How hot must a star get before it is considered to be a star?

From a physics perspective

From a physics perspective an object is a star when it is undergoing nuclear fusion, generally of hydrogen atoms at its core, this is regardless of its temperature!

A star is not determined by its temperature, it is instead determined by it's internal processes.

This does mean that if Jupiter began nuclear fusion it would be considered a star, albeit a minuscule one.

In this case it is a yes/no distinction of if an object is a star.

From an observational point of view once something is classified as a star there are 7 groups it can fall in to determined by its features.

Sourced From: http://en.wikipedia.org/wiki/Star#Classification

  Class Temperature  O:    33,000 K+
  B:    10,500–30,000 K   A:   7,500–10,000 K
  F:    6,000–7,200 K     G:   5,500–6,000 K
  K:    4,000–5,250 K     M:   2,600–3,850 K

Note: Three more classifications L T and Y have been added to the colder end of this list, but I am unsure of the cut off points so omitted them.

But strangely they are not classified by temperature but by their spectrum, it just so happens that their spectrum correlates to their temperature! The temperature spoken of here is of the photosphere of the star (where the photons begin free streaming), not its core (where photons are created from ongoing fusion reactions).

Dwarf stars have their own classification system prefixed by the letter D though.

Quote from Wiki article:

  White dwarf stars have their own class that begins with the letter D. This is further sub-divided into the classes DA, DB, DC, DO, DZ, and DQ, depending on the types of prominent lines found in the spectrum. This is followed by a numerical value that indicates the temperature index.


Does Polaris have a trail in the sky timelapse? Yes. Every star has a trail, but it's smaller if it's closer to the pole. Polaris is close to the pole but it's still 45 (arc)minutes away. I don't think it's visible with the naked eye; no star trail is visible with the naked eye as you only see one 'instance' of the sky. What perhaps would work, and is more feasible for stars closer to the celestial equator, is trying to see the movement of a star compared to an object on Earth, e.g. a tree, a building or the horizon.

The exposure times of the photos are different, as @J.Chomel notes in the comments. You can calculate the exposure time by looking at the arc length; in 23 hours and 56 minutes (a 'sidereal day'), a star will make a full circle.

The trails on the last photo have an arc length of about 8 degrees, corresponding to about 30 minutes; the first two photos are more like 9 resp. 5 hours. Also, the zoom factor (of the lens) may play a role; if a photo is zoomed in more, Polaris' movement is more likely to be visible. I'm not sure if this plays a role in this photos, though.


What is $\omega$ and $\tau$ in this celestial sphere

There seems to be very little to go on here.

"Zenith" and "Horizon" are easy to understand, it suggests that this is a projection of the sky (and ground) In that context "W" would appear to be "West", and you give that "S" is "sun". The horizon is curved. If you project the entire hemisphere the horizon would go straight across this isn't a complete hemispheric projection.

If we are looking West, then North would be to the right, and South to the left. The point P seems to be identified with NCP. The first guess would be that N of NCP means North, but this contradicts the reasoning that North is on the right. You mention SCP in the text, but it isn't in the diagram. The sun appears to be West, or just North of West, which is not an unusual but not impossible position.

The lines SW, WP and PS are curved. This may be to indicate that they are great circles on the celestial sphere. If so then $\omega$ is the spherical angle SWP and $\tau$ is the great circle distance SW, perhaps given as an angle with vertex at the centre of the sphere. Given three points on a sphere (in alt-az coordinates), it is a small exercise in spherical geometry to find the spherical distance and angles between them.

I'm not sure why you are being coy about the source of this. If you could link to the question from which this is taken the meanings of $\delta$ and H could be clearer. I guess the solution is an exercise in spherical angles, and projections from a circle to the plane


Measures for the degeneracy from protons?

Stars are electrically neutral. In the optimal case for proton degeneracy, a star is made of hydrogen, maximising the proton number density. The number density of electrons is exactly the same.

Since the Fermi momentum $p_F$ of a species is dependent only on number density$^1$, the Fermi momentum of the electrons and protons are the same. Pressure however$^1$, is an increasing function of $p_F/mc$ and since the mass of an electron is 1800 times smaller, the electron degeneracy pressure is always much higher than the proton degeneracy pressure.

Degenerate protons do exist. They form of order 1% of the density in the neutron fluid region (the bulk) of a neutron star. However, they contribute much less than a thousandth of the pressure and always lower than the degenerate electron pressure.

Actually, the form of the pressure equation (see below) does actually admit the possibility that the degeneracy pressures of electrons and protons become more similar as $p_F/mc \gg 1$ (aka ultra-relativistic). But by that time, the energies of the electrons and protons are easily sufficient to create neutrons and an equilibrium is set up so that neutrons always outnumber the protons by at at least 8:1 and therefore their degeneracy pressure will dominate.

(1) For those interested: The Fermi momentum is
$$p_{F} = \left(\frac{3}{8\pi}\right)^{1/3} h n^{1/3}\ ,$$
where $n$ is the number density of fermions.

Whilst ideal fermion degeneracy pressure is given by
$$P = \frac{\pi m^4 c^5}{3 h^3} \left[x (2x^2 -3)(1 + x^2)^{1/2} + 3
  \sinh^{-1}(x) \right]\ , $$
where $x = p_F/mc$.


Polaris consists of multiple stars,

α UMi Aa is the main star. It is a supergiant, and a cepheid variable. α UMi Ab and α UMi B are smaller (but still larger than the sun) and both are in orbit with α UMi Aa, (the former is close in, the latter is further out).

There are two further stars: α UMi C and D. These stars are thought to have formed before α UMi Aa, and so are not considered to be physically associated with α UMi A. There are other stars that appear nearby, one might be physically associated: J021006.2+891626 J021454.3+890852 are their catalogue numbuers. They are faint, but appear to be the same age as Polaris.

In conclusion, Polaris is known to have 3 stars: Aa, Ab and B. The stars C and D are not physically related to Polaris. There may be one or two more stars that are distant dwarfs and may share Polaris's orbit.

Reference http://iopscience.iop.org/article/10.1088/0004-6256/139/5/1968


Why aren't ground-based observatories using adaptive optics for visible wavelengths?

There's a pretty good discussion at this page.

There are several factors at work:

The smaller isoplanatic angle, as you note. This limits how much of the sky you can observe with AO, since your target needs to be within the isoplanatic angle of a bright enough references star. (Even with laser guide stars, there is still a need for a reference star for "tip/tilt" correction.) The difference in angular area on the sky means that the area of the sky that can theoretically be observed with AO will be about 20 times larger in the near-IR than in the optical, just from the difference in isoplanatic angle.
The effects of turbulence are stronger and have shorter timescales in the optical. This has three effects:

A. The corrective optics (e.g., deformable mirror) need to have more movable parts ("a near-perfect correction for an observation done in visible light (0.6 microns) with an 8-m telescope would require ~ 6400 actuators, whereas a similar performance at 2 microns needs only 250 actuators.") and need to operate on a faster timescale.

B. In addition to the electromechanical complexity, you'll have to do far more in the way of calculations to drive all those actuators, and on a shorter timescale. So the computing power required goes up.

C. In order to provide the inputs for the corrective computations, you need to observe the reference star on a much finer angular scale ("A large number of actuators requires a similarly large number of subapertures in the wavefront sensor, which means that for correction in the visible, the reference star should be ~ 25 times brighter than to correct in the infrared."). This limits how much of the sky you can do AO for even more: a star that might be bright enough in the near-IR to correct a region 20-30 arcsec wide isoplanatic patch won't be bright enough to correct the corresponding 5-arcsec-wide isoplanatic patch in the visible.
In order to make corrections, you need to observe the reference object in the optical. This is easy to do with a near-IR setup using an optical/IR beamsplitter: send the optical light to the AO equipment and send the near-IR light to the near-IR instrument. In the optical, you use an optical beamsplitter to send half the light to the instrument and the the other half to the AO equipment. This means that the AO equipment only gets half the light it would if it were used with a near-IR instrument, which makes it (even) harder to do the corrections.

Finally, there is an issue unrelated to the AO itself, which is that you need different science instruments depending on whether you're working in the optical or the near-IR. Optical instruments use silicon CCDs for detection; these are only sensitive out to about 0.9-1 microns. Near-IR instruments use different detectors (usually HgCdTe-based), which are good from about 1-3 microns. (Near-IR instrument also need a different design to reduce contamination from thermal emission from the telescope and optics for observations at wavelengths longer than 2 microns.) So in practice the choice has been: combine AO with a near-IR instrument and get good performance with affordable/feasible technology, or combine AO with an optical instrument and get very limited performance with more expensive (or even, until recently, unattainable) technology.

Nonetheless, there are some optical AO systems starting to appear, such as MagAO on the Magellan telescope (which has both an optical instrument and a near-IR instrument, and can correct for both simultaneously).


Can gravitational waves pass through a black hole?

No, gravitational waves cannot pass through a black hole.

A gravitational wave follows a path through spacetime called a null geodesic. This is the same path that would be followed by a light ray travelling in the same direction, and gravitational waves are affected by black holes in the same way that light rays are. So for example gravitational waves can be refracted by gravitational lenses just as light waves are. And just like light waves, if a gravitational wave crosses the event horizon surrounding a black hole it is then doomed to travel inwards to the singularity and can never escape.

There is one caveat to this. When we talk about a gravitational wave we generally mean a ripple in spacetime that is relatively small. Specifically it is small enough that the energy of the gravitational wave does not significantly affect the spacetime curvature. So when we calculate the trajectory of a gravitational wave near a black hole we take the black hole geometry as fixed, i.e. unaffected by the wave, and we compute the trajectory of the wave in this fixed background.

This is exactly the same approach as we use for calculating the trajectories of light rays. Since light rays carry energy and momentum then, at least in principle, they have their own gravitational fields. But for both the light rays and gravitational waves likely to exist in the universe the energy carried is too small to make a significant contribution to the spacetime curvature.

When you say in your question:

  I would presume that something interesting happens because of the way spacetime works near black holes

I would guess you are thinking that the gravitational wave could change the geometry near a black hole, but as described above typical gravitational waves don't have enough energy to do this. It would be reasonable to ask what happens if we give the wave enough energy, but the answer turns out to be that it no longer behaves like a simple wave.

Gravitational waves exist in a regime called linearised gravity where they obey a wave equation that is basically similar to the wave equation light obeys. If we increase the energy so much that gravity becomes non-linear (as if the case for black holes) then the oscillations in the spacetime curvature no longer obey a wave equation and need to be described by the full Einstein equations. For example it has been suggested, but not proven, that really high energy gravitational (or light) waves could interact with each other to form a bound state called a geon. I confess that I'm unsure how much work has been done studying oscillations in this regime.


Is there available space images taken by the south pole Gregorian telescope?

The South Pole Telescope (SPT) was funded, in part, by the NSF. This means their data must be made public as that is almost always a condition of funding by the NSF.

If you go to the SPT's homepage, they link to their Data Management Policy. Within this policy they state:

  The collaboration – in particular the P.I. and data managers at the University of Chicago – bear the responsibility for making data public. Public data products are hosted on University of Chicago machines and on the NASA Legacy Archive for Microwave Background Data Analysis (LAMBDA) server.

I went over to NASA's LAMBDA server and found that SPT data was hosted right here. Look to the left-hand side menu to see the links to raw data for the SPT-SZ and SPTpol instruments, individually.

The University of Chicago hosts their data right here (props to called2voyage). It seems that they provide publications and include any data used in that publication rather than just hosting all the RAW data like the LAMBDA server does. Use whichever is most convenient for you.


If oxygen was abundant in Neptune, would there be combustion?

"At high altitudes, Neptune's atmosphere is 80% hydrogen and 19% helium" (Wikipedia). No significant abundancy of free oxygen to react with.

A source of oxygen could easily made burn on Neptune, like a source of hydrogen on Earth.
Or take a sample of Neptun's atmosphere. It would easily burn in Earth's atmosphere.

Hydrogen oxygen combustion is sufficiently exothermic (483.6 kJ/mol of O2) to sustain burning even at the low temperatures on Neptune, provided the atmospheric pressure is high enough: With a gas constant of about 8.3 J / mol K, the gas can easily be heated up to the point of combustion by the energy released by the combustion.

Assuming oxygen would be provided, just sufficiently to burn up hydrogen and trace hydrocarbons:
After the explosion and condensation Neptune would get a huge ocean of water, later freezing to water ice, with an atmosphere of helium, and some less abundant gases like CO2, nitrogen, and argon, traces of water vapor. Traces of hydrogen and methane could survive the presence of oxygen. Hence some oxygen would remain in the atmosphere. Most of the CO2 would be resolved in the water ice.

The average density of Neptune would increase, it would shrink a bit, the core would be compressed and heated by adiabatic compression.

Before combustion Neptune's gravity is strong enough to hold an atmosphere of hydrogen, the easiest gas to escape. Hence the denser planet after combustion wouldn't improve its capability to keep an atmosphere significantly: The escape velocity of Neptune is 23.5 km/s, its surface temperature is below 100K. With Boltzmann constant $k=1.38\cdot 10^{-23}J/K$ and $E=3kT/2$ we get $$E=3\cdot 1.38\cdot 10^{-23}J/K\mbox{ or }4.14\cdot 10^{-21}J\mbox{ at }100K$$ for an average $H_2$ molecule with a mass of about $3.35\cdot 10^{-27}kg$. Hence for Neptun's escape velocity the kinetic energy of a hydrogen molecule is $$E=0.5\ mv^2=0.5\cdot 3.35\cdot 10^{-27}kg \cdot (23.5\cdot 10^3 m/s)^2 = 9.25\cdot 10^{-19} J.$$ Hence the energy needed to escape is more than 200-times the kinetic energy of the hydrogen molecules; no way to escape, 2000K or more would be needed to let escape some hydrogen over time (to get below a factor of about ten between the two energies). Photolyic dissociated atomic hydrogen would still need more than 1000K to escape.

The probability to catch asteroids or comets would be reduced due to the reduced radius, if the total mass is adjusted to the mass before combustion, since asteroids would miss instead of hit the planet.


What is the difference between the Yarkovsky effect and YORP effect?

The Yarkovsky effect is the thrust on a small object in space that has been heated by sunlight, created from radiant energy. The YORP effect is when the thrust produces rotation, likely due to uneven heating or differences in surface projections (IE: A mountain acts as a better solar collector than a flatter portion of a spherical object on the opposite side).

See: "The Yarkovsky and YORP Effects" (Feb 4 2015), by
D. Vokrouhlicky, W. F. Bottke, S. R. Chesley, D. J. Scheeres, and T. S. Statler:

  "The Yarkovsky effect describes a small but significant force that affects the orbital motion of meteoroids and asteroids smaller than 30 − 40 kilometers in diameter. It is caused by sunlight; when these bodies heat
  up in the Sun, they eventually re-radiate the energy away in the thermal waveband, which in turn creates a tiny thrust. This recoil acceleration is much weaker than solar and planetary gravitational forces, but it can
  produce measurable orbital changes over decades and substantial orbital effects over millions to billions of years.

  The same physical phenomenon also creates a thermal torque that, complemented by a torque produced by scattered sunlight, can modify the rotation rates and obliquities of small bodies as well. This rotational variant has been coined the Yarkovsky-O’Keefe-Radzievskii-Paddack (YORP) effect. During the past decade or so, the Yarkovsky and YORP effects have been used to explore and potentially resolve a number of unsolved mysteries in planetary science dealing with small bodies.".


Why can't neutron stars ignite and explode?

In a white dwarf, the dense matter is not in its lowest energy configuration. Energy can still be extracted from the white dwarf material by fusion, provided it can be ignited.

What exothermic nuclear reactions would there be that could take place in a neutron star? The bulk of the material is in the form of neutrons with a small number of protons and electrons. At these densities, that is the most stable equilibrium composition.

If a neutron star gains mass in a gradual way, then the most likely course of events will be that its radius will decrease (that is what happens in objects supported by degenerate matter) until it reaches a General Relativistic instability where its collapse to a black hole is inevitable (when $R$ is somewhere between 1.25 and 2 times the Schwarzschild radius). It is possible that neutrons may transform before that into additional hadronic degrees of freedom or into quark matter, but these are endothermic processes that suck kinetic energy out of the neutron gas and only hasten the collapse.


Need some sources on radiotelescope sensitivity

I'll try a quick estimate.

The strength of a radio source $S$ is often measured in Jansky's and 1 Jy is $1 \times 10^{-26}$ W/m^2/Hz. That means every square meter of collection area is exposed to $1 \times 10^{-26}$ per Hz of bandwidth.

Total power received would then be

$$P_S = S A \Delta \nu$$

where $A$ is the receiving area of the dish and $\Delta \nu$ is the bandwidth in Hz.

The noise limit of an RF receiving system is often expressed as an equivalent temperature. A typical number might be 50 Kelvin, which includes a receiver temperature, sky temperature, spillover (thermal radio emission radiated from the ground) etc. Though it seems counterintuitive, we can add temperatures as a proxy for adding powers because we are in the Rayleigh-Jeans regime where $\nu << k_B T/h$ as explained in this answer

You convert the noise temperature to a power in Watts:

$$P_N = k_B T \Delta \nu$$

where $k_B$ is the Boltzmann constant which is about $1.381 \times 10^{-23}$ Joules/K, and $\Delta \nu$ is again bandwidth in Hz.

Let's set the minimum strength easily detected to be equal to the noise, and equate the two powers:

$$P_S = P_T $$

$$S A \Delta \nu = k_B T \Delta \nu$$

$$S A = k_B T$$

$$S = k_B T / A$$

Note that the bandwidth drops out since both the thermal noise and most radio sources are fairly broadband. Even if the original line is fairly narrow, you'll tend set your minimum bandwidth of a specific measurement to cover the spectral region of interest.

Ballpark Estimate

The Arecibo dish has a radius of about 150 meters, so the area $A$ is about 70,700 square meters. A typical radio astronomy system temperature is 50 Kelvin (it can be significantly higher or lower depending on many factors, NASA deep space 70 meter dish liquid helium cooled front-end is 20K (see here and here), so plugging that into the equation above, we get $ = 9.8 \times 10^{-27}$ watts, which is (probably not coincidentally) almost exactly 1 Jansky!

Confirm that

According to Joeri van Leeuwen's Radio Astronomy lecture 5, The Radio Telescope at the bottom of page 79, the System equivalent Flux Density or SEFD of Arecibo is 3 Jansky, so a basic back of the envelope calculation gets us quite close!

Other radio telescopes will very roughly scale based on area, a smaller area dish will have a higher SEFD or signal level equivalent to the system noise.

Another example, VLA

A VLA dish is 25 meters in diameter, which is about 150 times less area. My back-of-the-envelope calculation would then be 150 Jansky.

According to Table 3.2.1 on the NRAO page Performance of the VLA during the Next Semester; Sensitivity at its lowest, a VLA dish's best SEFD is 250 Jansky. I've plotted the table below. Still checks out.


Finding a location using the direction of shadow, UT and date approximately

Planetarium software, such as Stellarium, or astronomical computation software such as pyephem can compute the exact position of the sun at any date, time and location. It is then simple trigonometry to calculate the length of a shadow:

if the sun is $\theta$ degrees above the horizon, a vertical building of height $h$ will cast a shadow on horizontal ground of length $\frac{h}{\tan(\theta)}$.


Can someone in a deep well see the stars in daytime?

No, because being in a deep well doesn't make the background dark.

The reason we can't see stars in the daytime is the atmosphere, which scatters sunlight in all directions such that along every sightline from the ground toward the sky you receive much more photons from the Sun than from a star.

That is, the photons that makes it impossible to see Vega, for instance, are the photons from the Sun that make their last scattering event at a molecule lying between you and Vega. Being in a well blocks the star- and sunlight from other directions, but lets through both star- and sunlight from Vega's direction.

Like this:


Why can't we see Saturn's phases from earth?

Phases are just different perceived illuminations of an object at different illumination and observing angles. If the observer is, with respect to the object, located in a similar direction as the light source shining on the object then you should expect to see the vast majority of the object illuminated, if the observer is located in the opposite direction you would see the object back-lit, and if you are at right angles observing the object with respect to the direction of the light source you would see the object half-illuminated.
Since Mercury and Venus are always inside Earth's orbit and move with different rates around the Sun as Earth does, the Earth (the observer) is able to locate itself at any angle with respect to the light source that shines on the observed planet. This means that you can see any phase of Mercury or Venus (except from a perfect 100% illuminated phase due to the body of the Sun blocking the view). Here you have an example for Venus:

Now think about what happens from the point of view of an Earth's observer for the exterior planets. The Earth will never have a chance to see the planet's back-lit side since there's no position in its orbit that would allow for this. As seen from the other planet, the Earth is always close to the Sun, so it can be seen almost exclusively during the day, which means that from the Earth you almost exclusively see illuminated regions of the planet.
Mars is the closest exterior planet, so the Earth manages to gain enough elongation to see a bit of the night side, but it is a tiny fraction of the disk as viewed from here. Here you have a picture of the phases of Mars as seen from the Earth:

This is even worse for far away planets. Saturn is so far away that from its vantage point, that the Earth is basically always close to the Sun (to the light source).  From the Earth, Saturn seems always fully illuminated, by an extremely small margin that allows to see a slim crescent of darkness in perfect conditions. Only with spacecraft like Cassini and Voyager have we been able to see what Saturn looks like from behind. The first time humanity did this was in 1980 with photographs like this one (from Voyager):

Before that we had never seen the night side covering more than a percent of the disk. Even the shadow of Saturn cast over the rings is nearly impossible to spot from the Earth. Look at this amateur photograph that shows precisely a bit of that shadow over the rings behind Saturn (bottom-right part):

That shadow is almost non-existent from Earth's point of view, and it's all because Earth's orbit is inside Saturn's and Saturn is far away from the Earth. The Earth is always so close to the line connecting the light source (the Sun) and the illuminated object (Saturn), that you shouldn't expect any more phases than "full" from here. Simple geometry.


How much do starlight and artificial light illuminate the Moon?

To reverse the question: if you were on the Moon, what sources of light would you see?

The Sun would be the major source of light.
Although the Moon's albedo is only about 12%, reflected light from the ground would be the second major source.
The Earth would be in a poor 3rd place. You could see the general shape of the continents, but not storms or any other light-emitting meteorological or biological activity (it's there, but much too faint to be seen from the Moon)
The other planets would be poor 4th place.
Starlight would be in a poor 5th place.

It would be similar to standing on Earth, except:

The Moon has no atmosphere, so:

the Sun would be brighter
there'd be no light diffusion, so you could see the Earth/planets/stars in the daytime
Meteors would not leave a fiery trail -- they would simply hit the moon silently. Asteroid impacts on the Moon would also produce no sound.

There wouldn't be a moon in the sky, but there would be an Earth instead.

You add a twist to the question by asking about past and future light:

In about 7.6 billion years, the Sun will expand to (probably) engulf the Earth and Moon, which will probably be the most light the Moon will see before it's reduced to nothingness.
The Moon possibly saw a lot of light when it was created through planetary collision. Not sure this counts, because the Moon didn't really exist until sometime after this.
Any other major source of light (such as supernovas) would be visible from Earth as well. That doesn't mean there haven't been any super-bright supernovas, but if they were bright enough and lasted long enough to generate more light than the Sun has during its lifetime, they would probably have a major impact on Earth (who knows, maybe that's why the dinosaurs died ), and there's really no evidence anything that liked happened.


How to calculate local position, knowing the RA and Dec of the center of an image, RPY of the camera, and precise time?

Picture was taken somewhere near 30°22'02.0"N 75°00'35.7"E
Bathinda, Punjab, India.
Answering your question about time - how it enters into the sight reduction: RA, all by itself, isn't useful. It's the angle between the plane of the meridian of a celestial location and the plane of the meridian of the first point of Aries. That angle is more or less fixed, but the first point of Aries, and its meridian, are continually changing, going through 360 degrees in a sidereal day.
Assumptions: pitch is the altitude of the celestial location (its angular distance above the horizon), and it's not corrected for refraction. Yaw is the azimuth of the location, CW degrees from north.
General method: knowing the time, find the Greenwich hour angle (GHA) of the first point of Aries at that time, by interpolating between hourly values in an almanac. The GHA of the celestial location is the GHA of Aries minus RA. The geographical position (GP) of a celestial location, is the point on the earth's surface where the celestial location is in the zenith. The GP I found after some number crunching was 3.56 deg N, 72.655 deg E. It's in the middle of the Arabian Sea. That GP, the observer's unknown location, and the north pole form a spherical triangle, which must be solved to get the observer's location fix. Two sides and an angle are known: one of the sides is 90 deg minus the latitude of the GP, the other is 90 deg minus the observed altitude of the celestial location. The angle is the 360 deg minus the azimuth of the location. It's one of those awkward spherical triangles where the angle is not the included angle between the sides. It's solved by first using the law of sines to find the angle at the pole then using Napier's Analogies to find the third side.
In this particular case, the celestial location GP is almost due south of the observer, so there's a way to make a simple estimate. The observer is roughly 26.9 deg north of the GP, a little less than 30.46 deg N, and a little east of the GP 72.65 E.
I have a few more things to say if anyone's interested. I'd appreciate a check on my numbers. It's very easy to make a mistake, which is why sight reductions in the pre-GPS era were done by two persons independently who'd compare their results. An unintentional grounding can ruin your whole day.


Were magnetic monopoles created before, during or after inflation?

Before. The point is that there were too many predicted, and no concept of inflation yet. Here is a reference that doesn't just gloss over it.

Here is the original paper by Guth, and physics.SE has a nice drawing of the same timeline shown on that page. (Original here)

You ought to try that physics.SE thread and continue there for more details.


What is the physics of the "spinning dust" contribution to Cosmic Microwave Background measurements?

"Spinning dust" is a mechanism proposed to explain a particular feature in the foreground emission of CMB; a bump around $\nu\sim20\,\mathrm{GHz}$.

Dust grains acquire charge through photoelectric emission and collisions with electrons and ions (Draine & Lazarian 1998). As zephyr comments, if the dust is a poor conductor, its charges will, in general, be unevenly distributed, causing small dust grains to exhibit an electric dipole moment $\mathbf{\mu}$. But the molecules making up the grains may themselves have a dipole moment, and even a perfect conductor
will, in general, have its charge centroid displaced from the mass centroid (Purcell 1975).

Collisions and radiation may cause the grains to start spinning, and in the presence of magnetic fields (which are very common in the interstellar medium), this spin in turn causes the particles to emit radiation with a power $P$, according to Larmor's formula which for a rotating dipole can be written as:
$$
P = \frac{2}{3} \frac{\mu_\perp^2 \omega^4}{c^3}.
$$
Here
$\mu_\perp$ is the component of $\mu$ perpendicular to the rotation axis,
$\omega$ is the angular frequency of the rotation, and
$c$ is the speed of light.

The emitted radiation matches the rotation frequency which lies in the (tens of) GHz-region, corresponding to wavelengths in the microwave region.

In contrast, the radiation from non-spinning dust will be thermal, thus lying in the infrared.

The paper you link to (Bennett et al. 2013) shows the difference in Fig. 22 (although thermal radiation peaks outside the observational range of WMAP): The spinning dust peaks around $\nu\sim20\,\mathrm{GHz}$, while the thermal radiation peaks around $\nu\sim2\,000\,\mathrm{GHz}$.

Note that spinning dust also emits thermal radiation, and in fact thermal fluctuations within the grains cause the charges to shift around quite rapidly, changing $\mu_\perp$ and $\omega$ (Hoang et al. 2015), thus effective smearing out the spectral lines.

A symmetrical dust grain will arguably be hit by particles/photons, on average, as much from one side as from another. A mechanism to accelerate an asymmetrical grain's spin is proposed by Purcell (1975). Below is Fig. 1 from his paper. A particle that strikes in a concavity is more likely to interact with the grains twice. If the gas it colder than the grain, it will be heated and leave the grain with a larger velocity than it entered with, causing the grain in the figure to start spinning counter-clockwise; it the gas is hotter (which is more often the case), it will cause the grain to spin clockwise.


Is nucleosynthesis responsible for the expansion of the universe?

Big bang nucleosynthesis did not happen "close to the inflationary period".
inflation happened at about $10^{-36}$s. Nucleosynthisis happened from about 10seconds.

In relative terms, the ratio of the age of the universe now, to 10 seconds is much much smaller than the ratio of 10 seconds to $10^{-36}$s. And, significantly, nucleosynthesis occurred after inflation, so can't be the cause of inflation.

As for dark energy, nobody has any real model for it. The test for any theory is how well it does empirically. As it stands you have advanced a pet theory, without any mathematical model or analysis.


Could celestial coordinates be encoded into three-word combinations?

Possible, yes. In fact since you can just use their algorithm. Sky locations expressed a RA and Dec map directly to coordinates expressed as Longitude and Latitude. The celestial sphere is no more infinite than the terrestrial one.

Each square seems to take about 0.1 arcseconds, which is very accurate, but it does raise a problem: The stars move. Barnards star has proper motion of over 10 arcseconds per year. This is an extreme example, but all the nearby stars move, so their addresses would change. The algorithm is designed not to give nearby locations similar names. So each year Barnards star would have 10 completely different addresses. Other stars would change address due to parallax. If you don't account for precession, all stars would be being renamed all the time.

Now the purpose of the three word addresses is to give addresses to the address-less, in a way that is memorable. Is there a need for addresses of sky locations? We have a cute way of naming larger regions of the sky (the 88 constellations) and stars have names (or catalog numbers) that don't change due to their real or apparent motion. If you need to refer to a location you can use the numeric coordinates.

So: this would not be useful, either for science (since the stars would keep moving address) or for outreach (the Constellations do very well for the whole sky)


"EW" stands for Equivalent Width. The equivalent width of a spectral line is essentially the range of continuum one would integrate over to get the same flux as the spectral line.

In your case, this is the Hα line.


Most of the communication on Earth and to-fro Space takes place over low frequency visible, infrared, micro and radio wave spectrum of the electromagnetic spectrum.

The immediate benefits of using low frequency EM spectrum which I think of are as following:

They are not blocked by Earth's atmosphere and magnetic field. On the other hand, cosmic rays which consists mostly of high energy electrons, protons and ß particles carries electric charge and magnetic field. Hence, they will be hugely affected by Earth's magnetic field. Maybe setting up receivers and transmitters at poles helps a little.

2.Speed:  EM waves travel at light speed, cosmic rays always travel at lesser speed which brings me to the next point.

Energy needs: The energy required to generate radio waves can be provided by low power solar panels. However, to generate cosmic rays one would need a particle accelerator. Something like a portable LHC, which I don't think is feasible. Moreover, if you want to generate faster cosmic rays then you have to put in more energy.

I can only think of these constraints at this moment. I don't know if sending information via cosmic rays carries any advantage.


Close, but not quite right - the blue light is indeed emission from CO$^+$, but it's from the CO$^+$ ions themselves, with no need for recombination to CO; that (ionized) molecule has a strong set of energy transitions around 425 nm (4250 Angstroms), which is in the blue part of the visible spectrum:

Spectrum of Comet C/2016 R2 (Pan-STARRS), Figure 2 from Cochran and McKay (2018).
There's more on the physics of comet tails from Chris Mihos here.


How many times brighter would full-earth-shine appear to people standing on the moon, than full-moon-shine does to people on earth?

How much brighter is full-earth-shine on the moon, than full-moon-shine on earth?

On average, it's about 41 times brighter -- in terms of luminosity, or a bit over four times brighter ($2.5 \log_{10} 41 \approx 4.03$) in terms of the logarithmic response of the human eye.
There are two factors that come into play in the luminosity calculation: The much larger size of the Earth than the Moon (a factor of over 13, the square of the ratio of the Earth's radius to the Moon's radius), and the much larger albedo of the Earth than the Moon (a factor of a bit over 3).
This is an average value. If the full Earth view from the Moon happens to be centered on a cloudless Pacific Ocean, that factor of three due to albedo becomes a factor of less than one. On the other hand, if the full Earth view from the Moon happens to be centered on a very cloudy Earth, that factor of three due to albedo becomes a factor of five or more.
The Moon's albedo (0.12) is roughly the same as that of slightly aged asphalt. The Earth's albedo varies a lot. The Earth's oceans are rather dark when seen from space, as dark as freshly laid asphalt. Clouds can have an extremely high albedo, up to 0.9, or 18 times the albedo of water.


Why does kerr black holes have a ring shaped singularity rather than point?

Firstly do not confuse the event horizon with the singularity. Wikipedia gives the formulae for the Kerr metric. There are a number of places where this formula appears to break down because you appear to be dividing by zero, essentially whenever $\Sigma = 0$ or $\Delta = 0$, in the notation of that page,  corresponding to the ergosphere and the event horizon respectively. There are two event horizons and two ergospheres but they all have the topology of spheres, one inside another (with some touching at the poles). However those are not singularities in general because you can choose other coordinates for space and time where the division by zero goes away. However you can't do this at points on the equator of the inner ergosphere, which makes those points a ring singularity. It is a ring with zero thickness (at least until quantum effects come in) .There is a helpful picture

Flow of liquids doesn't really enter the picture here. One can think of the properties of this metric as a "flow" of space-time, but that's an analogy at best.

There seems to be considerable doubt about what really happens inside a rotating black hole, or even, given that no information can escape from the black hole, about whether this is even a meaningful question.


If you are going to buy a telescope in the city let me suggest a large aperture and a reflective telescope.  I live in a city as well and I use a reflector telescope.  I did have trouble with some dimmer galaxies to start with but if you buy a light pollution filter you can see them a lot better.  Also take a series of images of the same object and then stack them together!  There are many free astronomy image stacking programs on the web.

Have fun!


Can I learn astrophysics and astronomy with my basic math?

Generally, astrophysics (and astronomy) is math-heavy. That should not discourage you, but rather act as a way of learning math: it is often easier to learn topics that you have a use for and are part of some personal project than just getting lectured about them.

Astrophysics is based on math because it is based on many forms of physics - mechanics, thermodynamics, nuclear physics, relativity theory and so on - plus chemistry. Mathematics is a common language that binds these fields together and allows you to combine them to build models to understand what is going on. Some parts just use some basic math, but in general calculus really is needed.

Mathematics in the more common "calculations with numbers" form also matters because it all starts with measuring things - where things are, how much they change, brightness and so on - and then we build and test our models and understanding from this. That also means that statistics is important for handling measurement noise and finding patterns in data.

One cannot really do astronomy and astrophysics without math. It would be just looking at things and making up explanations without actually checking if and how they hold true.


Collapse of a star below the mass of Chandrashekhar Limit

The Chandrasekhar mass is a theoretical maximum stellar mass that can be supported by ideal electron degeneracy pressure.

It was originally calculated by Chandrasekhar using polytropic equations of state for white dwarf stars of various compositions.

It is found that if the equation of state is $P \propto \rho^{4/3}$, which is appropriate for ultra relativistic electrons, then equilibrium is only obtained for $M_{CH} = 5.8 \mu_e^{2} M_{\odot}$, where the density becomes infinite and $\mu_e$ is the number of atomic mass units per electron in the gas.

Typical white dwarfs are carbon and oxygen with $\mu_e= 2$, and $M_{CH}= 1.45M_{\odot}$, but in the core of a supernova progenitor made of iron-56 it would be lower - around $M_{CH}=1.25M_{\odot}$.

In more recent years, the Chandrasekhar limit has evolved to colloquially mean the maximum possible mass for a white dwarf. The main corrections to the ideal case considered by Chandrasekhar are Coulomb corrections, the possible onset of inverse beta decay (electron capture) and the instability at a finite density predicted by using GR rather than Newtonian gravity. The latter probably sets the "Chandrasekhar mass" for carbon white dwarfs to be $1.38M_{\odot}$.

A carbon white dwarf that had a mass less than this could be supported by electron degeneracy pressure and would not collapse.

If your question is asking for a derivation of the Chandrasekhar mass, then I suggest you look at the stability analyses that are presented in a standard text on the subject like "Black holes, white dwarfs and neutron stars" by Shapiro and Teukolsky. An approximate result can be obtained using the virial theorem assuming a uniform density star.


If Jupiter is a gas-giant then why don't its features change?

Believe it or not, Jupiter isn't too consistent. Take a look at these pictures, the first taken in 2009 and the second taken in 2010:

and

Quite the difference, eh? Why?
Jupiter's atmosphere is made of zones and belts. Zones are colder and are composed of rising gases; they are dark-colored. Belts are warmer and are composed of falling gases; they are light-colored. The reason the two don't intermix is because of constant flows of wind, similar to the Jetstream. These winds make it hard for bands to mix.
There are two types of explanations for the jets. Shallow models say that the jets are caused by local disturbances. Deep models say that they are the byproduct of rotating cylinders comprising the mantle. At the moment, we don't know which explanation is correct.


Visibility of human activity on the moon

The video is hilariously wrong.

However, the principle of laser ranging is more or less right, and it does require the reflectors left behind by the astronauts on the Moon. It's just that the physics and technology involved are far beyond just pointing a toy laser at the Moon.

Project APOLLO (Apache Point Observatory Lunar Laser-ranging Operation) is actually doing this.

http://physics.ucsd.edu/~tmurphy/apollo/basics.html

You need a fairly large telescope to start with - both for collimating the outgoing light pulse, and for receiving as much of the reflection as possible. APOLLO uses a 3.5 meter telescope at the Apache Point observatory.

You need a laser that can generate a high-energy light pulse that is very short. The pulse is injected into the telescope's optical train and sent to the Moon. This is not a laser pointer; it's a high power Q-switched laser for research, a device the size of a refrigerator.

On the receiving end, you need a very good detector also plugged into the telescope. Of the many, many photons sent to the Moon in the pulse, only between 1 and 5 photons make it back down to the detector. You need a detector that can tell the extra 1 to 5 photons from the background noise of light coming from the Moon.

Using this technique, the distance from Earth to Moon can be measured with very high precision.

This is the APOLLO system in action:

Regarding observing human artifacts on the Moon with terrestrial telescopes, I wish it was doable, but it's not. Again the YouTube video is wrong.

The biggest telescope accessible to amateurs has an aperture (diameter) in the range of 1 meter or a little bit larger (the aperture of the largest amateur-owned telescopes currently). The resolving power of a telescope (the size of finest details discernible) depends on aperture - if aperture is measured in mm and the resolving power in measured in arcseconds, then the formula is:

resolving power = 100 / aperture

So a 1 meter telescope has a resolving power of 0.1 arcseconds.

The distance from Earth to Moon is 384000 km (3.84 * 10^8 meters). With a resolving power of 0.1 arcseconds, the size of the smallest detail discernible on the Moon is:

detail size = distance_from_Earth_to_Moon * arctan(resolving power)

or

detail size = 3.84 * 10^8 * arctan(0.1 arcsec) = 186 meters

Anything smaller than 186 meters would be blurred into a single dot. There's nothing we've done on the Moon that's as big as that. It's not possible to see traces of human activity with amateur-level telescopes, even with extremely large meter-class dobsonian telescopes. Even with professional telescopes, we just don't have the aperture yet to resolve that kind of detail.

However, satellites in orbit around the Moon, such as the LRO, were able to image traces of the Apollo missions. That's because they are a lot closer to the site.

http://www.nasa.gov/mission_pages/LRO/news/apollo-sites.html


Stan has essentially answered this in his comment, which I will attempt to spell out a little more laboriously.

The significant majority of our Sun's energy output comes from the proton-proton chain.  This was advocated by Eddington back in the 1920's, but at that time your basic concern was a very real and major problem.  Objects with like electrical charges repel each other.  In particular, protons will repel other protons since all protons have a positive charge.  What they knew of the sun then indicated that the core was far too cold for protons to overcome this repulsion—at least not even remotely close to the rate that was obviously necessary to produce a brightly shining sun.

With the development of quantum mechanics it was determined that a process known as quantum tunneling would give two protons a non-zero probability of "overcoming" this repulsion.  But not in the sense of they just suddenly gain enough energy (hence why "overcoming" is in quotations).  Instead, in the sense that the "fused together" state has the same energy as the "just about to repel each other before fusion can occur" state, and they just randomly switched from the latter to the former, despite every intermediate stage between the two of them requiring more energy than is available.  That this is possible is one of the many non-intuitive features of quantum mechanics, and I think it would be beyond the scope of this question (and site) to try to get much more precise.

Still, even this failed to explain why our sun was very obviously fusing atoms to the extent that it was.  If you fuse two protons together, you are left with an incredibly unstable state: the diproton.  As soon as one of these forms it pretty much immediately breaks up into two distinct protons.

In the late 30's Hans Bethe (a man who eventually won the Nobel prize and was a part of the Los Alamos team that developed the atom bomb) proposed that yet another random quantum mechanical event would save the day: beta decay, a feature of the recently discovered fundamental force known as the weak force.  In this situation one of the protons in the diproton undergoes a beta decay into a neutron before the diproton separates, at which point you have a stable deuterium nucleus: one proton, and one neutron.

That so many unlikely events have to occur for two protons to produce one deuterium is the reason why the sun's lifetime is approximately 10 billion years.  That the sun shines as bright as it does is, as you suggest, a sheer numbers game: there are truly prodigious numbers of protons in the sun's core, so even though the proton-proton chain is really unlikely, we have so very many chances at it that there is a great many successful fusion events.

Basically every other stellar fusion reaction proceeds much more rapidly than those that produce deuterium.  You may be familiar with the hydrogen bomb, which fuses its fuel very rapidly.  One of the key differences (and there are many) between the hydrogen bomb and a star like our sun is that the bomb is fusing together deuterium (and tritium) into helium, and is not using the proton-proton chain.  Stars have to produce large supplies of deuterium directly, and this is a lengthy process that consumes roughly 90% of a star's life—even for very massive stars, whose lifespans are orders of magnitude shorter than our Sun's.

(Note that very massive stars are rather more complicated beasts, in fact, but this is again beyond the scope of this question)


the initial astrometry calibration for a telescope

First you have enter your coordinates into the mount's software and then align it(e.g. 3 star alignment).After having done that, when you finish the observation you have to park the telescope.

Actually Colin's answer is really helpful.They keyphrase is "After installation set up".
The whole procedure should be explained in your telescope's manual.
If you tell us which model you have we might be able to help more.


Hypothetical CMB space telescope design problem, received power from extended thermal source versus receiver front end NEP?

The feed horn is an antenna which has an antenna pattern  Source
However, not all feeds are horns.  Here is an image of a parabolic dish with a Yagi feed:  Source
Ideally, you want the beam width of the horn antenna to cover the entire dish and only the dish.  So if the beam width of the horn antenna is, say, 30 degrees, you want the dish to subtend 30 degrees of the view from the center of the entrance of the feed.
Since the dish is parabolic, it can be described by the equation y=a(x-h)^2 Source.  Assuming the vertex of the parabola is at the origin, the feed antenna will be 1/(4a) units away from the vertex.  This is the focal length of the dish.  The aperture of the dish is its diameter.  Using the values from the above equation, the f-number of a dish antenna is (1/(4a))/x or 1/(4ax).
Now that we know where the focus is, we can compute the beam width of the feed required to cover the dish, then design a feed antenna meets that requirement.
Notice that the f-number doesn't contribute to the solution to the problem as we would expect if this were an optical system.  The most important thing is to match the beam width of the feed antenna to the radius of the dish, which is set by the value of "a".  So a feed antenna will receive the same amount of power regardless of f-number as long as the feed antenna is at the focus of the aperture.  (Whether one can design a feed antenna for a given "a" is another question.)


What are the reasons for referring to equinox J2000 or the equinox of the date?

Maybe you are interested in watching the occultation of a star by a solar system object. In which case you will need to compare their positions. Catalogues of stellar positions are most commonly given with J2000 positions.


Could there be a brown dwarf in our solar system?

The nemesis theory proposes that a low mass star or brown dwarf in highly elliptical orbit is a companion to our sun as a solution to the cyclical mass extinction problem (http://www.theatlantic.com/science/archive/2015/11/the-next-mass-extinction/413884/) and (http://www.space.com/22538-nemesis-star.html). Scientists noticed that some mass extinctions follow a seemingly cyclical pattern, so researchers proposed that the orbit of a small companion star (or a brown dwarf) was disturbing comets or asteroids as it approached aphelion and flinging them towards the earth.

As interesting as this would be, WISE's infrared scans of the sky have revealed nothing and there is currently no evidence supporting the theory. Furthermore the orbit of any companion would have to be very large and would therefore be very unstable and would likely be detectable. The Astrobiologist David Morrison stated of the theory: "(T)he Sun is not part of a binary star system. There has never been any evidence to suggest a companion. The idea has been disproved by several infrared sky surveys, most recently the WISE mission. If there were a brown dwarf companion, these sensitive infrared telescopes would have detected it."


Moon is a planet by "Clearing the neighborhood" condition?

According to the 2006 IAU definition of a planet, a celestial body must meet all 3 criteria to be considered a planet.  The Moon fails the first criteria "1. Is in orbit around the Sun".  The Moon orbits Earth, not the Sun, and thus it doesn't have a legitimate "neighbourhood" (solar orbit) to clear in that context.

Also to note, according to the Clearing the neighbourhood article, this criteria is contentious as it is vaguely written and Earth, Mars, Jupiter and Neptune have not "cleared the neighbourhood" of their respective solar orbits.


Have more recent LIGO/VIRGO gravitational wave measurements narrowed down the speed of gravity further?

To get these kind of measurements one needs a gravitational wave detection with a electromagnetic counterpart. The big problem is, that the now quite common BH mergers are not expected to produce a detectable electromagnetic signature.
Therefore those can not be used to measure the speed of gravitational waves. At least not to any precision.
What you need is a binary neutron star merger. During the destruction and following accretion of the two NSs, these produce photons over the whole electromagnetic spectrum.

Until now, there have only been detected two such mergers, GW170817 and S190425z.
As mentioned by @amateurAstro, GW170817 gave us the precise constraint added in the question. Judging from the citations in recent papers (e.g. this) the analysis by Abbott et al. 2017 is also still up to date.
For their constraint Abbott et al. assumed that the gamma-ray burst was emitted  between 0 and 10 seconds after the gravitational wave. If future simulations would be able to further constrain that range, this will also give us a more precise limit on the speed of gravitational waves.

The other one is S190425z. The problem here is that the only 1/4 detectors were working so almost a quarter of the sky had to be searched. It was also quite far away. So unfortunately no counterpart was found.

So until we have a better theoretical understanding of NS merger or/and we observe another one, the constraint by Abbott et al. 2017 stands.


Could we detect if Oumuamua asteroid wasn't an asteroid?

I think its a good question and has probably popped up in the back of many people's minds for milliseconds, both astronomers and lay people, though most of them would still be hesitant to admit it on an internet forum. And of course, our understanding of what is possible or even "normal" in the universe is vanishingly small.
edit 2:

Scientific American: Alien Probe or Galactic Driftwood? SETI Tunes In to 'Oumuamua
The Atlantic: Astronomers to Check Mysterious Interstellar Object for Signs of Technology
Breakthrough Initiative: Breakthrough Listen to Observe Interstellar Object 'Oumuamua'

From the NPR news item Astronomers Want To Know: Does This Interstellar Visitor Have A Message For Us?:

"Researchers working on long-distance space transportation have previously suggested that a cigar or needle shape is the most likely architecture for an interstellar spacecraft, since this would minimize friction and damage from interstellar gas and dust," the group said in its announcement.
"While a natural origin is more likely, there is currently no consensus on what that origin might have been, and Breakthrough Listen is well positioned to explore the possibility that 'Oumuamua could be an artifact."
"The possibility that this object is, in fact, an artificial object — that it is a spaceship, essentially — is a remote possibility," Andrew Siemion, a member of the initiative and director of Berkeley's Search for Extraterrestrial Intelligence Research Center, told The Washington Post on Monday.
"We don't want to be sensational in any way, and we are very realistic about the chances this is artificial," Yuri Milner, the Russian billionaire financing the project, told the Scientific American, "but because this is a unique situation we think mankind can afford 10 hours of observing time using the best equipment on the planet to check a low-probability hypothesis."
And they'll be checking on that hypothesis by scanning the object for possible artificial transmitters through a radio telescope at West Virginia's Green Bank Observatory.

edit 1: A few days after you've asked the question, a similar thought was uttered by the BBC correspondent Roland Pease in the Science in Action episode Interstellar Asteroid. Talking with astronomer Alan Fitzsimmons of Queens University, Belfast (who's image of X was featured in a NASA "Astronomy Picture of the Day"), Pease says:

You say ‘cigar-shaped”, that’s to me “rocket-shaped”. Is there any way you would know that this is an interstellar rocket?

So it looks like you're in excellent company!

The asteroid was discovered and has been observed mostly (if not completely) by it's reflection of radiation from the Sun. The Sun is roughly a blackbody emitter and most of the radiation is in the visible. Because of the dimness of the light even at maximum, there isn't much signal once dispersed spectroscopically.
There is a spectrum published in ArXiv Palomar Optical Spectrum of Hyperbolic Near-Earth Object A/2017 U1 on 2-Nov-2017 by Joseph R. Masiero though! Despite the low light level (V ~ +21) and expanded size due to poor seeing at low elevation resulting in a signal-to-noise ratio of order unity, the 5 meter Hale telescope (200 inch Mt. Palomar) was used to try to obtain some spectroscopic information. Ten 300-second runs (total of 50 minutes of collection) were used to produce the heavily binned spectrum below. I think those are 1-sigma error bars, so the spectrum doesn't say much more than "slightly red-ish".

See also Ye et al. 1I/2017 U1 (`Oumuamua) is Hot: Imaging, Spectroscopy and Search of Meteor Activity for further background:

The moderate spectral gradient of ‘Oumuamua indicates that its surface is devoid of ultrared material that is common on outer solar system objects like KBOs. Various classes of KBOs are typically very red in color, with a
spectral gradient $\gtrsim (\text{100 nm})^{-1}$, likely due to the irradiation of organic material resulting from the bombardment of energetic particles (Brunetto et al. 2006). The less reddish color of ‘Oumuamua suggests that the object was either formed close to its original central star, or has lost its ultrared material due to close approach(es) to its original or other stars. It is difficult to say which scenario is more likely due to the chaotic nature of small body dynamics.

According to Sky and Telescope's Update on `Oumuamua, Our First Interstellar Object:

The object had already dimmed to 20th magnitude by the time the Pan-STARRS 1 telescope swept it up, so observers worldwide had to scramble to get major facilities to gather more observations before the one-of-a-kind interloper becomes impossibly faint.
One lucky break came immediately after the MPC's initial announcement. Observer Joe Masiero (Jet Propulsion Laboratory) just happened be in the middle of a run with the 5-m Hale Telescope at Palomar Observatory, and he quickly obtained a spectrum that's reproduced below and explained in more detail here. There's no obvious absorption that would signify the presence of particular compounds or minerals.
But the gentle slope, increasing toward the red (right) end, suggests that `Oumuamua is not a particularly good match to objects in the Kuiper Belt (which tend to be much redder). Instead, it's the kind of spectrum that a rocky surface would exhibit after being "weathered" by long-term exposure to space radiation.

So I'd say that the answer to your question is "No", there's very little besides its orbit and light curve and approximate color that has been or probably could have been detected. With more warning, and serious interest, it's possible an observation campaign could have been organized to get more aperture area and time to make a better spectrum, but since S/N only scales as the square root of those, a lot of other observations would have had to have been rescheduled to make it happen.


How would seasons at the equator be if the Earth were tilted 90 degrees?

I assume you mean "what if the Earth's equator was perpendicular to it's orbit?" in which case the seasons would be greatly exaggerated at the poles and day and night would only really mean something for about half the year if you're nearer the equator. Temperature differential would be much more significant than it is now.

Unless you mean that the axis of Earth's rotation is exactly perpendicular with respect to it's orbit, in which case there would be no seasonal variation whatsoever. (Not including Malkovitch cycles and solar cycles)


First observation that the Sun and Jupiter (and friends) move around a common barycenter?

This idea would be a direct reduction from Newton's laws of motion.  In his Principia he states "Hence the common centre of gravity of the earth, the sun, and all the planets, is to be esteemed the centre of the world".  So that would be the first evidence of the Sun moving outside of a geocentric universe.  Of course he wouldn't have known about Uranus and Neptune at the time.


Why hasn't the CMB already receded away faster than the speed of light?

Let me try this from a building blocks point of view.

Imagine the Universe is basically uniform, that is, it's pretty much the same everywhere at any given time, or, at any specific age.  We think this is essentially true.

Next, imagine that the Universe is ridiculously, impossibly large, much larger than we can see, because when we look at things very far away, we also look back in time and we can only see as far as the universe is old.   We can't see most of the Universe because most of it is much too far away.   Given the observed expansion, we'll never see most of the universe.  The light from objects too distant will never reach us.   We assume there are countless galaxies out there, similar to the galaxies we can see, but we'll never see them.

Lets skip the very early parts when the observable universe was the size of a baseball, or the size of the solar system, because that happened long before the cosmic background radiation and it's not relevant to your question, and we can begin in the, lets say, around 370,000 years of age, when the Universe was much less spread out than it is today.

At that age, about 370,000 years after the big bang, the Universe was still hot everywhere, kind of like fire thinly spread out everywhere.   There were no galaxies or stars as they hadn't coalesced yet.  Just spread out clouds of fiery gas still spread out everywhere, but beginning to clump together under gravity.  Basically just hydrogen and helium nuclei and free electrons, hot enough to be in a plasma state, not too different than flame, and flame is opaque.   It's not fully opaque like a solid object, but flame diffuses light, see more details here.

so, 370,000 years of age, the entire universe was kind of a glowing fog.  You couldn't see very far in any direction.  You could maybe see an AU or a few hundred AU before the opaque plasma blocked your view if you happened to be around at that time, but basically everywhere you looked, you'd just see the heat-glow of the hot universe.

This all changed around the time when the universe was 379,000 years old, during a time called recombination.   The universe expanded enough and became cold enough that most of the free electrons rejoined with the atomic nuclei and the plasma became atoms.   As a byproduct of this, the universe went from a glowing fog to transparent for the first time.   This didn't happen all at once, it was more of a transition into transparency, but for the purposes of this question we can think of this as happening at 379,000 years.

But, because when we look in any direction, the farther away things are the older they are, so at the edge of this transparency, there's always going to be the cosmic background radiation, that was true shortly after the 379,000 year recombination and it's been true ever since.   What has changed, is the expansion of the universe keeps stretching the wavelength of the CMB and it keeps moving further away as we move forward in time.

What a person would actually see, when the Universe was 380,000 years old, is the cosmic background radiation in a sphere around them, about 1000 light years distant and in visible light, because the expansion wouldn't have affected it much over just 1000 years.   The CMB is kind of like a sphere of time in addition to a sphere of background radiation.   It's the furthest thing we can ever see in the EM spectrum, about 379,000 years younger than the age of the universe.

The thing to keep in mind about this sphere is the universe is much bigger than we can see.   The glow was everywhere, but the glow we see at any given time is the age of the universe minus 379,000 years old, it's a kind of observable universe bubble, past which we can't see.

The first stars didn't show up until about 180 million years later and the first galaxies, another 20 million years after that, so it was a long time between recombination and the first stars.  If a person was around 180 million years after the big bang, they could observe the first stars forming, but only stars that were close to them.  Because looking further out looks back in time, so the first stars anyone could observe would be in their neighborhood, but over time, we saw stars form further and further away, but never as far as the CMB.

But what we can see doesn't change what the largely uniform universe did.  Stars and Galaxies began to form everywhere, both inside and outside our visible edge of the CMB.   Every star and every galaxy is younger than the CMB, but we can infer, even though we'll never see them, that a great many stars and galaxies formed outside or our visible edge of the universe because the universe is much bigger than we can see.

That's kind of a step by step overview, perhaps more wordy than it needs to be.   Let me know if that helps.


Why does black hole draw off some mass from orbiting star instead of sucking it?

Black holes don't "suck things in", they pull on things gravitationally with their mass, same as any other object. You might as well ask why the Sun doesn't suck the Earth in. The natural thing is for objects to orbit, not collide with, stars or black holes.


Cycles that are longer than the year, but less than 1000 years

The Moon does not follow the ecliptic precisely like the Sun does; rather, it follows a path inclined of about 5.14°. The two points where the ecliptic and the Moon's path coincide are called 'nodes', and they travel around the ecliptic in a period of 18.66 years. This is a combination of the orbital period of the Moon, also called the sidereal month, about 27.32 days, and the slightly shorter draconic month, the average period between the moment where the Moon traverses a node, which is 27.21 days.

Once every 18.66 years, the ascending node coincides with the vernal equinox, and the Moon will reach declinations of about +28° and -28° (much more than the Sun, which reaches ±23°); this is called a 'major lunar standstill'. 9.33 years later, the opposite is true, and the Moon will only reach ±18° during a 'minor lunar standstill'.

According to Wikipedia these moments have had historical significance, all the way back to the Bronze Age:

  This time appears to have had special significance for the Bronze Age societies, who built the megalithic monuments in Britain and Ireland. It also has significance for some neopagan religions. Evidence also exists that alignments to the moonrise or moonset on the days of lunar standstills can be found in ancient sites of other ancient cultures, such as at Chimney Rock in Colorado and Hopewell Sites in Ohio.


Putting mass-luminosity relation and Hertzsprung-Russel diagrams together leads us to a mass-age relation; so how do stars lose their mass over time?

There is an alternative interpretation: Massive stars burn their fuel much faster than tiny ones. Hence massive stars are short-lived in comparison to tiny stars.

The lifetime of a star is proportional to about the inverse of the cube of its mass.
The Sun's lifetime is about $10^{10}$ years, hence for stars with 10 solar masses the lifetime is about $10^7$ years, and for stars with 0.1 solar masses it's about $10^{13}$ years.

But stars also loose mass, e.g. by stellar wind and coronal mass ejections. Heavy stars may loose mass (parts of their envelope) e.g. by pulsation or radiation pressure during their red giant or their Wolf-Rayet phases, before they explode as a supernova.


How many minutes does a sundial change every day?

You need to look up "the equation of time" - this is a correction to the time reported by the sun on any given day - the apparent solar time - (and measured by your sundial) and the standard civil time - the mean solar time - it varies throughout the year and the maximum your sundial will be out is about 15 minutes - but the difference and the rate of change varies. http://en.wikipedia.org/wiki/Equation_of_time

If your sundial is merely for educative/demonstration purposes, this might not matter so much of course - as the sundial will surely demonstrate that the sun's shadow can be used to tell the time.


Does anyone know why three of Jupiter's largest moons orbit in 1:2:4 resonance?

When the Galilean moons formed, they weren't in resonance with each other. All of them were in slightly smaller orbits than they are now. Over time after their formation, Io's orbit slowly moved outward due to tides from Jupiter. This is the same effect that is causing our moon to slowly move away from the Earth (at about the same rate your fingernails grow). It goes like this. The Moon's gravity causes tides to form in Earth's oceans. This bulge of water gets carried forward with Earth's rotation, because the Earth is spinning faster than the Moon is orbiting. Because the moon is still attracting the bulge, it causes a drag on the Earth, slowing its spin. At the same time, the bulge is attracting the moon, causing it to go faster in its orbit. As the moon speeds up, its orbit gets bigger. So essentially, the Earth's spin energy is getting transferred into the Moon's orbital energy. The same thing happens with Jupiter and Io, with a bulge in Jupiter's atmosphere causing Io's orbit to get bigger.
As Io's orbit expanded, its 'year' got longer, until it approached a 2/1 resonance with Europa. Once they reached resonance, they got 'locked in', their mutual gravity acting on each other reinforced it. Io was still raising tides on Jupiter, though, and its orbit was still trying to expand. As Io's orbit kept expanding, it gave a gravitational kick to Callisto on each pass, expanding both of their orbits until Callisto reached a 2/1 resonance with Ganymede. This is where the inner 3 Galilean moons got their resonance. The orbits are still expanding, but much more slowly because with each one you add on, it gets harder to transfer the energy. Given enough time, all 4 Galilean moons would probably reach resonance, although the sun will die before that happens.
I may have some of the minor details wrong here, but this is the story as I understand it.
EDIT:  Seven years later, When will Callisto be in orbital resonance with the rest of Jupiter's big moons? was asked.  The newest answer for when the first four moons will be in orbital resonance is much sooner than the Sun will die.  Specifically, about 1.5 billion years from now.


What are these objects on the night sky generally between 8:00 and 11:00 PM?

Your objects sound like satellites to me. Why do you think they are not satellites?

Where I live at 40 degrees north latitude, these are the paths that we see the satellites moving (just a rough estimate: not a scientific count)
• 90% are in a polar orbit, so they generally move south to north or north to south (different parts of the same orbit). These generally follow lines of right ascension.
• 9% are inclined 50 to 70 degrees to the equator, so they cross the sky from NW to SE or SW to NE (different parts of the same orbit)
• 1% go across the southern half of the sky from west to east .
•0% go from east to west because there are very few satellites in such orbit.

As Jan mentioned, what is your latitude? That can affect which objects you see and where you see them.

Also, how long are they visible? They ones you see with your unaided eyes will cross the sky in 4 to 7 minutes although the time they are visible can be less. You will see more in the eastern half of the sky after sunset than in the western half because of the phase angle.


Stable polar solar orbit with the Earth continuously observable

As ganbustein says, this is not too difficult to imagine.  The simplest case (approximating with circular orbits and only the Sun, Earth and Satellite) would have the satellite orbit orthogonally to the Earth with a 1 year orbit.  The Satellite will pass the Earth orbit plane in two places, call these "down crossing" and "up crossing" points.

To minimize Earth - Satellite iterations, keep them 90 degrees apart.  Have the Satellite over the north solar pole when the Earth is at the "down crossing" point.  Then when the satellite gets to the "down crossing point" the Earth will be furthest from the Satellite plane.  When the earth is at the "up crossing point" the satellite will under the Suns south pole. And so on.

This would not be completely stable when we include Jupiter and the mutual interactions, but I think they should be small, allowing this to work generally.  If someone "does the math" and proves me wrong, I will accept that.


Would an exoplanet without axial tilt have no seasons?

There are several factors that cause seasons.  In approximate order from least to most speculative:

Axial Tilt:  This is by far the significant factor in determining what the season is on Earth.  As you may know, axial tilt affects the seasons because the light is "spread out" when it falls on an angled surface.  Note that tilt causes different seasons in different hemispheres.  An untilted planet could still have seasons for any of these other reasons.
Eccentricity:  The Earth's orbit is very close to circular, so eccentricity has next to no effect on causing its seasons (and in fact, IIRC, currently Earth's eccentricity's tiny effect is actually opposing its seasons).  An exoplanet with a high eccentricity could easily have planet-wide seasons due to varying distance from its sun.
Star Systems:  As a commenter mentioned, another factor might be how many stars a system has.  If such a solar system is stable (which, surprisingly, it might be, although I can't find the article at the moment), you could get seasons depending on where the stars are in their mutual orbits.  A binary star system, for example, would have periods of winter when the stars align with the planets (so one blocks the other) and periods of summer when the stars are side-by-side.
Star Variation:  (From a comment): believe it or not, the sun actually has "seasons" (here) that are on the order of a decade or so.  These don't have much effect on Earth's climate since on the sun they mainly cause magnetic disturbances, but on an exoplanet the star's magnetic disturbances could be relatively larger, which would lead to significant temperature/brightness variations that would translate into seasons.
Internal Energy:  Planets could have varying/cyclic sources of energy within them, just like solar masses.  In our own solar system, Jupiter's seasons are partly due to this.  On a terrestrial planet, variations might come from, for example, volcanic activity.  Negative feedback loops of this sort will exhibit oscillations.  These sorts of changes might not count as "seasons" for you, though, as they occur over larger timescales.
Stored Energy: Earth's seasons are complicated by the oceans, which greatly affect the intensity of the seasons, especially in the Americas.  In particular, there are various cycles that work like subseasons underlying the main ones.  The Pacific Decadal Oscillation, for example, affects seasons years apart.  I don't see why an exoplanet couldn't have similar cycles that, unlike in the case of Earth, are the main factor in determining seasons.  This could be stored gravitational energy too; for example in a tidally-locked exoplanet/moon system.

If you mean a true Earth-like analogue (in the sense of being exactly-like-Earth-in-every-way-except-the-axial-tilt), then no--it would not have noticeable seasons, since none of the above factors really apply.


What is the main source of opacity in sunspots?

In Ignjatovi´c et al. (2014), ion-atom absorption processes are discussed. They suggest that these processes are of equal, if not greater, influence on the observed sunspot opacity as absorption processes in $\mathrm{H}(1s)+\mathrm{H}^+$ and $\mathrm{H}(1s)+\mathrm{X}^+$ collisions (where $\mathrm{X}$ is one of the metal atoms $\mathrm{X} = \mathrm{Na, Ca, Mg, Si, Al}$). The main scattering processes discussed within the paper follow the photo-dissociation of $\mathrm{H}_2^+$ and $\mathrm{HX}^+$ by UV photons in the spectral region ($110$ $\mathrm{nm}$ $\lesssim \lambda \lesssim 230$ $\mathrm{nm}$).

I highly recommend reading through the paper—it is very interesting.


What happens if gravity shuts down for only one second?

I expect stars would belch mightily, neutron stars and black holes would explode.

Stars are only held at their steady state size by the opposed forces of gravity and radiation, so eliminating gravity for 1 second will cause expansion, followed by collapse as gravity kicks in again.
A Neutron star's compactness gives it a surface gravity of up to 7×10^12 m/s², and expanding from 20km to ~70 billion km (or 300000km when c limit is considered) over the course of 1 second constitutes an explosion.
Black holes tend to have higher gravity values at the event horizon (same ref) so the result would be even more cataclysmic.


If gravity precedes the formation of a solar system, where did the mass come from that caused the gravity?

The start of it all is usually something called a Giant Molecular Cloud, a particular kind of nebula which is denser than usual and cooler than usual.  A GMC will typically be 10x to 1000x the mass of the Sun.

That mass is enough to cause the outer layers to fall in on the inner parts and the GMC starts to collapse and shrink. (Diffuse though it is, the mass exerts the same force on the outer edges as it would if the entire mass was concentrated at the center.)

The cloud collapses and frequently fragments into a bunch of smaller collapsing blobs centered on especially dense sections of the cloud. At this point it looks a lot like the Oriion Nebula. The young stars light up and blow away the remaining bits of the GMC, and in the end, you have a cluster of young stars with planetary disks looking something like the Pleiades.

There never was a central mass -- none is needed.  The GMC collapses under its own self-attraction.


Have I nearly found the event horizon of a black hole?

This is the Newtonian model of gravity. It is a very good model, it is used for accurate calculating the motion of objects in the solar system to a very high degree of accuracy.
However, for very strong gravitational fields you need to use Einstein's model, which accounts for things like the constant speed of light for all observers. I'm not going to go into mathematical details (not least because they are far to hard for me!)
Now you can calculate the gravity of a sphere and you find that at a radius of $r=2GM/c^2$ something strange seems to happen, nothing can travel from a radius of less than this to a radius of more.  It is the event horizon.  This can only happen in Einstein's theory of gravity.  Newtonian theory has no "speed limit" so there can be no event horizon.
Now it so happens that this is the same radius that Newtonian gravity would predict to have an escape velocity $v_e = \sqrt{\frac{2GM}{r}}$ equal to the speed of light.  That is mostly a coincidence(there was a recent question about this, that I haven't been able to find), but a neat one.
You seem to be comparing the Newtonian acceleration due to gravity with the speed of light.  The units of acceleration and speed are different, so the comparison is not valid.


Why haven't detected life in the universe?

To quote a favourite movie of mine

  Pardon me sir, but it's a big-ass sky.

Our current method of detecting life (electromagnetic signals) is by targeting and scanning stars indivdually using radio telescopes.

This method is pretty costly because we don't have a huge amount of them and they can be better used for other (proper) research purposes.  Basically, it costs a lot to monopolise a telescope's resources by repeatedly pointing it at stars without a reasonable hope of a result.  Understandably, astronomers want to keep their funding by producing results rather than nothing.

The current estimates is that we can stumble upon signals by about 2040

Source

  By 2040 or so, astronomers will have scanned enough star systems to give themselves a great shot of discovering alien-produced electromagnetic signals, said Seth Shostak of the SETI (Search for Extraterrestrial Intelligence) Institute in Mountain View, Calif.

But again, this is pure guesswork.

A related question here on SE asks this same question from another direction

how far away could we detect that Earth has life?

One of the answers, quoted in part:

  A "blind" search could look for radio signatures and of course this is what SETI has been doing. If we are talking about detecting "Earth", then we must assume that we are not talking about deliberate beamed attempts at communication, and so must rely on detecting random radio "chatter" and accidental signals generated by our civilisation. The SETI Phoenix project was the most advanced search for radio signals from other intelligent life. Quoting from Cullers et al. (2000): "Typical signals, as opposed to our strongest signals, fall below the detection threshold of most surveys, even if the signal were to originate from the nearest star". Quoting from Tarter (2001): "At current levels of sensitivity, targeted microwave searches could detect the equivalent power of strong TV transmitters at a distance of 1 light year (within which there are no other stars)...". The equivocation in these statements is due to the fact that we do emit stronger beamed signals in certain well-defined directions, for example to conduct metrology in the solar system using radar. Such signals have been calculated to be observable over a thousand light years or more. But these signals are brief, beamed into an extremely narrow angle and unlikely to be repeated. You would have to be very lucky to be observing in the right direction at the right time if you were performing targeted searches

I'd really recommend reading the rest of this answer, and other answers for this question.

So, our "noise" isn't really strong enough to be detected over long distances above the greater (and stronger) chatter of the natural universe.

While we could engineer a stronger broadcast message and hope for a reply, it's a long, long project and unlikely to be funded.


How to separate binary and optical stellar pairs?

If they are both in the Gaia DR2 catalogue, then a physical binary should have components with similar parallax and proper motion.


Is it difficult to see DSO in your eyepiece?

It's a pretty broad topic. It depends a bit on the type of DSO. Speaking in general, most DSOs lack brightness, and so to observe them you need two things:

lack of light pollution
lots of aperture

Light pollution

Observing from the city, DSOs are challenging. The further away you are from city lights, the better your views. Deserts, national parks, sparsely populated areas are all good. Driving at least 1 hour away from the city usually provides very noticeable improvements. Light pollution maps help in finding places with a dark sky.

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

http://www.lightpollutionmap.info/

Aperture

The bigger the aperture, the better you can see DSOs. If you're a DSO hunter, you must be focused on increasing aperture before you do anything else. What helps here is having an instrument with a good aperture / cost ratio, such as a dobsonian reflector.

There's essentially no limit here - more is always better.

It should be noted that a larger aperture will always perform better than a smaller one, even when light pollution is very heavy - but of course it's best if both aperture and dark sky cooperate.

Other factors that may help:

Filters

These are the most over-utilized and over-hyped accessories, especially for beginners (most people are too much focused on filters and not enough focused on things that really matter). That being said, some nebulae do look better in some filters.

Dave Knisely is one of the foremost experts in the amateur astronomy community in terms of DSOs and filters. Read anything he writes on the subject, e.g. this long article:

http://www.prairieastronomyclub.org/resources/by-dave-knisely/filter-performance-comparisons-for-some-common-nebulae/

In a nutshell: If you can only get one DSO filter, get a UHC. If you can get two, also get an OIII filter. Of course, dark sky and large aperture should always come first. Don't bother with filters if you're observing from the city using a small aperture.

Filters are useful for nebulae. They don't help with galaxies, star clusters, individual stars, etc.

Books

Find a book called Turn Left At Orion. It's a fantastic introduction for beginners to DSOs. It will help you find the easier targets, and essentially open up the sky for you. Later you'll be able to figure things out on your own.

http://www.amazon.com/Turn-Left-Orion-Hundreds-Telescope/dp/0521153972/

It should be noted that the entire Messier catalog can be observed in a relatively small telescope even in the city - and all Messier objects are DSOs. I've seen the Dumbbell Nebula (M27) and the Ring Nebula (M57) in 6" of aperture in a very light-polluted place (Silicon Valley). So, no, it's not that hard to observe DSOs.

Of course, if the sky is dark and/or the aperture large, all DSOs look better. This is especially true of globular clusters, such as M13. They already look pretty spectacular in any aperture, even in a small scope - but the view is mind boggling in a very large dob.


How do we know that light is redshifted/blueshifted and not the original light of a star/galaxy?

If you had a simple slit spectroscope, and looked at an incandescent light, you'd see a smear of light with red on one end and blue on the other. This is because the filament is producing light by glowing from being heated.If you looked at one of those orange colored sodium vapor street lamps, instead of a smear of color, you'd see a group of lines. This light is produced by ionizing the gas.The lines represent specific frequencies of light coming from the lamp. You could add a horizontal scale and find that the lines represent a specific frequency in the light spectrum. If the street lamp was coming at you at a significantly high speed, the individual lines would be shifted in frequency towards blue, but would still have the same pattern. Conversely, if the light was moving away from you, you'd again see the same pattern of light lines, but their frequencies would be shifted toward red.This is what's being measured when the spectra of stars and galaxies are measured: not just what the color looks like, but whether the spectra of things like hydrogen, helium and iron are shifted in frequency towards red or blue.So, it's the specific line patterns produced by these elements when they are ionized in stars that helps us identify them. Comparing the light frequencies of locally ionized elements to the frequencies coming from distant stars that tell us if the stars themselves are approaching or receding.


Why does the planet Saturn have numerous (62) moons compared to the rest of the planets in the Solar System?

Saturn and Jupiter have many moons for quite a few reasons, one of the main ones being that they have an absolutely immense gravitational pull. During the early stages of the formation of our solar system, there would of been many planet-like objects floating around which our gas giants would have attracted. Furthermore, these planets are so far out in the solar system water would if frozen (which explains Saturn's rings of ice). Infact, we can show that the ice can form moons by looking at some of the moons of Uranus, some of them are half made of ice!

A few of the outer moons of our planets are captured asteroids. Phoebe, which is a moon of Saturn, is believed to have been a captured asteroid.

I haven't heard anything about Saturn having more moons than jupiter.


What's more, stars are so small to our eyes that they look like "point sources," meaning that each resolution element in our eyes gathers light from a much larger region of the sky than a distant star can fill.  If you were to look at the Sun from, say, Uranus, and you have good eyes, it would still hurt almost as much because you could still resolve the Sun at that distance (sharp eyes can resolve objects 1/25 the width of the Sun).  That means whatever spot on your optic nerve the sunlight is focused to would be getting the same heat that it gets now-- there would be fewer such spots if you were at Uranus, but a burning spot still hurts.

But any distance larger than 25 astronomical units or so, you would start to not be able to resolve the Sun so even that single spot on the optic nerve would be mixing in light from the dark region around the Sun.  At $10^4$ times farther away than Uranus, the distance mentioned in the other answer, it means the sunlight would get diluted by the darkness around the Sun $10^8$ times larger than the Sun, accordingly reducing the heat that spot on the optic nerve receives by $10^{-8}$.  Far enough away, you can't even see there is a star there.


Do the gases in the Earth's atmosphere affect the color of a lunar eclipse?

From the way you have formulated your question, it seems to me that you are aware of the light diffusive process named Rayleigh Scattering and why it explains the various tints and colours of the sky depending on the Sun position. Other readers can refer to https://en.m.wikipedia.org/wiki/Rayleigh_scattering.

We are used to our atmosphere and as such often discuss the RS merely on geometrical bases, as I shortly did just above.

However, the intensity of RS does not only depends on the inverse of the fourth power of wavelength, but also on the size of the scattering particles and specifically on the square of the molecular polarisability.

As such, changing the nature and composition  of the atmosphere does effect the intensity of RS. An example is SO2, which tends to intensify the yellow tint of sunsets in polluted area.

This should answer the first part of your questions. Let us move to the second part, specifically using the example atmosphere that came to your mind, chlorine as a green gas, i.e. molecular chlorine Cl2.

In this case, the gas is not colourless. Different respect to nitrogen and  oxygen, chlorine does absorb in the visible part of the electromagnetic spectrum.
It absorbs the colours opposite to green in the colours wheel, to keep things simple as possible.

In an inherently coloured atmosphere the relative importance of scattering will be less or at least variable depending on parameters such as atmosphere thickness, density etc. Let me ignore RS in such a toxic atmosphere :)

What is important now is that when you point your eyes up to the sky, the green colour you see is not that one whose propagation is hindered, but rather the one that can propagate (given that we are ignoring RS). In other words, the light will be always richer of green colour.

To make the final answer simple let us look at the moon when the only light that hit it is that coming throughout the planet atmosphere.

With Earth, the Moon is bloody as for RS has removed the blue violet part of the solar spectrum.

A hypothetical change on Earth atmosphere composition will also have a minor impact, as we have seen that the polarisability of the molecules counts.

About Chloro, the planet having chlorine atmosphere. During the eclipse,  its moon will have a greenish tint, or likely appear green. In this case the moon is lit by sunlight depleted of the wavelengths absorbed by chlorine.

To resume, the colour of the moon is always that less affected by the planet atmosphere but the mechanisms and the final result differ.

When the atmosphere is intrinsically colourless, the colour of the moon is somehow "complementary or so" respect to that of the daily planet skydome. When the atmosphere absorbs, the moon and the sky dome will tend to the same colour.

Note: I did assume that sunlight can pass the intrinsically coloured atmosphere.  Again depending on various parameters, this is of course not given. Most likely we won't see the Moon. We shall really care of our atmosphere.

Also the intrinsic properties of the moon do not change. But this is even more trivial.


How to calculate the time of Ekadashi days?

It is not clear that it can be calculated in a simple way

The length of the lunar month varies between 29.3 and 29.8 days, due to perturbations from the sun. Since ekadashi is based on the actual astronomical moon, you would need to know the length of the current synodic month in order to work out when ekadashi is.  You can predict the motion of the moon very accurately, but the calculation is beyond this answer.

However on average, each lunar month is 29.53 days, so ekadashi is the day which is 259.86 hours and 614.22 hours after each new moon. The calculation is
24 hours/day * 11/30 * 29.53 days

And New moon last occurred on 4th Feb at 21.03, you can add 29.53 days to find the next new moon etc.

The irregular motion of the moon is why the Christians choose to use an "average moon" and an "average sun" for their calculation of Easter.


Do all the objects in the universe exert force on all other objects?

Yes - this is the formula:

$$F = G\frac{m_1m_2}{d^2}$$

Using this equation, we can say that all atoms in the universe exert force upon eachother. One carbon-12 atom has a mass of $1.660538921(73)\times10^{-27} kg$. That's a crazy small mass.

Now let's say that these two atoms are 100,000,000 light years apart. That's $9.461\times10^{23} m$, which is a very long distance.

Now, if we plug these values into our equation, we get that the force is: $1.709191430132 \times 10^{-59} N$

That's a very, very small amount of force. But it's still force.


If they mention times of night then you have to figure out to what extent the time zone (BST/GMT vs the US time zones) makes any difference, but it shouldn't make more than an hour difference either way for an object that is moving sidereally with the rest of the night sky.

The main difference would be that the USA is at lower latitudes than the UK. Therefore there could be instances where an object with a small or southerly declination is better seen from the USA than the UK. Conversely, more of the sky is circumpolar from the UK.

Finally, if you are trying to see a specific event at a given time (UT), then of course it may well make a difference. The Sun could be up in the UK and down in the USA, or vice-versa.


Are stars near the ecliptic the only stars from which Earth would appear to transit the Sun?

Yes. Heller and Pudritz 2016 define an Earth transit zone (ETZ) within 0.26° of the ecliptic and a restricted ETZ (rETZ) half as wide.
They also list 82 Sun-like stars in the rETZ and within 1 kiloparsec of the Sun.
Sheikh et al. 2020 report on a search for artificial radio signals from 20 of those.