What good is the Einstein de-Sitter model? Why is the EdS model still taught and used to predict the age of the universe if ours is not an EdS Universe. Does this model have any pros?

The scale factor of the universe depends on the matter and energy density in a way that depends on which component is dominant.  This is discussed in the question How does the Hubble parameter change with the age of the universe? on the Physics Stack Exchange, and from that I have taken this graph showing how the scale factor changes with time for our universe:

For a (cold) matter dominated universe we get a dependence:

$$ a(t) \propto t^{2/3} $$

and for a dark energy dominated universe we get an exponential dependence of $a(t)$ on $t$. Just by looking at the graph you can see that the $t^{2/3}$ dependence was good up to about half the current age of the universe and after that the exponential growth becomes increasingly dominant.

So you are quite correct that the Einstein de Sitter model is not a good description of our universe right now, but it is a good description of the first half of the lifetime of the universe. Furthermore, because when we view distance regions of the universe we are viewing them as they were in the past it turns out that all the parts of the universe we can see beyond a redshift of about $Z = 2$ are still well described by an Einstein de Sitter model.

I would take issue with your statement that the Einstein de Sitter model is still taught and used to calculate the age of the universe. The most precise calculations of the age of the universe come from the Planck observations of the cosmic microwave background, and these most certainly do not use the Einstein de Sitter model. As for being taught, if you do a general relativity course you will be taught the FLRW metric and Friedmann equations. The Einstein de Sitter model is just a special case of the more general result.


Two species of dark matter? At this point in time, evidence for the existence of dark matter has accumulated in many ways:

it affects galactic rotation curves
plays a major role in cosmology, and the evolution of structure in the universe
is predicted in copious amounts by gravitational lensing on a wide range of scales
influences the dynamics of galaxy clusters

to name a few.

There are many known candidates for dark matter particles: WIMPs, axions, WISPs, neutrinos, etc (in fact, even bricks, though some other considerations would exclude them).

The question then is:
Why do we expect that only one type of dark matter particles is responsible for phenomenological dark matter?

For example, $\Lambda$CDM cosmology, the standard cosmological model, requires dark matter to be cold (slow, non-relativistic), which is used to constrain the possible properties of dark matter particles. However, this doesn't actually imply, that dark matter is cold for all the astrophysical systems. For example, galactic halos could be made of warm dark matter, and halos of dwarf galaxies could be made of cold dark matter.

One might of course say that one-species model is the simplest one. The counter-argument would be that in reality there well may be many species. This in turn might have profound implications for astrophysical models.

To summarize the question: Is there any good reason, preferably supported by observations, to think that only one species of dark matter is present in all the models currently used?

Hot dark matter would be made from very light, fast moving particles. Such particles could not possibly be gravitationally bound to any structure, but rather would be dispersed all across the universe.

But dark matter is always "found" (or "inferred") either gravitationally bound to some visible structure (e.g. weak lensing detection of dark matter associated with colliding galaxy clusters / flat rotation curves of spiral galaxies / abnormal velocity dispersion in galaxy clusters) or not associated to anything visible but nevertheless forming clumps (weak lensing detection of galaxy clusters previously unseen). That is why dark matter is thought to be cold.

Additionally, there is a clear distinction between both types: there is not such thing as dark matter that is "not too cold but not too hot either" (see footnote as well). Dark matter is either made of particles with less than ~10 eV (hot dark matter, made of light particles, mostly dispersed everywhere) or particles with more than ~2 GeV (heavier, slower particles gravitationally bound to some structure). Both limits are found when imposing the maximum amount in which the candidate particles (neutrinos or something more exotic) can possibly contribute to the actual value of the density parameter due to matter in our expaning Universe.

Thus, either DM appears gravitationally bound (cold DM) or dispersed (hot DM), and both types are clearly distinct (10 ev vs 2 Gev). Observations favour the first case. However, Cold Dark Matter is not the ultimate solution, and still faces some problems.

Regarding the possibility of mixed solutions, many of them have been already ruled out. Microlensing has ruled out the possibility of unseen compact objects (brown dwarfs, stars, stellar black holes) in galactic halos, in our galactic neighbourhood as well as in the extragalactic domain. Ordinary matter (stones, bricks, dust) cannot possibly be, otherwise they would become hot and re-radiate. Any exotic mix of known particles doesn't work.

All we think we know is that DM must be made of some heavy particles yet to be discovered. In order to introduce a more complex model (e.g. different types of particles depending on the structure they appear attached to) one needs a justification (i.e. some predictions that better agree with reality) and nobody has been able to do that yet.

Remark Note that Dark Matter particles, either from the hot or the cold type, cannot possibly "slow down" and clump too much (e.g. forming planets) because they don't interact electromagnetically like ordinary matter, that is why DM is said to be collisionless. Wherever infalling ordinary matter forms any structure (e.g. protostars or accretion disks), a very important part of the process is thermalisation, i.e. the redistribution of energy of the infalling particles by means of numerous collisions. This cannot happen with Dark Matter.


Geographical Point of the Sun I want to draw a Sun in a 3d scene, so I figured the easiest way would be to calculate the latitude and longitude of the point perpendicular to Sun position and then just move it "up".

As a starting point, I want to calculate it's equatorial coordinates: declination and right ascension.

I followed instructions from this page, also cross-referencing with this one

Here's my code (in Swift):

import Foundation

class Astronomy {

  static func julianDays(since: Date) -> Double {
    return since.timeIntervalSince1970 / 86400 + 2440587.5
  }

  static func daysSinceJ2000(for date: Date) -> Double {
    return julianDays(since: date) - 2451545
  }

  static func sunAxialTilt(for daysSinceJ200: Double) -> Double {
    return 23.4393 - 3.563e-7 * daysSinceJ200
  }

  static func sunEquatorialPosition(on date: Date) -> (declination: Double, RA: Double, distance: Double) {
    let daysSinceGreenwichNoon = daysSinceJ2000(for: date)
    var meanLongitude = 280.460 + 0.9856474*daysSinceGreenwichNoon
    var meanAnomaly = 357.528 + 0.9856003*daysSinceGreenwichNoon
    meanLongitude = meanLongitude.rangedUpTo360
    meanAnomaly = meanAnomaly.rangedUpTo360
    let eclipticLongitude = meanLongitude + 1.915*sin(meanAnomaly.radians) + 0.020*sin(2*meanAnomaly.radians)

    let distanceToEarth = 1.00014 - 0.01671*cos(meanAnomaly.radians) - 0.00014*cos(2*meanAnomaly.radians)

    let axialTilt = sunAxialTilt(for: daysSinceGreenwichNoon)
    let rightAscension = atan2(cos(axialTilt.radians)*sin(eclipticLongitude.radians), cos(eclipticLongitude.radians))
    let declination = asin(sin(axialTilt.radians)*sin(eclipticLongitude.radians))

    return (declination: declination.degrees, RA: rightAscension.degrees, distance: distanceToEarth)
  }
}

extension Double {
  var radians: Double {
    return self * .pi / 180
  }

  var degrees: Double {
    return self  * 180 / .pi
  }

  var rangedUpTo360: Double {
    let multiples = (self/360).rounded(.down)
    return self - 360*multiples
  }
}

However, it does not seem to be working correctly. As of the time I write this article, the result is declination -16.98772292697088, RA -44.80674952684117. Compared with https://theskylive.com/sun-info the declination is about right (-17° 03’ 38”), but right ascension is way off:  20h 59m 43s, which is 314.93 degrees according to this site

What went wrong?

Edit:
As it turned out nothing was wrong, I can get to the second part: calculating the geographical point. (the title suggests it after all)

AFAIK for latitude, I can use declination as is and for longitude, I
need to calculate Greenwich Hour Angle(GHA)

After examining several articles, I found these equations:

SHA[°] = 360 – 15 · RA[h]
GHAAries[°] = 15 · GST[h]
GHA = SHA + GHAAries

For my case, I suppose I may use RA as is, without multiplying its hour component by 15. What keeps me a bit puzzled is GST:

May I use Greenwich Mean Sidereal Time instead if I don't need the calculations to be super precise?
Is there any "GST for dummies"? All the calculations I found rely on some floating second offset, I found this table but have no idea how to use it.

If we draw a circle and draw a 315 degree angle on that circle, in the counter clockwise direction, we would roughly get the same point as when we draw an angle of 44.8 degrees in the opposite direction (the equivalent of drawing -44.8 in the same direction) because 315+45=360. There therefore does not seem to be an error in your results... Visually the two angles are the same. As for the slight inaccuracy, it may be due to floating point precision errors or measurement errors.
If you want positive right ascension, all you need to do is add 360 if it is negative. Before doing that (this is general advice, it may not be important to your code), you ought to subtract the decimal part, get the remainder on dividing by 360, and re-add the decimal part.  Edit: The difference between GST and GMST is called the equation of time (for Grinich), and it does not grow beyond 16 minutes, so I think you can use it for estimates, but then there would be no point of rounding your results to the tenth decimal or so. I honestly don't know if there is a "GST for dummies" (normally I wouldn't post an incomplete answer, but the question was edited), but using GMST should suffice if the 16 minute error margin is acceptable.


Does Pi change when gravity changes Will Pi have the same value on the surface of say Jupiter, compared to the value of PI on Earth. Or does Pi vary with gravity?

$\pi$ is just a number. It doesn't vary with anything.

The ratio of the circumference to the diameter of a circle varies with geometry, so in the general case, this definition doesn't hold. But there are other definitions, such as
$$
\pi = \int_{-1}^{1}\frac{dx}{\sqrt{1-x^2}},
$$
or
$$
\mathrm{The\,smallest\,positive\,number} \, x \, \mathrm{for\,which}\, \arccos(2x) = 0.
$$
The latter definition can also be expressed a power series, which is independent of geometry.


Did NASA/JPL get "waning" and "waxing" backwards in this video? The NASA JPL video What's Up: September 2019 Skywatching Tips from NASA says (per both the closed captions and audio):

  We’re in a several-month period right now when the new Moon falls right around the end of each month. This means we get to enjoy lovely waning crescent moons at dusk for the first few days of each month, and delightful waxing crescents in the predawn sky near the end of each month.

Are "waning" and "waxing" used correctly here or is it just the opposite of what's said?
Would the video be correct if "waning" and "waxing" were replaced by "setting" and "rising"?

I think they did make a mistake.  The moon waxes after a new moon and wanes before a new moon. To wax means to grow bigger or stronger (and is cognate with "waist"!). To wane means to become smaller or weaker (cognate with "waste" and "vaccum").

The moon that sets shortly after sunset in the evening is a waxing moon. The moon that rises shortly before sunrise in the morning is a waning moon.


Using the Sun as a Gravitational Lens Can the Sun be used as a gravitational lens to achieve better telescopic viewing? Can this effect be practically used to view celestial objects?

Yes, it would be possible. There are two roads here:

Visible light
In case of detecting light in the range of visible wavelength, perhaps you would consider that rare occasion when there is a solar eclipse.
And it may also be possible at times of 'early' dawn and 'late' dusk.

Invisible light (outside visible range, beyond the IR and the UV)
Now, other than that, we have telescopes for radio, infra-red, UV rays and X-rays which wouldn't require us to look into the 'eyepiece'. Astronomers have detectors for that and light received is stored in 'pixels'.

You might also want to look at a similar question in Physics-SE here.
As pointed out by Jeremy in a comment below and also pointed out here, it is actually a far fetched idea. You need to be really far away from the sun to use it as a G-lens. The telescope has to be around 50 billion miles away and to cover that distance within one generation, probably we haven't achieved that technologically advanced level yet.


The life course for a massive star from birth to death using the HR Diagram Could you explain to me the life course for a massive star (30-40 solar masses) from birth to death using the HR Diagram (by showing key events in its life)?

I won't primarily explain the H-R diagrams, because I think focussing on some of the underlying physics is essential for understanding, especially stellar nucleosytheses.
For simplification let's assume the star initially consists of nothing else than ordinary hydrogen and traces of carbon and nitrogen.

Nuclear fusion of hydrogen forms helium; fusion of helium forms carbon; fusion of carbon leads to heavier elements like neon, also oxygen, sodium, magnesium; neon decays to oxygen; oxygen fuses to silicon and others; silicon fuses stepwise with helium to iron. Each of these phases of burning needs higher temperature, and it releases energy. The phases can be subdivided in first core burning, then shell burning.

Earlier phases last long enough to be be visible from outside, resulting in motion within the HR diagram. Heating from phase to phase generally results in an overall expansion of the star.
The last phases last only a short time, too short to propagate effects to outside.

Fusion of iron consumes energy. Therefore the star gets a problem after the core consists of iron. Outer pressure cannot be stopped by fusion and further heating: The core collapses to a neutron star or a black hole, while the outer part is blown off by a supernova.
Neutron stars usually rotate rapidly, and cool down over very long periods of time.

After this overall framework you may go deeper into details of the phases.


Could Venus be a source of Earth's apparent overabundance of water? I've watched documentaries about the solar system, where it is suggested that Venus once had oceans of liquid water similar to those that cover most of Earth today.  Venus is now in a period of runaway greenhouse effect and has lost its oceans.

I've always wondered whether the apparent overabundance of water on Earth could be partially explained by our planet sweeping up the water, pushed into Earth's path by the solar wind, after it had been liberated from Venus.  I know one theory on where our water came from is a bombardment of comets - but I've always wondered whether some could also have arrived after Venus heated up.

How stupid is this as an idea?

edits:-

10/12/2014 - article from the bbc - Rosetta results: Comets 'did not bring water to Earth'

20/09/2017 - article from www.spacetelescope.org - Hubble discovers a unique type of object in the Solar System which includes an interesting note - "[2] Current research indicates that water came to Earth not via comets, as long thought, but via icy asteroids."

No idea is stupid per se.  But in order to place an answer to your question, please consider Venus as a comet. It once had, in that image, a coma (a tail) pointing outwards from the sun, made of evaporated water pushed away by solar wind. How frequently is the Earth exactly inside that coma?

As a comparison, when the Earth strikes a real comet coma we see meteor showers that last no more than one day. So if we consider that Earth may go through Venus imaginary coma once a year, we have an upper limit of 1/365 of Venus' water captured by Earth. This, without taking into effect that Venus had not have a real coma, that evaporation and expelletion of water from Venus was not a so directional process, and surely other factors I can not think about just now.

Besides, do not consider comets the main source of water on Earth. You need to count also with the water vapour generated on volcanoes and that generated on acid-base reactions among Earth's rocks.


Would a telescope that uses Fresnel lenses be more practical than using regular lenses? If we were to make telescopes with Fresnel lenses instead of regular lenses, would it be more practical?  Would this also allow for much bigger telescopes to be made?

Here is an example of the type of lens I am talking about from https://commons.wikimedia.org/wiki/File:Fresnel_lens-1.svg

The main advantage of a Fresnel lens is its reduced mass compared to a normal lens. Its disadvantages include difficulty of manufacture and diffraction from the discontinuities in the aperture, these I suspect are killers given the practicality of the alternatives, reflecting telescopes, to a Fresnel lens refactor.

Diffraction will give reduced contrast in the image, already noticeable in a reflector with partially obstructed aperture, but may well be worse with a Fresnel lens. But the difficulty in manufacture may well be the real limit given the relative ease of manufacturing large mirrors.

While I'm at it; it seems as thought making a achromatic Fresnel multiplet might be quite fiddly.

We might also note where there is no alternative, systems analogous to Fresnel lenses are used. The example I have in mind are grazing incidence x-ray telescopes.


If the moon wasn't receding from Earth, what would be the impact on the weather and tides? Moon is slowly receding from Earth, which means that after its formation it was much closer to the Earth than now. How would weather look like if it wasn't receding at it would be now as close as at the beginning?

Surely, the tides would be much higher, but how big would be the difference? I've also watched document stating that the hurricanes would be much stronger. Was there any simulations, how much stronger they would be?

Tidal amplitudes roughly scale inversely with distance to the fourth power (this comes from the tidal force scaling as $r^{-3}$ and the Earth's potential scaling as $r^{-1}$). At its formation, the Moon was approximately seven Earth radii away from Earth, or about ten times closer than it is today. Thus, the amplitude of the tide on Earth would have been approximately $10^4$ times larger than it is today. As the typical ocean tide is 1 m, this means ocean tides would have been 10 km high on Earth! Of course, the collision that formed the moon turned the surface of the Earth molten, so really we're talking about a 10 km tall molten tide, which is far less scary...(kidding).

Imagine all the noise the ocean makes when it crashes on a beach. That noise requires energy to produce, and is partially driven by the action of raising/lowering the ocean via tides. Now imagine how loud a 10 km wave crashing would be. This is why the Moon receded quite quickly from the Earth shortly after forming, the tides enable the dissipation of a lot of orbital energy.

As for the hypothetical effect of this supertide on the present-day Earth's weather, the Earth's troposphere (which contains pretty much all of the Earth's weather phenomena) extends 10–20 km from the ground. Considering the height of the ground is changing by a similar amount due to the tide, I would imagine the effects on the weather would be quite dramatic.


Estimating the Rate at which Moon is Moving Away from the Earth Here it has been discussed why the moon is moving away from earth at a rate of 1.5 inches per year. Is there any way we can estimate the value (e.g, 1.5 inches per year) mathematically?

This is such a complex problem that I think your chances of getting an accurate answer are almost nil.

First, the equations are likely to be highly complex if you want to be accurate down to the tenth of an inch. For example, you'd need to model the exact shape of the Earth due to the Moon's tides, most likely accurate down to the meter or so. Scientists have done this on average, but finding how these types of models minutely change on an hourly basis is near impossible. The data just doesn't exist.

Second, filling in the variables you'll have in your equations is likely going to be impossible because many won't be known. For example, do you know the flexure/elasticity/tensile strength/etc. of the entire Earth's crust at every point on Earth over time? My guess is you don't. You'd need to know that though to do really accurate calculations.

The only way we know the recession rate of the Moon is through measurements. Crude models might get you crude order of magnitude numbers, but to match the precision of observations would require models that are far more informed than we currently are.


Which constellation would appear to revolve around Denab,Vega or Thuban when axis of the earth would point toward them? We have been calling the star by looking at which we can make out the direction of the North Pole the North Star. This was the only method at the disposal of the sailors during the days when they did not have any compass with them. When the axis of the earth would point toward Denab, Denab would become the North Star. When it would point toward Vega, Vega would become the North Star. Likewise, when the axis would point toward Thuban, Thuban would become the North Star. When we look at the Polaris from the earth, the Big Bear appears to be revolving around it. So I am curious to know which constellations would appear to be revolving around the Denab, Vega and Thuban when the axis of the earth would be pointing toward them due to wobbling of the earth.

When we look at the Polaris from the earth, the Big Bear appears to be revolving around it.

Ursa Major is relatively close to Polaris in angular distance, so this motion is easier to see.  But in fact all the constellations appear to revolve around the axis.  The closer they are, the tighter the arc and the easier this motion is to see.  If they are far away and you are not at very high latitudes, they will be obscured at some point by the horizon and the motion is hidden.

So for your question, look at a starmap and the closer a constellation is to the future axis, the more it will behave like Ursa Major today.


The big bang and our expanding universe The Hubble telescope can see the light from the big bang.My question is if our universe is expanding that would make everything that's in front of our solar system be in the future? And also can Hubble see ahead of our solar system and does the bang big light ends behind our solar system?

There are a couple of misunderstandings here. Let's take one at a time:

  The Hubble telescope can see the light from the big bang.

The Hubble Space Telescope (HST) doesn't actually see light from the Big Bang. HST has several instruments on board, both for imaging and spectroscopy, but they all operate in the infrared, optical (i.e. "visible to humans"), and ultraviolet wavelength range. When you mention "light from the Bing Bang", I suppose you're thinking of the cosmic microwave background radiation (CMB), which is light emitted 380,000 after Big Bang. This is "the closest" we can get to Big Bang (yet). HST is unable to detect microwaves; instead we have other telescopes for that, the most recently-launched being Planck.

  My question is if our universe is expanding that would make everything
  that's in front of our solar system be in the future?

The expansion of the Universe is one of the most difficult concepts to get one's head around. But firstly, Big Bang wasn't an explosion, hurling matter outwards from a central point in an otherwise empty Universe. Galaxies, on average, don't travel much through space. Rather, they lie relatively still in space, but space itself expands. The distances between galaxies increase all the time, and previously, before galaxies were formed, the distance between atoms and other particles increased. An often-used analogy — which you shouldn't take too far — is a balloon with dots painted on. The dots are fixed on its surface, and when you blow it up, the distances between the dots increase despite the dots still being fixed.

Secondly, your question seems to confuse space and time. Even if the expansion were an explosion, we would travel through space, and yes, you could say that we travel through time as well, but saying that the future is "everything in front of our Solar system" doesn't really make sense (to me, at least). What's in front of you when you travel through space is just more space; you can look at it before you get there, but that doesn't mean you look into the future (in fact you look at the past, since the light you detect with your eyes has spent some time traveling toward you).

  And also can Hubble see ahead of our solar system

It depends what you mean by "ahead". It can definitely see things outside the Solar system. The Solar system consists of one star and it satellites (planets, comets, asteroids, etc.) out of the few hundred billions that make up the galaxy we call the Milky Way. Our Galaxy is itself only one out of at least some hundred billion — and quite possibly infinitely many — galaxies that float around in our Universe. And yes, HST can see many of these. But if by "ahead" you mean "into the future", then the answer is no.

  and does the bang big light ends behind our solar system?

As mentioned above, the Big Bang light — or the CMB — was emitted shortly after Big Bang. That means that it has been traveling for 13.8 billion years through space. Light travels one lightyear per year, but since the Universe is expanding, the distance that the CMB has traveled is more than 13.8 billion lightyears; in fact it's some 46 billion lighyears. This is the most distant "thing" we can see, way, way beyond the Solar system which for comparion is of the order of one lightyear large.


Why does the distance between Sun and Earth stay the same? Well, we all know Sun gravity , makes the  Space time curvature and space pushes the earth towards the Sun. But my question is "why the distance between the Sun and Earth is Same? Earth should have moved closer to Sun and obviously Earth should fall into the Sun."

Thanks,
Selva

While the Sun and Earth attract each other, they cannot fall into each other because of angular momentum conservation. In a central field (where the force is acts in the direction of the distance vector and depends on distance only), the specific angular momentum vector $\boldsymbol{L}=\boldsymbol{r}\times\boldsymbol{v}$ is conserved ($\boldsymbol{r}$ is position and $\boldsymbol{v}$ the velocity). In particular $L=|\boldsymbol{L}|=rv_t$ (with $v_t$ the tangential velocity: the component of velocity perpendicular to the direction Earth-Sun). The specific orbital energy $E=\tfrac{1}{2}|\boldsymbol{v}|^2-GM/r$ is also conserved and must be negative for a bound orbit (such as Earth's). Combining these two we have
$$
E = \tfrac{1}{2}v_r^2 + \frac{L^2}{2r^2} - \frac{GM}{r}
$$
with $v_r$ the radial component of velocity. Since $v_r^2\ge0$, but $E<0$, not all radii are reachable. In particular there are two radii at which $v_r=0$: the apo- and peri-apse of the orbit.

In fact, for the Earth, the peri- and apo- apse are quite similar, and the orbit is nearly circular. The gravitational pull of the Sun is very nearly balanced by the centrifugal force due to the rotating orbit.

In GR, the picture doesn't really change much. Non-Newtonian (GR) effects are tiny in the Solar system and completely unimportant for Earth's orbit.

Note that the picture of pushing by ripples in space time (vs. pull from Sun's gravity) doesn't make a difference either. The point is that just because you're pushed some way doesn't imply that you're falling that way. All bodies also have something called inertia which makes them want to follow their path rather being pushed around. In case of a circular orbit (similar to Earth's around the Sun), gravity and inertia balance in such a way as to create a circular orbit: inertia want's to go straight, gravity want's to pull/push you into the Sun, but the result is something between those.


When did the first annular eclipse happen? A few hundreds of millions of years ago, the moon was closer to the Earth than it is today and hence was of a bigger apparent size. This made every solar eclipse either total or partial. Over the course of time, due to tidal forces between the moon and the Earth, the moon slowly drifted far from Earth decreasing its apparent size when viewed from Earth and hence causing annular eclipses to happen.

That would mean there must have been some day in history when Earth witnessed its first ever annular eclipse. Is there an estimate on when this happened?

A quick and dirty estimate is as follows.

An angular eclipse occurs first if the Moon is at apogee and Earth at perihelion. Assuming that these two events coincide (which happens every ~8.85 years owing to lunar apsidal precession), an angular eclipse requires that ($a$ denoting orbital semi-major axis and $R$ radius)

$$
      \frac{R_\supset}{(1+e_\supset)a_\supset} < \frac{R_\odot}{(1-e_\oplus)a_\oplus}
$$

or

$$
      a_\supset > \frac{R_\supset}{R_\odot} \frac{1-e_\oplus}{1+e_\supset} a_\oplus\approx 348300\,\mathsf{km}
$$

where I assumed that the orbital eccentricities, Lunar and Solar radii, and Earth's semi-major axis didn't change (over the period considered here) and are given by
$e_\supset=0.0549$, $e_\oplus=0.0167$, $R_\supset=1737.1\,$km, $R_\odot=695510\,$km, and $a_\oplus=149.6\times10^6\,$km.

At the present we have $a_\supset=384400\,$km. If we assume an average annual change by 22 mm, we find a time of

$$1.6\times10^9\,\mathsf{years},$$

though it could have been more recent if the annual change of $a_\supset$ was larger some hundred million years ago (which may well have been the case).

In a similar way, one may also estimate when the last full Solar eclipse will occur. This time, the Moon is at perigee and Earth at apohelion, when

$$
      a_\supset < \frac{R_\supset}{R_\odot} \frac{1+e_\supset}{1-e_\oplus} a_\oplus\approx 400800\,\mathsf{km}
$$

which last occurs in $430\times10^6$ or $745\times10^6$ years, depending on whether the current lunar recession rate of 38mm/yr or the long-term average of 22mm/yr is used.


Why is the Sun's density less than the inner planets? The density of the Sun is $1410~\frac{\text{kg}}{\text{m}^{3}}$ and Mercury's is $5430~\frac{\text{kg}}{\text{m}^{3}}$, but shouldn't the Sun be denser? Because when the Solar System was forming, there was a big disk of debris, and depending on the density of the debris it went closer or further from the centre, which then formed the planets, but the Sun is in the centre, and it's less dense than Mercury, why?

The sun isn't the same density all the way through.

According to MSFC's solar interior page, the core density at the centre of the sun is a whopping 150,000 kg/m$^3$. Surrounding it the radiative zone is around 20,000 - 200 kg/m$^3$ (already less dense than water). Eventually at the edge is the convective zone - the density at the part that we see is much less dense than our own air...

So although the Sun's average density isn't very remarkable, the core is the densest place in the solar system.

(Sun cross section from Wikipedia.org)


Why do some electromagnetic waves continue travelling while others disappear? Cosmic background radiation emitted when the Universe was very young still exists. But my wifi signal seems to disappear a short distance from my apartment. Why?

Electromagnetic radiation will continue to travel until it is absorbed.  Some of your wifi signal is escaping to space where it may continue traveling for a very long time.
However, the strength of your wifi signal will degrade with distance according to the inverse square law.

So if you double the distance between your device and the wifi transmitter, your signal will be reduced by 3/4. When you move far enough away, eventually your signal will degrade enough that your device will no longer be able to decode the signal.  So the radiation doesn't "disappear" but instead, your device will just be able to pick up less of the energy.  Of course, if you separate the device and transmitter by a concrete wall, some of the signal will actually be "disappearing" as it is absorbed by the wall.
The CMB (Cosmic Microwave Background) on the other hand, was not emitted from a point source, and is not subject to the inverse square law.  Instead, as wikipedia points out, the CMB

is very nearly uniform in all directions, but the tiny residual
variations show a very specific pattern, the same as that expected of
a fairly uniformly distributed hot gas that has expanded to the
current size of the universe.

So one can imagine that wherever we are in the universe, we can point an antenna in any direction and get about the same amount of signal (excepting Doppler shift) from the CMB. Of course, if we point our antenna towards existing matter, we will expect some or all of the CMB to have been absorbed in that direction. In many cases, astronomers can tell a lot about the composition of gas in absorption based on the narrow frequencies of the CMB that was absorbed. Back in the early 2000s, I spent quite a bit of time looking for cosmic formaldehyde in absorption lines of the CMB near the center of our galaxy, as an indicator for early star formation.


What is the difference between a depleted comet and an asteroid? When a periodic comet flies by the Sun enough times, it can become depleted of its volatile elements. Once it is depleted, what distinguishes it from an asteroid, other than records of past sightings showing that it used to have a tail when it got close to the Sun?

Extinct comets are a sub-category of asteroids.
Minor solar system bodies are classed as "Comets" if they show a coma, or "Asteroids" if they don't.  A depleted comet that has never shown any coma would be categorised as an asteroid.
It is hard to determine if an asteroid is an extinct comet. Some may be distinguished from other asteroids by its spectrum. F and B type Asteroids with low albedo may be extinct comets;  comets are very dark.  Asteroids with eccentric orbits at a high inclination are also often suspected of being extinct comets.
There is another meaning of "depleted", it means that the comet (with a coma, and perhaps a tail) is low in one or more species of molecule, which suggests it formed in an atypical location, or has been modified by some process since its formation.  For example, C/2016 R2 is rich in CO and N2, but depleted in HCN. (Which suggests that it formed in a region of space at ~50K, so much closer to the sun than the outer Oort cloud.


RA expressed in degrees Using the Python package Skyfield I thought that I could get the RA expressed in degrees simply by doing ra.degrees, but I get the following message

this angle is usually expressed in hours, not degrees; if you want to use degrees anyway, then please use the attribute _degrees

I tried a few things, but cannot find a solution. Could anyone help me?

The text is trying to suggest that you do ra._degrees. How could the message be phrased better to make clear that the suggestion is to change the attribute name you are asking for from .degrees to ._degrees? Would a leading . in front of the name help?


Do changes in the orbit of the Earth contribute to global warming "Researchers have largely put to rest a long debate on the underlying mechanism that has caused periodic ice ages on Earth for the past 2.5 million years – they are ultimately linked to slight shifts in solar radiation caused by predictable changes in Earth's rotation and axis."

How do we know that the amount of solar irradiation has increased since the ice age. Do current changes in the orbit path still cause the average temperature to still be increasing until another such event causes another orbit change?

This appears to be the article.

They found in 1976 that the axial tilt and precession changes matched the geologically known ice age time periods. (see this pic for precise details)

The graphs say the world will get a tiny bit warmer for a few hundred years and then colder, if it weren't for CO2 being at the highest level for many million years:

Figure 1:  Orbital Parameters:  Eccentricity, Precession and Obliquity- click for larger image

How is it measured? It's mostly straight mathematics, say if the earth is 10% nearer to the sun it casts a larger shadow into outer space and collects more solar radiation.

This figure is developed from A.L.Berger, 1978, Long Term Variations of Daily Insolation and Quaternary Climatic Changes, Journal of the Atmospheric Sciences, volume 35 (12), 2362-2367.

M.F.Loutre and A.Berger, 2000, Future Climate Changes: Are we entering an exceptionally long interglacial?, Climatic Change 46, 61-90

The effects of galactic cosmic rays on the atmosphere (via cloud nucleation) and those due to shifts in the solar spectrum towards the ultraviolet (UV) range, at times of high solar activity, are largely unknown. The latter may produce changes in tropospheric circulation via changes in static stability resulting from the interaction of the increased UV radiation with stratospheric ozone. More research to investigate the effects of solar behaviour on climate is needed before the magnitude of solar effects on climate can be stated with certainty.


Why does this twilight sky flat field have a grid of dark pixels? The image below is a small region of a twilight flat field. A grid of darker pixels with 728x36 cell dimensions can be seen.

The camera used was the SBIG STL-6303 according to the FIT headers, which uses a Kodak KAF-6303E CCD according to the online data sheet.

The CCD dimensions, these cell dimensions, and all the binning options are all perfectly divisible, so I'm assuming this artifact is due to the structure of the CCD.

I plotted the mean pixel value of each row to make sure this wasn't an artifact from the image viewing software. The plot clearly shows a drop every 36 pixels.

So, I'm hoping for some confirmation: is this decreased sensitivity due to the structure of the CCD?

If so, what about the CCD structure actually causes this?

Yes, it's some sort of CCD structure originated from the manufacturing process. CCD chips with large area were produced by chunks of several hundred pixels. see http://heasarc.gsfc.nasa.gov/Images/astroe/about/xis_config.gif

And sometimes position offsets between these chunks can be detected.

from Wilken et al., 2010, MNRAS, 405, L16


Will new stars stop forming at some point of time? New stars keep forming in the universe thanks to all the nebulae. Now, we need Hydrogen to form stars and there would a time when all the hydrogen will get exhausted, and no more star formation will take place, theoretically.

Will there practically be such a point of time? I guess there is no place where hydrogen is replenished after star formation. Or is there a feedback cycle?

Cosmic GDP has already crashed, as Peak Star was ~11 billion years ago.

According to Sobral et al's prediction, the future star production by mass will give only 5% of the stars in the universe today, "even if we wait forever." More theoretical predictions, such as this one, suggest that nebulae will run out of hydrogen on the order of $10^{13}$ years, while star formation will occasionally happen due to collision of brown dwarfs until somewhere on the order of $10^{14}$ years.

Of course, hydrogen itself may have a finite lifetime. The half-life of a proton is experimentally known to be longer than $10^{34}$ years, but it may still be quite finite.


What does it mean when there are two plus-or-minus signs for a measurement? As far as I know, the plus-or-minus sign is used to indicate the statistical margin of error of a measurement. But what does it mean when there are two plus-or-minus signs in the same measurement?

Some examples from Wikipedia:

The mass of M87* is given as (6.5 ± 0.2stat ± 0.7sys) × 109 M☉ (as measured by the Event Horizon Telescope).
The measured distance to Sagittarius A* is given as 7860 ± 140 ± 40 pc.

The first error is due to the precision of the measurement, the second is an uncertainty associated with methodology used.


Why is M87's relativistic jet bent? The end of the relativistic jet that emerges from M87 looks curved like a hook. See here:

Data from the VLA array show more of the jet: at the far end, the jet makes a complete 180 degree turn backwards toward the galaxy it emerges from.

What forces the jet to make such a drastic turn around?

I think the most common explanation for this kind of structure is ram pressure from the galaxy's motion through the surrounding intracluster medium (ICM). In the rest frame of the galaxy and jet, the ICM is blowing past with a velocity of several hundred km/sec, and this is enough to divert the flow of the jet. In this picture, the jet isn't moving back toward the galaxy; this is a projection effect (the inner jet is thought to be pointed toward us, only 30 degrees or so away from our line-of-sight to the galaxy).

A second factor may be that the point where the jet seems to bend is where its smooth, relativistic flow breaks up due to the Kelvin-Helmholtz instability. The initial small kink visible at the right-hand edge of the HST-WFPC2 image (upper right) is probably a sign of the instability starting to disrupt the jet. Once the smooth flow of the jet breaks up, it is likely to slow down, making it more easily diverted by the ram pressure of the ICM.


What causes a star to become a pulsar? What processes does a star undergo to become a pulsar? Does it take a very specific star with a certain set of qualities such as "Just the right mass, diameter, and composition," or is it a freak accident that certain stars live out their remaining life as a pulsar?

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

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


How do we estimate the dimensions and mass distribution of a galaxy's dark matter halo, beyond the effective limits of its visible matter? Current research suggests that the Milky Way's dark matter halo may extend up to about 10 - 15 times the size of the visible galaxy, and is believed to be roughly spherically symmetrical. But I don't understand how we deduce that.
In the case of a spherically symmetrical matter distribution, such as the classic "planet with a hole drilled through it", the matter within the radius of the object acts as a single mass at the centre, and the matter outside that radius makes no net contribution to force or motion. So if we somehow stood in such a shaft drilled through the Earth, and had no idea how far above us the shaft extended, we couldn't deduce the mass of any thin shell of matter at distance D from Earth's centre, because the net contribution would cancel out whatever that mass was. So we couldn't conclude much about Earth's higher-up mass distribution, or where this became effectively zero.  (We also couldn't deduce anything from depth if we couldn't measure ambient pressure, which might help.)
By analogy, I don't understand how we can draw conclusions about the extent of matter outside a star or region's galactic orbit.
So how can we deduce the mass distribution, or effective extent/size, of a galaxy's dark matter halo, in a "thin shell" region of space far from the galaxy's centre, if there are no visible stars of that galaxy to provide evidence of the local force of gravity at distances beyond the visible galactic extent?
My present understanding, for info:
I understand in overview some of the principles about a galaxy's dark matter halo, including why we believe dark matter is present and such a large proportion of a galaxy's mass, the need for extra mass to explain how a galaxy retains cohesion/rotation, and the fact that unlike visible matter, dark matter is limited in how dense it can get (Virial theorem? of which I think I may understand the conclusion and recognise the name, but nothing more?). I also understand that given a star (or galactic region's average) motion, we can deduce how much mass lies within that orbit, but I would naively expect a symmetrical "thin shell" of matter outside that orbit to have zero net contribution? But while this seems applicable, I'm less sure.

A rotation curve cannot be measured beyond the point at which no visible matter can be seen.
Rotation velocity measurements can come from clouds of hydrogen gas, distant planetary nebulae or globular clusters and satellite galaxies. The latter are numerous and extend to about 400 kpc from the centre of the Milky Way. The exponential scale length for the Milky Way's luminosity is a few kpc, so indeed observing these objects does explore the Galactic potential, and hence the matter distribution, to many times the size of the visible matter disc.


Does this photo show the "Little Dipper" and "Big Dipper"? There are two clusters of stars that I always thought were the Big Dipper and Little Dipper. But after looking at images of the Little Dipper and Big Dipper online, I am not too sure if that's what they are. Also, I found out that the front edge of the Big Dipper is supposed to be aligned with Polaris (or the North Star), which is part of the Little Dipper, but that's not the case with these clusters of stars that I see.
So last night I took this photo:

I circled the clusters of stars that I thought were the Little Dipper and Big Dipper. I would like to know if either of these clusters are indeed the Little Dipper or Big Dipper. If either of them are not the Little Dipper or Big Dipper, what are they? And where should I look in relation to these two clusters to find the Little Dipper and Big Dipper?

As mentioned, these are the Pleiades, and the belt of Orion.  These are visible in the South at this time of year. The Big and Little Dippers are in the North, so turn around.  The best way to find them is a map:

I've marked the approximate edge of your photo, with Orion, the V of the Hyades and the small cluster called the Pleiades.  The big dipper is much bigger. It is usually quite easy to find, in the North East,  In this map it is labeled "Ursa Major" which means "great bear".
The little dipper is a rather faint constellation. Even its brightest star, Polaris, is only second magnitude.  You can find it by tracing from the pointers in Ursa Major (I've coloured them green). It is the only moderately bright star in that region of the sky. Once you have found the Pole star, it is possible to make out the faint stars that make up the rest of the Ursa Minor, the little dipper.


If the planets only attract, then how can they stay in their orbit? Gravity is related to mass, and gravity is an attracting force on every body. Then what happens when bodies only attract each other..without repulsion they may strike the sun. I think there must be a repulsive force.

Another question arises, if only attraction is there, then why does the moon rotate elliptically around the earth?

And Mars comes closer to earth after 3 years?

Gravity does not repel, but there is another "force" at work here. The angular momentum of the original protoplanetary disk is conserved. In other words, just because the cloud condensed into planets doesn't mean it is going to stop spinning.

When we orbit artificial objects, we speed them up until they reach a velocity that sustains orbit--so again there is another force at work, just not gravity. Gravity keeps objects in orbit (rather than letting them fly off into space). Imagine if you spin a yo-yo around in the air. The tension of the string is like gravity, keeping the yo-yo in orbit, but if you let go of the string then the other "force" (its momentum) will cause it to fly off.

(Technically, momentum is not a force. It is the product of the mass and velocity of the system--a conserved quantity, but the object required a force to bring it up to that velocity. In the case of a planet, it is the astronomical forces that acted on the protoplanetary disk. In the case of the yo-yo, it was your applied force to spin it.)


Have any co-orbital exoplanet pairs been discovered (and not subsequently retracted)? For this question, I think a good working definition of co-orbital configuration would be two bodies orbiting around a third much larger body in a 1:1 resonance and where neither mass is negligible. In other words, you can't really say that either one is at the other's $L_3$ because they are interacting with each other more as equals.

I found this article in the New Scientist in my first on-line search, but it has a retraction at the top.

Just for example, here is a theoretical analysis of stability in an ArXiv preprint on the subject: Dynamics of two planets in co-orbital motion.

With the caveats above (which I think are fairly standard for the term "co-orbital planet") have any been discovered?

edit: Have any new candidates been at least identified or proposed?

One example is Kepler-91b; early observations included strange light curve behavior. While follow-up observations seem to confirm that it is in fact a planet, rather than some other object, Placek et al. (2015) suggested that there is a Trojan "planet" ($M\sim.025\pm.019M_J$) at one of Kepler-91b's Lagrange points - either L4 or L5.

The authors believe that this explanation explains the light curve data much better than the single-planet model, as the unexplained dimming could be the result of a "secondary transit" by the Trojan planet, rather than a "primary transit". However, predictions of the daytime temperature of the secondary planet fall squarely about 5,000 K, which is close to the temperature of the surface of the Sun. Albedo calculations are also higher than expected (0.49).

Therefore, the authors state that the light curve may therefore be a false positive, and Kepler-91b does not exist - and nor does the hypothetical Trojan planet.

Trojan co-orbital planets have been the main focus of theoretical and experimental research so far; see Giuppone (2012) for more information.


Official Definition of Satellite? So I had an argument with a friend, which was about whether the Moon is a planet or a satellite.

IAU 2006 Resolution B5 gives definitions on what it is to be a planet, but there is a vagueness, as I perceived, is that it does not define satellites, while using it to define dwarf planets.

There is a supplementary Q&A sheet, and it uses the barycenter to define a satellite, but it is implied, not explicitly defined.

My questions follow:

Should the supplementary Q&A sheet considered as legitimate official definitions, or should it be considered as a one way of interpreting the Resolution B5?
If Q&A is legitimate, then why is the definition of the satellite is not included in the Resolution B5? If not, is the official definition of the satellite nonexistent currently?
Are there any source or archive of the official or widely accepted definitions of glossaries in the fields of astronomy?

In fact there is no official definition of a satellite.

Issac Asimov makes a good argument that the Moon is a planet since its orbit is convex around the Sun for its entire orbit, unlike the other moons of the solar system which are concave when in opposition and convex (to the Sun) when in conjunction.

Additionally, the idea that a body is a satellite of another body based on the barycenter being inside one body is flawed. Consider moving the Moon further or closer from/to the Earth. In which direction would you move it to make the two a "binary system" as opposed to a primary/moon system. In fact, though naively one would consider two closer bodies to be a binary system, moving the Moon closer puts the barycenter closer to the center of the Earth. Yet moving the moon further away moves the barycenter outside the Earth. Would you consider the Earth/Moon system as a binary system if the Moon were 30% further from the Earth?

Thus, we see that all proposed definitions for a satellite are flawed and have corner cases which make a precise definition contentious. This is the same problem as we had when trying to define a planet. No definition will nicely fit all the corner cases that Nature will give us.

Note that this answer was translated into English by my father, who is the one typing it. I'm eight years old and don't use SE unsupervised. A good portion of this information I learned just now because of this question, so thank you to the OP.


Why was Pluto's resonance not discovered until 1965? Over here, I was shocked to learn:

  The Neptune–Pluto system lies in a 3:2 orbital resonance. C.J. Cohen and E.C. Hubbard at the Naval Surface Warfare Center Dahlgren Division discovered this in 1965.

Note: there is no cite for it.

I'm pretty sure the orbit of Pluto, including it's period, was calculated many years before 1965, to say nothing of Neptune. So how is it possible that astronomers missed the 2:3 resonance? Was orbital resonance just not a conceived concept until the sixties?

This much, I can answer:

  Was orbital resonance just not a conceived concept until the sixties?

Laplace recognized orbital resonance long before then, and he died in 1827.  I looked, couldn't find a precise year of publication.

https://en.wikipedia.org/wiki/Orbital_resonance#Laplace_resonance

Now, 3/2 resonances were perhaps suggested/discovered quite a bit later, I'm not sure.

Some mathematics on the subject in the following link, a little bit over my skill-set, but it makes the point of stable resonance.

http://www.applet-magic.com/solaresonance.htm

Also, the link points out that the the first resonance discovered, Io, Europa, Ganymede the 1-2-4 resonance, have ratios of 2.0076 and 2.0143, or, 0.4% and 0.7% off resonance.

I'm tempted to say that small variations like that fall within a stable zone, but I'd basically just be guessing.   I can't say for sure.

Pluto was discovered in 1930 and Charon, it's rather large moon, not discovered till 1977.   Perhaps in the 1930s, Pluto was no longer cutting edge physics and all the big brains were looking at Quantum physics, Hubble's discovery and other galaxies, implications from Einstein's discoveries, such as black holes and Nuclear bombs . . . but again, I'm only speculating.   It does seem that a discovery of the 3/2 resonance should have come before 1965 and I think it's a good question.

Not to take this too off subject, but I love the story of how Neptune was discovered.   Short summary here:   http://coolcosmos.ipac.caltech.edu/ask/146--When-was-Neptune-discovered-

Longer one here: https://en.wikipedia.org/wiki/Discovery_of_Neptune

My favorite potential orbital resonance is the Jupiter/Mercury one, which could, eventually, pull Mercury away from the sun.

https://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Mercury.E2.80.93Jupiter_1:1_perihelion-precession_resonance

The Jupiter/Mercury resonance is described as "coincidence" in the article, where as, Pluto's 3/2 with Neptune and Jupiter's 3 inner moons are (I think), probobly not coincidental resonance but some form of stable orbits.

Hope you don't mind my amateur/hobbyist answer.


How do we know the radius of the universe is 47 billion light years? Hubble deep field image at maximum resolution shows a few of the furthest galaxies very small & faint, only just visible, but suggests that there is more to see beyond if only the resolution were high enough. I know the age of the universe is 'only' about 14 billion years, but if Hubble can only show galaxies 13.2 billion light years away, how do we know the radius of the universe is 47 billion light years?

The observable universe's edge, by definition, is the farthest part of the Universe that light could have reached us from. Let's say light from the observable universe's edge has just reached us. The longest it could have possibly travelled for would be 13.8 billion years, since the Universe is only 13.8 billion years old.

As such, one might assume that the light came from 13.8 billion light years away, and hence the observable universe's radius should be 13.8 billion light years. The problem is that this overlooks a key phenomenon: the expansion of the universe.

The location from where the light had been emitted has grown increasingly distant from us. In other words, by the time the photon has reached us and we are able to see the object emitting it, the object has moved much farther than when it emitted the photon. This is why we are able to see objects of distances far beyond 13.8 billion light years.

So to work out the actual numbers, let's suppose an object emitted a photon at the beginning of the universe $t=0$, and it just reached us today ($t_0 = 13.8 \ Gyr$). Evidently, this object lies at the edge of the observable universe, so finding its distance allows us to find the observable universe's radius.

Let's consider a given time $t$ in the past. Suppose our photon travelled for a small interval of time $dt$; it follows that the distance it just travelled away from its source would be $dx = c \ dt$. Due to the expansion of the universe, that $dx$ will expand over time; compared to today, it was a factor $a(t) = \frac{dx(t)}{dx(t_0)}$ smaller (this is called the scale factor at time $t$). Hence, the distance it gained from its source grew to be $dx(t_0) = \frac{dx(t)}{a(t)} = \frac{c \ dt}{a(t)}$ by today.

At every time $t$, it increased by an amount $dx$. So if the photon reached us at time $t_0 = 13.8 Gyr$ after it was emitted, then the total distance it gained from its source is the sum of these changes:
$$r=\int_0^{t_0} dx(t_0) = \int_0^{t_0} \frac{c}{a(t)} dt \approx 46 \ Gly$$
Which is therefore the distance of the object from us.


How much telescope do I actually need for astrophotography? I've done the 'sparkly curtains behind landscapes' form of astrophotography for years with DSLRs, lately full frame Nikon. I know what I'm doing with the camera, and want to take the next step into incompetence with telescopes to point at stars with a camera attached, but I'm getting conflicting advice on what kit I need.

One group are saying 'basic motorised Newtonian will be fine' - e.g.

https://www.harrisontelescopes.co.uk/acatalog/skywatcher-explorer-130m-telescope.html#SID=34,
https://www.firstlightoptics.com/celestron-astromaster-series/celestron-astromaster-130eq-telescope.html.

Another are saying 'that's not powerful enough, you need a cadiotropic goto like

https://www.picstop.co.uk/goto-telescopes/celestron-nexstar-evolution-6-schmidt-cassegrain-goto-telescope.html?wgu=217_60210_15449648240362_d6add8be16&wgexpiry=1547556824&source=webgains&siteid=60210.

They're obviously very different bits of kit...! And I can see both arguments TBH. Either one is trying to waste my time with inadequate kit, or another to waste my money into his pocket with overspecified kit.

Help....?

Electronic viewing is filling the area between traditional eyepiece viewing and Astro Photography. Electronic images may be viewed live, near live and with digital processing on the fly as the image is viewed or processed after the images have been collected.

The cameras have lower resolution but higher sensitivity than DSLR's and astro photo cameras. Short exposure time are the norm. Imaging/processing time is generally measured in seconds and minutes instead of hours. Stacking algorithms have been integrated into the viewing software. While tracking is still necessary the software is much more forgiving. People are viewing/imaging DSO's with Dobsonians.

I use Mallincams in my observatory. They are used with a 10" SCT, a 5" SCT and several Canon DSLR lenses.

One has the choice of going big and beautiful or small and good enough is good enough.

Take a look at the Mallincam io forum* and judge for yourself. They have a photo section with thousands of images and answers to the questions those interested might have.

* You'll need to request membership first


In an area with lots of light pollution, which clear sky clock parameter is most important? I live on the periphery of a large light dome, looking towards the West. I get about 1/3 of the sky towards mountains that is darker. According to ClearDarkSky.com I live in an area with artificial to night sky brightness ratio of about 9.00, Sky Brightness mags / arcsec2 V Band of about 19.5, and Bortle Scale of about 5.

On the Clear Sky Clock (also at ClearDarkSky.com) I see the following sky values:

Cloud Cover
Transparence
Seeing
Darkness

Which of these values is most important in a light polluted area, especially when looking west in my case to the darker sky portion?  I'm thinking it is transparency as low transparency may result in more scattering of light pollution and thus more pollution?  But perhaps moon phase beyond a certain portion matters too, especially if the moon is in the west.

I have a large Dobsonian telescope and wonder which of these is going to have the most beneficial effect when viewing near the city?

I would agree that transparency is the more important item, but it does depend on what you are observing. Deep sky objects such as clusters, nebulae, and galaxies will benefit from the darker skies by the better transparency. The Sun, Moon, planets and double stars will benefit more from better seeing than better transparency (but my guess is that you are not using a large dob for the Moon and planets.)


"Who saw" the binary neutron star merger first? What was the sequence of events? (GRB/GW170817) I'm trying to read Multi-messenger Observations of a Binary Neutron Star Merger the "cast of thousands" OPEN Access ApJ letter 848:L12 (59pp), 2017 October 20 https://doi.org/10.3847/2041-8213/aa91c9 and get a feeling for the sequence of events that took place as the gravitational waves and gamma ray burst reached Earth at about 12:41 UTC 2017-Aug-17.

There seems to be five instruments involved in the first detection and direction determination; LIGO-Hanford and LIGO-Livingston, VIRGO, Fermi-GBM, and INTEGRAL. The first three are gravitational wave detectors and the last two are gamma-ray telescopes in Earth orbit. Figure 2 of the paper (part of which is shown below) provides a densely packed infographic of the early observations. In the top left one can see an inset that spans from twelve seconds before the merger during the gravitational wave (GW) ramp in frequency, to six seconds after, where the majority of the gamma ray burst (GRB) is detected.

Somehow the combination of the GW and GRB set in motion a sequence of events that triggered a world-wide observation campaign to look for the event in all the remaining electromagnetic spectrum from radio through visible and UV to X-rays. Neutrino data streams were checked as well.

Question: I'd like to ask about the sequence of events, the alerts, and the rapid automatic and manual analysis of the GW and GRB data which triggered the alerts. Which detector or combination first "saw" the event as some kind of flagged event? Did one trigger a rapid analysis of the other? Were these automated alerts triggering software to re-analyze, or SMS text messages to thousands of cell phones triggering everyone to sit down at their work stations?

below: Figure 2 (partial) showing the timeline seconds before and hours and days after (logarithmic scale). The GW and GRB data were used (see Figure 1) to start the search for the rest of the electromagnetic search.

below: Figure 1 showing the localizations made from different sets of GW and GRB detectors.

The initial Fermi trigger can be found here, and the following sequence of alerts that were sent out by the LIGO Scientific Collaboration/Virgo Collaboration (LVC) and various electromagnetic observers following-up the event can be found in the GCN circular archive here. This doesn't quite give the whole story of the time line of events, but is a good start and the linked "Multimessenger" paper is fairly comprehensive in giving the complete story.

As shown in the figure, the signal arrived first in the gravitational wave detectors, with gamma-rays arriving at Fermi and INTEGRAL approximately 1.7 seconds after the observed merger time. However, the online analysis software on-board Fermi was quickest off the mark in detecting the burst and generating an automated trigger just 14 seconds after the signal arrived (I don't know any more of the internal details of Fermi triggering, how the alerts get sent out to people, or what subsequent manual interventions are required). Within a little over 7 minutes the online automated software searching for gravitational wave signals using compact binary coalescence templates (see, this paper and this paper) had produced a candidate using data from the LIGO Hanford detector alone (LIGO Livingston data had been automatically vetoed by the software due to the presence of a glitch [Figure 2 of this paper], and Virgo data hadn't yet propagated to the location where the analysis was run) - this automatically notified (via email or text) a number people in the LVC that something interesting had happened. Less than 10 seconds after the gravitational wave candidate had been recorded, an automated code called RAVEN (see, e.g., Section 4.1 of this paper) noted the temporal coincidence between the Fermi trigger and the gravitational wave candidate. After being notified of the gravitational wave trigger, various people within the LVC initiated a conference call and started manually looking at the data and saw the obvious chirp-like signal in time-frequency representations of the data. Roughly 33 minutes after the gravitational wave trigger, and 40 minutes after signal arrival, it was decided (people were involved) to issue an announcement (the first entry in the GCN list here) that there was a joint Fermi gamma-ray burst trigger and gravitational wave trigger.

Regarding signal arrival times at the various gravitational wave detectors: it arrived first at Virgo, followed by the LIGO Livingston detector, and finally at the LIGO Hanford detector.


How can 16 Psyche be the core of a protoplanet? The recent approval of a NASA mission to study the asteroid 16 Psyche brought to my attention that Psyche (about 200 km in diameter) is thought to be the exposed iron core left over from a protoplanet (theorized as being about 500 km in diameter). My question is: what is the mechanism by which a body of that size can accrete, differentiate as thoroughly as Psyche apparently did, and then completely lose its mantle? I have a basic understanding of planetary differentiation and I had the impression that the migration of iron and nickel to the core took a long time, and that by that time major impacts capable of shattering large bodies must have been rare.

If this hypothesis is correct, the body from which 16 Psyche formed was a 500km planetesimal (about half the diameter of Ceres) It would have been destroyed in a collision with another body. The other body would have had to have been large enough (300km) to complete fragment the asteroid. The problem with this hypothesis is that no other pieces of this planetesimal have been found (they should still have similar orbits as Psyche).

Another hypothesis is that Psyche suffered from a series of smaller impact, that fractured and disrupted the parent body, but did not destroy it. This would give Psyche a composition similar to the stony-iron mesosiderite meteorites.

The Psyche mission is intended to answer these questions.


Historical estimates of the density parameter From the reference in this answer I learned that our current estimate for the density parameter (i.e., the density of the universe divided by the critical density, which determines the shape of the universe) is 0.9993 +/- 0.0019. But I’m also interested in the history: how our estimates of this parameter have changed since the first attempts at measuring it were made.
I have tried to find this information myself, and once again I have failed. Can anyone please help me?
I don’t need a complete year-by-year account – a few historical highlights will do. (But I do need sources.)
I have also asked this question on Physics StackExchange, but there have been no responses. I suppose this is a better forum to ask.

I found an article by Wendy L. Freedman Determination of cosmological parameters which contained the following graphics

The corresponding references should be listed in the references of that paper
References

The online article is also published as Invited Review given at the Nobel Symposium, Particle Physics and the Universe, Haga Slott, Sweden, August, 1998. Published in Physica Scripta, Vol. T85, 37-46, 2000.
http://hyperphysics.phy-astr.gsu.edu/hbase/Astro/denpar.html offers a value of 1.02 +/- 0.02 and dates back to the 2000s


How to choose portable, yet powerful telescope Due to the nature of my job, I have to spend log periods of time (4 - 6 months each), at different countries across Europe, Middle East and Far East. I have a 12 Inch Dobsonian at home, but I rarely spend 1 month a year, at home. Travelling with the Dob is out of question due to its weight and I don't always have access to low lying windows or a rooftop when abroad, making a dob pretty useless. I really want to have at one, at least 8", portable option while I'm travelling. Celestron NexStar 8 SE is looking good to me at the moment. It seems to have half the weight of an 8" scope with equatorial mount and from what I've heard, it can be used table-top, without tripod. I can also carry the OTA in hand luggage, to avoid damage to optical components. Can someone confirm if

(i) Does 8SE work on tabletop without tripod?
(ii) How durable is the mount? Can it survive transportation in checked baggage, if placed in its original packaging?
(iii) Any other suggestions for a scope that is at or above 8" and can be carried around easily when travelling internationally?

I was reluctant to do an answer because I do not own a NexStar 8 SE, so I can't speak from direct experience. But this information may prove valuable, so here it is.

The smaller NexStar scopes (like the 4) can be used tabletop-style. The scope is small enough so that the base is sitting just fine on a flat surface.

The 8 might be a little too big for that - I'm not 100% sure. However, if that's the case, there's a simple remedy. Take a piece of plywood or plastic, bigger than the base, thick enough to be rigid, and improvise a way to attach it to the scope base (use bolts / nuts with the same thread as the ones on the tripod). The wider area should stabilize it, and the extra piece of material is flat and easy to carry.

At your local home improvement store they surely carry plywood conveniently cut into round pieces of various diameters. Just get one of those, get the appropriate size bolts and washers, and that should be it. Maybe put 3 pads on the bottom of the plywood so it doesn't sit on the bolt heads. Get a spray can of shellac or urethane and give it a nice protective coating when you're done.

The NexStar mount should withstand the challenges of travel pretty well. It's the optics I'm more worried about.

As others have said, most tables are too wobbly for astronomy. Use a sturdy table, or some surface that doesn't move around, like the wide edge of a concrete balcony or something.

Another option is a collapsible dobsonian. Not the pseudo-collapsible ones like the big SkyQuest series, but a true collapsible scope like the Sumerian dobs. A 16" dob built this way folds down into the volume of a regular travel suitcase. A 10" collapsible dob is like a small suitcase. These are built for the specific purpose of being taken on long trips away from home.

http://www.sumerianoptics.com/

The 8" Alkaid Sumerian scope, at 800 mm focal length, is a pretty small dob already. The poles are probably pretty short. Should be easy to transport, and doesn't require much space when fully unfolded; you may actually have to place it on a stool or something.

Obsession Telescopes also makes dobsonians of all kinds. Their Classic series are bulky when packed, but the Ultra Compact series is basically a true collapsible dob which might be worth considering:

http://www.obsessiontelescopes.com/

You don't need a table for these scopes, since they are sitting on the floor. Table wobbliness issues are eliminated. :)

There are equatorial platforms made for all these scopes, so you could even use them for photography.

I do not own Sumerian or Obsession telescopes, but the feedback I hear through the grapevine is good. These are quality instruments that could be used as a general-purpose telescope even at home.

You could also take the DIY route and build a collapsible dob yourself, if you're so inclined. Many have done this, with great results.


What irradiance does the Moon receive during a total lunar eclipse? As we know, the total solar irradiance normally received by the Moon is about 1361 $W/m^2$ (the same as the top of the Earth's atmosphere). Of course, this changes during lunar eclipses, but the Moon doesn't go completely dark even during total eclipses (it's usually just a dim red). It still receives light through the Earth's atmosphere. My question is how much?

I've tried googling this, but all the search terms I could think of just gave me a bunch of unrelated astrology results. Surely people have measured this.

The worst case irradiance for any part of the Moon during a total eclipse is fine for my purposes, but I wouldn't mind a paper with an intensity curve over time if there are any.

Edit:
This is not a duplicate of "Red Moon a Characteristic of all Total Lunar Eclipses?" That question is asking if all lunar eclipses are red and/or if the amount of redness varies. I'm asking how much light hits the Moon (per second per area) during a total lunar eclipse. Furthermore, none of the answers there even mention irradiance.

The following is more a series of observations than a full answer, but I hope other people can fill in the details.

The apparent magnitude of the full moon is -13. The apparent magnitude of the moon during a lunar eclipse will differ from eclipse to eclipse (correlated to the Danjon scale value), but let's suppose it is 0 for a certain eclipse. (Note that this magnitude is different from the (pen)umbral magnitudes mentioned on the NASA website; these magnitudes can be computed and do not depend on the Earth's atmosphere.)

This means the full Moon is normally reflecting $100^{13/5}$ times more light than during this eclipse. I'm not sure if the albedo of the Moon is different for red light than it is for normal sunlight, but if not, that means the answer is $\frac{1361}{100^{13/5}} \frac{W}{m^2} \approx 8.6 \frac{\text{m}W}{m^2}$.


Are the compact objects in supernova remnants isolated neutron stars? It seems all the central compact objects correlated with supernova remnant(SNRs) are isolated, not in binaries, right?

I find only one exception. A source is claimed to be a compact binary which correlates with SNR G308.3-1.4.ads link

My question is, generally, are the compact objects in SNR always isolated?
And observationally, are the compact objects in SNRs neutron stars?

There are plenty of hard-x-ray binaries that involve a compact object with a companion.  Some should be in supernova remnants, but I don't know how many.  Are you sure there is a problem?


What decides the shape of a galaxy? As I know, most galaxies are spiral, some elliptical, and a rare type is ring-shaped.

What causes a galaxy to become a specific shape (i.e. ring vs spiral)?

The Hubble tuning fork classifies galaxies based on whether they are elliptical or have spiral arms, and whether there is or not a bar.

A historical aside

Edwin Hubble thought that galaxies started out as ellipses and evolved into spiral galaxies. This is why ellipticals are also called early-type galaxies, and spirals are spirals are called late-type galaxies.

As we studied galaxies more, we figured out that Hubble was wrong on that point, because elliptical galaxies typically have older stars and spiral galaxies still have new stars forming in them.

Elliptical galaxies

In elliptical galaxies, the velocity dispersion of the stars is high: the stars have random orbits around the center of mass of the galaxy. This is seen to be evidence for the fact that elliptical galaxies are the result of collisions and mergers between many galaxies.

When two non-elliptical galaxies merge, their planes are rarely aligned. So when many galaxies merge together, like what happened in the early universe, you end up with stars moving in many different directions. This is how a galaxy can end up being elliptical.

Spiral galaxies

The spiral arms of a galaxy might come to be through two different mechanisms. It is possible that some spiral arms are caused by the first mechanism, and others by the second.

First mechanism: The arms of a galaxy are density waves.

Picture a traffic jam: all the cars are slowed down on a portion of the highway, but the cars that are in the jam aren't always the same ones. Each car reaches the traffic jam, slows down, gets to the end of the jam and speeds back up. This is what the stars are doing in the spiral arms: a star reaches the arm, slows down because of the higher density, gets to the end of the arm, resumes its normal speed. As a result, the arms spin slower than the individual stars.

Second mechanism: The arms appear as star formation propagates through the galaxy.

The leading edge of the arm is where star formation occurs. The trailing edge is where stars die (so are no longer visible). In this case, there is not more matter in the arms, it simply is more visible.

As stars form in one region, the largest ones cause shock waves to spread through the interstellar medium, igniting the formation of new stars. This causes the star-forming region to propagate through the galaxy. Combined with the differential rotation of the galaxy, this makes spiral arms appear.

Other shapes

You mention ring galaxies in your question. These are possibly due to one galaxy having gone right through another.

Many galaxies have bizarre shapes (e.g. the Antenna Galaxies) that are caused by interactions between galaxies.


Age of the universe and time dilation Given our knowledge and the standard cosmological model, we estimate that the age of the universe is about 13.7 billion years old.

How much sense does it make to talk about the age of the universe as a whole?

We can observe time dilation so we know time passes differently for observers moving at different speeds (special relativistic time dilation) or located at different distances from centres of strong gravitational fields (as explained by general relativity). Is the estimated age of the universe a property of the universe or of us as its observers?

Would a different observer perceive the age of the universe to be different? Do different parts of the universe have different ages? Do the principles of special relativity even apply when thinking on a universal scale?

The answer is yes time dilation does affect how much time an observer experiences since the big bang until the present (cosmological) time.

However there is a certain set of special observers called comoving observers, these are the observers to which the Universe appears isotropic to. For example we can tell the Earth is moving at about 350 km/s relative to a nearby comoving observer by measuring the anistropy in the cosmic microwave background (in fact it causes a relatively large anistropy, the pictures of the CMB you see from WMAP, etc are with this anistropy factored out).

One special property of a comoving observer in the Universe is that they maximize the age of the Universe, i.e. no other observer can experience more time since the big bang than a comoving observer (in case you ask there is no observer that minimizes the time since the big bang). When we talk about the age of the Universe we are talking about the age of the Universe from the pov of a comoving observer.


As the Universe expands, do things like planets expand also, or just the space in between? Just what the question says. As the universe expands, and space expands, do objects like planets ever expand? Does the space INSIDE of those objects expand?

Things that are bound by other forces will not be seperated by the expansion of the Universe. For instance, planets do not expand due to cosmic expansion. The solar system does not either because it is held together by gravity. However, the space between galaxies or between galaxy clusters does expand.


Why are saturn rings so thin? I wonder why saturn rings are so thin ? Rings A, B and C are approximately just 10 meters thick which is extremely thin compared to the size of Saturn. What are the explanations for that ?

Thanks a lot in advance for your answers.

Not that reddit is the most reliable of sources, but here's a Q&A with a response that is at least consistent with orbital mechanics.

  [Q} Oh. But why doesn't [the original particle set] just scatter
  around the planet, creating a big "bubble"?

  [A]Let's say there's a particle that moves around in an inclined
  orbit, so it has both a horizontal and vertical component to its
  velocity. If it collides with something, it'll either get more
  velocity, or it'll cancel out. If the vertical component is cancelled
  out, its orbit doesn't change, it just becomes less inclined. If the
  horizontal component is cancelled out, it'll fall to a different
  orbit. If you've got lots of particles going in different directions,
  their vertical components will cancel out, while those not going in
  the predominant horizontal direction will collide with others, lose
  their velocity, and fall inward. The net result will be a ring.

  [Q]That makes sense in two dimensions. But it doesn't explain why the
  rings are thin.

  [A]If the rings are thick, it means the particles still have a
  vertical component to the orbit, each being slightly inclined in a
  different way so as to make up thickness. They would have to move very
  precisely for them not to collide and cancel out.


Centre (projected) of a galaxy cluster A galaxy cluster consists of somewhere around a few tens to several hundred galaxies.

To measure the position of a galaxy cluster, i.e. it's RA and DEC, do astronomers always assume it to be its Brightest Cluster Galaxy BCG's position?

It has been shown by Johnston et al. 2007 that BCGs are not always found in the centre of a cluster/halo. If so, does the assumption of taking the position of a BCG to be the position of the galaxy cluster still hold true?

Short answer, no, they don't.

Longer answer, it's complicated. There are, in essence, five different measures for the centre of a galaxy cluster, based on different physical properties of the cluster, and occasionally not in agreement with each other.

BCG, the brightest cluster galaxy. It should sink to the bottom of the gravitational well, but that's only the case if the cluster had enough time to evolve. Of course, galaxy clusters are, cosmologically speaking, quite young objects and still in the process of forming, so this is not a given.
Mean position of the cluster galaxies. Works for spherically symmetrical clusters with lots of galaxies. Problems: Poisson noise due to limited number of galaxies, identification problems (is this galaxy really a part of the cluster), asymmetry.
X-ray: the hot intra-cluster gas emits X-ray radiation that we can observe - more and hotter gas means more and harder X-rays and if the gas is in equilibrium, the highest intensity radiation comes from the centre. Unfortunately, the gas is typically not in equilibrium, there are cooling flows, streams, cold cores and other things that make the measurement less straight-forward.
(Weak) gravitational lensing. Measures the total mass of the cluster and doesn't care about ugly gas physics. Problems: Shape noise (measurements are based on mean ellipticity of background galaxies), resolution / smoothing issues with the mass map created from averaging over lots of background galaxies.
Sunyaev-Zel'dovich effect. Again the hot cluster gas, but this time we have inverse Compton scattering of photons in the cosmic microwave background on the electrons in the gas. The problems with gas physics should apply here, too.

So, there are other methods to find the centre of a galaxy cluster. This paper suggests that ususally the weak lensing centre is offset with respect to both X-ray and BCG. I am not aware of anyone using the mean position of the galaxies as a measure for the cluster centre.

This paper evaluates the cluster centre offset between measurements in X-ray and the SZ. This is interesting because they both trace the hot intra-cluster gas.


What is it that distinguishes one atmospheric layer from another? The atmosphere of a planetary body (assuming it has an atmosphere) is described as being made up of distinct layers.

For example, Earth, Saturn and Jupiter all have a stratosphere and a troposphere.

What is it that defines where one layer stops and another starts?

What is it that defines where one layer stops and another starts?

Temperature. More specifically, it's whether temperature rises or falls with increasing altitude.

In the troposphere, temperature generally decreases with increasing altitude, at an average rate of 6.4 °C/km (the environmental lapse rate). This decrease stops at the tropopause, the boundary between the troposphere and the stratosphere. The ozone layer is in the stratosphere, making temperatures in the stratosphere increase with increasing altitude. While this boundary is a bit fuzzy, it is still very real. It takes an incredibly strong thunderstorm (think hurricanes, storms powerful enough to spawn tornados, and very tall and strong storms in the Inter-Tropical Convergence Zone) to penetrate that boundary.

Temperatures in the stratosphere stop rising at the stratopause, the very fuzzy boundary between the stratosphere and the mesosphere. Like the troposphere, temperatures in the mesosphere fall (and fall very sharply) with increasing altitude. Thermodynamics makes the boundary between the stratosphere and mesosphere very different (and not nearly as clear) as the boundary between the troposphere and the mesosphere.

The boundary between the mesosphere and the thermosphere is similar to the boundary between the troposphere and stratosphere. Temperatures rise with increasing altitude in the thermosphere. This change from falling temperatures in the mesosphere to rising temperatures in the thermosphere makes for a rather stable boundary.

Something else happens near that boundary between the mesosphere and the thermosphere. Long-lived gases are fairly well-mixed in the troposphere, stratosphere, and mesosphere thanks to turbulence. Gases in the thermosphere and exosphere act more like a bunch of individual particles rather than a gas. Unlike the dense layers below, gases in the thermosphere and exosphere are differentiated. The makeup tends toward lighter and lighter particles (e.g., helium and hydrogen) with increasing altitude. Eventually, all one finds are hydrogen and helium. These are the gases that escape from the atmosphere.


Why bother looking for Earth-like planets? Is it really necessary to look for Earth-like habitable zone planets, when it is likely that the "humans" to reach that planet will be bio-engineered or entirely mechanical by the time we would be exploring "in person?"

The possibility of exploring exoplanets is so remote it is not a factor. We are interested in Earth-like planets as there is only one type of planet that we know can support life, and that is the Earth-like ones.

We are interested in Earth-like planets as they may provide a habitat for alien life. The question of "does extra-terrestial life exist" is of intrinsic interest.


The universe is dying vs the universe is ever expanding How does the idea that the universe is dying (trumpeted as some big new revelation the past few days on several new sites I keep seeing pop up on google news) square with the idea that the universe is ever expanding?

Both are true.

The universe is "dying" in the sense that stars eventually run out of hydrogen, and there aren't infinite amounts of hydrogen in galaxies to replace old stars. This will take a very long time, but eventually all stars will burn out and new stars will stop forming. This is what the study recently in the news is saying.

The universe is ever expanding in the sense that distances between galaxies which are not bound together gravitationally is expanding all the time. If you look at very distant galaxies, they are all receding into the distance.

Combine these two facts and the result is that the universe will become more and more "rarefied" (distances between galaxies will keep expanding), and also all existing stars in galaxies will burn out over trillions of years.