When does a solar eclipse become noticeable? For a human viewer located on Earth, is there an approximate apparent magnitude at which a solar eclipse becomes apparent in the form of a noticeable change in ambient brightness?
Assume normal atmospheric conditions, etc.
Note, I am not interested in the noticeability of the Moon covering the Sun, but whether someone outside would think, "Hey, it looks darker than usual today!"

It's hard to say because the situation is very similar to this question:
How much of the Sun's disk must be covered for a visible shadow to be cast?
The major problem lies in the sensitivity of the human eye, which is able to detect illumination changes sooner or later. Basically, there are people, who might not know about the ongoing solar eclipse until 98% obscuration! This is driven by the aforementioned human eye sensitivity and the surrounding scene, which keep seemingly the same level of illuminance for an observer.
I did a few observations both in solar and lunar eclipses and I have basic advice for you: try to remember the brightness of your scene at the very beginning and watch carefully how it changes against objects illuminated by other light than sunlight.
If, for instance, you are watching the eclipse outside, but close to the building, which rooms have a light turned on, you can try to detect this artificial light on the surface underneath this window (especially when it's placed in the shadow).
However, I assume, that for this purpose you will use a camera. In your device, you can set fixed image parameters and take the snaps at the same time intervals. Afterward, you can see how they vary from each other as the eclipse progresses.
See what I did for the lunar eclipse

The solar eclipse is analogous.
In the detection of the "visual start" of the solar eclipse, some amateur measurements can be helpful. You can install some light measurement applications on your phone. Some examples of use and pros & cons you can find here:
http://www.mkrgeo-blog.com/light-around-us-and-how-to-measure-it/
The smartphone headed towards any part of the sky should show how the level of light in your area changes as the eclipse progresses. With help of the images mentioned above, you will be able to detect the "beginning of visual eclipse visibility". Another example (also for lunar eclipse) you can find here:
http://www.mkrgeo-blog.com/remote-observation-of-the-total-lunar-eclipse/
and for a solar eclipse is here:
http://www.mkrgeo-blog.com/deep-partial-solar-eclipse-in-scotland/
http://www.mkrgeo-blog.com/light-level-measurements-during-total-solar-eclipse/
with visual examples I did myself

while also with artificial lights, which are highly recommended during observation such this. I used car dimmed headlights, but streetlamps are undeniably the best, because you can see also their light reflected from the ground, when gets darker.

http://www.mkrgeo-blog.com/light-level-measurements-during-total-solar-eclipse/
Bear also in mind the cloudiness. The observation above was carried out under the mostly overcast sky. In this case, clouds reflect the sunlight, hindering it from further way down to the ground. In turn, it's darker than it could be under clear skies. When the eclipse occurs along with some thunderstorms passing above your observation place, then it's extremely hard to define, when the eclipse-driven conditions started to be noticeable, as very thick clouds can reduce the sunlight up to 1000 times. Read more details here:
http://www.mkrgeo-blog.com/light-scattering-in-the-earths-atmosphere-part-3-clouds-haze-and-surface/


Why isn't there a large gap in the number of captured moons Jupiter and Saturn have? Given that Jupiter is roughly 3x as massive compared to Saturn and is located right next to the asteroid belt, it feels that Jupiter should easily have a larger amount of captured moons. However currently, due to a recent survey of Saturn discovering 20 new moons, Saturn has 57 observed captured moons compared to Jupiters 71.

Also summing the mass of the 6 largest captured moons for each planet gives:
- Jupiter: 7.844*10^18kg
- Saturn: 8.372*10^18kg

Why isn't there a larger gap between the number and mass of Jupiter and Saturns captured moons?

Note:
While Jupiter has a very large amount of Trojan asteroids, the main focus of the question is moons directly in orbit around the main planet.

Gravitational mass is not the be-all and end-all. Two other metrics are the Hill sphere and gravitational sphere of influence. These are widely used metrics that describe the volume over which a planetary body dominates gravitationally over than of the Sun. Saturn has a larger Hill sphere and a larger gravitational sphere of influence than does Jupiter.

This suggests that Saturn might have many more moons than does Jupiter. Saturn's much greater distance from the Sun compared to Jupiter's suggests that discovering a new Saturnian moon is over three times harder than is discovering a new comparably sized Jovian moon. This might be why the ratio is as close as it is.


Why do we not send diggers to Mars if we think there's underground water? A friend of mine asked me this question:

  If we have a guess that there may be underground water on Mars, why do
  we not send rovers with diggers?

I do have a few ideas, but I want alternative views by the community.

Sending anything to Mars is expensive. Sending a machine capable of drilling to the required depths to find subsurface water would be prohibitively expensive with current rocket payload limits.

There are also far more accessible sources of water on the surface - lakes of ice in craters, water rich minerals, and even an equatorial frozen sea .

There is also water at the surface of Mars on the poles, but the conditions here are much harsher than nearer the equator, and there is less sunlight, reducing the capability of any solar powered rover.

The subsurface water is not necessarily that far below the surface, but it is also not a torrent; more likely trapped as ice particles in between the rock or in secondary minerals.


What documented comet has/had the highest specific orbital energy? After reading this answer I wondered if there have been document comets with orbits on non-return, hyperbolic trajectories as they approached the Sun.

This means that they would not be gravitationally bound to the sun, so either they came from outside the solar system, or they managed to gain energy through a collision, or close pass with a large planet like Jupiter.

The quickest way to find the comet with the "highest" orbit is to ask for the one with the largest specific orbital energy, the total energy (kinetic + potential) divided by the mass, (Joules/kg). For a bound orbit the value would be negative, for an unbound, non-returning orbit it would be positive.

Question: What documented comet has/had the highest specific orbital energy? It should be an identified object with a known or measured orbit, not just a generic, nameless "edge-of-the-Oort Cloud placeholder."

That is a very good timing for this question.

There is a comet with the name C/2017 U1 (PANSTARRS), that was just discovered a few days ago (on 18th October) which has an orbital eccentricity of 1.18, which is by far highest eccentricity recorded so far.

The eccentricity describes how strongly "elliptical" an orbit is and refers to the variable e in the formula of Kepler's orbit. 1.18 makes the orbit a hyperbolic one, meaning it is not an actual orbit, but has higher velocity than the required escape velocity and is basically just passing by in our solar system. And it's the first recorded comet so far that has an hyperbolic orbit.

This comet is the first candidate for an interstellar comet ever in recorded history. It is already pretty certain that it originated from interstellar space, but not official yet.

Quote from minorplanetscenter.net:

  Further observations of this object are very much desired.  Unless there
  are serious problems with much of the astrometry listed below, strongly
  hyperbolic orbits are the only viable solutions.  Although it is probably
  not too sensible to compute meaningful original and future barycentric orbits,
  given the very short arc of observations, the orbit below has e ~ 1.2 for
  both values.  If further observations confirm the unusual nature of this
  orbit, this object may be the first clear case of an interstellar comet.

Sources:

 http://www.minorplanetcenter.net/mpec/K17/K17UI1.html
 https://en.wikipedia.org/wiki/C/2017_U1_(PANSTARRS)
 https://imgur.com/t/science_and_tech/vM4oh


How do we measure one complete orbit? This is being asked from a lay-person point of view. However, please do not hesitate to give more detailed answers that require prior knowledge - if these answer the question more accurately.

My confusion currently comes from the fact that everything in the universe is moving in some way or another - so any reference point we use would also be moving/not completely reliable.

As such, how can we accurately determine that we have completed an exact 360 degree rotation around the sun (and as such, measure exactly how long it took)?

To be clear, this is asking about how we can determine that we have completed a single 360 orbit of our sun, and not the timing aspect of how we know how long a second is. If this situation is different for measuring the orbits of other bodies around their star - it would be great to expand on those differences as well.

When one asks the question, "How is a year defined?", there are actually a few different answers one can provide. What you're implicitly asking, even if you don't know, is how one defines a Sidereal Year. In effect, the Sidereal Year is the time necessary to complete one, full, 360 degree revolution around the Sun. This is opposed to the Tropical Year (or else the Solar Year) which is the time necessary to travel just slightly less than a full 360 degrees (due to things like precession and the definition of the Tropical Year).

So your question boils down to, how is a Sidereal year determined? In practice, this is often done by choosing a specific starting point and noting where the Sun is with respect to the stars in the Sky. Then, it will be an entire sidereal year (and a 360 degree revolution) before the Sun returns to the same position as previously observed. Once you've observed the Sun in the exact same position you know a full revolution has occurred of the Earth around the Sun. It should be noted that the motion of the stars over a year is almost completely inconsequential. The stars are so immensely far away compared to their own motion, that their apparent motion (otherwise known as the proper motion) is basically nothing over a year. In this way, we consider the stars "fixed".


Do black holes produce thermal radiation? Do black holes produce thermal radiation, as expected on theoretical grounds? Does this radiation contain information about their inner structure, as suggested by gauge–gravity duality, or not, as implied by Hawking's original calculation? If not, and black holes can evaporate away, what happens to the information stored in them (since quantum mechanics does not provide for the destruction of information)? Or does the radiation stop at some point leaving black hole remnants? Is there another way to probe their internal structure somehow, if such a structure even exists?

Hawking radiation is the black body radiation that black holes emit. It is a well established theoretical result but it has never been observed.  For all currently known astrophysical black holes it will be totally insignificant and will never be observed.  The reason for this is that the equivalent temperature of the black body is inversely proportional to the mass of e black hole.  For all currently known black holes the temperature would be a tiny fraction of a degree above absolute zero.

Quantum mechanics is a probabilistic approach to physics, and deals with all kinds of probabilities and uncertainties. There is one such uncertainty of time and energy, given by (Δt) (ΔE) ≥ ℏ/2

One of its predictions, which has been experimentally confirmed, is that a particle may take an energy E from “nothing” and yet not violate the Law of Conservation of mass-energy if it “returns” the energy within a time given by (Δt)≥ ℏ/(2E).

One of its many implications is that random vacuum fluctuations constantly take place everywhere, i.e. even if a certain region of space has nothing or is purely vacuum, it can still constantly generate and annihilate particles due to the E-t uncertainty.

fast forwarding

When such a fluctuation causes a particle and its anti-particle to be generated on the edge of the event horizon of a black hole, one of the two enters the black hole and the other escapes. Now for returning the energy to the system, the particle relies on crashing into its anti-particle causing an “annihilation” and conversion of mass to energy. In the scenario I described, this can never take place since the pair of particles have been separated effectively forever. The particle that enters the event horizon has been trapped in an infinite potential gravitational well, so can be thought to have negative energy. So anti-intuitively (wait, what part of all this is even intuitive!), the particle left lurking outside the absorbs some energy from the black hole, (thus conserving mass-energy) gets converted into a “classical” photon and escapes the vicinity of the black hole.

Thus little by little, such events cause a black hole to effectively “evaporate” and diminish.

The apparant paradox about the fact that what happens to the information that goes into a black hole is a matter of debate. No one knows exactly how this Black Hole Information Paradox can be solved, however certain speculations are provided by Kip Thorne and Stephen Hawking, which include Firewall theory and Ads/CFT.


If an earthquake happens on Mars, is it still an earthquake? Or are seismic phenomenon named differently when they happen on other celestial bodies? If so, what are they called?

As pointed out in the comments, some do prefer to use different terms for quakes on other bodies, and you can see here that "marsquake" does have scholarly usage. That said, there is also scholarly usage for "earthquake" in this context. If you think about it, this makes sense given that one definition of "earth" is "the substance of the land surface", i.e. the ground. This definition of "earth" is agnostic to which terrestrial planet the land surface is on.


Do ASKAP and ALMA have "fast dump" interferometric modes? Can they see and perhaps report Fast Radio Bursts in real time? Chatterjee et al 2017 The direct localization of a fast radio burst and its host nicely shows a VLA detection of a repeating FRB in 5 millisecond steps nicely demonstrating the dispersion slope matching that of FRB 121102 and when de-dispersed providing a precise epoch for the particular event that could be potentially matched to other simultaneous observations while simultaneously providing the spatial resolution of an interferometric radio telescope array.

VLA fast-dump observations.
The VLA fast-sampled interferometric data were recorded with 5 ms integration time, 256 channels, and bandwidth 1024 MHz centred at 3 GHz.

Kumar et al. 2019 Faint Repetitions from a Bright Fast Radio Burst Source show ASKAP measurements of repeating FRB 171019 with 2.62 and 1.26 millisecond resolution. However the array was operated in fly's eye mode with each dish pointed in a different direction, so if I understand correctly, interferometric resolution from the array was not available.
I haven't found any reports of or references for ALMA FRB detections.
Questions:

Do ASKAP and/or ALMA have "fast dump" interferometric modes? Or do they have at least some way to extract dispersion of transient events at the millisecond level?
Can either see and perhaps report Fast Radio Bursts in real or near-real time?

Figure 1 VLA detection of FRB 121102. (a) A 5-ms dispersion-corrected dirty image shows a burst from FRB 121102 at MJD 57633.67986367 (2016 Sep 02). The approximate localization uncertainty from previous Arecibo detections (3' beam FWHM) is shown with overlapping circles. (b) A zoomed in portion of the above image, de-convolved and re-centered on the detection, showing the ~0.1" localization of the burst. (c) Time frequency data extracted from phased VLA visibilities at the burst location shows the $\nu^{-2}$ dispersive sweep of the burst. The solid black lines illustrate the expected sweep for
DM = 558 pc cm^−3. The de-dispersed lightcurve and spectra are projected to the upper
and right panels, respectively.

Figure 2. Dynamic spectra for both repeat bursts detected at GBT and ASKAP FRB 171019 dedispersed at their optimal DM. From the left: repeat burst 1 (resolution = 1.31 ms), repeat burst 2 (resolution = 2.62 ms), and ASKAP FRB 171019 (resolution = 1.26 ms). For each burst, the top panel shows the flux density averaged over frequency channels.

Short answer: No and yes.

Do ASKAP and/or ALMA have "fast dump" interferometric modes? Or do they have at least some way to extract dispersion of transient events at the millisecond level?

ASKAP has FREDDA which searches in total-intensity filterbanks (aka dynamic spectra) and triggers the voltages for imaging or localizing the FRB. general disclaimer that I am not 100% sure but do know about their searchpipeline
ALMA operates in a whole different frequency regime ($\sim 230$ GHz).
Dispersion delay scales $\sim \frac{1}{\nu^3}$ where $\nu$ is the observing frequency. So you'd need ridiculously high DMs to produce a sweep there. You could say there is almost no dispersion at these frequencies. There is also the question if FRBs are produced in that regime. The highest frequency any FRB (repeating or one-offs) has been detected is around $8$ GHz (FRB121102, Gajjar+2018 etal).
Having said all of that, the essence of what you are suggesting is captured in realfast which runs commensally on the VLA and does fast de-dispersed imaging at around $1-4$ GHz. Meaning, it produces an image on the milli-second timescale for many dispersion sweeps and then searches for bursts. There is also an AI running which vets many of the candidates it gets to test. All of this is done in realtime.
Check out this realfast detection of a periodic FRB detected last April.

Can either see and perhaps report Fast Radio Bursts in real or near-real-time?

These pipelines have the motivation to save the raw data and so employ realtime searches which issue triggers and then slices of the raw data are saved. There is AI running which decides if the trigger is to be issued or not. And even with it, there's a lot of False positives.
Reporting FRBs is a different story. Usually, confirmed FRBs are reported in Astronomer's Telegram (ATel), or Transient Naming Server (TNS). There is no one unified network (like LIGOs-VIRGO-FERMI) which shares triggers because of various reasons.
Or, you could choose to do follow up observations and then report.


What are the significant assumption of the Inflation Theory? Can it still be considered theory ?, because I can't find it's major or significant assumptions.

The most significant assumption is that very shortly after the Big Bang, the universe expanded very rapidly for a short period of time. Much faster than normal expansion due to Big Bang. It was an exponential expansion and occurred around ${10}^{-32}$ seconds after Big Bang and lasted for tiny fraction of second. The universe continues to expand but not at that exponential rate. Yes it is still a theory and is accepted by many scientists.


What does a black hole elements consists of? What are the elements of a black hole it just says what a black hole is? There must be elements that the sun consists of, right?

If your question is "What elements are in a black hole?"
The answer is: The gravity is too great for there to be "elements" (atoms or molecules) inside or near the surface of a black hole. The extreme gravity tears apart everything and compresses it.
Some distance away (especially far away, here on Earth) the black hole doesn't alter matter, but as things get closer they fall at increasing speeds to the surface, becoming a part of the black hole. Everything becomes crushed, becoming elementary particles; what atoms are made of.
NASA's webpage: "NuSTAR Probes Black Hole Jet Mystery" explains that plasma, x-rays, and eventually light can leave a black hole at the poles. See also: "An elevation of 0.1 light-seconds for the optical jet base in an accreting Galactic black hole system" (26 Oct 2017) which says:

"Relativistic plasma jets are observed in many accreting black holes. According to theory, coiled magnetic fields close to the black hole accelerate and collimate the plasma, leading to a jet being launched. Isolating emission from this acceleration and collimation zone is key to measuring its size and understanding jet formation physics. But this is challenging because emission from the jet base cannot be easily disentangled from other accreting components.
Here, we show that rapid optical flux variations from a Galactic black-hole binary are delayed with respect to X-rays radiated from close to the black hole by ~0.1 seconds, and that this delayed signal appears together with a brightening radio jet. The origin of these sub-second optical variations has hitherto been controversial.
Not only does our work strongly support a jet origin for the optical variations, it also sets a characteristic elevation of <~$10^3$ Schwarzschild radii for the main inner optical emission zone above the black hole, constraining both internal shock and magnetohydrodynamic models.
Similarities with blazars suggest that jet structure and launching physics could potentially be unified under mass-invariant models. Two of the best-studied jetted black hole binaries show very similar optical lags, so this size scale may be a defining feature of such systems.".

So, what goes in is not what comes out.
See also this article about AGN: "The largely unconstrained multiphase nature of outflows in AGN host galaxies" (28 Feb 2018).
 Black Hole-Powered Jet of Electrons and Sub-Atomic Particles Streams From Center of Galaxy M87
Astronomical evidence

Relativistic jets may provide evidence for the reality of frame-dragging. Gravitomagnetic forces produced by the Lense–Thirring effect (frame dragging) within the ergosphere of rotating black holes combined with the energy extraction mechanism by Penrose have been used to explain the observed properties of relativistic jets. The gravitomagnetic model developed by Reva Kay Williams predicts the observed high energy particles (~GeV) emitted by quasars and active galactic nuclei; the extraction of X-rays, γ-rays, and relativistic e$^-$—$\,e^+$ pairs; the collimated jets about the polar axis; and the asymmetrical formation of jets (relative to the orbital plane).

See also:

The Penrose process (also called Penrose mechanism) is a process theorised by Roger Penrose wherein energy can be extracted from a rotating black hole. That extraction is made possible because [some of] the rotational energy of the black hole is located not inside the event horizon of the black hole, but on the outside of it in a region of the Kerr metric spacetime called the ergosphere, a region in which a particle is necessarily propelled in locomotive concurrence with the rotating spacetime. In the process, a lump of matter enters into the ergosphere of the black hole, and once it enters the ergosphere, it is forcibly split into two parts.
...
Although momentum is conserved the effect is that more energy can be extracted than was originally provided, the difference being provided by the black hole itself. In summary, the process results in a slight decrease in the angular momentum of the black hole, which corresponds to a transference of energy to the matter. The momentum lost is converted to energy extracted.
The maximum amount of energy gain possible for a single particle via this process is 20.7%. The process obeys the laws of black hole mechanics. A consequence of these laws is that if the process is performed repeatedly, the black hole can eventually lose all of its angular momentum, becoming non-rotating, i.e. a Schwarzschild black hole. In this case the theoretical maximum energy that can be extracted from a black hole is 29% its original mass. Larger efficiencies are possible for charged rotating black holes.

Click to animate.
An unrelated effect is described as Hawking radiation:

Physical insight into the process may be gained by imagining that particle–antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles. As the particle–antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles lowers the mass of the black hole.

That shouldn't be confused with the Hawking-Penrose Singularity Theorem either.


What is the temperature of the solar atmosphere (the corona) and how is it measured? The temperature of the Sun's atmosphere, also referred to as the solar corona, is known to be hot even hotter than the Sun's surface. What is its temperature and how is it measured? How does it compare to other stars and how does it vary across the Sun?

This is a rather broad question and this will not be a fully comprehensive answer.

There is no single temperature to the solar corona. The coronal temperature varies by an order of magnitude from place-to-place. It is hottest ($\sim 10^{7}$ K) in magneticloops undergoing flares, which tend to be anchored in low latitude regions. It is coolest (a bit less than $10^{6}$ k in coronal holes which are usually at high latitudes. It is hard even to come up with some average number, because it is not altogether clear how the various pieces should be weighted in any average. It is also cyclic - in that magnetic activity, and average coronal temperatures vary with all the other magnetic phenomena on the Sun with a roughly 11-year period. If you weight it by emission measure, which is density squared times volume, then the solar corona is between $10^{6}$K at solar minimum to about $3\times 10^{6}$K at solar maximum.

The temperature of coronal plasma can be measured using X-ray spectroscopy. If you assume that the plasma can be represented by some equilibrium temperature, then you can make a prediction of what the X-ray/EUV spectrum will look like. To some extent the spectrum will also depend on the composition of the plasma and also on its density, but these produce comparatively minor differences compared to changes in temperature and can be simultaneously estimated in most cases.

For stars you essentially do the same thing but with worse data. Here you are getting some sort of average temperature for the whole corona - there is no spatial resolution. X-ray satellite observatories like Chandra and XMM-Newton have grating spectrographs and detectors with energy sensitivity that can record X-ray spectra from stars. You then use software like XSPEC along with an optically thin plasma model, such as that of Mewe-Kaastra-Liedahl, (and not forgetting to include absorption by the interstellar medium) to get the best fitting temperature for the measured spectrum.

As far as how the Sun compares with other stars, an excellent starting point is the review by Guedel (2007). A short summary would be that the temperature of a stellar corona depends on it's rotation rate - the faster it rotates, the hotter it is. Because (single) stars lose angular momentum and hence rotate more slowly as they age, then both X-ray luminosity and (average) coronal temperature decrease with age.

Telleschi et al. (2005) studied solar analogues at a variety of ages. They found a relationship that the average temperature
$$\langle T \rangle = 1.16\times10^{7} P_{\rm rot}^{-0.48}\ {\mathrm K},$$
where the rotation period $P_{\mathrm rot}$ is in days.

Given that rotation period and age are linked by a "gyrochronology" relation roughly of the form $P_{\rm rot} \simeq 27 (t/4.5\ {\rm Gyr})^{1/2}$, for $t \geq 100$ Myr, this means that
$$\langle T \rangle = 2.4\times10^{6} (t/4.5\ {\rm Gyr})^{-0.24}\ {\mathrm K}.$$
At ages below 100 Myr, the coronae "saturate" at the luminosity and temperature of a 100 Myr star.

In general, stars with photospheric temperatures cooler than the Sun tend to have hotter coronae, whilst those with warmer photospheres have cooler coronae. Stars with spectral types A basically don't have coronae, and hotter O and B stars have coronal winds that are heated by completely different mechanisms to the solar corona.


Are the radial spokes in Saturn's rings reliably visible via ground-based telescopes In the mid 1970's, Franklin and O'Meara saw persistent "radial spoke-like features" in the rings of Saturn, that should not have existed due to the differential rotation of the rings.  A publication on this observation was rejected by a journal apparently on the grounds that the phenomenon was considered to be illusory (c.f Sciparelli/Lowell's Martian canals?).

  From "Seeing in the Dark: How
  Amateur Astronomers Are Discovering the Wonder" By Timothy Ferris.

My question is, how reliably were these features visible from the ground-based telescopes of the day?  Could sufficiently strong evidence for the existence been obtained before images from space probes?  Was the rejection of the paper reasonable in the historical context in which the paper was submitted?

Bryan (2007) gives a number of reasons why O'Meara's discovery was largely discounted:

As you stated, the behavior was entirely inconsistent with Keplerian predictions of motion.
While O'Meara was able to reproduce his findings, no other independent observers could.
The detections were done entirely visually, rather than with numerical measurements.
There had been previous observations (i.e. in the 19th century), but in different places in the rings (the A and C rings, not the B ring). These have never been confirmed, and it is still believed that these were actually illusions.

Reliable observations from ground-based telescopes, especially those of amateurs, have arisen within the past decade or so. Since 2007, 67 "candidate" observations have been made of the spokes, many during the 2009 Saturnian equinox (keep in mind that the spokes may be a seasonal phenomenon). One additional factor that made observations difficult was that the spokes seemed to have vanished, even from space probes, in the years after the Voyager observations, making it impossible for ground-based telescopes to see them.


Is it possible to create a traditional clock solar system? Is it possible to create the following system of satellites?:

Place a satellite that orbits the some object X, once per day. Call it D.
Place a satellite that orbits D, once per hour. Call it H
Place a satellite that orbits H, once per minute. Call it M.
Place a satellite that orbits M, once per second. Call it S.

If possible, what are some values for the weights, rotational speed, and radii that would work?

Also, does the universe have a limit on how many layers of satellite systems it permits?

Is it possible to create the proposed system of satellites?

Yes (in theorey). Hierarchical multiple systems (like the one proposed in the OP) tend to be stable if the periods differ by $\sim 5$ or more and the orbits are near-circular. So, such a system could be stable. A slight problem may be the commensurability of the periods (the fact that their ratios are rational numbers), which implies orbital resonances. However, these resonances are rather weak (the lowest ratio is 1:24), but they may mess up the system in the very long term (a detailed analysis would be required to find out). In practice, it will be extremely hard to set-up such a system, even with advanced space engineering.

What are some values for the weights, rotational speed, and radii that would work?

Since this is a strongly hierarchical system, the masses $m$ and semi-major axes $r$ (=radius if the orbit is circular) can be worked out from Kepler's third law
$$
    \frac{2\pi}{T} = \Omega = \sqrt{\frac{G(m_1+m_2)}{r^3_{1-2}}}
$$
for each sub-system. Giving
\begin{align}
   \frac{4\pi^2}{1\mathrm{sec}^2} &= \frac{G(m_S+m_M)}{r^3_{S-M}} \\
   \frac{4\pi^2}{1\mathrm{min}^2} &= \frac{G(m_S+m_M+m_H)}{r^3_{SM-H}} \\
   \frac{4\pi^2}{1\mathrm{hour}^2} &= \frac{G(m_S+m_M+m_H+m_D)}{r^3_{SMH-D}} \\
   \frac{4\pi^2}{1\mathrm{day}^2} &= \frac{G(m_S+m_M+m_H+m_D+m_X)}{r^3_{SMHD-X}},
\end{align}
where $r_{ABC-D}$ denotes the distance (more precisely: the orbital semi-major axis) between object $D$ and the centre of mass of objects $A$, $B$, and $C$. Since these are 4 equations for 8 unknowns (4 masses and 4 distances), there is some freedom in the design of your 'clock'.

Does the universe have a limit on how many layers of satellite systems it permits?

So far, our formula only considered gravitational forces and neglected all else. Moreover, post-Newtonian effects (general relativity) have been ignored. As long as these assumptions remain valid (orbital speeds $\ll c$, orbital distances $\gtrsim\,$cm, solid objects with sizes $\ll$ distances so that tides are neglible, no electric or magnetic fields, vacuum), there is no limit. However, in practice, there is always some elctro-magnetic field and never perfect vacuum...


Why does the neutral hydrogen velocity have this characteristic behavior in the galactic plane? The recently released map of galactic neutral hydrogen density and velocity is really beautiful. The work has been relased by the HI4PI collaboration (HI = neutral hydrogen, $4\pi$ = complete spherical coverage).

My question is fairly simple. Looking at the image near the galactic plane, why do I see near the center a green bump on the left and a blue bump on the right, then near the edges of the image a much stronger blue bump on the left and a green bump on the right. In other words, there is a major and a minor peak in the positive radial velocity gas, and a major and a minor peak in the negative radial velocity gas.

Does this have some simple geometrical explanation?

above x2: neutral hydrogen density/velocity map from here with inset indicating color scale expanded.

above: for those who can not distinguish the green and blue in the image, this plots the average green (solid) and blue (dashed) intensity within the central equatorial band.

There is a nice article at phys.org with a video, which can also be seen in YouTube. It shows the intensity of HI as a cut in radial velocity is slowly scanned through the data set.

https://www.youtube.com/watch?v=Q2mgpsTFuV8

Since the stackexchange YouTube option is not turned on in astronomy stackexchange, here is a GIF of screenshots at -30, -20, -10, 0, 10, 20, and 30 km/sec:

The reason is detailed in depth in this pdf, which contains the following diagram:

Some key quantities:

$R_0$: Distance from the observer to the center of the Milky Way
$R$: Distance from target gas to the center of the Milky Way
$V_0$: Velocity of the observer with respect to a certain reference frame
$V$: Velocity of target gas with respect to the same reference frame
$l$: Angle between $R_0$ and $\mathbf{SM}$, the distance to the gas

When measuring the radial velocity, we look for the projection of $V$ onto our line of sight. Some geometry results in the relation
$$V_r=V_0\left(\frac{R_0}{R}-1\right)\sin l$$
for the radial velocity $V_r$, assuming $V\approx V_0$ (which is not always true!). The $\sin l$ term, in addition to the asymmetrical rotation of the Milky Way, leads to the peculiar data you asked about.


Is Jupiter as opaque as it looks? Although it is impossible to tell what is and isn't false color (aside from the adage that it probably is false).  It's inarguable that no picture of a gas giant shows any depth or significant transparency.  Of course this could just be because a single camera lacks depth perception, and since the planets are gas almost completely gas, there are no underlying boundaries to detect any transparency.

In particular what is the visibility inside the various conditions on Jupiter?  Could you see 1,000 km in every direction, or is it more like a constant fog or dust storm (with frozen ammonia instead of dust)?  What if you brought your own light source?

What you see are clouds, big clouds, and these are as opaque as the cumulus clouds on Earth. The pale bands are ammonia clouds and the darker bands are ammonium hydrosulphide clouds. The ammonia clouds form above the hydrosulphide clouds In the polar regions you can see deeper, down to the water clouds that lie beneath the ammonia clouds.
There is a layer of atmospheric haze that is above the cloud layers, it is easiest to see in the polar regions.
The difference between the layers is on the order of 50-100km.  If you were in a cloud, your visibility would be pretty limited and as you descended through 50km of cloud, it would get pretty dark. But like on Earth, there would be gaps and clear bands, between the layers of cloud.


Jupiter, Saturn, Uranus, Neptune system GM I'm modelling the solar system. I'm using data from Nasa.
To do this I'm using the barycenter of the planets Jupiter, Saturn, Uranus, Neptune. However the website fails to give the Gravitational Parameter $\mu$ for these barycenters, even though it does do so for Pluto's barycenter.
To make matters more confusing, the $\mu$'s given at Wikipedia are slightly different, though I assume these are just out of date.

So I'd like to know if there's an as up to date resource as possible available for these parameters. Paywalls should be OK as long as they are of major publications.
$$\frac{\mathrm{AU}^3}{\mathrm{M}_\odot\cdot\mathrm{day}^2}$$
-form also good.

In the JPL report on ephemerides DE430 and DE431,
on page 49, table 8 gives the mass parameters they used for the major planets.
They cite the
IAU 2009 system of astronomical constants.
There may be newer estimates, but to verify your own simulation against HORIZONS, you might as well use the same constants they did.


How to calculate the time and distance of closest approach of a star? How should I approach this problem?

A star has a motion on the sky of 0.78 arcsec/yr in declination and 1.33 arcsec/yr in right ascension.
Calculate the velocity of the star in the plane of the sky in km/sec.
The star also has a radial component of its velocity, coming towards earth
at 4.3 km/s. What is the time and distance of closest approach?

I know how to do 1. but I am not sure how to approach 2.

This is not really an astronomy so much a basic Maths problem.

Put the Earth at the origin and the star at $\vec{r}(t)$ with a position $\vec{r_0}$ at $t=0$. Define your axes such that the star is in the z direction and its tangential motion components are along the x and y axes.

$$\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k},$$
where these components are the two proper motion velocities and the radial
 velocity.

Assume no acceleration, so that
$$\vec{r} = \vec{r_0} + \vec{v} t $$

Use Pythagoras to get the distance to the Earth using the components of $\vec{r}$ at any time $t$ and then find the time at which this is a minimum (and if you are unsure how to do that $\rightarrow$ Maths SE).


Does Hawking radiation in fact bring mass into the universe? As I understand virtual particles pairs, e.g. electron and positron spontaneously pop into existence for tiny period of time, then they annihilate.

When virtual electron and virtual positron annihilate we should see two photons emitted to opposite directions, right? If so, shouldn't we detect photons appearing from vacuum for every such pair?

Also, when a pair of virtual particles pops near event horizon of a black hole one particle may be captured by the black hole, while the other escapes it. Both particles have mass and energy. Thus caught particle brings the black hole its mass. Then how the black hole loses mass through Hawking radiation? Shouldn't it gain mass instead?

The black hole loses mass in order to generate the Hawking radiation. The energy carried by the radiation is exactly proportional to the mass lost by the black hole according to Einstein's formula E = mc^2

No new mass is brought into the Universe this way.

EDIT:

(Keep in mind the explanation below is not rigorous. As Stephen Hawking himself has said, this is more like a mental model, good enough to visualize things.)

Virtual particles are called virtual because they don't really "exist" in the common sense. They have mass and they have energy. Due to Heisenberg's uncertainty principle, it's okay if particles with a total energy E almost "exist" for a very short while, as long as they return to nothingness afterwards.

The production / destruction of virtual particles could happen all the time, anywhere, not just at the edge of a black hole. This is fine, as long as each particle only seems to "exist" a short time. Conservation of energy is not violated.

However, at the edge of a black hole, something different happens. You have a pair of virtual particles popping up into existence. Soon after, one of the particles is captured by the black hole, while the other particle moves away from it. Since the captured particle is virtual, it doesn't really have mass and/or energy. No new mass is added to the black hole.

Now the two particles cannot re-annihilate each other. So you have one lone particle moving away from the black hole. It's not a virtual particle anymore, it's a real particle. But it has energy and perhaps it has mass also. Where does the energy come from?

The energy required to make the new particle real comes from the gravitational field of the black hole. The black hole is losing mass in order to power the creation of real particles moving away from it.

Again, don't get stuck in details. This is not what really happens. This is more like a metaphor.


Does a flat universe mean not finite and circular? I am trying to wrap my head around what a flat universe means; Trying to imagine why angles of triangles in spherical volumes are different.

Because my brain can't do that well and I became a bit annoyed by it, I started to think about a hypothetical circular one dimensional universe (a circle) - placing myself on the circle.

Without access to a "detached" view from the second dimension, just being able to look back and forward, I don't see any possible way to determine whether my universe is circular or not - apart of seeing my own back - if light already traveled around the whole circle.

So, thinking about two dimensions (a sphere, I guess), I still can't imagine to see any kind of circle, curvature or special angle, because I imagine the curves would lift into a "detached" third dimension which is not there from my point of view?

I don't understand how one would see curvature without access to a detached view of the higher dimension, could someone relieve me from this headache? - Does a flat universe actually mean that it is non-circular?

Yes, a "flat" universe is infinitely large, and hence "non-circular". I think perhaps you are confusing a flat and "closed" geometry.

Flat universes

A flat universe is a universe where your good old high-school math works as expected: the angles of a triangle sum up to $\pi$ radians (180 degrees), the circumference of a circle is $\pi$ times its diameter, parallel lines never meet, etc. In a 2D analogy, this corresponds to the geometry of a table; hence the name "flat".

Closed and open universes

But "flat" isn't the only possible geometry of space. You may, in principle, imagine any crazy geometry you like. However, we like to believe that the Universe is homogeneous and isotropic, i.e. that it looks the same everywhere and in all directions. Indeed, it seems to be on sufficiently large scales. If this is true, then there are only two other possibilities, namely "closed" (or "spherical") and "open" (or "hyperbolic").

Non-flat geometries are difficult to visualize in 3D, but the 2D analogy of a closed universe is that of a sphere, like the surface of a ball. On the 2D spherical surface, the angles of a triangle sum up to more than 180º, the circumference of a circle is less than $\pi$ times its diameter, and lines that start out parallel eventually meet. Moreover, if you travel far enough, you will get back to where you started.

The same is true in 3D spherical space. You may even, in principle, see your own back, although that would require either 1) that you start looking exactly at the Big Bang, in which case the light from your back will reach exactly when the universe collapses in a Big Crunch, or 2) that you have a somewhat fine-tuned amount of dark energy to slow down the collapse, without making it expand too fast so the photons never reach you.

Measuring the curvature in theory

Visualizing the 2D sphere is easy when it's embedded in a 3D space. But to describe how large its curvature is, you actually don't need the embedding 3D. You can just do a geometric measurement and compare your result to what you'd expect in flat space.

The quantity usually used to describe the magnitude of the curvature is $K$, which for a 2D sphere equals $1/R^2$ where $R$ is the radius of the sphere. If you draw a triangle of area $A$, and measure its three angles $\theta_{1,2,3}$, then the curvature is given by
$$
K = \frac{\sum_i \! \theta_i - \pi}{A}.
$$
This formula is valid for spaces of any numbers of dimensions (higher than 1D).

For instance, consider a triangle going from Ecuador, to Gabon, to the North Pole, and back to Ecuador. In this case, the sum of the angles is 270º, or $3\pi/2$, and the area of the triangle is $1/8$ of Earth's surface area, i.e. $4\pi R_\mathrm{Earth}^2/8$. Hence, the curvature is
$$
K = \frac{3\pi/2 - \pi}{4\pi R_\mathrm{Earth}^2 / 8} = \frac{1}{R_\mathrm{Earth}^2}.
$$

There are several other mathematical ways to measure the curvature. For instance you can parallel transport a vector along two difference routes; only in a flat space will they end up pointing in the same direction. In the example above, transporting a vector pointing from Ecuador to Gabon along the shortest path (red path in the figure below), and transporting it via the North Pole (green path), gives a difference of 90º.

So, to see the curvature of a 2D space in a higher-dimensional, flat space requires that theis space be 3D. But as described above, seeing the curvature is not required in order to measure it.

Furthermore, to see the curvature of a 3D space in a higher-dimensional, flat space, actually 4D wouldn't be enough — you would need 6 dimensions. And to see a 4D space (e.g. spacetime), you'd need 10 dimensions. But just as in the 2D case, there's no need for it, and to my knowledge (but please correct me if I'm wrong) no-one claims the existence of such higher dimensions.

Measuring the curvature in practice

In practice, this it not how we measure the curvature of our Universe. Rather, we use the Friedmann equation which gives the relation between the expansion rate $H$ of the Universe, the density parameters $\Omega_i$ of its various components (dark matter, normal matter, dark energy, and radiation), and its curvature.

The density parameters are defined as the densities $\rho_i$, divided by the critical density $\rho_c$, which is the density needed for a flat universe. If the total density $\rho_\mathrm{tot}$ is greater than $\rho_c$ — i.e. if $\Omega_\mathrm{tot} \equiv \sum_i\Omega_i > 1$ — the universe is closed.
Although "curvature" is not really a component of the Universe (such as e.g. matter), it may be parameterized as a corresponding density, and thus, in the Friedmann equation, appear as a term $\Omega_K \equiv 1 - \Omega_\mathrm{tot}$:
$$
\frac{H^2(a)}{H_0^2} =
\frac{\Omega_\mathrm{r}}{a^4} +
\frac{\Omega_\mathrm{M}}{a^3} +
\frac{\Omega_K}{a^2} +
      \Omega_\Lambda.
$$
Here, subscripts $0$ refer to the values measured today, whereas in general the values are measured at a time when the scale factor — the "size" of the Universe — is $a$. Subscripts $r$, $M$, and $\Lambda$ denote radiation, matter (dark+normal), and dark energy.

Hence, measuring the densities and expansion rates (preferably at various epochs given by $a$) will tell you $K$. This can be done in several different ways, e.g. by observing supernova brightnesses, the cosmic microwave background, the large-scale structure of the Universe, gravitationally lensed galaxy clusters, etc.

Luckily, the various more or less independent methods return roughy the same answer. And this answer seems to imply that our Universe is very close to flat; within the margins of the errors, it is consistent with being exactly flat (Planck Collaboration et al. 2018). Hence, for most practical purposes we may regard the Universe as flat. There was, however, recently a paper by di Valentino et al. (2019) claiming evidence for a positively curved (i.e. closed) Universe. But they considered only one out of Planck's many data sets, so it seems to me a bit like cherry-picking.


Longitude of the Ecliptic I originally went to math.stackexchange.com and they suggested I try here as well.

I'm trying to figure out how to calculate times of moonrise and moonset for a given position on the earth on a particular date. I'm using Vallado's Fundamentals of Astrodynamics and Applications because there is an example in the book that lays out a simplified process. I'm stuck at the point where I'm trying to calculate Longitude of the ecliptic for the Moon. The formula is below where T = -0.013634497.

  λEcliptic = 218.32° + 481,267.8813T + 6.29Sin(134.9 + 477,198.85T) - 1.27Sin(259.2 - 413,335.38T) + 0.66Sin(235.7 + 890,534.23T) + 0.21Sin(269.9 + 954,397.70T) - 0.19Sin(357.5 + 35,999.05T) - 0.11Sin(186.6 + 966,404.05T)

The expected answer is -0.8412457°. However, I'm unable to figure out how to get this answer. My calculations are below:

  1) 218.32° + 481,267.8813T = -6343.525

  2) 6.29Sin(134.9 + 477,198.85T) = 5.963779

  3) -1.27Sin(259.2 - 413,335.38T) = -0.900843

  4) 0.66Sin(235.7 + 890,534.23T) = -0.292285

  5) 0.21Sin(269.9 + 954,397.70T) = -0.126871

  6) -0.19Sin(357.5 + 35,999.05T) = 0.138211

  7) -0.11Sin(186.6 + 966,404.05T) = 0.054722

Simply adding or subtracting gets me -6338.688731. How can I get the value -0.8412457° from this. My trigonometry is rusty and I'm not sure how to get the right answer. Any help would be greatly appreciated. Thanks.

There's a mistake in the book (see page 280 on google books): the values for $\lambda_{ecl}$ and $\phi_{ecl}$ should be swapped. Your result is correct:
$$
-6338.688^\circ\text{ mod } (360^\circ) = -218.688^\circ,
$$
which is the value that is erroneously listed under $\phi_{ecl}$.


Theoretically, what is the biggest optical telescope that may exist? Recently, I read yet another news about E-ELT. It will have 39.3-metre-diameter segmented primary mirror. And I was interested in the next question:
Theoretically, what size of the primary mirror (single/multiple/segmented) can have a telescope on Earth for observing at optical wavelengths? And why? I mean what are physical limitations exists?

And the same question about space (not on Earth)?

Update:

On the advice of @TildalWave, to make this question answerable let's make a few adjustments:

Primary mirror should be segmented (or it variations) like on E-ELT.
Suppose we have a large (several square kilometers), flat surface high above sea level.
We have to build telescope for observing at optical wavelengths.

I know that, there is concept of OWL with 100 metre-diameter segmented primary mirror.

But what about 500 metre-diameter or 1000? Is it possible in theory?

It's complicated.

Until late-20th century, we've tried to make bigger and bigger monolithic telescopes. That worked pretty well up to the 5 meter parabolic mirror on Mount Palomar in California in the 1940s. It kind of worked, but just barely, for the 6 meter mirror on Caucasus in Russia in the 1970s. It did work, but that was a major achievement, for the twin 8.4 meter mirrors for the LBT in Arizona in the 2000s.

We've learned eventually that the way to go is not by pouring larger and larger slabs of low-expansion glass. It is generally accepted that somewhere just below 10 meters diameter is about as large as possible for monolithic mirrors.

The way to go is by choosing to make smaller mirror segments (1 meter to a few meters in diameter each) and combining those into a tiled mirror. It's somewhat harder to carve the asymmetrical parabolic (or hyperbolic, or elliptic, or spherical) reflecting curved surface in a segment like that, but it's far easier to manage thermal and cooling issues when you have to deal with smaller solid objects.

Each segment is mounted in an active mirror cell, with piezo actuators that very precisely control its position. All segments must combine into a single smooth surface with a precision better than 100 microns (much better than that in reality). So now you have a large array of massive objects, dynamically controlled via computer, each with its own vibration modes, each with its own source of mechanical noise, each with its own thermal expansion motions, all of them "dancing" up and down a few microns on piezo elements.

Is it possible to orchestrate a very large system like that? Yes. The 100 meter OWL was considered feasible technically. From the perspective of keeping the mirrors aligned, an even larger structure should be doable; the computer-controlled actuators should overcome most vibrations and shifts up to quite large distances.

Like you said, the real limits are financial. The complexity of such a system increases with the square of the diameter, and with complexity comes cost.

The entire discussion above was about "filled aperture" telescopes: given a round shape of a certain diameter, it is filled with mirror segments. For a given aperture, this design captures the largest amount of light.

But the aperture does not have to be filled. It can be mostly empty. You could have a few reflecting segments on the periphery, and the center would be mostly void. You'd have the same resolving power (you would see the same small details), it's just that the brightness of the image would decrease, because you're capturing less light total.

This is the principle of the interferometer. The twin 10 meter segmented Keck mirrors in Hawaii can work as an interferometer with a baseline of 85 meters. This is effectively equivalent to a single 85 meter aperture in terms of resolving power, but obviously not in terms of image brightness (amount of light captured).

The US Navy has an interferometer in Arizona with mirrors placed on 3 arms in a Y shape, each arm 250 meters long. That gives the instrument a baseline (equivalent aperture) of several hundreds of meters.

U of Sydney has a 640 meter baseline interferometer in the Australian desert.

Interferometers cannot be used to study very faint objects, because they can't capture enough light. But they can produce very high resolution data from bright objects - e.g. they are used to measure the diameter of stars, such as Betelgeuse.

The baseline of an interferometer can be made extremely large. For terrestrial instruments, a kilometers-wide baseline is very doable now. Larger will be doable in the future.

There are talks about building interferometers in outer space, in orbit around Earth or even bigger. That would provide a baseline at least in the thousands of kilometers. That's not doable now, but seems feasible in the future.


Does the critical density of the Universe also include Dark Energy? It is often said that it seems that the Universe is quite close to critical density (for example https://astronomy.swin.edu.au/cosmos/C/Critical+Density).
Does this include Dark Energy as well (apart from Dark matter and Baryonic matter)?

Yes. Quoting values from wikipedia (which in turn cites Ade et.al. in Astronomy and Astrophyics 517), the contribution of matter (both Dark and visible matter) is
$$Ω_\text{mass} ≈ 0.315±0.018$$
The contribution of photons and neutrinos is small, and within the boundaries of error of the other terms:
$$Ω_\text{relativistic} ≈ 9.24×10^{−5}$$
And the contribution of Dark Energy is:
$$Ω_Λ ≈ 0.6817±0.0018$$
$$Ω_\text{total}= Ω_\text{mass} + Ω_\text{relativistic} + Ω_Λ= 1.00±0.02$$


Usefulness of the hour angle (HA) The Hour Angle (HA) is measured west from the observer's meridian in hours, mins, seconds.

How is the Hour Angle used to help plan observations during the night at a telescope?

Imagine an hour which goes counter-clockwise. On the left, there are the numbers from 0 to 12, and on the right, there are the same, but the negative numbers.

Now imagine yourself standing in the middle of such dial. If you have HA = 0, you know that the star is on the meridian. If HA = 6 or HA = -6, then the projection of star onto the celestial equator is on the horizon. If you have HA = 12, then the star lies on the opposite side of the meridian. The same is valid for decimal values.

Look also at this simulation for demonstration of the hour angle, and at the Wikipedia page about this.

But beware! The altitude isn't the same as hour angle. HA is based on equatorial system, but the altitude is based on the horizontal system. The altitude tels you the height of the star, but the HA tells you the position of the star in the fictional dial.

The right ascension is useful because the star is (less or more) standing still in it. RA is for the distant star constant. But HA with the combination of declination tells you more presentable coordinate system. If you know RA and if you want to know, where the star is, you have to know date and hour, and you can't imagine the position easily. But if you know HA, you know it easily, you just look at the upper dial. (Note that we are looking toward the North in the upper image.) For example, do you know where the RA = 0 is located now? Of course not. And do you know where the HA = 0 is located now? Yes, you can just look at the dial. HA = 0 line is also called the meridian. Also, you can easily convert from HA to RA and conversely using current sidereal or astronomer time. (Look at the Wikipedia article.)

Short answer: You can easily imagine hour angle (HA), meanwhile, you cannot imagine the right ascension (RA) so easily.


Sun and moon's orbit in ecliptic coordinates using skyfield I am new to using skyfield, is there any doc or help file that can show me on how to get the orbit of Sun and moon in ecliptic coordinate for a particular date and time. This is a follow up question of this question

I cant help you with skyfield, but i usually use JPL Horizons Web interface. No installation required, you can also print it in a text file if you want:

https://ssd.jpl.nasa.gov/horizons.cgi

Otherwise i found the documentation for skyfield:
https://rhodesmill.org/skyfield/toc.html

And if nothing of that works, i made a little astropy script for you

from astropy import units as u
from astropy.coordinates import SkyCoord, EarthLocation, AltAz, get_body
from astropy.time import Time
import numpy as np

# Create 1000 Timepoints between Time 1 and Time 2 (one year later)
t = np.linspace(2451545, 2451545+365, 1000)

pointlist = []

#Loop through this times
for tn in t:
    # For every timepoint, create an astropy_time object
    astropy_time = Time(tn, format="jd")
    # Get Planet (as string, "earth", "moon", "mercury" etc. in aequatorial coordinates
    planet_aequatorial = get_body("moon", time = astropy_time)
    #Transform to Barycentric True Ecliptic (relative to the center of mass of the solar system).
    planet_ecliptic =  planet_aequatorial.transform_to("barycentrictrueecliptic")

    # Add a point to the orbit. Every point is described as (longitude [deg], latitude (ecliptic coords), distance (km))
    pointlist.append([planet_ecliptic.lon.deg, planet_ecliptic.lat.deg, planet_ecliptic.distance.km])
    print(planet_ecliptic.distance.km)
    # So pointslist is a 2D array. The rows are all the 1000 points of the orbit

# In every point there is 3 columns for [Long, Lat, Distance]
print(pointlist)

# You can also save the result with
pointlist = np.array(pointlist)

np.save("results.npy", pointlist)


What happens to the information on the event horizons of two merging black holes? What happens to the hairiness/information on the horizons of two black holes if they collide?

After Hawkin the information of the matter which has fallen into the blackhole is encoded on the surface of the event horizon. What happens to the information of both blackholes when the event horizons merge?

How black holes encode or destroy information is an open question, which RichS touched on.  However, while he provided an answer consistent with the No Hair Theorem, I will provide an answer derived from the holographic principle.  I'd like to stress that both are equally valid, since we (as of yet) do not know enough about black holes.

In terms of the holographic principle, the information about the matter that formed the black hole is encoded in some manner (micro gravitational fluctuations maybe?) on the two-dimensional 'surface' of the event horizon.  It has been theorized that one could reconstruct this information by measuring the outgoing Hawking radiation, since this process decreases the radius of the black hole, hence the surface area of the event horizon, and subsequently the amount of information.  I would liken this to burning a book and then trying to reconstruct the book by measuring the properties of the ashes and radiated light.

When two black holes merge, they form a black hole of smaller mass than the combined masses.  Once more the total surface area of the event horizons has decreased to that of the event horizon of the new black hole, so information must have been 'radiated'.  Black hole-black hole mergers are not believed to have any optical counterpart (burst of light for sake of simplicity) and the merging process is derived from relativity, so Hawking radiation is not a component of consideration.  Where could the information have gone?  Well, luckily enough these mergers do radiate gravitational waves, now proven by LIGO's recent discovery.  Thus the information, if it does in fact encode itself on the event horizon surfaces, could be radiated/lost by the gravitational waves created during the merger.

EDIT: The above description sounds very 'hand-wavey', so I will expand on proposed theoretical method behind it.

Gravitational radiation is generated by the changing quadrupole moment, caused by the two in-spiraling black holes.  However, just as for the case of electromagnetic radiation emitted by oscillating charges, the quadrupole moment contribution is only one part of the greater multipole expansion of the oscillating masses.  Oscillations of the the event horizon would then cause deviations of the system from the simple quadrupole approximation, and result in gravitational radiation from the high-order multipole terms.  This radiation falls off with distance faster as one moves to higher-orders, making measurement of these contributions much more difficult.

Of course this is just one proposed solution to the black hole information paradox.


What are the problems that need to be taken in consideration when trying to find planetary position by using jpl ephemerides? I'm currently trying to determine the position of the sun and the moon from jpl ephemeris DE200 which is referred to the dynamical equator and equinox of 2000 and uses tdb time. I am using the j2000  system ECI.The things that trouble me are the following:

1.Is the moon position given with respect to Earth or with respect to the barycenter? Also when we find the sun position by using the earth-moon barycenter position and lunar coordinate , will the sun position be with respect to Earth?

2.If I'm using UTC time  , how exactly to obtain the TDB or TAi, since it's oscillating?( After that I want to obtain the JD for TDB)

3.If I'm using  J2000 ECI coordinate system , do I need to take in acount the effect of nutation and precession? And how to do that?

You don't mention what you are trying to access this with but I would strongly recommend AstroPy for handling the ephemeris and handling the co-ordinates and time systems. For your questions:

The Moon and the Earth are both given relative to the Earth-Moon barycenter (center of mass). Separately you can get the vectors from the Solar System Barycenter to the Earth-Moon Barycenter. So if you want SSB->Moon you will need to do two calls to the ephemeris and do vector addition.
It depends how accurately you want things and how fast the thing you want is moving. You can probably use Terrestrial Time (TT) with the ephemeris and you want be horribly off for anything other than Moon. The offset TT-UTC (currently 67.184 seconds) consists of two parts; a variable part from UTC to TAI, currently 37 seconds, which consists (mostly) of the number of leapseconds and a fixed part (32.184 seconds) from TAI to TT. TDB varies around TT by about 2 milliseconds and it's somewhat periodic (dominated by Earth's orbit ellipticity and Jupiter and Saturn) but doing it properly requires a 791 term expansion (see e.g. http://star-www.rl.ac.uk/docs/sun67.htx/sun67ss154.html#Q1-158-946)
Yes, if you want to do anything related from the Earth such as predicting positions of something in the sky, you will need to take the frame bias, precession and nutation into account along with Earth rotation/sidereal time to align the celestial reference frame (ICRS/J2000) with that of the Earth (ITRS).

I recommend USNO Circular 179 (http://aa.usno.navy.mil/publications/docs/Circular_179.php) which has a lot of details of how to do this in the new IAU framework. Meeus's book is very good (I have it also) but is based on the old IAU1984 equator and equinox of date methods.


How big would the Milky Way be if all the stars were emptied into one sphere? Neglecting the fact that this ball of gas would just collapse on itself - I'm curious if there's an agreed-upon measurement that takes into account the volume of all the ~100 billion stars to predict the diameter of a milky way where all the gas is in one orb. If not, is it safe to just use the volume of our Sun (which is an "average" star) and multiply by 100 billion or is there a more accurate data set to include in the calculation?
Asking for a 6-year-old :)

Obviously the answer depends on what the density of such a sphere was!
The only sensible way I can interpret this question is to ask what is the combined volume occupied by all of the galaxy's stars and then get a spherical radius from this.
A modal star in the galaxy has a mass of around 0.3 times the Sun and a radius of 0.3 times the Sun. Since the stellar mass distribution is quite peaked at that value, we could do worse than simply assume that we have about $10^{11}$ stars like this, since stars of higher and lower masses are less common.
In which case, the radius containing a similar total volume is just $0.3\times 10^{11/3}\ R_\odot= 1400R\odot$.
The problem with this calculation is that although the masses of stars are quite peaked, their radii are not. Some evolved stars individually have radii that are a significant fraction of the combined radius figure. Thus although the figure above might be good for the combined radius of all main sequence stars it certainly isn't a good number if you wanted to include all giant stars, with the densities they have now.


How are shockwaves in a star that cause supernova created I know that during the collapsing of a star 8 times the mass of the Sun at the end of its life, a shockwave is generated that blows the star up. However, I do not understand how it is exactly generated. I would appreciate an answer.

You can use a rubber ball analogy.
At the end of a massive star's life the core consists mostly of inert nuclei of iron and nickel. The core is supported by electron degeneracy pressure - a quantum mechanical effect that prevents the electrons from having low kinetic energies. Their high kinetic energies are what provides the pressure.
If the core grows too massive and hot then an instability takes place - either it becomes energetically feasible for the electrons to combine with protons in the nuclei to produce neutrons, or the nuclei get photo-disintegrated and free protons combine with the electrons. Either way, the prop that has been holding up the core is removed and it collapses dramatically and suddenly - almost in free-fall.
A new equilibrium may be possible when the densities get so high that the free nucleons are "touching" - separated by around $10^{-15}$m. At this point there is a strong repulsive nuclear force that comes into play. This may be capable of stabilising the collapse and ultimately forming a neutron star, but not before the collapse has proceeded well past the equilibrium point. As such, there will be recoil, like a squashed rubber ball, and it is this recoil that powers the shock wave - this has become known as a "core bounce".
The mechanism by which the shock transfers energy to the envelope is still the source of debate. However it is strongly inferred that neutrinos are involved. These are produced in enormous quantities through the combination of protons and electrons during the collapse and end up possessing about 99% of the gravitational potential energy released by the collapse. But the matter is so dense that they cannot immediately escape. They are trapped for several seconds and this provides an enormous pressure that can propel material outwards.


Were effects of a planetary magnetic field reversal observed on other planets than Earth? From geological records in rocks and minerals we know that the magnetic field of Earth changed its polarity multiple times in the history. See Geomagnetic reversal.

Was a similar process of a planetary magnetic field reversal observed on other planets?

I suppose that the record could be in rocks directly found on the planet (i.e. Mars), on its moons or in asteroids.

There are such evidence in the case of Mars. Observations from Mars Global Surveyor show evidences of crustal magnetization. In particular, this magnetization has extensive, east-west trending linear features in Terra Cimmeria and Terra Sirenum. These are probably reminiscent of magnetic features associated with a reversing dipole.

There are no such evidence in the case of Venus: Venus has no intrinsic magnetic field today (contrary to the common belief, its inability to maintain a dynamo activity is either due to its chemical and physical conditions unable to form a solid core, either du to an early complete core solidification; it is not due to its slow rotation), and could not have maintained a remanent crustal magnetization, because its crust would have been to hot during its dynamo activity (above the Curie point).

Sources:

Acuna et al. (1999)
Luhmann & Russell (1997)


Does a parabolic trajectory really exist in nature? It seems it is very difficult to have e=1 perfectly in nature. The final state (being captured or running away) of a celestial body with a parabolic trajectory,
is determined by minor perturbation?

No, but nor do elliptical or hyperbolic trajectories.  These are features of a model we have for gravity, not part of nature.
The modelling process involves constructing a mathematical framework that describes a natural system and allows for predictions. It is often said that "no models are correct, but some models are useful".  For gravity, one common model is the Newtonian model.  In the Newtonian model, you can set up two particles in such a way that one follows a parabolic curve (in some inertial frame) Parabolic trajectories exist in the Newtonian model.
In nature, there are more than two particles in the universe and so there are no parabolic, elliptical nor hyperbolic trajectories.  Moreover, the Newtonian model is known to only approximate gravity (even General Relativity is probably an approximation, gravity is probably some kind of quantum interaction. But the details are not yet known)
Nevertheless.  The Newtonian model is a very useful model for solar system dynamics.  Long-period comets, falling from the distant Oort cloud, will have orbits that are very long and are well approximated by parabolas. (but only approximated, as comets, in particular, have significant non-gravitational accelerations due to reaction forces from outgassing, in addition to gravitational perturbations of the planets, and effects from the non-spherical shape of the sun)


Will Earth rotate backwards in the future? Hello recently I heave heard the earth will eventually start rotating in the opposite direction (ie sunrise from the west) . I am not sure if this is true, but a lot of sites are claiming it to be true, they say that the earth's rotation is slowing down slowly, and it will eventually stop and start rotating in the opposite direction, please let me know if this is true or not as I do not have much knowledge in astronomy.
EDIT:
Some of the sites that claim this event to be true, explain it in this way -
they say
"The earth is slowing down every year, so much so that when it first started spinning each day was only 4 hours but now it's 24 hours, a day will come when the earth will stop rotating and it will behave like a spring and then it will start rotating in the opposite direction (they don't mention if humans will be alive till that day or not) "
Here are the sources that led me to consider the possibility of this event-
1 - https://archive.siasat.com/news/nasa-confirms-possibility-sun-rising-west-102184/#:~:text=And%20this%20weird%20phenomena%20of,rise%20from%20the%20west!!
2 - https://medium.com/@loadjunker/nasa-confirms-possibility-of-sun-rising-from-the-west-ad5ddd166c11#:~:text=And%20this%20weird%20phenomena%20of,rise%20from%20the%20west!!
3 - https://www.express.co.uk/news/science/597607/Sunrise-WEST-Scientists-warn-magnetic-poles-SWITCHING-North-become-south
4 - https://medium.com/swlh/if-the-sun-rises-from-the-west-92452a570f86

No, the Earth will not start to rotate in the opposite direction. Ever.
The reason Earth maintains its direction of rotation is conservation of angular momentum. Just like a moving body resists changes in velocity because it has linear momentum, a rotating body will resist forces that try to change its rotation state. Angular momentum can be moved around in a composite system (consider a figure skater pulling in her arms while spinning), but it doesn't disappear unless there is a lot of friction. Which space lacks.
There is a grain of truth in the confused articles that led to the question: Earth is losing some angular momentum over time. Most of this is due to tidal interactions with the moon. When the moon formed Earth rotated much faster and the moon was close. Mutual gravitation caused tides that made some of the angular momentum transfer from Earth to the moon, slowing the rotation of both (to the extent the Moon now only turns one face towards the Earth, tidal locking) and giving the moon more angular momentum in its orbit, making it move outwards at a very slow rate. There are also tides from the sun (and other planets) but they are too small to matter here.
In reality, the sun will become a red giant in 5 billion years and expand, ending the story of the Earth and moon (ironically in this context, likely by tides dragging in Earth into the sun when it gets close enough to the surface). But if we ignore that, what would happen?
As the Earth's rotation is slowed by lunar tides, eventually it would become tidally locked to the moon in about 50 billion years. The moon would always stay over the same spot, visible only from one hemisphere. But Earth, like the moon today, would also continue to rotate around its axis. It is just that the "day" would be more than a month (about 47 days), corresponding to one full rotation of the Earth-moon system. But the direction would be the same as now, since the moon is moving around the Earth in the same direction as Earth is rotating.
At this point there would still be some loss of angular momentum due to solar tides, so over very long time periods (about 50 billion years more) the Earth and moon would slowly orbit closer and closer. Eventually they would merge into an (again) somewhat fast spinning body (still moving in the same direction!)
In the really long run tidal forces would slow this rotation so the big planet would be tidally locked to the sun. But again, the one-year rotation of the planet would be in the same direction as now.
There is a lot of nonsense claimed on the Internet. Learn the basic laws of physics like conservation of energy and momentum, get used to do rough calculations, and you can usually see through it quickly.


Are there known centaurs which orbit between Jupiter and Saturn without crossing their orbits? According to Centaur minor planet

  Centaurs are small Solar System bodies with a semi-major axis between those of the outer planets. They generally have unstable orbits because they cross or have crossed the orbits of one or more of the giant planets

Of all the known ones which I've found and I've seen data, all of them overlaps the orbit of Saturn, with the exception of  (514107) 2015 BZ509 which doesnt overlap the orbit of Saturn but the orbit of Jupiter.

My question is, are there known centaurs which dont overlap the orbits of Jupiter and Saturn but which orbit between those 2 orbits? Or does the definition of Centaur mean they have to cross orbits and an asteroid orbiting between those 2 orbits would be called another name if exists?

The JPL Small Body Database lists 5 asteroids and 5 comets with q > 4.56 au and Q < 9.04 au.
Two of those asteroid orbits have large uncertainties due to few observations and may not actually meet your criteria. These are fairly well determined:

Asteroid 494219 has a highly inclined 14-year orbit which occasionally brings it within 2 au of Jupiter. It was discovered in 2012.
Asteroid 2000 GM137 has a 22-year orbit which occasionally passes within 2 au of Saturn. It was discovered in 2000, unobserved for a few years, and recovered in 2014.
Comet 29P/Schwassmann-Wachmann 1 was discovered in 1927, has a nearly circular 15-year orbit, and is known for frequent outbursts.
Comet 39P/Oterma was discovered in 1943 on an 8-year orbit. A 1963 encounter with Jupiter switched it to its present 20-year orbit, where it became dormant and was lost until 2001. Numerical simulations indicate a prior Jupiter encounter in 1937; it was likely also a Centaur before that.


Which telescope to choose and some general questions I'm playing with the idea to buy a telescope for years now and finally decided to do so. There are two which I consider:

The Celestron NexStar 127 SLT Mak and the Celestron NexStar 130 SLT

I already know the difference between a Refractor and Maksutov-Cassegrain and for which observations they're suited for best (and also that the Mak doesn't need collimating).

First question: Which would be the better choice for a beginner and eventually astrophotography?

Second question: Is it possible to buy just the tube later (so I can reuse the motors and the tripod)?

Third question: As I've read, the tripod is pretty wobbly, so is it possible to replace the tripod later on (or is there a norm for the mounts)?

Macs also need to be collimated, but much more rarely than Newtonians.
 They hold their alignment better (for more time).

For astrophotography you also have to think about the field of view each scope will give with the camera a that you will use and the fact that their mounts are not equatorial (less important if you are into planetary photography).

I see that the Mac has a Vixen dovetail, which means you can later use it on other compatible head mounts or put some other scope with the same dovetail type to the SLT mount.

It might be cheaper to wait more time and buy from the beginning the "right" equipment for your task, than buy something cheaper and then replace its parts.

Cheers!


Relation between angular resolution and aperture? The angular resolution of a telescope is given by $\frac{1.22\lambda}{D}$.

This should mean that the smaller the aperture D, the higher the angular resolution. But it is exactly the opposite - why?

The angular resolution only tells you your ability to distinguish ("resolve") two separate points of light. For example, your ability to distinguish the two distinct headlamps on a car from being a single light in the distance.

The angular magnification has nothing to do with the angular resolution: it is determined by the focal lengths of your primary and objective, and the eyepiece if applicable.


Does Ganymede have a day/night cycle? If so, what does it look like? Over on the SF & Fantasy Stack, I've asked a question about a curious watch that shows 15 intervals on its watch face instead of the usual 12 hours. That watch belongs to someone living on Ganymede, close to the end of the 21st century.

Why does Jet's old watch show 15 hours?

Does Ganymede have any day/night cycle, or any other cycle, that would explain this curious watch?

In analogy to earth, we should expect a 30-hour Ganymede-day. But Ganymede is in tidal lock, always showing its same side to Jupiter. Also, calculations show that the light reflected by Jupiter pales compared to that of the sun.

So I would assume that Ganymede's day/night cycle is sun based, possibly complicated by passing through Jupiter's shadow.

What do day and night on Ganymede look like?

Here's some pertinent facts:

Apparent magnitudes: Sun from Ganymede     = -23.11
                     Jupiter from Ganymede = -16.02 (full)
                     Difference            = - 7.09
                     Brightness difference = 2.512 E7.09 = 686 times brighter

Earth comparison:    Sun from Earth        = -26.74
                     Moon from Earth       = -12.74 (mean full)
                     Full moon/ Jupiter from Ganymede difference = 20.5 times

Ganymede facts:      Orbital period = 1 Ganymede day: 7.16 days
                     Orbital inclinations: .2 deg to Jupiter equator
                                          2.2 deg to ecliptic
                     Orbital eccentricity: .0011 (circular)
                     Orbital speed (ave.): 10.88 km/s

Jupiter facts:       Mean radius: about 70000 km

Summary:
If you are on the "far side" of Ganymede you won't see Jupiter (maybe a glow on the horizon from Ganymede's thin atmosphere). The sun will be 2.512 E3.63 = about 28 times dimmer than seen from Earth (which is still pretty darn bright). A day will last 7.16 Earth days.

The near side will be far more interesting. The sun will go through the same cycle at the same brightness as on the far side. However, Jupiter will "hover" in the sky (maybe a little wobble/ libration) and when full will be about 686 times dimmer than the sun in the sky. Jupiter will go through phases over the course of the 7.16 day cycle and be less bright of course as it approaches eclipse. The eclipse (by rough calculations) will last about three and a half hours. The planet will have an angular diameter of about 7.5 degrees (compare with 0.5 for sun/moon from Earth) and be about 20.5 times brighter than a full moon when Ganymede is between the sun and Jupiter.

A fun side note is that the sky won't be totally dark during eclipse. Light from the sun will reflect off aerosols and hazes in Jupiter's upper atmosphere making Jupiter "glow" in Ganymede's sky.


Finding Quintuple Planetary Alignments with SkyField I am working on a python script that will use the SkyField and SciPy libraries to find quintuple planetary conjunctions and their corresponding constellation location. Specifically I am looking for dates when the 5 visible planets were all in conjunction within the constellation of Aries. This occurrence should be exceptionally rare and I just need something to find if and when it happened in the last 13K years or so...

I found this SkyField solution here to find conjunctions.

I was able to modify the above solution to find quintuple conjunctions for the last 15000 years. At least I think I did. Here is my solution:

import scipy.optimize
from skyfield.api import load, pi, tau, position_from_radec, load_constellation_map

efile = "de431t.bsp"; # ephemeris to use
ts = load.timescale()

print("Loading ephemeris... \nIf this is the first time running this script, a 3.4Gb file will be downloaded.")

# Load Ephemeris
eph = load(efile)

print("\nPlanetary Ephemeris loaded.\n")

# Define planets
earth = eph['earth barycenter']
venus = eph['venus barycenter']
mercury = eph['mercury barycenter']
mars = eph['mars barycenter']
jupiter = eph['jupiter barycenter']
saturn = eph['saturn barycenter']

constellation_at = load_constellation_map()

# Every month from start year
t = ts.utc(-12999, range(12 * 15000))

print("\nCalculation plaetary locations. This may take a while...\n")

# Where in the sky were the Planets on those dates?
e = earth.at(t)

lat, lon, distance = e.observe(venus).ecliptic_latlon()
vl = lon.radians

lat, lon, distance = e.observe(mercury).ecliptic_latlon()
ml = lon.radians

lat, lon, distance = e.observe(mars).ecliptic_latlon()
mal = lon.radians

lat, lon, distance = e.observe(jupiter).ecliptic_latlon()
jl = lon.radians

lat, lon, distance = e.observe(saturn).ecliptic_latlon()
sl = lon.radians

print("Looking for conjunctions...\n")

# When was Mercury conjoined with the other planets?  Compute their difference in
# longitude, wrapping the value into the range [-pi, pi) to avoid
# the discontinuity when one or the other object reaches 360 degrees
# and flips back to 0 degrees.
relative_lon = (vl - ml + pi) % tau - pi
relative_lon2 = (mal - ml + pi) % tau - pi
relative_lon3 = (jl - ml + pi) % tau - pi
relative_lon4 = (sl - ml + pi) % tau - pi

# Find where all planets are within a degrees of one another...
conjunctions = (relative_lon >= 0)[:-1] & (relative_lon < 0)[1:] & (relative_lon2 >= 0)[:-1] & (relative_lon2 < 0)[1:] & (relative_lon3 >= 0)[:-1] & (relative_lon3 < 0)[1:] & (relative_lon4 >= 0)[:-1] & (relative_lon4 < 0)[1:]

# For each month that included a conjunction, ask SciPy exactly when
# the conjunction occurred.

def f(jd):
    "Compute how far away in longitude Venus and Mercury are."
    t = ts.tt(jd=jd)
    e = earth.at(t)
    lat, lon, distance = e.observe(venus).ecliptic_latlon()
    vl = lon.radians
    lat, lon, distance = e.observe(mercury).ecliptic_latlon()
    ml = lon.radians
    relative_lon = (vl - ml + pi) % tau - pi
    return relative_lon

for i in conjunctions.nonzero()[0]:
    t0 = t[i]
    t1 = t[i + 1]
    print("Starting search at", t0.utc_jpl())
    jd_conjunction = scipy.optimize.brentq(f, t[i].tt, t[i+1].tt)
    print("Found conjunction:", ts.tt(jd=jd_conjunction).utc_jpl())
    e = earth.at(ts.tt(jd=jd_conjunction))
    a, b, d = e.observe(venus).radec()
    vr = position_from_radec(a.radians, b.degrees)
    print("In constellation:", constellation_at(vr))
    print()

This seems to be working and the next step is to plug the dates into XePhem and see what they look like.

Can anyone confirm I am doing this right?

This is not an answer.

I computed all major planet conjunctions in DE431 to answer How to calculate conjunctions of 2 planets and you might be able to port what I did (using CSPICE) to skyfield.

Several of the more interesting conjunctions I found are here:

http://search.astro.barrycarter.info/table.html

including 5 and 6 planet conjunctions:

http://search.astro.barrycarter.info/STELLARIUM/stellarium-191.ann.png
http://search.astro.barrycarter.info/STELLARIUM/stellarium-092.ann.png

I note that the 6 planet conjunction is unique in DE431, but I refer to the 5 planet conjunction as "rare", so it's likely there's at least one other one.

I'm too lazy to look through my own results, but this might give you a start.


Intuitive explanation for why the Doppler effect (and red-shift) happens? I was looking for an intuitive explanation as to why the Doppler effect happens. I haven't found any, but this is what I thought:
-Waves emitted travel at a constant speed
-The source emits a wave
-If the source remained still, then in a given time, there would a large distance between the emitted wavefront and the stationary source
-But if the source starts moving, the speed of the source relative to the wavefront is higher.
-This means in a given time, the distance between the emitted wavefront and the source will be lower
-So when the source (while moving) emits a new wavefront, the distance between wavefronts will also be lower
Is this explanation wrong? It suggests that the Doppler effect with light will be barely noticeable, because even if a source starts moving, because the speed of light is so high, the source speed will still be small relatively, so there will be a small (basically insignificant) decrease in distance between wavefronts. So how can red-shift be so noticeable, unless recession speed is extremely high (so high that it is actually significant relative to the speed of light)?

Your intuition is correct - a moving source emitting wavefronts periodically will be closer to the previously emitted wave in the direction of motion, and farther from the previously emitted wave in the opposite direction - see the simulation here.
You are also correct that the size of the effect depends on the speed of the observer relative to the speed of the wave.  That is why we can easily experience the Doppler shift for sound in everyday life (e.g. the sound of a car when it is approaching vs. receding) but we never notice the Doppler shift for light.  The speed of sound is a lot slower, and thus a lot closer to the velocities of everyday life, so the effect is larger for sound.
While the shift for light is too small for our eyes to perceive (e.g. not enough to change an object’s perceived color), sensitive instruments can measure extremely small changes in wavelength.  For example, the spectrographs that astronomers use to make radial velocity measurements of stars for the purposes of detecting extrasolar planets (via the star’s motion as the planet orbits) can now routinely measure velocities with a precision of 1 m/s - a leisurely walking speed for a human!


Could a star become a planet? Could a star become a planet? I am asking this because the gas giants are ¨Failed Stars¨ and they are classified as planets in our solar system.

A star could become a planet, but this almost never happens. It would probably have to lose almost all of its mass early in its life, perhaps by a black hole. Then, the exposed core would have to be stopped from becoming degenerate and further mass loss would have to continue. It is possible, but it is not realistic and  will not happen in a human lifetime, even if you observed every star in the universe (including white dwarfs).
A brown dwarf definitely can become a planet. A brown dwarf early in its life acts like a star, fusing deuterium and reaching temperatures of almost 2000 K. But as it ages it cools down and stops fusing deuterium. Then it acts like a planet.
In theory, a brown dwarf directly could become a planetary-mass object through mass loss. This could happen from a giant impact or another body accreting from it.
Contrarily, a planet (brown dwarf) can become a star if it accretes enough mass. For example, the most massive brown dwarfs have masses of about ~80 Jupiters, while the least massive stars have a mass of about ~60 Jupiters (there is an overlap). Therefore, a "planet" could become a star under special circumstances.
As @User123 said in a comment, all stars start as a "planet," without burning hydrogen. While it accretes more material, fusion ignites in its core and it becomes a star.


How do tadpole, horseshoe co-orbital states arise and how are they stable? A recent paper in Nature "Planetary science: Reckless orbiting in the Solar System" (Morais & Namouni, 2017) presents the following series of four co-orbital states:

While I understand the development of the quasi-satellite and trisectrix states, I am unable to inuit the development of the tadpole and horseshoe co-orbital states.
From explanation provided in the article:

The tadpole and horseshoe shapes arise because the planet's
gravitational attraction alters the body's orbital path — the body
goes through a cycle of catching up with the planet and falling
behind, seeming to change direction from the perspective of the
planet.

How, when approaching the planet, does the body "fall behind" instead of continuing to accelerate toward the planet?
The only means that comes to mind is if the planet orbits faster than the body, but then I imagine it would

No longer be considered co-orbital (because it would have a different orbital period), and
Would eventually be overtaken by the planet.

How, then, do these first two co-orbital states work?
Citation: Nature 543, 635–636 (30 March 2017)
doi:10.1038/543635a

If the body is in front of the planet (relative to the planet's orbital motion) and a little further from the sun, it will orbit the sun slightly slower than the planet.

As it is slower, the planet will slowly catch it up. (It takes many "years" for the planet to get close to the body.)

As the planet catches up with the body, the gravitational effect of the planet will tend to pull back on the body, this causes the body to lose energy and begin to fall slighly towards the sun.

The slight fall towards the sun cause the body to move to a lower orbit (relative to the sun) and so speed up. It appears paradoxical, but it the fact that the gravitational drag of the planet on the body actually causes it to speed up.

The body speeds up and so begins to move away from the planet.

When the body is makes an angle of more than 60 degrees from the planet, the gravitational pull of the planet now has the opposite effect. It combines with the pull of the sun to cause the body to accelerate, and (apparently paradoxically) this acceleration causes the body to move to a more distant orbit, and slow down.

There are two things key to understanding this. First remember that the diagrams you show are drawn with a fixed planet. I.e. the "camera" is rotating at the same rate as the planet takes to orbit the sun, so the planet appears fixed. If you don't rotates the camera, you would see that the body actually has a nearly elliptical orbit, relative to the sun. Secondly you need to grasp the idea that, in orbit, drag doesn't slow you down. Drag causes you to fall towards the sun and speed up.

How did bodies get into such orbits? Basically by luck. They happened to have orbits that were close to Earth's orbit, and by a lucky effect of the gravity of the other planets they were dislodged from their regular orbit of the Sun to a tadpole orbit. Once there, they can remain in their orbit for tens fo thousands of years. Even so, only one object is known to exist in a tadpole orbit relative to Earth.


What is the name of our Solar System? what is our Solar System called rather than "the Solar System"?

I've found "Sol System" and "Monmatia", too, but is there more to it?

It's just called "the Solar System". (Plenty of places and objects have names like that; it's no different from "the Arctic" or "the Moon" or "the Sun".)

("Sol system" is an invention of science fiction writers; it has no general use outside some science fiction contexts. Anything else is going to be something similar, or general crackpottery of one kind or another.)


Is there a retrograde star that passes closer to the galactic center than our Sun? Our Sun is moving in a prograde orbit around the galactic center with a distance of about 8 kpc.  This paper says there are many retrograde stars in the galactic halo 10-20 kpc from the galactic center.  Do any of these retrograde stars have an extremely eccentric orbit that would cause them to pass closer to the galactic center than our Sun?

There's a whole stream of them: https://iopscience.iop.org/article/10.3847/1538-4357/aadba3
Named Phlegethon, this is a stellar stream that covers about 75⁰ of the sky, and is probably the remnants of a disrupted globular cluster.
It orbits the galaxy retrograde on orbits that take it between 4.9 and 19.8 kpc from the galactic centre. The orbits are highly non-Keplarian. The image shows the orbits with the sun marked as a yellow dot.

Other streams exist and may have similar shaped orbits.