Would a killer asteroid shattered into thousands of pieces produce the same devastation? If we blow up a global killer (in the 3 - 20km range) days before impact into enough pieces that none of the them exceeds, let's say, a 30m diameter and the impact pattern is dispersed over most of the athmosphere, would that be a possibility that the impact energy is soaked up with less catastrophic consequences than the hit of the asteroid in one piece? Or would the distributed energy heat up the athmosphere so much that it would kill higher life instantly or eventually over chemical/biological chain reactions?
Or would the early impacts heat up or disturb the athmosphere so much that later (in the range of seconds or minutes) impacts would experience not enough friction so that there would be a surface bombardement with millions of little pieces?

TL;DR version: Too big and way, way too late.

The dispersal can't be done, even at the lower end of that 3-20 km scale. Holsapple claims 5 kilojoule/kg are needed to disrupt and disperse a solid 1 km asteroid asteroid, with energy scaling with radius1.65. Disrupting and dispersing a solid 3 km diameter asteroid with a density of 3 g/cc would require a deeply buried two Tsar Bomba weapon. The energy needed to disrupt and disperse a 20 km diameter solid asteroid is beyond anything humanity can achieve.

This disruption and dispersal will not be uniform. It will instead create a few big chunks, perhaps 1/3 the diameter of the original body, a larger number of intermediate sized chunks, and thousands upon thousands of little chunks. Those big chunks are still civilization killers. The intermediate-sized chunks will make Tunguska look small, and the thousands upon thousands of little chunks -- we recently saw what a little chunk can do, over Chelyabinsk.

What if it's a rubble pile or a comet? While they're not quite as dense, they need much, much more energy to disrupt and disperse. Rubble piles and comets are very good at absorbing impact energy. Disrupting and dispersing a one km diameter comet is beyond the scope of human technology.

The time scale, two days, is extremely short. A decade of advanced notice is considered to be very short notice when it comes to diverting a one km impactor. Even larger ones require even more advance notice. A century's advance notice might be enough time to deal with a 20 km diameter object.

Holsapple, "About deflecting asteroids and comets," Mitigation of Hazardous Comets and Asteroids: Vol. 1, Cambridge Univ, 2004.


Which software is appropriate to integrate radial velocity with respect to time I'm a high school student and currently I'm trying to use Baade-Wesselink method to independently derive period-luminosity relation of Cepheid variables. I noticed that to get the linear radius variation of a Cepheid, I need to integrate its radial velocity over its period.

The problem is, I don't know which program/software I should use to plot the best-fit line going through the points of the raw data of radial velocity over the phase. Even more, the thing is I need to integrate that best-fit line and plot the integrated curve as well.

Can anybody recommend good software that I can use?

(I've tried the integration with a program called Logger Pro but it was not successful. )

I'm not sure what you are actually trying to do. Radial velocity isn't generally measured, it is computed from the projected velocity. But if I take you at your word, that your raw data is radial velocity, then that simplifies things.

If you know the period, $P$, and you know the $v$ AND if you know the theoretical relationship between them $v = f(p)$ then you can use linear least squares to compute the adjustable parameters of $f(p)$ by plotting $v$ (measured) vs $v$(calculated).

Any 'best fit' equation is able to interpolate values of $v$ so all you need to divide up the period into small intervals and approximate the area under the curve using its two end-points (it is rarely necessary to use parabolic fits that also use the mid-points of those small intervals).

Thus the problem of integration becomes one of simple geometry (you have a tall thin rectangle with a triangle on top of it for each interval). Let's think about a star with a 30 day period. That's ~40,000 minutes, so I'd probably use 5 or 8 minute intervals ...starting at minute 0 what's the area under the curve from t=0 to t=8 minutes? then from 8 to 16, all the way up to 43,200th minute. Sum of those areas is total area. An Excel spreadsheet can easily do that.

Actually, if you also use 5 minute intervals and compare the results you'll see how much better (hopefully NOT much) a smaller interval is...analytical integration is this taken to the limit of the interval approaching zero, but in real life, numerical integration is necessary for almost all 'interesting' problems.  Here are 2 links, fwiw:

https://www.physicsforums.com/threads/finding-periodic-best-fit-equation-for-data-set.715638/
http://www.aanda.org/articles/aa/pdf/2004/08/aah4718.pdf ← see page 537 and references noted there.
The Institute for Astronomy (IFA) at the University of Hawaii has several pdfs on BW that may be helpful


Could the merger of two rotating black holes create another that breaks the rotational speed limit? The black hole GRS 1915+105 was observed to be spinning at near the theoretical maximum spin rate.

Assuming you had two such black holes spinning clockwise, and also orbiting each other clockwise, what would happen if they merged? How could the new black hole avoid breaking the rotational speed limit? Where would the extra angular momentum go?

The simple answer is that no, black holes cannot merge into a new black hole with a spin greater than the maximum allowed spin. The reason being that such an act would cause the black hole to show us a "naked singularity". Black holes have an event horizon which shields their singularity from observation by the outside universe. As they spin faster, that event horizon shrinks. At some point, it completely disappears, causing the singularity to become "naked" and observable to the universe. Some believe that naked singularities cannot possibly form, mainly because if they could, causality and physics would break down. This concept is known as the cosmic censorship hypothesis.

I found a statement concerning this matter in Numerical Relativity: Solving Einstein's Equations on the Computer, which I believe is relevant (emphasis mine).

  Black hole spins that are aligned with the orbital angular momentum
  increase the binary's total angular momentum. If this total angular
  momentum exceeds the maximum angular momentum of a Kerr black hole,
  then the binary cannot merge until a sufficient amount of angular
  momentum has been radiated away. Quite generally, we expect binaries
  with black hole spins aligned with the orbital angular momentum to
  merge more slowly than binaries with spins that are anti-aligned.
  This effect, sometimes referred to as "orbital hang-up", has been
  explored with numerical simulations.

One such numerical simulation looked at aligned and anti-aligned mergers and found that the aligned black holes took much longer to merge and radiated away more energy, in the form of gravitational waves, before they merged.

There is another way to burn off some of that spin angular momentum too. When spinning black holes merge, they experience a "kick" in their linear momentum. In other words, they suddenly speed up in their motion through space. This kick is a result of converting some of the two distinct black holes' orbital and spin momentums into the merged black hole's linear momentum. Quoting the same book as above concerning these kicks:

  Most of the initial calculations focused on black hole spins that are
  aligned or anti-aligned with the orbital angular momentum. The
  resulting kicks are several hundred km/s in magnitude, easily
  exceeding the maximum kick of approximately 175 km/s found for
  non-spinning black holes.

In summation, two spinning black holes cannot merge into a black hole which is spinning faster than the maximal rotation. Gravitational waves and the linear momentum "kick" play an important part in helping binary black holes to lose spin energy, such that the merged black hole doesn't exceed a maximal rotation.

I think the really interesting question that should be asked is, how do two black holes know, before merging, that their merger cannot happen as it would violate the cosmic censorship hypothesis? How do they know they need to radiate away extra energy before they can merge? What is the mechanism that prevents them from merging? Only time can tell for now.


Why does earth not collide with the sun? Gravity is most accurately described by the general theory of relativity (proposed by Albert Einstein in 1915) which describes gravity not as a force, but as a consequence of the curvature of spacetime caused by the uneven distribution of mass.(source -wikipedia)

Why it is moving on the boundary of that curvature, but not fall into the middle of that curvature and collide with sun?

The primary reason the Earth doesn't fall into the Sun is that it has a very large tangential velocity with which it is able to maintain an orbit. The physics is the same for describing satellites which we launch into orbit around the Earth. I've heard it described as such: "The ISS is falling towards the earth, but is moving forward so quickly that it continually misses."

In short, if you were to take a mass and insert it with zero velocity somewhere in the solar system, it would inevitably fall in towards the sun. But because almost all bodies in our solar system were formed from the same large clump of rotating gas and dust which due to conservation of angular momentum during its collapse both sped up and flattened into a disk, all of the solar system bodies (Earth, other planets, asteroids, etc), have enough forward velocity to continue orbiting the Sun without being pulled to the center.

From a physics perspective, you can see that centripetal acceleration is governed by the inward force of gravity, tangential to the direction of motion of moving objects.

In fact, the Solar Parker Probe which just launched is attempting to do exactly as you describe --- fall into the Sun (or well, get close, to take lots of measurements). To do so is actually difficult, because the probe, launching from Earth, has to lose roughly 30 km/s in velocity it already has (because that's the Earth's orbital velocity).

How it plans to do this is a cool story in and of itself, which you might be interested in, here's a good summary: https://www.theatlantic.com/science/archive/2018/08/parker-solar-probe-launch-nasa/567197/


How to compare a Catadioptrics light asorbtion with a reflector? I used to have a 6" Dobsonian reflector, and the next telescope I buy, I want it to have the equivlant of about a 11/12" dobsonian reflector.

I'm looking at this Nexstar VX 9.25 , it says 235mm aperature, but is thats more than what a 235mm reflector would acheive right? How close is to a 11/12" reflector?

Comparing optical instruments is very difficult. There are numerous variables involved, the performance is not a single number, and the theory is complex. Perfect scenario to baffle amateurs. Also, it sounds like you may have been listening to the wrong rumors.

The architecture of a telescope (refractor, reflector, catadioptric) is one of the least important factors in the performance analysis, regardless of how you define "performance".

Don't forget that how you maintain and where you use the scope are very important factors. Collimation will strongly impact resolving power - so make sure it's always perfect. Light pollution will impact you much more than light absorption within the instrument. Etc.

In terms of amount of light gathered, aperture is the dominant factor. 2x the aperture is 4x the area, so it's obviously very important. The central obstruction may factor in a little, but the effect is small. Going by diameter, a typical newtonian will have a 20% central obstruction; a typical catadioptric scope will have a 40% c.o. But if you translate that into areas, the newtonian is at 4%, whereas the catadioptric is at 16% - both are very small.

In terms of resolving power, again aperture is king. Bigger aperture = more resolving power, no matter what the architecture. But poor collimation will quickly destroy your gains, so keep that scope collimated.

In terms of internal losses, the newtonian will do a bit better - it's two reflections and that's it. The catadioptric will have two or more reflections (depending on design), plus absorption / reflection on the diopters. Again, with modern instruments, the difference is tiny.

Overall, at the same aperture, in theory, the newtonian pure catoptric scope delivers a bit more light to the sensor, compared to a catadioptric scope. In practice, the difference is far outweighed by other factors and you should completely ignore it.

You should make the purchase based on more important things. What's the size of the usable field? What kind of adapters can you use? Does it provide enough back-focus? What's the aberration in different points within the field? Is the telescope too heavy for the driven mount? Is it long focal ratio, or short? (each kind has different applications) Finally but not least important, price?

It's more common to see catadioptrics used as astrographs. That's not to say that a newtonian reflector cannot do the job - but if you're still new to the field, just do what most people do and get a catadioptric, with a good mount.

Final note:

When doing prime focus photography, the brightness of the image is given by both aperture and focal ratio - but not all objects behave the same.

For stars and other point-like objects (smaller than the resolving power), brightness is driven by aperture. Bigger aperture = brighter stars.

For extended objects (bigger than the resolving power), such as nebulae, many galaxies, the Moon, the planets, etc., the brightness of the object is given by the focal ratio. An f/4 scope will make brighter nebulae in prime focus, compared to an f/8 scope. Aperture doesn't matter. It's just that, at the same focal ratio, the bigger scope will also make a bigger image in prime focus.

This is true only for prime focus photography. If you're doing eyepiece projection, or something else, different rules apply.


What did I photograph? Last night I was experimenting with my mobile phone shooting the night sky adjusting some settings like shutter time and ISO. One of the first pictures I thought was a complete failure, I only adjusted the shutter time to 30 seconds. But when I zoomed in on what I believed to be just some bright stars, I was amazed about what I saw.

Here's the original picture

I zoomed in on the center most bright star and saw this

Then I zoomed in on that star at the bottom just a bit right of the center and saw this.

So I'm wondering what these are? Did I shoot some deep space objects? Or is this just static from the long exposure? Or maybe just pixalating from the zooming? I'm really curious about your answers.

My camera was just pointing at zenit and my location was 50°57'35.5"N 5°05'10.3"E the time was 1:06am on august the 6th 2018.

Here's a screen with some more information about the photo.

A contrast stretch reveals stars down to magnitude 4 or 5.
The stars you asked about are Deneb (center) and Vega (bottom).
The constellation in the center and below is Cygnus; we also see Cepheus at upper right and part of Draco at lower right.

The stars appear as blobs 10-12 arcminutes wide due to several factors:

Rotation around the north celestial pole, to the right of the upper right corner of the image. In your 30 second exposure, Deneb and Vega drifted 5-6 arcminutes in the 5 o'clock direction.
Diffraction due to finite aperture. The width of the ring pattern is inversely related to the width of the aperture. These rings are consistent with a 2-3 mm aperture.
Optical aberrations such as coma, increasing toward the edge of the image.
Vibration as noted in another answer.


Why are telescope mirrors nearly flat? My understanding of telescope mirrors is that they are (generally speaking) a slice out of an imaginary sphere. The center of this sphere is the focal point of the telescope, where the detector is placed. The bigger the sphere, the further away the focal point will be.
With this in mind, why are telescope mirrors nearly flat? Would it not make sense to instead use the same diameter mirror, but with a smaller imaginary sphere so the detector doesn't have to be so far away?
I have made a crude image to demonstrate my point. The blue arc and point represent a more traditional "nearly flat" mirror, whereas the red arc and point represent a mirror that has the same diameter but with a smaller sphere and therefore closer detector.

The main reason is that the red mirror produces a smaller image in its image plane, which requires a smaller pixel spacing in the detector (or a stronger eyepiece) to get the same resolution.
In a Newtonian telescope the focal length cannot be too small:

(Image Krishnavedala)
Also, a more ideal shape is a paraboloid (not a sphere). Spherical mirrors are relatively easy to manufacture, and "nearly flat" ones are very close to paraboloids.


Distance between two events I would like to start from the very basic such as Pythagoras Theorem.
If I want to calculate distance between two points in a right-angled Triangle We use
$c^{2}= a^{2}+b^{2}$; Using this We are quite capable to calculate the distance between two points in a 2D space. For 3D  this formula is slightly modified and We use $ds^{2}=dx^{2}+dy^{2}+dz^{2}$. But for a 4D we used to calculate distance (rather interval between two events) through this equation
$\boxed{ds^{2}=dx^{2}+dy^{2}+dz^{2}-c^{2}dt^{2}}$
Can anybody explain this modification?(looking for mathematical motivation)

We could define a distance $ds'$ in four-dimensional space-time as
$(ds')^2 = (dx)^2 + (dy)^2 + (dz)^2 + (cdt)^2$
This would be a Euclidean distance. The problem is that experimental observation shows that observers in different inertial reference frames disagree when they measure this $ds'$ distance between the same two points (events) in spacetime. This suggests it is not a physically meaningful value, since it depends on how the observer is moving.
However, if we use the Minkowski distance $ds$ defined by
$(ds)^2 = (dx)^2 + (dy)^2 + (dz)^2 - (cdt)^2$
we find that observers in different inertial frames agree on the value of $ds$ (at least if we restrict ourselves to flat spacetime), so this is an objective value (independent of how the observer is moving) and so more useful.
An everyday analogy is that when talking about locations in England (say) it is more useful to use the objective measure "distance from the centre of London" than the subjective measure "distance from where I am right now".


What is the Northernmost Latitude of Saturn? My wife and I were watching a TV show last night, which takes place in Salem, Massachusetts circa 1693.  One character went to a particular place in the woods at a particular time to be exactly under Saturn.  I told my wife that Saturn could never be exactly above such a Northern latitude.

What is the Northernmost Latitude that Saturn could possibly be above?  By "above", I mean if you draw a line through the center of the Earth and through Saturn, the point on the Earth's surface that line passes through.

Also, how often would it reach that point?  Would it be every few years, or every few thousand years?

Salem is at 42.5° north. At the equator, the position of the sun varies by 23.5 degrees on either side of the vertical over the course of each year, due to earth's 23.5° axial tilt with respect to the ecliptic.

Saturn, however is not on the ecliptic. It departs by about 2.5° over the course of its 29.45 year, 9.5 AU orbit. So once every 29.5 years Saturn's angle to the ecliptic will add 2.5° to the max latitude the planet can be seen overhead.
That adds up to 26° as the max latitude Saturn can be seen directly overhead.

There's one more factor to account for though. Earth being 1AU out from the sun, it's only 8.5 AU from Saturn at close approach. That will add another 0.2° (by trig) to the max latitude at that time of year when earth is nearest Saturn. That gets us to the possiblity of Saturn appearing directly overhead at up to latitude 26.2°: Miami, Jaipur, or Dubai, not Salem


Do celestial objects need to be big to have liquid water on their surfaces? I mean no asteroid, planetoid that I am aware of has water on its surface. It is way more common to see ice in it. So I figured that the size of the celestial body has something to do with the cycle of water. Does this theory hold true? Do exoplanets with water have to be similar size than Earth? To have water does it has to be a rocky planet?
I mean some moons have water inside their crust (I think). Some planets seem to have had water before, like Mars.

Liquid water can't exist in a vacuum. If there is no pressure, then the boiling point will drop to the freezing point and so there will either be ice or water vapour.
And if the world is "small" then its gravity won't hold on to any water vapour, and it will be lost to space.  The Earth can have liquid water because its gravity is strong enough to hold onto water vapour, and provide vapour pressure to raise the boiling point to 100 degrees, which is hotter than the temperature due to the sun.
A small world can have a sub-surface water layer, as the ice above it will stop the water from boiling off into space. Enceladus has such a layer, but if Enceladus was a moon of the Earth, the sun would have melted the ice and the water would have boiled off long ago.


How are the masses of celestial bodies determined? I was scrolling through the list of moons of Jupiter and I wondered how astronomers where able to determine the masses of those objects. When I tried to google it I meanly found results of people deriving the mass of Jupiter from the masses of its moons using Kepler's laws. So the mass of celestial bodies with satellite(s) can be estimated this way. But what about the rest? Or are they all derived from the trajectories of man-made satellites passing near them?

Newton's analysis of Kepler's laws lets you determine the ratios of the masses of everything in the solar system with quite high precision, but it doesn't let you determine the actual mass.  The actual masses were one of the large unsolved problems of physics for almost a century, until 1798 when Cavendish's work with a torsion balance (described by him as "weighing the world") determined the density of Earth, from which the masses of everything could be computed.


How to calculate shape of orbit when projected onto the plane of the ecliptic I'm trying to generate an 'overhead' view of the orbits of the moons of Neptune, similar to Figure 4 in this paper, using values given here. I initially tried using both the Argument of Periapsis, and the Longitude of the Ascending Node separately but neither seems to give the correct results. After a bit more reading I now think I will need to combine both these values to derive the orientation of the major axis of the ellipse, but I can't find the formula anywhere. Am I correct? Could someone show me the correct calculation?

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


How wide is the Moon's umbra and penumbra at EM-L2? Measuring the total width of the Moon's umbra and penumbra at EM-L2, 64,700 km, how wide would the shadow be?

The angle between the inner and outer edges of the penumbra is the same as the Sun's apparent angular diameter.
At 1.0 au from the Sun, that's 32 arcmin or 0.0093 radian.
At the distance from the Moon to EM-L2, this angle spans 0.0093 * 64700 km = 602 km.
The Moon's diameter is 3474 km, so the umbra diameter would be 2872 km and the penumbra diameter would be 4076 km.


Orange light that rises quickly and moves of to the righ I have recently noticed an orange "star" (or what I presume to be a star), rise above the hill line behind my home. It rises later than every other star and rises within a few hours while the other stars are stationary! It moves off to the right when half way in the night sky this happens every night when clear for the past few months. I also believe I'm looking south. It's really bugging me. Could anyone identify this object from my description?

The object is (most likely) Mars! It's been rising in the south as of late. When it will rise of course depends on your location...for me it rises pretty close to sunset. Planets appear in an essentially constant position relative to the background stars over the course of the night, and as with the stars, the movement you see is due to the rotation of the Earth. Over the course of weeks, however, you can see the planets change position relative to the stars due to their motion in their orbit around the Sun. We don't see this wandering of position with stars - at least, not in an amount or on a timescale we can appreciate - because they're so far away.

Keep an eye out for other bright objects in the sky that appear to move faster and (unless conditions are poor) don't really twinkle, especially near sunset before the stars properly come out - Mars probably isn't the only planet you can see! Jupiter, Saturn and Venus are often visible as well.

Check out this site if you want to see what planets are visible from your location and when.

https://www.timeanddate.com/astronomy/night/

ETA: Corrected explanation of planetary motion vs. motion of stars based on James' comment.


How do I download PSRCHIVE and Tempo2 for pulsar timing on Mac Mojave? I am fairly new to pulsar astronomy, and I need to download PSRCHIVE and then Tempo2. I have followed the instructions online here:

PulsarAstronomy.net

I'm also pretty new to linux commands/Mac terminal in general, so I think I might be missing something obvious.
So far I have untarred the tar file, which gave me a directory 'psrsoft' in Downloads.
I also attempted to open a psrsoft bash profile, but I don't think it worked because none of the psrsoft commands worked.
The instructions mention '$PSRSOFT_DIR' a lot, and I am not sure where that is supposed to be or what it represents.
Right now, my goal is to get the package installation options listed and then from there install Tempo2.

This is what psrsoft contains:

Downloads/psrsoft>bin: pkg_files, psrsoft, psrsoft-selfupdate, psrsoft-shell, psrsoft-update-index, psrsoft.py, shell_files, usr

Downloads/psrsoft>config: profile.example

Any help would be greatly appreciated.

The documentation is somewhat confusingly written. It refers to a $PSRSOFT_DIR as a shortcut to where you unpacked the tarball, which everything is done relative to but doesn't require this shell variable to be set. So there is 1 thing to do and then you can do 1 of 2 things.

The mandatory thing (as it says in the docs) is to copy the psrsoft/config/profile.example to psrsoft/config/profile and edit if necessary (I didn't need any changes but I'm running on pretty vanilla CentOS/RHEL 7; MacOS X may well be different). Then your two options are:

You can then manually "substitute in" for the unset $PSRSOFT_DIR variable e.g.  run /path/to/Downloads/psrsoft/bin/psrsoft psrchive when the docs say to run $PSRSOFT_DIR/bin/psrsoft psrchive
You can set PSRSOFT_DIR in your user's shell startup files such as ~/.bashrc or ~/.bash_profile (~ is a shortcut to your home directory which works on Unix/Linux and I assume MacOS X) by adding export PSRSOFT_DIR=/path/to/Downloads/psrsoft/. You should then be able to run the commands as shown in the documentation.


Road map for a faster spaceship New Horizons, speediest yet, is currently running at about 15 km/sec relative to the sun.  What is the roadmap for future ? How is it possible to reach much faster speeds ?

Here's a Wikipedia page discussing various theoretical methods of propulsion. It has some very interesting ideas.

The one that piqued my interest was the Alcubierre drive. For all intents and purposes it is a warp drive - well, as close as science can get to the fictional engine of The Enterprise anyway. It has been nicknamed the warp drive, I believe, by NASA scientists.


Why is there billowing dust in nebulae? Why do some nebulae look like the billowing clouds? Wouldn't that require something like a viscosity in empty space?

This is a very complex answer. It has to do a lot with gravity and electromagnetism. Nebulae are made of particles that act according to those forces and give you the illusion of viscosity. Those particles are actually moving. Take a look at this simulation it might give you an idea: Simulation


Problems with FITS headers I'm working with some NEOWISE images that seem to have incomplete header tags. I've been advised by the SAO (because I'm using DS9 to view them) that I have

"a partial PC matrix defined, yet it does not conform to the FITS WCS standard. With a combo of CDELT and PC keywords, the CDELT keywords hold the scaling factor, the PC keywords, the rotation matrix, which should be normalized. If you want the rotation matrix to also contain the scale factor, use CD keywords".

Here's one of the headers I need to fix:

SIMPLE  =                    T / conforms to FITS standard
BITPIX  =                  -32 / array data type
NAXIS   =                    2 / number of array dimensions
NAXIS1  =                 1459
NAXIS2  =                  903
WCSAXES =                    2 / Number of coordinate axes
CRPIX1  =                729.5 / Reference pixel for axis 1
CRPIX2  =                451.5 / Reference pixel for axis 2
PC1_1   =     -0.0333333333333 / Coordinate transformation matrix element
PC2_2   =      0.0333333333333 / Coordinate transformation matrix element
CDELT1  =                   1. / [deg] Coordinate increment at reference point
CDELT2  =                   1. / [deg] Coordinate increment at reference point
CUNIT1  = 'deg     '           / Units of coordinate increment and value
CUNIT2  = 'deg     '           / Units of coordinate increment and value
CTYPE1  = 'GLON-AIT'           / Projection for axis1
CTYPE2  = 'GLAT-AIT'           / Projection for axis1
CRVAL1  =                   0. / [deg] Coordinate value at reference point
CRVAL2  =                   0. / [deg] Coordinate value at reference point
LONPOLE =                   0. / [deg] Native longitude of celestial pole
LATPOLE =                  90. / [deg] Native latitude of celestial pole
RADESYS = 'ICRS    '           / Equatorial coordinate system
BAND    = 'W1      '           / Wise Band
END

I've already managed to fix a couple of the invalid tags, but apart from these minor edits, this is unaltered from the file that my research supervisor gave me. The FITS standard here gives descriptions of each of the tags, but I'm not sure how to implement the SAO's advice. Can anyone help me fix the tags?

You are missing the PC1_2 and PC2_1 elements of your pixel transformation matrix. If your coordinate transformation is a rotation, you would use

PC1_1 = cos(t)
PC1_2 = sin(t)
PC2_1 = -sin(t)
PC2_2 = cos(t)

for the PC matrix. The pixel scale (in degree/pixel, according to CUNITi) goes into CDELTi.

As pointed out, you can combine the PC and CDELT steps into a single CD matrix.

See here and here for more details.


Why do we think that there is no two-solar-mass black hole? We think that the mass boundary between neutron stars and stellar mass black holes is  around three solar masses.

The maximum mass of the neutron stars now is two solar masses and we may find a 2.6 solar mass neutron star in the future. That is based on the equation of state.

The stellar mass black holes may have different origins. Even the smallest stellar mass black hole originates in SN explosion,  we do not know the details of the explosion.

Why do not we think a two solar mass BH, or even a smaller BH, could exist?

Why do we think that the companion of Taylor-Hulse binary is a neutron star instead of a BH?

As you said in your question we only know about Black Holes creating in Super Novae, which lead to relatively big BH. The reason they can't be smaller in these cases it that they are created when the core of the star reaches a critical mass that overflows both the electron and the neutron degeneracy. Lower masses can be hold by those degenerate mass states and won't collapse any further.

Theoretically BH with masses less than the solar one could exists if there was some method capable to compress it's mass reaching radii smaller than their Schwarzschild radius.


Hypothetical trans-Neptunian planets orbiting the Sun The planet HD 106906 b was found at 650 AU from its home star, a star not so different from the Sun. Fomalhaut b is also a young planet that is orbiting very far away from its star.

These two planets were discovered mainly because they are young enough to still be hot enough to be seen in the infrared. I think that old planets in very high orbits would have temperatures below 10K, and thus be black and invisible.

Some people agree with the hypothesis that, in our Solar System, somewhere out beyond Neptune, some non-dwarf planets might be orbiting the Sun. The existence of HD 106906 b and Fomalhaut b suggests that this is possible. What are the odds that we could find at least one of these planets in our Solar System? If they do exist, how could we find them?

If such a planet exists, WISE should observe it. WISE is an infrared satellite that imaged the entire sky. In particular:

it was able to detect anything with a temperature above 70-100 K, whereas the coolest known exoplanets are in the 100 K range (see histogram below, taken from the Exoplanets Catalogue);
it was able to detect objects larger than 1km up to 3 AU from the Sun, or objects of 2-3 Jupiter masses in a distance up to 7-10 light-years (so roughly up to 440,000 to 630,000 AU, see here and here);

so I would say that its existence is quite ruled out now.


How do we define temperature in outer space? I was recently reading an article on space.com about : What's the Temperature of Outer Space.

They said :

  Some parts of space are hot! Gas between stars, as well as the solar wind, both seem to be what we call "empty space," yet they can be more than a thousand degrees, even millions of degrees.

But here is my question - "How do we define temperature in outer space?"

We perceive temperature as the average translational kinetic energy associated with the disordered motion of atoms and moleculules, so how can the temperature be so high at some places, if there isn't any (or very, very few) molecules out there in vacuum?

First of all, the medium they are mentioning is far from empty even though it is much less dense than what we do know on Earth (see this question for more details).

Then, the problem with temperature in astrophysics is that the medium you study is in general far from thermodynamic equilibrium (thermodynamic equilibrium means that there are no flows of enery or matter within the system and with the outside of the system).

Most commonly, however, we can use another flavor of equilibrium to estimate temperature. This is the kinetic equilibrium: since most collisions are elastic (meaning that the energy is conserved), particles velocities will follow a Maxwell-Boltzmann distribution. Depending on the ionization fraction (how much is your medium ionized?) this temperature will either the kinetic temperature of electrons or the kinetic temperature of hydrogen.

Then, if you want to understand these high temperatures, you have to take into account the photoionization rate and the photoelectric heating (a process in which you eject electron from the matter of the studied medium due to UV absorption). The thermal state of the medium then depends on the equilibrium between the energy absorption by the matter and the re-emission of this energy in thermal radiation. In this framework, the energy that can be absorbed comes from the interstellar radiation field (ISRF). In the solar neighborhood, it is dominated by six different types of radiation: galactic synchrotron emission of relativistic electrons (electrons accelerated to tremendous speed in the galaxy), the cosmic microwave background, the infrared emission of the matter heated by the light emitted by stars, emission from ionized plasma, the light emitted by stars itself, and lastly X-rays from hot plasma.

References:

Klessen & Glover (2016)
Ferrière (2001)


Equinox sunrise sunset direction Looking at sunrise and sunset data around 2017 vernal equinox (because of GMT timezone I took London) https://www.timeanddate.com/sun/uk/london

Date  Sunrise      Sunset        Day Length  Solar Noon
16    06:12 (92°)  18:07 (269°)  11:54:55    12:09 (37.0°)
17    06:09 (91°)  18:08 (269°)  11:58:53    12:08 (37.4°)
18    06:07 (90°)  18:10 (270°)  12:02:52    12:08 (37.8°)
19    06:05 (90°)  18:12 (271°)  12:06:50    12:08 (38.2°)
20    06:03 (89°)  18:13 (271°)  12:10:48    12:07 (38.6°)
21    06:00 (88°)  18:15 (272°)  12:14:47    12:07 (39.0°)
22    05:58 (88°)  18:17 (273°)  12:18:45    12:07 (39.3°)

We see that 12-hours day happens around 17-18 of March, however the Vernal Equinox takes place on 20 March 2017, 10:29 GMT.

I understand that due to refraction in the Earth's atmosphere and due to the fact that sunrises and sunsets are defined by the top of the Sun's disk, sunrises are "earlier" and sunsets are "later", so the 12-hours day indeed should happen before the equinox.

But why the directions of sunrise and sunset are east--west on that same day? The above factors should not influence directions, should they?

Due to the atmospheric refraction the Sun rises earlier and sets later as you correctly wrote. Refraction however is perpendicular to the horizon. So if the apparent sun rise is earlier, than the true (actual) position of the Sun will be more to the north east (smaller azimuth) than when the true sun rise occurs. As refraction is perpendicular the apparent sun rise will also be more to the north east and hence the azimuth will have a lower value (88-89° instead of 90°).

NB. This is of course only true from the northern hemisphere.

NB2. The Sun is not to scale in the figure below.


How to understand orbit definitions I've been reading about satellite observations (such as this one).
The writer uses terms such as these to define the orbit:

1 41584U 16036A   16167.96105997 0.00000000  00000-0  00000+0 0    07
2 41584   7.5055 353.7008 0046333  41.2140 319.1375  1.00195548    05

rms 0.004 deg      from 9 obs June 14.70 - June 16.79  (2.09 day arc)

What does each variable refer to, and what's the easiest way to view these on a map (either online or in an app of some kind?)

These are known as
two-line elements
(see also nasa.gov,
satobs.org).
On both lines, the first field after the line number is a NORAD ID,
which you can use to query a database such as
N2YO.
The second line gives geocentric
orbital elements
for the epoch given in the first line.
The last line looks like an error estimate by observation analysis software.

The first convenient visualization service I found was
Orbital Predictor,
where I naively pasted the above TLE data with a name less appropriate than "Mentor 7" or "2016-036A."


Acronyms of catalogs in SIMBAD database How to find acronym of specific catalog in SIMBAD database? For example, I want to select
stellar names, object types and all binary stars from "Stellar effective magnetic fields. II." catalog (Bychkov+, 2009; J/MNRAS/394/1338) via SIMBAD script page. How to do this?

It's roughly like this:

set limit 100
format object form1 "%IDLIST(1) : %OTYPE(S)"
query sample (otypes = '*i*') & (bibcode='2009MNRAS.394.1338B')

To get the list of the bibcodes choose
"Other" / "Catalogue collection"
from the menue bar.
The screen should look like

Then click the icon "Click to display the menu" at the left border:

Here you find the bibcode:


Can a star fall in a super massive black hole without getting destroyed? This is the first of many similar questions I have to build up to a specific question or scenario that I want to explore and find an answer to.

Yes, it is easily possible, but it depends critically on how massive the black hole is. A simple calculation will suffice.

If we take a Newtonian approximation for the tidal acceleration across a star of radius $R_*$ as it reaches the event horizon
$$ a_{\rm tidal} \sim 2\frac{GM_{\rm BH}R_*}{r_s^3}\ ,$$
where $r_s = 2GM_{\rm BH}/c^2$ is the Schwarzschild radius.

The gravitational acceleration at the surface of the star, due to its own mass is just
$$a_* = \frac{GM_*}{R_*^2}\ .$$

If we take the ratio of the two, we can assume the star will survive being ripped up before it crosses the event horizon if this ratio is less than $\sim 1$. i.e. when
$$\frac{a_{\rm tidal}}{a_*}= \frac{c^6 R_*^3}{4G^3 M_{\rm BH}^2 M_*} <1$$

The tidal acceleration reduces markedly for high mass black holes. We can rearrange this inequality to give us the minimum black hole mass for survival:
$$ M_{\rm BH} > \left(\frac{c^6 R_*^3}{4G^3 M_*}\right)^{0.5} = 1.5\times 10^{8} \left(\frac{M_*}{M_{\odot}}\right)^{-1/2} \left(\frac{R_*}{R_{\odot}}\right)^{3/2}\ M_{\odot}$$

To first order, this is almost independent of the type of star swallowed. It depends on the inverse square root of the average stellar density, which doesn't vary a lot.

This means that a star could fall into a black hole like that in M87 ($M_{\rm BH}\simeq 6\times 10^9 M_{\odot}$), but  would not survive to the event horizon of Sgr A* ($M_{\rm BH}\simeq 4\times 10^6 M_{\odot}$).

I think in practice this calculation can only be accurate to factors of a few. I have not taken into account shearing forces that would be present for a rotating black hole, or the compression due to tangential tidal forces, or any heating effects from hot gas near the black hole, or interaction with an accretion disk. A full hydrodynamic simulation is needed (and perhaps has been done?).

Also note, this calculation is just for getting past the event horizon. The tidal forces grow as $r^{-3}$, so any normal star will be ripped up shortly after this, even for the most massive black holes we know of.

Please note: This answer is written from the point of view of the falling star. A distant observer will not see the star (or anything else) cross the event horizon.


Calculate apsides without knowing eccentricity How would one go about determining the apsides of an orbit, knowing only the current distance between the 2 bodies, the orbital velocity, and the period? Every formula I've seen requires eccentricity, but conveniently, every formula for eccentricity requires the apsides. Surely there must be some way to find one without knowing the other?

You mention this is a system in which a vessel is orbiting a planet. In that case it's fair to assume the mass of the vessel is negligible compared with the planet mass and
the planet will sit in one of the focal points of the ellipse.

You can then use Kepler's second law to find the semi-minor axis as
$$ \frac{1}{2} r v = \frac{\pi a b}{P} $$
where $r$ the current distance and $v$ the current velocity, $P$ is the period and $a$ and $b$ give the respective semi-major and semi-minor axes.

From $b$ you find the eccentricity as
$$ b=a \sqrt{1-e^2} $$

And the apsides follow as
$$ \mathrm{periapsis} = a ( 1 - e ) $$
$$ \mathrm{apoapsis}  = a ( 1 + e ) $$


What would the cost be of visiting an asteroid? I'm trying to think through what the cost would be of an unmanned mission to a nearby asteroid.  To me it seems like the high-level costs would be the fixed cost of the "spaceship" itself (including any scientific or mining equipment required to carry out the mission), the cost of getting into orbit which would depend on the ship mass, and the fuel to get there and return (which would be a function of fuel cost, ship mass, how fast you want to get there, and the mass of anything collected at the asteroid site that would be returned).  Is there anywhere that this is discussed or has anyone thought this through further?

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


Milky Way mass fraction breakdown review I'd like to know a breakdown of the mass of the Milky Way, by the major categories. So, something like

Stars       ... 3%
Cool nebula ... 7%
Gray matter ... 12%
Etc
Etc

This order-of-magnitude knowledge that everyone should know.

(Of course, when talking of the galaxy, we can, obviously, think of either just the visible disk-bulge, or perhaps, a greater unit of the galaxy sitting in a "bulb" (of - perhaps - dark matter (?), gas (?), dust (?), halo stars (?) or something else): a brief clear answer in a few sentences would expose these concepts.)

Surprisingly, this information is nowhere to be found in one place as an overview.

I appreciate that the largest chunk may be dark matter (is that right?); but I don't even have a ballpark idea for the rest. It's the sort of knowledge that is obvious to professionals, but surprisingly not outlined in any of the usual popular science sources.

I should clear this up: the Milky Way isn't a completely well-defined body. Galaxies shouldn't be thought of as a single body — they are more like a collection of various material and stars. That said, I can only give you a wonky estimate of what might be the composition.

By "Milky Way", I'm assuming you mean only the Milky Way's disc and halo (and not everything that orbits the Galactic Center, like the satellite galaxies). That alone is hard to define, but let's get a rough estimate. The entire mass of the Milky Way is a bit uncertain, but it falls between $0.8–1.5×10^{12}M_☉$. Let's average that at about $1.2×10^{12}$.

Sagittarius A*, the black hole at the center of our galaxy, has a mass of roughly $4,000,000M_☉$. That would make it only ~0.0003% of the galaxy! A few sources state that dark matter composes about 90% of the Milky Way's mass.
So it has a mass of approximately $1.08×10^{12}M_☉$.

If we want to dissect the rest, we'll have to make generalizations. From some sources like this one, about 75% of MW stars are M type stars, 15% are F-K stars and the remaining 10% are mostly degenerate stars. Knowing that there are 100 billion stars in our galaxy, we can take their average masses and kind of conclude that 3% of the Milky Way's mass is from stellar objects.

Regarding the mass of gases in the Milky Way, this source states that both atomic gas and molecular gas would individually make up 0.25% of the Milky Way's mass. Combined, we get 0.5%. The rest is buried in a sea of contradictions and mystery.

That said, here's the best breakdown I could make:

Sagittarius A*: 0.0003%
Dark matter: 90%
Stellar objects: 3%
Gases and nebulae: 0.5%
Everything else: 6.5%

Again, it is extremely difficult to break down the mass of the components of a galaxy. This is the best you can really do without combining tremendous generalizations with shoddy math and sources.


Why does Io cast a hard shadow on Jupiter, but the Moon casts a soft shadow on Earth? Io (source):

The Moon (source):

It's due to the larger relative apparent size of the Sun. When the source of light is a point source the shadow is harder, and when it is extended it is softer.

Jupiter is approximately 5 times more distant from the Sun than the Earth, so the Sun is approximately 5 times smaller in the sky.

*Source: University of North Carolina CS


What star is directly north of the suns north pole? If Polaris is Earth's North Star as it is directly pointed at by our North pole, what would be the North Star of the Sun?

The North Pole of the sun points towards the constellation Draco. There is no bright star close to the North Pole of the Sun. The closest fairly bright star is Delta Draco, "Altais". A magnitude 3 star.

The other marks on the image are the "North Pole of the planets' orbits" The Earth's rotation is inclined by 23 degrees from its orbit, so the Earth's rotational North Pole is not the same as the pole of its orbit.


How can I use Polar Alignment to find objects in the sky? I have a six inch telescope on an equatorial mount and I've been struggling to use the mount properly. For a couple months now, the only strategy I've used is just 'point and look.' I've watched some youtube videos on polar aligning a telescope, but none of the videos ever tell me how I can use polar alignment to find other objects after I've aligned. Is polar alignment even the right method of finding objects? If so, how can I use the degrees and minutes on my RA and DEC knobs to point directly at an object?

If a mount is polar aligned, it becomes very easy to find objects that have the same Right Ascension or Declination as a bright star. It works like this

1) you align the mount to the pole

2) using a star map, or an app, you try to find an object that has the same R.A. or declination as a bright star. As an example, use Messier 4, which has almost the same declination as Antares.

3) you point the telescope at Antares. Lock the declination axis.

4) Look through the finder, or through the scope using the eyepiece that gives you the biggest field of view. This most often this is the eyepiece that gives you the lowest magnification.

5) move the scope to the west, it rotates around the RA. axis. Messier 4 should now enter the field. Move slowly, and move back to Antares when you get lost.

6) When you see Messier 4, lock the RA. axis too, and center Messier 4 in the field using the fine adjustment controls.

7) There are a couple of other bright objects you can try. I'm not familiar with the Southern sky, but for the Northern sky, Messier 13 can be found starting at eta Her(cules), and moving in declination.

8) a bit harder is Messier 57 in Lyra, you need to start a bit to the north of gamma Lyr, or a bit to the south of beta. Another easy target is IC 4665, which is very close to Beta Ophiuchi (Cebalrai), almost but not quite to the north.

9) Other objects to try are Messier 5 starting with Beta Librae (Zubeneschamali), and Messier 10 starting at epsilon Ophiuch (Yed Posterior).

10) The next step is to try and find objects by using multiple steps, alternating between moving in RA and in declination. You would move to a star or better, to a recognizable star pattern, and continue from there.

11) This way of star hopping works best if the angles moved are not huge. Think 10, 15 degrees. As you get more experienced you will start using the stars in between as waypoints so you can move bigger distances.


Why does gravity increase in star formation? When a star ignites ( ie. fusion starts ), the star maintains its form by balancing gravity's inward pressure, and radiation's outward pressure.

I get that the fusion of hydrogen atoms releases energy... fine...

How does gravity keep it together if the mass is lessening as a result of fusion( mass being converted into energy from fusion) while gravity is weakening( as mass lessens )?

Wouldn't the radiation overpower the force of gravity and tear the star apart?

The Sun's luminosity is $3.8\times 10^{26}$ W. Application of mass energy equivalence tells you it loses mass at a rate of 4.25 million tonnes per second as hydrogen turns into helium.

This is practically nothing as far as the structure of the star goes. Over its lifetime, the Sun has lost about 0.03% of its mass in this way.

Radiation pressure is a feature in stellar evolution calculations. It is almost negligible in the solar interior (at the 1% level compared with thermal pressure). However, it does become more important in more massive stars with hotter interiors and higher luminosities.


What is the distance between the solar systems towards the center of galaxy? The distance from our Solar System to the nearest is roughly 260,000 AU. I am wondering what is the minimum distance between two Solar Systems. Since the center of our galaxy is supposed to be more dense than the spiral on which our Solar System is situated, I am guessing that the distances between adjacent Solar Systems would be smaller. Please give quantitative figures and references to peer-reviewed journals if possible.

We don't really know what the statistics are for the frequency of solar systems is as a function of stellar density and many such systems in dense regions may be disrupted by close encounters with another star.

What I surmise you want is an estimate of the density of stars as a function of Galactocentric radius.

In the solar neighbourhood, there are 378 known stars within a distance of 10pc of the Sun, according to the RECONS project. This gives a stellar density $n = 0.09$ per cubic parsec, which includes stars of all masses and white dwarfs. However, this includes multiple systems. If we ask what the density of stellar systems is, then it more like $n = 0.075$ per cubic parsec.

The average distance between stars is tricky to calculate because of the issue of multiple systems. About half of stars are in multiple systems, with a companion at distances of anything from a fraction of an AU to $10^5$ AU, though the distribution peaks at around 1000 AU.

The average distance between systems is about $n^{-1/3}$, which is about 2.36 pc (or 7 light years).

The density of stars increases dramatically towards the Galactic bulge and then again towards the centre. However, since we would not actually detect all the very low mass stars that we can see in the solar neighbourhood, to some extent stellar density estimates rely on an assumption that the mass distribution of stars is similar throughout. There are huge uncertainties here; a lot of detail is discussed in the review by Genzel et al. (2010).
The consensus seems to be that there are of order a million stars in the central parsec, giving an average separation of 0.01 pc, or 2500 AU.

This density falls by several orders of magnitude in the first 10pc. The bulge of the Galaxy has a radius of about 2 kpc and a mass, mostly in stars, of about $2\times 10^{10} M_{\odot}$. If the average star has a mass of $\sim 0.3 M_{\odot}$ (as in the solar vicinity), then the stellar density in the bulge is about 2 per cubic parsec, with an average stellar separation of 0.8 pc (or 3 light years).


Do telescopes harm while observing sun through them? I don't know whether we can observe a sun through a telescope. But if we can then will it harm our eyes and if such telescopes exists what is their configuration?

Observing the Sun through a telescope is very dangerous, whatever the telescope you use, if you don't use the appropriate tools.

A telescope a basically a light collector: its purpose is to collect all the light that is arriving on his primary mirror and focus it on a point.

You may have already tried to make the Sun light converge through a little magnifying glass (which works quite similarly to a telescope). If you make it converge on something like paper, it will simply burn it. This can eventually be the cause for big fires (example, a glass bottle in a dry forest, concentrating the Sun's light on leaves). Here is an example to convince you: http://www.youtube.com/watch?v=N7nJ3wIxt3o

A telescope works the same, except that the collecting surface is usually bigger, so it will be more effective at burning ... your eye for example (astronomers make this joke that you can observe the Sun with a telescope twice in your life: once with your left eye, once with your right eye). Or, for bigger telescope, it could even start a fire, or burn instruments...

To observe the Sun, you will need a proper filter that you will put usually "before" the telescope. You may also observe it indirectly on small telescope by projecting the image of the Sun.You can see examples here: http://www.skyandtelescope.com/observing/objects/sun/Observing_the_Sun.html.


Is there any telescope on Earth that can see the lunar rovers on the moon? If I have the right numbers, it seems to me that even Hubble telescope might barely be able to make out a carcass of a blue whale on the surface of the moon, which puts objects as small as the lunar rovers or the American flag left there by Apollo out of range. Hubble is often lauded as the best telescope we have built. I know a large part of that is because it is in space where it's free from all of the interference of the atmosphere, but is there any ground-based observatory with better resolution? Is there any place on Earth where I could point a scope at the Sea of Tranquility and see what we left there over 40 years ago?

The largest optical wavelength telescope that we have now is the Keck
  Telscope in Hawaii which is 10 meters in diameter. The Hubble Space
  Telescope is only 2.4 meters in diameter.

  Resolving the larger lunar rover (which has a length of 3.1 meters)
  would require a telescope 75 meters in diameter.

Information extracted from The Curious Team answers: Are there telescopes that can see the flag and lunar rover on the Moon?

Update: In a comment to my answer, @Envite mentions Astronomical Interferometers which are:

  ...an array of telescopes or mirror segments acting together to probe
  structures with higher resolution by means of interferometry. The
  benefit of the interferometer is that the angular resolution of the
  instrument is nearly that of a telescope with the same aperture as a
  single large instrument encompassing all of the individual
  photon-collecting sub-components.

It also must be noted that probably this wouldn't be the best idea for watching the moon rovers since:

  The drawback is that it does not
  collect as many photons as a large instrument of that size. Thus it is
  mainly useful for fine resolution of the more luminous astronomical
  objects, such as close binary stars.


Could a moon orbit two planets at once? Let's say you had a binary planet system of two earth-sized worlds with about the same mass. Could a stable moon orbit around the barycenter of both planets? If so, is there a limit to the mass of such a moon? Would it have to orbit in a Lagrangian point?

The Pluto-Charon system is similar to this.

It's unlikely that you'd have two planets of nearly the same mass because some variation is far more common, but a moon orbiting a binary planet system is certainly possible.   It's easier if it's not that close to the sun, for the same reason that it's difficult for Mercury and Venus to have moons.   A larger hill-sphere and less solar gravitation would make your system more stable in the long term.

And the Moon couldn't be in a Lagrange point in your system because, for starters, L1, 2 and 3 are never stable and the potentially stable Lagrange points, L4 and L5 require a large central planet and a significantly less massive object, (less than 1/26th the mass) orbiting the central planet.   The L4 and L5 points are to the orbiting object.  That doesn't work in a binary system of 2 objects of similar mass.

Any working Lagrange point scenario you'd have two moons and one planet, not binary planets and one moon.    You could, in theory have two moons at Lagrange points to each other orbiting a binary planet system but that would be adding even more improbabilities, but theoretically possible and stable for some time.


How do we know background radiation does not come from undiscovered galaxies? Each year while observing voids, astronomers discover more and more unexpected galaxies within them.

How do we know the cosmic microwave background radiation is not the weak light from distant galaxies?
How do we know cosmic microwave background radiation is not stars' light reflected by dark objects?

Assuming you mean the Cosmic Microwave Background ...

The CMB is entirely in a very narrow range of the microwave part of the spectrum. Galaxy emissions are all over the spectrum.

The CMB comes uniformly from all over the sky. It's like looking at painting that is all one color. Galaxy light looks like pinpoints on a dark background. (Nearby galaxies are not pinpoints, but those at cosmological distances look that way.)

The CMB matches the exact wavelength we expect for light that is heavily redshifted from the moment of last scattering. That moment was hundreds of millions of years before galaxies ever formed. Light from oldest galaxies are at a redshifted wavelength from 500 million years after the last scattering.

If the CMB was due to early starlight reflecting off dark objects, you would see a very different spectrum. It would contain many absorption lines as the dark objects absorbed some wavelengths better than others. Also the reflected starlight would be across a wide range of frequencies, unlike the CMB which has a very narrow range.


What Is The Great Attractor? I've heard that there is something called the Great Attractor which seems to have some kind of gravitational influence on objects in the Universe.  Is this some kind of theoretical notion posited in order to help answer a question in Astronomy, or is there an actual object which is greatly attracting?

Or is this an obsolete notion?

No; it's real - observations of the movement of galaxies indicate that there is an unusual concentration of mass (probably not a massive "object" of course, but totalling as much as tens of thousands of galaxies) in a place around 200 million lightyears from our galaxy.

It's hard to study since the view in that direction happens to be blocked by our own galaxy, but given that it has been studied for decades, it seems unlikely that the phenomenon turns out to be based on some sort of methodical error.


What observational constraints are there in detecting the presence of volcanism on exoplanets? This question is somewhat related to my earlier question How are the compositional components of exoplanet atmospheres differentiated?, but this about a specific surface-atmospheric phenomena - volcanism.

Using our solar system as a rough analogue (where asides from Earth, Venus, Io and Triton have active volcanism, probably more), volcanism should be not uncommon amongst exoplanets given the right conditions.  Obviously, we could probably only detect either the massive shield, plateau volcanism, or prolonged volcanism from multiple vents.

What observational constraints are there in detecting the presence of volcanism on exoplanets?

Exoplanets are too far away to send satellites or to image them directly. So there is no way to go there and say: there it is a volcano. My guess is that we have to guess from what we know working in the solar system. And we might just get a statistical probability that the planet is active.

I would say that there are two cases:

If the planet is rocky and near enough to the star or a giant planet, you might expect volcanism caused by tidal effects, similarly to what happen on Jupiter's Io moon. In few words: the gravitational force of the nearby big planet/star deform the planet and this deformations are converted to heat due to friction. If the gravitational forces are large enough, you can end up having volcanism
If you are able to measure the spectral lines in the planet atmosphere you could try to identify signatures of gas or dust typically associated with volcanic activity. I don't know what you would want to look for, but my guess is that you want to search mostly in the infrared.


Sun's Stellar Class at Birth? The Sun currently has a stellar class of G2V. However, since its birth, the Sun has brightened significantly. What would the Sun's class have been when it was born ~4.6 gyr ago?

Spectral type is determined by (and is an indicator of) photospheric temperature and surface gravity. The classification of G2V applies to a star with a photospheric temperature of about 5800K and a surface gravity of about 100-1000 m/s$^{2}$.

The Sun's temperature and gravity have not changed greatly since it first appeared on the "zero age main sequence" - it has got a little bigger and more luminous and a little bit hotter (by about 170 K- Schroder & Smith 2008) - so maybe it was a G3V-G4V star at the zero age main sequence (Mamajek 2018).

However, there were much bigger changes during the 20 million years or so before the Sun reached the ZAMS. It was cooler and briefly much larger. This would be reflected in a spectral classification that was something like K5IV- M0IV; indicating a subgiant-like gravity and a temperature of about 4000K. As it aged, it would then shrink and become hotter, passing through the G5V-K5V spectral types before reaching the main sequence.

If you can synthesise these two diagrams in your head, you can see what I mean.

Pre main sequence evolution in the Hertzsprung-Russell diagram

Spectral types on the Hertzsprung-Russell diagram


Jupiter FM - What are practical and inexpensive ways for the amateur detection of signals from Jupiter, especially of the transit of her moons? What modifications to a standard AM/FM or shortwave radio are needed in order to be able to detect radio-wave signals emitted from Jupiter?

Would it be possible to detect the transit of the major moons and even minor planets (maybe even Ceres and Mars) using this method?

@Arne is right in his answer about two things, that the most suitable frequency for Jovian amateur radio is 20.1 MHz, and that this is a 15 meter wavelength. However, the antenna can actually be half the wavelength, and amateur radio astronomers have had good results listening to all kinds of Jovian sounds, including detecting occultations of its many moons as they produce change in frequency due to the Doppler shift when passing in front of it, amplification due to echo of Jupiter's own radio wave signature when close beside it from the observer's vantage point, and other effects causing radio wave frequency and amplitude changes with a simple dual dipole array that might look something like this:

        The Jove dual dipole array. Dipoles are suspended between PVC masts. Signals from the dipoles go to a power combiner and        then to the receiver. Source: Amateur Radio Astronomy Projects — Radio Signals from Jupiter (PDF)

Now this dual dipole antenna is more or less a simple two-piece of roughly half the wavelength of coaxial cable, stripped of insulation at the dipole ends and run dipoles parallel to each other at roughly 6.1 meters (20 ft) apart suspended over PVC masts. That is perfectly acceptable for nearly any amateur radio astronomer in both space and the price of the parts needed. There are also ways to help oneself with a lot smaller antenna, which I'll mention a bit later. Let's first clear the frequency range discussion with this really informative quote from Radio Receiver for Jupiter webpage (based on NASA's Radio JOVE project):

  The peak of Jupiter signals occur around 10 MHz. Still this frequency
  is not that suitable, as it is very near to Ionospheric cutoff. The
  best suitable frequencies are in between 18 to 22 MHz, as the chances
  of getting emissions are more. In practice 18.7 MHz, 20.1 MHz, 22.3
  MHz are common. The frequencies above 30 MHz are not suitable because
  of less strength. The frequency of 20.1 MHz is used for this project,
  as the probability of getting emission is high. So the entire receiver
  is designed considering 20.1 MHz as the operating frequency.

Antenna is of course merely one part of the story. Second thing that is required is the receiver. NASA is sponsoring Radio JOVE project with an assembly manual for roughly 100 electronic components and pieces of hardware worth RJ1.1 Receiver (Radio Jove 1.1 Receiver) that can be almost completely assembled out of parts ordered from Radio Shack (it even includes RS part numbers). I'll add a photograph and a few other links, and you'll have to be on your own then:

   Self-assembled Radio Jove Receiver's front plate with two turning knobs for volume and tuning. Source: KB0LQJ Observatory

Some relevant links for building Radio Jove (or Jupiter FM, if you like, it will be your own receiver so name it as you wish), starting with already mentioned ones:

Amateur Radio Astronomy Projects — Radio Signals from Jupiter (PDF)

  Ten years ago a group of (mostly)
  University of Florida graduates working at
  NASA conceived an educational outreach
  program known as Radio Jove. The idea was
  to build an inexpensive radio telescope kit
  suitable for detecting signals from Jupiter.
  The Jove receiver (Figure 2) is a simple
  direct-conversion design operating over a
  few hundred kilohertz range centered at
  20.1 MHz.

NASA's Radio JOVE project (PDF)

  The site is currently down due to US government shutdown, so here's a Google cached version that sadly doesn't come with in-document images

KB0LQJ Observatory Amateur Radio Astronomy — Radio Jove observations

  For my home observatory, I started with the Radio Jove receiver from NASA’s Radio Jove Project. This was a fairly easy-to-build kit with great directions, not just for the receiver but also the antenna setup and installation.  Unfortunately, I don’t have enough space to put up a phased antenna array.  Also, I’m bordered by power lines on the north and south sides of my property.   Also, since I’m in an urban area, I knew that I’d pick up a lot of noise.   Not to worry.   Solar observations are also quite interesting, and since the Sun is such a good signal source (especially so far this year), I opted for an antenna in my attic.

Radio Receiver for Jupiter

  This website contains the Technical details of the antenna and receiver used for receiving natural radio emissions from Jupiter at 20.1MHz. The antenna and receiver discussed in this site is based on the design given by NASA Radio Jove Program.
  Natural radio emissions from Jupiter or from Sun are detected by using a dual dipole array as an antenna and with a sensitive receiver. The RF voltage developed at the antenna terminals is amplified with the RF amplifier and converted in to audio frequencies using a mixer. The audio signal thus generated is recorded on PC through a sound card in ‘wav’ format. Also a strip-chart recorder software is available to generate the strip-chart of data coming through sound card.

So there are indeed many ways to build your own antenna and receiver out of affordable and easily obtainable electronic parts, and some of these websites listed above will help you through the process of self-assembly, even providing some tricks how to do it easier, like for example assembling a smaller, in-house antenna array.

Now, one other thing that all these websites mention is also using various PC software that lets you analyze through computer's sound card received Jovian radio sounds, but since there's many different and free solutions out there, including NASA's own that runs on Windows PCs, I'll let you discover those on your own. Here's one page listing many links, to get you started. More computer savvy of you might even write your own software for this purpose, this is Stack Exchange afterall.

And if anyone is wondering what such self-assembled radio receivers and antennas are capable of, here's a link to a collection of various Jovian sounds in the radio wavelengths on Astrosurf.com, and another one as a collection of solely amateur radio astronomy sounds of Jupiter and its moons (scroll a bit down to the list of recordings). And here's a short description of various sound types one might hear:

Chorus (quote source: The University of Iowa Department of Physics and Astronomy)

  Chorus consists of brief, rising-frequency tones that sound like the chorus of birds singing at sunrise, hence the name "chorus" or "dawn chorus." Chorus at Earth is generated by electrons in Earth's Van Allen radiation belts. Once generated, the chorus waves affect the motions of the electrons through a process called a wave-particle interaction. Wave-particle interactions disturb the trajectories of the radiation belt electrons and cause the electrons to hit the upper atmosphere.

Decametric noise storms (quote source: Radio-Jupiter Central on RadioSky.com)

  The emissions we can hear are often referred to as decametric noise storms, because the waves are tens of meters long. Okay, it is possible to hear Jupiter from 15 to 38 MHz, but what are the optimal frequencies?  The consensus seems to be that 18 MHz up to about 28 MHz is a good place to listen.  A good rule would be to pick the lowest frequency in this range which was not being hindered by ionospheric refraction.

Whistlers (quote and image source: Wikipedia):

  A whistler is a very low frequency or VLF electromagnetic (radio) wave generated by lightning.1 Frequencies of terrestrial whistlers are 1 kHz to 30 kHz, with a maximum amplitude usually at 3 kHz to 5 kHz. Although they are electromagnetic waves, they occur at audio frequencies, and can be converted to audio using a suitable receiver. They are produced by lightning strokes (mostly intracloud and return-path) where the impulse travels along the Earth's magnetic field lines from one hemisphere to the other.

Auroral Radio Emissions  (quote source: Wikipedia)

  Auroral radio emissions from giant planets with plasma sources such as Jupiter's volcanic moon Io could be detected with radio telescopes.

And so on. The last two on the list are probably a bit of a stretch for amateur astronomers, with auroral radio emissions average frequency range at 100 to 500 kHz, and whistlers cutoff at usually around 30 kHz, and both requiring too big antennae, but I wouldn't bet against it being possible with smaller antennae either, although bow frequency cutoffs might prevent identification of actual aural events. But there are many other sounds to listen to both from Jupiter, as well as it's many moons, mostly though the bigger and closer ones.

Good luck tuning in to your own Jupiter FM and happy hunting for rare radio frequency events!


Definition of Rotation versus Revolution Let there be a planet for which the duration of a solar day is equal to a year around the sun such that:

1 "day" = 1 "year" = 100 Earth days (2400 hours or 8640000 seconds)

Then, would we say that the rotation of the planet is 0 or equal to 100 days?

A planet that is not rotating with respect to the stars is not rotating. In your case this would give it a solar day length of 100 Earth-days. Such planets are not likely to exist in reality.

This is in contrast to the actual case when a body is tidally locked, such as our moon.  The moon rotates once per (lunar) month at exactly the same rate as it orbits the Earth.

There are different ways of defining "length of a day".  There is the (average) time from midday to midday (this is 24 hours, or one solar-day) or there is the time it takes for the stars to return to the same position that they were the previous night. This is the "sidereal day" and is about 23hours 56 minutes. The Earth takes 23hours 56 minutes to rotate (and not 24 hours)

For your planet, the sidereal day length is undefined and the solar day is  100 Earth-days. We would say that the planet is not rotating.


How many Supernovae have been Imaged in the Hubble Ultra-Deep Field? The Hubble Ultra-Deep Field is an image showing about 10.000 galaxies:

2
Image Credit: NASA

Consider Victor Buso's spectacular discovery of an early supernova in NGC 613 (named SN 2016gkg):

4
Image Credit: Carnegie Institution for Science/Las Campanas Observatory/UC Santa Cruz

This got me wondering about how many supernovae the Hubble Ultra-Deep Field may contain without us knowing because we would need a light curve or reference image to identify one with certainty. Some supernovae also would be only a pixel or less and impossible to identify.

Is it possible to calculate a meaningful lower bound?

Supernovae probably occur about once per century in our own (quite large) galaxy.

If we were to assume similar star formation rates and a similar stellar birth-mass distribution in other galaxies and that the galaxies in the HST-UDF were similar to our own, then we could also assume that the supernova rate was similar.

The UDF was taken over about 4 months, or 1/300 of a century. So one might expect a supernova to occur in about 1/300 galaxies - so about 300 of the 10,000 galaxies observed will have hosted a supernova during the period that the image was built up.

Of course, even if this figure is correct (see below), many of these may have been missed because the galaxies were too far away or the supernovae were obscured by dust, or because they brightened and faded between observations (there was a 6 week gap in the observations). On the other hand, the UDF was imaged later by HST with other instruments, leading to further opportunities to identify supernova candidates.

The calculated number could be awry in many ways. Most galaxies are smaller than the Milky Way, so one might expect fewer supernovae. On the other hand, most of these galaxies are far away and the star formation rate was likely higher in the past, leading to more supernovae. On the other hand, the supernova rate will be higher in some types of galaxies (Starbursts), but lower in others (ellipticals). Without a very detailed demography of the types, metallicity, masses and ages of the galaxies, and a detailed understanding of how the supernova rate varies with these parameters, your question is difficult to answer.


How could a 20 inch space telescope "be able to make out Earth-size planets" orbiting Alpha Centauri? The NY Times article Telescope to Seek Earthlike Planet in Alpha Centauri System describes a proposed effort to put a telescope into orbit fairly quickly with a narrowly defined purpose - to look for planets in the Alpha Centauri system, which is about 4.37 lightyears away from Earth.
The funding, design, assembly and space deployment fairly quickly is made possible in part by the narrowly defined purpose and the compelling public interest in this particular question.
I got out my box of envelope-backs and did the following. Since there is no air, there's no need for adaptive optics so  the standard diffraction limit applies. For an Airy disk:
$$ \theta \approx 1.22 \frac{\lambda}{d} $$
The article describes the telescope as washing machine-sized, with an aperture of 20 inches, or about 50 cm.  If I choose a visible wavelength of 500nm, I get a resolution of about 1.22 microradians ($\mu$R). 4.37 lightyears is about 4.13E+13 km or 2.76E+05 AU.
That makes the airy disk about 0.34 AU at Alpha Centauri. The actual point spread function of a telescope depends not only on the theoretical diffraction of the aperture, but diffraction from secondaries and supports within the aperture, on aberrations due to imperfectly shaped optical surfaces and scatter and haze due to nano-roughness of mirror surfaces. (see below)
Could a 20 inch telescope and associated experiment package be put together quickly that could actually resolve (or "make out" as the article says) an Earth-like planet in the Alpha Centauri system from space? If so, what is it exactly that it could do that can't be done from any existing telescope in orbit or on Earth?

above: "A rendering of the compact exoplanet imaging telescope Project Blue plans to launch into orbit and aim at Alpha Centauri. Credit Project Blue" from here.
It jumps out that the secondary is mounted off-axis, which can have mechanical and optical advantages (e.g. diffraction) but would probably limit the image quality to a narrow field of view, which might be OK for a single-target telescope if there are suitable calibration targets nearby.

A nice discussion of the point spread function of the Hubble space telescope can be found here: 20 years of Hubble Space Telescope optical modeling using Tiny Tim (Krist et al 2011, Proc. of SPIE Vol. 8127 81270J-1), and the reduction of scatter and haze from mirror nano-roughness by using refractive optics (lenses) is addressed by the DRAGONFLY instrument. See also What (actually) is the " deprojected half-light radius" of this almost-all-dark-matter Galaxy?

I'm not familiar with the design of the ProjectBlue telescope, but I think you have answered your own question.

The habitable zones for Alpha Cen A and B, are approximately centred at 1.25au and 0.7au. Both are at a distance of 4.37 light years.

1au at 4.37 light years, subtends an angle of 0.74 arcseconds. If working at blue wavelengths (the aim appears to be to find a pale blue dot, so let's say 400nm), an Airy disk of 0.74 arcsec is achieved with a telescope with a diameter of just 13cm. To stand a chance of resolving planets in the habitable zone, you would need a telescope a little bigger than this.

However, a bigger problem is separating the signal from the star and planets when they are barely resolved. The Airy disc is just the first minimum in the diffraction pattern. If the star is around 10 billion times as bright as the planet (ball park figure) then it would still swamp a planet that was just resolved.

Thus you need to build a bigger telescope to increase the angular resolution. And you need to build superb optics and a coronagraph to stop the light from the star swamping your detector. But this costs more money, the telescope has more mass, needs a bigger launch vehicle etc. So I guess a 50cm telescope is their compromise.

Is it achievable? I don't know. The best telescopes on Earth achieve almost diffraction-limited angular resolution using adaptive optics. However the contrast they can obtain for exoplanet imaging is compromised by the atmosphere to some extent. If you are looking for a "pale blue dot", then this can't be done from the ground because adaptive optics systems do not currently work at such wavelengths.

It is quite possible that Earth-like planets in the habitable zone of Alpha Cen could be discovered from the ground in the next few years using improved radial velocity measurements, but not by imaging.

An interesting question is why this can't be done now with HST? The proposed telescope will spend two years doing this. Even if HST could cut this to months, I am not convinced that detecting a faint smudge around the nearest star has the necessary science pay off to get the time, especially as there is a fair chance of failure. There may not be planets in the habitable zone or there may be sufficient zodiacal dust in the system to make the measurements impossible due to scattered light.

Further research suggests that the coronagraph on the HST ACS camera would only give around 7-8 magnitudes (factors of a thousand) contrast at angular separations of an arcsecond. So nowhere near what is required. This new telescope must have radically improved optics and a more advanced coronagraph.


What reason is there to doubt the existence of the hypothesised planet Tyche in the far distant solar system? Please note, I do not necessarily believe in its existence, just after a scientific (hence, non-Wikipedia) reasoning to doubt the planet's existence, other than the "we would have seen it" argument.  I am, nor are the articles below, referring to the alleged 'Nemesis' planet, as in the question What are the current observational constraints on the existence of Nemesis?

According to the article Astronomers Doubt Giant Planet 'Tyche' Exists in Our Solar System (Wolchover, 2011), the planet Tyche is a hypothesised massive planet

  by John Matese and Daniel Whitmire of the University of Lousiana-Lafayette, is not new: They have been making a case for Tyche since 1999, suggesting that the giant planet's presence in a far-flung region of the solar system called the Oort cloud would explain the unusual orbital paths of some comets that originate there.

Further information is found in the scientists' research paper Persistent Evidence of a Jovian Mass Solar Companion in the Oort Cloud (Matese and Whitmire, 2010), describe the hypothesised planet to be

  of mass  1 − 4 M(Jupiter) orbiting in the
  innermost region of the outer Oort cloud. Our most restrictive prediction is that the orientation
  angles of the orbit normal in galactic coordinates are centered on
  , the galactic longitude of
  the ascending node = 319 and i, the galactic inclination = 103 (or the opposite direction)
  with an uncertainty in the normal direction subtending  2% of the sky.

They posit that the elongated orbit of Sedna is a result of the presence of this planet.

So, the question, what reason is there to doubt the existence of Tyche?

Generally it would be that peers of Matese and Whitmire looking at the statistics and viewing that the results are not conclusive enough to indicate the existence of Tyche.  There can also be alternative reasons to explain the results mentioned in the paper other than a Jupiter sized planet in the Oort Cloud. As was mentioned in the comments, it is almost impossible to prove the non-existence of something.

At this time, the results from WISE are the best bet for locating Tyche if it exists.  As it will not be visible in the visible light spectrum but more likely in the infrared and the WISE mission scanned the entire sky which means that it won't have been missed due to looking in the wrong place.

http://www.jpl.nasa.gov/news/news.php?release=2011-060

http://www.nasa.gov/mission_pages/WISE/main/index.html#.UnJb3xCJ5fs

There isn't any hard science to show that Tyche doesn't exist.  It is skepticism that this is the explanation for observations to date.  These are the same reasons that Nemesis isn't thought to exist, the data isn't strong enough to support its existence.


Why is the main sequence often called 'zero age' main sequence? The main sequence stage for a star begins after it stars burning hydrogen into helium. This is often referred to as the zero age main sequence.

This is confusing since stars have already lived several million years in their pre-main sequence stage.

Where does this come from? Is it a historical designation that just stuck?

The zero age main sequence (ZAMS) is the theoretical locus of points in the Hertzsprung-Russell diagram where the luminosities of young stars (at a range of masses) are mostly supplied by the nuclear reactions that have begun in their cores.

Prior to this, the objects are known as pre-main sequence stars and most of their luminosity arises from the release of gravitational potential energy as they contract.

The exact definition (whether it is 99% or 90% of the luminosity) does vary, but since the nuclear reactions ramp up quite quickly this doesn't materially affect any quantitative analysis.

In any case, the age of a star is not measured from the ZAMS. The reason for this is that what is of physical interest is the time since the star formed and that must include the pre-main sequence lifetime. Whilst this is very short (maybe 1 million years) for stars more massive than the Sun, and certainly shorter than the total main sequence lifetime, it can be much longer (100 million years) for a low-mass M dwarf. This is also the reason that the ZAMS is a theoretical construct. Stars born at the same time will never be all on the ZAMS at the same time. By the time low-mass stars reach the ZAMS, more massive stars will have evolved away from the main sequence!

As to the origin of the term, I suspect it comes from the early computer/analytic models of stellar evolution. There you would likely miss out the pre main-sequence phase and start your model with "zero age" at that point where it achieves equilibrium with the luminosity supplied by fusion and with a pristine He/H ratio in its core. Thus in reality, stars of different mass would have different ages at this point, but from the point of view of a theoretical modeller, it would be their starting condition and hence a "zero age".


What do the calendar rounds in Tomás Filsingers "Map of the Universe 2.000" mean? Recently I did the Ravensburger puzzle "Map of the Universe 2.000 - Celestial Chart and Calendar Rounds" by Tomás Filsinger:

Tomás said the last time he visited this site in 2019, that he finished a new, improved map. I recognize some of the described features in the puzzle, e.g. the "60 galactic years since the big bang".
I am just trying to analyse the multiple rings of calendars, like the meaning of the blue and orange triangles, the tiny colored lines with squares (I think they are planets, but how do they relate to the calendar?), the red to blue-colored ring, why the grey (on the innermost ring) and yellow constellations (as part of the firmament), and the blue months and the grey months are shifted against another, why there are stars on the outer ring ... well almost most of what is going on in this exquisitely beautiful, but (at least for me) quite complex chart.
I asked the publisher, but he was not able to explain it.
If someone might be interested, I could send details of the map.
Perhaps Tomás is reading this, like in the other post, and could explain his map himself?
p.s.: I included a detail photo of the puzzle).
You can see several things here (from bottom to top, with my comments in parenthese; please forgive my choice of terms, I am a star map rookie):

grey triangles with the galactic years, divided in 12 steps (why?), and small diamonds with major events, but on another scale (10, 1.000, 2.150 years: why?). There must also be some kind of horizon, because there is the Southern Cross, and Alpha Centauri.
a dotted line, with a white dot for every 20 blue dots (dont know, must be something regarding planets, there is a tiny Venus on the right and the weekday is Friday).
4 lines with jovian planets (I think), and their movement over the year (I think).
a red dotted line, for Mars (I think).
the days of the respective month, with orange triangles (why are they pointing at the dotted lines?).
the months in different languages, with their meteor showers.
a white dotted line, for Venus (I think, text is "263 days as evening star").
another very tiny line (Mercury?)
days of week, in different languages, with some objects.
a band with a spectrum (which?), red to blue, here green, with missing lines (which?).
signs of the zodiac.
degrees of longitude, including hours of the day (why?).
sky map, with day of month ("16, "17", why?), moon phases, yellow dashed line (which?), ecliptic, celestial equator, and of course many objects and constellations.

And: how is it all connected to another ;-)
There are some more details, but these are the main constituents and corresponding questions. The booklet Tomás provided for downloading in the other post should help a bit, and if you refer me to other sources because this is all standard stuff, that would be fine for me.
Looking at my questions at the start of the post I already see that I answered some of my former questions myself, but much is left :-/

The star map is centered at the north celestial pole.
A thin red circle marks the celestial equator.
The Roman numerals around the edge of the star map are hours of right ascension, measured along the equator and subdivided into minutes.
Sidereal time runs slightly faster than solar time and corresponds to the right ascension of celestial objects currently crossing the overhead meridian.
The hour labeled XIV here is more commonly designated 0h.
The months in the blue outer ring show when the Sun appears along those parts of the ecliptic, marked on the star map with a white dotted circle.
Numbers at intervals along that circle indicate the approximate day of the month when the Sun passes those points.
Meteor showers are marked when the Earth encounters those meteoroid streams, e.g. the Orionids in late October.
The orange triangles pointing outward for each day may be mostly decorative.
As you inferred, the thin circles on either side of the month ring are for the planets in order of their distance from the Sun.
I would expect the number of dots punctuating each of these lines to match that planet's synodic period in days, indicating the planet's average rate of apparent motion along the ecliptic.
Depending on the Earth's and the planet's relative positions in solar orbit, this motion may at times be faster, slower, or retrograde.
The gray ring in 12 segments shows the signs of the tropical zodiac, 30° blocks of longitude measured along the ecliptic.
They are named after constellations which were in the same parts of the sky ~2000 years ago; precession since then has changed that alignment.
Some sign boundaries correspond to the equinoxes and solstices.
The Sun passes those points about 70% of the way through those calendar months.
Since the ecliptic is oblique to the equator, ecliptic longitude is slightly condensed around 0h and 12h right ascension and slightly expanded around 6h and 18h.
This chart may instead show ecliptic and equatorial coordinates aligned all around for aesthetic reasons.
The days of the week are named after the Sun, Moon, and bright planets in Latin and related languages.
Other than dividing the circle into 7 equal parts, their placement on this chart may be arbitrary.
The rainbow-colored ring shows the spectrum of the Sun, with thin dark lines where the electrons of gas atoms and ions absorb photons of particular wavelengths.
Astronomers use this information to identify elements in the atmospheres of stars and planets.
They also use displacements of spectral lines to measure radial velocity.
The dashed yellow circle in the outer part of the star map, tangent to the ecliptic at 18h right ascension, is the circle of -23.5° declination.
The corresponding circle of geographic latitude is called the Tropic of Capricorn.
South (outward) of -35° declination, the star map shows a few highlights in addition to Crux and α and β Centauri.
These include the Large and Small Magellanic Clouds and the globular cluster ω Centauri.
Crux may be plotted a few degrees east (clockwise) of its actual position for aesthetic reasons.


Long term effect on rocks of high pressure and temperature? The mean surface temperature of Venus is about 460 C and the pressure is about 92 atmospheres. Assuming this situation (or similar) has prevailed for millions of years what changes (if any) would there have been in the structure and chemistry of the rocks on the surface of Venus?

Wikipedia(Venus) http://en.wikipedia.org/wiki/Venus

Consider not only the physical attributes of the Venusian atmosphere, but its composition, from the NASA Venus fact sheet:

Major:       96.5% Carbon Dioxide (CO2), 3.5% Nitrogen (N2)
Minor (ppm): Sulfur Dioxide (SO2) - 150; Argon (Ar) - 70; Water (H2O) - 20;
             Carbon Monoxide (CO) - 17; Helium (He) - 12; Neon (Ne) - 7

The 1st and 3rd most abundant components - carbon dioxide and and sulphur dioxide, at the pressures and temperatures of the surface would result in:

  The reactions probably include the deterioration of silicates by carbon dioxide to produce carbonates and quartz, as well as the deterioration of silicates by sulfur dioxide to produce anhydrate calcium sulfate and carbon dioxide.

From the Venus Wikipedia page

A study reported in the paper Venus Geochemistry: Progress, Prospects, and New Missions (2009), focusing on the basalt-atmosphere interaction, determined that the conditions on Venus is roughly the equivalent of:

  a metamorphic environment of ‘amphibolite
  grade,’ ‘hornfels facies;’ the paucity of H2O
  in Venus’ atmosphere (30±15 ppm) prevents formation
  of amphibole

The metamorphic facies for Earth that are used in the above quote are shown below:

Image source

The implications of the surface-atmosphere chemistry, pressure and temperature would, according to the Venus geochemistry article (linked above), result in the formation of:

  anhydrite from essentially
  all the Ca in basaltic minerals and rocks. Iron in the
  rocks and minerals will likely react to either oxides or
  sulfides, depending on the oxidation state of the atmosphere.
  The fate of alkalis is not clear – our calculations
  yield only alkali feldspar (albite), but alkalis
  could be partitioned into ionic (carbonate-sulfate)
  melts. Some Venus basalts are so rich in potassium
  that phases besides feldspar may be present (e.g.,
  leucite, nepheline-group, or sodalite-group).