Physics Stack Exchange by Kyle Oman.
References

Goldreich et al. (2004) (early N-body simulations)
Krivova & Solanki et al. (2003) (early N-body simulations)
Meech et al. (2016)
Nesvorny & Morbidelli (2012) (more recent results)


Angle formed between a satellite, a ground station, and then another Satellite I am trying to check my work on calculating the angle between two sats and a ground station observing those two sats. I have the az and elevation angles to each sat from the ground station. What would be the equation to get the angle Sat1-GS-Sat2?

I know that it uses the law of Cosines, but not sure if I was using the correct equation.  Here is what I am using:

Angle = acos((abs(sin("V_Sat_1_EL")*sin("V_Sat_2_EL")))+(cos("V_Sat_1_EL")*cos("V_Sat_2_EL")*cos(abs("V_Sat_2_AZ"-"V_Sat_1_AZ"))))

Thanks

This figure from
Wikipedia:Spherical law of cosines
applies if the ground station is at the center of the sphere,
u is the ground station's zenith, and v and w are the apparent positions of the satellites as seen from the ground station.

Then a and b are the satellites' zenith angles, i.e. 90° minus their elevation angles.
C is the difference between the satellites' azimuths, and c is the angle you want.

Using the cosine rule for sides
$$\cos c = \cos a \cos b  + \sin a  \sin b  \cos C$$
and the rules for sines and cosines of complementary angles,
the expression in the question looks correct if the absolute value operations are omitted.


What (actually) is the " deprojected half-light radius" of this almost-all-dark-matter Galaxy? The recent news of the Ultra Diffuse Galaxy (UDG) Dragonfly 44 is an excellent example of what could be termed 'observe different' thinking. The dragonfly telescope is noted not for the size of its collective aperture, but for the absence of the diffracting effects of secondary mirrors and surface roughness that limit the contrast of dim objects in conventional telescopes when brighter sources are nearby. See here and here and here.

above: image of a Dragonfly refractive array telescope from here. Image: P. Van Dokkum; R. Abraham; J. Brodie

above: The Dragonfly 44 ultradiffuse galaxy from here. "Dragonfly 44 is very faint for its mass and consists almost entirely of dark matter. (Pieter van Dokkum, Roberto Abraham, Gemini Observatory/AURA)"

Once identified, the radial velocities of the stars in Dragonfly 44 were measured using DEIMOS on the Keck II telescope, in order to determine a value for the mass for the dim, ultra-diffuse galaxy.

I started reading the ArXiv article but quicky got bogged down in the abstract. The very exciting result is that the luminosity and therefore total number of stars is much smaller than what one would expect from the mass obtained from the radial velocity measurements, suggesting that it is made almost entirely of dark matter. I wanted to see if I could understand how the mass was calculated, but I got stuck on the phrase deprojected half-light radius.

Could someone just outline how this calculation is done, and what that phrase actually means?

  Recently a population of large, very low surface brightness, spheroidal galaxies was identified in the Coma cluster. The apparent survival of these Ultra Diffuse Galaxies (UDGs) in a rich cluster suggests that they have very high masses. Here we present the stellar kinematics of Dragonfly 44, one of the largest Coma UDGs, using a 33.5 hr integration with DEIMOS on the Keck II telescope. We find a velocity dispersion of 47 km/s, which implies a dynamical mass of M_dyn=0.7x10^10 M_sun within its deprojected half-light radius (my emphasis) of r_1/2=4.6 kpc. The mass-to-light ratio is M/L=48 M_sun/L_sun, and the dark matter fraction is 98 percent within the half-light radius. The high mass of Dragonfly 44 is accompanied by a large globular cluster population. From deep Gemini imaging taken in 0.4" seeing we infer that Dragonfly 44 has 94 globular clusters, similar to the counts for other galaxies in this mass range. Our results add to other recent evidence that many UDGs are "failed" galaxies, with the sizes, dark matter content, and globular cluster systems of much more luminous objects. We estimate the total dark halo mass of Dragonfly 44 by comparing the amount of dark matter within r=4.6 kpc to enclosed mass profiles of NFW halos. The enclosed mass suggests a total mass of ~10^12 M_sun, similar to the mass of the Milky Way. The existence of nearly-dark objects with this mass is unexpected, as galaxy formation is thought to be maximally-efficient in this regime.

The half light radius is the radius from within which half the luminosity emerges.

"Deprojected" means that the authors must have fitted some model to the 2D distribution of light, which can then be mathematically deprojected to give them a 3D model for luminosity as a function of radius, that they can then integrate to give a number for the half light radius.

In section 3, the authors explain that they have done this by fitting a "Sersic profile" to the surface brightness distribution https://en.m.wikipedia.org/wiki/Sersic_profile  The Sersic profile actually has the 2D half light radius as one of its parameters. But if you imagine looking through a ball of stars, this 2D measurement of the half light radius is an underestimate of the true 3D half light radius, because the surface brightness profile is more sharply peaked than the 3D stellar density distribution that produces it.

The authors appear to approximately correct (deproject) this by multiplying the half light radius by 4/3. They also make a small correction for the non-sphericity of the galaxy.

The deprojection factor depends (slightly) on the $n$ index of the Sersic profile and needs to be found by doing a numerical integral. The details can be found in the appendices of Wolf et al. (2010. http://arxiv.org/abs/0908.2995 ), who also provide expressions to directly estimate mass from the projected half light radius and the line of sight velocity dispersion.


What is the unit used for star brightness due to skyglow? What is the name and unit of the quantity in which star brightness due to skyglow is measured? How can it be measured from RAW image files of stars?

The brightness of a diffuse area is it's luminance, and can be quoted in units of $cd\,m^{-2}$. Measuring it would require knowledge of the properties of the camera: the sensitivity (at various wavelengths) of the ccd, the aperture, length of exposure and so on. This could probably be found using calibration images, in which a known luminance is imaged and then the brightness of the skyglow is compared.

The brightness of a star is usually given as a magnitude which is related to the amount of electromagnetic energy received from the star per second, in a particular range of wavelengths.

It's unclear what you mean by "brightness due to skyglow". The brightness of stars is not due to skyglow, but excessive skyglow can mask dimmer stars and diffuse objects.


How do we convert other waves of the EMS to visible light? So I just had a quick question and I was wondering how we convert other waves of the electromagnetic spectrum to visible light waves. For example we use X-ray's or radio waves to see things in space like nebulae that we wouldn't normally be able to see with just visible light waves. But how do we convert those waves to visible light so we can actually see them.  I don't know if I'm staying this right but another example that just came to me is pictures of the Sun we see that was taken with radio waves, we see them,  which must mean that they were covered into visible light some how, so I was just wondering how, thanks in return =)

It's just as simple as taking the flux at some wavelength (just a number) and using this number to represent a visible intensity.

If you only have one wavelength then you can only get a monochrome picture. However, if you have flux information at more than one wavelength, let's say three, you can use the flux at the longest wavelength to represent red (r), the middle wavelength to represent green (g) and the shortest wavelength to represent blue (b).

Put this together and you have 3 numbers representing a visible rgb signal that can be used to create a picture. Of course there may be lots of fiddling that goes on with the colour balance and the contrast to produce an effective picture, but this is the basic process.


What is the maximum Clear Sky Light Pollution rating to see the 2018 Perseid Meteor Shower? I'm talking about the Light Pollution ratings on this page: http://www.cleardarksky.com/csk/faq/2.html

I live in an area which is a 7 on the Bortle Scale ("red" on clearsky.com), but within 2 hours of an area with a rating of 3. I am trying to determine how far I need to drive in order to get a proper viewing.

What is the maximum rating that will enable me to view the Perseid Meteor Shower tonight (2018)?

The darker the better.

There are meteors of different brightnesses in the Perseid shower, including some fireballs that can be seen in any (cloudless) sky.

But many of the meteors are dimmer, and the quoted rate of 60 per hour assumes dark skies.  As there is no moon tonight, you would benefit if you can get to a dark sky location.  If you can see the milky way, you can see the Perseids.

I am under a blanket of cloud, but have seen a few while out running on Wednesday.


The sun "burns" hydrogen and even has "campfires" on it, but has anyone calculated a rate of actual chemical burning on the Sun? Answers to Hydrogen burning vs Hydrogen fusing explain that in astrophysics "burning" generally refers to nuclear fusion or at least nuclear reactions1, and information at

Why didn't we see those campfires on the Sun until now?
What could be the cause of solar "campfires"?

explains that those are astronomical nicknames for very small bright bits of plasma recently discovered on the Sun, not actual fires.
But in the Sun's atmosphere there may (or may not) be opportunities for chemical reactions. Certainly in some cool stars the atmosphere even makes dust.
Question: The sun "burns" hydrogen and even has "campfires" on it, but has anyone calculated a rate of some actual chemical burning on the Sun? Any oxidation reactions at all? If not our Sun, is oxidation simulated for some other stellar atmospheres?

1it's hard for me to think of two protons "fusing into a diproton", later beta-decaying for example; feels more like a nuclear reaction, but maybe that's just me.

I just became curious enough to search a few keyphrases.
This article talks about hydrogen molecules acting as energy "sink" in the sunspots - much like water phase changes on Earth create climate features. The heat capacity, mass density, adiabatic properties of the molecular hydrogen are pretty much different from the atomic one and these are important for the convection.
Hydrogen changing between atomic and molecular form is clearly a chemical process.
The article talks even about water molecules there. Given the high temperature and low pressure, they are expected to be pretty much short-lived and oxygen constantly moving between water, hydroxyl ion or radical and oxygen atom or ion.
Related matter here

The formation of H2 can significantly alter the thermodynamic properties of the sunspot atmosphere and may play a significant role in sunspot evolution. In addition to the survey observations, we have performed detailed chemical equilibrium calculations with full consideration of radiative transfer effects to establish OH as a proxy for H2, and demonstrate that a significant population of H2 exists in the coolest regions of large sunspots.

Looks like chemical processes not only do happen on the Sun, but are of great importance for the photospheric features we see.
Do these qualify as "burning" is a matter of oppinion.


Expansion and Gravity If space is undergoing accelerated expansion then regions of space are becoming inaccessible to us even at light speed. If gravity is bounded by light speed then inaccessible regions of space no longer contribute gravitational influence on us. If true wouldn't gravity waves... tiny variations in gravity given sensitive enough equipment... be all around us as matter crosses the threshold and becomes inaccessible to us? Does this violate conservation?

Joshua, your question combines a number of complex but related astrophysics issues, so let me try to untangle them as best I can in layperson's terms, with links where you can find more detailed information.

What you're describing is known as the cosmological horizon, which defines the limit beyond which a particle's light can never reach us because of the expansion of the Universe. While gravitational waves ("GWs") are unrelated to electromagnetic waves (what we call light), it's believed GWs are propagated at the speed of light (the universal limit for how fast information can travel), so the cosmological horizon would apply to gravitational waves as well.

However, gravitational waves are different from gravity.

To put it overly simply, the general theory of relativity says that mass distorts both space and time, and gravity is what we observe of this curvature of spacetime. Gravitational waves, on the other hand, are "ripples" in that spacetime curvature caused by non-symmetric acceleration of mass.

GWs are indeed "all around us", but they are extraordinarily tiny - Albert Einstein himself doubted they could ever be detected. Our most sensitive equipment, LIGO, looked for the very largest GWs for eight years (2002-2010) and found nothing. It wasn't until a significant enhancement came on line just three years ago (Sept 2015) that the first GW was detected - from the collision of two black holes with a combined mass equivalent to 65 Suns! - and since then we've only detected 11. GWs are really hard to detect!

Since gravity is an intrinsic geometric property of spacetime, it exists even beyond the cosmological horizon: it's inseparable from the Universe as a whole. All parts of the Universe contribute to the curvature of the Universe, no matter how far away one point is from another. There's no "threshold" for gravity.


Is it correct to assume that the relationship between an objects mass and size, is what decides if it’s a black hole or not? I’m asking this question in reference to a scenario I saw posted a couple of days ago, where it’s stated that if you would shrink earth to the size of a cherry, it would become a black hole.

Is it also correct to assume that changing the size of earth while maintaining the mass, will not affect objects which are further away than the event horizon?

I’m asking this in reference to the same scenario as above. It is also said that changing the size of the earth, so that it becomes a black hole, will have no effect on the orbit of objects around the earth - e.g. the moon.

I’m trying to understand this, with an image of how much the earth will bend spacetime. Both as a earth-sized earth and as a cherry-sized earth.

Yes. A black hole is in some ways a very simple object. It has mass, charge, angular momentum.

These three values determine the size of the event horizon. If the Earth's mass were compressed to less than the size of a cherry it would be inside its event horizon and so would be a black hole.

The gravitational field of the Earth out at the moon would be (almost) unchanged. The moon would orbit an Earth-mass black hole in the same way that it orbits the Earth now.  The small differences are due to irregularities in the shape of the Earth, which wouldn't exist if the Earth were collapsed.

Above the Earth's surface the curvature of space would be (almost) the same if the Earth were to be collapsed. If a rocket were to use its thrusters to hold itself stationary 6400km above the black hole, the inhabitants would experience 1g of gravity, just as we do standing on the surface, 6400km from the centre of the Earth.

If you go far enough below the Earth's surface, gravity weakens, because there is mass above you.  If you collapsed the Earth, then gravity (and curvature of spacetime) would increase all the way to the black hole.


Light Cone Explanation Not sure if this question belongs here, or on Physics. Would someone please describe the "anatomy" of a light cone and provide an intuitive example?

I have read the Wikipedia article (http://en.wikipedia.org/wiki/Light_cone) and I think I have a basic grasp of the concept. However, I still have questions: is a cone emitted every time there is a disturbance in the electromagnetic spectrum? Does every electron changing energy states emit a light cone? I understand the "future" light cone (I think). However, what of the "past" light cone? Is that what is observed when detecting redshifts?

Thank you for your answers/comments. They have helped my understanding of the light cone concept.

Think of dropping a pebble into a pond, and the circular pattern the resulting wave makes. Now take a photo every second, and stack these one on top of another, and you have a "cone" of sounds waves. The oldest image of the wave is also the biggest, and furthest back in time.


Do planetary rings have geometric bounds? Are there any bounds on where a planetary ring can form and maintain orbit relative to the planet?

Do they have to be in a circular equatorial orbit?

Is there a min or max altitude, say with respect to the Roche limit, geosynchronous orbit, or other points in space?

The Saturn's Phoebe ring that has an orbital inclination 173° to the ecliptic so it is actually in a retrograde orbit and is tilted 27° to Saturn's inner rings, shows that there clearly isn't any limit to orbital inclination. It is feeding off the Saturn's moon Phoebe (possibly due to micrometeorite impacts) indicating that as long as the planetary ring has a source of its materials with already inclined orbit, the ring will follow it (preserve the angular momentum of its originating body). That of course doesn't exclude polar orbits, but we have yet to see any rings like that. Theoretically though, their orbit is just as stable as any other as long as it doesn't intersect other gravitational disturbances, like Lagrange points, or paths of other celestials that would stop it forming a complete ring with accretion (something that is only partially done by Iapetus intersecting the Phoebe ring). Majority of planetary rings might be made of planet's own materials though, and would then naturally follow its own rotation, more or less (depending on their formation) and again preserving angular momentum.

Phebe ring is also extremely huge, stretching from calculated 59 to 300 Saturn's radii (observed by NASA's Spitzer Space Telescope in the infrared range between 128 and 207 Saturn's radii, more can be read in this Emily Lakdawalla's blog post), attesting that as long as the ring's materials would not reach escape velocity or be gravitationally attracted more by another celestial body (beyond the L1 point) there isn't much of a limit in their size either. In theory, if we for example imagine some rogue brown dwarf that isn't gravitationally bound to any solar system and floats freely in the interstellar medium not close to any stars, the ring size would only be limited by the interstellar medium's own pressure. At least until the brown dwarf in questions roams closer to some stronger gravitational influence and loses its ring, that is.

So the maximum size of the ring would in theory indeed be limited only to the L1 point (where gravitational attractions of two massive bodies cancel each other). As for the minimum distance, that depends on the size of particles that the ring is made of, stretch of planet's atmosphere, and strength of its radiation pressure that would be negating gravitational attraction. So it is difficult to give any fair minimum value. If we take a fast spinning planet (e.g. dwarf planet Haumea) as an example, one that would spin so fast that it would actually lose some of its materials due to it reaching escape velocity at the planet's equator, and if we assume this process can be sustained for long enough period, then there isn't any minimum as the planetary disk would be essentially touching the planet's surface. Eventually, the planet would lose much of its own rotation, similar like a fast spinning figure skater would by stretching her arms, by shifting some of its mass into larger radius, and the planet would cease replenishing disk's materials.


If we lived in a multiverse, what would our universe most likely then be named? We live in the Milky Way Galaxy, but we don't just call it "The Galaxy", because we know there are multiple different galaxies. So if we lived in a multiverse, it wouldn't make much sense to keep calling our universe "The Universe".
How are names for things like this created, and what would a likely name for our universe be?

New objects usually get some preliminary name or number according to one or more naming or numbering conventions (e.g. HR numbers for stars). If there aren't too many objects of interest, they are named by the discoverer(s) (e.g. Hubble volume), an institution (e.g. PANSTARRS), an occasional nickname liked by the public (e.g. black hole), or by a majority in a commitee or an election (e.g. Hydra).

For our universe I can only speculate; probably a notion of the respective multiverse theory would become part of the name, e.g. "Herman brane", if we take a brane cosmology as an example.


Why does Saturn have both moons and rings? From my understanding, a ring can form around a planet when a moon gets too close to its Roche limit, and gets ripped appart by the planet's gravity pull. That makes sense to me, but I don't understand why Saturn has both moons and rings at the same place.

I know the F ring is supposedly created by Enceladus' rejections, but I don't get why Pandora and Prometheus are not ripped appart as well as the ancient objects that formed the ring they are in.

I couldn't find specific explanations for this. Does it have any link to the moon's density?

You're right that density is the important thing here. The Roche limit is the distance from the main body $d$ such that
$$d=1.26R_M\left(\frac{\rho_M}{\rho_m}\right)^{\frac{1}{3}}$$
where $_M$ denotes the main body and $_m$ denotes the satellite.

As you can see from the chart on the Wikipedia page, Pandora and Prometheus are both at least one and a half times the Roche limit from Saturn. Therefore, they're in no danger of being ripped apart any time soon.


Why does the Sun show different colour at midday and at morning? As we see every day in the morning the sun shows a red-orange colour but at midday has a blue-white like colour why?

Is it concerned with the distance from the earth or about intensity? How do you see it?

Sunsets are red because the sky is blue.

Air, while it appears colorless actually scatters light just a bit. This is not the same as absorbing the light, but more like how light is scattered by fog.  If you are driving through fog with your headlights on, part of the light from their beams bounces off the fog particles and lights up the fog well outside the beam and some of it even bounces multiple times coming back to your eyes.

Air is much clearer than fog, but it does scatter light just a bit. What's more, it scatters blue light more than green, and green more than yellow, and yellow more than red.  So while the Sun's light in space is quite white, as it passes through the atmosphere on a sunny noonday a lot of the blue light gets scattered, while very little of the red does.  That blue light bounces many times and is the source of the blue sky light.  The loss of that blue light makes the sun appear redder.

At noon, the light comes down from high overhead and passes through a minimal amount of atmosphere and only some of the blue light is scattered. At sunset (and sunrise) the light comes in at a long, low slant and passes through much more of the atmosphere, and nearly all of the blue and even most of the green is scattered (and much light is absorbed, too), leaving mostly red and orange light left.

Hence the reddened sun.

See Why Is the Sky Blue? and Why Is the Sunset Red? for a bit more including some diagrams.


What are the so called clouds of dust and gas made of? We constantly read about "clouds of dust and gas" surrounding a (or being the initial state of a) star, but what elements constitute such dust and gas?

The clouds of gas and dust that form stars are usually what are called Molecular Clouds and Giant Molecular Clouds (GMCs).

The "Molecular" means that most of the atoms are combined into molecules rather than being free atoms.  So the hydrogen is mostly H2 and not H.  This is really important, since a cloud comprised of molecules can radiate away the heat of collapse much easier than a cloud of atoms can. If that heat isn't radiated away, the cloud won't collapse.

As far as the contents go, clouds int he solar neighborhood today have a composition very similar to that of the outer layers of the Sun: around 90% (71%) hydrogen, 9% (27%) helium .08% (1%) oxygen, .04% (0.5%) carbon and lesser amounts of things like nitrogen, silicon, magnesium, neon, iron and sulfur.  Note that these are atom% followed by (mass%).  See this for a table.

Astronomers would call it 90% (71%) hydrogen, 9% (27%) helium and less than 1% (2%) "metals".

Much earlier in the universe's history the percentage of hydrogen was higher and the percentage of everything else was lower.   Since the Sun is a relatively new star, the gas it condensed from was "polluted" by the gas emitted by generations of older stars which enriched it with helium and "heavy metals".


Transverse Comoving distance I don't understand the link between the comoving distance and transverse comoving distance : how can they be equal ?
Here an example of definition that I have found :
Angular Diameter Distance :
The angular diameter distance $D_{A}$ is defined as the ratio of an object's physical transverse size to its angular size (in radians). It is used to convert angular separations in telescope images into proper separations at the source. It is famous for not increasing indefinitely as $z$ -> infinity; it turns over at $z \sim 1$ and thereafter more distant objects actually appear larger in angular size. Angular diameter distance is related to the transverse comoving distance by
$D_{A}=\dfrac{D_{M}}{1+z}(17)$
Some clarifications would be fine.
EDIT 1: @benrg If I have well understood, the transverse comoving distance is simply equal to the comoving distance between object that has emitted at redshift "z" and us that we reveive now this light, isn't it ? If yes, why complicate the things by using the qualified word "transverse" into "transverse comoving distance" ?

Angular diameter distance is the reduced circumference of the circle, centered at our location, on which the object was located when it emitted the light (or the reduced area of the sphere if you prefer). If the universe is spatially flat then this is the same as the metric radius of the circle. In general it's related to the radius by the function $S_k$ defined here, where $k$ is the Gaussian curvature at that time.
Transverse comoving distance is the same thing scaled up by $1{+}z$, i.e., the reduced circumference of the circle on which the same object is located at the present cosmological time, if it had no net peculiar motion in the mean time. It's therefore equal to (radial) comoving distance if the universe is spatially flat.


What is the mnemonic reason behind b being galactic latitude? (in the Galactic Coordinates frame) I'm not sure if this is a question that has been posted before, and I'm also not sure if the answer is really mnemonic. If that's the case, I'd like to understand why we assigned $b$ to latitude anyways, so my brain can remember it more easily.
Thanks!

Länge und Breite are the German words for longitude and latitude. Thus 'b' seems like a natural choice for latitude and 'l' for longitude. A century ago when traditional choice of variable names were chosen by vote of feet, a reasonable amount of research papers were still published in German or people with some form of fluency in German. At the same time it's distinct enough from the variable names used in other coordinate systems which also have longitude and latitude (like the terrestrial where $\lambda$ and $\theta$ are commonly used), so the cause for confusion is reduced - even when you don't know German.


How to calculate the heliocentric velocity of an object? My task is to calculate the heliocentric velocity of an object to correct for wavelength in a spectrum. I have the RA, Dec and observation date, and I know that the  NED velocity correction calculator(https://ned.ipac.caltech.edu/forms/vel_correction.html) will do the job, and I found out the equation here (https://ned.ipac.caltech.edu/help/velc_help.html#notes), but honestly I don't have a clue where these parameters come from. Could some one give me some hints on the reasoning behind it? Thanks!

The heliocentric velocity $V_\mathrm{H}$ of an object is its velocity wrt. the Sun. When you measure an object's velocity, you measure it in the reference frame of Earth, which revolves around the Sun with ~30 km/s (varying a bit from aphelion to perihelion), so convert to $V_\mathrm{H}$ you need to know the time of the year of the observation (unless the line of sight toward your object is exactly perpendicular to the ecliptic plane), as well as the angle between the line of sight, and the line of sight toward the Sun. This involves a number of sines and cosines that you can find in e.g. Barbieri (2006).

If you further want to convert from $V_\mathrm{H}$ to the reference frame in which the Milky Way's center is at rest, the Local Group is at rest, or the Cosmic Microwave Background is isotropic (the "cosmic" frame), then you use the formula you link to, adding a similar term as described above, but instead using the velocity (i.e. speed and direction) of the Sun wrt. to the given frame.


Why were solar constant measurements before TSIS-1 all about 0.3% high? Phys.org's Solar energy tracker powers down after 17 years says:

"The big surprise with TSI was that the amount of irradiance it measured was 4.6 watts per square meter less than what was expected," said Tom Woods, SORCE's principal investigator and senior research associate at the University of Colorado's Laboratory for Atmospheric and Space Physics (LASP) in Boulder, Colorado. "That started a whole scientific discussion and the development of a new calibration laboratory for TSI instruments. It turned out that the TIM was correct, and all the past irradiance measurements were erroneously high."
"It's not often in climate studies that you make a quantum leap in measurement capability, but the tenfold improvement in accuracy by the SORCE / TIM was exactly that," said Greg Kopp, TIM instrument scientist for SORCE and TSIS at LASP.

Question: What was it about either the past measurements or their analysis or calibration that made their measurements of the Sun's output about 0.3% high? What TSI's improvement primarily instrumental, or due to better calibration?

From the abstract of Kopp & Lean (2011) "A new, lower value of total solar irradiance: Evidence and climate significance":

Scattered light is a primary cause of the higher irradiance values measured by the earlier generation of solar radiometers in which the precision aperture defining the measured solar beam is located behind a larger, view‐limiting aperture. In the TIM, the opposite order of these apertures precludes this spurious signal by limiting the light entering the instrument.

(emphasis mine)
For example, in the case of the ACRIM instruments used on the Solar Maximum Mission, the Upper Atmosphere Research Satellite, and ACRIMSat:

Notably, for the ACRIM instrument NIST determined that diffraction from the view‐limiting aperture contributes a 0.13% signal not accounted for in data from the three ACRIM instruments [Butler et al., 2008]. This correction lowers the reported ACRIM values, resolving part of ACRIM's difference with TIM. In ACRIM and all other instruments, the precision aperture used to define the measured solar beam is deep inside the instrument with a larger view‐limiting aperture at the front, which, depending on edge imperfections, in addition to diffraction can directly scatter light into the absorbing cavity. Additionally, this design allows into the instrument interior two to three times the amount of light intended to be measured; if not completely absorbed or scattered back out, this additional light produces erroneously high signals. In contrast, the TIM's optical design places the precision aperture at the front so only light intended to be measured enters (Figure 4a).

The paper goes into a bit more detail about the testing done to determine the reasons for the instrumental differences.


Could discoid galaxies be expanding? I understand that astronomers once thought that the material in the disc of a galaxy was moving around the galactic centre (where most of the mass was thought to be) in roughly circular orbits. The circular motions would be explained by the combination of inertia and gravitational acceleration towards the centre. This follows the Kepler/Newton model which describes/explains the orbits of planets around the Sun.  In a Kepler/Newton system the tangential or transverse velocity of a low-mass orbitting object depends on the radial distance of the high-mass object from the centre.  Objects further out must have lower velocities so that the weaker centrally-directed acceleration at that distance can keep the object moving in a roughly circular track.

In the centre of the galaxy the velocities are low and increase rapidly with distance away from the centre.  This is understandable as the mass of the central bulge of the galaxy  is not concentrated at the centre, it is spread out over a relatively large volume.

However it is now well-known that the measured velocities of visible material in the disks of discoidal galaxies do not vary as expected for a Kepler/Newton system.  At  a certain distance where the velocities would be expected to start decreasing (as per curve A) they either level out or continue increasing at a slow rate (as per curve B).

The current explanantion is that a large mass of Dark Matter extends throughout the galaxy in such a way as to produce non-Keplerian behavior.

QUESTION

But isn't there another possible explanantion of the non-Keplerian velocity profile namely that the ordinary detectable material (e.g. gas, dust, stars) in the disk is not completely gravitationally bound to the galactic centre and is actually gradually moving outwards along a spiral path?

No, for several reasons.

there are no radial velocities in the required range observed, for example in the Milky Way galaxy.
the reservoir of stars moving out from the centre would quickly be drained, so you would need to magically generate them at the centre -- this is much worse than postulating dark matter.
you'd expect huge amounts of stars and gas that have moved out to large radii.

The only serious alternative to dark matter is some modification to Einstein's/Newton's theory of gravity. There are several candidates, but none is really convincing (though arguably, dark matter is not too convincing either).


What would stargazing be like at the edge of our galaxy? Let's say, for the sake of speculation, we acquire the technology necessary to travel to the outer edge of one of the spiral arms of our galaxy. We go to a star in this "outer rim" and land on a planet orbiting a star there. What would we see when our side of the planet is facing outward from the galaxy? (Assume no atmospheric interference.) The answer will, of course, depend on the galactic coordinates of the star we choose to visit. Provided a particular set of coordinates, how much do we know about what we could see with the naked eye or through some form of optical instrument? I'm more interested in our ability to answer the question, rather than the specific objects we'd see (although they're pretty interesting too). You're welcome to select an arbitrary but still useful and appropriate boundary as the edge of the spiral arm.

Here is a render of what an observer might see near the edge of our galaxy. The "edge" itself being rather fuzzy, let's assume that we are in an area of the edge dense enough where a few occasional stars are still visible. (As far as we know, intergalactic space contains objects and stars, but so few that seeing even one naked-eye star in the sky there would be uncommon)

Looking at the sky towards the opposite side of the Milky Way, one could see the Magellanic Clouds, and a small quantity of naked-eye stars:

The Milky Way side of the sky would be interesting to look at:

Now, as interesting as these artistic renditions may be, there is very little that we know about actual objects in these regions. Seeing individual stars there is quasi-impossible from here; we have a general idea of how dense some regions of the galaxy are, but for the rendition part here, it is mostly procedural based on some general facts and statistics.

The software used for these renders is called Space Engine and contains all catalogued objects, and it procedurally generates objects to fill in the gaps based on what we know about space.


What are gravitational waves actually? I read about the LIGo experiment recently, but i didn't understand their significance. Although, it's clear to me that the gravitational waves do exist and are caused by large accelerating bodies in space.

but how are they significant to Humans as we do not have any mechanisms for harnessing these waves and Don't know what use they may be of.

It's a "ripple" in spacetime. Imagine the traditional picture of planets as marbles sitting on a sheet--heavier bodies push the sheet down farther and deeper. If you place a bowling-ball on one end of the sheet, the entire sheet is affected, but not instantly. It takes an amount of time for the sheet to be moved--this movement can be seen as a wave, with a wavefront, a speed, etc.

In reality, massive gravitational bodies don't magically appear in space, (that we know of... maybe God is planning to drop a giant marble in our Solar System) but gravitational waves are still generated by any change in placement of mass or any acceleration (Einstein's theories that show that gravity and acceleration are indistinguishable).

So you get gravitational waves caused by movements of planets around the sun, etc. That's going to be relatively smooth though, so hard to detect. What we really need is something that creates sharp waves--two giant black holes which are orbiting each other at an incredibly fast RPM is perfect since they're massive and will create a lot of high/low waves as they rotate that we can then detect.

As we improve the detection technology, we may be able to use gravitational waves to detect all sorts of things besides black holes, it just depends on how much we can measure. The nice thing is that you don't need "line of sight" to measure with a gravity wave. So if God drops a giant marble on the other side of the Sun, we could detect it even though we couldn't see it.


What does "Normalized to Si 10 to the power 6" mean? I'm writing a space game that I want to give a more "realistic" feel. When looking for a reference as to distribute elements in planets (for mining, etc) Instead of finding the more usual percentage tables, it seems that all elements are presented in a "normalized to Si 10 to the power 6", and furthermore, it seems that the data is presented as number of atoms, instead of actual mass. I've been searching this for days, and even the "parts per million/billion/trillion" tables sometimes just jump for percentages, where is is again no clear if what is being referred to are masses or number of atoms. Could someone clarify this, or better, point me to a mass percentage table of elements in a solar system/galaxy for reference? Thanks in advance.

The table you referenced is normalized so that the abundance of Silicon (Si) is exactly $10^6$, and all the other numbers are related to that.

"Abundance" in this table refers to the number of atoms.

Please note that this table is for the entire Solar System, taking into account the huge amounts of Hydrogen and Helium in the Sun. It will not be appropriate for creating Earth-like planets.


Can we (theoretically) spin the black hole so strong that it will be broken apart by centrifugal force? I can't imagine the forces involved in black holes' lives. So please, help me to find out, if it is possible or not to destroy black hole in this specific way.

Can we (theoretically) spin the black hole so strong so it will be broken apart by centrifugal force?

For a Kerr-Newman (rotating, charged, isolated) black hole of mass $M$, angular momentum $J$, and charge $Q$, the surface area of the event horizon is given by
$$A = 8M\left[M^2 + (M^2-a^2-Q^2)^{1/2} - Q^2/2\right]\text{,}$$
where $a = J/M$. An extremal black hole occurs when $M^2 = a^2 + Q^2$. Beyond that, if the black hole is even more overspun or overcharged, is an "overextremal" Kerr-Newman spacetime, which wouldn't really be a black hole at all, but rather a naked singularity.

Thus, I interpret your question as asking whether or not a black hole can be be spun up to the extremal limit and beyond, so as to destroy the event horizon. It's very probable that it can't be done.

Wald proved in 1974 that as one flings matter into a black hole to try to increase its angular momentum, the nearer to an extremal black hole it is, the harder it is to continue this process: a fast-spinning black hole will repel matter that would take it beyond the extremal limit. There are other schemes, and though I'm not aware of any completely general proof within classical general relativity, the continual failure of schemes like this is well-motivated by the connection between black hole dynamics and thermodynamics.

For example, the Hawking temperature of the black hole is $T_\text{H} = \kappa/2\pi$, where
$$\kappa = \frac{\sqrt{M^2-a^2-Q^2}}{2M\left(M+\sqrt{M^2-a^2-Q^2}\right)-Q^2}$$
is the black hole's surface gravity. Thus, even reaching the extremal limit is thermodynamically equivalent to cooling a system to absolute zero.


Precautions when retrieving material originated from space I'm assuming that when retrieving a meteor for study and later display, there are precautions taken against possible radiation coming from it. Is my assumption correct and are there other precautions taken such as protection from biohazards? (e.g. is it not possible for a meteor to carry a foreign viral agent?)

Meteorites are composed of rock and iron. They are not radioactive, at least no more so than any other stone.  A banana is probably more radioactive than a meteorite.

You do not need protection for foreign viral agents either. Meteorites are stones that have been orbiting in space, usually for billions of years. They do not contain any living things. The have also passed through the atmosphere so fast that their surface can begin to melt from the friction.

You can pick up a meteorite with you unprotected hands

For pedants: A meteorid is "space stone". Meteor is the flash of light in the sky as a meteoroid passes through the atmosphere. Meteorite is the stone from space that you find on the ground. It is the meteorite that you retrieve for study.

There is a suggestion that ancient meteorites bought the ingredients for life to Earth, or even seeded life on the planet, but it is speculative as we don't have the evidence to prove it.


Luminosities and Apparent Brightness of Two Stars of which one is 8 times the distance of the first star Suppose two stars have the same apparent brightness, but one star is 8 times further away than the other. What is the ratio of their luminosities? Which one is more luminous, the closer or the further star?

Welcome to Astronomy Stack Exchange!
Let the star which is farther away be A, and the fainter one be B.Since this perceived brightness is something called flux density (sometimes referred to as just flux), which the following formula holds its own for: $F = \frac{L}{4\pi r^2}$
($L$ is the luminosity, and $r$ is the distance) it would mean the luminosity ratio can be calculated in the following manner:
$F_A = F_B$
$\frac{L_A}{4\pi r_A^2} = \frac{L_B}{4\pi r_B^2}$Substituting for $r_A = 8r_B$:
$\frac{L_A}{64r_B^2} = \frac{L_B}{r_B^2}$
$L_A = 64L_B$
The farther star is much more luminous than the closer one.
P.S. This is only if there is no extinction, if it is present, I cannot make an estimate without more details.


By convention, when observing an exoplanet, is the observer's line of sight (on Earth) perpendicular to the plane of sky? This is just a quick question. In this picture of the orbit of an exoplanet, is the observer's line of sight perpendicular to the plane of reference, which is in the +z direction on the diagram? In the case when the inclination equals to $90^{\circ}$, it is ideal for us to measure the radial velocity. Is that right?

Yes, the reference plane for exoplanet orbital inclination is our sky plane.
From the NASA Exoplanet Archive documentation:

  The Observed Inclination is the orbital inclination with respect to the plane of the sky....
  0 degrees correspond to an orbit in the plane of the sky, face on with respect to our line of sight.
  90 degrees corresponds to an orbit that is edge-on with respect to our line of sight, perpendicular to the plane of the sky.

If you query the archive for exoplanets with discovery method "Transit," you get inclinations mostly between 85° and 90°.

If the only available observation is a fluctuation in the star's radial velocity, then the planet's mass m and orbital inclination i are unknown, but a minimum mass m sin i can be computed.
The actual mass is greater unless i = 90°.

The orbit diagram from Wikipedia can apply to extrasolar systems with one change: the reference direction is celestial north instead of ♈.


What is a low mass Black Hole? Is a Low Mass Black Hole also called a Stellar Mass Black Hole?
Can Low Mass BH be also called Primordial BH?

(If low mass BH are Stellar Mass BH-What is their size in terms of the mass of the sun?)

Stellar Mass black holes are (generally) black holes which form from collapsed stars. They get their name because the mass of the black hole is in the order of the mass of stars. I don't believe they would fit your definition of "Low Mass," as they are believed to be fairly standard in size.

Primordial black holes are theoretical black holes which were believed to have been created during the early moments of the universe. My understanding is that these would be a subset of quantum black holes, which also seem to be called micro black holes.

See this related question, in which the answer covers types of blackholes.

I'm not quite sure what the lower-end masses of observed black holes are, but in theory we could say that the lowest mass of a stellar black hole is just beyond the upper limit of mass for neutron stars, as above that limit they would be collapse into a stellar black hole. This is known as the TOV limit, and is approximated to be about 1.5-3 solar masses1, so we could say that the smallest stellar black holes could have a mass of about 1.5-3 times that of the Sun.

Note however, that this doesn't mean that the Sun is nearly massive enough to become a black hole. The mass of the resultant stellar black hole is only a small portion of the original mass of the star which created it.

1 I. Bombaci (1996). "The Maximum Mass of a Neutron Star". Astronomy and Astrophysics 305: 871–877.


What's in the center of a Galaxy? Ok, so it's supposed that there is a black hole in the center of any galaxy, that makes my head explode because:

There is light in the center of the galaxy, but it is supposed that black holes, also attract light.
For a black hole to exist there should be a star explosion, so what size does the star needed to be to create such a galaxy.
Why does the black hole doesn't eat us, I mean what objects are pulling us away from the extreme gravity of the black hole.

I'm not god, I didn't study Astronomy, I'm just a kid fascinated by the Universe. (This question should be dumb for anyone who studied Astronomy)

But if you think my question is dumb, (which isn't) or if you have any recommendation for further asking, let me know!

At the center of our galaxy is a powerful radio source named Sagittarius A*, which is believed to be a super massive black hole (SMBH). This blackhole would contain far more mass than your run-of-the-mill supernova remnant. Our galaxy is believed to contain a SMBH containing the mass of likely a bit above 4 million times the mass (Gillessen) (2) (Ghez) of our Sun. For reference, I don't think we've ever discovered a star more massive than 600 times that of our Sun.

It's also important to understand that while many people look at blackholes as mystical or all-consuming, they actually have to follow the same rules as everyone else in the stellar neighborhood. The stars that make up our galaxy don't fall into the black hole for the same reason our planet doesn't fall into the Sun. Our star orbits black hole, our star system's velocity in equilibrium with the attractive force of the galaxy's center of gravity. This should hopefully resolve point 3.

For point 1, we should make clear that the 'black' part of the black hole is only true once you cross the event horizon. This is the case because at this point the escape velocity to escape the gravity of the black hole because greater than the speed of light. The light that isn't within the event horizon and is moving away from it is free to escape. So we can see light around it.  But why is there so much light? Well, as it happens there are a lot of rather young and large stars in this area. It is not completely understand why this is the case. Lots of stars, lots of light! There are other factors that can contribute to this too, such as there simply being lots of stars between us and the center, not just in the center itself. The accretion disk of a black hole can also be exceptionally bright. Hopefully that clears up part 1.

Now for part 2. As far as I know, we don't really have any way of determining exactly where our SMBH came from originally. Black holes aren't necessarily formed just from a supernova event, there are a handful of other ways they can be created in nature. What is apparent, however, is that SMBHs contain far too much mass to be from a single star. It has probably consumed plenty of other black holes to grow to what it is now.

One interesting and notable difference between the comparison of a star system and a galaxy is the distribution of mass. While our Sun is believed to contain 99.8% of the mass of our solar system, the SMBH at the center of the Milky Way is not nearly as massive as the total mass of the Milky Way. The ratio can vary a lot, and there are some galaxies which are believed to host no SMBH at all.

Gillessen, Stefan et al. (23 February 2009). "Monitoring stellar orbits around the Massive Black Hole in the Galactic Center". The Astrophysical Journal 692 (2): 1075–1109.

Ghez, A. M. et al. (December 2008). "Measuring Distance and Properties of the Milky Way's Central Supermassive Black Hole with Stellar Orbits". Astrophysical Journal 689 (2): 1044–1062.


Create Position-Velocity diagram from a velocity field Can someone explain to me how I can generate a Position-Velocity (PV) diagram from a galaxy's radial velocity field? I understand that usually PV diagrams are generated across the major axis of the galaxy, but I am not quite sure why they appear as a distribution. For example:

Such diagrams are usually called rotation curves, and show the velocity of stars/gas in a disk galaxy viewed edge-on along the line of sight (LOS) as a function of distance from the center.

That is, the $x$ axis gives the distance of the stars from the center ($x=0$) of the galaxy, and the $y$ axis gives the measured velocities at a given distance. Along a given LOS you'll find not a single velocity, but a (Gaussian-like) distribution of velocities, both because there is a certain scatter in velocities, but also because you only measure the component of the velocities projected onto your LOS, while both speed and direction change along the LOS, as seen here:

If they all had the same velocity along the LOS, the plot you show would be a thin line; instead it is smeared out.

Due to mass being concentrated in the center, there's a steep increase in velocity at low $x$. If there were only the mass you can see, the velocity would decrease at larger $x$ due to the larger distance from the central mass. The reason it stay more or less constant is thought to be the extra, invisible mass known as dark matter.

The reason the left side of the plot is similar to the right, but mirrored not only across the $y$ axis but also the $x$ axis, is that on this side, the stars move toward you, while on the right side they move away from you. The whole galaxy evidently moves away from you at $500\,\mathrm{km}\,\mathrm{s}^{-1}$, since this is the velocity at $x=0$.

So, to recap: How do you generate such a plot? You measure the distribution of velocities along many lines of sight in the galactic plane, from one side ($-R_\mathrm{gal}$) to the other ($+R_\mathrm{gal}$). For instance, at $x=14\,\mathrm{kpc}$, you'd measure $V=675\pm12\,\mathrm{km}\,\mathrm{s}^{-1}$:

In practice, images like this is made by using a grism which disperses the light according to its wavelength, together with a slit placed along the galactic plane, such that you block the light from the rest of the field. This is called slit spectroscopy, in contrast to "normal" spectroscopy where you block all light except for a little hole placed on some object (e.g. a star), or integral field spectroscopy, where you get the whole spectrum behind every pixel in an image.


Why are there so many seemingly blue-shifted galaxies in deep space NASA released this image of deep space. The way light shifting vis-a-vis universe expansion is supposed to work is by making the furthest galaxies appear mostly red-shifted. But NASA's image shows somewhat of an even balance between blue and red shift. Why is that?

You cannot gauge the redshift of a galaxy by looking at a false colour image. The images taken through different filters are stacked and colourised to suit. You can say that the blue galaxies are indeed bluer than the red galaxies, but there is no absolute scale with which to judge redshift by eye.

Secondly, there is no detail in the NASA web page, but the ACS and WFC3 cameras have near infrared capabilities. So I would think that this image is a visual false-colour image of information that extends well redward of what the eye can perceive. So even the things that look blue might have a spectrum that peaks at redder wavelengths, whilst anything that looks red might actually be infrared!

However, beyond this, in order to judge what the redshifted appearance of a galaxy would be, you would need to know what the galaxy looked like with zero redshift. That is, light that is emitted in the ultraviolet could be redshifted into the visible part of the spectrum. It is entirely possible that visual colour of a redshifted galaxy would not change very much at all if the galaxy emitted lots of UV light in its frame of reference or equally, if it emitted lots of very red light that was then redshifted out of the telescope's sensitivity range.

Finally, there could be some genuine astrophysics going on. Many distant galaxies are bluer than nearby galaxies because they are undergoing intense star formation. Massive star forming regions emit copious UV light that is redshifted into the optical.

Finally, finally! Many of the galaxies in the picture will be quite close and will not be very redshifted.


How big and how far away would the moon be if earth was the size of Jupiter? If the earth were the size of Jupiter, what would its 'moon' be in both relative size and distance? I did some quick figuring and think I roughly figured out that if placed at the position of the sun, Jupiter's moon would be Neptune in both relative size and current distance from the sun. That seems very distant and I'm questioning my calculations. The moon is 27% as large as the earth and is ~27,700 times the earth's diameter distant. Jupiter is ~87,000 miles in diameter. That would make it's moon ~23,500 miles in diameter and ~2.41 billion miles distant. Neptune comes relatively close to those results. It would be slightly larger than our moon (30,600 miles in diameter vs. 23,500) and slightly farther away (2.8 billion miles away vs 2.41) but still close for comparison purposes.  That puts the moon relatively MUCH more distant from the earth than what I would have thought off the top of my head.

The basic numbers are:

Radius of Earth 6370km
Radius of Moon 1740km
Earth-moon distance 384000km and
Radius of Jupiter 71500km

So the scale factor is 71500/6370 = 11.22
The radius of the moon scaled is 1740×11.22 = 19500 km, and its distance becomes 384000×11.22 = 4310000 km
For comparison, Neptune has a radius of 24600 km and a distance from Jupiter of that is always more than 3700000000 km
The size of the "moon" is comparable to Neptune, but the distance from Jupiter to Neptune is nearly 1000 times greater than the distance to the "moon".


Why aren't the farthest objects close to each other? The last years we find some objects like this : MACS0647-JD which are more than 13 billion light-years away from us.

This means that the picture we get was created only some million years after the Big Bang.

Why can we see objects like this in all directions? Shouldn't those objects be close to each other?

the big bang did not happen in one place, it happened everywhere, and since then space has been expanding. so the stars that where formed shortly after the big bang where everywhere in space, but space was smaller. therefore you can see them everywhere if you look 13 billion light-years away, because they where everywhere.


Calculate depth and duration of exoplanet's transit using Python, astropy or lightkurve I am creating a program for the analysis of exoplanets in Python using Astropy and Lightkurve libraries. I have light curve of specified star and I would like to calculate depth and duration of planet transit. There is example for Kepler 10 bellow with these steps:

Load target pixel image for Kepler 10,
Convert it to flattened time series,
Calculate period using periodogram,
Create folded light curve.

import lightkurve as lk

target_pixel = lk.search_targetpixelfile("Kepler-10", quarter=1).download()
lightcurve = target_pixel.to_lightcurve(aperture_mask=target_pixel.pipeline_mask).flatten()
periodogram = lightcurve.to_periodogram(method="bls", period=np.arange(0.5, 10, 0.001))
period = periodogram.period_at_max_power

folded.scatter()
print(period)

The output is (I marked the depth and duration in red):

Now, how can I calculate duration of transit (approximately from -0.15 to to 0 on x-axis)? Interval from -0.5 to 0.5 should be whole period (0.837 d), so transit duration should be +/- 0.13 d.

And what about transit depth? I was thinking about:

min = np.min(folded.flux)
median = np.median(folded.flux)
median / min - 1 # 0.00038

However, the minimum point is not the one in the middle of the transit due to noise, so transit depth may not be accurate.

Calculation of both, depth and duration, is usually done not on the raw data but derived from a fit to the data.

In your last three lines of code you also calculate the average / medium over all data while you should calculate the uneclipsed mean or median flux only for the non-transit time (with using median it possibly has only a tiny influence, yet it might).

As a first and crude step, I'd de-noise the data by applying a floating average filter over the data; you will have to test for its width and see what gives you best results: you don't want to average out features, but you want to average-out noise.

The better approach is to not smooth but actually fit a physical model to the data which takes into account the typical light curve behaviour of a transit. For an implementation I can point you to pytransit (reference paper). (Are you sure you are not re-inventing the wheel?). See also this paper by Maxted and Gill for a comparison of a few algorithms.


5 sec flare near Arcturus Hello fellow amateur and professional astronomers,

On July July 17, 2017, I was at an observing session in Avrig, Sibiu county, Romania (45.7288° N, 24.3784° E). At approx 11:30 pm I pointed my green laser pointer (class III, wavelength 532 nm +/-10, YL-Laser 303) towards Arcturus in order to help my colleague align the SW AZEQ 5 mount within the 3-star procedure.

While the laser beam was on, at approx 1 arcminute beneath and to the left from Arcturus, with my naked eye, I observed a point of light (white/blueish) lighting up to the point which I estimate to be close to the visual magnitude of Jupiter or slightly higher for appprox 5 seconds, and then the "flare" faded out to black. As there was no "regular" star-like remanant, I was left to conclude that the initial flare did not originate from a star.

I ruled out a reflection of the laser beam from a satelite (as too far for my laser's power), a plane or a bird (due to the white/blueish light I saw, its fixed position and apparent distance to my eye), as well the possibility of a meteorite burst (as it had no apparent trajectory, but it appeared as a fixed point).

I am so intrigued and curious as to what possible optics or physics could have been involved and explanations or ideas you may have to explain the phenomenon?

I searched online for others' possible similar observations without success.

My first thought is that you saw a flare from a satellite, meaning a favorable reflection off of a solar panel or something. I use the software "Heavensat", and it shows the satellite Nextsat (Norad number 30774) passed very close to Arcturus at 23:40 pm from your location (45.7288° N, 24.3784° E). Note that I am assuming that your time is 3 hours ahead of UT (Universal Time).

I do not know if Nextsat typically flares, although reflections from satellites are common. Nextsat is in a low altitude orbit, so its motion would normally be detectible. Perhaps the duration of the brightening was shorter than you estimate (so that the motion was not detectible), or it was so unexpected that you did not notice the motion.

My second thought would be a head on meteor like you suggested.


Could the Moon naturally capture a large asteroid and keep it as a subsatellite? For how long? (Edited at James K's recommendation to stay on-topic for this SE.)

From my reading on this SE and from Googling I've gotten conflicting reports on the viability of subsatellites this close to the Sun, but a naive application of the Hill Sphere equation suggests that if it got captured at all, a large asteroid could theoretically orbit the moon in a stable orbit.

What I don't know is this:

For how long? Longer than a couple hundred thousand years, such that humans on Earth wouldn't have known anything else? Longer than a million years? Probably not longer than a billion, due to the tidal effects decaying the asteroid's orbit, but is that a bad assumption?
How eccentric would the asteroid's orbit be? Would it be unlikely for it to orbit with negligible eccentricity, whether on the ecliptic or not? If so, what kind of eccentricity/orbital inclination would be reasonable?

I'm going to discuss only our moon, to keep this on topic.

This sort of orbit could be stable on the sort of timescales that you want.

Objects in Low Lunar Orbit will likely suffer from the irregular gravity of the moon. There are "mass concentrations" as a result of ancient meteorite impacts. These "mascons" distort the orbits of objects in low orbit, and cause them to collide with the moon in a relatively short time. However at 24000km the gravity field is smoothed by your distance from the surface. You should not have problems with mascons at this distance, at least not over 100,000 years or so.

The Hill sphere of the moon extends to about 60000km, so your orbit is well within the Hill radius, and so can be stable.

The moon spins slowly, so tidal effects would tend to cause your asteroid to lose altitude, and eventually hit the moon. The timescale of this would be longer than a few 100,000 years

Getting your asteroid in orbit is very tricky. You would need to hypothesise double asteroids transferring momentum between each other and the moon, and one being lost while the other ends up in lunar orbit.


Occurrence of Venusian transits As transits of Venus come in pairs, each separated by 8 years, wouldn't that imply an accurate ratio between Earth's and Venus' revolution period? I've found the ratio to be about 0.681. I presume one could see a resonance of 2/3 in that number, but as other resonances in our solar system as for example Thethys and Enceladus (4:3) are far more accurate, could it be "just" coincidental? And if so, why the regularity in the transits?

Quoting http://eclipse.gsfc.nasa.gov/transit/catalog/VenusCatalog.html

"When a transit of Venus occurs, a second one often follows eight years later. This is because the orbital periods of Venus (224.701 days) and Earth (365.256 days) are in an 8 year (2922 days) resonance with each other. In other words, in the time it takes Earth to orbit the Sun eight times, Venus completes almost exactly thirteen revolutions about the Sun. As a result, Venus and Earth line up in the same positions with respect to the Sun. Actually, the two orbital periods are not quite commensurate with each other since Venus arrives at the eight year rendezvous about 2.45 days earlier that Earth. After the third eight-year cycle, Venus arrives too early for a transit to occur."


How to attach binoculars on starwatcher tracker? I am a complete newbie on trying to figure out how to attach the binocular on the starwatcher tracker.
I am having problem with the angle. This is the issue I am having:

I align the skywatcher tracker to the north star (Polaris), (I am at 40 degree longitude)

I attach the L shaped bracket on the skywatcher tracker. This L shaped will be able to support my binocular.  Once I attach the binocular which it sits parallel on the short leg of the "L" shaped bracket i.e. parallelly aligned to the pole star.

Now the problem is as as follows: I have the polar axis. Also consider a plane which is perpendicular to the polar axis passing through my center of my tripod. I am able to view objects on the side of the plane which has the north star. But I am unable to view any objects on the opposite side. If I turn my binocular around it will have towards the ground instead of the sky.

Am I missing an attachment here, or do I have something wrongly configured? Any help or comments is much appreciated.
Thanks in advance.

photo added. Say the binocular is focused at the north star. But the object I am trying to watch is behind the plane which is defined by the green L-bracket in the photograph. the binocular can rotate about the polar axis (which is perpendicular the plane defined by the green L-bracket)

I have two degrees of freedom without disrupting the polar alignment. I can rotate along the x or y axis. If I try to rotate anything using the parts in the circle I will loose polar alignment.
In order to observe something right of the plane (defined by: on which the y axis lies that is the y-z plane): First I will have to rotate the binocular along the y-axis by 180 (but that makes it point to the ground). Second I will have to rotate along a z-axis to point it to the sky -- which I have no controls to do.

I own such a mount. You can turn it to wherever you want by loosening the right ascension (the black “knob” that’s touching the white part just below right of the greenish dovetail plate. You can also loosen the declination with the “knob” that’s around the mounting bolt.