Why are the stars in my image so green (false color image, HSO palette)? This is my first try in false color imaging. The target was IC5070 - the pelican nebula. I took around 3 hours of data (around 1h for Ha, SII and OIII). Then after stacking did an LRGB combination in PixInsight using Ha for red, SII for green and OIII for blue. For luminance I used the stacked Ha image.
Now my question is, why are most of these stars so green? It would suggest that they were really bright in the silicon spectrum right? If so wouldn't that mean that those stars are extremly silicon heavy?
(Ignore the vertical blue-green line, its a stacking artifact resulting from a camera defect on those pixels)

I do some narrow band imaging and believe the most likely cause is you have stretched the SII channel to far. Most likely it is the weakest of the three. Also, I never use a luminance channel on narrow band. But if you must, I would sum the three channels and not just use a single channel.


Could a contracting Universe create the redshift effect observed by Hubble? What if we just can’t see far enough and the redshift actually shows the galaxies speeding up towards a gigantic black hole to where everything is converging, repeating a cycle that took place countless times “before” the big bang?

Complete text in this link: http://www.therealebook.com/Astronomy/Redshift_ContractingUniverse.html

Even in a contracting Universe, there is a region around the observer where all the galaxies are moving away; the galaxies that are moving towards the observer could be too far to be seen. If we can demonstrate this, we could challenge the conclusion that the Universe is expanding, based on the redshift effect observed in the “nearby” galaxies.

Let’s start with a simpler representation: 1 dimension only (1D). Assume that all the galaxies move along a single line. We could represent an expanding and a contracting Universe as per the images below:

In Situation 1, we have the currently accepted model, where galaxies move away from each other, at increasing speed, which creates the redshift observed by Hubble.

In Situation 2, we assume that in the contracting Universe the galaxies have a speed that is inversely proportional to the distance from the center (as in a sink flow where the particles accelerate towards the drain), galaxies 1 and 3 will display a red shift effect, because they are moving away from each other, while galaxies 4, 5 and 6 will appear with blueshift effect. However, in order to see the blue shifted galaxies it is necessary to see beyond the center of the Universe, which would be not possible either because of the distance or because of the gigantic black hole in the center traps light from galaxies 4, 5 and 6 preventing it from reaching us at galaxy 2.

In fact, assuming that from galaxy 2 we cannot see beyond galaxy 1, it is impossible to distinguish between situation 1 and 2.
With this we have demonstrated that for a 1D Universe it is impossible to determine if it is expanding or contracting, simple because we cannot see far enough. What about a 2D space?

Consider two galaxies (1 and 2); consider that they form a angle beta, relative to the center of the universe (assuming that there is one); Consider that galaxy 1 is at the distance r from the center; consider that galaxy 2 is at a fraction of r (say xr) from the center.

We can calculate the distance between galaxy 1 and 2 as follows:

From the equation above we can calculate alfa and from  we can calculate d.

We also suppose that the speed along the radius can be calculated with the equation v = c / r, where c is constant.

Therefore we can calculate the position of the two galaxies after a period of time t, and measure the distance d between them to verify if d increases (redshift) of decreases (blueshift).

We can see that no matter what the fraction x is, there is always a minimum value for the angle beta, such that any galaxy within this angle will be moving away from the reference galaxy. Now assume that the universe is so big that we cannot see galaxies beyond beta. In this way, all that we can see is redshifted galaxies!

This means that even for a contracting universe, depending on how far we can see, it is possible to detect only redshifted galaxies. Does not this disprove the theory of expanding universe based on redshift?

Aside from being, I suspect, totally incompatible with GR, the model fails to match observations.

The universe appears to be homogeneous and isotropic. In the model you suggest, Hubble's law would be different depending on which direction you were looking.

In particular, looking away from the black hole, galaxy redshifts would increase to some asymptotic value, whereas looking towards the black hole they would increase to infinity.

If the argument is that the central black hole is so far away that we cannot measure this asymmetry (i.e. well beyond the observable universe, given the homogeneity of the cosmic microwave background - but then does the model have an explanation for the CMB at all?), then the model also needs to explain why galaxies would move in this particular way, since they would not "feel" the influence of the black hole.


What class of elements are providing the many electrons needed to make the H- ion? Continuous absorption by the H- ion reaches its maximum in the cool atmospheric layers of G stars.  At these temperatures, most of the hydrogen atoms are neutral; but what class of elements is providing the many electrons needed to make the H- ion?

Easily ionised ones like sodium and potassium. Not much from lithium because that is rare.

You don't need many free electrons. The fraction of H$^{-}$/H is very small, something like $10^{-7}$ in the solar photosphere. And indeed the number densities of sodium and potassium to hydrogen are of that order.


Position of a star on the HR diagram Does the position of a star on a HR diagram depend on the magnitude system/color index used?

I made a CMD of a particular cluster using HST WFC3/UVIS data, and when I plotted $F275W vs F275W - F606W$, a particular star (at a given RA and dec) was present near the blue straggler region, but now when I plot $F438W vs F438W - F814W$, this same star (at the same RA,dec) was found near the start of the RGB.

Why is this happening?

Does the position of a star on a HR diagram depend on the magnitude system/color index used?

Yes, of course. But note that HR diagram refers to a luminosity vs. effective temperature plot, while you're dealing with a color-magnitude diagram (or CMD).

  Why is this happening?

First of all, you should check that it's not your fault. If you see unphysical things, like the horizontal branch going beyond the red giant branch, you know you've messed up. This can happen with bad calibrations. But I think you're safe, it's not your case.

Then, I assume that the dependence of magnitude on filters is trivial.

Now, how colors change with filters?
Let's start in first approximation. Start imagining the Planck's function (here you go), and assume your filters are monochromatic. Recall that color is the difference of two magnitudes, i.e. \begin{equation*}m_1-m_2= 2.5\log_{10} I(\lambda_1)/I(\lambda_2)\end{equation*} where $I$ is the intensity and $\lambda_1>\lambda_2$. Now take the points on the Planckian corresponding to your two wavelengths, i.e.  $I(\lambda_1)$ and $I(\lambda_2)$, and draw a line between these two points. If the slope of the line is negative (resp. positive), then the ratio of intensities is less (resp. more) than 1, and the color is negative (resp. positive). If the ratio is 1, then slope and color are 0. So color is somehow proportional to the Planckian slope between the two $\lambda$s. Here comes the point: color is most sensitive when the peak of the Planckian is near the wavelength of the filters. If the peak is too far from the filter wavelengths, then the Planckian slope becomes almost constant, and the color loses its sensitivy to temperature. That's the reason why in some CMD you see the bluer part of the horizontal branch suddenly going down: the color saturates and doesn't catch the temperature variations. First take-home message: different colors probe different ranges of the Planckian, and hence are sensitive to different ranges of effective temperature.

But I think we didn't hit the point. Going to the second-order approximation, we need to consider the whole width of the filter. This fact brings in a lot of effects. For instance, color excess starts depending on the star's spectral type. That is, hotter stars are more reddened, because reddening is more effective at shorter $\lambda$s (e.g. the famous Whitford's law), so early spectral-type stars emit most of their flux at short $\lambda$s. Another effect is that broad filters are highly affected by absorption lines and molecular bands, and I think this is your case. You have photometry in two broad UV filters, i.e. $F275W$ and $F438W$, of a putative blue straggler, so a source with a lot of blue flux. Moreover, the OH absorption band falls right in the $F275W$ band, while several CH and CN bands fall into the $F438W$ band. So your colors are not only proxies for effective temperature, but also for chemical abundances. Notoriously, the UV filters have been used to separate multiple stellar populations in the CMDs, exploiting their different abundances. So, second take-home message: reddening depends on spectral type, but most importantly colors can depend on chemical abundances.

Finally, as other answers have pointed out, the position of a star on the CMD can be changed by other effects, such as variability.


How can we tell how many exoplanets a star has? Part of this answer (discussing the moving center of mass of our solar system) explains that this movement is one way we know a star has planets:

  Bonus: We use this phenomenon to find planets outside the Solar System! If a distant star is observed to 'wobble' or oscillate about it's mean position, we can use that data to infer the presence of one or more exoplanets, and calculate their mass.

It makes sense to me that we could tell that a start has some planet(s) in orbit around it, based on how they affect the star's position. But is this movement really precise and predictable enough to pin down how many exoplanets, and how large they are? Can we produce a "map" of a distant solar system based on this movement?

If needed, we can even restrict the discussion to planets that are "big enough" to detect. For example, when performing this exercise on our Sun, would we be able to accurately predict at least the four gas giants, even if we couldn't get Mercury?

What you are describing is a basic signal processing problem. The doppler shift that one observes is due to the motion of the star in the system around the system's centre of mass. The star will be influenced by the gravitational pull of each of the planets in that system, each of which exerts a gravitational pull that increases with the mass of the planet and decreases with the orbital radius of the planet.

The overall motion of the star will be the sum of the effects of all the planets. Importantly, the effect of each planet will have its own amplitude and will be periodic with a period equal to the orbital period of that planet.

Let's imagine that each planet is in a circular orbit (elliptical orbits are more complicated, but the principle is the same). Each planet would cause a circular motion in the star about the centre of mass of the planet-star pair, leading to an observable doppler signal which has the form of a sine wave with a period equal to the orbital period of the planet. The amplitude of that signal will increase with the mass of the planet and increases with decreasing orbital radius.

The overall signal is the sum over all the planets in the system. Fortunately the decomposition of this signal back into its individual components is a well-trodden problem in physics, electronics and many other fields and is known as Fourier Analysis. Whether you can successfully recover the original signals from each of the planets depends on how long you observed the system (ideally you want to observe for longer than the longest orbital period) and the amplitude of the signals compared with the noise in your observations.

In general it is easier to recover high-mass planets with short orbital periods and more difficult to recover low-mass planets with long orbital periods.

The image below might be helpful. It shows the track of the solar system centre of mass compared with centre of the Sun over a period of several decades. Notice how the Sun executes a complex trajectory (with respect to the solar system centre of mass) that is mainly caused by the orbit of Jupiter, but then there are smaller, superimposed, signals caused by the smaller planets. In principle, if you observed for longer than the period of Neptune and had a detector which gave perfect measurements, you could reconstruct how many planets there were in the solar system, what their orbital periods were (and then from Kepler's 3rd law, what the planet-star sepration was) and what their masses were (multiplied by the inclination of their orbits with respect to the line of sight of observation, which is generally an unknown in doppler measurements).

In terms of what we could currently see if we observed the Sun as a star: basically we would (assuming we observed for 20 years) detect Jupiter quite easily with a doppler amplitude of about 13 m/s. We would also see that there was a drift in Jupiter's signal due to the influence of Saturn, but we would have to observe for >Saturn's orbital period in order to confirm the presence of Saturn, its orbital period and mass. The inner planets produce an amplitude that is too small to be visible using the technology currently available. e.g. The Earth would produce a doppler wobble of amplitude $<8$ cm/s, but the current precision of doppler measurements is limited to about 50 cm/s.

The doppler amplitudes in m/s due to each of the planets (assuming we view them edge on) are:

Mercury <0.01
Venus    0.08
Earth    0.08
Mars    <0.01
Jupiter  12.5
Saturn   2.6
Uranus   0.28
Neptune  0.26

Thus with current technology, only Jupiter and Saturn are detectable.

Below I simulate what the doppler signal due to these two planets would be. I hope you can see that the overall signal consists of the superposition of two sinusoidal signals with different periods and amplitudes. Myriad computational tools are available to do the fourier decomposition to establish these.


Is it possible for stars not to rotate? I'm just curious about that the existence of non-rotating stars. Is it physically possible for a star not to rotate at all? Does magnetic braking eventually stop the stellar rotation?

No, this is not possible.

During the stellar formation, some angular momentum will always be present. And any "braking" effects (magnetic, relativistic, tidal etc.) will become weaker as the rotation slows down. So the rotation will never completely stop, because any forces slowing down the rotation will weaken as well.


Is seeing the apollo moon Landers via earth telescope that hard? There are interferometers, which are combining multiple telescopes together, with 0.001 arcsecond resolution. With changes, couldn't these observe the Landers? Is the reason why nobody is seeing the Landers related to effort and not actual technical limitation? Besides atmosphere interference, brightness, time use and other general issues.

Is seeing the apollo moon Landers via earth telescope that hard?

Yes, it still is.

If the resolution of an interferometric telescope is really 0.001 arcseconds, that translates to 4.85E-09 radians. At a perigee of 362,600 km that's 1.8 meters.

The top part of the Lunar Modules is about 4.2 meters wide which is a little more than double the resolution, so if there is some good contrast then it might appear in a reconstructed interferometric image.

However, those optical interferometers are not designed for imaging of extended objects. They only have a few telescopes and so are mostly limited to resolving more than one point-like object from each other, such as a double star or a star and an exoplanet.

With so few telescopes in an optical interferometer, a complex image like lunar terrain would be difficult or impossible to reconstruct without a model. For example, the images of the accretion disk of a black hole by the event horizon telescope required some input modeling to interpret the interferometric data.

Near sunrise or sunset however, the shadow of a lunar lander will be much larger. The lander is about 7 meters tall, so the shadow could be 4 or more meters wide by several tens of meters long. Since the Moon's orientation, as well as that of the lander and the local terrain is extremely well documented, then at each moment you could estimate what the shadow would look like in size, shape, and orientation. With that model, an interferometric image based on a few telescopes could be used to confirm the shadow.

However, using a model like that won't appease conspiracy theorists.

 Source


Converting from $\mathrm{W \: m^{-2}}$ to $\mathrm{Jy \: km \: s^{-1}}$ Suppose I observe a spectral line with rest frequency $\nu_0$ and integrated flux density $F = 10^{-26} \mathrm{W/m^2}$. Is the following line of reasoning correct?
$1 \: \mathrm{Jy} = 10^{-26} \: \mathrm{W/m^2/Hz}$ therefore $F = 1 \: \mathrm{Jy \: Hz}$. This can be interpreted as a peak flux in $\mathrm{Jy}$ multiplied by a linewidth in $\mathrm{Hz}$. Thus to convert to $\mathrm{Jy \: km \: s^{-1}}$ I just need to multiply the linewidth by $c/\nu_0$ to convert it from $\mathrm{Hz}$ to $\mathrm{km \: s^{-1}}$. Hence, $F = c/\nu_0 \: \mathrm{Jy \: km \: s^{-1}}$ (where $c$ is in units of $\mathrm{km \: s^{-1}}$).

You are essentially correct, but here's a more in-depth explanation:
Both units are a measure of the total flux $F$ of a light source in some interval, e.g. a spectral line or a broader region given by a filter. The flux is obtained by integrating a spectrum’s flux density, i.e. the flux per "bin". A spectrum is a distribution of photons as a function of the photons' energy or, equivalently, wavelength or frequency. The bins are then whatever you choose on your $x$ axis.
Typically you will use frequency $\nu$ for low-energy photons (radio to microwaves, with units of e.g. MHz or GHz), wavelength $\lambda$ for mid-range photons (with units of µm for infrared, and Å for optical to UV), and energy $E$ for high-energy photons (keV for X-ray and MeV gamma-rays). The flux density can then be denoted $F_\nu$, $F_\lambda$, and $F_E$, respectively, although the latter notation is not common; these photons are often so infrequent that one simply uses "counts", i.e. the absolute number of individual photons.
Note though that astronomers often are quite slobby and uses the word "flux" when they actually mean flux density (as well as for intensity, radiance, etc.).
Doppler broadening
But there is another very common way to show a spectrum, namely as a function of $\mathrm{km}\,\mathrm{s}^{-1}$. If you consider a spectral line from some electronic transition, it is emitted at a particular frequency (almost a delta function). But because the atoms emitting the photon do not lie still, but have a random distribution of velocities $v$, they will broaden the line. To first order, an atom moving with a velocity $dv$ with respect to the systemic velocity Doppler-shifts its emitted photon by an amount
$$
d\nu = \frac{dv}{c}\nu_0,
$$
where $\nu_0$ is the line center frequency, and $c$ is the speed of light. This means that we can also show a spectral line centered on $0$, with positive and negative velocities on the $x$ axis corresponding to higher and lower frequencies.
This is an informative way to show physical processes such as temperature, turbulence or bulk motion of the gas.
Units
Although astronomers tend to like cgs units, we also dislike huge and tiny numbers with too many powers of tens, plus we're in general not very consistent. Historically, flux in optical and UV light is typically given in cgs ($\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}$), but wavelengths are measured in Ångström, not cm, so flux densities are measured in $\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Å}^{-1}$.
On the other hand, radio astronomers tends to use SI units for flux ($\mathrm{W}\,\mathrm{m}^{-2}$) and flux density ($\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}$). However, since typical astronomical sources are extremely faint, radio astronomers often measure flux densities in jansky (Jy), where
$$
1\,\mathrm{Jy} = 10^{-26}\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}.
$$
Converting from W m–2 to Jy km s–1
Because of the fear of small or large powers of ten, and because as described above spectra can be shown as a function of both frequency and velocity, you can actually integrate you flux density in jansky over velocity, and thus the total flux of radio sources is periodically measured in $\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}$. With $F_\nu d\nu = F_v dv$, combining the equations above the relation is (beware of the similarity between $\nu$ and $v$)
$$
\begin{array}{rcl}
\int F_\nu d\nu & = & \int F_\nu \frac{\nu_0}{c} dv\\
\int \frac{F_\nu}{\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}} \frac{d\nu}{\mathrm{Hz}} & = & 10^{-26} \frac{\nu_0/\mathrm{Hz}}{c/\mathrm{km}\,\mathrm{s}^{-1}} \int \frac{F_\nu}{10^{-26}\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}} \frac{dv}{\mathrm{km}\,\mathrm{s}^{-1}}\\
\frac{F}{\mathrm{W}\,\mathrm{m}^{-2}} & = & 10^{-26} \frac{10^9\nu_0/\mathrm{GHz}}{3\times10^5} \frac{F}{\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}}\\
\frac{F}{\mathrm{W}\,\mathrm{m}^{-2}} & = & \frac{1}{3\times10^{22}}\frac{\nu_0}{\mathrm{GHz}} \frac{F}{\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}}
\end{array}
$$
For example, for the CO(1→0) line with a rest frequency of $\nu_0 = 115.3\,\mathrm{GHz}$, to convert from flux in  $\mathrm{W}\,\mathrm{m}^{-2}$ to flux in $\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}$ you multiply by $2.6\times10^{20}$.

You can read more about conversions in this document on the low-resolution spectrometer SPIRE on the Herschel telescope. Note though that this conversion differs from mine by a factor of $\pi$. I think this is because in frequency the instrumental resolution of SPIRE broadens a delta line to a sinc function which is normalized to $\pi$, not unity.


Relativistic effects in stellar dynamical systems I am curious, if anybody knows of any stellar dynamical systems/environments, where relativistic effects could play a dynamical role on the motion of these stellar systems? As a subquestion - are there any known important weak, but cumulatively strong effects?

In other words, when can relativistic effects invalidate the applicability of N-Body/Collisionless Boltzman/Gas/.. models based on newtonian gravity.

From these systems I would like to exclude the simplest well known case of compact binaries.

Stellar clusters around supermassive black holes are systems in which relativity likely plays a role. Currently, only bright stars can be seen in our own galactic center because there is a ton of neutral gas between us and the galactic center that obscures it. As a result, we only have a few "test particles" out of the many stars that actually orbit the black hole at close distances.

Nevertheless, measuring relativistic precession may be possible for a star with one of the closest-known pericenter distances to Sagittarius A* (the central black hole in our galaxy), S2, potentially within the next few years once enough data has been collected.

As to how relativistic effects can affect dynamics of the cluster, the precession induced by general relativity can suppress resonant interactions, including three-body resonances such as the Kozai. Depending on if these sorts of resonances are important compared to other relaxation processes, the relaxation time can increase significantly, resulting in the cluster evolving more-slowly over time. This can affect things such as the rate of mass segregation, tidal disruptions, and production of hypervelocity stars/S-stars.


How much radiation would something need to output to sufficiently affect the CMB? Let’s say that I had some device capable of outputting insane amounts of radiation, across the entire EM spectrum. If I wanted to affect the CMB to the point that, were Earth-based scientists to know that the device existed, they would be unable to gather any useful data from the CMB regarding the early universe, how much radiation would this theoretical device have to output? Peaking in what frequency? Where would be the optimal place to put it? Or multiple copies of it?

If someone who actually knows stuff about the CMB is scratching their head thinking I’m nuts, I probably am. I don’t know much about the CMB, except that it’s random EM waves that are leftovers from the Big Bang, peaking in the microwave frequency range. So if what I’m describing is absolutely implausible across all frequencies and amplitudes, I would appreciate if you could explain why, given my low level of knowledge on the topic.

In the event that for certain waves what I’m describing is possible, I would be curious to know if there’s actually anything in the known universe that could do it.

It is not very clear what you mean. There are many cosmic sources of microwave radiation that emit so much energy and/or are so close to us that they confuse measurements of the cosmic microwave background. For example "compact sources" such as giant molecular clouds, supernova remnants, H II regions; "diffuse" sources due to dust in the galaxy and in our solar system. In order to get a clean signal, these foreground sources have to be removed from the data. There are also more subtle cosmological effects caused by lensing and gravitational time dilation caused by the large scale structure in the universe.

There are many ways and techniques that are used to characterise and remove these backgrounds (e.g. see Ade et al. 2013).

The most confusing background source would be one that is diffuse, spread over the whole sky almost uniformly and with a spectrum that is very similar to the cosmic microwave background. The source that most closely matches these properties is dust in our own solar system. An imperfect subtraction of this may have led to controversy over possible large scale anisotropies in the CMB- for example see Schwarz et al. (2015); who discuss the problems of removing the signatures of solar system dust.


Asteroids in langrangian Points 4 & 5 There are Asteroids "trapped" in Jupiters Langrange points 4 and 5 called trojans and greeks. Are there any asteroids in the earths L4 and L5? Have we seen asteroids in Lagrange points of the earth and moon system? Why are only L4 and L5 stable and L1, L2 and L3 require corrections to remain at those positions?

There is one known earth Trojan - http://en.wikipedia.org/wiki/Earth_trojan

The reason L1, L2 and L3 aren't stable is covered here:   https://physics.stackexchange.com/questions/36092/why-are-l4-and-l5-lagrangian-points-stable

The reason the earth has so few L4 and L5 asteroids is cause other planets like Jupiter and Venus tend to destabilize the orbits - same reason why  Jupiter has so many.


How can a singularity rotate? So I've heard that all black holes that have been observed rotate to some degree. But if it has zero radius, how is it even possible for the black hole itself to be spinning? Or is it just all the junk inside the event horizon that's rotating?
Maybe it spins with infinite angular velocity and a finite angular momentum? Please explain how this works.

There are multiple solutions to general relativity which allow for multiple different types of black holes. The "normal" black hole you see most people talk about, with a zero-volume, point singularity, is known as the Schwarzschild black hole. If the black hole is spinning, the Schwarzschild solution no longer applies and you're talking about a different type of black hole. This new type is referred to as a Kerr black hole (other types include the Reissner-Nordström and the Kerr-Newman).

In a Kerr black hole, the singularity at the center is still zero-volume, but it is no longer a point. Instead it is a disk of zero height, often referred to as a ring singularity or "Ringularity". The angular momentum of the spinning black hole is then the angular momentum around the axis of rotation passing through this ring.


Has someone ever captured footage of the collision of a meteoroid and a meteoroid? For a video game I'm looking for footage of a collision of meteoroids or maybe asteroids, not planets. Even though I have searched for a while I couldn't find anything. So has anyone ever captured footage?

Deep Impact, the NASA comet probe, not the movie, may be as close as you can get to what you want now.
Here's a video of the action when 370 kg of copper hits a comet at 10.2 km/s.
There may be better video out there, but the movie trailers really mess up the search.


Determine spectral type of star from its properties How can I determine spectral type of star, if I know all its another properties? For Example, Rigel A has spectral type B8Ia.
Rigel has temperature 12 500 K. According to Harvard spectral classification, spectral class of this star is B (10 000-30 000 K) and subclass is 8 (12 000-14 000 K), so B8.
Are there any exact boundaries for Yerkes spectral classification as well? How can I tell if a star is super giant (I), bright giant (II) or giant (III)? And then, how can I tell if it is bright (a), normal (ab) or faint (b)?

There are no exact boundaries in temperature, luminosity, surface gravity etc. for spectral classes because the classification system works in a different way - it is fundamentally an empirical system, with classification based only on the appearance of features in the spectra.
The Yerkes or Morgan-Keenan (MK) system is based only on a set of standard stars and their spectra, and the type of any other given star (that is not one of those standards) is based only on where that star’s spectral lines fit into the two-dimensional matrix of standards. The introduction of this Annual Reviews article by Morgan and Keenan spells out the principle.
Of course those lines reflect physical effects
Iike temperature or luminosity (really primarily surface gravity), but those physical properties are not the basis of the classification.  And there are other factors that can change the underlying properties while preserving the appearance of the spectral lines.  For example in the intro of this paper, Gray discusses how stars of the same luminosity class can have different luminosities due to rotation.
Divorcing the classification from the physical properties is an interesting philosophical approach, but one that has stood the test of time even as our understanding of stars and their physical properties has improved in the past 70 years.  If you have time, I encourage you to read the first couple of pages of the review article linked above - the quote from Dimitri Mihalas (who was a leading theorist on stellar atmospheres) is especially interesting, I think.


Why does the Moon stablize Earth's axis? In an academic study titled 'The Moon and the Origin of Life on Earth'  it is said that

  'If the Moon did not exist, the orientation of the Earth’s axis would not be stable, and would be subject to large chaotic variations over the ages. The resulting climatic changes very likely would have markedly disturbed the development of organized life'.

Since I don't have enough physics and mathematics knowledge in order to understand why the Moon stabilizes the Earth's axis, can anybody explain in simple words this phenomenon?

One way to look at this problem is to consider angular momentum. The Earth spins around its axis, and has therefore some angular momentum by itself. The angular momentum is proportional to the mass of the Earth, to its squared radius, and to its angular velocity. But the Earth is not alone; it has the Moon rotating around that adds angular momentum to the Earth-Moon system. And, even though the Moon is not as massive as the Earth (it is about 100 times less massive), nor rotates very fast around the Earth (and therefore a lesser angular velocity), it has a large orbit (about 300 000 km) and overall, it adds to the system an amount of angular momentum comparable to that of the Earth itself.

Now, think of a spinning-top: the faster it rotates (and thus the larger the angular momentum), the more stable it is. It is the same for the Earth-Moon system: without the Moon, the angular momentum of the Earth itself would be such that gravitational perturbations could be sufficient, in the long run, to perturbates significantly its axis (exactly like the spinning-top: if it does not rotate very fast, a small perturbation will rapidely increases and the spinning-top axis will start to oscillate more and more). But with the Moon, the global angular momentum of the system is larger, and it is therefore harder to pertubate sufficiently the system to get it oscillate strongly.

Sources:

For those who wants the dirty details, you can have a look at Laskar et al. 1993.


Light distorsion/ilusion when looking at bright objects What's the name of the phenomenon when you look at a bright object (e.g. the moon) and see some perpendicular "beams" going top-down and right-left (bare eye observation, also slightly visible on iPhone pic)?

https://ibb.co/j2b4AF
see the link for a picture

It's a bunch of different phenomena, somewhat related, but still distinct.

Naked eye

Since you mention naked eye observations - seeing a small amount of short spikes around bright objects is normal. The causes can be multiple: diffraction of light in/around your eyelashes, dirt/secretions on the cornea, just regular aberrations in the eye's optical system, or in extreme cases it might actually signal vision problems.

Your iPhone picture

It appears that you're taking the picture through some fabric or some other kind of rectangular network-shaped material. In this case, the spikes are due simply to reflection of light on the individual threads. If you removed all fabric and kept only one thread without moving it at all, the light would reflect on it exactly in the same place.

Telescopes

Many reflector telescopes show a pattern of cross-shaped spikes around stars. Those are due to diffraction. Those telescopes have a secondary mirror supported by an X-shaped element called the "spider". Light is diffracting on the spider and creates a similar pattern of spikes around stars.


Does a black hole become a normal star again? Stars which exceed the Tolman-Oppenheimer-Volkoff limit can become a black hole.

What happens to star after it becomes a black hole?
Does it regain its status of star?

There's several ways this question could be answered, but they all come down to an emphatic "no" - a black hole will not return to being a main sequence star. The simplest way to see this is probably that a black hole has a much higher entropy than a star or even another type of stellar remnant of even vaguely similar mass and so there simply could not exist a spontaneous process by which a black hole develops back into a star.

A black hole once formed will stay a black hole, however it is believed that the Hawking process will lead to the black hole eventually evaporating. The time scale for evaporation though of a stellar remnant black hole is mindbogglingly long and they will not evaporate until long after stellar formation has ceased in the Universe.


How many solar masses is the central black hole in the Whirlpool Galaxy? I am having trouble finding the mass of the central black hole in the Whirlpool Galaxy M51a/NGC5194, it is stated that its companion galaxy NGC5195 has a supermassive black hole at its centre that is 19 million solar masses but I can't find details on the Whirlpool Galaxy's central black hole, is its mass known?

It is not well known, but a paper by M. Brightman et al gives a value of $10^{6.3\pm0.4}$ or between 8 hundred thousand and 5 million solar masses, while noting that this estimate is lower than previous estimates which had $10^{6.95}$ or about 9 million masses.
It seems that, although the galaxy is face-on to us, its black hole is viewed from the edge.  This means that the black hole itself is hidden behind a torus of relatively cool dust and gas. [ESA] which is visible as a dark streak in the image of the central part of the galaxy
There is much that we don't understand about the M51a black hole.  It is active, but not as active as would be expected, given the interaction between the two galaxies in M51. There is a second dark streak making an "X" the cause of which is unknown, and, as suggested by the wide error margins above, the mass of the black hole is not well constrained.


How did the object CO-0.40-0.22 get its name, and how is it distinct from CO-0.40-0.22*? There seems to be three things described in the recent Nature Astronomy paper Millimetre-wave emission from an intermediate-mass black hole candidate in the Milky Way:

Molecular cloud CO–0.40–0.22 with an extremely broad velocity width, near the centre of our Milky Way galaxy.
a compact object with a mass of about $10^5 M⊙$ is lurking in this cloud.
a point-like continuum source called CO–0.40–0.22*

Question: What are the objects CO–0.40–0.22 and CO–0.40–0.22* and what is the difference between the two. How did the originally discovered CO–0.40–0.22 get its name?
The news of the paper is described in the Science item Long-rumored midsized black hole may be hiding out in the Milky Way.
There is a nice writeup in Gizmodo for lay scientists: Astronomers Think They Discovered The Missing Link in Black Hole Evolution
The paper is not paywalled and can be read here: https://www.nature.com/articles/s41550-017-0224-z.pdf

Abstract:
It is widely accepted that black holes with masses greater than a million solar masses (M⊙) lurk at the centres of massive galaxies. The origins of such ‘supermassive’ black holes(SMBHs) remain unknown, although those of stellar-mass black holes are well understood. One possible scenario is that intermediate-mass black holes (IMBHs), which are formed by the runaway coalescence of stars in young compact star clusters merge at the centre of a galaxy to form a SMBH. Although many candidates for IMBHs have been proposed, none is accepted as definitive. Recently, we discovered a peculiar molecular cloud, CO–0.40–0.22, with an extremely broad velocity width, near the centre of our Milky Way galaxy. Based on the careful analysis of gas kinematics, we concluded that a compact object with a mass of about 105 M⊙ is lurking in this cloud. Here we report the detection of a point-like continuum source as well as a compact gas clump near the
centre of CO–0.40–0.22. This point-like continuum source (CO–0.40–0.22*) has a wide-band spectrum consistent with 1/500 of the Galactic SMBH (Sgr A*) in luminosity. Numerical simulations around a point-like massive object reproduce the kinematics of dense molecular gas well, which suggests that CO–0.40–0.22* is one of the most promising candidates for an intermediate-mass black hole.

An earlier paper on the object (Oka et al. (2015)) explains that CO-0.40-0.22 is a "high-velocity compact cloud". The first discovery of such an object was two decades ago (Oka et al. (1998)), when CO 0.02-0.02 was found. The naming convention used for that object was
"CO" + Galactic longitude + Galactic latitude

For instance, CO 0.02-0.02 has latitude $0.02^{\circ}$ and longitude $-0.02^{\circ}$; CO-0.40-0.22 has latitude $-0.40^{\circ}$ and longitude $-0.22^{\circ}$. Naming based on coordinates is used for many other radio sources.
The * in the name of the continuum point source is, I believe, in continuation of the tradition set by the discovery of Sagittarius A*, the point source in Sagittarius A. It is often necessary to distinguish between the two. Quoting from Goss et al. (2003),

Eight years after the discovery, one of us (Brown) invented the name Sgr A* to distinguish the compact source from the other components in the galactic center and to emphasize the unique nature of this source. Brown (1982) proposed a model of Sgr A* consisting of twin precessing jets with a period of 2300 years. The model has not stood the test of time but the name immediately was accepted. As an example, the VLBI results discussed by Lo et al. (1985) uses the name Sgr A*; the review article by Lo (1987) also uses this nomenclature.
Bob Brown provides the following rationale for the name: “ Scratching on a yellow pad one morning I tried a lot of possible names. When I began thinking of the radio source as the “exciting source” for the cluster of H II regions seen in the VLA maps, the name Sgr A* occurred to me by analogy brought to mind by my Phd dissertation, which is in atomic physics and where the nomenclature for excited state atoms is He*, or Fe* etc.”

The difference between CO-0.40-0.22 and CO-0.40-0.22* is simple: The former is a high-velocity compact cloud, and the latter is a point radio source inside it, possibly an intermediate-mass black hole. The formation, evolution and interactions of the two are likely tied together, but, as with Sagittarius A and Sagittarius A*, they need to be distinguished.


Why don't we see "the milky way" in both directions? We're (basically) in the middle of an arm of our galaxy. That is to say, we're sitting in the middle of a dense disc of stars.
It would seem to me that. You should see:
the thick line of the milky way all around you, i.e. on that plane, in all four directions.
(Additionally - sure - in the particular direction of the galaxy center, you would additionally see the huge central bulge.)
However: this does not seem to be the case: when you look at the milky way from the vicinity of our solar system, you basically see it "in one direction".
What am I misunderstanding?  How come the sky object "the milky way" is well-known as only a lump/strip in one direction, rather than, a lump/strip that goes right around us?
--
Put it this way...
Quite simply does anyone have any photography of the anti-galactic point?  (Near "Auriga" right?)  Does it show any "milky way band" going through it?
If not why not? Looking outwards, we are still looking through ~30k lightyears of the dense disk we are sitting in.

I'll turn my comment into a full answer.

To put it simply, we actually do see the Milky Way all around us, even in the diametrically opposite direction from the galactic core. You can see this in the image below which is a full sky image I took from APOD.

If you look at the edges of the disk in that image, you're looking at what is actually the edge of our galaxy, opposite in direction from the core. In effect, this is precisely the image you requested because it does contain the part of the sky which contains the anti-core part of the galaxy. It is certainly not as bright, but there are still stars and dust out there. In fact, if you look very closely, you still see lots of dark splotches that mask distance stars and galaxies, indicating there is dust there.

I think the issue you may be having is that you expect many more stars to be within the outer regions of the disk than there actually are. The stellar density profile for our disk is roughly exponential, meaning there are literally exponentially more stars near the core than at the edges. If this means anything to you, the scale length for the exponential radial density profile is ~4 kpc.

To really get a good understanding of the stellar distribution, take a look at Jurić et al. (2008). They looked at (~48 million) stars from the SDSS and analyzed the stellar distribution across our galaxy (that is visible to us). You should find figures 10 through 18 of particular interest, however I'll present part of figure 16 here.

This image shows the (logarithmic) density of stars as a function radius from the galactic core. The varying shades of grey indicate varying heights above the galactic plane (numbered in parsecs). The dashed lines are various exponential decay models with differing scale heights. You can see that these star densities, even within the limited radial ranges covered by the SDSS fall off by an order of magnitude! Hopefully this helps you appreciate the significant difference between the core brightness/visibility and that of the galactic edge.

Here's a silver of Andromeda spectacularly demonstrating the falloff: even though galaxy photography tends to suggest to the casual eye an evenly-dense plate:

This synthetic image referenced by Andy, from the Tycho Catalog Skymap also shows the situation clearly.


Did universe have finite hydrogen According to this video stars at end of time  and many others on fate of universe they all agree on one  thing that stars will eventually convert  all the hydrogen into heavier elements and eventually there will be no light , i understand the documentaries are over simplified but is it really the case that universe had/has finite amount of hydrogen to start with and there is no known process that recycling hydrogen back into universe

Are you sure it's true that all hydrogen will eventually be processed?   I can see that star formation rate should eventually slow to a point that there would be essentially no more light being created, but that's not the same thing.  Right now, most of the hydrogen in the universe is in the space within clusters of galaxies, and some of it falls into galaxies and some gets ejected, so you might imagine that all of that will eventually get processed.  But there is also hydrogen in between galaxy clusters, in the "cosmic web", and since the universe is expanding quite rapidly, it seems quite possible to me that much or most of that gas will never fall into a galaxy, but instead just end up stranded in between the expanding clusters.

What's more, our models now include the "cosmological principle" that treats the universe as homogeneous, but that's just the universe we can see.   We have good evidence that it extends way beyond what we can see, so we have no way to know (or even to expect) that the homogeneity is a truly global property of the entire universe (i.e., not just the "observable" universe).  On scales we cannot probe, the structure might be fractal, meaning that there could be pockets somewhere doing very different things, and there might always be stars forming somewhere.  Another possibility is that there could be a "landscape" of universes, which is hard to say if they count as the same universe or completely different ones, but it does mean there could be in existence hydrogen elsewhere in the "landscape" even after our universe has none.   So it's a tricky question to answer if "all the hydrogen" will eventually be processed, but I suppose you could look at all the hydrogen in the observable universe-- but you still have to wonder if the gas between the clusters ever gets into a galaxy.


Ninth planet - what else could it be? So recently it was announced that there might be a ninth planet out there.

Some people (from NASA) say the odds are 50% that it is a real planet, since it hasn't been discovered by telescopes yet.

Let's say there is no ninth planet. Is there another explanation for those orbits that require a big mass (8 times earth mass) out there?

Could it even be a very small black hole (a few centimeters in radius) or maybe a clump of dark matter?

What else could it be?  Very simply, it could all be a coincidence...

What they have done is noted that objects in the outer solar system have their orbits grouped in interesting ways. Someone suggested this might indicate there was a heavy planet (or core of a giant protoplanet) that influenced their orbits in this way; the recent announcement is the result of examining this idea in detail.  It is likely to be caused by gravitational influence of such a body.

Of course the TV and newspapers have picked this up and said a planet HAS been discovered. But there might be nothing at all; the loose clustering could all be a coincidence - though that seems, statistically, unlikely at the moment.

Here is the paper for anyone who wants to check.  (Beware, lots of mathematical terminology.)


Why Can’t we make a star If we know that the reason a star burns without oxygen and fuel is that it converts hydrogen into helium and we have these two elements on earth then why can’t we make a star on our planet

It's very easy to create a star: you just put an amount of hydrogen gas quite near to each other and gravity will do the rest.

The problem is that the amount of gas you need for that is so large that even replacing the whole earth with hydrogen atoms is far from enough: you need far more hydrogen atoms for that and as such you can't do something on the earth when you need more space than the whole content of the earth :-)


What is $\omega$ and $\tau$ in this celestial sphere In the celestial sphere below, what is the $\omega$ and $\tau$ physically? I mean, In the normal world, how do we see them? for example, we can see the altitude of the sun (a) by examining the shadow of a pillar and comparing it with its height, How do we see $\omega$ and $\tau$ in real world?

S is the sun and W is the west.

Also  I want to know how do they effect on the shadow of wall which is built in direction of "$SCP$ to $NCP$".

There seems to be very little to go on here.

"Zenith" and "Horizon" are easy to understand, it suggests that this is a projection of the sky (and ground) In that context "W" would appear to be "West", and you give that "S" is "sun". The horizon is curved. If you project the entire hemisphere the horizon would go straight across this isn't a complete hemispheric projection.

If we are looking West, then North would be to the right, and South to the left. The point P seems to be identified with NCP. The first guess would be that N of NCP means North, but this contradicts the reasoning that North is on the right. You mention SCP in the text, but it isn't in the diagram. The sun appears to be West, or just North of West, which is not an unusual but not impossible position.

The lines SW, WP and PS are curved. This may be to indicate that they are great circles on the celestial sphere. If so then $\omega$ is the spherical angle SWP and $\tau$ is the great circle distance SW, perhaps given as an angle with vertex at the centre of the sphere. Given three points on a sphere (in alt-az coordinates), it is a small exercise in spherical geometry to find the spherical distance and angles between them.

I'm not sure why you are being coy about the source of this. If you could link to the question from which this is taken the meanings of $\delta$ and H could be clearer. I guess the solution is an exercise in spherical angles, and projections from a circle to the plane


First Telescope SkyWatcher Heritage 5" Tabletop Dobsonian Telescope I have been star gazing for awhile with a pair of binoculars and now I am ready to step it up a bit and purchase a telescope.

I have been looking at the SkyWatcher Heritage 5" Tabletop Dobsonian Telescope.

Would this be a suitable first telescope , I hope to be able to make out the rings of Saturn using stock the Eyepiece.

Or would this be considered overkill and something like the Celestron 21035 70mm Travel Scope be more suitable.

No telescope that fits your budget and observing habits is overkill.
As you may have read, the bigger the aperture, the more light it collects, and the finer detail it can resolve in steady air.

A 5-inch f/5 reflector (650mm focal length) is good for wide-field views of deep-sky objects and easy to transport to a dark site.
It should show the rings of Saturn clearly with a 10mm eyepiece (65x magnification); a shorter eyepiece could double that without losing clarity.
However, with such a short optical tube assembly, you might either operate mostly on your knees or put the mount on something less sturdy than the ground.
Also if the secondary mirror is exposed as shown here, it's vulnerable to dew, and stray light can enter the eyepiece and reduce contrast.

If you're primarily interested in planets, a 6-inch f/8 reflector (1200mm focal length) could be an even better choice.
A 10mm eyepiece gives 120x magnification, the primary mirror is less sensitive to collimation error, and the secondary mirror can be smaller for slightly better contrast.
On the other hand, you might want an additional eyepiece longer than 25mm for better views of certain faint objects.


Pyramid shaped observatories There are some pyramid shaped observatories around the world, like the Solar Lab Pyramid in the Canary Islands or The Pyramid International Laboratory in Nepal.

I was wondering, why a pyramid? Because of its relative stability, or is there something else?

below: Solar Lab Pyramid in the Canary Islands.

below: The Pyramid International Laboratory in Nepal.

Both are solar observatories, and so have different requirements from conventional telescopes.

The sloping sides may allow for smoother air flow around the building, as heating during the day causes air to rise, which affects the resolution of the image. Also the pyramid is structurally strong (useful when you have a building on top of a mountain) and attractive, echoing she shape of the mountains. The Nepali observatory uses the flat walls to mount photovoltaic cells.


Dfferent distances to comet 67P Earlier, I was looking for the current distance from Earth to 67P/Churyumov-Gerasimenko, in order to do a "back of envelope" calculation of signal delay. Not important, but I was somewhat puzzled to find quite a large discrepancy between values reported by different sites, and would love to understand why.

ESA's Where is Rosetta page currently gives the distance as around 719m km, which, if I'm using it correctly, is backed up by JPL Horizons tool, which gives me a delta of around 4.81 AU. Which is great.

But when I look at The Sky Live's 67P live tracker, it is reporting around 941m km, which their comet info page gives as around 6.3 AU. Am I missing something blindingly obvious?

I suspect that The Sky Live got their comet IDs mixed up.
The orbital elements show perihelia in Aug 2015 and Jan 2022,
aphelion 5.69 AU, period 6.45 years, consistent with other sources e.g.
Minor Planet Center.
However, the distance chart
shows a perihelion in mid 2018, aphelion about 11 AU, period about 16 years,
clearly not the same object.
Report the bug and they'll probably fix it.


How has the Earth's orbit changed over hundreds of millions or billion of years? First, I know that modeling orbital mechanics of 8 planets is hard, but there are some theories out there, for example, Jupiter is thought to have moved in towards the sun then started moving away.   Article

and Uranus and Neptune may have switched spots Article

Is there any pretty good evidence on how the Earth's orbit has changed over time.   I remember reading some geological evidence that a year used to be longer, implying that the Earth used to be farther form the Sun, but I've since been unable to find that article, and for purposes of this question, lets count a day as 24 hours even though a day used to be quite a bit shorter hundreds of millions or billions of years ago.   - footnote, I've still not been able to find that article but it occurs to me, it could have been counting shorter days, not longer years - so, take that part with a grain of salt.

Are there any good studies out there on how many 24 hour days were in a year, 100, 300, 500, 800 million years ago?  or 1 or 2 billion years ago?  Either geological or orbital modeling?   Preferably something a layman can read, not something written by and for PHDs?

Or any good summaries, also encouraged.    Thanks.

I also found this article, but it seems more theoretical than evidence based.   http://www.futurity.org/did-orbit-mishap-save-earth-from-freezing/

What governs the Earth's orbital period is its orbital angular momentum and the mass of the Sun. Two events have certainly changed the Earth's orbital period (a) whatever collision formed the Moon and (b) the continuous process of mass loss from the Sun. A third possibility (c) is that tidal torques from the Sun have increased the angular momentum of the Earth.

Given that (a) probably happened sometime in the first tens of millions of years and likely did not alter the Earth's angular momentum greatly  - it depends on the speed, mass and direction of the impactor and the amount of mass lost from the Earth-Moon system - I will ignore it.

(b). It seems, from observations of younger solar analogues, that the mass loss from the early Sun was much greater thanthe modest rate at which it loses mass now via the solar wind. A review by Guedel (2007) suggests a mass loss rate over the last 4.5 billion years that increases as $t^{-2.3}$ (with considerable uncertainty on the power law index), where $t$ is time since birth, and suggests an initial solar mass between 1% and 7% larger than it is now.

Conservation of angular momentum and Kepler's third law means that $a \propto M^{-1}$ and $P \propto M^{-2}$. Therefore the Earth's orbital period was 2-14% shorter in the past due to solar mass loss, but has been close to its current value for the last 2-3 billion years.

If the solar wind power law time dependence is very steep, then most of the mass loss occurred early, but the total mass loss would have been greater. On the other hand, a lower total mass loss implies a shallower mass loss and the earth spending a longer time in a smaller orbit.

(c)
The tidal torque exerted by the Sun on the Earth-Sun orbit increases the orbital separation, because the Sun's rotation period is shorter than the Earth's orbital period. The Sun's tidal "bulge" induced by the Earth applies a torque that increases the orbital angular momentum, much like the effect of the Earth on the Moon.

Quantifying this is difficult. The tidal torque on a planet from a the Sun is
$$ T = \frac{3}{2} \frac{k_E}{Q} \frac{GM_{\odot}^{2} R_{E}^{5}}{a^6},$$
where $R_E$ is an Earth radius and $k_E/Q$ is the ratio of the tidal Love number and $Q$ a tidal dissipation factor (see Sasaki et al. (2012).

These lecture notes suggest values of $k_E/Q\sim 0.1$ for the Earth and therefore a tidal torque of $4\times 10^{16}$ Nm. Given that the orbital angular momentum of the Earth is $\sim 3\times 10^{40}$ kgm$^2$s$^{-1}$, then the timescale to change the Earth's angular momentum (and therefore $a$ and $P$) is $>10^{16}$ years and thus this effect is negligible.


What's the difference between grism and grating? In spectroscopic observations, sometimes you meet grism, sometimes grating.

Both of them could cause light dispersion, but what is the difference?

Transmission gratings on their own introduce chromatic aberration.  This is because they change the effective focal length and do so as a function of wavelength.  The chromatic aberration can be eliminated by introducing a prism of the correct dispersion.  Known as a grism, the grating/prism combination provides an unaberrated image at the 0th order in addition to the spectrum, although the spectrum has fairly low resolution.


Why in my area does NEOWISE appear so dim and small? So, I was always amazed by images of NEOWISE around the Internet like this:

Or this:

Yet, all my efforts to see it or photograph it in its full glory were fruitless...

22.07.2020 22:01
Here you can barely see it at the bottom-left of the image...

24.07.2020 22:00
Here is the zoomed version... Very very blurry. The point is that I couldn't even see it with my "naked eye" (well I couldn't see it even if it were much brighter, because I use glasses, and that's automatically not "naked eye", but yea). The maximum that I could see was just some smudge. With 5x binoculars I could see it, again super dim and blurry. I decided to purchase a telescope specially for this, and although I could see Jupiter's moons and Saturn's ring quite clearly, I had no success with NEOWISE. Both at 56x and 140x zoom the comet seemed super blurry and dim. Although stars around it were in good focus, no matter how I tried to focus the comet itself, it always seemed out of focus...
BTW, these 2 shots were taken at the closest approach to Earth around 2
So I thought, maybe it's because of light pollution? I live in a not-super-giant city, but a light pollution map showed some amount of pollution still. So I went to a light grey area (only one step brighter than black) on the map. This, however didn't change much at all...

25.07.2020 22:32
The comet is still very dim and small. I couldn't find it with the "naked eye" reliably. Its size is almost the same as surrounding stars, yet on other people's photos it appears like 10 times bigger than the stars around it!
What could be the reason for this? If not light pollution, maybe it's that my location is very bad for viewing it? I'm located in southern Ukraine, Kherson city.

There's one main reason - and maybe some other reasons as well:
The primary reason is the date the photos were taken you compare your photo with. Two weeks ago (10th July 2020), the brightness of the comet was around 1 - 2 magnitudes - that is comparable to the brighter stars and it was readily visible with the unaided eye on onset of nautical twilight. A week later already (17th July) it moved further from the sun and it was only around 3.5 magnitudes (personal guestimate from comparison with surrounding stars like some from Canis Minor) - thus it needed already a dark sky to be nicely visible. Now is already another week past and its heliocentric distance increased further, its brightness dropped to something comparable to the stars just barely visible with the naked eye. Conversely its activity also reduced dramatically with increasing heliocentric distance. As it meanwhile has past its perihelion as well as closest approach to Earth, it's meanwhile both effects adding up: geometry and much less material to scatter the sunlight due to reduced activity.
See this animation for a trajectory.
While the pure physics above are the main reason for the difference, other reasons might be the amount and skill of post-processing done on images. People might have stacked several photos in order to gain a higher noise-to-contrast ratio and applied a logarithmic brightness curve or similar.


What could be the utmost lowest temperature in the universe/multiverse? What could be the utmost lowest temperature in the universe/multiverse?

What temperature means...

Temperature is the measure of the energy of particles. The higher the temperature, the more energized the particles. The more energy particles have the faster they move around. This is the particles' kinetic energy that is rising. As it rises, the particles will begin using up more space. Moving particles need more space. In a vacuum this can be measured as pressure which is the stress of thermal expansion on the closed system. In an open system the matter will expand freely. As the particles increase their speed they also move more erratically so the entropy, or measure of disorder, will also increase.

The coldest...

Now understanding all that, what would the lowest temperature, or energy state, be in the universe? The answer is a state of no energy, 0 K, or absolute zero in the Kelvin scale. It is −273.15° on the Celsius scale and −459.67° on the Fahrenheit scale. At this temperature, which is impossible to occur by only thermodynamic means, the particles are completely still and entropy drops to 0.

Temperature reference points...

The surface temperature of the sun is 5,778 K.
Water boils at 373 K.
Water freezes at 273.15 K.
The moon’s darkest craters that never receive sunlight are 33 K.
The cosmic microwave background fluctuates around 2.8 K.
The Boomerang Nebula, the coolest natural place currently
known in the universe, measures 1 K.
Absolute zero is 0 K.


Right Ascension of North Celestial Pole My scope simulator is pointing to the celestial north pole.
My position is 41º 27' lat and 2º 15' long.
At 12h 43m local time (11h 43m UTC) the coordinates read from the scope are:
RA: 16h 14m 25s
DE: 90º 00'00"
There seems to be a constant 4h 31m difference between North pole RA and UTC time.
What the relation is between UTC, my location and celestial North Pole Right Ascension ?
I am making all kind of operations trying to relate these numbers to no avail.
Sorry if it's too stupid a question. I try to create a model in my brain of how celestial movements work and at the moment I only manage to explode a few neurons. Getting a clear answer will help me get the puzzle together

The north celestial pole doesn't have a right ascension. It's like the north terrestrial pole not having a longitude. Asking for it creates an exception. In any case, the RA of a point, other than the poles, isn't dependent on time or the observer's location. Your simulator has an error in its error handling.


Are there ever any simultaneous transits of both Mercury and Venus as seen from the Earth? Transits of Mercury happen fairly frequently due to its short period, but transits of Venus are less frequent.  I've looked over the data available to me and found that there have been transits of both planets in the same calendar-year in 1631 (Nov & Dec) and in 1769 (Jun & Nov).

What would really be interesting is if the two planets transited the Sun at the same time! With the rule being, that the second transit begins before the first transit completes.

Can we figure out if they have ever so transited, or if they ever will so transit?

EDIT: As it turns out, I'm not the first or even the second person to run calculations like this:

https://www.fourmilab.ch/documents/canon_transits/
http://www.solexorb.it/SolexOld/Simtrans.pdf

Meeus' work (second link) mentions the 13425 CE event in "Table 1. Simultaneous and near-simultaneous transits of Mercury and Venus, years 1 to 300,000"

Within the limits of DE431 (7 May 13201 BCE to 7 May 17091 CE), there is no
time at which both Mercury and Venus transit the Sun.

The closest we get to this:

On 16 Sep 13425 CE at 11:57pm UTC, Venus starts transiting the Sun. This
transit ends the next morning (17 Sep 13425 CE) at 7:30am.
Less than 9 hours later, at 4:27pm, Mercury starts transiting the
Sun. This transit ends at 10:26pm.

The program I used to compute this:

https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-13227.c

The list of transits I computed while solving this:

https://github.com/barrycarter/bcapps/blob/master/ASTRO/mercury-transits.txt.bz2
https://github.com/barrycarter/bcapps/blob/master/ASTRO/venus-transits.txt.bz2

Although I believe this answer is correct, Stellarium does not agree with
me, and HORIZONS doesn't compute positions past 9999 CE, so don't put too
much faith in this answer, since there's no good way to confirm it. I
believe that I'm correct and Stellarium is wrong this far in the future, but
it could be the other way around.

Even if my calculations are correct, the uncertainty in calculating the
relevant positions (Sun, Merucry, Venus, Earth) this far in the future is
high. On their own transit pages, NASA only computes Venus transits from
2000 BCE to 4000 CE, and Mercury transits from 1601 CE to 2300 CE, even
though they could've made the same calculations I made from 13201 BCE to
17091 CE:

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

This suggests NASA isn't confident enough of Mercury/Venus (and Earth/Sun)
positions to predict that far in the past or future, so my results may be fairly inaccurate.


Origin of terms "white dwarfs" and "red giants"? Can someone explain my why "white dwarfs" and "red giants" were so named? And maybe that's a silly question but why not in the opposite way?

@EDIT
Is this somehow co-related with Stefan–Boltzmann law?

The terminology is related to Planck's Radiation Law. We have (from Wikipedia):
$$
{\displaystyle B_{\lambda }(\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}{\frac {1}{e^{\frac {hc}{\lambda k_{\mathrm {B} }T}}-1}}}
$$
where $ k_B $ is the Boltzmann constant, h the Planck constant, and c the speed of light, $ \lambda $ is the wavelength, and T is the temperature.

A typical white dwarf is very hot, with $ T > 10^5 K $ even. Here let's assume it is about 50,000 K. A typical red giant is of lower temperature than the sun. Let's take it to be about 4,500 K (it will usually be a bit lower). Then we can plot the Spectral Irradiance of both the sources and compare them (I had an older Matlab code).

(The y-axis is $ W m^{-2} sr^{-1} nm^{-1} $.)

For the white dwarf we see that the peak occurs at a significantly lower wavelength than visible light (at about 0.58 nm). Most white dwarfs are hotter so the peak would shift further to the left. Thus, we expect the white dwarf to appear white with a bluish tinge as blue is still marginally higher than the other wavelengths.

For the red giant, the peak occurs at about 640 nm, which is near the red end of the spectrum. Red giants colder than this would have peaks shifted to the right. So they appear orange-red.

As @AtmosphericPrisonEscape already mentioned, dwarf and giant relate to the physical size as compared to the original star in the main sequence.


How can GAIA's accuracy for position, parallax, and annual proper motion for a given star be the same number? Per Wikipedia's Gaia (spacecraft); Objectives:

  Determine the position, parallax, and annual proper motion of 1 billion stars with an accuracy of about 20 microarcseconds (µas) at 15 mag, and 200 µas at 20 mag.

As @RobJeffries painstakingly points out to us in comments below this answer when we talk about parallax as a unit, it is referenced to a baseline of 1 AU. It doesn't refer to the actual cyclic motion seen in the sky due to the orbital motion of whatever is moving around the Sun at its own particular distance.

But the accuracy of both "position" and "annual proper motion" have nothing to do with 1 AU and are measures of positional accuracy of the telescope and how its astrometric data is analyzed.

How can all three of these accuracies simultaneously be 20 µas for 15 magnitude stars?

Parallax and proper motion are determined from a series of position measurements taken over the course of (for Gaia DR2) 22 months. A "5-parameter" astrometric model is fitted to these position measurements, consisting of a sky position at some epoch, a parallax and a proper motion in each of the celestial coordinates.

The precisions of each of these parameters improves as the number of measurements increases, but parallax and mean position precision are basically unaffected by the timespan of the observations (as long as they are taken over more than a year), whereas proper motion precision is improved by a longer timespan given the same number of observations.

If a star had no proper motion, then a basic parallax measurement would be the difference between two (extreme) positions on the sky, and the parallax precision would be around $\sqrt{2}$ of an individual position precision. The mean position of the star is something like the average of the two measurements, so might have a precision that is a factor 2 better.

In practice, you have many more than 2 observations and measuring the parallax and mean position is like trying to fit a sine wave (with a DC offset), of known period (1 year), where the mean position is the DC offset and the parallax is the amplitude.  In such cases, the mean position has a variance of $\sigma^2/n$, where $\sigma$ is the uncertainty in a single position measurement and $n$ is the number of observations. It took me a while to track it down, but it can be shown that the variance of your best estimate of the amplitude is also $\sigma^2/n$, when the frequency is known and $n$ is large (e.g. https://pdfs.semanticscholar.org/df9b/0c876b488be088fd416387866cf8499fa438.pdf). Thus we expect the mean position and parallax to have similar uncertainties and for these to get smaller as $\sqrt{n}$.

Proper motion precision is different. If we imagine a set of measurements where parallax can be ignored (e.g. at the same time every year), the proper motion precision is akin to the precision in the gradient of a line fitted to two or more points (with time as the x-axis). The precision improves with both the number and precision of the position points and with the timespan of observation.

The answer to your question is therefore coincidence. You have picked on a magnitude of star for which the typical individual position measurement precision, typical number of measurements and typical timespan lead to a rough coincidence in the precision of the 5 parameters when proper motion is expressed in units of per year. For other magnitudes or for stars observed for an atypical number of times or duration, that isn't the case.

Quoting the abstract of Lindegren et al. (2018)

  For the sources with five-parameter astrometric solutions, the median uncertainty in parallax and position at the reference epoch J2015.5 is about 0.04 mas for bright (G < 14 mag) sources, 0.1 mas at G = 17 mag, and 0.7 masat G = 20 mag. In the proper motion components the corresponding uncertainties are 0.05, 0.2, and 1.2 mas yr−1, respectively.


What resolution is needed to see a quasar through the galactic center? The center of the galaxy is densely packed with stars and obscured by a whole lot of dust between us and it. For those reasons, groups that study the motion of stars around the super-massive black-hole in the center of the Milky way need to use big telescopes at near infrared wavelengths. Questions of dust obscuration aside, what resolution would a telescope need to have to pick out a background quasar looking through the center of the Milky Way?

SDSS QSOs start at around 8th magnitude (Vega) in 2MASS photometry ($H$ and $K_s$ bands), and rise like a power law as flux goes down Here's a quick and dirty plot of SDSS QSO survey made from SDSS data release 10 made using topcat. I would convert the graph to counts per solid angle, but I'm too lazy to look up the survey area of the SDSS spectroscopic survey right now.

There's a quick and dirty example of the effect I'm describing in Figure 2 of the Ghez et al. (2008) paper Rob Jeffries linked to below. The difference between the red points and blue points is both exposure time (more photons) and higher Strehl ratios (a measure of how close to diffraction limited the image is and, hence, a proxy for resolution given the same optical system).

Generically, you could say that the question is: how do we determine the confusion limit in images with point sources? This is just specifically applied to the field of view on the sky with the highest density of resolvable point sources available. Concretely, say we wanted to resolve $22^{\mathrm{nd}}$ magnitude sources in views of the galactic center, what resolution is required to do so?

Getting to the full answer to the more concrete question "how deep would we have to look to see a background quasar" would require additional information: size of field of view, and dust extinction along the sight line. So, for simplicity, I'm assuming that if you're resolving $22^{\mathrm{nd}}$ magnitude sources in $K$ band, you're probably seeing a quasar. Thus we only need the projected density on the sky of sources brighter than that cutoff, roughly, and some description of how resolved sources need to be from the background brightness to detect them to answer the question.

Whilst awaiting clarification, I'm going to assume that your question is asking about calibrating the reference frame within which the motions of stars at the Galactic centre are measured.

The process is described by Ghez et al. (2008); does not involve observing quasars at infrared wavelengths; and the study does not suffer as a result.

The Galactic centre adaptive optics images are 5-10 arcseconds across, but contain thousands of stars. Only a fraction of these are close to the Galactic centre. The ensemble can be used, iteratively, to define a relative reference frame by minimising the displacements of thousands of stars, excluding those very close to Sgr A* and those with large motions.

This is really all that is required to do the orbital analyses of the black hole.

To put the coordinates in the International Celestial Reference System, a wider field is observed that contains a few infrared-bright giant stars, that are also bright maser sources at radio wavelengths. The positions of these are used to bootstrap the coordinates onto the ICRS, and this is important if, for example, we are interested in the relative motion of the Sun around the Galaxy.

The maser sources have ICRS positions by virtue of their radio source positions versus the radio coordinates of distant (but still bright and point-like) background quasars.The issue of background sources, extinction and confusion does not arise at radio wavelengths.


Why doesn't the Moonrise appear red when viewed from orbit? After searching through many videos, I finally found one that shows, from the perspective of low earth orbit, the Moon transiting the horizon. It is actually a Moonset, but it shouldn't make a difference.

The Moon does not appear to change its color, neither gold nor red.

This contrasts with my understanding of atmospheric effects. Here on Earth, the Moon does appear more yellow (and when halfway or less above the horizon, even red) during Moonrise. The atmosphere tends to scatter blue light, leaving only yellow and red behind.

So why isn't this happening in the video? From orbit, we should be seeing the Moon through even more atmosphere. But I can't detect a trace of red or yellow in the Moon.

Granted it's not the best quality video that it could be. It could be more zoomed in. But it was the only video I could find that showed the Moon transiting the horizon from space.

In this video there is blue diffused sunlight present in the atmosphere.  At any point where you might expect blue light to be diffused from the Moon, there is also blue light diffused from the Sun being added back into what you see. To know for sure, you would need to have a recording from night.

The density of air could also be a consideration. The perspective in the video shows the moon from above cloud clover passing below cloud cover. The further the light passes from the Earth's surface doesn't only result in less distance spent in the atmosphere, but also the density of air throughout the entire passage.


Can one see a black hole? Can one see a black hole on earth?

If a black hole does not emit light, how can one take a picture of the black hole itself?

There's some discussion that the image composed by the Event Horizons Telescope is really just an accretion disk. To "take an image" of something, you need the light reflecting off of the surface of the object within frame. The "light" in the image is really just color-mapped ...
Can one see a black hole on earth?

I think, that's a bit semantics, and a question of what do you call 'I see something': If you see the shadow cast by an object on a bright background - do you see the object? Do you see Mercury or Venus when it passes in front of the disk of our sun? Similarly, if you look an object through an atomic force microscope - do you see the object, or do you just see some force indicator giving a value?

So yes, in this general sense, you see something when you have an instrument to detect it which allows to deduce properties of the object. So you also can see an object when you just see its shape in front of a bright background like a black hole within its accretion disk - even when all light you see comes from the latter.


How much mass does an object in space need to keep a human on its surface? Assume there were a roughly spherical object in space like a meteorite or a comet.

If I weighed 80kg on Earth, how much mass would be required for an object in space so that I could stay on its surface without "flying away?" It's not necessary have the same gravity as Earth, but I am wondering what minimal mass the object would require for it to have meaningful gravity for someone to remain on the surface.

Any object with mass (even you) has gravity. The mutual attractive force between two objects is given by the formula
$$ F = G \frac{M_1 M_2}{R_{12}^2}, $$
where the two mass are $M_1$ and $M_2$ and $R_{12}$ is the separation of their centres of mass.

So to answer your question we need to define some sort of parameter that specifies what you mean by "meaningful gravity".

One way of doing this would be to demand that the forces due to tides from the Sun are greater than the gravitational force holding you to the body in question. Even here though we need to make an assumption about how far away the object is from the Sun - let it be $r$. Now let your mass be $m$, the mass of the body be $M$, the radius of the body be $R$, and the mass of the Sun be $M_{\odot}$.

To stay attached to the object when you are at the "subsolar point" - i.e. the body rotates so that you are closest to the Sun, then we require
$$G \frac{M m}{R^2} > G \frac{m M_{\odot}}{(r-R)^2} - G\frac{m M_{\odot}}{r^2}$$

We can assume that $r \gg R$, so cancelling out $Gm$ and performing a binomial expansion on the middle term, keeping only the first two terms of the expansion:
$$\frac{M}{R^{2}} > \frac{M_{\odot}}{r^{2}}(1 + \frac{2R}{r}) - \frac{M_{\odot}}{r^{2}}$$
$$ \frac{M}{R^{2}} >  2M_{\odot}\frac{R}{r^3}$$
Thus this sets a limit on the density of the object in question
$$ \frac{3M}{4\pi R^3} = \rho > \frac{3}{2\pi} \frac{M_{\odot}}{r^3}$$

At $r=1\ au$, this means the density only needs to exceed $3 \times 10^{-4}$ kg/m$^3$, which will be satisfied by any solid body. NB: This limit would be a limit where you would actually be pulled from the surface by the Sun's tides.

A more stringent requirement might be to ensure that you can't jump off the surface. On Earth, an average person might be able to jump vertically about 50cm. Using the equations of uniform acceleration (SUVAT), we know $u^2 = 2g s$, where $g$ is the gravitational acceleration, $s$ is the height jumped and $u$ is the initial upward velocity. This tells us that you can jump upwards at about 3m/s. Assuming this is the same on any other body (its hard to say how well you could jump in much lower gravity) you could equate this to the escape velocity of the object, given by $v_{esc} = (2GM/R)^{1/2}$. Thus this gives a constraint of $M/R > u^2/2G$.

If we set a realistic density of $\rho = 5000$ kg/m$^{3}$ for an asteroid, we can replace $M$ with $4\pi R^{3} \rho/3$, to say that you would be able to jump off an asteroid if it were smaller than:
$$ R < u \left(\frac{3}{8\pi G \rho}\right)^{1/2}$$
For the numbers discussed this means $R< 1.8$ km and $M<1.2\times 10^{14}$ kg.

Many more details at
https://physics.stackexchange.com/questions/46318/is-there-a-small-enough-planet-or-asteroid-you-can-orbit-by-jumping

Incidentally, as you will have noticed, the result doesn't depend on your mass.


When Betelgeuse goes supernova, will the explosion hit any neighboring stars? There is evidence Betelgeuse might go supernova "soon". Are there any stars in the vicinity of Betelgeuse that are close enough that they will be hit by the explosion of this supernova? What would be the effect on stars near Betelgeuse?

A supernova "kicks" out the other star if it is in a binary pair, but Betelgeuse is not a binary, so this doesn't quite answer the question.

There have been 2 proposed spectroscopic companions, though never confirmed, traveling with Betelgeuse. These are thought to orbit the red supergiant. So when Betelgeuse eventually explodes, they would  certainly be blasted by radiation, atmospheres could be damaged or completely destroyed, and would lose their host star sending them on a new trajectory through space.  But with Betelgeuse having been ejected from its birth place and  traveling with no verified companion, its more likely that the surrounding galaxy would be left unchanged, aside from  supernova remnant left behind.
https://en.m.wikipedia.org/wiki/Betelgeuse has more information if you still have questions.


Is a three body gravitating system doomed to collapse? Suppose we have two gravitating bodies, which are rotating around each other. They are bodies and are affected by deformation caused by tidal forces. Moving tidal waves suck energy from the axial rotation of these bodies. After some time their axial rotation will be synced with orbital rotation and tidal waves will be stopped. Like Moon orbital rotation around Earth is synced with its axial rotation. Maybe I missed some additional effects but it seems to be a quite accurate assumption.
Except for some special cases, if we have three orbiting bodies, their orbital rotation cannot be synced with axial rotation altogether.  Tidal waves in such a system will exist and suck energy indefinitely. And the only source of energy is potential body energy. Will they fall?
I understand that such effects should be very small; it hardly can be measured in the solar system, because the variation of mass of planets and the Sun causes a much higher effect.
But anyway are there any measures that support the existence of such an effect for something? Or is there theoretical proof that this effect does not work for some reasons I missed?

Since you are talking about real bodies with some extent, yes. The case for point-particles is somewhat different.
Any dissipative system will (by definition) gradually lose energy, and that means that the positions and velocities of the constituent bodies will be confined to (hyper)surfaces in phase space with less and less energy, $E=V(\mathbf{x})+K(\mathbf{v})$. Potential energy $V(\mathbf{x})$ decreases with distance between the bodies and there is a minimum possible separation giving a minimum potential energy, and kinetic energy is always positive $K(\mathbf{v})>0$. So as $E$ declines the distances and velocities have to decline accordingly. Typically things getting close move faster, with kinetic energy increasing as potential energy decreases, but since there is a limit to how much potential energy can decrease before things collide there will eventually be a collision (or, in systems allowing it - not the case here - losing all kinetic energy and ending up in perpetual rest).
(Point particles can cheat in Newtonian gravity by performing weird and physically unrealistic collision or oscillating orbits since they can reduce their potential energy arbitrarily far, see the Painleve conjecture)
In practice dissipation effects are tiny in planetary systems, and for most planets their orbits will instead be disrupted by rare stellar encounters, chaotic resonances with other planets, or mass-loss from the star when it leaves the main sequence. Planets near stars may be dragged in because of tidal effects, but this matters most if one is a few stellar radii away. In the truly long run gravitational radiation will force any planet to spiral in, but the timescales are staggeringly long.
The number of bodies does not change things, but for many-body systems ($N\geq 3$) some bodies can gain enough energy to permanently escape while the rest merge.


Is dust the only reason why the Martian sky is so bright? The atmospheric pressure on the surface of Mars is 0.00609 atm on average and 0.012 atm at most in the Hellas Basin. On Earth, at these pressures, which are found at altitudes about 30-35 km (19-22 mi), the sky is very black and full of stars. On the Martian surface however the sky is still bright (namely orange) and no stars/planets are visible in daytime from what is considered the sea level (except the Sun and maybe the Earth). Is this due to the many dust in the Martian atmosphere only or are there other reasons too for why the sky is not black but still orange at these pressures?

It is very likely due to dust particles. The color of Martian sky is described to be yellow-brown "butterscotch" color during daytime (sometimes pinkish red) and blue during rising and setting of sun (very different from what happens on Earth). The typical color is because as the dust particles in the air are large relative to the wavelength of visible light, it absorb blue light and effectively act as mirrors by scattering the remaining wavelengths. Also, the presence of 1% by volume of magnetite in the dust particles intensify the color. Twilight lasts a long time after the Sun has set and before it rises due to that.
At times, the Martian sky takes on a violet color due to scattering of light by very small water ice particles in clouds.
For more information, read this excellent paper: The color of the Martian sky and its influence on the illumination of the Martian surface
Additional sources:

http://www.webexhibits.org/causesofcolor/14C.html
http://www.mkrgeo-blog.com/what-is-the-colour-of-martian-sky/

For more on Mars Twilight see:

At Mars' equator how long does twilight last?
How bright is the Martian landscape during the long (up to two hour) twilight?
How much more quickly does night come on Mars compared to Earth? Will it be easier to see Mercury there?
How bright is twilight on Mars compared to Earth?
Mars twilight photos


What kinds of stars would it be possible to observe in M31 with a small amateur telescope? I was wondering what kinds of stars would be visible in M31 and other nearby galaxies using a moderate (~8") telescope and a CCD camera. This would be a limiting magnitude of around 17.

I have heard of LBVs (Luminous Blue Variables). However, I would like to know what spectral class of stars would most likely be visible (would it be mainly OBA type stars)?

The context of this question is in the search for extragalactic transiting exoplanets. Current amateur detectors have the capability to detect planets with host star apparent magnitudes of 14. I was wondering if it might be possible to find exoplanets in M31, M33 or even the LMC/SMC with these limits. An advantage to searching for planets in M31 would be the very limited search region--an amateur astronomer would only have to perform photometry on the M31 region over several months or years to find a planet, rather than having to search a much larger region.

Are there any other issues with detecting transiting planets around very luminous stars? Would the percent dip in brightness be far less for a Jupiter sized planet? How would this affect our ability to detect extragalactic transiting exoplanets?

LBV stars are very powerful, very large and bright, and even though they may be near the end of their life, still young - since they don't survive for long. These are problematic.

Their powerful solar winds will clear the region around them, meaning that planets are less likely to form close to the star. It is much less likely for a LBV star to be aligned correctly to allow a transit to be visible from Earth. A hot Jupiter close to such a star would rapidly lose mass due to the extreme temperature.

If a transit does occur, because the star is so large, it obscures a small proportion of the light, so a the change in brightness is small. And of course they are intrinsically variable themselves, so detecting the very small signal is harder.

Since they are still young so planets may not have finished forming by the time they go supernova.

We haven't discovered any planets orbiting LBVs in the Milkyway.

Most stars that have known planets are F,K and G spectral type, this is partly sample bias (we look for planets around sun-like stars) But such stars would not be visible in M31 using amateur equipment.

An 8in telescope has a limiting magnitude of about 14, (E. Hubble was observing Cepheids with a mag of 16+) but you are already  close to the limit of the telescope, and distinguishing the very small changes is consequently is harder. Given the difficulties, it is perhaps unsuprising that we haven't found many planets more that 10000 ly away.


Attach a visible light telescope to the outside of the ISS Visible light imaging of the heavens is limited on Earth primarily due to the atmosphere. I know the ISS is moving at 17000mph, but given the delays to the James Webb telescope, is there any merit in attaching a space-rugged visible light telescope to the outside of the ISS? Could it ever approach the visible light performance of Hubble?

A self-sufficient orbiting telescope is basically Hubble mkII and would never get off the ground, literally and metaphorically

Hubble was expensive because it was state-of-the-art, requiring development of many new systems. The systems it needed to function as a standalone satellite (compared to being attached to the ISS) were cheap by comparison (reaction control, power, communications) because they mostly didn't require new development.

So a Hubble successor attached to the ISS would not be much cheaper than a free-flying one.

Any telescope attached to the ISS has to deal with:

high vibration levels which degrade performance
outgassing which degrades performance
severe pointing constraints because it's surrounded by a large structure
bandwidth constraints because it has to share a transmitter with ISS operations
end of mission in 5 years when the station reaches the end of its life.

The advantages of attaching to the ISS:

easier servicing,
you get the option to store lots of data on the ISS and bring it down on a capsule. Downside is a latency of several months.

IMO you'd be better off building a standalone space telescope.


At what wavelengths can black hole Sagittarius A* be observed from Earth? The supermassive black hole in the center of the Milky Way is blocked by dust and gas clouds as seen from where we are. Which wavelengths are blocked and which are most undisturbed for observation of Sagittarius A* and its immediate neighborhood? Does the matter between us change for example polarization at some wavelengths or cause disturbances other than blocking the light? Could galactic cosmic rays, heavy ions, from it be detected if they were emitted, or would they be deflected by magnetic fields?

By the way, at what distance from us is the most dust located? Is it concentrated in one region or are there several dust clouds in the way?

From Genzel et al. (2010), here's part of Fig. 7.7.1:

This is part of the spectral energy distribution of Sagittarius A*, a flot of $\nu$ (frequency) vs. $\nu L_{\nu}$ (frequency times luminosity). For comparison, visible light is in wavelengths from $\sim4\times10^{14}\text{ Hz}$ to $\sim8\times10^{14}\text{ Hz}$, which happens to be around the bottom of the trough of nonthermal electron emission. Right away, this makes submillimeter and millimeter wavelengths good candidates, leading to studies using Very Long Baseline Interferometry. Likewise, infrared emissions are a good target, and so the Spitzer Space Telescope, for instance, has been used. X-ray flares of up to $\sim10^{36}\text{ erg/s}$ also occur from time to time,1 so that part of the spectrum is sometimes used to observe this activity. Finally, of course, Sagittarius A* is a very strong radio source, and it was initially observed in radio wavelengths (and it still is!).

As Fish & Doeleman 2010 write,2

  Interstellar scattering, which varies as $\lambda^2$, dominates over intrinsic source structure at longer wavelengths, and the emission from Sgr A* transitions from optically thick to optically thin near $\lambda = 1\text{ mm}$ (Doeleman et al. 2001).

This means that the visible part of the spectrum is dimmed even more. Combine that with relatively low emission at those wavelengths, and you have a pretty poor target for optical telescopes, and a pretty good target (all other factors considering) at other wavelengths, especially near the emission peak.

1 Flares also are visible in other wavelengths, but I mention X-rays here because Sagittarius A* is typically much less luminous in X-rays.

2 I believe they mean $\lambda^{-2}$. Power laws for scattering in the area vary, but they are generally given between as $\lambda^{-1.5}$ and $\lambda^{-2}$. At any rate, the index may differ over different regimes.


Why can't we build a huge stationary optical telescope inside a depression similar to the FAST? So first we'll have to find a natural depression or we should create one. I understand you can only see a single portion of the sky since it can't be moved, so my money is on creating an artificial basin in an area where directly upwards it has the best view of the sky.
Imagine if we can build a 1000 m optical telescope like that. We'd be able to see exo-planets directly. We'd be able to see ancient galaxies. Even a 200 m one like this would do wonders. Are there any proposals?

The surfaces of telescopes need to be configured to a fraction of a wavelength. If one is working in the FAST wavelength range of 10 cm to metres, then that is a relatively straightforward engineering problem. However, the difficulty and expense ramp up enormously when the working the wavelength is 500 nm. This would entail the construction of literally millions of precisely configured mirror segments to tile the inside of the structure and then some way of pulling them all into position and alignment (and keeping them that way) with a precision of 10 nm or so.
It helps that you aren't going to move the primary, and you can still track objects for 1-2 hours by having instrumentation at a movable focus. However, a second big problem is that the air above the mirror is turbulent and the size of the turbulent cells is much smaller than the size of the telescope. To take advantage of its large size, in terms of angular resolution, the individual parts of the telescope have to be individually controlled/deformed at a continuous rate of $\sim 100$ Hz. It is probable that to get the tenfold increase in resolution implied by the much larger diameter of this telescope would require higher frequency and more precise adjustment.
This sort of engineering is going on for effective mirror diameters as large as 39-m, but comes at a billion-Euro cost. Scaling up to something ten times bigger would almost certainly be a two order of magnitude increase in cost. So although there may be no showstopping physics, it would cost about as much as all other astronomy research combined.
Large interferometers, with sparse arrays of telescopes, are a much more likely proposition but come with their own very difficult problems.