What are the objects that are visible from a city?

We can't know this without knowing what the sky is like for your city. However, you can determine what the limiting magnitude is for your location, then you can set planetarium software to only display catalogued stars that are brighter than this limit. You can do this for naked eye objects, but you could also do this using your telescope to find the limiting magnitude for your equipment also.


How to calculate the true length of day on a celestial body?

If you know the length of a sideral day and the length of a year, you can determine the length of a solar day.

From Wikipedia:

After one year, there will have been one more solar day than sideral day. Therefore :

solar day = sideral day / (1 - (sideral day/orbital period))


When was the martian dichotomy first observed?

It looks like the Mariner 9 era.
Based on a review paper by Watters et al. ("Hemispheres Apart: The Crustal Dichotomy on Mars"):  "The north-south asymmetry ... was clear from the first global image mosaics of Mars returned in the 1970s by Mariner 9 (Mutch et al., 1976) and the Viking Orbiters."  Other papers cite a 1973 paper by Hartmann ("Martian cratering 4, Mariner 9 initial analysis of cratering chronology") as one of the first papers to describe it, but it’s still based on Mariner 9 images.
I suppose in hindsight one could look at earlier flyby imagery and see that there is a dichotomy, but with 23 images from the first flyby, that's hardly representative of the surface and not enough to draw such conclusions.  Similarly, one could probably look at Earth-based telescopic observations from before the Space Age and retrofit observations into a north-south dichotomy paradigm, but I don't think one could legitimately claim earlier discovery of it.
So, 1971-1972 (Mariner 9 entered orbit in late 1971).


This appears to be the geocentric latitude ($\psi$), which is the angle between the equatorial plane and the point on the surface of the ellipse. It can be calculated from the geodetic latitude (also known as the geographic latitude) ($\phi$) and the eccentricity of the ellipse ($e$):
$$\psi(\phi)=\tan^{-1}\left(\left(1-e^2\right)\tan\phi\right)$$
If $e=0$, then the two lattitudes are the same, because the focus is at the center of the ellipse.

Image courtesy of Wikipedia user Wmc824 under the Creative Commons Attribution 3.0 Unported license.

It might make more sense to use this latitude to describe the position of an orbiting object, but the vector from the focus to the object's position $(r,\theta)$ is perpendicular to the tangential component of the object's velocity vector, meaning that the geodetic latitude may be better.


The creator seems to be referring to flare stars. Flares may be magnetic in origin, like various manifestations of the Sun's magnetic field. These flares can be quit luminous across the electromagnetic spectrum, including in the x-ray wavelengths.

Some red dwarfs are flare stars, though flare activity may be largely restricted to a small part of a red dwarf's lifetime. However, during this period, life on a planet orbiting close to a red dwarf could be in serious danger.

The reference to convection may be that red dwarfs are generally fully convective, unlike other stars. Magnetic fields in stars are generated via the dynamo effect; in red dwarfs, full convectivity can play a huge role in generating strong magnetic fields, and thus stellar flares.


What produces gravitational waves with "periods between about 100 - 8000 seconds"?

Any binary system produces gravitational waves at twice it's orbital frequency, i.e. with periods of half it's orbital period. So binary systems with periods between 200s and 16000s will produce such waves.
We can use Kepler's third law to say something about these:
$$ a = \left(\frac{GM}{4\pi}\right)^{1/3} P^{2/3},$$
where $P$ is the orbital period, $M$ is the total mass of the binary system and $a$ is the orbital separation.
For a binary with $M\sim 1M_{\odot}$ and $200<P<16000$s, then $0.11 < a < 2.00\ R_{\odot}$. Since normal stars of mass $\sim 0.5M_{\odot}$ have radii that are similar to this, then the stars would probably need to be stellar remnants (white dwarfs, neutron stars or black holes) except right at the longest period end, where it might be possible to observe W Uma binaries. More massive binaries have separations that increase as $M^{1/3}$, but the radii of normal stars increases more like $M$, so this conclusion is even firmer at larger masses.
It could be possible to have a compact binary involving a low mass star plus a compact object - perhaps a Roche lobe filling one, so as well as "double degenerates", the long period end of this range would include Cataclysmic Variables and Low Mass X-ray binary counterparts, with orbital periods of a few hours. Here is a prime example Time domain astronomy and fastest eclipsing binary ZTF J1539+5027 (+20 mag, 6.91 minutes): How to measure its minimum brightness?
Of course gravitational wave strain goes as something like $M P^{-4/3} d^{-1}$, where $d$ is the distance. These binaries are much longer period than the (presumably rare) massive, merging black holes seen so far and so probably need to be close, in our own Galaxy, to be detected.
e.g. LIGO was capable of detecting $M \sim 30 M_{\odot}$ merging black holes, with $P \sim 0.02$ s at distances of a billion light years. A similar strain amplitude would be produced by a $M\sim 2M_{\odot}$ binary with $P= 200$ s at a distance of 300 light years.


It appears to me that the full Spizter frame should normally be 256x256 but, if observing a very bright source or when requiring high temporal resolution, a different observing mode is performed wherein only 32x32 pixels are read from the CCD. However, what happens is that the exposure times are short and a total of 64 frames are observed in total, creating a data cube of 64x32x32. If you want, you can treat each individial 32x32 frame as its own image, or else stack them all together for the full image.

Full details can be found here.


Ultima Thule is a contact binary comprised of two formerly separate solid objects that gently1 (probably <5 km per hour) came together.  If the objects had been molten when they came together they would have formed a spherical object.  Since the lobes are distinct the objects must have been solid.

As for "understand[ing] how these two elements merged together", Ultima and Thule are believed to be have been two objects that formed separately through accretion.  At some point the two (solid) objects collided at a very low velocity of "one or a few miles per hour"1.  Since the escape velocity at point of impact is about 16 miles per hour2, gravity alone would be sufficient to keep the two lobes together1.

1 This topic was discussed in the 2019-01-02 New Horizons press conference.
The presentation talking about formation mechanism starts here (the portion discussing approach speed is ~80 seconds later).  A discussion of how strongly bound the lobes are starts here.

A note about the video links above: The official JHU APL video linked to above is missing the first few minutes.  If the video is updated to include the complete press conference then the links above will point to the wrong times.  The first two links are to Jeff Moore's presentation and the third is him fielding a question from Bill Horowitz (CBS News).

2 To quote the source:

  Modeling it as two spheres, one 16 km across the other 12 km, with a density the same as water, ... [a]t a distance of 8 km from the center of mass, the escape velocity ... would be about ... 16 miles/hour.


Recognising lunar month only by looking the sky

Not easily.

The Lunar Islamic calendar is based on a strict succession of months, with each month based only on the position of the moon, and no intercalated months or months extended to 31 days in order to keep the cycle of months aligned with the solar year. The start of the months are determined by the moon, the names of the months are not defined by any astronomical feature.

So to determine which Islamic month it is you need to know the actual amount of time that has passed. Now the positions of the planets change day by day, and by careful observation of the planets, and given a detailed ephemeris of their positions, one could work out the exact date, and therefore the Islamic month.

This is similar to determining the day of the week by astronomical observations, since the start of the day is based on the motion of the sun, but the name of the day is not defined on any astronomical feature.


Rendering asteroid orientation as seen from Earth

From the paper by M. Kaasalainen, J. Torppa, and K. Muinonen, equation (1) is:
$$ \vec r_{ast} = R_z(\omega(t-t_0)+\phi_0) R_y(\hat\beta) R_z(\lambda) \vec r_{ecl} $$
To reverse this equation you must reverse the order of the rotations and the rotation angles:
$$ \vec r_{ecl} = R_z(-\lambda)R_y(-\hat\beta)R_z(-\omega(t-t_0)-\phi_0) \vec r_{ast} $$
From your description, you reversed the order of the rotations but not the rotation angles.


How are shockwaves in a star that cause supernova created

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


A factor of cos(declination) in the formula of apparent of drift velocity of stars

All stars have to cover an angle of 360° in one sidereal day.

is close but incomplete; we should really say that right ascension covers 360° in one sidereal day.
Let me state that analogy: Imagine that you are near the north pole. You start walking with the velocity of 5 km/h to the east. In a short amount of time, you are at the beginning. You went through all the time zones and geographical longitudes. Now you fly to the equator and do the same. Now you need a lot of time to go around the Earth. Change of geographical longitude is same in both cases. But the angle, measured from the center of the Earth, is different - at the north pole 0° and at the equator 360°.
With that analogy, you can understand the stars movement.
Let me give the last proof: has the north pole star same angular speed as Rigel?


In theory it is possible to do time travel. There are several possibilities from theoretical point of view. Two aspects are there, travelling to the future and travel back to past. Although many scientists including Stephen Hawking said that travelling to the past is not possible (as there are several paradoxes including "the grand father paradox") that results in some weird phenomena (like in grandfather paradox, if you are allowed to travel back in time and you went to the time of your grandfather's youth and kill him then you are not going to born in the first place in future, thus a contradiction). But travelling to the future is seems possible. There are several ways to do that, thanks to The Theory of Relativity. As Einstein said, time flows at different speeds in different part of the universe, faster in Earth as compared to the near vicinity of a black hole, due to the stronger gravitational pull of the black hole then the Earth. So if you travel near to a black hole and keeping a safe distance and starts revolving around it for few hours and return back to Earth, then you will find the many decades has been passed here, due to the different speed rate of time in the two places, so essentially you are in future (but alas you cannot go back!). Another way is to travel at a very high speed (almost near the speed of light) in a spaceship for few years also take you to the future back on Earth, this is because time slows down in case you speed up. And many more concepts of time travel exists involving Wormhole and others.

Grandfather paradox
Time travel
See this


Was the Sun's gravitational lensing observed in other solar eclipses than the one in 1919?

Yes, observations of this kind are within the technical scope of amateur astronomers.  Several groups succeeded in replicating the experiment during the 2017 eclipse that crossed the USA.

For example Donald Bruns measured deflections of 2.8 arcseconds of multiple stars.

Nasa published a "How To" page for anyone wanting to test GR themselves.


How dense would planet earth have to be to have the same gravity as Jupiter?

There are at least two interpretations to this problem:

Per Wikipedia, Jupiter's surface gravity is $2.528$ times Earth's. Thus,
if the Earth were $2.528$ times denser, it would have the same surface
gravity as Jupiter. The Earth's current density is $5.514$ grams per
cubic centimeter, so the new density would be $2.528 \times 5.514$, or about
$13.9394$ grams per cubic centimeter. This assumes we change Earth's mass, but not its radius.

@Rob_Jeffries answer assumes the Earth's mass remains constant and the radius changes. If the radius shrinks by a factor of $2$, the volume decreases by a factor of $8$, and the planet becomes 8 times more dense. The surface gravity increases by $4$, since it depends on the radius squared. In general, shrinking the planet's radius by $x$ will increase the density by $x^3$ and the gravity by $x^2$. If we want the gravity $2.528$ times higher, we choose $x = \sqrt{2.528}$ or right around $1.590$. This makes $x^3$ equal to about $4.019$. Multiplying that by Earth's current density of $5.514$, we get $22.16$ grams per cubic centimeter, pretty close to what Rob got.

So, you can't really change the density without changing anything else: either the mass or the volume must change.


Questions about convolving/deconvolving with a PSF

Convolution is not a uniquely invertible process in the presence of random noise in your image. Deconvolving a noisy image can give misleading results, even if you have perfect knowledge of the PSF.

In general, when you are fitting models to data, it is far better to compare the models and data in the observational space of the data, where the uncertainties are reasonably well understood.


Is the North star going to stay in the North sky?

The relative direction to an object depends mainly on two things:

The location of the observer and the object
The coordinate system you use to determine direction.

So the first part of that concerns where we are compared with Polaris (and by extension of your question, where the other stars are as well).  Yes the stars and other components of the galaxy all have certain motions.  But this motion is not large compared to the distances between stars.

Most of the stars we can see in the sky are quite nearby, and are orbiting the galaxy at similar speeds.  The speed of the solar system around the galaxy is around $230 km/s$.  The relative speed between our sun and nearby stars is much less than that.  If we assume that most stars are farther than $5ly$ from us, then we can calculate how long it might take to make a $1\%$ change in distance.
$$t = \frac{d}{v}$$
$$t = \frac{4.7x10^{23}m}{2.3x10^5m/s}$$
$$t = 2.06x10^{11}s$$

That works out to over six thousand years.  There's a lot of caveats with that number, but it gives you a good idea that for the most part it takes too long for stars to move relative to each other for you to notice during your lifetime without very careful observation.

The second point is that things can move if we change what we mean by "north" or any other direction.  North, both on earth and in astronomy, is defined by the rotational axis of the earth.  But the axis changes its direction with respect to other stars over time.  Besides some smaller scale wobbles, the largest change to the direction of the axis is due to precession and has a period of about 26,000 years.  This change will take "north" from a direction that passes relatively near Polaris in a large circle around the sky.  With such a long period, again this isn't something to be seen by individuals without detailed measurements.

So the answer is that people alive today will be able to always think of Polaris as the pole star, but this will be less true for future generations as changes are easily visible on the scale of thousands of years.


How long would an occultation by a TNO last?

From the observer's point of view 1 million km away, the TNO's apparent angular motion is

$$\mathrm{\frac{8~m/s}{10^9~m} = 8 \times 10^{-9}~rad/s = 0.00165~^\circ/h}.$$

Assuming that the observer at 18 billion km = 120 au is in a circular orbit around the Sun, the orbital period is
1203/2 = 1320 years, making the Sun appear to move the other way at

$$\mathrm{\frac{360~^\circ}{1320~y} = 0.27~^\circ/y = 3.1 \times 10^{-5~\circ}/h},$$

so the TNO's apparent motion relative to the Sun is

$$\mathrm{0.00165~^\circ/h + 3.1 \times 10^{-5~\circ}/h = 0.00168~^\circ/h}.$$

The Sun would be totally occulted for

$$\mathrm{\frac{0.126~^\circ - 0.004~^\circ}{0.00168~^\circ/h} = 72.6~h}$$

and partially occulted for

$$\mathrm{\frac{0.004~^\circ}{0.00168~^\circ/h} = 2.4~h}$$

at each end.


How can we map masses orbiting the outer Solar System by determining the orbits of planets and smaller objects?

Iorio (Preliminary constraints on the location of Telisto/Planet Nine from planetary orbital dynamics) has suggested transmissions from the New Horizons probe (currently beyond Pluto) could be used to test for the influence of the proposed new planet, and provide constraints to its possible position.

It is estimated that the position of New Horizons could be affected by the proposed planet, of the order of 100 metres or more by 2019. It is stressed that this is a preliminary proposal:

  Moreover, the present investigation should be considered just as a preliminary sensitivity analysis which, by no means, is meant to replace a full covariance study based on extensive simulations of the ranging data from New Horizons and their reduction with accurate models including also PX itself, which is outside the scopes of this paper. Nonetheless, the possibility of using the telemetry of New Horizons seems appealing, and would deserve further, more detailed and dedicated studies which will, hopefully, prompted by the present one.


You're actually describing the twin paradox.
That's a little beyond Special Relativity, since the astronaut accelerates, and deaccelerates. Since Special Relativity treats only inertial frames of reference, accurate treatment would require some extension towards General Relativity.
But you may consider two astronauts, one travelling with constant velocity towards Earth, the other one with the same (by amount) velocity away from the Earth, and then apply the formula for time dilation of Special Relativity to both astronauts, as an approximation, and to avoid the twin paradox.
Then you get the almost ubiquitous factor $\sqrt{1-\beta^2}$ of Special Relativity, with $b=v/c$ the ratio of the relative velocity to the speed of light. For $\beta=0.1$, hence 10% the speed of light, we get a factor of

$\sqrt{1-\beta^2}=\sqrt{1-0.1^2}=\sqrt{1-0.01}=\sqrt{0.99}\approx 0.9949874371$.

Divide 10 years by 0.9949874371, and you get 10 years plus 18.4 days on Earth.

For $\beta=1$ this division would fail, implying a break-down of the formulas of Special Relativity. That's why travelling with the speed of light doesn't work for astronauts respecting Special Relativity.


Notifier about visible ISS passes and/or Iridium flares for Ubuntu PC

Here is something that I wrote for my personal purposes.

Fetches iridium flare details based on your heavens-above credentials and displays a notification in Ubuntu.

Github

I had one for ISS passes as well. Will find it and push it to the repo soon.


What is the highest angle the moon ever makes above the horizon at the North Pole?

${R_e}/{R}$ is about 0.017 radian or 1°.
Using your trigonometry and these values:

Lunar orbit inclination = 5.15° (varies between 5.0° and 5.3°)
Lunar perigee = 362600 km, apogee = 405400 km
Earth polar radius = 6357 km
Ecliptic obliquity = 23.44°

I get 27.7° at perigee and 27.8° at apogee.
At that altitude, atmospheric refraction is only about 0.03°.
The highest apparent altitude I found with JPL HORIZONS was 27.91° on 1987-09-15 around 17:15 UT.

Due to nodal precession,
the Moon doesn't reach ±28.6° geocentric declination every year, only near a major lunar standstill every 18 or 19 years, e.g. in 2006 or 2025.
Near a minor lunar standstill, e.g. in 2015, the Moon's declination is limited to ±18.3°.


Why do objects burn when they enter earth's atmosphere?

You'll often hear that it's because of friction, but that's often not the main factor. For larger objects it's more likely the pressure they create.

In both cases the reason is the enormous speed, often tens of kilometers per second. When a larger object enters the atmosphere at these speed the air in front of it gets compressed so much that it becomes extremely hot. (Think of pumping up a tire; you're also compressing air and you can feel the valve becoming hot.) The compressed air will often disintegrate the object in the air, and then the debris may burn because of the heat. This is exactly what happened to the asteroid above Russia last year: it exploded with an enormous flash in the air, and left little traces on the ground.

This happens on other planets as well, if they have a sufficiently dense atmosphere. In 1994 the comet Shoemaker-Levy crashed into Jupiter. It disintegrated before entering Jupiter's atmosphere due to the strong gravitation, but when the fragments entered the atmosphere they could easily be seen lighting up as they burned up.

edit
Remember the Space Shuttle? It had heat resistant tiles on the bottom of the craft to protect it from burning when it entered the atmosphere, even though its speed is only a fraction of a meteorite's speed when that enters the atmosphere.
During the last launch of the Space Shuttle Columbia some material from the external fuel tank damaged this heat shield, and upon re-entry the heat and the highly pressurized air under the craft could enter it, causing the craft to disintegrate and kill all crew.


How do the orbits of Nu Scorpii and AR Cassiopeiae work?

I checked the Multiple Star Catalog for data on these systems. It has hierarchical information on Nu Scorpii and AR Cassiopeiae, which I also used for this answer. I'll try to explain the orbital structure of both systems.
Nu Scorpii
Think of Nu Scorpii as two subsets of stars, Nu Scorpii AB and Nu Scorpii CD, orbiting each other with a period of 105,000 years. AB contains four stars:

Nu Scorpii Aa, Ab and Ac: Aa and Ab are a binary system (period: 5.5 days) orbited by Ac (period: 8.3 years).
Nu Scorpii B: This star orbits Nu Scorpii A - the three stars above - with a period of 675 years.

Nu Scorpii CD contains three stars:

Nu Scorpii C: C is a single star.
Nu Scorpii Da and Db: Da and Db are a binary system (period: unknown), which together orbit C over 3,100 years.

AR Cassiopeiae
AR Cassiopeiae is more complicated; it contains three main subsets:

AR Cassiopeiae AB: The contains three stars: Aa and Ab, a binary orbiting each other every 6 days, and B, orbiting A every 545 years.
AR Cassiopeiae CD: These pair of binary stars orbit each other once every 2,200 years, and orbit component AB in 438,000 years.
AR Cassiopeiae FG: This is another binary, with a period of 58,000 years. It orbits AB every 381,000 years.

Multiple star systems are usually nested like this - binaries and triple systems inside more binaries. I would image it's rare to have any other setup, like four stars orbiting together, because such a system would be highly unstable. You really need to nest the stars, with large separations between components (i.e. the distance between Nu Scorpii AB and CD must be much larger than the distance between Nu Scorpii A and Nu Scorpii B).


Calculating and mapping the best visibility of celestial occurrence

Partial answer to

Is it possible i.e. with Python or other (I am only basic with soft)?

Yes!
Use Skyfield.
As long as you can muster a little Python you can search for events, make topographical maps of event times and elevations, or even put it in a loop and find optimized locations.
The nice thing about learning to use Skyfield is that you can then do all your event searches, plots, predictions etc. this way in the future, and it's a great way to improve Python skills as well.
Skyfield also has a Github page where there are discussion and issues pages that are both very active.
You can also view other Stack Exchange questions tagged with Skyfield here and in Space Exploration SE.

See also:

Sunset on [Mountain] from [Viewing Point]
How to determine when Sun/Moon is aligned to two location on Earth's surface?
Where can I find the positions of the planets, stars, moons, artificial satellites, etc. and visualize them?
Inverse of the sunrise equation - finding locations with a given sunrise time on a given day


Excluding a rectangular region from an image in FITS

Usually pixel coordinates.
What I would do is to let DS9 make a region file, as your documentation site says it supports them.

To do so:

Open file in ds9
From region->shape select the kind of region you want.
Click on edit->region
Draw the shapes you want (just click on the img)
Click region->save. In the coordinates box choose physical pixel coordinates.

Probably that should work... Good luck!


About the formula for taking in count of the observer altitude in sunrise/sunset formula

It's simple geometry, you are calculating the downward angle you need to look relative to the horizontal, from a given elevation, so see the horizon. To lowest order, that angle simply gets added to the angle of rotation of the Earth to make the limb of the Sun reach the horizon.

To calculate that angle, let's measure all distances relative to Earth radius, and say you are at altitude $x$ above the surface. Draw a straight line to the horizon, where that line makes a right angle to the radius of the Earth (it is tangent to the horizon at that point).

So now you have a right triangle with hypotenuse $1+x$, and a side of length $1$, so the angle between the direction you are looking, and the direction to the center of Earth, has a sine of $1/(1+x)$. That means the downward angle we want is however much less that angle is from a right angle. Let's call the downward angle we want $y$, so that $y \ll 1$ if measured in radians, and then what we have from the above is $\sin(\pi/2 - y) = 1/(1+x)$.

Using a trigonometric identity then says $\cos(y) = 1/(1+x)$. But since $x$ and $y$ are $\ll 1$, $\cos(y)$ is close to $1 - y^2/2$ and $1/(1+x)$ is close to $1-x$, so we can say $y$ is close to the $\sqrt{2x}$. That's where the formula comes from, the rest is just unit conversion from radians to degrees and from the radius of the Earth to the altitude in meters.


Many models shown in books or television show a very populated asteroid belt but in fact the belt is mostly empty.

To answer your question, the inclination of the asteroids vary a lot going from 0° to 40° although most off them are in between 0° and 30°; See The orbital element distributions of real and modelled asteroids. So yes it would be 3 dimensional.


In India, what is the best time to view the geminid meteor shower in 2013

They should be visible any time at night (note that longitude has little to do with observing meteors), but there are many things that could thwart your observing.

Bright lights: In a big city you will see far fewer meteors. In a big enough city, you won't see any. On that note, if you keep checking a mobile device with a backlit screen, you probably won't see anything.
The Moon: The Moon will unfortunately be up until something like 4 AM local time (assuming I plugged numbers in correctly to the US Naval Observatory's table). Just as with cities, this will cover up smaller meteors. Just try not to stare directly at the Moon lest you lose night vision entirely.
Limited view of the sky: The key to watching meteors is to see a large amount of the sky at once. Never use binoculars or a telescope. The idea is to get comfortable by laying down (yes, this is important - people who crane their necks get tired after 20 seconds and then stop looking up!) and just look at as much sky as possible. Your eyes naturally catch and follow movement, so you'll instantly focus on a meteor once it comes into view.

As for where to look, the answer is "everywhere." Meteor showers are named after the radiant - the point in the sky from which the trails seem to originate if you trace them backward far enough. This shower originates in the constellation Gemini, which isn't terribly recognizable, but can be found by locating Orion:

Move from his belt to his right arm (the one indicated by the very red star Betelgeuse) and keep going until you come across two reasonably bright stars. That is Gemini. It is important to keep in mind, though, that meteors streak across other parts of the sky. Just because they seem to originate from Gemini doesn't mean they're actually visible there. Knowing the radiant just helps with knowing what direction to expect the streak to be going.


This is all subjective. Ask three different people about eyepieces and you'll get three different opinions. My answer is not the right answer, it's just my view as an amateur who has a case full of eyepieces.

The main difference between the less expensive and the more expensive eyepieces is in the quality of the view towards the edges.
Almost all eyepieces will give good views in the centre of the view. It's towards the edges that the view starts to degrade.
Wide field eyepieces need more lenses to provide a well-corrected view to your eye. But the more glass you add, the more surfaces you have which can reflect light back and away from your eyes, so this glass needs to be well-polished.
This is why the expensive eyepieces cost as much as they do. To provide a wide field that's well corrected at the edges needs lots of quality glass, and a good QA process.
Finally, there are definitely diminishing returns when paying more money. The difference between a €50 eyepiece and a €200 eyepiece will be more apparent than the difference between a €500 and a €1000 eyepiece.

I see three main groups of eyepieces:

€1 to €199
€200 to €399
€400+

The first group consists of a lot of different brand names all selling pretty much the same eyepieces.
Generally speaking, a factory in China makes a batch of eyepieces, and different brands buy these eyepieces and pay the manufacturer to put their brand names on the eyepieces they buy.
The brands who pay more money probably get to pick the better eyepieces from a production run.
Or take the case of Explore Scientific, who are partnered with Jinghua Optical Corporation (JOC), which suggests a closer relationship and explains why some of Explore Scientific's eyepieces (the 92°, 120°, and 3" 100°) aren't available under other names.
The eyepieces in this group are mostly all much of a muchness, but there are some gems to be found (the 8.8mm 82° offered by ES and Meade, or the Orion E-series 7-21mm zoom).
Also, it's perfectly possible to buy a €50 eyepiece and get lucky with a great manufacturing sample that gives eyepieces ten times its price a run for their money.

The middle group begins where I see a noticeable increase in optical quality.
The Baader Morpheus 76° eyepieces are at the lower end, and are an example of some fine "forever eyepieces", as you put it, which won't break the bank. (They're slightly more expensive in the USA than Europe, I believe, which might make the Pentax XW a better option if you're in the Americas)
APM have some good eyepieces in this price range, too, such as their 100° series (which many experienced observers have said are the second-best 100° eyepieces available today) and their 12.5mm 84° eyepiece.
Pentax's 70° XW range are well-regarded, with eyepieces from 3.5mm to 40mm.
Also Tele Vue have their "budget" Delite 62° eyepieces in this price range.

The expensive group contains the Tele Vue Delos 72°, Nagler 82, and Ethos 100° ranges.
Right at the top are the Tele Vue Apollo 11mm and the Nikon NAV HW with 102° FoV, which comes in only two focal lengths - 12.5mm and 17mm - but comes with a screw-in lens to give an extra focal length, allowing for 10mm and 14mm.

So, having said all of that, what would I recommend?
Low power I would recommend the APM UFF 30mm. Other options are the Pentax XW, Explore Scientific, and Tele Vue Nagler. (You could go up to 40mm if your pupil dilates to 6.5mm in the dark).
By the way, at low magnifications a FoV of ~70° is wide enough, so don't feel that you have to go for an expensive ~80° eyepiece here.
Medium power I would recommend the Baader Morpheus 17.5mm. Another option is the APM Hi-FW 12.5mm. (I have both of these, and am therefore biased towards them, but they are my most-used and favourite eyepieces)
There are many different options in this range. Anything from 10mm to 20mm would be good.
You might even want to wait a while before buying a medium power eyepiece. Get the low power and high power ones and use them for a while, and you'll get a much better idea of what you'll want out of a medium power eyepiece.
High power I would recommend the Baader Morpheus 6.5mm. Other options are the APM 100° 7mm, the Pentax XW 7mm, the aforementioned Meade/ES 8.8mm.
There are quite a few different options in this range, too.
(At high magnifications the main thing you'll want to consider is minimising "time to nudge": when observing a planet you want to be able to have it drift across the field of view for as long as possible before you have to nudge your telescope to bring it back into view.
For this you'll need a high FoV eyepiece - 82° is a comfortable minimum these days, now such a relatively high FoV is widely available at reasonable prices.)

This answer has become a lot longer than I thought it would. And I haven't even mentioned zoom eyepieces yet! (You could get a zoom. And a Barlow. And a low power eyepiece.)
Another option is to buy a set of SVBONY eyepieces and use them for a while, and gradually replace your most-used focal lengths with better quality versions.
But I'll wrap up. To conclude: don't take my word for it! Get some more points of view. Read some reviews. Spend some time with the astronomy.tools FoV calculator to plan the focal lengths you want to have in your eyepiece collection.
(And if you decide to go for the expensive eyepieces then look at the second hand market - people generally take good care of their expensive eyepieces, and some brands maintain good resale value)


As far as I can see, this is just a faint star. Here is some data from the USNO B1.0 catalogue:

USNO B1.0 star

B1.0 ID: 0940-0181715

RA: 10h10m11.13s +/- 69 milliarcseconds

declination: + 4 03' 12.3" +/- 195 milliarcseconds

Epoch: 1977.9

Proper motion in RA:  12 +/-       2 milliarcseconds/year

Proper motion in dec: 6 +/-       6 milliarcseconds/year

Photometry:

Band mag   Cal Survey  Emulsi field  S/G   xi    eta

B1: 14.36   1  POSS-I  103a-O 550     4   -0.05    0.03

R1: 13.33   1  POSS-I  103a-E 550     1    0.05    0.01

B2: 14.07   2  POSS-II IIIa-J 782     9    0.09    0.26

R2: 12.79   2  POSS-II IIIa-F 782     8   -0.07   -0.02

I:  12.37   2  POSS-II IV-N   781         -0.04   -0.30

"Cal" = calibration: 0 means calibrated via bright standard stars on
the plate; 1 = via faint standards on the plate; 2 = faint standards on
adjacent plate; 3 = faint standards two plates away,  etc.
  "S/G" = star/galaxy index. 11=object looked starlike, to 0=fuzzy/galaxy-like.
  xi, eta = displacement of the star's position on this plate,  in
arcseconds,  relative to the mean solution.

This USNO-B1.0 data was downloaded from a VizieR server.


Probability of planets in binary star systems?

There are only about a dozen circumbinary exoplanetary systems known - the first discovered was Kepler 16b.
The statistics of circumbinary planets is a relatively unstudied area. One of the major uncertainties is the inclination distribution of the exoplanets. If they are confined to the orbital plane of the binary system, then transiting circumbinary planets are easier to detect and so their observed frequency translates to a lower actual frequency than if their orbits are scattered around the binary orbital plane. Armstrong et al. (2014) use the Kepler survey results and find that if circumbinary exoplanets are coplanar with the binary, then the frequency of giant exoplanets within 300 day orbits is about 10%, and consistent with the frequency of similar exoplanets around single stars. On the other hand if their orbital inclinations were random with respect to the binary, then the exoplanet frequency could be a factor of 5 higher.
On the other hand we do know that circumbinary planets around very short period binaries (those with periods less than 7 days) are rare Fleming et al. 2018). This probably has something to do with orbital evolution of the binary system in these cases.
Exoplanets that orbit around one of the stars in a binary system are better studied. If the stars are widely separated then they have little influence on each other or on planetary systems that orbit within say 1/10 of the stellar separation. Exoplanets on orbits wider than about 1/5 of the separation are expected to be unstable.


How poor was our tally of objects that could produce potential extinction-level events back in 1998?

It was fairly poor.  Wikipedia has a list of large, potentially hazardous asteroids, and 9 on that list were known before 1998:

2201 Olijato
1620 Geographos
4183 Cuno
1981 Midas
3122 Florence
3200 Phaethon
4486 Mithra
4197 Toutatis
(4953) 1990 MU

The largest of these, 3200 Phaethon, is 5.8 km, so considerably smaller than the Yucatan impactor, but big enough to do plenty of damage if it does eventually get nudged into our orbit
At that time we hadn't observed (53319) 1999 JM8 a 7 km asteroid, so it is not inconceivable that we might not have observed a 10 km asteroid (similar in size to the low end of estimates of the Yucatan impactor).
Now Armageddon has a "Texas sized asteroid", and Texas is huge, 1300 km from Brownsville to the Rita Blanca grasslands.  It is considerably bigger than Ceres. There is essentially no chance that an asteroid the size of Texas would have gone unnoticed by NASA, indeed it would probably have been a naked eye object and known by the ancient civilisations.


How can we avoid needing a leap year/second?

Leap years exist for two reasons:

There are not an integer number of days in a year.
People perceive a need to keep the seasons where they are on the calendar.

Given the above, there is no way to avoid leap years, or something similar. Defining the calendar year as being a fixed number of days (e.g., 365 days) would result in the seasons shifting by one day per four years.

Leap seconds exist for two reasons:

The length of a day as measured by an atomic clock is not constant.
People perceive a need to keep midnight at midnight, noon at noon.

Given the above, there is no way to avoid leap seconds, or something similar. Defining the day as being a fixed number of atomic clock seconds (e.g., 86400) would result in your clock and the Sun disagreeing on mean local noon, but by a very small amount.

That said, there are serious proposals to eliminate leap seconds. Some people such as those who use UTC to timestamp financial transactions do not like them. So far, those proposals have been rejected. The standard response is that it's not UTC that's broken; it's using of UTC in a context where it shouldn't be used that is broken. If you need a monotonically increasing time scale, use TAI or GPS time instead.


Yes, they do.
Bars in massive galaxies undergo a vertical "buckling" phase about one or two billion years after they form; when this settles down (after about a billion years), the result is that the inner 1/3-2/3 of the bar is vertically thick, with the outer part of the bar still as flat as the rest of the galaxy disk.
When seen side-on, the thick part of the bar looks like a peanut (double-lobed), often with a hint of an X-shape; when seen at an intermediate angle, it looks more rectangular and "boxy". This is the situation for the Milky Way, in fact. (When seen with the bar end-on, the result looks very round, like a stereotypical spherical bulge.)
You can also see this when the galaxy isn't perfectly edge-on: some bars in galaxies inclined at 45 or 60 degrees (where 90 degrees = edge-on) have rectangular interiors with thinner, offset outer regions; this corresponds to the vertically thick/peanut-shaped inner part of the bar and the vertically thin outer part.
We've even seen the signature in a couple of galaxies of a bar in the middle of forming the peanut-shaped bulge (the "buckling phase").
(In principle, you could get a boxy bulge from some kinds of minor mergers, where a small galaxy has merged with a larger galaxy. But if these exist in real galaxies, they're much rarer than the boxy bulges that are part of bars.)


Given sunrise, noon, sunset, longitude and latitude can I calculate the "hight" or "ascension" of the sun?

To the precision that you need, you can calculate the azimuth A and altitude h of the Sun from the following equations:
$$
\tan(A) = \frac{\sin(H)}{\cos(H)\sin(\phi)-\tan(\delta)\cos(\phi)}\\
\sin(h)=\sin(\phi)\sin(\delta)+\cos(\phi)\cos(\delta)\cos(H)
$$
where

A is the azimuth measured from due south (positive towards the west)
H is the hour angle which is the number of degrees before (negative angle) or after (positive) solar noon. Calculate this as 15*1.00274*(time of day - time of noon).
$\phi$ is the latitude (positive for northern hemisphere)
$\delta$ is the declination of the Sun. This can be calculated from the second equation by assuming the altitude of the Sun h=0 when it is rising or setting, and H=15*1.00274*time from sunrise to solar noon. The assumption here is that the time from solar noon to sunset is the same.
h is the altitude of the Sun above the horizon.

All of the above quantities are in degrees. You may need to convert those to radians to perform the trigonometry calculations.

For a more accurate calculation, you would calculate the position of the Sun (its right ascension $\alpha$ and declination $\delta$) and Greenwich Mean Sidereal Time (GMST) based on the date and time throughout the day. Then using the latitude and longitude, you would calculate H, A, and h for various times of the day.


Why do X-ray binaries such as GRS-1915 have active and quiet states?

Such objects are surrounded by an accretion disk, that is a disk of gas, the gas being accreted onto the central object. However, at this stage of its life, the accretion disk around the black hole of a binary system like GRS-1915+105 is strongly depleted. Therefore, the accretion rate is not sufficient to support a coutinuous burst. Matter fills the outer disk until a critical density is reached, that triggered an outburst. Mass from the companion star is lost by a Roche lobe overflow mechanism.

References:

X-ray binaries in general: Remillard & McClintock 2006
For the outburst X-ray transients in particular: Dubus et al. 2001


As you said, the mass of the Moon is 1.2 percent that of the Earth.
Now, if you mean the gravitational acceleration at the surface, it is calculated like this $G\frac{M}{R^2}$, where $M$ is the mass, and $R$ is the radius of the celestial body. The moon's mass is a hundred times smaller, but the radius is four times smaller, meaning its surface gravity will be $100/16 \approx 6$ times smaller. Considering the factor is mass over radius to the power of two here, density alone does not help you determine the ratio of surface accelerations. On a side note, the moon's density is around 60 per cent that of the Earth.


Can astronomical radio sources be used as a verifiable randomness beacon?

I see two problems.  1) Any particular source in the sky is above the horizon for a limited region of the Earth at a given time, so one would need to relay the signal around the globe.  2) It is hard to think of an astronomical source that is truly random all of the time.  Astronomical sources are physical systems reacting to internal and environmental changes that are causal, so they all tend to be partially quasiperiodic to some degree or become temporarily nearly periodic.  So it depends on just how random you need them to be, but I doubt any astronomical source is dependable for this purpose.


How does a Solar Transit give more accurate determination of Earth-to-Sun distance?

The problem of determining the solar parallax (and inversely the Sun's distance from Earth) by classical optical astronomy has been in most periods since the 17th century dominated less by principle than by the great difficulties of controlling and reducing the measurement errors of the very small angles involved (the solar parallax is only about 8.794" of arc). A good reference source for the earlier period through the 18th century is Albert van Helden's "Measuring the Universe" (1985, Chicago UP).

The method of observing planetary transits across the Sun's disk in order to determine the solar parallax was arguably a method of clear choice mainly or only for the 18th-century solar transits of Venus, which took place only in 1761 and 1769.

The measurement advantage in observing the Venus transits (relative to a difficult direct solar-parallax determination) arises largely because Venus at inferior conjunction is only about a quarter the distance of the Sun, thus its parallax in that configuration is not far off four times as large as the Sun's. What was to be measured directly was the parallax of Venus itself rather than that of the Sun. The size of the parallax angle to be measured was one of a number of factors that it was hoped would allow a smaller relative error in the observational determination. Conversion from the parallax of Venus to that of the Sun was done by using the comparatively much-better-determined relative scale of solar-system distances.

(By contrast, Mercury transits, though more frequent in occurrence than those of Venus, only allow measurement of Mercury's parallax, which is not too much bigger at inferior conjunction than the Sun's own parallax itself, thus offering less advantage than a Venus measurement.)

By the time of the 19th-century Venus transits (again only two, in 1874 and 1882) there was an accumulated experience of the practical measurement difficulties, some of which (like the 'black-drop effect') had not been expected before the 18th-century transits were actually observed. The difficulties meant that advance expectations of measurement accuracy from the 18th-century Venus transits had proved to be over-optimistic.  There was some remaining optimism that the measurement difficulties could be addressed, and expeditions were mounted to observe the 19th-century Venus transits, but still the results were not as much better than before as had been hoped. (A useful reference source for this part of the history is J Ratcliff (2008) "The Transit of Venus Enterprise in Victorian Britain".)

In the meantime, competing methods of measuring the Sun's parallax began to be devised, and a more successful example was indeed an Earth-planet distance: to Eros, a minor planet whose orbit brought it much closer to the Earth than Venus' distance at solar transit. The crucial aim again was to measure by parallax a distance that should be as close as possible (and well-determined on the relative solar-system scale of distances, like distances of Venus and other planets). The measurements in the Eros case at its close approach to Earth were done by reduction of photographs and of the positions of Eros relative to selected reference stars as seen from a number of observing stations (A R Hinks, 1904, 'Reduction of Photographs of Eros, with a Determination of the Solar Parallax' (http://adsabs.harvard.edu/abs/1904MNRAS..64..701H)).

From the 1960s onwards, the traditional optical methods of determining solar and planetary distances using solar transits and close approaches of minor planets were entirely superseded by ranging methods, one of the earliest examples being radar-signal reflection from Venus (reviewed along with the Eros measurements e.g. by Eckstein (1963),  http://adsabs.harvard.edu/abs/1963AJ.....68..231E). More recently many other range measurements have  been obtained e.g. to spacecraft placed in orbit around solar-system bodies.


Are the stars in all galaxies orbitting around a middle?

Stars aren't really orbiting the center of a galaxy, as much as they're orbiting the gravitational potential of the galaxy. The galaxy doesn't need to have a black hole in the center for the stars to orbit, and even if it does (which in fact it does most of the time), the black hole does not dominate the gravitational potential, except in the most central part (see sphere of influence of a black hole).

All matter in the galaxy contributes to the gravitational potential, but since dark matter comprises 5/6 of the mass, this is the most important contributor. Only in the hypothetical case of a perfectly symmetrical distribution of masses, would the black hole and the center of rotation coincide. And since the paths of the stars are ellipses and not perfect circles, the center of the potential will not lie in the center of their orbits, but in one of the focal points. Furthermore, galaxies are dynamical and the potential continually changes, in turn changing the exact orbit of a star.

Galaxies form due to the collapse of a dark matter halo, and the gas that follows  along. Such a halo will in general have a non-zero angular momentum, which is partially conserved during the collapse (they do lose a significant amount, for instance through minor and major merging; see e.g. D'Onghia et al. 2006).

As in the case of a forming star, a forming planet, an ice skater, or you on an office chair, reducing the radius means increasing orbital speed. This is the origin of the rotation.


Does the loss of mass create an observable change in a comet's orbit?

This problem was studied in Yu & Zheng (1995), who evaluated the effects of the change in the Sun's mass over time and the change in a comet's mass over time, for the case of Shoemaker-Levy 9, which had recently crashed into Jupiter. Given their mass model for the comet (Equation 6), they found that the Sun's mass loss created an increase in semi-major axis of about 8.5 centimeters per year, whereas the comet's mass loss created an increase in semi-major axis of about 10,000 kilometers per year.

Several things to note:

The comet orbited Jupiter prior to breaking up, so it did not come as close to the Sun as most short-period comets.
Mass loss rates can change over time, depending on the distance from the Sun.
Shoemaker-Levy 9 should not be considered a normal comet, in any sense, given its orbit and eventual destruction.

10,000 kilometers a year, though, is nothing to sniff at. Over the course of an orbit, that can be quite a lot - although keep in mind that longer orbits involve much larger semi-major axes - and I'd argue that it should be observable, given the correct calculations of how the orbit should evolve over time.


I ended up making my own small code that you can find here:
     https://github.com/crhea93/AstronomyTools/blob/master/Astrometry/Correction.py

Essentially, after inputting at least 3 matching target (found by eye in DS9 for example), the program will calculate the mean offset and apply it to the target image. The output is a new fits file of the target image with the proper correction.


How to find period of a pulsar from a given time sampled voltage data?

To understand the basics of algorithms like that, I find it's often best to look at what amateur practicioners are writing (as opposed to diving directly into scientific papers). The description of the folding algorithm at joataman.net is thorough and pragmatic.

A great in-depth book on the topic is Lorimer/Kramer: Handbook of Pulsar Astronomy, Oxford University Press.

For the actual work, I would suggest the sigproc package by Duncan Lorimer. Its home is at Sourceforge, but there are forks all over the place because the original software is a bit quirky when run on modern Linuxes.


Why does Earth have more gravitational force than the Moon if I stand on it?

The Earth doesn't cover the hole, the Earth is the hole. Gravity is an attractive force, so were you not standing on the Earth, the Earth's gravity would cause you to accelerate towards it. As you are standing on the Earth, you feel this acceleration as your weight, just as you would feel pressure if you were to push against a wall.

To take a more complex view, we would say that you are being attracted to the local center of gravity, which is located within the Earth. Take your bridges example, it doesn't matter that something obstructs you from reaching the center of gravity. You will feel the attraction of gravity regardless. If you jump while standing on a bridge, you would land just the same as if you jumped while standing on the ground (though please try not to miss the bridge). The graphic is meant to illustrate the strength of the force based on the mass of the attractive object and distance.

If we were to put you on a more massive object, which would have a steeper curvature, you would experience this as increased weight, which is why our lunar explorers bounced around on the moon instead of walking.


The "no hair" theory of black holes states that a black hole formed from the collapse of normal matter would be exactly the same as a black hole formed from dark matter. You could even form a black hole from photons, if you could get enough of them close together at the same time.

What originally collapsed to form the black hole has no relevance to whether Hawking radiation is given off. A black hole formed of dark matter WIMPS would behave exactly the same as a black hole formed from normal matter.

Dark matter can be converted into "energy" (I'll assume that you mean light, ie photons), in the same way that normal matter can: for example by antimatter annihilation.

The actual formation of black holes, however, requires them to be mostly made of normal matter, since the matter has to lose energy as it falls together, and to do this it must interact with itself (essentially this is friction). As dark matter does not seem to interact with itself or with normal matter, it would not form a major part of a black hole, even in a galaxy dominated by an excess of dark matter. The supermassive black hole may be exceptionally small in such a galaxy, as there is less suitable matter to fall into it.


spider image in Samuel Oschin Schmidt Telescope DSS imagery

Those are almost certainly "ghosts" caused by the internal reflection of light from very bright stars. (That is, the light is reflecting off the insides of the camera, filter, etc.) The pattern is an image of the "entrance pupil" of the telescope, which in practice means the primary mirror + the blockage caused by the secondary mirror and its support structure (hence the four thin dark "arms").
There's a possibly useful technical discussion in this paper, which among other things shows that the off-center nature of the pattern is due to the bright stars being off-center within the field. (If they're in the center, then the ghost has a symmetric, centered pattern.)
See this question and its answers for the general idea when your entire image is out of focus (i.e., all the stars will look like that). In the case of these DSS images, the ghost reflection images are out of focus because the multiple internal reflections change the total path length of the reflected light, while the rest of the image is focus for light following the expected internal path.
The nature of a wide-field or all-sky survey -- like the DSS -- it that you generally try to image all the accessible sky. This will inevitably include some regions with very bright stars in them, even if you might avoid them for more focused observations (e.g., if you're studying type X objects, you might leave out those which happen to lie very close to a bright star).


Understanding expected TESS data structure

An insightful option is to open in a python IDE using Astropy. I find it useful because you can manipulate the files as your learn their structure.

It is also possible to use matplotlib in conjunction with Astropy and then basically have a FITS file viewer that you can code yourself.

Something like:

import astropy
import matplotlib.pyplot as plt
from astropy.visualization import astropy_mpl_style

    def open_fits_file(fits_path):
        for file in Path(fits_path).iterdir():
            #Not really using the with, but it'll manage closing the file...
            with fits.open(file) as opened_fits:
                #printing header infos for that file
                opened_fits.info()
                plot_image(fits.getdata(file, ext=2))

Then to plot the image:

def plot_image(fit_image_data):
    #astropy has better plot settings for this use
    plt.style.use(astropy_mpl_style)
    plt.figure()
    plt.imshow(fit_image_data, cmap="gray")
    plt.colorbar()
    plt.show()

I used that to plot the pixels  from these simulated data linked in the question.

As mentioned by Mike G, HEASARC as a list of viewer, including FV (fits viewer). It works fine and it has some built-in support to link up with other astronomy database. It has an apollo-era feel in terms of appearance, which is nice... I guess?


How would Jupiter's brightness relative to our sun seem to a remote observer (observing from a remote star)

The absolute visual magnitude of the Sun is about 4.8. (This is the Sun's brightness viewed from 10 parsecs).

The brightest visual magnitude for Jupiter is -2.7, when it is about 4 au from the Earth. Using the usual formula we can therefore calculate the absolute magnitude for Jupiter as 25.9. (Note that this is way brighter than the faintest objects seen for instance with the Hubble Space Telescope; but as Jupiter would be separated from the Sun by only 0.5 arcseconds, a telescope with even better angular resolution than the HST would be needed to pick it out of the glare from the Sun).

Thus to first order Jupiter is 21 visual magnitudes fainter than the Sun (a factor of 250 million) . The situation is slightly worse than this because as seen from outside the solar system, Jupiter could not be fully illuminated, because it must have an angular separation from the Sun to be visible at all. A half-illuminated Jupiter would be at least a magnitude fainter (actually a factor $\pi$ for a Lambertian reflector).

But all is not lost. If we we to switch to the infrared, the situation is far more favourable. The approximate ratio of Jupiter/Sun brightness improves from about $1.4 \times10^{-9}$ in the visible to about $2.8\times10^{-8}$ at a wavelength of 10 microns (Traub & Oppenheimer (2010). But this is still a 19 magnitude difference.

The technology to achieve these sorts of contrasts is not quite there yet, even if we were viewing from nearby stars.  An example of what is currently possible can be seen with the discovery of a few Jupiter mass planet around the star GJ 504 by Kuzuhara et al. (2013). The star-planet contrast was about 15-16 magnitudes at near-infrared wavelengths of 1-4 microns and with a star-planet separation of 2.5 arcseconds (equivalent to Jupiter as seen from only 2 parcsecs).


Can you find your own location in space by measuring angles between stars?

The related observation in astronomy is stellar parallax:
the observer knows the Earth's position relative to the Sun, measures tiny annual oscillations in the direction to another star, and estimates the distance to the star.
A star at a distance of 1 parsec (pc) appears to shift by 1 arcsecond (4.85 μrad) per astronomical unit (au) of lateral displacement.
Since parallax is inversely related to distance, the relative error of a parallax-based distance estimate increases with distance.
For any given star, the position uncertainty ellipsoid is shaped like a needle pointing toward the Sun.

Aboard an interstellar spacecraft with limited observational facilities, it may not be practical to measure angles between nearby stars directly.
Instead you might individually measure their positions relative to several distant reference stars near them in the sky.
For best results, you would need to account for the stars' proper motion over the duration of your journey.

After doing that data reduction, you can compute angles APB, BPC, APC fairly precisely.
However, the Sun distances SA, SB, SC have significant uncertainties which propagate to other quantities computed from them, e.g. distances AB, BC, AC.
You can estimate distances PA, PB, PC by the law of sines, but these are uncertain by similar amounts.
Alternatively you can estimate PA, PB, PC by comparing apparent magnitudes to absolute magnitudes, which are subject to the same uncertainties.
Then you can find the intersection of spheres around A, B, C to estimate position P, but still no more precisely than distances SA, SB, SC.
P is not so much a point as a probability distribution cloud which expands as you move away from the Sun.

In the real world, our most distant spacecraft Voyager 1 is only 145 au (0.0007 pc) away.
Within about 400 times that distance, or 1/5 the distance to α Cen, the Sun is the brightest star in the sky by far and would remain your primary point of reference.


What is the Age Difference Between Earth & Jupiter?

It is generally thought that Jupiter "formed" first via the core accretion process. An icy core forms first and then sweeps up the gas around it. This process is generally thought to take from a few million to ten million years. After that, the solar system gas was dissipated and Jupiter was basically as it is today, but larger and hotter.

Earth formed on a longer timescale in the maelstrom of rocky planetesimals that made up the inner solar system. An essential part of the process was the collision that formed the moon, which probably occurred between 30 and 60 million years after the protosun began to form. Further heavy bombardment of the Earth that may have significantly changed its surface and atmosphere continued for some tens of millions of years after that.

The wikipedia page on planet formation in the solar system looks ok to me
https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System
I may add some more academic references subsequently.


What would we find if we could travel 786 trillion light years in any direction

I'm afraid the only correct answer to this question is that nobody knows.

Cosmology is based on a number of simplifying assumptions (sometimes called principles) - the most basic of which is that the universe is "isotropic and homogeneous on large scales". Isotropic means that is (or looks) the same in any direction; homogeneous means that what we see does not depend on our position in the universe.

This "cosmological principle" is reasonably well tested and would certainly make sense if the universe were very much bigger than we are capable of seeing. The most distant thing we can "see" in our universe is the cosmic microwave background (CMB), the light of which has been travelling for about 13.6 billion years. However, if we were to think about a parcel of gas that emitted the CMB and where it is now, we could calcuate that it is about 46 billion light years away due to the expansion of space in the universe. Thus the "edge of the observable universe" is about 46 billion light years.

What we can say, is that on these scales the Universe is very isotropic, and homogeneous, because the CMB is very smooth and looks the same in all directions to a very good approximation (but see below). This suggests that if you were to travel 46 billion light years in any direction, that you would probably have a very similar view to what we have now (that is, if you were able to miraculously travel instantaneously; if you travelled more slowly then of course the universe would continue to get older).

You are talking about travelling more than a factor 10000 further. On these scales we do not know anything outside of speculation. The inflationary model for the big bang might suggest that our observable universe is just a tiny part of a much greater entity and that within this very much larger entity everything will be homogeneous and isotropic at a given cosmic time. However, the recent very precise Planck measurements of the CMB, suggest there may be small, but real, anisotropies on scales as big as the observable universe, which would then suggest that any extrapolation by a factor of 10000 might be wishful thinking.


Well, they know according to the Metonic Calendar created in ~430 B.C.

To know more about click here.


Astrophysical black holes deviating from Kerr black holes?

Evidence of "new physics" seems to be lacking (in my research into this question.) Black holes can be described fully by known consequences of general relativity as far as our ability to observe them.  There are known deviations from the Scharwchild metric, for example in black holes in which the accretion disc is inclined relative to the spin of the black hole, and in which the effects of frame-dragging can be observed as the accretion disc transtions to the spin plane of the black hole.
High energy cosmic rays are a poorly understood phenonomum. They seem to be charged particles (ie protons) that have been acclerated to very high velocities (much higher than 99% c) by black holes.  But the particular mechansim that does this is unclear.  A paper, https://arxiv.org/pdf/1904.04654.pdf, proposes that high energy protons may be an consequence of black holes with a Kerr-Newmann metric, that is a rotating charged black hole; as even a small charge would have a singificant effect on protons in the neighbourhood. This could be interpreted as evidence that some black holes have a Kerr-Newman spacetime metric.


How small could an orbital system be in our solar system?

If we only consider gravity, one answer may be found using the Hill sphere. This is the distance the gravity of a body dominates over the sun: $$r_H \approx a \left(\frac{m}{3M_\odot}\right)^{1/3}$$ where $a$ is the semi-major axis, $m$ the mass and $M_\odot$ the sun's mass.
Now, an actual body has some nonzero density $\rho$ and $m=(4\pi/3)\rho r^3$. If the Hill sphere is inside the body there will not be any orbits around it (they would be dominated by the sun's gravity). So, we get the equation $$r = a\left(\frac{(4\pi/3)\rho r^3}{3M_\odot}\right)^{1/3}$$ which simplifies to
$$\rho = \frac{9M_\odot}{4\pi a^3}.$$ Objects less dense than this have Hill spheres inside themselves: at 1 AU this density is  $4.3\cdot 10^{-4}$ kg/m$^3$ (a thin gas), while at 0.1 AU it is 0.4255 kg/m$^3$ - about a third of sea level air density.
For hydrogen atoms, if we calculate the density for a 25 picometer atomic radius I get a density of 25,570 kg/m$^3$ (in actual hydrogen gas the atoms are spread out way more). Hence there the Hill sphere argument actually allows them to orbit each other!
In practice this does not happen. The orbital period at (say) 3 atomic radii is $\sqrt{4\pi^2r^3/Gm}\approx 3.4$ hours and the binding energy is $1.5\odot 10^{-27}$ J. This is $4\cdot10^{-5}$ of the thermal energy of the cosmic background radiation: even if there wasn't any sunlight or other radiation from inside the solar system it would jostle the atoms enough that they would split.
This suggests an apparent way of answering the question: if the binding energy $Gm/r$ is less than typical disrupting energy the orbit will not be possible. Actually calculating the forces is non-trivial (there are many kinds, from Jupiter's gravity to solar heating) and weaker forces can sum up over time. Knowing the disruptive background also just gives an upper bound for $m/r$, one could have smaller orbits.
So the true answer will be given by how small dense objects we are willing to consider, and (as the other answer points out) local forces. In the solar system the most relevant may be electromagnetic charging due to the solar wind: if the objects are metallic and close they can even attract each other if they have the same charge (!). Things like magnetic fields, infrared radiation and solar wind will play a role, making the true answer somewhat undefined.


What are angular sizes of thin and thick disks in our Galaxy?

The sun is usually taken as the center of the galactic coordinate systems. Our solar system is well within the thin disk of our galaxy (population I stars). So taking the headline question literally, we see the thin disk as well as the surrounding thick disk in almost every direction, meaning any galactic latitude and longitude.

The radius interval, where we find mostly stars of the thin or thick disk, varies with angles. Along the galactic plane the radius interval is across the galaxy (up to about 30 kilo parsecs, varying a bit with direction, because we are not in the center of the galaxy) for the thin disk, containing the sun as origin; perpendicular to the galactic plane it's about 1 kilo parsec to both directions for the thin disk, followed by a respective radius interval from 1 to 3 kilo parsecs for the thick disk, if I follow the Wikipedia articles referenced in the question.

The region dominated by the thick disk is the farther away from us the closer we look along the galactic plane.

There are certainly no sharp boundaries between the star populations. Actually the thick disk penetrates the thin disk, but it's much sparser populated.

The Gaia project may give a much more detailed answer about the distribution of stars within our galaxy in the next 5 to 10 years.


The constellations change slowly, over thousands of years, due to the proper motion of the stars with respect to the Sun.  The main cause of these motions is the orbit of the stars around the centre of our Galaxy (with or without dark matter).  This video shows these changes for the constellation of Ursa Major over 400,000 years.

Another way the constellations can change is indeed through stellar evolution $-$ either the birth or death (or brightening/dimming) of stars.  However, the timescale for stellar evolution for naked-eye stars is typically hundreds of millions to billions of years (the Sun's life expectancy is around 12 billion years), so that this effect is much slower (and less well predictable).


What solar features would I see at 10x magnification?

You will definitely see some sunspots at 10x; in exceptional cases sunspots are big enough to be visible with the naked eye through the sunset haze near horizon, but many sunspots are visible in a small instrument like a 10x binoc with proper filtering.

The Sun is at low activity right now, but still a few sunspots are showing up once in a while. And the activity should increase gradually in the future.

Planetary transits - yes. However, the next transit of Venus will take place in a century or so. Mercury transits more often, and the filters will allow you to see it - a tiny dot on the face of the Sun. A bit higher magnification might be beneficial, but anyway even at 10x you should see the planet, even if it's rather small.

BTW, I recommend solar filters based on the Baader solar filter film (just google it). They are neutral-density (do not change the color of the image), and are recommended and have been tested with instruments from small binocs to dobsonian telescopes. Make sure the filters are firmly attached to the instrument, and are free of defects.


VSOP for exosatellites? (Io, Titan, Ariel, etc)

Yes, there are several I know of for the Galilean moons, less so for other satellites.  One very good place to start is from the same organization that created the VSOP series, now called IMCCE.  Here is a link to their satellite ephemeris.  According to the Explanatory Supplement to the Astronomical Almanac, most of the other moons of Jupiter behave to chaotically to create a similar theory and numerical integration is required to produce accurate results.


It depends entirely on the circumstances of black hole formation.

The remnant may be a stable neutron star, but if it accreted enough matter later on (by fallback in a supernova, or from a binary companion), then it may collapse. This could happen at any time after formation.

If the core of the collapsing star forms a proto-neutron star above the TOV limit of stability, then no force can prevent black hole formation. It will happen on the cooling timescale of the proto-neutron star at the centre. Once it is "cold" then the collapse would occur on a free fall timescale - which is about a millisecond in these circumstances. The cooling timescale of a proto neutron is governed by how long the copiously generated neutrinos can be trapped in the core, which could be as long as(!) a few seconds


How come the Sun does not make the Moon crash on the Earth?

The Sun's gravity does perturb the Moon's orbit but more subtly than you imagined.
The Moon's perigee and apogee migrate eastward in an 8.9 year cycle called apsidal precession.
Also the plane of the lunar orbit, inclined about 5.1 degrees to Earth's orbit, shifts westward in an 18.6 year cycle called nodal precession.
Lunar theory also addresses various effects on shorter timescales.

If the Moon were 4 times farther from the Earth (outside the Earth's Hill sphere), then the Sun would destabilize the lunar orbit.
Even so, the Moon would most likely drift into an independent orbit around the Sun, with little risk of collision with the Earth.


The Lyman-Werner background (LWB) is the part of the "meta-galactic" background radiation consisting of photons that are able to photo-dissociate molecular hydrogen ($H_\mathrm{2}$), but not ionize atomic hydrogen (HI). These photons are in ultraviolet, with energies ranging from $11.2\,\mathrm{eV}$ to $13.6\,\mathrm{eV}$, originating mainly from hot, massive stars.

Part of the reason this radiation field is interesting particularly in the early Universe is that a sufficiently high intensity will prevent gas from cooling. In order for gas to form stars, it cannot be too hot, since a hot gas cannot collapse to form structure. A gas with a high metallicity is able to cool through the many different atomic transitions in the metals, but in the early Universe, where stars haven't yet produced the metals, the most important coolant at these temperatures is $H_\mathrm{2}$.

It also has implications for the formation of supermassive black holes in the early Universe, since once you form a large amount of stars, the feedback from the massive stars (both through radiation pressure and heating of the interstellar medium) makes it difficult to have gas collapse directly into the supermassive black holes.

In simulations, you often have to assume a value of the LWB, since performing the "true" radiative transfer is computationally very expensive. Hence, these authors apparently have experimented with two different fields of $100\,J_\mathrm{21}$ and $1000\,J_\mathrm{21}$, respectively. Here, the term $J_\mathrm{21}$ doesn't have anything to do with $21\,\mathrm{cm}$ radiation (which is in the other end of the spectrum), but is simply the intensity in units of $10^{-21}\,\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}\,\mathrm{sr}^{-1}$.


How do we get the angular size of a crater without lens?

Most lunar craters are too small to resolve with the naked eye; I would measure a published photograph. A crater near the limb appears as an ellipse whose minor axis is foreshortened but whose major axis measures about the same as if the crater were centrally located. Surface curvature has little effect on this measurement because most craters are small compared to the lunar radius.


The contents of potential "message" in the Wow! signal lost?

I'm not an expert in this, but it's a fun little blip in the history of SETI.  Pretty much the only blip I think.

  I understand the incredibly high signal strength it entails.

I wouldn't call it "incredibly high".  It peaked at 30 times normal. source, and that's inside the "waterhole" a frequency range where the background radiation is the lowest in the cosmos.  30 times a normally very low signal is still pretty small, I think.  It was unusual, no question, but looking at the sheet of paper most often reprinted in articles about this, each space is 12 seconds, there are a handful of visible spikes.

Just on that one printout there's a 3,3,2,space,7 and a 3,6,1,3 and a 4,3,4,4
The Wow, 6,E,Q,U,J,5 or 6,15,26,30,20,5 compared to those numbers is maybe 6 or 7 times larger than some fairly common spikes.   Statistically speaking, that's high, but not enormous.  It's still an unexplained oddity, but it wasn't an "incredibly loud" signal, just unusually loud.

In 1977 they couldn't turn the dishes, so they had to wait 24 hours to scan that part of the sky again and 24 hours later, and every day since, and no similar signals have ever been found, so yeah, it's a mystery.

The way they had it set up, they couldn't tell which dish either, so there's 2 tiny slices of the sky that it might have come from, or it could have (theoretically) come from a ship of some sort transmitting it's location back to it's home planet, and the earth just passed through (or it could have been a mouse chewing on a wire, or a Russian prank) . . . it could have been lots of things.

In 1980 they set up the ability to have the dishes track a specific part of space, which would have come in handy in 1977, but new new signals of that strength were ever heard.

And, you're right, the big ear wasn't designed to record.  As I understand it, the system was designed to eliminate background signals, so if both dish's caught similar wavelengths while pointing at slightly different parts of the sky the duplication could be eliminated.  It was what one received that the other didn't that was printed on paper, but no way to tell which dish did the receiving.

Some cool pictures and write-ups about the Big ear here and here.  It was a monster of engineering for the 1950s when it was designed but it's primitive by today's standards.   The Wow signal has no triangulation to verify if it was from outer-space vs a closer event, and they couldn't turn their equipment to track it.   It was initially designed to look deep into space, by the way, not to record alien signals.   Seti began using it in 1973 after it was somewhat obsolete for more traditional astronomical pursuits, but throughout the 60s it was one of the better observatories.  Another problem, it didn't detect a wide range of radio waves, just a tight band, which is good for looking for a signal, but less good for eliminating rare stellar events that might have spanned a broader range of radio-frequencies.   Some possibilities here for stellar explanations and without triangulation, the possibility of rare atmospheric reflection of some kind seems possible too.

The 6 data points looks like a bell curve and that's exactly what we should expect if we pass through a tight signal.  It's not what I'd expect if a signal was directed at us, but passing through a tight beam would explain the bell curve of the spike.  A black hole in that part of space might explain it too.

  Is there a possibility that the burst could have carried information,
  however unlikely? It could have carried a one-time "message"?

The "one time message" idea strikes me as very unlikely because I don't see any point to it.   I also don't think a lot of information could be contained in 72 seconds, but some could, but there's no point in sending information for just 72 seconds and just once.  The point of sending information is so the person on the other end receives it.

But if it was Technology based and not natural in origin, Radio-waves can carry a lot of information, otherwise radios and antenna televisions wouldn't work, but the information is contained in signal variation usually in either amplitude or frequency, like these images below

or this:

Source

The big ear measured total signal variation between the disks every 12 seconds, so it wouldn't detect Amplitude or Frequency variations that would happen in much shorter intervals than 12 seconds.  We simply have no way of knowing.   Another possible method to transmit a information could perhaps be polarization variation in a consistent signal, but big ear couldn't pick that up either.

The problem is, we simply don't have enough information to know.  It's like somebody snapped a terrible picture of something that might have happened.   You can look at the picture forever, but it will never give you any new information.

More on it here, and I liked this article here.  Hope that wasn't too wordy or too much already known information.


Size of the Universe before the Big Bang

Nobody knows what the big bang was, but using the laws of physics as we understand them, and using what has been observed studying quantum mechanics and very high energy collisions, a lot of what's defined as moments after the big bang are basically working backwards and deduction.  Quantum physics models explain very accurately why we have the hydrogen/helium ratio that we see and why the cosmic background radiation formed.

The simple answer to how it was possible to get all that stuff into such a small space is two fold.   One, when it's that dense, it's no longer matter, but energy, and two, it was that small for an incredibly short period of time - too short for gravity to win, which, when you have that much mass in that small a space, you'd think gravity would win, but the primordial grapefruit expanded faster than the speed of light and gravity is limited by the speed of light, and perhaps limited in other ways by quantum gravity, but we don't know how to study that yet.

So you can have that a universe amount of mass in a grapefruit or watermelon size if it's that small for a very very short time, close to impossibly hot and faster than light expanding.

Somewhat different explanations could be that the Universe is infinite, or curved in on itself, in which case the gravity, if the young universe is uniform, balances out to zero everywhere until the uniformity breaks.   This is similar to the gravity at the center of the Earth.  With equal mass in every direction, gravity becomes zero.

The cosmic background radiation suggests that the young universe was very close to uniform and that, plus rapid expansion circumvents the "why didn't it become a black hole" problem.   Various shapes of the universe don't change the rapid expansion model, which is the most popular model today.

There are also models that involve additional dimensions which explain the big bang, but I think those have fallen out of favor, perhaps because extra dimensions have never been observed at CERN and because I've not read about that model recently, though if memory serves, Stephen Hawking put the 5 dimensional big bang idea in one of his books that he marketed to the general public.    Extra dimensions make all kinds of things possible, like putting all that mass in a small space.


Get orbital period of exoplanet from light curve using astropy.timeseries

The second light curve you show has no obvious periodic behaviour and I cannot see any sign of a planetary transit. The period-finding algorithm appears to be working correctly.

The planet (if it exists) is supposed to be one of the smallest planetary candidates found by Kepler and will have a barely detectable transit (depth of order 0.004%). The small light curve variations you see are either from the star or remaining instrumental artefacts. If you wish to verify or otherwise the results published by Morton et al. (2016), who show no light curves, you will have to do considerably more work to "detrend" the other variability in the light curve.


Why did the moon have a distinct reddish edge tonight?

Smoke (fire) is not the only explication - Dust is also a good candidate. I've noticed a drop in "seeing quality" even when observing deep sky objects or planets trough a telescope due to large amounts of dust in the air.

  The second reason for a red Moon is if there’s some kind of particle in the air. A forest fire or volcanic eruption can fill the air with tiny particles that partially obscure light from the Sun and Moon. Once again, these particles tend to scatter blue and green light away, while permitting red light to pass through more easily. When you see a red moon, high up in the sky, it’s probably because there’s a large amount of dust in the air.

(source: Red Moon - Universe Today)


Why hasn't the "9th Planet" been detected already?

Brown and Batygin, the authors of the paper on the possible planet, have a webpage addressing this.

A few reasons not already covered:

It moves quite slowly - the authors estimate 0.2-0.6 arc seconds per hour - so standard surveys may not notice the movement and fail to recognize it as a solar system object.

  Eris, which is the most distant confirmed object still known in the
  solar system, moves at a speed of 1.5 arcseconds per hour, which is so
  slow that it was missed the first time around. Most surveys of the
  outer solar system would not be able to find Planet Nine, even if it
  were quite bright, as they would just think it is a stationary star.

If the planet is near aphelion, it might be an order of magnitude further away than any major or minor planet we've found so far (excluding exoplanets, which are found by methods that don't apply in this case). The authors suggest an aphelion between 500 and 1200 AU. For comparison, Pluto is at 30-50 AU, while Eris at around 100 AU wasn't discovered until 2005. The potential 9th planet would be far larger than Eris, but is also likely to be much further away, and thereby fainter.
The WISE survey eliminated Saturn-sized planets within 10,000 AU, and Jupiter-sized planets within 26,000 AU. But the potential 9th planet is far smaller than those. WISE has also done a more sensitive search, which would pick up Neptune-sized objects, but that search has so far covered only a limited part of the sky.
The planet will be far harder to spot if it has the Milky Way in the background - there are too many stars potentially drowning out a faint object.

Here's the authors' summary:

Estimated orbit for the putative 9th planet. The horizontal axis is the right ascension. The colored segments are regions where it should have been found by existing surveys.
Illustration by Brown and Batygin, assuming fair use applies.

  The biggest unexplored territory is where, statistically, it is most
  likely to be: near aphelion. Sadly, aphelion is also very close to the
  Milky Way galaxy. Ugh.

  So where is it? Probably distant. 500 AU+. Probably fainter than 22nd
  magnitude. Very possibly in the middle of the Milky Way galaxy.

  Now go find planet nine.

More details on the authors' webpage: http://www.findplanetnine.com/p/blog-page.html

Finally, the gravitational dominance of the Sun reaches halfway to the nearest star. There's still plenty of unexplored territory for planets smaller than Saturn to hide in.

Note the log axis. We have a good map for the inner 50 AU, and are starting to find objects around 100 AU, but solar system objects might exist all the way to the outer edges of the Oort cloud.
Illustration from wikipedia.


How likely and severe is the threat of a gamma ray burst to earth?

I'll address WR104 first. The National Geographic article calls it a "potential threat." Yet that potential may be low. There are a slew of articles quoting astronomer Grant Hill on the subject. Hill studied the star and found that it looks like it isn't pointing straight at us. Its axis might be up to 45 degrees in another direction, meaning that we'd be fine if it underwent a burst.
From here:

It would appear the original Keck imagry may not have been as straight-forward as it seemed. Spectroscopic emission lines from the binary pair strongly suggest the system is in fact inclined 30°-40° (possibly as much as 45°) away from us.
So, Earth doesn’t appear to be in the firing line of WR 104 after all…

Here and here Hill is quoted as saying that the star looks like it could be pointed at us - which goes against the evidence he found! Finally, this is another example of Hill's cautious attitude: The evidence says we're good, but he's not making any assumptions.

Sabre Tooth, I know you said that you didn't want Wikipedia, but my primary goal here isn't to collect the bounty, so I'm going to reference it in this section. I hope you're okay with that. I may withdraw it, though, if Jonathan indicates that he doesn't want it.
Okay, back to the topic at hand. Wikipedia, of course, has a short tidbit on the frequency of gamma-ray bursts effecting Earth:

Estimating the exact rate at which GRBs occur is difficult, but for a galaxy of approximately the same size as the Milky Way, the expected rate (for long-duration GRBs) is about one burst every 100,000 to 1,000,000 years. Only a small percentage of these would be beamed towards Earth. Estimates of rate of occurrence of short-duration GRBs are even more uncertain because of the unknown degree of collimation, but are probably comparable.

'A small percentage' isn't too specific. The BBC article that is cited in this passage says the following:

Observations of deep space suggest that gamma ray-bursts are rare. They are thought to happen at the most every 10,000 years per galaxy, and at the least every million years per galaxy.

However, the 'small percentage' is never elaborated on.

That's all I have for this edit; more to come later. By the way, this Physics.SE question may interest anyone reading this. . .


Mmm... it's a pretty small scope using a fairly high power. The disc of Mars is not going to be very large (just a few seconds of arc across) so I don't think you should expect to see something that looks like a HST image! Mars is pretty red as a rule, so 'red light' is what you should be seeing because the disc is too small to show much detail.