Is there a star over my head? Say I'm standing up straight, and I draw a straight line from my core through the top of my head (perpendicular to the ground). What is the probability that that line intersects with a star?

EDIT: I'm not trying to exclude any stars. This should include stars which we've observed and stars which we haven't yet observed but can predict due to other things we've determined (like the overall star density of the universe). Also it should include all stars regardless of naked eye magnitude limit.

Summary

There's a 1 in 500 billion chance you're standing under a star outside the Milky Way, a 1 in 3.3 billion chance you're standing under a Milky Way star, and a 1 in 184 thousand chance you're standing under the Sun right now.

Big, fat, stinking, Warning! I did my best to keep my math straight, but this is all stuff I just came up with. I make no guarantees it's completely accurate, but the numbers seem to pass the sanity check so I think we're good.

Caveat the First: The numbers for stars other than the Sun are based on data with a great deal of uncertainty, such as the number of stars in the universe and the average size of a star. The numbers above could easily be off by a factor of 10 in either direction, and are merely intended to give a rough idea of how empty space is.

Caveat the Second: The numbers for the Sun and the Milky Way are based on the assumption that you are standing (or floating) at a random point on Earth. Anyone outside the tropics will never have the Sun over their head. People in the northern hemisphere are more likely to have Milky Way stars over their head, with the best odds being people near 36.8° N, because at that latitude straight up passes through the galactic center once a day.26

Note: You can mostly ignore everything in this answer and just look up the solid angle of the Sun to get the same result. All the other stars are really far away and very spread out. The difference in solid angle subtended is five thousandths of a percent more when we add the rest of the universe to the Sun.

Background

Let's try to get a somewhat realistic, hard number. To do that, we'll need some assumptions.

As pointed out in Michael Walsby's answer1, if the universe is infinite (and homogeneous2), there is only an infinitesimal chance of there not being a star overhead, which normal math treats as exactly zero chance. So let's presume the universe is finite.

Presumptions

Specifically, let's presume the universe only consists of the observable universe. (Look up the expansion of the universe3 for further information.)
Further, let's presume the contents of the observable universe are measured at their current (presumed) positions, not the position they appear to be. (If we see light from a star from 400 million years after the universe began, we would measure it as being about 13.5 billion light years away, but we calculate that it's likely closer to 45 billion light years away due to expansion.)
We'll take the number of stars in the observable universe to be $10^{24}$. A 2013 estimate4 was $10^{21}$, a 2014 estimate5 was $10^{23}$, and a 2017 estimate6 was $10^{24}$, with each article expecting the estimate to increase as we get better telescopes over time. So we'll take the highest value and use it.
We'll take the size of the observable universe7 to be $8.8\cdot10^{26} \text{m (diameter)}$, giving a surface area8 of $2.433\cdot10^{54} \text{m}^2$ 9, and a volume10 of $3.568\cdot10^{80} \text{m}^3$ 11.
We'll take the average size of a star to be the size of the Sun, $1.4\cdot10^{9}\text{m (diameter)}$ 12. (I can't find any sources for average star size, just that the Sun is an average star.)

Model

From here, we're going to cheat a bit. Realistically, we should model each galaxy separately. But we're just going to pretend the entire universe is perfectly uniform (this is true enough as we get farther away from Earth in the grand scheme of the cosmos). Further, we're going to start counting far enough out to ignore the Milky Way and Sun entirely, then add them back in later with different calculations.

Given the above presumptions, we can easily calculate the stellar density of the observable universe to be $\delta = \frac{10^{24}\text{stars}}{3.568\cdot10^{80} m^3} = 2.803\cdot10^{-57} \frac{\text{stars}}{\text{m}^3}$ 13.

Next, we need to calculate the solid angle14 subtended by a star. The solid angle of a sphere is given by $\Omega=2\pi\left(1-\frac{\sqrt{d^2-r^2}}{d}\right)\text{ sr}$ 15, where $\Omega$ is the solid angle in steradians16 (sr), $d$ is the distance to the sphere and $r$ is the radius of the sphere. Using $D$ as the diameter, that converts to $\Omega=2\pi\left(1-\frac{\sqrt{d^2-\left(\frac{D}{2}\right)^2}}{d}\right)\text{ sr}$. Given the average diameter presumed above ($1.4\cdot10^{9}\text{m}$), this gives an average solid angle of $\Omega=2\pi\left(1-\frac{\sqrt{d^2-4.9\cdot10^{17}\text{m}^2}}{d}\right)\text{ sr}$ 17.

At this point, we could set up a proper integral, but my calculus is rather rusty, and not very sharp to begin with. So I'm going to approximate the answer using a series of concentric shells, each having a thickness of $10^{22}\text{m}$ (about a million light years). We'll put our first shell $10^{22}\text{m}$ away, then work our way out from there.

We'll calculate the total solid angle of each shell, then add all the shells together to get the solid angle subtended by the entire observable universe.

The last problem to fix here is that of overlap. Some stars in the farther shells will overlap stars in the nearby shells, causing us to overestimate the total coverage. So we'll calculate the probability of any given star overlapping and modify the result from there.

We'll ignore any overlap within a given shell, modeling as if every star in a shell is at a fixed distance, evenly distributed throughout the shell.

Probability of Overlap

For a given star to overlap closer stars, it needs to be at a position already covered by the closer stars. For our purposes, we'll treat overlaps as binary: either the star is totally overlapped, or not overlapped at all.

The probability will be given by the amount of solid angle already subtended by previous shells divided by the total solid angle in the sky ($4\pi\text{ sr}$).

Let's call the probability a given star, $i$, is overlapped $P_i$, the solid angle subtended by that star $\Omega_i$, and the number of stars $n$. The amount of non-overlapped solid angle subtended by a given shell, $k$, is then $\Omega_{kT}=(1-P_1)\Omega_1+(1-P_2)\Omega_2+\ldots+(1-P_n)\Omega_n\frac{\text{ sr}}{star}$. Since we've said stars in a shell don't overlap each other, $P_i$ is the same for all $i$ in a given shell, allowing us to simplify the above equation to $\Omega_{kT}=(1-P_k)(\Omega_1+\Omega_2+\ldots+\Omega_n)\frac{\text{ sr}}{star}$, where $P_k$ is the probability of overlap for shell $k$. Since we're treating all the stars as having the same, average size, this simplifies even further to $\Omega_{kT}=(1-P_k)\Omega_k n\frac{\text{ sr}}{star}$, where $\Omega_k$ is the solid angle of a star in shell $k$.

Calculating Solid Angle

The number of stars in a shell is given by the volume of the shell times the stellar density of said shell. For far away shells, we can treat the volume of the shell as being its surface area times its thickness. $V_\text{shell}=4\pi d^2 t$, where $d$ is the distance to the shell and $t$ is its thickness. Using $\delta$ as the stellar density, the number of stars is simply $n=\delta V_\text{shell}=\delta 4\pi d^2 t$.

From here, we can use the calculation for solid angle of a shell (from Probability of Overlap, above) to get $\Omega_{kT}=(1-P_k)\Omega_k \delta 4\pi d^2 t\frac{\text{ sr}}{star}$.

Note that $P_k$ is given by the partial sum of solid angle for all previous shells divided by total solid angle. And $\Omega_k$ is given by $\Omega_k=2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right)\frac{\text{ sr}}{star}$ (from Model, above).

This gives us $\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right) \delta 4\pi d^2 t\text{ sr}$. Given that each shell is $10^{22}\text{m}$ away, we can substitute $d_k$ with $k 10^{22}\text{m}$. Likewise, $t$ can be substituted with $10^{22}\text{m}$. And we already calculated $\delta=2.803\cdot10^{-57} \frac{\text{stars}}{\text{m}^3}$ (from Model, above).

This gives us
$\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{(k 10^{22}\text{m})^2-4.9\cdot10^{17}\text{m}^2}}{k 10^{22}\text{m}}\right) 2.803\cdot10^{-57}\frac{\text{stars}}{\text{m}^3} 4\pi (k 10^{22}\text{m})^2 10^{22}\text{m}\frac{\text{ sr}}{star}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\left(1-\frac{\sqrt{k^2 10^{44}-4.9\cdot10^{17}}}{k 10^{22}}\right) 2.803\cdot10^{-57} 8\pi^2 k^2 10^{66}\text{ sr}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\,2.213\cdot 10^{11}\,k^2\left(1-\frac{\sqrt{k^2 10^{44}-4.9\cdot10^{17}}}{k 10^{22}}\right)\text{ sr}$

From here, we can just plug the numbers into a calculation program.

$\Omega_{T}=\sum_{k=1}^{k_\text{max}} \Omega_{kT}$

Where $k_\text{max}$ is just the radius of the observable universe divided by the thickness of a given shell. Thus $k_\text{max}$$=\frac{4.4\cdot 10^{26} \text{m}}{10^{22} \text{m}}$$=4.4\cdot 10^4$$=44000$

$\Omega_{T}=\sum_{k=1}^{44000} \Omega_{kT}$

Results

Because of the large numbers involved, it's difficult to just run this in a program. I resorted to writing a custom C++ program using the ttmath library18 for large numbers. The result was $2.386\cdot 10^{-11}\text{ sr}$, or $1.898\cdot 10^{-12}$ of the entire sky. Conversely, there's about a 1 in 500 billion chance you're standing under a star right now.

Note that we ignored the Milky Way and the Sun for this.

The C++ program can be found at PasteBin25. You'll have to get ttmath working properly. I added some instructions to the top of the C++ code to get you started if you care to make it work. It's not elegant or anything, just enough to function.

The Sun

WolframAlpha helpfully informed me the Sun has a solid angle of about $6.8\cdot 10^{-5}\text{ sr}$, or about 2.8 million times more than all the stars in the universe combined. The solid angle formula above gives the same answer18 if we provide the Sun's 150 gigameter distance and 0.7 gigameter radius.

The Milky Way

We could get an approximation for the Milky Way by taking its size and density and doing the same calculations as above, except on a smaller scale. However, the galaxy is very flat, so the odds greatly depend on whether you happen to be standing in the galactic plane or not. Also, we're off to one side, so there are far more stars towards the galactic center than away.

If we approximate the galaxy as a cylinder with a radius of $5\cdot 10^{20}\text{ m}$ (about 52000 light years) and a height of $2\cdot 10^{16}\text{ m}$ (about 2 light years), we get a volume of $1.571\cdot 10^{58}\text{ m}^3$ 20.

Current estimates of the galaxy's radius are closer to 100000 light years21 22, but I'm presuming the vast majority of stars are a lot closer than that.

There are estimated to be 100 to 400 billion stars in the Milky Way21. Let's pick 200 billion for our purposes. This puts the density of the Milky Way at $\delta = \frac{200\cdot10^{9}\text{stars}}{1.571\cdot 10^{58}\text{ m}^3} = 1.273\cdot10^{-47} \frac{\text{stars}}{\text{m}^3}$ 22, or about 4.5 billion times denser than the universe at large.

This time, we'll take shells $10^{17}\text{ m}$ thick (about 10 light years) and go out from there. But we need to re-organize the math into a spherical form, so we'll presume the galaxy has the same volume, but is a sphere. This gives it a radius of $1.554\cdot 10^{19}\text{ m}$ 24, or 155.4 shells. We'll round to 155 shells.

$\Omega_{T}=\sum_{k=1}^{155} \Omega_{kT}$

Using our formula from above (Calculating Solid Angle), we can start substituting numbers.

$\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right) \delta 4\pi d^2 t\frac{\text{sr}}{\text{star}}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{(k\cdot 10^{17}\text{ m})^2-4.9\cdot10^{17}\text{ m}^2}}{k\cdot 10^{17}\text{ m}}\right) 1.273\cdot10^{-47} \frac{\text{stars}}{\text{m}^3} 4\pi (k\cdot 10^{17}\text{ m})^2 10^{17}\text{ m}\frac{\text{sr}}{\text{star}}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\left(1-\frac{\sqrt{k^2\cdot 10^{34}\text{ m}^2-4.9\cdot10^{17}\text{ m}^2}}{k 10^{17}\text{ m}}\right) 1.273\cdot 10^{-47} \frac{\text{stars}}{\text{m}^3} 8\pi^2 k^2 10^{51}\text{ m}^3\frac{\text{sr}}{\text{star}}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\cdot\,1.005\cdot 10^6\,k^2\,\left(1-\frac{\sqrt{k^2\cdot 10^{34}-4.9\cdot10^{17}}}{k 10^{17}}\right)\text{ sr}$

Plugging this into the program gives $3.816\cdot 10^{-9}\text{ sr}$, which is $3.037\cdot 10^{-10}$ of the total sky. The odds you're standing under a star in the Milky Way are about 1 in 3.3 billion.

Solid Angle Totals

Solid angle is:

Sun, $6.8\cdot 10^{-5}\text{ sr}$
Milky Way, $3.816\cdot 10^{-9}\text{ sr}$
Universe, $2.386\cdot 10^{-11}\text{ sr}$
Total, $6.800384\cdot 10^{-5}\text{ sr}$ (the extra digits are basically meaningless, adding about five thousandths of a percent to the Sun's solid angle)
Milky Way plus Universe, $3.840\cdot 10^{-9}\text{ sr}$ (about 0.6% more than just the Milky Way)

References

1 Michael Walsby's answer to this question, is there a star over my head?. https://astronomy.stackexchange.com/a/33294/10678
2 A Wikipedia article, Cosmological principle. https://en.wikipedia.org/wiki/Cosmological_principle
3 A Wikipedia article, Expansion of the universe. https://en.wikipedia.org/wiki/Expansion_of_the_universe
4 A UCSB ScienceLine quest, About how many stars are in space?, from 2013. https://scienceline.ucsb.edu/getkey.php?key=3775
5 A Sky and Telescope article, How Many Stars are There in the Universe?, from 2014. https://www.skyandtelescope.com/astronomy-resources/how-many-stars-are-there/
6 A Space.com article, How Many Stars Are In The Universe?, from 2017. https://www.space.com/26078-how-many-stars-are-there.html
7 A Wikipedia article, Observable universe. https://en.wikipedia.org/wiki/Observable_universe
8 A Wikipedia article, Sphere, section Enclosed volume. https://en.wikipedia.org/wiki/Sphere#Enclosed_volume
9 A WolframAlpha calculation, surface area of a sphere, diameter 8.8*10^26 m. https://www.wolframalpha.com/input/?i=surface+area+of+a+sphere%2C+diameter+8.8*10%5E26+m
10 A Wikipedia article, Sphere, section Surface area. https://en.wikipedia.org/wiki/Sphere#Surface_area
11 A WolframAlpha calculation, volume of a sphere, diameter 8.8*10^26 m. https://www.wolframalpha.com/input/?i=volume+of+a+sphere%2C+diameter+8.8*10%5E26+m
12 A nineplanets.org article, The Sun. https://nineplanets.org/sol.html
13 A WolframAlpha calculation, (10^24 stars) / (3.568⋅10^80 m^3). https://www.wolframalpha.com/input/?i=%2810%5E24+stars%29+%2F+%283.568%E2%8B%8510%5E80+m%5E3%29
14 A Wikipedia article, Solid angle. https://en.wikipedia.org/wiki/Solid_angle
15 Harish Chandra Rajpoot's answer to a geometry.se question, Calculating Solid angle for a sphere in space. https://math.stackexchange.com/a/1264753/265963
16 A Wikipedia article, Steradian. https://en.wikipedia.org/wiki/Steradian
17  A WolframAlpha calculation, 2*pi*(1-sqrt(d^2-(1.4*10^9 m/2)^2)/d). https://www.wolframalpha.com/input/?i=2*pi*%281-sqrt%28d%5E2-%281.4*10%5E9+m%2F2%29%5E2%29%2Fd%29
18 Website for ttmath. https://www.ttmath.org/
19 A WolframAlpha calculation, 2*pi*(1 - sqrt(d^2 - r^2)/d), where d = 150 billion, r=0.7 billion. https://www.wolframalpha.com/input/?i=2*pi*%281+-+sqrt%28d%5E2+-+r%5E2%29%2Fd%29%2C+where+d+%3D+150+billion%2C+r%3D0.7+billion
20 A WolframAlpha calculation, pi * (5*10^20 m)^2 * (2*10^16 m). https://www.wolframalpha.com/input/?i=pi+*+%285*10%5E20+m%29%5E2+*+%282*10%5E16+m%29
21 A Wikipedia article, Milky Way. https://en.wikipedia.org/wiki/Milky_Way
22 A Space.com article from 2018, It Would Take 200,000 Years at Light Speed to Cross the Milky Way. https://www.space.com/41047-milky-way-galaxy-size-bigger-than-thought.html
23 A WolframAlpha calculation, (200*10^9 stars) / (1.571*10^58 m^3). https://www.wolframalpha.com/input/?i=(200*10^9+stars)+%2F+(1.571*10^58+m^3)
24 A WolframAlpha calculation, solve for r:  (4/3)*pi*r^3 = 1.571*10^58 m^3. https://www.wolframalpha.com/input/?i=solve+for+r%3A++%284%2F3%29*pi*r%5E3+%3D+1.571*10%5E58+m%5E3
25 My C++ program code on PasteBin. https://pastebin.com/XZTzeRpG
26 A Physics Forums post, Orientation of the Earth, Sun and Solar System in the Milky Way. Specifically, Figure 1, showing angles of 60.2° for the Sun, and 23.4° less than that for Earth. https://www.physicsforums.com/threads/orientation-of-the-earth-sun-and-solar-system-in-the-milky-way.888643/


About how many revolutions has the Earth made around the sun? The Earth is about 4.543 billion years old.
However, the length of a year can be changed by anything from an impact to a natural cycle of orbit changes to a slow drift toward or away from the sun to the tug of a nearby body.

So while the Earth is about 4.543 billion modern years old, it may have revolved around the sun a significantly different number of times.

What is a reasonable estimate for how many revolutions - factoring in known geologic events - that the Earth has made around the Sun?

This question touches on the topic of orbital changes but never really answers what I'm asking here.
The second answer on this Quora question seems to answer my question but I don't know if that source is reputable, and the actual calculations and reasoning are not given
This reddit discussion touches on the subject but has differing / inconclusive answers

Since the very early solar system, there have not been large-scale movements of the planets.

In the early solar system, while the planets were still embedded in a protoplanetary disk, there were large movements. (notably the hypothesised "Grand Tack" of Jupiter) However once Jupiter reached it's current position of 5.2AU it remained there, and the Earth formed then too.

There have not been any significant changes to the Earth's orbit since the Earth formed, otherwise there would have been significant heating or cooling. So the average year length was and is about 30 000 000 seconds, and Earth has made about 4.5 billion orbits of the sun since then.


Why is the great red spot shrinking? The Great Red Spot is an anticyclonic storm on Jupiter that is bigger than the Earth. Based on the Wikipedia article, it has been shrinking in size for the last 400 years, going from 41 to 16 thousand kilometers wide. What is causing this shrinking?
These articles (1, 2) talk about the spot shrinkage, but are rather vague on the causes. Both of them mention that we will likely learn more with the Juno probe. Now that we have data from Juno, do we know more about why the Great Red Spot is changing in size?

The Great red spot used to be bigger. Over the course of twentieth century, it lost half of its size and between 1996 to 2006, its area diminished by 15%. If the shrink rate is constant, then the spot will vanish within 20 years. However, computer simulation showed that Jovian weather is stable enough to sustain smaller red spots as Hubble space telescope showed two smaller red spots besides the great red spot indicating that small red spots will persist even if the Great red spot vanishes. The reason for this shrinkage is unknown but scientist suggested it has something to do with Jovian equivalent of global warming and climate change.
Unlike Earth, Jupiter does not have solid landmass so energy lost due to friction is out of question. This explains the long lasting nature of Jovian storms. The Great Red Spot is indeed long-lived, even by Jupiter standards. Researchers are able to pinpoint that Jovian storms that are located in cloud bands with the same direction of rotation tend to be longer lasting. So, the shrinkage of this spot might indicate that the spot is drifting from the clouds bands.
The Great Red Spot is confined by an eastward jet to its north and a westward jet to its south, confining the storm to a constant latitude. However, it has undergone considerable changes in longitude over time, and recent evidence suggests that its rate of westward longitudinal motion is increasing. The bands have also undergone little change in latitude over the time during which they have been observed. So, researchers suggested that this dynamic change of the cloud bands and drifting of the storm might be the reason for the change in the shape of The Great spot.
Although the reason has not been proved yet, researchers are trying to trace the evolution of the Great Red Spot, analyzing its size, shape, color, drift rate and storm’s internal wind speeds from the information available from spacecraft and hope they will provide an explanation in future.
Reference:

The Things that Nobody Knows: 501 Mysteries of Life, the Universe
and Everything William Hartston, Atlantic Books Ltd, 2011.
Celestial Calculations: A Gentle Introduction to Computational Astronomy, MIT Press, 2019.
https://cosmosmagazine.com/space/jupiter-s-great-red-spot-is-shrinking/
https://earthsky.org/space/jupiter-great-red-spot-getting-taller-as-it-shrinks
Historical and Contemporary Trends in the Size, Drift, and Color of Jupiter's Great Red Spot by Simon et.al., The Astronomical Journal, Volume 155, Number 4, 2018 (link)


Difference between zone time and local sidereal time? What is the key difference between LST and ZT?
Also, are UT, GMT and Greenwich Mean Solar Time the same?

Universal Time (UT) and Greenwich Mean Time (GMT) are based on the average (mean) position of the sun. The mean position of the Sun crosses the meridian every 24 hours. There are subtle differences between UT and GMT, but they are the same for most practical purposes. (The Wikipedia article for Greenwich Mean Time points out that the start of the day under GMT changed from noon to midnight sometime in the past. The Time and Date.com article GMT versus UTC describes GMT as a time zone and UTC as a time standard.)

Since our clocks are also based on the mean position of the Sun, the time in your time zone is a fixed number of hours before or after UT.

Local sidereal time (LST) is the time according to the stars at your longitude. Let's assume that star X is on the meridian at 0 hours (midnight). The same star will return to the meridian after 23 hours 56 minutes 4 seconds (approximately) of clock time. In other words, the star time runs almost 4 minutes fast each day. Therefore, there are two key differences between sidereal time and the clock time:

Sidereal time is different at each longitude. Clock time is the same for all longitudes within a time zone.
To convert from clock time to sidereal time, you need to know the time and date.


Is there a theory / equation showing whether or not two passing bodies will go into orbit around each other? I am looking for a theory that shows whether or not two passing celestial bodies will go into orbit.

I assume there would have to be a critical point where the gravitational attraction is stronger than the inertia of the moving bodies.

The bodies would then orbit until they collide, or be affected by a force to fall behind the critical point and escape orbit.

I am looking for a theory that states that objects past this critical point of gravitational attraction must either collide immediately, or go into orbit after falling within a certain distance.

If there are only two bodies, then they will never enter a mutual orbit. For two objects initially gravitationally unbound, in order to become gravitationally bound you must remove energy from the system. With only two bodies (that don't collide), this does not happen. They will accelerate toward each other, change directions according to how close they get, and then leave each other again with exactly the same total energy and momentum as before, but in general shared in some other ratio (for instance, if a small body encounters a large body, the smaller will gain energy and leave with a larger velocity).

On the other hand, if you have three (or more) bodies, one may get slung out with high velocity, thus extracting energy from the two others, which can then go in orbit. But alas, there's no equation for this; the so-called N-body problem has no analytic solution, and must in general be solved numerically.


Is it possible to know if you are travelling close to the speed of light if you have nothing to compare your speed with? This is a question that has been bugging me for some time. Let's see if I am able to make me understood!

Imagine you are travelling in a spaceship at 99.9999999999% the speed of light and you are somehow within a spaceship without windows.

My question is, is there ANY kind of experiment that would show you any time/space dilation effect as compared to being stationary? Is there ANY lab or thought experiment you could make or think that would tell you there are weird things happening with time and space outside your spaceship? I mean, would physics work differently than they would at home? Would radioactive atoms take longer to decay in a way you could measure and compare with your earth-based decay tables? Anything?

Or the only way to find out would be simply to step out of the spaceship and find out that life on Earth has gone extinct in the last couple of hours?

And another question related to the first. If you were travelling in the spaceship at that speed in the X-axis direction and you handthrew a baseball in that same direction, would that ball's mass or inertia explode asyntotically to practically infinity since you are increasing a fantastically high speed a tiny winy bit more? Or perhaps you could not threw the ball at all because of the huge mass increase being that close already to the speed of light? Or you wouldn't notice nothing special and a regular sized with normal weight baseball would impact the other side of the room at a relative speed to you of 60 km/h?

Velocity is relative. From your perspective, your velocity is zero. Without windows (or sensors) on your spaceship, you cannot know the difference. To measure your velocity, you have to define a reference point, eg a star, the average velocity of surounding stars or the galaxy center. If you have such a huge velocity relative to them, from you perspective, they are this fast and not you. So you can observe the time and space dilation of these objects, while your time appears to be normal. You ship is an inertial reference frame.

As for throwing the baseball: From your perspective, you have the speed zero, and the baseball eg. the speed 0.5c. A "fixed" observer, which observes you at 0.99c will measure the balls velocity at 0.997c. With those relative effects, speeds don't add up, only kinetic energy does. http://curious.astro.cornell.edu/about-us/139-physics/the-theory-of-relativity/special-relativity/1016-why-can-t-relative-velocities-add-up-to-more-than-the-speed-of-light-intermediate The increased mass is only a relativistic observation, resulting from the kinetic energy (relative to the observer).


Do we know the fluctuations and constant temperature depth of the moon? Disclaimer: I have no formal scientific training.

From what I understand, surface temperature effects on Earth extend only so far into the ground so that at a certain depth the temperature remains constant around the globe.

Has any attempt been made to identify/predict subsurface temperature fluctuations and/or constant temperature depth?

Note: The reason I am interested in looking under that particular rock, so to speak is that given the dark side/light side temperature difference on the moon I am wondering if we can predict the ideal location and depth on the moon to find a shallow surface layer that leads to a constant temperature in a range suitable for human/plant life without requiring machinery to preserve the temperature.

Related:

https://www.physicsforums.com/threads/temperature-for-3-or-4ft-underground.826193/

Do we know the fluctuations and constant temperature depth of the moon?

Yes, at some Apollo landing sites, it's been carefully measured! The astronauts make deep holes in the Moon's regolith and put in thermal probes that precisely measured temperature. These signals were radioed back to Earth by a solar-power transmitter, in some cases for years. These experimental packages were called ALSEP and included seismometers and passive retroreflectors and other "goodies".

I'm going to quote verbatim this answer from Space Exploration SE since it fits so nicely here.

  tl;dr: The number is often given as 252 +/- 3 K (or about -21°C or -6°F), so @BowlOfRed pretty much nailed it using first principles!

    Without atmosphere, equilibrium temperature at 1AU is about -17C. I would expect that to be the temp "near" the surface around the equator. Colder as the latitude increases.

  Reading the references, it seems that the diurnal oscillation (what Earthies might call "monthly" or Lunies might call "daily") in temperature decays with a 1/e depth of about 30 centimeters, so by the time one reaches 1 meter, the temperature "stabilizes" temporally, but of course will increase with depth since the Moon's interior is hot. You can see this in Figure 2 of the last paper ("Subsurface temperature histories, covering 3.5 years...") where the oscillations in temperature at different depths along the probe are plotted.

  Further, some of this data can already be found in @KimHolder's question Does the temperature near the surface of the Moon rise rather quickly with depth? who quotes as NASA News article Down the Lunar Rabbit-hole which then quotes Mark Robinson of Arizona State University, principal investigator for the LRO camera.

    "The tunnels offer a perfect radiation shield and a very benign thermal environment," says Robinson. "Once you get down to 2 meters under the surface of the Moon, the temperature remains fairly constant, probably around -30 to -40 degrees C."

  Why the difference between the "ground truth" actual temperature measurements discussed below and the quote of a quote?

  (image source also here) My guess is that the -30 to -40 number is a mid-latitude number, while the following concentrates on nearer to equator values. There's no physical way the temperature can drop by 10 or 20C in 1 meter of increasing depth, so the difference is more context and location.

  I'm not an expert on planetary science, but the oft-cited number for an average sub-surface temperature about 1 meter below the surface seems to come from the following papers:

  The Apollo 15 lunar heat-flow measurement, Langseth, M.G., Clark, S.P., Chute, J.L. et al. The Moon (1972) 4: 390. https://doi.org/10.1007/BF00562006

    Abstract:

    The heat-flow experiment is one of the Apollo Lunar Surface Experiment Package (ALSEP) instruments that was emplaced on the lunar surface on Apollo 15. This experiment is designed to make temperature and thermal property measurements in the lunar subsurface so as to determine the rate of heat loss from the lunar interior through the surface. About 45 days (1 1/2 lunations) of data has been analyzed in a preliminary way. This analysis indicates that the vertical heat flow through the regolith at one probe site is 3.3 × 10−6 W/cm2 (±15%). This value is approximately one-half the Earth's average heat flow. Further analysis of data over several lunations is required to demonstrate that this value is representative of the heat flow at the Hadley Rille site. The mean subsurface temperature at a depth of 1 m is approximately 252.4K at one probe site and 250.7K at the other. These temperatures are approximately 35K above the mean surface temperature and indicate that conductivity in the surficial layer of the Moon is highly temperature dependent. Between 1 and 1.5m, the rate of temperature increase as a function of depth is 1.75K/m (±2%) at the probe 1 site. In situ measurements indicate that the thermal conductivity of the regolith increases with depth. Thermal-conductivity values between 1.4 × 10−4 and 2.5 × 10−4 W/cm K were determined; these values are a factor of 7 to 10 greater than the values of the surface conductivity. If the observed heat flow at Hadley Base is representative of the moonwide rate of heat loss (an assumption which is not fully justified at this time), it would imply that overall radioactive heat production in the Moon is greater than in classes of meteorites that have formed the basis of Earth and Moon bulk composition models in the past.

  In-situ measurements of lunar heat flow (also here) in Soviet-American Conference on Geochemistry of the Moon and Planets, Langseth, M. G., and S. J. Keihm, NASA Spec. Publ., NASA SP-30, 283 – 293, 1977.

  The number is often given as 252 +/- 3 K (or about -21°C or -6°F) as for example in Latitude variation of the subsurface lunar temperature: Lunar Prospector thermal neutrons R. C. Little W. C. Feldman S. Maurice I. Genetay D. J. Lawrence S. L. Lawson O. Gasnault B. L. Barraclough R. C. Elphic T. H. Prettyman A. B. Binder
  First published: 29 May 2003 https://doi.org/10.1029/2001JE001497

  Some of the experimental and theoretical details behind the data and its analysis can be read in Revised lunar heat-flow values, Langseth, M. G., Keihm, S. J., & Peters, K., Lunar Science Conference, 7th, Houston, Tex., March 15-19, 1976, Proceedings. Volume 3. (A77-34651 15-91) New York, Pergamon Press, Inc., 1976, p. 3143-3171.


What would I see with a telescope of 70mm aperture , 300mm focal length and maximum magnification of 150x What would I be able to see with 70mm aperture telescope with 300mm focal length with maximum magnification of 150 x?

It's 70mm aperture, so the maximum useful magnification will be 165x.
If you have a 4mm eyepiece, you will have 75x magnification, and at that, you can expect to see rings of Saturn, moons of Jupiter in good conditions. Venus will show phases. I can't say much about Mercury.
You can see many surface features of the Moon. The view will be particularly good at 150x. You will have to use a 2x Barlow for that. Remember, the view will be good in good "seeing".
If you are in dark sky away from light pollution, you can expect to see many globular clusters and nebulas particularly, the Orion Nebula.


How do astronomers convert radio signals received by their antenna to images? I'm very new to radio astronomy and I am curious how an the signals received by a radio antenna are converted to data on a computer and then images? Is there a popular software to use for this, or do astronomers typically write their own code?
Thanks in advance!

There is no universal answer to this question. Instrumentation, data formats and software differ greatly between telescopes.

Most software packages in use are written by astronomers as open source and shared and modified widely within the community. But instead of listing popular packages, perhaps a description of the common way of plotting images is a better answer here.

Consider a very basic radio telescope: A dish that's movable in two axis with a simple radiometer (think "power meter for RF") as the receiver. During an observation, you record four-tuples of time, azimuth, elevation and signal strength. Plotting these is pretty straightforward: Convert the instrument coordinates to celestial coordinates, and plot the signal strength over these coordinates. You'll get a classical intensity distribution map (source). Those maps can be created using simple spreadsheet programs like Excel or custom code snippets.

Adding the capability to record spectra to your receiver makes the table you record your data in a bit more complex (since your last column is no longer a scalar value but a spectrum), and allows the creation of intensity-velocity-plots (source). From a software perspective, this is still a simple plot-points-in-2D excercise.

For anything more complex, especially array setups with multiple antennas and/or focal plane arrays, data reduction and plotting requirements are far more involved and highly depend on the actual setup. CASA and AIPS, mentioned in the comments above, are widely used packages, but there is a variety of custom code around.


How is the heat production inside moons compliant with Conservation of Energy? A Nova episode said that during stages of some moons' elliptical orbits, the moon is stretched, and friction causes the inside of the moon to heat up. I do not understand how this is compliant with the Law of Conservation of Energy - what energy would be lost to make the inside of the moon heat up?

The heat production inside moons is due to tidal heating. Due to this, the orbit of the moon would change.
The orbital and rotational energy of the moon is dissipated as heat either in the surface ocean or in the interior of the moon. This is especially seen in the moons of large planets, for example Io.
Io, being Jupiter's innermost moon, is pulled by Jupiter on one side and the other moons on the opposite side. Due to this, Io's orbit is highly elliptic. During this orbit, the shape of the Io changes continuously (the surface flexes by ~100m). This causes a large amount of friction in the moon.

"Tidal heating on Io" by Lsuanli - Own work. Licensed under CC BY-SA 3.0 via Commons.
Usually, the tidal heating leads to tidal dissipation and circularizes the orbit. However, it hasn't happened in Io because it is in orbital resonance with other moons of Jupiter.
If the orbital period of the moon is less than the rotational period of the planet, the moon is decelerated if the moons are out of the synchronous orbit, the opposite thing happens, conserving energy and angular momentum.
On the other hand, Europa, another of Jupiter's moons is tidally locked (it has same orbital period as the rotational period of the larger body) to the Jupiter, which causes tidal flexing on the surface.


What causes the duration of moon phases (the time the moon is visible above the horizon) to vary for an observer in the northern hemisphere? Why does each phase of the moon stay up a different number of hours per phase, so that, for example, January’s full moon was up twice as long as the new moon? Plus every month this bulge shifts and at some point the opposite phase is doubled. Three June full moon nights takes six March 1st quarter nights. Why does this happen?

The interval between moonrise and moonset varies for the same reason that summer days are longer than winter days.
How long a celestial object appears above the horizon depends on the observer's latitude and the object's declination (angle north or south of the celestial equator).
At full moon on 2019-01-21, the Moon's declination was +20$^\circ$, and there were 15 hours between moonrise and moonset at 40$^\circ$N latitude.
At new moon on 2019-02-04, the Moon's declination was -18$^\circ$, and moonrise and moonset at 40$^\circ$N were only 10h10m apart.

Another way to think of this is that when the Moon is in the first quarter, it's within about 5 degrees of where the Sun would be 3 months later; last quarter, the Moon is where the Sun would be 3 months earlier, etc.
A full Moon in June is up about as long as the Sun in December; a first quarter Moon in March is up about as long as the Sun in June.


How to see Saturn's rings through a pair of binoculars? I have a pair of 15x70 binoculars. I would not say what manufacturer to avoid being the subject of 'hold'.

The current angular diameter is 15.5 arcseconds. I am wondering if that condition is enough to be able to see rings. My eyesight is fair.

You will not be able to see Saturn's rings directly with a 15x70 telescope, but you will be able to notice that Saturn appears elliptical, not round. This is due to the shape of the rings around the planet, but you will not be able to resolve the hole between the rings and the planet (much less the Cassini gap).

Just as an impractical suggestion in the interest of giving to you a more complete answer, you might be able to mount a camera and use image stacking to see a smidge of ring and differentiate it from the sphere of the planet. If you do go this route, please post online about your setup and results, whether it succeeds or not.


Reference frames in relativity? I've recently been introduced to relativistic celestial Reference Systems (RS) such as the Geocentric Celestial Reference System (GCRS) and Barycentric Celestial Reference Systems (BCRS). So, for example we use GCRS to model near-Earth satellites or use BCRS to model probes far into our solar system. I understand the need for such RS i.e. to be in compliance with the theory of relativity - no favoured coordinate systems, limitation of the speed of propagation of information etc.

However, what are some layman definitions of advantages of such systems to really send the message home?

I notice that when describing $n$-body dynamics in our solar system for example, Barycentric coordinate time is used instead of proper time. This blows my mind!

It depends on your definition of "layman" but Patrick Wallace has a slide presentation on what changed and why with the new reference systems. There is also the USNO Circular 179 which covers all of this and is pretty readable.

In a lot of ways it boils down to the realization that the wobbly (in space and rotation) Earth was a lousy thing to base co-ordinate systems on. The new systems are what's known as "kinematically non-rotating" which mean they are fixed in space and don't move. This makes it much easier to make more precise theories (which were needed) to describe how things like the orientation of Earth's pole (the precession and nutation you may have heard about) move in the Solar System without having to reference it to something like the equator and ecliptic which are always moving about.


"Mysterious radio signals" - how could a geostationary satellite and Ross-128 line up with Arecibo? The Phys.org article 'Mystery' signal from space is solved. It's not aliens cites a summary of the conclusion "by Abel Mendez, director of the Planetary Habitability Laboratory at the University of Puerto Rico at Arecibo in a blog post Friday, revealing the true nature of the signals."

After further fueling speculation by summoning the world experts in the hunt for life elsewhere in the universe—The SETI Berkeley Research Center at the University of California—the team issued its conclusion.
"We are now confident about the source of the Weird! Signal," Mendez wrote.
"The best explanation is that the signals are transmissions from one or more geostationary satellites."
The signals only appeared around Ross 128 because it is located "close to the celestial equator where many geostationary satellites are placed," Mendez added.

The article also goes on to say:

Astronomers detected strange signals that seemed to be coming from a dwarf star about 11 light-years away, but have now determiend that the signals are interference from a distant geostationary satellite

Mendez is further quoted the Space.com article Weird Radio Signals Detected from Nearby Red Dwarf Star:

Each of these hypotheses has its issues, he said. For example, solar flares of the type that could be responsible generally occur at lower frequencies. In addition, Mendez wrote, there aren't a lot of other objects in the Ross 128 field of view, "and we have never seen satellites emit bursts like that."

and in the Space.com article Not Aliens: Weird Radio Signal from Star Likely Has Duller Explanation:

"This explains why the signals were within the satellite’s frequencies and only appeared and persisted in Ross 128; the star is close to the celestial equator, where many geostationary satellites are placed," Mendez added. "This fact, though, does not yet explain the strong dispersion-like features of the signals (diagonal lines in the figure); however, it is possible that multiple reflections caused these distortions, but we will need more time to explore this and other possibilities."

Wikipedia lists the star Ross 128 at at declination of about +0° 48', and the latitude of the Arecibo telescope to be about +18° 21'. Satellites in geostationary orbits are in a circular ring approximately coincident with the equator at a distance from the center of the Earth of only about 42,000 km. From the Northern hemisphere, geostationary satellites will appear several degrees south in declination. (See this interesting answer.)
For Arecibo, I calculate about -3.2° declination.
QUESTION: How could a radio telescope have picked up satellite signals from a satellite at about -3.2° declination while observing a star at +0.8° declination? And why wasn't this immediately ruled out - surely radio telescopists know about satellites by now, especially the crowded ring of geostationary satellites! Aren't there standard procedures to rule these out before making announcements?
Could the half-angle acceptance really have been 4 degrees? Since the brown red dwarf is extremely close to the solar system, could proper motion put it 4 degrees away from the value listed in Wikipedia?

above: "The signal that seemed to emanate from the red dwarf star Ross 128, as detected by the Arecibo Observatory in May 2017 (enclosed in the red frame)." Credit: PHL @ UPR Arecibo  From here.

The original team suggested three main possibilities: "(1) unusual stellar activity, (2) emissions from other background objects, or (3) interference from satellite communications"  Clearly radio astronomers know about satellites.

They noted "in the absence of solid information about the signal, most astronomers would think that [radio interference or instrumental failures] would probably be the most likely explanation."

The satellite hypothesis has some problems it "does not yet explain the strong dispersion-like features of the signals (diagonal lines in the [weird] figure [in the question])" But on a balance of probablities this is the most likely explanation.

There are satellites in the region of Ross 128:

Location of Geostationary satellites operating between 4-8 GHz in the same region of the sky as Ross 128 (yellow dot).
Credit: Enriquez et al. (SETI Berkeley), http://seti.berkeley.edu/ross128.pdf

Your analysis of the location of a satellite, based on the geometry of satellite orbiting over the equator is not supported by this image.

The observers did not claim to have received a SETI signal. Indeed if you look at their original report they note that "aliens" is at the bottom of their possible explanations. The signal looks different from typical satellite bursts. They hoped the signal was astronomical in nature so there may have been a little wishful thinking.

So the most likely explanation is that we are observing multiple reflections from geostationary satellites transmitting on the 4-8 GHz band.

All quotations from the press release from the planetary habitability observatory.


Shape of neutron stars I've heard that the more an object spins, the less of a true sphere it is.
Using this logic most of neutron stars would be far from spherical,in general what shape are most neutron stars?

I don't think you'll find a single agreed shape for a rotating neutron star, not least because we don't have an agreed single model for the equation of state of the material in a neutron star (which is more complex than the name suggests).

I found one openly available paper (I'm sure there are more) which will give you a rough flavor for the complexity of modeling the shape of neutron stars.  As you'll see the difficulty of no single model for an equation of state (EOS is the shorthand typically used) is just one issue.

I think "ellipsoid" should be considered as an approximation, although it's not something I'd consider written in stone.

Remember that to be useful a paper has to provide not just a model for what the shape might be, but also someone has to provide a way to measure this, which is challenging.  I think one of the hopes for the new era of gravitational wave astronomy is to be able (eventually) to make more useful and measurements that help us investigate the interior of neutron stars.

So this is an open question, I think.

@Rob-Jeffries asked a question in comment about typical numbers for the deformation, and I answered in comment but comments can be removed by the system, so I'm adding that information as an edit :

In the first section of the paper I linked to they do quote fractional deformations as being typically $10^{−5}$, perhaps $10^{−4}$ in special cases and in extreme cases up to $10^{−3}$. However another paper gives an analysis based on crustal rigidity and a very small deformation for a particular neutron star. The paper I initially liked to describes an upper limit based on gravitational wave considerations, I think, rather than a general analysis.


Is there any residual oscillation left from the Moon rotation? The Moon is tidally locked to Earth - a long time ago it was spinning, but after a time, the spin stopped - two factors stabilizing it: unequal mass distribution creating an equilibrium position where there's one "preferred" axis by which it faces Earth - an attractor, and the dynamic tidal forces acting as stress upon its structure, dissipating the energy - a friction, reducing the rotation speed.

So, at one point in time, it would no longer complete a full spin, but instead reverse its turn direction, start spinning in opposite direction, with maximum spin speed around the time where the "preferred" axis faces Earth, then slowing down again, until its spin stops, and reverses direction again - each time the angle smaller as the dynamic forces act against the direction of rotation - but also, with the rotation angle and angular velocity dropping, the value of the dynamic forces dropping.

There is a significant apparent wobble caused by the orbit's eccentricity, and possibly some actual continuous wobble, as the result - with the attractor axis missing Earth as result. But is there any long-term oscillation left over from the times when the Moon was spinning - a harmonic motion around its original spin axis, that would likely not be in tune with the orbital period? (I imagine the period of these oscillations would be very long; after all it's about reversing Moon's spin back and forth, by a force that isn't all that strong.)

This is called "physical libration". The Moon’s physical librations and determination of its free modes (2011) estimates it using the Apollo mission retroreflectors. Their result is very small. They note that the "damping times for these two modes are estimated as $2$x$10^4$ and $2$x$10^6$ years", so it is a mystery why it is even as large as their small result. They say:

  Some possible excitation mechanisms have been explored in the past without satisfactory explanation. It has been shown that a recent meteoroid impact is an unlikely source of such excitation Peale (1975). Eckhardt (1993) proposed an excitation process related to a resonance crossing of the longitude normal mode (of 2.9 years) and a close forced frequency. During the evolution of the lunar orbit, the free and forced frequencies change slowly and can cross. However, the mechanism excites only the libration in longitude mode. Yoder (1981) proposed an alternative mechanism, based on turbulent fluid core interaction, to excite the wobble mode. The new determination of the amplitudes of the free librations invites new investigation of their excitation mechanisms.

I did not search to see whether anyone has taken up their invitation.


meaning of p-wave charmonia What exactly does p-wave mean when referring to particles, like charmonia states.

For example, see the following reference:

http://arxiv.org/abs/hep-ph/9208254

My doubt is - p wave in scattering means l=1. But how is that used to label a particle state?

Thanks.

You're correct that "$p$-wave" in this context means that the charmonium has orbital angular momentum $L=1$. The principal observable effect is that two-particle states with even $L$ have even parity, while those with odd $L$ change sign under parity.

If you were to skim the Particle Data Group's list of charmonia (or the $c\bar c$ section of the short list) you would find that each particle has a value listed for spin, $J$, and parity, $P$. The lightest few states are

$\eta_c(1S)$, with $J^P = 0^-$. This is a charm-anticharm bound state where the two quarks have opposite spins and the angular momentum $L=0$. (The total parity is negative because the intrinsic parities of the $c$ and $\bar c$ have opposite sign.)
$J/\psi(1S)$, with $J^P=1^-$. This has the same parity as the spinless $\eta_c$, but a unit of spin. So it must be that here the two quarks have the same spin and $L=0$.
$\chi_{c0}(1P)$, with $J^P=0^+$. The change in parity suggests that this state must have a different $L$ than the $\eta_c$ and $J/\psi$, and it's the first with a $P$ in its name. There's also a $\chi_{c1}$ and $\chi_{c2}$ with $J^P=1^+, 2^+$, respectively. That suggests these are the three spin combinations of a charm-anticharm pair with spin $S=1$ and orbital angular momentum $L=1$.

There are also some charmonia with $(2S)$ and $(2P)$ in their names. There don't appear to be any definitively labeled $d$-wave charmonia.


If a star were travelling at 99% of the speed of light, directly at Earth would we see it? Stars being flung from black holes in the news made me question if something were heading directly for us at the speed of light, would we even know?

Update: Since it's impossible for a star to travel at the speed of light, would we see it coming if it were traveling at 99% the speed of light? and if so, how much warning would we have?

Processes energetic enough to accelerate a star to 99% of the speed of light are not known. A close interaction and ejection from a multiple black hole system could perhaps provide enough energy, but such close interaction with black holes would probably rip the star apart before it could reach such excessive speeds. Hypervelocity stars are known. They have speeds less than about 1% of the speed of light.

A star travelling through space at such speeds would also have problems.  It would be slamming into interstellar gas. I have no idea if it could even survive, or would it be eroded by the interaction with the thin gases in interstellar space.

A star at that speed would be significantly blue-shifted. This would make its spectrum very odd.  A small dim star could be visible, catalogued and observed to have a very odd spectrum at a distance of about several hundred light-years. It is likely that we would first observe such a star several hundred years before it got close to us. However such high speed objects probably don't exist.


TRAPPIST 1 and planet statistics? From what I understand, the TRAPPIST 1 system was discovered using the transiting method looking at 20 ultra-cool small stars. I'm not sure how many have been observed since this study.

What I wonder is, what would we expect to see if all 20 had planets? I'm not sure how many planetary systems in this and future similar studies we would expect to see edge on. What about radial velocity measurements?

Did we just get lucky with TRAPPIST 1?

The probability of observing a planet transit is approximately $(R_p + R_s)/a$, where $R_p$ and $R_s$ are the planet and stellar radius respectively and $a$ is the semi-major axis of the planet's orbit. This assumes that planetary orbits are circular and randomly oriented with respect to our line of sight to the star (for which there is little or no counter-evidence).

In the case of Trappist-1, it has an estimated radius of $R_p=0.11R_{\odot}$, the planetary companions have radii and semi-major axes of (in units of Earth radii and au respectively)  (1.09,0.011), (1.06, 0.015), (0.77,0.021), (0.92,0.028), (1.05, 0.037), (1.13, 0.045), (0.75, 0.059).

Using approximation above, then the transit probabilities are 0.051, 0.037, 0.026, 0.020, 0.015, 0.012, 0.009 respectively.

Now these probabilities cannot be treated as independent of one another, since it is quite likely (though this is still the subject of research) that planets will naturally tend to inhabit the same orbital plane (the Trappist-1 planetary orbit inclinations are within 0.2 degrees of each other - flatter than the solar system). Thus you are much more likely to see second and subsequent transits in a system with one transiting planet than you are in a star picked at random. On the other hand, a system with lots of planets that are not quite in the same orbital plane gives an increased chance of seeing one of them transit.

Either way, you can see from the probabilities above that seeing one and only one transiting system of this type from 20 targets, even if all of them had similar planetary systems,  is not at all unexpected.

Radial velocity studies of such systems are difficult, but currently being attempted. Trappist-1 has an apparent J magnitude of 11.4, which is very challenging for current telescopes and instrumentation to get high resolution spectroscopy.


When you travel fast around a black hole, do you experience high speed yourself? When an object in the space comes close to a massive body (like a black hole or a neutron star), we see from Earth that its speed significantly increases while it is close to the body. As we know, this is due to space distortion around massive objects known as gravity. But also, gravity causes time dilation, so an observer near a black hole will experience less time spans between same events than a far-away observer. So, the question is: if we travel on an asteroid and come close to a black hole, will we experience an increase in our speed, or the time dilation will compensate space distortion and we'll feel the same movement speed as before (i. e. no acceleration)?

If you jump out an aeroplane, do you experience high speed? Many people would say you do, since your velocity relative to the ground increases.

Note the importance of the phrase "relative to the ground" since to describe speed you always need to say what you are measuring the velocity relative to.

A simple choice is to measure your velocity relative to a distant observer. In this case then, yes as you pass close to the black hole your velocity relative to the distant observer increases. As you approach the event horizon, you velocity relative to the observer will approach the speed of light.

Depending on how close you pass the the event horizon, you will experience significant time dialation, so that more time will pass in the observer's frame than in yours. This means that if you manage to escape the gravitational pull of the black hole, and return to the observer, you will have found that the observer has aged compared to you.


How did Fermi manage to see neutron star merger? When the GW170817 neutron stars merger was observed by LIGO/Virgo, the Fermi gamma-ray telescope observed the event 2s after the merge. How did it know where and when to look? It must take some time to rotate the satellite and I guess that evaluation of data from LIGO/Virgo also isn't instantaneous. Or is it? And does it automatically send suspicious signals to the Fermi operation centre to point it to the right location?

The Fermi instruments have a very wide field of view; the gamma ray burst detector covers the whole sky not occulted by the earth with low angular precision, and the Large Area Telescope covers about 1/5 of the entire sky with arc-minute precision.  I don't know if the LAT happened to be pointing in the right direction (I guess ~20% chance?) but the burst detector would at least be able to identify the timing of the gamma ray signal from the collision from nearly anywhere in the sky.


How to check what telescopes were looking at a certain portion of sky at a particular time? I have some data that I collected from observations at a specific RA, DEC and I am trying to figure out what telescope surveys were looking at that portion of the sky at that particular time? I am currently trying to use NAVO's python notebooks to query databases but I'm struggling.

Any help would be greatly appreciated!

The areas of the sky covered by the major Near Earth Object (NEO) surveys are reported to the Minor Planet Center. You can plot visualizations of that sky coverage data using the sky coverage form where you can filter by depth, date and survey. The raw data is available, after the surveys give permission for it to be released, from the raw data page. This provides a download link to a tarball of all the pointing data so far, broken down by survey site and then a file per night. The description of the pointing fields is also shown on that page; hopefully it shouldn't be too hard to use a combination of e.g. astroPy's Table and 'SkyCoord` to read and parse the data.


How do stellar temperatures vary? The temperature of the surface of the Sun (photosphere) is between 4500° - 6000° Kelvin. Inside the core, it's around 15.7 million degrees Kelvin.

In other types of stars (neutron stars, white dwarves, etc.), what is the temperature of these areas (though many do not have these same layers) and how to they compare to the temperatures of the sun?

This question has two parts:

Surface Temperatures

A very useful diagram which shows surface temperatures, and also gives you the temperature of any star you can observe is the Herzsprung-Russell Diagram, this one from le.ac.uk.

As you can see, the yellow of our own sun places it in the 4.5 kKelvin to 6 kKelvin, as noted in the question. This temperature is down towards the lower end of average. The main sequence, where most stars are, tops out about 20 kKelvin, and there are some up towards the 40 kKelvin region - they aren't shown here as they are much rarer.

White dwarves are a little hotter than our sun - between 6 kKelvin and 10 kKelvin.

Neutron stars are way off the main sequence - young ones can be over 1 MKelvin!

Core temperature:

Internally, the core temperatures are dependent on the mass of the star. In our sun, energy is delivered via the proton-proton chain mechanism, which occurs up to about 20 MKelvins, whereas more massive stars can use the Carbon-Nitrogen-Oxygen cycle - which happens from about 15 MKelvins upwards.

The differences are mainly down to convection and radiation differences - this extract from Wikipedia's Main Sequence page describes this in some detail:

  Because there is a temperature difference between the core and the surface, or photosphere, energy is transported outward via radiation and convection. A radiation
  zone, where energy is transported by radiation, is stable against
  convection and there is very little mixing of the plasma. By contrast,
  in a convection zone the energy is transported by bulk movement of
  plasma, with hotter material rising and cooler material descending.
  Convection is a more efficient mode for carrying energy than
  radiation, but it will only occur under conditions that create a steep
  temperature gradient. In massive stars (above 10 solar
  masses) the rate of energy generation by the CNO cycle is very
  sensitive to temperature, so the fusion is highly concentrated at the
  core. Consequently, there is a high temperature gradient in the core
  region, which results in a convection zone for more efficient energy
  transport. This mixing of material around the core removes the
  helium ash from the hydrogen-burning region, allowing more of the
  hydrogen in the star to be consumed during the main-sequence lifetime.
  The outer regions of a massive star transport energy by radiation,
  with little or no convection. Intermediate mass stars such as
  Sirius may transport energy primarily by radiation, with a small core
  convection region. Medium-sized, low mass stars like the Sun have
  a core region that is stable against convection, with a convection
  zone near the surface that mixes the outer layers. This results in a
  steady buildup of a helium-rich core, surrounded by a hydrogen-rich
  outer region. By contrast, cool, very low-mass stars (below 0.4 solar
  masses) are convective throughout. Thus the helium produced at the
  core is distributed across the star, producing a relatively uniform
  atmosphere and a proportionately longer main sequence lifespan.


gravitational time dilation multiple sources I wondered how time would be dilated for an object in the middle of 2 black holes, however
https://en.wikipedia.org/wiki/Gravitational_time_dilation only provides a formula for one source of gravity

i tried approximating it by multiplying the time dilation factors, however when i used 1 single source with twice the mass the results differed from 2 objects with half the mass quite drastically

the examples:
mass $1,00E+036$kg distance $3000000000$m
time:$0,7105905766$

mass $2,00E+036$kg distance $3000000000$m
time:$0,099387802$

approximation: $0,7105905766*0,7105905766=0,5049389676
$
which is off by a factor of 5 and would be even more off the closer the distance gets to the schwarzschild radius

so how can the time dilation of multiple objects be approximated ?

In the static weak-field approximation, the metric is approximated in terms of the the Newtonian gravitational potential $\Phi$ as:
$$\mathrm{d}s^2 = -\left(1+2\Phi\right)\mathrm{d}t^2 + (1-2\Phi)\mathrm{d}S^2\text{,}$$
where $\mathrm{d}S^2$ is the metric for Euclidean $3$-space. A more general situation requires one to solve a multi-body problem in general relativity, although the parametrized post-Newtonian formalism provides a canonical approximation scheme. So the completely general answer to your question is "it's very difficult problem of numerical general relativity that doesn't have a neat analytic solution."

However, at large distances from slowly-moving objects (i.e., large compared to their Schwarzschild radii, we can approximate a multi-body time dilation relative to a stationary observer at infinity as just given by the sum of their gravitational potentials:
$$\frac{\mathrm{d}\tau}{\mathrm{d}t} = 1 - \frac{1}{c^2}\sum_k\frac{GM_k}{r_k} \text{.}$$
Note that this is consistent with the Schwarzschild formula for large radial coordinates, since by the Taylor-MacLaurin expansion,
$$\frac{\mathrm{d}\tau}{\mathrm{d}t} = \sqrt{1-\frac{2GM}{rc^2}} = 1 - \frac{GM}{rc^2} + \mathcal{O}\left(\frac{R^2}{r^2}\right)\text{,}$$
where $R = 2GM/c^2$ is the Schwarzschild radius. This also means that simply multiplying the factors is a pretty good approximation as long as those factors are close to $1$.


North node, South node correct understanding What is the correct concept of the North node and South node? There is some conflicting information on the internet.
Which one of these two is correct?
1. Sourced from some Astrology material - How can North and South be 180-degree opposites in this since these are on one side of the orbit curve?
https://astrologyquestionsandanswers.com/2016/10/12/major-and-minor-chords-the-moons-nodes-part-two/

https://www.spiritdaughter.com/blogs/align-with-magic/a-shift-in-perspective-the-north-node-moves-to-gemini

http://astrologyclub.org/eclipses-and-the-moons-nodes/

2.From Wikipedia source - This seems incorrect too since there is no "separate" Moon orbit as per around the Sun
https://en.wikipedia.org/wiki/Lunar_node

http://www.antidotebrooklyn.com/celeste-blog/2019/4/21/riding-the-tail-of-the-celestial-dragon

The first few diagrams show the lunar nodes coinciding with the first and last quarter phases, which is only the case every 173 days or so.
When the nodes are aligned with the new and full phases, eclipses can occur:

In this diagram from timeanddate.com,
our point of view is north of the ecliptic, the ascending node ☊ is in the 5 o'clock direction, and the descending node ☋ is in the 11 o'clock direction.
Nodal precession moves them retrograde (clockwise from north) about 19.4° per year, an 18.6 year cycle mostly due to the Sun's gravity.


Laser Colimator vs Cheshire eyepiece Both the laser collimator and cheshire eyepiece are tools used in astronomical telescope collimation. What are the main advantages/disadvantages of each? Which one might be best suited to start with for telescope newbies?

Why not both? (if you can afford them)

Both will do the job, but each has its strengths.

Cheshire eyepiece (or, more properly, Cheshire/sight tube combo, which is the popular combination nowadays):

It's easier to use when the scope is grossly out of alignment. Like, when you put it together for the first time, or after it dropped on the ground and rolled downhill for a minute. But can you do that with a laser? Yes.

It's easier to align the secondary mirror, and it tends to produce dead-on results for this purpose. But can you do that with a laser? Yes.

It is not immune from precision issues. Most parts of the Cheshire are usually well aligned (because they are machined on a lathe), but the crosshairs on the sight tube may or may not be perfectly centered. But can lasers also suffer from alignment problems? Yes.

Recommended Cheshire / sight tubes: the ones made by Catseye.

http://www.catseyecollimation.com/

They make much more than just Cheshire eyepieces, and all their tools are very high quality.

Laser collimator:

It's faster and arguably easier to use for normal daily collimation before you begin your normal observing sessions. But could you use a Cheshire instead? Yes.

A laser is only as good as its own alignment precision. A misaligned laser will not collimate your scope properly. But do Cheshire sight tube combos suffer from precision issues with their crosshairs? Yes.

Recommended laser collimators: the laser collimator and the tuBlug combo made by Howie Glatter. Unfortunately Howie has passed away recently, so either wait until someone picks up where he left off, or find a used laser/tuBlug combo somewhere on eBay or classified ads.

http://www.collimator.com/

Howie's lasers are essentially guaranteed to have zero collimation errors from factory. From anecdotes told by users, these devices tend to keep their perfect collimation even after being dropped to the ground repeatedly.

The tuBlug is not mandatory, but it makes primary mirror alignment much easier.

Keep in mind that you also need a precision center mark on the primary, for these methods to work well. Hopefully it's been placed there precisely from factory; otherwise there are ways to re-center it.

How to verify if the laser is centered:

Place it in the focuser. Do not tighten the screws too much, allow the laser to rotate in the focuser a little bit.

Turn the laser on. Watch where the spot hits the primary. Now rotate the laser in the focuser. If the spot describes a circle as you rotate the laser, then the laser is misaligned.

Different method: Make a V-shaped trough from 2 pieces of wood or aluminum. Placed the through horizontally on a sturdy table. Put the laser on the trough. Turn it on, projecting the spot on a distant flat object, like a wall or fence across the street.

Now rotate the laser in the trough and watch the spot. Does it stay put? Then the laser is fine. Does it move in a circle? Then the laser is misaligned - find the adjustments screws (if it has them) and adjust it until it's perfectly square.


Why aren't brown dwarfs the dark matter? There seems to be not enough matter to account for the fact that the speed of some stars located far from the Galaxy center is almost the same speed as those stars found nearer to the center of the Galaxy, therefore it seems there should be even more matter: so this missing matter is termed dark matter. There is a kind of stellar object that isn't massive enough to produce its own light but is many times more massive than Jupiter, a.k.a brown dwarf. Since it isn't bright enough to be detected by our deep space telescope nor the earth bound counterparts, they should be numerous in galaxies as I believe these objects sits in the middle range between planets and stars. However it seems the scientific community had ruled out this potential dark matter candidate without scanning the unobservable Milky Way galaxy as well as billions other with different light spectrums. Why aren't brown dwarfs considered candidates for the mysterious dark matter?

Two reasons.

We know from looking at galaxy rotation curves and the motion of galaxies in clusters and from gravitational lensing, that the amount of "dark matter" is some 30% of the density of the universe. But on the other hand, estimates of the abundances of deuterium, helium, tritium and lithium produced in the big bang indicate that only 5% of the density of the universe can be in the form of normal matter (e.g atoms, protons, neutrons etc). As brown dwarfs are made of these things, they can only contribute to this small percentage and cannot be responsible for most of the dark matter (the same is true for anything made of normal, "baryonic matter").
Astronomers have counted how many brown dwarfs there are, both in the local neighbourhood and in star clusters (see for example Andersen et al. 2008;  Kirkpatrick et al. 2011; Burningham et al. 2013; . In other words we can see brown dwarfs - they emit most of their light in the infrared part of the spectrum, and surveys like SDSS, 2MASS and WISE have uncovered them in their thousands. It turns out that there is about 1 brown dwarf for every 4 more massive stars. So, although numerous, they contribute a very tiny fraction of the normal matter in our Galaxy. In addition, microlensing experiments towards the Galactic bulge and Magellanic clouds suggest that although there are lots of brown dwarfs elsewhere in our Galaxy (e.g. Alcock et al. 2000; Novati et al. 2008), there are not enough to make much of a contribution to dark matter, and so the results from our solar neighbourhood appear representative of the Galaxy as a whole.


What is 'non-detection' of a Supernova? This may have a simple answer but I can't find it.

I've come across this term in a paper I was reading,  (Walmswell and Elridge, 2011) which mentions 'non-detection' of Supernovae, and while I can wrap my head around detection, I was lost at 'non'.

I tried finding a definition but to no avail. Would anyone be able to give one?

From a quick skim of the paper, it appears they using data from a previous paper by Smartt et al. (2009). That paper compiled a sample of observed supernovae (SNe) for which there are good images of the parent galaxies taken before the actual SNe. In some cases, Smartt et al. found plausible progenitor stars for the SNe; those are the "detections". Cases where they couldn't find a plausible progenitor on pre-SN images are the "non-detections".

(The authors of the paper you are talking about appear to be suggesting that Smartt et al. underestimated the masses of progenitor stars because they failed to account for extra local dust extinction from the more massive red supergiant stars.)


Can Milky Way be seen in South hemisphere only? I heard something(long time ago) that Milky Way can be seen in South hemisphere only, is it true? If so, what is the reason?

That is not true. The band of the Milky Way can be seen from any latitude on the planet. It just requires the right time of year and a dark sky.


What is the LSST's plan to address frequent satellite trails in data? update: From Nature.com News article: SpaceX launch highlights threat to astronomy from ‘megaconstellations’:

But an upcoming, cutting-edge telescope could be in bigger trouble. The US Large Synoptic Survey Telescope (LSST) will use an enormous camera to study dark matter and dark energy, asteroids and other astronomical phenomena. It will survey the entire visible sky at least once every three nights, starting in 2022. Because the telescope has such a wide field of view, satellites trailing across the sky could affect it substantially, says Tony Tyson, an astronomer at the University of California, Davis, and the LSST’s chief scientist.
He and his colleagues have been studying how up to 50,000 new satellites — an estimate from companies’ filings with the US government — could affect LSST observations. Full results are expected in a few weeks, but early findings suggest that the telescope could lose significant amounts of observing time to satellite trails near dusk and dawn.

In the 2011 Sixty Symbols video Spy Satellites (from Deep Sky Videos) after about 01:40 amateur astrophotographer Nik Szymanek says:

One thing I find when I’m out here imaging, by pointing the telescope at that part of the sky sort of south east at this time of night (early evening) there does seem to be a lot of satellites picked up by these images, and even on this 60 second exposure you can see we’ve got another satellite image there.
Well I mean satellites can be a problem for a deep sky imager in that if you’re using a telescope that has quite a wide field of view and you’re taking fairly long exposures, like five minutes or ten minutes, then at some point during that exposure it’s quite likely a satellite will be recorded passing in front of the image.
In fact a recent image that I’ve taken of the Andromeda galaxy with a refracting telescope actually recorded five satellite trails in a five minute exposure, so that gives you a good idea just how crowded the skies are with all these satellites are in orbit.

To me "using a telescope that has quite a wide field of view... taking fairly long exposures like five or ten minutes..." is a good description of how the Large Synoptic Survey Telescope (main site) will operate. Granted the exposures may be closer to 15 seconds than 5-10 minutes, but on clear nights it will be imaging almost continuously for high throughput. The FOV is about 3.2 degrees, so "wide field" applies.
Question: Has a plan or strategy been outlined or established to reduce the number of bright satellite trails that appear in LSST perpetual sky surveys, and/or to process images to remove the trails or at least flag pixel regions that contain likely trails?
Nearly every satellite and other large object in low Earth orbit is catalogued and it's position versus time can be predicted to a few kilometer accuracy (7km/400km is about 1 degree for example) and the object being in sunlight or dark can be likewise calculated, so one could conceivably adjust the exposure sequence to avoid some of the brighter ones and predict where they are likely to appear in data. So I'd be surprised if nobody has addressed this in a formal way yet.

  LSST from here

Has a plan or strategy been outlined or established to reduce the number of bright satellite trails that appear in LSST perpetual sky surveys, and/or to process images to remove the trails or at least flag pixel regions that contain likely trails?

Indeed LSST will be very affected by satellites in LEO. The strategy is to make several observations of the same area to have enought data to cover for the regions photobombed by the satellites and the artifacts generated by them in the sensor.

From the LSST webpage:

  The first group of Starlink satellites are sufficiently bright during dawn and dusk (when LSST would be surveying) that the trail would exceed sensor saturation, generating uncorrectable artifacts in the data. If instead these satellites were painted flat black making them a factor of 25 fainter, satellite trails should be less of a challenge for LSST due to its specific design.In that case LSST's frequent imaging of the same region of sky will provide enough data to correct for unsaturated satellite trails or other anomalies.


Would a future hole in the Cosmic Microwave Background show us the edge of the universe? If my understanding is correct, the Cosmic Microwave Background always come from our event horizon. By event horizon I mean the edge of the observable universe.

If the universe is finite does it mean that after some time (maybe in billions of years?) there will start being a hole in CMB. As our event horizon let us see CMB from regions further and further away, isn't there a time where it will reach the edge of the universe?

If this is the case, based on the known number of atom and mass of the universe, is it possible to calculate when this will happen?

First, let me clear up a misunderstanding:
Particle horizon
The "edge" of the observable Universe is called the particle horizon, and lies roughly 47 Gly (billion lightyears) away. It is always receding, both because the Universe expands and because light from increasingly large distances eventually reach us. In comoving coordinates (the coordinate system that expands along with the Universe), it asymptotically reaches a finite size (some 63 Gly), but in physical coordinates it increases forever$^\dagger$.
Event horizon
The event horizon is the boundary between the part of the Universe that contains the galaxies (and other stuff) from which light emitted today will be able to reach us, and the part where it won't. Currently, it is roughly 17 Gly away, but his horizon is always shrinking because the expansion of the Universe accelerates. In other words, stuff outside the event horizon is so far away that the expansion carries it away too fast for the photon amitted today to ever reach us.
Thus, at Big Bang (BB), the event horizon was 63 comoving Gly away, because that was the maximum distance from which we would ever receive information.
The CMB will always lag behind the particle horizon
The farther away we look, the more we look back in time. If there were galaxies a second after BB, and if we had the technology to look that far, we would see them right after BB, and they would be 47 Gly away. However, galaxies weren't born until several hundred Myr (million years) after BB. And now to your question: The CMB wasn't emitted at BB, so it doesn't origin from a region that is now 47 Gly away. Instead, it was emitted 380,000 yr after BB, and it was emitted from all of the Universe, in all directions.
This emittance wasn't instantaneous, but almost (took something like 10,000 yr), so let's assume for now that it was. Let's call that time $t_\mathrm{CMB}$. In that case, because light travels at $c = 300\,000\,\mathrm{km}\,\mathrm{s}^{-1}$, if you were present one second after $t_\mathrm{CMB}$, you would see the CMB that came from all regions that were 300,000 km away. That is, the CMB would be the light from a thin shell of radius 300,000 km.
Two seconds after BB, the CMB would be the light from a shell of radius 600,000 km, and so on. However, because the Universe is expanding, it is actually at little more than 600,000 km in radius.
And now, 13.8 Gyr after BB, and hence 13.7996 Gyr after the CMB was emitted, this shell is not almost 14 Gyr away, but in fact 45 Gly away which is almost, but quite, at the particle horizon.
Because of the expansion, the distance to the CMB shell always increases. But so does the distance to the particle horizon, and because the CMB was emitted after BB, the particle horizon will always "be ahead" of the CMB. As the CMB shell recedes, it becomes more and more redshifted, slowly fading into darkness but never disappearing. This is irrespective of whether the Universe is finite or infinite, except in the possible, but rather fantastic, event that the Universe is larger than the ~47 Gly we can see today, but smaller than the 63 comoving Gly that is the maximum size we will ever see.
Finite or infinite?
We don't know if the Universe is finite or infinite, but if the un-observable part of it looks like the observable Universe, then it is infinite. What we can say is that, at least in the observable Universe, the CMB was emitted everywhere in space, and there is currently no reason to believe that it was different elsewhere (although it could bee, if it expanded differently in other parts). To the best of our knowledge, space is homogeneous, meaning that (on average) there are as many baryons, photons, dark matter particles, and bicycles in one place as there is in another. In other words, to the best of our knowledge, there is no edge. Even if it's finite, it is difficult to (and most, if not all, astronomers don't) imagine an edge; rather it is thought that it would "curve back on itself", just as the surface of a ball is finite in area, but doesn't have an edge.
2D analogy
As an analogy, imagine living on a flat, inifinite Earth, populated evenly by people with loud voices. At some point in time, everbody yells "CMB". First you'll hear your neighbors yell, after 1 sec you'll hear those living 340 meters away, after 1 minute those living 20 km away, after a year those 10 million km away, and so on and so on forever.
Now imagine the same on a round Earth. After 16 hours, you would hear the folks living at your antipode, and if I understand correctly, this is when you expect to "hear a hole in the sound". But now you would start hearing the ones that are closer, but whose sounds have been traveling more than halfway around Earth. After 32 hours, you'd hear your own yell. And so on.
But, our Earth analogy is also expanding. If it were expanding at a steady rate, we would eventually hear our antipode friends. You might think that if it were expanding fast enough, their sound would never reach you, but as explained by the ant-on-a-rubber-band puzzle, that's not the case).
However, our fictional Earth is not only expanding, but in fact this expansion is accelerating. And in that case, there is a limit to how far away you can hear people.
Back to the real Universe, there is a limit to from how distant regions you may receive CMB photons (or any other information).
Moreover, if the Universe were finite, and were not accelerating, you might think that you would eventually see the CMB that was emitted from our own neighborhood and had been traveling around the Universe.
However, it can be shown that, in this case, the Universe will reach a maximum size, start collapsing, and end in a Big Crunch exactly at the time where you expect to see your antipode.
Only in a finite Universe with a rather special mix of gravitating matter and accelerating dark energy might it be possible to delay the collapse or the eternal expansion to a point where we eventually look back at ourselves. But that's sort of fine-tuning.

$^\dagger$Unless the Universe contains a yet unknown substance that is able to halt the acceleration. This is entirely possible, but in this answer I consider only a well-behaving Universe that acts according to current knowledge.


Most important feature of a telescope I was taking an astronomy test, the following question came up:

  The most important function of the astronomical telescope is:

  Resolving power
  Light gathering power
  magnifying power
  focal length

The focal length obviously means nothing here, as well as the magnification (that can always be done later). As far as resolving power and light gathering power go though, they both seem to be "equally" important, in that without the first, there is no detail, and without the second, there is no image.

It does seem that exposure time can compensate for a lack of light gathering power (what matters is the product between the two), but nothing can compensate for resolving power.

My professor seems to think the answer is resolving power. Why is this? Is this some well known idea in astronomy?

Well the angular resolution of a telescope depends on two things: 1) The wavelength of light you're looking in, and 2) the diameter of the primary lens/mirror.

$$ \theta = 1.22 \frac{\lambda}{D}$$

where $\lambda$ is the wavelength of light and $D$ is the diameter of the lens.

Light collecting power is simply how many photons you can collect in a particular wavelength in a given amount of time. Essentially, the bigger the bucket (or lens/mirror), the more rain you can catch. You can sort of get around light collecting power by increasing your exposure time. But, if the size of the object you're trying to observe is smaller than your telescope's resolving power, I don't believe there is really much you can do about it. Being limited by the resolution of your telescope is called being diffraction limited, but most telescopes on Earth are usually seeing limited due to the atmosphere's smearing effects.


Eclipse comparison: May 20 2012 vs Aug 21 2017 as observed from San Francisco, CA I'm preparing for the solar eclipse coming in August of this year, and want to make a decision on whether it will be worth my while to travel north from San Francisco.

As observed from San Francisco, when compared to the May 2012 eclipse, will there be more of the sun obscured by the moon in 2017, or less?

This is going to be the eclipse of the century for the USA. To make the most of it, you need to be on the path of totality (see map) to see a total eclipse. If you stay in San Francisco, you will only see a partial eclipse. You need to make arrangements now if you need somewhere to stay.

The gray band indicates the path of totality. Inside it, you will see a total eclipse, and will be able to see the Sun's corona in all its glory. If you are outside the path of totality, you will only see a partial eclipse (and no corona). The closer you are to the blue centerline, the longer totality will last. The black dot near Nashville indicates the location where totality will last the longest (2 minutes and 40 seconds).

During totality (but only totality), it is safe to observe the Sun's corona with the naked eye. At all other times, you will need to use a protective (solar) filter if you want to look at the Sun directly, or project an image of the Sun onto a screen.

www.eclipse2017.org


What happens for a "closed" universe without any content? Let an universe with no content and positive curvature.
Friedmann-Lemaître equation
$$H^2=\frac{8\pi G}{3}\left(\rho_m+\rho_r+\rho_{\Lambda}\right)-\frac{k\, c^2}{a^2},$$
where $a$ corresponds to the scale factor of Friedmann-Lemaître-Robertson-Walker Metric, and the $\rho$ to the density of the contents,
will become $$H^2=-\frac{\, c^2}{a^2},$$ as positive curvature (closed universe) corresponds to $k=+1$.

So, as $H=\frac{\dot{a}}{a}$ with $\dot{a}=\frac{da}{dt}$,

$$da=\pm i \, c \, dt$$

$$a(t)=\pm i \, c \, t + cst$$

what am I doing wrong ?

I see that you are using the convention that a has units of distance and is like the radius of curvature of the universe, and you are getting that either a or t would need to be imaginary.  So you can interpret that in two ways, either you say that is impossible, and conclude that it is impossible for an empty universe to be closed, or you attribute some physical meaning to imaginary radius of curvature or imaginary time.  I'm not sure what that significance would be, but what seems clear is there is something bizarre, or even bogus, about claiming that an empty universe can have positive curvature.

Another way to see this is to interpret your equation as an equation for k via H=Kc/a, where K^2 = -k.  If you further demand that nothing be imaginary, this requires that k be either 0 or -1.  Empty universes are flat or open, or else they are imaginary in some way.


Which atom is a Neutron star made of? I understood that everything was made of Atoms.

  An atom is the smallest constituent unit of ordinary matter that has
  the properties of a chemical element. Every solid, liquid, gas, and
  plasma is composed of neutral or ionized atoms

However with Neutron Stars;

  the basic models for these objects imply that neutron stars
  are composed almost entirely of neutrons, which are subatomic
  particles

Does this mean neutrons can exist outside of an atom, and that a Neutron star isn’t comprised of an element that is recognisable on the periodic table?

Yes neutrons can exist outside the atom (or nucleus). In free space a neutron will beta decay into a proton, and electron and an anti-neutrino on a timescale of 10 minutes. However, in the dense interiors of a neutron star, the electrons form a degenerate gas, with all possible energy levels filled up to something called the Fermi energy.

Once the Fermi energy of the electrons exceeds the maximum energy of any possible beta-decay electron, then beta decay is blocked and free neutrons become stable. This is what happens inside a neutron star and you end up with mostly neutrons with a small fraction perhaps a few per cent electrons and protons.

In the outer parts of the neutron star, the protons and neutrons can still arrange themselves into nuclei (but not atoms), but these nuclei are extremely neutron-rich (they would not normally exist in nature) and are only stabilised against beta decay by the process I described above. The very outer envelope may consist of completely ionised iron-peak element nuclei and there may be an ultrathin (few cm) layer of recognisable ionised hydrogen, helium and carbon (e.g. Wynn & Heinke 2009).

Once the density reaches about $3 \times 10^{16}$ kg/m$^3$ it becomes more favourable for the neutrons and protons to organise themselves into "macro-nuclei" - long strings and sheets of nuclear material, known colloquially as nuclear pasta.

At higher densities still, the pasta dissolves into a soup of mostly neutrons with about 1 per cent protons and electrons.

The diagram below (from Watanabe et al. 2012) shows roughly how these layers are arranged. It should be stressed that this is based on theoretical modelling, with the theory becoming less certain the further into the neutron star you go. Testing these ideas involves nuclear and particle experiments, observations of pulsars, of neutron star cooling, of X-ray bursts, mass and radius estimates in binary systems, pulsar glitches, etc., etc. None of the details have been observationally confirmed beyond dispute, but the basic picture below fits what we know. In particular, the the crust and the n,p,e fluid regions are well understood in theory. The details of the nuclear pasta phases are still the subject of a lot of theoretical work, as are the details of superfluidity in the interior, and what happens in the very central regions (solid neutron core, extra hadronic phases, boson condensation, quark matter) is still theoretically difficult and observationally untested except perhaps to say that the softest equations of state have been ruled out by the existence of $2M_{\odot}$ neutron stars.


What to do after first year of amateur astronomy? I have practiced astronomy as a hobby for a while now. I have an entry-level telescope (6-inch Newtonian). In addition to observing the night sky, I have been studying physics related to optics, astronomical distance & temperature measurement, gravity, etc.

My plan is to next get a CCD camera, and do some astrophotography. It should keep me busy the next winter. However, I feel I should try to find some kind of project where I could gradually make progress. Otherwise, after learning all the basics, and seeing everything that can be seen with the smallish telescope I have, I might gradually start to lose interest.

What would you recommend as the "next stage"?

Astrophotography is always a good step to take after just observation. That way, you can see things through your telescope that you cannot see with the naked eye,  and also share your findings.

Another thing you could do is start on a different part of astronomy, maybe getting a hydrogen alpha filter or a solar telescope to observe the Sun, which again could be combined with the astrophotography.


Why does e.g. Hubble's secondary mirror not block part of the picture? The SOHO observatory has an opaque object in the light path that blocks the direct light from the sun making it possible to observe the corona.
Hubble and many other telescopes have a big primary mirror in the bottom of a tube which bounces light on to a smaller secondary mirror at the opening of the tube. Why doesn't that other mirror cause a circular area of "no data" in the middle of the resulting images?

A particular point in the image is formed by the convergence of all the rays of light that leave from that point in the object, and bounce off the mirror.  As long as there are some rays that can do that, then the only effect of the secondary is to block some of the rays, dimming those parts of the image slightly.


Are there stars with an average density greater than the central density? Are there likely to be stars for which the average density is greater than the density at the centre? Intuitively, I would say not, as density tends to decrease outwards, but ideally I would appreciate a clear mathematical justification for this.

$$\frac{dP}{dr} = - \rho g,$$
is the equation of hydrostatic equilibrium, where $\rho$ and $g$ are the local density and gravity, $P$ is pressure and $r$ is the radial coordinate. This can be rewritten as
$$\frac{d\rho}{dr} \frac{dP}{d\rho} = -\rho g.$$
Since $\rho$ and $g$ are positive numbers, the pressure gradient is negative.
For all types of matter $P = f(\rho)$, such that $dP/d\rho$ is also positive (i.e. pressure increases with density).
Thus $d\rho/dr$ is negative and the density increases towards the middle.
EDIT: This argument only obviously works if you ignore the dependence of pressure on other variables - such as the temperature or number of particles per unit mass in your gas. It takes more work to establish that $dP/d\rho>0$ in a more general case. For example you might consider a situation where a sufficiently negative temperature gradient might counterbalance a positive density gradient.
i.e. if $P = f(\rho, T)$ then
$$ \frac{dP}{d\rho} = \frac{\partial P}{\partial \rho} + \frac{\partial P}{\partial T} \frac{dT}{d\rho}$$
and $dP/d\rho<0$ if
$$ \frac{dT}{d\rho} < -\frac{\partial P/\partial \rho}{\partial P/\partial T}$$
Leaving this as a placeholder for the moment.