Is a General Relativity approximation available in Universe Sandbox (1 or 2)? This answer points out that including General Relativity in a simulation can be computationally intensive and may not be present in Universe Sandbox.

However, answers to How to calculate the planets and moons beyond Newtons's gravitational force? discuss approximations that go most of the way toward including main GR effects in a typical solar system simulation where the effects are small.

Are there any options to "turn on" an approximation to GR in Universe Sandbox and/or Universe Sandbox 2? Or plans to add it later? And to double check, is it absolutely certain that only Newtonian mechanics is used currently for propagating orbits?

As noted by @MikeG in this comment, the Universe Sandbox 2 FAQ addresses the question, but not in a satisfying or helpful way:

Does it account for relativity?
No, the physics in Universe Sandbox² is currently only Newtonian.
Why? The short answer is that you need a supercomputer to accurately simulate general relativity. (emphasis added)
Jenn, astrophysicist and Universe Sandbox² developer, explains more in a blog post: "General relativity requires simulating the spacetime itself. That is, taking your simulation space, discretizing it to a hi-res 3-D grid and checking the effect that each and every point in that grid has on all neighboring points at every timestep. Instead of simulating N number of bodies, you are simulating a huge number of points. You start with some initial data of the shape of your spacetime and then see how it evolves according to the Einstein equations, which are 10 highly non-linear partial differential equations."

Why do I say this is not very satisfying or helpful?
Because it's quite easy to nearly completely account for General Relativity (GR) effects in the orbits of solar system bodies just by adding some terms to the ODEs that are integrated to propagate orbits.
Answers to the question How to calculate the planets and moons beyond Newtons's gravitational force? (including mine) show how this is done all the time. I reproduced what's normally done in Python rather than in a compiled language for transparancy, it would be much faster if included in a reasonable simulator implementation.
As explained here, to the Newtonian acceleration given by:
$$\mathbf{a_{Newton}} = -GM \frac{\mathbf{r}}{|r|^3},$$
You can just add this expression:
$$\mathbf{a_{GR}} = GM \frac{1}{c^2 |r|^3}\left(4 GM \frac{\mathbf{r}}{|r|} - (\mathbf{v} \cdot \mathbf{v}) \mathbf{r} + 4 (\mathbf{r} \cdot \mathbf{v}) \mathbf{v} \right),$$
To get close, you certainly don't need to evaluate these GR terms for every pair of objects. Interaction between the Moon and Ceres for example could still be Newtonian.
All approximations are approximate. The FAQ's suggestion that they don't do it because you need a supercomputer to be accurate is slightly misleading. Using this straightforward approximation is much better than not using it in this context.
Yes for Gravitational Waves, collapses, and other effects "requires simulating the spacetime itself. That is, taking your simulation space, discretizing it to a hi-res 3-D grid and checking the effect that each and every point in that grid has on all neighboring points at every timestep" but just adding the major GR effects to the propagation of solar system bodies certainly doesn't need anything more than  whatever computer you are already running your copy of Universe Sandbox 2 on already.
EDIT:
The FAQ answer continues:

We are, however, interested in adding in a few features which would address some effects of relativity. One example is setting gravity to travel at the speed of light, instead of instantaneously taking effect as it currently does. You can read more about these in Jenn's blog post: Gravitational Waves & Universe Sandbox².

I had thought so as well, but according to @DavidHammen's thoughtful answer this is probably not so. I haven't tested this yet, but since this comes from a reliable source, I tend to believe it. Retarding gravity alone, without these other terms, seems to be the wrong thing to do in this case.


How long would an occultation by a TNO last? A trans-Neptunian object, 1 million km from the observer, with an angular diameter of 0.126° occults the Sun (angular diameter of 0.004°) and the TNO and the observer are moving in the same direction, in the same plane.  The TNO is moving 8m/sec with respect to the observer, how long would the occultation last?

Assume the observer is 18 billion km from the Sun.

From the observer's point of view 1 million km away, the TNO's apparent angular motion is

$$\mathrm{\frac{8~m/s}{10^9~m} = 8 \times 10^{-9}~rad/s = 0.00165~^\circ/h}.$$

Assuming that the observer at 18 billion km = 120 au is in a circular orbit around the Sun, the orbital period is
1203/2 = 1320 years, making the Sun appear to move the other way at

$$\mathrm{\frac{360~^\circ}{1320~y} = 0.27~^\circ/y = 3.1 \times 10^{-5~\circ}/h},$$

so the TNO's apparent motion relative to the Sun is

$$\mathrm{0.00165~^\circ/h + 3.1 \times 10^{-5~\circ}/h = 0.00168~^\circ/h}.$$

The Sun would be totally occulted for

$$\mathrm{\frac{0.126~^\circ - 0.004~^\circ}{0.00168~^\circ/h} = 72.6~h}$$

and partially occulted for

$$\mathrm{\frac{0.004~^\circ}{0.00168~^\circ/h} = 2.4~h}$$

at each end.


Why are galaxies disk shaped? I always see galaxies in "disk" shape in pictures. It is like, there is a rotational movement on a plane and the system is balanced by the pulling-in gravitational force which prevents the stars to leap out of the galaxy. What is the physical reason behind this? I have never seen a star in disk shape. A star is nothing but a much much smaller non-solid mass revolving around an axis, just like the galaxies. Does size matter? What makes the difference? Why doesn't a typical galaxy form in a spherical shape rather than a disk shape?

Spiral galaxies may be the most commonly depicted probably because we find ourselves in one, however, galaxies come in many sizes and shapes. Wikipedia's page on Galaxy morphological classification explains the most common classifications.

Answering the question on the title:

  The basic answer to your question is Conservation of Angular Momentum.
  Astronomers have been interested in this question for a long time and
  are still writing papers on the subject.   Even if you assume that at
  the time of the big bang there was no net angular momentum in the
  universe, one would expect that there were local fluctuations.  When
  galaxies initially formed by gravitational attraction these net local
  amounts remained.  Now angular momentum is conserved so when the
  initial collection of stars and gas in one galactic volume start to
  collapse the net angular momentum has to stay the same.  Since angular
  momentum $\vec{L} =m\vec{v}\times\vec{r}$,  for a given $v$ its easier to conserve $L$ if the $r$ is
  large.   In the orthogonal direction (along the axis of the accidental
  angular momentum) there is no such hindrance to collapse and so the
  collection turns out to be disk shaped.   By the way there has been
  several studies on the net angular momentum of a large collection of
  galaxies coming from the Sloan Digital Sky Survey and the answer is
  pretty close to but not quite zero.  Stay tuned.

Extracted from an University of Phoenix Department of Physics article


How to determine the mass of a galaxy? I am having a slight disagreement with my professor.

We can measure the mass of a distant star cluster from:
A.Its color
B.Its radius
C.Its metallicity
D.Smearing of lines in its total spectrum

My thinking was:
Law of Conservation of Energy

Kinetic Energy = Gravitational force

1/2*m*v^2 =  G* m*M/r

v = sqrt(2GM/r)

So it is dependant on both velocity and radius.
Answer choice D would give you the velocity but I responded with answer choice B, radius.
After talking it over with my professor, she responded with:

If we think about it conceptually, the dominant parameter is the velocity.  In other words, you can get a meaningful estimate of the cluster mass from the velocity dispersion without knowing the radius (because all clusters are the "same" size), but not vice versa.
So I'd say it still seems pretty clear that the best answer is the velocities.

Is she right or is the question worded poorly?

I think she should have clarified the question by putting velocity and radius in the same choice.

By looking at her answer, it seems reasonable to think that velocity is more important. However  without knowing the radius we cannot determine the mass of the stellar system. She argued this idea by saying that the size is approximately known for stellar systems. But I don't think that is a solid argument.

Theoritically we should know both R and V to calculate M.

Note: I did not understand why you wrote kinetic energy= gravitational force ? Thats not possible.

In stellar bounded system you should use virial theorem. By stating that $-2<E>=<U>$  and from there you get

$$M=v^2R/G$$


Hypothetical planet jump: Will it perturb star systems? In a more speculative corner of SE somebody was dreaming up a story of planets traveling between star systems through wormholes.
I was wondering how much the appearing/disappearing of a planet the size of Earth would perturb the orbits of the other bodies in a solar system like ours. After all, the bodies in the system all interact, and the current situation appears to me as the delicate result of an evolution during which bodies collided, were ejected or were nudged into specific (dynamic) "places" which are stable, for example because they resonate with or are far enough away from others.
How much havoc would the sudden appearance of an Earth-sized planet in the habitable zone wreak on this balance? Both long-term and short-term (on astronomical scales; short-term may be 100 years) disturbances are interesting. Would it directly perturb the orbits of nearby planets (in our case Earth and Mars) enough to be prohibiting? For example, even relatively minor changes of the orbit can have an impact on climate.
Could it lead to catastrophic events much later, for example because some Trojan bodies are pushed out of their Lagrange neighborhood and start marauding the Solar System, increasing the chance of catastrophic impact events on Earth or an inhabited Mars?

There is essentially no room in our habitable zone for another Earth-sized planet.  The exact timescale for instability would depend on the details of where you put it, but the system couldn't survive long term.
Most investigations of the long-standing question of solar system stability (without any additional planets) find that the system is stable in the long term with about 99% probability, e.g. this page by Scott Tremaine and this discussion on Wikipedia, but that there is some chance of instability even with the current set of planets, and no room for another.
It appears that many extrasolar planetary systems are "full" as well, although selection effects make it hard to know whether there are planets we aren't seeing in systems that aren't yet known to have multiple planets.
The "fullness" of these solar systems suggests that it may be that even more planets form initially, but then some are ejected until the system reaches a stable configuration.
Interestingly, in our own inner solar system, the place where there is the most dynamical space for another planet is inside the orbit of Mercury, say with a semimajor axis of 0.1 AU or so.  That might seem counterintuitive, since the spacings between Earth (1 AU), Venus (0.7 AU), Mercury (0.4 AU), and a planet at 0.1 AU would all be about 0.3 AU, the same in absolute terms.  But what matters most for stability is the ratio of the semimajor axes, so Mercury to an inner planet would be a factor of 4, whereas Earth/Venus is only 1/0.7 = 1.4.
Such a planet isn't in the habitable zone, and it likely wouldn't be stable on very long timescales with our current configuration of other planets, but we do know that many (perhaps even most) extrasolar planetary systems have planets at those very small orbital radii - see the image below, made from data from the NASA Exoplanet Archive.  There are strong selection effects there (close-in planets are the easiest to detect with both the transit and radial velocity methods), but the high frequency does tell us that such planets are relatively common and can have stable orbits.


Force exerted by space debris on satellites Hypothetically, if space debris were to collide with the currently orbiting satellites, how much force would they exert, and how much damage would they make?
Will the satellites be able to withstand that much force, especially if they don't have Whipple shields?
Since space debris travel at tremendous velocities, I am guessing the force is pretty large too. How do you calculate the force? Is it a simple $\rm{F=m \cdot a}$ calculation?

The force depends on the details of the collision. Depending on how hard the materials involved are, and the angles and everything, it might be a larger force for a shorter time, or a shorter force for a larger time.

What are relatively predictable is the impulse -- force times time, essentially (or the integral of force over time, if you want to be exact) and the energy released by the impact. For a small piece of debris these are determined by the mass and velocity of the debris with respect to the satellite.

Relative velocities are typically a few km/s, so, 1 mg piece of debris (a paint flake or something like that) will deliver about $10^{-3} kg m/s$ of impulse, and about $10--100J$ of energy.  The impulse will change the orbit of the satellite (typical mass perhaps 1 ton) by very little. Perturbations from tides and other effects will be bigger. The energy is more of a concern -- depending on what it hits something may get hot, or have a hole made in it.

A much larger piece of debris (say 1g -- a small screw) carries 1000 times the impulse and energy and may will wreck a satellite (although if it hits a solar panel or something it may just make a hole and go on through). If the satellite survived, then it the change to its orbit would be noticeable, but not a huge deal.


Is it possible to explain the difference between synodic month and sidereal month by degrees? Is it possible to explain the difference between synodic and sidereal month by degrees - in the following way?

Just recently I learnt about these two types of months, and I want to see if I can understand the difference by degrees method. Assuming the orbit's moon is 360 degrees, the sidereal month starts from the point where 1st degree to (include) 1st degree, while the synodic month starts from 1st degree to the 0.5 degree (after full orbit's moon, because the angular size of the moon is 0.5 degree). Is it possible to understand why it that's way?

A better diagram is something like this:

Day 0. The Moon is aligned with the Sun (New Moon) and some star X behind the Sun (of course :-), as shown by the arrow at 1. (Neither the Sun nor the star X are shown in the figure.)
Day 27.32. One sidereal month after figure 1. The Moon has travelled through 360 degrees orbiting around the Earth. (During that time, an arrow from the Earth through the Moon is pointing at different stars.) After one sidereal month, the Moon is again aligned with star X, as shown by the solid arrow in figure 2. (The arrow in 1 and 2 are parallel because the star is very far away.) Because the Earth has moved in its orbit around the Sun, the position of the Sun (shown by the dashed arrow in figure 2) has changed relative to figure 1. The Moon is still a few days prior to New Moon.
Day 29.53. One synodic month after figure 1. The Moon is again aligned with the Sun (New Moon), as shown by the solid arrow in figure 3. The direction to star X is shown by the dashed arrow which is parallel to the solid arrows in figures 1 and 2.

Note how the arrows pointing to the Sun all point to the center of the Earth's orbit while the arrows pointing to star X are parallel.

You are probably interested in the angle between the two arrows in position 3; that is, how far has the Moon moved in its orbit from the position of the sidereal month to the position of the synodic month. The average angle = (360 degrees)/(27.32 days) * (29.53-27.32 days) = 29.12 degrees.


Have we observed any rogue/wandering stars? Do we know (have we observed and cataloged) any rogue star, being not part of galaxy, but drifting somewhere in inter-galactic space?

I know that determining if a star is a part of galaxy or not is a matter of definition, but let's take those, that inter-galactic space is where the gravitational influence of any galaxy isn't dominating.

Different types of observations have been made:

hypervelocity stars: there are stars in the halo of our Galaxy (it is kind of the suburbs of the Milky Way) with such velocity that they are escaping it, and aiming to the intergalactic medium (IGM). There not yet fully in the IGM, but the're on their way. (Reference)
molecular clouds in the IGM: there are molecular clouds detected in the IGM, that are forming stars. Stars are not directly observed, but these star forming molecular clouds are proof of their existence. (Reference)

It's hard to do better, because it is hard to observe directly stars outside of the Milky Way.


What's a good model for the vernal equinox year? Can it increase forever? Duncan Steel (2002) makes an interesting argument that the notion of year is (1) somewhat arbitrary and, more interestingly (2) one notion of year can increase while another decreases.

His examples are tropical year (which decreases over time, at least in the linear Newcomb model) and the vernal equinox year which seems to be currently increasing.

Actually looking at the data that his point is based on, namely Meeus and Savoie (1992), we can find another measure that is decreasing, namely the solstice year(s), either June or December; the autumnal equinox year is also decreasing:

Steel actually uses the fact that the  vernal equinox year is increasing as an argument that the Gregorian calendar is "getting better" in a sense that it's closer to hitting its target year length. (The opposite is the case if one consider say either of the solstice years... or the tropical year, which is the point that a lot of other people make, e.g. Moyer (1982).)

What I really want to ask here: assuming a simple linear decrease model for the tropical year (e.g. Newcomb), will the trends of the equinox and solstice years reverse at some point in the future?

(And if it's not too much extra to ask:) What are the implications of the more complex models for the tropical year (e.g. cubic VSOP) on the evolution of the equinox and solstice years?

The main factors that affect the length of the March equinox year (*1) over time can be divided into three components:

Changes in the length of the solar day
Changes in the length of the mean tropical year
Difference between the mean tropical year and March equinox year

Of these, the change in the length of the solar day is most significant in the long term.

The length of the solar day

The length of the solar day is continuously increasing at a rate of roughly 1.75 milliseconds per century on average. This means that if we measure any type of year in terms of solar days, its length will decrease simply because our measuring unit is becoming longer; this comes to about 0.64 seconds per century, or 7.4 millionths of a day per century. In particular, calendars count solar days, which means that any calendar with a fixed average year length will inevitably become inaccurate.

Change in the mean tropical year over time

The change in the mean tropical year over time (as measured in atomic time) is mainly an oscillation of period around 41,000 years and amplitude around 15 seconds, which is linked to the obliquity cycle. The long-term average (over millions of years) is relatively stable in the vicinity of 31,556,924 seconds (365 days 5 hours 48 minutes 44 seconds).

Combined with the above, the mean tropical year when measured in solar days decreases over time at a slightly varying rate.

Difference between the March equinox year and mean tropical year

The difference between the March equinox year and mean tropical year depends on where the vernal equinox is in Earth's orbit, and is thus affected by the precession of Earth's orbit. In short, when the March equinox is near perihelion the March equinox year will be longer than the mean tropical year, and when it is near aphelion the March equinox year will be shorter (as a reminder, perihelion is when the Earth is closest to the Sun, and aphelion is when the Earth is furthest away). (*2)

Kepler's laws tell us that the Earth moves faster in its orbit when it is nearer to the Sun; this leads to variations of a few days in the lengths of the seasons. Due to apsidal precession, the aphelion and perihelion move slowly over time; this movement causes the variation to show up when computing the March equinox year. Specifically, as the apsides move forward in Earth's orbit over time, a tropical year is slightly less than one orbit relative to the apsides (one such full orbit is known as an anomalistic year). As a result, a small chunk of the orbit is left out, and the duration of that chunk depends on where it is relative to the apsides i.e. shorter near perihelion and longer near aphelion. (*3)

Since perihelion is currently in January and advancing towards the March equinox, the March equinox year is increasing relative to the mean tropical year; the same holds for the June solstice. The rate of this increase will slow down as perihelion approaches the equinox. Once perihelion passes the March equinox in around 6000 CE the difference between the March equinox year and the mean tropical year will decrease; at the same time the June solstice year will be increasing at its fastest rate, and the September equinox year will begin increasing relative to the mean tropical year as the perihelion advances towards that equinox.

Note that the size of this effect will be influenced by the obliquity of Earth's orbit; as the obliquity decreases over the next few tens of thousands of years, the magnitude of these differences will also decrease.

Results

When we combine these effects, we find that each equinox and solstice has a period of relative stability alternating with a period of fast decrease. In stable periods, the increase due to apsidal precession cancels out the decrease due to lengthening of the day, while in fast decrease periods both effects add together.

It just so happens that the March equinox has been in such a stable period for thousands of years. The March equinox year will stop increasing around 3000 CE, then begin to decrease. By that time, the June solstice year will be increasing, while the September equinox year will continue decreasing for a long time, and the December solstice year will be in the middle of its fast decrease period.

All of this is most easily seen in the graphs on Irv Bromberg's excellent page.

(*1) Of course, the same factors apply for any other year based around a fixed solar longitude, such as the June solstice year or the Besselian year.

(*2) This is largely a reminder for myself -- I mixed them up while writing this post.

(*3) If Earth's orbit was in the opposite direction for some reason, this would be reversed, with a tropical year being longer than an anomalistic year and an extra chunk of orbit being added.


Enceladus; why use the words "geysers", "jets", and "plumes" interchangeably? Background, Europa:

In an earlier question How can “Geysers” on Europa reach heights of 100km? I brought up the use of the word "geyser" in association with the reported confirming observations of water vapor plumes on Europa. Europa size and surface gravity are nearly as large as those of Earth's moon. With an escape velocity of about 2,025 m/s it was hard for me to understand how an actual geyser could exist within the ice, a geyser so powerful that it could be shooting liquid water from the surface at 2 kilometers per second.

Part of the problem here is that I still can't be absolutely sure there is a clear definition of what is or isn't an actual geyser on solar system bodies other than Earth. On Earth, while a geyser is usually powered by geothermal conversion of water into steam, it pushes additional liquid water up to and above the surface as liquid. I believe that other sources of pressure may still result in proper Earth geysers.

Most scientific references I found referred to the vapor as plumes or jets, the word geyser was used in more popular and general interested articles.

It seems one likely explanation is a venting of water vapor (gas) from a particularly shaped vent can accelerate the molecules to an average velocity that is several times the speed of sound by converting random thermal motion to directed vertical motion, as described in a U.T. Austin thesis by Jared James Berg. Simulations there showed that this jet phenomenon can result in some water vapor reaching a substantial fraction of escape velocity, and rise to 100 to 200km altitudes. The model also addresses the formation of solids as ice crystals within the plume.

About Enceladus:

I have just come across the paper How the Geysers, Tidal Stresses, and Thermal Emission Across the South Polar Terrain of Enceladus are Related, (Porco, DiNino and Nimmo (2014), Ast. J. 148 (3), 45), an in depth study of sources of water vapor from Enceladus. I saw that both geysers and jets were mentioned in the abstract, but at the end of the second paragraph in the introduction I ran into the very interesting sentence:

  (Throughout this paper, we use the terms “jets” and “geysers” interchangeably.)

The escape velocity on Enceladus is only about 240 m/s, nearly an order of magnitude lower than that on Europa, which means much less sub-surface pressure would be required to push liquid water high above the surface into near space. So I can understand that the term geyser might be more likely to be appropriate for these plumes than it would be for the plumes on higher surface gravity Europa.

Question:

I would know why the words "geysers" and "jets"/"plumes" would be used interchangeably. Is this because it hasn't been confirmed yet what they are, and the authors don't wish to show favoritism by choosing one or the other, or are they synonymous and can always be used interchangeably, or si there some other reason?

above x2: Figures 3a and 3b from Porco, DiNino and Nimmo (2014).

Question:
I would know why the words "geysers" and "jets" would be used interchangeably. Is this because it hasn't been confirmed yet what they are, and the authors don't wish to show favoritism by choosing one or the other, or are they synonymous and can always be used interchangeably, or is there some other reason ?

A jet is a generic term to describe the ejection of material.  You'll see it used all over the place in physics (and astronomy).  Jet's can be any form of matter ( solid, liquid, gas, plasma ).
A geyser is a specific type of geological process and, as I understand it, means (at the least) a liquid is being vented (typically water).  The exact process is, of course, not known for certain for non-terrestrial bodies.  I don't think using the term "geyser" is a statement of certainty, but would be used to indicate the general type of event.
I think you're more likely to use "jets" if it's not properly understood and "geysers" when you're either proposing a mechanism that's geyser-like or it's generally accepted that they're geyser-like already.  But I don't think that's written in stone.


When did astronomy first discover that the stars are bigger than the moon? We take it for granted these days that the stars are unimaginably bigger than the planets and the moons. But when you look at the sky, it does not appear this way.

The moon looks bigger and brighter than the stars. This leads me to wonder when astronomy first discovered the relative size of the stars.

Did people believe that the moon was bigger than the stars for a long time?
If so, where was this believed, and up until when?
What discoveries or observations led people to conclude that the stars are larger?

When did astronomy first discover that the stars are bigger than the
  moon?

Your choice of words "discovery" implies evidence, not just hypothesis.  The discovery that stars were like our sun wasn't possible until the distance to some nearby stars was measured and with that, a brightness comparison could be made between stars and our sun.   That discovery was made in 1838.

The first person with that theory goes back to 450 BC.   Short summary and longer summary.

Quote:

  Anaxagoras, who lived in Athens, Greece, around 450 BC (about 2450
  years ago), thought that the Sun and stars were fiery stones, that the
  stars were too far away for their heat to be felt, and that the Sun
  was perhaps more than a few hundred miles in size. With that
  Anaxagoras was, as far as we know, the first one to suggest that the
  Sun is a star. His ideas were met with disapproval and he was finally
  imprisoned for impiety, because his ideas did not fit the prejudices
  of the time.

So, while he suggested that stars were like our sun, his size was way off and it's unclear how large he thought the Moon was, or even, if he had a good idea how large the Earth was.   Erastosthenes' estimate came later.

Aristarchus of Samos came up with a much better, but still quite inaccurate, measure of the Sun and Moon's size and he recognized that the Moon was smaller than the Earth and the Sun much larger.  It's unclear what he thought about stars though.

And much more recently:

  Christiaan Huygens of Holland lived from 1629 to 1695. He determined
  the distance to the star Sirius, assuming that that star was as bright
  as the Sun and appeared faint only because of its great distance. He
  found that the distance to Sirius must be very great. At this time,
  then, the idea that the Sun is a star was considered seriously by
  scientists.

So the idea that stars were like our sun was popular and perhaps even accepted, for over 100 years before the size and brightness of stars was first accurately measured.

  Did people believe that the moon was bigger than the stars for a long
  time?

This is hard to say.  Aristarchus of Samos ideas were less popular than Aristotle's  (who came a little earlier) and Aristarchus' theories weren't central to Ptolomy's model which became the model for over 1000 years.  I've often wondered what might have been different if Aristotle and Aristarchus could have exchanged ideas.  Aristotle might have learned something and re-written his model, but Aristarchus was born too late.

Ptolomy, who's model was accepted by the Roman Catholic Church, didn't pay much attention to sizes and, in fact, his model doesn't account for size well at all.  It's one of the ways that his model breaks down, but his basic model had the Moon closer to the Earth than the Sun, so presumably he recognized that the Sun was larger than the Moon and it's not unreasonable to guess that Ptolomy might have said the planets are smaller than the moon.  The stars, however, were for the most part, not addressed.

If your question is, in particular, what Aristotle, Ptolemy up through scholars in the middle ages thought about the size of stars in comparison to the moon, then, I have to say that I don't know what was assumed about the size of the Moon and stars, though Giovanni Bruno was burned at the stake for suggesting stars were like our sun.  It's entirely possible most scholars thought stars were smaller than the Moon from the time periods between Aristotle to Kepler.  If someone knows more specifics, I encourage them to answer.

Not long after Galileo and Kepler, it seems to have become more generally accepted that stars were large and very distant and the Moon was relatively small by comparison.


What celestial mechanics parameters define a lunar occultation of Venus visible from Earth? As a carry on from the following thread
Are lunar occultations visible to the naked eye?
Is it correct that the following parameters must hold true if we use Moon occult Venus as an example

[LongMoon = LongVenus = RAMoon = RAVenus] AND [DECMoon = DECVenus]
Moon and Venus must both be visible, i.e. not too close to the Sun

I expect both of these events will need to occur above the horizon (ascendant) for the location right?
Am I missing anything? Are there agreed parameters for degrees above the horizon for visibility?
Like eclipses, such events wont be observed everywhere on earth. Is there an orb convention that captures when occultations can be observed on earth? Is there a way to identify where on earth the total occultation can be observed?
Thank you for any guidance with this.

A lunar occultation of Venus happens when light from Venus can not reach the observer because the Moon blocks it.
The simplest model for this would be to draw a line in 3D from Venus to the observer and detect when that line passes through some part of the Moon. However that doesn't take into account the time it takes for the light to travel from Venus to the observer for example.
Angular separation
If you have apparent positions in RA and Dec as seen from the observer, then those take care of several corrections.
In that case, the Moon will occult Venus from the viewpoint of the observer when the angular separation between their centers is smaller than the angular radius of the Moon, which is very roughly 1/4 of a degree but changes a lot during the month because the Moon's orbit is elliptical.
Above horizon
You need to make sure they are far enough above the horizon to be visible to the observer. If the location has a flat, dark horizon, perhaps only a handful of degrees is sufficient. If in an urban setting then it may need to be 10 or 20 degrees to clear the lights and buildings, or more.
Visibility (when above horizon)
To be visible by eye is a tough call. Venus is always roughly magnitude -4 (-3.8 to -4.8) and so if it were daytime and passing behind the dark edge of the Moon, it might be visible in the daytime to a careful observer with excellent eyesight since they know exactly where to look for it. Looking through a cardboard tube to block other light might help, and certainly as you mention the farther the angular separation from the Sun at the time the better! But this is pretty rare and many people would not see it easily.
With binoculars or a telescope it will be a lot easier to see in the daytime, since magnification reduces the surface brightness of extended objects (like the Moon and the sky) relative to unresolved objects as Venus would be in a relatively low magnification view.
If Venus were passing behind the bright edge of the Moon, this might also be difficult to see by eye. The surface brightness of the moon is about -6 magnitude per square arc minute (the apparent size of Venus for less-than-perfect vision) which is way brighter than what Venus looks like, so it's going to be a real challenge to see this situation by eye.
But again with binoculars or a telescope this will be a lot easier.


Are we already moving faster than light? Okay, so the speed of revolution of earth around sun is 30 km/s[1], speed of revolution of sun around milky way's center is ~200 km/s[2], speed of rotation of milky way is 828,000 km/h[3] i.e. 230000 m/s. Since speed of light is 300000000 m/s, so on an even bigger scale (bigger than clusters) are we already moving faster than the speed of light? Or am I totally wrong?

Short Answer

NO.

Long Answer -

The Earth "wobbles" on its polar axis.  This motion is not relevant (IMO) to the calculations needed to return to earth from interstellar space as it is an "in situ" motion, but it exists.
The Earth revolves {spins} on its axis. For these calculations we will use the equator with a circumference of approx. 25,000 miles. One rotation of Earth is approx. 24 hours. Again this is "in situ" motion so not relevant to space travel, but adds to our "stationary chair" model
25000/24 =1041.7 MPH
The Earth is orbiting the Sun once a year. The circumference of the Earth's orbit is approx.  607.6  million miles {or 940 million kilometers}. One year is approx. 365 days
365 days X 24 = 8760 hours
607,600,000/8760 = 69,360.73 MPH

  So far we have basically 3 Directions of Motion {Wobble, Spin and
  Orbit} and a combined speed of 1041.70 + 69,360.73 = 70,402.43 MPH for
  a person sitting in a chair at the Equator.

The sun {and hence the solar system} is moving towards the constellation Hercules, namely to the star Lambda Herculis at 12 miles per second {or 20 kilometers per second} which is 43,200 MPH
The Solar system is also moving upwards, at 90 degrees to the plane of the Milky Way, at 4.34 miles per second or 15,624 MPH. But we are actually leaving the Galaxy, out about 50 light years now and will be moving out to 250 light years before it reverses. Details of the mechanics of this are explained in the link below. We also crossed the Galactic plane 2 million years ago.
The Solar system is orbiting around the Galaxy at an "estimated" speed of 124 miles per second {or 200 kilometers per second} which is 446,400 MPH. The way that figure has been calculated can be found at the link below.
So our "armchair Astronaut" is now moving through 6 different directions and a combined speed of approximately 574,585 MPH

  69,361 MPH Spin and Orbit    43,200 MPH Towards Lambda Herculis
  15,624 MPH Perpendicular to Galactic Plane  446,400 MPH Orbiting the
  Galactic Center {or Galactic Spin Rate}  574,585 MPH  or 256862.478 m/s Speed of Earth
  within Our Galaxy

light's= velocity = 3*10^8 m/s
So the ratio = 256862.478 / 3*10^8 = 0.00085620826


Why did it take five years to "figure out" how to use astrometric calibration sources to deblur LOFAR images? Quanta Magazine's The Hidden Magnetic Universe Begins to Come Into View is a fascinating review of a rapidly evolving field in astronomy.
It contains some statements:

In their paper last year, van Weeren and 28 co-authors inferred the presence of a magnetic field in the filament between galaxy clusters Abell 399 and Abell 401 from the way the field redirects high-speed electrons and other charged particles passing through it. As their paths twist in the field, these charged particles release faint “synchrotron radiation.”
The synchrotron signal is strongest at low radio frequencies, making it ripe for detection by LOFAR, an array of 20,000 low-frequency radio antennas spread across Europe.
The team actually gathered data from the filament back in 2014 during a single eight-hour stretch, but the data sat waiting as the radio astronomy community spent years figuring out how to improve the calibration of LOFAR’s measurements. Earth’s atmosphere refracts radio waves that pass through it, so LOFAR views the cosmos as if from the bottom of a swimming pool. The researchers solved the problem by tracking the wobble of “beacons” in the sky — radio emitters with precisely known locations — and correcting for this wobble to deblur all the data. When they applied the deblurring algorithm to data from the filament, they saw the glow of synchrotron emissions right away.

The filament between galaxy clusters Abell 399 and Abell 401 is also discussed in How is it determined that the X-ray and radio intensity come from a magnetic field bridge between two clusters of galaxies?
Question: Why exactly did it take five years for the radio astronomy community to "figure out" how to improve  the calibration these LOFAR measurements and apply deblurring algorithms? Answers to

Would Adaptive Optics be Useful in Radio Astronomy?
What is precipitable water vapor in millimeter-wave radioastronomy and how is it measured?
What can be learned from low frequency radio astronomy available outside of Earth's ionosphere?
In the 1950's how were radio-astrometric positions with portable dishes so precise they could be assigned to their dim optical counterparts (Quasars)?
What exactly is interplanetary scintillation; what was the Interplanetary Scintillation Array looking for? Did it successfully observe any?

suggest that correcting for spatial variations in water vapor or other effects is a known thing in radio astronomy. However discussions at How large does refraction become in radioastronomy? suggest that the ionosphere becomes increasingly important at low frequency (the "LOF" in LOFAR) so perhaps this was part of the challenge?

The point of the paper is not that no calibration has been used previously. It points out that they now use a more refined method and set of parameters to calibrate the data.
That in general is usual procedure: the more data you gather, and the longer a mission runs, the better you understand your instrumentation and all external factors which influence the data you gather - and especially the sensitivity towards different systematic and instrument errors. Thus with data becoming available you only have the people to work on the problems, often Bachelor, Master or PhD students. Thus prior to instrument availability and detailed data analysis you can only do a coarse calibration based on general knowledge of the problem. Five years is a reasonable and good time interval two have one to two generation of students work on it and get an good overall picture which considers all aspects you mention (and probably more) in the necessary detail and assess their contribution in a way which satisfies scientific scrutiny including the review process.


Can BY Draconis variable stars sustain any form of complex life? I wanted to ask this, because while I was working on two fictional planets back on the Worldbuilding SE, one of them being habitable, which orbit the 61 Cygni A star of the 61 Cygni binary star system, I realised something: 61 Cygni A is a BY Draconis variable star.

Here is the Wikipedia article link for more information.

My problem is: can a BY Draconis variable K-class star like 61 Cygni A permit the development of complex life, unlike its partner, 61 Cygni B, is a flare star that can barely get life past the primordial stage, if any planets are there? Or does it have some problems in that regard? If so, what are those problems that a habitable world in that star system could have, and other star systems of the BY Draconis variable star category?

Life could easily form underground, where flares don't reach.  Never underestimate the power of 10^(googol) possible locations in the universe for living things to come into being.   No reason to limit to organic, or CO2 / O2 connected, or photon-dependent, life forms.


The RA and Dec of lunar poles The moon faces the earth with one fixed side. However there is a small vibration.

So I am not sure the south and north poles of the moon is fixed. Even if they are fixed points on the moon, the pole axis should point at different locations in the sky.

What are the general positions of the two points in equatorial coordinate system of our earth?

Here's a NASA movie showing Libration: http://svs.gsfc.nasa.gov/cgi-bin/details.cgi?aid=4000

As given in the details here, the position angle is the angle between the Moon's north and celestial north. This changes because of the inclination of ecliptic to equator + inclination of moon's orbit to ecliptic and plus, there might be some contribution from the inclination of moon's rotation axis to its orbital plane. Note that all these changes are apparent. The pole need NOT physically shift at all for these effects.

Having said that, whether the pole does shift, requires one to correct for all those effects and the remaining effects would be cause by some precession or nutation of the lunar poles, which is likely to be negligible given that it is tidally locked to the earth.


What is the furthest star for which we can measure the diameter? If a star is close enough that we can measure its parallax and that it doesn't appear point-like, then we can deduce its diameter. Can we measure the diameter of all the stars for which we can measure the parallax, or is the angular diameter a limiting factor?
How far out is the diameter of a star measurable?

It's neither the angular diameter or prallax precision that is the limiting factor, but the fact that it is difficult to get the interferometric measurements for faint stars.

State-of-the-art angular diameters are measured by infrared interferometry (e.g. with the CHARA array - Gordon et al. 2019). The most precise measurements of angular diameters have an uncertainty of about 17 microarcseconds in that paper. That means the smallest angular diameters that can yield
 a solid detection of the disc of a star are about $\theta = 50$ microarcseconds. But such measurements can only be achieved for stars down to about $V \sim 6$.

We can write down an equation for the relationship between $\theta$, the radius of the star $R$ and its distance $D$.
$$ D = \frac{2R}{\theta} = 186\left(\frac{R}{R_{\odot}}\right) \left(\frac{\theta}{50\ \mu{\rm as}}\right)^{-1}\ {\rm pc}$$

This means that if a $1R_{\odot}$ star has its distance known more precisely than the $\sim 30$% error in the angular diameter, then it's radius can be measured out to distances of about 200 pc (with a precision of about 30%). This is easily achieved - a $1R_{\odot}$ solar-type star at a distance of 200 pc has a brightness of $V \sim 11$ and has a parallax uncertainty in the Gaia catalogue of only about 0.1 mas (mostly a systematic error at present), meaning the distance is known to about 2%. Thus, parallax uncertainty is not the limiting factor. However, a 6th magnitude solar-type star would be at $D \sim 20$ pc. Much further away than this and the source is just not bright enough to get an angular diameter measurement with CHARA.

On the other hand, a large red giant star, with $R \sim 200 R_{\odot}$ could in principle have its radius measured to 30% out to a distance of 37 kpc with the angular resolution available. Such a star would not however have a measurable parallax at present. The parallax would be 27 microarcseconds, which should just be resolvable in the final Gaia data reduction in a few years time. However, the star would also have $V \sim 14$ and is way too faint for the current interferometric capabilities of CHARA. A 6th magnitude red giant is at $D \sim 500$ pc, a distance at which both its angular diameter and parallax are precisely measurable.

So the answer is that the limitation is not the angular resolution, or the parallax. It is the brightness limit for the interferometers that are used to do the measurements. This limitation means that you can measure the diamters of solar-type dwarfs to a few 10s of pc, and the diameters of the largest red giants to about 1 kpc.

Update: There are other interferometers that can be used. The VLTI/Amber interferometer in Chile utilises larger telescopes than CHARA and can in principle work to fainter magnitudes. A paper by Chesneau et al. (2014) measures an angular diameter for HR5171A (a red hypergiant) to be $3.39 \pm 0.02$ mas, at a distance of 3.6 kpc.
https://www.aanda.org/articles/aa/full_html/2014/03/aa22421-13/aa22421-13.html

Although much further away, this huge star is bright enough ($V \sim 6.5$) that an angular diameter measurement can still be achieved. Note though that the limitation here is still the brightness of the object, not its distance or angular size.


Why is Proxima Centauri called Proxima Centauri? It seems very logical, to call the star, closest to earth, "Proxima" ("Proximus" is Latin for "close" and the word for star, "stella", is feminin, hence "Proxima"), but there's one thing I don't understand: in order to call a star ±"The closest one", you must be sure that in future, no closer star will be found.
This is not so simple: lots of very close stars, like Barnard's star, Kapteyn's, ..., are very weak and even not visible to the naked eye.
So, how can we be that sure that Proxima is indeed the nearest one? Have astronomers indeed examined the complete sky in order to prove this?

So, how can we be that sure that Proxima is indeed the nearest one? Have astronomers indeed examined the complete sky in order to prove this?

The minimum absolute magnitude for a star is, according to this well-known diagram about +16. Proxima Centuari is at the very low end of this scale with an absolute magnitude of +15.60. Still, it has an apparent magnitude of +11.13. That means that if a star is closer to us, it needs to be +11.5 or brighter. Every object of that magnitude or brighter has been categorized, and a hypothetical close star would've stood out due to its large parallax. So it's safe to say there are no closer stars ... at this moment.

  you must be sure that in future, no closer star will be found.

Stars move (they rotate around the galaxy center), so in the (far) future, there might be a star closer to use than Proxima Centauri, just like sometimes Venus is the planet closest to Earth, and sometimes Mars. So it's possible that Proxima might not be the nearest star in the far future. This is not something we have to worry in our lifetimes, though :)


Why was Dawn's (the spacecraft) flight path circular? In comparison to the twin Voyagers, Dawn's flight path circled the sun in an anti-clockwise manner. The map of its path shows its targets, Ceres and Vesta were in a trajectory that was close to a straight path, if Dawn had taken a clockwise direction, with a close flyby of Mars. I can't find a reason to show it was studying the space around Lagrange point 3 (in relation to Earth), as according to this flight path map. So what was the reason for a longer direction?

Dawn was not exploring L3. But instead it explored both Vesta and Ceres. In order to do so in a cost-saving manner, Dawn was also a mission demonstrator for an interplanetary mission equipped with ion propulsion.

Because of the low specific impulse of ion propulsion, Dawn cannot fly in a straight line, as this would imply a high initial velocity and the spacecraft couldn't deaccelerate in time. Therefore, Dawn started with low $v_0$ at Earth, and raised its orbit slowly and continuously, creating the spiral you see.


How does the earth's plane of orbit (ecliptic) vary over time? Does the earth's orbital plane change with time relative to the distant stars? Does the Sun follow the same path through the zodiac (ignoring proper motions of the zodiacal stars)?

The position of the ecliptic does change, due to perturbation of other planets, but only very slowly.

The invariant plane is the plane of angular momentum of the whole solar system. This is dominated by the orbital momentum of the planets, particularly Jupiter, and since other stars are much too far away to perturb the solar system, it is invariant with respect to distant stars.

The ecliptic does vary. It is currently decreasing by about 1.4 degrees per 10000 years, as indicated by this graph, (from Wikipedia). This variation is one hundredth of the variation in the Earth's rotational axis.


Can a planet with a hydrogen atmosphere have a water ocean with dissolved oxygen? Reading up on rogue planets (i. e. planets not orbiting a star), it turns out that a rogue Super-Earth with a sufficiently dense hydrogen atmosphere could retain its internal heat and have oceans/lakes of liquid water. What I'd like to know is whether those oceans could contain dissolved oxygen in equilibrium. Please advise if this should be best posted to Chemistry...

I'm going to say no, or at least, very unlikely.   Short answer, no sunlight, no oxygen formation.    Longer answer below.

A water ocean on a large Rogue planet with a thick enough atmosphere is possible, at least for while.  If there's enough heat coming from the surface or under the oceans through volcanism, perhaps with it being a a younger planet with higher internal heat and more radioactive elements, and/or if the atmosphere was thick enough then liquid oceans are certainly possible.  You run into problems with too think an atmosphere, because the pressure and perhaps temperature would be enormous where liquid water could form.

But if we assume somewhat Earth like, with extensive volcanic activity to maintain enough surface heat, Oxygen is more unlikely for a few reasons.   The hydrogen atmosphere would prevent significant amounts of free Oxygen from ever being present and without oxygen in the atmosphere, disolved oxygen in the oceans would be temporary if it was formed there, not equilibrium.

Oxygen is also very reactive.  It readily binds or reacts with all kinds of common elements like hydrogen, methane, Iron, Silicon, Magnesium, Aluminum, Sulfur and inside the Earth, trapped Oxygen wouldn't escape as a gas but it would likely bind with silicates pretty much permanently.

Oxygen in it's pure form is rare on planets because a whole lot of it needs to be produced before it's abundant enough to begin collecting in the atmosphere (or in liquid form if the planet is cold enough . . . but that's not relevant with a liquid water question).  Photosynthesis is a great way to produce Oxygen from water and CO2 but it only works with sunlight as the process is very energy intensive.   On a rogue planet without sunlight, I can't think of any way to produce any statistically relevant amounts of oxygen inside the oceans.

Plants on Earth produce oxygen in abundance because of free energy from sunlight and easy access to CO2 and H2O, but without all that free energy, I don't see it happening.

Extremophiles come in may types, so I don't want to say it's impossible, but life that we know of relies on chemical energy and the laws of chemical energy are pretty straight forward and the production of free oxygen as a by product is difficult without a significant supply of energy.

Extremophiles, loosely speaking, eek out a living on not very abundant chemicals from volcanic gas escaping into the ocean floor or in other difficult situations.    There's many different kinds and more possibilities if you factor in Astrobiology theories, but I have a tough time seeing a method where oxygen is released.   There's plenty of heat in ocean vents, but heat isn't easily converted into usable energy by bacteria.  Chemical energy and exothermic chemical reactions to create stored energy, or photosynthesis is generally how the life we know creates and stores it's energy.   The chemical energy required to release oxygen in significant amounts seems very impractical without significant sunlight.

Now, granted, the biology and chemistry behind what extremophiles can produce, gets pretty complicated and is closer to biology and perhaps chemistry than astronomy (though some might say astrobiology is part of astronomy), but in addition to that, exploring what chemical processes might be possible would get very long and above my paygrade so, there's a chance it's possible, but I feel pretty confident in saying your scenario is unlikely.


What does the BinTableHDU store? I'm just getting started with very primitive analysis of FITS data, and I have a 'raw' FITS file which I don't know how to get different wavebands of data from (if this is even possible).

Filename: casa_raw.fits
No.    Name         Type      Cards   Dimensions   Format
0    PRIMARY     PrimaryHDU      33   ()
1    EVENTS      BinTableHDU    793   14621280R x 19C   [1D, 1I, 1I, 1J, 1I, 1I, 1I, 1I, 1E, 1E, 1E, 1E, 1J, 1J, 1E, 1J, 1I, 1I, 32X]
2    GTI         BinTableHDU     28   2R x 2C      [1D, 1D]

Could anyone explain what exactly this information conveys?

This shows that the file contains 3 Header-Data Units (HDUs; you can read more about this here). The headers contain meta-data. The primary HDU (0) has some 'cards' (basically keyword/value pairs most of the time, or comments) but no data. The next HDU (1, called 'EVENTS' in your file) is a binary table which contains 19 columns. Finally, the last HDU (2, also called GTI) contains another table with 2 columns. You can access the data in one of the columns of the EVENTS table using for example:

from astropy.io import fits
hdulist = fits.open('myfile.fits')
hdulist['EVENTS'].data['colname']

where colname is the name of one of the columns. You can find out the column names by looking at:

hdulist['EVENTS'].data.names


how to connect two eyepieces together with the x3 Barlow lens how to connect two eyepieces together with the x3 Barlow lens

I have 10mm,25mm, and x3 Barlow. Is there any way to connect them together to archive the maximum magnification of 105mm?

You will get the maximum magnification by using the 10mm eyepiece with the barlow lens.  You cannot connect 2 eyepieces in the way you suggest, but even if you could add the 10 and the 25 to get a 35mm, that would give lower magnification, as I explained in a previous answer.


What is the lyman-werner background? I've seen several papers that discuss the Lyman-Werner Background (LWB) when modeling star and galaxy formation in the early universe. One in particular (https://arxiv.org/abs/1803.04527) even varies the strength (?) of the LWB using the units $J_{21}$. I assume the subscript is referencing the 21cm band, as opposed to a more generic use of $J_{c}$. But I am equally confused as to what the flux in terms of J is referring to. I've found and read on Lyman-Werner photons but going from these photons to a variable "background" is where I'm having a disconnect.

The Lyman-Werner background (LWB) is the part of the "meta-galactic" background radiation consisting of photons that are able to photo-dissociate molecular hydrogen ($H_\mathrm{2}$), but not ionize atomic hydrogen (HI). These photons are in ultraviolet, with energies ranging from $11.2\,\mathrm{eV}$ to $13.6\,\mathrm{eV}$, originating mainly from hot, massive stars.

Part of the reason this radiation field is interesting particularly in the early Universe is that a sufficiently high intensity will prevent gas from cooling. In order for gas to form stars, it cannot be too hot, since a hot gas cannot collapse to form structure. A gas with a high metallicity is able to cool through the many different atomic transitions in the metals, but in the early Universe, where stars haven't yet produced the metals, the most important coolant at these temperatures is $H_\mathrm{2}$.

It also has implications for the formation of supermassive black holes in the early Universe, since once you form a large amount of stars, the feedback from the massive stars (both through radiation pressure and heating of the interstellar medium) makes it difficult to have gas collapse directly into the supermassive black holes.

In simulations, you often have to assume a value of the LWB, since performing the "true" radiative transfer is computationally very expensive. Hence, these authors apparently have experimented with two different fields of $100\,J_\mathrm{21}$ and $1000\,J_\mathrm{21}$, respectively. Here, the term $J_\mathrm{21}$ doesn't have anything to do with $21\,\mathrm{cm}$ radiation (which is in the other end of the spectrum), but is simply the intensity in units of $10^{-21}\,\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}\,\mathrm{sr}^{-1}$.


Looking back in time by looking further away I have been pondering the question of how we can look back to the state of the early universe for some time. So I came here and found, to my delight, that several such questions have been asked, such as this one. Fortunately the answers to that (and other questions) has helped immeasurably, but I have one lingering area of confusion:

I often hear, in cosmological TV documentaries (such as Horizon in the UK) words such as "The further away we look, the further back in time we are looking. Our technology has now advanced to the point we can see objects so far away they began emitting their light just [n] after the big bang" [this is not a quote from a programme but my synthesis of what I have heard].

My question is this: I understand the limitation of c and light taking finite time to reach us from distant objects. Having now understood that inflation caused the early universe to "move" so far away we are only now receiving its light; How can today's technology evolve into tomorrow's technology and enable us to see "older" light as we can only see what is arriving at the Earth right now. I.e. if a time-traveller turned up today bearing a better telescope from the future, I should not be able to use it to see further away and therefore back in time, because it is dependent on the age of the light arriving at my location.

The only solution I can think of is that due to tricksy inflation rate effects, the light of successively further objects is arriving at, essentially, the same time (when looked at over a universal age). But that doesn't seem to make sense and even if it did how would we sort the older light from the newer light from the same object if it is all arriving at once (notwithstanding red shift, which I do (hopefully) understand)...

Before the advent of telescopes, we could only look back in time from a few years (for nearby stars) to a few thousand years (for the most distant stars visible to the unaided eye). In addition to this, a handful of galaxies are visible without binoculars, letting us look back a few million years.

The first telescopes allowed us to see much farther, like hundreds of megaparsec, so we could look back hundreds of millions of years. As technology improves and our telescopes become better, we are able to observe more and more distant objects, and thus look further and further back in time.

The farther away an object is, the less light we receive, and hence the bigger telescope and better detector we need in order to see it. So, being able to look further back in time than previously is not about the light from more distant objects eventually reaching us, but simply a matter of developing the tools necessary to see it.

However, even the most distant galaxy we will ever see (which are at earliest a few hundred million years after the Big Bang, i.e. roughly 13.5 billion years ("Gyr") ago), are not farther away than the cosmic microwave background, which was emitted 380,000 years after Big Bang, i.e. ~13.8 Gyr ago, and which we have been able to detect since the mid-1950's. If we ever manage to build an efficient neutrino detector, we may be able to look back to seconds after the Big Bang, when these particles decoupled from matter and began traveling freely through space.


Determining the extinction by counting stars? Hi I'm a highschool student studying astronomy at school.
I learned that one can determine how much a luminosity of a star is reduced by counting stars, but I can't understand how it works. My textbook has the following equation:

$\Delta m=2.5 \log(\frac{N}{N_0})$ or also written as $\Delta\log N=0.4\Delta m$
, where N is a number of stars and m is the magnitude.

The textbook says that by comparing the number of visible stars from the inside of an interstellar cloud with that of stars from the outside of the cloud, one can determine how much the luminosity is decreased using that equation.

I have no idea how the equation is created. I searched it on google but I only found this pdf: http://rpi.edu/dept/phys/observatory/obsastro9.pdf
Can anybody explain it for me?

The general idea is that the more extinction there is, the less you'll be able to see stars. So there is a direct relation between the decrease in flux observed for any individual star and the number of stars you can see in that general direction. So the surface distribution of star counts is an excellent tracer of optical extinction, and $\Delta m = 2.5 \log\frac{N}{N_0}$ is a fairly good approximation in many cases.

This relation between $N$ and $\Delta m$ is not derived mathematically, but rather observed empirically (here or here, for instance), under the assumption that the density of stars is the same in the direction of reference and in the direction of interest. This means that the extinction is measured with another method (for instance looking at how much specific emission lines are absorbed), counting the stars, and then fitting a curve that will give the relation between star count and absorption.

But why don't we just use the other methods directly, might you ask! Well the other methods require data such as high-enough resolution spectroscopic data that aren't always available for the region of the sky you are studying. Then star counting gives you a close enough guess without being prohibitively expensive.


Are there Dark Stars In the visible universe, we have a lot of matter to energy conversion going on in stars.  Is there an equivalent structure that is converting dark matter to dark energy?

Dark matter has different properties than ordinary matter. It cannot collide, and thus it alone cannot coalesce to form a star. Furthermore, dark matter and dark energy are unassociated properties. There is not currently any reason to assume dark matter can be converted to dark energy.

That said, there is a strictly hypothetical entity referred to as a dark star, which does involve dark matter. While composed mostly of ordinary matter, this entity is hypothesized to include a high concentration of dark matter, generating heat and preventing fusion of the ordinary matter. If dark stars existed, it is likely they existed in the early universe and did not survive to the present. They are modeled to be large clouds--4 to 2000 astronomical units in diameter--of hydrogen and helium, with a surface temperature low enough to make them invisible to the naked eye.


How to test direct sun viewing glasses I purchased a 10 pack of glasses for viewing the upcoming eclipse. They show to be ISO 12312-2:2015 certified. How can this be tested and verified?

The American Astronomical Society (AAS) answers this question on their How to Tell If Your Eclipse Glasses or Handheld Solar Viewers Are Safe page.

Basically they say that since anyone can claim ISO 12312-2 compliance even if they haven't been certified, you should make sure you obtained your eclipse glasses from a vendor whose ISO 12312-2 compliance has been verified by the AAS Solar Eclipse Task Force:

  We used to say that you should look for evidence that they comply with the ISO 12312-2 international safety standard for filters for direct viewing of the Sun. But now the marketplace is being flooded by counterfeit eclipse glasses that are labeled as if they're ISO-compliant when in fact they are not. So now we suggest that you make sure you get (or got) your eclipse viewers from one of the suppliers listed on our Reputable Vendors of Solar Filters & Viewers page.

They say that checking eclipse glasses to see if they meet the ISO 12312-2 standard would require a spectrophotometer, which most people do not have access to:

  Unfortunately, you can't check whether a filter meets the ISO standard yourself — doing so requires a specialized and expensive piece of laboratory equipment called a spectrophotometer that shines intense UV, visible, and IR light through the filter and measures how much gets through at each wavelength.

So they have assembled a task force to put together a list of manufacturers and vendors of eclipse glasses that they have verified to be ISO 12312-2 certified (their list is longer than NASA's list of 5 manufacturers):

  The AAS Solar Eclipse Task Force has been working diligently to compile a list of such vendors, now posted on our Reputable Vendors of Solar Filters & Viewers page. We've checked manufacturers' ISO paperwork to make sure it's complete and that it comes from a recognized, accredited testing facility, and we've personally examined manufacturers' products. We've asked manufacturers to identify their authorized resellers, and we've asked dealers to identify the source of the products they're selling. Only when everything checks out do we add a vendor to our listing. If we don't list a supplier, that doesn't mean their products are unsafe — only that we have no knowledge of them or that we haven't convinced ourselves they are safe.

They also give advice on how to tell if your eclipse glasses are not safe to use:

  How can you tell if your solar viewer is not safe? The only thing you can see through a safe solar filter from a reputable vendor is the Sun itself. If you can see ordinary household lights through your eclipse glasses or handheld viewer, it’s no good. Safe solar filters produce a view of the Sun that is comfortably bright (like the full Moon), in focus, and surrounded by black sky. If you glance at the Sun through your solar filter and find it uncomfortably bright, out of focus, and surrounded by a murky haze, it’s no good.


How do people measure the distance between the Earth and The Moon? How do people measure the distance between the Earth and The Moon?
What methods were used? Where can I find the official data?

There are many ways and I'm not entirely sure who you mean with "how do people measure the distance" (does this exclude space observatories like e.g. Clementine, probes currently in lunar orbit like e.g. Lunar Atmosphere and Dust Environment Explorer a.k.a. LADEE, or any other currently available technology?), but one interesting and extremely precise way is by laser ranging, pointing them towards one or more of the retroreflectors that were left on the surface of the Moon by the Apollo landers, in what is known as the Lunar Laser Ranging Experiment. Since they are laid on the Moon's near side that is always pointed towards the Earth, these retroreflectors are available for measurements to any properly equipped researcher, no matter which country they're from.

By measuring the time it takes for the light to reflect back, we can infer distance by simply multiplying the c (speed of light in vacuum) with the time taken between the light signal being transmitted and received and then dividing all of it by two to get a single leg of the distance. I'll let you read the linked to Wikipedia page on LLRE for more information, but if you find that too boring, here's a YouTube video on Mythbusters: Moon Hoax Retroreflectors that explains it from the practical perspective.

As for results, again quoting that same Wikipedia page on LLRE:

  Some of the findings of this long-term experiment are:

  The Moon is
  spiraling away from Earth at a rate of 3.8 cm (about 1.5 inches) per year. This rate
  has been described as anomalously high.
  The Moon probably has a
  liquid core of about 20% of the Moon's radius.
  The universal force
  of gravity is very stable. The experiments have put an upper limit on
  the change in Newton's gravitational constant G of less than 1 part in
  $10^{11}$ since 1969. The likelihood of any "Nordtvedt effect" (a
  composition-dependent differential acceleration of the Moon and Earth
  towards the Sun) has been ruled out to high precision, strongly
  supporting the validity of the Strong Equivalence Principle.
  Einstein's theory of gravity (the general theory of relativity)
  predicts the Moon's orbit to within the accuracy of the laser ranging
  measurements.

So these distance measurements are so precise they're a lot more interesting and reliable than merely telling us how far away the Moon is. For one, it also blows any Moon landing hoax theories right out of the water. Links to many more official results can be found in references and external links sections of Wiki on LLRE.

There is also a plethora of other ways to measure the distance to the Moon, with different precision and equipment requirements, but I will let others add some other methods in their answers. If however you had some specific "people measuring the distance" equipped with specific measuring devices in mind, don't forget to mention that in your question. ;)


Could the Philae comet lander be recharged by laser? Could the Philae comet lander be recharged with a laser (or possibly many lasers) from earth?  My thought would be to aim one or more powerful lasers from earth at the Philae lander to recharge it through its solar panels.

Another thought would be to heat up a small part of the comet with one or more lasers, possibly causing the comet to slowly rotate into a position where the sun light falls onto the lander.

Would either of these ideas be feasible?

Could the Philae comet lander be recharged with a laser (or possibly many lasers) from earth?

No.

You could make a laser beam with a pretty low divergence - meaning a beam that remains tight and parallel over a very long distance. However, "a very long distance" is nothing on the cosmic scale.

The Lunar Laser Ranging experiment is using high-power lasers focused by telescopes, shooting a beam into the Moon. The light is reflected back to Earth by mirrors placed there by the Apollo program.

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

The problem is, even with all the tight focusing at the origin, of the initial 10^17 photons in a pulse, only 1 makes it back to the detector. It's very hard to tell that something makes it back at all, let alone charging any solar panels with it.

The comet and Philae are much, much further away than the Moon. You couldn't possibly project any useful energy density at that distance. Not with the current technology.


Why are ice giants not considered suitable for life? Ice giants are currently not considered suitable for life due to temperature, pressure and composition.  This paper gives some parameters for life in regards to temperature, pressure and composition:

Temperature: $-15\ ^\mathrm{o}C$ to $122\ ^\mathrm{o}C$
Pressure Upper Limit: $1,100\ \mathrm{atm}$, and
Composition to Include: liquid water, $C$, $N$, $S$, and $P$.

I'm trying to find out what the conditions of the hot dense ammonia/water mantle border with the hydrogen/helium atmosphere are, in regards to temperature and pressure. We know the mantles are under very high pressures, have liquid water (though in a likely different state due to pressures), C, N and S. What are the pressures and temperatures at this mantle border?  Specifically in reference to Uranus and Neptune.

A basic approach would be to estimate the air-pressure at the top of the mantle.

Uranus' surface gravity is a bit less than Earths, but, somewhat counter-intuitively, it's atmosphere is light enough that it's gravity increases as you move inside the planet towards the top of the mantle.  (the same is true for Earth).   For ballpark estimate, lets say the gravity works out to be about the same as Earths, because I don't want to calculate, but it might be a bit more than that.

The Mass of Uranus' atmosphere, by Wikipedia is about 0.5 times the mass of Earth.  It's surface area is about 16 times Earth (4 times the Radius), but the surface area of the Mantle would be less, about 80% of the radius which works out about 10.2 times the surface area.

If you spread 0.5 Earth masses over 10.2 times Earth's surface area, you get a pressure between 50,000 and 60,000 atm.   Some 50 times the upper limit you provided in your question.   That's probably a limiting factor for most "ice giants".   There's simply too much hydrogen and helium atmosphere to allow liquid water at 1,100 ATM.   Ice giants might create their own hydrogen by heat and pressure on Methane, creating diamond like structures and free hydrogen.

The heat below all that atmosphere (about 20% of Uranus' radius) would be outside your criteria too.  I couldn't find any good estimates for Uranus' mantle's temperature.   There's too many unknowns about why Uranus is as cold as it is on it's surface, so estimates on it's mantle temperature seem pretty tentative and unavailable, but what little I know about how temperature rises inside a planet, I don't see any way the top of Uranus' mantle would come close to the temperature criteria you gave.  Not at a depth of 20% of Uranus' radius.

Consider this chart from Uranus' wikipedia at 200,000 mbar Uranus temperature is approaching 340 K.    200,000 mbar is about 200 atm.   The mantle begins, by my ballpark estimate at about 50,000 atm.   The temperature wouldn't be close unless the planet had a very long time to cool down, perhaps tens of billions of years.

I could run ballpark estimates on Neptune too, but I suspect the results would be similar.

For an ice giant to meet your criteria, it would need to have comparatively very little hydrogen and helium compared to Uranus.  Perhaps if an ice giant was close enough to it's star that it had a hot enough surface temperature to lose much of it's hydrogen and helium through Jeans escape, and then, it had a long time to cool down, it might get the right criteria of liquid water at the proper range of temperature and pressure.  I don't see any other scenario where it would work.


Earth and ferromagnetism Earth's core is a giant liquid iron ball actually. If I know well, the magnetic field of our planet (that protects the surface from some particles comes from the Sun) can exist because as Earth rotates, the liquid iron core also rotates, and it generates a magnetic field.

However, ferromagnetism has strict conditions. How can the liquid and hot core still be ferromagnetic? How is it possible for Earth to have a magnetic field generated by a body that is not even solid?

Actually, the core has two parts. The outer core is liquid, while the inner core is solid.

As explained in the Wikipedia article about Earth's magnetic field:

  The Earth's magnetic field is mostly caused by electric currents in the liquid outer core, which is composed of highly conductive molten iron.


How many satellites does the Milky Way have? The small and large Magellanic clouds are galaxies that are orbiting our own galaxy. How many such "galactic satellites" do we know of?

Starter answer:

Wikipedia's Local Group lists the following items as galaxies, dwarf galaxies, or possibly galaxies. You should refer back to the Wikipedia page, but here is a summary:

Galaxies: Sagittarius Dwarf Galaxy, Large Magellanic Cloud, Small Magellanic Cloud

Dwarf Galaxies: Ursa Minor, Draco, Carina, Sextans, Sculptor, Fornax, Leo I, Leo II, Ursa Major I, Ursa Major II

Disputed Dwarf Galaxy: Canis Major

other: several additional ultra-faint dwarf spheroidal galaxies

However: There is a much longer list of "galaxies" (over 60 items) Wikipedia's Satellite galaxies of the Milky Way

Hopefully another answer will be posted discussing just how galactic these need to be to be called galaxies, and if they are all known to be gravitationally bound to the Milky Way or just in its proximity.


Do moons have moons? Have we discovered any natural satellites of natural satellites of planets or dwarf planets? Even very small, or relatively short-lived - e.g. ringlets around Saturn's moons, some meteorites orbiting Jupiter moons, or something to orbit Charon? Or is the Star-Planets-Moons the deepest naturally ocurring orbital recursion level?

I don't think there are any in the Solar system. We do have around 250 asteroids with moons. Rhea's ring seems to be the only exception.

Edit: Originally I said "a moon with a moon would be an unstable system, due to the gravitational influence of the planet." @Florian disagrees with this. However, the answer is more complex than the Hill sphere alone.

At first approximation, the Hill sphere gives a radius in which orbits around a moon could be stable. Our Moon's Hill radius is 64000 km.

For our own Moon, we know that most low orbits are unstable due to mascons: mass concentrations below the surface which make the Moon's gravitational field noticeably uneven. There are only four inclinations where an object orbiting the Moon avoids all mascons and would be stable: 27º, 50º, 76º, and 86º.

High orbits above the Moon aren't all safe either: above 1200 km and inclinations of more than 39.6º, Earth's gravity disrupts the satellite's orbit. Note that these orbits are comfortably within the Moon's Hill sphere.

There are stable orbits at high inclinations and high eccentricity:

As for other moons in the solar system: most of them are smaller and orbit around larger planets, so their Hill spheres are small, and the planet's gravity will disrupt much of the volume inside the Hill sphere too.

Moons below the limit where their gravity is strong enough to make them spherical, will have problems with uneven gravitational fields. Mascons may also be present.


Are there naturally radiation shielded areas on Mars? (shadowed by terrain formations) Mars' thin atmosphere and lack of magnetosphere exposes its surface to lots of Solar UV radiation and cosmic radiation. But Mars also has some dramatic altitude gradients in for example Valles Marinaris. Are there any areas which in this way are protected from A) Solar UV radiation and flares and coronal mass ejections, and/or B) substantially protected from cosmic radiation?

If you want protection from radiation on Mars, you'd have to either live in caves, or else build some kind of thick dome overhead, or live behind thick walls/ceiling. Cosmic radiation will reach you anyway as long as the sky is visible, so deep valleys can only help you so much.

The only way to stop hard radiation is to place a lot of "stuff" (thick mass) between you and the source. A pile of dirt would suffice in many cases. The geologic layers on top of a cave would work well. I guess a dome could also work if it's thick enough. Habitats with thick ceilings (and walls) could also be an option. Have a large water reservoir? Put it on the roof of the habitat: voila, instant radiation shield.

The main concern is the stuff coming in from directly overhead; near horizon even Mars' atmosphere is thick enough to reduce radiation a little. So prioritize the ceiling.

I guess at the base of a hill with some overhang above your head you'd get some shielding, but again - is the sky visible? That's the main concern. Also, areas under an overhang may or may not be safe (stable) long term.

Find a very large boulder (the size of a house or bigger), much taller than your habitat, and place the habitat right next to it - you've cut 50% of radiation right there.

For a shorter term stay, a simple habitat with thin panels would be enough - not very different from the interplanetary vehicle after all.


Why do we use solar filters instead of simply reducing aperture of the telescope? Everywhere I read that to view the Sun in the telescope, one has to use a solar filter. The filter must be handled carefully to avoid any holes in it and even mustn't be stretched so as to avoid reducing its reflectivity in some spots. Otherwise it becomes dangerous due to possibility of overexposing/burning observer's retina.

But, if all we need is to reduce illuminance, and covering part of the telescope's aperture won't obscure the image (since the aperture is far from being in focus), why not simply reduce the aperture to a very small hole? Photo cameras do this all the time, why wouldn't this work for telescopes? Or does a solar filter block IR/UV light, in addition to reducing visible illuminance?

Suppose we want to make the Sun appear only as bright as the full Moon does in a 5-inch telescope.
A difference of 14 magnitudes is a factor of 400 000; we can either add a filter transmitting 2.5 millionths of the light (including UV and IR), or reduce the aperture diameter by a factor of 630.
We can achieve that by covering the 5-inch aperture with either a large #14 welding shade, a similarly dark solar filter, or a mask with a 0.2 mm pinhole.

However, angular resolution is inversely related to aperture.
For a wavelength of 500 nm and an aperture of 0.2 mm, the Rayleigh criterion gives 1.2 * (500 nm / 0.2 mm) = 0.003 radian = 10 arcminutes, about 1/3 the angular diameter of the Sun.
Even large sunspots would be blurred beyond recognition.
A 1.0 mm hole would give 2 arcmin resolution but make the Sun too bright for direct viewing.


If the tilt of the Earth changed, would the tropics latitude change too? I wanted to know if the tilt of the earth were changed to 30 degrees would the tropics of cancer and Capricorn also both change to 30 degrees?

Yes, the Tropics of Cancer and Capricorn are defined as the highest and lowest latitudes where the sun is directly overhead at least one day a year. The sun's highest position appears to move up and down on the sky over the year because of the Earth's tilt, so if the Earth's tilt were greater or lesser, so would be the extent of the Tropics. (Side note: I may be wrong, but I think the Tropics were named after the Zodiac sign the sun is in when it reaches its highest or lowest point.)


Orion Constellation request This isn't a question so much as a request. I'm looking for an image of the Orion constalletion I could use, but I would like to have a very high resolution HD image, whereby a lot of the deep sky objects can be seen to high detail other than just the 7 hourglass stars of the belt and four corners. Would love to have perhaps the Orion nebula (Messier 42), the Horsehead Nebula (near ζ Orionis), the emission nebula NGC 2174, Messier 43 and Barnard's Loop.

The Astronomy Picture of the day for the 23rd November 2015 has all the neblosity around Orion, in fantatic detail. The result of 212 hour exposure.


How to calculate position of an unknown star knowing positions of some other stars from an image? Suppose I know the coordinates (right ascension and declination) of the stars marked with red. How should I calculate the coordinates of an unknown star, marked with yellow?

Searching over the web, I found a technique called least squares plate constants. I guess plate constants dates back before CCD cameras. Unfortunately there was only little explanation. However, if I understood the matter correctly, I should assume the stars are on a surface of a sphere imaged to a plane? Please help :)

Also, how should I calculate the distances between the stars?

The original image is from Wikipedia.

First, let assume your image geometry is homogeneous, and has no peculiar distorsion in either direction.

Second, let assume you have the resolution of your image: the number of arcseconds / pixel.

Now, take one 'red-cross' star, call it A. It will be the origin of the triangle we will draw. Name your 'yellow-cross' star B. Now, take a new point, called 'C', that is at the same pixel-y coordinate of A, and the same pixel-x coordinates of B.

Drawing lines between A, B and C gives you a right triangle. You cannot simply apply flat geometry equations, since you are on the celestial sphere.

Hence, one must compute the 'Bearing' angle between A, B and C. See for instance here: (A bearing is an angle, measured clockwise from the north direction).

Here is a little piece of code you should be able to read:

double adjacent = sqrt(pow(B.x-C.x, 2) + pow(B.y-C.y, 2)); // dBC
double opposite = sqrt(pow(A.x-C.x, 2) + pow(A.y-C.y, 2)); // dAC

double theta = atan2(opposite, adjacent) * ONE_RAD_IN_DEGREES;

theta is the bearing angle, here expressed in degrees, thanks to the conversion constant ONE_RAD_IN_DEGREES. atan2 is the Arc-Tangent function that takes care of which quadrant you are in (it computes arctangent(opposite / adjacent), correcting for the quadrant, see the wikipedia article for instance).

Now, depending on whether you have East to the left or not (astro images have East to the left usually), you need to correct your angle. Again this little piece of code:

BOOL eastLeft = <true or false>

if (B.x < A.x && B.y > A.y) {
    theta = (eastLeft) ? theta : 360.0 - theta;
}
else if (B.x < A.x && B.y < A.y) {
    theta = (eastLeft) ? 180.0 - theta : theta + 180.;
}
else if (B.x > A.x && B.y < A.y) {
    theta = (eastLeft) ? theta + 180. : 180.0 - theta;
}
else if (B.x > A.x && B.y > A.y) {
    theta = (eastLeft) ? 360.0 - theta : theta;
}

Now, we have the correct theta value. Now, compute the distance (below, in degrees) between A and B, and call it delta.

Assuming the R.A. and Declination of A are called lambda1 and phi1, you can compute the R.A. and Declination of C, lambda2 and phi2, using the formulae given in here, under the section "Destination point given distance and bearing from start point".

In my code:

double phi1 = declination_A * ONE_DEG_IN_RADIANS;
double lambda1 = rightAscension_A * ONE_HOUR_IN_RADIANS;

double delta = degrees * ONE_DEG_IN_RADIANS;
double theta = bearing * ONE_DEG_IN_RADIANS;

double phi2 = asin(sin(phi1)*cos(delta) + cos(phi1)*sin(delta)*cos(theta));
double lambda2 = lambda1 + atan2(sin(theta) * sin(delta) * cos(phi1), cos(delta) - sin(phi1) * sin(phi2));

with the usual meaning of trigonometric functions (sin is sine, asin is arcsine, etc).


How does the Earth move in the sky as seen from the Moon? I just want to be sure I am visualizing this correctly, because it seems odd. The Moon is tidally locked to the Earth but there are wobbles to its motion due to libration. So from a point on the surface of the near side of the Moon, the Earth would always be near the same place in the sky? It would describe a small circle or a side-to-side wobble over the course of a month, but never move far from that point?

That would seem very strange, like it was a gigantic stage prop or something. We are so conditioned that everything rises and sets (except for a few stars near the poles).

Still new at stellarium but here are some quick capture gif  lasting one month. Sorry about the quality- limited to 256 colors for smaller gifs. Date on lower left corner. By the way the sun is of course the brightest and i use it as reference for recording (start record when sun is in frame then stop when it appears again in the same position which is roughly one month)

Location on Moon : Sea of Tranquility

You are looking straight up

Yellow lines are azimuth

second picture:

Zoomed view of earth

One whole day 24 hours (give or take a few minutes)

Location on Moon: Sea of Tranquility

You are looking straight up

Yellow lines are azimuth (gif itself rotated to approximately match the orientation of the 260 degree azimuth line in the first picture)

Answer for Emilio Pisanty comment on whether the oscillation is detectable by eye

stellarium

Field of view 60 degrees (default view when stellarium first opened)

Picture 3

Shows one end of the oscillation near the 200 azimuth line (see yellow arrows)

other end of ellipse near 185 azimuth line

you can use Gemini (lower right corner) for reference (see yellow arrows) from head to crotch of one Gemini twin would be a good reference for how wide the oscillation is (not sure if your sundial can detect the difference)

picture 4

shows other end of the oscilation near the 185 azimuth line (see yellow arrows)

picture 5 shows my stellarium location settings on the moon


Does the rotation of galaxies precess? Does the axis of a galaxy as a whole "wobble" the same way that the Earth does?  If so do we have any idea how much or how fast the Milky Way is doing so?

To expand a bit on FJC's answer: galactic disks are not rigid bodies, and so we should not expect them to rigidly precess.

Most disk galaxies do show warps at large radii (including our own), in both the interstellar gas and the stars. From observations of warps in gas disks, we know that the azimuthal angle of the warp often changes with radius: i.e., the tilt of the disk changes as a function of radius.[*] Since the angular rotation speed changes as a function of radius,[**] this suggests the tilt could precess at different rates for different radii.

Unfortunately, since there isn't really a consensus on what causes or maintains warps, it's very hard to go from observing a warp to predicting what the underlying precession rates might or might not be. (Although one can argue that the precession rate can't vary too much with radius, otherwise the warp would rapidly "wind up" and lose its coherent warped shape, and we wouldn't see warps as often as we do.)

See here for some discussion of the theoretical issues.

[*] This is a kind of precession, although it's a change in the orientation of the rotational axis with radius rather than a change with time (as in the Earth's precession).

[**] Even if the circular velocity is approximately constant, as it often is in the outer regions of galaxies, the increase in the circumference as a function of increasing radius means the angular velocity goes down.


Why is the color of the moon so similar to the clouds? Why the moon looks like the cloud-structure on the blue sky among the clouds?

The Moon is a dark gray in color -- it reflects light poorly but uniformly. (Its color is about the same  as fresh asphalt!  It's albedo is about 12%)  Clouds are (usually) light gray with an albedo ranging from 10% to 90%.

Both the Moon and the clouds shine by reflected sunlight, so they're both the same color. (Clouds usually have a greater surface brightness owing to their lower albedo.)


Could there be Earth-like planets in binary or trinary systems? Could Earth-like (suitable for Human habitation) planets exist in systems with 2 or 3 suns, have any been found?

With more than a single star, isn't it likely that the habitation zone would be too irregular, gravitational forces be irregular/strong and numerous other issues?

Firstly, a few planets have been discovered around binary or trinary systems, mostly Neptune sized or larger.  An example is are the two planets described in the NASA article "Kepler-47: Our First Binary Star 2-Planet System", where they are described as:

  Now Kepler mission has discovered Kepler-47b and 47c, the first transiting circumbinary system — multiple planets orbiting two suns. To compound the excitement of the discovery, one of those planets is in the binary system's habitable zone (where liquid water may exist)

(A representation of the Kepler 47 system is pictured below)

The second planet discovered in the system, Kepler 47c, is within the binary stars' habitable zone, but is slightly larger than Neptune and is likely to be a gas giant - but that raises the potential for terrestrial satellites.

Another example is the Saturn sized Kepler 16b reported in the article "Weird Exoplanet Discovered Orbiting Two Stars" (Klotz, 2011), which orbits 2 stars that are smaller than our sun, thus has habitable zones closer to their parent stars, in this case, the equivalent of the Venus' orbit.

In terms of the habitable zones around multiple star systems, the article "Habitable Binary Star Systems" suggests a major advantage of a binary (indeed a trinary) system is that the combined energy between the stars could indeed extend the habitable zones, particularly,

  Low-mass twins could make the best hosts, because their combined energy extends the habitable region farther away than would exist around a single star.

However, temperatures would vary significantly and the orbits would be irregular, and in simulations reported in "Double-Star Systems Can Be Dangerous for Exoplanets" (Wall, 2013) results in such disruptions that one or more of the planets therein could be ejected and sent hurtling into interstellar space.  But such perturbations are likely to take a considerable amount of time (millions of years) to fully occur.

This is also discussed in "Exoplanets Bouncing Between Binary Stars" (Moeckel and Veras, 2012), where the planet in such as system is described as 'bouncing' between the binary stars, then resuming a relatively regular orbit.

The recent discovery of a terrestrial-like planet in the Alpha Centauri system, as reported in "Discovery! Earth-Size Alien Planet at Alpha Centauri Is Closest Ever Seen" (Wall, 2012) - despite this planet not being in the habitable zones of either star, this does raise the hopes for the possibility of Earth-like terrestrial planets existing in the habitable zones of other multiple star systems.

There is also the possibility that an Earth sized moon could orbit the larger planets such as those in the Kepler systems mentioned earlier.


Have we ever observed a large meteor hit the Moon? Since the moon has no atmosphere, its surface has many more meteorite impacts than Earth's does.

Image source

Have we ever observed a meteor strike the surface of the Moon?

Yes, we have, and the impacts occur very frequently according to "Bright explosion on the moon" (NASA):

  For the past 8 years, NASA astronomers have been monitoring the Moon for signs of explosions caused by meteoroids hitting the lunar surface. "Lunar meteor showers" have turned out to be more common than anyone expected, with hundreds of detectable impacts occurring every year.

On March 17, 2013 a significant impact occurred that could have been observed from the Earth.

Updated information about lunar impacts is found at the Marshall Space Flight Centre's Lunar Impacts page.  Below is a map of the impacts that occurred on the lunar surface between 2005-2008, all of which, according to the article "Amateur Astronomers see Perseids hit the Moon, could have been seen from the Earth (with a telescope):

There have been instances throughout our history when, so is believed, people on the ground have witnessed lunar impacts. One of the more famous ones was in 1178 AD, when monks witnessed fire coming from the moon; However, according to the article "Historic lunar impact questioned", the accuracy of this historical account is questionable.