Mass, Radius, Colour, Size, Type of a Star from the Hipparcos Catalog Using data from the Hipparcos Catalog download file, is there a way to determine the Visual or Apparent Magnitude from the data within?

Can I obtain the Mass, Radius, colour, temperature, type (dwarf, sun-like, hypergiant, etc) of the star from the catalog?  Which column relates to which item i'm looking for?  Any examples?

The apparent magnitude (in V ~ visual) is in column 42-46. Other magnitudes are BT magnitude (columns 218-223) and VT magnitude (columns 231-236). These are magnitudes recorded with different (colour) filters.
The colour ($B-V$) can be found in columns 246-251, expressed in magnitudes. This is the difference between V (visual) magnitude and B (blue) magnitude also known as the colour index. Blue stars have a low colour index ($B-V<0$) as the star will be brighter in the blue (B) passband (filter), and will therefore have a low B magnitude, while red stars will have a higher $B-V$ colour index.

Mass, Radius, and Temperature are not in the catalogue. Mass can only be directly determined for multiple star systems and requires a lot of observations per multiple star. Radius is determined from stellar models. And Temperature can be determined from the spectrum of the star. A crude indication can be gotten from the colour index of the star. (see for instance this webpage).

Finally the type can be found in the star's spectral type. The spectral type is in columns 436-447. These spectral types are not always easy to read but they consist of a spectral class (e.g. O, or F type stars) and the luminosity class (and some other information). The luminosity class gives the type of class in roman numerals. Below is the list from the wikipedia page on stellar classification.

0 or Ia+ (hypergiants or extremely luminous supergiants). Example:
Cygnus OB2#12 (B3-4Ia+)[12]
Ia (luminous supergiants). Example: Eta
Canis Majoris (B5Ia)[13]
Iab (intermediate luminous supergiants).
Example: Gamma Cygni (F8Iab)[14]
Ib (less luminous supergiants).
Example: Zeta Persei (B1Ib)[15]
II bright giants. Example: Beta
Leporis (G0II)[16]
III normal giants. Example: Arcturus (K0III)[17]
IV subgiants. Example: Gamma Cassiopeiae (B0.5IVpe)[18]
V main-sequence stars (dwarfs), Example: Achernar (B6Vep)[15]
sd (prefix) subdwarfs. Example: HD 149382 (sdB5)[19]
D (prefix) white
dwarfs.[nb 2] Example: van Maanen 2 (DZ8)[20]

The Sun's spectral type is G2V, which means that the spectral class is G2 and the luminosity class is V, which means that the Sun is a main-sequence star.

An example:

The following record is for the star Arcturus (α Boötis):

H|       69673|H|14 15 40.35|+19 11 14.2|-0.05|1|G|213.91811403|+19.18726997|*|  88.85|-1093.45|-1999.40|  0.92|  0.42|  0.74|  0.60|  0.52|-0.52| 0.16|-0.13|-0.07| 0.05|-0.01| 0.03|-0.12|-0.03|-0.31|  7| 2.01| 69673| 1.629|0.010| 0.286|0.009|*| 1.239|0.006|G|1.22|0.02|A|*| 0.1114|0.0020|0.015| 70|*| 0.09| 0.14|       |U|2| |14157+1911|H| 1| 2|C| |A|AB|198|  0.255|0.039| 3.33|0.31|S| | |124897|B+19 2777 |          |          |1.22|K2IIIp      |X

The apparent visual is(V) magnitude $-0.05$. The colour index is $B-V = 1.239$, which means that it is probably a red K type star with temperature $T\approx 4000 \textrm{K}$ (from figure 2 on this webpage). The spectral type is K2IIIp, i.e. the spectral class is K2 and the luminosity class is III, a normal giant star.

An indication for the Radius can be found from the Hertzsprung-Russel Diagram in which radii are added (see Figure here). The luminosity of Arcturus can be calculated as follows:
$$ d = \frac{1}{\varpi}$$
where $d$ is in parsec and $\varpi$ is the parallax in arsceconds. The parallax is in columns 80-86 and is $\varpi = 88.85mas = 0.08885^{\prime\prime}$ for Arcturus. The distance is, therefore, $d=11.25\textrm{pc}$. You can then calculate the absolute magnitude ($M_V$):
$$ M_V = m_V -5(\log{d}-1)$$
where $m_V$ is the apparent visual (V) magnitude. The absolute magnitude of Arcturus is: $M_V=-0.05 - 5(\log{11.25}-1)=-0.31$.
The luminosity is then given by:
$$M-M_{\textrm{sun}} = -2.5\log{\left( \frac{L}{L_{\textrm{sun}}}\right)}$$
where the absolute magnitude of the Sun is $M_{\textrm{sun}} = 4.83$ and $L$ is the luminosity of Arcturus. This gives:
$$\log{\left( \frac{L}{L_{\textrm{sun}}}\right)} = 2.056 $$
$$L = 114  \textrm{L}_{\textrm{sun}}$$
From the Hertzsprung-Russel diagram we get a radius for Arcturus of about $R \approx 10  \textrm{R}_{\textrm{sun}}$.


If two white dwarfs collided, would they become a star? Would the temperatures during such a collision be able to ignite nuclear fusion, bringing the dead star back to life? If so, would it only be able to fuse for a short time before running out of fuel, or would it turn into a full-fledged star?

The answer to your question is both yes and no, depending on the circumstances.

Two white dwarfs colliding would likely yield a Type Ia supernova, assuming the combined mass exceeded the Chandrasekhar limit ($\sim1.4$ solar masses). The unstable object resulting from the collision could not be supported by electron degeneracy pressure; when the temperature rises rapidly due to the collision, there is nothing to stop it (compare this to thermal pressure in a "normal" star, which can make the star expand or contract to compensate for changes in temperature).

The rise in temperature triggers fusion, which then increases the temperature, which triggers more fusion . . . and so on and so on, in a very short time span. This is the same process, more or less, as in an accretion-powered type Ia supernova. The resulting explosion destroys the object, ejecting matter into space.

As with normal Type Ia supernovae, this sort of collision would likely happen in a binary system, with two white dwarfs in close orbits losing energy to gravitational waves and spiraling in (the odds of two unrelated white dwarfs colliding is very, very low). I'm not sure how luminous this event would be in the gravitational wave spectrum; likely fainter than a neutron star-neutron star inspiral but still strong. The speeds of the components could be quite fast, meaning that a lot of energy would be released in the collision.

Things are slightly more complicated if the combined mass is less than the Chandrasekhar limit. An example system that is predicted to undergo a collision is SDSS J010657.39-100003.3. The total mass of the two white dwarfs is $\sim0.6$ solar masses, certainly below the limit. In roughly 37 million years (Kilic et al. (2011)), they will collide and form a subdwarf - fusing helium, interestingly enough (compare this to the carbon/oxygen fusion in Type Ia supernova, which in those circumstances is unstable). Similar scenarios could play out in other systems.

This sort of a star would not be a "normal" star in the sense that it wouldn't truly be on the main sequence - just a bit below it. It would be an O- or B- type subdwarf, less luminous than main sequence stars of the same spectral type. I saw a couple of articles on these objects claiming that they would be stars just like the Sun - it's not true! Given that white dwarfs are highly depleted in hydrogen - remember, they are the degenerate cores of old stars - there can be virtually no hydrogen burning, even in a shell. Core helium burning is the only option.

On a final note, one explanation for the formation of R Coronae Borealis variables is the collision of two white dwarfs. I don't think we have solid evidence yet for preferring this over other models (e.g. something involving hydrogen loss from the outermost shell), but if it is the correct pathway, then the answer to your question could be a definitive yes. Note that this would require intermediate-mass white dwarf progenitors.


360 degree image of the whole sky I look for an image that I know exists.
It is an image showing the whole sky, showing all stars and galaxies bright enough to be visible at the resolution, projected to the same shape that is commonly used for the microwave background / CMB.
I am sure the image is well known and has a name.
Can you give a reference to it?

There are night sky images in high resolution available from ESO, NASA and potantially others. They are already projected, openly licensed, and can directly be used, e.g. as skyboxes or -domes in computer graphics.
https://www.eso.org/public/images/eso0932a/
https://svs.gsfc.nasa.gov/3895


How much time does it take for a meteor (shooting star) to touch the ground after its combustion? I know that meteors that We see as shooting stars travel tens of thousands of kilometers per hour and are tiny as a grain of sand, but after hitting the atmosphere and burn I suppose that its ashes would touch the ground at some point. How long does it take? Can it be measured or observed?

I think the best model we could use for this is nuclear test fallout, whose dispersion was monitored quite extensively.  It's very, very dependent on weather patterns at altitude.

For example, the Castle Bravo test mushroom cloud reached at altitude of about 40km and hence hit the Jet Stream.  In addition to this the wind speeds at lower altitudes were higher than ground speeds and this was not checked for, so the combination of factors spread the fallout downwind about 280 miles (about 430 km) in about 16 hrs.  This was one reason for the irradiation of many people.

What these tests and other nuclear incidents show us is that material ejected into the high atmosphere can remain there a long time, even circling the globe like traces from Fukushima.

In theory trace particles from a meteor could remain in the upper atmosphere for days, months, even years.  Most would be expected to fall within days, however.


How can many stars be formed from the remains of one supernova? A supernova is the explosion of a single star; so how is it that thousands of stars can "be born of" that one explosion (presumably only using the unspent fuel / lighter elements of the original)? Does that also mean the stars coming from a supernova cannot possibly become supernovae themselves?
For example, according to Wikipedia, the Rosetta Nebula has about 2500 including a couple of O-type stars!

approximately 2500 young stars lie in this star-forming complex, including the massive O-type stars HD 46223 and HD 46150

Edit to reduce confusion and risk of reinforcing my false premised path of questioning: I somehow had the idea that nebulae came from single supernova explosions and that all nebulae were planetary nebulae. After research, that was an absurd understanding on multiple levels.

Stars don't "come from" a supernova.
Stars come from the interstellar gas in the galaxy, particularly where it is more concentrated into nebulae.
This gas is mostly hydrogen and helium, but it is "enriched" with heavier elements from old stars, including from stars that have exploded in supernovae. Over the billions of years since the galaxy formed, there have been many supernove which have gradually increased the amount of heavier elements in the interstellar gas.
The shockwave from a supernova can cause the gas to become squeezed and more dense, and this can trigger the gravitational collapse that leads to star formation, but the gas for those stars doesn't come from one particular supernova.
As a nebula collapses into many stars, some (the largest) will be large enough to supernovae (which can trigger subsequent periods of star formation).


MOOG 2019 and gfortran I've been trying to install MOOG Nov2019 (utexas, Wiki) on my desktop running OpenSUSE 13.1. I have the SM and X11 library paths copied into the makefile, and I am 3-sigma confident that they are the correct pathways. I'm using cc as the C compiler, and gfortan for Fortran. When I run the makefile, I get a string of errors that looks like this:
gfortran -Wall -g -std=legacy Abfind.o Abpop.o Abunplot.o Batom.o Begin.o Binary.o Binplot.o Binplotprep.o Blankstring.o Blends.o Bmolec.o Boxit.o Calmod.o Cdcalc.o Chabund.o Cog.o Cogplot.o Cogsyn.o Correl.o Crosscorr.o Curve.o Damping.o Defcolor.o Discov.o Doflux.o Drawcurs.o Eqlib.o Ewfind.o Ewweighted.o Fakeline.o Findtic.o Finish.o Fluxplot.o Gammabark.o Getasci.o Getcount.o Getnum.o Getsyns.o Gridplo.o Gridsyn.o Infile.o Inlines.o Inmodel.o Invert.o Jexpint.o Lineinfo.o Lineabund.o Linlimit.o Makeplot.o Molquery.o Moog.o Mydriver.o Nansi.o Nearly.o Number.o Obshead.o Oneline.o Opaccouls.o OpacHelium.o OpacHydrogen.o Opacit.o Opacmetals.o Opacscat.o Params.o Partfn.o Partnew.o Plotit.o Plotremember.o Pltabun.o Pltcog.o Pltflux.o Pltspec.o Pointcurs.o Prinfo.o Putasci.o Readobs.o Rinteg.o Setmols.o Smooth.o Specplot.o Stats.o Sunder.o Synpop.o Synspec.o Synth.o Tablepop.o Taukap.o Total.o Trudamp.o Ucalc.o Vargauss.o Vmacro.o Voigt.o Wavecalc.o Weedout.o Writenumber.o -o MOOG -L/usr/local/lib/X11  -lX11 \
        -L/usr/local/pkg/sm/2.4.36-F90/lib -lplotsub -ldevices -lutils

Abunplot.o: In function `abunplot_':
Abunplot.f:(.text+0x624): undefined reference to `sm_defvar__'
Abunplot.f:(.text+0x64e): undefined reference to `sm_window__'
Abunplot.f:(.text+0x66c): undefined reference to `sm_limits__'
Abunplot.f:(.text+0x6c6): undefined reference to `sm_ticksize__'
.
.
.

Running the nm command in the sm library yields:
> nm /usr/local/pkg/sm/2.4.36-F90/lib/libplotsub.a | grep ylabel
0000000000001cc6 T sm_ylabel
                 U sm_ylabel
0000000000000dae T sm_ylabel_
                 U sm_ylabel

Based on the readme for MOOGStokes (https://www2.mpia-hd.mpg.de/homes/deen/Software/MoogStokes/MoogStokes.README.txt), it looks like this error arises from how gfortran and SM handle underscores. I've tried using the gfortran flags listed in the MOOGStokes readme, but the problem persists. Are there other flags I should use, or is there another way to resolve this?

The SM developers seem to have anticipated such issues.
In your nm output sample,
T sm_ylabel is the native C function, and
T sm_ylabel_ is its wrapper to be called from Fortran.
gfortran usually appends single underscores to external symbol names.
If the MOOG object files were in fact built with the gfortran options in the question, the symbol references should match the wrappers in the SM library.
The double underscore suffixes may be left over from a prior build attempt with -ff2c.
When you change compiler options, remember to make clean before rebuilding.
If that doesn't lead to a solution, and nothing else is using SM,
the SM manual suggests
that you could rebuild it with this in its options.h:
#define FORTRAN_APPEND __


Accretion of in-falling material for a young main sequence star I'm reading material that is seemingly contradictory.  Some sources indicate that the evolution of a protostar to a main sequence star is characterised by a stellar wind that precludes the accretion of further in-falling material.  That is, the (young) star now has a constant mass.  However, other sources suggest that material may continue to accrete for a (brief) period after the protostar has become a main sequence star.

Can someone please confirm the actual process?

At the end of the protostar phase a vigorous outflow from the star develops called the T-Tauri Wind and this could cut off accretion.  Eventually, it develops into a normal star and the strong wind dies down.  Material that was not totally blown away could then continue to  fall down and be accreted


Would it be possible to discover an asteroid belt around another star? With existing or planned technology (e.g. the ELT or the JWST), would it be possible to infer the presence of a ring of asteroids similar to our own asteroid belt? Or would any signal be drowned out in the noise?

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

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

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


On what scale does the universe expand? According to the theory (or my understanding of it), the universe is not only expanding, but speeding up.

If the galaxies are moving apart, are the solar systems within them also moving apart from one another (& the planets within them), if they are, won't the Sun be moving away from the Earth?

Does the expansion of the universe occur at all scales or just large (inter-galactic) scales?

The short answer is: gravitationally bound system will not be ripped by the accelerated expansion.

A longer answer: The current standard model ($\Lambda$CDM model) says that everything started with the big bang. It released a lot of energy that pushed the Universe and is since then expanding. While it's expanding, gravitational attraction has been working to create the structures that we see today (planets, stars, galaxies, cluster of galaxies).

As the Universe expands, the density of matter and radiation within the Universe decreases. At about 5 billions years after the big bang, these densities became smaller than the density of Dark Energy (Dark-energy-dominated era). The strange things about Dark Energy are that, 1) its  density remains a constant regardless of the expansion of the universe, and 2) it has anti-gravitational properties - it pushes rather than pulls. Once Dark Energy came to dominate the density of the universe it began to dominate its overall dynamics, and has caused the expansion of the universe to accelerate.

But in structures which have already formed, gravity is much stronger than dark energy and so they don't feel it.


3D Positions of Nearby Stars I have come across a number of star catalogs that list stars by right ascension and declination, along with other data such as magnitude.

Is there a star catalog that lists the 3-D position of stars (e.g. includes distance from the sun, or otherwise specifies a 3-D position)?

One of the most popular ones known as HIPPARCOS lists parallax which is indicative of distance.


Why is this HST WFPC2 image so much fuzzier than the WFC3 image of the same field? I was looking at the comparison of 2007 and 2015 HST images of the "disappearing" star N6946-BH1 in Phys.org's article Collapsing star gives birth to a black hole. The 2007 images is from the WFPC2 while the one from 2015 is from WFC3.
I wondered why the WFPC2 image is substantially fuzzier than that from WFC3. I looked at the WFPC2 manual for cycle 17 and found it has the "L-shaped" detector. Three quadrants have CCD elements with pixel spacing of 0.1 arcsec, the fourth "planetary" array has a pixel spacing of 0.046. However it seems more complicated because - if I understand correctly - there are different focal ratios used for the two.

The WFPC2 field-of-view is divided into four cameras by a four-faceted
pyramid mirror near the HST focal plane. Each of the four cameras
contains an 800x800 pixel Loral CCD detector. Three cameras operate at
an image scale of 0.1″ per pixel (F/12.9) and comprise the Wide Field
Camera (WFC) with an “L” shaped field-of-view. The fourth camera
operates at 0.046″ per pixel (F/28.3) and is referred to as the Planetary
Camera (PC).

However, while the WFC3 manual for cycle 23 gives the pixel spacing as

Table 5.1: WFC3 Detector Characteristics
0.04″ per pixel

I didn't find a focal ratio.
I'd like to understand why the 2007 image is so "gaussian fuzzy". If it were only because of larger pixel spacing it would look more blockish or pixellated. Does this mean it was using a different optical path with lower optical resolution? Is the 2007 image from the WFC part at F/12.9? Does the WFC3 use F/28.3?
I also don't understand the optical reason why different paths would have different optical resolutions. This would not come from different Airy disks from simple diffraction - the angular size would be unchanged.

above: cropped sections from the same image shown below.

above: From here.

As Rob Jeffries suggested, in the WFPC2 observations the progenitor star fell located on one of the wide-field ("WF") chips of WFPC2. (I verified this by downloading one of the preview images from the Mukulski Archive.) This was purely by chance, since the WFPC2 observations were taken years before the discovery of the failed supernova. It's good luck that the region of the galaxy where the failed SN took place was imaged by WFPC2, and only slightly bad luck that it wasn't in a region imaged by the planetary-camera ("PC") chip.

WFPC2 is/was an array of four equal-sized CCD chips, each with 800 x 800 15-micron-wide pixels. However, the optics were set up so that 3 of the chips (the WF chips) were imaged using a focal ratio of F/12.9, while the fourth (the PC chip) was imaged separately with a focal ratio of F/28.3. In essence, a magnified optical beam was focused on the PC chip, giving a scale of about 3.0 milli-arcseconds per micron, or 0.0455 arc seconds per pixel. Meanwhile, the WF chips received a scale of 6.67 milli-arcseconds per microns, or 0.097 arc seconds per pixel. This undersamples the actual resolution of the telescope (which has a point-spread function with a full-width half-maximum of about 0.09 arc seconds in the I-band).

The UVIS (UV-visual) part of WFC3 has two 2k x 4k CCDs (with, I think, 15 x 15 micron pixels) and a focal ratio of F/31, which works out to 0.0395 arc seconds per pixel.

So in practice, WFC3 will always resolve things better than the WF chips of WFPC2 can, because its pixels can properly sample the telescope resolution and the WF pixels cannot.

You can see the difference in the upper "progenitor" (WFPC2) images, compared with the lower "2015" (WFC3/UVIS) images from the actual paper:

Note the stronger visible pixelation in the WFPC2 images (top two panels), due to the fact that the WF pixels are larger in angular size.

I suspect any other apparent differences in the press-release images are probably due to the processing used to create the color image. In particular, the side-by-side (WFPC2 vs WFC3/UVIS = "before and after") color image from the press release shows both images at the same spatial scale, as part of a single image file, which means that the same pixel density is present throughout the combined image. But since each original WF pixel is about 2.5 times the (angular) size of a WFC3 pixel, the WFPC2 image had to be rebinned to match WFC3 pixel scale (1 WF pixel = 6.25 WFC3 pixels) before the combined (JPEG/PNG/TIFF) color image could be made. The interpolation used for this rebinning process may be what introduced the extra smoothing in the WFPC2 image.

[Edited to add comment about rebinning used to generate color press-release image.]


Supernova explosion nearby A nearby star has crossed Chandrasekhar's limit. How do we come to know when it will undergo supernova explosion?
Or has it already undergone supernova explosion?

The Chandrasekhar limit applies only for white dwarfs. Stars on the main sequence (or even off the main sequence) can easily surpass it, but if a white dwarf's mass is greater than the Chandrasekhar limit ($1.39 M_{\odot}$), it will undergo some sort of collapse.

First, though, in response to

  Or has it already undergone supernova explosion?

White dwarfs are not formed from supernovae, but from stars that (relatively) gently shed their outer layers to form planetary nebulae. So no, the stellar remnant will not have gone supernova already.

Back to the main question. If a white dwarf exists, its mass is less than the Chandrasekhar limit. To surpass that, it needs to gain mass. This can happen in binary systems where the white dwarf accretes matter from the more massive star (generally a giant). The white dwarf can then undergo a Type Ia supernova and explode. The light curve will look a bit like this:

A second (but far less likely) scenario is that two white dwarfs merge. The result? A Type Ia supernova and, possibly, a neutron star.


Is this the Florence asteroid? I saw this blazing thing in the sky from Chennai, India at around 21:00 IST. Is this by any chance the Florence asteroid ? If not, what is this ? I'm pretty sure that this cannot be a firecracker because it was huge with large flames and it actually lasted in sky for at least 30 seconds till it reached heights where commercial aircrafts fly.

NOTE: This object was moving from the ground and upwards towards the atmosphere.

Congratulations, you have seen a fireball. An especially bright meteor. It was bright and fast. This is a rock, a very small asteroid, hitting our atmosphere and "buring up". It is also possibly a satellite or space junk re-entering. Re-entering satellites look move slower than asteroids. You can see fireball videos on youtube, as well as re-entries, to compare.

This event was also witnessed from Puducherry, the details seem to match up quite well. The time is slightly out, but the location is spot on. You can add your report at http://fireballs.imo.net/members/imo/report_intro

No, its not Florence, which will not be entering our atmosphere, and so will not appear to be burning, it will be a single point of light. It will be moving much slower, and won't be visible with the naked eye.


Expansion of the Universe Please forgive my ignorance. I dare say that these questions arise from me trying to understand the nature of the Universe using Newtonian physics. There are a few things that really bother me when topics such as red shift and expansion are discussed.

1) We can view a Galaxy that is 30bn light years away because the light we are seeing it as it was 13bn years ago, and at that time it was only 13bn light years away. So, it has travelled 17bn years further in the last 13bn years? Faster than the speed of light?

2) 13bn years ago, it was 13bn light years away. That is around the time of the big bang. How did it manage to get that far away so soon after the Universe began.

3) How is the known Universe even as big as 40bn light years, after only 13bn years in existence.

The current explanation is that "space" is continuously created in the intergalactic void, "pushing" the galaxies apart. From what i understood ( i am also ignorant ), the current rate at which "space" is created is expressed by the Hubble's constant 67.15 ± 1.2 km/s/Mpc ): for every million parsecs of distance from the observer, the rate of expansion increases by about 67 kilometers per second, which means that the distance between 2 points that are 1 megaparsec apart grows with 67km every second; between 2 points that are 2 megaparsecs apart with 134km/s, and so on so the relative speed between 2 points that are 5000 Mpc apart would be 335000 km/s which is more than the speed of light. In fact, two objects at this distance would not move, but 335000km worth of "extra space" would be "created" between them every second.

However in the last week there was an article depicting the shape of the universe, and it is hard for me to understand how it got to look like this : http://ifreepress.com/wp-content/uploads/2014/09/laniakea.png

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


Have more recent LIGO/VIRGO gravitational wave measurements narrowed down the speed of gravity further? This answer to How precise are the observational measurements for the speed of gravity? says:

...in 2013 a Chinese group built a model using Earth's tides that helped them narrow it down.

... [T]he speeds of gravity are from 0.93 to 1.05 times the speed of light with a relative error of about 5%. This provides first set of strong evidences to show that the speed of gravity is the same as the speed of light.

This is so far the most accurate measurement I've seen.  See the paper for more.
In the near future, LIGO may be able to provide more accurate measurements by comparing the distance among detectors and the delay of observation.

In case the links break, the papers are:

"Observational evidences for the speed of the gravity based on the Earth tide" TANG KeYun et al. Chinese Science Bulletin, February 2013 Vol.58 No.4-5: 474-477 doi: 10.1007/s11434-012-5603-3
"Bounding the speed of gravity with gravitational wave observations" Neil Cornish, Diego Blas, Germano Nardini 2017,  https://arxiv.org/abs/1707.06101

update: As noted in @amateurAstro's comment and linked well-sourced answer the time between a gravitational wave detection and X-ray burst of GW170817 and GRB 170817A constrains the difference between the speed of gravity and the speed of light to be "...between $-3 \times  10^{-15}$ and $+7 \times 10^{-16}$ times the speed of light..."
So my updated question is:
Question: Have more recent LIGO/VIRGO gravitational wave measurements narrowed down the speed of gravity further?

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

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

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

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


Is the measurement of distance and position of remote celestial bodies accurate? Considering that light is affected by gravity, how accurate are measurements of distant stars and galaxies?
When light passes through objects with great mass, such as Jupiter size planets, stars, or black holes, shouldn't it change its course and result in a deviation of trajectory? Does the light we see with telescopes when observing the many celestial bodies in the sky pinpoint the real location of such bodies in the past, or was its past location divergent from the point in space indicated by the light reaching us today?

Large masses can bend light, but space is largely empty.  The light from distant stars and galaxies rarely passes close enough to another star or galaxy to have deviated. On the few occasions when it does, it is special and notable.
For example, the Einstein cross looks like four quasars in a (very small) square, with a galaxy in front of it.  In fact it is four images of one quasar, the light having been bent by the gravity of the foreground galaxy.  In this case the image of the quasar is split up and moved by a few thousanths of a degree, because there is a very exact alignment of a quasar and a galaxy.
Such examples are rare.  For nearly everything else the light has travelled in a straight line through flat and empty space. The light we see pinpoints the location that the object was when the light was emitted.
An exception to this is that our own sun (and to a lesser extent the other planets) create local distortions. In very high accuracy measurements this can be taken into account. But the distortions are very small and as we know the location of the sun, they can be fully taken into account.
Although the light travels in straight lines, it can be very hard to measure the distance to stars and galaxies.  Often there is considerable uncertainty on the distance of astronomical objects. But this is not a result of gravity, it is just because measuring distance is hard.


Source of mirror reflectivity data I am trying to find a source of mirror reflectivity versus wavelength for the typical protected overcoat Aluminum mirrors used in astronomy. I need this to cover the whole visible range (300-1200nm) and ideally machine-readable. I've only been able to find part coverage of the wavelength range or single spot measurements at a specific wavelength.

The best I've managed is this page at Edmund Optics which has (small) plots but no table or downloadable file.

The spectral reflectance curve depends on the coatings applied.  For 'scopes used by people there is no point in maintaining reflectivity below 450 or above 700 nm.  You idea of "visible" range is, I fear somewhat optimistic.  You may be better off with an uncoated Al mirror (spectral curves easily found on the net) and giving up some signal in the photopic range.  -- or with a nice Ag overcoat, perhaps.
edit
Per request, some data on uncoated Al.
From   laserbeamproducts,

REFLECTIVITY OF PURE UNCOATED ALUMINIUM AT NORMAL INCIDENCE

Wavelength            % Reflectivity
248nm                      92.6
400nm                      92.0
532nm                      91.6
633nm                      90.7
800nm                      86.8
900nm                      89.0
1um                          94.0
3um                          98.0
10.6um                     98.7
20um                        99.0
100um                      99.4

An image published (by the author) at Researchgate


What is the difference between precession and axial tilt? I'm a right that the difference is that precession indicates the direction of the inclination of the Earth'axis and the tilt is the quantity of this inclination? So that when there is no tilt there can't be any precession?

And are the inclination degrees of the precession and the tilt related or the same?

It seems that the precession causes different possible polarstar, but is the tilt also have effect on this?

Perhaps this above it not right, so what else is the difference? Is perhaps the precession just the changing of the tilt?

The Earth orbits the sun. The plane in which it orbits is called the Ecliptic plane, and perpendicular to the plane is a direction, known as the ecliptic pole. The direction of the Ecliptic pole is towards the constellation "Draco". The Earth's North pole doesn't point in the same direction. It points towards the pole star "Polaris" in Ursa Minor. The difference between these two poles is the axial tilt (it is about 23.5 degrees)

Over time the direction of the North Pole changes, it moves in a circle around the Ecliptic pole, taking about 26000 years to complete one circuit. This is precession. The amount of tilt doesn't change much since the North Pole moves in roughly a circle. The tilt varies over a few degrees with a period of about 41000 years. The variation in tilt is not precession.

So axial tilt and precession are not the same thing. One is the angle between the ecliptic pole and north pole, the other is the motion of the north pole relative to the ecliptic.


What will future populations of stars be like? As stars slowly using up hydrogen and ejecting heavier elements into space, the future stars will form from nebulae that are more metal and helium rich.

The question is: how does the increased ratio of helium and metals affect star formation and stellar evolution?

Does more helium and metals in the core at the beginning means the star begins it's life in an advanced state so it dies faster?

(Adams & Laughlin 1997) discuss the effect of increasing metallicity in the future. A higher metallicity increases the stellar burn rate since the density increases but
the higher opacity reduces it a bit. The total effect is nonlinear; in their model the maximal lifespan happens for $Z\approx 0.04$ and beyond that it declines. Whether there are newer and better models of this I do not know.

They also point out that the maximum stable stellar mass decreases as metallicity increases, while the mass of the minimum mass mainsequence star declines, both as a result of opacity effects.

Overall, the metallicity growth is likely to continue across the stelliferous era. While extragalactic gas may fall in, it merely adds to the hydrogen pool and continued star formation. What could stop the metallicity growth is quenching due to a galactic merger or active galactic nucleus that blows away enough gas to stop further star formation.


When will Callisto be in orbital resonance with the rest of Jupiter's big moons? The four Gallilean moons of Jupiter (from innermost) are Io, Europa, Ganymede, and Callisto.  Io, Europa, and Ganymede are in a 1:2:4 orbital resonance. Callisto's orbital period is 16.689 days, which is not much more than twice Ganymede's period of 7.1546 days.

If Jupiter's orbit around the Sun was stable, when could we expect Callisto to come into a 1:2:4:8 orbital resonance with the Jupiter's other Gallilean moons?
Related: Does anyone know why three of Jupiter's largest moons orbit in 1:2:4 resonance?

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


How can the observable universe be so small if there are so many stars in it? The observable universe has a radius of about 46.5 billion light years. That’s big but I just wonder how that can be big enough to fit in everything we know exists in the universe. There are hundreds of billions of galaxies and each galaxy has billions of stars.

Plus there are vast spaces between galaxies. And galaxies tend to cluster up into local groups, so there would be even bigger gulfs of empty space between these local groups.

If the observable universe was only 46.8 billion light years, and there were this many and galaxies stars that need to fit inside, wouldn’t everything be much closer together?

If you know the total number $N$ of galaxies, and you know the size $V$ of the (observable) Universe, then you can calculate the number density $n = N/V$ and mean distance (of the order $1/n^{1/3}$) between galaxies, and convince yourself that it actually works out fine.

However, the reason we know how many galaxies are in the observable Universe in the first place is the other way round:

Observe an average number density of galaxies (which of course varies quite a lot from dense clusters to voids).
Calculate the size of the Universe by integrating the Friedmann equation (using as input observed values of the densities of the constituents of the Universe; dark energy, dark matter, gas, stars, radiation, etc.).
Finally, multiply the two numbers to get the total number.

Note that when observing the number density, we must not look too far away from the local Universe, since this means looking back in time to a period where the number density was different from present-day's value.


Shouldn't we not be able to see some black holes? Whenever there is a picture of a black hole, (even though they are artist depictions, this is how they are described) most are shown kind of in a flat disk like this.

My question is, why do we see the black hole itself? If it is sucking in material from all directions, then shouldn't the material be kind of covering the black hole, not just spinning around it to be sucked in from only the sides. Especially in the case that a star would be sucked in, then wouldn't all of its light & matter in the accretion disk be encasing the black hole?
In other words, why is the accretion disk flat, and not a sphere surrounding the black hole? So if this were true, we shouldn't be able to see some black holes, only their accretion disks around them.

(unrevised/old post: If a black hole is spherical and has infinite density, then it would be sucking in objects from all directions because the gravity would be the same all around. In other words, the event horizon should be all around it!

I understand everything would orbit the black hole a little before 'falling in', but since the amount of matter and light is so large, shouldn't it encase the black hole, thus making it covered in a ball of light? The event horizon would be a shell so we could not see some black holes, which would mean there could be billions more out there.

As for the other black holes which we have seen/proven, if the information above is true, then maybe they are the result of black holes with poles with low force/gravity/magnetivity/energy. Possibly they could be a disk of some sort or have an unequal distribution of density, either of which could lead to the flat event horizon, maybe even both.

Poles on a black hole would also help account for the quasars shooting out of some. Since the gravitational pull would not be as great from the poles of a black hole, then this would allow the quasars to beam out (otherwise they would have to be going faster than the escape velocity, which is the speed of light).

Maybe the quasars are even the black holes that I was talking about earlier, and this is why you would see no event horizon, only a shell of bright light.

I am no expert in astronomy but if any of these theories sound plausible, please let me know! Just looking for some explanation, I thought of this earlier and couldn't figure it out, I could be completely wrong though. Thanks!)

Well it doesn't exactly work the way you describe it. Matter doesn't just fall into a black hole.

A black hole still has a finite mass, which means other matter orbits its just like it would a comparable object of the same mass, like a star. In fact the gravitational disturbance it causes in this way is usually how we determine that there are, for example, supermassive blackholes in the center of galaxies.

Now the light you speak of is an accretion disk. When a star orbits a black hole, the star can get torn apart due to gravitational variation in the star as a consequence of the nearby black hole. Matter from the star starts orbiting the black hole, but it cant just do that in any direction. Due to conservation of angular momentum (the star itself was rotating) it is forced to rotate in one specific circular path, thats what we call the accretion disk and is the reason it doesn't become a huge sphere of light. As for light itself, it can't orbit, it either gets deflected or it gets trapped inside.


What kind of effects can maintain Uranus' rings' eccentricities? The recent paper Thermal Emission from the Uranian Ring System has been in the news recently. The introduction mentions:

The ε ring, the brightest and most massive of the narrow rings, was shown to maintain an appreciable eccentricity (e = 0.00794) and an azimuthally-varying width; the ring is five times wider and ∼2.5 times brighter in reflected sunlight at apoapsis than at periapsis (French et al. 19881; Karkoschka 2001a2). However, many fundamental parameters about the ring system remain unknown, including the filling factor, composition, thickness,
1French, R. G., Elliot, J. L., French, L. M., et al. 1988, Icarus, 73, 349, doi: 10.1016/0019-1035(88)90104-2
2Karkoschka, E. 2001a, Icarus, 151, 51, doi: 10.1006/icar.2001.6596

I've read in Wikipedia's Rings of Uranus that several rings are expected to be fairly young, and there may be interaction with some small moons, but I am surprised that the rings could maintain a significant eccentricity, and do not circularize or smear out due to collisions.
Question: What kind of dynamical effects can maintain Uranus' rings' eccentricities?
Uranus itself has a huge J2 and so any ring that isn't perfectly equatorial should have an eccentricity-dependent precession. The paper states that the apoapsis of the rings is five times wider than the periapsis, so there's definitely a population of different eccentricities and semi-major axes within the ε ring.
List of rings and their orbital parameters

A relevant paper is Papaloizou & Melita (2004) "Structuring eccentric-narrow planetary rings" which starts off promisingly with the following:

  The nature of the dynamical mechanism that maintains the apse alignment of narrow-eccentric planetary rings is one of the most interesting and challenging problems of Celestial
  Mechanics.

  According to the leading model (Goldreich and Tremaine 1979) the self-gravity of the
  ring counter-acts the differential precession induced by the oblateness of the central planet.
  Using this hypothesis, a prediction of the total mass of the ring can be made, which, in
  general, is not in good agreement with the inferred mass of the observed eccentric rings in
  the Uranus system

It then goes on to describe subsequent refinements taking account of additional effects including particle interactions and perturbations that bring the predictions more in line. The approach described is given in the introduction:

  In this work we build, from first principles, a simple general continuum or fluid like
  model of a narrow-eccentric ring. The eccentric pattern in the ring can be described as
  being generated by a normal mode of oscillation of wave-number m = 1 which may be
  considered to be a standing wave. Dissipation can be allowed to occur due to inter-particle
  collisions leading to a viscosity which would lead to damping of the mode. However, this
  global m = 1 mode can also be perturbed by neighboring-shepherd satellites which can
  inject energy and angular momentum through resonances. In this way losses due to particle
  collisions may be balanced. It is that process that is the focus of this paper. Other possible
  mechanisms, such as mode excitation through self-excitation through viscous overstability,
  that could arise with an appropriate dependence of viscosity on physical state variables (see
  Papaloizou and Lin 1988, Longaretti & Rappaport 1995), are beyond the scope of this paper
  and accordingly not investigated here.

The specific case of the ε-ring of Uranus is described in section 10.2, where the main satellite forcing considered is due to the 47:49 second-order mean-motion resonance between the ring and Cordelia.

The paper goes into quite a bit of mathematical detail on the various processes involved.


Is there a Vela Nebula? The Vela pulsar is a neutron star from a supernova that exploded 10,000 years ago. It spins 11 times a second. The Crab Nebula along with the central Crab Pulsar formed 1000 years ago. This pulsar spins 30 times a second. If there's a famous Crab Pulsar with a nebula, is there a Vela Nebula in which the Vela Pulsar is within?

The Vela Pulsar (PSR J0835-4510 or PSR B0833-45) is a radio, optical, X-ray- and gamma-emitting pulsar associated with the Vela Supernova Remnant in the constellation of Vela.

First line of the Wikipedia entry on the Vela pulsar.
It's obviously much more spread out and diffuse than the Crab nebula because it occurred more than 10 times as long ago.
Here it is as an optical image, roughly 6 degrees across and almost entirely filled by the supernova remnant. (The arrow marks the position of the Vela pulsar, which is distinctly off-centre.)

The first picture on the Wikipedia page is

which shows the pulsar wind nebula around the Vela pulsar. This again has an analogy with the Crab pulsar (and looks similar) and is due to the interaction of relativistic particles emitted by the pulsar with their surroundings and the strong magnetic fields. The picture is taken in X-rays and is only a few arcminutes across.


Which Saturn satellite passes closest to Saturn's rings and at what distance? I was using Stellarium to watch Saturn from its moon Pan and I saw the rings were very close to this moon. Now Saturn's rings extend for a large distance so several moons see them from close. I was wondering, is it know which Saturn moon passes closest to Saturn's rings and at what distance?

Pan, Daphnis, and various other moonlets, I would argue, are inside the rings.

If you explicitly discount the Encke gap (which Pan orbits in) and the Keeler gap (which Daphnis orbits in) as being part of the ring system, Daphnis would be your answer, as it is a ~8 km object in a 42 km gap. (for comparison, Pan is a ~35 km object in a 325 km gap)

Really, your answer depends on what you want to consider a moon. There are many objects small enough to be classed as moonlets (several hundred metres across) embedded in the rings, and presumably countless more of smaller size.


minimum size for a black hole to capture a photon If a photon is moving past a black-hole, what is the minimum size of the black-hole that will capture the photon and prevent the photon from going onto orbit around the mass? Do black-holes exist that have a stream of photons orbiting them?

There isn't really a minimum size. All black holes have an event horizon from which nothing, not even light can escape. The (Schwarzschild) radius of this event horizon is 3km for a black hole of a solar mass and scales linearly with mass. There is a region just outside the event horizon called the photon sphere, within which light can briefly orbit the black hole, but not in a stable fashion. It will either fly off in some other direction or spiral into the black hole.

You can check out this Physics SE question to see a derivation of the impact parameter (the effective radius at which light will be absorbed) and you will see that it is a small numerical factor ($\sqrt{27/4}$ for a non-rotating black hole) times the Schwarzschild radius (the event horizon of a non-rotating black hole), which is indeed proportional to the black hole mass.


Why isn't Ultima Thule spherical, while its two component parts are The two component parts of the Ultima Thule contact binary are approximately spherical. Why have the same forces that made the two parts spherical when they were separate not made the whole thing spherical after they joined?

The explanation is that as Ultima and Thule were forming, they collected out of small hunks of debris and piles of accumulated rubble.  The key thing is that once U and T became the biggest piles of rubble around, any new rock or pile of rubble that gently approached them would be much smaller then U and T and hit and gently bounce a bit and, on average, settles down in a low spot. The collisions are mostly pretty gentle and even rubble heaps probably mostly stay stuck together because the force of gravity isn't enough to overcome the cohesion of the material.

U and T grow and are -- roughly -- spherical since the things colliding with them and sticking to them are much smaller than they are, there are lots and lots of them, and they have a small tendency to end up in low spots. So a rough sphere emerges.

When U and T finally collide to produce Ultima Thule, it's another gentle collision, and the same thing happens: Gravitational forces are too small to overcome cohesion, so U and T stay pretty much the shape they were in and you have a body which is two spheres stuck together at the point of collision.

U and T formed spherical because they formed out of many much smaller pieces which are probably still pretty distinct, but which are too small to see -- yet.  UT formed out of two big pieces which are also still pretty distinct, but much easier to see..


Number of actuators in adaptive optics It is well known that diffraction resolution of a telescope depends on the exterior radius of aperture: One can, for instance, use just a narrow outer ring of a parabolic mirror and the resolution will be the same as it were the whole mirror (obviously, intensity will be much much smaller).

Now, take such a ring telescope, say, 3cm wide ring which is 100 meters in diameter.

Suppose we are using adaptive optics. How many actuators are needed to reach the maximal possible resolution?

I assume that spacing between the actuators should be of an order of Fried radius r0 (i.e. one actuator about every 10cm along the circumference of the ring). I.e. number or actuators should be of order D/r0, where D is the aperture diameter.

Is it correct?

It is well known that diffraction resolution of a telescope depends on the exterior radius of aperture: One can, for instance, use just a narrow outer ring of a parabolic mirror and the resolution will be the same as it were the whole mirror (obviously, intensity will be much much smaller)

The premise of your question is false. I tried to convince you to separate it out as a separate question, but to no avail. So I'll answer it here instead.

From comment:

  I personally made computation of the telescope transfer function for annular aperture and found that support of the function in the fourier space is the same as for circular aperture. This is an elementary excercise.

To help illustrate that a narrow annular aperture has much worse resolution than the full aperture, I'll do a quick comparison. The first plot is the diffraction pattern (scale is +/- 0.5 arcseconds) for a clear aperture of 2 million wavelengths (e.g. 1 meter radius with  1 micron wavelength), and the second is for the same aperture with a 0.9 meter obstruction.

I use Airy disks and Babinet's principle to subtract the E-field of the obstruction before squaring. For this symmetry there's no need to do an explicit Fourier analysis, the Airy disk is the Fourier transform of the circular aperture step function. (see also)

This is for an annulus with outer radius 1.0 and inner 0.9. If it was 1.0 and 0.99 it would look much much worse!

  Now, take such a ring telescope, say, 3cm wide ring which is 100 meters in diameter.

That would give you a giant splat pattern of roughly the same size as a simple 3 cm diameter telescope would.

Python script:

from scipy.special import j1
import matplotlib.pyplot as plt
import numpy as np

def get_E(a, lam, theta):
    ka  = (twopi/lam) * a
    x   = ka * np.sin(theta)
    return a * (2. * j1(x) / x)

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads        = 180/pi, pi/180

arcsecs = np.arange(0, 0.5, 0.0001)[1:]
theta   = (arcsecs/3600.) * rads

r1, r2  = 1.0, 0.9
lam     = 1E-06

arcsecs   = np.linspace(-1, 1, 1000) # arcsec (skips zero to avoid singularity)
Xas, Yas, = np.meshgrid(arcsecs, arcsecs)
Ras       = np.sqrt(Xas**2 + Yas**2)
theta     = Ras * pi / 3600. / 180.

E1, E2    = [sign*get_E(r, lam, theta) for r, sign in ((r1, +1), (r2, -1))]
asec      = np.linspace(0, 1.0, 500)[1:]
th        = asec * pi / 3600. / 180.
e1, e2    = [sign*get_E(r, lam, th) for r, sign in ((r1, +1), (r2, -1))]

Etot = E1 + E2
etot = e1 + e2

if True:
    plt.figure()
    plt.subplot(2, 2, 1)
    plt.imshow(E1**2, cmap='gray')
    plt.title('r = 1.0, lam=1E-06', fontsize=16)
    plt.subplot(2, 2, 3)
    plt.imshow(Etot**2, cmap='gray')
    plt.title ('r1, r2 = 1.0, 0.9, lam=1E-06', fontsize=16)
    plt.subplot(2, 2, 2)
    plt.plot(1E+06*th, th*e1**2) # weight by th for azimuthal (phi) integration
    plt.xlabel('urad', fontsize=14)
    plt.title('r = 1.0, lam=1E-06, integ theta')
    plt.subplot(2, 2, 4)
    plt.plot(1E+06*th, th*etot**2) # weight by th for azimuthal (phi) integration
    plt.title('r1, r2 = 1.0, 0.9, lam=1E-06, integ theta')
    plt.xlabel('urad', fontsize=14)
    plt.show()


When would we detect a tiny meter size natural satellite in a geostationary orbits? A natural, tiny (meters-size, maybe 10.000kg mass) natural satellite could be trapped in a geostationary orbit. I wondered for quite some time:

When and how are we able to detect these satellites?

I assume the distance of 35.700km of way too far for naked eye detection. So the next realistic opportunity would have been Galilei who first used telescopes for scientific observations of the sky. Could he have detected such an object?
Now even if he had enough resolution, he certainly didn't systematically cover the whole 4$\pi$ of the sky (especially as the satellite could be syncronous with the other side of the earth, and he could never see it).
Then, would it be at the times of Hubble (because of the Mount Wilson Observatory and other similarly powerful telescopes)? Would it at the times when quality equipment for hobby astronomers became cheap and therefore widespread enough (to cover huge areas of the sky)? Or would we - until today - not be able to detect such objects?
To answer this question, one needs to consider both the technical ability as well as the area of the sky that is covered.

tl;dr: At distances far enough from Earth that the motion with respect to the stars was slow, a serendipitous survey photographic plate from a large enough telescope might catch a trail, and in a doubly serendipitous situation it might have been a short exposure, duplicated on the next night, an Earth orbit suspected, and a hunt for a second Earth satellite begun.
However starting in the 1960's and 1970's radar and visual scans for artificial satellites in Earth orbit would have found this natural satellite in Earth orbit if it were low enough.

I'll start with @CarlWitthoft's 5 meter asteroid refer to this answer and especially this answer. Two equivalent equations to get the absolute magnitude of an asteroid are:
$$ H = C - 5 \log_{10} D - 2.5 \log_{10} p_V$$
where $H$ is absolute magnitude, $p_V$ is albedo, D is in km, and $C$ = 15.618, and
$$M_{Abs} = 5 \left(\log_{10}(1329) -\frac{1}{2}\log_{10}(\text{albedo}) -\log_{10}(D_{km})\right).$$
A 5 meter diameter asteroid with an albedo of 0.1 has an absolute magnitude of +29.6.
The apparent magnitude from this answer:

Knowing the absolute magnitude of an object, you calculate the apparent magnitude $m$ using:
$$ m = M_{Abs} + 5 \log_{10}\left(\frac{d_{SR} \ d_{RE}}{1 \ \text{AU}^2 O(1)}\right), $$
where $d_{SR}$ and $d_{RE}$ are the Sun-Roadster and Roadster-Earth Sun-satellite and satellite-Earth distances, each normalized by 1 AU, and the factor $O(1)$ is the phase integral, of order unity, taking into account the angular difference between the direction of illumination and the direction of viewing. In an order of magnitude calculation, this only becomes really significant when the body moves between the Sun and the viewer. See https://en.wikipedia.org/wiki/Absolute_magnitude#Solar_System_bodies_(H).

Let's pick two distances. One is the geostationary distance where the satellite would appear to hover above the observer, probably drifting up and down in a roughly annelema shape because the Earth's oblanteness will eventually tilt the orbit. See Geostationary orbit; Orbital stability. It will have a distance to Earth as close as 36,000 km.
The other is a low Earth orbit, but high enough that it will not decay due to drag too soon. Call it 1000 km altitude, or a circular orbit with a 7378 km semi major axis.
Plugging all of this into the equation above, I get:
     orbit                   closest distance      visual magnitude
Geosynchronous altitude         36,000 km                +20.6
Low Earth Orbit                  1,000 km                +16.7

In low Earth orbit the apparent magnitude is almost as bright as Pluto, but it's going to be moving pretty fast. $\sqrt(GM/a)$ gives 7350 m/s, at a distance of 1000 km that's about 0.4 degrees per second. Any large telescope that's being used in astronomy will be tracking the motion of stars, or close to that, so it will be fast +17 magnitude track rather than a dot, and last only a fraction of a second. That probably wouldn't expose a photographic plate or if it did it would be dismissed as an artifact, meteor, or scratch. Visually it wouldn't be noticed.
At GEO type distances and +20.6 magnitude, the object would move about 0.25 degree in a minute, so it might also be captured on a photograph, but by the time the plate was developed it would be impossible to know when it appeared in a long exposure. However if the exposure (say at the Hale 200 inch telescope) were short, it really is possible that at a short-term trajectory on the celestial sphere could be considered. The problem is that nobody would suspect that it was in Earth orbit, and they would extrapolate to a heliocentric orbit and never find it again.
If the plate happened to be a series and there was another exposure of the same patch of sky the next night, then they would see it again and get pretty suspicious that it was in Earth orbit.
However, in a post-Sputnik cold war era radar and optical searches of the sky for objects in Earth orbit became particularly interesting.
So I would say that satellite surveys (both optical and radar) in the 1960's and 1970's would be the first likely candidates to find this 5 meter, 0.1 albedo satellite.

For some insight on optical tracking, see the two videos linked in Are commercial communications satellites in GEO being constantly monitored by telescopes?. Currently these links will take you to a new tab with the YouTube video:
https://www.youtube.com/watch?v=8ebIAUjFfZM
https://www.youtube.com/watch?v=4FXX1kSNljU
If you would like to see them here, then leave a comment or answer or vote at Interest in looking into adding the YouTube viewer?.


Comet Neowise c2020 f3 will return in ~6800 years, how can we be sure Earth will be in the right place to see it then? I have read in several places that the orbital period of comet Neowise c2020 F3 is about 6800 years. In several popular media sources it is said that we will be able to see the same comet after ~6800 years.
How is it known that the Earth will be at exactly the right place in order for it to be visible?

The time of the orbit is not well known but the inclination can't change as much (small changes in velocity due to outgassing can significantly change the orbital period, but can't change the inclination much. So, while we have a good idea of it's orbital track close to perihelion, we just don't know when it will be there.
We don't know where the Earth will be in its orbit when it returns, but a comet is typically visible for several months. It is possible that the Earth will be particularly badly placed, but even then it will be visible
The Earth doesn't need to be in an exact position for the comet to be visible. It may not be a naked eye object next time, but it should still be visible, no matter where Earth is on its orbit.


Solar Telescope Projector Screen? [& intro telescope for solar eclipse] I don't know very much about telescopes; I used one a few times as a kid, but that's about it. I understand the general design & use of reflectors & refractors, but I don't know enough to decipher focal length, etc. So, I don't feel comfortable trying to DIY my own contraption, that won't be nearly as good, assuming it works at all.

I'm looking to buy a minor telescope (a few hundred dollars [US] or less) to view the American Total Solar eclipse on Aug. 21, 2017; and possibly reuse for the Mercury transit in 2019. I'm a decently well off computer programmer, so blowing $200 (& some gas to drive up-state) for a one off just to experience the Solar eclipse is worth it to me, even if I don't use it again.

I spent a few hours wandering around the internet for something, and I found a few styles of "Projection Screens/Scopes" that are really cool, but I can't find anywhere to buy them. The first is an image that several websites used as their article's banner image, but none say what it is or where to get it.

I found another one, where a guy hacked together a projector scope from an old 8mm film projector.

I've found many DIY tutorials to roll-your-own out of dixie cups & Pringles cans, but as I said before: I'd rather not waste my time fumbling with that, since I'd probably screw it up; I'd rather pay money to get one that's good quality & done right.

I've wanted to buy a big light bucket for years now, so I can start looking at nebulae & planets, but never got around to it; mostly because I'm not sure what to buy. I've been looking at the "Orion SkyQuest XX14i Dobsonian" for a while, but never pulled the trigger. Given that the eclipse is only 3 weeks away, I figured something smaller would be a faster choice in the short term.

I found the "Celestron EclipSmart 50 50mm f/7.2 Alt-Az Solar Telescope" that comes with a backpack storage bag, for \$100. It has a built-in (i.e. non-removable) solar filter, & a collapsible tripod. But that seems pretty cheap; is \$100 for a telescope like that normal, or is that a poor quality trap for laymen like me?

Do yall have any advice on something to get for the solar eclipse? And is there anywhere to get a professionally made projector screen like that, so I can watch the multi-hour process from a chair rather than bending over for 2 hours? (p.s. I'm 6 foot 3 inches tall) Also, if possible, being able to record video of it on my Samsung Galaxy S6 would be cool, but not required if that's a big deal.

Thanks for your help,

-Yurelle

For your purposes, I'd really recommend just purchasing the EclipSmart. The telescope itself is of reasonable quality. If you have a camera tripod, bring it with you to view the eclipse$-$the stock tripod is so flimsy as to be useless. If you dont have one, buy one. Additionally, you should invest in different eyepieces. The telescope comes with a 20-mm EP, which will provide relatively low magnification. I'd recommend getting a 6-mm EP, and possible an intermediate one as well.

Clear skies!

P.S. I would recommend against getting the XX14i as your first telescope. I think instead you should get a smaller EQ, in the 4-6-in bracket. Having to manually find your target will teach you much about the skies, and having to do so with a smaller aperture will humble you. I worked a 4" Alt-Az for a decade as a kid before moving to a 4.5" EQ for a year, and then finally an XT12i as a highschool sophomore. Having to find my targets in my relatively bright skies made the sights I see through my 12" that much more incredible. I think allowing yourself the same experience will be a decision you won't regret. Don't fall for aperture fever!


What are the dark spots in this Hubble image of the Crab Nebula? I recently ran into this NASA Hubble image of the Crab Nebula in this article talking about its core. What I am interested to know though is: what are these round dark spots appearing in the image (upper right, lower center and lower left for example)? "Holes" in the Nebula? Objects in front of the Nebula, visible against its illuminated gas? Blemishes in Hubbles lens(es)?

CLARIFICATION: I am asking about the small pixel like dots. Not the large "holes" from this question.

They are most likely dense clumps of dust absorbing whatever wavelength of light this image was captured in. It's hard to say just how big they are as I can't find any scale for this image, but generally nebulae are lightyears massive, with the Crab Nebula according to wiki being 11 light years. So whatever they are, they're still pretty big.

Edit: Just look at the horizontal band of darker lumps, you can tell that they are between us and the brighter region. The band would be of the same material as the bright green cloud, just less energised as it is further away from the star.


How to calculate the argument of perihelion for exoplanets with incomplete orbital elements? I'm trying to simulate the Kepler-444 exoplanetary system using data from this website. The problem is that there is no argument of periapsis parameter for the Kepler-444 system, so while I think I'm getting the correct shape of the orbits, their periapsises and apoapsis are all aligned in the same direction, as you can see in the screen dump below.

So my question is how, or if, I can get the periapsis to be correctly aligned with the data that is available on this website for these planets (unfortunately I can't link to a query, but you can search Kepler-444 and the planets should pop up in the table)? The way I'm doing it now is that I take the semimajor axis and the eccentricity data from the website I linked to, and then I calculate the periapsis for each planet and then I set the x axis to be that value while I let y and z be zero. Once there, I use the vis-viva equation to calculate the velocity vectors for the planet, and then I iterate the simulation forwards in time and get the shapes of the orbits. With the solar system planets, I take the argument of periapsis and the inclination of the orbit and rotate the position and velocity vectors that I derived accordingly, and this way I end up with orbits that have the right shape and their periapsises are correctly aligned, too. Naturally I don't get the right position of a planet in its orbit at a given time using this approach, but for now that is not important; I just want the orbits to have the correct shape and to have their periapsises correctly aligned.

I know very little about orbital mechanics, so any help would be much appreciated so that I can learn something and improve the quality of my simulation.

Ford et al. (2008) give the following relationship for eccentric transiting planets:

$$\frac{t_\mathrm{D}}{P} \simeq \frac{R_\ast}{\pi a \sqrt{1-e^2}}\sqrt{(1+r)^2 - b^2} \left(\frac{d_\mathrm{t}}{a}\right)$$

Where $t_\mathrm{D}$ is the total transit duration, $P$ is the orbital period, $R_\ast$ is the stellar radius, $a$ is the semimajor axis, $e$ is the eccentricity, $r \equiv R_\mathrm{p}/R_\ast$ is the ratio of planetary radius $R_\mathrm{p}$ to the stellar radius, $b \equiv d_\mathrm{t} \cos i / R_\ast$ is the impact parameter, $i$ is the inclination and $d_\mathrm{t} = a(1-e^2)/(1+e \cos \omega)$ is the distance between the star and the planet during the transit. By inverting this you would be able to estimate $\cos \omega$ from the parameters listed on the site, which unfortunately won't give you a unique value (and if you're unlucky, the various uncertainties may conspire to give you a value $\left|\cos \omega \right|>1$).

In the case of Kepler-444, the website gives a link to the paper where the orbits were published, Campante et al. (2015). Their table 4 tabulates both $e \cos \omega$ and $e \sin \omega$ for the planets. Unfortunately it is not possible to simply determine the value from these estimates without having access to the full distribution of samples, so this isn't entirely helpful.

The good news is that the statistical significance of the non-zero eccentricities doesn't appear to be too high in this case, so you can probably treat the orbits as circular and not worry too much about it: tightly-packed systems like this usually have very low eccentricities otherwise they would become unstable.


Pinhole Projector: Can I use a specific solar optical eye piece to focus the Transit of Mercury? The title pretty much explains this one. Can I use a specific solar optical eye piece, or other optical device, in conjunction with the pinhole projector method to focus the Transit of Mercury?

Would anyone have any specific recommendations for specifications, etc if it is possible to achieve this...

The whole point of a pin hole projector is that it doesn't need any optics just a pin hole.

If you are after magnification then that is related to the distance you project the solar image behind the pin hole, with a projection distance of 1000mm the solar disc will have a diameter of about 9mm. Increasing the distance will increase the size of the image but also reduce its brightness. Increasing the size of the pin hole will increase the brightness but reduce the sharpness of the image.

I suppose you could use a handheld magnifying glass to look at a small image but would not personally recommend it as unless you are extremely careful there is always a risk of fire or even blinding yourself or someone else when using a lens near an unfiltered solar observation. There is a famous story of an experienced astronomer setting his beard on fire because he left the cap off his telescopes finder.


How can I find out which known stars lie within or close to M8 I'm looking for a way to find stars within a certain region of space, the Lagoona Nebula in this case, based on what we currently know. Are there some databases that allow you to make such queries?

Please note that I'm not looking for stars that can be seen as being near M8 from earth, but for a list of known stars that are thought to actually be near or part of M8 itself.

Have a look at the following references - found relatively easily using the following query in the NASA ADS site.

UBVRI and Hα Photometry of the Young Open Cluster NGC 6530
This paper by Sung & Bessell (2000) contains an extensive photometric catalogue. The main table can be queried interactively.

The Lagoon Nebula and its Vicinity
This is a review article by Tothill et al. (2008), from which you can get pointers to most of the significant literature (pre-2008) on the region.


How would water waves behave in partial gravity? Would an "indoor" pool of water on the Moon or on Mars behave differently than on Earth? Would the waves caused by a splash be higher or lower? Would they propagate faster or slower? Would boyance be different? And what would it be like to shower in slowly falling water?

Would an "indoor" pool of water on the Moon or on Mars behave differently than on Earth?
Depends on what you are observing.

Would the waves caused by a splash be higher or lower? The waves would be higher.

Would they propagate faster or slower? They would propagate at the same rate.

Would boyance be different? No, because buoyancy depends on the displacement of mass.

what would it be like to shower in slowly falling water? How can anyone here answer this since only a few people (<0.001%) have experienced this.


Are the Trappist-1 planets close enough to be tidally heated? According to the NASA press release, the Trappist planets are close enough (only a few million kilometers) that "the tidal forces between the planets are not negligible".  The speaker says that this could cause ocean tides on the planets.  Are they close enough that the tidal forces could heat the interior of the planets?

This is a complicated question that would really require a full physics simulation and better knowledge of the system to accurately answer. But let's try a few back of the envelope calculations to see what we get.

Calculating tidal forces from TRAPPIST-1c on TRAPPIST-1b

I'm going to calculate the tidal effects of TRAPPIST-1c on TRAPPIST-1b (simply because, a priori, this seems likely to be where the strongest tidal heating will be induced). See the figure below which describes the parameters.

The tidal force of 1c on 1b is defined as the differential force of gravity across 1b, that is, the difference of the force of gravity on the side of 1b facing towards 1c and the force of gravity on the side of 1b facing away from 1c. Mathematically, we get.

$$F_{tide,c-b} = F_{g,-R_b} - F_{g,+R_b} = \frac{GM_bM_c}{(D-R_b)^2} - \frac{GM_bM_c}{(D+R_b)^2} = 4GM_bM_c\frac{DR_b}{(D^2-R_b^2)^2}$$

We can presume that $R_b << D$ (for this case $R_b/D = 1\%$) and reduce this to

$$F_{tide,c-b}(D) \approx 4GM_bM_c\frac{R_b}{D^3}$$

But this isn't enough to determine the amount of tidal heating that may occur. Tidal heating only occurs when the tidal force changes. It is this constantly changing tidal force which results in tidal flexing of the planet and thus creating heat through tidal friction. Fortunately, for these two planets, the tidal force will be changing since $D$ will be constantly changing. So let's calculate $F_{tide}$ for the two extremes where these planets are as close as possible and as far as possible and difference them.

$$\Delta F_{tide,c-b} = F_{tide}(0.004\:\mathrm{AU}) - F_{tide}(0.026\:\mathrm{AU})$$

If I plug in numbers to this, I find that

$$\Delta F_{tide,c-b} \approx 3.7\times10^{20}\:\mathrm{N}$$

Okay, but what do we do with this number? It's somehow a metric of the change in tidal forcing that 1c imparts on 1b, but is it negligible? To determine this, we have to compare it to something. Let's compare this to the tidal forcing that TRAPPIST-1b would receive from the star.

Calculating tidal forces from TRAPPIST-1 on TRAPPIST-1b

I've already set up the math, so we don't need to go over that again. But first, let me discuss where this tidal forcing is actually coming from. Quoting an article from space.com, the paper's author, Gillon, states:

  Because the seven alien worlds orbit so tightly, they're probably all tidally locked, Gillon said. That is, they likely always show the same face to their host star, just as Earth's moon only shows the "near side" to us.

As I said above, the only way to produce tidal heating is to have changing tidal forces. These planets are likely synchronized and always present the same side to the star. It's marginally possible these planets aren't perfectly tidally locked, but rather have some higher spin-orbit resonance. That is, their spin orbit resonance might not be 1:1 (as it would be if they were tidally locked) but instead could be something like 3:2 (which is what Mercury has). I'll ignore that complication though and just assume 1:1 resonance. So if they're tidally locked, they can't experience differing tidal forces through their own rotation. Instead, the differential tidal forcing comes from the orbit's ellipticity. Sometimes the planet will be closer and sometimes it will be farther, causing a differential tidal force on TRAPPIST-1b from the star as it orbits. This is exactly what occurs in the tidal heating of Io. Let's calculate $\Delta F_{tide,*-b}$ by using the varying distances TRAPPIST-1b will have from the star. I found that TRAPPIST-1b will orbit between $0.0101\:\mathrm{AU}$ and $0.0119\:\mathrm{AU}$1. This means the differential tidal force is:

$$\Delta F_{tide,*-b} \approx 4GM_bM_*R_b\left(\frac{1}{(0.0119\:\mathrm{AU})^3} - \frac{1}{(0.0101\:\mathrm{AU})^3}\right) = 1.8\times10^{23}\:\mathrm{N}$$

Is planetary tidal heating non-negligible?

The back of the envelope calculations show that the differential tidal force on TRAPPIST-1b from TRAPPIST-1c is about $0.2\%$ of the differential tidal force due to the star. Whether you consider this negligible or not is up to you. I'd personally consider it a pretty small effect and say that most of the tidal heating these planets experience comes from the star itself.

Could the inter-planetary tidal heating still contribute to the tidal heating of the planets enough to heat the interior?

This is a remarkably hard question to answer and I can't even really do a back of the envelope calculation without making wild, unjustifiable assumptions. The calculations above simply determined the maximum tidal force variation over time. That doesn't tell us anything about how much tidal heating this may induce though. That requires knowing more about the planet itself, in particular the planet's Love numbers which define the rigidity of the body and thus how easy it is to stretch via differential tidal forces. You can vary your tidal forcing as much as you want, but if your planet is pure iron (and thus very rigid) you're unlikely to have as much of an effect as if it were primarily silicate (and thus much less rigid). The paper produces the plot below which defines the potential constituents of each planet. This would be a first step in determining the planet rigidities, but as you can see from the error bars, it would be highly uncertain.

Overall, and this is entirely opinion based and from my calculations above, but I'd say the chances of the inter-planetary tidal heating having significant effects on the interior heat of of these planets is negligible. More than likely the biggest contributing factor is radioactive decay, followed by tidal heating from the star (but this is amplified by the eccentric orbits induced by planetary gravitational perturbations)

1 Note that this calculation involves using the eccentricity and the paper only provides an upper limit. These distances then represent an upper limit as well and the final answer will also be an upper limit. It may be less.

Values used in calculations:

$G = 6.67\times10^{-11}\:\mathrm{m^3 kg^{-1} s^{-2}}$
$M_b = 5.075\times10^{24}\:\mathrm{kg}$
$M_c = 8.239\times10^{24}\:\mathrm{kg}$
$M_* = 1.604\times10^{29}\:\mathrm{kg}$
$R_b = 7.34\times10^{6}\:\mathrm{m}$


How big do comets get? If my understanding is correct, Chury, the best explored comet so far is rather on the smallish side as comets go. Most comets observable with naked eye were so because they flew close to Earth, not because they were so big. I wonder, though, how big are the biggest ones.

Let's get it in two variants:

How big is the biggest known comet? (mass, diameter)
Is there an estimate, or a theoretical limit on how big a comet can be? (for whatever reason, e.g. becoming a planet, breaking up due to tidal forces or whatever)

I suppose size limit of the comet is still under process but astronomers have found a few large ones though!
The comet Holmes which appeared in 2007 became brighter and also swelled up in size. Its nuclear size is only 3.6km (2.2 miles), but it released so much dust and gas so that it expanded to about 869,900 miles (1.4 million kilometers) i.e larger than the sun.
The largest comet in terms of nuclear size was discovered in the year 1729, comet Sarabat which had a nucleus of about $100 {\rm km}$.


What are the present, and potentially the future roles of data scientists in Astronomy? I am a data scientist and curious about their roles in modern Astronomy, both now, and in the near future. What are type of roles and projects where data scientist are most active in Astronomy? What are the basic solutions provided by data scientists in astronomy. Are there data scientists as principle investigators, or are they at least team members in research efforts?

Astronomer is moving towards the big(ger) data era because of many sky survey technologies. The coming ones include, e.g., LSST, JWST, and WFIRST.

By the meaning of survey, it normally means observing the whole sky over a few days, and keep repeating over and over. Also, since most of the surveys are imaging technologies, every pixel in an image is important.

So, you can imagine how much data is incoming per day, and how much data science is important in many fields of astronomy (and cosmology). So, data science is, kind of, taking over when it comes to analyzing data from surveys.


What are the Earth-like features of Titan? This NASA site says that hidden beneath the smog of Titan are Cirrus like clouds and This space article describes how Titan's atmosphere is similar to Earth's too.

What other features does Titan boast that make scientists so ready to boast that "Titan may well be the most Earth like planet we know of besides Earth itself"? (paraphrased)

Answers should focus on the features of Titan itself, obviously a similarity is that it orbits the same Sun, but this type of thing is not what I'm after. I am mainly searching for geological or atmospheric similarities that Titan shares all the way down to potential similarities in the composition and flow of the planetary core.

To expand a bit on Undo's answer, there are some geomorphological features on the surface of Titan that resemble similar features on Earth, suggesting similar processes.

For example, according to this NASA image comparison, some lake features have features that are similar to groundwater based salt pans on Earth. Image below:

The feature on the left is Ontario Lacus on Titan compared to the Etosha Pan of Earth on the right.  It should be noted, that the Titan feature is considerably larger than the Earth analogue.


How to calculate planetary position over vast time spans Ignoring the turbulent beginnings of our solar system I wonder if there is a meaningful way to calculate the relative position of the planets of our solar system 100 million years into the future using the standard parameters
N /  i  /  w /  a /  e   /  M / +- pertubations , {assuming no unannounced black hole or neutron star zips through our solar system in the meantime.}
How far can the orbital calculations be projected into the future before error accumulation make these calculations completely meaningless?

Jacques Laskar, at IMCCE, has been working on describing planetary motions on long term time scales.

Quoting and translating a paper by him:

Motion in the solar system is chaotic, meaning that trajectories diverge exponentially: an error on the location of a planet gets multiplied by 10 every 10 million years
This means that a 15 meter error on the initial location of the Earth yields an error of 150 meters after 10 million years, and 150 million km afer 100 million years. Long term predictions involving a single trajectory are impossible.

For predictions over longer periods, it becomes necessary to study multiple possible trajectories. Laskar and Gastineau have published results of a simulation of the solar system involving 2501 variations of the location of Mercury. In this simulation, "initial conditions  differ by only 0.38 mm in the semi-major axis of Mercury". Some of those trajectories involve "possible collisions of Mercury, Mars or Venus with the Earth".


What is the physics of a gas mass subject to gravity in space? There are many formulas for atmospheric pressure on earth, but how does gas behave in free space?

I am thinking about why stars form. I am guessing that the gas density will influence pressure, as well as gravitational force. So both factors must determine if a given gas mass in a given volume in space will collapse into a star.

So a more well defined question would be: what is the relation between mass and volume in space (let's think about hydrogen only) so that this mass will collapse into a star (or at least forms a spherical body)?

A second question is: given a mass and volume so that the gas collapses into a spherical body, what is the pressure vs. distance from the center of the sphere?

Quite interesting I think!

Any ideas?

The formula for the change of pressure with altitude is the same-you need hydrostatic equilibrium.  This means that at constant temperature and over small height differences (so gravity can be considered constant) the pressure falls exponentially with the scale height.  You have $\frac {dp}{dz}=-{\rho g}$  You can compute the local $g$ by taking everything closer to the center as a point mass and ignoring all the mass outside the current radius.  You can consider the compression adiabatic, but need to use the appropriate value of $\gamma$ in $PV^\gamma=k$ for the local temperature.  Calculating the radiation is hard unless there is enough dust around, in which case the grains can be black bodies.  If you assume rotational symmetry, you have a 2D problem, which you can model on a grid and step through.


What record do we have of the length of supernovas? recently I heard an astronomer on the radio claim that the supernova we have observed in our own galaxy lasted about 6 months, while in distant galaxies they last about 7-8 months, due to relativity and the expanding universe. He didn't give any references. Just to be clear, I don't mean to imply that he implied the actual duration was different, but that it only appeared different because of expansion.

I've found records of supernova, but only 1 mentioned how long it lasted, the first recorded one was said to last 8 months according to wikipedia.
Can someone point me to a list that contains both the location and duration of supernova's .

I would expect that the further away a supernova is, the "longer" we observe it will be active because of an increased red shift.  Objects in the universe that are further away are generally going away from us faster than objects that are close, hence the increased red shift.  This increase in time due to redshift is very similar to the Doppler effect, where a siren is high pitched (and faster) when approaching you (e.g. blue shift with light), and low pitched (and slower) when going away from you (e.g. red shift with light).  The faster the siren is moving away, the more the effect (both lower pitch and longer time) will be increased, likewise with light.

Here are some sources that may help with what you are looking for (although I haven't been able to find a list exactly like you are looking for):

http://www.wolframalpha.com/input/?i=supernova

http://en.wikipedia.org/wiki/Supernova#Light_curves

http://en.wikipedia.org/wiki/Cosmic_distance_ladder#Type_Ia_light_curves

http://curious.astro.cornell.edu/question.php?number=49


Wavelengths for observed objects -- emitted or observed? The WISE craft (Widefield Infrared Survey Explorer) surveys the sky in 4 wavelength bands; 3.4, 4.6, 12, and 22 $\mu$m. When it's observing an object with an estimated redshift (given by

$z = \frac{\lambda_{obs} - \lambda_{emit}}{\lambda_{emit}}$

from Wikipedia) of, say, 0.373 (the lowest redshift for one of the objects I'm studying), would the wavelength it's seen in be its observed wavelength? I'm interested in calculating the wavelength actually emitted by the object.

If the redshift is $z$, then the wavelength of light you observe is at a wavelength that is $1+z$ times longer than the wavelength emitted in the frame of reference of the galaxy.

In this case $z=0.373$, so when you observe light centered on say the 3.4 micron band, then the light was emitted at a wavelength of 3.4/1.373 microns.


Estimating Galactic Mass of Andromeda So I need to estimate the mass of the andromeda galaxy using:
$$ M = \frac{v^2r}{G}$$
where $v$ is the rotational velocity, $r$ the galaxy's radius and $G$ Netwon's gravitational constant. I'm told the following:

Distance (m) = 2.403×$10^{22}$
Radius (arcmin) = 12250.40
Velocity (km/s) = 300

I went so far as to convert the radius:
$$ \frac{12250.4}{60*57.3} = 3.563\ rad$$
then, calculated the radius:
$$0.5*3.563*2.403\times10^{22} = 4.2757\times10^{22}$$
Then subbed into original formula to get:
$$\frac{ 4.2757\times10^{22} \times 300^2}{G} = 1.7595\times10^{36} kg$$
It should be $2\times10^{42}kg$. where have I gone wrong?

You have gone wrong because the radius of Andromeda is not 12250.40 arcminutes.
It is difficult to be more helpful. Possibly the units are arcseconds?

Neither do I see why you have inserted a factor of 0.5 in your second equation.
or why you have inserted $300^2$ into the third equation, when $v$ is given in km/s, so it should be $(3\times 10^5)^2$.


How do I adjust the sunrise equation to account for elevation? I have been working with the sunrise and sunset formula, using the equations from Wikipedia. The problem is that, in my usage thereof, it seems that it only works for sea level; if I start with coordinates that correspond to any point above sea level (or below, for that matter), it will still output the sunrise and sunset for my coordinates at sea level.

For example, today (21/01/2018), in Denver, CO (~1mi above sea level), sunrise was at 07:15:28. Yet my equations keep yielding 07:15:41, which is exactly on-target, were Denver at sea level. How do I account for the fact that higher altitudes experience sunrise earlier than lower altitudes of the same coordinates?

Higher elevations by themselves do not change the time of sunrise.

The horizon at sea level is 90 degrees from the zenith. The horizon at Denver is still an angle of 90 deg from the zenith if you assume that the ground is "flat". In other words, even though you are 1 mile high in Denver, you cannot see the horizon at 0 mile elevation.

If you are on a hill or mountain and looking down so that the distant horizon is more than 90 degrees from the zenith, then sunrise would occur early. If you are in a valley and the sun rises behind a hill or mountain, the sun rises later.

So you need to know the elevation of the observer and horizon (or obstacles). From that you can calculate the altitude of the sky in that direction. From that you can calculate the time of rising.

(edit Jan 24) For points that are close to each other, simple trigonometry can be used.
$$\tan(\theta)=y/x$$ where y is the difference in elevation between points 1 and 2, and x is the distance. Sunrise occurs when the altitude of the Sun is $\theta-sun's\;radius-refraction$.

For more distant points with latitude (lat) and longitude (long), solving for the sides and angles of a triangles gives an approximate solution. (I am sure there are more precise formulas that account for the Earth's shape.) From the figure below,
$$a=radius\;of\;Earth+Elevation1$$
$$b=radius\;of\;Earth+Elevation2$$
$$\cos(\gamma)=\sin(lat1)\sin(lat2)+\cos(lat1)\cos(lat2)\cos(long1-long2)$$
$$c=\sqrt{a^2+b^2-2ab\cos(\gamma)}$$
$$\cos(\alpha)=\frac{b^2+c^2-a^2}{2bc}$$
$$\theta=\alpha-90$$
and sunrise occurs when the altitude of the sun is $\theta-sun's\;radius-refraction$. Naturally, you need to check all locations along the line of sight between the observer at point 1 and the obstruction at point 2 to find the highest obstruction.


Creating a basic "fixed" solar system to host a 3D space travel simulation I'm trying to simulate a virtual/imaginary "solar system" in software--just a hobby project for now. Unfortunately this has made me realize exactly how much math I've forgotten since college.

Complete accuracy isn't important, and I'm not looking to do anything sophisticated... no need for multi-body calculations, just a star and a planet will do. I just want something that a lightweight simulation or a game might do.

I'll have a configuration file with some basic starting parameters for each planet. At the start of the simulation, time=0 seconds, all planets would be lined up in a row (planning on just placing them on the same axis at their perihelion distance from the star). I'd like the planets to move around their orbits at the correct speed given the parameters supplied. So if I decide to set time = 1e10, I need to be able to calculate the x and y positions of each planet in their orbit at that precise time.

I'm flexible on WHICH parameters need to be provided. So if this becomes easier using some other method of defining the orbit, I'm open to it. Right now I'm thinking perihelion, mass, and time elapsed.

(Since this is a 3D simulation, eventually I'd also like to work inclination in, but I'm trying to start off simple for now, so inclinations will all be zero.)

Here's some sample parameter data below in case it helps provide a better example.

Star mass = 2.00e30

Planet mass = 3.30e23

Planet perihelion = 5800000

(Ex. Where would this planet be, in X/Y coordinates, at time=x?)

If someone posts a formula I can just plug in I won't complain. But if I could just be pointed in the right direction for what to learn or refresh my memory on, I'd like to try and figure this out myself. So far I've been looking at Kepler's 3rd Law equations but I'm not sure if that's a good place to start. Feeling a little guilty that I seem to have forgotten virtually all math I learned in college just over a decade ago, and I kind of want to reverse some of that decay.  Hopefully this makes sense!

The easiest approach is to do a numerical integration of the equation of motion of your system. For that you need to remember that the gravitational force of $a$ on $b$ is $$\vec{F}_{a\rightarrow b} = -G\frac{M_a M_b}{r_{a\rightarrow b}^{3}} \vec{{r_{a\rightarrow b}}}$$
where $\vec{{r_{a\rightarrow b}}} = \vec{x_b} - \vec{x_a}$ and that
$$ M\vec{a} = \sum \vec{F}$$
where $M$ is the mass of your object, $\sum\vec{F}$ is the sum of the forces applied to the object and $\vec{a}$ is the acceleration.

To do the simulation, you then have to follow this numerical scheme

set some initial conditions to your problem: initial position $\vec{x_p}$ of the planet and $\vec{x_\star}$ of the star and initial velocities $\vec{v_p}$ and $\vec{v_\star}$.
compute the force $\vec{F}$ of the star on the planet and the accelerations of the planet and the star:
$$ -M_\star \vec{a_\star} = M_p \vec{a_p} = \vec{F_{\star\rightarrow p}}$$
update the speed of the planet and the star using the respective acceleration:
$$ \vec{v}(t+\Delta t) = \vec{v}(t) + \Delta t\times \vec{a} $$
update the positions of the planet and the star using their respective velocity:
$$ \vec{x}(t+\Delta t) = \vec{x}(t) + \Delta t\times \vec{v} $$
recompute the distance between the planet and the star $\vec{r_{{\star\rightarrow p}}} = \vec{x_p} - \vec{x_\star}$ and go back to 2.