What would happen if a wormhole gets swallowed by a black hole?

The concept of a wormhole is, of course, highly speculative.  None the less, such a phenomenon could take many forms - most simple something which to outside observers looks like a blackhole on each end of the wormhole.  Alternatively, a traversable wormhole (see for example Morris & Thorne 1988; which has the same basic topology but no horizon), could still behave for the most part* like a normal, massive object.

In either case, if one end were 'swallowed' by a blackhole, it is unclear to me whether it would remain stable (i.e. remain a wormhole).  In the case of a non-traversable wormhole, if it remained stable, it would continue to look like a blackhole on each end --- one end of which now merged with another blackhole, and subsequently more massive.  A traversable wormhole on the other hand could (extremely hypothetically) still remain traversable if there were enough exotic matter, or it could retain the horizon from the blackhole it merged with and be no-longer traversable.

All wormhole metrics that I've seen are more or less symmetric, and thus either have no horizon, or horizons on both ends... I don't think it would be possible to have a horizon on only one side --- allowing something to fall into an apparent blackhole on one side, and come out of a wormhole on the other.  Even so, however, remember that a blackhole isn't a magic vacuum --- and wouldn't 'suck in' the 'entire universe'.

*A traversable wormhole is believed to require some sort of 'exotic matter' in the neck, which may behave somewhat differently...


Supergalactic and Equatorial coordinates

Equatorial coordinates have their equator and poles at the equator of the earth and the poles at the earth each projected onto the sky.

Supergalactic coordinates on the other hand have their equator in the same plane as the "Supergalactic plane", which comes from the distribution of nearby galaxy clusters. And the poles just perpendicular to the the plane.You can kind of see this preferred plane in the image below.

One is therefore just a rotated or tilted version of the other. To be precise the north poles of Supergalactic coordinates lies at approximately RA = 18.9h, Dec = +15.7° in Equatorial coordinates.

See also here here for a comparison of common celestial coordinate systems.


How do ASKAP's focal plane phased array feeds interact with the entire array phasing?

The phased array feeds (PAFs) on ASKAP are indeed located at the focal plane. The ASKAP PAFs have 188 receivers, and from this digital beam-forming will create 36 independent beams on the sky. Some more information can be found from the CSIRO. Imaging is then done by scanning the dish (or letting the sky roll overhead). The scanning can then be done much faster than with a single receiver.

Its important not to confuse these independent beams with purpose of the array itself. The main reason for using an array is interferometry. We use an array for this purpose to have multiple baselines (the distance between the antennas). Having multiple baselines allows the telescope to 'see' structure various angular resolutions. It's true that you can use the array in a 'fly's eye' mode, and then every dish is providing 36 independent beams. In this mode even larger areas of the sky can be scanned quickly. This is useful for searching for transient events such as fast radio bursts.

So, the phasing within the PAF needs to be correlated and the beams formed digitally, and then the signals from each dish need to correlated. A huge task indeed! The computing breakdown for ASKAP is given here. A new supercomputer (Pawsey) has been built in WA for the purpose of processing ASKAP data.


Which telescope's model is being shown to Albert Einstein in this file footage? What event might this be?

I'm pretty certain it is the 200inch Hale telescope at Palomar Observatory. There is an image galley of the 200 inch telescope. According to Richard Preston's book First Light: The Search For The Edge Of The Universe, Einstein attended the completion of the telescope truss, yoke and horseshoe bearing at Westinghouse Electric and Manufacturing Company in South Philadelphia (page 41) so it may have been this event that the footage is from.

Searching for "celluloid model of Hale telescope" in the Hagley Digital Archive will yield information such as

  This celluloid mount was part of a 1/32 scale model of Hale Telescope that was constructed by Westinghouse Electric Corporation engineers to test design utility and points of stress.

and images such as these:


On (minuscule) dark matter production in supernovae

The most favoured WIMPS at the moment are probably neutralinos, see
http://en.wikipedia.org/wiki/Neutralino

These particles are purely hypothetical at the moment.
The mass estimates in the above Wikipedia article for the lightest neutralino range between
10 and 10,000 GeV, meaning that the production rates in SNs will be much lower than with an assumed 1 GeV.
Higher production rates should have already been detected at LHC.

Hence from the non-detection (in the form of energy loss) of WIMPS at LHC an estimate of an upper bound of the production rates in SNs should be possible.


Leap seconds don't correct for the problem of the calendar years cumulatively drifting out of sync with the seasons, which is what the first source means with 365.2425 being inequal to 365.242374. Instead, the purpose of leap seconds is to correct the problem that 24 * 60 * 60 = 86400 seconds is not the length of the day, and the reason we don't get leap seconds nicely on schedule is that the slowing of the Earth's rotation is nonuniform.
What the second source means when talking about the slowing of the Earth's rotation, is that when the day gets longer, the length of the tropical year in days gets shorter, which means that the relative error between the mean calendar year and the tropical year gets even worse.
The cumulative error in the calendar is not corrected by any means currently, and it will take a long time for the error to be meaningful in the sense of impacting everyday life, so it is not viewed to be a real problem now. This might change in a couple of thousand years, or then people even then might value the algorithmic stability of an unchanging calendar more than the spring equinox's occurence near March 20th.


What is the air pressure in the heliosphere (Sun's atmosphere)?

I am going to assume that what you mean here is "what is the pressure" in the solar wind? There is no "air"!

The solar wind is pretty complex, consisting of a "fast wind" observed predominantly at high solar latitudes and a slower more variable wind a low latitudes. Both components essentially consist of an expanding stream of protons, electrons plus a small fraction of helium nuclei.

I think all the information you want is contained in this paper by Ebert et al. (2009), but I will attempt a broad-brush summary.

The fast wind has a typical proton number density of $n_p=3$ cm$^{-3}$, a temperature of $T=2\times10^{5}$ K and a speed of $V=700$ km/s. Note that "temperature" is a slippery concept here since the velocity distribution of the particles is non-Maxwellian. However, ignoring this and assuming the protons are matched by numbers of electrons, then the "thermal pressure" $P_{th} = (n_p + n_e)kT \simeq 10^{-11}$ Pa. However, this is negligible compared with the dynamic pressure $P_d = \rho V^2$, where $\rho$ is the mass density given roughly by $\rho = n_p m_p$. Thus $P_d \simeq 2\times 10^{-9}$ Pa (i.e. about $2 \times 10^{-14}$ bar).

The "slow wind" is quite variable, but mean values could be $n_p = 6$ cm$^{-3}$, $T=8\times 10^{4}$ K and $V=400$ km/s. Here again, the thermal pressure $P_{th} \simeq 10^{-11}$ Pa is dwarfed by the dynamic pressure $P_d = 1.6 \times 10^{-9}$ Pa.

The thermal pressure component falls as distance from the Sun cubed, ie. as $P_{th} \propto R^{-3}$. This is because for mass conservation, then a sherically expanding wind must have a density that falls as $R^{-2}$, but it is experimentally found that the proton temperature also falls as $\simeq R^{-1}$.

The dynamic pressure does fall almost as $R^{-2}$. This is because there is very little radius dependence for the proton velocity (i.e. the protons do not slow down, right out to the heliopause - as measured for instance by Voyager 2).


How well supported are the theories of black-holes containing their own universes?

More and more of modern theoretical physics deals with unphysical models and ideas. The only basis we have of predicting what is happening beyond an event horizon, ANY event horizon, is what we understand about what happens around us, on this side of that horizon. Feel perfectly safe in making up anything you wish about what the Physics is like beyond the event horizon, we'll never be able to contradict you. We'll never be able to falsify any claim made for events on the other side.

Purple people eaters? Why not? Ghosts and Gods? Sure. 1+1 = 3? Yup, that too. The event horizons in nature are the natural limits to our Universe. The edge, if you will. Unless and until we are able to observe consequences of the claims you make (called verifiable predictions) you're not dealing with Science, at least not what I regard as Science.

Right now, we know of 4 types of edges to our Universe:

The edge at ~1E-44 seconds after the hypothetical Creation Event
The event horizon of any black hole
the event horizon at about 15-20 billion light-years distance (at which space is expanding at greater than the speed of light and
the edge of our Observable Universe (from which no signals will ever reach us).

Physics is about what goes on on this side of these "edges". Metaphysics, philosophy, religion, opinion are what you may use to deal with those things beyond, but don't call it physics or science.

P.S. I am sympathetic to the idea that there may be an (as yet unknown) physical law which prevents singularities from actually forming. In fact, we have only weak indirect evidence that event horizons actually exist, and absolutely none that singularities form. Generally, when a physical theory and its mathematical expression leads to infinities, we change the theory - see renormalization.

You want to know about multiverses or what is beyond the event horizons of black holes? OK. I want to know about the color of the nail polish that the tortoise carrying the Universe on its back uses. Both are equally valid areas of inquiry. See for example https://en.wikipedia.org/wiki/Multiverse#Proponents_and_skeptics ... and links found there.

I should also mention that you know an article which talks about events prior to time zero (which I call the Creation Event) are metaphysics and/or religious. None of them, I repeat: none of them are based on known Physics. (Why? because our known physics breaks down at about 10E-44 seconds, extrapolating back to t=0 isn't possible.) Sure they make an interesting story, but they should be regarded in the same way as any fiction. You get to choose which you like and which you don't without any regard for ever having to justify your choices.

Much of human knowledge is about narratives which allow us to understand our place in the Universe. It is part of our human nature to be curious and seek out narratives that make sense of our world. The set of scientific narratives differ from this set because the goal is different. They doesn't require our understanding (for instance, Feynman said: "I think I can safely say that nobody understands quantum mechanics." (1965) Little has changed since then.), they only require our ability to apply them to make/verify observations.

Creating a satisfying narrative does not require the scientific method(s) be enforced. But ignore them at your peril.


None of those stars can go supernova, so the question is rather moot.  If you look at the classifications, the most luminous is Sirius A (an A sequence star even) you can get an idea of its mass. If you look at your source page, and link to the explanation you see that A stars range from 1.4 to 2.1 stellar masses. In order to go supernova though, you need about nine solar masses.  The nearest supernova candidate is IK Pegasi (HR 8210), located at a distance of 150 light-years (and that's a Type Ia). The closest Type II candidate that I can think of is Spica, which is 260 light years away. Although this list doesn't include Spica, and has the closest Type II as Betelgeuse, at 640 light years away.

In his book, Death From The Skies, Dr. Phil Plait (a professional astronomer, writer, lecturer, etc.) covers exactly what would happen to earth if a star nearby did go supernova, and it isn't pretty. Basically, it would strip us of our ozone layer and turn the upper layer of our atmosphere into a dirty brown smog layer, let in a lot of UV radiation that would devastate plankton and plant life, and be very bad for all life down on the surface. However, for a Type II, they need to be around 25 light years or closer to affect us.  Dr Plait's writing style is very much geared for a layman reader, and explains things in a very conversational tone, so I highly suggest that you get the book and read it. It will tell you exactly what to expect, and also tell you exactly why it's nothing to worry about.

All that said, there are two things in his book that you may want to consider as more likely than a supernova. A GRB from Eta Carinae, or an asteroid impact. The latter we may actually have the technology to eliminate as a concern though, so we're not doing too badly!


Why is moon light not the same color as sunlight?

The light from the moon is light being reflected from the sun.

This is at least one reason you should not expect the Moon to have the same color.  Sunlight hitting an e.g. blue object would appear predominantly blue and similar.  So the color we might expect to see from the Moon is going to be adjusted by the color properties of the Moon's surface.

As it happens lunar regolith is essentially gray in color (i.e. neutral), so it more or less reflects the spectrum it receives.

  The sun, in space, is white

No it is not.  There are a couple of reasons.

The Sun is an exceptionally bright object (for us) and will saturate our color vision so that it seems like it's white when we look at it directly.

The Sun is actually a yellow star.  It's spectrum peaks in the (visual) blue-green range.  There are stars that have a red or blue color.

White is a tricky one.  White is a perceived color.  There is no "white" wavelength in the EM spectrum.  White is a balance between the three different color sensors your human vision uses.  The Sun itself is not white because the color combination "white" we're balanced for is based not on the light from the Sun in the vacuum of space, but on the light that reaches the surface of the Earth and also what is reflected from the Earth (which is part of the ambient lighting we normally experience).

  why is moonlight white through the atmosphere?

It seemed a nice clear sky last night around my neck of the woods and the Moon looked yellow to me.  This will vary with your location and atmospheric conditions, but also with your personal perception of color.


Relation between absolute magnitude of UV and star formation rate

The UV luminosity of a galaxy can be calculated, given a stellar population. This population, in turn, can be calculated given an initial mass function (IMF), i.e. the distribution function of stellar masses. In this case, the UV luminosity should be linearly proportional to the star formation rate (SFR), sometimes written $\Psi$.
The UV is primarily emitted by young stars with lifetimes of $\lesssim10^8\,\mathrm{yr}$. Assuming that the SFR is constant over longer time scales, Kennicutt (1998) calculates in his seminal paper (with ~5000 citations) for a Salpeter (1995) IMF (equally seminal; ~6700 citations) with mass limits 0.1 and 100 $M_\odot\,\mathrm{yr}^{-1}$ that
$$
\frac{\Psi_\mathrm{Sal}}{M_\odot\,\mathrm{yr}^{-1}} = 1.4\times10^{-28}\frac{L_\nu}{\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{Hz}^{-1}}.
$$
were $L_\nu$ is the emitted UV luminosity density, i.e. luminosity per frequency bin. At a (luminosity) distance $d_L$ from the galaxy, the observed flux density $f_\nu$ is then
$$
f_\nu = \frac{L_\nu}{4\pi d_L^2}.
$$
The flux density is related to the apparent (observed) UV magnitude $m_\mathrm{UV}$ as (Oke & Gunn 1983; ~1900 citations despite their +/- error in their most important equation)
$$
m_\mathrm{UV} = -2.5\log \frac{f_\nu}{\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-1}\,\mathrm{Hz}^{-1}} - 48.6.
$$
Distance modulus and dust extinction
The apparent magnitude is related to the absolute magnitude $M$ through the distance modulus
$$
\mu \equiv m - M.
$$
Furthermore, the above relations don't take into account the fact that some of the emitted light is absorbed by dust. Dust extinction increases the observed magnitude (because magnitudes are backwards) by a number $A_\mathrm{UV}$.
That is, the apparent UV magnitude is
$$
m_\mathrm{UV} = M_\mathrm{UV} + \mu + A_\mathrm{UV}.
$$
Typically, the extinction $A_\mathrm{UV}$ isn't measured explicitly (because UV measurements are difficult); rather the extinction is measured in two optical bands, e.g. $B$ and $V$. The extinction $A_\lambda$ at a given wavelength $\lambda$ can then be estimated as
$$
A_\lambda = k_\lambda E(B-V),
$$
where $E(B-V)\equiv A_B - A_V$ is the color excess, and $k_\lambda$ is given by the assumed extinction law, e.g. a Cardelli (1989) law or a Calzetti (2000) law.
The role of the initial mass function
The Salpeter IMF is old, and at that time we didn't have too good data for the low-mass stars (which are faint), so he just fitted a single power-law. With better data, it was eventually found that there were less low-mass stars than previously thought, i.e. it takes less stellar mass to produce a given UV luminosity. A popular choice today is the Chabrier (2003) IMF (~5100 citations), which gives roughly a factor of 1.8 more UV than the Salpeter IMF per star formed, on average. That is, the conversion factor is 1.8 times smaller.
With a Chabrier IMF, you hence get that the relation between absolute magnitude of UV and star formation rate (which is what you're asking) is
$$
\frac{\Psi_\mathrm{Cha}}{M_\odot\,\mathrm{yr}^{-1}} = 7.8\times10^{-29}\,\times\,4\pi d_L^2 \times10^{-(M_\mathrm{UV} + \mu + A_\mathrm{UV}+48.6)/2.5},
$$
where $d_L$ is measured in cm.
Simpler relation
The above formula resembles the one you give, but is a bit strange in that it both contains $\mu$ and $d_L$, which are just two different measures of the same thing. Using the definition
$$
\mu = 5\log\left(\frac{d_L}{10\,\mathrm{pc}}\right)
$$
we can — after a bit of algebra — express the definition of the Oke & Gunn (AB) magnitude as
$$
M_\mathrm{UV} = -2.5\log L_\nu + 51.6.
$$
With this, we can can write the relation as
$$
\frac{\Psi_\mathrm{Cha}}{M_\odot\,\mathrm{yr}^{-1}} = 7.8\times10^{-29}
\,\times
10^{-(M_\mathrm{UV}+A_\mathrm{UV}-51.6)/2.5},
$$
or even
$$
\boxed{
\frac{\Psi_\mathrm{Cha}}{M_\odot\,\mathrm{yr}^{-1}} = 3.4\times10^{-8}
\,\times
10^{-(M_\mathrm{UV}+A_\mathrm{UV})/2.5}.}
$$


Why are telescopes positioned in Lagrange points?

tldr;

  L2 is a very stable thermal environment as well as good instantaneous sky visibility and high observing efficiency.

The main reason space telescopes are placed in an L2 orbit is because L2 is a stable thermal environment. Telescopes in Earth orbit can receive sunlight and earthlight in different directions, meaning that the telescope would have to shield in most directions if it wanted to keep itself cool. At L2, the Sunlight and Earthlight come from the same direction, meaning the telescope has to only shield itself from that direction. This is very important for infrared telescopes, as heat radiation acts as interference to measurements. Of the 6 space telescopes going to L2 listed, 3 are infrared telescopes.

For example JWST, one of the infrared telescopes, has to cools some of it's detectors to 7 kelvin to operate properly.

JWST L2 Halo Orbit

The other main benefit from being in L2 is that being so far away from Earth means that there is a much greater available field of view as the Earth blocks a very minor portion of the sky. This increases sky visibility and observation efficiency.

Other Benefits:
- Ease of communication
- Delta-V requirements low (2-4 m/s per year)


Well, your problem is that you have a sphere. To compensate for the polar declination skew, you just calculate $$\sin^{-1}(\mathtt{rand})$$
Where $\mathtt{rand}$ is a number in an evenly divided -1 to 1 range. Alternatively, you can use another trigonometric function based on what you have available, but you get the idea.

RA can obviously be divided evenly.


Why wasn't CHEOPS data taken during passage through the South Atlantic Anomaly downlinked in this case, resulting in gaps in photometry?

The high density of energetic charged particles causes problems for any instrument system transiting through the South Atlantic Anomaly. Here, for example, is a description from the Hubble Space Telescope WFC3 Instrument Science Report 2009-40 (Barker et al. 2010, "WFC3 IR SAA Passage Behavior") of its effects on HST:

The South Atlantic Anomaly (SAA) is the region where the Van Allen
belts are closest to the Earth and is distinguished by elevated levels
of high-energy particle radiation (protons and electrons) which
adversely affects the scientific instruments of low-earth orbit
satellites such as HST. Short and long term effects may include: high
cosmic ray (CR) hit rates, high dark and background levels, residual
glow and count rates, damage to the detector lattice, reduced charge
transfer efficiency, flipped memory bits, and Single Event Upsets that
can suspend the operations of HST instruments.

Below is a figure from that report, showing three 159-second exposures taken by the infrared channel of WFC3 with the shutter closed. The left and right exposures were taken before and after transiting the SAA (the left image's brighter background is Earth light leaking into the camera). The middle image was taken deep inside the SAA. They note that the cosmic-ray incidence is about 50 times higher in the middle image.

It's standard to not do observations when passing through the South Atlantic Anomaly. From the HST Cycle 29 Primer's "Orbital Constraints" section:

No astronomical or calibration observations are possible (except for
some specialized and infrequently executed WFC3 observations) during
spacecraft passage through the SAA because of high background induced
in the scientific instruments and FGSs.

(While certain types of observations with the WFC3 instrument may be permitted, the document also notes that some instruments can't be used on an orbit that passes even through the edge of the SAA, so different instruments can evidently be affected more or less strongly.)
My guess (knowing very little about the details of how CHEOPS observes) is that "no data were downlinked" means either:

The telescope kept taking data during the time it passed through the SAA, but the data were not downlinked because it was known they would be too noisy to use;
"No data were downlinked" is a compact (and maybe slightly awkward) way of saying, "No data were downlinked because the telescope was not taking data during that time".

Edited to add: A quick look at the most recent CHEOPS Observers Manual reveals this discussion of the SAA

... a target is typically observable whilst flying through the SAA,
however the data taken will show an increased number of glitches and
noise spikes due to the impact of high energy particles. Data taken
during IOC showed that images acquired during SAA passage are not
amenable to photometric reduction due to the huge number of cosmic ray
hits. Therefore, the images taken during SAA passage are not
downlinked to ground in order to save bandwidth.

(So it's #1 of the two possibilities I suggested.)


Can a star emit heat but no visible light?

The minimum temperature of an object classed as a "star" is something like 2700 K. Such an object, although emitting the bulk of its radiation in the infrared, would still emit something like a few per cent of its energy at visible wavelengths.
Note that the visible and near infrared spectra of cool stars are nothing like blackbodies, so if the calculation is important then you would need to look carefully at empirical or model spectra.
Lower temperature objects do exist - these are called brown dwarfs. Some have (a subset called Y dwarfs) been found that are only a few hundred kelvins, so a planet could orbit not far above their surface and receive a kW per square metre of infrared light.
A plausible scenario might be a brown dwarf of spectral type T7 (a T-dwarf), with a temperature of about 1000 K (e.g. Burgasser et al. 2002). Such an object has a luminosity of only $\sim 10^{-5} L_\odot$, so a planet would have to orbit at about 0.003 au (about 0.5 million km) to receive the same bolometric flux as does the Earth from the Sun. Such objects cannot be described as "invisible" at visible wavelengths, but a typical $R-J$ (red minus near infrared colour) is about 9 or 10 - indicating that there is thousands of times more flux in the near infrared than there is in the red part of the spectrum. An example of the spectrum of such an object is shown below (GL229b is a T7 brown dwarf). Note how the spectrum is nothing like a blackbody.

Edit: There is likely a limit to how close a planet can get to its parent body, depending on a comparison of its density with the density of "star" it is orbiting. Too close and it will break up due to tidal forces, but for an Earth-like planet, whch would be less dense than a brown dwarf or of similar density to a very low-mass star, then that limit is either about a stellar radius above it or very close to the surface respectively.
Edit 2: NB cool white dwarf stars might be considered, but the oldest, coldest ones in the Galaxy are still around 3500 K and their spectrum encroaches significantly into the visible band. It is possible though that the most massive white dwarfs ($>1 M_\odot$) in old binary systems could be cooler than 3000 K, though no definitive detection has been made (e.g. Kaplan et al. 2014).


How big would the asteroid belt planet be?

The mass of the asteroid main belt is estimated at 4% the mass of our moon according to Wikipedia so any object formed from the aggregation of that mass would not be a planet.

It would be the size of a very small moon.

Even if all the asteroids in the solar system were combined, the total mass would be below a third of the moon's mass.


We have apparent magnitude of stars, planets, DSOs.

That's true, and these are calculated from the total integrated brightness across the apparent size of the object for an observer on Earth.

What about the apparent magnitude of Milky Way?
Can it be estimated?

There are two options if one wants an apparent value:

Try to get the total integrated brightness of the milky way as seen from Earth, either integrated over the whole celestial sphere or by choosing a place on earth, season and time of night and using that part, and converting that to a magnitude
Switch over to a surface brightness approach, i.e. convert a brightness per unit solid angle to a magnitude per unit solid angle.

While option 1 can be executed in a straightforward way, and might be useful if one is trying to photograph a scene on Earth lit by the Milky Way in a similar way to how one might light a scene by blue sky brightness by blocking out only the Sun, it's not very helpful if one asks how bright the Milky Way appears, because it is huge compared to a tiny fuzzy nebula or planet, stretching across the whole sky rather than being a barely-resolved spot.
Option 2 is what's normally done for a large extended object. This large table of apparent magnitudes only includes objects of small angular size, the largest being that of the Sun or Moon (each roughly 30 arc minutes in diameter as seen from Earth).
Wikipedia's Sky_brightness; Relative contributions explains some of the concepts required for this kind of calculation.
                      Surface brightness [S10]
Airglow                     145
Zodiacal light               60
Scattered starlight         ~15

and S10 is the brightness of a 10th magnitude star smeared over the solid angle of one square degree.
So to "Can it be estimated" my answer is yes!
From Wikipedia's Milky Way; Appearance:

The Milky Way is visible from Earth as a hazy band of white light, some 30° wide, arching across the night sky.[38] In night sky observing, although all the individual naked-eye stars in the entire sky are part of the Milky Way Galaxy, the term "Milky Way" is limited to this band of light.

and

The Milky Way has a relatively low surface brightness. Its visibility can be greatly reduced by background light, such as light pollution or moonlight. The sky needs to be darker than about 20.2 magnitude per square arcsecond in order for the Milky Way to be visible.41
Crumey 2014 open access in MNRAS: 41Human contrast threshold and astronomical visibility also readable in arXiv

Further reading here in Astronomy SE related to apparent magnitude and smaller objects in contrast to something as large as the Milky Way:

What is the difference between albedo, absolute magnitude or aparent magnitude?
Venus' magnitude during inferior conjunction
How (the heck) was this photo of Venus at inferior conjunction (between us and the Sun) taken?
Can you see the difference with the naked eye whether Venus is at the other side of the Sun or at the Earths' side?


how far away can close stars diverge with time?

Only one force counts on galactic scales: gravity.

Stars like Alpha Centauri are orbiting in the galaxy, as is the sun. Both stars are moving in a similar direction and at roughly the same speed, relative to the galactic centre. The distance to such stars will vary slowly, but both stars will remain within the galaxy, and the stars will stay fairly near each other for some time yet. The direction of stars doesn't change suddenly, but the gravity of the galaxy will pull the star's path into a roughly circular orbit within the galactic disc.

High-velocity stars are still affected by the gravitational pull of the galaxy. The galactic escape velocity is over 500km/s (1800 ly per million years) Stars can have sufficient speed to escape from the galaxy's gravitational pull. Such stars can escape from the galaxy into intergalactic space. They will be decelerated by the gravity of the galaxy, but if their speed is enough, then gravity won't be enough to pull them back. Stars that are moving more slowly than that will pass into the galactic halo, before being pulled back by the gravity of the galaxy. They will then usually orbit in the galactic halo, with an orbital period measured in hundreds of millions of years.


Do all planets rotate in the same direction in relation to each other?

This Q can be split into two:

Do all the planets in a stellar system always rotate around the star[s] in the same direction as the star[s] itself/themselves?

Question #1 is a duplicate of Why do (most of) the planets rotate counterclockwise, i.e. the same way the Sun does?. As you mentioned, calling it counter-clockwise is simply due to the convention of North on top.

How did the convention of North being the top of maps and globes get established?

Question #2 is about earth science history or the history of cartography. I encourage you to ask that as a separate Q either on the Earth Science Stack Exchange or on the Geographic Information Systems Stack Exchange.

In many cultures, East was/is put on top, as a nod to the rising sun. I imagine if the convention had come from the Southern hemisphere, South or East would have ended up on top.


This answer is conjecture, but typically when an object has a nickname that identifier is used frequently as an alternative to the index number. In that case, you would expect a brief search through a page or two of Google to find at least one instance of the object referred to by that name. As there is none, it is likely there is not a common nickname for IC 1101.


Theoretical sudden appearance of a black hole in BBC drama show *spoiler warning*

Firstly the show is set in London and everyone seems amazed when the image below rises from the dusk yet clearly it's been in existence for the time taken to create the spiral (hours at least?) so nations east of the UK would have seen it for that time and reported on it globally surely?

And we've kind of a large number of space based satellites that do nothing but monitor the Sun and all the people monitoring them would certainly spot this minor event.

  My real question is more around is there any potential scenario where a black hole of that size could develop suddenly and get to that size quickly enough for that image to be a surprise ?

About the only way I can think of is someone creating a Kugelblitz, although it's pretty far fetched.

Maybe a very freak collision of two high mass objects head-on, but that would create far worse problems - essentially an absolutely huge explosion and we would not be looking at the result quietly afterward.

  My understanding, very much as a scientific-layman, is that a black hole of that size would have to have the mass of at least one of our suns to get to that size (probably more) - is that a valid assumption ?

There's a minimum size for the formation of a black hole by gravitational collapse of a star and it's larger than our Sun's mass.

There is, however, no theoretical minimum for the size of a black hole if you can make one some other way.

  If so where has the mass come from ?

Ask the writers. :-)

  Also wouldn't the sudden existence of something else with at least the same mass as our sun suddenly appearing at the centre of our solar system have enormous gravitational impact on everything in the system, including Earth?

Yes that's a major issue.

Any black hole large enough to start siphoning off the Sun's outer layers would need a significant mass and it would have a significant impact on our orbit - and do so more or less immediately.

Our orbital speed is enough to maintain our almost circular orbit around the Sun.

With something as massive as a small star added close to the Sun, our orbital speed would no longer be enough for a circular orbit and we'd be in an elliptical orbit about the Sun-BH pair.  It's more complicated than than that, but that's the broad strokes.

But that ellipse would be with our current distance as an aphelion, whereas we'd have a smaller perihelion distance.  So we'd get warmer.  And it would not take much to raise temperatures here uncomfortably.  Very roughly if we were to have a closest approach of 0.72 AU (Venus's current orbit) we'd find our temperatures rise by perhaps 24 degrees Celsius, which would be enough to make the planet unlivable for us - ( maybe 60 C at the equator and enough to melt significant parts of the polar ice and permafrost ).  And as this change would happen every six months as we orbited moving from close in to far out, we'd be permanently shifting climate.  And that ignores greenhouse effects - we could end up with a runaway greenhouse effect that raised temperatures to lethal levels everywhere.

  We'd have massive volcanic/tectonic impact wouldn't we?

That's a hard one to call.  This would depend on how far apart they Sun and the BH were and how massive the BH was.

But we'd have lots to worry about.


How do people measure the distance between the Earth and The Moon?

There are many ways and I'm not entirely sure who you mean with "how do people measure the distance" (does this exclude space observatories like e.g. Clementine, probes currently in lunar orbit like e.g. Lunar Atmosphere and Dust Environment Explorer a.k.a. LADEE, or any other currently available technology?), but one interesting and extremely precise way is by laser ranging, pointing them towards one or more of the retroreflectors that were left on the surface of the Moon by the Apollo landers, in what is known as the Lunar Laser Ranging Experiment. Since they are laid on the Moon's near side that is always pointed towards the Earth, these retroreflectors are available for measurements to any properly equipped researcher, no matter which country they're from.

By measuring the time it takes for the light to reflect back, we can infer distance by simply multiplying the c (speed of light in vacuum) with the time taken between the light signal being transmitted and received and then dividing all of it by two to get a single leg of the distance. I'll let you read the linked to Wikipedia page on LLRE for more information, but if you find that too boring, here's a YouTube video on Mythbusters: Moon Hoax Retroreflectors that explains it from the practical perspective.

As for results, again quoting that same Wikipedia page on LLRE:

  Some of the findings of this long-term experiment are:

  The Moon is
  spiraling away from Earth at a rate of 3.8 cm (about 1.5 inches) per year. This rate
  has been described as anomalously high.
  The Moon probably has a
  liquid core of about 20% of the Moon's radius.
  The universal force
  of gravity is very stable. The experiments have put an upper limit on
  the change in Newton's gravitational constant G of less than 1 part in
  $10^{11}$ since 1969. The likelihood of any "Nordtvedt effect" (a
  composition-dependent differential acceleration of the Moon and Earth
  towards the Sun) has been ruled out to high precision, strongly
  supporting the validity of the Strong Equivalence Principle.
  Einstein's theory of gravity (the general theory of relativity)
  predicts the Moon's orbit to within the accuracy of the laser ranging
  measurements.

So these distance measurements are so precise they're a lot more interesting and reliable than merely telling us how far away the Moon is. For one, it also blows any Moon landing hoax theories right out of the water. Links to many more official results can be found in references and external links sections of Wiki on LLRE.

There is also a plethora of other ways to measure the distance to the Moon, with different precision and equipment requirements, but I will let others add some other methods in their answers. If however you had some specific "people measuring the distance" equipped with specific measuring devices in mind, don't forget to mention that in your question. ;)


The equator (Earth's rotation plane) and ecliptic (Earth's solar orbit plane) cross at the vernal equinox ♈, the zero point of both right ascension and ecliptic longitude.
Relative to the ecliptic, the Earth's north pole is tilted toward ecliptic longitude λ=90°, where the Sun appears at the June solstice.
Constellations in that general direction are highest around midnight in December and January: Auriga, Orion, etc.
The Earth's south pole is tilted toward ecliptic longitude λ=270°, where the Sun appears at the December solstice.
Constellations in that part of the sky are highest around midnight in June and July: Sagittarius, Ophiuchus, Hercules, etc.
Even though axial precession changes the celestial poles' orientation relative to the stars in a cycle of about 26,000 years, their ecliptic longitudes with respect to the equinox of date are constant.
The north and south ecliptic poles precess by much smaller amounts and have constant right ascensions α=18h and α=6h respectively, relative to the equinox of date.
Stellarium might help you visualize all this.
In the View window, Markings tab, you can ask it to plot the equator, ecliptic, and celestial and ecliptic poles of date.


Our universe the surface of a 4-dimensional sphere?

The surface of the 4-dimensional ball (called 3-sphere) is a slice through the universe as a whole for a fixed cosmic time. This slice describes just three spatial dimensions. The observable universe is a tiny part of this 3-sphere; hence it looks flat (3-dimensional Euclidean space) up to measurement precision (about 0.4% at the moment).

Adding time makes the universe 4-dimensional. The observable part is similar to a 3+1-dimensional Minkowski space-time. The universe as a whole may be a de Sitter space-time. A de Sitter space-time is the analogon of a sphere ( = surface of a ball), embedded in a Minkowski space, instead of an Euclidean space, but it's not literally a sphere.

If time would be taken as an additional spatial dimension, the de Sitter space would ressemble a hyperboloid of revolution, if embedded in a 5-dimensional hyperspace.
The difference to an Euclidean space is due to the different definition of the distance: In a 4-dimensional Euclidean space the distance between two points is defined by
$l = \sqrt{\Delta x^2+\Delta y^2+\Delta z^2+\Delta t^2};$ for a 3+1-dimensional Minkowski space it's $l = \sqrt{\Delta x^2+\Delta y^2+\Delta z^2-\Delta t^2}.$ For simplicity the speed of light has been set to $1$.

This model of the universe as a whole can hold, if it actually originated (almost) as a (0-dimensional) singularity (a point); but our horizon is restricted to the observable part, so everything beyond is a theoretical model; other theoretical models could be defined in a way to be similar in the observable part of the universe, but different far beyond, see e.g. this Planck paper.


Why does radio astronomy offer higher resolution images than optical?

The relation you cited holds for a single telescope. But, as also noted in the lecture you linked
"One thing that is possible in radio astronomy is to use interferometry, which combines the signals from an array of antennas as if they were all part of the same aperture. That means that the resolving power of a radio telescope is not just what it would be for each antenna alone, but is as for a single telescope the size of the separation between antennas."

This basically means you use multiple telescopes - which in case of radio astronomy are called antennas for historical reasons - and combine their measurements into a single one. To get an intuition of how big of arrays we are talking about, google images of ALMA, LOFAR (which has stations in several countries) or SKA.

In principle other wavelengths, such as optical, can also be observed using interferometry, but it is technically much more complicated to build such instruments. On top of that, optical light is the one most affected by the atmosphere and it is hard enough to take it into account for one telescope, combining such effects to create an image from multiple ones is a very daunting task.


Is a spotting scope or binoculars a better choice for astronomy?

The main point is that it all depends on the optics quality. Nikon is a good brand, but $300  for a 50mm binocular is way too much. Remember the golden rule while buying any kind of telescope or binocular: Aperture matters, not the Magnification. In a binocular, say "7x50" models, you get 7x magnification with 50mm aperture, which restricts how far you can see. But the Celestron 80mm scope with a 20x magnification is perfectly fine. It would be advisable if you can get hold of a good tripod, get it. The beauty of the heavens is unimaginable; they shall unfurl as you explore. A 80mm scope will let you see even Dark Sky Objects like Nebulas and Distant Galaxies.

In a nutshell, it would be best for you at this stage to grab the Celestron 20x80 scope, which is both better and less costly.


How bright are geostationary satellites due to reflected sunlight?

Have a look at Is it possible to capture geostationary satellites with DSLR?

See the excellent advice in this answer to Are any geosynchronous satellites visible with the naked eye?

No, but they are easily seen with a small telescope on a sturdy mount. March and September are the best times. Use an app to help you. My favorite way is to keep M11, the Wild Duck Cluster, in view with a medium power eyepiece. Every few minutes, a "star" will slowly track through the southern edge!

It's a fun and instructive exercise to calculate the declination of an equatorial GEO orbit from a given latitude!

GEO satellites can be quite large, and while they are said to be +10 to +14 magnitude they can flare far brighter due to flat surfaces, both solar panels and thermal radiators. From http://www.satobs.org/geosats.html

Typically the satellite will be in the mag. +11 to +14 range (or dimmer), but brightening by several magnitudes when the geometry is favourable (around mag. +5 to +6 is not untypical). One satellite is reported to have briefly been visible to the naked eye at mag. +3!

See Richard Huziak's (Saskatoon Centre (contact info in link) excellent tutorial Observing Geostationary Satellites reprinted from Saskatoon Skies:

...I found three other satellites, all between magnitudes 13 and 13.5, each a degree or two from the last! Chances are that one of the satellites is Canada’s Anik C1, although I could not identify it for certain.

From Sky and Telescope's How to See and Photograph Geosynchronous Satellites

Many geosynchronous satellites shine between magnitudes 10–12, so you can spot them in telescopes as small as 4 inches. They're also easy to photograph. High ISOs and fast, low light lenses aren't necessary, just a camera capable of a several-minute-long time exposure — long enough for the stars to trail, so you can easily tell them apart from the satellites. Set your shutter speed to "B" and ISO at 400. You can hold the shutter button down with your finger, but a shutter release cable is much better and vibration-free. Use a 100–200-mm telephoto lens, focus sharply, and expose for 2–4 minutes. When you enlarge the image, you'll should see long trails and a line of pinpoint dots — satellites!

See and enjoy the videos of GEO satellites colliding with space junk linked in Are commercial communications satellites in GEO being constantly monitored by telescopes?
From this answer to Are there any satellites in geosynchronous but not geostationary orbits? we can see that while many objects in GEO have a well station-kept equatorial orbit, some are allowed a small inclination and others can have quite a large one!

Large communications satellites can have large flat surfaces that can dramatically increase their visibility at certain oriatiantations. These include solar panels that are often articulated to always point at the Sun, and large thermal radiators that have high thermal emissivity (are dark) in thermal IR yet highly reflective (mirror-like) in visible to avoid solar heating. For more on that see answers to What are these very large, square panels on Inmarsat 5? and What is the function of this array of what looks like mirrors on TESS? and some cool photos of spacecraft in question.

For some math, see answers to At what magnitude would adjacent large geostationary communications satellites see each other? and When would we detect a tiny meter size natural satellite in a geostationary orbits?  and Calculating the apparent magnitude of a satellite

For a truly weird looking communications satellite bound for GEO see Why do these satellite antennas look so weird? (strangely shaped with little white dots…)

From How do commercial broadcast satellites in GEO produce such carefully shaped signal footprints? you can see thermal radiators that are highly reflective in the visible (and near IR)

From How will GOES-R simultaneously point some instruments down at Earth and others sunward?


If Planet Nine indeed exists, how large a impact event there should be to be seen from Earth?

We can compare the brightness of a comet hitting planet 9 with the 1994 impact of Shoemaker-Levy 9 / SL9 on Jupiter. It was seen by Hubble, and Galileo which was orbiting Jupiter at the time.
View from Galileo of SL9-W, a 49 second event involving a 350m comet fragment at 215,000km/h:

info from NASA:

_Galileo RADIOMETER

brightness of L-impact > that of G-impact by 20 percent   * G impact about 15 percent of total light from Jupiter at
maximum

_CCD (late 1980's tech, 800x800) time resolution 2.5 sec

data returned for K, N, V and W impacts   * W-impact clearly seen as light point with 1 percent of total luminosity of jupiter, and afterwards the fireball  roughly constant during 49 secs at 15 percent of total Jupiter intensity.

_UV Spectrometer at 292 nm

G-impact seen at 07:33:32; 20 percent increase in total
Jovian brightness   * secondary impact at 07:34:36, 1/4 as intensive as G
itself

The fireball events from SL9 reached 3200km above the cloud deck the total energy released was 60,000,000 Megatons of TNT.
Jupiter is 5 Astronomical Units away and Planet 9 is estimated at ≈500AU away, so an SL9 equivalent impact would be 100 times further away and 100x100 times less bright than SL9 because of the inverse square law, corresponding to 13.6 apparent magnitude difference compared to SL9, which was 20% of Jupiter's brightness.
Perhaps i'm wrong, i'm suggesting a maximum visual magnitude of 13.4 / 11.2 for an SL9 event of comets ranging from 350m / 3.5km for planet 9.
Asteroids are probably very rare at 400AU, otherwise a processor could check historical records for matching flashes of light where Planet 9 would emit single or multiple flashes ranging from 5 to 20 apparent magnitude, conforming to a characteristic signal of a 5-10 second rise, a 30 to 120 second constant brightness and a fade.


How long would a photon take to reach us if it's emitted from a galaxy receeding at c?

Actually, this is a perfectly good question if interpreted in a reasonable way, i.e., interpreted as asking how long will light take to get to us if emitted at comoving age 13.8 billion years from a galaxy whose comoving distance is currently increasing from us at rate c (and the Hubble law with H=70 tells us that comoving distance is 4.3 Gpc or 14 billion LY).  That is not an infinite time, as that is not the definition of the Hubble sphere.

The answer to how long it would take, less than infinity, depends on the cosmology model, and is a little hard to calculate because cosmology calculators (such as http://www.astro.ucla.edu/~wright/DlttCalc.html) tend to focus on light reaching us now-- as that's what we see.  For example, that calculator gives that a galaxy that is receding from us at rate c now would have had to emit light 9.25 billion years ago for us to be seeing it now, but that's not what was asked.

As mentioned, it is not easy to calculate the time it would take light emitted at the Hubble sphere now to get to us, but it is not infinite.  The Hubble sphere would only be the edge of what we could ever see if the Hubble constant were constant with time (as in an accelerating universe completely dominated by a cosmological constant), but even though the expansion does seem to be accelerating, it is not yet completely dominated by a cosmological constant (assuming that is what is causing the acceleration).  Hence, we would indeed eventually see the light emitted at the Hubble sphere now, but it might take a very long time, perhaps 50-100 billion years but that's just a guess without doing a real calculation with a full cosmological model.


All these objects are rocky, also the icy ones given the low temparatures at their usual orbits. Your observation is also correct, that mutual collisions at usual orbital velocities in today's solar system tend to be catastrophic among the remaining small objects (comets, asteroids, KBOs etc); that is even true when you consider that usual collision velocities are smaller than orbital velocities as most objects rotate prograde and only objects in similar orbits collide with approx. differential Keplerian velocities. The objects don't grow anymore today.

The collisions you likely refer to occur during planet formation. Planet formation happens concurrent to star formation inside the star's accretion disk which evolves into a protoplanetary disk around the pre-main sequence star. This protoplanetary disk consists of both, gas and solid particles (mass ratio ~100:1) so that motion of the objects is not in vacuum but influenced by the gas - the smaller the more. The outcome (sticking, bouncing, fragmentation) of mutual collisions depends on what the particles look like (how porous or compact) and what is their relative size which defines for small particles their collision speed. You can extend these physics to larger particles until they reach a size at which their motion decouples from the gas. Then likely processes kick-in which combine the motion of gas and interaction with collective behaviour of these 'pebbles' (like disk turbulence, streaming instability which forms zones or vortices in the protoplanetary disk in which the small bodies can amass and gently collide to form big bodies, etc), so that these pebbles agglomerate to larger bodies, the so-called planetesimals which are of a few km-size like today's comets.

These bodies are big enough that they are only little affected by gas - the bigger the less drag compared to their mass they experience and thus will rotate with Keplarian speed around the being-born star - and catching up with smaller bodies in the same or very similar orbit, so that they gather them to form even bigger bodies. Thus a few 'lucky winners' will experience run-away growth, collecting the smaller bodies in their orbital distance which increases even more the bigger they become as gravity will start to play a role. The star enters main sequence and the emerging solar wind will push out the gas of the disk so that a debris disk, thus big particles, planetesimals and protoplanets remain, entering the phase of late heavy bombardment where many small bodies collide with the bigger bodies, effectively clearing especially the inner part of the solar system from many small bodies (as the collision timescale is particularily short)

Many, many details have been left-out here, of course and in parts it may even be over-simplified.

So in essence: what you see in the videos only works because there is still gas which slows down things - and because stuff is in similar orbits and thus their relative velocity is small so that gravity is NOT overcome by the shattered bodies when they collide.


How to find star location knowing latitude and LST

Welcome !. I think you still need the co-ordinates of the star that you are interested in, its Right Ascension (the sky equivalent of longitude on the Earth) and the Declination (equivalent to latitude). Then as you say, you can use the hour angle (HA) formula:

HA = LST - RA

The Hour Angle tells you how many hours west or east of the meridian your star is (and can be further converted into an azimuth).

For the altitude of the star, you need a bit of trig:

altitude = arcsin(sin(latitude) * sin(declination) + cos(latitude) * cos(latitude) * cos(HA))


Is our central black hole actually at the CG of the galaxy?

In simple terms: yes, the Milky Way's supermassive black hole (SMBH) is at the center of the galaxy, we know approximately where the center is (but not terribly precisely), and we should expect the SMBH to be there.

You can define the Galactic Center using the orbits of stars and gas: what is the (average) center of their orbits? You can also use the spatial distribution of the same (which is probably a bit closer to what you're thinking of as a "center of gravity" or "center of mass"). This is never going to give you an exact answer, and since the galaxy is not perfectly symmetric or unchanging, there will always be some uncertainty and variation.

There's a 2D coordinate definition, which is the point on the sky where Galactic latitude and longitude both = 0. This is intended to approximate the direction toward the GC, but since it was defined back in the 1950s, no one is worried if modern attempts to measure the (3D) center of the galaxy might be slightly different.

As far as we can tell, yes, the SMBH is at the Galactic Center. If a SMBH did form outside the central region of a galaxy, dynamical friction would make it spiral into the center of the galaxy (and damp any radial orbital oscillations as well). This would also cause a SMBH from another (smaller) galaxy that merged with the Milky Way to spiral into the GC and merge with the SMBH there; see my answer to this question.

The SMBH is not the "main player" in our galaxy: it has a mass of about 4 million solar masses, which you can compare with the total stellar mass of about 50 billion solar masses. (And a dark-matter mass of roughly one trillion solar masses.) Nonetheless, there is evidence that SMBHs "co-form", in some sense, with their host galaxies. In particular, the mass of SMBHs is fairly tightly correlated with several properties of the host galaxy (e.g, the mass of the "bulge", or the velocity dispersion of stars in the inner part of the galaxy).

[Edited to fix a typo.]


Instantaneous gravity in equation of motion for ephemerides calculation

Would this not mean that information travels infinitely fast and thus contradict Relativity?

Look more closely at equation 27 in the referenced document. I'll simplify this as
$$\boldsymbol{\mathrm a} = \left(\sum_{B \ne A}\frac{GM_B\,(\boldsymbol{\mathrm r}_B - \boldsymbol{\mathrm r}_A)}{r_{AB}^{\,3}} \Bigl(1 + \text{other terms}\Bigr)\right) + \text{another term} + \text{yet another term}$$
Assuming Newtonian gravity, the "other terms", "another term", and "yet another term" all vanish, simplifying the equation to
$$\boldsymbol{\mathrm a} = \left(\sum_{B \ne A}\frac{GM_B\,(\boldsymbol{\mathrm r}_B - \boldsymbol{\mathrm r}_A)}{r_{AB}^{\,3}} \Bigl(1 + 0\Bigr)\right) + \boldsymbol{\mathrm 0} + \boldsymbol{\mathrm 0}$$
or just
$$\boldsymbol{\mathrm a} = \sum_{B \ne A}\frac{GM_B\,(\boldsymbol{\mathrm r}_B - \boldsymbol{\mathrm r}_A)}{r_{AB}^{\,3}}$$
All of those extra terms in equation 27 in the referenced document are due to a linearization of general relativity assuming that distances are very large compared to the Sun's Schwarzschild radius and that velocities are very small compared to the speed of light.


A Sky Map of Gravity - what would it look like?

The rough answer is: just like the sun makes it hard or impossible to see planets and stars during the day, it dominates the gravity sky.

But there are interesting patterns there if we view the sky using a logarithmic scale.

The above plots are a combination of the gravitational force of the sun, moon and planets, the stars in the Hipparchos catalog, the gravitational field of the Milky Way, and the nearby galaxies. Each component have a somewhat different effect.

The solar system

The force from the sun is shockingly big: 0.0059 N (per kilogram - I will omit that henceforth). At first this might appear impossible since we never notice the force, but remember that the setup for the question is a fixed point in space rather than the free-falling orbit of Earth. Because of the orbit this force exactly cancels.

Note that for spherical objects we only need to calculate the force from the direction of their centre of mass, not for their full extent. Hence regardless of what resolution I use for my map planets and stars will only change one box.

The force from the moon is $3.3198\cdot 10^{-5}$ N, a factor of 179 less. (Why do we experience tides from the moon more than from the sun? Because the gradient for the gravitational field is steeper for the nearby moon than for the remote sun). The other planets have much less effect, from $10^{-6.8}$ N for Jupiter down to $10^{-9.5}$ N for Neptune (and Pluto at $10^{-13.4}$).

For positions I used the Matlab Aerospace toolbox planetEphemeris function. Incidentally, I might well have messed up some of the many different coordinate frames here, so don't use these maps to navigate. The important thing for this question is the magnitudes of the forces anyway.

Stars

I added the field of the stars in the Hipparchos catalog, but they do not show up even on my logarithmic map because they are overpowered by the galaxy at large. If we just plot their contributions to different boxes we get the following map:

Note that the peak force in each box is on the order of $10^{-14}$ N: even Pluto matters more than nearby stars. This map also shows an annoying problem: if we just sum the contributions in each 3.6 times 3.6 degree rectangle in the coordinate system the polar rectangles will be narrow and have smaller area; hence there are fewer stars included and less force. Dividing by $\cos(\delta)$ removes this bias and shows that the stars indeed pull on us roughly equally in all directions.

(I used a rough estimate of their masses: assuming them all to be on the main sequence I calculated their luminosity from their absolute magnitude, and then estimated the mass as $M\approx M_\odot (L/L_\odot)^{1/3.5}$.)

The Milky Way

I modelled the Milky Way using the approximate values and formulas in Carroll & Ostlie (p. 884), essentially a neutral gas disk, a thin disk, a thick disk, a central bulge, a stellar halo plus a dark matter halo. The disks for example have density scaling as $$\rho(r,z)=\rho_0 e^{-r/h_r}e^{-|z|/h_z}$$ while the dark matter halo has the Navarro-Frenck-White profile $$\rho(r)=\frac{\rho_0}{\frac{r}{R_s}\left(1+\frac{r}{R_s}\right)^2}.$$ I integrate these functions along a ray extending from the part of the sky I am interested in. Note that the integral of the force will be $$F=\int_0^\infty \frac{G\rho(\mathbf{x}_0 + \mathbf{d} r)}{r^2} r^2 dr = G \int_0^\infty \rho(\mathbf{x}_0 + \mathbf{d} r) dr$$ since it is really an integral of an ever widening pyramid of area $r^2$: the total force is only dependent on the total amount of mass there. Since these density fields change slowly I ignore lateral variations. Again there should be a correction for declination, but it does not make a great difference in the picture.

The end result is a fair amount of force, especially along the galactic plane. This causes "galactic tides" in stellar orbits and can sometimes rival the force from remote planets in the solar system. Again, we do not notice most of the gravitational field because the sun is free-falling in the galaxy.

Sag A* does not show up in this image because it is not very massive compared to the full galaxy. The force is just $9.3678\cdot 10^{-15}$ N.

Nearby galaxies

I used the NEARGALCAT catalogue to estimate the effect of nearby galaxies. This is in many ways the most dodgy estimate of all. While we know positions well we have weak data on actual masses. I used the estimated masses whenever they existed, and a linear fit between angular extent and mass derived from those to impute masses for galaxies lacking values. I am a bit uncertain if the mass estimates are just photometric or include dark matter; if not including DM, they should be an order of magnitude higher.

Typical forces are on the order of $10^{-19}$ N, although heavier galaxies might pull us with $10^{-15}$ N. They are hence on par with the nearby stars: totally overpowered by the galactic gravitational field and the solar system.

Technically, since galaxies are extended non-spherical objects, I should integrate over areas here. But there are limits to my energy.

In short, the gravity sky has nontrivial patterns if one has gravity-eyes with a logarithmic response.


What happens to a white dwarf past the Chandrasekhar limit?

It's not completely clear what you are asking, but if this is a multi-choice quiz, then the only option that could be correct is (a).

(b) Is not correct, because a white dwarf that just passes the Chandrasekhar mass is comfortably below the maximum mass that is supportable by a neutron star. So neutronisation followed by neutron degeneracy pressure and the strongly repulsive nuclear force between neutrons at small separations ought to be capable of preventing black hole formation.

(c) Is not correct because by definition, type II supernovae result from the collapse of massive stars. Observationally, they are distinguished by hydrogen absorption in their spectra, but since a white dwarf will contain little if any hydrogen, then this is not possible.

(a) Might happen. As mass is added to the white dwarf it will become smaller and denser. It is possible that nuclear reactions (carbon fusion) might begin. Because the white dwarf is supported by temperature-independent electron degeneracy pressure, then the nuclear reactions take off at constant density, but the increasing temperature leads to runaway nuclear reactions that detonate the star as a type Ia supernova.


If you have a date and time you can go to Flightradar24 and recreate all of the planes that flew over the area. An oncoming large body jet can look a lot like a yellow star if it has its "headlights" turned on, and if it is coming in your direction and descending, it can look surprisingly motionless. Since the light is focused forward, it can suddenly dim if the plane changes direction.

The lovely NATS video A Day Over the UK shows a substantial amount of East/West air traffic in that area, including several flights off of the main corridor.

However if you watched it for 20 minutes, this can obviously be ruled out.

Venus and Mars were too close to the horizon at 21:00 GMT in the UK, but I recommend you play with in-the-sky.org a bit. Choose Planetarium. It should know your approximate location based on your internet access so as long as the location is roughly correct the only think you need to do is choose a date and time to recreate, and enjoy! I hear the site works on phones nicely as well.

above: Screen shot fro in-the-sky.org.


Given that gravity bends the path of light, how confident can we be that galaxies are where they seem to be?

The widely known phenomenon termed gravitational lensing is believed
  to curve space, thereby affecting the path followed by light and other
  electromagnetic waves.

and gravitational waves! In fact, the paths followed by light and other massless particles (for the lack of a better word) are called null geodesics and they have in common that $$\mathrm ds^2 = 0, $$ i.e. the 4-distance they experience is zero.

  This effect is associated principally with galaxies and galaxy
  clusters, as it requires considerable mass. But in theory any star or
  star cluster, or black hole, can have this effect to some degree.

One could argue that the Shapiro delay is of the same nature, only temporal rather than spatial. You can observe the Shapiro delay from the sun using radar probes!

  What can we tell for sure about the position of external galaxies, if
  we are uncertain that their light is actually coming to us along a
  straight path:

It is by definition! The path that light takes is straight, even if it seems bent to us in 3-dimensional space. Therefore, galaxies appear where they are because that's where they are. It's just that we project our idea of Euclidean (i.e. unbent) space onto the sky!

I know what you want to ask though: How can we be sure that the sky isn't curved in some insane way somewhere and we're actually looking somewhere else? We do assume that on the whole, space is Euclidean.

Well, on the small scale this might be the case: We might be looking at stuff whilst it is "really" somewhere else (note that it is really there, the "really somewhere else part" is just us imposing our coordinates on the sky!!).

But we know beyond reasonable doubt that the Universe, as a whole, is more or less Euclidean. At least up to redshift z~1000, i.e. about 300,000 years after the Big Bang. We can observe the Cosmic Microwave Background, and it tells us something about the curvature in the sky. We know that there's structures there which have a certain angular size, as we know that they had 300,000 years to form and we know the speed at which they formed. So we can predict their typical size, and it turns out that the angle under which they appear would be bigger if the universe were oblate/hyperbolic and smaller if the universe were closed. They are just right for a flat universe. So we know that on big scales, space is indeed Euclidean. It might not be the case for individual stars, galaxies, etc. but on average, it works. And cosmology / gravitational lensing usually works with averages and statistics.

  Can a galaxy appear to be in more than one place in the sky?

Totally. There's a plethora of cases of multi image lens-source systems. You can usually sort them out by looking at their spectra and checking their time-variability.


Does the new moon appear at the same time around the world or in one region last

If the new moon is measured by moonrise then it's possible that it could land on a different day in different longitudes.   Or, if it's not measured by day, but by astronomical position, it's still possible that it would fall on a different day in different timezones.    180 degrees of longitude should equal roughly a 50% chance of landing on one day before or after (2 days apart isn't possible).

30 degrees of longitude should be roughly 1 chance in 12 of landing on a different day.    More detail is in the possible duplicate question.

Map of timezones.  While the lines aren't precisely straight, each line should correspond to about 15 degrees longitude.


Why is liquid water considered a requirement for life?

The reason for which water is needed for Life as we know it (and we have not been able to successfuly imagine a different model from that of DNA that has ability to self-replicate, react to the ambient, and evolve), is that water is the "universal solvent".

That means that most chemical reactions needed for life happen on a liquid environment, more specifically a water-based environment for most of them.


How can a star be "thrown out of the Milky Way" by Sagittarius A*?

It's very similar to the question noted in the comments which has a nice mathematical answer.

But some comments in your question are incorrect and worth addressing.

  As far as I can see, the escaping star can at most end up with the
  kinetic energy of the other star on top of its own, accelerating it by
  a factor of 2–√ if the stars have the same mass.

The combined kinetic energy remains unchanged, unless there's a collision or some kind of relativistic interaction, but that's not really important.    It's easier to think of a gravitational slingshot interaction between a larger object (like Jupiter) and a smaller object like a Voyager spacecraft.   The kinetic energy and momentum exchanged is equal between the two, but the acceleration to the larger object is negligible but the acceleration to the smaller object can be up to 2 times the velocity of the larger one plus the initial velocity, or 2U + V, which is more than enough to escape both the central black hole and the entire galaxy when the initial velocities are high enough.

See this question for more math.

And you wrote this:

  A "slingshot effect" is impossible because the central black hole is
  stationary relative to the galaxy.

This isn't relevant, because it's not the central object's velocity that matters but the velocity of the stars orbiting the central black hole, which can be very high.   When two stars orbiting the central black hole pass close to one another, then a gravity assist happens and the two stars exchange velocity and momentum.

This can be a hard thing to picture.   It's often described as bouncing a ball off a moving train, the train's velocity (times 2) can be added to the ball's initial velocity.   The problem with that image is that there's a point of impact where velocity is reflected and that moment never happens in an orbit.    The way I like to think about it, is, imagine you're standing on a station and a train is approaching and the train is carrying a very heavy object - say a neutron star, and you throw a ball towards, but just behind the neutron star.

You throw the ball at 100 kph and the train is moving towards the station at 100 kph.   The neutron star on the train sees the ball approaching it at 200 kph, it then swings around the Neutron star's gravity and flys away from the Neutron star at 200 kph, because orbital speeds are conserved in the hyperbolic orbit.

But if the ball is now flying away from the train at 200 kph and the train is moving towards you at 100 kph, the ball, inevitably is traveling in your direction at 300 kph, or 2U + V.

  Does that mean that this effect necessarily involves binary systems
  with very different masses so that more energy is available for
  transfer to the lighter star? Or are there other effects at play?

This is an interesting question and the math is a bit messy so maybe somebody else could do it.   Two stars of equal mass and opposite directions could accelerate each other and one would fly away from the black hole and the other would fly towards the black hole.    I think it's easier with a larger and smaller object where the smaller objects is accelerated and the larger one only moved a tiny bit, but I see no reason why two equal mass objects couldn't exchange significant velocity to each other, changing both trajectories.   That's a good question whether that would be sufficient to achieve escape velocity for one of them.


Convert dates to julian day number in astropy?

The accepted string formats for date-hms are:

Format  Class    Example argument

fits    TimeFITS    ‘2000-01-01T00:00:00.000’
iso     TimeISO     ‘2000-01-01 00:00:00.000’
isot    TimeISOT    ‘2000-01-01T00:00:00.000’

https://docs.astropy.org/en/stable/time/index.html

You can either change one of these, or you can define your own format by deriving a class for the format from the TimeFormat class.
https://docs.astropy.org/en/stable/api/astropy.time.TimeFormat.html#astropy.time.TimeFormat


How to calculate the orientation of planets at current epoch

The IAU WGCCRE (Working Group on Cartographic Coordinates and Rotational Elements) seem to be the authorities on this.
Table 1 in their 2015 report gives the following information for the major planets:

α0, δ0 are the ICRF right ascension and declination of the planet's north pole. If the planet's axial precession is known, these are expressed in terms of T, the number of Julian centuries since epoch J2000.0.

W is the rotation angle eastward from reference point Q to the planet's prime meridian at time d, where Q is the intersection of the planet's equator and the ICRF equator at RA α0 + 90°, Dec 0°, and d is the number of days since epoch J2000.0.

They define a planet's north pole as the one pointing north of the solar system's invariable plane.
This means that W decreases with time for Venus and Uranus.
Their figure 1 appears to indicate Q's RA as 90° - α0, in conflict with the text. I believe the text is correct.
When you are ready to validate your calculations, the S&T Mars Profiler shows which side of Mars faces Earth at a given time.


How does the Sun's gravity have so much force and pull on the solar system? How does it scale?

Many questions can be answered using the vis-viva equation:
$$v^2 = GM\left(\frac{2}{r} -  \frac{1}{a} \right)$$
which gives the velocity of an object in a Keplerian orbit at distance $r$ from a body of mass $M$ and with a semi-major axis $a$. $G$ is the gravitational constant. And for convenience and accuracy, the product $GM$ or standard gravitational parameter for the Sun and for the Earth are 1.327×1020 and 3.986×1014 m3/s2.
For circular orbits set $r=a$ and get
$$v^2 = GM/a.$$
The orbit's circumference $C=2\pi a$ and the time for one orbit (period) is $T=C/v = C=2\pi a / v$ so
$$T^2 = 4 \pi^2 \frac{a^3}{GM}$$
The mass of a sphere is
$$M = \frac{4}{3} \pi R^3 \rho$$
and we're going to keep the density of the Sun fixed at 1408 kg/m3 which is only 41% higher than water! (see At what depth below the Sun's surface does the density reach that of water?) So:
$$T^2 = 4 \pi^2 \frac{a^3}{GM}$$
$$T^2 = \frac{3 \pi}{G \rho} \left(\frac{a}{R}\right)^3$$
or
$$T = \sqrt{\frac{3 \pi}{G \rho}} \ \ \left(\frac{a}{R}\right)^{3/2}$$

punch line: So the period is going to be one year, i.e. about 365 days whether we use the current values for $a$ and $R$ or scale them up or down by any factor!!
In other words, while:

in plain English, in fact a sun-density sized soccer ball will keep a grape-sized object (of the same density as the planet it represents) in the same scale orbit with the same orbital period. This does in fact all scale down.

...is nearly correct. If the Sun were a 22 cm diameter ball with the same 1.4 g/cm^3 average density, and the sesame seed-sized Earth was 47.4 meters away with a diameter of 2 millimeters and the same average density of 5.5 g/cm^2, then it would orbit the soccer ball-sized Sun once a year, unless there were external forces pulling on it from other astronomical object.
Alternatively you could keep the Sun and Earth and all the planets the same sizes and distances but make them a hundred times less dense, and the orbital periods would be $\sqrt{\text{100}} = $ 10 times longer.

This is actually variant of the rule of thumb that the period of a low orbit around a spherical body is just inversely related to the square root of density. So a dust particle orbiting a 1 meter diameter spherical chunk of "average Earth" will orbit in about 90 minutes just like the ISS orbits around the whole Earth in about 90 minutes.
But you can always replace a spherically symmetric mass distribution with a smaller spherically symmetric mass distribution, (even a point).
Not the same, but similar to what's discussed in this answer to Delta-V required for lift-off from a planet/ asteroid


Are black holes expected to contain the same ratio of dark matter to regular matter as the rest of the universe?

(Short answer: No, scroll to the last point.)

It is irrelevant to an external observer whether the matter that fell into the black hole was dark matter or baryonic, by the no hair theorem. The only properties of a black hole from our point of view are mass, electric charge and angular momentum. (But of course we don't understand quantum gravity.)
From the point of view of matter which has fallen into the black hole, nothing special happens upon crossing the event horizon. This means that dark matter stays dark and baryonic matter stays baryonic when viewed from inside the black hole.
There is some controversy about how much dark matter exists in the universe. This recent article, for example, indicates in the abstract that more accurate modelling of galactic rotation curves could eliminate a large percentage of the expected non-baryonic dark matter. (Note as @pela indicated in the comments, that this author's papers have not been peer reviewed and could be suspect.) Obviously, the amount of dark matter in the universe would greatly affect the question's answer. I should note that the controversy is mostly composed of a small number of vocal scientists who appear disproportionately in the media. Following the mainstream news science sections, I get the impression that the death of dark matter seems to be announced once a month or so.
The formation of supermassive black holes is poorly understood. One hypothesis is that they may form by successive merger of stellar mass black holes. As there have recently been gravitational wave observations of such mergers, and as candidates for intermediate mass black holes have also been observed recently, I will assume here that this is how they form and that supermassive black holes are therefore made of roughly the same stuff as stellar mass black holes.
Black holes lose most of their mass during the formation process. It is important to always keep in mind whether we are talking about the mass of the stellar core which collapsed to form the black hole (this is often the "mass" of a black hole that is referred to when speaking about e.g. the minimum size black hole that can form from core collapse) or the mass of the black hole as seen by a distant observer after the supernova.
Dark matter particles can not lose much orbital energy by interacting with other matter nor by radiation, therefore will remain in orbit around a black hole rather than falling in, unless they happen by unlikely chance to hit it near the event horizon. This paper indicates that simulated supermassive black holes derive no more than about 10% of their mass from dark matter.

However, it must be said that some scientists suspect that dark matter is made of primordial black holes in the first place. There is also the theory of MACHOs (Massive Compact Halo Objects), that dark matter is composed of large compact bodies such as black holes, but it is believed by most that this theory can not account for the dark matter in the universe.


Fingers of God effect for galaxy clusters

I'm going to break your question down into two sections.

  Do galaxy clusters have measurable peculiar velocities?

The answer to this depends slightly on what is meant by "peculiar velocity". In the most general sense, the peculiar velocity of some object is the velocity it has with respect to some standard of rest. In the cosmological sense, the peculiar velocity of a galaxy in a cluster is simply any motion that galaxy may have which is not attributable to the Hubble flow. Primarily this peculiar velocity is a result of that galaxy's orbit around the cluster barycenter.

With that in mind, it should hopefully be easy to reason out that clusters themselves will have peculiar velocities. It is highly unlikely that a cluster will be "fixed" in space and only move according to the Hubble flow. A quick google search revealed Bahcall & Oh 1996 which cites an rms one-dimensional galaxy cluster peculiar velocity of $298\pm28\:\mathrm{km/s}$ (based on 22 clusters). While this is smaller than a typical galaxy's peculiar velocity, it is certainly measurable and capable of being distinguished from the Hubble flow (at least for reasonable redshifts).

The real difficulty here is not whether the cluster has a non-negligible peculiar velocity, but rather, is that peculiar velocity even measurable due to many of the potential limitations. While the value of the peculiar velocity of a cluster is certainly "high enough to be taken into account" it can often be difficult to measure such a velocity with any reasonable error (a point that Bahcill and Oh discuss).

  Can these galaxy cluster peculiar velocities produce the "Fingers of God" effect?

To answer this you have to consider a very particular point I made in the last section which was that for galaxies the primary contribution to their peculiar velocity is their orbital motion around the cluster barycenter which can result in peculiar velocities on the order of $10^3\:\mathrm{km/s}$.

When trying to map out a particular cluster in redshift space, you can often get "Fingers of God" in your maps which are a result of the cluster appearing elongated towards the observer and due entirely to the orbital peculiar motion. An example of such an effect is below, with the red circles added by me to indicate these fingers of god.

Galaxy clusters on the other hand are not gravitationally bound to other galaxy clusters! This means any peculiar motion a cluster has is simply its general motion through space. Since the fingers of god are a result of orbital peculiar motion, it stands to reason that galaxy clusters will not have such a motion and as such clusters won't produce these fingers of god. You'll find that more or less, the peculiar motion of clusters will simply move cause them to be randomly re-positioned without any real "structure" that you see in the fingers of god for galaxies.


Is there a general term for epicycles, deferents, and eccentrics in Ptolemaic astronomy?

Orbit can also mean "the usual course of one's life or range of one's activities" so I would find Ptolemic orbit quite acceptable if you hesitate to use just orbit when the path isn't elliptical.


Parent stars of our Sun - Where are its remains?

We have no way of identifying where the Sun was born, what the surrounding environment was, or where the Sun's siblings are right now. This is easy to see from some bare numbers: the Sun's current orbital period around the Milky Way is some 250 million years, and it's been around for some 4.5 billion years, making for some 20 orbits around the galaxy since its birth. There's a lot of stellar-movement dynamics that can happen over those twenty orbits, and any siblings have long parted ways. Similarly, any stellar remnants that are close to us right now are close because we're passing by, not because we were born in their neighbourhood.

As such, whatever the source of the heavy elements in the solar system, it's several billions of years too late to figure out any individual sources.

That said, it is important to emphasize that there is no such unique source to begin with. As explained in the answer you linked to, there are plenty of such predecessors:

  There are grains of material trapped inside meteorites that consist of solids that were already present in the pre-solar material. These are important because these grains were thought to have formed in individual stellar events and their isotopic compositions can be studied. These tell us that the Sun formed from material that has been inside many different stars of different types.

Moreover, if you want to place those predecessors, you are fighting against the internal mixing of the interstellar gas on a galactic scale:

  Mixing in the interstellar medium is reasonably effective. The material spewed out from supernovae and stellar winds 5-12 billion years ago has had plenty of time to mix throughout the Galaxy before the Sun's birth. Turbulence and shear instabilities should distribute material on galactic length scales in a billion years or less.

This means that the Sun was not born out of the ashes of its neighbours. Instead, it was born out of the ashes of stars that might have died on the opposite side of the galaxy, several billion years before the Sun's birth, whose ashes then got thoroughly distributed by the mixing of the interstellar medium.


I'll touch on a few of these because I disagree with some of those ideas.

  There are approx. 2000 planetary systems discovered, and all of them,
  except the Solar System, have angular momentum mostly in a star

There's a few problems with that argument because planets close to their star are easier to identify.  We don't have a complete picture of the solar systems around other stars, so we don't know the full answer to this.

  Earth is the only planet in the Solar System that has continental
  crust and oceanic crust

Again, we don't know.   I think it's interesting that Earth has entire continents of mostly granite and Mars/Venus have almost no granite and maybe that is a byproduct of oceans and water mixing in the crust, or, maybe something else, like the giant impact . . . but we don't know enough about exoplanets to say how common continents of granite might be.    As for plate tectonics, there's some models about that which suggest the size of the rocky planet is a factor.   I think plate tectonics might be fairly common, at least based on models.

  Earth had an Lagrangian satellite (near Earth), comparable to Mars in
  size, which is itself unlikely, since Lagrangian satellites are
  typically much smaller

I have a big problem with this statement because during planetary formation there were perhaps a few dozen planetesimals.    That a planet might have a lagrange point minor planet when planet formation is very young doesn't seem that strange to me, and we also don't actually know if Theia was in a Lagrange orbit, it's often speculated that it was, but nobody really knows.    If Theia was in a Lagrange point, it was only there for about 1/50th of the age of our solar system.   Not all together that strange.

  then some unknown power brought out that satellite from Lagrangian
  point

Orbital perterbations could have done it.   No need for anything so exotic as a wandering or, as I like to call it "just passing through" star.

There are reasons to think that life friendly Earth-like planets might be rare.   Earth's relatively stable orbit, though Earth did spend significant periods frozen, it's still mostly been in the right distance from the Sun for a few billion years.   That requires a very stable orbit and the right size star - too big, the star would age too fast.  Too small, red dwarfs get colder over time, not warmer.   I think a star that grows larger and more luminous works well with a photosynthesis planet which removes CO2 over time.

The perfect combination of events to allow life to grow (it took a billion years just to make the first nucleus), might be rare, but I think events like the giant impact would still happen from time to time.

The James Webb space telescope might provide more information on this subject.  I think for now, it's more speculative and the information that can be observed is very limited.

I disagree with some of the arguments but I agree with the overall premise that planets with Earth's combination of seemingly just right circumstances - oceans of water but not covered by water, continents of granite that basically float over the basalt, an atmosphere that's thick enough to protect the planet but thin enough that it can be changed by life on the planet over time, making a partially Oxygen atmosphere possible.  A magnetic field, plate tectonics and pretty impressive long term orbital and axial tilt stability and relatively few globally catastrophic meteor impacts.     Earth got a lot right and that combination of stuff that went right might be pretty rare.    But conclusions based on assumptions are limited.   We need better telescopes to get real answers and some bigger and better telescopes are on the way.


Why is true and average anomaly of planet important?

Given one body orbiting one another, the radial distance between the two and the true anomaly are the polar coordinates of the orbiting body. This wouldn't be all that useful if there was no way to predict where the orbiting body will be at some point in the future.

The mean anomaly along with Kepler's laws of motion do just that. Kepler's first law says that the planets orbits about the Sun are ellipses. His second law says that the rate at which a planet sweeps out area (with the Sun as the central point) is constant. His third law says that the period of a planet's orbit depends on the semi-major axis length but not eccentricity.

This means that a planet in an eccentric orbit and a planet in a circular orbit, both with the same semi-major axis length, will, on average, exhibit the same motion. In particular, their periods will be exactly the same. It's easy to predict where the planet in a circular orbit will be at some point in the future because that planet's true anomaly is a linear function of time. Calling this the mean anomaly, this linear relationship means that
$$M(t) = M(t_0) + (t-t_0)n$$
where $n$ is the mean motion:
$$n = \frac{2\pi}{T} = \sqrt{\frac{\mu}{a^3}}$$
Here, $T$ is the period, $a$ is the semi-major axis length, and $\mu$ is the constant of proportionality in Kepler's third law.

What's needed is a bridge between the mean anomaly and the true anomaly that makes the true anomaly satisfy Kepler's second law. This bridge is the eccentric anomaly. Mean anomaly and eccentric anomaly are related via Kepler's equation,
$$M = E - e \sin E$$
Here, $E$ is the eccentric anomaly and $e$ is the eccentricity. Note that by this equation, it's trivial to calculate the mean anomaly given the eccentric anomaly and the eccentricity. Simply plug the values in the right hand side and evaluate. What's wanted is a means to calculate the eccentric anomaly given the mean anomaly and the eccentricity. Unfortunately, there is no close-formed inverse of Kepler's equation in the elementary functions. Fortunately, there are many ways to solve Kepler's equation for the eccentric anomaly. One easy way is to use Newton-Raphson iteration.

What's still needed is a relationship between eccentric anomaly and true anomaly. This relationship can be represented in closed form:
$$\tan\frac\theta2 = \sqrt{\frac{1+e}{1-e}}\tan\frac E2$$


Models are uncertain

I'm guessing that you are referring to age estimates of very old stars. The age of a star can be determined from certain observables, among one is its absolute luminosity, which in turn requires a precise measurement of its distance. Even today, the uncertainties involved yield estimates with mean values longer than the age of the Universe, which of course is impossible. For instance, the age of the star HD 140283 was found as late as in 2013 to be $14.46\pm0.8$ billion years (Gyr). This is not really in conflict with the estimated age of the Universe, $13.798\pm0.037$ Gyr; all this tells us is that our methods are still not perfect, but that HD 140283 was formed shortly after the Big Bang.

Time dilation is small

General relativity predicts that time evolves more slowly in a graviational field. This is with respect to an observer outside the gravitational field, and since there is no external observer wrt. the Universe, the global time of the Universe can be said to be the "right" time. You're right that in principle the total mass of the Universe makes time run slower than in a hypothetical, empty universe, but almost anywhere in the Universe, the gravitational field is way too weak to cause any noticeable dilation. Close to massive objects, time does run slow. On the surface of a star, the time dilation factor is roughly $(1 - GM/rc^2)^{-1/2} = 1.000002$. Inside the star, the factor is even closer to unity, because only the mass between the center and you contributes, so the farther you dive into the star (kids, don't try this at home), the less the time dilation. Only if you compress the star, so as to squezze it into a neutron star of a black hole, will you get a larger time dilation. For instance, at the "surface" of a black hole, time stops (but again, only wrt. an external observer; for the person at the surface, time evolves as expected)


Python: Find brightest pixel in fits-file and ignore brighter artefacts

Following Mick's comment, I found an astropy-function - DAOStarFinder - that does what I was looking for. It scans an image for light-sources and one can set threshhold for a detection, e.g. the "roundness", "sharpness" or brightness of an object have to be between [-0.05,0.05], [0.6,0.8] or >5*standard_deviation, respectively. I must say, that I have the impression the roundness and sharpness measure don't work consistently and sometimes values for the same object change enormously from one frame to the other. But DAOStarFinder gives you also a rough estimate of the flux density of the object, which I could use for sorting the sources and filtering out the relevant one. Here is a link to example script and to the documentation of DAOStarFinder. And here the example script explicitly:

>>> from astropy.stats import sigma_clipped_stats
>>> from photutils import datasets
>>> from photutils import DAOStarFinder

>>> hdu = datasets.load_star_image()
>>> data = hdu.data[0:400, 0:400]
>>> mean, median, std = sigma_clipped_stats(data, sigma=3.0, iters=5)
>>> daofind = DAOStarFinder(fwhm=3.0, threshold=5.*std)
>>> sources = daofind(data - median)

>>> print(sources)


In my answer to Color of planets I estimated the colour of the Earth as #2f6a69 (based on averaging the colour in various images of Earth, and adjusting for albedo), but noted that it is the hardest of the planets to give a single colour to.

If you were on the near side of the moon with no sun the cones in your eyes would not be fully active: you would be have Mesopic vision. Also your eyes would adapt to the blue light conditions. As a result you probably wouldn't notice blueness of the light much.


If the axial tilt of Uranus (97°) was caused by a giant impact, then why do its moons orbit around its equator at the same tilt (≈97°)?

The obliquity of Uranus is a tricky one to explain, the issue with keeping the satellites in the equatorial plane is one of the problems. Another issue is that giant impacts would be expected to randomise the rotation, yet Uranus has ended up with a rotation period similar to Neptune despite its tilt.

As noted by Morbidelli et al. (2012), tilting Uranus from zero obliquity with a single giant impact runs into the problem that while the circumplanetary disc can realign with the planet's equatorial bulge after the impact, the sense of its rotation would be retrograde. This does not match the present-day system, where the satellites are prograde.

Rogoszinski & Hamilton (2020) investigate the possibility that Uranus was tilted via resonances between its spin, orbit and the other giant planets. This is the mechanism that appears to be responsible for Saturn's obliquity, via a resonance with Neptune. The process would likely be gradual enough to bring the satellites along with it as they tend to align with the equatorial bulge, while avoiding the rotation period randomisation of a giant impact. In the case of Uranus they find that with the right initial conditions, Uranus can be tilted to about 70°, requiring a much smaller impact to push it into the current configuration, so you could avoid having to reverse the sense of rotation of the circumplanetary disc. This has the further advantage that there were likely to be more Mars-sized objects than Earth/super-Earth sized impactors, so such collisions are less improbable.

In their follow-up study, they investigate whether the right initial conditions are compatible with the Nice model for the formation of the Solar System. Unfortunately getting to the current situation through the spin-orbit resonance mechanism appears to take too long.

Currently there does not seem to be a particularly satisfactory scenario to explain the rotation of Uranus, but it does appear that the process of tilting the planet was more complex than a single giant impact, and the satellite system is one of the indications that the simple scenario is unlikely to be correct.