Shadows of Light = Space or Dark Matter? Perhaps my title is misleading. but here is my understanding and I'm looking for confirmation.

You cannot have dark without light.

The only thing faster than light is darkness (i.e. shadows)

Stars are in the galaxy and are inherently bright.

Does this make "dark matter" mere shadows of extremely bright objects?

Is that also why it has to get dark before we can really see the majority of stars (subjective experience)?

Update:

With all due respect to user with answer below and thank you for taking the time to answer... I think this can be expanded on.

I have provided a source for my claims: youtu.be/JTvcpdfGUtQ

Check out the video, and it more explains my understanding of things...

Why I don't accept answer below:

Can you provide sources for your claims?

Also I didn't say you can't have DARK MATTER without Light , I said you can't have Dark, as in Darkness. I think you're getting hung up on Dark Matter which is a big part of my question, but not the entire question.

Also, are you claiming that all darkness in space is dark matter? I don't like to assume but for scientific purposes (not an accredited scientist) would we not agree that the measures made on dark matter were on a sample, and not the entire universe?

So to further question this, do shadows not exist in space?

You cannot have dark without light.

Not true in the case of dark matter (for the general case, see below). Dark matter is called "dark" because it appears that it doesn't absorb or emit electromagnetic radiation - light! It can interact with light via gravitational lensing, but dark matter particles have no electromagnetic charge (we think) and so are considered "dark".

  The only thing faster than light is dark (i.e. shadows)

This is true, as the video explains.

  Does this make "dark matter" mere shadows of extremely bright objects?

This is the meat of the question, and the answer is a clear "no".  First, any "bright objects", while capable of casting a shadow of something else, will emit electromagnetic radiation.

Okay, so why can't there just be non-extremely luminous objects casting a shadow? One reason is that the dark matter simply isn't where these shadows should be. Many galaxies have a dark matter halo that extends far beyond the galactic plane. Any light sources emitting light (that is then shadowed) would have to be on the opposite side of the galactic disk. This is simply not the case.

The second, more convincing, reason is that dark matter has mass. In fact, that's the reason the idea was first conceived (to account for anomalies in galactic rotation curves)! Shadows don't have mass; therefore, dark matter cannot be explained by shadows.

  Is that also why it has to get dark before we can really see the majority of stars (subjective experience)?

No. I don't know for sure, but I believe that that's because the Sun's light is still blocking out the light of distant stars.[citation needed]

Addendum after edit to question

  Can you provide sources for your claims?

I think that they're pretty well sourced! The first section can easily be checked via Wikipedia (which I assumed you had seen). The rest seems to be well-sourced, too (aside from the last bit, which is, admittedly, unsourced but only tangential).

  Also I didn't say you can't have DARK MATTER without Light , I said you can't have Dark, as in Darkness. I think you're getting hung up on Dark Matter which is a big part of my question, but not the entire question.

Okay, I may have misinterpreted this.

Darkness is, by definition, the absence of light (electromagnetic radiation). So no, you don't have to have light for there to be darkness.

  Also, are you claiming that all darkness in space is dark matter? I don't like to assume but for scientific purposes (not an accredited scientist) would we not agree that the measures made on dark matter were on a sample, and not the entire universe?

I never said that, but you did. Your original question asked if dark matter could be shadows. The answer is no.

  So to further question this, do shadows not exist in space?

They exist. Any object blocking a light source casts a shadow:

This happens to have been used in the video, so it shouldn't be new to you.


Is the universe simply a mass of atoms in 'void'; what does it look like? I've always been stumped with the question of what lies beyond the universe, and I've come to the firm conclusion - void (i.e. completely empty).  Does it make sense then to think of the universe as a finite 'blob' of atoms (the completely full) flying around each other in nothing?  Furthermore, and more importantly, does it have a shape like a sphere?

It is not known. The universe has a finite age so we cannot see beyond the speed of light times that age (and taking the Hubble expansion into account as well). This means there is a 'visible universe' which defines a boundary of what we can observe. Information does not travel faster than light.

The whole universe is bigger than the visible universe, but we don't know how much bigger (it could even be infinite). However, as the universe contains all space ánd time, the stuff outside the universe does not contain both (try to wrap your head around that!)

As I mentioned it could be infinite, we could live on the three dimensional surface of a four dimensional ball (like an ant on a balloon) or our universe could be embedded in a higher dimensional multiverse (like string theory tells us).

We probably will never know because we cannot see past the visible universe.


Sky in the very far future / past Motivated by this question about a stranded submarine in the Jurassic I would like to know:

Is there anything remotely left of our current skyline if we go back or forward millions of years so that the stars rotating around the milky way covered a measurable distance ? I would expect that the proper motion of a star causes an irreversible drift out of the original position. Are there any stabilizing factors which may allow that some star configurations are still discernible ?

The sun still rises in the East and sets in the West. So you quickly identify the cardinal directions.

By observing the point around which the stars rotate each night you can find the altitude of the pole, which gives you your latitude. You can't find your absolute longitude, but with good time keeping you can find your longitude relative to your starting point. The planets are still in the same orbits. You might find that Saturn doesn't have rings (it's not certain how old the rings are)

The Galaxies would still be in the same positions, The Milkyway, Andromeda and the Magellanic clouds are all visible. Many more with a moderate telescope, The Milkyway would look different, in detail, the LMC and SMC would be in different positions.

However, the stars in the background have all changed. There are no constellations that have not undergone radical changes. There would be no stars that you recognise: Stars are visible either because they are close (and so have a large proper motion) or powerful and short-lived. All the stars you know would either have moved, or not be born yet.


Why are Cassegrain telescopes shorter than Newtonian reflecting telescopes? Many websites say this is because the effective focal length of the objective is increased by making the secondary mirror convex. This allows a Cassegrain telescope to be shorter than a similarly powered Newtonian.

What I don't understand is that if the focal length is made longer, how come the telescope doesn't have to be made longer too? Could someone explain how the focal length is increased and how this leads to shorter telescopes?

Basically, the light has a longer way to go after hitting the secondary mirror in a Cassegrain than in a Newton telescope.

In a Newton telescope (top image), the light hits the primary mirror and goes through the telescope tube once. In a Cassegrain telescope (bottom image), the light hits the primary mirror and goes through the entire length of the tube twice.

The focal length of a telescope is the distance between the telescope's primary mirror and the point where the light rays come together in focus. This is how far the light has to go after hitting the primary mirror before reaching the eyepiece.

Since the light goes a longer distance within a Cassegrain telescope than a Newton telescope of the same length, the focal length of the primary mirror can be made longer.
Conversely, for two telescopes with the same focal distance, the Cassegrain will be shorter than the Newton telescope.

You can imagine the Cassegrain telescope folding the focal length in two, whereas the Newton telescope leaves the focal length stretched out flat.

In this image, light has a longer path to go in the bottom (Cassegrain) telescope than in the top (Newton) one. Even though they are the same size, the bottom telescope has a longer focal length.

This would make it seem that a Cassegrain can have a focal length twice the length of the tube, whereas a Newton can have a focal length roughly the length of the tube.
In fact, Cassegrain have an extra trick up their sleeve to increase the focal length, and can have a focal length equal to five times the length of the tube.

The secondary mirror of a Cassegrain telescope is convex, and has its own focal length. The effective focal length of the primary mirror + the secondary mirror can thus be made longer than that of the primary mirror alone.


How does gravity interact with a photon? If photon is mass-less and gravity can interact only with matter, then how  does gravity alter the trajectory of light?

It is simply not true that gravity can only interact with mass. Rather, any long-range spin-2 force interacts with all energy-momentum equally, and it source is the stress-energy-momentum tensor. That is one way to state the equivalence principle.

Note that a massive object in its own rest frame has an associated energy $E = mc^2$, which under ordinary conditions is usually much higher than any stress terms (including pressure) or momentum (outside the rest frame). Thus one can usually pretend that gravity couples to mass, but it isn't so--rather, the gravitational charge is energy, and the entire stress-energy-momentum tensor couples to the gravitational field. This is analogous to how for electromagnetism, there is electric charge, but the electric currents also make a difference.

In the weak-field limit of general relativity, one can consider the fact that light is gravitationally deflected twice as much as a naive Newtonian prediction would be (for an object at the speed of light under Newtonian mechanics, that is) as the fault of light putting a pressure in the direction of its propagation equal to its energy density. But in general, the motion of a test particle is determined by its four-velocity and the geometry of spacetime only, and the light would just be a special case of having a 'lightlike' four-velocity.


what is the time mentioned after date (1962-07-03+02:00) the +02:00? i was retrieving a some personalities birthday date through computer programming

when i retrieve a person's birthday the result was

1962-07-03+02:00   [this is tom cuise's birthday for information]

In the above result i could understand his birthday is on 1962-07-03 ,

But what is +02:00 , is that the location of birth or GMT ?

for more information the data is retrieved from the URL

http://dbpedia.org/page/Tom_Cruise

sorry to ask this question here , I asked this question because it deals with time and i believe astronomers can help with this

As indicated in the comments, the +02:00 is the ISO 8601 syntax for the time zone. Specifically, it indicates a time zone 2 hours east of UTC.

whois dbpedia.org indicates that the site is hosted in Berlin, Germany, which is consistent with the +02:00 time zone.

The fact that the data includes the date and time zone but not the time is probably an error. This particular datum 1962-07-03+02:00 is Tom Cruise's birthday; in fact he was born on that day in Syracuse, New York.

A data representation that includes a date and time zone but no time is peculiar, and I'm not sure that it conforms to the ISO 8601 standard. The description in the Wikipedia article seems to imply that a time zone can only follow a time. I haven't checked the standard itself.

A representation of just the date: 1962-07-03 would make more sense, and would be more useful. This is probably a bug in the software that runs dbpedia.org.


Why is the height of a satellite greater than the apogee of its orbit? I read in the German Wikipedia for the Indian Technology Experiment Satellite, that the apogee (Apogäumshöhe) is 567 km, while the height of the orbit (Bahnhöhe) is 572 km. I had always thought that the apogee is the furthest point of the orbit from the Earth and I wonder how can there be a height greater than that.

You are right!

Apogee is higher than Perigee, except when the orbit is perfectly circular when they're equal.
Semi-major axis will be half-way between the two.
572 km doesn't make sense.

As pointed out in comments, expressed like this they are the distances from Earth's center minus a fixed equatorial radius of the earth, in this case about 6378.137 kilometers. From the numbers in Wikipedia for peri and apo you'd call this a 567 x 515 km orbit. It's a simple way to express roughly the size and shape.
The linked de.Wikipedia page says
Umlaufzeit:       95,3 min[1]
Bahnhöhe:         572 km
Bahnneigung:      97.7°
Apogäumshöhe:     567 km
Perigäumshöhe:    515 km

[1]  Bahndaten nach [TES](https://www.n2yo.com/satellite/?s=26957). N2YO, 18. November 2017, abgerufen am 18. November 2017

I don't know what Bahnhöhe means, google translates it as "track height", so I've asked What exactly would a distance labeled Bahnhöhe mean in the context of a satellite's orbit around the Earth? in German SE.
While the data might be static, the mentioned n2yo link says:
NORAD ID:           26957
Int'l Code:         2001-049A
Perigee:            513.2 km
Apogee:             564.0 km
Inclination:        97.7 °
Period:             95.3 minutes
Semi major axis:    6909 km

and looking it up in Celestrak gives the two line element set or TLE as
TES
1 26957U 01049A   20269.88561907  .00000195  00000-0  15464-4 0  9995
2 26957  97.7413  31.7408 0036778 246.8588 112.8763 15.11565287 39121

The TLE is also shown in n2y0.
Here's a link direct to Celestrak for this object. They are always changing, the orbits of Earth satellites are never really fixed.
Atmospheric drag tends to lower them and Earth's lumpy gravity causes them to precess, just for starters.
Many satellites have small thrusters to raise their orbit after atmospheric drag lowers it, so we can't expect numbers in static Wikipedia data fields to be accurate.


What exactly is this VLA radio image of Jupiter showing? This VLA image is beautiful, and instructive, but I don't understand it fully.

This version is from an article in Gizmodo where the credit reads: A radio image of Jupiter captured by the Very Large Array. Image: Imke de Pater, Michael H. Wong, Robert J. Sault and which cites the recent article in Science: Peering through Jupiter’s clouds with radio spectral imaging. However, that paper does not actually show this image.

Another version of the image can be found in Berkeley News: New radio map of Jupiter reveals what’s beneath colorful clouds and gives more information how the image was produced.

  A radio image of Jupiter from the VLA at three wavelengths: 2 cm in blue, 3 cm in gold, and 6 cm in red. A uniform disk has been subtracted to better show the fine banded structure on the planet. The pink glow surrounding the planet is synchrotron radiation produced by spiraling electrons trapped in Jupiter’s magnetic field. Banded details on the planet’s disk probe depths of 30-90 km below the clouds. This image is averaged from 10 hours of VLA data, so the fine details seen in the other maps are smeared here by the planet’s rotation. Image by Imke de Pater, Michael H. Wong (UC Berkeley), Robert J. Sault (Univ. Melbourne).

I would like to know more about how the image was processed. What does it look like before the uniform disk is subtracted - is the radio emission from the region around the planet actually as strong as the planet disk itself, as this image suggests? If there is a reference for this image I'd also like to read it!

In the mean time, it really is beautiful!

The image is sometimes paired with the following image, taken here from Science News.

Since this image resolves lateral detail, I'm guessing it might be the same 10 hours of data, but processed with a moving frame and corrected for distortion as it wraps around during the rotation of the planet, while the global image above is smeared by rotation

Was it? Where can I read more about this?

A uniform disk has been subtracted to better show the fine banded structure on the planet.

This means the surface of the planet is far brighter (in the raw data) than the surrounding magnetosphere. The planet itself has been darkened considerably in the image, otherwise it would be a pale circle and show little or no detail.

The press release from the VLA's site itself doesn't even show the wide field image however; it concentrates on the image showing detail of the spot itself. It mentions:

  an innovative data-reduction technique to "unsmear" the data from many hours of observing

There is an earlier 2006 paper by some of the same researchers showing earlier attempts, with a much more coarse radio map of Jupiter. This goes into more detail about the earlier image processing techniques though (requires good knowledge of Fourier analysis etc.)

(Note: I can't read the current Science paper in full, it's behind a paywall.)


Transit of Uranus from Neptune I would like to help find sources or calculations for the following dates found in Wikipedia (in French). The next Uranus transit from Neptune will happen :
1. October   38172
2. August   111551
3. November 124067

Would you please help me find sources or calculations stating these three dates?

The only one who seems legit is the first date, October 38172. It appears somewhere in Transits, by Jean Meeus (1989), and some say that also in the SOLET webpage, but I can't find it either. Taking into account that, if true, it comes from a book from 1989 and probably is not very accurate, I wouldn't take that date as an exact thing.
The other dates I can't find anything.


App that shows what's passing in sky? I'm not an astronomer but I enjoy looking up at the sky.

So tonight I saw something moving in the sky. It looked like a star, and it was definitely not a plane (I checked in flightradar).

Is there an app, similar to flightradar for planes, that will tell me what's traveling above me in the sky?

Chris Peat maintains a very sophisticated database of satellites and their visibilites at http://heavens-above.com, and there is an app available.


Are the stars outside of the galactic plane in the galactic halo? The majority of the stars we see in the sky, like Pollux, are outside of the galactic plane. That means that all those stars we see are not in the galactic disk, and therefore are in the galactic halo, whereas Capella, for example, is where?

The galactic disk, as Riley Jacob wrote, has a definite thickness. It's actually composed of a thin disk $\sim0.3\text{ kpc}$ thick and a thick disk $\sim1\text{ kpc}$ thick, at least (McMillan (2011) has models with data from the Sloan Digital Sky Survey). There's also a central bulge that is even thicker, as the following diagram (from here) shows:

Pollux is $\sim10\text{ pc}$ away from the Solar System, which is about 1% of the thickness of the thick disk (Capella is barely farther). Essentially, it's in the same plane as the Solar System; that's an insignificant distance.

There are stars in the thin disk, thick disk, and halo, which compose different populaitons based on metallicity. In the thin disk are Population I stars, which are high in metals. In the halo (and thin disk, to some extent) are Population II stars, which are lower in metals and on average older than Population I stars. The Sun is a Population I star in the disk. There may also be a sub-population of Population II stars in the thick disk.

Halo stars certainly do not compose the majority of stars in the Milky Way galaxy. Most stars are contained within the thin disk, thick disk, and bulge.


Different full moon face on different places on Earth same time I have theoretical question.
Let's assume that it is full moon and two different humans are looking at it at the same time. One human is for example in London and one in Beijing. It doesn't really matter where exactly, important is that they are almost on opposite side of Earth but they both see the moon at the same time.
I assume they see the moon from different angle. Is it possible that one of them see different part of the moon because of this angle. It could be just the border, tiny part.

Could you please tell me, how big is this tiny part of the moon which one see and the other don't? Few meters? Kilometers?

Thanks everyone for help.

Yes, they do see a slightly different face of the Moon.

It's easy to calculate how different. If they're on nearly opposite sides of the Earth so that one sees the Moon just rising and the other sees the Moon just setting, they are seeing it from about 1.9 degrees apart. (A triangle with a base of 8000 miles (the distance they are apart) with sides of 240,000 miles (the distance to the Moon) yields an angle of 1.9 degrees (8/240 radians).

So they'll be viewing the Moon from 1.9 degrees apart and will see a full moon displaced by 1.9 degrees, or about 36 miles on the surface. So one will see 36 miles of Moon more on the Lunar east side than the  other and the other will see about 32 miles more on the Lunar west side.


Expresion of comoving distance I have a simple question :
How to prove the following relation :

The comoving distance to an object at redshift $z$ can be computed as

$$r(z)=\dfrac{c}{H_{0}} \int_{0}^{z} \dfrac{\mathrm{d} z}{E(z)}$$
from the relation :
$$r(t)=\int_{0}^{t} \dfrac{c\mathrm{d} t}{R(t)}$$
I tried to use with the definition : $1+z= \dfrac{R_{0}}{R(t)}$ but I can't conclude.
Any help is welcome.
UPDATE 1 : @Tosic's demonstration seems to be correct. But the factor $R_{0}$ is not disappearing. Indeed, If I do :
$$\dfrac{\text{d}(1+z)}{\text{d}t} = \dfrac{\text{d}z}{\text{d}t} = -\dfrac{H(t)}{R(t)}\,R_{0}$$
which implies :
$$\int_{0}^{z} \dfrac{c\text{d}z}{H(z)} = \int_{0}^{t}c\text{d}t\dfrac{R_{0}}{R(t)} = R_{0} \int_{0}^{t}\dfrac{c\text{d}t}{R(t)}$$
How to get rid of the factor $R_{0}$ ? Since if I multiply the comoving coordinate $r(t)$ by $R_{0}$, I get the cosmological horizon (the limit of observable universe if I integrate up to $z=1100$), don't I ?

According to this, the Hubble constant for redshift z is $H_0E(z)$. Meaning we need to prove that $$\int_{0}^{z_0}\frac{cdz}{H(z)} = \int_{0}^{t_0}\frac{cdt}{R(t)}$$
Take the first derivative of both sides of your second equation to obtain, by the chain rule, and the equation (this is the definition of the Hubble constant) $H = \frac{R(t)'}{R(t)}$ the following:
$$\frac{dz}{dt} = -\frac{1}{R(t)^2}*R'(t) = -\frac{H(t)}{R(t)}$$
That the redshift is zero for the current time, and some value greater than zero for some time before the current time should explain the minus sign. After multiplying with $c$ and placing the small changes on both sides this becomes $$\frac{cdz}{H(z)}=-\frac{cdt}{R(t)}$$, which looks like it can be integrated to obtain what we need to prove (the integral from z to 0 is the one from 0 to t).This proof is not very formal, but it's the best I can do, so I hope it is somewhat correct and that someone will give a more detailed answer if this is not good enough.


Looking for planetary position calculation In Hindu astrology, for creating horoscopes (birth charts), I am looking for calculations of planetary positions. So, how could I calculate planetary positions according to given date, time and place?

If you're only interested in the positions, you'll probably be best off just using a program instead of diving into celestial mechanics. Depending on what you exactly want with "planetary position", here are three different approaches to finding a planet's position at a given time:

Stellarium is a free program to visually solve for the position of a planet, and it gives some precise data for the positions of the planets too, but it needs to be installed and run on your computer so it won't be the quickest way to find the position.
If you're looking to get a planet's coordinates on ecliptic and/or equatorial planes, you can let this web app do the calculations for you for a given time.
Also, a quick Google search came up with this web app meant for working with zodiacs.


Is it possible to do moon sighting in advance for 5 years with 100% accuracy? It is Ramzan and the usual heated disagreements between clerics here in Pakistan as to the sighting of the crescent. This time the minister of Science and Technology (who is not a scientist himself btw) has intervened and formed a committee comprising of meteorologists and our space agency. This committee is tasked to form a 5 year lunar calendar, specially one that will fix the issue of moon sighting of major religious months and events such a Ramzan, Hajj (pilgrimage), Eids etc.

My question is that: Is it scientifically possible to predict (simulate...) in future with 100% accuracy as to the location of the moon? My surface reading of Lyapunov time suggests that the solar system can be predicted with accuracy for 5 million years.

I would like to know your understand on this matter.

Also some commenters mentioned that this question might be a duplicate of an other one. I would like to add that there are school of thoughts and in one the condition is that the crescent is to be observed from the surface of the Earth. So I think this question is unique in that sense that it includes that component too.

Note: this answer was posted under duress; though I mentioned in a comment under the question that I was composing an answer, several users have decided to close the question out from under me. Therefore I've put this together a little hastily. It's late here and I'll come back in the morning to address any questions or requests for clarification, and probably add some more information.

  Is it scientifically possible to predict (simulate...) in future with 100% accuracy as to the location of the moon?

Nothing is 100% accurate or possible. But it is highly likely that the apparent position of the Moon from a given geographic position on Earth can be predicted to very high precision over the next five years. Each aspect:

Geometry
Geography
Time
Meteorology/atmospherics

is predictable based on well-understood scientific principles which have been tested against a substantial extent of data.

Without a specific specification in the question, I can present these aspects of the problem and give an example of the precision to which they can be predicted.

Geometry

We have records of bits and pieces of the Moon's motion for several thousand years from accounts of eclipses, and several hundred years with much more precision with respect to time and to other stars from records of lunar occultations of stars as seen through telescopes.

Combine this with laser ranging measurements of the moon which fit theoretical predictions to centimeter accuracy over a period longer than forty years, and it seems that the relative positions of the Earth and Moon are probably going to continue to be within a few centimeters for the next five years as well.

You can read further about that in
 - Progress in Lunar Laser Ranging Tests of Relativistic Gravity
 - Lunar laser ranging: the millimeter challenge

Now the position of the Earth relative to the Sun is a different matter. If you want to read about how these calculations are done you can have a look at this answer to How to calculate the planets and moons beyond Newtons's gravitational force? These days the movements of the Sun, Earth, planets, and other bodies with enough mass to affect their motions substantially are know extremely well.

Let's look at one of the major JPL Development Ephemerides. In the abstract of The Planetary and Lunar Ephemerides
DE430 and DE431 there are a few sentences that give us some idea of what's discussed in this pretty in-depth report:

  The planetary and lunar ephemerides DE430 and DE431 are generated by fitting
  numerically integrated orbits of the Moon and planets to observations. The present-day
  lunar orbit is known to submeter accuracy through fitting lunar laser ranging data with
  an updated lunar gravity field from the Gravity Recovery and Interior Laboratory (GRAIL)
  mission. The orbits of the inner planets are known to subkilometer accuracy through fitting radio tracking measurements of spacecraft in orbit about them.

Geography

The visibility of a crescent moon from a given spot on Earth needs more than the positions of the center of masses of the Moon, Earth, and Sun. We need the position of the viewer. These days we can get GPS coordinates of the site, but those coordinates are fixed to the Earth, and the Earth rotates around its axis and its axis nutates and wobbles.

That motion is pretty small on the scale of five years, you can see the image below showing the motion of the Tropic of Cancer each year.

Time

The rotational speed of the Earth increases and decreases in a somewhat unsteady way, and this is why the placement of leap seconds looks a bit uneven. However, the fact that we are talking about one second makes it clear that the uncertainty over five years is likely to be a handful of seconds.

Still, this may be the single biggest risk in the accuracy of the prediction five years into the future!

There could be a major seismic event, magma movement, or big chunk of ice that falls into the ocean. Specific details about how these can affect the rotational speed of the Earth (and the exact direction of its axes) need to be addressed in detail in separate questions, but this could be an upset.

Of course we would know about it, and a request for an updated five year report could be issued. The time error wouldn't suddenly jump to some big number, it would slowly start accumulating, and there would be time to recalculate before it became significant.

Atmospherics and Meteorology

This is the hardest one. Refraction from the atmosphere can move the apparent position of objects significantly, up to a fraction of a degree. But the amount of refraction is fairly well predictable. It depends a little bit on the Weather because index of refraction's deviation from 1.0 depends almost linearly on the density of the air, and slightly on the humidity. But this is a very small correction on top of a small correction.

Lunar Libration image from here

Lunar Laser Ranging images from here

 Source

Source


How well is Earth’s motion through the universe quantified? If I had a time travel machine that could drop me into any point in space and time, could I accurately predict earth’s “position”?  I know position is an improper term.  Would it be possible to know exactly where to land so that you are at the same spot on the surface of Earth as when you “departed”?

Surprisingly, we will soon be able to do this well enough to go back several hundred million years and place ourselves, if not in the solar system, inside the Galaxy and roughly in the right part of the Galaxy.   It turns out that we know, with high accuracy, the Earth's motion with respect to the microwave background radiation, which provides a well defined frame of reference for motions.  Therefore, we will be able to work out which direction and about how far the MW has moved with respect to this universal frame and separately the Earth with respect to the center of the galaxy. It requires solving for the growth of what is called the 'microwave dipole velocity' over time.  Some of this velocity develops from nearby objects like the MW and the nearby groups of galaxies.  So, we need to know their masses and relative motions a bit better than we presently do.  But, by far, most of the velocity is from objects quite far away, and this allows us to use the simpler and well understood 'perturbation' techniques. The biggest uncertainty for the Earth's position might be the interactions with local stars back in time and what effects they had on perturbing the Suns orbital motion about the center of the galaxy.  Again we need to know better the masses of local stars and their relative velocities to calculate the Earth's trajectory back in time.  As it happens, the space telescope GAIA, which is about to make its second release of data, will greatly improve our knowledge of the motions and, indirectly, the masses of all these stars and nearby galaxies.

So, if you have got that time machine working, we are nearly ready to go!


Would a high albedo reflective substance cool down Venus? If we were to put a highly reflective substance into Venus's upper atmosphere, could we cool it down?  I am envisioning adding a highly reflective fine dust or gas, possibly lighter than "air" to the atmosphere.  I believe to have the proper effect, it should also reflect infrared light, which causes heat.  Apparently Venus already has a high albedo of .75 (reflects 75% of light): http://www.universetoday.com/36833/albedo-of-venus/ .  However, if we were able to raise that higher, even up to 99%, would that cool it down, perhaps even making it habitable (temperature wise), or possibly even making it ice cold?

Eventually, yes.

Interesting information about Venus: Venus is hotter than Mercury, despite being nearly twice as far from the Sun. Earth, despite being further from the Sun, receives more
energy from the Sun than Venus, due to Venus's very high albedo.

As you might guess by this information, the major factor that keeps Venus hot isn't how much energy it receives from the Sun, but rather its inability to rid itself of energy. While adding a cloud of some high albedo substance to the atmosphere may effectively prevent the Sun from heating Venus, it would also likely prevent infrared radiation from leaving Venus, thus keeping it warm.

As you know, Venus already has a very high albedo. It's thick atmosphere prevents energy from leaving the planet, which is what caused the temperature to rise far above the Earth-like temperatures we believe it once enjoyed. This is caused by IR radiation being emitted from the surface and being reflected back down towards the planet by the atmosphere more detail, which plays an essential part in the greenhouse effect.

In my opinion, a more effectual method in terms of making it habitable would be to find a way to drastically lower the albedo of Venus to allow all the excess energy to escape.


How do supernova explosions reconcile with light speed constraints? I have learned about the various types of supernovae, and both Types Ib and II involve the extremely rapid compaction or explosion of a star's core.

How does the "information" that (respectively) the Chandrasekhar limit or the degeneracy pressure have been reached propagate through the star with enough speed (yet below $c$) that the resulting collapse remains symmetric and happens in a very short time?

Interestingly enough, the limit on the time needed for such a change to propagate through the star isn't from the speed of light, but from something called the dynamical timescale, $t_{\text{dyn}}$:
$$t_{\text{dyn}}\sim\frac{R}{v}\sim\sqrt{\frac{R^3}{GM}}\simeq\frac{1}{\sqrt{G\bar{\rho}}}$$
where we've assumed that the speed of a pressure wave, $v$ is on the order of the escape velocity of the star. $R$ is the star's radius, $M$ is its mass, $G$ is the gravitational constant, and $\bar{\rho}$ is its mean density (density is not constant throughout a star).

$t_{\text{dyn}}$ can vary greatly among different stars, depending on their characteristics. It can be on the order of seconds to minutes to hours to days, and it can determine the time it takes for parts of the supernova to take place. It's a lower limit, though; large dynamical changes can take more time than $t_{\text{dyn}}$, just not less. For more information, see Section II.B of these notes.

We have to ask ourselves, what precisely is $v$? I waved the specifics away as a "pressure wave", but is there more to it than that? Unfortunately, we still don't have great models of what goes on inside a supernova - I'm sure you've heard that before - but we do have inklings of the process, which I touched on a bit here.

Essentially, as the core collapses, matter at the inner regions begins to fall in, reaching extremely high densities. Some strange things happen here, including the production of neutrinos. At first, degeneracy pressure slows the collapse of the inner star, and a shock wave forms. The so-called "supernova bounce" happens, as this wave travels through the outer layers of the star.

The shock wave is further energized by the release of neutrinos, created via electron capture (note that some have already escaped the outer layers):
$$e^-+p\to n+\nu_e$$
This adds to the energy of the shockwave, which eventually blows away the outer layers of the star. Therefore, we can see how the dynamic timescale relates to the shock wave and the time of the resulting explosion.

The exact processes causing the bounce and shock wave are not yet fully understood; this is just a basic picture.


How to estimate uncertainty of measurements of equivalent widths? I'm measuring equivalent widths of absorption lines using a spectrum of a star. I make two or three measurements of each line by making reasonable gaussian fits of the line with IRAF's splot tool. Then I calculate the mean of the measurements, which serves as my final equivalent width estimate.

What is a good way of estimating the uncertainty of this measurement?

My current method

I'm currently using half of the range for the uncertainty. For example, if I made two measurements 10 and 16 mA (milliangstrom), then the mean is 13 mA and uncertainty is 3 mA. This gives the estimate of equivalent width to be 13±3 mA. Do you see any problems with this method of estimating uncertainty?

Yes there is a problem. You seem to be trying to derive an uncertainty in the measurement of EW by doing repeated measurements of the same data?

This can only give you the uncertainty associated with your measurement technique (i.e. where you define the limits of the line and how you set the continuum level) - the systematic error you might call it (although there can be other systematic errors inherent to EW measurements, like whether you subtracted the sky or scattered light in your spectrograph correctly for example).

What it does not do is evaluate the uncertainty in the EW caused by the quality or signal-to-noise ratio of the data itself. You might assess this using some rule-of-thumb formulae for a Gaussian line, e.g.
$$\Delta EW \simeq 1.5 \frac{\sqrt{fp}}{{\rm SNR}},$$
(eqn 6 of Cayrel de Strobel 1988) where $f$ is the FWHM of the spectral line (in wavelength units), $p$ is the size of one pixel in wavelength units and SNR is the signal-to-noise ratio of the data in an average pixel. Or you could take a synthetic spectrum and add some artificial noise to it with the appropriate properties and measure the EW of several randomisations of the same spectrum, taking the standard deviation of your EW measurements to indicate the EW uncertainty for a particular level of signal-to-noise ratio.

If this statistical uncertainty is not negligible, then you would then need to add it to any systematic uncertainties associated with your analysis of the spectrum. As far as the latter is concerned then your suggested method does give some indication of what that error might be, though I suspect it will overestimate the 1-sigma uncertainty.


Where does the energy of light go, when it red-shifts? When talking about the expansion of the universe, it is said that it can be proven by the red-shifting of light.(As we would need higher than lightspeed to get this redshift by the Doppler effect)

I am an amateur, so I am not sure I am correct, but here is what I think.

Redshifting increases the wavelength of the light. higher wavelength = lower frequency = less energy.

So, if my assumptions are correct, where does this energy from the light go?
If not, where did I make an incorrect assumption?

The problem is that conservation of energy is a slippery concept in General Relativity. There are arguments back and forth but most people accept that conservation of energy is only a local law - it applies only to a local inertial frame and cannot be applied to the universe as a whole. However in an expanding universe it is very difficult to identify any inertial frames and certainly not ones that encompass a cosmologically significant volume.

What this means is that if you make a local "box" small enough that it is not affected by the expansion of the universe, then energy conservation will apply. But of course in such a box, a photon would enter and leave with the same energy because the box is unaffected by the expansion of the universe and so there would be no redshifting of the photon.


Has this image of the Draco Constellation been altered Have the stars in this photo of the Draco Constellation been magnified so as to help me see the Draco or not?

Also I really like [this image][2] of Orion; if someone knows of a site with a lot of such images that are in a user friendly, easy-to-find format, I would appreciate it.

The image of Draco has definitely been altered.  Most of the stars in Draco are quite faint, even the ones making the dragon shape, whereas in the image they are all shown as bigger/brighter than the stars in Ursa Minor (the quadrilateral in the middle with 3 stars trailing down to the lower left.  The last of the 3 is polaris, which is in reality brighter than all the stars in Draco.

The same is true for Kocab.
The image of Orion is typical of probably thousands taken by amateur astronomers.  If you want to find others, try a search engine.  There may be some specific sites that collect all or most of the constellations, but I'm not aware of them.


Do more massive stars become larger or smaller white dwarfs? I wonder whether e.g. stars more massive than the Sun (but below the mass where they'd go supernova) would become larger or smaller white dwarfs than a white dwarf Sun. Since Procyon B is larger than the more massive Sirius B, I suppose the more massive the star the smaller its remnant white dwarf would become. Am I right and if so, why is it like this?

More massive stars have a more massive core and produce more massive white dwarfs. The relationship between the initial mass of the main sequence star and the final mass of the white dwarf is monotonic, but not linear. The Sun is expected to produce a white dwarf with a mass of around $0.5 M_\odot$, whilst a $8M_\odot$ star is expected to produce a carbon/oxygen white dwarf of mass $\sim 1.1 M_\odot$. It is possible that some stars with $8-9M_\odot$ could result in a slightly more massive oxygen/neon/magnesium white dwarf.
The difference between the initial and final masses is accounted for by mass loss in the red giant and asymptotic giant branch phases. Note that white dwarfs close to the Chandrasekhar limit are probably not produced by the normal evolution of a single star and that stars with initial mass $>9M_\odot$ probably end their lives with a supernova of some kind, that either destroys the star or leaves behind a neutron star or black hole.
White dwarf stars are supported by electron degeneracy pressure. This form of pressure, does not depend on temperature (to first order), but does depend on density. To work out how big (in terms of radius) a white dwarf of a given mass is, one uses the density-dependent pressure to solve the hydrostatic equilibrium equation and one finds that for lower densities (which means $<10^{10}$ kg/m$^3$ in this context!), the pressure $P$ is proportional to density $\rho^{5/3}$, and this results in a radius $R$ that is proportional to mass $M^{-1/3}$. That is, a more massive star has a smaller radius. The increased weight of a more massive star requires a bigger pressure gradient to support it and this is provided by a bigger central pressure and smaller radius.
The mathematics of this is a very standard piece of bookwork involving "polytropes" (where the pressure is proportional to density to some power) and I won't produce it. A simple argument is the following. The hydrostatic equilibrium equation is
$$ \frac{dP}{dr} = - \rho g,$$
where $g$ is the gravity due to the mass within radius $r$.
If we work in proportionality, then $P \propto \rho^{5/3} \propto M^{5/3} R^{-5}$. Thus $dP/dr \propto M^{5/3} R^{-6}$.
On the right hand side, $\rho \propto MR^{-3}$  and $g \propto MR^{-2}$.
Thus the hydrostatic equilibrium equation tells us
$$ M^{5/3}R^{-6}\propto M^2 R^{-5}$$
and so
$$ R \propto M^{-1/3}.$$
As a more massive white dwarf gets smaller, the density goes up as $\rho \propto M^2$ and the degenerate electrons become relativistic. This drives the pressure equation towards $P \propto \rho^{4/3}$. This leads the radius to decrease more steeply as a function of mass and it asymptotes to zero radius and infinite density at the Chandrasekhar mass.


How small would you have to crush an object for it to become a black hole? I was wondering how much you had to crush an object for it to become a black hole. Recently I learned that anything could become a black hole (even you) if it were crushed down small enough, for example our sun would have to be crushed to the size of Manhattan to become a black hole (According to a video I saw). I did a little research and found out that something would have to be crushed down to Schwarzchild radius meaning all of that objects mass would have to be crushed into that objects radius. But if you say crush the sun down to the size of Manhattan, that is much smaller than the sun's actual radius. So I haven't really gotten a clear answer and I'm a little confused, thanks for clarifying =)

Well, I wasn't going to answer but the other two answers are wrong, or at least incomplete. If you wish to make a black hole from a stellar-sized object, then there is no need to compress it to as small as the Schwarzschild radius (though that would certainly work and would certainly be the answer for smaller objects with negligible self-gravity). Instead, you just need to compress it to a size at which it cannot be possibly supported by any plausible equation of state against further gravitational collapse. It turns out that this is somewhat bigger than the Schwarzschild radius and hence the density required is considerably lower. More details below.

There is a radius, larger than the Schwarzschild radius at which a neutron star, quark matter, whatever its equation of state, cannot be supported against collapse.

There are limits imposed by causality and General Relativity on the structure of compact stars. In "Black Holes, White Dwarfs and Neutron Stars" by Shapiro & Teukolsky, pp.260-261, it is shown, approximately, that even if the equation of state hardens to the point where the speed of sound equals the speed of light, that $(GM/Rc^2)<0.405$.

The Schwarzschild radius is $R_s=2GM/c^2$ and therefore $R > 1.23 R_s$ for stability. This limit is reached for a neutron star with $M \simeq 3.5 M_{\odot}$. A more accurate treatment in Lattimer (2013) suggests that a maximally compact neutron star has $R\geq 1.41R_s$.

If the equation of state is softer, then collapse will occur at smaller masses, and higher densities but at a similar multiple of $R_s$.

Thus it is not necessary to compress matter within $R_s$ to form a black hole.

The picture below (from Demorest et al. 2010) shows the mass-radius relations for a wide variety of equations of state. The limits in the top-left of the diagram indicate the limits imposed by (most stringently) the speed of sound being the speed of light (labelled "causality" and which gives radii slightly larger than Shapiro & Teukolsky's approximate result) and then in the very top left, the border marked by "GR" coincides with the Schwarzschild radius. Neutron stars become unstable where their mass-radius curves peak, so stable neutron stars are always significantly larger than $R_s$ at all masses.


Is there, in fact, any close-up photography of Saturn's rings, showing individual pebbles/rocks? After reading this excellent but difficult answer, https://astronomy.stackexchange.com/a/18407/13071

I've been thinking how difficult it is to get a sense of the distance between and the size of typical items of dust, pebbles, rocks and/or anything else in Saturn's rings.

In short, have any photographs been taken by spacecraft, more or less "in" the rings, close enough to actually see individual dust, pebbles and/or rocks?

(As I understand it, the rings are on the order of 10m thick.  So I'm asking, have any photos been taken from that order of distance from the rings?  What's simply the closest viewpoint ever taken?)

Failing that, is there perhaps an artist's rendition somewhere of "what you'd see from 100m above the rings looking straight down at them?"  What would be in that frustrum - nothing?  Lots of fist-size rocks you could see? Or?

It's completely unclear to me if each (say) pebble-size piece is inches apart, meters apart, or for all I know 100s of km apart?

The rings are about 3% solid in the densest parts, but this translates to a separation between 30cm particles of about 1metre.

There are no images Because approaching the dense part of the rings would be extremely difficult. A probe would be travelling at orbital speeds, and even a small percentage difference in speed would translate to a large absolute difference. As the probe would also be in orbit, it would be extremely difficult to find an orbit that doesn't collide catastrophically with ring particles.

Cassini did travel through one of the gaps in the ring system during Saturn orbital insertion. It survived 100000 dust impacts in 5 minutes.

Nature has some artist's impressions of the ring particles.


Are any other planetary bodies warming up as well in the Solar System? Do we have a few decades worth of surface temperature data to compare?

The short answer to your question is "decades", only for Mars, and only a two Mars decades (about 4 Earth decades)

I don't know if your question was related to climate change, as there were a number of extremely unscientific arguments made that Pluto and Mars are getting warmer, "and there are no SUV's on Mars" . . . so, maybe it's not the CO2 but those claims weren't science.

Silly articles and bad arguments aside, global temperature on other planets is a legit scientific question and there were scientific articles that said Pluto is getting warmer (2002, story below) and Mars is getting warmer (2007, also below), so there is some truth to this.   I'll give an answer as best I can.  Each planet deserves it's own bit of research, so this will be longish.

Pluto

Here's an article on Pluto, the title is iffy but the article is good.   It's from 2002, so it's long before New Horizon's fly-by, it's even before New Horizon was launched, but "pluto is getting warmer" was a real scientific story in 2002, and Mars, in 2007 was real science too, but some of the conclusions drawn, loosely based on those two stories were ridiculously bad.

The gist of the article is, we can't see Pluto's atmosphere well, even with Hubble, but they were able to measure some aspects of Pluto's atmosphere by watching what happens when a star passes behind Pluto and that was measured twice, in 1988 and again in 1997.

Pluto's Perihelion was in 1989 and most everyone figured that the little ball of ice/dwarf planet would get colder after 1989 when it passed it's closest point to the sun and stated moving further away but the readings said otherwise, and that was one of those really cool "oh, wow, things are more interesting than we thought they were" moments.  Pluto's atmosphere had apparently gotten thicker and it was about 2 degrees warmer.  Not what anyone expected.

The problem is, the sun couldn't make Pluto 2 degrees warmer while it moved further away without making the Earth 20 degrees warmer at the same time, so, clearly, something else was at play and it wasn't the sun warming Pluto.  Generally speaking, the warmest periods aren't necessarily precisely timed to closest to the sun, and there's a lag time caused (on Pluto) by sublimation of the frozen gases on it's surface, which thickens the atmosphere adds trace amounts of Methane and can warms the planet past Pluto's perihelion, just as on the Earth, June 21 isn't the warmest day on the Northern Hemisphere, nor December 21st the coldest, and 12:00 noon isn't the hottest time.  Could that seasonal "lag time", apply over a number of Earth years on Pluto?   It's plausible.

But we should remember, that's just two data points, which isn't nearly enough.  Or, 3 data points now with New Horizons, but it's still just 3 data points spanning 27 years on a planet where one year is 248 earth-years, so, there's a lot we don't know about seasons on Pluto.

New Horizons got better data on Pluto's atmosphere and frozen surface than we've ever had before, and what I remember reading about it, though I'm very far from well read on the subject, is that Pluto has a much more dynamic atmosphere and surface than anyone expected.

Gas Giants

you asked:

  Do we have a few decades worth of surface temperature data to compare?

Surface temperature isn't well defined on gas giant planets because we can't see the surface.   Also (I think all 4, I know for sure Jupiter and Neptune), they give off more heat internally than they receive from the sun.  Temperature on the visible surfaces / upper atmospheres of the gas giants has more to do with circulation than anything else.

Mars

There's a number of articles on Mars warming, but there's still a lot we don't know about Mars' climate.   It's not too difficult using modern technology and with Mars' thin atmosphere, to read surface temperature on Mars by measuring the infra red light that comes from it's surface.  The surface temperature on Mars was first measured in 1920 (see 4th paragraph from the top), and NASA says here that they've been taking accurate surface temperature readings on Mars for 30 years (article from May 2007), so it's pretty safe to say we have about 38-39 Earth years (20 or so Mars years) of reasonably good global temperature readings on Mars and a few decades more than that with less accuracy.

There's many problems with reaching any kind of conclusion on Mars based on that data though.  Mars has significantly greater eccentricity than Earth, so it may be prone to larger temperature swings than Earth and just 20 Mars years of data is too few.

Winter to Summer on Mars' enough CO2 can freeze to reduce it's atmosphere by as much as 25% and that can drive temperature changes and mars is also prone to planet wide dust storms that can last months.  These can have the effect of blocking sunlight and after settling, darkening the ice caps, which lowers albedo.  Like Pluto, Mars' atmosphere, climate, temperature and orbit are dynamic and there's still a lot we don't know about how they worth together to form a predictable climate on Mars.

As an FYI, dust storms are nothing like in the movie the Martian, but they can cause a visible haze over the planet that can obscure viewing it's surface.   A nice short article on Mars' dust storms here and here.

I think it's kind of cool that Mars forms an annual "cloud", not unlike how parts of Earth form annual snow and ice covers in winter that thaw in Summer, but we don't know nearly enough about Mars, even with 20 Mars years of observation, to make good long term predictions on Mars' climate.

Venus

Venus, as far as I know, pretty much never changes.  It's the same temperature, Winter, Summer, Pole, Equator.  There might be some slight variation of Venus temperature depending on elevation, like the gas giants, but I'm not sure.   Venus traps heat so effectively that is changes very little, in fact, I've never seen one article mentioning Venus' surface temperature changing as much as 1 degree.   Venus reflects most of the sunlight that hits it away anyway, so it would be less prone to solar variation anyway.

I won't do Mercury cause I don't know enough about it, but that's 7 out of 8 anyway.

. . .

If your question is about having a good understanding of climates on other planets, then decades of surface temperature data wouldn't be enough even if we did have it.  Centuries of accurate surface temperature data, you could begin to recognize patterns, for the inner 4 planets anyway, but decades isn't enough to differentiate between a baseline, a natural fluctuation or a transition period.

If anyone tells you there are no SUV's on mars, so how could Mars be warming, get away from them as fast as you can.  They're not interested in science.


How to calculate geocentric transformation to heliocentric coordinates? I have Sun data (lon+lat+distance) in geocentric coordinates:

but I want to visualize in heliocentric way. How to calculate that?

Earth's position as seen from the Sun is directly opposite the Sun's position as seen from Earth, at the same distance.
In ecliptic coordinates,
$$ \begin{align}
l_\oplus &= \lambda_\odot \pm 180^\circ \\
b_\oplus &= -\beta_\odot \\
r_\oplus &= \mathit{\Delta}_\odot
\end{align} $$
The heliocentric position of the Sun is always at the origin ($r=0$) by definition.


Why can't we see Saturn's phases from earth? In the Astronomy Picture of the Day from August 8, 2020 ("Crescent Saturn"), the caption states

From Earth, Saturn never shows a crescent phase

Why is this? We can observe phases on planets such as Venus which have an apparent size a lot smaller than Saturn. Saturn is not tidally locked to the sun (only Mercury is).
So why does the sunlit face of Saturn always face Earth?

Phases are just different perceived illuminations of an object at different illumination and observing angles. If the observer is, with respect to the object, located in a similar direction as the light source shining on the object then you should expect to see the vast majority of the object illuminated, if the observer is located in the opposite direction you would see the object back-lit, and if you are at right angles observing the object with respect to the direction of the light source you would see the object half-illuminated.
Since Mercury and Venus are always inside Earth's orbit and move with different rates around the Sun as Earth does, the Earth (the observer) is able to locate itself at any angle with respect to the light source that shines on the observed planet. This means that you can see any phase of Mercury or Venus (except from a perfect 100% illuminated phase due to the body of the Sun blocking the view). Here you have an example for Venus:

Now think about what happens from the point of view of an Earth's observer for the exterior planets. The Earth will never have a chance to see the planet's back-lit side since there's no position in its orbit that would allow for this. As seen from the other planet, the Earth is always close to the Sun, so it can be seen almost exclusively during the day, which means that from the Earth you almost exclusively see illuminated regions of the planet.
Mars is the closest exterior planet, so the Earth manages to gain enough elongation to see a bit of the night side, but it is a tiny fraction of the disk as viewed from here. Here you have a picture of the phases of Mars as seen from the Earth:

This is even worse for far away planets. Saturn is so far away that from its vantage point, that the Earth is basically always close to the Sun (to the light source).  From the Earth, Saturn seems always fully illuminated, by an extremely small margin that allows to see a slim crescent of darkness in perfect conditions. Only with spacecraft like Cassini and Voyager have we been able to see what Saturn looks like from behind. The first time humanity did this was in 1980 with photographs like this one (from Voyager):

Before that we had never seen the night side covering more than a percent of the disk. Even the shadow of Saturn cast over the rings is nearly impossible to spot from the Earth. Look at this amateur photograph that shows precisely a bit of that shadow over the rings behind Saturn (bottom-right part):

That shadow is almost non-existent from Earth's point of view, and it's all because Earth's orbit is inside Saturn's and Saturn is far away from the Earth. The Earth is always so close to the line connecting the light source (the Sun) and the illuminated object (Saturn), that you shouldn't expect any more phases than "full" from here. Simple geometry.


Are there known objects rotating at close-to-light speeds? For example, $0.1c < v < c$? Is this actually possible?

Background: is that true that the black hole in the center of the Milky Way rotates once per 20 minutes?

A particle travelling in a circle of radius $R$ at velocity $v$ has an acceleration of $v^2/R$. So either $R$ is very large, or a very strong force is acting to stop the particles at the surface of the object flying away. We see no very large objects (galactic clusters, for instance) rotating this quickly, so we need to look for strong forces.

One option is gravity. We know of sub-millisecond pulsars, which we believe are neutron stars with a radius of perhaps 10 km, rotating 1000+ times per second. This gives a surface velocity at the equator of about $60\,000 km/s$ which is $0.2c$. So those meet your conditions.

Defining how fast the surface of a black hole is rotating is difficult on several levels, but black holes do have angular momentum, and there is a dimensionless number relating that to their mass (or equivalently radius) which can be loosely thought of as the fraction of light-speed at which the event horizon is moving. This paper measures a value of about 0.44 for that parameter for the black hole in the centre of the milky way.

At the other extreme, this stackexchange answer suggests that the electrons in a hydrogen atom rotate about the nucleus at something a bit less than $0.01c$. This scales up linearly with the charge on the nucleus, so the inner electrons in a lead or uranium atom are in some sense rotating at close to light-speed.

Finally, the particles in a large particle accelerator, rotate around the centre of the accelerator at velocities very close to that of light.


Why are these objects moving at Vastly Different Speeds along the same orbit? UPDATE: The on-line simulation seems to be working beautifully now! I'd recommend anyone to go back and take another look to enjoy both the mathematical beauty of orbital mechanics, and the aesthetics of the solar system visualization!

New York Times article Visualizing the Cosmic Streams That Spew Meteor Showers links to a solar system viewer that allows one to visualize aspects of meteor-shower-inducing comet orbits by Ian Webb.

Now, astronomers and engineers have created an animation that lets you witness the entire journey. Using data from the Cameras for Allsky Meteor Surveillance, a network of about 60 cameras pointed at the sky above San Francisco Bay, researchers have recorded more than 300,000 meteoroid trajectories since 2010. They plan to use the data to confirm more than 300 potential meteor showers that scientists have observed, but not verified.
“Each dot that you see is a shooting star that was captured by one of our cameras,” said Peter Jenniskens, an astronomer at the SETI Institute and NASA Ames Research Center in Silicon Valley, Calif., who runs CAMS. His interactive transforms meteor showers like the Geminids and the Orionids into shimmering rivers of space rocks. Viewers can pinpoint the moment the Lyrids or Eta Aquarids light up the night by watching when their streams intersect with the Earth’s orbit, shown in blue. There is even an option to see all the meteor showers at once, making it look like a meteor hurricane.

I need some help understanding what I'm looking at. When I first open it up I believe I see a number of objects following the same general hyperbolic (or very elliptical) orbit. What bothers me is that some seem to zip past very fast, some seem to crawl slowly, and some are moving at intermediate speed. If they are all associated with one primary comet, and follow similar paths in this large (roughly 10 AU) view, shouldn't they have at least roughly similar velocities?
Edit: I've snapped screenshots from 30-10-2017 to 04-11-2017 at a speed setting of 0.005 and made two GIFs below The second one is annotated with a red, green, and blue arrow indicating three objects (slow, medium, and fast) following nearly the same orbit with at least a factor of 10 different speeds between them.
I have replayed viewing from different angles, they follow nearly the same orbit in 3D, all the way around the sun and back into space - it is not related to a particular view.
I believe this to be completely unphysical!

 Red, Green Blue (slow to fast)

Looking at the simulation, and the gifs attached here, some things are clear. The dots cannot represent actual meteoroids in their orbits, as the 6 orbital elements fully determine the state vectors of a body. You can't be in the same orbit, at the same place, at the same time and have very different velocity. Thus the dots don't obey Kepler's third law.

It is not clear that they obey the second law either (that equal areas are swept out in equal times) They do seem to move faster at perihelion but perhaps not fast enough.

The graphic doesn't seem to show the "lumpiness" of some meteor streams. Notably the Leonids, which have a very intense storm, roughly every 33 years. Other streams are also lumpy which doesn't seem to be apparent in this graphic.

What it does show is the variation in orbit of the meteoroids that form a meteor shower. This is where (I guess) Nasa was involved. From the observation of a meteor (ideally several observations of the same meteor to triangulate its position) it should be possible to work out its orbit before it fell into Earth's gravitational field and hit the atmosphere. Observe enough meteors and you can get an idea of the distribution of orbits: the mean and standard deviation in the inclination, eccentricity, semimajor axis, ascending node and angle of perihelion. From these, you can get a distribution of elliptical orbits. There may be some modelling work do be done: we can only observe those meteorids that hit the Earth, but we can suppose the they are distributed all around the parent body. But from our sample of meteoroids that hit the Earth, we can model the distribution of those that never will.

The dots are drawn on these Keplerian orbits, but the speed at which the move on the orbit doesn't seem to represent the actual speed of meteoroids in space. Rather it is a way of illustrating the variation in elliptical orbits, without drawing a solid ring for each orbit.


Are there any predictions of what galaxies exist in the Norma cluster/ Abell 3267? I am working on a map for a science fiction story and using what information is available I have a good idea of locations of nearby galaxy groups and clusters and where the known supermassive black holes are.
As the Norma cluster is a big area of debate and due to that and other areas being obscured by our own galaxies, observing what is in those areas is difficult. Going off Wikipedia this cluster has a binding mass of 1E15 solar masses, which is the same as the Virgo cluster and only one galaxy is mentioned which is ESO 137-001.
Is this mass based of our galaxies motion in that direction so mass behind the cluster (some believe the Shapley supercluster) is giving the value we have or is that a known mass of galaxies in that cluster? If so can we predict if it has a similar number of large galaxies and supermassive black holes as the Virgo cluster or is there other predictions for the Norma cluster?

VizieR lists two catalogues of Norma cluster members:

J/MNRAS/383/445 Woudt, P.A. et al. (2008) "Radial velocities in the Norma cluster (A3627)"
J/MNRAS/396/2367 Skelton, R.E. et al. (2009) "NIR Ks photometry of Norma cluster (A3627)" (this one contains angular diameter information)

The paper associated with the first of these, "The Norma Cluster (ACO 3627): I. A Dynamical Analysis of the Most Massive Cluster in the Great Attractor" describes the methodology used for estimating the cluster mass from their measurements of the radial velocities (in particular their dispersion) of the galaxies. From the paper:

For the determination of the dynamical mass of the Norma cluster, we have used both the virial theorem ($M_{\rm VT}$) and the projected
mass estimator ($M_{\rm PME}$), see equations (21) and (22) of Pinkney et al. (1996). The use of the biweight velocity centroid and scale (Beers et al. 1990) in the virial theorem (instead of the velocity mean and
standard deviation) leads to a more robust mass estimate ($M_{\rm RVT}$).
The latter is more robust against the effects of contamination by the
inclusion of possible non-members in the analysis. The projected
mass estimator (Bird 1995), on the other hand, is sensitive to the
presence of (spatially separated) subclusters due to its proportionality to the projected distance between galaxy $i$ and the cluster centroid
($R_{\rm \perp,i}$) (see equation 22 in Pinkney et al. 1996).

They also note that the masses they derive are consistent with masses estimated from the X-ray emission by Böhringer et al. (1996) and Tamura et al. (1998).
As regards what types of galaxy these are, unfortunately neither of the catalogues provide morphological types directly. On the other hand, they do include for many of the objects a reference to the WKK98 catalogue (Woudt, P.A. & Kraan-Korteweg, R.C. 1998).
WKK98 is also available on VizieR as J/A+A/380/441 and does include the morphological type.
You can use the VizieR query interface to join the tables (or you can download the tables in their entirety via the FTP page on VizieR and do the joining on your local machine, e.g. via the various lookup functions in your spreadsheet software of choice), which will give you a list of the morphological types of the objects.


Brightest Radio Source In the Universe I would have assumed that the brightest radio source in the Universe is a quasar of some kind (perhaps 3C 273) given that the average pulsar has luminosity of $~10^{40} \text{Watts}$ and this is the brightest (optically) and is radio loud.

My professor seems to be of the opinion that the Arecibo message is louder, but that does not seem to pass the straight face test, given that it was only transmitted at a power of only $1\text{MW}$, and only a very small fraction of the pulsars output (a mere $1/10^{34}$th) would be sufficient to outshine the message.

Is one of these the brightest radio source in the universe (excluding the Big Bang if it matters), or is it something else entirely?

Additionally, is Watts even the right unit to be measuring this in?

Update: Here is the text of the question (it was a true/false, and the answer is true):

In 1975, Frank Drake and Carl Sagan sent a message in to the Universe, directed to extraterrestrials, from the radio telescope at Arecibo in Puerto Rico that, for a brief time, made the Earth the brightest radio object in the Universe.

1MW of transmitted power is very modest by RADAR terms, to give one example. Thus, you probably need to establish what exactly your professor means by "brightest" (may it be a specific wavelength or point in time, or some other definition). Otherwise, there's no reason to believe that Arecibo transmission was any special or outstanding from radio engineering stand point, while there were such nice toys operating at the very same time:

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


Does wavelength affect redshift caused by the metric expansion of space? Does wavelength play a role in cosmological redshift? Do we see certain wavelengths affected less by expansion or notice any delay in arrival of certain wavelengths?

How accurate are any observations?

Standard cosmological models predicts that the cosmological redshift and the speed of light are wavelength-independent. This result is confirmed observationally e.g. by Ferreras & Trujillo (2016), who used $\sim500\,000$ SDSS galaxy spectra down to a precision of $\Delta z \sim 10^{-6}$ and $\Delta z \sim 10^{-5}$ for galaxies at $z<0.1$ and $z>0.1$, respectively.


Do stars of a galaxy change their positions relatively to each other? Complete astronomy noob over here who would be happy if he get a simple answer (and who is also aware that this may be not possible)...

I've learned from a tv documentary that the stars at the edge of the galaxy are not traveling more slowely than the ones closer to the center.

Does this also mean that all stars in a galaxy do not change their positions relatively to each other?

To put it simply: If I note the relative position of a star compared to our sun, and then do it again 10 (100 / 1000) years later: Are the coordinates of that star identical or will the star be in a different relative position?

I would assume that the star is in a completely different (relative) place given the speed star systems are travelling because the galaxy is rotating AND the "fact" (read: I don't know if this is really a fact) that the solar system is oscillating through the galactic plane.

Stars do in fact move relative to one another within galaxies of all types. The orbital period of stars in a typical spiral galaxy (at around the same distance as the Sun is from the center of the Milky Way) is on the order of hundreds of millions of years. For the Sun it's something like 230 million years (source). A particularly old person (~100 years old), will have been around long enough to see something like one-tenthousandth of 1% of the orbit of stars in this region of the galaxy! Most of these stars simply do not move much relative to us in that amount of time.

The exception to this view of the motions of the stars is the Sagittarius A* (which is not a star at all, but is in fact a supermassive black hole at the center of the Milky Way). The period of stars traveling around it are on the order of 10's of years.

Also, to give you a rough idea of how galaxies look dynamically, here's a link to a particularly nice galaxy simulator implemented in the browser (created by Adrian Price-Whelan).


Are meteors and meteorites considered "Small Solar System Bodies"? The difference between meteorites, meteors, and meteoroids is one of altitude relative to a celestial surface: in space, it's a meteoroid; in the atmosphere, it's a meteor; and on the surface, it's a meteorite. (See here and here for more information.)
The International Astronomical Union (IAU) defined the term Small Solar System Body (SSSB) in 2006 with Resolution B5 as

(3) All other objects³,except satellites, orbiting the Sun shall be referred to collectively as "Small Solar System Bodies".
[Footnote 3] "These currently include most of the Solar System asteroids, most Trans-Neptunian Objects (TNOs),comets, and other small bodies."

And, although this official definition clearly states, "All other objects ... orbiting the Sun", I'm not sure how verbatim this definition is meant to be treated.
What I mean by this is, well, for example, in Law, all definitions are, by default, unless previously stated, treated in an context of exactness. But although science is oftentimes based on specificity, it is not always, and the degree thereof in this context is thus not wholly clear to me.

Your question is really well answered by this article at the Encyclopedia Britannica website, written by Prof. Ed Tedesco of the Planetary Science Institute.

Two quotes that resonate well are...

The small bodies populate the solar system in vast numbers and include ... the fragments of these bodies—commonly called meteoroids—over a continuum of sizes down to microscopic grains known as interplanetary dust particles or micrometeoroids.

And...

The term small body is sometimes restricted to objects that can be observed telescopically while still outside Earth’s atmosphere. In practice, this results in a lower limit of a few metres on the diameter of a small body.

So in general you can assume that anything other than the Sun, planets, minor planets, and their moons is a small solar system body. But in practice it makes sense to limit the minimum size for a SSSB, and the "observable telescopically" caveat provides that limit so that we don't have to include single molecules or one grain of sand in the list.


Are all satellites thought to be the result of collisions with the planets? The leading hypothesis for how the Moon was formed was that another object collided with the Earth, throwing off a large lump of matter which formed into the moon.

Is this thought to be the case with all satellites? e.g. Phobos, Deimos, Charon, Titan, Europa, etc. ? If not, then why do we need such a dramatic "origin story" for the Moon but not the others? Why do we not just assume the Moon was formed the same way as all satellites?

There are three main formation scenarios for planetary moons.

The giant impact hypothesis: The satellite forms as a consecuence of an impact between the planet an a large planetesimal. The Moon is an example, and one of the arguments is that the chemical composition of the Moon matches that of Earth with a significant accuracy suggesting that it is a in part our planet and in part the original impactor (Theia). We also know that the Moon has been getting farther away from Earth because we have evidence that it gained orbital potential energy by absorbing it from the rotational energy of Earth. We know this because days were not 24 hours long a few million years ago and we can keep track of those changes in the rotational period of Earth using rings in fossillized corals (those have rigns like the tree rings but ones that generate on a daily basis). We then can see that the moon was extremely close to the Earth some billion years ago (we have more evidence of this from the fact that tides were huge in those times and led geological evidence of daily floods across the recently formed planet). If you keep going back in time you see that the Moon was basically emerging from Earth. There are many more evidences of this scenario for our moon.

The accretion scenario: The satellite coalesced from a disc of material around the new-born planet (just as the planet accreted from the protoplanetary disc), the so called circumplanetary disc. As an example we have the four galilean moons around Jupiter (Io, Europa, Ganymede and Callisto). Since the disk was relatively flat the moons formed in the same orbital plane, also they move in the same direction as the planet rotates (which makes sense as both are generated from the same material spining with a certain angular momentum). This is the most frequent scenario for large moons. Our moon couldn't form like this because the expected size of the circumplanetary disc was in no way as massive as our moon is today (Earth is a tiny planet and it has a huge moon in relative terms).

Capture scenario: The satellite formed elsewhere in the Solar System as an independent minor body. With time some dynamical interaction might have led the object close to a planet and both got gravitationally binded. An example of this is Triton, Neptune's largest moon. The retrograde orbit is inexplicable in terms of the accretion scenario and the energy needed for a gian-impact scenario to work on Neptune is too large. Triton was captured (we think it formed as another planetesimal in the Kuiper Belt as it shares many chemical features of Pluto and other objects of the region). There are not so many moons on Neptune probably because they dissapeared (crashing into the planet or been ejected) just as Triton arrived to the system and dynamically destabilized their orbits. Another clear example are the tiny irregular satellites of Jupiter. This scenario is very difficult to immagine for Earth since capturing a massive moon like ours and making the orbit circular would have been a feat in terms of how precisely tunned would the orbit insertion parameters had to be. The giant-impact scenario leads to the current situation in simulations for a larger range of impact parameters, thus is much more probable in statistical terms.

There are some less frequent and some speculative scenarios also:

Eyecta fragments from other moons: Some satellites might have their origin on other satellites. A large impact might eject material into orbit. An example would be Hippocamp (a Neptunian moon) which is now regarded to be a fragment stripped out from Proteus (a larger moon).

Lagrangian/Trojan moons: This is similar to the circumplanetary disc scenario but here the accretion in the disc of the planet is further stimulated in certain regions because of a moon that formed a bit earlier. An orbiting body can generate five equilibrium points (Lagrange points) by sculpting the gravitational landscape. Two of those equilibrium points (L4 and L5) are stable equilibrium points; thus they are like gravitational traps where matter can accumulate until a new moon is formed. As a potential example we have Telesto and Calypso in the Saturnian system. They both lie on the L4 and L5 Lagrange points of Tethys (a much larger moon with great gravitational influence). They might have formed as regular objects and then got trapped on the equilibrium points or they might have actually formed there as matter coalesced on those gravitational traps.

Sprayed into existence by cryovolcanism from another moon: It sounds ridiculous because it is an hypothetical scenario, I just think might happen somewhere in the universe. Take a look to Enceladus (a large active moon of Saturn). Enceladus has water plumes and jets that shoot material from its interior into space through craks in the icy crust (because tidal stresses heat the interior like a pressure cooker and the pressure gets liberated that way). The entire E-ring on Saturn was created by orbiting ice and dust grains sprayed by Enceladus. We know that the E-ring has a mass of nearly $12\cdot10^8\; kg$ and we know that it is so far from Saturn as to make possible the coalescence of this matter (tidal forces wouldn't disrupt it: see Roche limit). Thus it is possible that the material from the ring could make a moon with the density of Aegaeon (another saturnian moon) and with a diameter of just $162\; m$ (a third of the size of Aegaeon). Enceladus prohibits this because of the gravitational influence it excerts on the E-ring, it also prohibits it to grow (the ring would be more massive if the material was not continuosly reabsorbed by Enceladus). But if Enceladus migrated to another orbit in relatively short timescales then at least I think is possible for this to happen. The new moon greated from the material of the ring could then end up coliding with Enceladus since both would probably still interact chaotically. The material once sprayed out of its interior would had returned to home.

Centrifugal breakup scenario: This is also hypothetical but we think this happens a lot on asteroids. Lightweight moons could be ruble piles just as many comets and asteroids are. Losely bounded material with little cohesion. If the moon starts to spin faster and faster (because of some mechanism like the Yarkovsky-O'Keefe–Radzievskii–Paddack effect), it could eventually break into two pieces due to the extreme centrifugal forces (described as seen from the co-rotating reference frame of the original moon). It is believed that asteroids like $1999 \;KW_4$ splitted in two because of this effect. I don't see any reason why a moon couldn't do the same, forming a new independent moon in the process.

Moon made of chunks of other disrupted moons: As crazy as it seems, it is one of the formation hypothesis regarding the formation of Miranda (one of Uranus satellites). Miranda's surface is so complex and varied that some speculate it might have formed as several pieces that orbited Uranus came toghether gently. These pieces might have been chunks from other moons or might have been chunks from an earlier iteration of Miranda itself, fragmented after a disruptive event. The geology on each chunk would have evolved independently until they re-assembled together. But this is also quite speculative.


What is the mass limit in a stellar accretion disc? I became curious about the maximum mass in a star's accretion disc while watching an episode of Star Trek involving a Dyson Sphere. I wondered if some maximum amount of stellar material would limit natural and artificial structures around any system of one or more stars. What affects these limits? Is there a logarithmic/linear/exponential correlation between stellar mass and maximum accretion mass?

The accretion disc is formed by material in orbital motion against a central body, which can be a star.

The size, mass and other characteristics are usually determined by the central object, in this case the star.

In general, the protoplanetary accretion discs are the largest ones (with the largest mass) and as the age on the central star increases, the average size decreases.

This is because as the age of the system increases, more mass is drawn into the system and less is left out in the disc itself.

Once the central star crosses the Chandrasekhar limit, it will become unstable and become a supernova and once it passes the Tolman–Oppenheimer–Volkoff limit it will become a black hole.

The physics of black hole accretion disc is entirely different and is possible to derive a relation between the black hole mass and size of the accretion disc. The relation appears logarithmic.

Any natural or artificial structure around a star would feel the gravity of the star and consequently, it is the mass of the star that determines amount of the counteracting force (rotation, radiation pressure etc.) required to maintain stability.

For example, as the stellar mass increases, the rotation speed must increase if the orbit is maintained at the present position or the orbit must become larger. In case of stellar radiation pressure is used to counteract gravity, as the mass (and gravity) increases, the system becomes unstable (as the ratio of radiation pressure to gravity is constant).


A rough analytical expression for the Milky Way's radial mass distribution? I found the image below in Space.com's article This 3D Color Map of 1.7 Billion Stars in the Milky Way Is the Best Ever Made, although it is not the map mentioned in the title.

The caption for this image reads:

  This radial velocity image shows the movement of 7 billion stars. The colors run from blue (stars moving at 50 km/s toward us) to red (stars moving 50 km/s away from us). The white color shows when, on average, the stars are not moving in the line of sight with respect to us. Stars lagging behind as they orbit the center of the Milky Way appear to be traveling away from us, and those speeding up appear to be traveling toward us. Credit: ESA/Gaia/DPAC

If you imagine a band along the galactic equator the dominant velocity shows two positive and two negative "peaks", with a zero crossing in the direction of the galactic center.

Purely for fun I wanted to see if I could reproduce this behavior with a simple calculation based on a 2D calculation assuming circular motion and a radial density distribution $\rho(r)$ which I could then use to figure out a rotational velocity distribution $v(r)$, bun I swiftly realized that I have no idea what the density profile would look like.

For the purposes of this simple exercise, what would be an analytical expression that roughly matches the Milky Way's radial density profile, projected on to its equatorial plane?
For spherically symmetric distributions, Newton's Shell theorem allows one to treat all mass inside a sphere defined by an orbit's radius as if it were at the center, and to ignore all mass in the shell outside of that radius. Is there anything like an analog to this for a radial distribution within a plane?

For the purposes of this simple exercise, what would be an analytical expression that roughly matches the Milky Way's radial density profile, projected on to its equatorial plane?

The simplest example astronomer's work with is the radial density profile of a Single Isothermal Sphere (SIS). It is called this because it is spherically symmetric (and thus applicable to a 2D plane for your purposes) and all objects orbit with the same velocity (and thUs have the same "temperature", hence isothermal). The density profile takes the form:

$$\rho(r) = \frac{v^2}{4\pi Gr^2}$$

where $v$ is the rotational velocity. Note that you may see other formulations which use $\sigma_v$ rather than $v$. In this case they're using the velocity dispersion which is slightly different than the rotational velocity.

Other, more realistic density profiles have been found by running simulations of the Universe and matching functional equations to the density profiles of the resulting galaxies. Such popular results are the NFW profile and the Einasto profile.

The NFW profile is a two-parameter function, given by

$$\rho(r) = \frac{\rho_0}{\frac{r}{R_S}\Big(1+\frac{r}{R_S}\Big)^2}$$

where $\rho_0$ and $R_S$ and two, halo-dependent parameters.

The Einasto Profile is again a two-parameter model given by

$$\rho(r) \propto \exp(-Ar^\alpha)$$

where $A$ and $\alpha$ are configurable parameters.

  For spherically symmetric distributions, Newton's Shell theorem allows one to treat all mass inside a sphere defined by an orbit's radius as if it were at the center, and to ignore all mass in the shell outside of that radius. Is there anything like an analog to this for a radial distribution within a plane?

The Shell theorem for gravity does not extend to a 2D ring. However, I will say that when talking about the orbits of stars in galaxies, the mass of stars outside a star's orbit are generally considered negligible. The primary reason for this is that it is Dark Matter which comprises most of the mass of a galaxy and contributes the most to defining a star's orbit in a galaxy. The Dark Matter halo is often assumed to be spherically symmetric, in which case Newton's Shell theorem does apply and the mass you're concerned with in determining a star's orbit, is the mass of the Dark Matter halo interior to the star's orbit.


Is there any difference between parallel universes and multiverse? Nowadays we come across with these two terms - parallel universes and multiverse. I wonder if this is the same thing or its two entirely different principles because the names are a bit confusing to me. Can anybody please shed some light on this?

I'd like to see the notion "parallel universe" restricted to universes similar to each other in some sense. This applies best to the many-worlds interpretation of quantum theory. Two universes in this sense can be almost identical, just different in a couple of quantum states.

Two such parallel universes could be interpreted (in the simplest case) as two elements of a Hilbert space. A Hilbert space defines a metric as a measure of distance between two elements (here applied to universes).
This may be seen in some analogy to points (universes as analog) in an Euclidean plane (many-worlds multiverse as analog). Here the notion "parallel" can make some sense. At least, it's possible to compare the items in some way (by the distance between two points/universes).

Talking of universes as parallel in the very general case of items in a multiverse, when there is no similarity at all between these items, would be a bit too lax in my eyes.


When and how was the length of a Mars year determined? I guess it is known for a long time that Mars has an orbital period of around 687 days.

But when and how was this determined?

There are two answers to this, because the period "687" was known long before it was recognised as an "orbital period".

The Babylonians knew by 600 BCE that Mars took 79 years to go through all 12 constellations 42 times, giving a period of 79/42 × 365.25 = 687 days.  But they wouldn't have thought of this as an "orbit" (in the modern sense) since they would not have thought of Mars as being a physical object. Similar knowledge was known in India and China.  It is relatively easy to determine, and just requires the intent to measure and the ability to make careful records.  This value was known wherever people had writing, an interest in the stars, and a stable enough society.

The understanding of the value as being "the time for Mars to orbit the sun once" depends on having a heliocentric, so is due to Aristarchus, Copernicus, Kepler, or Newton (depending on what you mean by "know that Mars orbits the sun")


Question on the singularity theorem I have just started studying Cosmology and we have been asked to prove that in an expanding FRW Universe which obeys the strong energy condition: $$\rho + 3P >0$$
Then there must exist a Big Bang singularity. I can see that this condition implies $$\ddot{a}/a \leq 0$$
which when you plot against t gives an always decreasing rate of expansion over time. This when plotted then shows that if you extend the curve far back enough will cross the t axis at some finite t-value representing a Big Bang singularity. My question is why can't the universe begin with some finite non-zero a value? is there some physical reason that we expect the Universe to begin with a=0?

It does not have to start from the big bang. There are also different universe models such as the big bounce. Where the universe has an infinite past, such that it expands and then contract.
See
https://www.quantamagazine.org/big-bounce-models-reignite-big-bang-debate-20180131/
https://www.wired.com/story/what-if-the-big-bang-was-actually-a-big-bounce/
For more detailed explanations you can look some cosmology textbooks


What types of signals is SETI (and other such efforts) looking for? This seemed to be the closest group for this question.

Every few months there are a few articles asking "Are we alone in the universe" which focus on the Fermi Paradox, the Drake Equation, and end by talking about SETI and its lack of progress. As I understand it SETI is looking for signs of technological civilizations beyond our solar system, both in our galaxy and beyond through radio astronomy.

My question is about what would count as a "sign" of a technological civilization, and is there a maximum distance beyond which such a signal would be indistinguishable from other radio emissions from stars and galaxies?

What signals do SETI receive?
SETI uses radio telescopes to look at large portions of the sky across a range of wavelengths. They believe that we are most likely to find alien civilisation buy looking for microwaves.

How would we know if they were from aliens?
There are radio signals from many different sources bouncing around the universe, many of which we don't quite understand yet. However, natural sources of energy tend to produce a range of wavelengths. SETI are hoping to find "narrow-band signals", which  are strong signals at a single wavelength. These (as far as we know) only come from devices made to produce them, so only a technologically advanced species would emit these signals.

Is there a maximum distance for detecting these?
Yes, these signals would decrease with distance. It is not so much that they become indistinguishable form natural sources, as they are a different type of signal, but as the waves travel they spread out, like dropping a pebble in a pond, so the signal is much weaker as you get further away. Also, space is not empty. If the waves are emitted from a planet in a galaxy, there will probably be lots of dust and gas (interstellar medium) in the galaxy which will absorb the radio waves, so it might be like looking for car headlights on a foggy day - we will see something vague and fuzzy but it will be much weaker if there is more matter between us and them.

If you want to find out more about SETI they have a website with FAQs and news. If you want to get involved there is a project called SETI@home which uses spare processing power on your home computer to analyse the data they are receiving and search for these narrow-band signals.