Can you see something active in the sky apart from satellites? Can there be amateur time-domain astronomy? What I mean is an event unfolding that is viewable by naked eye or telescopes, and doesn't take comparing days of footage to see moving pixels. Apart from satellites, is there something that moves or happens so quickly that humans could sense it within a reasonable time? The whole sky is like a still painting, I'd love to see parts of it being "animated". Things on my mind are:

Crescent moon occulting a star (revealing or hiding it)
Solar and Lunar eclipses

Or more exotic things like:

Stars exploding (catching that as it happens, is it even possible?)
Other planets' moons moving a reasonabl/detectable amount in 30 minutes perhaps
Slow pulsar blinking (they are still super fast afaik)
Star changing its brightness and not blinking due to haze/turbulence

If it moves or flashes it isn't astronomy, it is meteorology or technology.
There are only a few exceptions to this:  Meteors are an atmospheric phenomenon, and a meteor will appear to move rapidly across the sky. But because they "come from space" and occur well above the clouds they are often considered to be part of astronomy.
As you note, eclipses and occultations happen quickly enough for the changes to be visible. The moons of Jupiter do move notably during the night, and when one moves into or out of shadow, it can appear or disappear over the course of a couple of minutes: easily noticeable.
A supernova could brighten quickly enough for the variation to be visible over a night's observation:  You wouldn't see it suddenly appear, but over the course of a night it could appear brighter at the end of the night than at the beginning.
Similarly, Algol will fade from magnitude 2 to 3.5 over a few hours. It isn't quick enough to notice the change in the moment, but it is quite clear if you are observing over a few hours.
Pulsars are too dim to be seen with normal equipment even the bright nearby crab pulsar has a magnitude of 16.5 (and at one flash in 33 milliseconds, it is too quick to see by eye)
GRB 080319B was an exceptional object. It was a gamma-ray burst. It gave a flash of gamma rays that lasted a little over a minute.  If you had happened to know exactly where to look you could, (marginally with the naked eye, but easily with binoculars) have seen the optical counterpart to the gamma-ray burst.  This was created by the formation as a massive early star collapsed to a black hole producing a jet of energy that happened to point our way.
The sun is changing and on a small scale it does change on the scale of minutes, but it is hard to see any movement at that scale with basic equipment.  If you can get your hands on a Hydrogen-alpha solar telescope you will, be able to see prominances that change over a period of hours.
Rarely an asteroid will pass so close so as to be visible.  Apopsis will have a close approach in 2029 and will appear as a slowly moving star.
Jupiter is active at radio frequencies.  You can tune in to Jupiter with the right equipment and hear it changing at rates from a few fractions of a second to a few seconds, see e.g. radiosky.com > Jupiter Central
These are exceptions.  In general, it is rare for anything so big and powerful that it can be seen over a distance of many light-years to be able to change fast enough that those changes can be noticed by our eyes.


Why does star formation cease? The universe is expected to have a time when star formation slowly ceases. However, this seems bizarre as there is always an abundant amount of matter in the Universe. Massive amounts of hydrogen exists in several galaxies, but they are deemed dead. Any gravitational disturbance can send the gas into a protostar, and even if there weren't hydrogen left, enough helium or carbon can turn a star into a star that fuses heavier elements, until the universe is depleted of such and mostly elements heavier than iron.

So why does star formation cease when these is still abundant hydrogen around? Is there some sort of interaction in these hydrogen supplies that prevents star formation? Is the gas kept at a temperature that is too hot for formation? Or have I made a misconception?

Star formation in a galaxy can end for a number of reasons:

Ram pressure affecting a galaxy falling into a cluster can strip away gas, which can end most star formation in the galaxy itself.
The winds of massive stars (as well as supernovae shocks) can eject cold gas via radiation pressure.
Mergers with other galaxies can also eject gas, although these interactions often lead to an increase in star formation, briefly.

As far as I'm aware, how strongly each of these phenomena contribute is still an open question, but we do know that they either eject gas or heat it up (and therefore make it unsuitable for star formation, which requires cold gas), thereby slowing or even effectively ending star formation in a given galaxy. A gas cloud is supported against gravity by thermal pressure (and occasionally magnetic fields). Clouds that are hot have higher thermal pressures and therefore can resist gravity; cold clouds cannot. Cold, dense clouds are therefore the best places for star formation (more gravity and less pressure).

As for the amount that can be ejected, the exact amount depends on a) the initial gas quantity, b) the mechanism, and c) likely a large number of other free parameters, so I don't think I can answer that for the general case.

Regarding if stars can form with little hydrogen: If hydrogen was ejected, so would other elements, so it's not as if all the hydrogen would be blown away and helium, lithium and others would remain. Hydrogen-poor stars exist, but they're older, evolved stars, not young ones.


What is the bluest narrow band filter mounted on ground based telescopes? If we want to check a QSO's lyman alpha narrow band image, could a ground based telescope do that?

I mean under the circumstance that its lyman alpha shifts to 3300A.

You probably will have difficulty imaging Ly$\alpha$ from a z=1.7 quasar at 330nm.

Says P. Veron: The lower limit [z=1.8] is set by the ultraviolet transmission of the atmosphere... between 3300 and 3600Å.


Which camera body is best introduction for beginner amateur astrophotographer? I currently own an old 1980s Celestron eight inch orange tube schmidt cassegrain telescope with tripod. I am seriously looking to dive into some astrophotography.

I also own an olympus pen epl3 camera. It's great for photos during my hikes, but not quite as versatile as an DSLR. Unfortunately, I know of no software out there which can tether this camera to my laptop. I don't have any great ideas on remotely controlling this camera to avoid shaking my telescope.
I'm thinking of getting a DSLR at some point anyway.

Is there a camera body that anyone would recommend for me?

When I first started out, I used my Olympus pen e-pl1, so you're doing great with the e-pl3. It takes higher resolution photos (330 vs 300 dpi) than my wife's Nikon D5300, so your camera should work just fine. I still whip this one out every now and again just because of its simplicity.

The one on the right is the adapter that I use regularly. I mounts up easily to each of my telescopes. Works great and has held up nicely the last (almost) decade.

To avoid the shake, set as long of a timer as you can! I'm not sure if the newer pen cameras have the option for the tethered remote, but mine definitely does not.

As for the software, I wrote some python codes that I use for image processing, and DS9 is the interface that I use regularly, however good old photoshop (pre-everythingisstupidandcloudbased) really works wonders if you're on a budget and have or "find" a copy laying about.

p.s. the adapter is from telescopeadapters.com


Where can I find the image of MY Camelopardalis? In this news article, it says

  The image captured by the telescope at the Calar Alto Observatory in southern Spain, reveal the physical properties of each of the stars including temperature and shape.

I Googled to see the actual image but unfortunately all are graphical images.So can I see the actual image as captured by the Calar Alto Observatory?

The work on the recent paper was done using spectra and light curves, not via direct imaging.  There may not be a good image to view.  I checked the Hubble Legacy Archive and could not find a Hubble image of the object.


Where should I look at if I had a 25m parabolic radio reflector? I am most interested in (exo)planets with atmospheres, see e.g. my question
Requirements to resolve position of Jovian Whistlers up to magnitude of Red Spot with amateur radio equipment?  So if I had a 25m radio dish in my backyard, would I have the chance to observe anything useful about (exo)planetary atmospheres by pointing it at one of the known exoplanets with atmosphere?
Another cool thing to observe would be a fast radio burst. Where should I point my reflector to increase the chance to see one?
References

How big a dish do I need for radio astronomy?
What kind of things I could "see" with an amateur radio telescope?
In German, with some English summaries: Observations at Astropeiler Stockert until 2019, see also Wikipedia about Astropeiler which happens to be an amateur radio observatory with a 25m antenna. I am not associated with them (yet).

On the fast radio burst side of things:
The FRB distribution is isotropic so far, which isn't surprising - you'd expect cosmological sources of this sort to be more or less everywhere in the sky. This means that you could point your telescope at random and still have a roughly equal chance of detecting one - albeit an extremely low chance. Fortunately, in the last couple of years we've discovered that at least a small subset of FRBs repeat; pointing your telescope at one of those known repeaters could substantially increase your odds of a detection.
I don't think there's been a substantial effort to look for FRBs solely with small dishes, but it's been attempted with a 25-meter dish. Spitler et al. 2018 used the 25-meter Stockert radio telescope, operated by the amateur astronomy group Astropeiler Stockert e.V., in conjunction with main observations by the 100-meter dish at Effelsberg. They were looking at FRB 121102, the first known repeating FRB. While Effelsberg detected three bursts from that source during their observations, Stockert turned up nothing - not overly surprising. As it turned out, the bursts Effelsberg picked up had flux densities much lower than you'd need for a reasonable signal-to-noise ratio with Stockert. There have been a couple other non-detections with small dishes, including the Nanshan 26-meter dish (Men et al. 2019).
On the plus side, you might have better options than FRB 121102, depending on your latitude. The ASKAP folks recently detected a bright repeat burst from the source FRB 20201124A, and say that the first burst could have been detected at SNR > 10 with a 20-meter telescope (with adequate specifications, of course). So depending on your precise setup, you might be in business - assuming, of course, you've got the proper setup.
I know less about exoplanet radio emissions, but I'd say your odds are probably much worse, partly because this is an area which has only had success even more recently than repeating FRBs. There was a detection from Tau Boötis this past year (Turner et al. 2021), and some previous tentative signs from elsewhere, but we have yet to get follow-up observations confirming things. You could point your telescope at Tau Boötis and hope for the best, but I wouldn't get your hopes up. Even that signal wasn't exactly strong.


Will the JWST be affected by dust at L2 (gegenschein?) Gegenschein is a "faint brightening of the night sky" at the anti-solar point.  A naked eye limiting magnitude of about 7.6 might enable an observer to make out gegenschein.

The Wikipedia article on gegenschein suggests that interplanetary dust at the Sun-Earth L2 point might be responsible for gegenschein.

The James Webb Space Telescope will also be at the Sun-Earth L2 point.

Does this imply that the same interplanetary dust that may be responsible for gegenschein will affect the JWST?

The Wikipedia article cites "Zdeněk (1962)" for the statement that the dust responsible for the Gegenschein has a possible concentration at L2. I haven't been able to obtain that paper, but I can't really see why that would be the case, since L2 is not dynamically stable.

However, the dust consists of millimeter-sized grain (see e.g. this APOD image), which is quite large for dust. Such large grains probably has a highly anisotropic "phase function", with a high preference for backscattering. Since when you look from Earth toward L2 you have the Sun in your back, you will thus see an increased brightness compared to other directions, even though the density of that dust is the same.

If this is the case, then placing JWST in L2 is no problem (except when it looks exactly away from the Sun, but that would be a problem anywhere in the ecliptic).


Do black holes really exist? Black holes from my knowledge are created from the collapse of a star’s own gravity crushing itself, Now my question is well two questions, is it really infinitely dense to where space-time can be distorted to the point where it just stops? if so, this would mean objects caught in the event horizon would freeze and time would be stopped how is the black hole itself not effected by it’s own mass wouldn’t it just cease to exist in the first place?

The literal answer to your question is "yes". We can see clear astronomical evidence of black holes of a variety of sizes in this and other galaxies. It is possible of course that all that evidence is actually the result of some unknown physical process(es) and not black holes, but it would be a remarkable coincidence that it looks so much like we would expect black holes to look.

So I'll take your question as asking for a way to understand what black holes are. The best answer to that question is basically just equations -- solutions to the governing equations of general relativity that have event horizons, but I'm going to assume that is not what you want here.

So we are left looking for a better intuitive explanation or understanding, bearing in mind that this is at best inexact, and at worst misleading. If you have a lot of matter in a small region of space, and you look at it from the outside you see that the light coming from the surface of the region is red-shifted and (which is basically the same thing) that clocks on the surface seem to run more slowly.  As you pack the matter in more tightly this effect becomes more pronounced. But because the clocks are running more slowly, whatever process is packing the matter in also runs more slowly. Eventually, as seen from the outside the clocks on the surface approach a complete standstill, and you will receive a final few extremely red-shifted photons from the surface and then nothing. What you have is now a black hole. From your point of view it is "frozen" and all you can see of it from the outside is the gravitational field surrounding it (and maybe an electric field as well). No information can ever leave it (I'm ignoring Hawking radiation and quantum mechanics in this answer, which is reasonable for big black holes).

Relativity being relativity, things can look quite different from another point of view. If you are falling freely into the black hole (and it is big enough that tides don't rip you apart) everything seems fine, as in this simulation Light from the rest of the universe is blue-shifted and increasingly distorted by gravity, but locally everything is OK. However, no matter what you do, within a finite time (a few days for the largest black holes we know about) you will reach a singularity -- a place where the equations of general relativity break down, and some other physics must presumably apply. We don't know what happens there, but we do know that no information about it can come back out of the black hole, so we can't send in any kind of probe to tell us about it.


period of an eclipsing binary I am confused about the definition of the period of an eclipsing binary, a binary configuration where one component crosses the other across the line of sight. When people say that an eclipsing binary has a period of T years do they mean that the period of the individual component is T years?

The period (at least in physics) is defined the time an oscilating system needs to get back to its starting point (for a sinus curve its 2*pi).

Now when they say an eclipsing binary has a period of T years, it means that its brightness varies as seen from earth when they are eclipsing (thats how we detect eclipsing stars and planets by the way), and that the curve of its brightness is like an oscilation that gets back to its starting point every T years.

The meaning of this is, that this system consisting of 2 stars will be exactly as you see it now again in T years.


What is the detection threshold of gravitational waves for LIGO? Since now two neutron stars have been detected merging via gravitational waves, I was wondering what is the current detection threshold that the LIGO detectors can achieve.

Considering that the first observed objects were two black holes with a combined mass of more than 60 solar masses and they now detected two neutron stars with a combined mass of only about 3 solar masses I was wondering what was the threshold that these detectors can actually detect.

Obviously there are much larger stars out there which orbit each other, but their size and distance from each other make gravitational waves too difficult to detect. So what masses and at what distances can we expect to be detected in the future?

I'm afraid this is not straightforward

The amplitude of the gravitational wave strain signal from a merging compact binary (neutrons star or black hole) is
$$h \sim 10^{-22} \left(\frac{M}{2.8M_{\odot}}\right)^{5/3}\left(\frac{0.01{\rm s}}{P}\right)^{2/3}\left(\frac{100 {\rm Mpc}}{d}\right),$$
where $M$ is the total mass of the system in solar masses, $P$ is the instantaneous orbital period in seconds and $d$ is the distance in 100s of Mpc. $h \sim 10^{-22}$ is a reasonable number for the sensitivity of LIGO to gravitational wave strain where it is most sensitive (at frequencies of 30-300 Hz).

So you can see that to increase the observability you can increase the mass, decrease the period or decrease the distance.

But here are the complications. LIGO is only sensitive between about 30-300 Hz and the GW frequencies are twice the orbital frequency. Thus you cannot shorten the period to something very small because it would fall outside the LIGO frequency range and you also cannot increase the mass to something too much bigger than the black holes that have been already seen because they merge before they can attain high enough orbital frequencies to be seen. (The frequency at merger is $\propto M^{-1}$).

A further complication is that the evolution of the signals is more rapid at lower masses. That is - the rate of change of frequency and amplitude increase rapidly with total mass. That is why the recent neutron star merger was detectable for 100s by LIGO, whereas the more massive black hole mergers could only be seen for about 1 second. But what this means is that you have fewer cycles of the black hole signal that can be "added up" to improve the signal to noise, which means that higher mass sources are less detectable than a simple application of the formula I gave above would suggest. A further complication is that there is a geometric factor depending on how the source and detectors are orientated with respect to each other.

OK, these are complications, but the formula can still be used as an approximation. So if we take the GW170817 signal, the total mass was about $2.8M_{\odot}$, the source was at 40 Mpc, so at frequencies of 200 Hz (corresponding to a period of 0.01 s) you might have expected a strain signal of about $3\times 10^{-22}$.  This gave a very readily detectable signal. The discovery paper (Abbot et al. 2017) says the "horizon" for detection was approximately 218 Mpc for LIGO-Livingston and 107 Mic for LIGO-Hanford. As the source was much closer than these numbers then it is unsurprising that the detection was strong.

Taking the formula above and a fixed orbital period of 0.01 s, we can see that the horizon distance will scale as $\sim M^{5/3}$. So a $10 M_{\odot} + 10 M_{\odot}$ black hole binary might be seen out to $218 \times (20/2.8)^{5/3} = 5.7$ Gpc (this will be an overestimate by a factor of a few because of the issue of the rapidity of the evolution towards merger that I discussed above).

A more through and technical discussion can be read here, although this is a couple of years out of date and LIGO's reach has been extended by about a factor of five since these calculations were done.


Would an object shot from earth fall into the sun? Would an object shot from earth fall into the sun?

If an object is shot at 107,000 km/h via rocket or otherwise, in the opposite direction to our orbit about the sun, it will be traveling at 0 km/h relative to the sun. The moon is not close enough to the object to have a significant force for the purposes of this question.

Will this object start accelerating towards the sun or will it somehow fall into another stable orbit?

Could it instead get trapped in the L4 Earth-Sun Lagrange point?

Assume that a spacecraft is instantaneously accelerated at the Earth's surface (disregarding the atmosphere for simplicity). We'll consider this from the Sun's reference frame; in other words, the Sun is stationary and the Earth is moving around it.

The spacecraft is accelerated to a velocity which is precisely equal and opposite to the orbital velocity of the Earth around the Sun, making it completely stationary in the instant after the acceleration.

What happens next? Well, we can consider the forces acting on the spacecraft:

The Earth's gravity causes a force in the direction of Earth.
The Sun's gravity causes a force in the direction of the Sun.

The stationary spacecraft is therefore going to accelerate towards the Earth and towards the Sun. Since the Earth is moving away quickly on its orbital path, the gravitational force is not enough to pull the spacecraft back into an Earth orbit; however, it will nudge the spacecraft into an elliptical orbit.

To demonstrate the situation, I have created a small simulation which can be viewed in a desktop browser. Click here to try the simulation. (You can click "View this program" to check the code, and refresh the page to restart the simulation.)

The simulation is physically accurate (ignoring the effects of other planets), but the spheres have been enlarged for easy interpretation. The Earth is represented as green, while the Sun is orange and the spacecraft is white. Note that, while the spheres representing the spacecraft and Sun intersect, the distance between the two physical objects is always larger than 3.35 solar radii.

This screenshot shows how the spacecraft has been pulled into an elliptical orbit by the Earth:

Finally, we could consider a more realistic scenario where the spacecraft is accelerated until it reaches zero velocity (again, in the Sun's reference frame) at a certain distance from the Earth. At the instant it reaches zero velocity, the engine is stopped.

In this case, the result is essentially the same: there are still forces exerted by the Earth and the Sun, so an elliptical orbit will result. The further the rocket is from the Earth when it reaches zero velocity, the more elliptical the orbit. If the Earth is so far away that its gravity is negligible, the spacecraft will fall directly towards the Sun.


Why are rocky and small planets nearer to the Sun whereas big, gas giants are farther? Planets, including Mercury, Venus, Earth, and Mars are rocky planets and are small compared to gas giants. And all these planet are near to Sun. But mass is directly proportional to force. So why don't any planets lie between rocky planets. And if gas giants lie farther always they why does Pluto lie after all the gas giants?

The planets all have differences based on where they are and their compositions. The outer gas giants are so large today because of how quick they were to accrete material. In a protoplanetary disk, once an object gets massive enough, it can attract others through gravity, making the process much quicker than granules sticking together through electromagnetic forces or random collisions. The reason the outer planets were able to get massive much quicker than the inner planets is that there is more material with larger circumferences and that material was able to stay a solid rather than become gaseous.

With a larger mass, the outer planets attracted material quicker and the process was self-perpetuating. In addition, these gas giants had the opportunity to collect more gas that the sun did not end up taking as they are farther out. Hot Jupiters can confuse scientist because of how close they are to their host star when they could not have formed there.

The Kuiper Belt and Oort cloud are also two interesting topics in planetary formation and are most likely relics of planetary formation. This would result from the material that was created around that area, but also material that was flung out by gas giants to larger orbits. The Oort Cloud could also have been formed from interactions of the sun with neighboring stars but we do not have direct observation of this cloud yet.


Finding Asteroid Geocentric Longitude and Latitude I started using the software that comes with Jean Meeus book Astronomical Algorithms. I am now using the updated code from naughter(dot)com/aa.html.

I am able to calculate the heliocentric longitude and latitude for an asteroid using the orbital elements. Below is for the asteroid Juno. I can't find any source code or instructions for converting the asteroid coordinates I calculated below to geocentric longitude and latitude. Can anyone recommend a method to find asteroid geocentric longitude and latitude? Thanks.

double Longitude = 0.0; //West Greenwich
double Latitude  = 51.4826; //North Greenwich
double TZOffSet = 0.0;

//Asteroid Juno Elements;   Using epoch J2000
CAAEllipticalObjectElements elements;

//Calculation Julian Date  2017, 9, 25
elements.JDEquinox = JD;

elements.T = (JD - 2451545.0) / 36525.0;

//semi-major axis of the orbit
elements.a = 2.6700912;

//eccentricity of the orbit
elements.e = 0.25498122;

//inclination of the plane of the ecliptic
elements.i = 12.98211;

//Longitude of ascending node
elements.omega = 169.91138;

//longitude of perihelion
elements.w = 248.10807;

//-------------------------------------------------------------------------
CAAEllipticalObjectDetails details = CAAElliptical::Calculate(JD, elements, true);

1) Heliocentric Ecliptic Longitude  = 215.6670202
2) Heliocentric Ecliptic Latitude = 9.377841131
3) True Geocentric RA = 13.90415220
4) True Geocentric Declination = -3.802673727
5) True Geocentric Distance = 4.136531368
6) True Geocentric Light Time = 0.0238906126
7) Astrometric Geocentric RA = 13.90397025
8) Astrometric Geocentric Declination = -3.80217202
9) Astrometric Geocentric Distance = 4.13653445
10) Astrometric Geocentric Light Time = 0.023890630
11) Elongation = 26.74464206
12) Phase Angle  = 7.928218201
//-------------------------------------------------------------------------

This can be accomplished through the following steps :

compute the position of the Earth in the same heliocentric frame,
express both positions (Earth and asteroid) in heliocentric rectangular coordinates,
change the origin of the frame to get the location of the asteroid in geocentric rectangular coordinates,
convert the position of the asteroid in geocentric latitude and longitude.


How can I calculate the apparent position of stars as seen from a location on Earth at a specific time? I'd like to be able to generate star maps using Python, similar to what is done at YourSky. I believe this is possible with Astropy and the details provided in this question. I can import stars from Hipparcos and use this tutorial to map stars using a Mollweide projection.
What calculations are needed to adjust for location on earth and time of year? Is the Mollweide projection the appropriate projection to use, or would something else be better?

Also related: Correct units to use while plotting mollweide projection of sky chart with RA and Dec values, matplotlib

There is doubtless a way to do it in AstroPy, but in case you’re interested, the Skyfield library both supports the operations of loading the Hipparcos catalog and computing apparent coordinates. Here is its overall guide to positions, with running examples in the text:
https://rhodesmill.org/skyfield/stars.html
And here's a more elaborate and realistic example star plot, using the stereographic projection, which is my favorite when doing astronomical work:
https://rhodesmill.org/skyfield/example-plots.html#drawing-a-finder-chart-for-comet-neowise
In Skyfield you will account for location and time of year by using a specific time and location when generating your apparent positions, a maneuver that's included in this example:
https://rhodesmill.org/skyfield/examples.html#at-what-angle-in-the-sky-is-the-crescent-moon


Does gravity bend light, and how much time does it take for light to cross gravity of a Black Hole? This is very confusing to me, so the question might be same. Please go through it, if required i will add more text.

According to Einstein's theory of general relativity, photons are not technically affected by large gravitational fields; instead space and time become distorted around massive objects and the light simply follows this distorted curvature of space.

My questions are:
1) For example: 1 hour near a black hole, is equivalent to 100 Years on Earth. If the light travels through this distorted space and time around a black hole, how much earth time does it take for light to cross the black hole's gravity?

2) We say the Universe is 14 Billion Years old, and its diameter is 91 Billion Light Years approximatly. My assumption is, these age and diameter of universe is based on the measurement of the light travel (Light is one of the factor in many).

3) So, if first (1) and second (2) points are right, how can we say the Universe is created 14 Billion Years ago (if light takes 100 Earth Years to cross over the black hole's gravity)?

You are quite right: Einstein's theory says the curvature of space is locally deformed. The essence of this is captured in the spacetime 'metric', a mathematical tool that tells us what space looks like and, derived from this, what is meant by 'a straight path', which photons take. If there is no source of gravitation present, the path of a photon will be what you know intuitively as a straight line. However, for some mass concentration (e.g. a black hole, as you say), this path will be bended such that the mass concentration acts as a lens. This is immediately clear from this image from the CFHTLenS survey:

Crucially to your question I think, you must remember that photons do not experience time and their speed is equal to $c$, the speed of light. Photons are not unaffected by their movement through a gravitational field, mind you: but this shows up as a gravitational redshift (for a time-varying potential), rather than a time delay (other than the slightly elongated path, perhaps).

Besides that, the age of the Universe is typically not quite measured as you say, but rather through parameter estimation in e.g. the cosmic microwave background. The effect of gravitational lensing needs to be taken into account for that, but not in the way you presume.

Also, remember that an extreme gravitational field such as that of a black hole is relatively rare, and even if it did delay the photon for a 1000 years that is still a tiny fraction of the age we would, according to your way of thinking, infer; the photon would have to encounter a LOT of such black holes for this 'effect' (which does not occur) to have a huge impact.

As a general point, you seem to have some (Interstellar-induced?) misconceptions about the universe, but I think that's not for me to address here.


What happens if a black hole is produced inside another black hole? Similar question here.

The question I linked got me thinking.

It is known that a black hole doesn't let anything escape its event horizon, not even light. Thus, naturally, anything that passes the event horizon would get "eaten".

Now, assume a supermassive black hole(small Roche limit, large event horizon) was in the path of a high eccentricity, massive, dying star. Assume the star is travelling in a straight line at 5% the speed of light, and the black hole is stationary. The star would be swallowed by the black hole, and also swallowed intact. Now, the star goes supernova inside the black hole, but since its mass is very high, it too, creates a black hole. What would happen?

Would the supermassive black hole gain the mass of the star and things would go as normal for an outside viewer(everyday common things: black hole swallowing objects)? Would the black hole inside the larger one pull the larger one in? Or something else?

Thanks!

Max0815

From the no-hair theorem, black holes can have only three properties: mass, angular momentum, and electric charge. From this we'll see the original, supermassive black hole gain the mass, angular momentum and electric charge of the ingoing star. The black holes will merge (already merged actually, since one is inside the other). The singularities will merge too (all world lines, i.e. paths a particle can take, lead to the big singularity in the supermassive black hole).

From the outside, we will see a more massive black hole with different angular momentum & electric charge, and that's it. We can't see the star go supernova because anything that happens inside the event horizon is invisible to us.


Why does the Earth have a tilt of ~23°? Is there a reason that the Earth has the tilt that it does (~23°)?
How do we know which way is supposed to be 0°?
Does this tilt have major consequences on the planet?
Has it changed and will it change in the future? (If so, would there be any consequences?)

First up, the tilt is exactly 23.45 degrees.
The reason for Earth's tilt is still not yet really proven, but scientists at Princeton stated on August 25, 2006 that planet Earth may have 'tilted' to keep its balance. Quote:

By analyzing the magnetic composition of ancient sediments found in the remote Norwegian archipelago of Svalbard, Princeton University's Adam Maloof has lent credence to a 140-year-old theory regarding the way the Earth might restore its own balance if an unequal distribution of weight ever developed in its interior or on its surface.
The theory, known as true polar wander, postulates that if an object of sufficient weight -- such as a supersized volcano -- ever formed far from the equator, the force of the planet's rotation would gradually pull the heavy object away from the axis the Earth spins around. If the volcanoes, land and other masses that exist within the spinning Earth ever became sufficiently imbalanced, the planet would tilt and rotate itself until this extra weight was relocated to a point along the equator.

Same goes for the lack of proof about the exact consequences on the planet or if it'll happen again. All that is still being researched and debated, but those Princeton scientists have thrown in some interesting perspectives which you'll discover when you visit the link I provided and completely read what they wrote.
Besides "consequences" for Earth as a planet, it should be noted that the tilt of Earth is the reason why we have seasons. So even when we yet have to find out what the consequences for Earth (as a planet) have been and/or will be, we do know that the tilt surely has consequences for all beings that live on planet Earth… the seasons that influence us all: summer, fall, winter, and spring.
And - last but not least - as for your question how they know where 0 degrees would be, you just have to look at the Earth's rotation around the sun. Take a 90° angle of the planet's orbit and you know where 0° would be. But I guess a picture explains more than a thousand words, so I quickly created the following graphic:


How to get the longitude/latitude from solar zenith/azimuth? How do I get the longitude/latitude from zenith/azimuth?

Or how can I position my sun in a 3d application?

code extracted from here

public static AzimuthZenithAngle getSolarAzimuthZenithFromLatLon(float lat, float lon) {
   GregorianCalendar time = new GregorianCalendar(TimeZone.getDefault());
   return PSA.calculateSolarPosition(time, lat, lon);
}

public static Vector3 getSolarPosition(float radius, AzimuthZenithAngle az) {
    double a = az.getAzimuth();
    double z = az.getZenithAngle();

    float X  = (float) (radius*Math.sin(z)*Math.cos(a));
    float Y  = (float) (radius*Math.sin(z)*Math.sin(a));
    float Z  = (float) (radius*Math.cos(z));

    return new Vector3(X, Y, Z);
}

this is my code, but it's not working very well.

Well let's see:

Local Standard Time Meridian (LSTM)

The Local Standard Time Meridian (LSTM) is a reference meridian used for a particular time zone and is similar to the Prime Meridian, which is used for Greenwich Mean Time.
The (LSTM) is calculated according to the equation:
$$
LSTM = 15^{o}.\Delta T_{GMT}
$$
where $\Delta T_{GMT}$ is the difference of the Local Time (LT) from Greenwich Mean Time (GMT) in hours.

Equation of Time (EoT)

The equation of time (EoT) (in minutes) is an empirical equation that corrects for the eccentricity of the Earth's orbit and the Earth's axial tilt.
$$
EoT = 9.87 sin\left(2B\right) - 7.53cos\left(B\right) - 1.5 sin\left(B\right)
$$

Where $B = \frac{360}{365}\left(d - 81\right) $ in degrees and d is the number of days since the start of the year.

Time Correction Factor (TC)

The net Time Correction Factor (in minutes) accounts for the variation of the Local Solar Time (LST) within a given time zone due to the longitude variations within the time zone and also incorporates the EoT above.
$$
TC = 4\left(Longitude - LSTM\right) + EoT
$$
The factor of 4 minutes comes from the fact that the Earth rotates 1° every 4 minutes.

Local Solar Time (LST)

The Local Solar Time (LST) can be found by using the previous two corrections to adjust the local time (LT).
$$
LST = LT + \frac{TC}{60}
$$

Hour Angle (HRA)

The Hour Angle converts the local solar time (LST) into the number of degrees which the sun moves across the sky. By definition, the Hour Angle is 0° at solar noon. Since the Earth rotates 15° per hour, each hour away from solar noon corresponds to an angular motion of the sun in the sky of 15°. In the morning the hour angle is negative, in the afternoon the hour angle is positive.
$$
HRA = 15^{o}\left(LST - 12\right)
$$

Declination angle:

The declination angle denoted by $\delta$, varies seasonally due to the tilt of the Earth on its axis of rotation and the rotation of the Earth around the sun. If the Earth were not tilted on its axis of rotation, the declination would always be 0°. However, the Earth is tilted by 23.45° and the declination angle varies plus or minus this amount. Only at the spring and fall equinoxes is the declination angle equal to 0°.
$$
\delta = 23.45^{o} sin\left[\frac{360}{365}\left(d - 81\right)\right]
$$

where d is the day of the year with Jan 1 as d = 1.

Elevation angle:

The elevation angle is the angular height of the sun in the sky measured from the horizontal.
$$
α=sin^{−1}[sin\delta sin\phi + cos\delta cos\phi cos(HRA)]
$$
Where $\delta$ is the declination angle, $\phi$ is the local latitude and HRA is the Hour angle.

Azimuth Angle:

The azimuth angle is the compass direction from which the sunlight is coming. At solar noon, the sun is always directly south in the northern hemisphere and directly north in the southern hemisphere.
$$
Azimuth = cos^{-1}\left[\frac{sin\delta cos\phi - cos\delta sen\phi cos(HRA)}{cos\alpha}\right]
$$

Where $\delta$ is the declination angle, $\phi$ is the local latitude and HRA is the Hour angle.

Zenith Angle:

The zenith angle is the angle between the sun and the vertical. The zenith angle is similar to the elevation angle but it is measured from the vertical rather than from the horizontal, thus making the zenith
$$
Zenith = 90° - \alpha
$$

Where $\alpha$ is the elevation angle.

Note that your input parameters are going to be:

Longitude
$\Delta T_{GMT}$ is the difference of the Local Time (LT) from Greenwich Mean Time (GMT) in hours
LT local military time in hours
$\phi$ the local latitude
d the day of the year


Is nucleosynthesis responsible for the expansion of the universe? Is it just a coincidence that the two major expansionary periods occur close to periods of nucleosynthesis?

Big Bang nucleosynthesis occurred very close to the inflationary period.
Supernova and stellar nucleosynthesis created heavier elements as the stars died, and the universe underwent increased acceleration.

Further thoughts:

The metaphor for the expansion of the universe is often described as an inflating balloon, but could it be equally valid to think of expansion as matter falling into its own gravity well, away from infinity minus other matter within its gravity field. The effect being that matter radiates space itself as gravity pulls itself away from infinity. Just as heavier elements create a stronger gravity field, heavier elements would accelerate away from infinity minus other matter within its gravity field, at a faster rate.

Eventually, on a cosmic scale, atomic decay would release the confined energy and the energy would return to infinity, creating the appearance of a contracting universe that restarts the Big Bang.  If this idea is reality it would create a geodesically complete cosmology.

Big bang nucleosynthesis did not happen "close to the inflationary period".
inflation happened at about $10^{-36}$s. Nucleosynthisis happened from about 10seconds.

In relative terms, the ratio of the age of the universe now, to 10 seconds is much much smaller than the ratio of 10 seconds to $10^{-36}$s. And, significantly, nucleosynthesis occurred after inflation, so can't be the cause of inflation.

As for dark energy, nobody has any real model for it. The test for any theory is how well it does empirically. As it stands you have advanced a pet theory, without any mathematical model or analysis.


Hydrogen and oxygen in space chemistry tells us that when mixing hydrogen and oxygen the result is an explosion and water.

my question is: does this happen on astronomical scale? for example, did they ever observe planet-size hydrogen cloud collide with planet-size oxygen cloud to produce a huge explosion and planet made primary of water?

Hydrogen and oxygen only react when there is sufficient energy. For instance, the autoignition temperature of hydrogen at 1 atmosphere is 536 °C. This is why you can do that experiment with mixed hydrogen and oxygen in a balloon, that only explodes when you touch the balloon with a lit taper.

Space is cold. Molecular clouds have temperatures in the tens of degrees Kelvin. In the rare (if not impossible) case that a molecular cloud of pure oxygen did exist, and it encountered a molecular cloud of hydrogen, there wouldn't be enough energy for an explosion.


Does every object in the Universe have gravity? Space has no gravity, why? My science teacher told me that every object in Universe has gravity but space is a part of Universe too, so,  why does the space have no gravity?

A quick search up on google you would find this:

  Gravity causes every object to pull every other object toward it. Some
  people think that there is no gravity in space. In fact, a small
  amount of gravity can be found everywhere in space. Gravity is what
  holds the moon in orbit around Earth.

Why is this? That's because:

  Outer space is the closest known approximation to a perfect vacuum. It
  has effectively no friction, allowing stars, planets and moons to move
  freely along their ideal orbits. However, even the deep vacuum of
  intergalactic space is not devoid of matter, as it contains a few
  hydrogen atoms per cubic meter.

If there's matter in space, thus this makes this statement valid.

  Yes, everything that has mass will cause a gravitational pull on other
  objects.


What is the name of meteoroids which hit Moon, Mars, or pretty much anything that isn’t the Earth? A meteorite according to the Oxford English Dictionary is

  A piece of rock or metal that has fallen to the earth's surface from outer space as a meteor.

But, meteors impact other planets.  Are they still meteorites?  Or is there another word for them.  And, if they are still called meteorites, what is the threshold for an object to be impacted upon by a meteor or meteoroid in this case, and still have the meteoroid be called a meteorite.  Are meteoroids impacting Pluto or Pallas meteorites?

We expect the OED to be authoritative so I am just curious if they have let us down or if the clever space brains have a taxonomy to differentiate them.

Oxford English Dictionary is an authoritative source, but it's aimed at the general public. I think no astronomer will frown upon you when you're using the terms meteoroid and meteorite to describe a small body impacting another body in the solar system. As an example, Wikipedia doesn't impose the limitation that it must hit Earth:

  A meteorite is a solid piece of debris from an object, such as a comet, asteroid, or meteoroid, that originates in outer space and survives its passage through the atmosphere to reach the surface of a planet or moon.


How can we learn the inclination of the orbiting plane of the exoplanets using radial speed method? I heard about the radial speed method to discover exoplanets a few years ago, however, these questions kept confusing me. If we know a planet star of mess $m$ is orbiting a star by interpreting the  radial speed of the central mass, how can we tell wether this is a even more massive object whose orbiting plane has a larger inclination? Further more, even if we know the angle of our sight and the axis of orbit, can we determine the direction of orbit?

Thanks in advance.

I assume you are talking about the radial velocity method.

Analysis of a radial velocity curve yields the minimum  mass of a planet $M \sin i$, where $M$ is the true planetary mass and $i$ is its orbital inclination (90 degrees would mean the orbital plane is in our line of sight).

In general, with no other information that's all you can say. In principle $M$ could have any value $\geq M\sin i$.

Of course increasingly small values of $\sin i$ are increasingly unlikely and the average $\sin i$ of a randomly oriented population is $\pi/4$.

If you see a transit then you know that $i$ is close to 90 degrees and can in fact estimate $i$ from the transit duration.

If you measure the star's rotation period and can estimate it's projected equatorial velocity from spectral line broadening, then you can estimate it's inclination to the line of sight. One could then assume that the planet orbits in the equatorial plane of the star and use the same inclination for its orbit.


How long until we cannot see any stars from other galaxies? Since all the galaxies are moving further away from each other at a faster and faster rate, eventually we won't be able to see stars from other galaxies. How long until we can't see stars from other galaxies?

First of all, we can't see stars from other galaxies (with a few exceptions, Cepheid variable stars for example are regularly used to determine distances to nearby galaxies). As it currently stands, we can only see stars from our own Milky Way (in which I'm including the large and small Magellanic Clouds). Type Ia supernovae aside, it might be possible to see Cepheid variables in the Andromeda galaxy, but anything further is almost certainly not possible.

The relevant number when it comes to the distance one could possibly see in the universe is the comoving distance to the cosmological horizon. This horizon defines the boundary between what can be seen and what cannot be seen, simply because the universe is not old enough for particles to have traveled that far (yes, even at the speed of light; in fact, if we're talking about what we can observe, the photon is the particle we care about).

The definition of the comoving distance is:

$$\chi = \frac{c}{H_{0}} \int_{z=0}^{z=z_{Hor}} \frac{dz^{'}}{E(z^{'})}$$

where the function $$ E(z^{'}) $$ contains your choice of cosmology (the following is for flat universes only):

$$E(z^{'}) = \sqrt{ \Omega_{\gamma} (1+z)^{4} + \Omega_{m} (1+z)^{3} + \Omega_{\Lambda} }$$

This distance can be extrapolated out into the future (given your choice of cosmology), and ultimately will tell you what the distance to the cosmological event horizon will asymptotically reach. This is the distance at which no object beyond could ever come into causal contact with you.

Another simpler way to calculate the particle horizon is to calculate the conformal time that has passed from the beginning of the universe, and multiply it by the speed of light $c$, where the conformal time is defined in the following way:

$$\eta = \int_{0}^{t} \frac{dt^{'}}{a(t^{'})}$$

where $a(t^{'})$ is the scale factor of the universe, and its relationship to time depends specifically on your choice of cosmology. Again, if the ultimate value of this horizon is desired, you would need to integrate this to infinity.

SUMMARY: I'm having a hard time finding the exact number of the comoving size of the cosmological event horizon, though I'll keep searching. If I can't find it I suppose I'll have to do the calculation when I have some free time. Almost certainly it is larger than the size of the local group of galaxies, making our night sky largely safe from the fate of the universe. Anything further than this horizon, however, will monotonically become both fainter and redder. I also think that it does asymptotically reach some value, rather than eventually increasing infinitely or decreasing passed some point in time. I'll have to get back to you on that.

Here's a nice review of distance measurements in cosmology.


How fast is Neptune getting brighter? When was was this first noticed and reported? This answer to Compute Planet's Apparent Visual Magnitude reports the new work on planetary magnitudes, and contains an intriguing blurb about Neptune:

Neptune keeps getting brighter. No one knows why.

So I'd like to ask: How fast is Neptune getting brighter? When was was this first noticed and reported?

TL;DR: There was apparent 11% increase of Neptune brightness during 1980 and 2000. This could be due to multiple reasons. Recent observation suggested the reason to be change in the amount and brightness of the banded cloud mostly in the planet's southern hemisphere due to seasonal variations and thus affect overall brightness of the planet.

Neptune's visual magnitude stayed relatively constant before 1980 and after 2000, but during this period, there was an unexpected rise in the brightness, about 11%.  Astronomers have proposed various ideas to explain this peculiar variation but none are convincing. But the more promising explanation would likely to be seasonal change due to tilt and cloud formation variation.

A team from Wisconsin made three sets of observations of the full rotation of the planet in 1996, 1998, and 2002. The images showed progressively brighter bands of clouds encircling the planet's southern hemisphere. The findings were consistent with observations made by G.W. Lockwood at the Lowell Observatory, which show that Neptune has been gradually getting brighter since 1980. They believe that cloud bands are getting brighter because the Sun is warming the atmosphere in the south apparently changing the brighness of Neptune.

Neptune's near-infrared brightness is much more sensitive to high altitude clouds than its visible brightness. The recent trend of increasing cloud activity on Neptune has been qualitatively confirmed at near-infrared wavelengths with Keck Telescope observations from July 2000 to June 2001 by H. Hammel and co-workers which could provide an explanation to this brightness change.
Tilt could also be a reason. Hubble images reveal a real increase in Neptune's cloud cover consistent with seasonal change is the apparent absence of change in the planet's low latitudes near its equator. This do has a reason. Since Neptune is inclined at a 29-degree angle, the amount of solar energy received determines the season and this seasonal variation might affect the brightness. When the Sun deposits heat energy into the atmosphere, it forces a response like heating the atmosphere that lead to constant moving and condensing of cloud cover.
Another reason could be internal heat source of Neptune which may contribute to the planet's apparent seasonal variations and ultimately affect the planetary brightness.
Current visual magnitude of Neptune is 7.88 but it could reach 8.0 in the near future.
For more information, refer to this paper, "The Secular and Rotational Brightness Variations of Neptune"
Other references:

https://www.sciencedaily.com/releases/2003/05/030516082529.htm
https://skyandtelescope.org/observing/measuring-planet-magnitudes/
https://www.ssec.wisc.edu/media/Neptune2003.htm


If the earth stopped rotating, would we really fly inertially at 1000mph? I heard this on Prof. Neil Tyson's StarTalk podcast (this episode) and had a question about it. In a Q&A session, a listener had asked what would happen if Superman really did stop the earth from rotating and rotated it in the other direction. Prof. Tyson's response was that if the earth, which is rotating at approx. 1000mph about its axis, is stopped, everything that is not anchored on earth will move the opposite direction at 1000mph and we'd all go crashing into things at that speed.

I suppose that is just simple inertia: if a body moving at 1000mph is stopped abruptly, everything that was moving with it but not attached to it would indeed attempt to continue in the same direction at 1000mph. However, the earth is rotating rather than moving in a straight line. Essentially, gravity is strong enough to ensure that we are not thrown off into space by the centrifugal force from the earth's rotation (we aren't holding on to our dear lives as the earth turns at 1000mph). So in a sense, aren't we anchored to earth by gravity? And therefore, doesn't it mean that if the earth stopped abruptly, we'd just stop with it? (I'm thinking of say a centrifuge with test tubes attached to it. Though as I think about this, I feel that we'd sway at the point of attachment instead of stand still, were the earth to stop).

Can someone help resolve this for me?

You would be sliding, because the friction between you and Earth would be too weak, or else you wouldn't survive the acceleration, if the rotation of Earth would be changed abruptly. That's like crashing into a wall with several hundred mph. This is with the exception of close to the poles.

Your weight wouldn't change much. On a smooth, plane, and horizontal surface you would slide (due to your inertia). Your velocity you would be sliding depends on your geographic latitude. At the equator it would be at the maximum. At the poles it would be zero. More precisely the velocity would be proportional to the cosine of the latitude (latitude 0 at the equator, and (+/-) 90 degrees, resp. $\pi/2$ at the poles).

Even more precisely, the speed you would be sliding would be $$2\cdot 465.1~ m/s \cdot cos ~\varphi,$$ with $\varphi$ the latitude.
Factor two, because superman accelerates to the opposite direction. Earth's equatorial rotation velocity is $465.1~m/s$.


Why are there no Oe or Ae stars? The unique characteristics of Be stars are mainly because of their circumstellar discs.
I wonder what makes Be stars special that only they have unique characteristics?
Which kind of stars have discs generally?
Why are there no Oe and Ae stars?

Well, there might be.

I strongly disagree with your statement that circumstellar discs are what differentiate B(e) stars from other related stars. In fact, there are four primary criteria for stars that satisfy "the B(e) phenomenon":

Strong emission lines in the Balmer series
Emission of lines of certain (low ionization) metals such as Fe II
"Forbidden emission lines" in the visible part of the electromagnetic spectrum such as O I
Hot circumstellar dust

The dust is not the key to the B(e) phenomenon, though it can change the spectrum slightly.

HAeB(e) stars are a class of stars (and a subclass of B(e) stars) related to Herbig Ae/Be stars, which have not yet left the main sequence. There are additional criteria that a star must satisfy to be considered an HAeB(e) star, too. These stars can be of spectral type A, meaning that you can have B(e) stars that are not of spectral type B, if you use a certain definition.

This answer is based largely on Lamers et al. (1998), a fascinating paper on the classification of B(e) stars.


JPL DE Documentation I'm currently trying to write a C++ library that makes accessing binary data from a JPL ephemeris file easier. So far I sifted through the web, trying to find a good and thorough documentation about what is actually stored in there and how to work with the data.

Does anybody know of a PDF/Book or something similar, that explains anything of that sort?

Thanks in advance!

Be aware there are differing formats. SPK is the newer format - the older text format seems to be deprecated.

It's worth having a look at the documentation on this jplephem python library as it explains the format reasonably well. If you do use SPK, you'll need to read all of the documentation in NASA's SPK repository

I'm not sure if this is from the same author as this library - worth a read as it discusses the FITS format.


Galaxy behind the Ring Nebula Would anyone be able to reference the galaxy in the background of this image of the Ring Nebula:

It's obviously not a Messier object, I'm more confident in it being in the New General Catalogue.

As Andy pointed out, the spiral galaxy is IC 1296. You're correct that it's not in the Messier Catalog, but it's obviously in the revised Index Catalogue. NED and SIMBAD have some more information on it:

Morphological type: SBbc, as classified according to the Hubble sequence. It's a spiral galaxy with two clearly visible arms.
Redshift: 0.017075 $\pm$ 0.000070, according to Marzke et al. (1996).
Distance: 66.6 Mpc to 68.9 Mpc after corrections, depending on your distance definition of choice (angular diameter, luminosity, etc.). Large uncertainties arise when calculating the difference from redshift. For comparison, Andromeda is approximately 0.78 Mpc away, and the Ring Nebula is $10^{-3}$ times that distance.
Apparent magnitude: 14.80 $\pm$ 0.30, at visible wavelengths. For comparison, the Ring Nebula's apparent magnitude is 8.8.


Can I learn astrophysics and astronomy with my basic math? I want to learn astrophysics and astronomy, I'm very interested in it because there are many questions that popped up in my head, and really curious about things, but I'm afraid of a little bit on math, but I'm a fast learner, I can understand things quickly and easily especially if it is translated on my native language, I have learned basic trigonometric functions, special angles, trigonometry, algebra, quadratic equations, and some basic math but not calculus, will proceed on that later, and I have learned basic physics, until vertical dimension, but still afraid a little on math and equations, how can I counteract that fear? I mean I really want to learn astrophysics but I'm afraid of numbers,  can you give me tips? And can you suggest me some book? Or can I learn astrophysics without interacting too much with math? Btw, I'm just taking advancement because I'm really curious and interested in astrophysics, I'm still at high school, hope you can help me.

Generally, astrophysics (and astronomy) is math-heavy. That should not discourage you, but rather act as a way of learning math: it is often easier to learn topics that you have a use for and are part of some personal project than just getting lectured about them.

Astrophysics is based on math because it is based on many forms of physics - mechanics, thermodynamics, nuclear physics, relativity theory and so on - plus chemistry. Mathematics is a common language that binds these fields together and allows you to combine them to build models to understand what is going on. Some parts just use some basic math, but in general calculus really is needed.

Mathematics in the more common "calculations with numbers" form also matters because it all starts with measuring things - where things are, how much they change, brightness and so on - and then we build and test our models and understanding from this. That also means that statistics is important for handling measurement noise and finding patterns in data.

One cannot really do astronomy and astrophysics without math. It would be just looking at things and making up explanations without actually checking if and how they hold true.


Can I sense a bright star pointing an eight foot antenna towards it? If I connect an eight foot Yagi or other comparable sized antenna to my oscilloscope and point the antenna at a bright star will I see a voltage on my oscilloscope?

I am not interested in turning the voltage into an image just wondering if I would see a voltage increase when it is on a bright star. I’d like to know your thoughts before I take the time to build the antenna. I’m thinking about in the 25cm range. I’ve heard that’s an active area. My oscilloscope will read down to about 20 millivolts.

Stars are too dim for amateur radio equipment. There are two possible radio sources that you can detect: the sun and Jupiter.

Jupiter is particularly interesting as interactions between Io and its magnetic field produce beams of radio waves that sweep past earth every 10 hours.  These are detectable in the amateur range, at about 20 MHz.

Nasa make a kit for detecting these radio signals, or it is possible to use a ham antenna, but of course it must be cut for the frequency of operation. The Nasa kit uses a phased dipole antenna which must be set up in a field or similar as the antenna is about 7m long.

Stars are not very good radio sources. Supernovae remnants such as Cassiopeia A or the Crab nebula are much brighter at radio wavelengths. Most supernovae are too distant to be powerful radio sources; radio supernovae are rare. A local supernova would be a radio source but we haven't observed a supernova in the milky way for several hundred years.


In which direction is the sun travelling? Relative to this simple drawing, can someone explain to me,

in which direction is the sun and its solar system travelling,
which star (irrespective of distance) is closest to that direction,
and in which direction the milky way center is.

The direction of solar motion is referred to as the solar apex. It is at around RA=18h28m and Dec=+30d in the constellation of Hercules (you can see a map on the page linked to). Note that this is the direction in which you would see the "star wars" effect of onrushing stars diverging from the apex point. i.e. It demonstrates the sun's motion with respect to the local stellar population.

On your 2-dimensional cartoon of the Earth's, the circle is the ecliptic plane and the Earth's northern hemisphere (at winter solstice) is tilted away from the Sun at an angle of  23 degrees. The celestial coordinate system defined above is geocentric and the Sun is at 18h -23d at the winter solstice. The apex point is therefore towards  the Sun but 57 degrees above it.

Look at the map below (produced by Christian Ready), you can judge where the solar apex is with respect to the Sun and ecliptic plane (green line), which is roughly the path (drawn in your diagram) that the Earth takes around the Sun.

The very bright star Vega is reasonably close to the solar apex direction.

The Milky way centre is at RA=17h45m, Dec=-29d in the constellation of Sagittarius - i.e. almost where the Sun is in December. In Galactic coordinates (see Best approximation for Sun's trajectory around galactic center? ) the Sun moves at about 10 km/s towards the Galactic centre, about 5 km/s faster than the average circular speed (which is itself about 220 km/s) of local stars in the tangential direction and 7 km/s upwards out of the Galactic plane.

Thus the motion of the Sun with respect to the Galactic centre, as opposed to the local stellar population, is basically at $\sim 230$ km/s towards a point that is slightly less than 90 degrees around the Galactic plane and slightly above it. This would be at 9h30m -50d in the constellation of Vela. I think that is into the page and down about 30 degrees in your cartoon.


What is the most asymmetric known planet? Planets are more or less spherical, can you tell which one is the most asymmetric? Do they figure out such a property from an eccentric behaviour, or what?

I have a simple technical question: an asymmetric body gets a torque in a gravitational field, is the rotational energy/acceleration to be subtracted from the regular energy/acceleration? In other words: does an asymmetric planet get less radial acceleration than a perfect spherical one?

What is the most asymmetric known planet?

That depends on what you mean by asymmetric, on what counts as planet, and on what parts of a planet are to be considered / excluded. The dominant feature of many planets is that they aren't quite spherical due to their rotation. If that counts then Saturn is the winner. Saturn's poles are 5900 km closer to Saturn's center of mass than are points on Saturn's equator.
That oblateness is however exactly the shape an object such as Saturn is expected to take on. It means that Saturn is in an equilibrium shape. Moreover, it has a very definite axis of symmetry, the rotation axis. So that's probably not what is meant in the question as being asymmetric. All of the giant planets are quite symmetric by this standard. We need to look to the terrestrial planets.
One concept that might be of aid is the center of mass - center of figure offset of each of the four terrestrial planets. Do the oceans count? Do the icecaps count? Excluding the oceans but not the ice leads to COM-COF offset of about 2100 meters. Including both leads to a COM-COF offset of about 800 meters, about the same as Mercury and Venus. Mars however has the Earth beat by far. Various estimates place the Mars COM-COF offset between 2.5 km and 3.3 km.
Another approach is to look at how lumpy the planet's gravitational field is. Mars once again is the winner in this regard.

I have a simple technical question: an asymmetric body gets a torque in a gravitational field, is the rotational energy/acceleration to be subtracted from the regular energy/acceleration? In other words: does an asymmetric planet get less radial acceleration than a perfect spherical one?

No. In a system of particles bound by central forces such as self-gravitation, Coulomb forces, and chemical bonds, the acceleration of the system's center of mass is the net sum of the external forces acting on the individual particles, divided by the total mass.


Is it possible to tell if a certain extrasolar planet produces its own magnetic field? I know by observing the dimness of a star it is possible to calculate an exoplanet's distance from the star and its mass by how much the star wobbles. However, is there any way to accurately determine if said planet has any magnetic field through observation alone directly?

There are three approaches with which people have looked for this, and not yet been too successful:

A transiting planet planet with a strong magnetic dipole and/or very strong host star winds might produce a visible signal when the magnetospheric bow-shock passes in front of the star as well. The idea is that at the bow-shock the streaming hydrogen would be heated up, ionize and produce a detectable ultraviolet photometry signal upon recombination at a characteristic distance (the stand-off distance) from the planet.
HD189733b was for a time the first planet to have measured its magnetic field in that way, however the discovery was retracted because the ultraviolet opacities that the group used to determine the bow-shock position were wrong.
It turns out there is no signal from the bow-shock.
The second idea is to look at a hot Jupiter that has an evaporating atmosphere. The shape of the ionized hydrogen tail in the magnetosphere will be influenced by the strength of the planetary dipole. But then, the amount of hydrogen that is in the planet's shadow is also a function of the magnetic field strength.
Hydrogen that comes into the planet's shadow will recombine with magnetospheric electrons due to the lack of direct irradiation from the star and produce a signal in the Lyman-$\alpha$ recombination line.
Going further, only the particles that stream away from the planet, and towards us will be part of this process. Thus the blue-shifted side of the Lyman-$\alpha$ line can inform us about the strength of the magnetic field.
So far the theory. The practice is that our last remaining UV-spectrometer in space that could perform such a thing (mounted on the Hubble) is even only barely capable of observing this process. But it has been done for the planet HD209458b
The data quality is lousy, and the interpretation that has been done in Kislyakova et al. 2015 has more parameter than datapoints and is therefore questionable.
But if you choose to believe the authors, then this planet has a magnetic moment of $0.1$ Jupiter magnetic moments.
The third possibility that could yield real results are radio observations with the LOFAR array. LOFAR observes at the edge of observable frequencies from Earth (at very low frequencies the ionosphere interferes with radio observations) at the for the moment highest achievable spatial resolution. The idea here is to look for analogues of Jupiters decametric radiation, which is simply synchrotron radiation produced when ionized particles start gyrating around a magnetic field line.
This already has been tried at higher frequencies / different observatories with the planets around $\epsilon$ Andromedae  and $\tau$ Bootes, as yet to no avail.


Betelgeuse and sun classification In this video, between 4 and 6 minutes, the lecturer describes M type stars as being less massive, smaller, and redder than G type stars.

He then gives Betelgeuse as an example of an M type star and mentions that the sun is a G type star.

However, Betelgeuse is more massive than our sun, and larger than our sun.

How can I reconcile these facts?

If you take a look at the Hertzsprung-Russell Diagram that helps us to classify stars

(c) Wikimedia commons

(while this one uses data gathered from the Hipparcos satellice, roughly 100.000 stars) we see, that there are 2 main types of red stars: Giants and M-dwarves. The latter are stars on the main sqeuence (they have stable hydrogen-burning in their cores), while the giants are stars that have left the main sequence due to aging. They appear red as they bloat up to bigger radii compared to their main-sequence life which also makes them cooler, and thus redder.

Betelgeuse is now one of those giants, and he's actually more massive than the sun (somewhere between 7 and 20 solar masses, according to wikipedia), so there is no contradiction with M-Dwarves also being red.

Please let me know if you desire to know more details on this.


What state of matter are nebulae in? After doing research it said that nebulas are made of gas and dust. Are they actually gases and solids, or are they plasma?

Nebulas are not static objects - they evolve.

Initially a nebula might contain mostly (as an example) hydrogen, helium and some other elements in a cloud, but over time that material will clump together to form larger and/or denser material.  Eventually parts of some nebula can even condense enough to form a star and it's planets.  Our solar system was (probably) once part of a nebula.

So there's no single answer to your question.  They can contain most forms of matter.


Why can I see the dark side of the moon? Last weekend (29/30 April) the sky in my area (Central Europe) was pretty clear and the Moon was nicely visible. Funnily, one could not only see the sunlit area of the Moon, but also the dark side* could be easily recognised throughout the whole evening from blue hour until dark night. I attached two photos of the Moon that should illustrate the effect. The pictures (sorry for the poor quality) needed to get a little overexposed to capture the illuminated dark side, but with plain eye the effect could be well seen.
We were speculating about the reasons why the non-sunlit area of the Moon could be seen.

diffraction of sunlight through the atmosphere can be neglected as the Moon doesn't have an atmosphere.
reflection of light that in turn was reflected from earth (Sun -> Earth -> Moon -> observer) sounded a little unlikely to me
a psychological effect that kind of mentally completes the shape of the Moon, because we know it should be circular, sounded ok to me. But could be debunked as the dark side stays visible if one covers the bright side on the photo?

So I guess the  sunlight has to reach the dark side of the Moon somehow differently.

What could be provide enough light to make the dark side visible?

* not the far side of the Moon, but the non-sunlit area of the Moon

You can generally see the unlit side of the moon when a considerable amount of sunlight is reflected off the earth. This reflected sunlight illuminates the unlit side of the moon. This is referred to as earthshine, and a decent explanation can be found at timeanddate.com. I seem to recall reading somewhere (but now can't find a reference) that it is more prominent when it is very cloudy on earth near where the moon is overhead.


rotation curve in the innermost part of galaxies I read in Sofue's articles (1999) that he used a method to investigate the bulge of galaxies to obtain rotation curve of this part of galaxies. So my question is how could Vera Rubin et al. obtained the rotation curve in their article (1978), especially the steep increase in the begenning of the curve since in Sofue's article he seems to say it was rather difficult, at that time, to investigate galactic bulges ?

So they are talking about different scales. Rubin is finding the rotation curve on kpc scales. And Sofue is talking about how hard it is to get the rotation curve in the central less than a kiloparsec. In the center you cannot use the same techniques as Rubin did.

Rotation curves of galaxies are most often measured by the gas speed. The speed of the stars can be used, but owing to dynamical heating over time, the stellar rotation speed does not indicate the true rotation speed of the galaxy as well as the gas speed.

When looking for the speed of something, astronomers will look for the redshift or blueshift of the stars or the gas which they are observing. So, they need to identify a gas, which will be present in all galaxies, which has a well known and easy to observe "line" in the spectra which will allow a velocity determination by seeing how red or blueshifted it is compared to that emission line in a laboratory.

Two such lines are H$\alpha$ and molecular CO (the one mentioned in that paper you've linked).

The H$\alpha$ line is an optical line at 656.28 nm. The CO lines are radio waves at 2.6 and 1.3 mm.

It looks like the Rubin+1978 paper took spectra of the H$\alpha$ lines, and from Figure 2, it looks as if they've collected spectra from various radii from the center. If you do it to many different points inside the galaxy at different distances from the nucleus, you can see how fast they are moving relative to each other.

That steep increase at the beginning of the curve usually occurs for around the extent of the bulge, a few kiloparsecs (as in this image from Rubin+1978).

It looks like they had a resolution of about a kpc, which would not tell you much about the center, except that the rotation curve is very high, very quickly (you assume the very center is 0 rotation):

Sofue+1999 seems to be talking about even more central than that, like the "central few hundred pc region" which will require a new technique.

A large limitation, as mentioned in the Sofue+1999 paper, is that HI is often obscured in the very centers (<0.5 kpc) of these galaxies by dust. Galactic dust, in general extincts shorter wavelength light much more effectively than long wavelength light. In this way, looking at the galaxy in longer wavelengths allows you to see through the dust. The HI line is much more easily erased by galactic extinction than the CO line at a longer wavelength. So CO makes it easier to look into the galactic central hundred parsecs than H$\alpha$. You can see this in the galactic extinction curve:

The CO line is in the radio band (mm range), a type of astronomy that was just picking up speed in the 1960s and 1970s. Having been predicted in the 1800s by Maxwell’s equations, scientists were unable to observe any radio signals from the Galaxy until a serendipitous observation of Sagittarius A* by Karl Jansky at Bell Labs in the 1930s.

However, radio astronomy comes with a problem of resolution, the resolving power of a telescope is basically:

$\Theta = 1.22 * \frac{\lambda}{D}$

Where lambda is the wavelength of the observation and D is the diameter of the telescope. And, for reference, the distances to two galaxies from that rotation curve above are:

NGC4594 = 9.45 Mpc
NGC7217 = 15 Mpc

At the distances of these galaxies, we need to be looking in the right wavelength to penetrate, and we need enough resolution to get the rotation speed at maybe 10 parsec scales. This resolution, at 15 Mpc, is maybe 0.1 arcseconds.

$3600 * arctan(\frac{10}{15e6}) * \frac{180}{\pi} = 0.1375 arcseconds$

To get that resolution observing at the 2.6 mm you need for the CO line, you need a:

$1.22 * 3600 * \frac{180}{\pi} * \frac{2.5e-3}{0.1375} = 4581.48$

4.5 km telescope… Hmmmmm. A bit of a problem considering the unfortunate tale of the 300 foot telescope.

Then, computers to the rescue. Interferometry is an idea whereby one can use two telescopes to increase their resolving power. When combining their signals, they have an effective diameter of their separation. Of course, they are not equivalent to a single dish telescope of that diameter, because they collect far less total starlight.

The technique was successfully utilized scientifically as early as the 1920s, to resolve the star Betelgeuse. But it was difficult to use this technique in it’s modern form, combining tens or hundreds of telescopes at once without the advent of computers which could handle the calculations in the 1960s and 1970s.

And now, the final piece. Interferometry grows much more difficult the smaller your wavelength. Interferometry is only possible if amplitude and phase of the incoming wave is recorded and recombined precisely. So an optical interferopeter needs: more precisely polished mirrors, more precisely measured separations, stable signal information transport from the dishes (fiberoptics), and more computing power to run more fourier transforms. While radio interferometry was performed in the 1920s, it took all the way until the 1990s to do optical interferometry, and it's still far from routine (see the Large Binocular Telescope).

So, to summarize:
1) To get a rotation curve of a galaxy, we need a molecular line that we can get a redshift or blueshift from. HI (optical) and CO (mm) are good choices.

2) The very centers of galaxies are obscured by dust, which destroys optical signals, but allows mm to pass, so CO is the molecule of choice for rotation curves of the nuclei of these galaxies, as H$\alpha$ signal is poor in the centers.

3) However, for a mm telescope to be able to spatially resolve small parts of the galaxies at the large distances of 10+ MPC in this wavelength, we need a telescope a few kilometers wide.

4) We can solve this with interferometry, but only recently has the technology to do interferometry at this wavelength been accomplished.

The first interferometer observing at this wavelength, the SubMillimeter Array (SMA; eight, 6-meter telescopes), opened to guest astronomers for the first time in 1990. The SMA was still being expanded at the time of the paper you linked.

These days, the Atacama Large Millimeter Array (ALMA) is just now nearing completion, and is focusing on the same wavelength range only with huge resolution and surface area (66 telescopes from 7-12 meters). They are collecting CO rotation curves to such precision that they claim to be able to estimate the central black hole masses of galaxies to 5%.

Stuff:

https://en.wikipedia.org/wiki/Astronomical_interferometer
https://en.wikipedia.org/wiki/Aperture_synthesis
https://ned.ipac.caltech.edu/level5/Fitzpatrick/Figures/figure1.gif
https://www.nrao.edu/whatisra/images/300ft-before.gif
https://www.nrao.edu/whatisra/images/300ft-after.gif
http://www.atnf.csiro.au/outreach//images/wallpapers/atcamorningsml.jpg
http://www.almaobservatory.org/


Is there any planet out there which is half gaseous and half rocky ? is this possible? I have heard many times that this planet is gaseous like jupiter this planet is like super earth like rocky planet etc. so is there any planet which is mix of gase and rock ?

Probably.   Relatively little is known about exoplanets because they're very hard to get a good look at, but there's no reason why a rocky world couldn't accumulate enough ices and/or gas to also resemble a gas giant.   Now half rocky half gas giant (hydrogen/helium) might be rare.   Half rocky half "ices" is very possible and those have likely already been observed.  It helps to understand planet formation and elemental abundance.

There's likely a size limit to rocky planets with thin atmospheres because mass tends to accumulate and retain atmosphere and with hydrogen and helium being the most abundant gases in the universe, once a planet gets massive enough, it should collect and retain hydrogen and helium and begin to resemble Jupiter or Saturn, even if it started out as a rocky world.

Rocky worlds, need elements like Magnesium, Silicon, and Iron, often bound with oxygen. close to 90% of the mass of rocky worlds in our solar system comes from those 4 elements and 90% isn't a bad estimate for rocky exoplanets.

Based on our solar-system:

Iron makes up about 0.117%, Silicon 0.065% and and Magnesium 0.051%, for about .234% of the mass of the solarsystem.  Oxygen bound to those elements, using Earth's composition as an estimate adds about another 50%, up to about 0.35%.   Hydrogen and Helium make up about 97.6%, which leaves about 2% of the mass of the solar system in the form of ices and heavier gases, not hydrogen and helium.  Those are primarily water, ammonia, CO2 and Methane, with smaller amounts of other gases/ices.

These numbers are rough but good enough for an estimate.   In our solarsystem there's about 6 times as much ice and heavier gases by mass than there is rocky material and there's close to 40 times as much hydrogen and helium as the other elements.

Planets need to be quite large to hold hydrogen and helium.   Jupiter and Saturn are the only planets in our solar-system that are hydrogen and helium abundant.    Even the smaller gas giants, Uranus and Neptune have comparatively little hydrogen and helium.

So there's probably no such thing as a small gas giant.   Gas giants need to be large or they don't exist at all.  (I should make a footnote that as a gas giant loses atmosphere, you can get a small gas giant, but that would be in a transition phase).

I don't like the term "Ice giants", though it's often used to define Neptune and Pluto because they are primarily ices (water, methane, CO2, NH3), not primarily hydrogen/helium.     I don't like calling them "ice giants" when they can also be hot, so I prefer the term Neptune like planets or Neptunes.

Uranus (a Neptune type planet) is mostly made up of ices and heavier gases with an estimate of just 3%-10% hydrogen and helium.  Estimates of about 70%-90% of ice/not hydrogen/helium gas and the remaining, about 7%-20% more dense material.   That would be some other elements, perhaps some solidified carbon (diamonds), higher sulfur and lower Iron by percentage than Earth as well as some Earth like silicates. magnesium-oxides, Iron and Iron-oxides, though the high internal temperature might not leave much in the way of chemical bonds towards the center of the planet where temperatures reach 9,000 degrees C.

Neptune is similar to Uranus but more massive and more dense.  It's thought to have more water and perhaps a larger internal mantle, though like Uranus, the majority of it's mass is ices and heavy gases (not hydrogen/helium).

Neptune and Uranus have hydrogen in the form of Methane, Ammonia and Water, but I'm counting that hydrogen as part of their "heavy gases/ices".   They both have a comparatively low percentage of free hydrogen (3%-10%) making a clear distinction between Neptune like planets and gas giants like Jupiter and Saturn with which the majority of their mass is hydrogen.

It helps to think of how planets form.   Smaller planets have too little gravity to retain hydrogen and helium gas and solar-systems are mostly too warm for hydrogen to freeze, so planets are made up of solid material for the most part, either rocky or ices that clump and stick together during formation.   When they have enough gravity, then they can begin to hold onto an atmosphere and further from the star makes this easier.

The Moon Titan, as an example, is small to have an atmosphere, but it's quite far from the sun and largely made up of ices on it's surface, so as the ice on it's surface melts, it basically replenishes it's atmosphere.  Titan is still out-gassing, and it losing atmosphere, but it loses atmosphere slowly enough that it retains an atmosphere, but in a sense, it's in a transition, where it's atmosphere is outgassed.  When it runs low on surface ice that readily thaws, it should begin to resemble the icy moons of Jupiter.

So, back to your original question.   Many combinations are possible, some are not.   Baby Jupiters (the mass of Neptune) are probably unlikely, though a gas giant close to it's star that's lost a lot of it's atmosphere could resemble a baby Jupiter, but like Titan, that would be transitional.

Rocky Jupiters are unlikely because hydrogen outnumbers rocky material by over so much, over 200 to 1 in most of the milky-way.   Now, some solarsystems are likely more "metalic" than others, so that ratio will have some variation, but it's once there's enough mass for a gas giant to form, hydrogen is likely to be the abundant element and Helium #2 and any rocky core would be dwarfed by the hydrogen and helium.

A rocky Neptune however, no reason why not.   A planet with 8 earth masses of Earth like material and 8 Earth masses of ices (and 5%-10% hydrogen) would basically be a rocky Neptune.   It would likely be a little smaller and certainly denser, but not all that different in outward appearance.

Water worlds is a common term that might qualify as a "rocky-neptune".   We don't know exactly what they are made of, but density estimates suggest a high percentage of water (and presumably CO2, CH4, NH3), which is the majority of Uranus and Neptune.

Here's a chart of exoplanets of less than 20 Earth masses.   I suspect there's a considerable margin of error in these estimates, but it more or less agrees with the rocky-neptune argument.   Planets with 2 or 3 grams per cc would be in that range.

See chart:  www.hpcf.upr.edu/~abel/phl/hec_plots/exoplanet_df.png

Source.

Baby Neptunes of just 1 or 2 Earth masses might be possible too, but they'd probably need to be quite cold and far from their star or they'd be in danger of losing their atmosphere.

Iceworlds like Pluto can get quite small, but Pluto has very little atmosphere.  Planets generally need to be fairly large to retain their atmosphere.   If they are too cold, that atmosphere freezes.   Titan, as I mentioned above, is in what could be called a slow transition where it's atmosphere comes from it's surface and it loses it slowly.   Ceres, based on it's density, is an icy moon like object too, though it's lost nearly all it's surface ice, but it probably has a lot of water/"ices" below it's surface.


How often is a blue moon? A blue moon is two full moons in the same month.  How many times will a blue moon occur in the average lifetime of say 70 years?

As months are not of fixed length it is only possible to calculate an average.

From Wikipedia:

The average frequency of a blue moon can be calculated as follows. It is the period of time it would take for an extra synodic orbit of the moon to occur in a year. Given that a year is approximately 365.2425 days and a synodic orbit is 29.5309 days, then there are about 12.368 synodic months in a year. For this to add up to another full month would take 1/0.368 years. Thus it would take about 2.716 years, or 2 years, 8 months and 18 days for another blue moon to occur.

So for a 70 year life span the average number of blue Moons would be 70/2.716 or 25.77.


Surface brightness of deep-sky objects and telescope specs I'm new to amateur astronomy, and I recently acquired my first small telescope. I've just learned a bit about surface brightness after an unsuccessful night of trying to locate a few nebulae that were beyond my scope's abilities, but I'm wondering if it's as cut-and-dry as I've understood so far.

The reason I'm asking is that when I look up the surface brightness of the Great Orion Nebula on Stellarium, it shows up at 13.07, a bit beyond the limiting stellar magnitude of my scope at 11.7. However, I am pretty confident that I was able to see the Orion Nebula with my scope on another occasion. It's a large enough object that I'm pretty sure I observed it, and the features looked correct (but faint and colorless) when compared to high-detail photos.

So to get to the point of my question, are there other things I should keep in mind when determining whether I'll be able to see an object with my scope? I would like to avoid more frustrating nights in the future by not picking impossible targets. I also recognize that as an amateur I might have seen something completely different than the Orion Nebula.

Although you have not given any information about your telescope, the Orion nebula is easy to see with almost any telescope, even under urban light polluted skies.

Surface brightness is almost always greater (ie fainter) than magnitude - so for example the total magnitude of the Orion nebula is about 4 (from memory) but you say surface brightness of 13.  The reason is that the area of the Orion nebula is quite large, at least when measured in arc seconds squared, and the eye/brain has some ability to integrate the light.

This works both ways: there are objects, eg M33 and M101 in my experience, which have a magnitude that would make you think they are easy targets for small telescopes, but because the brightness is spread over a large area, they are difficult to see from a typical urban or suburban location.  The important measure in these cases is to compare the surface brightness with that of the background sky.

It's quite a complex topic, and Roger Clark's book "Visual Astronomy of the Deep Sky" has some useful discussion of this, and much more.


What are the challenges in finding Earth-like planets via the radial velocity method? [I have re-asked this, because it was a good question by @Banyan, which was deleted whilst I was composing an answer.]
Most exoplanets that are found by the radial velocity ("Doppler wobble") technique are quite massive - Jupiter/Saturn-like objects.
What are the principle challenges in extending this technique to finding Earth-like exoplanets?

Assuming a circular orbit of a planet of mass $m_p$ around a star of mass $m_*$, with an orbital period of $P$ and we can assume that $m_* \gg m_p$ for the case of an Earth-like planet.
Newton's second law tells us that
$$ m_p a \frac{4\pi^2}{P^2} = G\frac{m_* m_p}{a^2}\ , \tag*{(1)}$$
where $a$ is the orbital radius.
The observed radial velocity amplitude of the star will be
$$ K \simeq \frac{2\pi a}{P} \left(\frac{m_p}{m_*}\right)\sin i\ , $$
where $i$ is the orbital inclination. Replacing $a$ using eqn. (1)
$$ K \simeq \left( \frac{2\pi G}{Pm_*^2}\right)^{1/3} m_p \sin i\ .$$
Putting in some typical numbers
$$ K = 0.09 \left(\frac{P}{1\ {\rm year}}\right)^{-1/3} \left( \frac{m_*}{M_\odot}\right)^{-2/3}\left(\frac{m_p}{M_{\rm Earth}}\right)\sin i\ {\rm m/s}\ . $$
This equation shows that to detect an Earth-like planet in an Earth-like orbit around a sun-like star we would need to detect a sinusoidal wiggle of amplitude 9 cm/s, with a period of 1 year. Things become more favourable if (i) $m_*$ is lower or (ii) $P$ is shorter.
Indeed Earth-sized (Earth-mass) planets have already been found around very low mass stars (e.g. a planet with $m_p \sin i = 1.35$ Earth masses in a 10-day orbit around the M-dwarf Ross 128; Bonfils et al. 2018).
The state of the art fibre-fed, temperature and pressure-controlled high resolution spectrograph at the moment is known as ESPRESSO at the VLT. It has a demonstrated precision capability of 25 cm/s and an instrument stability that is likely to be at the 10 cm/s level over several years (Pepe et al. 2020). ESPRESSO may be able to start picking up many Earth-mass planets, with orbital periods of up to 1 year around solar-type and less massive stars.
However, just achieving the necessary instrument stability is only part of the problem. The spectra of sun-like and lower-mass stars are not well behaved because of (i) convective motions in their photospheres and (ii) the presence of dark spots and bright plages that are inhomogeneously distributed over their surfaces. Both of these effects can caused velocity "jitter" that are an order of magnitude (or more) larger than 10 cm/s. The two approaches to deal with this are avoidance - where you try to choose targets that are old, quiet and magnetically inactive; or mitigation - where you try to work out what perturbations are being inflicted on your velocity measurements by looking for other signatures of the jitter like asymmetric spectral line profiles or by decorrelating with magnetic activity indicators (e.g. Wright 2017).


Generate East and North unit vectors tangent to a point on the celestial sphere, using vectors? I am stuck and a little embarrassed.

I'm trying to implement this answer. I have a unit vector $w$ pointing in the direction of a radio source on the celestial sphere.

$u$ and $v$ are the two other, mutually orthogonal unit vectors that point east and north from that point, tangent to the sphere.

Without back-converting to R.A. and Dec or using trigonometry, is there a simple way to generate $u$ and $v$ vectorially?

Slide 38 here may or may not be helpful, I think my description is sufficient by itself.

If $\mathbf{\hat{n}}$ points to the north celestial pole, then the eastward tangent vector is

$$\mathbf{\hat{u} = \frac{\hat{n} \times \hat{w}}{\|\hat{n} \times \hat{w}\|}}$$

and the northward tangent vector is

$$\mathbf{\hat{v} = \frac{\hat{w} \times \hat{u}}{\|\hat{w} \times \hat{u}\|}}$$

$\mathbf{\|\hat{w} \times \hat{u}\|}$ analytically should be 1 but can be slightly different numerically.