Python

I see that you are quite active in Stack Overflow but have no posts tagged with python yet, so maybe it's time to give it a try!

Run pip install skyfield
read the simple documentation: https://rhodesmill.org/skyfield/
calculate x, y, z positions in the solar system's coordinates with a single line of Python; e.g. `x, y, z = venus.at(time).position.km, or
use Topos to define a viewing location on Earth and the .altaz() method to get the position of each in the sky.

NASA JPL Horizions

This is the same basic source of information that Skyfield uses.

Go to the Horizons website at: https://ssd.jpl.nasa.gov/horizons.cgi
Follow the instructions I've written up in this answer.
You can also read the documentation there to specify topographic coordinates and get positions in the sky, and you can do that viewing from other planets and bodies as well!

Have fun!


What are the challenges for the building and data analysis of the CHIME telescope?

There are many radiotelescopes that consist of large arrays of antennas. We usually think of an array individual dishes and VLA and ALMA are two of the most commonly recognized.

But if you want to keep an eye on a huge patch of sky in order to try to catch a very rare event, you want a wide field of view more than ultrahigh spatial resolution. And if what you are looking for is super-hot science, you want to build your device quickly which means a modest budget.

So wide field, lower resolution means a smaller baseline, and CHIME is of order 100 meters on a side. but building 1000 small say 3 meter dishes or 1000's of log-periodic dipole arrays and trying to steer them all would be a mess.

So instead CHIME built four, simple, parabolic cylinders and told all those antennas to be nice to each other and share!

Each antenna picks up signals from a huge area of the cylinder. If you include all directions it's like 25x25 meter section. The antennas are spaced only about a wavelength or so apart, but that's plenty far to do interferometry with the signals in a computer.

The ArXiv preprint The CHIME Fast Radio Burst Project: System Overview goes into great detail about the design and optimization of the antenna, considering what it is supposed to do and the necessity to make it straightforward to build an have a reasonable budget.

  Figure 1. Photograph of the CHIME telescope on 15 September, 2016, looking North-West. The shipping containers housing the X-Engine and CHIME/FRB backend can be seen adjacent to the right-most cylinder. The receiver huts containing the F-Engine are beneath the reflectors and cannot be seen here. The DRAO Synthesis Telescope (Kothes et al. 2010) can be seen in the background. See Table 1 for detailed properties
  of CHIME.

The Field of View in the North-South direction (the axis of the parabolic cylinders) is about 110 degrees, but in the East-West direction it's of course very narrow, only 2.5 to 1.3 degrees.

I've borrowed the following text, images and links from my question How does Canada's radiotelescope CHIME manage polarization-dependent apodization of the fast f/0.25 optics?

Here's a picture of some of the antennas, looking down into the trough:

below x2: from The cloverleaf antenna: A compact wide-bandwidth dual-polarization feed for CHIME

Fig. 1. At left, the shape of each petal consists of two perpendicular straight
lines, two 45 degree circular arcs with radius R and one half an ellipse. W is
the major axis of the ellipse and L is the length from the intersection of the
straight sides to the outer edge of the ellipse. The shape is illustrated here for
the adopted values of gap, R, L, and W. Each of the four tabs shown at the
centre is connected to one side of a vertical microstrip transmission line and
in each case the full width of the adjacent petal is connected to the other lead.
At right, CST simulated currents for one linear polarization at 600MHz are
shown. Note the small asymmetry in the curent distribution near the centre
becasue of the tab geometry.

Fig. 3. A linear array of eight cloverleaf antennas installed at the focal line of
the CHIME Pathfinder at the Dominion Radio Astrophysical Observatory in Penticton, BC, Canada. The picture is taken through the wire mesh reflective surface (mesh spacing 19 mm) illustrating a photonsview of the antennas and ground plane. Notice that each feed has an image-feed in the ground plane, 1/2 λ away at the passband centre frequency. Notice also the four slots cut to remove dielectric material from the gaps between the petals.

below: From https://chime-experiment.ca/instrument

  CHIME consists of four adjacent 20m x 100m cylindrical reflectors oriented north-south. The focal axis of each cylinder is lined with 256 dual-polarization antennas, each of which receives radiation from a large swath of sky that nearly stretches from the northern horizon to the southern horizon. This gives CHIME its enormous field of view. The CHIME antennas are custom-designed to have good sensitivity from 400 to 800 MHz, in both linear polarizations. This gives CHIME its large frequency coverage. Signals from the antennas are amplified in two stages, using low-noise amplifiers developed by the cell-phone industry. This novel application of consumer technology makes CHIME affordable. The 2048 signals from these inputs (256 antennas x 2 polarizations x 4 cylinders) are fed to the F-Engine for the initial stage of digital processing.


Which is the least massive object?

Quoting Wikipedia,

  In the hierarchical, restricted three-body problem, it is assumed that the satellite has negligible mass compared with the other two bodies (the "primary" and the "perturber"), . . .

This is the case studied in Kozai (1962), specifically, the case of asteroids being perturbed by Jupiter. While not massless, the difference in mass is large enough that the asteroid's mass is negligible.

  How does the three-body system evolve? . . . Whose inclination and whose eccentricity?

Wikipedia is again rather direct in this, stating that the conserved quantity $L_z$ depends on the satellite's orbit's eccentricity and inclination:
$$L_z=\sqrt{1-e^2}\cos i$$
As the satellite's mass is negligible, it will not have any significant effect on its perturber.

  Can it become circular, eccentric, circular, eccentric?

This is basically asking if the eccentricity (and therefore inclination) can be described by some periodic function. Again, this is given in the Wikipedia article. A slightly different but just as simple xpresison is given in Takeda & Rasio (2005):
$$\text{Kozai Period}=P_{\text{perturbed}}\left(\frac{m_{\text{star}}+m_{\text{perturbed}}}{m_{\text{perturber}}}\right)\left(\frac{a_{\text{perturber}}}{a_{\text{perturbed}}}\right)^3(1-e_{\text{perturber}}^2)^{3/2}$$
In the approximation discussed above, in cases of extreme mass difference, $m_{\text{perturbed}}\to0$.

  The inner binary will lose energy during the whole process, right?

The whole thing is periodic, so no energy is lost.


I just want 28.88 deg = 28 deg 53' confirmed.

Yes, 0.88 degrees is equal to 53 arc minutes, so 28.88 deg = 28 deg 53'.

But the Yahoo source is off in its value. According to this reference, the maximum declination is closer to 28° 44' (28.73°).

Title: Extreme declinations of the moon.
Authors: Können, G. P. & Meeus, J.
Journal: Journal of the British Astronomical Association, Vol. 82, p. 192 - 193
Bibliographic Code: 1972JBAA...82..192K
Available on NASA Astrophysics Data System.

That reference gives a declination of 28° 43' 21" on Sep 15 2006. Using the Lunar Polynomials from the Astronomical Almanac Online, which permit you to calculate the declination to a far greater precision than you can know the time, I calculate a value of 28° 43' 22". Thus, I trust the article.

28° 43' 22" = 28.723 degrees (Sep 15, 2006). The maximum between 1920 and 2050 is 28° 44' 11" = 28.736 degrees (March 15, 1932).


Are mass extinction events more likely during meteor showers / passing through comet debris?

The American Meteor Society FAQ page notes that about 95% of observed meteors are probably of cometary origin. But they're apparently so fragile that they're only about 1/3 of observed fireballs, and 0% of known meteorites.

Also, since the Earth passes through streams of cometary debris several times a year, and yet there's been only one plausible mass extinction due to cometary or asteroid impact in the last few hundred million years (the one that did in the dinosaurs), it appears to be extremely unlikely for cometary debris to be a meaningful danger.


From this NASA FAQ page about Europa:

Theory and observation indicate that Europa's icy shell is around 15 to 25 kilometers (10 to 15 miles) thick, overlying an ocean approximately 60-150 kilometers (40 to 100 miles) deep. Support for this hypothesis comes from observations of pits, domes, and spots on Europa's surface. The size and spacing of the features suggests that they are due to churning within the ice shell, and theory suggests that such churning (called convection) can occur only if the shell is between greater than about 15 kilometers thick. Measurements of the height of domes on Europa (up to a kilometer or just over half a mile high) also suggest that the ice shell must be fairly thick for the domes to be so tall.

However, it seems like there has been a some disagreement on this matter:

Some scientists have argued that the ice shell might be thinner, only a few kilometers thick.

But, the text doesn't seem to elaborate on any exact figures, so I'll look for some more sources.
Here's a paper that looks at Europa's ice shell:
Constraining the thickness of Europa’s water–ice shell: Insights from tidal dissipation and conductive cooling
From the abstract:

...when considering the present thickness of Europa’s ice shell, if tidal heating is included at a global dissipation rate of ∼1 TW, the shell is found to be, on average, approximately 28 km thick. However, if this dissipative heating is solely restricted to the shell, the local rate of heating may vary significantly due to crustal compositional heterogeneities and it is shown that this process may, in turn, produce thermal maxima in the crust, which could lead to local melting and structural instabilities...

So, with the combined information from these sources, I would stand by a figure of somewhere between 15-28 km. The reason I don't go any higher is because I don't see an error bound or range of values in the paper's abstract - it just says "approximately 28 km thick" - and the full paper is unfortunately behind a paywall. As the paper's abstract says, there are local heating phenomena that could occur and cause the crust to vary in different places, which could explain why the figure isn't that precise. The reason I don't go any lower, on the other hand, is because that NASA FAQ didn't specify any exact values for their mention of "few kilometers thick", and didn't provide much explanation as to why those said scientists supported that idea.


Would space-faring vehicles in interstellar space be in pitch darkness, show reflections from the nearest star or be bathed in light completely?

The brightness of the spaceship follows (almost) the inverse square law, meaning twice the distance from the star, the brightness will be a quater.
In the middle of nowhere, but within a galaxy, it would look like in a moonless, and cloudless night, far away from any artificial light source.
It wouldn't be pitchblack, but much too dark to read a newspaper. You would see the stars in the background, and the fuzzy band of the Milky Way. The starship would occult the stars; that way you would see the black silhouette of the starship.

At 630.1-fold the distance of the Earth from the Sun (about 3.65 light-days), it would look like at full moon (derived from apparent magnitudes of full moon and Sun)

But maybe the starship is illuminated artificially.


is this really true that speed of light is same as we look from different parts of universe?

According to theory of relativity . Light speed is same for every observer . How can we prove it is right?

By doing experiments. Pela mentioned the Michelson-Morley experiment. That's what got Einstein thinking. That's why he came up with the theory of relativity. However other experiments will prove that the speed of light is not the same for every observer. It's the same for every inertial observer, because of the wave nature of matter. But see Is The Speed of Light Everywhere the Same? on the PhysicsFAQ website. It's written by Don Koks, a physicist in Adelaide. He says this:

"That the speed of light depends on position when measured by a non-inertial observer is a fact routinely used by laser gyroscopes that form the core of some inertial navigation systems.  These gyroscopes send light around a closed loop, and if the loop rotates, an observer riding on the loop will measure light to travel more slowly when it traverses the loop in one direction than when it traverses the loop in the opposite direction.  This is known as the Sagnac Effect.  The gyroscope does employ such an observer: it is the electronics that sits within the gyro.  This electronic observer detects the difference in those light speeds, and attributes that difference to the gyro's not being inertial: it is accelerating within some inertial frame.  That measurement of an acceleration allows the body's orientation to be calculated, which keeps it on track and in the right position as it flies".

Your measurement of the speed of light is not totally constant, it depends on your acceleration. If you were in a room in an accelerating rocket, you would be what Koks called the "global" observer, and you would say that light at the ceiling travels faster than light down near the floor.

  But how can we know that this applies to more than a local region?

By understanding general relativity. See the general relativity section of the PhysicsFAQ article:

"Einstein talked about the speed of light changing in his new theory.  In the English translation of his 1920 book "Relativity: the special and general theory" he wrote: "according to the general theory of relativity, the law of the constancy of the velocity [—Either Einstein or his translator obviously mean "speed" here, since velocity (a vector) is not in keeping with the rest of his sentence. People often say "velocity" when they clearly mean "speed".] of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [...] cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity [speed]  of propagation of light varies with position." This difference in speeds is precisely that referred to above by ceiling and floor observers."

  What if light travelled to us from a distance region where the speed of light was different or where the relativistic principle did not apply?

You wouldn't measure any difference, because you measure the speed of light here, not somewhere else.


How do astronomers distinguish the spectral lines?

The lines show as either peaks above the black body emission (for emission lines) or troughs below the black body emission (for absorption lines). Though stars more often have absorption lines and nebulae emission.


Is there a variation of ISO 8601 for julian calendar dates?

Astronomers use "Julian day numbers" to indicate time. The Julian day number is the number of days since Noon GMT on January 1st 4713BC (in the Julian calendar). A time is then a single real number. For example, 00:30:00.0 UT January 1, 2013 (Gregorian), is 2456293.520833 (wikipedia)

So except to specify the epoch, the Julian calendar is not in use by astronomers to specify dates. There is no international standard equivalent to iso8601 for representing dates in the Julian calendar.

Ancient dates are often expressed as years before present, where "present" is 1950. 1950 is chosen as it preceeds most of the atmospheric nuclear testing that dumped massive amounts of C14 into the atmosphere and made carbon dating only valid for years before 1950.


How big does the moon have to be to have the Earth tidally locked to it

There's a lot of variables for the timing, but according to the formula from Wikipedia, the time ratio is inversely proportional to the square of the mass.

If our Moon will tidally lock the Earth to it in 50 billion years, a moon 7 times the mass should tidally lock the Earth, everything else being equal, about 7^2 or 49 times faster.  About 1 billion years.

There's numerous variables, such as how large the oceans are on your planet, the density, the radius, how long the surface stays molten after the impact, the initial rotation rate and initial distance.  (that one is proportional to the 6th power).

If somebody wants to give a better and more numerical explanation, please feel free, but this should get you started.


Well the angular resolution of a telescope depends on two things: 1) The wavelength of light you're looking in, and 2) the diameter of the primary lens/mirror.

$$ \theta = 1.22 \frac{\lambda}{D}$$

where $\lambda$ is the wavelength of light and $D$ is the diameter of the lens.

Light collecting power is simply how many photons you can collect in a particular wavelength in a given amount of time. Essentially, the bigger the bucket (or lens/mirror), the more rain you can catch. You can sort of get around light collecting power by increasing your exposure time. But, if the size of the object you're trying to observe is smaller than your telescope's resolving power, I don't believe there is really much you can do about it. Being limited by the resolution of your telescope is called being diffraction limited, but most telescopes on Earth are usually seeing limited due to the atmosphere's smearing effects.


Does one need to take into account finite gravity speed in N-body simulations?

If you're asking whether it's sufficient to use a retarded (time-delayed) positions to calculate gravitational forces, then no, that would be much worse than Newtonian gravity. For example, that would predict that the Earth should spiral into the Sun on the order of about 400 years. See also answers to

Besides retarded gravitation, anything else to worry about when calculating MU69's orbit from scratch?
Is the influence of gravity instantaneous?

  Most small-scale N-body simulations (e.g., planetary systems, solar system, stellar clusters, ...) use classical Newtonian gravity.

This is true, but some simulations use relativistic corrections.

The post-Newtonian expansion works in order $\epsilon\sim v^2\sim U$, where $U$ is (the negative of) Newtonian potential, and including the first-order post-Newtonian correction to the N-body equations of motion forms the Einstein–Infeld–Hoffmann equations. Because they contain $3$-body terms, the simulation would now be $\mathcal{O}(N^3)$ instead of the Newtonian simulation $\mathcal{O}(N^2)$.

The post-Newtonian three-body equations are known to 2PN, while the two-body problem to 3.5PN. Effects beyond the first order can be important in e.g. binary simulations because gravitational radiation losses are seen at 2.5PN order. However, they are not translatable to an N-body problem.

A technique of making the EIH equations more tractable by making a second expansion that depends on the problem at hand can be found in Will (2014), which is applicable when most the relativistic dynamics are dominated by a few contribution (e.g., a supermassive central black hole in galaxy simulations).

  Most large-scale N-body simulations (e.g., galaxy clusters, ...) use general relativity.

As Walter says, most of them use Newtonian gravity on a expanding background. Not necessarily directly computer particle-particle Newtonian force, but perhaps modeling Poisson's equation through mesh methods, or perhaps some combination thereof (e.g., direct force calculations for close particles, mesh methods for larger scales).

  My question: if I were to make a large-scale N-body simulation using Newtonian gravity, would it be enough to "correct" that to first-order by modifying the equations of motion to take into account the fact that gravitational fields propagate at the speed of light?

If you discount cosmological expansion, it will simply be wrong. However, first-order corrections to the Newtonian gravity part of simulations do exist in the post-Friedmann formalism. I'm not really familiar with PFF, but according to Bruni et al. (2014), the leading-order correction to Newtonian gravity introduces a vector potential in the $g_{0i}$ term in the metric that can produce weak lensing effects but doesn't affect matter dynamics. Which is entirely sensible because that kind of gravitomagnetic effects should be suppressed by a further $v/c$ term for matter, just like magnetism.

For references on how low order GR corrections to Newtonian gravity simulations are often made, see answers to How to calculate the planets and moons beyond Newtons's gravitational force?

References:

Will, C. M. "Incorporating post-Newtonian effects in N-body dynamics". Phys. Rev. D 89, 044043 (2014) [arXiv:1312.1289]
Bruni, M., Thomas, D. B., Wands, D. "Computing general-relativistic effects from Newtonian N-body simulations: Frame dragging in the post-Friedmann approach." Phys. Rev. D 89, 044010 (2014) [arXiv:1306.1562]


Why can't stars be multicolored like gas giants?

Just rotation is the wrong tree to bark up on.
You see color variations on gas giants due to differences in composition, i.e. ammonia vs. sulfuric acid clouds on Jupiter, which are transported differently on the rotating planet in the up/downwelling bands.
Spots on stars originate due to very different physics. At the temperatures which are prevalent on stellar surfaces, molecules are mostly dissociated and ionized, we call this state a plasma, so no more colour effects due to ammonia or others. Star spots are concentrations of the local magnetic field, which is coupled to the plasma dynamics. The magnetic field pushes the gas aside and lets the surface cool locally. This makes dark spots that you see on stars like our sun.
There is a transition region in terms of surface temperature of 2000-3000K, at the masses of brown dwarves. Those failed stars seem to exhibit dark bands on their surfaces, which are thought to be due to absorption from exotic, high-temperature molecules like TiO and VO (Titanium and Vanadium oxide).


Is Sahara desert a good place to build radio/optical telescopes?

No, the Sahara isn't a good place to build telescopes.
The Atacama desert is used because it is at high altitude, which means that there is less atmosphere to get in the way. Other telescopes are located on mountaintops for the same reason.
The Sahara is mostly at sea level. It's also very hot, so you get lots of turbulence due to rising air, which distorts the image.
The Sahara is also undesirable for other reasons: no infrastructure, unstable regimes, lots of erosion due to sand storms.


How many GAIA Data Release 2 files are there?

There are 61,234 data files in the directory. There are three additional (non-data) files:

MD5SUM.txt
_citation.txt
_disclaimer.txt

If you include those, the total number is 61,237.


How can we tell that a short-period binary is tidally locked?

The phenomenon of tidal locking occurs primarily in short period binary systems with convective envelopes. The tides raise bulges in the stars and these interact with convective motions to dissipate energy and to synchronise the orbital and rotational periods of stars and circularise their orbits. The process is much less efficient in stars with radiative envelopes.

How can we tell observationally? Well, we have to show that the rotation period of a star is the same as its orbital period. The orbital period is easily found, either by measuring eclipses, in the case of binary systems with a high inclination, or by measuring the motions of the star(s) using the doppler shift in their spectra.

Measuring rotation periods is also not so difficult. Fast-rotating stars with convective envelopes have magnetic activity and starspots (analogous to sunspots, but covering a larger area). As the star rotates then the light from the star goes up and down depending on what starspots are on the visible surface. This modulation gives the rotation period. In eclipsing binaries it can be difficult to separate out spots from eclipses, but here one can use the known orbital inclination and known radii of the stars to see whether their projected rotation velocities (measured from the widths of spectral absorption lines) matches the predictions from a tidally locked orbit.

There are many known tidally locked bnaries. Any low-mass binary with an orbital period of less than a few days is almost certainly tidally locked, but the situation at orbital periods greater than about about 6-8 days becomes more complicated (e.g. Meibom et al. 2006). The plot below is taken from a review by Mazeh (2008), showing the rotation vs orbital periods from some small, but well-defined samples.

The absolutely classic, must-read, theoretical work on this is by Zahn (1989). The timescale for synchronisation depends on structural properties of the stars, their moments of inertia, the binary mass ratio and the ratio of their separation to their radii to the power of 6. It is this latter property that means tidal locking is nearly bimodal; binaries with periods shorter than some threshold (about 6 days) are usually tidally locked, but those with slightly longer periods almost never are.


Why are there no ISM clouds with temperatures between 100 and 6,000K?

Let $n$, $T$, and $x_i$ be the number density of hydrogen, the temperature of the gas, and $n_i/n$, where $n_i$ is the number density of the $i$th component of the interstellar medium. We can then write the criteria for thermal equilibrium as
$$n^2\Lambda(n,T,x_i)-n\Gamma(n,T,x_i)\equiv n^2\mathcal{L}=0$$
where $\Lambda$ and $\Gamma$ and the heating and cooling functions, respectively, and $\mathcal{L}$ is defined through those functions and $n$. If the equilibrium is unstable,
$$\left(\frac{\partial\mathcal{L}}{\partial S}\right)<0$$
for entropy $S$. This leads to different instability conditions, termed the isochoric and isobaric instabilities (Field (1965), $\text{Eq } 4a,4b$). These can be determined from the temperature, pressure, and density of the gas (also assuming that the gas can be approximated as an ideal gas).

In general, $\Lambda$ and $\Gamma$ are complicated to determine, although combinations of power-law and exponentially decaying factors may often be sufficient. An example curve that seems to pop up a lot as an example was calculated by Dalgarno & McCray (1972), Figure 2:

More accurate measurements have been made since then, but the general shape is still applicable. The warm neutral medium occupies the area near the sharp change at around $\sim10,000\text{ K}$, and the cold neutral medium occupies the area in the left of the diagram. Another way to visualize this is on a $\log P/\log n$ diagram like this one (from these slides, annotated from Wolfire et al. (1995)):

In reality, the two-phase model is an oversimplification, and the ISM has more distinct components. However, the isochoric/isobaric instabilities still limit the range in which clouds can exist in stable equilibria, and explain the dearth of gas in the relevant temperature range.

Let me elaborate on the terms isobaric and isochoric. In thermodynamics, it is sometimes convenient to assume that some thermodynamic variables remain constant in a certain situation. Isothermal processes occur at constant temperature; likewise, isobaric processes occur at constant pressure and isochoric processes occur at constant volume.

The equations for the two instabilities are
$$\left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\rho}<0\tag{Isochoric}$$
$$\left(\frac{\partial\mathcal{L}}{\partial T}\right)_{p}=\left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\rho}-\frac{\rho_0}{T_0}\left(\frac{\partial\mathcal{L}}{\partial\rho}\right)_T<0\tag{Isobaric}$$
In the first, we assume that the cloud is at constant volume, and since the total amount of matter in the system is constant, the (mean) density must also be constant. In the second case, we assume that the cloud is at constant pressure. Perturbations that lead to instabilities thus arise from perturbations of other thermodynamic variables.

As a final note on notation,
$$\left(\frac{\partial\mathcal{L}}{\partial A}\right)_B$$
means that we take the partial derivative of $\mathcal{L}$ with respect to $A$ while keeping $B$ constant. It's a common thermodynamical convention.


Stars do not get hot because of nuclear fusion, they become hot enough to sustain nuclear fusion and this process maintains their temperatures. Nuclear fusion actually stops a star getting hotter.

Protostars (before nuclear fusion) get hot because of a well known statistical relationship between the gravitational potential energy of a gas and the internal kinetic energy of the particles that make up the gas. [In an ideal gas, the kinetic energy of the particles is directly proportional to the temperature of the gas.] This is known as the virial theorem, which says that twice the summed kinetic energy of particles ($K$) plus the gravitational potential energy ($\Omega$, which is a negative quantity for a bound object) equals zero.
$$ 2K + \Omega = 0$$

Now you can write down the total energy of the system as
$$ E_{tot} = K + \Omega$$
and hence from the virial theorem that
$$E_{tot} = \frac{\Omega}{2},$$
which is also negative.

If we now remove energy from the system, for instance by allowing the gas to radiate away energy, such that $\Delta E_{tot}$ is negative, then we see that
$$\Delta E_{tot} = \frac{1}{2} \Delta \Omega$$

So $\Omega$ becomes more negative - which is another way of saying that the protostar attains a more collapsed configuration.

Oddly, at the same time, we can use the virial theorem to see that
$$ \Delta K = -\frac{1}{2} \Delta \Omega = -\Delta E_{tot}$$
is positive. i.e. the kinetic energies of particles in the gas (and hence their temperatures) actually become hotter. In other words, the gas has a negative heat capacity. But a hotter temperature usually means more radiation is produced and if the energy losses continue, then so does the collapse.

This process is ultimately arrested in a star by the onset of nuclear fusion. This replaces the radiative losses with nuclear energy and the star attains a quasi-equilibrium that lasts as long as it has nuclear fuel to burn.


What is in the brightest area of the night sky?

All quoted text in this answer is from image captions in the Wikipedia article on the Milky Way.

  360-degree panorama view of the Milky Way (an assembled mosaic of
  photographs) by ESO

  This magnificent 360-degree panoramic image, covering the entire
  southern and northern celestial sphere, reveals the cosmic landscape
  that surrounds our tiny blue planet. This gorgeous starscape serves as
  the first of three extremely high-resolution images featured in the
  GigaGalaxy Zoom project, launched by ESO within the framework of the
  International Year of Astronomy 2009 (IYA2009). The plane of our Milky
  Way Galaxy, which we see edge-on from our perspective on Earth, cuts a
  luminous swath across the image. The projection used in GigaGalaxy
  Zoom place the viewer in front of our Galaxy with the Galactic Plane
  running horizontally through the image — almost as if we were looking
  at the Milky Way from the outside. From this vantage point, the
  general components of our spiral galaxy come clearly into view,
  including its disc, marbled with both dark and glowing nebulae, which
  harbours bright, young stars, as well as the Galaxy’s central bulge
  and its satellite galaxies. As filming extended over several months,
  objects from the Solar System came and went through the star fields,
  with bright planets such as Venus and Jupiter. For copyright reasons,
  we cannot provide here the full 800-million-pixel original image,
  which can be requested from Serge Brunier. The high resolution image
  provided here contains 18 million pixels.

  Here is a schematic map of our POV in the Milky Way galaxy.

  Observed (normal lines) and extrapolated (dotted lines) structure of
  the spiral arms. The gray lines radiating from the Sun's position
  (upper center) list the three-letter abbreviations of the
  corresponding constellations.

  A "God's view" map of Milky Way as seen from far Galactic North (in
  Coma Berenices). The star-like lines center in a yellow dot
  representing the position of Sun. The spokes of that "star" are marked
  with constellation abbreviations, "Cas" for "Cassiopeia", etc. The
  spiral arms are colored differently in order to highlight what
  structure belongs to which arm. H II regions are marked as dots
  colored in the same color as their spiral arm. They come in three
  sizes, measured by the excitation parameter U: small - U > 200 pc cm-2
  medium - 200 > U > 110 pc cm-2 large - 110 > U > 70 pc cm-2

It turns out we are in an arm -- the Orion-Cygnus Arm. The much brighter part of the Milky Way from our POV is in the direction of the galactic center, but the actual nucleus around the supermassive black hole is obscured by dust. If it was visible, it would be quite bright. What we are seeing that is bright is mostly the pseudobulge, or galactic bar formation, in the middle of the galaxy. We are looking at the galactic bar almost end-on, so it resembles a sphere from our POV.


What is this bright "glow" in the center of galaxies?

If this is the case, then we should not see a "light" coming out from the center since light get's sucked in black holes.

You are overestimating the size and the capabilities of a supermassive black hole. Contrary to pop sci portrayals of black holes, black holes are not giant vacuum cleaners in space that suck up anything and everything close by. While the supermassive black hole at the center of the Milky Way is indeed very massive (about four million times the mass of our Sun), it isn't very large physically. It's less than a couple dozen solar diameters across. It also isn't that hungry, gobbling up perhaps the equivalent of four or so Earth masses over the course of a year.

On the other hand, the central bulge of a spiral galaxy contains several million stars in a fairly small volume. That central bulge is what you are seeing in those images. The supermassive black holes near the centers of those bulges gobbles only a tiny, tiny fraction of the light emitted by those millions of stars.


Did the Babylonians believe in the Heliocentric version?

In 150 BC a babylonian heliocentric model is documented. Heliocentric models were not uncommon historically, but Ptolemaios geocentric model was actually better at predicting the movements of the planets until Kepler came up with the idea of elliptic rather than circular orbits in the early 1600s. In the 1500s Copernicus hesitated to publish his heliocentric ideas because they did not improve predictions.


What is a hypernova and have we observed any?

A hypernova is just a really, really big supernova.  UMass has a (tacky designed) web page that explains it.

  A hypernova explosion typically has a mechanical energy output of ~ 10^53 ergs, or about a factor of 100 greater than a supernova.

Regarding aftereffect, the page says this:

  The age of the hypernova remnant NGC5471B is about 30 thousand years, while the age of MF83 is about 1 million years.

As best as I am able to determine, we have not had a chance to observe any hypernova candidates in the modern era, although APOD featured the same one as mentioned by UMass:

According to New Scientist, research suggest the upper limit to a star is about 150 solar masses. These are the stars that create the infamous supernovas.his is as derived from the Eddington Limit.  Although R136a1 is an example of a star at nearly double that.  It is believed that these monsters are what are responsible for hypernovas. There is much to be learned from these stellar monsters!


Could a near collision with small black hole account for the effects of the 9th planet?

To quote from Batygin and Brown (2016)

Trujillo & Sheppard (2014) point out that the Kozai mechanism allows libration about both ω = 0 as well as ω = 180, and the lack of ω ~ 180 objects suggests that some additional process originally caused the objects to obtain ω ~ 0. To this end, they invoke a strong stellar encounter to generate the desired configuration. Recent work (Jílková et al. 2015) shows how such an encounter could, in principle, lead to initial conditions that would be compatible with this narrative. Perhaps a greater difficulty lies in that the dynamical effects of such a massive perturber might have already been visible in the inner solar system.

So the short answer is that it is unlikely a single encounter with a massive object could cause the perturbations without leaving other traces in the dynamics of the solar system. Furthermore, stellar encounters are rare enough as it is - encounters with black holes should be even rarer since stellar evolution and the critical mass requirement of black hole formation dictate there should be far fewer black holes than stars in existence.
To me that means favoring an encounter with a black hole to be responsible for the "unexpected clustering in their respective arguments of perihelion" of Kuiper belt objects over other theories would be a violation of Ockham's Razor.
All that being said, due to the chaotic nature of orbital dynamics and the relatively short Lyapunov time of Mercury in particular, it is very difficult to rule something like that out. It may be the case that an encounter occurred long enough ago that the only remnant is in the perturbations of long-period Kuiper Belt objects. And that all other traces of its influence on the inner solar system have been obscured by its characteristic emptiness - it could in fact have even been the cause of that unusual emptiness.
It should be noted of course that this would not have been a 'small black hole' (also sometimes referred to as micro black holes) as has recently been theorized in place of a 9th planet, but a stellar black hole (roughly 5 or more times as massive as our sun).


Have Pluto or any of its moons been mapped?

Yes, but they're not very good and they're amazing.

The Wikipedia article for Pluto shows a low-resolution map of the surface, generated from Hubble images:

And the Wikipedia article for Pluto's largest moon Charon shows a low-resolution map of the Pluto-facing side of Charon (not to scale):

Larger image here.

Only the Pluto-facing side is shown because the map was generated from brightness variations during Pluto–Charon occultations.

Maps with similar resolution would show no details of any of the other moons, which are much smaller.

Expect much better images in a couple of years, likely including moons that haven't yet been discovered.

UPDATE : Thanks to user HDE 226868 for reminding me: it's been a couple of years, and we have those much better images now. Here's one from New Horizons via Wikipedia:

Since the question was about a map of Pluto, here you go. You can see the much lower resolution on the opposite hemisphere, which New Horizons saw only from a greater distance (it was a flyby mission).

We also have some good images of Charon:

And here's a map of Charon:

and some lower resolution images of the other moons, Styx, Nix, Kerberos, and Hydra.

My guess that New Horizons would discover new moons turned out to be wrong.


Having some weird issues with my tracking (Advanced VX)

Looks like you have a periodic error issue. The worm period on the AVX is about 9mins I think. The issue has a period of 90sec * 6 = 540sec (9min).


Why do Earth and moon move apart but binary black holes move closer?

Here is how the tides move the moon away from the Earth:

The moon orbits the earth, and there is a difference in gravitational force between the the side of the Earth nearest the moon, and the side far from the moon.

This difference in force tends to pull the Earth into a oval shape with its long axis pointing towards the moon.

But the Earth is also spinning, and this spinning moves the axis of the oval forward, so the oval doesn't point towards the moon, but a little ahead of it. So there is a bump on the Earth, and it is permanently a little in front of the moon. This bump has mass and it pulls the moon towards it, So the moon is being pulled forwards. The moon pulls the Earth back. So the Earth's spin is slowed, but the moon gains energy and moves a little further from the Earth.

The moon has also been slowed down to the extent that the same side always faces Earth, and it will remain locked like this.

Gravitational waves, on the other hand, imply an emmision of energy from the binary system, and as energy is lost, the black holes spiral in.

There are no tides on black holes because there is nothing there to be pulled into an oval shape. The event horizon is not a solid surface. The mass of a black hole is concentrated entirely at the singularity, there is no structure to be deformed into a bump.


Clearing the Neighbourhood: Definition of 'Orbital Region'?

The IAU gives no definition, it leaves the term somewhat vague.

Various calculations of the "planetariness" of solar system bodies use slightly different notions of the "neighbourhood"

Soter's $\mu$ considers bodies to be in the neighbourhood if they share a common radial distance (ie the orbits cross each other) and the orbital periods differ by less than an order of magnitude

Margot's $\Pi$ uses $2\sqrt 3$ times the Hill radius.

Stern and Levinson's $\Lambda$ in the original paper consider the neighbourhood to be bodies that cross the orbit of the planet, while not in an orbital resonance, to be in its neighbourhood, in a somewhat complex way that requires knowledge of the relative inclinations and and orbital velocities of bodies that cross orbits.

Changing the details of these values doesn't change the conclusion of the calculations: There is a clear and measurable difference between bodies like Mars and Neptune that are (to use Stern and Levinson's term) uberplanets and Ceres and Pluto that are unterplanets (or dwarf planets as termed by the IAU).


Why are distant galaxies not visible in the observable Universe?

Quick answer: Because they didn't entered our event horizon. Some never will. And some will move out of our event horizon - their last photons that'll be received here being sent right now.

Let's do some fact checks first:

  [...]the galaxies that are now at the 'edge' (not visible
  theoretically) must have been (at some point in time) at place around
  where the Earth is at now[...]

According to current calculations, the first galaxies may have formed around 200 million years after the Big Bang - older estimates went with the 400-500 MY range. For a long, long while, there were no stars to be seen. So if you go back in time you won't be seeing the same structures we see today.

Second, and that may be an awkward mental experiment, nothing else was occupying our place other than ourselves. I'll ask you to excuse the cliché, but the old balloon example is very apt to explain this:

As the universe expands, the distance between celestial bodies increases. Now, here's a way to put it: space is being generated in between the objects.

And not only that - while there's a limit on how fast you can move on the balloon's surface, it doesn't apply to the amount of space being generated.

As a direct consequence there's a bubble around us that basically works in the very same way as a black hole's event horizon does:

In a black hole, gravity is so strong that its pull on photons exceeds C (the speed of light, or 299,792,458 m/s);
On a fringe object, the amount of space generated by the expansion, per second, may exceed C; its photons will never reach Earth.

And it seems that we live in a accelerated universe - that is, the spatial acceleration ratio is actually going up. If that's correct, some fringe objects will  slip away form our event horizon, disappearing (from our point of view) into a sample of the heat death of the universe.

Scary, huh?


Why do these astronomers say that "Carbon chains with more than nine atoms are unstable"?

Unlike the saturated hydrocarbons in kerosene, carbynes are unsaturated carbon chains with alternating single and triple bonds.
Molecules containing such chains are called polyynes, e.g. the short cyanopolyyne HC5N:

H−C≡C−C≡C−C≡N

Those carbon atoms readily interact, and long chains (if they form; see comments) are more likely to crosslink or form cycles than to remain linear.
Tarakeshwar et al. suggest that iron clusters in "pseudocarbynes" inhibit this by bonding to some of the carbons.

Loomis et al. 2016 looked for cyanopolyynes in radio spectra of Taurus molecular cloud 1.
They got a good HC9N signal but did not detect the HC11N lines Travers et al. 1996 observed in the lab.


Why is blue light extincted more strongly than red?

There are two effects causing this:

The relevant quantity for determining whether or not a photon interacts with a particle is the ratio
$$
x \equiv \frac{2\pi r}{\lambda},
$$
where $r$ is a size of the scattering agent.

Rayleigh scattering

When $x\ll 1$, we're in the Rayleigh regime where the wavelength is so long compared to the particle that the electric field of the photon causes charges within the particle to oscillate with the same frequency, effectively resulting in dipole scattering. This has a very strong wavelength dependence, with the probability of interaction begin proportional to $\lambda^{-4}$.

Mie scattering

When $x \sim 1$, we're in the Mie regime, where the probability of interaction is largest. So if you had an ensemble of randomly sized particles with an equal amount of any size, and you shone a beam of light consisting of photons with the same distribution of wavelengths, you would scatter all photons with equal probability (unless of course your distribution extended to very small values of $x$).

However, as with many other things in the Universe — e.g. galaxies, stars, cities, lakes, and wealths — dust grains are not equally distributed in size, but has a distribution which declines as a power law from small to large size. In the case of dust grains, this power is very steep, with a power law slope of roughly $-3.5$; that is, the density $n$ of dust grains as a function of size drops as
$$
n(r) \propto r^{-3.5}.
$$

Hence, there are many more small dust grains available, casing the blue photons to be scattered much more often.

Larger particles / smaller wavelengths

When $x>1$, but less than $\simeq1.8$, the probability of scattering declines, meaning that in this narrow window, red light is scattered more easily than blue. Coincidentally, these values correspond to optical light and the dust on Mars, leading to blue sunsets, as discussed in this answer.

When $x\gg 1$ — i.e. when the particles become very large compared to the wavelength — we have "geometric scattering", all light is scattered equally. In astronomy, we call this "gray dust", and would be caused e.g. by smalles pebbles.

Scattering vs. absorption

The albedo $A$ of the dust grains — i.e. the probability that an interaction result in a scatter rather than an absorption — is also wavelength dependent, but not as strong. Typical values are of the order of $0.5$, but decreases toward zero for long wavelengths (e.g. Pei 1992)


Why does a particle feel no force at radii greater than itself?

"in a spherically symmetric distribution of matter, a particle feels no force at all from the material at greater radii, and the material at smaller radii gives exactly the force which one would get if all the material was concentrated at the central point".

This may be poorly worded. The idea is not that the distance from the center of the body must be smaller than the radius of the particle. Rather, those parts of the (uniform, spherical) body which are further away from the center than is the particle have all their respective gravitational attractions cancelled out.


How can we install a radar on radio telescopes like FAST or GMRT?

The article is quite informative. It is a summary of "a paper published in the journal Scientia Sinica Information" which appears to be Discussion on the requirements and feasibility of constructing China's near-Earth asteroids radar system. While it is written in Chinese the tables and figures alone are very informative.
They are going for a "planetary scale" or very long range system; out to 0.1 AU or 15 million km for an asteroid is ambitious, and what's really interesting it sounds like their transmitted radar beam will com from "four or five 35-metre (115-foot) diameter radio dishes" rather than one big one.
I've pasted some bits into google translate so far and it looks like they will try to have the option to use the multiple transmitting dishes coherently

(3) X 和 Ka 频段上行天线组阵技术. 在国家高技术发展计划支持下, 我国在上行天线组阵技术研 究方面也已经取得了突破, 成功实现了对地球同步静止轨道通信卫星在 C 频段 (发射频率 6 GHz) 的 3 个天线上行组阵技术验证, 达到了 80% 的合成效率. 后续需要针对频率更高的 X 和 Ka 频段上行天 线组阵, 重点开展上行链路相位延迟变化准确估计技术、时延和相位对齐的精确控制技术和大回路系 统标校技术等研究.

X and Ka band uplink antenna array technology. With the support of the national high-tech development plan, China has also made breakthroughs in the research of uplink antenna array technology, and successfully achieved the C-band communication satellites in geosynchronous geostationary orbit. (Transmit frequency 6 GHz) 3 antenna uplink array technology verification, achieved 80% synthesis efficiency. The follow-up need for higher frequency X and Ka band uplink antenna array, focus on accurate estimation of uplink phase delay changes Research on precise control technology of technology, time delay and phase alignment and large loop system calibration technology.

Rather than build one giant dish for transmitting like Goldstone Solar System Radar which is

...a large radar system used for investigating objects in the Solar system. Located in the desert near Barstow, California, it comprises a 500-kW X-band (8500 MHz) transmitter and a low-noise receiver on the 70-m DSS 14 antenna at the Goldstone Deep Space Communications Complex. It has been used to investigate Mercury, Venus, Mars, the asteroids, and moons of Jupiter and Saturn. The most comparable facility was the radar at Arecibo Observatory, until that facility collapsed. GSSR now stands alone.

...they will use several 35 m transmitting dishes and an even larger number and more widely separated array of receiving dishes.
Table 7, "China asteroid detection multi-bases radar system layout station composition" is a list of transmitting sites. Obviously not all can be coherent and they will be used in a variety of ways.

It's really hard to add a new multi hundred kilowatt transmitter to an existing very large dish like a 70 m DSN or FAST. You can read more about that in answer(s) to What is a Beam Waveguide dish and why does the Deep Space Network use them?. The big 70 m DSN dishes use the focus up between the primary and secondary; it is a real challenge to add more hardware there. The image below shows a DSN 70 m dish, for scale, the red lines in the dish itself are a safe walking path and going up each arm of the secondary reflector are stairways for people, not ants.
I think that the Chinese project is quite ambitious but it's a next step in technology, rather than a makeshift retrofit that would interrupt availability of currently very busy large dishes, and it seems that adding a transmitter to FAST is not an option. From this answer to How will the closure of the Arecibo dish impact deep space communications? (found here):

With the loss of Arecibo, Goldstone's DSS-14 now becomes the world's largest and most powerful radar dish.  (China's 500 meter FAST dish is larger, but has no transmitter and is purely passive.)  Sky and Telescope reports that "Arecibo offered 18 times the sensitivity of other existing facilities, such as NASA's Goldstone receiver."  It also states

Arecibo is also irreplaceable for scientists. Even though it’s technically the second-largest radio dish in the world (China’s Five-hundred-meter Aperture Spherical Telescope, or FAST, recently broke the record Arecibo held for decades), the observatory has unique capabilities, among them its radar. “FAST cannot do radar, it’s specifically incapable of doing active observation,” Springman explains. Because of that, FAST can’t take Arecibo’s place in planetary defense by characterizing asteroids and their orbits.

See also answers to

What will succeed the Arecibo Observatory?
Arecibo: Advantages of Giant Dish?
What are monostatic radar observations, and how will Deep Space Network's DSS-13 be used to observe asteroid 1999 WK4's flyby of Earth?

From Yahoo News China urged by scientists to create biggest radar system to save lives threatened by asteroid strike (similar/same article reprinted in several places, including intellasia.net, toysmatrix.com):

The proposed system would use four or five 35-metre (115-foot) diameter radio dishes in Kashgar, Xinjiang, to send powerful beams into space. Returned signals would be picked up by large antennas across the Chinese mainland in a number of cities, including Jiamusi, Beijing, Tianjin, Shanghai and Kunming.
The radar detection range would be expected to exceed 0.1 astronomical unit – or a tenth of the mean distance between the Earth and the sun.
The project “will fulfil the responsibility of our nation to the mankind,” said the team led by Li Haitao with the People’s Liberation Army’s Beijing Institute of Tracking and Telecommunications Technology in a paper published in the journal Scientia Sinica Informationis on Tuesday.
Continuing to rely solely on information provided by the Americans for a threat that could cause a global catastrophe would have a negative impact on China’s rising status in the international community, the researchers said.
The US had two planetary radars but one of them – the Arecibo Observatory in Puerto Rico – collapsed in December because of ageing, hurricane damage and a shortage of funding for repair and maintenance. This leaves the Goldstone Solar System Radar in a Californian desert as the last line of defence, and things could go wrong.
“A single failure of the system would leave us without an essential tool to ensure our protection. Such a failure occurred as recently as 2019-20, when the Goldstone transmitter was inoperative because of klystron (high power microwave vacuum tube) failures,” said professor Jean-Luc Margot, a planetary radar expert with the University of California, Los Angeles.
“I am glad to hear that China is considering building a radar system to study near-Earth asteroids,” he added.

From this answer to Why is the reflector on this millimeter-wave antenna spinning?:

Below are photos of one of the 70 meter Deep Space Network telescopes for talking to deep space spacecraft. This one is in the Goldstone complex. From the relative sizes in the image, the secondary mirror is close to 8 meters in diameter. Considering the size and mass of the secondary (those are stairs for humans on each leg, and the red lines in the dish are "safe paths for walking") other, more advanced techniques to scan electronically have been developed, but the concept is the same.

above: Photo credit JPMajor, creative commons CC BY-NC-SA 2.0. click for larger.

above: From commons.wikimedia.org click for larger.


How do we not know what's in the Solar System?

There could be another planet in the solar system that has not been discovered if that planet had a highly elliptical orbit with a single orbit taking many hundreds of earth years to complete. That planet could be in the part of its orbit further from sun and could stay in that area for a few hundred years, longer than we've had large telescopes.

At such a distance it would receive very little sunlight and the chances of visually detecting it would be very small. And we wouldn't know where to look either, which just compounds the problem.

Other more distant plants have been discovered but their discovery was due to the effects on the star or stars it is orbiting. We observe a planet passing in front of a star which briefly lowers the apparent magnitude of its parent star. This is a good clue of a planet in orbit. This is impossible in our solar system because an unknown planet would have to come between us and the sun. We would easily see it if that were the case.

We have observed some discrepancies in the orbits of some of the outer planets. This has led to the speculation of another planet. It is very difficult to determine the cause of these small discrepancies though. All bodies orbiting our sun are not only effected by the suns gravity but by the gravitational attraction of all other bodies (including all objects in the asteroid and Kuiper belts). This is a very intense calculation. This attraction constantly changes due to the motion of all the bodies.

Simulations have been developed to study this but must take a lot of computing time even with today's high speed computers.

I hope this helps explain the situation.


could dark energy be a force coming from the hyperspace outside the universe?

Nobody really has any idea what "dark energy" is. But there is a principle known as Occam's razor. It can be expressed in various ways, for example "When an observation can be explained in many ways, the one with fewest assumptions should be used."

To explain the properties of dark energy as being due to changing pressure of hyperspace, requires the assumption that "hyperspace" exists, that it has a property that pressurizes the universe, that this "pressure" is changing, and that affects the rate of expansion of the universe.

There are other hypothesis: a cosmological constant, or a as yet undiscovered force field. None have much observational evidence, but have the benefit of requiring fewer assumptions.

As Rob mentioned what you describe is similar to "brane cosmology", in which the universe is embedded in a higher dimensional space. These could influence our space, by controlling the scale of a cosmological constant, making it small, but non-zero. As with much of string theory, the experimenters are well behind the theorists.


What happens if a black hole is produced inside another black hole?

From the no-hair theorem, black holes can have only three properties: mass, angular momentum, and electric charge. From this we'll see the original, supermassive black hole gain the mass, angular momentum and electric charge of the ingoing star. The black holes will merge (already merged actually, since one is inside the other). The singularities will merge too (all world lines, i.e. paths a particle can take, lead to the big singularity in the supermassive black hole).

From the outside, we will see a more massive black hole with different angular momentum & electric charge, and that's it. We can't see the star go supernova because anything that happens inside the event horizon is invisible to us.


How to calculate shape of orbit when projected onto the plane of the ecliptic

You could convert the orbital elements into x,y,z and $ \dot{x}, \dot{y}, \dot{z}$ and then calculate a whole orbit. A source for converting orbital elements into cartesian coordinates is here.


What would happen if those gravitational waves were much stronger?

Based on a post by an Astrophysicist my understanding is as follows:

Gravitational waves would still be faint to detect irrespective of the size of the Black Holes. BTW aren't black holes meant to be more concentrated rather than bloated?
The distance from ground zero is what matters rather than the size of the colliding black holes and even if they were a billion times closer they would still not measure up to 1 mm.

There wouldn't be significant impact from the waves themselves rather the Black Holes are the ones that would do the damage.
For the shift to be 1 mm we would require a force enormously stronger than that of a gravitational wave.

Here's the link to the article: http://www.forbes.com/sites/briankoberlein/2016/02/13/could-gravitational-waves-ever-be-strong-enough-to-feel/#1605e7fa4aac

Disclaimer: All my points are a result of my learning from the above link rather than any native knowledge or original research that I profess to have done.


How do we know an event is a gamma-ray burst?

Yes, on the right track where the energy produced in a supernova can indicate if it was a LGRB or SGRB.

An indicative tools stems from the light curves of the remnants, both the afterglow brightness and it's region of explosion can indicatively be traced to how long the event lasted. Large gamma-ray remnants are typically associated with rapid star formation regions and are a more luminous remnant, where short gamma-rays are typically associated with little or no star formation regions and are a less luminous remnant.

It's not an exact science, as reproducing the effects and duplicating results have proven vastly difficult, however using distribution plots can definitively separate the differences and provide a theoretical model to work on.

Some minor information can be found at Wikipedia, however further reading and a more in depth view can be found at Cornel University Library. It is a bit of a read and attempts to explain the physics behind emission through mathematics, I'm not entirely sure whether is will provide the answer you are seeking.


Is Earth's moon the only one where a total eclipse of the sun covers the entire sun?

As planets get farther from the Sun, the Sun takes up a smaller part of the sky.  The Sun is about 31 arc-minutes when viewed from Earth, but just 6-7 from Jupiter and 3-4 from Saturn.   Less than 2 from Uranus and about 1 arc-minute from Neptune, not much bigger than Venus appears from Earth when Venus is visibly large in the sky and when Venus transits the Sun, viewed from Earth, its size is about 1 arc-minute, about the same size that the Sun appears from Neptune.  To get a sense of what size the Sun is from Neptune, look at a photo of a Venus transit.

Jupiter has 5 satellites capable of creating a total eclipse, the 4 Galilean satellites and the tiny but close Almathea, which is irregularly shaped, so it wouldn't be the same size as the Sun.

Saturn has 6. All are larger than the Sun when viewed from Saturn.

Uranus, because the Sun is quite small that far away, has 12 moons that can create a total eclipse.  All 12 are considerably larger than the Sun in the Uranus' sky but because of Uranus's nearly sideways axial tilt, solar eclipses are rare and only happen at the midpoints in its orbit every 42 years.

Neptune has 7, but due to its axial tilt and Triton's off-equator orbit, eclipses are rare.

Io probably casts the largest shadow.   Though Pluto and Charon cast a shadow over the largest percentage of the other object, about once every 120 years.   Two of Pluto's smaller moons, Nix and Hydra are large enough to block the Sun completely though eclipses from them are probably rare.

As far as I can tell, having checked, our Moon appears unique in our solar-system in being a nearly perfect fit over the sun.

Further reading if interested, though there's some variation in the answers, they only mention 4 of Jupiter's moons.


What's the farthest object as determined only by parallax?

The first data release from GAIA was in 2016 so fairly new. If you want to be certain you should take a look at it because as you say, GAIA will no doubt be the source of the most accurate parallax measurements in the near future. Speaking of accuracy, I believe that this is the essence of your question because it basically depends on how uncertain measurements you are willing to accept.
Basically, modern values of parallaxes are very small numbers, e.g. 0.1 milliarcsec which means a distance of

d(pc) = 1/(parallax in arc-seconds)

around 10,000 pc (one parsec is 3.26 light years). But the important point is the accuracy of these estimates. Clearly if your measurements is 0.1 milliarcsec, but the accuracy is 1 milliarcsec, it would be going out on a limb to postulate that you have measured a distance of 10,000 pc. So the real answer to your question is another question "to what accuracy?" So if you want something like 10 or 20 % accuracy you are more or less limited to our neighbourhood, say some 200 pc. If you check out ESA's website on stellar parallax, they will go up to 500 pc but it really depends on what level of accuracy your work requires.

If you check out the GAIA satellite then the aim is actually to go down to an accuracy of 24 micro-arc seconds so whatever object is measured to be farthest (with a given accuracy) is sure to replaced as the data from GAIA is analysed.

For fun, you might wanna look at this

also found at the GAIA website.. I think it provides a nice visual guideline to what distances we can expect (and it also nicely shows a ring at 500 pc - the limit of fairly accurate measurements from Hipparchos!)


Since the Universe is expanding, is it accurate to say that a galaxy is 5 billion light years away?

I'm not 100% sure if I'm understanding what your asking, but if your question could be rephrased as "how do we measure distances in an expanding universe?", then I can try to answer that.

Depending on what astronomers measure they use different distance measurements. For example the comoving distance between two objects takes into account the expansion, and so does not change with time. If you know the redshift of the galaxy, for example by measuring the spectrum, and have a cosmological model, then you can calculate the comoving distance. Here cosmological model means constraints on the amount of dark energy, matter (both dark and regular) and radiation. The current accepted model is that the universe is around 70% dark energy and 30% matter, most of which is dark (with negligible radiation). The percent here refers to the fraction of energy density. Note that these values change over time, mostly since dark energy is like a property of space and so increases as the universe expands.

For some more info see:
https://en.m.wikipedia.org/wiki/Distance_measures_(cosmology)

Note that light years is a definite measurement, and so we can use it a valid unit for all distance measures.


Are galaxies moving away with constant acceleration?

I think the correct answer is that nobody knows - it all depends on the behaviour, or equation of state, of the dark energy.

If the dark energy takes the form of a cosmological constant, then the energy density due to dark energy will end up completely dominant as the universe expands.

The acceleration of the scale factor $a$ is given by
$$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}\left(\rho + \frac{3P}{c^2}\right) $$
where $\rho$ is the energy density, and $P$ is the pressure. A cosmological constant has $P = -\rho c^2$, which leads to
$$ \ddot{a} = \frac{8\pi G \rho a}{3}$$

So, if dark energy dominates, then $\rho a$ increases with time, so the acceleration is positive and increasing with $a$ and hence time.

If matter dominates, which it did in the past, then $P \simeq 0$ and $\rho a \propto a^{-2}$, so the acceleration is negative, with a magnitude that becomes smaller.

The transition from deceleration to acceleration happens when $\rho_{\Lambda} + \rho_m + 3P/c^2= 0$. If we assume a cosmological constant with $P = -\rho_{\Lambda} c^2$, where $\rho_{\Lambda}$ is the dark energy density (which is constant), and $\rho_m$ is the matter density, where $\rho_m = \rho_0 a_0^{3}/a(t)^3$ and $\rho_0$ and $a_0$ are the present-day matter density and scale factor respectively. Thus
$$ \rho_{\Lambda} + \rho_m -3\rho_{\Lambda} = 0$$
and the transition point occurs when $\rho_m = 2 \rho_{\Lambda}$. But because $\rho_{\Lambda}/\rho_0 = 0.7/0.3$, then
$$ \rho_{\Lambda} = \frac{7\rho_0}{3} = \frac{7}{3}\left(\frac{a}{a_0}\right)^3 \rho_m$$
and so the scale factor at the time of transition from decleration to acceleration is found from
$$ \rho_m =  \frac{14}{3}\left(\frac{a}{a_0}\right)^3 \rho_m,$$
$$ a = \left(\frac{3}{14}\right)^{1/3} a_0$$
i.e. when the scale factor was 60% of the present day value. According to this cosmology calculator, this happened roughly 6 billion years ago.

Thus the universe was decelerating then accelerated. So clearly the acceleration now is increasing with time.

On the other hand, if for whatever reason, the dark energy started to "decay" then it might be that the acceleration would decrease.


The wikipedia page you linked to tells you that the solar system is gravitationally "chaotic", in part because the mass of the sun is not fixed over time.

But even more simply than that, focusing just on the gravity (ignoring loss of stellar mass, etc.), the solar system is an N-body problem.  We have 8 planets, a sun, and millions of asteroids, comets, and who knows how many individual particles gravitationally bound to our sun (plus ones that aren't and are just passing through the neighborhood, so to speak).  When you have more than 2 bodies, the solutions to the N-body problem are unstable.  What this means is that, say we describe the N-body problem with data $D$ (the "initial conditions", or a perfect description of the state of the system at some specific point in time).  With a given complete data set the (Newtonian) gravitational evolution of the system is completely determined (but so difficult to do we can only approximate it).  What instability means here is that if we have some other data set $D'$ that is only a little bit different from $D$, then the differences between the evolution from $D$ and $D'$ will become exponentially large over long enough time scales.  So what may seem like minor differences now will result in radically different looking solar systems in the long run.

Since all of our observations can never give exact values, but only a range of values, there is necessarily a bit of uncertainty in what the exact gravitational state of our solar system is.  We have very poor data on the exact asteroid and comet content of our solar system, and even planetary data has significant error margins.  All of this means that there are lots of justifiable picks for the data $D$, each differing by a small amount from each.  But due to the instability, eventually these data will produce radically different futures from each other.  Currently we can only predict the solar system's evolution up to a few million years or so (the exact value stated can vary wildly depending on how you opt to define and compute the Lyapunov time).  After that the evolutionary tracks become so disparate we can't really say we're predicting anything other than "it'll definitely do something".

One way or another, it is currently impossible for us to make any clear assertions about what the solar system will look like on a timescale of billions of years.  Maybe all 8 planets will still be there; maybe their orbits will be very similar, but maybe they'll have much different orbits; maybe several planets will have been ejected from the solar system.  At best we can observe a few things that lead certain objects to be most likely to undergo significant alteration.  For example, Jupiter and Mercury appear to have a certain orbital resonance right now which could ultimately lead Mercury to undergo a significant orbit change.  This may ultimately cause it to collide with another planet, or the sun, or be ejected from the solar system entirely.  But maybe it won't.  It's hard to say.


Can we actually predict an asteroid impact?

There's a wide variety of answers for how much warning.   Most of what the Near Earth Asteroid Program or focuses on is Earth Crossing Objects or Near Earth Objects.   Those pose the greatest risk of impact and they are also, somewhat conveniently, the easiest to track.

The hardest to track is a high eccentricity long period orbit that approaches from the sun side of the planet.   Chelyabinsk was a sun side asteroid (an asteroid that hits during the day generally approaches from the sun side.   Those are the hardest to see, and even though Chelyabinsk was too small to worry about (20 meters diameter estimate).  It's possible that it would have been observed by the luck of having a telescope pointed in the right direction had it approached from the night side of the planet.   Still, an object that size, it's unlikely that they'd have had more than a few hours warming.

The larger the object, the easier it is to see and once an object is observed for some time and it's trajectory has been studied, it's path can be predicted years, even decades ahead of time.   The warning time drops sharply for an object not yet observed.

Potentially Hazardous Objects are at least 100 meters across, sometimes 150 meters is used as a minimum.   25-50 times the reflective surface of Chelyabinsk, so an object that size could be seen a fair bit further ahead of time.

Finally, Planet X doesn't exist and if it did exist, it would have cleared it's orbital path by now and it wouldn't be sending asteroids and/or comets towards Earth once every orbit and predicting an exact date (Sept 23rd) from an unobserved planet X. That's just silly.

There is stuff out there in our solar system that we haven't seen.   Lots of stuff, maybe even a hundred or more unidentified dwarf planets, possibly another planet or two.  (Click here for cool diagram)

The thing is, orbits tend to clear out over time and our solar-system is over 4 billion years old.   Most of the stuff that had an orbit that was going to crash into the Earth did so long ago and Planet Nine, assuming it exists, has cleared out it's orbit and it rarely sends stuff flying towards Earth.

Space is also so big that Earth is, by comparison, a very tiny target, so an asteroid coming from the outer reaches of the solar system and hitting Earth is like shooting a hole in one from the space shutting.   Stuff is far more likely to miss than hit the Earth.

But to answer your question on an object coming from a distant orbit, sent our way by a theoretical planet, it's hard to say when we'd see it in that scenario if it's a brand new observation and it would also depend on whether it came towards the Earth on the sun side or the night side.  (Objects can hit the Earth from any direction), but sun side observations are the hardest to see.

The good news is, high eccentricity impacts of objects large enough to be considered potentially hazardous are rare enough that you are unlikely to be affected by an Asteroid Impact in your lifetime.   It's somewhere around 140 on the list of things you should worry about.   And low eccentricity orbits that could hit Earth, or NEOs, those are being carefully tracked.

The other good news is that Planet X (if it even exists) doesn't threaten Earth.  No object in distant orbit around our sun would be a threat.  In fact, having a planet out there would actually send more asteroids away than it would cause to fly in towards the inner solar system.   They would reduce, not increase the number of impacts, similar to Jupiter.

But a large object passing through might post an asteroid/comet impact threat. Gilese 710 could do that, but it won't pass close enough to have that effect for about 1.36 million years.

I don't want to make light of this.  Asteroid impacts do happen and Earth may have been hit by a very destructive asteroid or comet about 12,900 years ago (the Younger Dryas impact hypothesis) and a global changing event like that might happen every, 50,000 or 100,000 years or so, which means it's 100%, absolutely worth looking out and seeing as much as possible, because Earth will get hit by a large object again.    But such impacts are very rare on a human lifetime scale.

But I'm not sure I ever answered your question.  How much warning for a large object, say 150-250 meters?   If the trajectory makes it tough to see, maybe only a few hours.   If it approaches from the night side, maybe a few days or a few weeks.   Longer if someone gets lucky and has a big telescope pointed in the right place, but I'm kinda guessing.    As our technology improves, the warning period will increase.  (I invite correction, especially on this last paragraph).


Yes, there is no surface in the sense as we have on the Earth.
However, there is the Barometric High/Pressure formula. It says, that in an ideal gas of atmosphere, the density (and pressure) increases exponentially high the decrease of the height. Its cause is understable even without differentiation: all horizontal layers of the atmosphere holds the mass of all the layers over it.
The result is, that in the atmosphere of the Earth, it is a good estimation of the pressure, that it halves with every $\approx$ 5km of height. Thus, 5km height we have about half of our surface atmosphere density, 10km heigh we have about a fourth, and so on.
In gas planets or stars, where the whole body is gaseous, this barometric formula holds only until the gas loses its ideality, i.e. its density does not grow more linearly with the pressure. In layman terms, it means that its molecules are so close to each other, that doubling the pressure does not halve the volume any more.
This happens typically in very high pressures (some thousand bars). At this point, the gas is already so dense, that we can say, we are already "inside of the body". If the gas is radiating (for example, the thermal radiation due to the $\approx$ 6000K temperature of the Sun), it also means that it is not transparent any more, thus we can not see inside it.
While there is no surface on the Sun (Jupiter, etc), the height difference between the near-total vacuum, and between the point where the atmosphere is not an ideal gas any more, is surprisingly small because of the exponentiality of the barometric formula. For example, the photosphere of the Sun, is only about some hundreds of km high. This is the layer,

Which is not enough dense to be able to fade the region below it;
But it is already enough dense to be its light enough well visible.

While there is no solid surface, this some hundreds of kms can be considered as "the surface of the Sun", particularly if we compare it to its $\approx$ 1.4million km diameter.


Will all galaxies eventually become elliptical galaxies?

First of all, stars are not confined to the plane of a spiral galaxy's disk by friction with the interstellar medium. Stars are very massive objects and the interstellar medium is extremely thin (comparable to the best laboratory vacuums on Earth), so there is effectively no friction involved. (And if there were, it would cause stars to spiral into the center of the galaxy.) Collisions between gas clouds do tend to keep the gas confined to a thin disk.

In practice, an isolated spiral galaxy will gradually turn into a lenticular (a.k.a. S0) galaxy, not into an elliptical. As it uses up its gas in forming stars, star formation will then gradually cease, and the visible spiral arms will weaken and fade away; you will be left with a smooth disk of stars. (This process can happen more quickly in galaxy clusters, where the high-speed motion of a spiral through the hot, high-pressure gas of the cluster rapidly strips out the gas, leaving behind just the stars.)

Stars in spiral-galaxy disks do tend to increase their vertical motions a bit over time, due to the gravitational effects of minor mergers, spiral arms, and massive molecular clouds (not because of friction, but because the gravity of the molecular cloud shifts the orbits of nearby stars a bit). Ironically, this means that the thinning out of the interstellar medium will probably weaken the tendency of stars to increase their vertical motions, because there won't be any more molecular clouds to perturb the stellar orbits.

Given sufficient time, close encounters between pairs of stars will tend to randomize their orbits, and the lenticular galaxy will eventually evolve into something more like an elliptical galaxy. However, because the average distances between stars are so large, the frequency of close encounters is very low, and the whole process will take trillions of years.


Question from a non-astronomer /non-mathematician regarding rotation speeds

This is a bunch of questions, many of which have answers already, but they are linked.

  Most black holes are depicted with a disk of matter that spirals into the black hole. Does the mass and diameter of the disk have a bearing on the rotational speed, maximum rotational speed of a black hole?

It's much more the other way round. The size of the accretion disk is determined mainly by the mass of the black hole, and by the supply of new material passing close enough to the black hole to possibly get trapped in the disk. The rotation of the black hole only plays a role in the innermost part of the disk, where the strange effects of General Relativity cause the material of the disk to be dragged around.

  Also, do black holes have gravity?

Yes. Indeed they don't really have much else.

  Does the gravity of a black hole create gravitational waves much as the electromagnetic waves created around the earth?  Do these waves have any influence on the speed of the black hole's rotation. What about if the rotation if the black hole is not symmetric?

A black hole is only asymmetric for very brief periods after large objects fall into it (for example when two black holes merge). In this case it produces very large amounts of gravitational wave energy, rapidly shedding the asymmetry. This process is called "ringdown". A symmetrical spinning object does not radiate gravitational waves.

A very good source for seeing these kinds of things are the videos produced by the SXS consortium with the caveat that they are visualisations of aspects of what is happening, not images. There are far too many dimensions of data to show the whole story.


How did Bradley arrive at the +/- correct speed of light when he calculated it?

He didn't need to know the absolute position of the stars. The change of the appearent position  has been sufficient to get a good estimate.
For velocites small in comparison to the speed of light, the shift of the angle is still small.

For small angles the sine is proportional to the angle with 2nd order precision.
The first Taylor summands are
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - + ... $$

The 2nd order approximation $$\sin x\approx x$$ can be applied on stars perpendicular, or at least apearently perpendicular, to the plane of the Earth orbit, where the method is most accurate.

Hence the underlying reason, for which it works, is the smoothness of sine.

This argument can be adjusted to less optimal conditions.


Unfortunately, many cheap mounts have a specific, proprietary connection to the telescope tube rather than something like rings, standard tripod threaded attachment or dovetail, so replacing them is difficult or impossible.
If you are fortunate enough that your inexpensive scope has a standard 1/4" threaded attachment, you could replace the mount with an inexpensive camera tripod with probably better results.


Is it just the observable universe that is expanding?

There is no evidence to support the idea that some part of the universe (at a cosmological scale) is contracting. Obviously one could construct such a theory but there is currently no observational data to support it.

Until relatively recently there were three broad ideas about the Universe's expansion - that gravitational pull would slow the expansion but by never enough to halt it, that (a boundary case) there was exactly enough matter in the universe for expansion to halt at an infinite point in the future, or that eventually gravitational attraction would first slow and then reverse the expansion, leading to a big crunch.

In fact, though, the observational evidence suggests that the expansion of the Universe is accelerating - that there is some "dark energy" that is speeding up the expansion. This is now the generally accepted view, though arguments continue as to what this "dark energy" is.


Why is Enceladus's albedo greater than 1?

To answer this, one really has to understand how the geometric and bond albedos are defined. Let's start with the bond albedo since its simpler.

Bond Albedo

The Bond Albedo is just the fraction of energy hitting a surface that gets reflected. To simplify the process, let's say I shoot 100 photons of all the same energy at Enceladus. Of those 100 photons that hit the surface, 81 of them will be reflected back and 19 of them will be absorbed. That means the bond albedo will be $81/100 = 0.81$.

Simple right? As such, the bond albedo can never be greater than 1 because you obviously can't reflect back more photons than you received. This means that a bond albedo of zero implies no light was reflected and all of it was absorbed, whereas a bond albedo of one implies all light was reflected and no light was absorbed.

Geometric Albedo

The definition of the geometric albedo is a bit more complicated. The first thing you have to realize though is that it is not a straight zero to one scale. A geometric albedo of one doesn't mean all the light got reflected like it did for the bond albedo. So let's try to define the geometric albedo.

There are two important points about the geometric albedo that make it differ from the bond albedo.

For the bond albedo, we talked solely about the total amount of incident light and the total amount of reflected light. This says nothing about an observer of said light. We just have some magical way of knowing about every photon that hits and is reflected by your surface and that allows us to calculate the albedo according to the bond albedo definition. The geometric albedo on the other hand specifically takes into account your ability to observe the light from a specific vantage point.

The second major point is that the geometric albedo is a measure of how well your particular surface reflected the light back to you compared to a reference surface. Imagine now you have two surfaces, one is the surface of Enceladus, the other is your "reference surface". This reference surface is an "idealized" reflector which means that it reflects every photon that hits it (making it have a bond albedo of 1). The key here though, is that it reflects light isotropically. What I mean by that is that the light doesn't get reflected in any preferred direction, but rather gets reflected in all directions equally. So now you have your 100 photons hitting your reference surface and all 100 photons get reflected, but because they're all reflected in random directions, only 10 photons actually get to your camera/eye/detector. What can happen with Enceladus though, is that the surface has just the right properties such that 100 photons hit the surface, and 14 photons get reflected to your detector because the surface preferentially reflects light in a specific direction. It's sending more photons to your detector than the reference surface. The geometric albedo is the ratio of how much light gets reflected to your detector by Enceladus over how much light gets reflected to your detector by the reference surface. In this case, $14/10 = 1.4$.

One small, additional point, is that this definition relies on your detector being in the same direction as your light source. In other words, the geometric albedo is a measure of how much your surface can retro-reflect (i.e., reflect back to the source of the photons) compared to the reference surface. Technically, the geometric albedo attains a maximum value when your surface is made up of retroreflectors.


Keplerian Elements for Approximate Positions of the Major Planets

I believe you are right, this is extracted from the "Explantory supplement to the Astronomical Almanac (2006?)"  (although it might date back to the 1960s...)
For example a similar PDF document gives that citation.
The supplement is published in a separate binding from the main astronomical almanc and a new version is published each year.  Finding a copy may be challenging (its not going to be on the shelves of your local Waterstones).


Does Halley's Comet travel past the outer bounds of the Oort Cloud?

No. Halley's Comet has an aphelion of 35 AU, which is  far less than the believed boundaries of the Oort cloud.
Any object on an elliptical orbit  with a perihelion of 0.6 AU and an aphelion of 2,000 AU has a semimajor axis of about 1,000 AU, and orbital period of about 31,600 years, much longer than Halley's Comet's current orbital period of 75-76 years. To get out to 200,000 AU and back on an elliptical orbit, you'd be looking at an orbital period of more than 31.6 million years.
As mentioned by @NilayGhosh and @planetmaker, 1P/Halley likely started as an Oort Clout object, was perturbed into becoming a long-period comet, and probably  subsequently became the prototypical periodic comet through gravitational interactions with the solar system's gas giants at some point in the last couple hundred thousand years.
As such, though it is possible that Halley's comet may have been in an orbit whose aphelion went beyond the outer boundary of the Oort Cloud in its long-period-comet days, it certainly does not do so now.


Can Jupiter's nightside be classified as spectral type Y?

This Wikipedia page lists Jupiter as a disputed sub-brown dwarf because its mass reportedly is sufficient to qualify for one. Distance to another star like the Sun doesn't matter, nor does the visibility in the infrared spectrum either since every planet is emitting some infrared light. The fact that Jupiter emanates more energy than it gets from the Sun can't be accounted for either, since that's also a matter of distance from the Sun, and Saturn and Neptune do so too. But, as the article states, Jupiter obviously has enough mass to possibly count as a sub-brown dwarf. Sub-brown dwarfs are still planets because they don't undergo nuclear fusion unlike brown dwarfs and hydrogen-fusing stars and never did.
Edit: There is no reason to only classify the night side just because the day side is illuminated by a(nother) star. Spectral classifications are given to the entire object. Since sub-brown dwarfs reportedly have a mass of at least one Jupiter mass, you may want to classify Jupiter a Y9V sub-brown dwarf.


If the Universe is Flat, why does it look 3D from our View?

So I know that the universe is flat in a 4D point of view, but what about a 3D point of view?

You seem to be misunderstanding the term "flatness". In this context, it does not mean the universe looks like a piece of paper in any sense. It instead refers to whether space is warped.

  i'm just a kid who is really interested in the universe

I'll try to simplify as much as possible. I'm assuming you've learned that triangles have 180°. For the mathematics you'll be dealing with in middle school, high school and even most of college, you'll only have to deal with that.

The more accurate statement is that triangles have 180° when our geometry is not warped. When triangles have 180°, we say we are dealing with Euclidean geometry (named after Euclid, an ancient Greek mathematician). There are in fact cases in higher mathematics when this does not hold, which we call Non-Euclidean geometries. Just as a side note: Euclidean geometry was developed thousands of years ago. On the other hand, Non-Euclidean geometries only began to develop in the 1800s, so they're pretty recent.

Anyway, Euclidean geometry deals with flat spaces. For example, if you draw a triangle on a piece of paper, it will indeed have 180°. However, Non-Euclidean geometry deals with curved (warped) spaces. For example, if you draw a triangle on the surface of a ball (which is not flat), the triangle will actually have more than 180°. If you instead draw the triangle on a potato chip or saddle-shaped surface (also not flat), it will have less than 180°.

So with regards to the Universe: let's say we draw a triangle connecting three stars in outer space (we can actually use any object or simply points, but I'm using stars as an example). Will that triangle be less than, equal to or more than 180°?

According to our measurements, it will actually be equal to 180°; the Universe is indeed flat. This has major implications for physics and astronomy.

  i'm sorry (in advance) if this is a simply answered bad question.

Future advice: don't be afraid to ask questions. Yours wasn't bad and is pretty common.


Why the lows of the GRB 2014-04-27 in Andromeda?

It turned out to be not real GRB event, just a false alarm.

My understanding is that, since the source is actually a steady X-ray source, you have a high count rate level at the base:

I am not expert of GRBs, but this high-oscillation behavior seems a characteristics of GRBs: you can see few examples here.
In fact, the BAT time resolution is very high ($\sim100\mu s$), then you can resolve very tiny details of the source's timing event.

About the source variability, this is something more theoretical (so I can not be of much help), and there is no accepted theory, but qualitatively, in some models the gamma rays are produced by clumpy clouds of relativistically ejected material, which could account for the high variability indeed.

Other source: Wikipedia classification .


How did Jocelyn Bell Burnell discover the periodicity of CP 1919?

Yes - at that time, the chart was the source of data used for analysis. Jocelyn visually reviewed hundreds of metres of traces every night looking for quasars (the reason the Interplanetary Scintillation Array was built)
The wiggle looked different from expected signals, and in fact was misidentified as interference at first, but when she noticed it again, and again, she conducted further analysis, identifying its repetition rate of 1.337 seconds.
From the New Yorker:

Bell Burnell - who analyzed the roughly seven hundred feet of paper generated each week as galactic radio waves were recorded in inked peaks - noticed a faint signal arriving from one slice of sky. Then it disappeared. In November, she saw it again. By adjusting the speed of the recording device, she determined that the signal came in every 1.34 seconds, a regular beat against the background static of the cosmos.

All from visual measurement of the trace!


How accurate are NASA's eclipse predictions?

The path of totality, which eclipse maps show to be 70 miles wide, is actually narrower than that, by up to 1 mile, according to some experts.
An article by the Kansas City Star discusses this:

Those maps, provided by NASA and others, show a crisply defined, 70-mile-wide path of totality where the moon will block 100 percent of the sun. But they are not as precise as they appear, at least on their edges.
The southern edge of the path as shown on the maps could be off by as little as the length of a football field or as much as a half-mile, eclipse mapping experts say. Likewise for the northern edge, meaning the path of totality might be just 69 miles wide.
“This is an issue. This is really an issue, but it’s not advertised. … Yeah, all the maps are wrong,” Mike Kentrianakis, who is the solar eclipse project manager for the American Astronomical Society and who routinely consults with NASA, told The Star.
Xavier Jubier, a French engineer whose calculations have been used to create the interactive Google maps of the eclipse, confirmed to The Star by email that the actual path of the totality is slightly narrower than the 70 miles shown on current maps.
Ernest Wright, who created maps and other multimedia presentations on the eclipse for NASA, said he thought the map might be narrower by about 100 meters, slightly longer than a football field. He also said it’s possible that Kentrianakis is correct in his estimation that the path is narrower by a half-mile or more.

The article explains the reason the path of totality is narrower than the eclipse maps indicate is because they use a 41-year old value for the radius of the Sun, which is now known to be too small a value:

Wright explained that eclipse maps are made based on what is known about the relative sizes and positions of the moon and the sun. “We have really good information about the orbit of the moon, the positions of the sun, the positions of the Earth. All of that is really well nailed down,” Wright said. “In order to get more accuracy, we need to take into account the mountains and valleys on the moon, and the elevations on the Earth. And we’re starting to do that, as well.”
The size of the moon, in fact, has been measured to within a meter, and its position in the heavens has been measured to within a centimeter. “But the last sort of uncertainty might surprise you,” Wright said. “It’s the size of the sun.”
Jubier said that the current maps are accurate using the the 696,000-kilometer radius and other standards agreed upon in 1976 at a meeting of the International Astronomical Union. “This is perfectly accurate but we know it does use a solar diameter that is not large enough. Why don’t we change the value(?)” Jubier wrote. “Well simply because the IAU (International Astronomical Union) has not yet approved a new value. This is part of the research we’re doing and for which we’re looking for funding.”
He continued, “So technically speaking if the Sun is larger than the adopted IAU value, and we know it is, the eclipse path is necessarily narrower and our tools can simulate this, yet the standard maps for the public will still retain the currently adopted solar radius until a new value has been accepted."


What fills the new space created during expansion?

It's not the case that something needs to be put in there, it's just that space itself is expanding. So if the bit you are watching expand is void then it will still be void, only bigger - you don't get new space, you just get bigger space.