Diameter of Astronomical Object Using Magnification

You could use your image to estimate the angular size of Jupiter that night. To find the actual diameter of Jupiter you would also need to know the distance that Jupiter is from Earth, and since Jupiter and Earth are both orbiting the sun, this distance changes.

You can find the distance with software (Eg Stellarium) or using Nasa's Horizons web interface (You will need to use the column labeled "delta" which gives distance in astronomical units). To find the diameter you will need to do some trigonometry. (Diameter = distance $\times$ angular size in radians)

Alternatively you could use the published Diameter of Jupiter (70000km) and use your observations to calculate the distance.

To calculate the distance to Jupiter without knowing it's diameter requires more than a single observation. You will need to have:

A long enough series of observations to establish the orbital period and elements of Jupiter.
Knowledge of Kepler's laws of planetary motion
Knowledge of the distance of Earth from the sun, and to get this you can fund a expedition to the South Pacific during a transit of Venus.
A sharp pencil to do all the maths.

I'm being a little glib... The point is that it took more than 150 years of research between the broad acceptance of heliocentric solar system, and our knowledge of its scale, or the actual sizes of planets.


Is there any planet or satellite in the solar system in the process of getting tidally locked?

The Moon is in the process of tide-locking the Earth...
Tidal interactions between the Moon and the crust and oceans of Earth have been slowing down the Earth's rotation since the moon's formation. In the past 4-odd billion years, the Earth's rotational period has gone from about 6 hours to about 24. These tidal interactions are also responsible for the Moon receding from Earth, and given enough time, is projected to result in the orbital and rotational period of the mutually-tide-locked Earth-Moon system stabilizing at about 47 days long.
But the Sun might run out the clock...
How long is this tide-locking projected to take? My quick googling on the answer comes up with a number of different estimates, from 7.5 to 50 billion years.
According to Wikipedia: Tidal Locking:

Earth is not expected to become tidally locked to the Moon before the
Sun becomes a red giant and engulfs Earth and the Moon.


A singularity in physics is basically a big signpost saying "the theory you are using has probably ceased to be a good approximation of reality by the time you get here". An example is the infinite amount of short-wavelength radiation produced by a blackbody under the Rayleigh-Jeans law. This is obviously an incorrect description of reality and the issue ended up being resolved by the development of quantum theory: taking into account the corrections gives Planck's law, and the infinity goes away.

Inside a black hole, general relativity does predict singularities, and it seems to be fairly common to talk of them as though they are physical objects. However we should be cautious: even though general relativity is our best description of gravity so far, we already know that it does not describe the actual universe. The issue here is that general relativity is incompatible with the Standard Model of quantum field theory, which is our best theory of how physics works excluding gravity.

By the time you get to the region near where general relativity predicts a singularity, you're working with a very small region with an intense gravitational field. Quantum corrections are almost certainly going to be relevant here, and unfortunately we do not yet have a self-consistent theory of "quantum gravity" to describe what should happen in such a region and so far experiments (such as the ones being performed at the Large Hadron Collider) haven't given much of a clue as to how to proceed with developing one.

In a hypothetical universe where general relativity is correct then there would be a singularity (which is fortunately hidden from the rest of the universe by the event horizon), but we do not live in such a universe.


A good telescope for the viewing of Nebulae, Stars and Planets

This is a very common question, yet very hard to answer if you prefer a clear, concise, uncontroversial answer that applies to all situations. So I'm not going to do that.

Instead, I'm going to describe your main options, and let you choose. Be aware that you'll make the choice while still not knowing much about optics. So, in a sense, it will be just the beginning of a learning journey, and you'll have to keep learning and trying things out. Hope you like that sort of thing, because it's kind of mandatory in this hobby.

Also, there are two kinds of people in this field: those who enjoy using the instruments, and those who enjoy making the instruments. I'm 25% of the former, 75% of the latter kind (I make telescopes and mirrors, and I do some optics design). So that's the perspective you'll get.

First off, a few words about the objects you'll observe. There are two kinds of objects:

HIGH RESOLUTION, HIGH BRIGHTNESS

These are the planets, the Moon, many double stars, and the like. These are typically pretty bright, so the amount of light you capture is not essential. But these objects show high-resolution detail, so the resolving power of your scope is crucial.

You get more resolving power if the aperture (the diameter) of the scope is bigger. But there's a problem - seeing (the inverse of turbulence).

http://en.wikipedia.org/wiki/Astronomical_seeing

When air is turbulent, it will reduce the resolving power of your instrument. If your scope is 200 mm (8") or bigger, this will happen A LOT. If your instrument is 100 mm (4") or smaller, it's not so noticeable because you don't have that much resolving power to begin with.

There are ways to forecast seeing, because it depends on weather:

http://cleardarksky.com/csk/

So, for high resolution targets, the point of having a large instrument (like a huge dobsonian) is because once in a while seeing is so good (turbulence is so low) that you can actually use all its huge power - but most of the time you'll be using it at much less than its theoretical power, as if it's a 4" instrument or whatever the seeing is at the moment.

These objects are not affected by light pollution. You could observe the Moon from a brightly illumined parking lot in the middle of the city. You don't need to be in deep darkness while observing them - in fact, it's beneficial if you're not adapted to very deep darkness.

LOW BRIGHTNESS

These are most of the nebulae, star clusters, and galaxies. These are distant, faint objects. Their light is very faint, so it's essential you capture as much of it as possible. Again, bigger aperture is better.

These are the kind of objects that huge dobsonians (half-meter aperture, or more) are made for. In a very large scope, there's galaxies everywhere.

But again, there's a catch. These objects are affected by light pollution.

http://www.jshine.net/astronomy/dark_sky/

You need to take your scope far and away from any city, town and industrial complex (at least a 1 hour trip at highway speed). This is pretty hard to do in Europe; in the US, population density is low and this is somewhat easier. Anyway, just stay away from city lights and the view will improve.

A few of these objects will be visible from the city: the Andromeda galaxy, the M13 cluster, the Orion nebula, etc. But many of them look much, much better away from light polution. In any case, you should be able to observe the whole Messier catalogue (over 100 objects) from the city - but out in the boondocks these things look better.

http://en.wikipedia.org/wiki/List_of_Messier_objects

It's important that your eyes are adapted to very deep darkness when observing these things, so stay away from any source of light. Even the screen of your phone is far too bright. Street lights 500 meters away are too bright. And turn off that flashlight and keep it off. One second of exposure to a bright light destroys a dark adaptation that took 20 minutes to build.

Since these objects are so faint, the eye doesn't see a whole lot of high-resolution detail, so they are not affected by seeing (as far as the human eye is concerned).

Okay, now here are the main types of instruments that you're most likely to encounter:

NEWTONIAN REFLECTORS IN GENERAL (DOBSONIANS IN PARTICULAR)

Many people will say "just get an 8 inch dob as your first scope", and that's usually correct. This design gives the biggest aperture per price. Biggest bang for the buck. It scales up easily to very, very large sizes. It's the ideal instrument for observing the "faint fuzzies" (nebulae, galaxies). It's great for learning, because you can easily observe its guts. When seeing is good, a large dob will open up the sky for you - you can crank up the magnification a lot (depending on aperture).

On the flip side, due to the low price, you will see many dobs out there that are shoddily made.

You need to remember two things with a dob: collimation, and thermal management.

Collimation means keeping all the optics aligned. The mirrors will move a little, due to the way a dob is built, and they'll lose collimation. So don't forget to re-adjust it once in a while. In time, you'll get so good at it, you'll give it a two-minute quick adjustment every time you observe, and it'll be good to go; that's what I do.

This is why collimation is important:

http://legault.perso.sfr.fr/collim.html

This is a good introduction to newtonian collimation (which is a vast topic, you could write a whole book on it):

http://www.cloudynights.com/documents/primer.pdf

Thermal management means keeping the scope at the same temperature as the ambient air. If you don't do this, internal convection will destroy the image. This is why:

http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/

http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/

Small dobs (up to 6...8"), just take it outside at least 1 hour before you observe - or better yet, 2 hours. Let it "breathe". Dobs 10" or larger, you MUST use a fan on the primary mirror to force cooling.

REFRACTORS

These instruments are very convenient and very low maintenance. Take it out of the box, and it's good to go. It's collimated from factory, and it doesn't care about thermal issues. A refractor makes a great "travelscope", and a great grab-and-go scope.

On the flip side, they are the most expensive, in terms of how much aperture you get for the money you spend. And they are typically made in small apertures; most commercial refractors are not bigger than 6" or so; those that are bigger are extremely expensive. The small aperture limits the maximum performance of these instruments.

There is this myth out there that refractors are somehow magically better for planets (and other high-resolution targets), whereas dobs, even very large ones, are not so great. This is false; load up the designs in Zemax, do a basic performance analysis, and you'll see that aperture is the dominant factor, with everything else being secondary. But there are several reasons why this myth is around, and it's instructive to see what these reasons are:

refractors perform at 100% of their potential no matter what, right out of the box. Dobs require maintenance and fiddling to perform at 100%.
many dob owners simply don't do the required maintenance - they don't collimate their scope, and don't give it time to reach thermal equilibrium. Then of course it will perform miserably on planets. There are 12" dobs out there that perform worse than 4" refractors because of how out of collimation they are and how hot the mirror is.
many refractors out there are quite well made (they ought to be, given the higher price you're paying), whereas many dobs are cheaply made in a sweatshop. So you're not comparing apples to apples.

Bottom line: don't buy a refractor because you "want to watch planets". But do buy a refractor if simplicity of use and the low maintenance are very important to you (and you don't care much about price).

A decent refractor on an excellent mount can do pretty good astrophotography. More on that below.

CATADIOPTRICS (CASSEGRAIN, MAKSUTOV, DALL-KIRKHAM, RITCHIE-CHRETIEN)

Broadly speaking, these are for astrophotography (not always, but usually). But can you do visual astronomy with them? Sure. It's just that it's easier to optimize them for photo, with this design.

Usually, they come installed on a motorized mount. All but the very cheapest ones have a computer on-board; you just punch the name of the object into the remote control, and the scope will turn around automatically to face the object. Sounds cool, right?

Well, I am going to say something a little controversial here: this is not a good place to start.

First off, astrophotography has an extremely steep learning curve. It's very easy to take a little blurry photo of Jupiter that will impress nobody. It's very hard to take a great, high resolution photo like these:

http://www.acquerra.com.au/astro/

The skills and knowledge required to take that kind of photos are more easily acquired by doing purely visual astronomy, on a dob or whatever, for a couple years or more. Also, the equipment required for that sort of stunt costs many thousands of dollars - your cheap little Cassegrain cannot do it.

So why buy a small Cass? Well, if you have an overwhelming interest in astrophoto, and are prepared to deal with the difficult learning, then go ahead. But realize that the first images are not going to be very impressive, and the motorized mount will not teach you some valuable skills that you would acquire more easily pushing a dob around and keeping your eye plugged into the ocular for a while. Also, money spent on the motorized mount is money NOT spent on aperture, and the aperture is so important for overall performance.

For astrophotography, the mount is of huge importance; the scope per se is relatively less important (I'm generalizing). A so-so scope on a great mount is much better than the other way around. But a great GEM mount can be very expensive.

Well, that's it. Do some more research on your own, then make a decision. Don't forget that learning only begins with the moment you acquire the scope.

Above all, have fun, and keep looking up.

Clear skies to you!


What kind of a visual trajectory would an asteroid have if it were to hit Earth?

There is no simple answer to this question other than: you have to do the math.

Most often you would see the asteroid brightening more rapidly and moving across the sky faster as the days pass, but not necessarily.  It could arrive directly along our orbital path and just brighten but show no motion in the sky.  It could start out lit up fully (full phase) and then arrive with only partial lighting from the sun, so it does not necessarily brighten.

You generally would not know the phase, unless you calculate it, because you will not be able to resolve it until it gets really close.


How to compute Ascension / Descension cycle of the Moon


Computing the average hours of sunshine at given location&orientation

Between September and March (ie winter, assuming a Northern Hemisphere location) The sun rises in the South East, is Due south at midday and sets in the South west. Your South facing window would get sun for as long as the sun is up. The North West facing window would only get a glimpse of the sun around sunset (and it would then be low and weak)

Between March and September (Northern hemisphere summer) The sun rises in the North East, it is in the south between roughly 6am and 6pm, then sets in the North West. Your Southern window would get 12 hours of sun each day. The North West window would get some sun in the evening, and this sun would be low (weak but shining into the room)

Overall the South facing window would get a lot more sun than the North West window. The reason is that the sun is always above a point on the Earth that is between 23.5 degrees North or South of the equator. For anyone in the Northern Hemisphere, the sun is usually South of them.

A quantatative estimate of this could be done, but the actual values you got would depend on the Latitude and precise aspect of the two windows. However the result from SunCalc looks to be robust. The North West facing window will only get a little sun on summer evenings, and little or nothing during the winter.


If dark matter bends light, how do we know the stuff in the sky is where we think it is?

The local dark matter density is actually quite tiny, on the order of $\rho\sim10^{-19}\text{ g/cm}^3$ (see e.g. Bovy & Tremaine (2012)). This means that there is roughly $0.001$-$0.01M_{\odot}$ of dark matter per cubic parsec - a staggeringly small amount. 1000 cubic parsecs would contain about one solar mass of dark matter - and that's a cube 10 parsecs in length on each side! Now, the distribution of dark matter in galaxies is not homogeneous - it follows, roughly, a Navarro-Frenk-White profile, decreasing in density from the center of the galaxy - but on the scale of parsecs (and certainly in the Solar System), we can consider it to have roughly uniform density.

On small scales, then, we have approximate homogeneity and low density. This means that any gravitational lensing effects from dark matter should be extremely low or self-cancelling, arising only from inhomogeneities containing large clumps of dark matter. However, such clumps are unlikely to form solely through dark matter's interaction with itself (if we discount the MACHO hypothesis, which, as far as I know, is not currently favored).

On intergalactic scales, however, dark matter can have some effects. Weak lensing is a commonly-observed phenomenon in galaxy clusters, which may have extremely high fractions of dark matter. There are several techniques currently used to model the mass distribution of the lensing galaxy (see the KSB+ method) and to reconstruct the image and position of the original galaxy via deconvolution (see Chantry & Magain; a visual example is given here). I'm not too familiar with either technique, though, so I can't give you a good overview.

Even large-scale lensing has large mass requirements. zephyr pointed out that the foreground object that created the Einstein cross contained $\sim10^{10}M_{\odot}$ of dark matter (van de Ven et al. (2010)). That is enormous!


When will Sirius be closest to the solar system?

Sirius is moving towards the sun at 5.5 km/s, and according to Sky and Telescope, will reach its closest distance in 60000 years, at which time it will be 7.8 light years distant, and be at magnitude -1.64, only slightly brighter than it's current magnitude -1.46.

Calculations using the latest Hippacos results give a slightly different value: a closest approach of 8.18 ly in 46000 years, and consequently a slightly dimmer peak. However Sirius is not moving rapidly relative to the sun, and will not change its brightness greatly in the next 100000 years or so

It will be the brightest star in the sky for another 90000 years, and no star will match its current brightness for another million years (when delta Scuti will reach magnitude -1.84.


Robert Hooke's Micrographia from 1664 has a detailed discussion of "seeing"

The table [of contents], which is at the end of the book, explains that pages 230-232 discuss:

  that the Air near the Earth is composed of parts of differing density...this property produces the effects of waving and dancing of Bodies; and of the twinkling of the Stars


What is the angle between the equator of the sun, and the plane of the Earth's orbit

The plane that contains the orbit of the Earth is known as the "ecliptic". The rotation of the sun is tilted by 7.25 degrees to the ecliptic, and this value does not vary over time.

The rotation of the Earth is also tilted, by 23.45 degrees to the ecliptic, it is this angle that causes seasons.

An old paper Position of the Sun's axis describes how the Earth crosses the solar equator on June 4th and Dec. 6th each year.

Nasa has a factsheet, that has this and other information about our nearest star.


What will be the impacts on the planets' orbits from Gliese 710?

It'll have basically no effect on the planet's orbits, except perhaps on the theoretical planet 9 and even then, not very much.  As close as it's expected to get Gilese will till be too far away to affect the 8 planets much.

Gilese 710's mass is about 60% the mass of our Sun.

I've seen two estimates, of 13,000 Astronomical Units (AU) and 13,365 or about 77 light days or 0.21 light years at closest pass, and some estimates put a +/- margin of error at as much as 20% of that, but lets use 13,365 AU as an estimate.   For a star, that's exceptionally close but it's enormously distant for a planet.

Neptune, the most distant planet, averages about 30.1 AU from the Sun, so Gilese will be about 440 times further away from Neptune than Neptune is from the Sun.   That will give it a negligible gravitational effect of less than 1 part in 300,000 on Neptune's orbit.  That's, by crude calculation, a bit less than the effect Saturn has on Earth's orbit.  That's enough for a small perturbation, but you'd need some pretty high tech equipment to even detect it.   It's about 50 times less than the gravitational effect Neptune had on Uranus when it was noticed that Uranus' orbit was behaving oddly and that observation lead to the discovery of the planet Neptune.

So, basically, for the 8 known planets, the outer planets will get small pushes, noticeable by very meticulous observation only and the inner planets, essentially no effect.

It's effect on Planet 9 would be greater but still small.  The estimate for planet 9 at Aphelion is about 1,200 AU or 1/11th the closest pass of Gilese.  That's close enough for a noticable push.  Not enough for any major changes like pushing it's eccentricity to where it passes anywhere close to Neptune, or ejecting it from the solar system, but Planet 9 could be jostled somewhat, perhaps as much as a 1% change to it's orbital period and/or eccentricity, which for a planet's orbit, 1% is a large change over a single orbital period.

The bigger effect, as you mentioned, will be on more distant objects, several thousands to tens of thousands of AU from the Sun, or several times more distant than the theoretical Planet 9 at it's most distant.   Objects that distant have never been observed, but are theorized to be out there in vast numbers based on a handful of observed objects with very high eccentricity.  Oort cloud objects at those distances will get cast every which way, some cast out of the solar-system and a small percentage, pushed to higher elliptical orbits where they could pass through the inner solar system.   Our sun will do the same thing to Gilese 710, casting it's Oort cloud objects in every direction and each star will steal a number of orbiting objects from the other.

The initial effect, the inner solar system might pass through the outer edge of Gilese 710's outer oort cloud.  I don't think anyone knows how extensive it's Oort cloud is at 13,000 AU, but any initial increase in objects passing through the inner solar system will be from Gilese 710's Oort cloud.

Any of our Solar system's Oort cloud objects that Gilese 710 sends towards the inner solar system will take (roughly) between 50,000 and 400,000 years to reach the inner solar-system, so Gilese 710 will be perhaps still visible to the naked eye when the proposed rain of comets begins, but it will be long past being the brightest star in the sky if/when any objects it sends towards the inner solar system make it here.   The reason for this is that it moves much faster than any objects it's likely to send our way.   Gilese 710 will pass through the solar system at about 50,000 kph and any objects it sends towards the inner solar system, because they're so spread out and most are likely to just get a small push from the passing star, they will leisurely drift towards the inner solar system at roughly between 500 and 1,500 kph, requiring about 50,000 years for the first ones to begin to reach the inner solar system and become comets or perhaps crash into the Earth or Moon or other planets.

Also, fun fact.   At closest pass, Gilese 710 should be a hair brighter than Jupiter at Jupiter's brightest, nearly -2.8 apparent magnitude.  It will be the brightest star in the sky for ballpark 15,000 - 20,000 years, over that time-frame it will move across the fixed stars a good percentage of the 180 degrees of horizon.   In a human lifetime it might move a bit over 1 degree of arc at closest pass, so it won't be very noticeable that it's moving, but that's very rapid proper motion for a star.

(Too long?)

I've excluded the mathematical formulas cause I don't do the that well and it would get a little ugly.  I can't do the math of a 3 body system, so this is an approximation, but I my approximation should be in the ballpark.


Modern astronomers are not instructed on the constellations and pay little attention to them except as a naming convention for stars.  Positional information is nearly always either equatorial (ra,dec), or galactic (gl,gb), or supergalactic (SGL, SGB).  To improve the speed in which information on a region can be extracted from a database, a scheme called Hierarchical Triangular Mesh (HTM), is often used.  The sky is divided into three sections (0,1,2) and each of those is divided into 3 sections (00,01,02,10,11...) and this iterated until the level of finest size is reached.  One can address the finest region by using all of the digits or a larger region by dropping some of the final digits.


Why is the exoplanet HD 131399Ab so hot?

It's a giant planet and it's young. p.3 of the paper you reference gives a likely age of 16 million years.

The heat arises from gravitational contraction and it would be a similar temperature even if it were further away from the stars.

When a ball of gas contracts, the virial theorem establishes that half the gravitational potential lost is radiated away and half heats up the gas.

Jupiter would also have been much hotter in the past and even now emits more radiation than it receives from the Sun. The rate of contraction (and hence the planet's luminosity and temperature) decreases as the planet ages.

Direct imaging surveys for giant planets deliberately target young stars for this reason.


What is the farthest point of light that is emitted by a torch?

If you point a flashlight up, then the light will reach as far as the other ends of the universe, if one has a sensitive enough measuring device, so long as it is not absorbed. Of course, said device might require a lot of work, and a huge telescope. Lights can easily be seen by satellites, but it would take a huge telescope to see a single light from, say, the Moon, and even larger the further you go.

In the case of your moving light, I would suspect a laser of some kind. Lasers are focused enough where one could easily see a reflection from the clouds, and pretty focused as well. It could be a moving light source, such as a plane or balloon of some kind as well.


Is it possible to have a star orbiting around a brown dwarf?

Yes

If a brown dwarf count as a star in this case, the solution is as easy as a smaller brown dwarf orbiting a larger one.

If not, it is still possible if you have two brown dwarfs orbiting close to each other, both just too small to be red dwarfs, together out-massing an orbiting red dwarf, just large enough to be counted as a star.

While I have no example of such a system, the Alpha Centauri system has to closely orbiting stars, with a more distant star orbiting, and Luhman 16 is a system with two brown dwarfs orbiting each other.


Why do we observe spectral lines of specific energy?

When an interaction occurs that results in the absorption or emission of a photon, we are not dealing with stationary states. Stationary states are solutions of the time-independent Schrödinger equation, but the absorption and emission of photons is a time-dependent phenomenon.

Spectral lines in fact do not have a definite energy. They have a natural width associated with the time taken for the transition to occur.

The probability that a transition might occur resulting in the absorption or emission of a photon of a given frequency can only be tackled by considering a time-dependent Hamiltonian. What one finds is that transition probabilities become large for frequencies close to the those corresponding to energy differences between stationary states.


Yes, #1 is the correct answer.

✅ For small particles such as dust grains and molecules, light generally interacts most easily with particles roughly the same size as its wavelength. This is seen as the maximum in the figure below. That is, the particle’s cross section doesn’t depend on its exact shape, but rather on the ratio between the size and the wavelength of the incident light: As the particles become larger, you enter the regime of “geometric” scattering, where the cross section approaches the geometric cross section (e.g. $\pi r^2$ for a sphere etc.), whereas as the particles become smaller, you enter the “Rayleigh regime”, where the cross section drops proportionally to the wavelength to the power of $-4$.
Since observationally we know that interstellar dust affects optical
and, in particular, UV light significantly, we know that the dust
grains must be roughly the size of that light’s wavelength. More carefull modeling shows that a typical cosmic dust size distribution can be described by a power law with the probability of size $r$ decreasing roughly as $P(r) \propto r^{-3.5}$ (Mathis et al. 1977). In other words, for every dust grain of size $r$, there are 3,000 times as many dust grains of size $r/10$.

 "Normal" dust particles are in the range 1–100 $\mu\mathrm{m}$ (thanks for the sources @uhoh & @PeterErwin, unless they mean dust bunnies which are more like mm- to cm-sized. At any rate, cosmic dust is not much larger than normal dust, so #2 isn’t correct.

 #3 also isn’t correct. While it’s true that 98% of the interstellar gas is hydrogen and helium, dust is composed of the last 2%. That is, iron, carbon, silicon, oxygen, etc.

 Finally, while it’s true that blue light is absorbed or scattered more efficiently by dust than red light, dust doesn’t emit blue light. If it’s absorbed, the energy goes into heating the dust and is then later re-emitted as infrared light. So #4 also isn’t correct.

In the figure below you can see how the cross section of a(n idealized) particle depends on the ratio between its size $r$ and the wavelength $\lambda$ of the incident light.

Credit: Catslash, Public domain, via Wikimedia Commons with own annotations.


What would the cost be of visiting an asteroid?

This is something that has already been done, so we already have a point of reference. Hayabusa landed on the asteroid 25143 Itokawa in 2005. The mission was considered a success and a sample was returned to Earth. According to sources from the time, the total mission cost was around 170 million dollars and other sources put the development cost of the spacecraft at around 138 million dollars. There is also a planned Hayabusa 2.


Can gravity be amplified by other celestial bodies in its vicinity?

Gravity itself cannot be amplified but the forces are additive. The earth is held in orbit by the Sun's gravity but earth is also affected by the moon's gravity. Depending on their position the moon's gravity could be aligned with Sun's gravity, increasing the gravitational pull. Or the moon could reduce the effect of suns gravity if on opposite side of the earth relative to the sun.


Compton scattering of high energy radiation

The answer is pair production. Once photon energies exceed 1.02 MeV it is possible to spontaneously create an electron-positron pair in the presence of an atomic nucleus (to conserve momentum). In general for high energy photon interactions with matter you need to consider the photoelectric effect, Compton scattering and pair production. The former is more important at lower energies, the latter at higher energies.

The picture below shows the mass attenuation coefficients for Aluminium and Iron as a function of energy, with the contributions of these three processed identified.


What exactly is a "conclusive association" in Astronomy?

A "conclusive association" is an oxymoron; for example, "brawling love", "loving hate" (these first two are from Shakespeare), "act naturally", or "jumbo shrimp". An association is a vague term while conclusive is very specific and means proven without a doubt.
The underlying journal article does not claim the authors have provided a "conclusive association between the diffuse gas in the space between galaxies and the large-scale structure of the cosmic web predicted by cosmological theory." That is what the press release claims. Press releases are not subject to peer review. Press releases can, and oftentimes do, use over flowery speech.
The underlying journal article does claim that "statistics of absorption by the intergalactic medium (IGM) via spectroscopy of distant quasars support the model yet have not conclusively tied the diffuse IGM to the web." This is true. The implication is that the authors' model has conclusively tied the diffuse IGM to the web. The author do not explicitly make this claim in the journal paper. Suppose the authors had make this claim in early versions of the paper. If that was the case, peer reviewers would inevitably have said to remove this claim on the basis that "conclusive" is an extremely strong word in science.
Even if peer review had allowed that claim, it would still be dubious. Many scientific press releases claim that the research being reported is ground shaking and conclusive. Don't take press releases as fact. Only a tiny few scientific papers show something that is ground shaking and conclusive. Those that do do so (e.g., Einstein's annus mirabilis papers) are very few and very far between.


Mathematical astrophysics subjects for math graduate students

As a first step Newtonian Mechanics is very important. Both at an undergrad level and graduate level. So a calculas based first course is very important. Also very useful is a source on undergrad Atomic Physics and a course in Nuclear Physics. There are several others besides these. I would recommend the series of courses in a Halliday and Resnick type of text as a start. This will help set you up for many other courses in Physics. IMHO.


How to determine the age of a star using asteroseismology?

Asteroseismology effectively measures the sound speed inside a star by finding the characteristic oscillation frequencies of a star.
The sound speed depends on the composition because the pressure at a given temperature depends on the average mass of a particle in a gas.
As the star gets older, it turns its hydrogen into helium, changing the composition, the average mass per particle in the gas, and hence the average sound speed.
Asteroseismology tells you the age of the star.
The answer is somewhat model-dependent, so ideally you would have an effective temperature and surface gravity in order to constrain matters. Precisions of 10% of the main sequence lifetime are possible for stars of around a solar mass. Accuracy is hard to assess since we have an accurate age for only one star - the Sun.


How did Astronomers deduce that the Sun was not a ball of fire?

I think it's maybe not the case that there was a moment when the astronomy community conclusively rejected the ball-of-fire hypothesis; astronomers simply accumulated more and more evidence against it. If you want to put a rough date on it, you could put your finger somewhere in the middle of the 19th century, as by then, other ideas had taken hold.

Back in the classical period, Anaxagoras had proposed that the Sun was a heap of molten metal. I don't know whether this was widely accepted by his contemporaries. The idea of the Sun as a ball of metal or fire certainly persisted for some time, though perhaps largely for lack of any better ideas. We didn't even understand oxygen and combustion until the work of Lavoisier and others in the late 18th century, so detailed calculations were presumably out of the question for a millennium or two after Anaxagoras. I don't know when calculations of how long combustion could sustain the Sun were first done, but it appears to have been not more than several decades after the theory of combustion was developed.

Why? Well, we can say that by the middle of the 19th century, the predominant explanation for the Sun's luminosity was not the burning of coal but instead gravitational potential energy. By the 1860s, it was widely known that chemical reactions could only power the Sun for a few thousand years. We also now had a potentially viable alternative: a decade earlier, Hermann von Helmholtz had begun exploring the idea that gravitational contraction of some sort, by what we now call the Kelvin-Helmholtz mechanism, was the source of energy, with gravitational potential energy being transformed into heat$^{\dagger}$. Around the same time, Lord Kelvin suggested that meteors falling into the Sun provided the necessary energy, a similar mechanism to Helmholtz's. I believe astronomers continued with the contraction hypothesis through the turn of the century - I've seen an article written around 1900 to that effect.

However, during the early and mid- 1900s, quantum theory and nuclear physics were being developed, and the work of Eddington, Bethe and others would lay the groundwork for our current understanding of solar energy production. Previous models (including, finally, Kelvin-Helmholtz contraction) were now known to be insufficient because they allowed the Sun to shine for only thousands or millions of years, and geologists had established that Earth itself was much older than this. Fusion, on the other hand, allows the Sun to survive for billions of years - a timescale that matches up well with the age of the Earth. We also knew that hydrogen and helium were the dominant constituents of the Sun and other stars; while Wollaston and Fraunhofer had performed the first solar spectroscopy observations in the early 1800s, the true composition of the Sun was not accepted for more than a century, when Cecilia Payne made a detailed study of spectral lines.

$^{\dagger}$ While this does produce heat in various bodies, including T Tauri stars, it is not significant in most stars beyond that stage.


.2015: When last did both New/Full moon in a fort-night cause an Eclipse?

As written here:

  Another oddity of nature is that solar eclipses and lunar eclipses tend to come in pairs – a solar eclipse always takes place about two weeks before or after a lunar eclipse.

And here:

  Rules of Eclipses (Solar and Lunar) ... Eclipses tend to go in pairs or threes : solar-lunar-solar. Lunar eclipse always preceeded by or followed by a solar eclipse (two weeks between them)

So, seems like a totally common thing.

P.S. The previous pair was first lunar on Oct 8, 2014, and then solar on Oct 23rd.


How many satellites orbit their planet faster than the planet rotates?

Uranus rotates once every 0.718 days. 11 of its satellites have a shorter orbital period. These are inner satellites of Uranus which are roughly in the equatorial plane of Uranus. I don't quite understand what direction they orbit in compared to the direction in which Uranus spins.
Neptune rotates once every 0.671 days. 5 of its satellites have a shorter orbital period and orbit prograde.
Jupiter rotates once every 10 hours. 2 of its satellites have a shorter orbital period and orbit prograde.
Saturn rotates once every 0.4 days and none of its satellites orbit faster than that.
So Phobos is not alone in its case. Around Mars, Jupiter and Neptune (and maybe Uranus), there are moons that orbit so fast that they rise in the west and set in the east.


Matter and Antimatter Interaction in the early Universe

Matter and antimatter aren't perfectly symmetric, when following the "forward" direction of time. This is commonly called charge-parity violation.
It's assumed, that this asymmetry caused an excess of matter in the early stages of the Universe after the Big Bang, annihilating almost all antimatter, and leaving behind matter.


When we see the Sun, do we actually see its past?

Yes, you are right. We don't only see the Sun 8 minutes in the past, we actually see the past of everything in space. We even see our closest companion, the Moon, 1 second in the past.

The further an object is from us the longer its light takes to reach us since the speed of light is finite and distance in space are really big.


A magnetic field is only one of the requirements for an aurora (and given the signs of emissions from excited oxygen on Venus after solar flares, it may not actually be a requirement). There needs to be an atmosphere, too. In Mercury's case, there isn't much in the way of an atmosphere. The magnetic field still channels solar wind to the polar regions, where it causes sputtering of sodium. Given this is the result of the same magnetic channeling that produces the aurorae on Earth, you could think of this as Mercury's equivalent of aurorae. The sodium forms a faint glowing tail as it escapes the planet.


Why is Uranus able to support a regular satellite system?

The solar perturbations on most of the satellites of Uranus are on a very small scale indeed, which may explain the absence of the instabilities noted in the question.

A perturbational effect depends on the scale of the solar perturbing accelerations relative to the ordinary inverse-square attraction of the primary body. The scale factor (often designated $ m^2 $ in texts on the lunar theory such as Airy's 'Mathematical Tracts' and Godfray's 'Lunar Theory') increases with the separation of the satellite from the primary, it is also as the cube of the distance-ratio satellite-primary : primary-Sun (see detailed calculation below).

Taking as an example the most distant of the major moons of Uranus, Oberon at 583500 km:

The scale factor for the solar perturbing accelerations on Oberon in its orbit relative to Uranus is only about 1/5,000,000 of the gravitational acceleration towards Uranus felt by Oberon (calculation below).

Comparing the corresponding scale factor for Earth's moon and its solar perturbations, the factor is very much larger, close to  1/178, as is well-known.  The Moon has a fairly strongly sun-perturbed orbit relative to the Earth, but the solar perturbations on Oberon are more than four orders of magnitude smaller, really very tiny.

It is possible that the two tiny outermost satellites XVI and XVII of Uranus may be experiencing perturbations large enough to drive growth of their eccentricities, their eccentricities at 0.18 and 0.52  are very large compared with the major satellites, all with eccentricity < 0.004 (source, Astronomical Almanac 2016).

Detail of the approximate calculation:

The 'sampled' solar perturbing acceleration on a satellite is taken for present purposes to be represented by the perturbing acceleration on the satellite when it is close to quadrature with the Sun (as seen from the planet). In this configuration, the solar perturbation on the satellite is directed towards the planet, adding here to the ordinary inverse-square attraction of the satellite towards the planet.

With mass-constants of the Sun, S, and of the planet, P ;  planetary semi-axis a ; and planet-satellite separation d , plus an assumption that the mass of the satellite is very small relative to the other two:

1 ** the planet's accelerative attraction on the satellite is $P/d^2$ ;

2 ** the Sun's accelerative attraction on the planet is $S/a^2$ ;

3 ** and so also (to a quite close approximation) is the amplitude of the Sun's accelerative attraction on the satellite  $S/a^2$ .

4 ** When the satellite is in quadrature, the resolved part of the Sun's vector attraction on it that does not cancel with the Sun's attraction on the planet, i.e. the net perturbation on the satellite at that point, is given very nearly by the proportions of the triangle satellite-planet-sun:  the lengths of two of the sides are approximately a and the third d. In this configuration, the resolved part of acceleration #3 in the direction of the planet, to a close approximation, is thus  --

$ (S/a^2) .  (d/a) $.

The ratio of perturbing acceleration 4 to ordinary planetary attraction 1 is thus

$ (S/P) . (d/a)^3 $.

For the Sun and Uranus, S/P ~= 22902 ,

for the Sun and Earth+Moon, S/P ~= 328901 .

Uranus is approximately 19 au from the Sun to the Earth's 1, the au is  149597871 km, the mean planetocentric distance of Oberon is 583500 km and of the Moon 385000 km.

Using these figures, the ratio

'solar perturbation on satellite : planetary attraction on satellite'

comes to ~ 1/178 for the Earth and Moon, and  ~ 1/5,000,000 for Uranus and Oberon.

The tiny outer satellites of Uranus are farther away from it than Oberon is, in a ratio of about 20.8 for the outer (XVII). Thus the perturbations on it, as a proportion of the ordinary attraction of Uranus at its distance, is $(20.8)^3$ times larger than for Oberon, making the relevant perturbation scale factor as large as about 1/550 , still smaller than for the one for our Moon, but perhaps enough for the disturbing effects to be reflected in its higher eccentricity of about 0.52.

{Update:} It turns out that the orbits of the outer Uranian moons have actually been studied: Brozovic, M.; Jacobson, R. A. (2009), "The Orbits of the Outer Uranian Satellites", The Astronomical Journal, 137 (4): 3834-42. Just one of the (high-eccentricity) outer satellites was found to be disturbed by a Kozai resonance and may be in unstable orbit. It appears that the conditions for the effect are not fulfilled for the others.


How early could life supporting planets been formed?

You can get relatively high metallicity rather quickly in parts of the early universe -- especially some globular clusters and the centers of massive galaxies -- because star formation rates in those places were very high, which means lots of massive stars and several rounds of "massive stars form, go supernova and seed surrounding gas with metals, new massive stars form out of enriched gas" -- the timescale from formation to supernova can be as short as 2 million years for really massive stars.

So how high a metallicity do you need to form planets, and how early did (parts of) the universe get there?

One the one hand, we know of a star with a metallicity (iron
abundance) only a quarter of the Sun's ([Fe/H] = -0.63) with a
Neptune-mass planet, and Kepler has found roughly Earth-sized planets
around stars almost as metal-poor.
So you don't need very high metallicities to form planets.
On the other hand, there is evidence of gas around quasars enriched
to above solar metallicity by redshifts of 6 to 7 (about 800
million years after the Big Bang). There are also rather old stars in
our galaxy with high metallicities: for example, this paper estimates that
the stars in the globular cluster NGC 6258 are 10-12 billion years
old with near-solar metallicity.

So the short answer is that you could have planets forming within the first 1-2 billion years after the Big Bang, and quite possibly within a few hundred million years after the first stars formed. Because the metals would mostly be from core-collapse supernova (as opposed to the iron-dominated products of Type Ia supernovae), there would be an excess of so-called "alpha elements" -- particularly oxygen, magnesium, neon, silicon, sulfer, argon, and calcium -- relative to iron; I don't think you'd have to worry about missing too many life-critical elements.

These planets would not all be very good places for life, however, because they would be in dense regions with lots of stars and lots of star formation, meaning lots of supernovae going off nearby, and possibly an active quasar as well if we're talking about a galaxy center.


Who first hypothesized that the universe is accelerating?

Science isn't concerned with belief. The possibility of a dark energy term and therefore an accelerating universe is implicit in GR and in the consequent Friedmann-Lemaitre equations that govern the dynamics of the universe. When the supernova cosmology teams set out to measure the deceleration of the universe, the models that they fitted to their data explicitly allowed for the possibility that the universe might actually be accelerating.

However, such hypothetical universes had certainly been explored before. The De Sitter universe (proposed by Willem de Sitter in 1917) is an accelerating universe consisting only of a cosmological constant. Our own universe appears to be headed asymptotically towards being described as a De Sitter universe.

The idea of $\Lambda$ as a form of "dark energy" that had the potential to cause an accelerating universe was already put forward as an idea by Lemaitre in the late 1920s and early 30s. The picture below shows the evolution of the scale factor for families of solutions and was apparently made by Lemaitre in 1927 (see this science history blog about Lemaitre). The upper 5 curves correspond to a universe that is accelerating today.


Instrumental magnitude to "real" magnitude - Photometry with not enough data?

The easiest method of photometry is by comparing the measured instrumental intensity of the unknown target with one or more other known targets in the same field and then converting the difference into magnitudes.  This is called differential photometry, and is easier because you don't have to worry about things like atmospheric extinction.  The more comparison stars you use, the better your result.  There are tools that will automatically do this analysis and calculate your error bars.  I use MaximDL.

Visit http://www.aavso.org and use the Variable Star Plotter (VSP) tool to generate a chart for your star of interest.  This will give known magnitudes of comparison stars in the same field.  You can use those to determine the magnitude of your target.  While you're there, check out some of the FAQs and documentation on variable star observing, and submit your results to the online database.


Are Symbiotic and Vampire Stars the same type of star or different?

As the title of the research paper linked to in the ESO press release is "An incisive look at the symbiotic star SS Leporis", it is safe to conclude that they are the same type of star. My suspicion is that, since symbiotic stars transfer and accrete mass via either Roche-lobe overflow or stellar wind overflow, the term "vampire star" was coined as part of the press release as a "cute" way to imply that one star is "feeding" on another.


Considering that the moon’s gravitational pull on earth is so weak- we need special instruments to detect it- how does it cause the tides?

Gravity is an acceleration, and force = mass x acceleration.  Thus, if we want to detect the Moon's gravitational pull, we need either a giant mass or a really precise instrument.  Since the oceans are a giant mass, the force on the ocean is thus correspondingly large.  The fluidness of the ocean also allows it to move, even if the force per volume is small.
To expand on this, the tidal acceleration due to the Moon on the surface of the Earth is about 1 μm/s$^2$ (1), which means that the force due to the weight of a 1000 kg object will vary by about 0.001 N over the course of a day.  1000 kg of water is about 1 cubic meter, so every cubic meter of the ocean experiences that tiny variation in force as well.  Since the ocean has about 1.3 x 10$^{18}$ cubic meters (2) of water in it, you could say the tidal force is 1.4 x 10$^{15}$ N, which is an absolutely enormous force - translated into weight, it's more than the weight of all living things on the Earth (3).
There is also a corresponding acceleration from the Sun, about half as strong as the Moon, which also has some effect on the tides, resulting in a monthly very high and very low tide, known as the spring tide, and a somewhat averaged tide, known as the neap tide.  It might also help to realize that this 1/2 as strong force is what is also responsible for keeping the whole Earth in orbit around the Sun.


What is the probability of an astronomical body hitting you?

Comets are part of the meteor calculation.

Space junk is a minor concern. Most satellite and rocket bodies are fairly flimsy and are destroyed by the atmosphere. Sometimes chunks do come down. But by the time they reach Earth they are not falling fast enough to form a significant crater.

However being directly hit could be fatal.

  According to Mark Matney, a scientist in the Orbital Debris Program Office at NASA's Johnson Space Center in Houston, the odds that any of the 7 billion people on Earth will be struck by a piece of the soon-to-fall satellite is 1 in 3,200. "The odds that you will be hit are 1 in several trillion

In '97 a piece of debris from a Delta rocket hit Lottie Williams in Tulsa. It caused no significant injury. http://www.livescience.com/33511-falling-nasa-satellite-uars-risk.html


Would it be possible to discover an asteroid belt around another star?

It's currently very hard to impossible to observe asteroids around other stars.
Asteroids are bodies with a size of at least several metres to a few hundred kilometres in diameter. There's three main ways to detect bodies:

by their characteristic thermal emission
by obstructing the light of their central star
by reflecting light of their central start to us

Either of those three methods is extremely challenging for bodies of that size: they are not very numerous and their thermal emissions is in the mid-IR (depending on proximity of their host star) and is easily drowned in the emissions of their host star.
Obstruction of the direct light of their host star (transit method). This relies on their geometric cross-section which is very small. We can detect reasonably transits of fraction of 0.001 or possibly with quiet stars and good conditions down to even 0.0001. An asteroid will cause a smaller signal.
They are so small that any reflected light is also very hard to detect even with interferometric imaging methods. This has only been proven possible for some objects at quite a separation to their host star and which are much larger. The reflected light is also much smaller at those separations, and given their size so that we cannot make direct observations.
So in summary: we are currently at the edge of detecting planets of Earth size and are approaching Mars and Mercury with the transit method - both are considerably larger than asteroids.
Thermal emissions are mostly used for dust. Single asteroids have a much smaller total surface and thus are not big enough to cause a significant IR excess at their typical emission wavelength (even to indicate the presence of an asteroid belt with more than single object).
Detection of asteroids via astrometry is unlikely even in the near and mid-term future. We currently only have technology to detect (rather distant and larger) gas-giants around nearby planets. The best we have for astrometric detection currently is the Gaia satellite which has detected a 10 Jupiter-mass gas giant in a 22 year orbit as a companion to Beta Pictoris.


The physical processes of emission lines in cosmic nebula

Just a short answer, and likely others will fill in more details.

If there is ionization of some atoms, then generally there is recombination as well - you will have both processes going on, roughly in balance with each other.  Typically when an electron recombines with an atom, it does so into some excited state.  Then as it drops from that excited state into lower states, it emits one or more photons on the way down.  This is how ionization gives you emission lines - it’s the inverse process that matters for the emission.

To have emission lines, the cloud only needs to be hotter than the background behind it.  So in your sketch, an observer at the bottom would see an emission spectrum (as you’ve drawn), but an observer to the right would see an absorption spectrum (with the same lines) - from that perspective, more light is lost relative to the (hot) background source than is re-emitted in that direction by the cloud.

This link may help with some additional explanation of these ideas.


Astrophysical "unitised" version of the Gravitational constant

Useful formulations which have taken some time to dig out:

$G\sim{4.3\times{10^{-6}}}\,\mathrm{kpc}\,\mathrm{M}_{\odot}^{-1}\,\mathrm{km}^{2}\,\mathrm{s}^{-2}$

or

$G\sim{4.3\times{10^{-3}}}\,\mathrm{pc}\,\mathrm{M}_{\odot}^{-1}\,\mathrm{km}^{2}\,\mathrm{s}^{-2}$


If the Mars polar caps were melted, how deep would the oceans be?

The oceans would be smaller because Mars has lost most of its water. From Wikipedia's article on the Martian polar ice caps:

  Evidence that Mars once had enough water to create a global ocean at
  least 137 m deep has been obtained from measurement of the $HDO$ to $H_2O$
  ratio over the north polar cap. In March 2015, a team of scientists
  published results showing that the polar cap ice is about eight times
  as enriched with deuterium, heavy hydrogen, as water in Earth's
  oceans. This means that Mars has lost a volume of water 6.5 times as
  large as that stored in today's polar caps.

  The water for a time may have formed an ocean in the
  low-lying Vastitas Borealis and adjacent lowlands
  (Acidalia, Arcadia and Utopiaplanitiae). Had the water ever all been
  liquid and on the surface, it would have covered 20% of the planet and
  in places would have been almost a mile deep.

So if all the water melted (and the atmospheric pressure allowed liquid water) the ocean would cover roughly 6% of the planet, and be roughly half as deep as the original ocean. This crude estimate totally ignores the topography of the Martian surface. In the comments, David Hammen writes:

  Given the marked unevenness of Mars's surface, if both of its ice cap melted completely the resulting water would cover a fraction of its northern hemisphere, plus a few depressions such as the Hellas Planitia (8 kilometers deep!), but to a depth much greater than 137 meters.

Heavy water is slightly heavier than normal water, so it evaporates slower and has a higher boiling point. So in a mixture of normal and heavy water the normal water evaporates at a slightly faster rate, which increases the relative concentration in the unevaporated water.

That same article mentions that the total volume of both Martian polar caps is around 3.2 million cubic kilometres. Most of that ice is water ice, but some of it is frozen carbon dioxide (aka "Dry ice"). The article doesn't state the proportions, but it says

  During each year on Mars as much as a third of Mars' thin carbon
  dioxide ($CO_2$) atmosphere "freezes out" during the winter in the
  northern and southern hemispheres.
  [...]
  Analysis of data showed that if these deposits were all changed
  into gas, the atmospheric pressure on Mars would double.

Note that not all of Mars's water is in its polar caps, some may be frozen underground, especially near the poles, and some underground water may even be in a liquid state, due to a high salt concentration.

For further details, please see the linked article.


Factors included in the average density of planets

Whether of not you include the mass of the atmosphere doesn't matter much. For instance, the mass of Earth's atmosphere is roughly $10^{-6}$ of Earth's mass, corresponding to the uncertainty in the gravitational constant $G$ used to calculate the mass in the first place.

But the volume that you use for calculating the average density could potentially matter, depending on your definition of the height of the atmosphere. For Earth, the lowest-orbiting satellites are at a height of $\sim100\,\mathrm{km}$, so if you include that, your volume will be 5% larger. This number can be much lower if you define the height of the atmosphere to be the scale height (0.3%), or much higher if you include the thermosphere (1600%).

Because the height of the atmosphere is quite arbitrary, this is rarely if ever included in the average density of a (rocky) planet.


What are the advantages and disadvantages of a 2-inch eyepiece versus a 1.25-inch eyepiece?

When you look into the eyepiece, you've noticed that the image is round, framed by a black ring. That ring is called the field stop, and it's an actual round piece of metal or plastic inside the eyepiece. Its function is to limit the image to the size where the image quality in that eyepiece is acceptable.

If you were to remove the field stop, the image outside of it would start to look bad, and its edges would be fuzzy and barely usable. Also, most people prefer a neat, sharp edge to the field of view, so there's the esthetic element to consider also.

The size of the field stop is given by two parameters:

the focal length of the eyepiece, measured in mm
the apparent field of view of the eyepiece, or how wide the image appears when you look in the eyepiece, measured in degrees of angle

Longer focal length eyepieces will naturally have bigger field stops, keeping everything else in proportion. Eyepieces with a wider field of view will also require a bigger field stop.

IMPORTANT: the field stop, obviously, must be smaller than the diameter of the barrel of the eyepiece, otherwise it would not fit in. This is what dictates the lower limit for the diameter of the barrel. The barrel must be bigger than the field stop. You need to increase the barrel diameter when the field stop grows too big.

So, when you stick to short focal length eyepieces with a narrow field of view, the required field stop is pretty small, and it can easily fit into a 1.25" barrel.

But with longer focal length eyepieces, especially with modern eyepieces with a wide field of view, the field stop can grow pretty large, and eventually becomes bigger than the 1.25" barrel. That's when you need to move up to a 2" barrel. That's what drives the need for larger size barrels.

Long time ago, there used to be eyepieces with 0.965" barrel diameter, because back then opticians didn't know how to make wide field eyepieces. Then the field of view kept increasing, so opticians moved up to 1.25".

Eventually the newer 2" standard was adopted, as the field of view kept growing. At 32mm focal length, a Plossl eyepiece with a 52 degree field of view can fit in a 1.25" barrel, but a more recent 68 degree design requires 2".

Nowadays there's a new standard emerging, with a 3" barrel diameter, imposed by the very wide field eyepiece designs invented recently. At 30mm focal length, an eyepiece with an 82 degree field of view could use a 2" barrel, but if you demand 100 degrees FoV at 30mm, you need to move up to 3".

Of course, larger barrel diameters make heavier and more expensive eyepieces. So you need to mix and match. At smaller focal lengths, the 1.25" barrel works well. For longer focal length eyepieces, as you move up in terms of the size of the field of view, eventually you must switch to 2".

If you only use Plossl eyepieces, you can have all of them in the 1.25" format, because even at the longest focal lengths that make sense in practice, the Plossl field of view is narrow enough to not require a big field stop.

As a practical example, my collection of eyepieces currently is about evenly split between 1.25" and 2", with the 2" barrels being reserved for eyepieces over 18mm focal length. All my eyepieces have an 82 degree field of view.

Of course, with a bigger barrel the eyepiece is larger, there's more glass in it, it's heavier and it can be more expensive. This is why manufacturers try to use the smaller barrels when it makes sense, and switch to bigger barrels only when physics and math force them to take this step.

If you need a certain focal length combined with a certain field of view, and that requires a bigger barrel, then by all means use that size. It's kind of a mix-and-match situation.


For what parameter does the abbreviation MA stand for?

It stands for mean anomaly. The start of section 3 gives the epoch.

The epoch for all the fits in this work is
JD 2450602.093.


Is it possible that Mercury was originally the moon of Venus after a giant impact?

This was originally going to be a comment, but it ran too long, so I'm making it an answer.

Some models argue that the scenario of a satellite of Venus escaping like this is unlikely. Alemi & Stevenson (2006) have explored the possibility of a prior Venusian moon, starting from the assumption that Venus would not have been able to avoid a giant impact. Here's their sequence of events:

A large body collides with Venus in a similar manner to the proposed Earth-Theia collision.
Debris from the impact moves outwards into a disk surrounding Venus,
A moon coalesces from the disk, and begins to slowly recede because of tidal acceleration.
Another large body hits Venus. It reduces Venus's angular momentum, reversing its rotation.
The moon spirals into Venus as it undergoes tidal deceleration, finally colliding with it again.

One of the tricky things about testing this model is that the authors say that there would not necessarily have been drastic composition changes, meaning that it would be hard to analyze the planet's surface and see if there is evidence supporting the double impact hypothesis. So far, there have not been tests.

It is certainly true that Venus could have suffered other impacts - the model does not preclude that. There are a couple problems with Mercury arising from such a collision:

Other impacts could have ended up with the same result as the original moon.
The chances of many more impacts aren't too high.
Solar tides would likely have destabilized the orbit of any moon larger than a few kilometers in diameter (see Sheppard & Trujillo (2009)).
MESSENGER determined that Mercury has a high potassium/thorium ratio on its surface, which would seem to disprove any events involving extremely high temperatures, including any giant impact variant.

Of course, if we accept that Venus could have captured a moon, only the third objection remains - still a strong point against the survival of a satellite, even by itself.


Why is declination positive in the northern hemisphere?

After quite an extensive search of dictionaries and books on astronomical nomenclature I came across this article that has a comprehensive review of celestial coordinate systems. And upon reading page 84 (8/14) I came across the answer to your question.

It would appear that the reason declination is used, as opposed to inclination, is that declination was first interpreted as a "deviation from standard" not an inclination of angular distance from a given point as we know it today.

This "deviation from standard" was first used to describe the motion of the sun  along the ecliptic that "deviated" from the then widely accepted notion of all heavenly bodies following the line of the equator. It goes into more detail in the article but it would appear that when translated from Greek to Latin and then Latin to Arabic, the precise meaning of the word changed and evolved so that now the term declination refers to the angular distance of any celestial body.

Source:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1942PASP...54...77W&db_key=AST


How do we know that the intrinsic brightness of Cepheid variables corresponds to their period?

This relationship was discovered empirically by Henrietta Leavitt by comparing the apparent magnitudes of stars in the Magellanic clouds. Since these clouds are far away (200,000 ly) and relatively small (7,000 ly), the difference in distances to the 47 different Cephid variables she observed could account for only a small difference in brightness.  By plotting the brightness against the period she noticed that there was a relationship, cephids with longer periods were brighter.

As you guessed, modern measurements are made via parallax for close stars. The Hipparcos satellite compiled a vast list of parallaxes, and by measuring a large number of periods, we can calculate the relation.


You can see some of them with a naked eye. Many of the Messier objects, including galaxies and nebulae are observable even without any telescopes. As for others, if Charles Messier could catalog them in 1771 with a 4 inch refractor, you should see most using your 3.3 inch reflector too given good observation conditions, knowing where to point and using a properly collimated and otherwise maintained telescope.

Messier was from the Northern hemisphere and you seem to be closer to the equator judging by information you provide in your profile, so you shouldn't have problems seeing these objects either, and there are equally interesting objects also in the Southern skies that you could observe. Some of the easiest recognizable on the Northern skies are the Andromeda galaxy, or Messier 31, and the Orion Nebula, or Messier 42. You should also be able to discern quite a lot of detail of other astronomical bodies, including e.g. Pleiades, or Messier 45 which is also visible to the naked eye and many astronomers use to determine observational conditions and their visual acuity by.


How can we be 13.2 billion light years from another galaxy?

The distance you are referring to is the "light travel distance". It is the length of time that light has been travelling from the distant galaxy to us.

When the light was emitted, our Galaxy was much closer to that distant galaxy. What has happened in the meantime is that the space between the galaxies has expanded.

Furthermore we can ask where is that galaxy right now (where now is a time measure since the big bang) and it turns out that the galaxy would be much further away than 13.2 billion light years - more like 40 billion light years if you were to somehow measure that distance with a very long ruler.


How much time does the ISS take to repeat its orbit?

Objects in orbit pass over the surface above what's commonly called their ground track. For objects in low earth orbit, though they go around the earth every 1.5 or more hours, the Earth rotates beneath them so they don't trace a simple "circle" over the same points on land.

The orbit is around the Earth in a fixed plane, so we are likely (but not guaranteed) to see the satellite pass over 12 or 24 hours later.


Does the Sun's atmosphere have a scale height?

The exponential decrease in density comes out naturally whenever you have a gas in hydrostatic equilibrium. The scale height $H$ is then given by the balance between the kinetic energy of the particles due to thermal motion, $kT$, and the gravitational energy of the particles, $mg$. This is often a good approximation, both in planetary and stellar atmospheres, and even in galaxies.
That is,
$$
H = \frac{kT}{mg}
$$
where $k$ is Boltszmann's constant, $T$ is the temperature, $m$ is the average mass of the particles, and
$$
g = \frac{GM}{r^2}
$$
with $G$ the gravitational constant, and $M$ the mass inside the radius $r$.

On the surface of our Sun, $g$ works out to $274\,\mathrm{m}\,\mathrm{s}^{-2}$, 27 times as high as on Earth.

The average particle mass depends weakly on metallicity, and mostly on the ionization state of the gas, since the small mass of free electrons compared to that of atoms pulls the average down. For a fully ionized gas, the mean molecular mass — i.e. the mass in terms of the hydrogen mass — is $\mu \simeq 0.6$, whereas for a fully neutral gas it is (e.g. Carroll & Ostlie 1996)
$$
\mu \simeq \frac{1}{X + Y/4 + Z/15.5} \simeq 1.25,
$$
where $X$, $Y$, and $Z$ are the mass fractions of hydrogen, helium, and metals, respectively. I mistakenly wrote initially that the gas is fully ionized, but that's not true; it is only partially ionized, and the hydrogen is largely neutral.

Taking the average mass of a particle to be roughly equal to the proton mass $m_p$ (i.e. setting $\mu=1$), and taking the temperature to be $T = 5770\,\mathrm{K}$, the scale height is thus
$$
H_\odot = \frac{kTR_\odot}{GM_\odot m_p} \simeq 170\,\mathrm{km}.
$$
With $\mu=0.6$ you'd get $H\simeq290\,\mathrm{km}$, while $\mu=1.25$ yields $H = 140\,\mathrm{km}$.

Realistic density profiles

The above calculation are quite basic, assuming a completely isotropic Sun. But observations and more realistic models, both 1D and 3D, do indeed predict exponential density profiles, although with quite large variations across the surface (according to a Solar physicist colleague down the hall). I found this model from these lecture notes where the yellow curve shows the number density profile in the Sun's atmosphere.

Extracting the data and plotting on a log-linear scale shows reasonable agreement with an exponential decrease of scale height $H = 140\,\mathrm{km}\,\mathrm{s}^{-1}$:


Could a magnetars magnetic field have saved Mars atmosphere?

How would you get a magnetar close to Mars, within the solar system?

OK, let's leave that to one side, but the Earth's magnetic field is of  order $10^{-4}$ Tesla at strongest. As the dipole field of a magnetar diminishes as $r^{-3}$ and could be as high as $10^{11}$ Teslas at the surface of the magnetar, at a radius of 10 km, then the magnetar woud need to be at a distance of $10^{6}$ km to surround Mars with a field of $10^{-4}$ Tesla.

The problem is that the radiation coming from the magnetar in the form of charged particles accelerated from the magnetic poles would completely dwarf the charged particle radiation from the Sun at Mars.

So no, a nearby Magnetar would not save Mars' atmosphere - quite the reverse.

EDIT: A BOTE calculations: If we look at the Crab pulsar - not quite a magnetar, but close - it is calculated that 10% of the rotational KE being lost ends up in synchrotron raiation from charged particles. That's about $10^{30}$ Watts. So there is at least $\sim 10^4$ solar luminosities, just in the kinetic energy of relativistic particles.

Magnetars are also very strong and variable sources of hard X-ray and Gamma ray radiation. Typical magnetar bursts/flares can release $10^{7}-10^{9}$ solar luminosities of hard X-rays and Gamma rays over periods of 1-40 s. That should fry any atmosphere (or anything else) that was only $10^{6}$ km away!


pc is the abbreviation for parsec, a common unit of distance objects outside the Solar system. In order to compress the scales, they have also taken the logarithm (to base 10) of the distances in parsecs. So the scale in the figure on the x-axis in the radial (r) direction from the central black hole spans from 1e-5 to 100 parsecs. The z-axis, which is perpendicular or normal to the plane of the galaxy and the accretion disk, spans from 1e-5 to 1000 parsecs.


Has someone ever captured footage of the collision of a meteoroid and a meteoroid?

Deep Impact, the NASA comet probe, not the movie, may be as close as you can get to what you want now.
Here's a video of the action when 370 kg of copper hits a comet at 10.2 km/s.
There may be better video out there, but the movie trailers really mess up the search.


If you don't mind an oversimplified answer.   There is a relatively straight forward relationship.   The life of a comet (in orbits not years) depends on 3 factors.   It's size, it's composition and how close it passes to the sun.

Comets are often referred to as dirty snowballs but the ratio of different types of ice and how much "dirt" or silicate material varies from comet to comet.   Average density is about 0.6 g/cc.  That variation is a factor in how much light gets reflected right back off the comet, and how easily the surface melts, but to make this answer somewhat easier, I'm going to ignore composition and make a simplified assumption that all comets are made of basically ice.

With that in mind, how much a comet loses per pass around the sun can be simplified into loss of radius.   Think of a block of ice sitting in space, facing the sun.   The rate of sublimation would simply be a product of how close the ice was to the sun.  It's not far off to oversimplify melt rate in terms of loss of radius over a pass near the sun.

Halley's comet, for example has a fairly close pass to the sun, at about 0.587 astronomical units (closer than Venus).  The closer a comet passes to the sun the faster it sublimates.   Halley's is expected to last another 100-125 orits.  Source.

As mentioned above, there's some variation in terms of the density and different materials that make up a comet, but the primary factors are Perihelion and size of the comet.

Halley's has a diameter of about 8 KM. (roughly speaking, it's 16x8x8 km, ellipsoid shape).   8,000 meters over 100-125 passes suggests it loses about 60-80 meters per pass around the sun and if we estimate that the "dead comet" after 100 or so orbits retains 1-3 KM of rocky material, the ballmark melt-rate of Halley's is somewhere in the 30-60 meters per pass around the sun, or a meter every couple-few days by sublimation.   A comet with a perihelion at .8 AU or 1 AU would lose less radius per pass.

A comet that passes within 0.3 AU instead of 0.6 might lose about 4 times as much per day as it's hit by 4 times the solar energy per square meter, but it spends fewer days due to a faster orbit near the sun.   Somebody might be able to run the math based on highly elliptical orbits, perihelion and solar energy per square meter over the entire pass.   I'm afraid to try myself, but it would be something less than the square of the distance due to the faster orbital speed for smaller perihelion.

Wit this estimate in mind, Wikipedia says a sungrazing comet with a perihelion of about 0.055 AU and an initial radius of about 2-3 km could survive that pass and be about 1 km.   That would be a loss of 1,000-2,000 KM over the pass, but that's a very close pass.

I think that's the start of an answer to this question.  The closer to the sun, the more meters of radius the comet will lose per orbit and calculating the radius losing so many meters every pass gives a rough estimate of the life of the comet.   Sublimation can begin as far away as the orbit of Saturn, but at that distance the sublimation is very slow.


Yes it is. See this 2012 blog post by Paul Gilster:

https://www.centauri-dreams.org/2012/05/17/powering-up-the-antimatter-engine/

It says things like this:

  Antimatter in space is an idea that James Bickford (Draper Laboratory) analyzed in a Phase II study for NASA’s Institute for Advanced Concepts, for he had realized that high-energy galactic cosmic rays interacting with the interstellar medium (and also with the upper atmospheres of planets in the Solar System) produce antimatter. In fact, Bickford’s calculations showed that about a kilogram of antiprotons enter the Solar System every second, though little of this reaches the Earth. To harvest some of this incoming antimatter, you need a planet with a strong magnetic field, so Jupiter is a natural bet for Baxter’s scientists, who go there to forage.

But it also says Saturn's a better bet, and that Earth is handier, along with this:

  The problem with this — and this has been noted by The Physics arXiv Blog and Jennifer Ouellette in recent days — is that PAMELA could come up with only 28 antiprotons over the course of 850 days of data acquisition. There is no question that Bickford is right in seeing how antimatter can be produced locally. In fact, the paper on the PAMELA work says this: “The ﬂux exceeds the galactic CR antiproton ﬂux by three orders of magnitude at the current solar minimum, thereby constituting the most abundant antiproton source near the Earth.” But does the process produce enough antimatter to make local harvesting a serious possibility?

If you want some antimatter, buy a banana. I kid yet not. See the Symmetry article antimatter from bananas. Flip Tanedo says this: “While researching natural sources of antimatter, I discovered a curious article about a naturally occurring potassium isotope that, some fraction of the time, decays via positron emission. The conclusion was that: The average banana (rich in potassium) produces a positron roughly once every 75 minutes”. Better still, buy a whole bunch of bananas!


The orbits of the planets are coplanar (in the same plane) because supposedly during the Solar System's formation, the planets formed out of a disk of dust (ha, ha)  which surrounded the Sun. Because it was a disk, all in one plane, all of the planets formed in that one lousy plane as well.

Single rings and disks are common in astronomy. Jupiter's moons are coplanar too.

The common explanation is that orbits are unstable and they all eventually tend to get into the same plane and stay there. Wouldn't it be fun to have a planet orbiting the Sun at a right angle to the other planet's orbits? You'd never know when the damn thing's going to hit what.


Units of integrated flux density (irradiance) along longitudes

You could say "MJ in bins of x degrees in l" where x is the degrees spanned by a pixel.  Or, you could divide the fluxes by the steradians in each column and keep it MJ/sr or "Mean MJ/sr within longitudinal bins". Now the width of the bins is not important as long as they are small.


When will Callisto be in orbital resonance with the rest of Jupiter's big moons?

we tried to answer to this question in this article:
https://arxiv.org/abs/2001.01106
According to the current estimation of the tidal dissipation in the Jovian system, we expect that Callisto will be captured into resonance in about 1.5 billions of years, forming 1:2:4:8 resonant chain with the other Galilean moons.


How can there be 1,000 stellar ancestors before our Sun?

The Sun is actually a THIRD generation star. What I mean by this is that there are chemical elements in the Sun that were made inside another star, but that star itself can only have made those elements because it had material in it that must also have been made inside previous, second generation stars. Eventually we get back to the first generation stars, born out of primordial gas from the big bang that contained almost no heavy elements (those beyond helium) at all.

That is quite a mouthful, so let me explain using an example - barium.

There is barium in the Sun. We can tell that by looking at the spectrum and seeing absorption lines due to barium. But barium cannot be made in the Sun. The barium is made via the s-process, which involves the slow capture of neutrons onto the nuclei of iron-peak elements. This happens during the asymptotic red giant branch phase of stellar evolution, and the Sun has 6 billion years or so before it gets to that point. [NB: not even half the abundance of chemical elements beyond iron are produced by supernovae explosions$^1$.]

So, before the Sun, there must have existed a star - probably an intermediate mass star - which evolved to become a giant, made barium in its interior, then lost its envelope through a massive wind into the interstellar medium, and that material was incorporated into the protosun. Such stars (between, say, 2 and 10 solar masses) would have much shorter lifetimes than the Sun$^2$, so plenty of time for them to live and die before the Sun was born.

But wait a minute! That previous star must have already had iron-peak elements in its interior to act as a "seed" for the s-process production of barium. These were not and could not be made inside that star. They must have been made in a previous star, probably a massive star, that burned through all the nuclear fusion stages before exploding as a supernova, casting heavy elements, including iron-peak elements, into the interstellar medium. This previous star could also have had its own (metal-rich) ancestors, but ultimately as we go back in time we reach a point where the previous star was a first generation star, made from primordial H/He gas, with almost no heavy elements. These first generation (a.k.a. population III stars, just to be confusing) were probably very massive and short lived - a few million years. They would be born when the universe was a few hundred million years old and we can see no examples of them in our Galaxy today.

To try and define more precisely what I mean by "generation".

First generation - made from primordial big bang material.
Second generation - a star made only from the detritus of dying first    generation stars, enriched in heavy elements but lacking in primary s-process elements.
Third generation - a star made from material already enriched in heavy elements and including elements that are produced in the s-process inside previous second (or third) generation stars.

So that is why I claim the Sun can be classed as a "third generation star" - it contains atoms/nuclei that must have been inside at least two previous stars.

But you should not take this too literally. There are grains of material trapped inside meteorites that consist of solids that were already present in the pre-solar material. These are important because these grains were thought to have formed in individual stellar events and their isotopic compositions can be studied. These tell us that the Sun formed from material that has been inside many different stars of different types.

Stellar evolution and nucleosynthesis calculations tell us the same story. For example, whilst most of our oxygen was made in massive stars that underwent a core collapse supernova, such events do not produce that much carbon. The C/O ratio tells us that most of our carbon comes via the winds from intermediate mass AGB stars. Heavy elements like uranium may be dominantly produced in neutron star collisions, but others like barium and strontium are not.

The details of how many ancestors have contributed to the Sun has no simple answer. Much of the solar hydrogen and helium could be pristine; some will have been through more than one star. Heavier elements (bar some lithium) will have been through at least one star. The fact that we have s-process elements like Ba, Sr, La and Ce, which are formed by neutron capture onto iron-peak elements, tells us those have been through at least two stars.

However, these are vast underestimates. Mixing in the interstellar medium is reasonably effective. The material spewed out from supernovae and stellar winds 5-12 billion years ago has had plenty of time to mix throughout the Galaxy before the Sun's birth. Turbulence and shear instabilities, driven by the winds and supernovae from massive stars, should distribute material on galactic-length scales in a billion years or less (Roy & Kunth 1995; de Avillez & Mac Low 2003), though local inhomogeneities associated with nearby recent events can persist over $10^{8}$ years. If this is the case, then the Sun is the product of the $\sim$ billion stars that died before it was born.

The reason you are confused with your lifetime argument is that you have ignored the possibility of the Sun being made from stars that lived at the same time in different parts of the Galaxy. The material that they ejected near the end of their lives has just been thoroughly mixed up.

$^1$ The rest are produced by the s-process in intermediate mass AGB stars; through nova events on white dwarfs; or perhaps, in the case of the heavier elements, through the collision of neutron stars (see this Physics SE question).

$^2$ A rough expression for the lifetime of a star is $10 (M/M_{\odot})^{-5/2}$ billion years.


You're talking about the temperature gradients in the atmosphere of a star, that's where the emission and absorption lines are originating. This is a much more complicated matter than the star itself.

The temperature of a star must increase from its surface downwards.
This is a simple result of the fact that the stellar structure is given by pressure gradients balancing the force of gravity. We can express this via the law of hydrostatic equilibrium
$$\frac{\partial P}{\partial r} = -g(r)\rho  $$
which is a pretty good approximation to reality in most cases.

The pressure gradient then translates into a temperature gradient, as the pressure originates in the thermal motion of particles, formalized as the law of ideal gases
$$ P = \frac{\rho k_B T}{\mu} $$.

So wherever there's non-zero gravity and no other force in a gaseous medium, there will be temperature gradients.

In an atmosphere, things depend on how efficient the gas is at cooling. In very thin media, like in Earths or any planets / stars exosphere cooling is inefficient enough to make temperature gradients vanish.


We detect black holes mostly by their effects on nearby matter. The "radial velocity" method is important after x-ray observations suggest a likely black-hole.

Matter in orbit around a black hole can form an accretion disc. This disc will become very hot, and glow brightly in X-ray radiation. If we see X-ray radiation coming from a star we look at the radial velocity of the star, and use this to infer the mass of the orbiting object. Since a neutron star cannot exist at more than 3 solar masses, if the object is more than that it must be a black hole.

Compared to planets, black holes are rare. Only about 20 candidates are known in the Milky way.


This doesn't really answer your question, but may help you get started. If you run https://github.com/barrycarter/bcapps/blob/master/STACK/bc-stell-travel.pl and follow the instructions at https://github.com/barrycarter/bcapps/blob/master/STACK/README you can create a video similar to https://www.youtube.com/watch?v=_4ej5rfYm4A but higher quality (youtube lowers the quality of uploaded videos).

With the default settings, it's a ~6 hour Mach 1 flight from Menton, France to Rampart House, YT Canada starting at 2017-02-10 at 00:00:00 GMT at approximately 120x regular speed.

The only vaguely interesting things about the default flight:

Since the flight goes fairly far north, you can see Polaris rising in the Northern sky for most of the flight.
Since you traveling westwards fairly quickly, once you get far north enough, the stars go "backwards": they rise in the west and set in the east.


Okay, I think I know what Max Tegmark is talking about in the video. He is referring to the fact that, when you observe the cosmic microwave background radiation (CMB) — i.e. the radiation that was "released" when the Universe had expanded and cooled sufficiently to allow protons and electron to combine into neutral hydrogen without immediately being ionized again — you find that it has almost (to within $1$ in $\sim10^5$) the same wavelength everywhere, no matter in which direction you look, meaning that the temperature of the Universe was almost the same when it was emitted.

In a universe like ours, but without inflation, this is puzzling. The reason is that, using the observed expansion rate and densities of the universe, you can calculate that at early times such a universe would simply expand so quickly that it would be impossible to reach thermodynamical equilibrium on such large scales as the Universe we observe today. In fact, when we look at the sky, any two points separated by more than roughly 1 degree would never have been in causal contact, i.e. been able to exchange information. Thus, there would be no way for one of these points to know e.g. what is the temperature of the other points, not to mention the rest of the universe.

This puzzle is called the horizon problem.

Map of the CMB sky. The color coding shows the temperature of the radiation, and red is roughly $18\,\mu\mathrm{K}$ hotter than blue.

Credit: ESA/Planck Collaboration (plus my own annotations).

What Tegmark points at is that although we're told that "Big Bang happened everywhere", this doesn't make sense, because how can one part of this "everywhere" know about the other parts of "everywhere"?

One solution, obviously, is pure luck. After all, the regions on the sky are a little bit different. But this is a very unsatisfying solution, known as fine-tuning, which we loathe.

Enter inflation.

If you assume the our Universe was much, much denser in the beginning, but then went through a period of extreme (!) expansion, then in the very first tiny fraction of a fraction of a second, before this epoch which started when the Universe was $\sim10^{-36}$ seconds old, everything was so close to each other that a region much, much larger than the part of the Universe that we can see (the "observable Universe") could exchange information and reach thermodynamical equilibrium.

The reason that many people believe in this scenario, I think, is that it simultaneously solved two other puzzles, known as the flatness problem and the magnetic monopole problem (and that, so far, no better scenario has been proposed).

So, to recap, Tegmark is not saying that Big Bang didn't happen everywhere, but rather that, before Alan Guth proposed the inflation scenario in 1979, but after we first observed the CMB, this didn't make sense.


How are East and West defined on other bodies of our solar system?

Actually, it is a bit contentious. The IAU defined the North Pole of a planet to be that axis of rotation which lies north of the solar system's invariable plane. Which means that it is possible for bodies to rotate retrograde, and Venus and Uranus are considered to rotate retrograde with this definition. This definition is not universally accepted.

From early periods when solar maps were drawn, they have been labelled such that features appear on the "Eastern" limb of the sun and move to the "Western" limb, though this means that they are moving from Earth's East to Earth's West. This convention hasn't changed, though things have changed for the moon, just to confuse things.


LIGO: How can laser interferometry (wavelength >$10^{-7}$m) detect length changes of arms <$10^{-18}$ m?

The LIGO interferometer uses a homodyne detection technique. Basically, the light travelling in each arm of the interferometer is derived from the same laser source and is combined in the output channel and falls onto a photodiode.
The interferometer is operated so that when there is no gravitational wave (GW) passing through the instrument, the beams combine to produce a dark fringe (i.e. they are set to destructively interfere). There is a small offset from this, but basically the phase difference between the combiniing beams is close to $\pi$.
The phase difference caused by a GW, due to the changing length of one arm with respect to the other, can be derived as
$$
 \Delta \phi \simeq 2\pi \left(\frac{2hL}{c}\right) \left(\frac{c}{\lambda}\right) = \frac{4\pi}{\lambda} hL \ ,
$$
where $L$ is the length of the arms, $\lambda$ is the laser wavelength and $h$ is the strain amplitude of the fravitational wave signal. Actually, it is a little bit more complex than this, since the arms act as Fabry-Perot resonators which means the light effectively travels backwards and forwards many times in the arms (about 300 for LIGO, i.e. $L$ is effectively 1200 km).
For a typical dimensionless GW strain of $h \sim 10^{-21}$, $\lambda = 1064$ nm, then $\Delta \phi \sim 10^{-8}$ and is modulated at the frequency of the GW (typically 20-2000 Hz).
The problem then reduces to combining
$$ E_{\rm tot} = E_0\sin (\omega_l t) + E_0 \sin (\omega_l t + \alpha + \Delta \phi)\ ,$$
where $E$ is the electric field in each arm, $\omega_l$ is the angular frequency of the laser and $\alpha$ is the offset phase between the arms (close to $\pi$).
Using the identity $\sin a + \sin b = 2 \cos[(a-b)/2] \sin[(a+b)/2$ and squaring the total E-field to get an intensity:
$$I = 4E^2 \cos^2[(\alpha + \Delta \phi)/2]\, \sin^2[\omega_l t +(\alpha + \Delta \phi)/2] $$
Since $\omega_l$ is much greater than the GW frequency and much higher than can be sampled by any photo-sensitive detector, then the second term in the product above can be replaced by its time-average of $1/2$. If we now identify the total power $P_{\rm in}=E^2$  as the average input power to each arm of the interferometer and note that $\cos^2 (a/2) = (\cos(a)+1)/2$ and $\Delta \phi \lll 1$
$$ I = P_{\rm in} \left[1 + \cos(\alpha + \Delta \phi ) \right] \simeq P_{\rm in} \left[1 + \cos(\alpha) -\Delta \phi \sin(\alpha)\right] = 2P_{\rm in} \left[ \cos^2 (\alpha/2) - \frac{\Delta \phi}{2}\sin \alpha \right]\ .$$
It is the second term inside the bracket that contains the signal of the GW. That signal is proportional to the power in the interferometer and the phase difference between the arms. Note that although the signal-to-(shot) noise is mathematically maximised when $\alpha=\pi$, this would mean the SNR was 0/0 ! In practice there is always some other noise present so $\alpha$ is shifted a little bit away from $\pi$ - Fricke et al. (2012) suggests that $\alpha \sim \pi+ 6\times 10^{-5}$ is used.
The power input into each arm is about 600 W (the 100s of kW Steve Linton mentions in a comment is after accounting for the Fabry-Perot resonator, which I did above by talking about an "effective $L$").
In the absence of other forms of noise then photon counting (shot noise) becomes the limiting factor and is proportional to the square root of the power.
The output signal is the modulated GW signal discussed above which is recorded by detecting photons with the photodiodes. The response function that translates the photodiode signal into a strain is determined by acting on the test masses/mirrors with precisely calibrated lasers modulated at GW frequencies that can produce monochromatic phase shifts in the arm lengths.


How much are Neptune's and Uranus' orbits perturbed by Pluto and other KBOs?

Since (1) the mass of the Pluto/Charon system is very small compared with Neptune, and (2) their orbits are in 2:3 resonance, so the perturbation of Neptune's orbit would have been undetectable. University of Rochester has a good explanation of this.

  The Accidental Discovery of Pluto

  Later supposed perturbations of the orbits of Uranus and Neptune suggested
  the presence of yet another planet beyond the orbit of Neptune.
  Eventually, in 1930, a new planet Pluto was discovered, but we now
  know that the calculations in this case were also in error because of
  an incorrect assumption about the mass of the new planet. It is now
  believed that the supposed deviations in the orbits of Neptune and
  Uranus were errors in measurement because the actual properties of
  Pluto would not have accounted for the supposed perturbations. Thus,
  the discovery of Pluto was a kind of accident