Is there an objective difference between space expansion and reduction in speed of light I just read an article about the cosmological model of an inflationary universe. As you probably know it basically states that space itself is subject to an expansion over time.

Since there is no moving matter or energy involved, this might even happen at a "speed" faster than light. So far, so good and obscure. What strikes me is the principle that the speed of light as a fundamental constant can only be expressed as a function of space-time. Where do we know that the one is constant but the other can suddenly be variable?

Is there any reason why the point of view of an expanding space is preferred over, say, a reduction in the speed of light or an increase in the "speed" of time? Is there any objective difference or is it just the good old rubber metaphor being stretched (pun intended) too far?

In physics the "speed" of anything depends on the coordinate system you choose, since speed is measured as change in coordinate position in some interval of coordinate time. Even in the special theory of relativity, which doesn't take into account gravity and hence involves no spacetime curvature, the notion that the speed of light is always equal to the same constant (labeled $c$ in physics and astronomy) would only true in a special class of coordinate systems known as inertial frames, it is quite possible to define a "non-inertial" coordinate system in special relativity such as Rindler coordinates in which the speed of light does not have the same value $c$. In the general theory of relativity, which models gravity in terms of mass/energy curving spacetime, you can only have "local inertial frames" defined on very small patches of spacetime (specifically, the limit as the size approaches zero)--see this article on the "equivalence principle" for the conceptual details of how local inertial frames can be defined by observers in freefall measuring events in their immediate neighborhood (like an observer looking at events within an elevator in freefall). Such observers will always measure the local speed of light within a vacuum in their local region to be equal to $c$, regardless of the larger-scale properties of the spacetime they're embedded in, like the "expansion of space".

But if you try to define a global coordinate system on a large region of curved spacetime, this coordinate system is always a non-inertial one, so there is no guarantee that the coordinate speed of light in this coordinate system will be equal to $c$, and indeed the coordinate speed of light may vary from one region of spacetime to another depending on what coordinate system you choose (the equations of general relativity work in all smooth coordinate systems, as long as you define the metric correctly relative to your chosen coordinate system). In the basic model of curved spacetime in cosmology (the FLRW model), the simplifying assumption is made that matter is a sort of uniform fluid filling all of space, so that if you pick the right definition of simultaneity (multiple definitions are always possible in relativity due to the relativity of simultaneity), you will find that the density of this fluid is identical at every point in space at any given moment of cosmic time. This obviously isn't completely true to life, but it's expected that on large scales the density of matter is close to uniform at any given cosmic time, so it's seen as a reasonable approximation. The expansion of space basically means that the density of the fluid gets lower as time passes, and that if two objects are at rest relative to the local fluid in their immediate neighborhood, then the proper distance between them will grow with time (proper distance corresponds to what you would measure if you laid a bunch of short rulers end-to-end between the two objects at a particular moment in time, and then added up the distances).

As it happens, this cosmological model has a further nice feature (as discussed in the 'proper distance' link above which is based on this paper, see p. 99 of the 'Full Refereed Journal' link). The most "natural" coordinate system to use in this model is one in which the time coordinate corresponds to the proper time measured by a set of observers who have been at rest relative to the cosmic fluid since the big bang, and the spatial coordinate is such that the coordinate distance between any such observers at a given time corresponds to their proper distance at that time. If you use such a system, it works out that the overall coordinate velocity of any object can be broken down into a sum of two velocities:

The "recession velocity" at any given space, which is the velocity that an observer at rest relative to the cosmic fluid would be moving (the rate that their proper distance from the origin of the coordinate system is growing as a function of time, where we can assume the origin corresponds to our own location in space).
The "peculiar velocity" of any object which is not at rest relative to the cosmic fluid, which is just the same as the velocity of that object as measured in the local inertial frame of an observer at the same location who is at rest relative to the cosmic fluid. So, the peculiar velocity of a light ray must always be $c$.

So, if we know the recession velocity $v_{rec}$ at some distant location in space, then a light ray emitted directly towards us from that location will have an overall velocity $v_{rec} - c$ in this coordinate system, and a light ray emitted directly away from us will have an overall velocity $v_{rec} + c$. So from the perspective of this coordinate system, it makes sense to say as a shorthand that the light itself always travels at $c$, but space is also expanding away from us and this accounts for why the light is redshifted, and also why light originally emitted at distance $d$ won't necessarily take a time of $d/c$ to reach our own location. But this neat way of describing things is specific to both the cosmological model being assumed and the coordinate system used, things may not work out so neatly in other choices of spacetime or other coordinate systems. The only really general statement you can make about the speed of light is the one I mentioned earlier, that regardless of what global coordinate system you use and what the speed of a light ray works out to be in that system, it's always true that in a local inertial frame defined on a small patch of spacetime, light traveling through that patch always has a speed of $c$ as measured in that local frame.


What exactly are the "ν6 secular Sun-Jupiter-Saturn" and the "1:4 Sun-Jupiter" resonances? In the recent Acta Astronautica article The edge of space: Revisiting the Karman Line, Harvard-Smithsonian Center for Astrophysics astronomer, Space SE contributor and "inverse namesake" of asteroid (4589) McDowell Johnathan C. McDowell lists a number of different boundaries in the solar system in section 3.2. One of them is described as the following:

  The ν6 secular Sun-Jupiter-Saturn resonance which marks the conventional inner edge of the asteroid belt at 2.06 astronomical units (308 million km) from the Sun (26); it coincides with the 1:4 Sun-Jupiter resonance (27) and asteroid orbits near this resonance are unstable, soon perturbed to enter the inner solar system. Although there is no generally agreed definition, this location is a reasonable place to mark as the boundary between the inner and outer solar system.

What (the heck) are the ν6 secular Sun-Jupiter-Saturn and the 1:4 Sun-Jupiter resonances exactly?

The 1:4 (or 4:1) Jupiter resonance is a mean-motion resonance: an asteroid there takes 1/4 as long to orbit the Sun as Jupiter does.
Perturbations by Jupiter at recurring ecliptic longitudes alter the asteroid's orbital period, so asteroids do not remain long in this state.
The Kirkwood gaps in the asteroid belt include this and other mean-motion resonances with Jupiter.

The $\nu_6$ secular resonance (also called $g = g_6$) is more complex: in certain combinations of orbital period and inclination, an asteroid experiences perihelion precession at the same rate as Saturn (the 6th planet) does, resulting in a gradual increase of the asteroid's orbital eccentricity.


Examples of Equal Mass, Unequal Mass and Double Binary Star Systems Can anyone give me an example of a star (ideally those with a common name e.g. Rigel, Sirius, etc) of the following multiple star systems.

Equal Mass Star System, where two binary stars orbit round a central point of space.

Unequal Mass Star System, where a smaller star(s) orbits round a larger binary star.

Double Binary where two sets of binary stars orbit round a spot. (Regulus?)

Equal mass
Alpha Centauri, the nearest star system to us, is composed of a binary star and Proxima Centauri (making it a triple star system). Alpha Centauri A has a mass of 1.1 M$_{\odot}$, while Alpha Centauri B has a mass of 0.9 M$_{\odot}$.
If you count brown dwarfs, Luhman 16 are a good target, with each being about 0.04 M$_{\odot}$.
Unequal mass
Eta Carinae contains a Luminous Blue Variable (LBV) with a mass of 120-160 M$_{\odot}$ and a smaller (but still very massive) star with a mass of 30-60 M$_{\odot}$.
Sirius is okay, though with two stars of much smaller masses, 2 M$_{\odot}$ and 1 M$_{\odot}$.
Double binary
Mizar, in reality part of a six-star system (Mizar/Alcor), has four stars separated into two pairs. This  ensemble then interacts with the binary star Alcor.

You can find plenty of stars here.


Black holes at the center of galaxies At the center of massive galaxies there is a huge and massive black hole that radiates gamma rays. So at the center of the Milky Way there is also a huge and massive black hole and it also radiates gamma rays. Do these gamma rays reach our solar system? What is the age of black hole in the center of our galaxy? Does it still radiates gamma rays?

Black hole in the centre of our galaxy is not active (which means it is not feeding much matter) as compared to black holes in the centre of other galaxies. So it does not emit that much gamma rays as other supermassive black holes does but still emits some amount of gamma rays. Along with gamma rays cosmic rays are also emitted which are ward off by radiation pressure of our sun's heliosphere.


What did I photograph around 6 AM on April 27th, 2021? Supermoon plus comet? Facing SW just before sunrise on April 27, 2021 in Central Florida:
The timestamps on the pics start at 6:23 end at 6:27.  It was still visible and breaking up like a post launch trail that I would see in the east.  By the size, from beginning to end, looks like the direction was from/could be described as, from New Orleans toward Ft. Myers crossing between the moon and Gulf of Mexico.  Because its small early Northwest of moon and larger late south of moon.

Having seen the extra photos, this is certainly a cloud.
It is a high altitude cloud: a fragment of cirrus composed of ice crystals.  These are aligned roughly horizontally, and they are catching the light of the moon. This is why the region of the cloud is more highly lit than the parts to the side: a sun-pillar (or moon-pillar?) effect.
Its movement over the course of a few minutes also indicates a nearby object. It is not a body in space, as astronomical bodies are so distant that they appear to move only very slowly.  One very simple rule is "if it moves or changes, then it isn't astronomy, it's meteorology or technology" This isn't quite true, but very few astronomical bodies can be seen to change position over the course of a few minutes.


Is there a timekeeping word for the orbit of a moon? For a planet, we can colloquially refer to its period of rotation as a "day" and its period of revolution around its parent star as a "year." Some worlds have unique terms, such as Martian days being referred to as "sols", but the principle is the same: one word for rotation around its axis, another for revolutions around the star.

Is there an equivalent word for the period of a moon's revolution around a planet? On Earth it roughly lines up with a "month," but multi-moon planets would have different periods for each moon.

If I was in charge of timekeeping for a mission to Europa or Ganymede, what term would I use to refer to the period of time it takes for the moon to complete one orbit of Jupiter?

You should use one of the precisely defined terms of specific measures of orbital period, which are described on this Wikipedia page.

You might get away with just "orbital period" is you're not using it in a context that requires precision.

For length of day there's synodic day as well as sidereal period.  There's a discussion on the difference between these here.


What does the Sun look like from the heliopause? I calculated the angular diameter of the Sun at the heliopause to be 0.004°.  If that is correct, would the Sun appear as a disk, or like the pinpoint of light from a star?

How bright would the light from the Sun appear?  I understand at Pluto the Sun appears as bright as the full moon at night on Earth.  Assuming the heliopause is 3x distant from the Sun as Pluto, does that mean there would be 1/9th as bright at the heliopause.  What would that brightness translate to an everyday measure?

0.004° converts to about 14.4 arc-seconds.
That's within the range typical for Mars; 5-25 arc-seconds.
You'll see the sun as a point with the naked eye.
Heliopause is about 14 billion miles out. Absolute magnitude of sun is 4.83.
14 billion miles = 0.0007297pc.
Apparent magnitude will be -15.8543. Magnitude of the full moon, from earth, is about −12.74
So you'll see a point source of light, brighter than the full moon.


Brown Dwarf Cloud Models There has been research that suggests that the changing brightness of Brown Dwarves are due to bands of Iron/Silicate clouds. Several models have been made that supports that hypothesis. My question is: Which part of the model are the actual clouds? Are the clouds the bright orange bits? Or are those bright areas “breaks” in the clouds and a view of the Brown Dwarf below.

5 MB animated GIF: Brown Dwarf Cloud Model

Initial research proved to be wrong as intuition does not match up with expectations.

Anyway, the paper referenced by the OP seems to sum up the answer to the question pretty well:

  In  this  explanation,  the  dark  areas  of  our  map represent  thicker  clouds  that  obscure  deeper,  hotter  parts  of  the  atmosphere  and present a  higher-altitude (and  thus  colder) emissive  surface, whereas bright  regions correspond to holes in the upper cloud layers that provide a view of the hotter, deeper interior.

Extract from A global cloud map of the nearest known brown dwarf

Initially I assumed that the metallic clouds would reflect radiation from the nearby star, however as it turns out the hot atmosphere is emitting infrared radiation which affects the brightness much more significantly.

More on Brown Dwarfs at here.


whose gravitational pull is more stronger , Neutron Star or Black hole? We know neutron star is very densest in the universe and it weighs the thousand times of our Sun. And Black hole having strong gravitational pull not even light can escape From it . So , whose gravitational pull is greater ?

The black hole has greater pull.

If you stand beside a massive object (planet, star, black hole or anything else with mass) you will feel a gravitational pull, and unless you do something about it you will start to accelerate towards the object.

So the gravitational pull (or more precisely the gravitational acceleration) depends on two things: How massive the object is, and how far you are from its centre.

Neutron stars are all between about 1.2 and 3 times the mass of the sun, any larger and they become black holes. Objects that are believed to be black holes are know with masses of about 4 times the mass of the sun. A black hole can be of any mass, there is no known way for smaller black holes to form.

Now if you fix the distance (say 150 million km), then the gravitational pull depends only on the mass of the object. A neutron star can be at most about three times the mass of the sun, black holes are nearly all larger than that, so the gravitational pull of the black-hole is greater. But if you find a way to make a small black hole, then it would have lower mass and so less gravitational pull, at the same distance.

You can also look at surface gravity. Black holes don't have a proper surface, but you could use the event horizon instead. Here the answer is easy: The gravitational pull becomes infinite at the event horizon (from a certain point of view), Neutron stars have very strong, but not an infinitely strong pull so the gravitational pull of a black hole is greater than a neutron star.


What's the origin and culture of funny astronomical terminology? I'm not in the industry myself, but as an interested member of the public the terminology of astronomy seems a bit funny. Astronomers who today talk publicly about the interstellar medium say that the solar system is in the "local fluff" inside the "local bubble" near a "local chimney". Cosmology uses a term like "the dark ages" to describe an era in the history of the phenomena they study. But geologists instead prefer to use terms like "cenozoic" to name a period. Early on, astronomers also used pompous words and names from antique languages.

Was "Big Bang" (coined in 1949, I think) the term which started off this fluffy bubble blob bang type of terminology? Or were there precursors?
Is there a connection to sci-fi literature?
Are the "blobby bang fluff" words just a mass media ("reach out") construction which professional astronomers don't use at all even as quick references in conversations with colleagues?

Any causal claims or associations relating to language use are never going to be able to be proven, so the best answer to your overall question is: "nobody knows".

You can use Google's ngram viewer to visualise the appearance of a specific term or terms over time.

Perhaps astronomers generally have been better as a field in making their research more approachable by not pitching their discoveries in superior-sounding high-falutin sesquipedal pomposity. But it isn't universal; for every approachable 'Hanny's Voorwerp' there is an ancient-language-originating label like 'heliosphere'.

Geologists also use non-latin terms like "slip" and "fault" and "hanging wall". Is that to make geology more approachable for mass-media public reach-out PR?


Which kernel does HORIZONS get the Geometric Albedo from? I am new to using SPICE and the SPICE Toolkit.  My goal is to extract body radius, albedo, and distance from the sun of a number of solar system objects.  The tools provided by NAIF are great and I have been able to get the information items I need from HORIZONS.  I would like to use mice and the right selection of kernels to construct a table of body information.  I am unable to locate the geometric albedo.  It must be in there somewhere since HORIZONS reports it in its body summary (example below).  Please point me to which kernel(s) the information in the example is extracted from.

Horizons> 500

JPL/HORIZONS                500 Selinur (1903 LA)          2018-Apr-26 15:32:09
Rec #:     500 (+COV) Soln.date: 2018-Apr-05_11:16:59   # obs: 1803 (1903-2018)

IAU76/J2000 helio. ecliptic osc. elements (au, days, deg., period=Julian yrs):

EPOCH=  2455485.5 ! 2010-Oct-16.00 (TDB)         Residual RMS= .28963
   EC= .1453222754778464   QR= 2.232822954662918   TP= 2455876.0720004463
   OM= 290.0072894804399   W=  75.13840817439457   IN= 9.764544808931515
   A= 2.612473556522465    MA= 268.8351702646443   ADIST= 2.992124158382012
   PER= 4.22266            N= .233413627           ANGMOM= .02750885
   DAN= 2.46541            DDN= 2.65631            L= 4.9371039
   B= 9.434865500000001    MOID= 1.25316           TP= 2011-Nov-10.5720004463

Asteroid physical parameters (km, seconds, rotational period in hours):
   GM= n.a.                RAD= 20.414             ROTPER= 8.0111
   H= 9.3                  G= .150                 B-V= n.a.
                           ALBEDO= .202            STYP= n.a.

ASTEROID comments:
1: soln ref.= JPL#41, OCC=0
2: source=ORB

Select ... [A]pproaches, [E]phemeris, [F]tp,[M]ail,[R]edisplay, [S]PK,?,:

As barrycarter commented, codes_300ast_20100725
has ephemerides for 300 large asteroids, generated in 2010 and supposedly valid from 1800 to 2199.
For other asteroids, a Celestia developer forum shows how to how to request an SPK from HORIZONS's telnet interface.
However, they appear to use the SPK for ephemeris only, adding albedo in Celestia's own SSC file format.
In SPICE documentation, neither the tutorials nor the "required reading" seems to mention albedo.

A more direct approach to your goal might be the JPL Small Body Database Search Engine.
For example, if you query for numbered NEO asteroids with absolute magnitude H < 16, you get a table of 169 rows.
Diameter and albedo are known for some asteroids but not others, even though their orbits and rotational periods may be well established.


Where can I look up assigned origins of gamma ray bursts events? (GRBs) Gamma ray burst GRB170817A is believed to have originated from the merger of Neutron Star(NS)-Neutron Star(NS). From this article on the associated gravitational wave detection GW170817:

The gravitational wave signal, designated GW 170817, had a duration of approximately 100 seconds, and shows the characteristics in intensity and frequency expected of the inspiral of two neutron stars. Analysis of the slight variation in arrival time of the GW at the three detector locations (two LIGO and one Virgo) yielded an approximate angular direction to the source. Independently, a short (~2 seconds' duration) gamma-ray burst, designated GRB 170817A, was detected by the Fermi and INTEGRAL spacecraft beginning 1.7 seconds after the GW merger signal. These detectors have very limited directional sensitivity, but indicated a large area of the sky which overlapped the gravitational wave position. It has been a long-standing hypothesis that short gamma-ray bursts are caused by neutron star mergers.

Are there lists or compilations or reviews of all reported gamma-ray bursts (GRBs) together with an assigned or hypothesized event or source that produced it?

I found a database that contains all GRBs from April 1991-August 1991. However, it has 1637 registered GRBs so it would probably be enough for your purposes. Here is the link-Goddard Space Center GRB Archive. Also, just a fair warning, that website is incredibly old and very difficult to use, it took me nearly fifteen minutes to navigate to the actual database. Hope this helps!


How can black holes be sometimes so gaseous? One of the most massive Supermassive Black Hole observed is the one at the centre of the galaxy NGC1600 with a mass of 17 billion suns. It would have a density of ∼0.01kg/m^3, or one part in 100,000 times the density of water, or 1% the density of earth atmosphere at sea level.

But if there is so much gas why isn't the bh shrinking and becoming more dense as gravity would pull everything together?

A black hole is not a solid or gaseous body. It is a shape that space-time can take. Now the event horizon is sometimes treated as being the "edge" of the black hole, but there is nothing at this edge. The black hole can't shrink its event horizon, since the event horizon is not a physical object

There is no matter inside the black hole (for this reason the shape of space-time around a black hole is sometimes called a "vacuum solution" of Einstein's equations). The mass of the black hole is concentrated at the singularity (which, due to the curving of space time you can imagine as more like a point in time that you can't avoid if you cross the event horizon.

Since the radius of the black hole's event horizon is proportional to the mass, the volume of the black hole is proportional to the cube of the mass. If you double the mass, you increase the volume by a factor of 8.

Density is mass/volume so if you double the mass you change the density by 2/8=0.25, so the more massive a black hole is, the lower is the average density of the region inside the event horizon.

This does mean that if you did manage to find a gas cloud with 17 billion times the mass of the sun (much bigger than any known nebula - Orion Nebula M43 is only a few thousand solar masses) and you were to somehow compress it to a sphere with 0.1 atm pressure, it would undergo gravitational collapse and form a black hole.


Period coincidence of Planet Nine with furthest Kuiper belt objects A lead scientist of the Planet Nine theory said that "The period with which it goes around the sun is a rational multiple of the periods of all the furthest Kuiper Belt objects"

To what degree of precision can Konstantin Batygin make that kind of measurement? Has it been backed up by other physicists with a degree of certainty?

He also says that they think it's near it's aphelion "because it hasn't been seen, so it's far away", not because of the above period coincidence, why can't they say where it is more convincingly?

Does that mean that they have very strong mathematical evidence for Planet Nine?

I did a little digging and I found this article.

It's worth noting that the initial publication wasn't about resonances but the fact that all 6 objects came from the same side of the sky, from somewhat similar directions, which, given that they were the 6 most distant at the time, is statistically unusual and it was both similar direction and similar inclination to the plane, so the statistical improbability was high enough to investigate.  Orbital resonance wasn't the primary argument, it was tacked on later as a secondary.

From the article:

  Their analysis also offered suggestions as to what kind of resonance
  the planet has with the KBOs in question. Whereas Sedna’s orbital
  period would have a 3:2 resonance with the planet, 2010 GB174
  would be in a 5:2 resonance, 2994 VN112 in a 3:1, 2004 VP113
  in 4:1, and 2013 GP136 in 9:1. These sort of resonances are
  simply not likely without the presence of a larger planet.

  “For a resonance to be dynamically meaningful in the outer Solar
  System, you need one of the objects to have enough mass to have a
  reasonably strong gravitational effect on the other,” said the
  research team. “The extreme Kuiper belt objects aren’t really massive
  enough to be in resonances with each other, but the fact that their
  orbital periods fall along simple ratios might mean that they each are
  in resonance with a massive, unseen object.”

  But what is perhaps most exciting is that their findings could help to
  narrow the range of Planet 9’s possible location. Since each orbital
  resonance provides a geometric relationship between the bodies
  involved, the resonant configurations of these KBOs can help point
  astronomers to the right spot in our Solar System to find it.

now, I want to point out that it's them, not me who did the typos on 2 of the objects, I'm pretty sure (based on Wikipedia and the chart below) that it should be 2004 VN112 (not 2994 VN112) that has the 3:1 and 2012 VP113 (not 2004 VP113) that has the 4:1.

This chart below also has a typo, it should be 2010 GB174 not 2012 GB174.  I guess nobody proofreads letter/number assignments.   Correct names will help if anyone wants to run the numbers see how see how close the resonances come to each other.

As to the degree of statistical certainty?   If you're a good enough statistician, you can run the numbers.   3/2, it seems to me, would be the most common resonance based on Jupiter's abundant Hildas and the Plutinos and only 1 of the 5 potentially resonant objects has a 3/2, but 3/2 is also very far from the sun and hardest to detect, so that's not a deal breaker.

I'd guess that the degree of certainty from these orbital resonances is OK but not great - but that's entirely a guess.   I also don't know enough about astronomy to know how close numbers have to be to provide some evidence of orbital resonance.  Jupiter's resonant moons, for example, if you check their orbits, are off by 1 or 2 percentage points from each other.


Does the gravity oppose to the Universe expansion? I'm trying to understanding the very basic of the current cosmology.
I searched in the old questions but I found nothing that answers specifically to my questions.

As far as I understand, we have to think the "expansion" more like a "streching" of the space that makes room for "new" space. This is confirmed because the Universe is homogeneous and non-local.

First question: does the gravity contrast this expansion? In other words: does the space expand more in the areas with less gravity (i.e. among galaxies) and less in the nearby of stars and planets?

If we imagine to place a ruler between two stars, say at 10 ly, it would "extend" over time - even if we will still read 10 ly, right? But if now we place another ruler that was kept in our (huge) pocket, it will measure a longer distance. Is it correct?

That because the first ruler has expanded together the space between the stars, while the second one endured the force of gravity and was kept rolled up so it wasn't able to expand.

Expansion depends on the amount of mass

Yes, space expands more in regions with less matter — in fact this has been proposed as an alternative explanation to dark energy as the cause of the observed accelerated expansion: If by chance we happen to be located near the center of a "cosmic void", then nearby space expands faster than distant space, and since nearby space corresponds to recent times, this would be interpreted as recent accelerated expansion. This idea was popular some ten years ago (e.g. Conley et al. 2007; Wiltshire 2008), but isn't to my knowledge so hot anymore.

Similarly, in regions with more matter, space expands less. In fact, on scales as small as galaxies, and even galaxy groups, space doesn't expand at all. Gravity prevents galaxies from expanding, and keeps galaxies near each other from receding. On even smaller scales, electromagnetic (and nuclear) forces keeps objects like stars, planets, and bicycles from expanding.

Expansion is physical

For this reason, space between two stars doesn't expand. So, to address your question, let's consider instead two galaxies, separated by millions of light-years. However, fundamentally we cannot place a ruler between them and watch it expand. Electromagnetic forces would try to hold it together, fail, and the ruler would break.

Why is this? Imagine a rigid ruler, one megaparsec long (i.e. 3.26 million light-years). Since the Hubble constant is $H_0 \simeq 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, the region at the other end of the ruler recedes at $70\,\mathrm{km}\,\mathrm{s}^{-1}$. That is, space "drags" the two ends of the ruler away from each other at $70\,\mathrm{km}\,\mathrm{s}^{-1}$. If the ruler doesn't break, a local observer will see the end of the ruler whizzing by at high speeds.

Perhaps you might be able to construct a rod able to withstand this stress. If I were a solid state physicist I might be able to calculate the maximum possible length of a rod, given an optimal material, but I'm not. But the point is that, no matter what material you come up with it is not even practically impossible to make it arbitrarily long, but "theoretically" impossible. There is a fundamental limit to the length of a physical ruler, namely the length corresponding to the distance at which two regions of space recedes at the speed of light, i.e. $d = c/H_0 = 4400\,\mathrm{Mpc}$, or $14.4\,\mathrm{Glyr}$. If the ruler were longer than this and didn't break, then a local observer would see the end move by faster than the speed of light, which is not possible.

Luckily, there are other ways to measure distances, e.g. by "standard candles". And yes, if you measure the physical distance between the two galaxies at two different times, it will have increased in the meantime, physically.

(You could also imagine "freezing" space by magic, laying out 1-meter rods. In that case, second time you do it you will need more rods.)


Can an Earth-like planet survive if our Sun went Supernova? First, I would like to point out, yes, our sun does not have enough mass to be a candidate for Supernova, this is a scenario were our sun is though, and after the various life cycle stages of the sun and whether or not life managed to survive or not on the planet, The Sun goes of as a Supernova, it is 1 AU from an Earth-like planet, or equivalent to 1 AU from a massive star, would the planet still be standing or would it be obliterated from the face of the Galaxy?

Here are some of the links that I have been looking through but could not find an answer for:

https://www.physicsforums.com/threads/effective-destructive-range-of-supernovae.312925/
https://earthsky.org/astronomy-essentials/supernove-distance
https://www.popularmechanics.com/space/deep-space/a26483/supernovas-deadly-twice-as-far-away/

One way of estimating this is to look at how much energy could be received by the planet. At distance $d$ it takes up $r^2/4 d^2$ of the sky as seen by the supernova. So for a $10^{44}$ J blast that is about  $4.5444\times 10^{34}$ J.

The gravitational binding energy of Earth is about $2\times 10^{32}$ J. So we have about 227 times as much energy as is needed to separate all pieces of the planet to infinity. It is also a few thousand times the energy needed to heat up an Earth-mass of iron from 0K to vaporisation. So, yes, it looks like it could well obliterate the planet.

Whether it actually does so is a complex question of how protected the evaporating planet is by its plasma sheath during the explosion. Given that actual terrestrial exoplanets in close orbits at 2000K may lose an Earth mass per gigayear it looks plausible that at least slower scenarios can vaporise planets.


How do I understand a brown dwarf with a M-type spectrum? There are a large fraction of M dwarfs which are claimed to be brown dwarfs.

Why do we still use M-type and not create a new stellar type like L, T and Y? The main signature of M is TiO absorption bands. They appear in M-type brown dwarfs?

Is a spectrum enough to judge whether a celestial object is a brown dwarf or not? Or do we need to know its mass?

The earliest type is M3.5. It means M-type objects older than M3.5 are necessarily brown dwarfs? Or is there a safe value, say, M9, is necessarily a brown dwarf? A brown dwarf is necessarily colder than a red dwarf?

There should be more brown dwarfs or red dwarfs?

The reason we do not need to create a new stellar type is the spectra of brown dwarfs are more like red dwarf's spectra?

The spectral type of an object is almost entirely determined by the temperature of its photosphere. ie Saying something is type M3.5 is just a measure of its surface temperature. An M3.5 brown dwarf is at a very similar temperature to an M3.5 star.

Brown dwarfs begin their lives as hot balls of gas and gradually cool with time. They start off as M-type objects and then cool to become L-type and then eventually, T-type objects. Stars on the other hand, start their lives by cooling, but stabilise their temperatures at a mass-dependent value that corresponds to an M-type classification or even as cool as type L2 for the very lowest mass stars.

Given that stars and brown dwarfs appear to have been born throughout Galactic history, then a range of masses are possible for any spectral type. In order to use the surface temperature (or spectral type) as an indicator of whether something is a brown dwarf, then you also need to know its age. In general that is not known, unless the object can be demonstrated to be part of a cluster of stars (of known age).

In order to be a brown dwarf the mass should be below about $0.075 M_{\odot}$, but unless the object is in a binary system, this cannot be measured. So what people do is trust the models(!) and compare the luminosity (or estimated temperature) with model evolutionary tracks at a given age. This gives an age- and model-dependent mass that can be used to claim that an object is a brown dwarf.

The fact that brown dwarfs can have an M-type spectral class when they are young, but then move through L- to T-dwarfs as they get older is not a problem, it is a confirmation that the evolutionary models are (roughly) correct!

A plot might help (from the work of Burrows et al. (1997)). This shows the evolution of temperature with time. Each track represents a brown dwarf of a given mass (the tracks from top to bottom) are masses of 0.2 - 0.08$M_{\odot}$, representing stars (in blue). Then come masses of 0.07 to 0.015$M_{\odot}$ representing brown dwarfs (in green, that will never ignite hydrogen in their cores to any extent). Then below this are the "planetary mass objects" from 14 Jupiter masses to 0.3 Jupiter masses at the very bottom (in blue - the definition here is that these objects don't even ignite deuterium in their cores).

The horizontal lines mark the boundaries of the spectral type classifications (M at the top, L in the middle, T at the bottom).
Notice how the stars level-off in temperature (when they begin H-burning) but can have M-type or even L-type classifications, but the brown dwarfs cool throughout their lives. Even objects as low as $0.01M_{\odot}$ begin their lives as "M-type" objects.

To answer some of your more specific queries:
The warmest spectral type below which you can almost guarantee that an object is a brown dwarf is $\sim$ L2. Any object warmer than this could be a star.

A brown dwarf is always cooler than a low-mass star of the same age.

Current observational estimates are that there is about 1 brown dwarf for every 4 low-mass stars - the weighted mean ratio for stars between 0.08 and 1 solar mass to brown dwarfs from observations of several clusters is $4.3 \pm 1.9$, according to Andersen et al. (2008); the commonly adopted/assumed "Chabrier Initial Mass Function" has 4.9 stars for every brown dwarf.

The spectrum of a brown dwarf and an M-dwarf star at the same temperature are very difficult to distinguish. The M-dwarf is only slightly bigger and this might lead to small gravity-dependent differences. Lithium is not burned in brown dwarfs less than $0.06 M_{\odot}$, whereas it is burned completely in low-mass stars on a mass-dependent timescale. Thus an older (than say 1 billion years) M-dwarf star would certainly have burned its lithum, whereas a younger M-type brown dwarf of lower mass would still have lithium.


Reflector vs refractor in terms of magnification, which one wins? I'm thinking about getting a telescope and i'm debating between these 2 telescopes: meade and orion. Theyre both the same price so this question is not about the money, but about getting the most out of your money. I like the Meade 102 mm (or 4 inches) because it is a refractor, which means it is more lightweight and compact. The orion starblast 4.5 on the other hand is a reflector and therefore more clumsy.

My question is, is a reflector necessarily better than a refractor in terms of magnification and sharpness of image? The orion does have a slightly bigger aperture (half inch bigger) but I don't think that makes a huge difference. The question really is, a reflector vs a refractor, which one is more desirable in terms of magnification, and how much of a difference does it make?

Just judging from the reviews it would seem like the reflector (orion) wins the fight. For example, Saturn's rings according to some reviewers are "clearly visible" in the reflector, but not so with the refractor which most reviewers agree, Saturn will be very tiny and the rings barely visible.

So will I settle for magnification if I get the refractor instead of the reflector, or are they basically the same and in that case I would choose the refractor (meade) over the reflector since it is more compact, and is also easier for beginners like me?

Magnification is (mostly) done by the eyepiece. You can change the magnification by using a different eyepiece. Whether it is a reflector or refractor doesn't make much difference.

There is a limit to the useful magnification that can be done. At a certain point, you are just magnifying the blur. Larger apertures can resolve better, and so can allow greater magnification. Using more magnification also makes the view dimmer. A bigger aperture collects more light, so also allows for more magnification.

The choice between reflector or refractor is not about magnification. Lenses work by refraction, and this can create coloured fringes (which can be compensated for, but this increases the cost). Thicker lenses also absorb light, but so do mirrors to some degree.  Generally, bigger is better, but other factors (stability of mount, ease of use, ease of transport, computer control etc.) may be significant. "Magnification" is often not a good measure of quality.


How are the compositional components of exoplanet atmospheres differentiated? How are exoplanetary atmosphere compositional spectra distinguished from those of the parent star(s), from the composition of the planetary surface or any other factor?  Is it actually possible to determine the atmospheric composition precisely using this method?

Are there any specific examples of such an analysis having been performed on an exoplanet?

In theory one should just be able to determine the difference between the spectra during the star's eclipse of the exoplanet (starlight alone) and the spectra of the star and exoplanet together, but in reality equipment is not precise enough for this. To remedy this problem, the analysis is integrated over many eclipses. Some other calibrations are made as well which you can read more about in the paper A Spitzer Spectrum of the Exoplanet HD 189733b. This is called occultation spectroscopy.
There's another method used as well called transmission spectroscopy, which detects the change in light as the exoplanet passes in front of its star. You can read more about that in this paper: Infrared transmission spectroscopy of the exoplanets HD 209458b and XO-1b using the wide field Camera-3 on the Hubble Space Telescope.


How many stars are in the constellation Canis Minor? I've read Wikipedia and found a list of stars in Canis Minor, and counted the number of stars in the list. I came up with 56 stars, a surprising number. I did some more research and found that Canis Minor was just the two binary stars orbiting together: Procyon and Gomeisa. So I came to the conclusion that the answer could be found here somewhere. Please help and tell me how many stars the constellation Canis Minor is made of. Thank you.

You possibly are confused by these two entries in Wikipedia (click on the quotations to go to the original distinct entries:

Canis Minor contains only two stars brighter than the fourth magnitude, Procyon (Alpha Canis Minoris), with a magnitude of 0.34, and Gomeisa (Beta Canis Minoris), with a magnitude of 2.9.

and

This is the list of [56] notable stars in the constellation Canis Minor, sorted by decreasing brightness.

The first statement is about stars brighter than the 4th magnitude, of which there are two, and not about all the others. Those two stars are NOT a binary. Procyon is 11 lightyears away and Gomeisa is 170 lightyears away. Procyon itself is a binary, but for the purpose of general observation is considered to be one star.

The second statement is about "notable stars", which is rather arbitrary. The list is 56 stars long, but in fact is one were to map every star visible to telescopes, there would be thousands.

Another quibble: What perchance is the exact area in the sky that is assigned to Canis Minor? This is not precisely defined. See below.

~~~~~~~~~~~~~~~~~~~~~~ and ~~~~~~~~~~~~~~~~~~~~~~

Note the completely different boundaries.


Can I see comet ISON from Saudi Arabia? I live in western Saudi Arabia, I am trying to see the comet ISON, reading here and there tells that I can see it by naked eye somewhere next to the constellation leo, I used an app to tells me where constellation leo is but I could not see the comet.

Is it visible from Saudi Arabia? if so, in which direction (west, east, etc.) and what time is the best to see it?

I'm not far from you (Israel) and I am also waiting to see comet ISON.

Comets are very unpredictable, and comet ISON has not yet reached the critical part of its journey which will determine visibility for us. Comet visibility is usually due to reflection of the Sun's rays on the coma and tails of the comet, as the nucleus is to small to see directly. So ISON's visibility (and all comets for that matter) depend on how much gas, water, and other materials that it outgasses. This in turn depends on how much volatiles it has, how much the Sun melts them, how much the slingshot around the sun and tidal forces fracture the nucleus, and other factors. So the actual visibility that will occur is very difficult to determine in advance. That said, ISON has all the markings of this being its first trip around the Sun. Thus, we assume that it has many volatiles to release and we assume that ISON will be exceptionally bright.

In any case, ISON will be most visible when it is closest to Earth, near the end of December. It should be most visible in the early morning sky, I'm guessing at about 4:00 AM or so. This is because ISON will be close to the Sun (thus visible close to the time that the Sun becomes visible). Like all celestial bodies, it will rise in the East. Each day the comet will move farther from the Sun and thus will be visible earlier and earlier in the night, until it disappears from view probably sometime in mid January.

You are welcome to visit Beersheba and have a peak through my binoculars!


Why can't we observe the Oort cloud with a telescope? The Oort cloud is a hypothetical structure based on our observation of long-period comets. There are currently proposals to design probes to confirm the existence of the Oort cloud.

Now, sending a probe would have other benefits, but why can't we observe the Oort cloud with a telescope?

The angular resolution of the telescope really has no direct bearing on our ability to detect Oort cloud objects beyond how that angular resolution affects the depth to which one can detect the light from faint objects. Any telescope can detect stars, even though their actual discs are way beyond the angular resolution of the telescope.

The detection of Oort cloud objects is simply a question of detecting the (unresolved) reflected light in exactly the same way that one detects a faint (unresolved) star. Confirmation of the Oort cloud nature of the object would then come by observing at intervals over a year or so and obtaining a very large ($>2$ arcseconds) parallax.

The question amounts to how deep do you need to go? We can do this in two ways (i) a back of the envelope calculation assuming the object reflects light from the Sun with some albedo. (ii) Scale the brightness of comets when they are distant from the Sun.

(i) The luminosity of the Sun is $L=3.83\times10^{26}\ W$. Let the distance to the Oort cloud be $D$ and the radius of the (assumed spherical) Oort object be $R$.
The light from the Sun incident on the object is $\pi R^2 L/4\pi D^2$.
If we now assume that a fraction $f$ of this is reflected uniformly into a $2\pi$ solid angle. This latter point is an approximation, the light will not be reflected isotropically, but it will represent some average over any viewing angle.

To a good approximation, as $D \gg 1$ au, we can assume that the distance from the Oort object to the Earth is also $D$. Hence the flux of light received at the Earth is
$$F_{E} = f \frac{\pi R^2 L}{4\pi D^2}\frac{1}{2\pi D^2} = f \frac{R^2 L}{8\pi D^4}$$

Putting some numbers in, let $R=10$ km and let $D= 10,000$ au. Cometary material has a very low albedo, but let's be generous and assume $f=0.1$.
$$ F_E = 3\times10^{-29}\left(\frac{f}{0.1}\right) \left(\frac{R}{10\ km}\right)^2 \left(\frac{D}{10^4 au}\right)^{-4}\ Wm^{-2}$$

To convert this to a magnitude, assume the reflected light has the same spectrum as sunlight. The Sun has an apparent visual magnitude of -26.74, corresponding to a flux at the Earth of $1.4\times10^{3}\ Wm^{-2}$. Converting the flux ratio to a magnitude difference, we find that the apparent magnitude of our fiducial Oort object is 52.4.

(ii) Halley's comet is similar (10 km radius, low albedo) to the fiducial Oort object considered above. Halley's comet was observed by the VLT in 2003 with a magnitude of 28.2 and at a distance of 28 au from the Sun. We can now just scale this magnitude, but it scales as distance to the power of four, because the light must be received and then we see it reflected.
Thus at 10,000 au, Halley would have a magnitude of $28.2 - 2.5 \log (28/10^{4})= 53.7$, in reasonable agreement with my other estimate. (Incidentally my crude formula in (i) above suggests a $f=0.1$, $R=10\ km$ comet at 28 au would have a magnitude of 26.9. Given that Halley probably has a smaller $f$ this is excellent consistency.)

The observation of Halley by the VLT represents the pinnacle of what is possible with today's telescopes. Even the Hubble deep ultra deep field only reached visual magnitudes of about 29. Thus a big Oort cloud object remains more than 20 magnitudes below this detection threshold!

The most feasible way of detecting Oort objects is when they occult background stars. The possibilities for this are discussed by Ofek & Naker 2010 in the context of the photometric precision provided by Kepler. The rate of occultations (which are of course single events and unrepeatable) was calculated to be between zero and 100 in the whole Kepler mission, dependent on the size and distance distribution of the Oort objects. As far as I am aware, nothing has come of this (yet).


Is Jupiter still an anomaly? I remember a few years ago hearing that Jupiter was an anomaly in the landscape of exoplanets. Back then, most Jupiter-mass planets discovered were Hot Jupiters, orbiting very close to their host star.

In the last few years since I heard this, many new exoplanets have been discovered. Do we know now if there are many more Hot Jupiters and our own Jupiter is a rather rare occurrence, or if this was an observational bias and there are many more "Cold Jupiters"?

A recent study indicates that Cold Jupiters similar to Saturn and Jupiter greatly outnumber Hot Jupiters. The authors studied 18 years worth of data to find long-period exoplanets, that is planets far from their host star.

Cold Jupiters, being farther from their host star, have longer periods than Hot Jupiters. Therefore, they need to be observed over a longer time frame to see multiple transits. ETs would need to observe the Sun for 12 years before seeing two transits of Jupiter, and that still wouldn't be enough to confirm its presence.

Planets that are further out also induce smaller variations in the velocity of their host star, so the spectrum of the star is shifted by a smaller amount. Therefore higher resolution spectrographs are needed to detect Cold Jupiters using the radial velocity method.

Finding more Warm and Hot Jupiters at the beginning of the era of exoplanetary discovery was an observational bias due to the limits of the instrumentation available at the time, and the amount of time needed to find long-period exoplanets.


'Little h' usage in cosmological simulations I am running a cosmological simulation and am having some trouble putting things into code units. The physical distance units in my simulation are in terms of $\text{Mpc/h}$, where $h$ is the dimensionless Hubble parameter. This makes enough sense because, as noted elsewhere, simulations are often scale-free so it makes sense to factor out the $h$ dependence and make it explicit. This unit convention is causing me some confusion however. In one calculation I have to do during the simulation, I essentially (ignoring context which I can provide later) have to multiply the speed of light $c$ by an inverse distance $1/x_0$ which is given in units of $\text{Mpc/h}$.

In order to properly have the units cancel, I first put $c$ in units of $\text{Mpc/s}$ to get
$$
9.716 \times10^{-15} \text{Mpc/s}
$$
However, should I know factor out the $h$ dependence? These seems strange to me because in my mind, the value of the speed of light should not depend on the underlying cosmology I have simulated. On the other hand, I feel that I should not cancel units of $\text{Mpc}$ with units of $\text{Mpc/h}$. To make things concrete, let's assume I have a value of $h=.7$. Should I then take the quantity above and multiply it by $.7$ to yield
$$
6.802 \times 10^{-15} \text{(Mpc/h)/s}
$$

and use that result in my calculations? I think this situation confuses me because it doesn't involve measurements, where it is clear how $h$ can enter, and it involves a constant of nature, which should be independent of the assumed cosmology.

Let me know if more info is needed.

Partial answer; although the question is 17 days old, I hope that this guide to the dependence on h in cosmology can give you a good insight into this concept.

Quick guide to h dependence in cosmology

which links to Damn You, Little h! (Or, Real-World Applications of the Hubble Constant Using Observed and Simulated Data). The abstract says:

  The Hubble constant, H0, or its dimensionless equivalent, "little h", is a fundamental cosmological property that is now known to an accuracy better than a few per cent. Despite its cosmological nature, little h commonly appears in the measured properties of individual galaxies. This can pose unique challenges for users of such data, particularly with survey data. In this paper we show how little h arises in the measurement of galaxies, how to compare like-properties from different datasets that have assumed different little h cosmologies, and how to fairly compare theoretical data with observed data, where little h can manifest in vastly different ways. This last point is particularly important when observations are used to calibrate galaxy formation models, as calibrating with the wrong (or no) little h can lead to disastrous results when the model is later converted to the correct h cosmology. We argue that in this modern age little h is an anachronism, being one of least uncertain parameters in astrophysics, and we propose that observers and theorists instead treat this uncertainty like any other. We conclude with a `cheat sheet' of nine points that should be followed when dealing with little h in data analysis.


The role of gravity during planetesimal accretion As far as I understand, during the formation of a planetary disc, grains of dust stick together due to collisions and chemical bonding. Considering the size of the grain - say, 1 mm - and the speed - 1cm per sec - it would seem that gravity plays a minor role during accretion. My question is, to what extent is gravity a factor during accretion? To what extent is it related to fragmentation? In other words - how does gravity affect whether the fragments return and stick to the main body? Is there a size barrier, or is it dependent only on the speed?

The very first stage of planet formation involves purely inelastic (i.e., sticky) collisions between pairs of small particles of dust to make slightly larger particles of dust, pairwise collisions between these slightly larger particles to make even larger particles, and so on. So far, the only gravitation that comes into play is that of the central protostar and the disk as a whole. The gravitation of these little objects is very small.

Then some magic happens that makes the little rocks that result from this very first stage grow into planetesimals that do gravitate. How do little rocks (a few centimeters in diameter) grow into little planetesimals (a few hundred meters in diameter)?  This is the key open problem in the theory of planet formation.

The core of the problem is that objects the size of a spec of dust to an object the size of a small pebble move with the gas that forms the bulk of the disk. The outward pressure from the gas in the disk makes the gas and small objects move at a slightly less than orbital speed. Once an object reaches the centimeter scale or so, the pressure from the gas no longer provides the support needed to keep those objects moving at that slightly suborbital speed. The objects start to become separated from the gas. By the time an object reaches a meter or so in diameter, the gas is equivalent to a hurricane force headwind.

One consequence of this is that collisions become ever more violent as objects grow beyond a centimeter in size, and hence are more likely to break colliding objects apart than to make them coalesce. An even bigger problem is that that headwind makes centimeter- to meter-sized objects spiral into the protostar, and very quickly (a few hundred years from 1 AU). A number of proposals have been put forth to overcome this meter barrier problem. So far, none appears to have stuck. (Sorry for the pun.)

Once objects reach some critical size, the drag from the gas reduces in importance thanks to the square-cube law and the gravitational attraction increases. That critical size varies from hundreds of meters to tens of kilometers, depending on which paper one reads (experts vary). These objects are "planetesimals," and there are lots of them. These once again combine readily to make planetary embryos. From then on, its just a mopping up process as gravitation either makes these planetesimals / planetary embryos combine or ejects them from the star system.


How close would quasars have been to each other? My understanding is that quasars are the early phase of formation of galaxies.

At the time quasars were forming, the universe would have been smaller and therefore things closer to each other.

About how close would quasars have been to each other?

That's an image of two quasars, very close to each other, "separated on the sky by only about 70 thousand light years".

According to this paper the peak density should have been about $10^{-6}$ quasars per Mpc, resulting in a peak mean distance between two quasars of about 100 Mpc. (Density of $10^{-6}$ quasars per Mpc results in one quasar per cube of $10^6 Mpc^3$, hence on average one quasar each $\sqrt[3]{10^6 \mbox{Mpc}^3}=100 \mbox{ Mpc}$)

With about 1 million known quasars in the observable universe I get a mean distance of about half that distance. (With a considered light-travel distance of about $4 \mbox{ Gpc} = 4\cdot 3.26 \mbox{ Gly} = 13.04\mbox{ Gly},~~$  a little less than the 13.81 Gly light travel radius of the observable universe, the volume of a sphere with this radius is about $\frac{4}{3}\pi(4\cdot 10^3 \mbox{Mpc})^3=2.68\cdot 10^{11}\mbox{Mpc}^3$. With 1 million quasars, it's about one quasar per $268000 \mbox{ Mpc}^3$ or one quasar per $\sqrt[3]{268000 \mbox{ Mpc}^3}=64\mbox{ Mpc}$ in one dimension. Due to non-constant quasar frequency over cosmic time, the minimum mean distance should have been below this overall average distance. Using the light-travel distance considers the higher density of the earlier universe better than the comoving distance of 13.8 billion years after big bang. 4 Gpc are used as light travel radius to restrict roughly to redshift z < 7, the range where quasars have been observed. A redshift calculator can be found e.g. here, take $H_0=70.4$, $\Omega_m = 0.265$, $\Omega_{\Lambda}=0.728$, and $z=7$.)

The peak number density seems to be somewhere between a redshift of 2.0 and 2.5 (interpreting the paper). If it will turn out by future observations, that the peak number density has been earlier in the universe, we get much shorter peak mean distances.

So you have at least an order of magnitude: Somewhere between 150 and 300 million lightyears (between 50 and 100 Mpc, 1 parsec equals 3.26 lightyears) should have been the peak average distance between two quasars. Might be this estimate will shrink over time, as more (distant) quasars get known.


Lifespan of B-type subdwarfs According to the Extrasolar Planets Encyclopedia, the B-subdwarf Kepler-70 lost its helium envelope and became a blue-white subdwarf roughly 18.4 million years ago.

(sources: archived encyclopedia page, also the paper "A compact system of small planets around a former red giant star")

This has got me curious regarding the lifespan of these stars - 18.4 million years isn't that long in stellar terms!

Although I realise this will vary depending on e.g. mass of the star, is anything known about the length of time a B-subdwarf will remain in the subdwarf state before finally becoming a white dwarf? The information quoted above for Kepler-70 was a lot easier to find than for any other blue-white subdwarves.

In section 3 of this paper on the possible origin of two planets orbiting a B-type subdwarf, called KIC 05807616, and of which I asked a question a while ago, the survivability of the planets to the intense UV radiation of their host star is examined. Here is an excerpt from the summary:

  In section 3 we examined the survivability of the planets to evaporation by the UV
  radiation of the EHB star. Equation (3) for the evaporation rate implies that the inner
  planet will be completely evaporated within ∼ 10⁷ yr. This is shorter than the ∼ 10⁸ yr life duration on the HB.

KIC 05807616 is still on the horizontal branch (HB), like any other star of its kind, and will remain in that stage for approximately one hundred million years, according to the paper, until it finally becomes a white dwarf. I guess that's the average lifetime of the subdwarf B stars.

EDIT: I also found this paper, which is more specific with respect to the average lifetime of the subdwarf B stars, I quote from its introduction:

  An sdB stays on the EHB for roughly 10⁸ years and directly evolves along with
  the white dwarf (WD) cooling track after its core helium has
  been exhausted.


Are there any hot jupiters orbiting red dwarfs? Do we know of any hot jupiters that could be orbiting a red dwarf (or, more probably, orbiting a barycenter between the two)? Is this scenario even physically possible?

I found one "hot Jupiter" in the Kepler data (Kepler 45b). The star is a M dwarf with an effective temperature of 3820K. The planet has an estimated mass of 160.5 M(Earth) and radius of 10.76 R(Earth). This gives a density of about 0.8 g/cm2 which is consistent. The planet is located at approximately 0.03 AU from the star with an orbital eccentricity of 0.11. Also, it may be interesting that the star has a very high metallicity of Fe/H = 0.28. (This data is from http://kepler.nasa.gov/Mission/discoveries/ ).

It must be said, however, that M dwarfs only made up about 4.3% of the initial Kepler study which found 82 "Jupiter size" planets (as of my tally in April of 2015). Thus, there is an indication that there are less "hot Jupiters" around M dwarfs, but sample size is way too small. There does seem to be a correlation between stars hosting "hot Jupiters" and their metallicity (more common around higher Fe/H stars).

My gut feeling is that "hot Jupiters" in general are uncommon. They are just the easiest thing to find.


What causes a neutron star to become a pulsar? Leading on from this pulsar question, what criteria/processes make neutron stars become pulsars, do all neutron stars become pulsars?

It is thought that all neutron stars are pulsars when they are first born, as all spin extremely rapidly and have strong magnetic fields shortly after birth, arising from their dramatic collapse from a few Earth-sized massive star core into a city-sized neutron star. Conservation of angular momentum means that an initially slow-rotating core becomes a rapidly spinning neutron star, which also drives a dynamo action that generates a strong magnetic field.

The emission from a neutron star comes as a result of pair-production in its magnetosphere, which has an approximately dipolar shape and thus the strongest field strength at its poles. This field is rarely aligned with the spin axis, thus giving the "lighthouse" effect where the beam sweeps over an observer at twice the spin frequency. These pair-produced particles radiate energy via synchrotron emission (primarily in the radio), sapping energy from the magnetic field. As the neutron star ages, the magnetic field eventually decays below a critical point where electrons/positrons cannot be pair-produced. This is called the "death line," illustrated here and denoted by the word "graveyard" on a P-P_dot diagram:

After crossing this threshold, the pulsar loses its ability to emit radio waves. Pulsars only remain pulsars for the first ~10^7 years of their lives, and thus about 99% of neutron stars are no longer pulsars.

Additionally, we can only see a pulsar if its beam crosses over the Earth while the pulsar rotates. As the beam angle is usually tens of degrees, the total fraction of pulsars whose beams cross the Earth is approximately 10% the total that exist.

In the lower-left of the diagram you can see very short period pulsars, these are millisecond pulsars and have been spun up by accreting mass from a companion. These pulsars have very long lifetimes, but only represent a small subset of the neutron star population.


Why is it that night has a almost a bluish tint to it instead of red? From my understanding/been told, the color red has the ability to travel the farthest in our atmosphere, being the reason why we have red sunsets and such. But with this in mind, why is it that at night the moonlight we receive seems to have a slight blue tint to it, and since red can travel the best out of our visible spectrum, how come our moonlit nights don't have a red tint instead of a blue one?

The blue tint is an illusion caused by the wavelength sensitivity shift when switching from rods to cones as the light intensity decreases below certain level.

It is called the Purkinje effect.

Objectively, moonlight is not blue. In fact, it is even more yellow than sunlight! (comparing the same elevation, because obviously the true color can be distorted by the atmoshpere, too). The moonlight is not bright enough to fully enable the photopic vision and our red color sensitivity suffers, producing the blue moonlight illusion.


What are the 10 most abundant elements in the universe by number of atomic nuclei? I've seen websites that show the ratios of the 10 most common elements but they compare them by relative mass. I think it's more interesting to know the relative abundances of atomic nuclei because that is what you need to know to understand why the universe has the chemical makeup that it does. What are the 10 most abundant elements in the universe by number of atomic nuclei?

All right, so I took the first list on wikipedia listing the 10 most common elements by mass in parts per million, and did what Rob recommended and here's what I got.

https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Abundance_of_elements_in_the_Universe

Hydrogen - 739000amu(H)/1amu(H)=739000 H atoms

Helium - 240000amu(He)/4amu(He)=60000 He atoms

Oxygen - 10400amu(O)/16amu(O)=650 O atoms

Carbon - 4600amu(C)/12amu(C)=383 C atoms

Neon - 1340amu(Ne)/20.1amu(Ne)=66 Ne atoms

Iron - 1090amu(Fe)/55.845amu(Fe)=19.5 Fe atoms

Nitrogen - 960amu(N)/14amu(N)=68.5 N atoms

Silicon - 650amu(Si)/28.1amu(Si)=23 Si atoms

Magnesium, - 580amu(Mg)/24.3amu(Mg)=24 Mg atoms

Sulfur - 440amu(S)/32.1amu(S)=13.7 S atoms

So therefore, the 10 most common elements in the universe by atom, with the relative ratios between them, is;

1.Hydrogen (739000)
2.Helium (60000)
3.Oxygen (650)
4.Carbon (383)
5.Nitrogen (68.5)
6.Neon (66)
7.Magnesium (24)
8.Silicon (23)
9.Iron (19.5)
10.Sulfur (13.7)

If anyone sees a mistake that I made with my reasoning or calculations then please point it out.


Has the Lyman-alpha forest ever been used to test Arp's theory? Has the Lyman-alpha forest ever been used to test Halton Arp's theory that quasars are not cosmological but instead are ejected from relatively near-by galaxies?  If Arp was right, then the spectra of the quasar and the 'parent galaxy' should both show the same Lyman-alpha forest; if it only appears in the quasar spectra, then the quasar must be much further away.

All possible tests of the Arp model have been failed. High redshift quasars have Lyman alpha forests, low redshift galaxies do not.

The Arp model suggests that the Lyman alpha forests are somehow intrinsic to the quasar. There are many problems with this idea.


What is the highest point on Mercury? Continuation of "Gravity on Mercury's highest elevation?"

Because when you Google that, you see "Caloris Montes" which is actually a mountain range consisting of mountains 1-2km tall from base to peak (Wikipedia). However, the previous question noted a value of 4.48 km "from what is considered a sea level". But mercury doesn't have sea, does it? Moreover, this value is confirmed here:

This new model reveals a variety of interesting topographic features,
as shown in the animation above, including Mercury’s highest and
lowest points. The highest point on Mercury is at 2.78 miles (4.48 km)
above Mercury’s average elevation, located just south of the equator
in some of Mercury’s oldest terrain.

A certain discrepancy is noted for our planet Earth. Mt. Everest is considered the tallest mountain (8848 m) when considered from sea-level. But, when we consider the height from base to peak, Mauna Kea/Loa is the highest mountain (of volcanic origin) (Total height=10.2 km, 4.2km above sea level). But, Mercury doesn't have "a sea level". So, why this discrepancy?
Questions:

What is the highest point of Mercury? Why is it "Calores Montes" even if it is a mountain range consisting of many mountains? Even if that peak of height 4.48 km is considered the highest point, why doesn't it have a name (since it is the tallest point, it should be notable and should have a name)?
What should be considered the base for planets which doesn't have a "sea" (sea level is only possible for Earth and Titan)? Wikipedia says " peak elevations above an equipotential surface or a reference ellipsoid could be used if enough data is available for the calculation, but this is often not the case". Where is it not applicable? Doesn't it lead to discrepancy like for instance, Mercury?

*This video says something called Mt. Hermes as the highest point of mercury.

There are a couple of components to your question, so rather than directly answering, I need to discuss a few things in a more narrative form.
Mercury was first spacecraft-imaged in the 1970s, with only about 45% coverage.  From those images, some topography could be generated through stereogrammetry (two images taken with different viewing geometries that use parallax to reconstruct topography) and photoclinometry ("shape-from-chading").
The MESSENGER spacecraft orbited for several years, 2011–2015.  Its orbit was designed, in part, for global stereogrammetric coverage and it had a laser altimeter that could operate at high- and mid-northern latitudes.  As data were collected, early results were published.
The Wikipedia page you link to for Mercury's highest point, the Caloris Montes, was based on earlier results.  It goes at least as far back as 2015, I didn't check further.  Caloris Montes is named because it is an important feature on Mercury due to its association to the giant Caloris impact basin.  It's not named because it's a high point, specifically.  "Mons" is a singular mountain, while "montes" is used for a mountain range.
The article you link to is based on the full MESSENGER mission dataset, which was used to construct a new global mosaic and digital terrain model (available here).  From the new, better, global model, there is a higher point in the southern hemisphere than the Caloris Montes.  To me, it looks like a random point between some large craters, not anything especially important.  Therefore, it doesn't have a name.  Planetary nomenclature is used only when features are considered important and so need a name for easy referencing.  So, instead of saying, "That really big basin on Mercury," I can say "Caloris" and people will know what I'm talking about.  Similarly, as I noted above, the Caloris Montes are important as a mountain range, but a specific highest point among them is not.
As for a YouTube video saying the highest point is "Mt. Hermes," that sounds dubious to me with respect to anything official (for one, mountains on other bodies are "montes" after the Latin, if they have a formal name).  A search shows no one else is using this name.
Regarding "sea level," it's what Wikipedia said: You use a reference ellipsoid, usually.  The reference ellipsoid is made usually by fitting many different limb profiles to an ellipsoid, and/or via a laser altimeter.  The best fit is the ellipsoid.  Any deviations from that would be topography.  Mercury has plenty-enough data to make a reference ellipsoid, indeed you linked to one in the comments above.  Once these are published in the scientific literature, assuming they are generally accepted, the International Astronomical Union makes that a standard and publishes it in various reports.  They don't often update their standards, though I think the last one was in 2015.  So, it would not reflect the latest you found.
Practically all resolved solar system bodies have a reference ellipsoid that's used and published.  How good it is, though, is another question.  Asteroids, given their highly irregular shapes, for example, that were imaged by flyby missions would generally have a much more poorly constrained ellipsoid than, say, Earth's moon that has been observed through telescopes for over 400 years, orbited by numerous spacecraft, has retroreflectors on it, and has >8 billion laser altimeter points from the Lunar Reconnaissance Orbiter Laser Altimeter.


DSO in a 60mm telescope I am thinking about buying a telescope, but here in Brazil they are very expensive and hard to find where to buy. So, the better telescope I can pay for is a Celestron Powerseeker 60AZ. My main application would be to see nebulas and some star clusters, but I don't know if this scope can be used for this pourpose. Will I see big blurry images, faint lights (like a tiny fog on the darkness of the sky) or only its main stars?

*Note: I've found some images about planets in small scopes on YT, but I haven't found about DSOs (although, I know the images won't be so atractive like bigger telescope ones). Also, can I use the star diagonal as a barlow lens?

Specifications:

Optical Design: Refractor

Aperture:   60mm (2.36")

Focal Length:   700mm (28")

Focal Ratio:    12

Focal Length of Eyepiece 1: 20mm (0.79")

Magnification of Eyepiece 1:    35x

Focal Length of Eyepiece 2: 4mm (0.16")

Magnification of Eyepiece 2:    175x

Finderscope:    5x24

Star Diagonal:  1.25" Erect Image Diagonal

Highest Useful Magnification:   142x

Lowest Useful Magnification:    8.57x

Limiting Stellar Magnitude: 11.4

Resolution (Rayleigh):  2.32 arc seconds

Resolution (Dawes): 1.93 arc seconds

Optical Coatings:   Fully-Coated

Optical Tube Length:    711mm (28")

Barlow Lens:    3x

To know what you would see with a telescope, you can use a field of view calculator such as this one from astronomy tools. There, you just fill in the details of your optics, and you can get an idea of what you would be able to see with your telescope.

For a Celestron Powerseeker with the focal lengths and apertures you provided, and a 3x Barlow, the Crab Nebula might look something like this

The yellow line here is the edge of the field of view. You can also use the web app to compare several different configurations. This will give you an idea of how big the object will be in your telescope.

An other factor to take into account is how bright you will be able to see the object. This depends greatly on where you will be observing (near a city or far away from any bright lights). In any case, the contrast will not be as stark as in the above image.

For this, you first need to estimate the magnitude of the faintest objects observable with the naked eye where you expect to be observing. If you know how to recognize a few constellations, you can look up (in a sky atlas or online) the magnitude of the faintest stars you see. For instance, if you can see M13 in Hercules, then the limiting magnitude with your naked eye is higher than 5.8. However, if you can't see the entire wings of Cygnus, then the limiting magnitude is between 3 and 4.

Once you figured this out, you can use a limiting magnitude calculator. This will give you an idea of the faintest objects you can see with the telescope.


Whats the farthest distance a spacecraft has been detected (except by radio) As far as I understand it, one can not hide spacecraft, because the heat from the engine would be very visible.

Now I tried searching for the farthest distance an actual spacecraft has been detected besides its radio signals that are intentionally sent to us. Unfortunately I could not find any data on this.

OSIRIS-Rex has been spotted on its approach to Earth, at a distance of approximately 7 million miles (12 million kilometers) away, with a brightness of approximately 25th magnitude. The Large Binocular Telescope was used for this observation, this has a pair of 8.4-meter mirrors.


How does one measure velocities of far-off, bright objects As the title already says, I want to know how one measures velocities of far-off, bright objects, e.g. when the mean parallax drift isn't measurable with current apparatus (this means when there is only a "non-moving" picture of that object available)

I know that one can measure the redshift of spectral lines and correct them for gravitational redshifting if the distance is already known.
But then the relation between source-frequency $f_S$ and observed frequency $f_O$ is
$$ f_O = f_S \frac{\sqrt{1-\frac{v^2}{c^2}}}{1+\frac{v}{c}\cos(\alpha)}$$,
where $\alpha$ is the angle between line of sight and the velocity vector.

If one only has a "non-moving" picture of the object, there is no way to determine this angle $\alpha$ and therefore no way to get the velocity by means of redshift.

As an addition: At which distance does the mean parallax drift become unmeasurble with current apparatus?

In general the total velocity of a faraway object is not obtainable, only the component along the line of sight. However, if a star is not too far away, and its position on the sky is measured very accurately with an interval of some years, the velocity component perpendicular to the line of sight can be determined, yielding together with the redshift the total velocity. The recently launched Gaia spacecraft is expected to be measure the proper motion of millions of stars in this way.

In distant galaxies, this is not possible. If one is interested in, say, the 3D velocity dispersion of the gas, the 1D dispersion is measured, and assuming (fairly) an isotropic dispersion, this number is then multiplied by $\sqrt{3}$ to get the total velocity.

The distance to which parallax is measurable depends on the angular resolution of your instrument. For milliarcsec resolution, parallax measurements are good out to roughly 1 kpc. If you wait long enough, the motion around the galactic center increases your baseline such that larger distance can be probed.


What does "all stop" mean in reality? In science fiction, when a starship captain commands "all stop," the audience simply assumes that, like a terrestrial naval ship anchored to the ocean floor, the ship literally comes to a complete stop.

Does our current understanding of celestial mechanics provide a meaningful definition of a complete stop in space — for any object?

As mentioned in the comments, in relativity, "all stop" can only be in reference to some other object, or set of objects.  Still, I would argue there remains some interesting potential for the "all stop" concept in physics theory, because Einstein was motivated by Mach's principle when creating general relativity (regardless of whatever version of that principle is formally supported in that theory), and so we might suspect that "all stop" can have the theoretical local meaning of "at rest with respect to the center of mass of everything around it that is affecting it."

That notion is trickier in an expanding universe, where the concept of a center of mass gets replaced with the concept of "at rest with respect to the cosmic microwave background", which is normally treated as a kind of accident of history, rather than a theoretical truth.  Still, since the Big Bang model is often taken as asserting that the accidental history of our universe is actually relevant to the origin of our universe and its theoretical properties, as well as to the origin of the laws that our universe obeys, it is not a great stretch to imagine that an "accident of history" kind of preferred frame could still, nevertheless, have a formal status in the laws as a kind of true "all stop" frame-- albeit only locally.


How can I calculate how the debris of an object ripped apart at the Roche limit will spread out? Let's say I have a moon orbiting a planet at a distance $d$. Eventually, $d \leq d_R$, where $d_R$ is the Roche limit of the moon-planet system. I can figure out the mass of both bodies by experiment, as well as their radii, and thus their density (which is how I can calculate the Roche limit).

I'm curious as to how the debris will spread out, though, once the moon is ripped apart. Can I calculate the inner and outer radii of the newly formed ring, based on what I know? Or is there not a direct relationship between these radii and the initial conditions?

I think there are two key aspects to the answer.  1) Solid/rocky bodies should tend to collide before they reach the Roche limit.  2) When gaseous bodies reach the Roche limit (and undergo 'Roche-Lobe Overflow'), the dynamics are basically those of test-bodies and are fairly straightforward and well understood from binary stellar dynamics.  To expand on both:

1) Rocky Roche Limit.  If you are thinking about a gaseous donor (the object being disrupted), then this is an irrelevant point, but It sounds like rocky is what you had in mind.  To an order of magnitude, the Roche limit is the same as the Hill sphere, or the Tidal Radius (e.g. Rees 1988) --- which is simply the radius at which the density of the donor equals the average density of the primary in a sphere of that radius:

$$R_t^3/M \approx r^3/m \rightarrow R_t \approx r \left( M/m \right)^{1/3}$$

Rocky material varies very little in density (e.g. Iron meteorites are only about twice as dense as chrondritic ones), which means that the densities will only match very near the radius of the primary.  The tidal bulge in the secondary (donor) also makes it easier for the objects to collide before meeting this criteria.

e.g.

2) Gaseous Roche Lobe Overflow.  Basically every text on stars and star systems will have a section on mass transferring binaries which will describe the dynamics of mass transfer (this PDF is by Philipp Podsiadlowski who is a wizard of the field).  In the type of situation you are describing, like in (semi-)stable binary systems, it is a very gradual process where material is slowly syphoned off of the donor.  This material can either form an accretion disk (high angular momentum material) and gradually accrete onto the primary, or directly impact the primary (low angular momentum material).

e.g.


How to explain the equation 'Altitude of a star at culmination = co-latitude + declination'? Altitude of a star at culmination = co-latitude + declination.
I understand that the co-latitude is equal to the altitude of the celestial equator (since it is 90 degrees away from the NCP). I also understand that declination is the height of a star above the celestial equator.
But I am just unsure why you can directly link altitude and declination at the exact point of culmination?
And why is a star reaching its highest point as it crosses the meridian?
thanks for any help.

Unless the observer is at a geographic pole, the celestial equator is not parallel to the horizon but crosses it at an angle equal to the observer's colatitude.
A star's path across the sky is parallel to the celestial equator, with maximum altitude midway between the eastern and western horizons, i.e. on the meridian.

Rendered by Stellarium


Units and some equations I'm a grad student and I feel like this is an extremely elementary question that I should know, but I admit that I don't. Occasionally in science I find myself solving an equation for some answer where it feels like the units don't add up.
For instance, in a spatially flat matter-dominated universe with nonrelativistic matter, the accepted age of such a universe is:
$$t_0 = \frac{2}{3}H_0$$
where $H_0$ has units of ${\rm km} \cdot {\rm s}^{-1} \cdot {\rm Mpc}^{-1}$.
Why am I getting a time from this? I know it works, I know this is correct. This isn't the only example of this I've ever seen, but it is the freshest in my memory. It conceptually makes sense, but usually I'm in the habit of doing meticulous dimensional analysis and just wondered if there's some deep metaphysics or philosophy behind when that applies and when it doesn't.

The Hubble constant effectively has a unit of inverse time only: ${\rm s}^{-1}$:
It is usually given as ${\rm km}\cdot{\rm s}^{-1}{\rm Mpc}^{-1}$. Yet looking at it, km and Mpc are both units of length, so they cancel eachother out when converted to the same unit. Thus add the appropriate unit conversion from km to Mpc and it remains an inverse time only:
$$t = \frac{1}{H_0} = \frac{\rm Mpc}{60 {\rm km}\cdot{\rm s}^{-1}} = \frac{3\cdot 10^{22} {\rm m}{\rm s}}{60\cdot 1000 {\rm m}} = 5\cdot 10^{17}{\rm s} = 15{\rm Ga}$$
Add dimensionless factors to the equation to account for different cosmological models and inflation theories. This is anyway only a ballpark number - but an easy one to get for the age of the universe.


Why does the half moon appear to have a "nose"? Yesterday evening, the moon from my place looked like this:

Basically, instead of the crescent shape I am used to it looked like there was three tips: one at each extremity (as usual) and one in the middle between the two first ones. This picture shows the rough shape of the visible moon, with arrows pointing at all three tips:

What is the explanation of this?

Here's an image I captured from a Moon globe app on my iPad.

From your description, I think you may be referring to the southern rim of Mare Tranquillitatis, the Sea of Tranquility, perhaps extending a little into the Rimae Hypatia. This is an area which is lower than the surrounding mountains and craters, into which the shadow is encroaching.

Incidentally, Mare Tranquillitatis is the region where Neil Armstrong and Buzz Aldrin landed on the Apollo 11 mission!


How does the Gamma Ray Burst that occurred when 2 black holes merged compare to other GRB's? A Gamma Ray Burst was detected 0.4 seconds after the gravitational wave event, GW150914, caused by a black hole merger, and it was in the same part of the sky. It is uncertain whether that Gamma Ray Burst was associated with the black hole merger. The odds of a GRB being coincident (or just background noise) is 0.22%. That implies a 99.78% chance the black hole merger was related to the GRB. Later analysis suggests the GRB was just a background event that occurred in the same place in the sky at just 0.4 seconds after the black hole merger, and therefore not related.

While a GRB may have relativistic jets beaming out from it in opposite directions, research "excludes the possibility that the event is associated with substantial gamma-ray radiation, directed towards the observer." I interpret that to mean this GRB did not have relativistic jets, but was omnidirectional. (Not sure how astronomers came to that conclusion.)

Anyhow, whether it was coincident or not, I am asking about how the energy output of that GRB compares to other GRB's. The Wikipedia article in the link says the "energy emission in gamma-rays and hard X-rays from the event was less than one millionth of the energy emitted as gravitational waves."

How much energy does a typical GRB emit? What was the spectrum spread of a typical GRB? Over what amount of time does a typical GRB emit that energy? (How many seconds?)

How do the energy levels, spectrums, and time durations of other GRB's compare to the GRB associated with the black hole merger?

If it is similar to other GRB's, that would support the hypothesis this GRB was just a coincident event at nearly the same time and location in the sky.

If this GRB has different energy emissions and time duration than other GRB's, that would support the hypothesis it is truly associated with the black hole merger.

When you answer, please provide data, citations, or quotes from original research. I am not looking for unsupported speculation, but real analysis supported by real data.

The interpretation you suggest in the second paragraph is incorrect. It is understandable, since there is a debate in the literature - different papers come to potentially contradicting conclusions.

"Excluding a possibility that the event is associated with substantial gamma-ray radiation, directed towards the observer" simply means that no observable GRB is found, hinting that may be the original detection was caused by instrumental background: each detector has it's own instrumental backgrounds, only real events should be seen by all sufficiently sensitive instruments.

In principle, it is always possible that the merger generated beamed emission, directed somewhere else. No instrument should have seen it directly, and there is no simple way to know if it happened.

This upper limit is derived from the observation of another satellite (INTEGRAL/SPI-ACS, Savchenko et al 2016) than the original detection (Fermi/GBM, Connaughton et al 2016).
Also an alternative analysis of the Fermi/GBM data (Greiner et al 2016) suggested that no event can be found in the GBM data - their opinion is that it was a background fluctuation of some kind.

Right now, the teams who reported these conflicting results are working together, trying to come up with a consistent picture, which might be, in principle, a GRB with this or that properties associated with the GW150914, an unrelated GRB with some properties, or no detection whatsoever. This work is centered on cross-calibrating and comparing instruments, and is also useful to avoid these kind of uncertainties in the future.

One could try to characterize the spectral properties of this event, following the approach of the original Fermi/GBM team. But unfortunately, the measurement appeared to be in very unlucky conditions for Fermi/GBM (in bad direction). Which is why the signal was very weak (below what would be usually reported for a real GRB, though recently the attempts were made to decrease these thresholds, see Goldstein et al 2017 ), and spectral characterization is loose. You can look for some details to Veres et al 2016.
With these large uncertainties the spectrum is compatible to that of known short GRBs.

The luminosity estimate depends on the spectrum, but is seems to be at the lower end of the short GRB sample (see e.g. Wanderman et al 2015)
But because the uncertainties are large, the event, if real, might be unusual as well.

The INTEGRAL observation, non-detection, would imply much softer (perhaps, unusual for a short GRB) or/and a weaker burst, possibly incompatible even with the highly uncertain Fermi/GBM data.

Duration of this possible GRB possibly associated with GW event is the easy part and is about 1 s long, typical for a short GRB (Kouveliotou  et al 1993).


How can I calculate the orbital periods in a binary star system? I have two stars, with known masses and known orbital radius. How do I calculate the orbital periods of both stars?

If you need a rough estimation, you can sample the positions of the stellar objects from their acceleration, using Newton's law. A full picture is drawn on this Wikipedia page, but basically, given N stellar objects at position P(i), with their respective mass M(i):

$$
M_i.\vec{acc_i} = -G\sum_{n \neq i}^N \frac {M_i.M_n.(\vec{pos_i}-\vec{pos_n})}{\begin{vmatrix} \vec{pos_i}-\vec{pos_n}\end{vmatrix}^3 }
$$

You can then derivate acceleration into speed, then into position using a small enough time delta. With initial positions and speed (which are the trickiest and also funniest to choose), you can simulate a rough N-body system.

On the fun and simulation side, this is what Universe Sandbox aims to present (in advance, sorry for linking to a commercial store, I'm not related to this studio).

Lagrangian points are also interesting to look at when simulating.

However, to simplify things, you may like to think nearly all stellar objects "close" to the binary system would have been eaten by the massive couple over time, and then consider only the barycenter of the pair of stars.

EDIT: although OP was clear, my answer is for N objects. The simplification to N=2, with A and B being the positions of the two objects, results in:
$$
\vec{acc_A} = \frac{G.M_B}{AB^3}\vec{AB}
$$
$$
\vec{acc_B} = \frac{G.M_A}{BA^3}\vec{BA}
$$
And with an iterative process, once acceleration is computed for step k, using known initial conditions for speed and position:
$$
\vec{spd_{k+1}} = \vec{acc_k}.\Delta{t} + \vec{spd_k}
$$
$$
\vec{pos_{k+1}} = \vec{spd_k}.\Delta{t} + \vec{pos_k}
$$

EDIT2: well, another rough estimation, for the period duration (which I misunderstood as "trajectory" from the question). I'd use cylindrical coordinates as reference, for their clean spatial representation, and for the single-angle conversion:

$$
\left( \begin{array}{c} R_k \\ \theta_k \\ h_k \end{array} \right) = \left( \begin{array}{c} \sqrt{pos_{x,k}^2+pos_{y,k}^2} \\ \arctan \left( \frac{pos_{y,k}}{pos_{x,k}} \right) \\ pos_{z,k} \end{array} \right)
$$

...with adequate precaution. Introduce part of that conversion at each iteration, and wait for theta to complete a rotation.

(in advance, sorry for the weird notation, which I just tried to make consistent from my previous posts)


What is the probability that we are the first ever intelligent life in the whole universe? What is the probability that we are the first ever intelligent life developed all over the universe?

Everything starts somewhere, Is there any probability that we are the beginners to be part of intelligent life all over universe?!

We know that intelligent life exists on one planet. We do not fully understand the processes by which life (a) starts and (b) becomes intelligent.

Therefore, the probability of these things happening elsewhere ranges from 1 to as close-to-zero as needs to be to ensure we are the first and perhaps only intelligent life in the universe.

Therefore the answer to your question ranges anywhere from zero to one.

A more sensible, informed calculation could arise if we ever create life in the lab or if life with an independent genesis is found elsewhere (in the solar system or beyond).