I think the argument is the following. The central temperature can be estimated from a form of the virial theorem. At least dimensionally speaking,  total thermal energy $MkT/\mu$ (where $\mu$ is the average mass per particle) is proportional to gravitational potential energy $GM^2/R$. So $T \propto GM/R$.

But for a given mass, standard polytropic theory for a star, says that
$$M \propto \rho_c^{(3-n)/2n} G^{-3/2}$$
$$R  \propto \rho_c^{(1-n)/2n} G^{-1/2}$$
 where $n=3$ is an approximate polytropic index for the Sun and $\rho_c$ is the central density. See for example Does Eddington's variable polytropic index better fit data from the Standard Solar Model?

Putting this together we have
$$ T \propto GMR^{-1} \propto G^2 \rho_c^{-1/3}$$

The power generation in the solar core from the pp cycle goes approximately as $L \propto \rho_c^2 T^4$ and using the proportionality for $T$ derived above, we have $L \propto G^8 \rho_c^{2/3}$.

Finally, we note that for a $n=3$ polytrope, that $\rho _c$ and $M$ are independent, but that $\rho_c^{2/3} \propto G^{-1} R^{-2}$. So for a fixed solar radius we have $L \propto G^7$.


Which part of the milky way does the image show?

I wholeheartedly recommend astrometry.net for this sort of thing. It Just Works(tm); running on your image produced this output with absolutely no hints or guidance from me:

For avoidance of doubt, I have no association with astrometry.net.


How impermeable are the surfaces of the moon?

The moon's surface is made up of many layers.

The top few centimeters of the moon are generally composed of a layer of thin powder.  This is easy to dig into, and the Apollo astronauts easily scooped up this stuff.

Wikipedia - Lunar Soil

Below that layer is the bulk Regolith,  which contains a mixture of powder, grains, pebbles, and rocks - the remnants of billions of years of meteor collisions which have thoroughly mixed and distributed the material from collisions all over the surface.  The regolith ranges in depth around the Moon based on the age of the surface - the old highland and farside surface contains regolith up to 15 meters deep, while the mare regolith averages about 4 meters deep.

The regolith is not solid rock,  but it's very compacted and the grains are rough and billions of years of vibrations from impacts and quakes has caused the particles to interlock.  Therefore, it can be extremely difficult to dig into.  The Apollo 14 astronauts only managed to drive a heat probe 70cm into the regolith, using about 20 hard hammer blows to do so.  For later missions they had to use electric drills to get deeper, about 3 meters.

Lunar and Planetary Institute - Lunar Sourcebook, chapter 7

Apollo 14 - Soil Mechanics Investigation

Below the regolith is rock.  On the Mare, the rock is a very hard basalt (8 out of 10 on the hardness scale).  Getting through this is like cutting granite.  On the older highland regions, the most common material under the regolith will be anorthosite, which is less dense but it's still drilling through rock.

The mare basalts range in depth from 100m to about 3-4 km, with an average depth of .74 km.  Below that is old anorthositic crust.  The crust itself ranges from 10-70 km thick, with an average of about 35 km.  The crust on the near side is significantly thinner than on the far side.

Planetary Society - GRAIL maps of crustal thickness

If you want to drill deeper,  you're into the mantle, which is only partially melted near the core, and made up of roughly the same olivine basalt that makes up the floor of the mare, and goes down to a depth of about 1100 km.

The Moon - Internal Structure and Magma Ocean

Hope that helps.


When a star reaches the red giant phase, why does it become more opaque?

There are some answers in this lecture and this one although even there the author admits that the whole story is complicated, and not perfectly known.

It seems that shell burning and contraction of the no-longer-burning core of the star causes some expansion of the outer layers (this is described as the "mirror principle" and partly explained in lecture 12 (section 12.4). This causes the outer layers to cool, until H- ions start to form in them (around 5000K). These interact with light much more than neutral hydrogen atoms or bare protons (H+) which accounts for the increase in opacity.


How precisely is planetary tilt defined (the tilt direction, not just the angle)?

A bit too simply (but not too much), it's the angle between the planet's rotational angular velocity vector and the planet's orbital angular momentum vector.

Where it gets tricky is dealing with little wobbles and such. Planets don't quite orbit in a plane because of gravitational interactions amongst the planets. This means the instantaneous (osculating) orbital angular momentum vector isn't quite constant, both in magnitude and direction. There are a number of different mean orbital elements that smooth out most of those tiny wobbles. One of those mean orbital elements sets (I'm not sure which one exactly) is used to determine the  mean orbital angular momentum vector.

Planets don't quite rotate nice and smoothly because planets aren't perfect spheres and aren't rigid bodies. This gives the Moon, the Sun, and other planets a handle by which they can exert tiny little torques in the planet in question. The response of the planet to these torques is somewhat arbitrarily divided into two categories based on frequency. Very slow responses are called precession; faster responses, nutation. The non spherical nature of a planet means the planet undergoes a small torque-free nutation as well as the torque-induced precession, nutation, and polar motion. In addition to these, there are a number of terms (all small) that don't yet have a very good model behind them. This oddball terms, along with the torque-free nutation, are collectively called polar motion. The motion is fairly smooth if one ignores nutation and polar motion, and the effects are all small.

Precession, while very slow, can be quite large in magnitude. The rotational angular velocity vector that is used in determining the axial tilt incorporates precession but smooths out nutation and polar motion.


How will the Breakthrough Initiatives affect other astronomy observations?

I was fortunate enough to spend the past week at a workshop held at the Green Bank Observatory (GBO), so I can give you a partial answer based on how the Breakthrough Initiatives affects it in particular.
Telescope time
The GBO actually operates a selection of telescopes, of which the most widely-known (and largest) is the Green Bank Telescope (GBT). The GBT is the main instrument used by Breakthrough Listen at the observatory. During the second half of 2016, the GBT was oversubscribed by a factor of 2.6, meaning that only about one in every 2.6 proposals was approved outright (“filler” proposals will be accepted if short periods of time are available). This doesn’t mean that for every hour of telescope time, 2.6 hours worth of observations were proposed, although the ratio is close (2.3), but it does mean that the GBT is very much in demand - and those numbers have increased this year. The telescope is not in use on only two days annually, Thanksgiving and Christmas, although there are blocks set out each week for maintenance.
The answer to your first question, then, is yes - there is approximately indeed a zero-sum redistribution, because the telescope is extraordinarily busy. The actual data-gathering proportion may be slightly different because overhead times vary by the type of observations, and things like changing receivers or pointing at a different source do take time. However, it is true that Breakthrough Listen, like any other set of observations, takes time away from other scientists - about 20% of the GBT’s annual total observing time.
(As an aside, Breakthrough Listen operates a “backend” in the GBO computer room - a set of GPUs that analyze incoming data. There are plenty of backends for various receivers and instruments, and they do take up space - and power. We were told that the room the Breakthrough Listen computers are in wasn’t actually designed to hold that many computers, which did cause some initial cooling issues (which were quickly rectified) when these new units were installed. So it’s important to realize that the project takes up more resources than just observing time!)
Money
The financial system at Green Bank is actually a little complicated. In 2012, it was recommended that the GBT and the Long Baseline Array be removed from the NRAO, and this was implemented in 2016. In some ways, the transition is still happening - much official GBO information is still on the NRAO’s website - but the relevant point here is that in the 2017, the GBO only received about 66% of its funds from the National Science Foundation (NSF). The current agreement is that the NSF will gradually defund the GBO over the next few years.
This is why Breakthrough Listen is so financially crucial, and why it is getting so much publicity at the GBO. Without sources (and currently, without non-governmental sources) of funding, the observatory simply cannot operate. That said, the GBO is not beholden to people like Yuri Milner (the man behind the Breakthrough Initiatives) who have specific projects in mind - and indeed, that type of donor is quite the outlier. It does seek out donors and philanthropists who are willing to give money to simply ensure that the facility keeps going.1 If it can be guaranteed that these sources will continue giving without setting limits on what types of observing should be done, then things can progress as close to normal as possible.
The exact details of funding for the GBO are not (as far as I know) public, and I don’t want to speculate too much. However, I can say that the observatory is almost certainly not going to push ordinary astronomers out of the way for the sake of big projects - in the short-term. Long-term funding is uncertain, true, but that does not mean that it will be renamed the Breakthrough Listen Observatory any time soon. The number of people who have operated the GBT number in the thousands, and they don’t pay to use it. It was built using public funds and has operated largely using public funds for quite some time now. I think the GBO intends to keep operating like this, and will not turn to the alternative - subject to the interests of only a select few projects - unless matters become extraordinarily desperate. It’s possible, but not something to be immediately worried about.

1 One day last week, we noticed a helicopter landing at the airstrip. It turns out that the passenger was a VIP who was given a full tour of the telescope, guided by several observatory higher-ups. He was, as you can image, a prospective donor.


Is there any difference between M87's image and predictions?

The observation did indeed confirm general relativity, see e.g. Lisa Grossman wrote in October 2020 for ScienceNews: The first black hole image helped test general relativity in a new way

That iconic image, of the supermassive black hole at the center of the galaxy M87 about 55 million light-years away, showed that the shadow closely matched general relativity’s predictions of its size. [...]

Specifically, the researchers used the size of the black hole to perform what’s known as a “second-order” test of general relativity geared toward boosting confidence in the result. That “can’t really be done in the solar system” because the gravitational field is too weak, says EHT team member Lia Medeiros of the Institute for Advanced Study in Princeton, N.J.


Well, it's a relatively new field of study so I don't think there's any exact answers, but I've read up on this a bit out of interest.  Red Dwarfs have 2 main disadvantages.   One, which you've mentioned.  They're relatively cold compared to our star so the planet would need to be comparatively close and therefor, probobly tidally locked.   The other problem is red dwarfs tend to have very active sunspots compared to our sun, so the planet would need a crazy strong magnetic field to protect the planet.   Neither of those are "deal breakers" but perhaps less likely than a sun-like star.

A tidally locked planet wouldn't necessarily have a scorching hot side and a freezing cold side.  Modeling suggests that temperature would circulate.   If it had extensive oceans on the tidally locked side, for example, that could create a cloud cover and not cause intense heat.   You could also have the sun side be far enough away to be temperate and the star-side icy, or a livable section in a ring around the planet.

There's also the possibility that a tidally locked planet could have winds that we would find very difficult to adapt to.

I think you'd ideally have the sun side be the habitable side for photosynthesis.

On the flip side, if the star is too large, the star's life would probobly be too short for the planet to cool down enough to be habitable for us.  The life of a star is the inverse cube of the size, so a star twice the size of our sun would only have a main sequence of 1.2-1.3 billion years.  When you get a bit bigger than that, there's not enough time for a planet to cool off and a solar system to get past it's heavy bombardment period.

So I'd give a red dwarf a better chance of having a habitable planet than a sun 2.5 times or larger than our sun.   That, and red-dwarfs are the most common stars in the galaxy, so I think there's a probability that some red dwarfs could have habitable planets.


Has life been discovered outside of Earth?

No life has been discovered out of the Earth. It is still discussed that bacteria once existed at Mars.


Are there ever any simultaneous transits of both Mercury and Venus as seen from the Earth?

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

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

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

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

The closest we get to this:

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

The program I used to compute this:

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

The list of transits I computed while solving this:

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

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

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

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

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


Can I find exoplanets without a telescope?

No. With a probability nearly equal 1 you will not find any exoplanet with naked eye from Earth, even under perfect conditions like no light-pollution. The human eye simply does not have enough resolution to observe any of the closest exoplanets directly.
Neither of the methods of detection an exoplanet work without any additional tool (such as a telescope for any part of the spectrum).


Does the given radius of a planet/moon include its atmosphere?

I looked for a source without luck, so I'll just post cause I've read this answer before.   The diameter is the solid surface, not the atmosphere for rocky planets.   For gas giants it's the atmosphere up to 1 bar.

To determine Venus' angular diameter, you'd need to add something for it's atmosphere.   It's difficult to say how high it's atmosphere stops being reflective (and if anyone knows the details, please answer), but I would guess about 50 km where it's clouds reach to on it's sunny side.   You'd need to add more if you look at it through infra-red.   But I think it would be an error to add the full 250 km listed as the thickness of it's atmosphere and it's reflectivity probably trails off well before then.

A curious sidebar, Ganymede is listed as the largest moon by diameter, but because of it's atmosphere, Titan appears larger.  It's visibly larger on accurate moon charts.   Because of ice in it's upper atmosphere and low gravity, Titan's atmosphere remains reflective for a significant distance.


What are the correct symbols for vernal and autumnal equinox?

The Wikipedia article on astronomical symbols says:

  The zodiac symbols are also sometimes used to represent points on the ecliptic, particularly the solstices and equinoxes.
  Each symbol is taken to represent the "first point" of each sign.
  Thus, ♈ the symbol for Aries, represents the March equinox; ♋, for Cancer, the June solstice; ♎, for Libra, the September equinox; and ♑, for Capricorn, the December solstice.

The article cites Roy and Clarke, Astronomy: Principles and Practice.
This question copies an example usage of ♈ and ♎ from that book.
Stellarium uses the same symbols if you set View: Markings to show the equinoxes and solstices.

$\gamma$ is a poor substitute but clear enough if ♈ is unavailable.
Since $\Omega$ stands for an orbit's longitude of ascending node (angle ♈-☉-☊), it would be a confusing substitute for ♎.

"First point of Libra" is archaic but correct.
These names and symbols represent tropical zodiac signs, not IAU constellations.
♎ is fixed at ecliptic longitude 180° regardless of constellation boundaries.
Its location in Virgo this century doesn't change its name or symbol.


How do we know distant stars still exist

We don't know from sensory evidence.  As far as we can tell, no information ever moves faster than the speed of light, so we are always at least 450 years out of date for anything happening in the Pleiades.

The only evidence we have is inferential: As far as we can tell (and we've looked real hard at this) the world works the same both here and at the Pleiades.  And as far as we can tell (and, once again, we've looked real hard at this) stars like those in the Pleiades don't just disappear or go out in only 450 years -- it takes millions of years.

So, based on what we observe of them as of 450 years ago, and based on what we know of how stars work, we conclude that they are pretty much the same "today" -- though, really, what we're concluding is that when we observe them 450 years from now, they'll still look pretty much like they do today.

This is all inferential, but I'd bet just about anything on it.  (If only I were here in 450 years to collect!)


With respect to "eternally collapsing", they are probably referrring to the fact that in the reference frame of an external observer, gravitational time dilation prohibits matter from ever reaching the event horizon (the "surface") of the black hole.

Denoting the radius of the event horizon $r_\mathrm{S}$ (for "Schwarzschild radius"), time runs slower by a factor of $(1 - r_\mathrm{S}/r)^{-1/2}$ for an observer at a distance of $r$. As $r$ approaches $r_\mathrm{S}$, this factor goes to infinity, i.e. the observer will never reach $r_\mathrm{S}$

However, this is only for the external observer; the falling observer would cross the horizon and reach the center in a finite time.

With respect to "not being stable", you're probably right that they are referring to Hawking radiation which presumably makes the black hole slowly "evaporate". Near $r_\mathrm{S}$, pairs of virtual particles are being created and can be turned into real particles by the gravitational field. If they avoid falling into the black hole, energy is taken away from the hole, reducing its mass.


There are lots of apps you can download on your laptop / smartphone :

Stellarium : the one i use on my computer. Useful for beginners. You will see planets, some of the best known object, ...
Skymap for android : used it before, quite nice.
SkySafari : don't use it but some people told me it was great.

All of them are free. However, you can just search on google, there are plenty of online solutions.


Open problems in astronomy that an amateur (with a PhD in some other field) would have a chance of solving?

If you have good knowledge of software development and pattern recognition, there are several problems that you could assist in solving. Much of observational astronomy requires long time series data, and removing the noise from this data. I have just left the field where some colleagues are trying to develop some software to use image subtraction techniques to isolate individual stars in the centre of clusters. The centre of the cluster is typically more dense and harder to get clear measurements of each individual star to analyse.

Pattern recognition would be particularly useful in pipeline analysis, where a generic pipeline is used on large amounts of data to 1: find the types of stars one is interested in; and 2: extracting some interesting information on these stars. Machine learning techniques could also be used to assist in developing the general pipelines for more specific interests.

I am happy to put you in touch with a few people that could provide you with some specific problems that you might be able to assist in.


Are there collected data about the direction of rotation of black holes and the direction of the magnetic field?

For pulsars the relation is complex. The magnetic field tends to "freeze in place" after the star collapses, but I'm sure there are all kinds of perturbation mechanisms that may change its initial orientation.

For blackholes (the Kerr-Newman metric, rotating and charged black hole) the relation is very simple: the rotation axis of the BH and the symmetry axis of its magnetic field always coincide.

https://en.wikipedia.org/wiki/Kerr%E2%80%93Newman_metric#Some_aspects_of_the_solution

There is a simple and intuitive justification, although perhaps not rigorous. If the magnetic field was not aligned with rotation, the BH would appear to be "blinking". But a BH cannot "blink", nor can it otherwise send any signals to the external universe. Therefore, its magnetic field axis must coincide with its rotation axis.


What is the Galactic Latitude and Galactic Longitude? How is different from the latitude and longitude we use on Earth?

The Galactic Coordinate system is a longitude-latitude coordinate system that is used to define the positions of objects in space, most commonly objects within our own galaxy. It uses the center of our galaxy as the focal point (i.e., where we consider $(0^{\circ},\:0^{\circ})$), much like we use the position just off Africa as the focal point of the Earth's lon-lat system. You can see a visual representation of this system below.

The reference point is of course from our solar system. Longitude is measured in the plane of the galaxy, going positive as you sweep to the left (thus following a right-hand rule). The symbol $\ell$ is generally used for galactic longitude. Latitude is measured above or below the plane, going positive "above" the plane and negative "below". We define above and below based on the right hand rule as applied to our galaxy's rotation. The symbol $b$ is generally used for galactic latitude.

If someone says an object is at galactic coordinate position $(\ell,b) = (180^{\circ},\:+45^{\circ})$, then the object will be located directly opposite the galactic center and above the plane of the galaxy.

Aside from being a lon-lat system, just as the Earth's is, it does not share any features with Earth's lon-lat system. That is, there's generally no relation between the two and the definition of one is not affected by the other.


Is there a relation between mass, radius, and luminosity in deuterium-burning brown dwarfs?

Yes, there are monotonic relationships between mass and luminosity and radius on the "deuterium burning main sequence".

Deuterium "burning" begins when the core temperature exceeds just over $10^6$ K. This happens very early in the life of a contracting protostar and, because it is fully convective and thoroughly mixed, all the D is "burned" in less than a couple of million years. The process takes much longer (5-50 million years) in lower mass brown dwarfs. Thus only systems younger than 50 Myr could have a D-burning brown dwarf. Below about 13 Jupiter masses, the brown dwarf core never gets hot enough to ignite D.

D burning mimics hydrogen burning in that it is possible to stabilise the contraction of a protostar if the D burning rate can match that which would be released by gravitational contraction. The burning process then acts as a thermostat, keeping the core at roughly constant temperature, and the protostar/brown dwarf at constant luminosity, until all the D has been depleted. The big difference between D and H burning though, is that the initial D/H ratio is something like $2\times 10^{-4}$, so D burning does not last long.

The rate of gravitational contraction of a protostar/brown dwarf increases with mass. Hence the rate of D burning, and hence the luminosity, must increase with mass. The virial theorem used with an ideal gas approximation then tells us that since the central temperature is roughly proportional to mass/radius, and since the D burning reaction rate is very temperature sensitive ($\propto T^{12}$), the D burns at almost the same temperature, regardless of mass, and so the radius during D burning will be roughly proportional to the mass. This is only rough because the cores of such objects are close to degenerate and ideal gas approximations stop working.

The plot below (from [Tout et al. 1999]) 1) shows the H-burning (lower) and D-burning (upper) "main sequences" for stars as dotted lines in the Hertzsprung- Russell diagram. You can use this plot to estimate the luminosity and radius for a star of a given mass on the D burning sequence. Note the isochrones (line of constant age, running across the diagram, labelled in Myr). There is no convenient formulae that I am aware of.

For brown dwarfs I can't find anything similar. Below are cooling curves from Burrows et al. (1997), showing how luminosity changes with time. D-burning manifests itself as a plateau at a luminosity that increases with increasing mass, and with a duration that increases with decreasing mass. The brown dwarfs are the green curves.


The major axes of the planets are not aligned.

The major axis passes through the perihelion, the solar system barycentre and the aphelion. So by looking up the Longitude of perihelion we can find the direction of the major axis:

planet  Longitude of perihelion
M       77
V       132
E       103
M       -24
J       15
S       93
U       171
N       45

There is no evidence of alignment.


If thousands of exoplanets have been discovered, why haven't we discovered planet 9 yet?

There are two primary reasons Planet Nine (if it exists) is extremely hard to detect: it's very dim and its position is not well constrained.

Brightness - We are able to see planets in our solar system because they reflect light from the sun.  The amount of light they reflect depends on how large the object is, how reflective it is, and (most importantly) the amount of sunlight that hits it.  The hypothesized distance at which Planet Nine is orbiting is so far away that it will receive only a tiny amount of sunlight. At this distance, its apparent magnitude would be greater than 22.  In comparison, Pluto's apparent magnitude is about 15.  Magnitude is a logarithmic scale, so that means that Planet Nine would be $631$ times dimmer than Pluto, making it incredibly hard to see.
Unknown Position - The estimates of Planet Nine's characteristics are based on how it gravitationally disrupts the orbits of some trans-Neptunian objects.  Based on these observations, scientists can calculate where it is likely to be, but these calculations depend on its size.  Estimates of its mass range from 2 to 4 times the mass of the Earth.  And all of these are just estimates, giving us a relatively large patch of sky to search to find Planet Nine.

Combining both of these reasons, it is very difficult to search for Planet Nine.  We're looking for an incredibly dim object in a large area of the sky.

Now, when we look for exoplanets, things are actually a lot easier.  To detect exoplanets we hardly ever look for the planets directly.  We've only detected about 20 planets via direct imaging.  Instead, we look at the light from the star.  The stars are many orders of magnitude brighter than the planets, and are very easy to see.  All we have to do is see how the light from the star changes over time.  An early method detect shifts in the radial velocity of the star (its motion towards or away from us) using Doppler spectroscopy.  The more prevalent method today, the one that the Kepler mission used, is transit photometry.  We measure the light output of a star, and look for dips in its intensity, caused by the planet orbiting in front of the star and blocking some of the light.

Because we're looking at a star, we don't have the problems we have when looking for Planet Nine.  Stars are very bright, their positions are known very accurately, and we don't have to wait too long to make a detection.  Plus, surveys like Kepler are able to scan many, many stars at once.


Can an average person see stars from the bottom of a well or through cardboard tubes? Definitive answer required!

This article (Hughes 1983, "On Seeing Stars (especially up chimneys") from the Quarterly Journal of the Royal Astronomical Society seems like a pretty good account of the "phenomenon", including both classical and literary claims (e.g., in stories by Dickens and Kipling) and actual experiments. It appears that the Aristotle reference ("people in pits and wells sometimes see stars", from Generation of Animals) is not an eyewitness report by Aristotle himself -- and, given that Aristotle claimed men had more teeth than women, one should always be a bit skeptical about his claims; there is also the possibility that these were sightings of planets, which can sometimes be seen in the daytime. The Herschel reference is to an "optician" Herschel knew who claimed to have seen a star this way once or twice, not something Herschel himself verified.
In Section 2 of the article there are discussions of actual serious (and unsuccessful) attempts to test this, including numerous attempts by Alexander von Humboldt and well-described attempts by astronomers like J. Allen Hynek, who tried (with "several members of my astronomy class") to observe Vega from the bottom of an abandoned smokestack, and some measurements taken at the botton of a tall chimney which verified that the brightness of the sky was unchanged from that seen outside.
There are also some discussions of the sensitivity of the human eye. The conclusion I get from this is that it is barely possible that someone with good eyes and an extremely restricted field of view (much smaller than "from the bottom of a well" or a "cardboard tube") might be able to make out Sirius, but probably nothing fainter.
Edited to add: This article (Können et al. 2015) reports a daytime sighting of Sirius from the La Palma observatory. They note that there were able to do this because they had pointed their telescope at Sirius, and looked along the finder scope (knowing the offset), so they knew exactly where to look. (Even then, it was difficult: "Both of them saw Sirius, but it was difficult; as Wallis put it during his attempt: 'It comes and it goes.'") Knowing exactly where to look is important, because

the limiting magnitude of photopic vision decreases linearly and
rapidly with distance from the center of field of the human eye: at 2°
from the center the limiting magnitude has worsened by 1 magnitude;
at 4° by 2 magnitudes [16]. This means that for an unintentional
sighting, a star should be roughly 2 magnitudes brighter than the
limiting magnitude. Venus meets this requirement, but for Sirius there
is little hope for an unprepared sighting.

(Short answer: Generally speaking, you cannot see stars from the bottom of a well, chimney, mining pit, etc., during the daytime. And you certainly aren't going to see multiple stars from the bottom of a canyon, even if there's a lovely evocation of that idea at one point in The Lord of the Rings.)


Do the E and B modes of the CMB polarization have anything to do with electric and magnetic fields?

In this context, the $\kappa$ you are referring to is called the dimensionless matter density field. It is gravitational lensing jargon, and is usually just referred to as the 'convergence' field.

What you have written there (that a field can be split up into divergence-less and curl-less components) is generally speaking true for any field which can be expressed as the curl of another field $A$. @chris is correct in that they refer not to the electromagnetism you are probably used to hearing these types of things discussed in, but refer to the polarization states of the cosmic microwave background due to quantum fluctuations in the early universe. They are called 'E' and 'B' only because they are reminiscent of electric and magnetic field lines.

The big find was the measurement of B-modes. As a reminder, here is what these E- and B-mode patterns in the CMB would look like:

The big thing here is that E-modes are symmetric upon reflection, whereas B-modes are clearly not ($B>0$ maps into $B<0$). In order to produce B-mode polarization, you need two things:

1) Thompson scattering of photons by free electrons (present in the surface of last scattering as well as reionization).

2) A temperature quadrupole.

The exciting thing about the second item is that there are only a few things which can produce just such a quadrupole moment, and one of them is gravity waves! Gravitational waves physically warp the coordinates of space-time as it travels through space, which of course have existing particles, causing these types of distortions:

This is the source of the temperature quadrupole moments. Sides which are squeezed are hotter and sides which are stretched are cooler.

Gravitational lensing can also produce small amounts of B-mode like features in the cmb polarization, but I believe they occur on much smaller scales and can therefore be ruled out as the cause of the BICEP2 findings. Lensing tends to stretch things out perpendicular to the radial vector (in the $\hat{\theta}$ direction):

The reason why this matters is because lensing provides a mechanism for turning E-mode polarization (take the $E<0$ mode for example) into a B-mode type polarization. Most plots of the BICEP2 results show the effect of lensing at much higher multipole moment (i.e. - smaller physical scales), so I believe they have a good way of differentiating between the two types of secondary cmb anisotropies.


What is the distance of the Earth's orbit around the Moon?

The mass of the Earth is about 81 times that of the Moon and the distance between their centres of mass is typically 384 400 km, so the centre of mass will be about 1/82 of this distance from the centre of mass of the Earth, which is about 4700km.

The orbits are not too far from circular, so the Earth travels about $2 \pi$ times that each month, so a little under 30000km.


What to do after first year of amateur astronomy?

Astrophotography is always a good step to take after just observation. That way, you can see things through your telescope that you cannot see with the naked eye,  and also share your findings.

Another thing you could do is start on a different part of astronomy, maybe getting a hydrogen alpha filter or a solar telescope to observe the Sun, which again could be combined with the astrophotography.


Where can I find a catalog of all stars in the Milky Way?

Hipparcos, the predecessor to Gaia, has a dataset (http://cdsarc.u-strasbg.fr/viz-bin/Cat?I/239) with 3D positions for 100,000 stars. While we have much larger datasets of galactic stars, such as SDSS, finding the distances to stars is much harder. Parallax is the most precise technique for finding distances. Even for Gaia, we will only be able to measure distances to 10% precisions.

There is a "galactic coordinate system", but that stil has the sun at the center and the line between the sun and the galactic center as its reference. There are calculators that can convert RA and Dec to galactic coordinates (e.g. http://python4astronomers.github.io/astropy/coordinates.html). You can use that and the distance to place them in space.


Let's assume that every particle in the Sun quantum tunneled far away from us (there's a near 0% chance this will happen, but let's just assume). For the first eight minutes, nothing would change from our point of view. Gravitational waves propagate at lightspeed, so we would continue to orbit the Sun and see it there.

After eight minutes, however, the lights would go out. The Moon will no longer shine upon us (it receives its light from the Sun), so our only source of light will be that of the celestial objects. Mercury and Venus will have seemingly disappeared from our optical telescopes — they will no longer be illuminated by the Sun. It would take another 30 minutes to an hour for the reflected light of Jupiter to shut off, and soon all the objects in the Solar System will no longer be visible to optical telescopes.

According to Ahad (2004), the illumination by the rest of the Milky Way would be approximately 1/300 of that of a Full Moon. So we might be able to see around thanks to the night sky and our own man-made technologies. However, most plants will not be able to undergo photosynthesis and will die.

At this point, we would be traveling tangentially to our orbit around the Sun, as you mention. We would just keep going into the abyss; it's very unlikely the gravity of any object will pull us in. However, things will get quite cold. The Earth would radiate heat away at a moderate rate, and temperatures would be unsustainable in most parts of the world after a year.

Luckily, the Earth has a nice and toasty core that could provide life-sustaining heat in some places (like Yellowstone), so humans that migrate over there would likely be better off.


How to tell whether an optical spectrum is featureless or just with a low S/N?

Convert answer in comment to community wiki

You would need to do statistical tests. i.e. Is the data you have (given the known noise properties) consistent with a featureless spectrum? If not, then it has features. Of course, the lower the signal-to-noise or the weaker the features might be, then the harder it would be to rule out a featureless continuum.


What is this thin, square-ish object on asteroid Bennu? How big is it?

In this composite of the rotating asteroid you can see that it is not thin, but actually protruding:

This series of images taken by the OSIRIS-REx spacecraft shows Bennu in one full rotation from
a distance of around 50 miles (80 km). The spacecraft’s PolyCam camera obtainedthe thirty-six 2.2-millisecond frames over a period of four hours and 18 minutes.
Credit: NASA/Goddard/University of Arizona

It looks like an 'ordinary' boulder.
(You can see several more in that sequence)

Simply comparing the 'height' from the second picture with Bennu's diameter of 492m shows this 'boulder' to be 28 meters high.


In section 3 of this paper on the possible origin of two planets orbiting a B-type subdwarf, called KIC 05807616, and of which I asked a question a while ago, the survivability of the planets to the intense UV radiation of their host star is examined. Here is an excerpt from the summary:

  In section 3 we examined the survivability of the planets to evaporation by the UV
  radiation of the EHB star. Equation (3) for the evaporation rate implies that the inner
  planet will be completely evaporated within ∼ 10⁷ yr. This is shorter than the ∼ 10⁸ yr life duration on the HB.

KIC 05807616 is still on the horizontal branch (HB), like any other star of its kind, and will remain in that stage for approximately one hundred million years, according to the paper, until it finally becomes a white dwarf. I guess that's the average lifetime of the subdwarf B stars.

EDIT: I also found this paper, which is more specific with respect to the average lifetime of the subdwarf B stars, I quote from its introduction:

  An sdB stays on the EHB for roughly 10⁸ years and directly evolves along with
  the white dwarf (WD) cooling track after its core helium has
  been exhausted.


How to calculate the altitude of the Moon?

With a lot of difficulty if you want to do it numerically by yourself. The Moon's orbit is very complex.

Far and away the easiest method would be to use NASA's HORIZONS service for calculating ephemerides, which will get you very comprehensive and accurate information on objects at your specified times.

I don't think that HORIZONS can directly return the time that the Moon is at its maximum altitude. However, if you can write some basic computer code then it shouldn't be difficult to query the ephemerides with a precise (say 1 minute) time interval and find the maximum altitude from there.


Why objects are uniquely defined by their right ascention and declination?

Stars do not change their right ascension or declination values as the
Earth turns.

The position of the stars in the night sky (called azimuth and
elevation) will change, but that's because the lines of right
ascension also move.

However, right ascension and declination do not uniquely define an
object. Over time, stars move with respect to the sun (called "proper
motion"), and the direction of the Earth's north pole (currently
pointing near Polaris, the North Star) also changes (called
"precession").


Could Pluto and Charon have extra-Solar origin?

The surface of Pluto is not just young, but very young, and also differentiated. The 'heart', Sputnik Planum, may be currently active, with flowing glaciers, and nitrogen snow. This area may have be resurfaced during Pluto's perihelion. Your theory cannot account for the very young age of the surface.

Next capture of a rogue planet is not easy. It would need Pluto to interact with the outer planets, to slow it down to enter solar orbit (otherwise a rogue planet would just pass through the solar system and come out the other side) A captured rogue would be unlikely to have the relatively low eccentricty of pluto.

Finally, while we don't know how common rogue planets are, they are probably not common, and given the massive distances between stars, the chances of any rogue just happening to have passed anywhere near the solar system in the last billion years in the right direction to be captured to Pluto's orbit is very low.

Conclusion. Pluto formed around the sun. A capture theory does not explain the observed young surface of Pluto, and modelling suggests a capture would be very unusual.


How do we know that a natural satellite cannot be a hollow object?

I'd caution that absolute statements are not good things in science. When Sagan says a natural satellite cannot be hallow, it is an absolute statement, but there's no way to 100% prove it true.

That being said, everything we know about planet/satellite formation tells us that the chances of a satellite forming as a hollow structure (or becoming hollow afterwards) is pretty much nil.

There is still much debate on the specifics of planetary formation, but the general process seems to be that small rocks and particles start to form, then they coalesce into bigger chunks and slowly that planetesimal grows in size. This type of formation process does not allow for the center to be hollow since it is the center which forms first! You cannot form the "edges" of your hollow planet and have those edges come together.

Moons and satellites can potentially have a different formation path. Look at our own moon for example. It is hypothesized that a Mars-sized proto-planet crashed into the Earth, flinging out a large amount of material. Much of this material then coalesced into our current Moon. Again, this process doesn't really allow for making something which is hollow.

Aside from the difficulty in forming a hollow satellite, you also have the challenge of maintaining the structure. Gravity constantly pulls everything to the center of the body. If there's nothing at the center, there's nothing preventing the outer shell from collapsing. It is technically possible that the satellite could not collapse, but that would require it to be fairly small and be perfectly structured such that the structural integrity of the rock forming the shell could withstand the crushing gravitational forces. Really though, such an object would have to be intelligently designed as the chances of it forming perfectly like that are non-existent. There's no possible way an object like Phobos could maintain itself without collapsing if it were hollow.


We see sunspots, which are giant, planet-sized storms on the surface of the Sun. There are however many differences between sunspots, the Great Red Spot on Jupiter, and tropical cyclones (e.g., hurricanes) on the Earth. Tropical cyclones are low pressure systems fueled by evaporation of warm ocean water and sustained by the Earth's somewhat rapid rotation rate. The Great Red Spot is a high pressure system that is sustained by Jupiter's quite rapid rotation rate. Sunspots are low temperature systems fueled by the Sun's magnetic field and carried along by the Sun's rather low rotation rate.


How are younger objects outside our observable universe when the older CMB is an observer's temporal "edge"?

The part of the universe that we can see is a cone in spacetime (our past light cone) and there are young and old parts of the universe both inside and outside the cone.
Here's a picture from Ned Wright's cosmology tutorial that shows what's going on.

We're at the top center. The red lines are (a slice through) our past light cone. Ignore the wavy $\phi(x)$ line and imagine a straight horizontal line there instead. Below that line the universe was opaque, so our observable universe effectively ends at that line (just like our view of the sun ends at its surface). There are young and old galaxies, stars, etc. both inside and outside the cone.


How small a location is possible to identify when any visible star is at its zenith directly above?

Lets assume your observer has a good watch, so she knows the exact UTC time. But lets also assume that she has no special kit, and has no special training.

She might be able to estimate the location of "directly up" to an accuracy of about one hand width (at arms length) which corresponds to roughly 5 degrees.  To achieve better than this, she would need some tools for sighting the position of stars.

With an error of +- 5 degrees, the error on the Earth's surface of any calcuation based on her observation is also 5 degrees of the Earth's surface. That is roughly a circle of radius 550km (350 miles).

Such an observation could prove you were in Mexico and not Canada. But it couldn't distinguish Conneticut and Washington DC, much less a square mile.

To get 1 mile accuracy you would need measurements at least 350 times better, This is better than one-sixtieth of a degree. The most skilled observers did achieve this level of accuracy with naked eye equipment, but not from a single observation. It could be done with a telescope designed for measuring the position of stars, provided the operator was sufficiently highly skilled.

If your observer doesn't have a good watch, she is subject to the problem of Longitude: It is relatively easy to find Latitude just from astronomical observations. It is much harder to find the longitude.


The physical difference between high- and low-mass X-ray binaries is that the latter has a donor star that fills its Roche lobe. Usually, the compact object is more massive than the donor star. Mass transfer occurs via Roche lobe overflow.

In high-mass X-ray binaries usually the donor star is more massive than the compact object and mass transfer is via a stellar wind.

The reason that a disc may form (in either case actually, but almost always for a LMXB) is that there is too much specific angular momentum (angular momentum per unit mass) in the accreted material to allow it to fall directly onto the compact object (angular momentum must be conserved). Broadly speaking, a condition for the formation of an accretion disc would be that the specific angular momentum of the accreted material exceeds the specific angular momentum of material in a Keplerian orbit at the surface of the compact object. The formation of an accretion disc would be favoured by short orbital periods or slower wind speeds. Material can then only be transferred onto the compact object as angular momentum is transferred outward by (possibly) viscous torques.

This simple picture is immensely complicated by the presence of any strong magnetic fields.


The moon is so big that the processes that circularize and reduce the equatorial inclination would take much longer. The moon is big because of how it formed: a huge collision in the early solar system. (Unlike, say the Galilean moons that probably formed along with Jupiter, or Triton, that looks like a captured TNO)

The other fact that makes its orbit complex is that the moon is the closest moon to the sun, so solar perturbation is more significant than for other moons.  Solar perturbation is the main cause of the periodic variation in inclination and the various other forms of precession seen in the moon's orbit.

The sun also pulls the moon out of a equatorial orbit and into a roughly ecliptic orbit as the perturbations of the sun become more significant than those of the equatorial bulge. As noted elsewhere  at lunar distance, the sun provides more than 50% of the gravitational field.

So the reason the moon's orbit is complicated is that it does follow the laws of physics, and with three (or more) bodies, those laws have complex effects.


Doing Correct Calculations with Binary Star Systems

The semi-major axis is 4.3 arcseconds. At a distance of 3.51 parsecs, this corresponds to 15.09 au. Using your formula I get 40.3 years, which is close enough given the 2 significant figure precision of the semi-major axis.


Is the composition of stars in future made of more and more heavy elements?

The initial stars were made of hydrogen and helium.

These enriched the interstellar medium (ISM) with some chemical elements right across the periodic table, when massive primordial stars ended their lives as supernovae.

Subsequent generations of stars continue to enrich the ISM, if their lives are short enough.

So the general gist of what you suggest is true, but the rate of enrichment is very slow. The Sun was born 4.5 billion years ago, but sun-like stars born within the last few million years also have a metallicity consistent with the Sun.

Why is this? Well, star formation is quite inefficient; only about 10 percent of a collapsing gas cloud ends up in stars. Then a star only processes about 10-20 percent of its mass as nuclear fuel; and then only a fraction of this is returned to the ISM in the form of winds and supernova ejecta.

To get a more enriched ISM requires bursts of massive star formation in dense environments. Such conditions occurred early in the life of the Milky Way and perhaps will again when we collide with the gas-rich Andromeda galaxy.

In conclusion, star formation will continue into the future; the stars will very slowly get more metal-rich; there is nothing to prevent more metal-rich stars forming; perhaps the main consequence would be a small reduction in the maximum possible stellar mass (due to the increased gas opacity and consequent radiation pressure).


First off - good job for paying attention to collimation. For a newtonian telescope, it is as important as changing the oil on your car.

It would help to know a little bit more about the telescope. What make and model? What is the focal length?

EDIT:

Saw your comment below. It's an f/4 parabolic mirror, you're always going to have coma at the outer edge. But the center should be aberration-free if the scope is collimated.

Beware - it's an inexpensive scope, so depending on your luck some residual aberrations may remain even after perfect collimation. But that doesn't mean you should stop trying to collimate it. On the contrary, the better the collimation the better the performance, always.

  I observe Venus and Jupiter with several eyepieces but I got some "coma like" effect all the time with this telescope.

If the image in the center is fairly good, but you get "coma like" effects at the edge, then that is indeed coma (perhaps mixed with other aberrations). It is normal at focal ratios of f/5 ... f/4 or faster. If the focal ratio is around f/6 ... f/8 or slower, coma is not so easily visible, or not visible at all.

  But when I test this setup with laser collimator it show vast miss-alignment in both primary and secondary mirrors.

It is a very common issue with cheap laser "collimators" that the collimator itself is miscollimated. The laser only helps if it is very precisely centered, otherwise it is misleading. Unfortunately, many, many commercial lasers are woefully off-axis.

Plug the laser into the focuser. Tighten the screw only a little bit, so that the laser is sitting there semi-snugly, but you can still turn it around its own axis (rotate it in place in the focuser). Fire up the laser. Now gently turn the laser around, while watching the laser spot on the primary. What does the spot do?

If the laser is centered, the spot should never move as you rotate the laser in place. If the spot moves even a little bit (try to avoid rocking the laser tube sideways, as that would falsify the test), if the spot describes a circle on the primary as you rotate the laser in place, throw that laser in the garbage can.

(Ok, some lasers can be adjusted, provided they have the 2 or 3 adjustment screws on the collar, on the lateral part of the laser tube. Sometimes the screws are hidden under a plastic or metal ring. If that's the case, try and rescue the laser by carefully adjusting those screws. Otherwise dispose of the laser, or destroy it, since it's misleading.)

Other things to try

See if you can get better results with the Cheshire only.

Try and make a very simple collimation cap. It's basically a plastic cap with a tiny hole in the exact center, that's all. The procedure for using it is explained here:

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

A more detailed procedure is explained here:

http://www.garyseronik.com/?q=node/169

If you're patient, you could collimate by the stars. It's complex and time-consuming, but it's guaranteed to be 100% accurate. It is always a good idea to use a quick star test at the end of whatever other collimation procedure you use - think of it as a reality check. The laser may or may not work properly, but the stars never lie.

http://www.astrophoto.fr/collim.html

Laser collimators can be very good and very quick - provided they are centered very precisely from factory. I use Howie Glatter's collimators, but they are not cheap. If you could borrow a Glatter collimator it would show you what's really going on.

http://collimator.com/

Is your primary center-dotted with good precision? In many cases, mass-produced telescopes come from factory with an off-center dot on the primary, which is as misleading as a bad laser. Even the best laser will not work properly if the primary is not precisely center-dotted. But you can always re-center the dot with good precision. See here:

http://www.garyseronik.com/?q=node/168

(Always use a donut shape to center the primary, instead of a solid dot. The donut allows you to use laser collimators, unlike the solid dot markers.)

More thoughts on collimation below. These articles are written by very experienced amateur astronomers and telescope makers. Highly recommended to read these before you begin.

http://www.garyseronik.com/?q=node/165

http://web.telia.com/~u41105032/myths/myths.htm

Try something simple (like the collimation cap) before you open up the scope and do "major surgery" on it. Then do a star test. Does it look pretty good on the stars? Then you're probably doing something right. The stars are looking pretty bad no matter what? Then perhaps you need to open up the scope, re-center the primary, check if the mirrors are sitting properly in their cells, etc.

Final comments

If you use too much magnification, the image will not look very sharp, no matter what. Stay within 100x ... 200x while testing. It is pretty rare that atmospheric conditions and/or the quality of mass-produced optics allow good performance above 250x.

Always check the results in the center of the image. With most mass-produced optics, performance degrades at the edge, and there's not much you can do about it (other than replacing them with better optics).

Before you do anything, take the scope outside and let it "breathe" for 30 ... 60 minutes. After it reaches ambient temperature, its performance might get better. More details here:

http://www.garyseronik.com/?q=node/55


Newtonian reflector collimation question

Your scope is f/5. There will always be some coma aberration.
Always, the image quality is worse as you get closer to the edge of the field of view (FOV).
Personally, given that the fact that you will want to observe your targets at the center of the FOV, this collimation would be acceptable.

Happy photon hunting!


Why does exoplanets' naming start from "b" and not from "a"?

A  quick lmwtfy reveals the answer:

  In the WMC naming system, the brightest member of a star system
  receives the letter "A". Distinct components not contained within "A"
  are labeled "B", "C", etc. Subcomponents are designated by one or more
  suffixes with the primary label, starting with lowercase letters for
  the second hierarchical level and then numbers for the third.[2] For
  example, if there is a triple star system in which two stars orbit
  each other closely with a third star in a more distant orbit, the two
  closely orbiting stars would be named Aa and Ab, whereas the distant
  star would be named B. For historical reasons, this standard is not
  always followed: for example Alpha Centauri A, B and C are not
  labelled Alpha Centauri Aa, Ab and B.


Is there any evidence that the Gas Giant planets in our solar system are experiencing orbital migration?

There is no direct evidence of change of orbits of the gaseous planets of the Solar System. But one could wonder about the stability of the Solar System, and if such events could happen again.

Theory:

The Solar System is a chaotic system, as most of gravitational systems involving $N$ bodies. The KAM theorems (for Kolmogorov, Arnold and Moser, the three authors of the theory) show that, for non-degenerate Hamiltonian systems, quasiperiodic trajectories exist, that could ensure the stability of the system. But unfortunatly the KAM theorems cannot be applied to the Solar System, because the planets are too massive.

Simulations:

It leaves us with numerical simulations, that show that the Solar System is indeed chaotic, and that planetary collisions, ejection and migration can happen. But if it is true for the telluric planets, it is not the case for the gaseous planets: one reason could be that they are not much perturbated by the inner planets.

Sources:

Laskar (1994)
Laskar (2012)


How can something infinitely big have expanded from an infinitely small?

The universe started with "big bang" is a slogan for "There was a time when the universe was hot and dense, and this time may be considered to be the start of the universe, and the start of time."
The universe has (as far as we know) always been infinitely large.  That's not certain, but we have no evidence contradicting it.  The universe was never an infinitesimally small point. It was always infinite at every time t>0.
We know how much matter the big bang has produced in the observable universe, and we can model this. (We don't really know why it produced this much matter, but we have enough parameters in our models to get a universe like ours).
We also think that the universe is kind of the same everywhere.  Again, there is no proof of this, but nothing contradicts it, and so there would be an infinite amount of matter in the universe.
There was no "blast wave".  Because the whole universe was hot and dense at the same time.  The whole infinite universe, filled with an infinite amount of matter, which then expanded. (and yes, an infinite thing can expand, it just means that things within it get further apart.)
As for t=0 or for t<0  We have no idea.  I don't know if those are even things that we can talk about. It seems that we can talk about the state of the universe at time t=1 second.  But we can't talk about the state of the universe at t=0 seconds, at least, not with the mathematical and physical knowledge that we currently have. Time t=0 is a singularity.


tiny diamonds in meteorite "signature of a long-lost planet": why come to Earth in 2008?

Many comets are such lucky ones...
That's why we see them with their icy tails evaporating: They have only recently been scattered into the inner solar system.
Just like the well-studied comet 67P, you could ask "where has 67P been for all those billion years?". For at least this comet, we know the answer: It has been orbiting in the outer solar system, until a close encounter in 1840 and a few subsequent later ones with Jupiter brought its' perihelion closer to the sun, where it started its' cometary activity.

I would place my bet on the origin of those comets being in the outer solar system, because:

The nature article you cite mentions 20 GPa pressure, for which one needs at least a Mercury or Mars type body to generate. This is not necessarily a problem, because Mars-type bodies form quick and early (we know this from theory as well as data), within the first 3-5 Million years of the solar system, when gas is still there. This is about 10 times faster than an Earth-type body, which has 10 times the mass.
It is also thought that Mars-type bodies should form the basic building blocks for larger planets - thus there should be many more of them in the early solar system.
However even if there are many, now there are essentially none in the inner solar system - the asteroid belt contains roughly 0.1% of a Mars mass. The total mass of the Kuiper belt is estimated to be far larger, by factors of 10-100.
If a giant planet instability occurred, it requires a even much more massive Kuipter belt as trigger, after the gas disappears in the early solar system.

So, Summarizing: They're probably out there in the outer solar system, merrily orbiting along.


Yes, you could definitely see Jupiter as a round object if it's
angular diameter was 369 arcseconds. The limit for seeing something as
round appears to be a diameter of around 1 arcminute. Notes:

The most direct comparison is M16, the Eagle Nebula. According to
http://www.nightskyinfo.com/archive/eagle_nebula/

The M16 complex is visible to the naked eye on clear nights as a hazy patch

(and thus not a single point), and http://messier.seds.org/m/m016.html
notes that M16 has an angular diameter of about 7 arcminutes.

http://www.astropix.com/doubles/ notes that:

In the real world, folks with excellent eyesight can resolve down to
about 2 arcminutes. Folks with average eyesight can resolve about 3
arcminutes.

Of course, resolving a double star as two separate points of light
isn't the same as seeing something as a circle, but provides a good
upper limit.

And
https://en.wikipedia.org/wiki/Phases_of_Venus#Naked_eye_observations
notes that:

The extreme crescent phase of Venus can be seen without a telescope by
those with exceptionally acute eyesight, at the limit of human
perception. The angular resolution of the naked eye is about 1 minute
of arc. The apparent disk of Venus' extreme crescent measures between
60.2 and 66 seconds of arc,[4] depending on the distance from
Earth. Nevertheless it is possible for observers with extremely acute
eyesight to see a crescent Venus under ideal atmospheric
circumstances.

Note also that Venus appears circle-like (not point-like) when it transits the Sun, but the same is true of Mercury, so this may be a special situation that doesn't count.


Will I be able to see the Leonid Meteor Shower during mid night?

Leo starts to rise at about midnight, local time, so you should be fine, weather permitting.  Note also that you may be able to see meteors even before then, as Leo is merely that part of the sky from which the meteors appear to originate, but you will likely be able to see them some distance away from Leo.

I don't know where the Google rise time comes from.

EDIT; the link you added shows Leo rising at about midnight, confirming that the google source is incorrect, or not being correctly interpreted.


When did astronomy first discover that the stars are bigger than the
  moon?

Your choice of words "discovery" implies evidence, not just hypothesis.  The discovery that stars were like our sun wasn't possible until the distance to some nearby stars was measured and with that, a brightness comparison could be made between stars and our sun.   That discovery was made in 1838.

The first person with that theory goes back to 450 BC.   Short summary and longer summary.

Quote:

  Anaxagoras, who lived in Athens, Greece, around 450 BC (about 2450
  years ago), thought that the Sun and stars were fiery stones, that the
  stars were too far away for their heat to be felt, and that the Sun
  was perhaps more than a few hundred miles in size. With that
  Anaxagoras was, as far as we know, the first one to suggest that the
  Sun is a star. His ideas were met with disapproval and he was finally
  imprisoned for impiety, because his ideas did not fit the prejudices
  of the time.

So, while he suggested that stars were like our sun, his size was way off and it's unclear how large he thought the Moon was, or even, if he had a good idea how large the Earth was.   Erastosthenes' estimate came later.

Aristarchus of Samos came up with a much better, but still quite inaccurate, measure of the Sun and Moon's size and he recognized that the Moon was smaller than the Earth and the Sun much larger.  It's unclear what he thought about stars though.

And much more recently:

  Christiaan Huygens of Holland lived from 1629 to 1695. He determined
  the distance to the star Sirius, assuming that that star was as bright
  as the Sun and appeared faint only because of its great distance. He
  found that the distance to Sirius must be very great. At this time,
  then, the idea that the Sun is a star was considered seriously by
  scientists.

So the idea that stars were like our sun was popular and perhaps even accepted, for over 100 years before the size and brightness of stars was first accurately measured.

  Did people believe that the moon was bigger than the stars for a long
  time?

This is hard to say.  Aristarchus of Samos ideas were less popular than Aristotle's  (who came a little earlier) and Aristarchus' theories weren't central to Ptolomy's model which became the model for over 1000 years.  I've often wondered what might have been different if Aristotle and Aristarchus could have exchanged ideas.  Aristotle might have learned something and re-written his model, but Aristarchus was born too late.

Ptolomy, who's model was accepted by the Roman Catholic Church, didn't pay much attention to sizes and, in fact, his model doesn't account for size well at all.  It's one of the ways that his model breaks down, but his basic model had the Moon closer to the Earth than the Sun, so presumably he recognized that the Sun was larger than the Moon and it's not unreasonable to guess that Ptolomy might have said the planets are smaller than the moon.  The stars, however, were for the most part, not addressed.

If your question is, in particular, what Aristotle, Ptolemy up through scholars in the middle ages thought about the size of stars in comparison to the moon, then, I have to say that I don't know what was assumed about the size of the Moon and stars, though Giovanni Bruno was burned at the stake for suggesting stars were like our sun.  It's entirely possible most scholars thought stars were smaller than the Moon from the time periods between Aristotle to Kepler.  If someone knows more specifics, I encourage them to answer.

Not long after Galileo and Kepler, it seems to have become more generally accepted that stars were large and very distant and the Moon was relatively small by comparison.


Where goes original asteroids mass after impact?

All three.  The impact velocity will be huge and the kinetic energy of astroid extremely large. The asteroid will be pulverized into millions of fragments of asteroid. Many of these will be as small as dust particles and will be vaporized from the heat associated with collision. Hence some small fragments will evaporate will other larger fragments will be buried or blown away from impact site. The scatter field is an ellipse about the crater elongated along direction of motion of meteor.


Is a comet necessarily confined to the solar system?

It means that the sum of the kinetic and gravitational potential energies can be greater than zero. Such objects are not bound to the solar system. They fall towards the Sun on hyperbolic trajectories, picking up speed and kinetic energy as they do so. After rounding the Sun they then have enough speed to escape from the solar system again.

The origin of such comets is not known for certain - some may have come from other star systems. Some comets can enter the solar system on parabolic orbits - these have come from a long way out - perhaps the Oort cloud - and are just bound to the solar system. But interactions with other planets and the Sun can result in them gaining energy and leaving the solar system on unbound hyperbolic orbits.


The really short answer is ... you don't (use Alt/Az data).  You own a German Equatorial Mount that uses RA/Dec values and have no need of the Alt/Az coordinate system.
A slightly expanded answer is that use of coordinates is meant to help you find objects.  Realistically it would be challenging to do with your telescope mount because the RA/Dec setting circles on the mount are not highly accurate (and sometimes move when they aren't supposed to move).  More on this later.
Right Ascension (& Declination)
The Right Ascension and Declination system for objects in space is somewhat like Longitude and Latitude coordinates on Earth.  But since Earth spins, we can't use actual Longitude / Latitude values.
A object directly above Earth's North Pole has a Declination of +90°.  An object above Earth's South Pole has a declination of -90°.  A declination of 0° means the object lies along a path somewhere directly above Earth's equator.
Right-Ascension works like a clock (which is why it is expressed in hours and minutes) based on 24 hour day.  If you look at the setting circles for your telescope, you'll probably notice that the Right Ascension setting circles have two sets of values... one counting up, the other counting down.  Those are actually to allow for use in the Northern vs. Southern Hemispheres (a telescope in the Southern Hemisphere is aligned to the Southern celestial pole, but the motor drive needs to spin the telescope in the opposite directly as compared to the Northern Hemisphere.
There is a 00 hour 00 minute position in the sky.  It's "name" is the "First point of Aries" (a confusing name because the point is not in Aries ... for reasons I'll get to in a moment).
The point is defined based on the position of the Sun (relative to the background stars ... which of course we cannot see during the daytime) at the Vernal equinox (when the Sun crosses the equator from the Southern Hemisphere into the Northern Hemisphere on or around March 20 each year).
But since Earth's axis precesses very slowly (a cycle that takes nearly 26,000 years), the point moves.  When the point was originally defined, the Sun was in Aries... but a couple of thousand years later, it has migrated to Pisces ... and is actually very close to Aquarius.
So imagine if you could point your telescope at the First Point of Aries (the 00 hour 00 minute 00 second point) and then ... switch off the motor and just watch the sky drift along from East to West.  How long would you have to wait before the imaginary "longitude" value of the object you want to view in the night sky is in front of your telescope?  That time duration... is how the RA coordinate gets it's value.
With the exception of solar system objects (planets, comets, asteroids, etc.) objects in the universe seem as if they don't move.  They do move... but are so incredibly far away that it requires very precise measurements to detect any movement at all.  This means that, for the most part, we can assign them fixed coordinates that don't change (these coordinates are occasionally updated).  On the timescale of a single human lifetime, we don't notice the movement.
So this means, conveniently enough, that we can mostly get away with a simple look-up table to find objects and ignore their proper motion.
This is why the RA/Dec values do not change when you look up objects in those apps... because they don't actually change from hour to hour or day to day.  Many years are required to notice even very subtle changes ... and using very accurate instruments to take those measurements.
Also, because of the way this coordinate system works... the RA/Dec coordinates for these non-solar-system objects are the same for all observers, everywhere on Earth, regardless of date or time (well... regardless of date or time excluding very large amounts of time... such as thousands of years ... at which point their movements would change enough to notice.)
One little tiny nit... Polaris is not located at Earth's celestial North Pole ... but it is close.  It's roughly 2/3rds of a degree away from the pole.  But keep in mind that the angular width of the Moon (as observed from Earth) is just about 1/2° wide in the sky.  This means there is enough angular width between Polaris and the true Celestial North Pole... that you could fit the Moon in that space with a little room to spare.
For visual astronomy ... pointing your mount at Polaris is "close enough".  A computer driven telescope would miss it's "go to" targets ... but probably get close enough that those objects would appear in the Finderscope (and just need to be re-centered). Not a big deal.  (For astro-imaging using long camera exposures ... a more accurate alignment is needed.)
Altitude Azimuth
This coordinate system is much simpler to understand.  But it has a major drawback.
The azimuth is simply the compass heading... in which compass heading do you need to be facing in order to view the object, and... how high above the horizon line do you need to look?  The compass heading is the Azimuth.  The angle above the horizon is the Altitude.  Simple enough.
But suppose you are collaborating with someone about an object you are observing ... so you want to share coordinates with your collaborator.  But your collaborator is not nearby ... they live on another continent.  The Earth is spherical and the curvature of the Earth means that the Altitude and Azimuth you use ... isn't the same as the Altitude and Azimuth your colleague will need to use.  Even if they live along the same latitude and are simply in an adjacent time zone... their coordinates will be different enough to completely miss finding your object.
This is why Altitude/Azimuth values aren't useful in catalogs.  They change ... a lot.
The German Equatorial Mount
The benefit of this mount is that once you orient the mount so that it's major axis (the Right Ascension) is parallel to Earth's axis, then as the Earth rotates from West to East ... the telescope mount can be moved East to West.  If this mount were equipped with a motor and if that motor spins the mount at the same rate that Earth spins... then the telescope will remain fixed on the same position in the sky.
Also, adjusting the mount in the RA direction causes the telescope to move in a perfect east-west direction.  Adjusting the mount in Declination moves the telescope in perfect north-south direction.
This simplifies the process of tracking objects in the sky because once you target and center on an object, only the Right Ascension motor needs to operate to track an object.  With an Alt/Az mounted motorized scope, a computer needs to interpolate the movement of both axes in order to track an object.  The exception is... if that telescope happens to be located directly at Earth's North Pole or South Pole ... then an alt/az telescope works just as well as a German Equatorial Mount (RA/Dec).
Converting from one system to the other requires a bit of math -- but not particularly complicated.
Using RA/Dec to find objects
I previously mentioned RA/Dec (for non-solar-system objects) is mostly just a lookup-table of values that don't change in a very noticeable way within the span of a single human lifetime.
Computer-controlled "go to" telescope mounts have to be "aligned".  The mount is polar aligned (physically).  The computer will typically want to know your geographic coordinates as well as the date & time to have a good idea of what objects should be visible in the sky.  Most "go to" mounts will then ask that you accurately center the telescope on a couple of bright stars ... and use this information to create a "fix" on the entire celestial sphere.  From this ... the computer can now automate moving the telescope to any object you want to observe.
As your telescope mount is not not motorized and controlled by a computer, what point is there to have setting circles with the RA/Dec values?
It turns out you can (theoretically) use the setting circles as a kind of analog computer.
Suppose you want to locate some hard-to-find object.
Start ... by pointing the telescope to a known easy-to-find object.  A really bright star such as Sirius... or Vega... etc.
Look up the RA/Dec values for that really bright star.  Now adjust the setting circle values on your mount to match the values in the table.
Next, find the RA/Dec values for your hard-to-find object.  Then re-point your telescope'S Declination axis to the declination of the hard-to-find object.  Now do the same for the RA value.
Theoretically your telescope is now pointed at the hard-to-find object.  I say "theoretically" because large professional telescope used many years ago to find these objects (before telescopes were controlled by computers) had very large settings circles that were much easier to use.   Consumer telescopes usually have tiny setting circles that are sometimes rather frustrating to use (My personal experience with my own budget mounts were that the bits that were meant to be adjusted to match an object and then remain "fixed" as you move the mount ... often slipped and moved while adjusting the mount.  This made accurate pointing a fairly frustrating experience.)
This means "star hopping" is probably the preferred method to find objects (find the nearest easy-to-find object .. then move the mount only slightly to locate your hard-to-find object).


What was the greatest spatial extent of the Asteroid Belt prior to the Late Heavy Bombardment?

There have been some studies based on observations on the best preserved impact craters of that time in the inner solar system, namely those on Mercury and the Moon.  In the paper, Impact-induced compositional variations on Mercury (Edgard et al. 2014), state that based on the nature of Mercury's craters, they have developed a

  New dynamical simulations of the early evolution of the asteroid belt hint at the presence of additional
  asteroids in a region interior to the present-day belt, known as the “E-belt”.

Through their modelling and simulations, they have determined that the 'E-belt'

  The recent reevaluation of the early history of the inner Solar System suggests the primordial asteroid belt could have originally extended well into the Mars-crossing zone

The 'Hungarias' are thought to be a remnant of this belt - as shown in the diagram below

Image attribution:
By Agmartin (Own work) [CC-BY-3.0], via Wikimedia Commons


The furthest horizon in the Solar system

The distance depends on the diameter of the planet and of your height above the surface (such as on a mountain). The greater the diameter, the farther away the horizon will be.

You can see this in the figure below. $d_3 > d_1 > d_2$. On a large planet your horizon (at distance $d_1$) will be farther away than on a smaller planet with horizon distance $d_2$. But if you stand on a mountain on the same planet then the distance will be even larger ($d_3$).

You can calculate the height from Pythagoras' rule because we have a triangle with one unknown edge. The distance from the planet's centre to the point where you see the horizon is the radius of the planet and the distance of the eye of the observer is the radius of the planet plus the height of the observer above the surface.
$$(r+h)^2 = r^2+d^2$$
where $r$ is the radius of the planet, $h$ the height above the surface, and $d$ is the horizon distance. If we rewrite this equation we get:
$$d^2 = (r+h)^2 - r^2 $$
$$d^2 = r^2+h^2+2rh-r^2$$
$$d^2 = h^2 + 2rh$$

For Venus we have $r=6052\textrm{km}$ and $h=11\textrm{km}$ (Maxwell Montes), we therefore get a horizon distance of 365km. This is still smaller than the horizon distance from Olympus Mons. Olympus Mons is the highest mountain and all other rocky planets are smaller, the horizon distance from Olympus Mons is, therefore, the greatest horizon distance.

Whether the horizon distance in the Rheasilvia crater is greater depends on whether you would be able to see the crater edge from the middle of the crater. Vesta's radius is 262km. If you stand on the crater edge, which is 13km above the crater surface, you would only have a horizon distance of 83km. This means that from the centre of the crater you would not be able to see the crater's edge.

EDIT Above I assume that the 13km is the height of the crater edge above the centre of the crater as measured from the geodesic. If the height is the height above a flat (not spherical flat) surface than the horizon is indeed 505km (see comments below).


What's the fastest moving object in the universe?

The answer to this is surprising:

We are.

And many (if not all) other galaxies.

And they move faster than light.

See, the universe is expanding, at an accelerating rate. The fabric of spacetime itself stretches out, so that galaxies seem to move away from each other. The interesting thing is that relativity does not forbid these from moving away faster than light. While local space is flat and the local speed of light must be upheld, this need not hold at a global scale, so it is possible to have frames which move away from each other faster than $c$. Indeed, there are some galaxies that are moving away from us faster than light (the only reason we see them is that they used to be closer and moving at a slower speed). Any pair of galaxies that are 4200 Mpc away from each other (that is, with a redshift of 1.4), are moving away from each other faster than light in each other's frames (numbers stolen from the linked page).

Since the only consistent way to talk about motion is relative, one can say that we are moving away from other galaxies faster than light, since the reverse is also true. This can put galaxies in the bucket of the fastest moving objects in the universe. As for which is the fastest, I don't know, we would have to find a pair of galaxies which are the farthest apart (distance measured in the frame of the galaxy, of course), but since the universe is probably more than what we observe1, we can't pinpoint the pair of galaxies for which this is true.

For those who think that it is cheating2 to short-circuit the question with space expansion, there are other objects that go faster than light (they are not the fastest objects in the universe though), and these can be found on good 'ol Earth.

Electrons:

In nuclear reactor cooling pools3, we have a phenomenon known as Cerenkov radiation. Basically, emitted beta particles move faster than the speed of light in water. This creates an effect of similar origin as the sonic boom, where strong light emanates from the medium.

Saywhat? You think I'm cheating again2 by putting everything relative to the speed of light in a medium?

Alright, fine. Here are some fast objects that don't require space expansion to be fast, nor do they involve any trickery of semantics where the medium in which they are being measured is not mentioned. Many have already been mentioned by astromax.

Tachyons:These are particles which go faster than $c$ — this does not violate relativity as long as they never decelerate to subluminal speeds. However, there isn't much (any?) experimental evidence for these. A lot of BSM models do predict their existence, though. So there's still some cheating here, on to bradyonic matter:
Gluons: These are massless, and though they don't occur freely (except possibly in glueballs, though these most probably have mass) they do travel at $c$. But these can't move at any other speed, so again, this is slightly cheating. On to fermionic matter:
Neutrinos: Now these are viable candidates. The electron neutrino is known to have very, very little mass (we have an upper bound for it, which gives ), and as a result it can easily attain very high speeds. Put it in a gravitational field, and it goes even faster. If you want macroscopic objects, however:
Stuff spiraling around spinning black holes: Black holes have a strong gravitational field, and when rotating, they can impart angular momentum (lots of it) to nearby objects like accretion disks. Objects close to a black hole are accelerated to pretty high speeds. In fact, if an object is within the ergosphere, it moves faster than light from the point of view of certain frames of reference.
Stuff falling into black holes: From the faraway frame, an object speeds up and approaches the speed of light as it approaches the horizon of a black hole. Arbitrarily large speeds bounded by $c$ can be attained here.
Black hole plasma jets: Jets being flung out of black holes can get pretty fast relative to each other.

1. Due to cosmic expansion, there can be galaxies that are no longer visible to us. Some galaxies may never have been visible to us, if we start watching from when galaxies started forming.

2. I, for one, agree with you.

3. And other places where you have massive particles being emitted really fast into  a medium


Your intuition is correct, as far as classical dynamics is concerned. The main problem, I think, arises in that you are referring to a strictly quantum mechanical phenomenon - what is the physical singularity? In General Relativity (GR), we know these singularities exist, but we cannot really know more than that (aside from events that occur not too close to the physical singularity). But, it's deeper than that, because in GR we cannot get information out of a black hole, so we'd never really know. That's why people in the comments are referring to a "theory of quantum gravity" which would take over where GR breaks down (at the physical singularity).

For a non-rotating (Schwarzschild) BH, the physical singularity is at $r = 0$, but the coordinate singularity is at the event horizon, $r = 2M$.

For a rotating (Kerr) BH, the physical singularity is now a ring with no thickness but non-zero radius - this is the ringularity. We literally don't know what should happen here (need "quantum gravity"), but in GR we can't even know in principle, because if someone were to try to fall into the ringularity we would never receive any signals back from them (as far as GR tells us).

OKay, with all that out of the way, we can still make a similar comment like what you want to within the framework of GR:

Consider a Kerr BH (in Boyer-Lindquist coordinates) with dimensionless spin $a$ whose symmetry axis is along the $z$, so that the ringularity lies in the $x$-$y$ plane. One can show that at $z = 0$ and at $r = 0$ (where $r$ is not the physical radial distance), the equation of the ringularity is

$$ x^2 + y^2 = a^2$$

and so the radius of the ringularity is the spin. Therefore, as the spin of the Kerr BH increases, the radius of the ringularity also increases. This is the opposite of what your classical intuition leads to, but then again, this is a very nontrivial spacetime metric, and it exists on the fringe between classical and quantum physics.

  Why doesn't a singularity spin infinitely fast?

the ringularity is not spinning itself, rather it is defined by the spin of the BH. Mathematically, we can construct some kind of "spin" of the ringularity, but it wouldn't make much physical sense.

Now, if the BH was spinning infinitely fast, i.e. $a \rightarrow \infty$, then the ringularity would have an infinite radius!!! And THAT's certainly problematic (but interesting!). Fortunately, as far as we know, BH's cannot have a spin greater than 1 - the maximum spin of a Kerr BH is capped by the naked singularity hypothesis. You see, if BH's could have a spin greater than 1, then the ringularity would extend "outside" the event horizon, which is problematic because it would mean that we could observe the collapse of an object into infinite density. ;)


The first question anyone asks about a telescope is "what is the magnification?" It is almost always not the most important thing. Any telescope can magnify a million times, given a short enough eyepiece - the problem is, how good the image is.

For observing planets, the main thing is resolving power - the ability of the telescope to discern tiny details. The resolving power is limited by aperture, or the diameter of the primary lens or mirror. If the primary lens diameter is measured in mm, and the resolving power in arcseconds, then the limit is:

resolving power = 100 / aperture

Your 50 mm diameter lens would have a resolving power limited to 2 arcseconds in ideal conditions. Any details smaller than that would be blurry, no matter how much magnification you put into that thing. For comparison, Jupiter's apparent diameter is between 30 and 50 arcsec, Mars' between 3.5 and 25 arcsec, Saturn's between 14 and 20 (without the rings).

With that aperture you would see the rings of Saturn, and a couple equatorial belts on Jupiter. Venus shows a crescent when it's far from the Sun. Mars is a little orange disk when it's closest to Earth. I've looked at planets through 50 mm of aperture, and you can see quite a few things that way.

The Moon also looks interesting, you can see craters and mountains, plus it's available every month.

Because resolving power is limited by aperture, as you give it more and more magnification, at some point the image is just huge and bloated; big, but blurry. So there's a limit to the useful magnification - again, this depends on aperture. If the aperture is measured in mm, the formula is:

maximum magnification = 2 * aperture

Your 50 mm aperture would support up to 100x magnification before things get too blurry.

So how do you calculate magnification for a given instrument? It is the ratio between the focal length of the primary lens (or mirror) and the focal length of the eyepiece (the lens close to your eye):

M = F / f

For your 1000 mm focal length primary, you achieve 100x magnification with an eyepiece with 10 mm focal length. Anything shorter than that would just bloat the image uselessly. Eyepiece focal lengths that would make sense in your instrument would be between 10 mm and 120 mm.

Words of advice:

Make sure the lenses are perfectly aligned. When you build the instrument, have some way to adjust the direction of the primary lens, so as to align it with the main axis of the instrument. Some tweaking will be required - tilt the primary a fraction of mm, check the results, tilt again, repeat until the image looks best. This is called collimation and it's very important for the overall performance of the instrument.

Have some way for the eyepiece to move back and forth until you achieve perfect focus. This will have to be re-adjusted every time you observe. The adjustment is quite sensitive. A real focuser would help a lot, but this could be achieved also with two tubes sliding into each other, held in place by friction, like in an old pirate captain's telescope.

For astronomy, when the scope is focused at infinity, the distance between primary lens and eyepiece is the sum of their focal lengths. E.g. with a 1000 mm primary and a 10 mm eyepiece, the distance would be 1010 mm. This assumes a convergent eyepiece.

The telescope will not work well hand-held. The image will be jumping around too much. You will need some kind of mount. At the very least put the top of the tube on a fence or something. But astronomical instruments work best when they are on a rock solid mount. Even a broomstick hammered into the ground, with a loop on top for the tube, is better than nothing.

Whichever way you mount the primary lens, don't squeeze it too hard in the lens cell or tube. It will deform the lens and make the image bad. Some pressure on the edge is okay; lots and lots of pressure not okay. You could glue the edge of the lens on some kind of ring. Just use common sense.

This is totally doable. I've built a telescope just like that, many years ago. It's great to see what you can achieve on your own.

Good luck!


What is the scale of things you can see with gravitational lensing?

You are right that the stars seen on the sky are within the Milky Way. Only with a large telescope is it possible to resolve individual stars in other galaxies, and only for the nearest ones.

I don't know which sources you refer to, by I think perhaps you are confusing the different types of gravitational lensing. I cannot explain them better than the excellent review by astromax, but briefly, there are three types:

Strong lensing, where are foreground galaxy cluster (i.e. a group of $\sim$100–1000 galaxies) magnify and severely distort background galaxies,
Weak lensing, where clusters or individual galaxies distort the shapes of many background galaxies on the percent scale, which can only be seen statistically, and
Microlensing, where a single object inside the Milky Way happen to pass in front of another single object, also inside the Milky Way. These objects are usually stars or planets, and do not distort the images of the background objects, but merely increase the flux for a while. This effect has been used to find exoplanets.

Whereas type 1 and 2 are more or less static in a human lifetime, type 3 is an event that happens once only for a given set of stars (as seen from Earth).

Scales

Strong and weak lensing happen on very large scales, from a few hundred million lightyears, up to the order of the size of the observable Universe (e.g. Wong et al. 2014). While the lenses themselves are cluster of galaxies, and thus a few to $\sim10$ megaparsec across, the lenses object are typically individual galaxies. Gravitational lensing is most efficient when the lens is halfway between us and the background source.

Occurring in the Milky Way, microlensing, on the other hand, happen on the scale of a few kiloparsec, again with us-lens distance being of the same order as the lens-background object distance (see e.g. this Wikipedia list).


There is sound, but you cannot hear it.

1) There is no sound in space:
This shows how sound gets fainter in an evacuated glass jar. The same happens with an explosion (or a shout) in space - there is no air to transmit the pressure waves, so no sound.

2) There is sound in space:
The interstellar and even intergalactic medium is not completely empty, even though it is much more empty than the best vacuum we can create on earth. Nonetheless, there can be pressure waves in those media.

 (The bullet cluster, taken from NASA's APOD)

This image shows the mass distribution of two colliding galaxy clusters in blue, and the X-ray gas in red. The galaxies in the clusters have passed through each other without much disturbance, but the intra-cluster gas has collided. The red cone to the right of the centre looks a lot like the sonic boom familiar from aircraft.

For more regular sound waves (repeating pattern as opposed to the one-off of an explosion), you might find those in the interplanetary gas / solar wind of oscillating stars (RR Lyrae or Delta Cephei type stars). I am not sure if anyone has ever measured this.

At any rate, while human ears can pick up sound in the range of ca. $10^2$ to $10^4$ Hz, these sound waves would have a much much lower frequency. Also, the force exerted by these pressure waves on a human ear would be much too small to register.


What would a 220km diameter asteroid impact do to Earth?

Using the impact effects calculator
The energy before impact (making a some reasonable assumptions) is $3.35 × 10^{27}$ Joules = $7.99 × 10^{11}$ MegaTons TNT, most of which would be deposited in the Earth.
The impact effects calculator doesn't discuss the formation of antipodal mountains (this is not a well-understood process) but it certainly seems possible that there could be major antipodal effects. The initial crater would fill with lava, but there would not be a planet-wide magma ocean.
Would life survive? perhaps, life is surprisingly resiliant.  This would cause mass extinction, and probably wipe out much or all of complex life. But "until the sun expands to a red giant" gives us millions of years for evolution to find a way. But any answer would have to be speculative.
The average interval between impacts of this size is longer than the Earth's age.