What does the coordinate of time really mean?

First, imagine that you're in an inertial reference frame. We can think of this as a sort of lattice throughout space, with a synchronized clock at each lattice point in space. Now imagine two events, $A$ and $B$, which occur at space-time coordinates $(t_A, x_A, y_A, z_A)$ and $(t_B, x_B, y_B, z_B)$. The reference frame has (again, synchronized) clocks at $(x_A, y_A, z_A)$ and $(x_B, y_B, z_B)$. The coordinate time $\Delta t_{AB}$ is the difference in time measured by the clock at $B$ and the clock at $A$, so that $\Delta t_{AB}=t_B-t_A$. If $\Delta t_{AB}=0$, the events are simultaneous in that reference frame. However, an observer in a different inertial frame (say, in one with a boost $\vec{\beta}$ in the $x$-direction) might measure a different $\Delta t_{AB}$. This means that coordinate time is frame-dependent - it depends on the reference frame you're in.

Proper time is sometimes called "wristwatch time". Now, there's a single clock, and it measures the time between two events $A$ and $B$ on a single worldline. This proper time $\tau_{AB}$ does not depend on your reference frame, but it does depend on the path you take from $A$ to $B$ - that is, it's path-dependent.

One way to think of the difference is that coordinate time tells you something about one reference frame relative to another (because it's frame-dependent), while proper time tells you something about the worldline of the observer (because it's path dependent).


Has the iron core of Mars really solidified?

You're mostly missing an important point to do with the geothermal gradient.

The geothermal gradient used in that question, and applied to a $55\:\mathrm{km}$ depth is not consistent all the way through Mar's interior. They cite the Earth's geothermal gradient (see image below) and from that page you can see how the temperature profile changes with depth. There is a much stronger gradient towards the surface. Thus you can't naively apply the surface geothermal gradient through to the core. Rather, to get the core temperature, you'd need to integrate the gradient function.

Earth's Temperature Profile


How to convert Proper Motion in RA & Dec to Galactic Coordinates

The DEC conversion in the question is incorrect.
Try this:
double DEC = dec * Math.PI / 180.0;

positionrotation looks like a composition of two rotations bringing (1, 0, 0) to the RA and DEC of interest.
It seems correct, but something like this would be easier to verify:
using Microsoft.Xna.Framework.Matrix;

Matrix positionrotation =
    Matrix.CreateRotationZ((float) RA) * Matrix.CreateRotationY((float) -DEC);

The units in vector are mixed up.
According to these Gaia DR2 docs, radial_velocity is in km/s.
pmra and pmdec are in mas/yr; dividing (*) them by parallax in mas (1 au baseline) gives results in au/yr. Try this:
using Microsoft.Xna.Framework.Vector3;

// conversion to km/s
double au_yr = 1.49598e+8 / 3.15576e+7;

Vector3 velocity = new Vector3(
    (float) radial_velocity, // km/s
    (float) pmra / parallax * au_yr,
    (float) pmdec / parallax * au_yr);

Vector3 vel_icrs = Vector3.Transform(velocity, positionrotation);

If galacticrotation is from equation 3.61 in these Gaia DR2 docs,
// Matrix galacticrotation as above except
galacticrotation.M12 = -0.873437f;

Vector3 vel_gal = Vector3.transform(vel_ircs, galacticrotation);

(*) Actual distance in parsecs may differ from the reciprocal of measured parallax in arcseconds, which may be zero or negative.


OK, guessing rather wildly in the absence of solid information....

The press-release page is here; that provides a bit more information
than the news article, and includes two images of (different regions of) the cluster and a hint that the cluster is named MACS0940.

I think the main image is a color composite from HST imaging. It shows a galaxy cluster (the yellow galaxies). Also visible is light from an Einstein ring, the gravitational lensing by the cluster of a (single) distant background galaxy. The thin, white, ringlike streaks within the pink arcs are the stellar light from this galaxy (probably UV light from massive, recently formed stars); the broad pink arcs themselves are (I'm guessing) the Lyman-alpha emission as seen by MUSE. (So the image is a composite of both HST data and MUSE data.) Thus it is plausible to say that the Einstein ring is made up of stellar light from the background galaxy and (in pink) emission from gas in and around that same galaxy; the assumption is that said gas is mostly in a halo around the galaxy.

The second "magnified" galaxy referred to in the news article is probably in a different part of the cluster, and doesn't form a nice Einstein ring; it is presumably the four pink blobs in the second image on the press-release page.

This is the abstract for an HST proposal on which Claeyssens was a co-I; the data were taken on May 1, 2019, so it's plausible the image is from those observations:

  Recent deep integral-field spectroscopy observations using VLT/MUSE
  revealed two     high-redshift (z>4) lensed arcs in the galaxy cluster
  MACS0940. Notably, both lensed galaxies show highly extended
  Lyman-alpha emission (> 5 arcsec)... one of these galaxies forms an
  almost complete Einstein ring, with the Lyman-alpha arc reaching an
  unrivaled extent of 30 arcsec.


These are FITS binary tables not FITS images so the normal FITS image tools very likely  won't read them. You have a couple of options to examine and manipulate them:

Use the 'fv' tool which works on FITS images or tables (and integrates with ds9)
Use TOPCAT, a very powerful tool for examining, plotting and many other operations on all formats of tables (not just FITS ones)
Use Python and AstroPy Tables e.g.

from astropy.table import Table

table = Table.read('EN2_STAR_CHR_0102990364_20070206T133547_20070401T235934.fits')
table
<Table length=71042>
          DATE                DATEJD            DATEHEL       STATUS ...     BG    CORREC_RED CORREC_GREEN CORREC_BLUE
  yyyy-mm-jjThh:mi:ss    COROT JULIAN DAY   COROT JULIAN DAY         ... ELECTRONS ELECTRONS
        bytes23              float64            float64       int32  ...  float32   float32     float32      float32
----------------------- ------------------ ------------------ ------ ... --------- ---------- ------------ -----------
2007-02-06T13:35:47.000   2593.06626508258 2593.0705519515977     16 ... 373.35654        0.0          0.0         0.0
2007-02-06T13:44:19.000  2593.072191038021 2593.0764776164892      0 ... 378.81073        0.0          0.0         0.0


Does the Milky Way move through space?

Yes it does.

  I'm very fascinated with space, although I don't have a degree or any formal education, I'm still very in love with everything about it and want to learn constantly.

Good man Mike.

  One thing I ask myself is if our galaxy moves through space?

It does. When we look at the Cosmic Microwave Background Radiation we see a "dipole anisotropy" due to the motion of the Earth relative to it:

Image courtesy of William H. Kinney's Cosmology, inflation, and the physics of nothing

See Wikipedia for more:

"From the CMB data it is seen that the Local Group (the galaxy group that includes the Milky Way galaxy) appears to be moving at 627±22 km/s relative to the reference frame of the CMB (also called the CMB rest frame, or the frame of reference in which there is no motion through the CMB) in the direction of galactic longitude l = 276°±3°, b = 30°±3°.[82][83] This motion results in an anisotropy of the data (CMB appearing slightly warmer in the direction of movement than in the opposite direction).[84]"

627 km/s is quite fast. See this article, which says it's 1.3 million miles an hour. The speed of light is just under 300,000 km/s or 670 million miles per hour, so the Milky Way is moving through the Universe at circa 0.2% of the speed of light. Also see the CMBR physics answer by ghoppe which talks about the CMBR reference frame, which is in effect the reference frame of the universe.

  Or does it stay in a single location? If it does move, what causes it to move?

I'm afraid I don't know why it's moving. Perhaps it's because the Universe is full of things moving in fairly random directions. Like a gas.

  Hopefully the question makes sense, if not I can elaborate.

It certainly makes sense to me!

Edit 13/09/2017 : as Dave points out in the comments, there are other motions, including the motion of the solar system around the galaxy, which is circa 514,000 mph. (See the Wikipedia Galactic Year article). And the motion of galaxies isn't neat and tidy either.


In relation to Meteor Showers, what is meant by Lambda Omicron?

Not $\lambda \omicron$ or $\Lambda. \mathrm{O}.$ but $\lambda_⊙{}^\circ$.

This is the solar ecliptic longitude, the position of the sun relative to the vernal equinox, in degrees clockwise. This varies from 0 at the spring equinox, over 360 degrees. The Solar longitude is a way of giving the date of a meteor shower.

The effects of leap days mean that the date of the peak of a meteor shower will vary from year to year by a few days. However, the solar longitude at a meteor's shower's peak should remain the same from year to year. That is any variation in the solar longitude peak represents a real astronomical variation in the shower, not just an artefact of how we structure the calendar.


What's the largest non-spherical astronomical object in the universe?

The largest sub-galactic astronomical object (in volume) that we know of is the Carina Nebula, which is a non-spherical diffuse nebula. The Carina Nebula has a radius of about 100 parsecs.

Image credit: ESA

If you consider astronomical objects of the galactic scale, then the galaxy IC 1101 is the largest astronomical object (in volume) that we know of. From Wikipedia:

  The galaxy has a diameter of approximately 6 million light years, which makes it currently (as of 2013) the largest known galaxy in terms of breadth. It is the central galaxy of a massive cluster containing a mass (mostly dark matter) of roughly 100 trillion stars. Being more than 50 times the size of the Milky Way and 2000 times as massive

IC 1101 is on the left in the image below.

Here is an image from the Sloan Digital Sky Survey (SDSS) of IC 1101:

I'll leave it to you to decide if an elliptical galaxy counts as "spherical" or not.

If you don't count elliptical galaxies, the largest spiral galaxy is NGC 262, pictured below. It has a diameter of 1.3 million light years.


First spectroscopic detection of Uranus' rotation?

Slipher 1912 and
Lowell 1912
used a one-prism spectrograph on the 24-inch refractor at Lowell Observatory in this way.
Their 1911 observations of Uranus yielded a period of 10.8 hours, which made sense given the polar flattening previously observed by others.

They knew the obliquity of Uranus's satellites' orbits and assumed that its equator was in a similar plane.
This affected their observations in two ways.
First the solstice of 1901-02 presented an unfavorable pole-on aspect; after a negative result in 1903, Lowell waited until 1909 to try again.
Then Uranus required an unusual orientation of the spectrograph, so Slipher took special care to avoid instrument flexure bias.

Moore and Menzel 1930 credited Slipher and Lowell with the first such observation and obtained a similar result themselves.


If an earthquake happens on Mars, is it still an earthquake?

As pointed out in the comments, some do prefer to use different terms for quakes on other bodies, and you can see here that "marsquake" does have scholarly usage. That said, there is also scholarly usage for "earthquake" in this context. If you think about it, this makes sense given that one definition of "earth" is "the substance of the land surface", i.e. the ground. This definition of "earth" is agnostic to which terrestrial planet the land surface is on.


Mini black holes, a possible result of the LHC experiment at CERN

This result is due to Hawking radiation.

Due to random quantum fluctuations, particles are radiated from a black hole.

When the black hole has a mass less than the Moon, it will evaporate.

EDIT

I should add that any black holes created can only have the mass of the particles involved in the collision - two protons. So it's not going to do any harm before it evaporates. However, we would learn a lot from the event.


What if the two Black Holes spiraling around each other are evaporating via their Hawking radiation?

Gravitational waves are efficiently emitted by massive black holes orbiting each other - the power emitted increases with mass. Hawking radiation on the other hand is a process that increases with decreasing mass. As a result only very tiny black hole binaries would emit more power in Hawking radiation than they do in gravitational waves; at least towards the end of the inspiralling phase.
Details:
The characteristic timescale on which a black hole binary system spirals to merger is
$$\tau_{\rm GW} \simeq \frac{20c^5}{256 G^3}\left(\frac{a_0^4}{M^3}\right)\ ,$$
where $a_0$ is the separation, $M$ is the total system mass and I've assumed the binary components are of equal mass.
The characteristic evaporation timescale by Hawking radiation (again assuming each black hole is of mass $M/2$, is
$$ \tau_{\rm Evap} \simeq 640\pi \frac{G^2 M^3}{\hbar c^4}$$
For $\tau_{\rm Evap}$ to be shorter or even comparable with $\tau_{\rm GW}$, then
$$ 640\pi \frac{G^2 M^3}{\hbar c^4} \leq \frac{20c^5}{256 G^3}\left(\frac{a_0^4}{ M^3}\right),$$
which means
$$ \frac{M^6}{a_0^{4}} \leq 3.9\times 10^{-5} \frac{\hbar c^9}{G^5}$$
For black hole binaries, a reasonable value for $a_0$ would be a few times the Schwarzschild radius of the final merged black hole since most of hte power in gravitational waves is radiated in the final few orbits before merger, i.e. $a_0 \sim 10 GM/c^2$, so we can say
$$ M \leq 0.62 \sqrt{\frac{\hbar c}{G}} \sim 2\times 10^{-8}\ {\rm kg}\ ,$$
which is the Planck mass.
So the only merging black holes for which energy loss by Hawking radiation will play a significant role, close to the end of their inspiral, would have a mass of a few $\sim 10^{-8}$ kg or less. But the dynamics of such tiny black holes would not be controlled by gravity at all and they would evaporate in a fraction of a second.


Are features in the CMB correlated with features in the SDSS map of the most distant galaxies?

There shouldn't be any correlation. The CMB light that we see is from a spherical region in the early universe. Its homogeneity strongly suggests that the interior of the sphere was just as homogeneous, but we can't actually see CMB light from the interior. The galaxies that we can see formed from matter inside the sphere, and quite far from the edge. Between the CMB redshift ($z\approx 1090$) and the redshift of the most distant observed galaxy ($z\approx 11$), light traveled a comoving distance of roughly 14 Gly, around 30% of the radius of the sphere. Therefore only the very largest features of the CMB should be reflected in the distribution of observable galaxies.


Converting an RGB image to fits (astropy)

I haven't done much astronomical image processing before, but as this question is unanswered I'll give it a shot - hopefully to some avail. If the problem is more specific, a code sample/image sample would probably be useful for further diagnosis, but otherwise this example may help. It discusses the process of writing a 3-channel image to separate FITS images. I would try this first, and check if this output is also garbage - it might just be an issue with how you're using the HDUList.writeto() function. The relevant code is pasted below (with a couple of edits, as the example uses the Pillow Image class at first):

Split the three channels (RGB):

r, g, b = img[:, :, 0], img[:, :, 1], img[:, :, 2]

Write out the channels as separate FITS images

red = fits.PrimaryHDU(data=r)
red.writeto('red.fits')

green = fits.PrimaryHDU(data=g)
green.writeto('green.fits')

blue = fits.PrimaryHDU(data=b)
blue.writeto('blue.fits')

Also out of curiosity - why is your "numpy RGB array" storing values between 0 and 512? The standard for images is usually integers between 0-255 or floats between 0 and 1; but my guess is that astronomy requires a larger bit depth? In any case, you should also make sure the dtype of your numpy array is large enough for the desired bit depth - often images are stored with dtype="uint8" which ranges from 0 to 255.


Can the supernova remnant SN 1572 be observed by amateur astronomers?

As you say, SN 1572 is not very bright in the optical.  There are some Hα regions that have been observed with world-class optical telescopes, but they do not look like the X-ray and infrared images that you normally see.  In fact, images from the Palomar Optical Sky Survey 2 (with a limiting magnitude of ~22) do not reveal any nebular emission from this object at all:

Therefore, the optical emission must be fainter than 22nd magnitude.  I do not think there are amateur photographs of an object this faint and I believe a CCD on an amateur telescope would saturate from noise before revealing any of the nebular emission.


Could the earth's atmosphere have been partially stripped by a passing rogue planet?

Very interesting question. I'm okay with this question being here since you specifically ask for the effect of rogue planets on Earth's atmosphere. Thinking "astronomically" could be useful in coming to conclusions about whether or not your idea is even a possibility.

When I saw this I instantly thought of ice cores, since trapped gas bubbles could tell you about the composition of the atmosphere. Unfortunately ice cores go nowhere near that far back to be able to tell you about the composition of the atmosphere during the Mesozoic.

Now for the answer:

I can't really speak to the title of your question effectively, since I am unaware of any research involving rogue planets and atmospheric stripping, however, from an astronomy point of view, here is one other thing I would like to point out that may work to reduce the thickness of the atmosphere with time.

A small fraction of the gasses of the atmosphere exceed the escape velocity of Earth at any given time. Over time, many of these gasses get bombarded by high energy photons (as well as other particles) and fly out of Earth's gravitational potential well. If there are no other sources of these lighter gasses then there is a net loss.

Temperature does play a key role here (as the above plot suggests), but regardless, this process works to reduce the amount of atoms and molecules in the atmosphere (and it affects smaller mass atoms/molecules more so than heavier ones - this is why the Martian atmosphere is primarily carbon dioxide. Essentially it wasn't massive enough to be able to hold on to its lighter gasses, and over time the vast majority of them escaped).

If you had temperature information going back as far as the Mesozoic era, you could in theory run the clock backwards by running a simulation of the evolution of the atmosphere. I don't know how helpful this would really be (or how accurate), since there are potentially other things like volcanic outgassing and the bi-products of living things to complicate the situation. Quite honestly, I know little about these topics and so I couldn't tell you whether there is a net loss or gain of atmospheric gasses.

As far as rogue planets are concerned, I would find it rather unlikely that such a massive object would still be lurking around the solar system between about 252 to 66 million years ago. The early solar system was certainly more chaotic than it is now (for example, Giant impact hypothesis: $\sim 4.5$ Gya; the collision which caused Uranus' tilt: a few billion years ago), however it is not out of the realm of possibility.


Is it possible to view the Andromeda galaxy with the naked eye in Greece?

Assuming you mean Bortle class 4, then the Andromeda galaxy (M31) should be a fairly easy object with the naked eye. It won’t be particularly impressive, but rather will look like a hazy blob. On the other hand it is more than 2 million light years distant, so it’s impressive to be able to see it at all.


What's going on in this photo of Halley's comet's tail?

Negatives taken with various intstruments at Lowell Observatory on the morning of 1910-05-13 appear in a 1981 conference paper by Henry L Giclas.
The wide-angle images were taken with cameras piggyback on a telescope tracking the comet nucleus.
These exposures are naturally deeper in the center than at the edges; modern astrophotographers take flat fields to correct for this.
The stars in the upper right of the image in question were low enough to the horizon to be obscured for part of the exposure, making their trails shorter.
To make this more intuitive, rotate the image so that the tail points to the right and upward.

Comet tails generally have two parts: a smooth, curved dust tail driven by steady radiation pressure; and a straight, blotchy plasma tail (also called an ion tail or gas tail) driven by the fluctuating solar wind.
A coronal mass ejection can severely disrupt the latter, as the STEREO spacecraft observed with Comet Encke in 2007.

The 1910 image in question may show a tail disconnection event in progress. If so, the highlighted feature is the old plasma tail drifting away from the nucleus, and the new plasma tail is not yet large enough to be distinct.
Barnard 1914 includes three photos of such an event on June 6-7, 1910.


What percent of planets are in the position that they could be viewed edge-on from Earth? (and thus able to undergo transits)

Because there are so many planets out there!

There just happens to be an entire web page dedicated to calculating that answer.

  Transits can only be detected if the planetary orbit is near the line-of-sight (LOS) between the observer and the star. This requires that the planet's orbital pole be within an angle of $d_*/a$ (part 1 of the figure below) measured from the center of the star and perpendicular to the LOS, where $d_{*}$ is the stellar diameter (= 0.0093 AU for the Sun) and $a$ is the planet's orbital radius.

  This is possible for all $2\pi$ angles about the LOS, i.e., for a total of $4\pi d_*/2a$ steradians of pole positions on the celestial sphere (part 2 of figure).

  Thus the geometric probability for seeing a transit for any random planetary orbit is simply $d_*/2a$ (part 3 of figure) (Borucki and Summers, 1984, Koch and Borucki, 1996).

  For the Earth and Venus this is 0.47% and 0.65% respectively (see above Table). Because grazing transits are not easily detected, those with a duration less than half of a central transit are ignored. Since a chord equal to half the diameter is at a distance of 0.866 of the radius from the center of a circle, the usable transits account for 86.6% of the total. If other planetary systems are similar to our solar system in that they also contain two Earth-size planets in inner orbits, and since the orbits are not co-planar to within $2d_*/D$, the probabilities can be added. Thus, approximately $0.011 \times 0.866$ $= 1\%$ of the solar-like stars with planets should show Earth-size transits.

That's pretty freaking amazing! Kepler has been up there for a short while, and has a possible list of nearly 2000 planets just looking at about 150,000 stars for only a couple of years! So if only 1% statistically transit, that would mean that just randomly 1500 systems would have the correct orientation (given the results to date, that makes sense). And given that about 7500 stars were eliminated from consideration due to being variable of one sort or another... I think it would be pretty safe to say that pretty much every star out there has at least some sort of planetary body around it.


Is there a website that shows equivalent views through different telescopes?

Deep sky watch has an article that covers this
http://www.deepskywatch.com/Articles/what-can-i-see-through-telescope.html
It includes sample image of what can be seen in, for example, a medium telescope (150mm objective) with 180x magnification


Theoretical measure of distance to black hole?

Yes. If you could spatially resolve a black hole and measure the redshift of radiation at a given angular radius, then that would yield the distance to the black hole.

This works because the amount of redshift depends on the ratio of radius to Schwarzschild radius, but the latter is known if the mass is known. Thus the redshift yields the radius. The distance is then given by the radius divided by the apparent angular radius.

There will be complications due to gravitational lensing and any orbital velocity around the black hole, but this does not stop it working in principle.


When and how was the length of a Mars year determined?

There are two answers to this, because the period "687" was known long before it was recognised as an "orbital period".

The Babylonians knew by 600 BCE that Mars took 79 years to go through all 12 constellations 42 times, giving a period of 79/42 × 365.25 = 687 days.  But they wouldn't have thought of this as an "orbit" (in the modern sense) since they would not have thought of Mars as being a physical object. Similar knowledge was known in India and China.  It is relatively easy to determine, and just requires the intent to measure and the ability to make careful records.  This value was known wherever people had writing, an interest in the stars, and a stable enough society.

The understanding of the value as being "the time for Mars to orbit the sun once" depends on having a heliocentric, so is due to Aristarchus, Copernicus, Kepler, or Newton (depending on what you mean by "know that Mars orbits the sun")


Has the sun become way brighter the last years?

The Sun's luminosity is stable to 0.1% - it varies slightly in response to, or associated with, levels of solar magnetic activity (Solanki & Unruh 2012).

This variation would not be perceptible and is completely outweighed by the $\pm 3.4$% variation caused by the non-circular orbit of the Earth (the Sun is closest in January).

Obviously, if you live in the northern hemisphere, then the Sun is generally higher in the sky now (in May) than it was in December. Its direct light is less absorbed/scattered by the atmosphere. This is a much more perceptible effect. It could change the visual brightness by tens of per cent, and by even more in the short wavelength blue part of the spectrum. That is the only physics-based explanation I can offer.


A comet is usually characterized by its tail. A dead comet has lost all its ices and gases (responsible for producing this tail), leaving just a rocky core.

The Halloween comet is such a dead comet, in that it has no tail, but furthermore it resembles a skull, making it particularly relevant for Halloween.


Flamsteed designations are assigned by John Flamsteed. He was an English Astronomer.

Flamsteed Designations (Flamsteed Numbers) are numbers assigned to the stars for the purpose of identification and cataloguing.
Flamsteed designation contained 2554 stars.

Flamsteed designations for stars are similar to Bayer designations, except that they use numbers instead of Greek letters.

[ Bayer designation is a stellar designation in which a specific star is identified by a Greek letter, followed by the genitive form of its parent constellation's Latin name. The original list contained 1,564 stars.
For the most part, Bayer assigned Greek and Latin letters to stars in rough order of apparent brightness, from brightest to dimmest, within a particular constellation. Since in a majority of constellations the brightest star is designated Alpha (α), followed by beta for the 2nd brightest and so on.
But this was not the rule always followed by Bayer.
Somtimes by position of a particular star in that constellation, sometimes the order of rising up in the sky (the star rises up early will get earlier number), sometimes he follows mythological stories or data or sometimes he even uses his own way of assigning the number.]

In Flamsteed designations, Each star is assigned a number and the Latin genitive of the constellation it lies in.
e.g. 52 Cancri is star in the Constellation Cancer and assigned a number 52 according to star's right ascension. i.e. there are 51 stars in the Cancer constellation to the West direction of 52 Cancri star or there are 51 stars that rise before/earlier of 52 Cancri star.

The numbers were originally assigned in order of increasing right ascension within each constellation.
[Right ascension is the angular distance measured eastward along the celestial equator from the vernal equinox (it is one of the point of intersection in the space of two imaginary circles: Celestial Equator and Earth's orbit around the Sun. Our Earth is at this point on 21/22 March each year while orbiting around sun.) to the the position of the star. It is measured in degrees  but for simplicity and ease of use, mentioned in the form of hours. That the 360 circle is divided into 24 hours (1 hour = 15 degrees). ]

But due to the effects of precession (Top like rotation of Earth's axis of rotatation in the space mataining a constant angle of 23.5 degrees with verticle) they are now slightly out of order in some places.

Flamsteed catalogue don't cover all the stars visible from Earth with naked eye. Flamsteed's catalogue covered only the stars visible from Great Britain, and therefore stars of the far southern constellations have no Flamsteed numbers.

Reference/Information Source/Quoted From:

1) List of constellations for a list of constellations and the genitive forms of their names: https://en.wikipedia.org/wiki/88_modern_constellations

2)Bayer Designation: https://en.wikipedia.org/wiki/Bayer_designation

3)Flamsteed Designation: https://en.wikipedia.org/wiki/Flamsteed_designation


Hohmann transfer orbit (Earth-Jupiter system)

I want to calculate the following:

  The semi-major axis and eccentricity of Hohmann transfer orbit from Earth to Jupiter.
  The speed of the probe at Earth.
  The launch speed of the probe from Earth to reach Jupiter.
  The duration of the flight to Jupiter.
  The elongation of Jupiter at launch to ensure that the probe arrives at Jupiter.

$a_E$, $a_J$ are 1 and 5.2 AU, respectively. The Hohmann transfer ellipse touches both, so the major axis is 6.2 AU and the semi-major axis is half of that, or 3.1 AU as you wrote.
Definitely use the vis-viva equation whenever you can! The standard gravitational parameter $GM$ of the Sun is 1.327E+20 m^3/s^2. You can use that if you convert your distances to meters to get the velocity in m/s,

$$v^2 = GM\left( \frac{2}{r} - \frac{1}{a} \right)$$

or as long as you are orbiting the Sun you can stick with AU and years and use:

$$v^2 = \frac{2}{r} - \frac{1}{a}$$

which will give the velocity in AU per year (once you take the square root).
I'm not exactly sure what "The launch speed of the probe from Earth" means and how it is different from the speed of the probe at 1 AU. If it means the speed you'd fire it from the surface of the Earth from a gun if there was no air, then just add the Earth's escape velocity to the velocity from #2.
this answer should help you calculate the period of an orbit for a given semi-major axis . For the Hohmann transfer ellipse, use its semi-major axis to calculate its period, and then use half of the period for the duration of the flight from Earth to Jupiter.
I think you can figure this one out for yourself. Draw a picture of the 1 AU and 5.2 AU circles and the Hohmann ellipse that touches both. Put a dot where you want Jupiter to be when you get there, then move Jupiter backwards by a certain angle, which would be 360 degrees times the ratio of the flight time to Jupiter's period. If the flight time is 5 years (it's not) and Jupiter's period is 10 years (it's not) then you'd back Jupiter up by 180 degrees for its starting position when you launch from Earth. Then read how elongation is defined.


It's called sunglint. This can be problematic for Earth-observing satellites in low Earth orbit. Such satellites typically don't take a "picture". They instead continuously scan the Earth a line at a time. This means the sunglint moves with the satellite.

You can see this effect yourself while flying in an airplane. Little ribbons of rivers and lakes can appear to be on fire. Here's a nice short YouTube video that shows this perfectly.


New Horizons encountered Pluto when it was about 33.5 au from the Sun.

The solar diameter is $1.39\times 10^{9}$ m.

An astronomical unit is $1.496\times 10^{11}$ m.

The angle subtended by the Sun at Pluto (in radians) is the ratio of the diameter of the Sun to its distance from Pluto. Converting to degrees, minutes, seconds, gives 0.01589 degrees, or 0.95 arcminutes.
The atmosphere of Pluto is way too thin (much thinner than the Earth's) to make any significant difference to this at all.

The diameter of Pluto is $2.37\times 10^6$ m. In order for Pluto and the Sun to subtend a similar angle in an image, you just need to move far enough away from Pluto to make the same angle.

Thus
$$ D_P/R_P  \simeq (D_P + D_{PS})/R_S,$$
where $D_P$ would be the distance from New Horizons to Pluto, $D_{PS}$ the distance from Pluto to the Sun, $R_P$ the radius of Pluto and $R_S$ the radius of the Sun. Because I will show that $D_P \ll D_{PS}$ we can say that
$$ D_P \simeq \frac{R_P}{R_S} D_{PS} \simeq 0.057\ au$$.

New Horizons is travelling at about 14.5 km/s. So a simple bit of Maths tells us it would takes about 6.5 days to get into this position.

The picture I think you are talking about was taken on July 15th 2015 only 7 hours after the flyby. The Sun and Pluto do not share the same angular size in this photo. You are not seeing the Sun at all in this picture. It is eclipsed by Pluto completely. All that is being seen is sunlight refracted (or more likely scattered) through Pluto's atmosphere. The Sun acts more-or-less like a point source of light at more-or-less infinite distance.

The angle that the light needs to deviate in order to be seen through the atmosphere is about $0.19$ degrees (the half-angle subtended by Pluto at the Sun [negligible] plus the angle subtended by Pluto at New Horizons after 7 hours). If we were to model the atmosphere as a glass prism. The deviation angle is $\delta \simeq (n-1) \alpha$, where $n$ is the refractive index and $\alpha$ is the opening angle of the prism. If you let $\alpha$ be quite large (let's say 1 radian) to simulate rays coming through the atmosphere, then $n = 1 + \delta$, where $\delta$ is in radians. So $n\simeq 1.003$. NB: This assumes there is no scattering in Pluto's atmosphere, but I suspect that is not true and that my estimate of the refractive index is a large overestimate. In fact I'm sure it is an overestimate because this is larger than the refractive index of air on Earth. i.e. I am sure that what we are seeing is scattered light, not refracted light.


One property of distant objects, that follows from the limited speed of light, is that we are observing them in the past; the light that they emit takes time to reach us. For very distant objects, we are effectively observing the universe at a very early time.

Objects that are very far away, such as distant galaxies, are receding away from us, which causes the light that they emit to be doppler shifted. The recession velocity is proportional to the distance away, and in turn so is the magnitude of the redshift. This redshift allows us to put a number to the distance away; combining this with the speed of light limit allows us to determine how old the light we are observing from them is.

These two effects, the limited speed of light and redshift, effectively allow us to determine an evolutionary timeline for the universe, and place constraints on when certain objects appear.


Where is the habitable zone of a Hypergiant?

As in the case of Betelgeuse, it isn't really meaningful to talk about the habitable zone of a hypergiant. These are massive, short-lived stars that don't last long enough for terrestrial planets to form: their evolution is measured on timescales of only a few million years. Their progenitors are very hot O-type stars that are highly destructive to protoplanetary discs, eroding them with extreme ultraviolet and X-ray emissions. Even if you did manage to form a terrestrial planet there, there wouldn't be sufficient time for the magma ocean to cool before the star undergoes a supernova or collapses to form a black hole. The intrinsic heat of the planet would likely keep water as superheated steam in a rock vapour atmosphere. Furthermore, these stars undergo extreme rates of mass loss, which would make the space weather environment very hostile to maintaining an atmosphere over long periods of time. This is not an environment where you would find Earth-like planets or life.
As for the question of what would happen if the Sun turned into a hypergiant: it won't because (fortunately for us) it is not massive enough to do so. Such a scenario is in the realm of fantasy so it isn't really possible to say: once you're invoking wizards and magic sufficiently advanced AIs with the technology to do this then all bets are off.


How large were the deviations of Uranus's positon from predicted which led to the discovery of Neptune?

I also asked this question at the History of Science and Mathematics Stack Exchange and got two answers so far.
The answer by Ben Crowell makes a rough calculation that the displacement should be on the order of ten arc seconds.  It also has a plot of the variation in longitude of the observed and predicted positions of Uranus over time, which seem to differ by up to a few tens of arc seconds from predicted positions.
The answer by cktai states that according to his source the deviation between the calculated and the observed orbit was 30 arc seconds in 1835 and 70 arc seconds by 1840.
An arc second is one in 1,296,000, or 0.0000007, of a full circle.
At the average distance between the Sun and Uranus, about 19.22 Astronomical units or AU, a full circle is about 120.76271 AU. Since the orbital period of Uranus is 84.01 years, Uranus travels at an orbital speed of about 1.4374801 AU per year, or about 0.0039356 AU per day, or about 0.0001639 AU per hour, or about 0.0000027 AU per minute, or about 0.000000045 AU per second.
0.000000045 AU is 6.81433 kilometers, so the average orbital speed of Uranus should be about 6.81433 kilometers per second.  The average orbital speed of Uranus is listed as 6.8 kilometers per second.
Earth is at an average distance of 1.0 Astronomical Units from the Sun, so the distance between Earth and Uranus varies between about 17.33 and 21.11 AU, and the average distance is about 19.22 AU.
So since there are 360 degrees in a full circle, at the average distance between Earth and Uranus a degree of arc would be about 0.33545 AU, and an arc minute would be about 0.00559 AU, and an arc second would be about 0.00009318 AU or 13,939.61 kilometers.
Being off position by 30 arc seconds would equal being off position by 418,188.38 kilometers, and being off position by 70 arc seconds would equal being off position by 975,772.88 kilometers.  And that would be approximately the distance traveled by Uranus in about a day.
And I guess that gives some idea of the scale of the deviations which were considered a problem and which led to the discovery of Neptune.


"Early" and "late" nomenclature regarding O, B & K, M stars

What's wrong with your argument is that you are using modern theories of stellar evolution. The terminology was introduced in the context of the theory that stars are powered by the Kelvin–Helmholtz mechanism (heat generated via contraction due to gravity). In this model, stars would start out hot and gradually cool down as they evolved, hence hot stars are called "early-type" and cool stars are called "late-type".
This theory was shown to be inadequate because it can only power the Sun for tens of millions of years, while geological evidence indicated that the Earth had been around for far longer than this. The resolution was the discovery of nuclear fusion. However by this point the terminology of "early-type" and "late-type" stars had become established and it survives to this day.
(Incidentally the terms "early-type" and "late-type" are also used for galaxies, where once again they do not correspond to modern theories of how these objects actually evolve.)


How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it?

Humphreys & Larsen (1995) suggest, using star count information, a distance of $20.5 \pm 3.5$ pc above the Galactic plane; consistent with, but more precise than the Bahcall paper referred to by Schleis.

Joshi (2007) is more guarded, investigating some systematic uncertainties in the estimation techniques and ends up with distances between 13 and 28 pc above the plane.
This paper gives an excellent review of the topic in its first couple of pages.

The Sun moves at about 15-20 km/s with respect to a local standard of rest defined by the general motion of stars in our vicinity around the Galaxy. In three-dimensions, this "peculiar velocity" is $U=10.00 \pm 0.36$ km/s (radially inwards), $V=5.25 \pm 0.62$ km/s (in the direction of Galactic rotation) and $W=7.17 \pm 0.38$ km/s (up and out of the plane).  (Dehnen & Binney 1998)

The Sun executes oscillations around its mean orbit in the Galaxy, periodically crossing the Galactic plane. I borrowed this illustration (not to scale!) from http://www.visioninconsciousness.org/Science_B08.htm to show this oscillatory motion.
As the Sun is currently above the plane and moving upwards, and each cycle takes about 70 million years with an amplitude of 100pc (Matese et al. 1995), it will be roughly 30 million years before we cross the plane again.

EDIT: An update. A new analysis by Karim & Mamajek (2016) yields a distance above the plane of $17.1 \pm 5$ pc and they provide a meta-analysis of previous analyses that gives a median distance of $17.4 \pm 1.9$ pc. They also refer to a paper by  Schonrich et al. (2010) that gives the velocity away from the plane as $7.25 \pm 0.37$ km/s (though with an additional systematic uncertainty of 0.5 km/s).


Is it possible to determine an observer's position on Earth from a photo of the night sky and the time of the shot?

What can you tell from a picture of some stars?

At the very least, you need a recognisable asterism

If the exposure is too low or the light pollution too high to identify unambiguously an asterism of three or more stars, you're simply not going to be able to tell much from the photo.

An asterism on its own can only tell you where the picture wasn't shot.

If a star in a picture has a declination of $60^o$, you can rule out all shooting locations south of $-30^o$ latitude. A star like Polaris can only really be seen in the northern hemisphere, while Sirius can be seen everywhere south of Svalbard.

An asterism plus the horizon is sufficient for determining latitude

If the edge of land in the picture is sufficiently close to being the actual horizon (and not a range of mountains) and it is present alongside a recognisable asterism in the photo, you can calculate the latitude. This is true for any magnification, as you can use the angular distance between stars (a known quantity) to measure the angle to the horizon. I can't think of any special locations where you wouldn't be able to determine the latitude from the stars + horizon.

I haven't really done the math for how accurately one would be able to determine the latitude, but here are some considerations: There are two sources of error, the position of the stars and the position of the horizon. The star image may spread out over several pixels, and if the picture was shot through a standard camera and not a telescope, there would be an error, at a wild guess, of around ten arcminutes. The horizon can appear lower than $0^o$ depending on your distance above ground, but the effect is small. Looking out from Mt Everest, the horizon will appear a twentieth of a degree lower down than it ought to. I believe the horizon's exact position can also be obscured through atmospheric refraction.

All in all, the error in the angle between an identified star and the horizon equals the error in latitude of your identified position. For a 10 arcminute error, that corresponds to ten arcminutes of latitude, which is about 18 kilometres.

For longitude, you need global, not local time

If you only have local time, the photo could have been taken anywhere along that latitude. If you have some form of global time (UTC, GMT, whatever, it doesn't matter which as long as you know what it is) you can figure out longitude from there.


Around the worm hole would be a region with an intense gravitational field, and there would be gravitational lensing occurring, similar to that around a black hole. The difference would be that light that has passed through the worm hole would would be visible. So, a distorted view of the "other side" could be seen.

This physicsbuzz article has some images and discussion. Note that the wormhole would be a spherical region in space, and not a 2d "stargate"


Could a contracting Universe create the redshift effect observed by Hubble?

Aside from being, I suspect, totally incompatible with GR, the model fails to match observations.

The universe appears to be homogeneous and isotropic. In the model you suggest, Hubble's law would be different depending on which direction you were looking.

In particular, looking away from the black hole, galaxy redshifts would increase to some asymptotic value, whereas looking towards the black hole they would increase to infinity.

If the argument is that the central black hole is so far away that we cannot measure this asymmetry (i.e. well beyond the observable universe, given the homogeneity of the cosmic microwave background - but then does the model have an explanation for the CMB at all?), then the model also needs to explain why galaxies would move in this particular way, since they would not "feel" the influence of the black hole.


Why does the convective core in an intermediate to high mass star shrink?

Whether convection exists depends on whether the interior radiative temperature gradient reaches the adiabatic temperature gradient.

The interior radiative temperature gradient is proportional to the opacity and the outward energy flux, and inversely proportional to $T^4$. As the star evolves on the main sequence, the central temperature goes up and the opacity (e.g. Kramer's opacity goes as $T^{-7/2}$ goes down. You are also removing free electrons (combining them with protons to form He) which reduces the Thomson/Compton scattering opacity. This means the radiative temperature gradient goes down and can fall below the adiabatic gradient, meaning that energy transport reverts to being radiative.

It stays convective right at the centre for longest because that is where the radiative temperature gradient is still largest (driven by the extreme temperature dependence and high outward energy flux of the CNO cycle nuclear reactions).


Why are these telescope mirror makers not upside down? (Daniel K. Inouye Solar Telescope)


The first handful of hits on Google actually return incomplete and even wrong answers (e.g. "Because the Moon is much brighter" which is plain wrong, and "Because the Moon is closer" which is incomplete [see below]). So here's the answer:

As you mention, when light enters our atmosphere, it goes through several parcels of gas with varying density, temperature, pressure, and humidity. These differences make the refractive index of the parcels different, and since they move around (the scientific term for air moving around is "wind"), the light rays take slightly different paths through the atmosphere.

Stars are point sources

Stars are immensely far away, effectively making them point sources. When you look at a point source through the atmosphere, the different paths taken from one moment to another makes it "jump around" — i.e. it twinkles (or scintillates).

The region in which the point source jumps around spans an angle of the order of an arcsecond. If you take a picture of a star, then during the exposure time, the star has jumped around everywhere inside this region, and thus it's no longer a point, but a "disk".

…the Moon is not

The same is true for the Moon, but since the Moon (as seen from Earth) is much larger (roughly 2000 times larger, to be specific) than this "seeing disk" as it's called, you simply don't notice it. However, if you are observing details on the Moon through a telescope, then the seeing puts a limit on how fine details you can see.

The same is even true for planets. The planets you can see with the naked eye span from several arcsec up to almost an arcmin. Although they look like point sources (because the resolution of the human eye is roughly 1 arcmin), they aren't, and you will notice that they don't twinkle (unless they're near the horizon where their light goes through a thicker layer of atmosphere).

The image below may help understanding why you see the twinkling of a star, but not of the Moon (greatly exaggerated):

EDIT: Due to the comments below, I added the following paragraph:

Neither absolute size, nor distance is important in itself. Only the ratio is.

As described above, what makes a light source twinkle depends on its apparent size compared to the seeing $s$, i.e. its angular diameter $\delta$ defined by the ratio between its absolute diameter $d$ and its distance $D$ from Earth:
$$
\delta = 2 \arctan \left( \frac{d}{2D} \right)
\simeq \frac{d}{D}\,\,\,\mathrm{for\,small\,angles}
$$

If $\delta \lesssim s$, the object twinkles. If it's larger, it doesn't.

Hence, saying that the Moon doesn't twinkle because it's close is an incomplete answer, since for instance a powerful laser 400 km from Earth — i.e. 1000 times closer than the Moon — would still twinkle because it's small. Or vice versa, the Moon would twinkle even at the distance it is, if it were just 2000 times smaller.

Finally, to achieve good images with a telescope you not only want to put it at a remote site (to avoid light pollution), but also — to minimize the seeing — at high altitudes (to have less air) and at particularly dry regions (to have less humidity). Alternatively you can just put it in space.


What you describe is a mutual planetary occultation (or transit), and yes, it has happened—in 1737—and will happen again—in 2133 (both dates are for Venus/Mercury events; there are other possible instances of planets passing in front of each other; for example, in 2065, Venus will pass in front of Jupiter). I would suggest giving a look at https://en.wikipedia.org/wiki/Planetary_transits_and_occultations
What is interesting with Mercury and Venus is that, considering the respective positions of these planets as well as the Earth, it is technically possible for Mercury to pass in front of Venus! The last time it happened was in 796, and it won’t happen again at least until the year 2500. See More Mathematical Astronomy Morsels by Jean Meeus or the link above for more information as well as for the frequency at which mutual planetary transits occur.
A “near-miss“ will happen on 27 April 2022 at 19:11 UT, when Neptune will be very close to Venus—about 25″ depending on where you will be on the Earth at that moment…


What determines the configuration of orbits in a binary system?

Following the convention $\mathbf{r} = \mathbf{r}_2-\mathbf{r}_1$, $M = m_1+m_2$, in the center-of-mass frame we have, by definition,
$$\begin{eqnarray*}\mathbf{r}_1 = -\frac{m_2}{M}\mathbf{r}\text{,}\quad&\mathbf{r}_2 = \frac{m_1}{M}\mathbf{r}\text{.}\end{eqnarray*}$$
Hence, $\ddot{\mathbf{r}} = -GM\hat{\mathbf{r}}/r^2$ implies that the individual orbits are similar conic sections in this frame, and moreover in the bound case they are ellipses that share a common focus at the center of mass, with all three of the distinct foci being collinear and with the center between them. Although there is a degenerate case of a circular orbit.

Therefore, there's a geometrically simple condition for the intersecting configuration: there is intersection if, and only if, the distance to apoapsis of the larger mass (smaller orbit) is greater than or equal to the distance to periapsis of the smaller mass (larger orbit). Since a general ellipse in polar coordinates about a focus can be described by
$$r = \frac{p}{1+e\cos(\phi-\phi_0)}\text{,}$$
where $e$ is the eccentricity, and the individual orbits are proportional, we have intersection if, and only if,
$$\frac{1-e}{1+e}\leq \frac{m_1}{m_2}\leq\frac{1+e}{1-e}\text{,}$$
as the apsides occur when the cosine term is $\pm 1$.


It is an optical illusion.

We perceive nearby objects in 3d because we have two eyes. As we see objects from two different viewpoints, our brain can put the images together to make a 3d image.

Objects that are more distant perceived as 3d if they are moving fast enough for us to see them change appearance. Again it is not in the eyes but in the brain that the perception of depth occurs

The moon is a long way away, far too far for any 3d from our two eyes, and doesn't change fast enough for to see any motion. It is too small for us to see shadows or any other clues that we could interpret as 3d structure, so it looks like a small  flat shape.

Telescopes can make shadows visible, and time-lapse can make the motion of the moon apparent, and so add to the perception of a solid.


How can I calculate 27 tropical/solar year since my birthdate?

From the Wikipedia page on the Tropical year,

  The mean tropical year is approximately 365 days, 5 hours, 48 minutes, 45 seconds.

Starting from the Tropical year proceeding after your birth date, you can calculate 27 Tropical years forward by adding that approximate amount of time 27 times over.


The reason why we centre our galactic coordinates on the sun is that is where we are.

What actually is the purpose of a coordinate system‽ I'd like to be able to point to a body and say that it is at coordinates X, Y, Z (with uncertainties in each direction)

Now we know the direction of the centre of the galaxy, but we don't know it's distance with accuracy. If we were to use the location of the centre of the galaxy for the galactic coordinates we could not give accurate coordinates for any of the local stellar objects.

If we had a warp drive and could travel halfway across the galaxy before lunch then it might make sense to use a different set of coordinates. However, as things stand there is nothing wrong with being Earth centric.

I have mentioned the remainder of the question in comments.


Where is the Perseids meteor shower coming from?

The comet sheds material each time it get close to the sun. The sun heats this dirty snow ball, the ice evaporates and tears dust and smaller rocks with it.
These dust particles then follow a similar orbit around the sun as the comet. But as they are ejected with some velocity and react differently to the solar wind & radiation, their orbit is slightly different from their parent comet.
Different orbit means generally different orbital velocity and, over time, the particles spread out along the comets orbital path. They form a steam of material that constantly dissipates and is replenished each time the comets approaches perihelion.
And each time the earth passes through the part of the solar system this stream occupies, we see meteor.
Here is a nice interactive animation:
https://www.meteorshowers.org/view/Perseids


How can we ge to the exoplanet Proxima b?

It's habitability is uncertain.   Red Dwarf stars that small could be unkind to their planets and strip them of most of their atmospheres.   We'll get a better look at the exoplanets around Proxima Centauri when the James Webb Space Telescope is up and running.  It's scheduled to launch in 2019 and it should be able to tell us something about exoplanet's atmospheres, which would go a long way towards estimating any habitability.

Actually getting there, even with a small unmanned craft is a long ways away.   There's still significant technological difficulties with Breakthrough Starshot.  The lasers aren't set up yet for one.  For two, the aim has to be just about perfect, and three, we currently have no means of receiving a signal from a craft that far away.   We have to improve our communications tech.   Now, in theory, we know how to do it.   We could build giant satellite dishes on the Moon, for example, or orbiting in space, but that's very expensive.

Another issue with starshot is that they have no means of slowing down, so they'd pass through the solar-systems they were sent to examine in a rapid fly-by, perhaps 10% of the speed of light, with only a few hours within the solar-system.   Fly-by missions are still useful, and given the difficulty of transmitting data an orbiting mission in another solar-system might not make practical sense, because such a mission could easily gather more information than it could send back.    The recent New Horizons / Pluto mission was a fly-by and it gathered a lot of information, that took over a year to send back to Earth.

Rapid fly-bys shouldn't be dismissed though.   They can gather a lot of information that we have trouble seeing from Earth, even with newer and bigger telescopes.

More ambitious missions, and ships with their own fuel so they can break and remain in a distant solar-system are harder.   That requires much more equipment, much more mass and as a result, would accelerate more slowly.   Manned missions 4.2 light years away are so far out of reach that it makes little sense to talk about them outside of science fiction.   A manned mission to Mars is hard enough and that's over 100,000 times closer.

Starshot is many years away.   We might see Starshot launch in a decade or two - if it moves quickly.   Then it'll be another 2 decades or so for the ship to reach the Proxima system and another 4 years and change for any information to be returned.   35-45 years is probably on the ambitious estimate before we receive the first starshot data.

Most of the new information about exoplanets that we're going to get over the next few years will come from telescopes like James Webb and others that follow.


I would suggest that this:

  LST = GMST + longitude;

line is wrong. 360 degrees of longitude correspond to 24 hours, so it should be something like:

  LST = GMST + longitude/15.0 ;

Assuming these are all in floating point and you are subsequently handling the potential overflow of time into the next day correctly...


How can a planet have a 90° inclination?

As Wikipedia notes,

  Since the word 'inclination' is used in exoplanet studies for this line-of-sight inclination then the angle between the planet's orbit and the star's rotation must use a different word and is termed the spin-orbit angle or spin-orbit alignment. In most cases the orientation of the star's rotational axis is unknown.

As of now (7/26/2015), no other planets, smaller objects, or debris disks have been found around Kepler-452. This means that the situation is not as unusual as might be thought. If there were other planets in the system with orbital inclinations near 0°, then things would appear exceedingly odd. Why would this one planet be different than the others?

One possible scenario is that the star's rotation axis has been changed. Think about how Uranus's axial tilt is skewed. Nobody knows for sure what happened, though one popular theory is that it was hit at some point in time by another object. Granted, it would take a rather large object to have any effect whatsoever on the axial tilt of Kepler-452, but it is still possible.

The point of this idea is that perhaps it is the star, not the planet, which is unusual.

Several papers have been written by those much more knowledgeable than I, notably Crida & Batygin (2014) and Xue et al. (2014). Both refer to Hot Jupiter models, but the ideas behind them are still possible for other planets.

The former suggest

  Contemporary observational surveys suggest that a considerable fraction of solar-type stars are born as binary or multiple systems (Ghez et al. 1993; Kraus et al. 2011). Moreover, most stars form in embedded cluster environments (Lada & Lada 2003) where dynamical evolution can lead to the acquisition of transient companions (Malmberg et al. 2007). Recently, Batygin (2012) showed that the presence of a companion to a young star can force the proto-planetary disk to precess around the binary’s axis, so that the plane in which planets eventually form and the equatorial plane of the central star can differ.

They also suggest that encounters with other planets could be responsible for some cases of this (although an angle of 90° makes this seem rather improbable); a related idea that has been applied to Kuiper Belt Objects is the Kozai mechanism. The companion to this theory is that the star was perturbed by an encounter, possibly with a companion (as I discussed earlier).

Xue et al. discuss similar scenarios, focusing in part on "dynamic" interactions between planets. Direct tidal interactions between the star and the planet are also a possible scenario.


If you extrapolate, you'll find that those ~4 minutes per day add up to 24 hours per year.
The difference is due to the Earth revolving around the Sun at a rate of almost 1 degree per day.
For every 365$\frac{1}{4}$ solar days, there are 366$\frac{1}{4}$ sidereal days.

Civil time is based on mean
solar time,
in which the Sun crosses the
celestial meridian
every 24h 0m on average.
In sidereal time,
24h 0m is the time between successive meridian transits by any given star, equivalent to 23h 56m 4s of solar time.

Sidereal time also corresponds to the
right ascension
of stars crossing the meridian at that time.
If a star in Orion has right ascension 5h 30m, then Orion is on the meridian at 05:30 sidereal time, which occurs 2 hours earlier in the solar day per month.


what magnification is required to see jupiter, saturn, venus and mars with a 50mm refractor scope with f/12 ratio and 600mm focal lenth

Venus, Mars, Jupiter, and Saturn are visible with your eyes, so a telescope and magnification are not required to see them. Depending on the quality of the telescope, the magnification will show you some details.
Venus - you will be able to see the phase.
Mars - for several months every 26 months, you will be able to see some dark markings and a polar cap.
Jupiter - you will see the cloud bands and 4 of its moons.
Saturn - you will see its rings and moon Titan.

Enjoy the view!


It's generally dictated by how massive the star is. Remember what a pulsar is, it's a very rapidly rotating, highly magnetized neutron star.

Neutron stars are a category of objects which have masses between 1.4 and 3.2 solar masses. This is the end stage of stars which are not massive enough to form black holes (they're held up by neutron degeneracy pressure), but are massive enough to overcome electron degeneracy pressure (which is what prevents white dwarves from further gravitational collapse).


What observation can be expected on LIGO if any when Betelgeuse goes supernova?

Potentially a short (less than a second) burst of gravitational waves (GWs) would be detected. Much depends on asymmetries in the core collapse, since a spherically (or even axially) symmetric collapse would not produce GWs (e.g. Morozova et al. 2019). However, theoretical models suggest that the GWs start at low frequency (tens of Hz) and are associated with the core bounce at the end of core collapse. After about half a second the frequencies rise to kHz and are associated with the oscillations of the proto-neutron star at the centre. Since LIGO is sensitive from about 10Hz to 1kHz, then only the initial phase that lasts a few tenths of a second would be detectable, but snce GWs can travel through the star without much hindrance they will indicate the moment of core collapse quite accurately.

The theoretical calculations referenced above suggest GW amplitudes of $h_{+}d \sim 10$, where $h_{+}$ is the dimensionless GW strain and $d$ is the distance to the supernova in cm. Betelgeuse is at a distance of $\sim 6 \times 10^{20}$ cm (600 light years), so the strain at the detector would be $h_{+} \sim 2\times 10^{-20}$. LIGO is sensitive to strains of above $10^{-23}$ in this frequency range, so there should be no problem (according to these models) in detecting such an event.

If they were produced with detectable strength, then they should be received at the same time as ananticipated neutrino burst. The difference would be that the neutrino burst should extend over about 10 s. That is because the dense centre of a supernova is opaque even to neutrinos (but not gravitational waves) for that period and so they "leak out" over that period.

The electromagnetic signature of the supernova (i.e. a visible light increase) would be seen a few hours after that, since that is how long it takes for energy to be transferred to the envelope of the star.

NB: For all practical intents and purposes, over the distance to Betelgeuse, neutrinos, along with light and gravitational waves, travel at the same speed. i.e. The delay of the ($\sim  10$ MeV) neutrino burst due to their non-zero ($\
< 1$ eV) mass would be negligible. If the Lorentz factor is $10^7$, then the neutrino speed is the same as the speed of light to 1 part in $2\times 10^{14}$, which means they take an extra 0.1 ms to arrive from Betelgeuse compared to light speed.


If Mars were tidally locked to the Sun, what would the temperatures on its surface be like?

Tidal lock would mean only one side of Mars would be exposed to solar irradiation, so its temperature would be mostly defined by the equilibrium between the energy received from the Sun and the black-body radiation.

Because of the orbit eccentricity, solar irradiation reaching Martian surface varies between 493 and 717 W/m² (source). According to Stefan-Boltzmann law, the temperature required for the black-body radiation to match these levels will vary between 305 and 335 K (32°C an 62°C), respectively.

Such temperatures will be reached in the area directly facing the Sun (so the Sun is seen in zenith), and gradually decline towards the background radiation temperature of space (about 3K) when approaching the dark side (where the Sun is just barely above the horizon). Such temperature distribution is due to the angle of incidence, not to the atmospheric "filter" or lack thereof.


If the Earth was in Venus's orbit, would it be habitable?

A planet's environment is a complex interplay of factors.  If you dramatically change the insolation of a planet, it becomes very difficult to predict how the other factors which influence the planet's environment will respond.

Venus, for example, seems locked into its high-CO2 environment (which makes its surface hot) because its surface is too hot to allow for the formation of carbonates.  Even if it had water (also needed for the formation of carbonates), it would be continuously boiled away from the surface.

All planets - Earth, Mars, and Venus included - have their temperatures influenced by their albedo.  Albedos of planets are often strongly correlated with cloud cover.  Cloud cover generally increases as a wet planet warms, reflecting more light.  But atmospheric composition can overcome this balance, as in the case with Venus, which has a very high albedo but is still swelteringly hot on the surface.

Even if you get rid of the feedback factors like water and carbon dioxide, the planets have very different nitrogen levels - Venus at about 2 1/2 times that of Earth (enough to cause mild nitrogen narcosis on its own), and Mars almost nothing.  Really, Mars in Venus's position would be in a terrible state - it's been so stripped of its atmosphere even as far from the sun as it is, it would be all the more stripped being much closer to the sun.  Over geological timescales it'd become highly water stripped as well, as it'd be much easier for water to enter a gas phase, and the solar wind can very effectively strip water planets even as large as Venus, let alone Mars.

Venus in Mars' position would probably fare much better.  But it's difficult to say exactly how much.  Given that we can't say with great confidence exactly what set in motion the path that led to its current state today (thicker, more viscous lithosphere preventing crustal overturn? cause or consequence?), it's hard to say whether that would have happened had it been in Mars' orbit.  One curiosity if you took today's Venus and moved it to Mars' position: you may well cause an atmospheric collapse to a supercritical state. It's been theorized that even in its current orbit that's happened before (temperature runs counter to the formation of supercritical fluids).


Where is the center of the Milky Way located?

Distance estimates to the center of our galaxy vary around $r \approx$ 8000 $\pm$ 200 parsecs.
Its spherical equatorial coordinates are $\alpha$ = 17h45m40s, $\delta$ = -29$^\circ$00'28" (J2000).
Converting those to rectangular form, we have:
$$\begin{align}
X &= r \cos \delta \cos \alpha \approx -437 \pm 11 \ \mathrm{pc} \\
Y &= r \cos \delta \sin \alpha \approx -6983 \pm 175 \ \mathrm{pc} \\
Z &= r \sin \delta \approx -3879 \pm 97 \ \mathrm{pc}
\end{align}$$
As with most celestial objects, its direction is known much more precisely than its distance.
We can't pinpoint its 3D position, but we can narrow it down to a long, fuzzy ellipsoid pointing away from the Sun.


Calculate how long it takes for the sun to complete one orbit around the galaxy?

The time it takes for the Sun to complete one orbit around the center of the Galaxy is called a Galactic year. It is estimated to last 225 to 250 million terrestrial years.
Now if you know the speed of the Sun in the Galaxy and you know how far the Sun is from the center, you can estimate this period. Let's assume a circular orbit and ignore the (large) error bars, for the sake of simplicity. The Sun is roughly $8\,200$ parsecs away from the Galactic center. So the circumference of the Sun's orbit is $8\,200 \cdot 2 \cdot \pi = 52\,000 {\rm \,parsecs}$. You mention the speed of the Sun is $2.1\cdot 10^{-4} {\rm \,parsec/year}$. So to complete one orbit, it takes circumference/speed = $52\,000 / 2.1\cdot 10^{-4} = 245 \cdot   10^6 {\rm years}$, that is 245 million years. Pretty close to the actual value!


Is there another explanation, apart from the doppler effect, for the redshift observed in distant galaxies?

Tired light has been used as an explanation before, whereby light loses energy whilst travelling through space---a sort of drag effect.

I don't think anyone actually supports it nowadays, though.


What's the reason that we have a different number of days each month?

You make a great point. The reason behind the discrepancy between the dates is due to a complicated history behind it.

The calendar is based on the calendar created by ancient Romans, which is based on one Moon cycle. One lunar cycle is

  29.53 days.

    www.universetoday.com/20620/lunar-year/

which does not evenly divide into the 365.25 days of the year.

Note: The Romans borrowed the calendar from the ancient Greeks to develop the idea of a 10-month calendar that left approximately 60 days unaccounted for.

Consequently, the earliest ancient Roman calendars had months that were either 29 or 30 days to account for this discrepancy.

The Romans started using the 10-month calendar in 738 B.C. Their months were initially called:

Mensis Martius (March)
Mensis Aprilis (April)
Mensis Maius (May)
Mensis Iunius (June)
Mensis Quintilis (July, quin- meaning "five")
Mensis Sextilis (August, sex- meaning "six")
Mensis September ("seven")
Mensis October ("eight")
Mensis November ("nine")
Mensis December ("ten")

To account for the remaining $\sim60$ days, Mensis Ianuarius (January) was added to the beginning of the year and Mensis Februarius (February) to the end of the year during Numa's reign around 700 B.C. with a leap year every 4 years.

  Leap years were decreed by proclamation the pontifex maximus, in such years an extra month was added to the calendar.

    Quoted from PM 2Ring

The calendar stayed in that order until 452 B.C. when a small council of Romans, called the Decemvirs, moved February to follow January.

However,

  in the initial 36 years of its adoption leap days were added every 3 years due to a misunderstanding (Romans used inclusive counting),

    Quoted from PM 2Ring

so that caused a big problem. In fact, by the start of the reign of Julius Caesar, the previous calendar was off by a whole week!

Julius Caesar noticed this and thus modified the Roman calendar in 46 B.C. to make each month have either 30 or 31 days, with the exception of Februarius[February], which had 29 days, with the three year leap year fixed into a four year leap year. Quintilis[July] was later renamed Julius[July] in his honor. Likewise, Sextilis[August] later became Augustus[August] to honor Augustus (Gaius Octavius, the first Roman Emperor). Augustus[August] was also given an extra day (taken away from Februarius[February]), so that Augustus and Julius would have an equal number of days, representing their equal power. This is the Julian Calendar.

  This calendar was extremely accurate, but was still not accurate
  enough as it drifts by approximately 3 days every 400 years. So after
  a few hundred years, the drift becomes noticeable
  (i.e. the seasons don't properly match the calendar), which is
  important for farming, and therefore human survival.

    quoted from  CJ Dennis

Then a pope named Pope Gregory XIII modified the Julian calendar again in October 1582. This is the currently used Gregorian calendar. He made it so that

  Every year that is exactly divisible by four is a leap year, except for years that are exactly divisible by 100, but these centurial years are leap years if they are exactly divisible by 400.

    https://en.wikipedia.org/wiki/Gregorian_calendar

This calendar is the most accurate calendar currently used, and won't be a single day off until the year of $\pm 3200$ AD. It isn't perfect though. Compared to the tropical year, it is 27 seconds too long. This is...

  due to astronomical effects (the
  rotation of the Earth speeding up and slowing down, etc.) we can't be
  more precise than that.

    quoted from CJ Dennis

As for your other question, unless the plates taken had accurate dates of when recorded, it would be rather impossible to calculate precisely.


Try this link http://geo.pds.nasa.gov/missions/mgs/megdr.html
This should have the data you're looking for, though you may need to process it a bit to get it into a format hat will be usable.


What values can a color index variable get?

The only physical mechanism I can think of for quasars to have such a large difference in two bands would be at very high redshift ($z\gtrsim6$) where you have the Gunn-Peterson trough completely erasing everything blueward of 912 Å (restframe). But with typical magnitudes of $\sim20$ or so at these redshifts, a color of $20$ that would then mean that the magnitude in the bluest band would be of the order $\sim 40$, which no instrument is able to detect.

So effectively, there would be no detection on the bluest band, in which case the color would be infinite, or rather undefined.

In other words, I think those values must be errors.


What is the reason for magnetic braking?

The magnetic braking in contact binary stars AFAIK is assumed to be similar to the magnetic braking mechanism that spins down the rotation of single stars.

A stellar wind couples to the large scale magnetic field and maintains this coupling out to the Alfven radius (where the thermal pressure and magnetic pressure are roughly equal). At this point the wind decouples, stops co-rotating with the star and hence removes a certain amount of specific angular momentum.

The situation will be more complicated in a binary system (entangled, asymmetric fields etc.) but the basic principle will be the same. Many contact binaries have fast rotation and strong, dynamo-generated magnetic fields.


Welcome to StackExchange. Good question. Hubble's Law says that an object's velocity away from an observer is directly proportional to its distance from the observer. In other words, the farther away something is the faster it is moving away from us. The redshift tells how fast a star is receding from us and we can therefore get the distance. Hubble's equation states that $v\ =\ H_0\cdot D$ where $H_0$ is Hubble's constant.  It makes sense that the further away a star is the faster it has been moving.

The redshift is measured for a star and for small velocities relative to c it can be written that $z \approx \frac{v}{c}$. For larger speeds the equation is $z \approx \frac{H\cdot D}{v}-1$

A good general description is given here.


The Sun is immensely loud. The surface generates thousands to tens of thousands of watts of sound power for every square meter. That's something like 10x to 100x the power flux through the speakers at a rock concert, or out the front of a police siren. Except the "speaker surface" in this case is the entire surface of the Sun, some 10,000 times larger than the surface area of Earth.

Despite what "user10094" said, we do in fact know what the Sun "sounds" like -- instruments like SDO's HMI or SOHO's MDI or the ground-based GONG observatory measure the Doppler shift everywhere on the visible surface of the Sun, and we can actually see sound waves (well, infrasound waves) resonating in the Sun as a whole! Pretty cool, eh? Since the Sun is large, the sound waves resonate at very deep frequencies -- typical resonant modes have 5 minute periods, and there are about a million of them going all at once.

The resonant modes in the Sun are excited by something. That something is the tremendous broadband rushing of convective turbulence. Heat gets brought to the surface of the Sun by convection -- hot material rises through the outer layers, reaches the surface, cools off (by radiating sunlight), and sinks. The "typical" convection cell is about the size of Texas, and is called a "granule" because they look like little grains when viewed through a telescope. Each one (the size of Texas, remember) rises, disperses its light, and sinks in five minutes. That produces a heck of a racket. There are something like 10 million of those all over the surface of the Sun at any one time. Most of that sound energy just gets reflected right back down into the Sun, but some of it gets out into the solar chromosphere and corona. No one can be sure, yet, just how much of that sound energy gets out, but it's most likely between about 30 and about 300 watts per square meter of surface, on average. The uncertainty comes because the surface dynamics of the Sun are tricky. In the deep interior, we can pretend the solar magnetic field doesn't affect the physics much and use hydrodynamics, and in the exterior (corona) we can pretend the gas itself doesn't affect the physics much. At the boundary layers above the visible surface, neither approximation applies and the physics gets too tricky to be tractable (yet).

In terms of dBA, if all that leaked sound could somehow propagate to Earth, well let's see... Sunlight at Earth is attenuated about 10,000 times by distance (i.e. it's 10,000 times brighter at the surface of the Sun), so if 200 W/m2 of sound at the Sun could somehow propagate out to Earth it would yield a sound intensity of about 20 mW/m2. 0dB is about 1pW/m2 , so that's about 100dB. At Earth, some 150,000,000 kilometers from the sound source. Good thing sound doesn't travel through space, eh?

The good folks at the SOHO/MDI project created some sound files of resonant solar oscillations by speeding up the data from their instrument by 43,000 times. You can hear those here, at the Solar Center website. Someone else did the same thing with the SDO/HMI instrument, and superposed the sounds on first-light videos from SDO. Both of those sounds, which sound sort of like rubber bands twanging, are heavily filtered from the data -- a particular resonant spatial mode (shape of a resonant sound) is being extracted from the data, and so you hear mainly that particular resonant mode. The actual unfiltered sound is far more cacophonous, and to the ear would sound less like a resonant sound and more like noise.


I wrote https://github.com/barrycarter/bcapps/tree/a944a8d8315fd6169907f73e47698f030befe3bb/ASTRO/bc-zodiac.c to solve this. I may've tweaked bc-zodiac.c since then, so you'll need the exact link above (not the most recent version of bc-zodiac.c) to confirm/replicate my results. In the directory https://github.com/barrycarter/bcapps/blob/master/ASTRO/

The file "houses.txt.bz2" contains a list of all house changes for now +- ~15000 years in a mostly-English-readable format.
The file "house-states.txt.bz2" is a list of all "horoscopes"(house states for all planets) in machine readable format for the same period. Please see https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-zodiac.txt (which also contains the text of this answer) for details on how to read this file.

Important caveats below, but first, some notes:

You'll notice that lines like:

BCE 13200-JUL-12 09:48 SUN ENTERS ARIES PROGRADE -479626193452 S0+

suggest the vernal equinox was on July 12th about 15000 years ago (of course, no one actually used our calendar back then), which seems a bit off. Further, you'll notice:

CE 17191-FEB-23 00:27 SUN ENTERS ARIES PROGRADE 479386211293 S0+

suggesting the vernal equinox will occur on February 23rd (about a month early) 15000 years from now (assuming they continue using our calendar).

See https://astronomy.stackexchange.com/a/13009/21 for why this happens (my answer there has slightly different dates/times, because I used a less accurate model).

In the 11,100,017 days (about 30,391 years) covered, there are 6,359,874 horoscope changes, for an average horoscope change roughly every 1.75 days. Broken down by planet:

The moon changes house 4,875,276 times, about once every 2.3 days.
Mercury changes house 450,523 times (including retrograde house changes), about once every 25 days.
Venus changes house 384,971 times (including retrograde), about once every 29 days.
The Sun comes in third with 364,691 changes, about once every 30 days. A more exact number is once every 30.4367697585 days. When you multiply this by 12, you get 365.2412371020 days, which, as expected, is a good estimate of the length of the tropical year (the exact value given by http://hpiers.obspm.fr/eop-pc/models/constants.html is 365.242190402, which is 82 seconds different).
Mars changes house (including retrograde) 209,501 times, about once every 53 days.
Jupiter (w/ retrograde) 49,215 times, once every ~225.5 days.
Saturn (w/ retrograde) 25,697 times, ~432 days.

As expected, most of these house changes (6,283,141) are in the forward or "prograde" direction and very few (76,733) are in the retrograde direction. Of course, the Sun and Moon always move prograde, another reason the prograde number is so large.
Since there are 12 houses and 7 planets, it would seem there are 12^7 = 35,831,808 possible horoscopes. However, there are actually only 3,732,480:

Since Mercury's greatest elongation is 28 degrees, it must remain within one house of the Sun. Given the Sun's house, there are only 3 possibilities for Mercury's house (either the Sun's house or one of the two adjacent houses).
Since Venus' greatest elongation is 47 degrees, it must remain within two houses of the Sun, so there are only 5 possibilities given the Sun's house.
Combining these, we see there are 12*12*3*5*12*12*12 or 3,732,480 possible horoscopes.

Of these possibilities there are 2,565,570 (about 69% of the total possible) actual horoscopes. Of these,

969,067 horoscopes occur only once
The horoscope "4045424" (translated below) occurs 18 times (most recently in 1799 and next in 5391), but two others occur 17 times, 10 appear 16 times, and many appear 15 times, so it's only the most frequent by coincidence (probably just because of where DE431 happens to start and end).
"4045424" translated is Sun in Leo, Moon in Aries, Mercury in Leo, Venus in Virgo, Mars in Leo, Jupiter in Gemini, Saturn in Leo.

Note that this horoscope is probably only the most frequent because of how I chose the start and end dates (based on the DE431 ephemeris).

Important caveats:

My standard disclaimer: please do not rely on anything I do/post anywhere. I can and do make mistakes. Feel free to check my work and alert me to mistakes.
If you actually want to check my results, you will need to download kernels from NASA. See http://naif.jpl.nasa.gov/naif/tutorials.html for more details on how this all works.
Note that NASA can't predict planet positions (or even the Earth's own orientation) into the far future: they solve differential equations to predict planetary positions and then publish an approximation of those solutions. The further away we get from today, the less accurate these predictions are.
We have a pretty good approximation of how the Earth precesses, but we can't even predict THAT super accurately more than a few months in advance. Note that https://naif.jpl.nasa.gov/pub/naif/generic_kernels/pck/ shows the high precision precession kernel is only good through 6 Mar 2017, and even THAT is updated twice a week (see https://naif.jpl.nasa.gov/pub/naif/generic_kernels/pck/aareadme.txt for details). Since precession affects the location of the first point of Aries, this adds another uncertainty.
Precession doesn't change the ecliptic itself, but nutation does. We don't have a super-accurate model of nutation, and I am using the undocumented SPICE functions zzmobliq_ and zzwahr_ as per a private reply to my https://naif.jpl.nasa.gov/pipermail/spice_discussion/2016-January/000423.html
The calendar on https://cafeastrology.com/astrologyof2017horoscopes.html (and the 2016 equivalent) should agree with my results within +- 1 minutes. Remember to convert between ET on their site and UTC in my file.
http://aa.usno.navy.mil/data/docs/EarthSeasons.php can also be used to check some of my calculations (within +- 1 minute), recalling that, in the northern hemisphere:

"SUN ENTERS ARIES" (as in "first point of Aries") marks the vernal equinox.
"SUN ENTERS CANCER" (as in "Tropic of Cancer") marks the summer solstice.
"SUN ENTERS LIBRA" marks the autumnal equinox. As nearly as I can tell, there is no well-known associated phrase with this point in the zodiac.
"SUN ENTERS CAPRICORN" (as in "Tropic of Capricorn") marks the winter solstice.

If you use HORIZONS (http://ssd.jpl.nasa.gov/?horizons) to check my work, you'll notice my results are accurate to the minute for about +- 1000 years from now, but drift in accuracy after that. I'm not exactly sure why this is.
I used a "stepsize" of 86400 seconds (1 day), so if a planet changed house and then changed back in under a day, my program may miss it. Someone w/ more computing power may wish to re-run my program with a smaller stepsize.
The moon's position varies with your location on Earth. I used a geocentric model, giving the moon's position for a theoretical observer located at the center of the Earth.
I actually computed planetary barycenters (the center of mass of the planet and its satellites) because NASA doesn't provide planet position information for the entire 30,000 year period I computed (see http://ssd.jpl.nasa.gov/eph_spans.cgi?id=A for details).

For Mercury and Venus, this no effect (no moons), and the Mars' system barycenter is within a few inches of Mars' own center (Phobos and Deimos have very little mass). Jupiter is fairly massive, so its system's barycenter is fairly close to its own center, and even the effect Titan has on Saturn isn't that great.

The human-readable time in houses.txt.bz2 is the output of the C function http://naif.jpl.nasa.gov/pub/naif/toolkit_docs/C/cspice/timout_c.html (format ""ERAYYYY##-MON-DD HR:MN ::MCAL ::RND" modified slightly by Perl). They are reasonable for current dates, but not necessarily for past dates because the Gregorian Reformation of the calendar occurred at different times for different countries, and the BCE/CE time system didn't exist prior to 1 CE (and probably not even then). For precision work, see bc-zodiac.txt for details.
As above, I use "ARIES" to mean the "1st house", ie, the constellation that starts at the vernal equinox ("the first point of Aries", where the ascending ecliptic meets the celestial equator) and ends 30 ecliptic degrees of ecliptic longitude later.

This was reasonably accurate in 56 BCE, but is certainly not true today as the vernal equinox is in Pisces and drifts towards Aquarius, heralding the "Age of Aquarius".

In the 30,000+ year period covered by my program, the first point of Aries makes a complete circuit (and a bit more) of the Zodiac, even ending up in Ophiuchus for a while.

Of course, the constellation boundaries I mention above are those defined by the IAU. I don't know how (or if) the ancients drew their constellation boundaries.

I am leaving my old answer below, because NASA is a trusted site for astronomers, and WebGeoCalc and HORIZONS are important resources. The step-by-step method I show below can help people here make their own calculations or help others. I am in no way affiliated with NASA or the government of the United States, except as a citizen.

EDIT: As I should have guessed, people have already done this: https://cafeastrology.com/astrologyof2017horoscopes.html (and I'm guessing there are many many other sites too, just google "Mercury enters Aries" "Venus enters Taurus" type stuff).

I hope to provide a better answer later, but for now:

Visit http://wgc.jpl.nasa.gov:8080/webgeocalc/#CoordinateFinder and do this:

and click "Calculate". This will tell you every time Mercury has an ecliptic longitude of 30 degrees, the cusp between astrological Aries and Taurus.

If you do this for all 7 planets and every multiple of 30 degrees (12*7 = 84 times), you can use the results to see when the combination of constellations change. However, this is has several problems:

It's extremely inefficient
If you hover over "Solar System Kernels kernel set" above, it says the kernels are valid from 1950 to 2050. However, with the options I chose, you can only get results from 1966 to 2034, possibly because the ECLIPDATE frame is only defined in that interval (when I choose years outside this range, I get an unhelpful and inaccurate error-- you may get different results for other planets).
The results are in calendar times; however, this may be my fault. You may be able to get the results in a more numerical and useful format.

The "correct" way to do this is to use CSPICE (http://naif.jpl.nasa.gov/naif/tutorials.html) which lets you make calculations +- 10000 years from now.

Leaving that aside, let's check our results. If you scroll down a bit in the results window, you'll see:

Let's check the result for 2017-04-20 by visiting http://ssd.jpl.nasa.gov/horizons.cgi and entering these parameters:

to get these results:

As you can see, Mercury's ecliptic longitude crosses 30 degrees between 2017-04-20 and 2017-04-21, so our results are reasonable. Interestingly, Mercury is in retrograde at this time, so it actually crosses from Taurus into Aries, not vice versa.


To find the mass and radii of hill sphere of a planet after passing through a region of planetary nebula

I have come up with a technique to find a solution to this problem as shown in this paper
A Study to Show : Jupiter : Our Next Star
My 11th grade research paper 2018


Is the Milky Way seen as how it looked millions of years ago?

Most of the matter in the Milky Way is contained within a radius of 100,000 light-years from the galactic center. This encompasses the disk and most of the halo, including halo stars and globular clusters. It also includes quite a few (though not all) of the Milky Way's satellite galaxies. This is rather large - the Sun is only about 25,000 light-years from the galactic center - but not truly inter-galactic scales. Andromeda, the nearest large spiral galaxy, is 2.5 million light-years away, and so we see it as it was 2.5 million years ago. However, we see most of the objects in the Milky Way as they were no more than 100-200,000 years ago, usually much closer to the present day.

It's true that some galaxies are a couple million light-years across, but they are quite rare. In fact, the largest known spiral galaxy, Malin 1, is a mere 650,000 light-years in diameter. I should make it clear, though, that most of these galaxies are large because they have extremely large halos, consisting of gas, pockets of stars, and dark matter. Most matter is not contained in the halo. Additionally, it is not always clear where a galaxy ends and where the intergalactic medium begins.


How are neutrinos produced in blazar jets?

Neutrinos are typically produced in AGN jets through what we refer to as hadronic processes. Protons are accelerated to relativistic speeds and interact with nearby photons. Depending on the particular type of AGN, there are a number of possible sources of these photons:

Ambient light within the jet
Emission from the accretion disk
Photons from the broad line region surrounding the AGN
Emission from the nearby dust torus

As I understand it, contributions from the latter two sources are less prominent in blazars, although they may play an important role in the radio-loud flat spectrum radio quasars (FSRQs). Murase et al. 2014 have an excellent diagram showing the possible sources of this radiation:

The photons interact with the protons in the jet and produce other hadrons, typically pions. Neutral pions decay to a pair of photons, one key mechanism of gamma ray production in AGN jets:1
$$\pi^0\to\gamma+\gamma$$
However, it is quite possible for the proton-photon interactions to produce charged pions ($\pi^{\pm}$), which then decay to muons and muon neutrinos (Atoyan & Dermer 2002). The muons decay to electrons, while the muon neutrinos may be observed here on Earth. This appears to have been the source of the neutrino detected by IceCube in 2018, originating from the blazar TXS 0506+056.

1 The other main set of gamma ray production pathways in the jets are the leptonic processes, where electrons and positrons transfer energy to ambient photons via the inverse Compton effect; see Degrange & Fontaine 2016, which I found quite helpful when learning about those mechanisms.


Does the critical density of the Universe also include Dark Energy?

Yes. Quoting values from wikipedia (which in turn cites Ade et.al. in Astronomy and Astrophyics 517), the contribution of matter (both Dark and visible matter) is
$$Ω_\text{mass} ≈ 0.315±0.018$$
The contribution of photons and neutrinos is small, and within the boundaries of error of the other terms:
$$Ω_\text{relativistic} ≈ 9.24×10^{−5}$$
And the contribution of Dark Energy is:
$$Ω_Λ ≈ 0.6817±0.0018$$
$$Ω_\text{total}= Ω_\text{mass} + Ω_\text{relativistic} + Ω_Λ= 1.00±0.02$$


The formula in the wikipedia article explicitly uses integer division with round toward zero. Python's integer division uses round toward negative infinity (i.e., floor).
The wikipedia article formula repeatedly uses (m-14)/12. This evaluates to -1 for months 1 and 2 (January and February), zero otherwise. You can use the wikipedia article formula in python3 if you change every instance of (m-14)/12 in the wikipedia article formula to any of the following:

int((m-14)/12)The single / performs floating point division. Python's built-in int function rounds toward zero.
((m+9)//12-1)The inner (m+9)//12) evaluates to 0 for January and February (m=1 and 2), 1 for all later months.
(-1 if m<3 else 0)While more verbose, this does a better job of conveying the intent of the wikipedia article's (m-14)/12.