How does the Grand Tack Hypothesis explain how Jupiter formed inside the frost line? The Grand Tack Hypothesis states that Jupiter formed at roughly 3.5 AU from the Sun, then interacted with a large portion of objects in the Solar System before settling into its current orbit.

Yet as I found from researching for an answer to this question, giant planets generally form beyond the frost line - in fact, this explains why the giant planets are cut off from the terrestrial planets.

How does the Grand Tack Hypothesis get around this?

A direct cribbing of the wiki's page for the frost line:

  In astronomy or planetary science, the frost line, also known as the snow line or ice line, is the particular distance in the solar nebula from the central protostar where it is cold enough for volatile compounds such as water, ammonia, methane, carbon dioxide, carbon monoxide to condense into solid ice grains. This condensation temperature depends on the volatile substance and the partial pressure of vapor in the protostar nebula. The actual temperature and distance for the snow line of water ice depend on the physical model used to calculate it:

  170 K at 2.7 AU (Hayashi, 1981)
  143 K at 3.2 AU to 150 K at 3 AU (Podolak and Zucker, 2010)
  3.1 AU (Martin and Livio, 2012)

  Occasionally, the term snow line is also used to represent the present distance at which water ice can be stable (even under direct sunlight). This current snow line distance is different from the formation snow line distance during the formation of Solar System, and approximately equals 5 AU. The reason for the difference is that during the formation of Solar System, the solar nebula was an opaque cloud where temperature were lower close to the Sun,[citation needed] and the Sun itself was less energetic. After formation, the ice got buried by infalling dust and it has remained stable a few meters below the surface. If ice within 5 AU is exposed, e.g. by a crater, then it sublimates on short timescales. However, out of direct sunlight ice can remain stable on the surface of asteroids (and the Moon) if it is located in permanently shadowed craters, where temperature may remain very low over the age of the Solar System (e.g. 30–40 K on the Moon).

  Observations of the asteroid belt, located between Mars and Jupiter, suggest that the water snow line during formation of Solar System was located within this region. The outer asteroids are icy C-class objects (e.g. Abe et al. 2000; Morbidelli et al. 2000) whereas the inner asteroid belt is largely devoid of water. This implies that when planetesimal formation occurred the snow line was located at around $R_{snow} = 2.7$ AU from the Sun.

So in all of the three listed models, as well as the snow line inferred from asteroid properties, it seems that the 3.5 AU starting distance for Jupiter in the Grand Tack hypothesis remains comfortably behind the snow line (for water, at least, which is still an advantage over the inner planets).

Since it ultimately fell well within the snow line for a time, one would expect any exposed ices to rapidly sublimate.  I would guess that it made up for this by accreting the rockier planetesimals found in the inner parts of the Solar System, which could then shield the remaining ices.

This site offers a pretty good detailing of the basic structure of the Grand Tack Hypothesis.  The most digestible piece of information is at the very bottom, in the form of an AVI file that models the first few million years of the solar system.  The short answer to your question is "Jupiter is assumed to have already formed some fundamental core at the start of the hypothesis, then moves around accreting and scattering things through the inner solar system, until returning to the outer solar system to finish growing".  The initial inward migration is caused by interactions with the relatively thick gaseous disc (the majority of it will never become part of any planet), which creates drag on the planets/planetesimals and causes them to lose velocity and fall inward.

But the video's pretty cool, so I'm going to link it and talk about it just for its own sake.

Get the video, as produced by the paper Walsh et. al., 2011

You may find it best to run this at half speed or slower to really see what's going on.

Here's how it starts out.

The solid black orbits are the gas giant orbits, with Jupiter being the innermost one, and Saturn the one after that.  The dashed circle represents a 2 AU distance from the sun (modern day Mars is currently at about 1.5 AU, and the modern asteroid belt stretches roughly 2.2-3 AU).  This is noteworthy since the model is trying to explain why Mars is so small, and why there is so little material (relatively speaking) in the asteroid belt.  The numerous small blue dots are the icy comets/planetesimals that form right around the point Jupiter starts out, and points beyond.  The numerous red dots are the rocky asteroids/planetesimals characteristic of the inner solar system.

For the first several seconds we see Jupiter creep in towards the inner solar system.  You'll notice a lot of rocky planetesimals get scattered out into the outer solar system, while some of the icy ones are scattered to the inner solar system.  Indeed, it seems to be a general feature of this model that the total amount of material within approximately 1 AU undergoes a net increase.

Saturn is also slowly moving in, but not nearly as quicky as Jupiter is (Neptune/Uranus seem pretty stable, but do "pulse" in and out a bit).

In this image we see that Jupiter hits the 2 AU mark at ~3 seconds in, corresponding to a model time of about 68,000 years.  At this point you can see a dashed circle at about the 3.5 AU distance, corresponding to Jupiter's starting distance.  Saturn is about 4 AU out now, having started at over 5.

The third through fourth second are fairly stable-looking, with Jupiter moving in to about 1.5 AU but the other outer planets staying mostly put.

But in the 5th second (~94,000 years model time)...

We see Saturn make a very rapid move inwards.  This is the main reason I suggest playing it slowly, as you can very easily miss it and lose track of where each solid black orbit has come from.  By model time 98,000 years (still in the 5th second of the video time), Saturn has already come to within about 2 AU of the Sun; Jupiter is still about 1.5 AU, and there is a thick concentration of mostly rocky planetesimals within 1 AU.

By model time of about 120,000 years, Saturn and Jupiter are rebounding back out into the outer solar system, to much the same distances we see today, scattering most of the planetesimals that remained in the areas that would become Mars and the asteroid belt.  This leaves a fairly dense disc of material for Earth, Venus, and Mercury to form and grow from, while leaving significantly less for Mars, and lesser still for the asteroid belt.

The whole 10 second movie covers a model time of 600,000 years.

Walsh, K. J., Morbidelli, A., Raymond, S. N., O’Brien, D. P., Mandell, A. M. (2011). A low mass for Mars from Jupiter’s early gas-driven migration. Nature 475, 206-209.

Walsh, K. J., Morbidelli, A., Raymond, S. N., O’Brien, D. P., Mandell, A. M. (2012). Populating the asteroid belt from two parent source regions due to the migration of giant planets – ”The Grand Tack”. Meteoritics & Planetary Science 47(12), 1941-1947.


On the lack of observational records by European astronomers of the 1054 guest star This question relates to the Crab nebula, the Crab Pulsar and the known concurrent supernovae event as recorded by Chinese astronomers in 1054.

I'm wondering, I'm looking at the history of the discovery of the Crab pulsar...and I'm wondering what the reason is (or suspected to be) as to why European records show a discernible lack of observations?

Of the three known medieval supernovae (in 1006, 1054, and 1181 AD), only SN 1006 was definitely recorded by Europeans, and even then, was only discussed in any detail in a single European source (the annals of the monastery of St. Gall in Switzerland). Although there are multiple accounts (some quite detailed) of the supernova of 1006 in Arabic/Islamic records, there is only one known Arabic record of SN 1054, and none for SN 1181.

Since SN 1006 was by far the brightest of the three, the simplest explanation would be that SN 1054 was, like SN 1181, just not bright enough or long-lasting enough to merit sufficent widespread commentary in Europe. (Remember that China in the 11th Century was a much more advanced society than Europe of the same time, with a dedicated tradition of government-supported astronomical/astrological observations.)


When was the atmosphere of Venus first observed? Nowadays, atmospheres of transiting exoplanets are being characterized. We can measure the absorption of the star's light through the exosphere of exoplanets.

The same can be done closer to Earth, with the inner planets. In fact, this was done during the 2012 Venus transit.

When is the first time that a transit was used to study the atmosphere of Venus ?

The discovery is attributed to Mikhail Lomonosov, who inferred the presence of the Venusian atmosphere from observations of the transit of Venus in 1761, although this has been challenged by Pasachoff & Sheehan (2012) who argue that Lomonosov's observations do not match the phenomena observed during the 2004 transit, arguing that Lomonosov's telescope was not capable of detecting the Venusian atmosphere. They suggest that the strongest claim for detecting the atmosphere was by David Rittenhouse who observed the 1769 transit, although they assign the greatest credit to Johann Schröter (1796) who also took into account out-of-transit observations of the ring of light around the planet at inferior conjunction.

This challenge was itself challenged by Koukarine et al. (2013) who observed the 2012 transit of Venus using antique refractors, concluding that Lomonosov's experimental procedure was sufficient to detect the atmosphere and support the attribution of the discovery to Lomonosov.

An in-depth summary can be found in a review article by Vladimir Shiltsev in the Journal of Astronomical History and Heritage (2014).


Pinhole projector for the Transit of Mercury Very quick and simple one today. What would be the best/optimal pinhole size for a pinhole projector to observe the transit of Mercury on May 9th?

I want to get the optimum between resolution and brightness...has anyone any experience in this area?

I've just rewritten this answer - @MikeG caught a glaring error by pointing out a really basic handy relationship called the Rayleigh criterion.

\begin{align} {\theta}_R \approx1.22 \frac{\lambda}{D}. \end{align}

It's better to read the (or any) article, but very briefly, the angular resolution is roughly the ratio of the wavelength to the diameter of a circular aperture. You can apply this equally well to pinhole-only imaging, or to a system which images by focusing with curved mirrors or lenses.

Mercury's diameter is about 4900 km and since it will be on the line between the sun and the earth, the distance will be about 150,000,000 minus 58,000,000 or 92,000,000km. In that case the angular width of Mercury will be about:

\begin{align} {\theta}_{merc} \approx \frac{4.9 \times 10^3\ \mathrm{km}}{9.2 \times 10^7\ \mathrm{km}} \approx 0.000052 \ \mathrm{rad} \approx 0.0031°  \approx 11 \ \mathrm{arcsec}.  \end{align}

So to even poorly resolve Mercury as a dark fuzzy dot, you'd like the diffraction width to be equal or less than the angular width. If you set the two angles equal and let $\lambda$ = 580 nm, you get

\begin{align} D \approx 1.22\ \frac{5.8 \times 10^{-7}\ \mathrm{m}}{\theta_R}  \approx \frac{7.1 \times 10^{-7}}{5.2 \times 10^{-5}}\ \mathrm{m}   \approx 14\ \mathrm{mm}\ (minimum)   \end{align}

Since the light is essentially parallel, your geometrical resolution on the screen is about the diameter of the pinhole. To make a 5.2E-05 radian object 14mm on a screen the screen would have to be VERY far away:

\begin{align} L_{to screen} \approx \frac{D}{\theta_{merc}}   \approx  \frac{0.014}{0.000052} \approx 270 \ meters! \end{align}

You can try that, use a 14 or 20mm "pinhole" at 270 meters away, but I think the light will be far to faint to see. I once did something similar to see a solar eclipse. It may have been a 10mm "pinhole" but I'm sure I wasn't that far away. I used household mirrors to bring the light indoors to a very dark area, and it worked great.  But that was at most only abour 30 meters!

If you would really like to try it, here are some tests you can do ahead of time:

Search to see if someone has done it and given details
Try some simple experiments using the full sun. If the fuzziness at the edge of the sun's image really seems to be 300 times smaller than the sun's image, then that's a suggestion that it might work.
Check the internet for daily sun images and see if you can find a sunspot. Then test your setup and see if the sunspot suggests your resolution looks as good as 1/300.


Space curvature and its rotation As I learn, the gravity is caused by the spacetime curvature. Is there any gravitational/magnetic field that really exists around the Earth, or any other planets?

Since the planets/stars rotate, does this cause the space time curvature to twist in the direction of their rotation? If it does then the gravitational pull should also twist the the path of the objects falling(though we may not perceive the curvature of this path, we should be
able to calculate at-least theoretically)

Unlike gravitation, which is entirely a result of space-time curvature, magnetic fields do exist and there is a magnetic field around the Earth and many other planets.

(There is a deeper level of understanding of electromagnetism called Quantum Electrodynamics which explains how fields work in terms of more fundamental concepts, but unlike gravity, the E-M field does exist. (And there is hope -- so far unrealized -- that eventually all of the fields of nature plus gravity will be explained as aspects of one yet deeper thing.  But not today.))

Frame dragging -- the distortion of space-time by a rotating body -- does exist and the link suggested by @PM (en.wikipedia.org/wiki/Frame-dragging) is a good place to start.  Frame dragging has been measured by careful observation of satellite orbits, and possibly by astronomical observations.  The Gravity Probe B satellite measure a closely related effect.  As far as I know, the effect is too small to have been measured in the lab.

Frame dragging does indeed affect the path of a falling body, though for the Earth the effect is really, really small.  It would be large near a millisecond pulsar or a rapidly rotating black hole.  It's also large for merging black holes and is an essential part of the calculations of the gravitational waves emitted during the final rotations before merger.

The effect can be precisely calculated and agrees perfectly with the (still rather crude, alas!) experiments and observations.


Bad seeing on a clear night? I want to say first off, I am a complete beginner, starting out my interest in astronomy (so go easy!).

I have been able to see some nice sights on some nights, however, I had a strange experience the other night when viewing Saturn.

I had been viewing it the night before, using my highest magnification eyepiece and got some really nice views. It can be noted my telescope has a focal length of 1000mm and aperture of 100mm and I was using a 5mm eyepiece, which was really pushing the limits of the useable magnification, as far as I am aware.

So anyway, the next night had clear skies again, if anything maybe clearer. I was excited to see Saturn again, but this time once I had found it, I put in my 5mm eyepiece and just couldn't get it into the same focus I had the previous night. I googled it and somewhere said that although the skies can be good, the seeing can be poor.

What is meant by this statement and how am I able to tell if there is going to be a night with poor seeing in advance so I know which eyepieces I can use for that particular session? The app I use to check the skies only tells me if it is going to be clear so I don;t know what to be looking out for.

Seeing refers to turbulence in the atmosphere that distorts wavefronts passing through and results in the blurring of distant objects.

Unfortunately, there is no foolproof way of predicting what the seeing will be from night to night, though building up a database of local seeing and meteorological measurements might allow you to find some crude correlations using principal component analysis.


Can Summer Triangle asterism be used for navigation? I am wondering, whether is it possible to use vega-altair-deneb asterism for navigational purpose? Or if there is any geometric relationship of it with Polaris?

Any idea.
Thanks

Yes, it can be used, but it is not easy.

For the latitude, you usually need to know the Alt of Polaris. You can derive that if you know the Alt of two of the three, but the (spherical geometry) formulas for that are not simple.

For the longitude, you need Greenwich time and the RA of any star. Any star includes Vega, Altair and Deneb ;)


Sunrise and Moonrise time are same I checked an app and saw that today, the sunrise and moonrise time are almost same. The sunset and moonset time are also same.
Why does this happen?
I have always known that moon rises at night and sets at dawn.

That's called New Moon and happens roughly once a month.
It's when (viewed from earth) the moon, circling the earth, is in the same direction as the sun. So, by rotation of the earth, both become visible (sunrise/moonrise) and invisible (sunset/moonset) at roughly the same time.
Moon is then hardly visible for two reasons:

The back side of the moon is lit by the sun, the side facing to us does not get sunlight.
The moon is visibly rather close to the sun, so sunlight will blind the human eye when looking in that direction.

Two weeks later, we have Full Moon, the moon is opposite to the sun, it rises when the sun goes down and goes down when the sun rises, meaning that it's visible at night. And the side facing us gets full sunlight, so the moon becomes a very bright object on night sky.


What visible differences do Geminids have to other meteors Unlike all other meteor showers, which originate from comets, Geminids originate from asteroid 3200 Phaethon. Seeing a large number of meteors last night, I noticed they seemed to be much more green than other meteors, but is there anything else I should expect to see different to other meteor showers? Brightness, size, length of trail etc?

Well done on your observations, I've been completely clouded out.

You are quite right, the Geminids are noted for a variety of colours, and a good number of fireballs, though most observers note a predominiance of yellow meteors This is partly due to their composition and density (coming from a peculiar "rock comet") and partly to their velocity (35km/s) which is fairly fast.

There is a good chance of seeing some good trails with this shower. Brighter meteors and most fireballs will leave a trail, and the Geminids have more of these than most other showers.


Does gravity propagate? Gravity is sometimes described as a curvature in space-time. Due to relativity, doesn't this imply that gravity doesn't propagate?

A thought experiment. If you are moving toward a black hole, this could also be considered as if the black hole is moving toward you. If a black hole was moving toward you at the speed of light, and gravity propagates at the speed of light, you would experience something similar to a supersonic blast, except its a gravity blast, and it would allow you to remain unaffected until you arrived at the very centre of the black hole?

If you were able to turn around, just before you arrive at the center, and travel away from the black hole at the speed of light you would never notice it, even though you would be travelling at the same velocity as the black hole in the same direction?

Black holes distort the geometry of spacetime. So you need to take care about that.

Changes of gravity propagate with the speed of light. That's thought to occur for accelerated mass, e.g. a binary of black holes.

A second fundamental point: Speeds don't add up in an additive way, but in a subadditive way: speed of light + speed of light = speed of light.

If you or the black hole travel with the speed of light, you resp. the black hole needs to be massless at rest, otherwise your relativistic mass, or that of the black hole would be infinite.

Photons are massless at rest, so you're actually talking of light, or something which behaves rather similar to light in many aspects.

So at the end you're talking of light directed to a black hole and reflected by a mirror inside the black hole.

The light can get inside the black hole, but it cannot leave the hole after being reflected from the mirror due to the high curvature of spacetime below the event horizon.

If you are travelling towards a binary of black holes, you get blue-shifted changes of gravity.


Does the Giant Impact Hypothesis explain how the Moon circularized its orbit? I just read most of Wikipedia's article on the Giant Impact Hypothesis. Basically a large object impacts Earth and creates debris that soon coalesces into the Moon.

But there is something I did not find mentioned: How would the Moon get its circular orbit?

A debris field from any impact would at first follow a very elliptical orbit, with a perigee very close to the surface of the Earth and an apogee much farther away. In orbital mechanics, where you start is where you end up. In fact it doesn't even have to be where you start. Pick any point in the orbit and you will return there again. In other words, the original elliptical orbit doesn't change its shape.

...Unless there are pertubations or impacts that change your velocity. How exactly did that happen? Does the GIH just assume something else did that fortuitously?

I've also seen youtube video simulations like this. It's hard to tell if the camera angle is changing or if the orbital plane is unstable (how?), but you can clearly see the debris goes around and impacts the surface again and again. I.e., the orbit remains highly elliptical. I've never seen a simulation that shows the debris circularizing.

I also want to note that this question is highly relavent not just to the Moon, but many other things like orbital capture as well. I've never understood orbital capture theories as they all seem to leave out how the initial elliptical orbit later becomes circular, e.g., Triton.

I want to point out that the Moon's orbit isn't circular now.   A 0.055 mean eccentricity isn't that circular.

But, onto your question.   I think you're making a bad assumption on the "must have started very elliptic.   Individual objects that are ejected from a planet need to follow their orbital path.   So any object flung from the earth would need likely either escape the Earth or fall back into it because you can't launch something, with a singular push, into a circular orbit.   All the individual bits of debris should have had highly eccentric orbits.

But the formation of the Moon was much more complicated than that.   First, it was enough matter to have it's own gravitational field, basically influencing itself, and the forming moon would be the combination of all those individual objects, so it wouldn't necessarily be eliptic at all.   The eccentricity of billions of objects could have (and likely did) cancel each other out to a large degree.

Because the impact was off-center, it set the Earth spinning and most of the debris moved in the same direction around the Earth.   But the coalescing mass of material that didn't fall back to the Earth could have been in a reasonably circular orbit at formation.

We should also define what number constitutes "highly" elliptic.   The Moon is thought to have started out 3-5 Earth Radii distant.   Any material that passed closer than 3 radii would have been inside the Roche limit and had a hard time coalescing into the Moon.    This article suggests that the debris would have had a hard time being ejected further than 5 Earth Radii, though I'm not sure how that conclusion was reached.  But you have a few things going on.  Debris colliding with other debris, and some (presumably quite a bit) of debris falling back to the Earth and, mentioned above, the debris having a gravitational effect on itself.   Ultimately momentum needs to be conserved and calculating the formation of the Moon requires a super-computer, but in general I think the initial orbit could have been relatively circular.

If the Moon formed close to the Earth, which most models suggest it did and since it couldn't pass inside 3 Earth Radii and stay solid, those numbers put a limit on how eccentric the initial orbit probably was.   If we use the 5/3 Earth Radii as an estimate, Ra=5, Rp=3, then the eccentricity is 0.25, and that's an estimate for the upper limit of the initial eccentricity after formation.  It may well have been quite a bit less.

But for the sake of argument, lets give the newly formed moon an eccentricity of .25 or, maybe a little higher with a more distant apogee, but we have to keep the perigee at 3 Earth radii or greater.   Your question still stands, how did the Moon's initial eccentricity become somewhat circular.

The answer is tidal circularization.   I can't find a good article on it, but for many 2 body systems, tides circularize orbits.   It's well known that the tides on Earth, caused by the Moon's gravity rotate ahead of the Moon and this creates a tug on the moon that pushes (pulls?) the Moon into a higher orbit, further away from the Earth.   This same secondary effect circularizes the Moon's orbit because the closer the Moon is too the Earth, the greater the push.  The math behind this gets very complicated and it's above my paygrade to put numbers behind it and the Wikipedia article I've linked is very lacking, so I invite anyone to provide more details if they can. But Tidal circularization is a commonly accepted theory and given the strength of the tides early on, it's likely any eccentricity in the Moon's orbit circularized relatively quickly, at least, astronomically speaking.   (Tens of millions of years is a long time to you and me, but not to a stable orbit).


Why are the time zones calculated as 360°/24 and not 361°/24 or 360°/23.933? Background: I'm training to be a geography teacher. Currently I have practice lessons and I'll be discussing solar time and standard time with the class. Now I stumbled over an issue to which I could not find an answer:
We teach the students:

Sidereal day: In 23 h 56 min the earth rotates 360°
Solar day: In 24 h the earth rotates 361°

We also teach that in order to construct time zones (as a replacement of solar time), one divided 360° by 24 (to have a zone for each hour of the day), which results in 24 zones with 15° each.
Now my question is: Why does one mix the measures for sidereal day and solar day? Or put differently: Why doesn't one calculate 361°/24 or 360° / 23.933?

The Earth takes 23 hours 56 minutes to rotate once.  But that is not relevant to most people. Sure, the stars will be in the same position again after 23 hours 56 minutes, but the sun will not be in the same position.
It is far more important, for most people, to measure the time from noon to noon. And the average time from noon to noon is 24 hours. This is because the motion of the sun is a combination of both the spinning of the Earth and the orbit of the Earth around the sun. The orbital motion of the Earth adds four minutes. You should also teach the students

In twenty-four hours the sun advances 360 degrees. (solar day)

Time zones are based on clock time, which is based on the motion of the sun and not the motion of the stars.


How does 2015 RR245 compare to the other KBOs predicting "Planet 9"? 2015 RR245 was recently announced, in a highly eccentric orbit in the KBO. From what I understand, the so called "Planet 9" was predicted because these objects tended to be pointed in a particular direction. How is this object pointed relative to the other KBOs?

It was the first thing I thought too, but I would say it won't count in the statistics.  The earlier paper by Batygin & Brown was talking about bodies orbiting further out.

2015 RR245 orbits between 33.7 and 129 AU.  The "cluster" of bodies considered in the paper have aphelion distances from 150 AU upwards.  So I think it would be a mistake to try and fit this new body into the existing figures.


How to calculate the stellar visibility? Please could you let me know a (basic) formula for calculating at what time a star would become visible of an evening in clear weather conditions.  I presume the formula would involve the stars apparent magnitude, the location of the observer, and how far the sun has dipped below the horizon (so, the sky's limiting magnitude).  Working with these three parameters, I guess there must be a simple enough formula, however I cannot find it!  Could you help with this?  For example, assuming clear weather and good eyesight, how could I calculate at what time the star Deneb will become visible on the 1st May
2019 for a rural location?
I am asking this as I want to create an art piece where something happens each time a new star (or stars) becomes visible around twilight time....

Maybe that piece shouldn't consider necessarily individual stars rising from the East but entire constellations/parts of constellations. Trying to express the sky movement as a whole sounds really nice to me. There could be planets setting (like Mercury or Venus right after dusk in the West) or others rising (quite bright) Jupiter, Saturn Mars. Even large deep sky objects portrayed there could be amazing - like the Orion Nebula, Andromeda Galaxy, The Double Cluster, Pleiads, M44 Cluster, meteors, etc. I found a good expression of items of the night sky here.


Are there a lot of collisions between stars in the core of the galaxy? I'm reading on Wikipedia about Halo Stars that orbit the center of the galaxy at a high inclination away from the plane of the Milky Way.  It seems that at some point, these stars must dive back into the core of our galaxy at very high relative velocities.  This causes me to wonder...wouldn't this, combined with the greater density of stars in the core, cause a large number of stellar collisions in the galactic core?

The answer is yes. Stellar collisions are events that occur mainly in Globular clusters due to their high stellar density. The likelihood of these events is highly dependent on the stellar density. As we know the galactic centers concentrate large amounts of matter and, of course, thousads of stars which are mainly in the old red main sequence. However, recent observations hane shown as the existence of young stellar clusters located in the Galactic center of Milky in contrast with our models and predictions about the galactic structures. It is currently believed that in these clusters occur a lot of stellar collisions. For more information there is a related paper here.


What would be the obstacles to building a 6.5km space telescope besides cost? In the comments of another question on this Stack Exchange site, I talk with another user about the possibility of building an optical space telescope with a 6.5 km aperture.

Obviously, this is much larger than any rocket we have constructed, so it would need to be launched from Earth in a number of components and assembled in orbit in a similarly massive number of spacewalks, and the sheer scale of it would likely require active heat management systems to prevent heat expansion of the components from sunlight warping the mirrors or stretching the distance between the mirrors.

Are there any other obstacles that would prevent this project from being possible?

The basic issue with a large optical telescope is to keep all of the mirror in an extremely stable position relative to the rest of the mirror. Depending on the wavelength of your light you probably want to limit deviations from your perfect shape to less than about 100nm, so, you need to account for and deal with somehow any force which might act on one part of your mirror differently from another with enough force to bend the whole mirror by more than this tiny amount. At a glance, therefore you will need to deal with:

tidal forces from Earth, Moon and Sun (I think you can ignore tides from Venus and Jupiter, but I'm not certain)
Orbital perturbations arising because Earth is not spherical
temperature variations from moving in and out of Earth's shadow at different times and from sunlight and Earthlight (and maybe moonlight) hitting different parts of the structure differently
interactions of any conducting components of the mirror with Earth's magnetic field
flexing and oscillations of the structure arising from thrusts used to point the mirror

and probably a lot more that I haven't thought of.  These problems would become somewhat simpler if you put the telescope much further from the Earth, or even better much further from the Sun (say a very distant orbit around Neptune) but that would make assembling it by spacewalking impossible.


Where does Jupiter's gravitational force come from? Why don't Jupiters gasses fly away? Gravitational force is the attractive force between any two masses.

What about gaseous planets, where do they get their attractive force from?

While the big rocky planets don't have enough attractive force to keep enough gases in their gravitational field, Jupiter has lots of gases.

Does this means there is a concentrated mass inside, under which gravitational field the gases are kept 'in'? And if so, then why are gaseous planets not also considered to be rocky ones?

if they don't have any solid mass inside them is an incorrect conclusion.

Some have 'ordinary' solid cores, and the pressures in the centers of gas giants are so high, that they are even 'suspected' of compressing hydrogen into a 'metallic' state. Quoting from the Planetary core Wikipedia page:

Gas giants also have cores, though the composition of these are still a matter of debate and range in possible composition from traditional stony/iron, to ice or to fluid metallic hydrogen.

For Jupiter especially:

Jupiter has an observed magnetic field generated within its core, indicating some metallic substance is present. [... ]
Jupiter has a rock and/or ice core ten-thirty times the mass of the Earth [...]
Thermal contraction/evolution models support the presence of metallic hydrogen within the core in large abundances (greater than Saturn).

The earth's core has an average density of about 12,000 kg per cubic meter, Jupiter's core density is estimated at 25,000 kg per cubic meter (link). Together with its much larger radius than the earth (70000 km versus 6370 km), you will have a pretty massive core with a mass of from 12 to 45 times the Earth's mass, so there's plenty of gravitational pull there.

More to read: The outer planets: Giant planets: Interiors


Do all moons orbit their planets on the ecliptic? Do all moons orbit their planets on the ecliptic? Do they all follow this same process as the moon that orbits Earth, or is their some random distribution with some even having extreme polar orbits.

Of course not.

See Neptune's moons. Triton's orbit has a tilt of 157º from Neptune's rotation plane, which is in itself tilted 30º from Neptune's orbit, just 2º away from the Ecliptic.

Or look at http://en.wikipedia.org/wiki/Moons_of_Uranus


Strength of gravity during the big bang? I was watching a documentary on the big bang, one of the astronomers said that initially the four fundamental forces were combined as one, they then emerged to become: The strong and weak nuclear forces, the electromagnetic force and the gravitational force. He claimed that the first force to emerge was the gravitational force, he went on to say that had the gravitational force been weaker than it is everything would fly apart so no galaxies would form and if gravity was too large than we would end up with black holes everywhere, so the force of gravity needed to be just right, as it is now. I couldn't find anything backing up this information online or the reasoning behind this information, any help would be appreciated.

The truth is we do not have a working or widely accepted "theory of everything" that unifies gravitation with the other fundamental, and quantum, field theories. What we do have is strong evidence that the other field theories - electromagnetic, strong nuclear and weak nuclear - are unified at high energies (indeed the evidence is so strong that I doubt it is widely disputed at all).

We do have a theory of gravitation - general relativity - that appears to work well in "most" circumstances, though it also has a singularity (black holes) and there is a fundamental difference between it and the quantum theories about vacuum energy (the so-called vacuum catastrophe).

What we also know is that gravitation appears to much weaker than the other forces, though it can act over very large distances and so is fundamental in shaping the nature of our universe.

There are various arguments about how/why our universe is able to support life/is the way it is. Perhaps it is just one of many/an infinite number of universes that have a range of physical properties and we just happen to be in one that "works" for life - obviously there could be no life in a universe that had physical properties that didn't support life.

At another extreme perhaps we are all inside a computer simulation.

I'd suggest hunting down works by Max Tegmark or Brian Greene for more information.


What's the distance between Venus and Jupiter? What is the minimum, maximum and average distance of the Venus and Jupiter?

I only found 670,130,000 km and 670,198,000 km as an average distance (here and here).

The furthest distance Venus is from the Sun is 0.73 AU

The furthest distance Jupiter is from the Sun is 5.46 AU

The closest distance Jupiter is fro the Sun is 4.95 AU

So if we assume the orbits are concentric circles (they're not !) then :

Minimum distance = 4.95-0.73 = 4.22 AU
Maximum distance = 5.46+0.73 = 6.19 AU

And a simple average of these gets you an average distance of about 5.2 AU.

The problems with that simple estimate are :

The orbits are not circles, but ellipses
The orbits are at an inclination
A more detailed calculation for circular concentric orbits produces a tricky double integral with no easy solution
Over time (the only way to do a true average) the orbits would change from even "simple" ellipses.

A number of online sources quote the average distance as 4.48 AU.  This appears to be nothing more than the semi-major axis of Venus's orbit subtracted from the semi-major axis of Jupiter, which is probably an under-estimate of an average separation, as it neglects that they can be at opposite sides of the Sun sometimes and the same side at others.

Minor update :

For those interested in seeing the effect of a more detailed calculation I tried treating the orbits as concentric circles and doing the full integral on WolframAlpha.com .

The required integral in this case is (units are AU) :

$$L = \frac {5.2044}{4\pi^2} \int_0^{2\pi}\int_0^{2\pi} \left[ ( sin(v)-0.139 cos(u) )^2 + ( cos(v)-0.139 cos(v) )^2 \right]^{1/2}du\, dv$$

where $0.139$ is the ratio of Venus's orbital radius to Jupiter's and $5.2044$ AU is the mean radius of Jupiter's orbit.

And the result of this tedious integral $L=5.2293$ AU, which is not exactly much of an improvement on the much simpler $5.2$ AU given earlier.  But it at least dispels then notion on some online pages that it could be $4.48$ AU.

Last (?) minor update :

The above calculation just did a geometric average over the two orbit shapes, but made no attempt to consider the motion.  Well it occurred to me to revise that and allow for two bodies in a circular motion in the same plane.  Guess what ?  Same answer !

Well, almost.  In this case it's $5.2297$, which is an insignificant difference as we're already ignoring inclination and using circular orbits.  I'm inclined to put down the difference to a numerical calculation margin of error (the integrals do not have simple closed forms and would easily propagate errors, I think).

The integral boils to an average over the synodic period and is a simpler single integral which reduces to :

$$L = \sqrt{a^2+b^2} \frac 1 {2\pi} \int_0^{2\pi} \sqrt{1-2\frac {ab}{a^2+b^2} cos\theta}\, d\theta$$

I do prefer this result myself as it's taking account of orbital motion and over time, rather than purely geometric.


Magnification with a refractor Using a refractor telescope at 200x, how close will an object appear at four miles away? Is there a table out there enabling you to calculate apparent seeing by filling in known variables?

This is an interesting question!
Let's first ask; What does "200x" mean exactly?
If we set a photocopy machine to 1.5x or 2x we get a paper copy that we can set next to the original and compare them directly. With a ruler we can confirm that something that's 1 inch on the original is 1.5 inches or 2.0 inches on the copy.
But how big or how much closer something looks is a more subjective because our vision system is not flat. Our vision works with angles rather than distances.
For small angles like several degrees or so, we can use the small angle approximation fairly well, so let's go with that.
When a telescope is focused far away, near infinity, perhaps 100 times farther away than the length of the telescope itself, we can treat 200x as the angular amplification of 200.
Something in the distance (e.g. 4 miles away) that's 0.01 degrees wide will appear 200 x 0.01 = 2 degrees wide through the telescope.
If we move that object 200 times closer, it will also appear almost exactly 2 degrees wide.
So as long as we are considering things far away and small, we can say that something that appears 200x larger through the telescope would appear to be the same size as if it were 200x closer.
4 miles times 5280 = 21,120 feet. Divide that by 200 and we get 105.6 feet.
At 21,120 feet I would appear about 0.015 degrees tall. Viewed through a 200x telescope I would appear about 3 degrees tall. If I moved to 105.6 feet away, I would also appear about 3 degrees tall.
This works as long as we use the small angle approximation and our telescope is focused on things far away, say 100 times the length of the telescope or farther. The same 200x telescope would not necessarily provide exactly 200x angular amplification if they eyepiece were pulled out so far that one could focus on something across the room.


Rotation of far away and near by galaxies Do the far galaxies rotate the same as does the milky way and the near by galaxies?

It depends a bit on what you means by "far away" and "the same", but:
Galaxy formation
Galaxies form from collapsing and colliding clouds of gas and dark matter in the early Universe. The first structures began to form a few hundred million years after the Big Bang, with masses of the order of $10^5$ Solar masses (e.g. Mo et al. 2010). As time went, progressively larger structures built up in a hierarchical manner from smaller clumps, with more galaxies forming as time went, until galaxy formation peaked around 2–3 billion years after the Big Bang (e.g. Thomas et al. 2017).
Because the Universe expands, everything was closer in the early Universe. Hence, early galaxies formed under constant harassment by other galaxies coming from all sides. This made it difficult for galaxies to settle into an ordered, rotating disk. Indeed, some galaxies in these early epochs had all their gas blown out by merging with other galaxies and feedback from supernovae and quasar activity. Without gas it is difficult to settle into a disk, and to form new stars. These galaxies are seen today as elliptical galaxies without an overall rotation (e.g. De Lucia et al. 2006).
When did rotating disks form?
Galaxies that survived such a bombardement eventually tended to form a rotating disk. So when did galaxies start to rotate? This question is still not entirely resolved. Galaxies don't just suddenly start rotating, but rather progress from being dispersion-dominated (i.e. stars and gas follow randomly oriented orbits) to being rotation-dominated.
Previously, it was thought that (most) galaxy formation took place in a "hot mode", where accreting gas from the intergalactic medium is shock-heated to high temperatures ($T\sim10^6\,\mathrm{K}$), eventually "drizzling" down on the central galaxy (e.g. Rees & Ostriker 1977; Fall & Efstathiou). Consequently, disks would form quite late in the history of the Universe (at redshifts $z\lesssim1$, some 6 billion years after the Big Bang).
However, partly due to numerical simulations it was eventually realized that galaxies could accrete gas through cold, narrow streams of gas which reach the central galaxy faster, and is cold enough to form stars (e.g. Dekel & Birnboim 2006; Dekel et al. 2009). These streams also carry angular momentum which aid the rotation.
Simulations now routinely see rotating disks forming already ~1 billion year after the Big Bang (e.g. Grand et al. 2017; Pillepich et al. 2019). Observationally, disks at these high redshifts are detected (primarily I think) through carbon and carbon monoxide with large radio telescopes (e.g. ALMA), and rotation-dominated disks have been detected out to redshifts of at least $z=4.55$, corresponding to 1.3 billion years after the Big Bang (Fujimoto et al. (2020). The distance to this galaxy is 25 billion lightyears. I think the "record" for seeing rotation is Smit et al. (2018) at redshift $z=6.8$, seen only 800 million years after the Big Bang, at a distance of 28 billion lightyears, though its rotation is perhaps a bit up to interpretation.
A very recent (yet non-refereed) study by Kretschmer et al. (2021) find galaxies 1.8 billion years after the Big Bang to be rotation-dominated for several orbital periods, before being disrupted by mergers and counter-rotating streams.
Although these galaxies are not as evolved as the Milky Way, some of them already show not only rotation, but also sign of spiral structure. So I'd say that the answer to your question is "Yes, for the last 90% of the history of the Universe, some galaxies do, although most early galaxies don't."


Is there an alternative (potentially easier to understand) Euler diagram of the new solar system body classification? note: This question is asking for a representation of the astronomical classification scheme now used for solar system bodies as established (presumably) by the Working Group for Planetary System Nomenclature (WGPSN) of the International Astronomical Union or IAU. While graphical layout is mentioned as part of the question's motivation, the question is not about layout or graphics.

The Euler diagram shown in this helpful answer is hard for me to understand, I think mostly due to graphical layout. The diagram shows the mathematical relationship between various sets (in this case solar system body classifications) so it could be drawn with widely differing choices of layout and still retain exactly the same mathematical relationships.

I am wondering if there is some attempt here to indicate relative numbers or ratios by the degree of overlap? Look closely at the small overlap of Minor Planets with Comets. Is that meant to reflect in some way a relatively small overlap quantitatively? Did you notice it took a while to even make sure you were seeing the overlap correctly? This style of layout requires the use of color perception to help identify which label is associated with which contour while simultaneously trying to follow one line in a groups of perfectly parallel lines.

Have other versions of this Euler diagram representation of solar system body classification been published? A link to one or even inclusion of the diagram in the answer itself would be greatly appreciated!

above: Example of an Euler diagram in this Wikipedia article, from here.

above: Example of a somewhat easier to use and understand Euler diagram, from here, used in the same Wikipedia article.

Per the IAU (Resolution B5, adopted by the 2006 IAU General Assembly), there are five supposedly mutually exclusive categories of things in the solar system:

The Sun (only one), which is distinguished by the ability to fuse hydrogen into helium in its core.
The planets (eight, might possibly become nine if Michael Brown is correct), which are distinguished by having the ability to "clear their orbit" (a well-defined concept with numerical meaning) before the Sun turns into a red giant.
The dwarf planets (unknown number, most likely well over 100), which are distinguished as being bodies that don't qualify as the sun or as planets, but are sufficiently massive to more or less have pulled themselves into a roundish shape.
Small solar system bodies (unknown number, well into the millions) which are distinguished by not being distinguishable as any of the other categories.
The satellites of planets, dwarf planets, and small solar system bodies (unknown number, the number of which change over time as satellites are captured or rejected) which are distinguished by being in orbit about a planet / dwarf planet / SSSB rather than the Sun.

Contrary to that resolution, these boundaries are fuzzy rather than sharp, particularly when one looks outside of our solar system. (Note well: The IAU definitions pertain to our solar system only. There are currently eight planets in the entire universe, with the possibility of growing to nine if the conjectures regarding a largish planet beyond Pluto turn out to be correct.)

Consider the first boundary, that which separates planets from stars. This becomes fuzzy with the smallest red dwarf and the largest super-Jupiter. There's no clear-cut boundary here. In fact, there's a good amount of debate over this very issue. This boundary is quite sharp in our solar system, but it almost certainly is anything but sharp in star systems beyond our own.

The second boundary, the one between planets and dwarf planets (which, despite the name, are not planets) also has the potential for being fuzzy. In our solar system, that boundary is once again rather sharp. Any time you can come up with a boundary that is separated by five orders of magnitude you have a nice sharp classification boundary. However, given the hugeness of the universe, there is almost certainly some star system out there, somewhere, where this boundary is as fuzzy as can be.

The third boundary, the one between dwarf planets and small solar system bodies, is fuzzy in our own solar system. Objects less than 100 meters in diameter are, as a rule, anything but "round". Objects over a kilometer in diameter are, as a rule, quite "round" (when the shape is resolvable). Astronomers have found lots and lots of objects between 200 and 600 meters in diameter. Squint hard enough at the smaller ones and you might be able to convince yourself that they are "roundish". Focus intensely on the larger ones and you might be able to convince yourself that they are not quite round. There's no sharp boundary here, even in our own solar system.

The fifth category, satellites vs non-satellites, opens yet another can of worms. Planets (particularly the giant planets) can temporarily capture objects as satellites, only to lose them as satellites after a short time (tens or even hundreds of thousands of years is a "short time" in this context).

Note well: I'm not saying Pluto should once again be promoted to the status of "planethood". That very sharp boundary (at least in our solar system) says that it most definitely is not a "planet".

But what about asteroids versus comets, which is what started this series of questions? That boundary is ridiculously fuzzy. Fuzzy logic (a real and serious concept) doesn't come close to addressing how "fuzzy" this is. Centaurs and formerly active comets are the boundary cases. There is no boundary. There instead is a soft melding from one to another.


Why is the equinoxes only 2 days and not more? The earth is tilted at 23.5 degrees which means that in the solstices the north/south hemisphere is more radiated by the sun and the tilt to the sun is maximum.

In the equinoxes, the tilt is not towards the sun but to the "side" so the equator has direct exposure to the sun. As a result, we have 12 hours on day and night on the equinoxes.

What I don't understand is why is it only for these 2 days (equinoxes) that this happens. In the next day(s) I think the equator should have the same exposure to the sun as the equinoxes. Has the tilt changed so much?
I couldn't find any animation that shows the exact path of the tilt throughout the year.
Can someone please help me understand this?

First, don't think that equinox is when the day and night are the same. It is the moment, when the sun's declination is zero, or when the sun is directly over the equator. Let me explain ...
Earth is always rotating around the Sun, so declination of the Sun is always changing (except at the solstices when it stops for a minute and goes in the other way). As @JohnHolst has pointed out, Sun's declination is changing for 1 angular minute per hour (mark 1 mm on the 3 m of distance – this is nearly 1 angular minute). An equinox doesn't last for a day. It lasts for a moment when the Earth is totally aligned and straight with the sun:
(this figure shows the geocentric model, which is easier to represent and imagine)

Also, Wikipedia says:

An equinox is commonly regarded as the instant of time when the plane (extended indefinitely in all directions) of Earth's equator passes through the center of the Sun. This occurs twice each year, around 20 March and 23 September. In other words, it is the moment at which the center of the visible Sun is directly above the equator.

P. S.
You can look this simulation.


Why is time considered a (fourth) dimension? Since dimensions usually refer to space and we naturally acknowledge three of them, we also perceive time and we separate space from time ("space AND time"), why is it, that time has ended up in the same shelf as spatial dimensions. Is time really a dimension "by nature"? Or is it a concesus, shifting our natural understanding what dimensions are, for practical reasons, similarly to defining particles?

EDIT: by Is time really a dimension "by nature"? I asked if there are good intrinsic reasons, given by to label time as a dimension, as opposed to us labeling it as such, to make it more tangible, explainable.

(The speed of light) times (time) has the properties of a linear dimension.  Just like length, width, or height, it can be used to measure how far you can go.  The Lorentz contraction is consistent with applying the Pythagorean theorem to it and the other dimensions.


Based off of the study that found the universe has a positive curvature, how big would the whole universe be? This study found that the universe has a positive curvature https://www.nature.com/articles/s41550-019-0906-9. I didn't really want to buy it and see if it says how big the universe might be. I know it didn't have a low enough p-value to be considered confirmed. Is there a big margin of error for the whole size of the universe?

Surprisingly small!

(To me, at least.)

The paper referred can be found on the arXiv as Di Valentino et al. (2019).

As is customary with Planck results, the exact values of the cosmological parameters depend on how much confidence you put in auxiliary data, such as baryonic acoustic oscillation data (from surveys such as the 6dF Galaxy Survey, SDSS, and BOSS), supernova data, and Big Bang nucleosynthesis models. In this answer, I'm going to assume that the inferred value of the curvature parameter is given by their "$\Lambda\mathrm{CDM}$$+$$\Omega_K\!$" model, with their 99% confidence limits of $-0.007>\Omega_K>-0.095$, i.e. I'll use
$$
\Omega_K = -0.0438^{+0.0368}_{-0.0512}.
$$
The dynamics of the Universe is given by the Friedmann equations, which can be rearranged like this:
$$
\Omega_K = -\frac{kc^2}{R_0^2\,a(t)^2 H(t)^2},
$$
where $k=+1$ for a closed universe, $a\equiv1$ today, $H=H_0$ is the Hubble constant today, and $R_0$ is the radius of curvature, and $c$ is the speed of light. The value of $H_0$ depends somewhat on the assumed value of $\Omega_K$; here I'll use $H_0 \sim 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$.

I this case, we get that the radius of the Universe is roughly
$$
R_0 \sim 67_{-21}^{+100}\,\mathrm{billion\,light\text{-}years}.\qquad (99\%\,\mathrm{C.L.})
$$

Accepting the "mainstream" cosmological parameter the observable Universe has a radius of $46.3\,\mathrm{Glyr}$. This result will change somewhat in the case of a closed Universe, but I can't seem to identify the author's preferred set of cosmological parameters. If we just use this value nonetheless, it means that we are currently able to observe a volume fraction of
$$
f = \frac{V_\mathrm{obs}}{V_\mathrm{tot}} \simeq \left(\frac{46.3}{67_{-21}^{+100}}\right)^3 \sim 33_{-31}^{+67}\%
$$
of the total Universe, i.e. we see somewhere between a few percent and everything, but most likely "a third".


Is it possible to use Photolithography for telescope image sensor? The current largest digital CCD camera is the Vera C. Rubin Observatory, which has a whopping 3.2 gigapixels with a pixel size of 10 micrometers. In contrast, Agfa 10E56 holographic film has a resolution of over 4,000 lines/mm—equivalent to a pixel size of 0.125 micrometers.
It makes me think. Will it be possible to use a chemical image sensor to achieve a nanometer pixel size?
Between 1983 and 2006 the Digitized Sky Survey(DSS2) produced 1.1-gigabyte images (23,040 x 23,040) of 15 micrometers corresponding to 1.0 arcseconds using Photographic-plates. One of their challenges was storage space. They were forced to use a lossy adaptive algorithm to shrink the output size, during which data was lost.
But since 2018, we can produce 7 nanometers chips via the Photolithography process.
Can the same technology be used for image sensors? What can we do in 2021?

The pixel size is not a major constraint on the imaging capabilities of an astronomical telescope. The angular resolution of the image is limited by diffraction at the aperture of the telescope. For a reasonable pixel size, the telescope geometry can then be designed so that each pixel corresponds to a region of sky smaller than that angular resolution, so that making the pixels smaller would gain you very little.
Responding to a comment. A diffraction limited telescope with an aperture of $d$  observing at a wavelength of $\lambda$ has (at best) an angular resolution of $\lambda/d$ radians. So if it has an effective focal length of $l$, the smallest useful pixel size will be around $l\lambda/d$. The effective focal length depends on the whole optical setup and will usually be designed to make this size big enough.


How accurately can you tell the time using a photo of the stars? If I were to go out on a clear night, in an area with low light pollution, and take a photo of the stars, could an expert analyze the photo and figure out what date the photo is from? We can assume a powerful camera, but not reaching up to the strength of a telescope.

This question stems from a larger project. It would be very convenient for me if it were possible to take a photo of the stars to tell the time. Even more convenient if you could narrow it down to within 10 minutes, but if that's not possible I'm wondering how close we can get. Can you tell the year? The millennium? Do you need to know which location on Earth the photo was taken from? Does it help if you already know the year in which the photo was taken?

You can assume access to modern computing, existing databases, state-of-the-art software, etc. (no supercomputers)

Thanks!

To make a clock you need something that changes predictably with time, and stars are nearly constant, making them poor clocks.

The stars do move across the sky each night, but since the morning stars of winter become the evening stars of  spring, and the sky looks different in different locations you need to know the date of the image, the location of the photo, and the exact direction of the photo. That could give sub 10 min accuracy. Unfortunately while date and location may often be recorded with a photo,  exact direction is not, unless the horizon is visible in the image.

Without that information, and it is possible to 'get lucky'. For example, if there are two or more planets in the shot, their locations could be measured to give at least the date of the photo. Similarly if a rare event like a nova is visible that can offer dating information. If the shot happened to catch a bright meteor, and the date was known well enough, you could get almost 1 second accuracy. Variable stars might also give the chance of at least dating the image.

With no planets you can use the very slow proper motion of the stars. With a good enough telescopic shot of the night sky you can be quite accurate, but with a wide view of a camera you wont be able to get enough accuracy to give any useful date.

The exposure time of a typical camera is to short to capture any stars. If you just point and shoot at tye sky, you just get black. To capture images of stars  you would need a timed exposure.  A telescope would help too.

So if you photographer wanted to create an image which could be timed accurately, and they had time to prepare for it, or it was a lucky shot this might be possible, otherwise not.


How do we give directions in space? If I'm on a life boat, in the middle of the ocean, radioing another ship to come pick me up - I'd say something like "I'm 40 kilometers directly east of you."

In space, it's not that easy. How do you give a direction?

You need to define an agreed standard for orientation of axis and origin.

What co-ordinate system you choose as a standard depends on your purpose.

Here are some examples and links : Celestial coordinate systems.

Typical reference points are defined using very distant objects whose position can be measured accurately ( typically quasars ).


Can there be a 2nd possibility deduced from the fact that galaxies are moving away from us? I just read that according to the BBT the universe is expanding. This is evident from the fact that distant stars (and galaxies) are moving away from us. But it could also happen that the universe is in fact shrinking to a very small point and we being far away from the point are experiencing its pull less than those that are near it. It may be like a race where a very fast (in 1st place) , a moderately fast (in 2nd place) and a slow car (at last) are racing . To the moderately fast car it would seem that the other two are moving away from it,but actually they all are moving towards the same point -the end. So it may happen that earth is like the moderately fast car moving towards the collapse point at a different relative speed. Is this proposition possible? If not,why?

Galaxies are moving away from us in every direction.

If it were like your race, galaxies at our same distance from the central point will be slowly approximating. That is, we will note that galaxies nearer to the center than us will move away from us (since they are faster), galaxies in the opposite side of the sky will move away from us (since they are slower), and galaxies in the plane that lies between these two point will be slowly approximating to us (since we converge with them to the same point at the same speed).

So the answer is no, that is not a possibility.


When I read that a system is located x light years away, why is there no direction? Well, I just read an article about a galaxy with two black holes, and it stated that the galaxy is 1 billion light-years away. Why is there no direction given? If the Big Bang happened, and things are moving away from each other, we can still have at least two directions - away from the center, and to it. Is that a right way to think about space and our drift in space?

The galaxy's catalog name, SDSS J1126+2944, also gives its direction. This means the galaxy is at approximately 11h26m right ascension and 29 degrees 44 minutes north declination.

The "full name" of the galaxy, given here:

http://www.sci-news.com/astronomy/pair-black-holes-distant-galaxy-03546.html

is "SDSS J112659.54+294442.8", which gives its position even more precisely. A finder chart is here:

http://simbad.u-strasbg.fr/simbad/sim-id?Ident=SDSS+J112659.54%2B294442.8


How do disk-shaped galaxies then form at the sites of spherical galactic halos? Do the halo stars start orbiting or just sit in place? Phys.org's Astronomers detect large amounts of oxygen in ancient star's atmosphere mentions

"Stars like J0815+4729 are referred to as halo stars," explained UC San Diego astrophysicist Adam Burgasser, a co-author of the study. "This is due to their roughly spherical distribution around the Milky Way, as opposed to the more familiar flat disk of younger stars that include the Sun."

Wikipedia's Formation of galactic halos says

The formation of stellar halos occurs naturally in a cold dark matter model of the universe in which the evolution of systems such as halos occurs from the bottom-up, meaning the large scale structure of galaxies is formed starting with small objects. Halos, which are composed of both baryonic and dark matter, form by merging with each other. Evidence suggests that the formation of galactic halos may also be due to the effects of increased gravity and the presence of primordial black holes. The gas from halo mergers goes toward the formation of the central galactic components, while stars and dark matter remain in the galactic halo.
On the other hand, the halo of the Milky Way Galaxy is thought to derive from the Gaia Sausage.

"Halos... form by merging with each other" isn't profoundly instructive, but I'm guessing it means that there are small inhomogeneities and the coalesce into larger ones without a lot of rotation being involved (thus the spherical shapes). I suppose the part about the Gaia Sausage means that the Milky way is not a good example of a disk-shaped galaxy forming at a spherical halo.
Question(s):

But how do disk-shaped galaxies then form at the sites of the spherical halos?
Do stars in the halo just sit there without rotating, or perhaps slowly accelerate towards the gravitational center?

The key to understanding this is that dark matter and, effectively, stars are collisionless, whereas gas is collisional.

Structure formation

As written in the second quote, structure forms in a hierarchical, bottom-up fashion. That is, small halos of gas and dark matter (DM) collapse first on small scales, and these halos then merge to form larger halos. This is in contrast to what was previously thought, namely that galaxies formed in a so-called monolithic collapse from a huge, primordial cloud (Eggen, Lynden-Bell & Sandage 1962).

On the very smallest scales, thermal motion will tend to wash out structure and prevent collapse. The first structures to collapse are thought to have masses of some $10^5\,M_\odot$, but the exact value depends on your assumed initial conditions. During the collapse, these clouds then fragment to form stars.

Collisionless matter

Because DM is collisionless, it is difficult for it to "relax", i.e. settle down into a dense, "virialized" structure: A DM particle falling into the combined potential of gas and DM will tend to go right through through the center. However, because the potential isn't static, but contracts, the particle doesn't go as far to the other side, and eventually a DM halo will relax (thanks to Peter Erwin for making me aware that I underestimated this effect).

The same is effectively true for stars: Of course, stars can smash together, but the chance of two stars getting close enough for this to happen is minuscule because they're so far apart.

Collisional matter

The story is different for gas: The collapse is dominated by DM (because there's much more), but when gas gets dense enough, hydroforces become significant. The atoms will collide, electrons are excited, de-excite, and emit photons, removing energy from the system. In other words, gas can cool.

Formation of the disk

This causes the gas to basically smash together in the bottom of the potential well, and eventually settle down into the center of the halo. Although there are ways to shed itself of some of the initial angular momentum, it will usually maintain enough to form a rotationally supported disk, with centrifugal forces impeding the collapse in the plane of the rotation.

The halo

The first population of stars are formed before the disk has settled, and so have random motions around in the potential. The massive stars quickly die, leaving the longer-living, dispersion-dominated (as opposed to rotation-dominated) population of relatively low-mass, metal-poor stars, known as Population II stars.

That is, the stars in the halo do not just "sit there", but orbit the galactic center like the disk stars. However, the orbits are quite unordered.


How does dark SMBH evaporate? I read that Dragonfly44 is a galaxy mostly consists of dark matter, and it should contain a super massive black hole too. If there really is a dark SMBH, how does it lose mass? I think there is no way for dark matter to be converted into energy and vice versa, no?

The "no hair" theory of black holes states that a black hole formed from the collapse of normal matter would be exactly the same as a black hole formed from dark matter. You could even form a black hole from photons, if you could get enough of them close together at the same time.

What originally collapsed to form the black hole has no relevance to whether Hawking radiation is given off. A black hole formed of dark matter WIMPS would behave exactly the same as a black hole formed from normal matter.

Dark matter can be converted into "energy" (I'll assume that you mean light, ie photons), in the same way that normal matter can: for example by antimatter annihilation.

The actual formation of black holes, however, requires them to be mostly made of normal matter, since the matter has to lose energy as it falls together, and to do this it must interact with itself (essentially this is friction). As dark matter does not seem to interact with itself or with normal matter, it would not form a major part of a black hole, even in a galaxy dominated by an excess of dark matter. The supermassive black hole may be exceptionally small in such a galaxy, as there is less suitable matter to fall into it.


If a captured moon has prograde, circular and equatorial orbit, can we know it is captured? As we know, Triton has retrograde orbit , which is an stong evidence it is captured by Neptune. Suppose if a moon is captured, but results in prograde orbit, and the orbit becomes circular and lies on equator of the planet after billions of years, can we still know(or find) it is captured?

If a satellite has a prograde, circular(ish) orbit, it will certainly be much more difficult to determine if said satellite has been captured or not. However, there are other indications we could look for in determining if that body was captured or formed in place.
Composition
As a general rule, the composition of any given body in a stellar system is a function of its distance from the star. The reason for this has to do with the temperature of the primordial disk as a function of distance from the Sun. Early on, as all the objects in our solar system were forming, the began coalescing out of the stellar disk surrounding our Sun. However, in order for objects to solidify out of this disk, they had to be cool enough to do so. Very close to the Sun, it was of course extremely hot and molecules like water or methane could not precipitate into solids. Only materials like iron and silicate could do so. Much father out, where it is colder, things like iron and silicate can still solidify, but now its cold enough that you can pack on ice and methane. This process is evident by the fact that Mercury, our closest planet to the Sun, has a core of almost complete Iron which makes up 42% of the planet's volume (compared to the Earth's 17% by volume Iron core). A graph of this concept is shown below.

What this means for your question is that you can look at the composition of a given satellite and determine if that composition matches what we'd expect for an object forming at that location. If for example, you see a satellite around Saturn which has a lot of Iron and very little ice and methane, you can likely presume it formed closer in to the Sun and was captured by Saturn (how this might happen, I don't know but its just a simple example).
In fact, the Moon's nearly identical composition to the Earth is a strong indicator that our Moon formed in place (and from debris from Earth) rather than being captured.
Now of course measuring composition is no easy feat. You can determine something like that by analyzing any emissions the planet produces (such as with the water plumes from Europa) or else by flying a satellite very close to the body and measuring gravitational effects.
Cratering
This is a less reliable method, but can be telling. The number and distribution of craters on a body tells us a lot about its history. If we assume that the surface of a given satellite is not constantly changing (through some geological process) then the craters on the surface represent a record of past interactions with that satellite. One approach I can imagine is to statistically predict the number of craters that should exist on the body given that it formed at its current position, then compare to what we see. If the surface hasn't been changed for billions of years, then a difference in the expected cratering indicates the body hasn't formed in the way we thought it might and possibly formed elsewhere or under different conditions. I admit though that this method is much less telling and more of a guess than using composition.


Michigan's Pole Star Unfortunately, I'm not any good at figuring these things out myself so hope for help here. If the (true) north pole were over the Keweenaw Peninsula of Northern Michigan, what would the North Star be; a star, similar to Polaris that never sets? Or, what star would be directly overhead? I couldn't even pick out Venus or Mars in the night sky when I was younger and now, for personal reasons, I'm unable to go out and look. Any help would be incredibly appreciated. Thank you

It would depend on what time the Earth started to spin around Michigan. There is no way to answer this without saying what time, time of year, and which century (or at least millennium) it happens.

Edit: Hi Dayna. Thanks for the the time of year the Keweenaw Peninsula becomes the North Pole. One could be omniscient except for that and not know this. But we also still haven't ruled out a single star since your OP. With the time though we could tell you exactly what star. The decade is not important. What century or part of history it happens, yes but not the decade. (yes, astronomy is weird).


How to convert density from physical unit to comoving unit? I am learning co-moving units and I am interested to see how the conversion works. If Hubble constant is assumed to be 0.7, then I want to convert $1gm/cm^3$ into comoving unit. If I want to express length in co-moving kilo-parsec per hubble constant (${\rm ckpc}/{\rm h}$), instead of cm, then how does the conversion happen?

You're asking two different, but somewhat related questions: One has to do with a practical way of describing the expansion of the Universe; the other has to do with a way of dealing with our ignorance of the exact expansion rate of the Universe.
I will answer your specific question in the last paragraph.
Comoving and physical coordinates
A comoving coordinate system is one in which particles that don't move through space have fixed coordinates, even is that space is expanding (or contracting, or otherwise warping).
In contrast, the particles' physical coordinates are the actual distances you would measure between them if you froze space and started laying out measuring rods.
Expansion, scale factor, and redshift
Cosmological distances are often measured in megaparsec (Mpc), i.e. one million parsec, or roughly 3.3 million lightyears.
Our Universe expands, and we describe this expansion not by its size (because we don't know how large it is, or even if its finite), but by a scale factor $a$, which is defined to be equal to $1$ today. Thus, in the past, when everything was half the distance from each other compared to today, we had $a=0.5$. Because light moving through an expanding space is redshifted, light from a certain time when $a$ had a certain value, is redshifted ("stretched") by a certain factor. It turns out that if we observe some light to have been redshifted by a particular value of $z$, it must have been emitted at a time when the scale factor was $a = 1/(1+z)$.
Definition of comoving coordinates
Comoving coordinates are defined to coincide with the physical coordinates today. That is, if the physical distance to galaxy "CLASH 2882" is 3.4 Gpc, so is the comoving distance, always was, and always will be (save for a very small movement through space, but this can be neglected). The light we receive from CLASH 2882 is redshifted to twice its emitted wavelength, so $z=1$, and $a=1/(1+1)=0.5$. At earlier times, when $a$ was, say, $0.75$, $0.5$, and $0.1$, the physical distance to CLASH 2882 was 2.5 Gpc, 1.7 Gpc, and 340 Mpc, but its comoving distance was always 3.3 Gpc.
In other word, a comoving meter is only equal to a real, physical meter today, but was smaller in the past and will be larger in the future. To be explicit, we sometimes prepend a unit with a "c" or a "p", like this:
$$
1\,\mathrm{pMpc} \equiv \frac{1\,\mathrm{cMpc}}{1+z}.
$$
The use of comoving coordinates
Comoving coordinates are practical for instance when comparing properties of something that may or may not evolve with time, but also change with the size of space, e.g. densities. One example is the number density of galaxies. If galaxies had just been around forever, without evolving, then their number density would decrease with time proportionally to $1/a^3$. To study how galaxies form, evolve, merge, etc. it is helpful to factor this out, i.e. to use comoving coordinates. If the number density in comoving coordinates change, it must be due to some "real", astrophysical effect, not just because of cosmological expansion.
Another example is the density of neutral hydrogen, in $\mathrm{g}\,\mathrm{cm}^{-1}$. In physical units, if nothing happened this density would decrease as $1/a^3$, whereas in comoving units it would be constant. Nevertheless, when $a$ was about $0.1$ the global, comoving density dropped several orders of magnitudes, evident of a physically significant process referred to as the Epoch of Reionization.
Converting between physical and comoving units
So, if you want to convert between physical and comoving units, just remember that the comoving coordinates factor out the expansion of the Universe, and you have, for distances,
$$
d_\mathrm{physical} = a d_\mathrm{comoving} = \frac{d_\mathrm{comoving}}{1+z},
$$
and consequently, for densities,
$$
n_\mathrm{physical} = a^{-3} n_\mathrm{comoving} = n_\mathrm{comoving}(1+z)^3,
$$
Little $h$
The little $h$ commonly seen in length units (and units derived hereof) is a way of factoring out the exact magnitude of the expansion rate $H_0$ of the Universe. It is defined by
$$
H_0 = 100h\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}.
$$
That is, if $H_0=70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, then $h=0.7$, and if $H_0=67.81\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, then $h=0.6781$. The use of $h$ is something of a reminiscence from the times when we only knew that $H_0$ was of the order of $50$–$100\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, and factoring out the exact value allowed people assuming different values to more easily compare their results. It may be argued — and has in fact been (Croton 2013) — that today, where $H_0$ is known rather precisely to be around $70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, $h$ only acts to confuse. It does, however, seem to prevail, especially in numerical, cosmological simulations.
$h$ notation
The way you express your question, namely that you

want to express length in co-moving kilo-parsec per hubble constant

shows, in my opinion, a common misinterpretation of the use of $h$: The factor $h$ is not a part of the unit of some number; rather it should just be thought of as a number that is multiplied on the quantity. So, although the result is the same, I think it's better to write, say,
$$
d = 2300h^{-1}\,\mathrm{Mpc}
$$
than
$$
d = 2300\,\mathrm{Mpc}/h.
$$
$h$ in quantities
$h$ appears whenever a derived quantity depends on $H_0$. Because some quantities can be derived in different ways, you may actually encounter the same quantity multiplied by different powers of $h$, depending on how it was derived. For instance, deriving a galaxy's mass from its luminosity introduces two factors of $H_0$, so if its "true" (stellar) mass was, say, $M_\star=10^{10}\,M_\odot$,
then if you had assumed $h=0.7$ you'd write
$M_\star=4.9\times10^9h^{-2}\,M_\odot$.
But if the same galaxy's mass was derived from dynamics, you'd only have one factor of $H_0$, so you'd write
$M_\star=7\times10^9h^{-1}\,M_\odot$.
Converting to "$h$-units"
In general, to convert a number to "$h$-units" (bearing in mind that $h$ is not a part of the unit), you need to know how it was calculated. Derived distances, for example, usually scale inversely with $H_0$ (to the power of one), so it makes sense to measure in $h^{-1}\,\mathrm{Mpc}$, that is to divide the quantity by $h$. Thus, assuming $67.8\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ we may write the distance to CLASH 2882 as
$$
\begin{array}{rcl}
d & = & 3.3\,\mathrm{Gpc}\\
  & = & 2.3h^{-1}\,\mathrm{Gpc}.
\end{array}
$$
On the other hand, since derived times scale proportionally to $H_0$, they may be measured, e.g., in $h\,\mathrm{years}$ and thus, in velocities these two $h$ factors cancel, such that velocities are just measured in, say, $\mathrm{km}\,\mathrm{s}^{-1}$.
The answer to your question
We are now ready to answer your specific question of converting from cm to comoving kpc with $h$ factored out.
You need to know two things: The assumed value of $H_0$, and the redshift of the object you're considering. For the sake of this calculation, let's assume that $h=0.7$ and $z=0.1$, and that we want to write the diameter of a galaxy whose physical diameter is, say, $10^{23}\,\mathrm{cm}$ (a Milky Way-sized galaxy). Then
$$
\begin{array}{rcl}
R_\mathrm{phys.} & = & 10^{23}\,\mathrm{cm}\\
                 & = & \frac{3\times10^{-22}\,\mathrm{kpc}}{\mathrm{cm}} 10^{23}\,\mathrm{cm}\\
                 & = & 32.4\,\mathrm{kpc}\\
                 & = & 32.4 \times h \, \times \, h^{-1}\,\mathrm{kpc}\\
                 & = & 22.7 h^{-1}\,\mathrm{kpc}.
\end{array}
$$
and
$$
\begin{array}{rcl}
R_\mathrm{com.} & = & (1+z) \, R_\mathrm{phys.}\\
                & = & (1+0.1) \, 22.7\, h^{-1}\,\mathrm{kpc}\\
                & = & 25 h^{-1}\,\mathrm{kpc}.
\end{array}
$$
Similarly, a density of $n_\mathrm{phys.}=1\,\mathrm{g}\,\mathrm{cm}^{-3}$ could be expressed as$^\dagger$ $n_\mathrm{com.}=3.9 h^3\,\mathrm{g}\,\mathrm{cm}^{-3}$.

$^\dagger$ Note however, that if instead of particles per cubic centimeter you were to measure Solar masses per cubic Mpc, then depending on how you had measured the mass, there could be up to two factors of $h$ canceling some of the $h^{-1}$ in this expression.


Composition and Proton Flux from the Solar Wind I am taking a class on the space environment and I have a few questions about the solar wind.  A number of sources list the composition as (approximately) 4% He$^{++}$ and 96% H$^{+}$.  What exactly does this mean?  Does it mean by mass or by number of ions?

Also, when discussing proton flux due to the solar wind at Earth:  What exactly is included in the flux numbers quoted (e.g., one source I have says 2e12 m$^{-2}$s$^{-1}$).  Is that the flux due to only the H$^{+}$ and excluding the He$^{++}$ and other minor constituents?  Or, are they including all the protons from the other ions as well?

That's by number-- you get a lot more helium than that by mass.  Also, when they talk about proton flux, they just count the bare protons, not protons embedded in larger ions like helium.  Those are the kinds of things that should be made clear, but end up being common assumed conventions.


At what rate is the observable universe expanding? The observable universe is $4.4 \times 10^{26} m $ in every direction. I told this to my daughter and then added, "And guess what? It's a little bigger today than it was yesterday."

"Wow!"

But by how much? My first thought would be: one light-day, in each direction. But that can't be right. The boundary is receding faster than the speed of light.

Yesterday, a lonely photon ended its 13.8-billion-year journey with the collapse of its wave function (or whatever, let's not start on this) on a rock orbiting a smallish star toward the edge of a spiral galaxy in a relatively sparse region of the Universe.

Today, yet another photon met a similar fate.

How much further away was the 2nd photon when it began its trek?

How fast does the observable Universe grow?

The observable Universe is $r = 46.3\,\mathrm{Glyr}$ (billion light-years) in radius, so by Hubble's law, galaxies at that distance recede from us at a speed
$$
v_\mathrm{rec} = H_0\,r = 9.6\times10^5\,\mathrm{km}\,\mathrm{s}^{-1},
$$
or $3.2$ times the speed of light, $c$.

At the same time, light from ever-increasing distances will have had the time to reach us. Since light travels at the speed of light, this adds another factor to the edge of the observable Universe. That is, its distance recedes at $v_\mathrm{rec}+c = 4.2c$.

Hence, today the radius of the observable Universe is
$$
\begin{array}{rcl}
dr & = & (v_\mathrm{rec} + c)\,dt\\
   & \simeq & 4.2\,\mathrm{light\text{-}days}\\
   & \simeq & 10^{11}\,\mathrm{km}\\
   & \simeq & 730\,\mathrm{AU}
\end{array}
$$
bigger than it was yesterday ($dt=1\,\mathrm{day}$).

But remember that most of this growth does not mean that our observable Universe encompasses more galaxies; only that they're farther apart.

How far apart were two photons emitted, if they arrive one day apart?

In reality, we don't see photons from the Big Bang. The oldest photons we can detect are from the cosmic microwave background, but humans don't see those. But for the sake of the argument, consider two photons, both emitted right after cosmic inflation, but at slightly different distances from the place you will once stand.

Taking inflation to end when the Universe was $t\sim10^{-32}\,\mathrm{s}$ old, and the Universe has expanded by a factor of $a^{-1} \simeq 2.3\times10^{26}$ since then. Hence, what we today call the observable Universe was, at the end of inflation, only
$$
r_\mathrm{then} = ar\simeq10\,\mathrm{m}.
$$

That is, the photon you saw yesterday, was emitted 13.8 Gyr ago, only 10 meters from you.

To find the distance at which the photon you see today was emitted, we can calculate the fractional distance between the two regions emitting the photons — but now we shouldn't use physical coordinates, but instead comoving coordinates, because the part of the extra distance that is due to expansion shouldn't be taken into account. In comoving coordinates, the observable Universe increases a distance of $r_\mathrm{com} =$ one light-day per day, or a fraction
$dr_\mathrm{com}/r_\mathrm{com} = dr_\mathrm{com} / r = 6\times10^{-14}$.

In other words, the region emitting photon #2 is, and always has been, a fraction $6\times10^{-14}$ farther away than the region emitting photon #1.
If region 1 was 10 meters away, then region 2 was
$$
\begin{array}{rcl}
\frac{dr_\mathrm{com}}{r_\mathrm{com}}\times 10\,\mathrm{m} & \simeq &  6\times10^{-11}\,\mathrm{cm}\\
& \simeq & 0.6\,\mathrm{picometers}.
\end{array}
$$

(I originally forgot to calculate in coming coordinates, resulting in a slightly too large value.)

A note on the calculations

In the above calculations, I've used cosmological parameter data from Planck Collaboration et al. (2016), because those are implemented in Python's astropy module. The newer data from  Planck Collaboration et al. (2018) would change the result only a little bit. If instead of $10^{-32}\,\mathrm{s}$ you assume $10^{-33}\,\mathrm{s}$ for the end of inflation, as some models predict, you'd get slightly smaller Universe at the end of inflation (a factor $\sqrt{10^{-32}/10^{-33}}\simeq3$).


Dark energy and light red shifting When light is red shifted from distant galaxies, the photons have lost energy. When dark energy pushes objects apart, those objects have gained energy from a larger gravitational potential. Is the amount of energy that dark energy applies to push objects apart equal to the amount of energy lost because light from distant galaxies is red shifted?

Is the amount of energy that dark energy applies to push objects apart equal to the amount of energy lost because light from distant galaxies is red shifted?

No, dark energy existed prior to the expansion and the shift towards blue or red is based on the direction of movement of the emitter relative to the observer, so the amount isn't equal.

Sources:

Wikipedia: Dark Energy - "Change in expansion over time":

  "Recent results from the Hubble Space Telescope Higher-Z Team indicate that dark energy has been present for at least 9 billion years and during the period preceding cosmic acceleration.".

Wikipedia: "Integrated Sachs–Wolfe effect":

  "The integrated Sachs–Wolfe (ISW) effect is also caused by gravitational redshift, but it occurs between the surface of last scattering and the Earth, so it is not part of the primordial CMB. It occurs when the Universe is dominated in its energy density by something other than matter. If the Universe is dominated by matter, then large-scale gravitational potential energy wells and hills do not evolve significantly. If the Universe is dominated by radiation, or by dark energy, though, those potentials do evolve, subtly changing the energy of photons passing through them.

  There are two contributions to the ISW effect. The "early-time" ISW occurs immediately after the (non-integrated) Sachs–Wolfe effect produces the primordial CMB, as photons course through density fluctuations while there is still enough radiation around to affect the Universe's expansion. Although it is physically the same as the late-time ISW, for observational purposes it is usually lumped in with the primordial CMB, since the matter fluctuations that cause it are in practice undetectable.".

Wikipedia: "Theories of Dark Energy":

  The Equation of State (EoS) of Dark Energy for 4 common models by Redshift:

              A: CPL Model   B: Jassal Model   C: Barboza & Alcaniz Model   D: Wetterich Model

This question was closed on our Physics.SE site: "Could some Red and Blue shifts be the result of light passing through “dark matter”? [closed]", while this: "Redshifted Photon Energy" was not.

See also:

"New Aspects of Photon Propagation in Expanding Universes" (Oct 6 2016), by H.-J. Fahr and M. Heyl.
Wikipedia: "Friedmann equations":

  "The Friedmann equations are a set of equations in physical cosmology that govern the expansion of space in homogeneous and isotropic models of the universe within the context of general relativity.".

Wikipedia: "Cosmic Expansion History":

  Since the densities of various species scale as different powers of $a$, e.g. $a^{-3}$ for matter etc., the Friedmann equation can be conveniently rewritten in terms of the various density parameters as

  $$H(a)\equiv {\frac {\dot {a}}{a}}=H_{0}{\sqrt {(\Omega _{c}+\Omega _{b})a^{-3}+\Omega _{\text{rad}}a^{-4}+\Omega _{k}a^{-2}+\Omega _{DE}a^{-3(1+w)}}}$$

  where $w$ is the equation of state of dark energy, and assuming negligible neutrino mass (significant neutrino mass requires a more complex equation). The various $\Omega$ parameters add up to 1 by construction.

As you can see the contribution to shift (either way) is very small compared to the amount of dark energy present. The dark matter particles interact with each other and other particles only through gravity and possibly the weak force.


What would happen to the solar system if Earth's mass was removed? This question is inspired by the "Stargate SG-1" episode "The Road Not Taken". In that story a device is used to shift the entire Earth to another dimension to defend against an attack.

I'm not asking about the science of whether that is even theoretically possible but rather the effect on the rest of the objects in our solar system if it did happen. Since this was apparently a short shift (likely days at the most) I would imagine there might be little overall effect to most things but it occurred to me the Moon might immediately start drastically altering its position relative to the now massless Earth (there was no indication that the effect extended to the Moon but given that ships very close to Earth were unaffected it would seen that the Moon was also unaffected) so that when the Earth returns to "normal" it may no longer be in Earth's orbit or worse, Earth may rematerialize partially within or completely around it.

There would be three effects to consider, one very short-term, one medium term and one long-term.

The medium-term effect is that the Moon has an orbital velocity around the Earth of about 1 km/s.  If the Earth suddenly disappeared, the Moon would just keep on going in a straight line at 1 km/s.  This would have the effect of putting it into a Solar orbit quite different from the Earth's. It would have a different inclination to the ecliptic (the Moon's orbit is inclined to the ecliptic by 5 degrees), different period and different eccentricity. (How different depends on exactly where in the Moon's monthly orbit the Earth disappeared.)

The effect wouldn't be enough to cause the orbit to intersect that of Venus or Mars right away, but perturbations on the new Planet Luna would potentially be very different from those on Earth and it might evolve fairly quickly (centuries) to a Mars-crossing or Venus-crossing orbit with an eventual collision.

It is highly unlikely that it would be very near the spot where Earth reappears.

The long-term effect is smaller: The Earth is part of the generally stable Solar System and contributes small perturbations to all the planets. It is possible (but I suspect unlikely) that the disappearance of the Earth and the subsequent lack of the Earth's perturbations would result in significant changes to the Solar System over many centuries to tens of thousands of years.

The very short-term effect is that the disappearance of the Earth's mass would result in a speed-of-light propagation of an abrupt change in the curvature of spacetime.  I.e., gravitational waves.  I don't know enough to calculate if these waves would have any noticeable effect within the Solar System. My inexpert guess is that it's plausible that they would.