When the Moon formed and early on, did it orbit over the Earth's equator?

Yes, the Moon was likely formed in an orbit over the equator. I'm going to recycle my answer to a different question about the Moon's orbit as most of it is relevant to this one as well.
The main concept here is the Laplace plane, which is the plane about which a satellite's orbit precesses. In general, the inclination of a satellite with respect to this plane gets damped to small values. Close to the planet, the main influence on the Laplace plane is the equatorial bulge of the planet, further away the Sun (and to a lesser extent, the other planets in the Solar System) becomes a more significant influence.
The transition between satellite orbits aligned with the planet's equator and satellite orbits aligned with the planet's orbit occurs around a distance called the Laplace radius $r_{\rm L}$, from Nesvorný et al. (2014), which can be thought of as the scale of a planet's regular satellite system. This distance is given as:
$$r_\mathrm{L}^5 = J_2' R_\mathrm{p}^2 a_\mathrm{p}^3 \left(1-e_\mathrm{p}^2 \right)^{3/2} \frac{M_\mathrm{p}}{M_\odot}$$
where $R_{\rm p}$ is the planetary radius, $a_{\rm p}$ is the planet's semimajor axis, $e_{\rm p}$ is the planet's orbital eccentricity, $M_{\rm p}$ is the mass of the planet and $M_\odot$ is the mass of the Sun.
$J_2'$ is the quadrupole coefficient of the planet $J_2$, adjusted for the presence of $n$ equatorial inner satellites:
$$J_2' R_\mathrm{p}^2 = J_2 R_\mathrm{p}^2 + \frac{1}{2}\sum_{i=1}^n a_i^2 \frac{m_i}{M_\mathrm{p}}$$
In the case of the Moon, there are no inner satellites so $J_2' = J_2$.
The angle between the planet's spin axis and the Laplace plane $\phi$ is given by:
$$\tan 2\phi = \frac{\sin 2\theta}{\cos 2\theta + 2r_\mathrm{L}^5/a^5}$$
Where $\theta$ is the planet's obliquity and $a$ is the semimajor axis of the satellite's orbit. For $a \ll r_{\rm L}$, the denominator becomes very large, resulting in $\phi \approx 0$. For $a \gg r_{\rm L}$, the $2r_\mathrm{L}^5/a^5$ term becomes very small, and the equation becomes $\tan 2\phi \approx \sin 2\theta / \cos 2\theta$, giving $\phi \approx \theta$, which is the behaviour described earlier.
Using the $J_2$ value of Earth of 1.08263×10-3 from the NASA Earth Fact Sheet, the Laplace radius works out to 53600 km, or 8.4 Earth radii. Note that this may not necessarily have been the exact value billions of years ago, due to differences in the Earth's shape caused by rotational spindown, deformations due to ice ages, etc.
The Moon likely formed closer to the Earth than this distance, so would have been aligned with the equator. As it migrated outwards across the transition region, the orbit would have undergone rapid changes in orientation, which would also have an effect on the Earth's obliquity, potentially allowing the Earth to transition from a high-obliquity state which might explain the Faint Young Sun paradox (Jenkins, 2000) to the less-severe obliquity we observe today.
Sarah Stewart-Mukhopadhyay created a video of a simulation of the migration of the Moon across the transition region, with the Earth initially in a high-obliquity state.


How would the sky look if Earth orbited a red giant at a safe distance?

Rayleigh scattering happens at all wavelengths, but the scattering cross section goes as $\lambda^{-4}$.

On Earth, the atmospheric optical depth to Rayleigh scattering is very small at red wavelengths, so hardly any red light is scattered, even at sunset when the Sun is viewed through a thick atmospheric layer. On the contrary, there is sufficient optical depth to scatter some blue light, even if it arrives from the Sun at zenith. Some numbers are that the optical depth at zenith, from sea level is about 0.36 at 400 mm (blue) and ten times smaller at 700 nm (Bucholtz 1995).

However, the spectrum of light that is being scattered is very different in the case of a red giant. The solar spectrum peaks at about 500 nm and is about a factor of two less intense at both 400 nm and 700 nm. A red giant has a spectrum that peaks at around 900 nm (in the infrared), and the flux is about 100 times lower at 400 nm and two times lower at 700 nm (which is why they are called red giants).

If Rayleigh scattering was all that was going on, and the total flux incident at the top of the atmosphere was the same, then the scattered spectrum from the red giant illumination would be quite different. The overall amount of scattered red light would be about the same as in the solar case, but the amount of scattered blue light would be reduced by about a factor of 50. The net effect would be that the sky was much darker, and rather than being dominated by blue light, would actually have a redder spectrum (what colour this would be perceived as, I'm not sure).

But Rayleigh scattering isn't the only thing going on. The optical depth to scattering can be dominated by particulates in the atmosphere at wavelengths above 600 mm. This scattering is much less wavelength dependent, depends on the size distribution of the particles and is much stronger for small scattering angles. I think that this would enhance the relative redness of the scattered light a bit more, but given that the incoming flux at 700 nm is similar to that of the Sun, it wouldn't increase the sky brightness.

In summary, I think the sky would be much darker (factor of 50) and would have a much redder spectrum.


When we say the sun takes 230M years to orbit the galaxy, what is this in relation to?

The Sun's velocity is known (with uncertainties) with respect to the Galactic centre (GC), as is the Sun's distance from the Galactic centre. These measurements have a long tortuous history, which I am not going to attempt to summarise here. However, the motion of the Sun with respect to the GC is established by measuring the proper motion of the Sgr A* source with respect to the assumed zero proper motions of distant radio quasars (see for example Backer & Sramek 1999; Reid & Brundthaler 2004).

The astrometry frame of reference now is known as the International Celestial Reference Frame and is defined by hundreds of compact, extragalactic radio sources.

If you then assume the Sun's orbit is circular and that Sgr A* is at the GC, then the job is done - the relative velocity divided by the distance to the GC gives the angular velocity $\Omega$, and the orbital period is $2\pi/\Omega$ . If you have a model for the Galactic potential then you can do a bit better.


Do main sequence stars in binaries transfer mass by Roche lobe overflow?

Yes. An example would be the Cataclysmic Variable stars (CVs), where the donor star (often in the main sequence), loses material via Roche Love overflow onto a compact white dwarf companion.
The other obvious example (though less common) are the Low Mass X-ray Binaries (LMXBs), where the Roche Love overflow is onto a neutron star or black hole.
Nelson & Eggleton (2000) did a theoretical study of Case A mass transfer (where the donor is hydrogen burning star) and found several subcategories where all the mass transfer takes place whilst both stars are on the main sequence (labelled AD, AR and AS). In all three cases, the endpoint is a contact binary of the W UMa type, where both stars fill their respective Roche lobes.
W UMa stars are very common, but that is because their lives are long, whereas the period over which mass is transferred between the progenitors is very short by comparison, so objects "caught in the act" will be rare. The basic picture is that they start out as short-period, deteached binary systems. But as they evolve, two things happen. (i) If they are of low-mass then dynamo generated magnetic activity can result in angular momentum loss which brings them closer. (ii) Normal stellar evolution on the main sequence leads the stars to get bigger. The two combine and lead to the primary star filling its Roche lobe, transfering mass and accelerating the process towards a W UMa status.
The exception is the AS case, where the transfer occurs on the "nuclear timescale" (e.g. the Sun's nuclear timescale is 10 billion years) and the mass transfer is very slow.
Thus semi-detached main sequence systems are either rare or not doing much, which makes them hard to spot. Nevertheless, they are found if you look in the literature. e.g. Deb & Singh (2011) identify several in a sample of contact or close-to-contact eclipsing binaries, from their distinctive light curve shapes. Other authors refer to these as Pre-Contact W UMa Binaries (PCWBs) and have catalogued some tens of objects (Samec et al. 2013).


Why are telescope mirrors made of glass?

you are partially correct. Glass is used for several reasons. It is a very stable material and will hold its shape well for thousands of years. Glass can also be polished to a high degree of accuracy without having defects.  Another major reason is the expansion properties are very favorable as well. Glass doesn't corrode and is easier to mold into shape than metal mirrors. Glass is relatively cheap and readily available.

Some useful references are found here ,here and here.


Can a material other than glass be used for making telescope mirrors?

Front-surface glass mirrors weren't developed until the 1850s.
Before that, the most common material was speculum metal, a high-tin bronze.
Notable examples include Isaac Newton's 1668 prototype and William Herschel's late 18th century telescopes.
Since speculum metal tarnished easily, it was common to have a spare mirror to continue observations while the other was being reconditioned.
Reflectivity and thermal characteristics were not great either.


Converting the log (Stellar Luminosity/Solar Luminosity) and the log of the plus and minus variation

You're completely right, except that the authors, as far as I can see, don't specify whether the central value (i.e. $\log L = 2.01$, or $L = 102$) is the most likely value (i.e. the mode), or the average, or the median. My guess is it's the average, but especially for logarithmically distributed values, my opinion is that the median would makes more sense to quote, since in that case you'll get the same result whether you take the median of $L$ or of $\log L$.


Tunnel vision while viewing through a reflector

This is completely normal behaviour. What you are seeing is the edge of the projection angle for the lense. This is probably sort of what you're seeing? :

Typically this is more noticable with cheaper eyepieces as the individual elements are smaller, and have a smaller field of view.


what is the highest throughput astronomy project in pixels?

If you refer to "data volume" instead of real image pixels, it would be the Millenium Run (25TB of data):

http://www.mpa-garching.mpg.de/galform/virgo/millennium/

  The Millennium Run used more than 10 billion particles to trace the
  evolution of the matter distribution in a cubic region of the Universe
  over 2 billion light-years on a side.


Could a closely orbiting pulsar be used to study a black hole?

Yes. And we are actively searching for these systems, e.g., with the Pulsar timing array and "soon" with instruments like SKA.

From the Astro2020 Science White Paper on Fundamental Physics with Radio Millisecond Pulsars:

  A broad class of alternative theories invoke mediation of gravity through both tensor and scalar
  fields, whereas general relativity relies strictly on a tensor-field description. A key prediction from
  tensor-scalar theories is dipolar gravitational-wave radiation in compact binary systems with large
  differences in component binding energies (Eardley, 1975).

And:

  Future improvements to tensor-scalar tests will also come in the discovery of pulsars tightly
  orbiting stellar-mass black holes. Shao & Li (2018) recently estimated that a small but detectable
  pulsar/black-hole binary population – between 3 to 80 such systems – resides within the Galactic.
  disk

And about super-massive black holes:

  A key science goal for future radio-astronomical observatories is the discovery and timing of radio
  pulsars in orbit around the supermassive black hole residing in the center of the Milky Way galaxy
  (e.g. Bower et al., 2018). Recent projections have shown that an entirely new class of tests can
  be achieved with pulsar/black-hole orbits shorter than 1 yr in period, even if only one pulsar is
  discovered and yields low timing precision (Liu et al., 2012). These tests will directly probe mass
  and spin properties of the Galactic-centre black hole, as well as the validity of the storied “nohair” theorem (e.g. Will, 2008) with measurements of mass, spin and the quadrupole term of its
  gravitational potential.

So, you see, these source will be great for testing GR in general and studying the structure of BHs (or whether there is any) in particular.

Btw, such gravitational time delays, are often called Shapiro delay .


Can a black hole reach a limit where it can no longer attract more matter?

There is no known upper limit for the mass of a black hole.
A black hole stops attracting matter, when there is no matter within reach.
In this case the (macroscopic) black hole is assumed to evaporate by Hawking radiation over very long time scales.

The estimates for the currently (as of early 2014) known most massive black holes (ultramassive black holes) are below about 40 billion ($4\cdot 10^{10}$) solar masses. A black hole of that mass would have a Schwarzschild radius of about 780 times the distance Earth-Sun, or about 4.5 light-days.

(Hypothetical) micro black holes are assumed to be able to evaporate faster than they can accrete matter.


How strong are the Lagrange Points of Earth?

Because The triangular points (L4 and L5) are stable equilibria, provided that the ratio of M1/M2, moon 0.166 g and earth 9.8 g  so 0.0169 g in L4 and L5


Here's a rough sketch of the Milky Way and the Andromeda galaxy,* showing their approximate sizes and distance from each other to scale:

What the picture (hopefully) illustrates is the incredibly vast gap of empty space — around 2.5 million light years, to be exact — between the galaxies, each of which has a diameter of only(!) around 100 thousand light years or so.

While both galaxies are shaped pretty much like "fuzzy whirlpools" of dust and gas (and, apparently, dark matter), with no precisely defined sharp outer edge, they are still pretty unambiguously separated.  While we may not be able to point to a specific line in space and say that "the Milky Way ends right here", it's still clear that the Milky Way is over here, while the Andromeda galaxy is over there, and there's a huge gap of pretty much absolutely nothing at all in between.

(Of course, there is some stuff even in intergalactic space, including some very diffuse gas, a little bit of dust and even the occasional stray star.  Still, compared to the galaxies themselves — which, from a human viewpoint, are already pretty full of empty space — the intergalactic medium can be pretty well described as empty.)

*) The little specks in the picture around each major galaxy are meant to represent their smaller satellite galaxies, such as the lesser and greater Magellanic clouds. Their relative positions and distances, alas, are probably not very accurate.


Which is the most early type star with a planet discovered by radial velocity method?

HD 103774 b, detected by radial velocity, orbits a star of spectral type F5V according to The Extrasolar Planet Encyclopeadia.

HD 102956 b, detected by radial velocity, orbits a spectral type A subgiant, the star of the "earliest" type harboring a planet detected by radial velocity, I could find (as of May 2014).

The entry for
Kepler-88 c around a spectral type B star (according to the catalog as of May 2014) appears to be incorrect, or encodes something else, the star should be classified as spectral type G6V.


Why does light from objects take time to reach us?

For the very simple reason that light travels very fast, but not infinitely fast.

It takes light about 1 nanosecond to travel 30cm. That is so fast that you don't normally notice it. However it does take some time for light to reach your eyes, even from just across the room.

Objects in space are a lot further away. It takes light about 1.2 seconds to reach the moon, and as you note it takes light 8 minutes to come from the Sun. The reason is that the sun is 150 million km away, and it takes about 8 minutes for light to go that far.

If light did have infinite speed, it would be a weird universe. Effects would happen at the same time as causes, so the basic nature of cause preceding effect would be broken. The "speed=frequency*wavelength" formula for light would be broken. Refraction and diffraction wouldn't exist. A photon would be emitted and absorbed at the same time. Could something that is absorbed at the same time as it is emitted actually exist?


How could I determine what areas of the Milky Way will be visible?

Have you tried to use Stellarium?
You can insert the date of the observation and see how the sky will look like from your location.
 There is also setting to make the Milky Way be seen more intense.


How did TESS scientists choose which stars to observe at a "2 minute cadence" in each field?

A combination of algorithm and by committee.

The majority of targets are bright, low-mass dwarf stars. The primary goal of TESS is to discover planets by transit, and these targets offer the best chance of that. The way this is done is described by Stassun et al. (2017). This paper summarises how an input catalogue is constructed (from many data sources) and used to estimate a luminosity, distance, temperature, mass and radius for each star. The catalogue is substantially complete for stars capable of being observed by TESS, since it is based on the 2MASS all-sky catalogue.

The candidate target list is designed to meet the priorities of the NASA TESS mission - to find small (rocky) planets in $<10$ day orbits using the transit technique. A subset of targets (near the ecliptic poles) are observed for longer, and there, there is a possibility of finding small planets in longer orbits - up to 120 days. The steps that are taken to prioritise this list are (i) remove giants -- too big for appreciable transits to be detected; (ii) avoid crowded sources -- the TESS point spread function is quite large; (iii) the prioritisation of small stars (which give a large transit signal) and of bright stars (which yield more signal-to-noise in the light curves). The final priorities are artificially boosted for those stars near the ecliptic poles  (the continuous viewing zone) and reduced for stars near the galactic plane, where the de-reddening calculations are more uncertain.

However, space in the target list was also reserved for targets selected by the "community".
There were a number of working groups setup (by science area - I was in the open star clusters group).

There was then an open call for proposals last September to bid for which stars to include in the 2-minute cadence list. The bids were mainly from the working groups. A committee (of astronomers) then met to decide which bids/targets were meritorious.


Why does M101 search string have RA variable from 0 to 360 and Dec variable from -90 to +90 in the Hubble Legacy Archive?

I think that somehow that string “M101” got inserted into the search box but has nothing to do with your query results.   For example, if you click the “M101” link right below the box it will insert that into the search box without changing your search results until you click “Search”.
The images shown in your search output are labeled with their object names, RCS0327-1326.
If you’re still getting odd results, the editing your question to show your full URL from your query would be helpful.


How to install the Barlow lens on this telescope?

Based on the length of the optical tube (which doesn't appear to be roughly 1 meter even though you mention it is a 1000mm scope) this is probably a "Bird Jones" telescope and not a typical Newtonian reflector.

  A Jones-Bird reflector telescope (sometimes called a Bird-Jones) is a mirror-lens (catadioptric) variation on the traditional Newtonian design sold in the amateur telescope market. The design uses a spherical primary mirror in place of a parabolic one, with spherical aberrations corrected by sub-aperture corrector lens usually mounted inside the focusser tube or in front of the secondary mirror. This design reduces the size and cost of the telescope with a shorter overall telescope tube length (with the corrector extending the focal length in a "telephoto" type layout) combined with a less costly spherical mirror. Commercially produced versions of this design have been noted to be optically compromised due to the difficulty of producing a correctly shaped sub-aperture corrector in a telescope targeted at the inexpensive end of the telescope market.

Also see Astrowiki

The "Bird Jones" designs use a spherical primary mirror and then use (or attempt to use) a corrector installed in the focuser assembly.  It also acts as a 2x Barlow.  A 1000mm "Bird Jones" design scope would really be roughly 1/2 meter in length.

The optics are usually not great (not as good as a Newtonian) and not a good candidate for using a Barlow (because now you're basically putting a Barlow on a Barlow).  The image will get faint and very fuzzy.

Have you tried using the scope with just a 10mm and no additional Barlow?

Remove the eyepiece (and Barlow if you've inserted one) from the focuser.  When you look through at the secondary mirror... do you see any "glass" in the path (other than the secondary mirror?)  If yes, then you have a "Bird Jones" design telescope and not a typical Newtonian reflector.

While you are certainly able to try using a Barlow, I suspect you wont be happy with the results no matter what you do.  I suspect a Barlow with a 10mm eyepiece would be pushing the magnification well beyond the practical limits of the optical system of that telescope.


The currently accepted answer is not relevant for finding the distance to a star like Proxima Centauri.

Here's how parallax works. You measure the position of a star in a field of stars that are (presumably) much further way. You do this twice, separated by 6 months. You then calculate the angle that the star has moved against its background stars. This angle forms part of a large triangle, with a base that is equal to the diameter of the Earth's orbit around the Sun. Trigonometry then tells you what the distance is as a multiple of the distance from the Earth to the Sun. [In practice you perform many measurements with any separation in time and combine them all.]

The "parallax angle" is actually half this angular displacement, and a star is said to be 1 parsec away if the parallax angle is 1 second of arc. So 1pc is 1 AU/$\tan (\theta) = 3.08\times10^{16}$ m. The larger the parallax, the closer the star.

The Gaia satellite is currently mapping the entire sky and will estimate tiny parallaxes
with precisions of $10^{-5}$ to $10^{-4}$ arcseconds (depending on target brightness) for about a billion stars.

Parallax - as illustrated at http://www.bbc.co.uk/schools/gcsebitesize/science/21c/earth_universe/earth_stars_galaxiesrev4.shtml

Now in reality, it is a bit more difficult than this because stars also have a "proper motion" across the sky due to their motion in our Galaxy relative to the Sun. This means you have to do more than two measurements to separate out this component of motion on the sky. In the case of Proxima Centauri the motion against the background stars due to proper motion is larger than the parallax. But the two components can clearly be seen and separated (see below). It is (half) the amplitude of the curved motion in the picture below that corresponds to the parallax. The proper motion is just the constant linear trend with respect to the background stars.

HST images of the path of Proxima Centauri against background stars. The green curve shows the measured and predicted path of the star against the background field over the next few years.

Parallax measurements work best for nearby stars, because the parallax angle is larger. For more distant stars or those without a parallax measurement, there are a battery of techniques. For isolated stars, the most common is to attempt to establish what type of star it is, either from its colour(s) or preferably from a spectrum that can reveal its temperature and gravity. From this one can estimate what the absolute luminosity of the object is and then from its observed brightness one can calculate the distance. This is known as a photometric parallax or spectroscopic parallax.


Discovery in Astronomy vs one in Physics - do they differ in required burden of evidence?

There is no central authority in science. There is no council that sets the standards. The criteria for a discovery are the same: You publish your findings, and your peers accept your results.

There is the 5 sigma rule in particle physics. Perhaps you were thinking of this. But that is not an official rule, instead it's a convention among particle physics. And it's not applicable to every field.

Things are never cut and dry. What does "Peers" mean. What if only most are convinced. Does it count? Discoveries are usually judged in retrospect. After the dust has settled and the debate s are over.

//edit

Why is the bar set so high in physics?  When "discovering" a particle there is no way to "see" it, instead you observe a mass of data and see a statistical discrepancy. Cern produces masses and masses of data, and it is sifted for anything unusual. And with so much data, the chance of seeing something unusual is actually quite high. (imagine searching for repetitions of '9' in the digits of pi - if you search far enough you can be sure to find a string of 6 9s even though the chance of a random string having 6 9s is very low)

With a mass of data, and the opportunity to repeat experiments it makes sense to set the bar very high.

Compare with an observation, such as the "chirp" at Ligo. We can't repeat it at will, the data is there: something happened that is consistent with a black hole merger. No other theory has been proposed that can explain the observations. The observation is less dependent on a statistical finding after running multiple experiments, but a single direct observation.

In fields in which a result could be explained by chance, then a statistical analysis is done, and published. This contributes to the quality of a finding, and so the number of people you will convince. And convincing your peers is the only criteria that counts in the end.


What cluster of stars is this with a "dark donut" to one side?

OK, having (finally) actually looked at the video, it's clear that Szymanek is looking at the center of M33. There is in fact a nuclear star cluster in the center of that galaxy; not knowing the field of view or the resolution, I can't tell how much of the central condensation is simply the unresolved, smeared-out nuclear star cluster (which is small -- you really need something like HST to resolve it) and how much is that plus the surrounding central region of the disk.

The "dark donut" is undoubtedly the diffraction shadow due to a dust grain on the filter or other optical element in front of the detector; see the "Dust Rings" subsection of this page for an example and brief discussion.

  Dust on either the filter, the window protecting the CCD, or any of the corrective optics will leave little donut shapes on an image like the one below. They appear as rings because the dust grains lie on optical surfaces above the focal plane so when they cast a shadow on the CCD, it is out of focus. Astronomers can measure the size of a dust ring and tell exactly where in the optics the grain of dust lies. Dust grains on the CCD itself leaves little dark spots. Calibration frames completely remove dust spots and rings from images.


Is there a name for a planet and its moons/satellites?

This question has been asked before on the Space Exploration page.
In summary, the term used is system, e.g. the 'Jupiter system'.


How much does the stellar radius change if seen through different filters?

The "radius" of the Sun is basically where the optical depth to radiation at any particular wavelength is about 1 (or some times 2/3 is used). This is known as the photosphere and is from where the light that we measure comes from.

The "thickness" of the photosphere - i.e. the depth over which the optical depth changes from being negligible to very large is perhaps a few 100 km at optical wavelengths, compared with $7\times 10^{5}$ km for the radius of the Sun. So the exact definition is important at the level of about 1 part in a 1000.

The light coming from the Sun in a "band" is a mixture of light coming from a variety of depths, depending on the exact wavelength. So your question is a bit ill-defined. If we are looking a the wavelength of a prominent absorption line, then the photosphere is higher up in the atmosphere than if we are looking at a  continuum wavelength. The difference is as I say, at most a couple of hundred km.

It's a bit more difficult to define a solar radius in terms of X-ray or radio emission. The Sun is surrounded by a corona that emits radiation at both of these wavelengths, though that plasma is "optically thin" to the radiation. If we were to shine a beam of X-rays or radio waves at the Sun, they would be totally absorbed perhaps about 1000km above where the visible photosphere is defined in something called the chromosphere.

However, the Sun's outer atmosphere (chromosphere and corona) does not have a plane-parallel geometry; it is shaped to some extent by the magnetic field and you can find spicules and prominences that are optically thick that stick up to greater heights in the atmosphere.


In outer space, a vacuum isn't what sucks in matter to a black hole. That may be the case for your household vacuum cleaner, but everywhere in outer space is essentially a vacuum. Instead, it's the black hole's powerful gravitational pull.

However, black hole can have energy. One type that comes to mind is angular kinetic energy. This is the energy that a spinning object has - in this case, a rotating black hole (see also Kerr black hole and Kerr-Newman black hole). Black holes can also have translational kinetic energy - the energy of an object in motion. Also black holes have gravitational potential energy - potential energy due to gravity - and black holes with electric charge can have electric potential energy.

So black holes do have energy, although perhaps not in the form you thought.

Keep asking good questions!

I hope this helps.


Will a new planet form if Jupiter's influence on asteroid belt will diminish in a few billion years?

Sometime after 4.5 billion years, the sun will change into a red giant star and expand to about 1 AU. Some people think it will be beyond Earth's orbit. (In any case, Earth's orbit would have changed and likely to have struck the surface of the expanded sun.) The asteroids would experience a stronger solar wind and their orbits would become more elliptical, so Jupiter's influence would be significant.


Does the sun cross other spiral arms in its movement around the galaxy's center?

What is a spiral arm?

The reason that the Sun, in principle (but see below), may cross spiral arms is that galactic spiral arms are not rigid entities consisting of some particular stars; rather they are "waves" with a temporary increase in density. An often-used analogy is the pile-up of cars behind a slow-moving truck: At all times, all cars are moving forward, but for a while, a car behind the truck will be moving slow, until it overtakes and speeds up.

Similarly, stars may overtake, or be overtaken by, the spiral arms. Inside a certain distance from the center of the galaxy called the corotation radius ($R_\mathrm{c}$) stars move faster than the arms, while outside they move slower.

Since the stars and the interstellar gas follow the rotation of the arms for a while once they're inside, the density of the arms is higher than outside, but only by a factor of a few (e.g.
Rix & Rieke 1993).

When interstellar gas falls into the potential well it is compressed, triggering star formation. Since the most luminous stars burn their fuel fast, they will mostly have died once they leave an arm. Hence, what we see as spiral arms is not so much the extra stars, but mostly due to the light from the youngest stars which are still inside the arms. Since most luminous also means hottest, their light peaks in the bluish region — hence spiral arms appear blue.

Origin of the spiral structure

At least the most prominent spiral arms (especially grand designs) are thought to be created by these long-lived, quasi-stationary density waves (Lin & Shu 1964). The reason that the density waves exist in the first place is not well-understood, I think, but may have to do with anisotropic gravitational potentials and/or tidal forces from nearby galaxies (e.g. Semczuk et al. 2017).

But in fact even small perturbations may spawn gravitational instabilities that propagate as density waves. In computer simulations of galaxy formation, even numerical instabilities may cause this, so the fact that your simulated galaxy has spiral arms doesn't necessarily mean that you got your physics right.

When the luminous and hence massive stars die, they explode as supernovae. The feedback from this process, as well as that exerted by the radiation pressure before they die, may help maintaining the density waves, at least in flocculent galaxies (Mueller & Arnett 1972). Perhaps this so-called Stochastic Self-Propagating Star Formation may also initiate the density waves (see discussion in Aschwanden et al. 2018).

The rotation speed of the material in the galactic disk is roughly constant with distance from the center (this is mainly due to the dark matter halo hosting a galaxy). Hence, stars close to the center complete a revolution faster than those farther away. In contrast, the spiral pattern rotate more like a rigid disk such that, in an intertial frame, the pattern can be described by a constant angular speed $\Omega_\mathrm{p}$ throughout the disk. However, note that spiral arms are transient phenomena; they appear and disappear with lifetimes of the order of (a few) Myr (e.g. Grand et al. 2012; 2014). Sometimes you also see multiple spiral patterns propagating with different velocities.

The Sun in the Milky Way

Note: This section first contained errors based on dubious values for the angular speed of the spiral arms, as pointed out by @PeterErwin and @eagle275.

In the case of our Sun, we happen to be located very near the corotational radius $R_\mathrm{c}$; we sit at a distance of $R_0 = 8.32\,\mathrm{kpc}$ from the center of the galaxy (Gillessen et al. 2017), while
$R_\mathrm{c} = 8.51\,\mathrm{kpc}$ (Dias et al. 2019).

Using Gaia data, Dias et al. (2019) find a pattern angular speed $\Omega_\mathrm{p} = 28.2 \pm 2.1\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{kpc}^{-1}$. At the location of the Sun ($R_0$) this implies a pattern speed of $\simeq235\,\mathrm{km}\,\mathrm{s}^{-1}$. This is only a little bit slower than (and in fact statistically consistent with) the Sun's velocity of $239\pm5\,\mathrm{km}\,\mathrm{s}^{-1}$ (Planck Collaboration et al. 2018). If the spiral arms were "permanent"*, the timescale for the Sun crossing an arm would hence be gigayears.

However, because as described above they're quite transient, we might once in a while overtake a spiral arm. I haven't been able to find firm evidence for whether we have or haven't crossed any arms; Gies & Helsel (2005) argue that we have crossed an arm four times within the last 500 Myr, but base this on matching glaciation epochs with passages through spiral arms (and admit that this requires a lower but still acceptable pattern speed).

The article you link to

I now wrote to Jesse Christiansen (who the linked article quotes) and asked her if she knows whether or not we are moving in and out of spiral arms; she replied within roughly 8 seconds, tagging Karen Masters who chimed in even faster — they both agree that this is an ongoing debate with no conclusive evidence.

Anyway, the article seems to have misunderstood the tweet from Jesse Christiansen. In her animation she shows the journey of the Sun, but shows the galaxy itself as being static, which she did on purpose to keep it simple. Hence, you see the Sun traversing the arms unnaturally fast.


Why are the stars in my image so green (false color image, HSO palette)?

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


How do proto-planetary nebulae gain momentum?

The angular momentum comes from the chaotic/turbulent collapse of the parent molecular cloud. The collapse is turbulent because the initial conditions in the parent cloud are not spherically symmetric. This leave the proto-stellar clumps with non-zero angular momentum even if the parent cloud started with zero angular momentum. The sum of the angular momenta of all the material of the original cloud after fragmentation is still that of the cloud before fragmentation.

Some other/external triggering events for molecular cloud collapse can impart non-zero angular momentum to the cloud as a whole, but this is not necessary for the the individual fragments to end up with angular momentum.


Chris, you are actually on the verge of understanding how special relativity works. You're very close. You only need to take one extra step.

  to state that all speed is relative to an object is to essentially say, "speed does not exist, but is only relative to an observer"

This is correct. Absolute space, as in newtonian physics, does not exist. Same for absolute time. Both are just relations between objects, and between events.

As long as you hang on to the notions of absolute space and absolute time, relativity cannot be understood. If you let go of them, relativity actually seems natural.

Einstein simply noticed this:

A) All previous experiments (like Michelson-Morley) have shown that whenever you measure the speed of light, you always get the same result, no matter what.

B) Maxwell's equations for electromagnetic fields (such as light) include the speed of light, and the value of that speed is frame-independent. Which means, they always show the speed of light to be the same, no matter how the observer is moving.

Both A and B tell the same thing: speed of light is always the same. But how can that be, when different observers are moving through absolute space at different speeds?

The answer is - absolute space is not a thing. Space is just a background for a special type of relation between objects, called "distance". Time is just a background for a special type of relation between events, called "duration". But neither space nor time are things, separate from objects and events. Only when you put together a whole lot of "distance" type of relations, space seems to appear.

The really important thing is speed of light. It's a fundamental constant of nature, like the Planck energy, etc. This universe is built in such a way that speed of light and other constants like it remain the same no matter what.

But space and time are derivative notions. They are not primordial realities like the speed of light. So they depend on the observer. For me, the bundle of distance relations we call "space" looks a certain way. For you, the same bundle looks a bit differently. This is okay, because space is not absolute; it's derived from a lot of other things.

When you throw the flashlight, the speed of light from it is the same for everyone. But how can that be, since different observers are moving in different ways? Simple: they all see the bundles of distance relations differently; those differences are such that speed of light always remains the same.

And that's special relativity in a nutshell.


"LinAlgError: SVD did not converge in Linear Least Squares" with eleanor.py

I installed eleonor and ran your code - similar but deleted the line:

eleanor.Update(sector=1)

Because it was giving me an error. And it runs without issues.
Attached I post my output. I’m guessing you must have some problems with the packages themselves. I can't pinpoint what, but my naive advice would be to create a conda environment (I can try and help if you need). In this new and empty conda environment, you install the packages that you import (as you did before), then give it a run.


Why can I see the whole moon during various non-full-moon phases?

The dark parts of the Moon are partially illuminated by "Earthshine". That is sunlight which is reflected from Earth to the Moon. Just like the ground on Earth is lit up a bit by the Moon at night, so is the ground on the Moon lit up a bit by the Earth in the lunar night. And even more so because Earth is much larger in the lunar sky, than is the Moon in our sky. And the surface of Earth is actually more reflective per area unit than is the surface of the Moon.


Tricky to say for sure, but I would imagine it comes about from measurements of the luminosity and inference of the black hole mass in such systems.

The most extreme objects radiate at the Eddington luminosity, where gravitational forces on matter falling into the black hole are balanced by radiation pressure from the heated material closer in.

If infalling mass is converted to luminosity at a rate of
$$ L = \epsilon \dot{M} c^2,$$
where $\dot{M}$ is the mass accretion rate, $L$ is the luminosity and $\epsilon$ is an efficiency factor, which should be of order 0.1; then the mass accretion rate at the Eddington limit is given by
$$ \dot{M} = \frac{4\pi G M m_p}{\epsilon c \sigma_T} \simeq 1.4\times 10^{15}\frac{M}{M_{\odot}}\ {\rm kg/s},$$
where $M$ is the black hole mass, $m_p$ the mass of a proton and $\sigma_T$ is the Thomson scattering cross-section for free electrons (the major source of opacity in the infalling hot gas).

The biggest supermassive black holes in the universe have $M \simeq 10^{10}M_{\odot}$ and thus the Eddington accretion rate for such objects is about $1.4\times 10^{25}$ kg/s or about 2.3 Earths/second or 140 Earths per minute. The difference between this estimate and the one on the wikipedia page could be what is assumed for the biggest $M$ or that $\epsilon$ is a bit smaller than 0.1 or indeed that the luminosity could exceed the Eddington luminosity (because the accretion isn't spherical).

Perhaps a simpler way to get the answer is to find the most luminous quasar and divide by $\epsilon c^2$. The most luminous quasar ever seen is probably something like 3C 454.3, which reaches $\sim 5\times 10^{40}$ Watts in its highest state. Using $\epsilon = 0.1$ yields about an Earth mass per second for the accretion rate.

So perhaps the number on the wikipedia page is a little exaggerated.


Without some reference to compare to, time passing more slowly makes no sense at all. But every clock has its own proper time that measures time along its own worldline.

In cosmological models, the cosmological time is the proper time of a certain kind of ideal observer: one comoving with the Hubble flow. In other words, imagine space filled with observers for whom the universe looks as close to the same in every direction as possible (zero CMBR dipole anisotropy). The cosmological time is what's measured by their clocks.

As such, there is no time dilation... because we defined our time coordinate as that which is measured by the comoving clocks no matter how small or dense the universe is. You can see this in the standard form of the FLRW metric:
$$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\left[\frac{\mathrm{d}r^2}{1-kr^2} + r^2\,\mathrm{d}\Omega^2\right]\text{,}$$
where nothing happens to the time component no matter what the scale factor $a(t)$ is.

How could we compare the rate of clocks now with rate of clocks then? We could have the past clock emit some pattern, say a light at a specific frequency, and then later catch and measure it. If we find a different frequency, we could interpret this as evidence of time dilation. But we could also fix the clocks as our time standard and interpret the frequency shift as the result of a changing scale factor.

We could also arbitrarily rescale the time coordinate any way we like, but there's no point in doing so because it wouldn't change the time experienced by comoving clocks.

  are you saying that we've decided to look at it as an increasing scale factor where clocks are fixed, instead of time dilation - but in fact both are just different point of views?

That was phrased too poorly, so let me clarify. If you're watching a far-away supernova, the collapse and explode process will take a different amount of time in the supernova's own local rest frame than is seen by you. Taken alone, it's up to you whether you interpret this fact as time dilation of your clock relative to the supernova or a stretching of wavelengths due to cosmic expansion, or any combination of both.

However, what we have decided is to use a frame in which the homogeneity and isotropy of the universe is directly manifest, in which case we're not "looking" at a single supernova but all around us. For a general FRW spacetime, we cannot re-interpret cosmological redshift as due entirely to time dilation without spoiling those conditions.

The reason is that homogeneity and isotropy picks out a sequence of spatial hypersurfaces, i.e. a sequence of "nows", snapshots of the universe at instants of time. Then generally the distances between different galaxies are different at different times.


Thanks for your replies. Yes it appears what I saw was Elon Musk's Starlink satellites. I didn't even know they existed, what a sight! Thank you. Chris


Yes, it is a thing.

Weakly Interacting Massive Particles (aka WIMPS) are thought to come in matter and anti-matter forms and have a self-annihilation cross-section in order to produce the "correct" amount of dark matter (in relation to photons) that we measure today.

As the opening paragraph of the wikipedia article on WIMPS states - there are a number of experiments searching for the photons associated with Annihilation of WIMP anti-WIMP pairs, from which it appears that there is expected to be a neutrino and a gamma ray signature or even the presence of anti-matter (positrons) that might be detected in cosmic ray experiments.

Of the other dark matter candidates - whether sterile neutrinos have an anti-particle depends on whether they are Majorana fermions or not. The hypothetical axion I think is its own anti-particle (a chargeless boson) and there are no annihilation-type experiments for their detection that I can find.


Are sulfur and phosphorus also created in the nucleosynthesis of big stars?

It is not true to say that the CNO cycle produces these elements. It is more like a catalytic chain that aids the conversion of hydrogen to helium. Thus pre-existing CNO nuclei are required and all the reactions do is change the balance of these elements because the reaction steps in the chain have differing timescales.

Sulphur is mainly produced by alpha capture (fusion) onto. carbon and oxygen via neon, magnesium and silicon. This requires the high temperatures found near the cores of massive stars (more than 8 solar masses) and occurs fairly late in their lives, not long before the type II supernova that casts a lot of the processed material into space.

Phosphorus is not part of this "alpha chain" (https://en.m.wikipedia.org/wiki/Alpha_process) because its mass number is not divisible by 4 and it has an odd atomic number. However, it is also expected to be produced in massive stars via neutron capture onto isotopes of silicon (see https://arxiv.org/abs/1112.3824 ). This occurs in the neon and oxygen burning shells, late in the lives of massive stars.


What happens to photons that don't interact with something?

Your statement "most of the universe is composed of mostly empty space" is very vague. In reality, the Universe is embedded in diffuse background photon fields, from low energy (like the so-called cosmic microwave background which is relic from the big bang) to very high energy (from the extreme compact objects, like AGNs). Of course, there is also unknown dark matter. It seems that dark matter probably only takes part in weak interaction. Thus, it probably won't interact with photons (the same argument with cosmic background neutrinos). The chance of photon-photon interaction and photon-electromagnetic-fields interactions depends on the energy of the photons. For high energies, it is not low as you expected. Even if a photon gets lucky and doesn't interact with the other particles, still it won't reach the edge of the Universe. The reason is that the Universe is expanding.


LIGO gravitational wave chirp signal frequency

There's a detailed graph of it here:

https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves

What is the most precise reference for your query is the Time-Frequency graphs in black and green, also called waterfall graphs, spectrographs, sonograms and so forth.

On the Y axis, the width of the green smudge is about 30-50Hz, That's the frequency-uncertainty for the TF graph for every wave of the roughly 10 waves that make up the signal.

That's the actual real measurement that LIGO returned, and there are idealized versions here with an uncertainty range of about 5Hz.

If you are interested in sound theory, you can research about Wigner-Ville TF graph, which is the most precise sound analysis mathematics known, and which uses quantum uncertainty estimates to minimize the quantum uncertainty of placing single audio waves in time at a given known frequency. If you have java JVM you can compute a more deeply processed TF graph using sonogram 4 visible speech demo version.

As an synth-reverse-engineering sound scientist, I have to say that the precision of the wave is incredibly clear, it's the same as recording an individual bird chirp from 20 meters away on a clear microphone. Gravity actually has sound, like a bird call, except it's not carried by air compression it's carried by space time at c.

https://losc.ligo.org/s/images/omega_sample.png


How does the Earth move in the sky as seen from the Moon?

Still new at stellarium but here are some quick capture gif  lasting one month. Sorry about the quality- limited to 256 colors for smaller gifs. Date on lower left corner. By the way the sun is of course the brightest and i use it as reference for recording (start record when sun is in frame then stop when it appears again in the same position which is roughly one month)

Location on Moon : Sea of Tranquility

You are looking straight up

Yellow lines are azimuth

second picture:

Zoomed view of earth

One whole day 24 hours (give or take a few minutes)

Location on Moon: Sea of Tranquility

You are looking straight up

Yellow lines are azimuth (gif itself rotated to approximately match the orientation of the 260 degree azimuth line in the first picture)

Answer for Emilio Pisanty comment on whether the oscillation is detectable by eye

stellarium

Field of view 60 degrees (default view when stellarium first opened)

Picture 3

Shows one end of the oscillation near the 200 azimuth line (see yellow arrows)

other end of ellipse near 185 azimuth line

you can use Gemini (lower right corner) for reference (see yellow arrows) from head to crotch of one Gemini twin would be a good reference for how wide the oscillation is (not sure if your sundial can detect the difference)

picture 4

shows other end of the oscilation near the 185 azimuth line (see yellow arrows)

picture 5 shows my stellarium location settings on the moon


Why doesn't Earth's leading tidal bulge cause the Moon to start spinning in the opposite direction than its original spin?

why doesn't Earth's leading tidal bulge (encircled in the green circle 1) pull on the moon's tidal bulge

It does....

leading to a force, and consequently a torque

There's nothing here that will give an ongoing torque.
Because of the Earth's high rotational speed and lag in response to forces, the earth's bulge stays displaced from the direction of net gravitational pull.
You have drawn the moon's bulge directly in line with the Earth's center.  But in fact it would be pointed at the net gravitational pull direction, which includes the earth's bulge.  Since the moon's rotation does not carry its distortion away from that line, there is no net torque in either direction.  (I'm ignoring libration and long-term changes in rotation due to orbit raising).
So, the Earth's bulge is continuously displaced due to rotational mismatch, while the moon's bulge is not.  If the moon's bulge were directed at earth's center, then there would be a temporary torque and the moon's orientation (in this non-inertial frame) would appear to pendulum back and forth, not a consistent torque that would get it spinning.  (Again, I've completely ignored libration effects and assumed a circular orbit for simplifications).


The classification of stars using spectral class is a very useful classification when considering the properties of (the atmosphere of) a star at that moment. If you consider the different stages in human development (embryo, fetus, infant, toddler, etc...), for instance, here one person also continuously changes its class.
So it is not strange/uncommon/forbidden to change a classification of a thing whether it is a human or a star.

It also does not mean that other classifications are not useful. It depends on how you want to use the classification. In literature, concepts such as low-mass star, or Sun-like star are also used.

Classification according to the evolutionary stage (i.e. the type of fusion going on) of a star is also very common.

See the image below from ATNF in which you can see how stars of different zero-age main-sequence mass move along a colour-magnitude diagram. If we follow the track of a star of one solar mass then we get the following subsequent evolutionary stages:

MS - Main sequence stars (spectral type V) fuse hydrogen into helium in the core. When the hydrogen in the core is depleted, the star becomes an RGB star.
RGB - Red Giant Branch stars (V->II/III) fuse hydrogen into helium in a shell around the core while the core is inert helium. At some point (Helium flash) the core temperature becomes high enough to start fusing helium into carbon and oxygen, the star then becomes an HB star.
HB - Horizontal Branch stars (II/III) fuse helium into carbon/oxygen in the core. A shell with inert helium surrounds the core, which in turn is surrounded by a hydrogen burning shell.
AGB - Asymptotic Gian Branch stars (I) have a carbon/oxygen core with several inert or fusion shells around it.

These types (MS/RGB/HB/AGB) are classes in a very useful and common classification scheme.

In the end it just depends on what you want to do with your classification scheme.

NB. The classification of the Sun changes from G2 to M because the outer envelope and atmosphere expand enormously, causing the atmosphere to cool down. A cooler atmosphere means that the Sun changes from a yellow (6000K) to a red (4000K) star. The wavelength, and therefore colour, at which most energy is transmitted depends on the temperature of the star's atmosphere.


How hard is it to find the Sun's "sisters?"

Here are the problems/issues:
Most stars are born in clusters/associations but a cursory investigation of cluster demographics with age reveals that the vast majority of clusters do not survive to old age. The majority either are never gravitationally bound to begin with or become unbound in the first 10 Myr.
The Sun was likely born in a cluster of $10^3-10^4$ stars (Adams 2010). If it dispersed more than 4 billion years ago, then its members would have had time to spread all around the galaxy by now. That is because although the members would be kinematically coherent to begin with, the velocity distribution of stars in the Galactic disc becomes "heated" as stars scatter in the potential of giant molecular clouds and pass through spiral arms (Wielen 1977). They attain a significant velocity dispersion which means they could now occupy positions over a broad swathe of the Galactic disc and be displaced above/below the disc by $\sim 200$ pc. If we argue that the Sun's siblings could be in an annulus, with width $\pm 1$ kpc and thickness 400 pc, then if the solar cluster was $10^4$ stars, then the nearest sibling is expected to be at a distance of 160 pc.
Another way of looking at this: The density of stars in the Galactic disc is around 0.1 per cubic parsec compared to the expected density of solar siblings of $2.4\times 10^{-7}$ per cubic parsec.
Thus only about 1 in 400,000 of the local stars might be a solar sibling.
How could they be found?
Cluster members would have a similar age and a similar chemical composition.
We know the age of the Sun precisely and accurately. We don't have that information for any other star. Age estimates of field stars are imprecise, model-dependent and are of indeterminate accuracy. In the best cases - solar type stars and a little more massive - asteroseismology and the HR diagram position might give an age to about $\pm 0.5$ Gyr. But you need good data to do that and we don't have asteroseismology for the nearest 400,000 solar-type stars (in order to find ONE solar sibling). Even if we did, it would only narrow the candidates down by a factor of 10.
What about chemical abundances? There are now big spectroscopic surveys that have observed large-ish numbers of stars. Again, the most robust data comes for stars like the Sun, so that a differential abundance analysis can be done. The more numerous cooler stars have spectra which are more difficult to deal with and there may be systematic abundance errors when comparing with the Sun. Giant stars are bright, with narrow spectral lines, which is good, but their abundances may have been modified during their evolution. Thus we are limited to looking at solar-type field stars. No spectroscopic surveys (yet) have detailed spectra of 400,000 potential solar siblings.
Chemical abundance may also not be that discriminatory. In the solar neighbourhood, stars have an abundance dispersion of about a factor of two , centered quite close (a bit below) the solar metallicity. Good quality spectroscopic analysis can give the metallicity to 10%, so enough to resolve the distribution, but it would only whittle down the candidates by factors of a few.
Very high quality spectra could look at the detailed abundance mixture, but is only available perhaps for a few thousand stars. The majority also have an abundance mixture that is not very different to the Sun. Thus whilst the evidence is that the detailed abundance dispersion in a cluster is much smaller than the dispersion in field stars (Paulson et al. 2003), it does not look like the Sun is that unusual or has any unique chemical "markers" (Bensby et al. 2014). Some hope may be offered by the detailed mix of s- and r- process elements; the former may be age sensitive (e.g. Jofre et al. 2020), while the latter might reveal some peculiarity associated with contamination by nearby supernovae in the Sun's natal environment. At the moment, although candidate solar twins can be found, it is unclear by how much that narrows down the one in 400,000 figure, especially when you consider that abundance peculiarities may also be imprinted by later accretion events or some process associated with planet formation (e.g. Melendez et al. 2012) or that the abundance dispersion in a cluster may not be exactly zero (e.g. Liu et al. 2016).
In summary, the space density of solar siblings is likely to be very low compared with unassociated field stars. The properties of those siblings are not that unusual compared with typical field stars and we lack precise enough measurements of age and chemical composition to have anything more than a  list of candidate solar twins at the moment. It's like looking for a needle in a huge pile of other needles.


What would people on Earth see during a close comet pass?

Hear and smell: Nothing. The comet and its coma are well outside the atmosphere, no sound or scent can travel through space.

The coma of the comet could be 30000km across, bigger than the moon, and the comet is half the distance to the moon, so the coma could appear about 20 times larger than the full moon. However, it would be very dim. Although its total brightness might be quite high, the surface brightness would be still very low. It is likely that only the central part of the coma would be visible, as a hazy blur, with no clearly defined edge. Some jets might be visible towards the nucleus. In a dark sky it would look quite dramatic.

The nucleus would be only about 4 arcseconds across, however it is close enough to be visible as a star-like body in the centre of the fuzzy nucleus. Telescopes would reveal the shape of the body, but telescopes on Earth would show few details. The comet would slowly track across the sky, clearly moving relative to the stars if you observe it for any length of time. The exact speed would depend on its orbit, but it wouldn't be moving like a meteor, the encounter would take a few hours.

The tail, which might be visible in dark sky locations, would be pointing away from the sun, rather than behind the direction of travel.


First off, you're much more likely to see parsecs, light years, or astronomical units than Planck lengths in astrophysics.

However, for those who like to work in a system of natural units, the Planck length is the obvious choice for the unit of length. This choice has nothing to do with quantizing spacetime. It instead results from a desire to make conversion factors such as the speed of light $c$, the reduced Planck constant $\hbar$, the Boltzmann constant $k_b$, and the gravitational constant $G$ all have a numerical value of one or some other nice number such as $
\frac{1}{4\pi}$.

One way to look at this: Compare the conventional units used in the US and a few other backward countries. These countries have to use $F=kma$ rather than the $F=ma$ used in the metric system. The $k$ is a conversion factor that results from having those conventional units being an inherently inconsistent system. From a modern perspective, force is the product of mass and acceleration; no conversion should be needed.

That said, that the mass-energy equivalence equation expressed in metric units is $E^2=\bigl(m c^2\bigr)^2 + \bigl(pc\bigr)^2$ is a sign that the metric system, too, is inconsistent. With natural units, this expression simplifies to $E^2=m^2+p^2$. Energy, mass, and momentum all have the same units; no conversion is needed.


Maximum disturbance to space time of gravitational waves, and will this affect humans?

Gravitational waves cause a stretch and squeeze in space-time. It has two polarizations Plus and Cross polarization. This serves as an axis for the stretch and squeeze when a gravitational wave hits a body. However, the order of stretching and squeezing is actually really small. It changes space-time by an order of $10^{-21}$m. That's around the diameter of a proton. There are Numerical relativity simulations where you can see for yourself how much the change in space-time occurs for a certain pair of mass collisions.

Check out the Waveforms for SEOBNR, Taylor etc. You might get a good idea.

FYI : Question belongs in physics.stackexchange


Red shift is usually measured for galaxies rather than individual stars. Unless a star has just gone supernova, it's usually not bright enough to be seen even w the world's most powerful telescopes at the distances where cosmological redshift comes into play. Hubble's law operates over large distances; the expansion constant being 67.8 km/sec per megaparsec (3.3 million light years)
Andomeda galaxy (M31) is 2.54 million light years away, and although some individual stars are visible at that distance, the galaxy has a net blueshift of 0.001001 (303 km/sec) due to its peculiar (i.e. non-Hubble) velocity.
Further out is where you start to see redshifts in excess of peculiar motion; and there you need to start using standard candles of some sort.
Cepheid variables when visible, Supernova, globular cluster brightness functions, are all used, as well as some spectral methods: Cosmic distance ladder

Supernova measurements go the furthest out, highest redshift, but are still a bit of a can of worms as far as interpretation goes.


Are stars expected to become dimmer before a supernova?

The connection between the dimming and a putative supernova relies on the interpretation that the decrease in luminosity may be due to circumstellar material, ejected in the years/decades/centuries immediately preceding a supernova. There are several mechanisms that could lead to this sort of mass loss (see slides 24-25), including

gravity-wave driven envelope loss (in red supergiants, during neon/oxygen core burning)
pulsations due to pair-instability in the days to decades prior to the supernova
turbulent eruptions à la luminous blue variables

These then lead to Type IIn supernovae, with narrow lines arising from interactions with the previously ejected circumstellar material. If the dimming is due to extinction by circumstellar dust from eruptions due to any of these mechanisms, it could mean that a supernova is imminent on a timescale of days to years.


Why does the Sun have different day lengths, but not the gas giants?

It's a matter of how "day" is defined.

Wikipedia's article on Jupiter cites this IAU/IAG paper for the length of a Jupiter day.  In it, footnote (e) of table I has the following:

  The equations for W for Jupiter, Saturn, Uranus and Neptune refer to the rotation of their magnetic fields (System III)

The radio emissions of the gas giants have well-defined periodic variations.  These variations are caused by the rotation of the magnetic fields of those planets, and are evidence that they have a reasonably coherent core of some sort that's rotating at a uniform speed.  The periodic variations then represent the rotation speed of that object, which is taken as the rotation speed of the planet.

We're reasonably certain the Sun doesn't have a coherent core.  Measuring the variation of the magnetic field doesn't show a well-defined period, and doesn't provide a useful definition of the Sun's rotation speed.


Is moisture harmful to a Newtonian Reflector?

The tube itself is not that much of a problem, just wipe it with a soft cloth. The screws should be stainless steel, surface rust can be wiped off as well.

But moisture, which contains humic acids from plants and pollen as well as mold damage the reflective surfaces of mirrors. If it is an expensive one, leave it outside or in a shelter until the next day and store it with the main mirror pointing down.  Don't expose it to a temperature change of -10 to +20°C. Do not clean optical surfaces mechanically if your unsure about the coating and the cleansing stuff. Wiping the aluminium surface will scratch it if it has no protective coating.

If it is a particular expensive bucket and you're in an area that experiences much moisture (where i live, water sometimes runs off the tube, really), consider heaters. Or try another day :-)

Amateur astronomer forums like e.g. cloudynights.com have much more info on this.

Hope that helps.

Edit: also: How do I remove fungus from a telescope mirror?


How can I calculate the orbital periods in a binary star system?

If you need a rough estimation, you can sample the positions of the stellar objects from their acceleration, using Newton's law. A full picture is drawn on this Wikipedia page, but basically, given N stellar objects at position P(i), with their respective mass M(i):

$$
M_i.\vec{acc_i} = -G\sum_{n \neq i}^N \frac {M_i.M_n.(\vec{pos_i}-\vec{pos_n})}{\begin{vmatrix} \vec{pos_i}-\vec{pos_n}\end{vmatrix}^3 }
$$

You can then derivate acceleration into speed, then into position using a small enough time delta. With initial positions and speed (which are the trickiest and also funniest to choose), you can simulate a rough N-body system.

On the fun and simulation side, this is what Universe Sandbox aims to present (in advance, sorry for linking to a commercial store, I'm not related to this studio).

Lagrangian points are also interesting to look at when simulating.

However, to simplify things, you may like to think nearly all stellar objects "close" to the binary system would have been eaten by the massive couple over time, and then consider only the barycenter of the pair of stars.

EDIT: although OP was clear, my answer is for N objects. The simplification to N=2, with A and B being the positions of the two objects, results in:
$$
\vec{acc_A} = \frac{G.M_B}{AB^3}\vec{AB}
$$
$$
\vec{acc_B} = \frac{G.M_A}{BA^3}\vec{BA}
$$
And with an iterative process, once acceleration is computed for step k, using known initial conditions for speed and position:
$$
\vec{spd_{k+1}} = \vec{acc_k}.\Delta{t} + \vec{spd_k}
$$
$$
\vec{pos_{k+1}} = \vec{spd_k}.\Delta{t} + \vec{pos_k}
$$

EDIT2: well, another rough estimation, for the period duration (which I misunderstood as "trajectory" from the question). I'd use cylindrical coordinates as reference, for their clean spatial representation, and for the single-angle conversion:

$$
\left( \begin{array}{c} R_k \\ \theta_k \\ h_k \end{array} \right) = \left( \begin{array}{c} \sqrt{pos_{x,k}^2+pos_{y,k}^2} \\ \arctan \left( \frac{pos_{y,k}}{pos_{x,k}} \right) \\ pos_{z,k} \end{array} \right)
$$

...with adequate precaution. Introduce part of that conversion at each iteration, and wait for theta to complete a rotation.

(in advance, sorry for the weird notation, which I just tried to make consistent from my previous posts)


Telescope size to detect Ceres (Newtonian)

Ceres varies between mag 9.3 and mag 6.7 which are below the nominal naked eye limiting magnitude of ~5. The gain in limiting magnitude for something like 10x50 binoculars (~4.5 magnitudes) would bring Ceres even at its faintest above the limiting magnitude of about 9.5.


What do cosmologists mean when they talk about "the running of the spectral index"?

The spectral index $n_s$ describes how the clumpiness of stuff varied on various scales just after cosmic inflation. The primordial power spectrum $P(k)$ of the fluctuations (where the wave number is $k=2\pi/\lambda$ with $\lambda$ being the physical scale), is predicted by many inflationary models to be:
$$P(k) \propto k^{n_s-1}.$$
If $n_s=1$, the fluctuations are scale invariant.

If $n_s$ is not a constant, but instead changes with $k$, that is if
$$\frac{dn_s}{d \ln k} \neq 0,$$
it is called a "running spectral index". And in fact it seems that that $n_s$ does chance with $k$ (see e.g. Cherny et al. 2014).

The term "the running of the spectral index" refers to the quantity $dn_s\,/\,d\ln k$.


Is the influence of gravity instantaneous?

The first question as stated has a rather trivial answer:

  "If the sun magically disappeared, instantly, along with all its influences, how long would it take its gravity to stop having an effect on us?"

Since the Sun's gravity is among its influences, it would instantly stop having an effect on us. That's just part of the magical situation, and doesn't even involve any physics. A bit more interesting is the question without the bolded part.

In general relativity, changes in the gravitational field propagate at the speed of light. Thus, one might expect that the magical and instant disappearance of the Sun would not affect earth for about eight minutes, since that's how long light from the Sun takes to reach Earth.

However, this is mistaken because the instant disappearance of the Sun itself violates general relativity, as the Einstein field equation enforces a kind of local conservation law on the stress-energy tensor analogous to the non-divergence of the magnetic field in the electromagnetism: in any small neighborhood of spacetime, there are no local sources or sinks of stress-energy; it must come from somewhere and go somewhere. Since the magical instant disappearance of the Sun violates general relativity, it does not make sense to use that theory to predict what happens in such a situation.

Thus, the Sun's gravity instantly ceasing any effect on the Earth is just as consistent with general relativity as having any sort of time-delay. Or to be precise, it's no more inconsistent.

  My big question, now, is: "How do we know it's instant?"

It's not instant, but it can appear that way.

  We can't possibly move an object large enough to have a noticeable gravitational influence fast enough to measure if it creates (or doesn't create) a doppler-like phenomenon.

We don't have to: solar system dynamics are quite fast enough. An simple calculation due to Laplace in the early nineteenth century concluded that if gravity aberrated, Earth's orbit would crash into the Sun on the time-scale of about four centuries. Thus gravity does not aberrate appreciably--more careful analyses concluded that in the Newtonian framework, the speed of gravity must be more than $2\times10^{10}$ the speed of light to be consistent with the observed lack of aberration.

This may seem quite a bit puzzling with how it fits with general relativity's claim that changes in the gravitational field propagate at the speed of light, but it's actually not that peculiar. As an analogy, the electric field of a uniformly moving electric charge is directed toward the instantaneous position of the charge--not where the charge used to be, as one might expect from a speed of light delay. This doesn't mean that electromagnetism propagates instantaneously--if you wiggle the charge, that information will be limited by $c$, as the electromagnetic field changes in response to your action. Instead, it's just something that's true for uniformly moving charges: the electric field "anticipates" where the change will be if no influence acts on it. If the charge velocity changes slowly enough, it will look like electromagnetism is instantaneous, even though it really isn't.

Gravity does this even better: the gravitational field of a uniformly accelerating mass is toward its current position. Thus, gravity "anticipates" where the mass will be based on not just current velocity, but also acceleration. Thus, if conditions are such that the acceleration of gravitating bodies changes slowly (as is the case in the solar system), gravity will look instantaneous. But this is only approximately true if the acceleration changes slowly--it's just a very good approximation under the conditions of the solar system. After all, Newtonian gravity works well.

A detailed analysis of this can be found in Steve Carlip's Aberration and the Speed of Gravity, Phys.Lett.A 267:81-87 (2000) [arXiV:gr-qc/9909087].

  If he was wrong, how do we know it's not?

We have a lot of evidence for general relativity, but the best current evidence that gravitational radiation behaves as GTR says it does is Hulse-Taylor binary. However, there is no direct observation of gravitational radiation yet. The connection between the degree of apparent cancellation of velocity-dependent effects in both electromagnetism and gravity, including its connection with the dipole nature of EM radiation and quadrupole nature of gravitational radiation, can also be found in Carlip's paper.


The problem with trying to form a black hole with dark matter is that dark matter can only weakly interact (if at all) with normal matter and itself, other than by gravity.

This poses a problem. To get dark matter concentrated enough to form a black hole requires it to increase its (negative) gravitational binding energy without at the same time increasing its internal kinetic energy by the same amount. This requires some sort of dissipative interaction between dark matter and normal matter (or itself).

The following scenario should make this clear. Suppose we have a lump of dark matter that gravitationally attracts another lump of dark matter. As the two approach each other, they accelerate and gain kinetic energy. The kinetic energy gained will be exactly enough to then separate them to a similar degree to which they started, unless some dissipative process takes place.

An example is to suppose that dark matter is weakly interacting massive particles (WIMPs). WIMPs are gravitationally drawn towards the centres of stars. If the weak interactions happen sufficiently frequently then it might be possible for them to accumulate in stars, rather than shoot through and out the other side.

It has been hypothesised that black holes could be made like this near the centre of a Galaxy, seeded by dense neutron stars. The density of neutron star matter, combined with the enhanced density of dark matter near galaxy centres could result in dark matter accumulation in the neutron stars, leading to the formation of black holes.

Once a black hole is formed then any dark matter that enters the event horizon cannot emerge regardless of what kinetic energy it gains in the process. However, there is still a problem. Material in orbit around a black hole has less angular momentum the closer it orbits. To pass inside the event horizon requires the dark matter to lose angular momentum. Normal matter does this via an accretion disc that can transport angular momentum outwards by viscous torques, allowing matter to accrete. Dark matter has almost zero viscosity so this can't happen.

So building a supermassive black hole from a smaller seed would be difficult, but forming small black holes out of neutron stars might be easier. It has been proposed that a relative lack of pulsars observed towards our own Galactic centre could be due to this process.


What is the LSST's plan to address frequent satellite trails in data?

Has a plan or strategy been outlined or established to reduce the number of bright satellite trails that appear in LSST perpetual sky surveys, and/or to process images to remove the trails or at least flag pixel regions that contain likely trails?

Indeed LSST will be very affected by satellites in LEO. The strategy is to make several observations of the same area to have enought data to cover for the regions photobombed by the satellites and the artifacts generated by them in the sensor.

From the LSST webpage:

  The first group of Starlink satellites are sufficiently bright during dawn and dusk (when LSST would be surveying) that the trail would exceed sensor saturation, generating uncorrectable artifacts in the data. If instead these satellites were painted flat black making them a factor of 25 fainter, satellite trails should be less of a challenge for LSST due to its specific design.In that case LSST's frequent imaging of the same region of sky will provide enough data to correct for unsaturated satellite trails or other anomalies.


star atlas with absolute magnitude and spectral type

BSC5p may be a nice database to start with.

The parallax (check according box) and vmag will provide a basis for the calculation of the absolute visual magnitude. The absolute visual magnitude may be lower (star may be absolutely brighter), if the extinction isn't negligible.

1/parallax is a good estimate for the distance. Take this distance as an approximate for the luminosity distance, if you can assume negligible extinction. Now calculate the visual magnitude of the star at 10 parsec distance to get the absolute visual magnitude.

Example: Alp1Cen (HR 5459) has vmag -0.01, and parallax 0.751 arc seconds. Therefore the distance is about 1/0.751 = 1.33 parsec. We get absolute visual magnitude
$$ M=m-5\cdot (\mbox{log}_{10}D_L - 1)=-0.01-5\cdot (\mbox{log}_{10}1.33 - 1)=4.37.$$

An overview of absolute magnitude and spectral types.

More detailed discussion, and tables can be found in this paper.

More tables.

This book seems to contain a table with values as of 2009, which is cited here.


What liquids can be found in the void space?

No liquid can be completely stable in a vacuum, since all liquids have some non-zero vapour pressure, and so will evaporate at some rate. However some liquids have an exceptionally low vapour pressure, and so can be used in a vacuum.

The vapour pressure of silcone fluid DC705, which is used in diffusion pumps is 2.6e-8, and it is designed to function in a high vacuum.

If a location could be found at which it was warmed sufficiently to remain liquid (in deep space it would just freeze, whereas too close to the sun and its vapour pressure would rise) it could remain in a liquid state for some time.  Not indefinitely but it could be stable for a while.


On analogies between gas and stellar systems

The analogy is rather weak and not really useful.

So-called collisionless stellar systems (those for which relaxation by stellar encounters has no appreciable effect over their lifetime), such as galaxies, can be described by the collisionless Boltzman equation, but never settle into thermodynamic equilibrium (only into some dynamical or virial equilibrium). Thus, the only other systems with somewhat similar behaviour are collisionless plasmas.

Sound, turbulence, viscosity etc are all effected by close-range collisions (not mere encounters) between the molecules. These also maintain thermodynamic equilibrium and a Maxwell-Boltzmann velocity distribution. Stellar systems have none of these processes and their velocities are in general anisotropically distributed and don't follow a Maxwell distribution.

Gases are in some sense simpler to understand, because their dynamics is driven by local processes and because statistical methods are very useful. Stellar systems are driven by gravity, i.e. long-range non-local processes, and intuition from the physics of gases is often very misleading (for example, a self-gravitating system has negative heat capacity -- this also applies to gas spheres, such as stars).

Note also that the number of particles in a gas is much much larger ($\sim10^{26}$) than the number of stars in a galaxy ($\sim10^{11}$), though the number of dark-matter particles may be much higher.


The effects of solar radiation are the main suspect for clearing out the vulcanoid region of whatever was there to begin with. Radiation pressure will tend to blow small dust out of the region. Larger objects tend to get cleared out due to the Yarkovsky effect and the YORP effect.
The Yarkovsky effect occurs as a result of temperature variations across a rotating object and the time lag for regions of the object's surface to heat up and cool down as they move into and out of the sunlight. This affects the distribution of the radiated photons (which carry momentum), and over time will act to shift the asteroid's orbit. Vokrouhlický et al. (2000) estimate that this effect would clear kilometre-sized objects out of the vulcanoid region on a timescale of a few billion years.
A further effect that is relevant is the YORP effect (short for Yarkovsky, O'Keefe, Radzievskii and Paddack, see also the question What is the difference between the Yarkovsky effect and YORP effect?). This affects the rotation of irregular objects like asteroids and can result in them spinning up to the point where they break apart. This is thought to be a significant contributor to the population of binary asteroids. In the vulcanoid region, this would be a mechanism for breaking up larger asteroids into fragments small enough for the Yarkovsky effect to rapidly clear them out of the vulcanoid region: Collins (2020) puts this right in the title: "The YORP Effect Can Efficiently Destroy 100 Kilometer Planetesimals At The Inner Edge Of The Solar System".

... The YORP effect destroys Vulcanoids by spinning them up so fast that the gravitational accelerations holding components of the body together are matched by centrifugal accelerations, this causes the body to rotationally fission. i.e break apart. We calculated the timescale of this fission process for a parent Vulcanoid and for each of their subsequent generational fragments. We show that objects with radii up to 100 kilometers in size are efficiently destroyed by the YORP effect doing so in a timescale that is much younger than the age of the Solar System...


Would gravitational waves be subject to external gravitational perturbations?

Firstly, gravitational waves (GWs) are not an echo - we measure the direct signal.

The process you describe here is known as gravitational lensing, the deviation of (usually) light rays due to massive objects between the source and the observer.  This also applies to GWs.  The result will be similar - the direction of the waves can be changed, resulting in a distorted image of the source and possibly a 'brighter' image (stronger signal) due to 'gravitational focussing'.

However, LIGO and Virgo do not image GWs, they detect point sources.  Hence, there is no distorted image, but the 'luminosity' of the signal may be affected, resulting in a (somewhat) misjudged distance (although they use a 'luminosity distance', which is not necessarily the same as a current physical separation).

Just as in the case of light, a black hole could trap some of the GW signal that would have reached the observer (gravitons that fall directly into the black hole), if it is directly between the source and the observer, but the gravitational focussing makes it likely that the resulting signal will be stronger rather than weaker.


Do we know if we are overtaking or falling behind our arm?

The answer to the question depends on precisely where you are in the Milky Way. According to density wave theory, which appears to describe the formation and evolution of spiral arms quite well, the arms move relative to some static inertial frame at a global pattern speed $\Omega_{gp}$. At some defining radius $R_c$, known as the corotation radius, stars move at $\Omega_{gp}$. Inside $R_c$, stars move quicker and thus move ahead of the arms; outside $R_c$, stars move slower and thus "fall behind" the arms.

There is uncertainty surrounding the values of $R_c$ and $\Omega_{gp}$ in the Milky Way in all its spiral arms - especially since local stellar kinematics differ in each of the arms, and so too may $R_c$ and $\Omega_{gp}$ - but measurements place the Solar System slightly inside the corotation radius, meaning that we are moving ahead of the arms.

I have data from a review by Gerhard (2011). Here are some basic average parameters, for the Milky Way as a whole:

Bar corotation radius ($R_{CR}$): $3.5$-$4.5\text{ kpc}$
Spiral arm corotation radius: $8.1$-$8.4\text{ kpc}$
Bar pattern speed ($\Omega_b$): $50$-$60\text{ km}\text{ s}^{-1}\text{ kpc}^{-1}$
Spiral arm pattern speed ($\Omega_{sp}$): $17$-$28\text{ km}\text{ s}^{-1}\text{ kpc}^{-1}$

This also goes along with a local mean speed of stars in the solar neighborhood - the Sun included - of $\sim220\text{ km}\text{ s}^{-1}$.

Hence for the three questions: (1) The pattern is rotating clockwise as seen in that diagram; (2) It rotates about once every 300 million years; (3) and at our radius, we're moving a little faster than the pattern so we are overtaking it.


Why does the Sun move slower during the solstices?

All points on the celestial sphere execute a full circle every (sidereal) day, but the speed of a point with nonzero declination is slower than a point on the celestial equator because it's traveling on a small circle, not a great circle. This is exactly the same as how a point on the terrestrial equator travels at a higher speed than a point not on the equator. Eg, a point with a latitude or declination of 60° travels at half the speed of a point on the equator, because cos(60°) = 0.5

At the solstices, the Sun is on either the tropic of Cancer or Capricorn, so it has its maximum or minimum declination, approximately ±23.4°. So its speed is cos(23.4) $\approx$ 0.9178 relative to a point on the celestial equator, or about 8% slower, as Mike G mentioned in that comment.

Here's a diagram from an answer about latitude speed by SF. on the Space Exploration Stack Exchange:


Can IR sensitive cameras read signals from Venus's surface emissions?

Yes. A thermal image from the surface can be detected, even by amateurs.

The ground is heated to about 700K, and thermal emission at a wavelength of 1 micron is just about observable by skilled amateurs, with the right equipment. Mountains, which are cooler by about 30K, can be seen at this wavelength

From orbit, more details can be seen. Akatsuki used the 1 micron band to observe the surface in 2015

At other wavelengths, cloud patterns are revealed. Galileo took images at 2.3 microns during its flyby. They show sulphuric acid clouds at 50km above the surface (10km below the visible cloud layer) and at about 270K, with gaps exposing the lower atmosphere, glowing at 500K.


What are the color progression of a typical sunrise?

These steps are in there for credit

I used Stellarium to simulate a sunrise, and took screenshots of the progression. Each screen shot was 2 seconds apart, and stellarium was accelerated to 10 minutes a second.

Then, using imagemagik to crop batch those 13 images with this command:

mogrify -crop 1280 1024 *.png

(those numbers are the size of my screen. I cropped to get rid of my second screen)

Next I put the images into Color Thief and got the pallet of each image. This is the result of those 13 images:

Those are the colour pallets. You can pick whatever colours you want from them with a colour picker, such as KColorChooser.


Why (the heck) is the basic angle of GAIA 106.5°?

I did not find anything conclusive on the web, and I am yet waiting to receive feedback from ESA's contact page. In a German speaking IT forum there is something a bit less vague than the FAQ cited in the question:

The reason for the 106.5 degrees is not easy to see. It has to do with the fact that we are (also) doing an astrometric solution on a great circle and for this all angles 360x(n/m) [n, m = small integers] have to be avoided because otherwise one would have a congruence state that would hinder the solution. Values other than 106.5 degrees would also be possible. 90 degrees would be ideal for parallax determination, but unfortunately 90=360x(1/4).

This said, why not simply choosing 91° as $\gcd(360,91)=1$ (with $\gcd$ being the greatest common divisor)?
Maybe it is really only a numerology of the engineers, just like the Mars parachute Morse message. An argument for that would be that Gaia also has also exactly 106 CCD sensors. Or it is simply a logistical issue on how to mount things on the spacecraft. I am really curious what the official ESA statement will be.


Is it possible for a moon to only be out at night?

If the Earth were alone in space and the hypothetical moon had negligible mass, you can calculate the distance at which the period would be one year using:

$$ a = \left( \frac{GM_E T^2} {4 \pi^2} \right)^{1/3} $$

where $GM_E$ is the standard gravitational parameter of the Earth (about 3.9860E+14 m^3/s^2) and $T$ is the period (about 365.2564 x 24 x 3600 sec). You should get about 2.2 million kilometers.

It turns out that because the Earth is orbiting around the Sun, the added gravitational effect results in a location a bit closer, about 1.5 million kilometers from the Earth, where a low mass object would tend to remain at the far side from the Sun of the Earth for a while.

This location is one of the five Lagrangian points, and has the designator Sun-Earth L2. There is a similar point L1 between the Earth an the Sun as well.

However, this is not a stable orbit, even in a perfect circular restricted three-body problem (CR3BP or CRTBP) scenario, the least little displacement would cause it to eventually wander away into it's own, somewhat chaotic orbit around the Sun within a few years. The reality of the more elliptical orbits of Solar System bodies plus gravitational perturbations would only reinforce this instability.

So find a way to give the hypothetical moon a propulsion system to do station keeping maneuvers to keep it there, and you've got almost what you are looking for. The problem is that this point is in the same neighborhood as the point where the Umbra of the Earth's shadow disappears.

Since the Earth's orbit is elliptical, the point where the umbra terminates moves a bit, and since there is no exact definition of a Lagrange point's location for a messy, realistic Solar System, the question of whether a satellite in the vicinity of L2 would be in partial or total eclipse is debatable.

If you already have the propulsion to do station keeping, you could possibly move it in to the umbra and maintain it there as well.


Perhaps one way of answering this is to note that the (approximately) dipole magnetic fields of a magnetar fall off radially as something like $r^{-3}$. So at the magnetar surface the fields could be of order $10^{11}$ Teslas (at a maximum), but these are reduced by a factor of $\sim 8$ at 2 magnetar radii, a factor of $\sim 27$ at 3 magnetar radii etc.

Now given that the radius of a magnetar is going to be $\sim 10$ km, then the fields die away pretty quickly.

So let's say the Earth orbited a magnetar. The field strength at $1.5 \times 10^{11}$ m is going to be $\sim 10^{11} \times (10^4/1.5\times10^{11})^3 =  3 \times 10^{-11}$ Teslas - much smaller than the Earth's magnetic field.

To put this in context. The field at the surface of a small neodymium magnet are of order 1 Tesla and they don't disrupt your atoms when you pick them them up. To get a field this strong from a magnetar you would have to get to within about 50,000 km of it. The synchrotron radiation field would have fried you before then and any planet would have been ripped apart by tidal forces about an order of magnitude further away. The Roche limit given by
$$ R_{Roche} \sim 1.26 R_1 \left(\frac{\rho_1}{\rho_2}\right)^{1/3},$$
where $\rho_1$ is the density of the neutron star ($\sim 3 \times 10^{17}$ kg/m$^{3}$), $\rho_2$ is the density of the Earth ($\sim 5500$ kg/m$^3$) and $R_1$ is the radius of the neutron star (10 km). This yields $R_{roche} = 480,000$ km.


Because the uncertainties in the number of counts detected in each pixel are considered to be independent. That means some of them are higher than the "true value" (the count rate you would measure if you observed for an infinitely long time), whilst others are lower. To some extent these independent uncertainties will cancel, and the net result is that for normally distributed errors the correct protocol is what you have described.

Perhaps a good way of seeing this is that suppose you took 100 independent measurements of the same thing, each with its own, roughly equal uncertainty. If I asked what the uncertainty in the average was, you wouldn't just add up all the errors and divide by 100 because that would give an uncertainty that was identical to that of an individual measurement. Instead you would do the quadrature sum of the uncertainties and divide by 100, which reduces the uncertainty in the average by a factor of $\sqrt{100}$.


Why is the SETI survey limited to only Red Dwarf Stars?

Sun-like yellow dwarf stars are generally thought to be the best places for intelligent life to develop, if only because 100% of known life is in orbit about a yellow dwarf.  Larger, brighter stars don't last long enough for intelligence to evolve and red dwarf stars were thought to be too prone to solar flares.
However yellow dwarf stars are relatively rare, and the nearby ones have already been investigated.  Red dwarfs are far more common and so there are far more red dwarf stars in our neighbourhood, and perhaps solar flares are not as much of a problem as we previously thought.  So current investigations are focussing on those red-dwarfs that had previously been excluded.


Formation of elements in the Sun other than helium

The Sun is currently turning hydrogen into helium. There are no other nuclear reactions taking place at any significant rate in the Sun. The Sun will not start to make heavier elements until it reaches the tip of the red giant branch in about 7 billion years time.
The elements heavier than helium that are present in the Sun were almost all made inside other stars. These stars lived and died before the Sun was born 4.5 billion years ago. The nucleosynthesis products from these previous stars were recycled in stellar winds, supernovae, kilonovae and novae, and then mixed into the turbulent interstellar medium.


How can only 1 magnetic pole of the Sun change at a time?

The Sun is not a point object, in fact it occupies a huge swath of space. Nor is it solid, rather the Sun is mostly fluid. Therefore, there is not one single "Sun's magnetic field" but rather the aggregate of hundreds, thousands, or perhaps trillions of magnetic fields, each of which changes in complex ways and are affected by (and affect themselves) nearby fields. Thus, the aggregate field does not always resemble the simplification to one single large "Sun's magnetic field".

Note that the page the OP links to has a nice diagram which explains the "two south poles" phenomenon. During the process of the Sun's poles swap, for a short time the magnetic positive 'pole' shifts to the Sun's equator and both geographic poles have magnetic negative poles. Then the once-magnetically-negative geographic pole becomes positive and the equator no longer acts as a magnetic pole. I've never seen this mentioned anywhere else, and I'm sure that it is an oversimplification, but it does explain the "two south poles" statement.