What is the gaseous-to-rocky ratio of exoplanets? Around the Sun, there are as many gaseous planets (Jupiter, Saturn, Uranus and Neptune) as there are rocky planets (Mercury, Venus, Earth and Mars).

Do other stars have similar gaseous-to-rocky ratios among their planets? For any given stellar system, are there typically as many gaseous planets as there are rocky planets? Or do most stars have either mostly gas giants or mostly rocky planets, making our own Solar system an exception? Perhaps we need more data to conclude, since we often have mass or radius, but not both for exoplanets?

>  Do other stars have similar gaseous-to-rocky ratios among their planets? For any given stellar system, are there typically as many gaseous planets as there are rocky planets?

With the current instruments and methods we have, we can only access certain population of exoplanets. Terrestrial-mass exoplanets are quite hard to find hence we have more discoveries of Jupiter-mass exoplanets, which does not mean that in reality, we have more chance to have a formation of Jupiter-mass, it just means that they are easier to detect so we detect more. Therefore it is hard to conclude and make statistics right now.

Or do most stars have either mostly gas giants or mostly rocky planets, making our own Solar system an exception?

This is more linked to planetary formation. I invite you to read about the 'core accretion model'. It explains very well the formation of terrestrial and gaseous planets in the proto-planetary disk. But in general, both can form and our Solar System is not an exception, we just don't have the instruments to find a system like it for now.
For example, an alien with the same instruments as us that would study the Sun with the radial velocity method would only detect Jupiter, the other planet would be invisible to it.


How long would it take to reach the edge of the reachable universe? How long would it take to reach the current edge of the reachable portion of the universe, with the following bounds in mind:

We figure out a way build a space ship to provide a constant 1 G acceleration for an extended period of time.
We accelerate until we are halfway there, then decelerate past this point (so we stop when we get there.
Our target is about 15 billion lightyears away.  As per http://curious.astro.cornell.edu/about-us/104-the-universe/cosmology-and-the-big-bang/expansion-of-the-universe/616-is-the-universe-expanding-faster-than-the-speed-of-light-intermediate galaxies past 4,740 megaparsecs are no longer reachable. Using this site http://www.convertunits.com/from/megaparsecs/to/light+years to convert megaparsecs to lightyears means this "unreachable" point is currently 15,459,812,396.5002 lightyears away.  Let's simplify to 15 billion lightyears, which is still supposed to be reachable.
We somehow work out all the details of life support, collision with particles at relativistic speeds, etc...

I would like to know the time the traveler experiences, and the time that everyone else (e.g. those on earth) experience.

Jonathan's answer is essentially correct, but as Rob Jeffries comments, he doesn't take into account that the Universe is expanding during the journey.

The edge of the observable Universe is 47 billion lightyears (Gly) away. Even if you are a lightbeam, you cannot reach that point. The farthest you can go if departing today is roughly 5 Gpc, or 17 Gly, but this journey would of course take infinitly long (or else it wouldn't be "the farthest you can go"). This distance is probably what the linked article is referring to (I didn't read the article; it's very, very long).

So, in order for the answer to be any fun, you have to freeze the Universe, using magic, which is what Jonathan's calculator is doing. Here I'll just provide the analytical solution: In that case, the proper time $\tau$ (i.e. the time experienced by the traveler) to reach a distance $x$ when traveling at a constant acceleration $a$ is
$$
\tau = \frac{c}{a} \cosh^{-1} \left( \frac{ax}{c^2} +1 \right),
$$
where $c$ is the speed of light. If you wish to decelerate after having reached halfway, you just divide $x$ by $2$ and multiply the result by $2$.

If you plug in the $x=15\,\mathrm{Gly}$ you request, you get roughly 45 years. To get to the edge of the Universe at 47 Gly actually only takes a few years more. The reason for this is simply that traveling at 1G gets you to (almost) the speed of light in only a couple of years, and hence you experience (almost) no time, no matter how far you go.

The time experienced for the Earthlings for the traveler at constant acceleration is given by
$$
t(\tau) = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right),
$$
which works out to 15 Gyr for the 15 Gly, and 47 Gyr for the observable Universe. The reason is simply that the traveler, from the point of view of the Earthlings, extremely fast reaches a speed which is almost the speed of light.


How to watch at best this week's planet alignement? I read somewhere that counting from today, five planets will be aligned during almost two weeks IIRC. The planets are : Mars, Jupiter, Saturn, Mercury and... us. I know Jupiter and Mars can be seen with the naked eye, but I would really like to catch a glimpse of Saturn.

Assuming one does not own a telescope, but small binoculars, how can one maximize his chances of seeing at least Mars, Jupiter and Saturn ?

All these planets are naked eye objects, and except for Mercury, are easy to see. Venus is also part of this grouping, Meaning that all 5 planets known to ancient astronomers are on the same side of the Earth. (and the sun is on the other)

The planets are not exactly "aligned", just they are all visible in the sky, spread over a wide part of the south eastern sky, a little while before dawn.

Sky and Telescope has some nice guides to what you can expect. Mercury is always the hardest, as it is always in the twilight. Saturn is an easy sight, appearing as a slightly yellow star. Binoculars will show its moon, Titan.

Best time is around the first week of Feb about 45 min before dawn.


How to Interpret Lomb Scargle periodogram lomb scargle periodogram:

import numpy as np
import pandas as pd
from matplotlib import pyplot as plt
from scipy import signal
from astropy.time import Time
from astropy.timeseries import LombScargle

m5_data = pd.read_csv('tmp')
m5_data.head()

     date    sales
0  2011-01-29  32631.0
1  2011-01-30  31749.0
2  2011-01-31  23783.0
3  2011-02-01  25412.0
4  2011-02-02  19146.0

"""Converting dates to MJD """
tmp_str = m5_data.iloc[:,0].astype(pd.StringDtype())
t_date = np.array(tmp_str.values,dtype = 'str')
t_date = Time(t_date, format='isot', scale='utc')
t_date.format = 'mjd'

y = m5_data.iloc[:,1]
m5_ls = LombScargle(t_date, y)
m5_frequency, m5_power = m5_ls.autopower()
plt.plot(m5_frequency,m5_power)
plt.show()

What does the x axis represent? Is it frequencies in days or is it frequencies in 1/day?
So the spike at 2 does that mean there is a period of every two days or does it mean there is a period 1/2 days, which is a fractional day and I do not understand how I could get fractional days when I only have 1 observation per day.

The overall goal is to find if the data is periodic. Given this data is sales of a product from a store I would expect that if the data is periodic that the periods would be 7days(weekly), 30days(monthly), 182days(semi annually)
So the spike at 2 does not make sense to me.

The x-axis of a periodogram is usually a frequency, measured in inverse time units. It is possible to plot the inverse of frequency, in which case the axis would be non-linear.
In your case, the x-axis looks linear, and the the code you present appears to be using something labelled as a frequency on the x-axis. Therefore I would assume (though I would read the documentation for the functions you are using) that the x-axis is frequency. The units will be based on whatever units are in your input. i.e. It is just the inverse of whatever you are giving it for the times - in this case, $d^{-1}$.
If your sampling rate is once per day, then you are getting no reliable information at all about frequencies $>1$ d$^{-1}$. The stuff you should be concentrating on lies between $0\leq f < 1$ d$^{-1}$. Even in this range you won't have good fidelity on signals with $0.5<f<1$ d$^{-1}$ because they are undersampled. You can also expect to get aliasing signals in the range $0\leq f <1$ that are caused by true signals (including those with higher frequencies) aliasing with the 1 d$^{-1}$ frequency of your sampling.
It looks to me, by eye, like you might have a genuine signal at $1/7$ d$^{-1}$, corresponding to a weekly variation. There may be a weaker signal at much lower frequencies that might correspond to a monthly signal. The 2 signals either side of 1 day look like aliases of the weekly signal with the sampling frequency.
I would highly recommend reading VanderPlas 2018 for a thorough discussion of the Lomb-Scargle periodogram and its interpretation.


Observable universe equals its Schwarzschild radius (event horizon)? The estimated age of the universe is 14 billion years.

The estimated Schwarzschild radius (event horizon) of the observable universe is 14 billion light-years.

What are the ramifications?

There's an error in your source, but even if there weren't, it wouldn't mean that the Universe is a black hole (see below): The Schwarzschild radius of the observable Universe is not equal to 13.7 Glyr. Wikipedia cites some random, non-refereed paper that uses the Hubble radius as the radius of the observable Universe, which is too small by a factor of $\gtrsim3$.

The Schwarzschild radius of the Universe

Although the age of the Universe is indeed ~13.8 Gyr, its radius is much larger than 13.8 Glyr, because it expands. In fact, the radius is $R \simeq 46.3\,\mathrm{Glyr}$.

The mean density of the Universe is very close to the critical$^\dagger$ density $\rho_\mathrm{c} \simeq 8.6\times10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$.
Hence, the total mass (including "normal", baryonic matter, dark matter, and dark energy) is
$$
M = \rho_\mathrm{c}V = \rho_\mathrm{c}\frac{4\pi}{3}R^3 \simeq 3.0\times10^{57}\,\mathrm{g},
$$
and the corresponding Schwarzschild radius is
$$
R_\mathrm{S} \equiv \frac{2GM}{c^2} \simeq 475\,\mathrm{Glyr}.
$$

The Universe is not a black hole

Even worse! you might say. If our Universe is much smaller than its Schwarzschild radius, does that mean we live in a black hole?

No, it doesn't. A black hole is a region in space where some mass is squeezed inside its Schwarzschild radius, but the Universe is not "a region in space". There's no "outside the Universe". If anything, you might call it a white hole, the time-reversal of a black hole, in which case you could say that the singularity is not something that everything will fall into in the future, but rather something that everything came from in the past. You may call that singularity Big Bang.

The error in the source

You will see questions like this many places on the internet. As I said above, they all assume the Hubble radius, $R_\mathrm{H} \equiv c/H_0$ for the radius. But this radius is well within our observable Universe, and doesn't really bear any physical significance. In this case, the age in Gyr works out to be exactly equal to the radius in Glyr, by definition.

So, what does that tell us?

Nothing, really. Except that our Universe is flat, i.e. has $\rho\simeq\rho_\mathrm{c}$, which we already knew and used in the calculation.

That is, setting $R = R_\mathrm{H} \equiv c/H_0$,
$$
\begin{array}{rcl}
R_\mathrm{S} & \equiv & \frac{2GM}{c^2}\\
             & =      & \frac{2G}{c^2} \rho V \\
             & =      & \frac{2G}{c^2} \rho \frac{4\pi}{3}R^3 \\
             & =      & \frac{2G}{c^2} \rho \frac{4\pi}{3} \left(\!\frac{c}{H_0}\!\right)^3 \\
             & =      & \frac{8\pi G}{3 H_0^2} \rho \frac{c}{H_0} \\
             & =      & \frac{8\pi G}{3 H_0^2} \rho  R,
\end{array}
$$
so if $R=R_\mathrm{S}$, we have
$$
\rho = \frac{3H_0^2}{8\pi G},
$$
which is exactly the expression for the critical density you get from the Friedmann equation.

$^\dagger$The density that determines the global geometry of the Universe.


Mapping illumination to moon phase (font) icon I use algorithms that give me the moon phase, age and illumination (and more) for a given timestamp and location. I am struggling to map that information to a (font) icon of the Moon Phases font. Each icon is an approximation because it ignores the rotation. Still, I'd like to map illumination (I guess?) to icon as precisely as possible.

As you can see above there are 13 characters/icons per half cycle; 26 characters a-z. However, new/full moon are represented by the numbers 0 and 1. Particularly there's a gap between m and n where the new moon is in reality. The font uses a-z and A-Z to distinguish between northern hemisphere and southern hemisphere (again, I guess).
Mental model
Illumination 100%                              0%                              100%

Moon         full                              new                             full

              |---------------------------------|--------------------------------|

Character     1            a - m                0             n - z              1

Assuming my mental model is correct this means (for the northern hemisphere)

13 characters plus 0/1 to distribute across the 0-100% illumination range
pick 0 if the illumination is <= 100/15 i.e. <= ~6.66%
pick 1 if the illumination is >= 100 - 100/15 i.e. >= 93.33%
if waning (waning gibbous, last quarter, waning crescent)

round(illumination/100 * 15) + 97 to find character index for a-m

if waxing

round(illumination/100 * 15) + 110 to find character index for n-z

Does that sound about correct?

While you have "illumination" it would seem to be easier just to map "moon age" (in days) to a character
The average synodic month is 29.53 days, and you have 28 characters, so calculating age*28/29.53 and then rounding to give a number between 1 and 28 (or 0 and 27). Then it is just a matter of adding 97 if that number is in the range 0-13, or 36 if that number is 14... etc.
Slightly better would be to find out the actual length of the synodic month (if you can calculate moon age and illumination this should be possible) and use it in place of 29.53.  The actual length varies from about 29.2 to 29.8 days depending mostly on the position of the Earth's orbit


Do star systems stripped from galaxies have different expected behavior? Another way of stating this question: Does the fact that a star system (or comparable concentrated mass) exists in a galaxy affect its long-term evolution, in terms of expansion, contraction, nebulation, etc.?

This question was inspired by a view of the "Umbrella Galaxy" where previous interactions have torn streams of stars far from the galactic disk.

The local environment is very important to star formation because it requires material as well as something to perturb the nebula to start collapsing and forming star(s). Therefore, both the type of galaxy as well as the location in the galaxy are very important to creating stars.

However, once a star moves from protostar onto the main sequence, external influences don't much affect its development. This is how we can identify stellar streams because the stars come from a similar population when they are created, but some tidal forces cause them to spread out around the galaxy.


How can astronomers pinpoint the location of the source of a neutrino? In the popular press, in recent months, we have heard a lot about high-energy neutrinos from far outside our solar system reaching our detectors....
But I wonder...
If a single neutrino from a great distance randomly, rarely, smacks a xenon atom in a detector, how the heck can they 'triangulate' it's direction?

You correctly state that neutrinos do not interact too often. The physical parameter describing that is the effective cross-section. So what you observe in a detector is not the neutrino itself, but secondary particles, e.g. muons. Colloquially put, you may regard anything with high mass (density) in between the neutrino source and your instrument (to detect mainly the secondary particles) as detector for the neutrinos. This might be a large chunk of ice (like for the icecube experiment), a mountain range (as for the Gran Sasso lab), or even a complete planet, if the incident neutrinos originate from a star located on the other side of Earth at the moment of observation. The latter is possible since neutrinos e.g. from the Sun may pass through the whole planet without interacting with its matter (at night-time for the scientists working for the detector).
Let me try to explain with a drawing I took from a 2018 blog entry of the Antarctic Muon And Neutrino Detector Array (AMANDA). The red dots are on the ice surface, the vertical lines are the boreholes into the Antarctic ice shield and the little black dots are the PMTs, the Photomultiplier Tubes which detect the Cherenkov light generated by some fast muons, in the picture on the top of the blue cone. The readings of the PMTs (detecting the Cherenkov light) are displayed as color-code, I assume the color scale somehow relates to the time passed since the trigger event.

So how do we now know where the neutrino came from? We can use particle physics to derive the secondary particle's velocity (vector) and from that where the neutrino actually came from. This is not trivial and requires quite a bit of computer power, but it is possible.


Why is $H_\delta$ prominent in type A stars? I understand factually that H$\delta$ lines are most prominent in type A stars and less so in more extreme types of stars on the H-R diagram. However I was wondering the reason for why they are not prominent in more extreme type stars such as type O or type F. For type F stars I believe (this may not be correct) none of the hydrogen is in the 4th excited state (or very few are) so there aren't many H$\delta$ balmer lines. However I am very unsure of why the EW H$\delta$ is low in type O stars. Can someone please explain this, Thanks!

H$\delta$ absorption is formed when hydrogen in the level $n=2$ is excited to $n=6$.

To get strong H$\delta$ absorption lines you need large amounts of hydrogen in the first excited state $n=2$ and a radiation field that contains large numbers of photons with an energy equal to the difference between the $n=6$ and $n=2$ states.

These requirements are satisfied in stars with photospheres between about 8,000K and 15,000K, and we call these early-F and A-type stars.

At cooler temperatures most of the hydrogen is in the ground state, there are few photons with sufficient energy (3eV) to be absorbed and the H$\delta$ line weakens. At much higher temperatures (in O-stars for example) then none of the hydrogen is in the $n=2$ level, and in fact most of it is ionised.


Ancient greeks know about leap day? I'm reading about Antikythera machine and its functions.
So I read about solar cycle and moon cycle and how they used 19 solar years to compare with 235 lunar cycles.

So this is my question: They know about leap day at the time (~ 200 BC)?

Well, they know according to the Metonic Calendar created in ~430 B.C.

To know more about click here.


Possible Reason for the "NASA UFO sighting: Alien ship 25 TIMES size of Earth spotted near the Sun" news (does it fake or ...?) I have this news:
NASA UFO sighting: Alien ship 25 TIMES size of Earth spotted near the Sun
and tried to find the NASA site link by this link, but not found that news , so like to know what could be the reason of this situation, does it fake?
Thanks.

The Express is notorious for low-quality "science" stories.
In this case, someone who wanted to see a UFO looked at images from the Solar and Heliospheric Observatory until he found a feature he could say something sensational about.
One might as well say, "I like turtles, that cloud looks like a turtle, therefore turtles come from the sky."
The mention of hollow sun "theory" is another red flag.
SOHO and its successor the Solar Dynamics Observatory have recorded many legitimate phenomena including coronal mass ejections and Sun-grazing comets.


Could we make "tiny black holes"? To study the physics of black holes (What we presume as really dense matter) could we not simply simulate that here on earth? What if we took 1 hydrogen molecule and just crushed it with enough force that I'd fall apart and then crush it even more? I'm curious what would happen.

"Really dense" is a bit of an understatement for small black holes.

A Black hole the mass of the Earth would have a diameter of about 18mm.

And it doesn't get any easier if you start smaller. The density of small black holes is bigger than large ones:

A black hole the mass of a human would have a diameter that is no only smaller than a proton, but a billion times smaller than a proton. Consider the difficulties in crushing something that small.

A black hole the mass of a proton...

The notion that you can make a black hole by "crushing matter very very hard" is many magnitudes beyond what is realistic.

It is, perhaps, possible that tiny black holes could be created in very high energy particle collisions.  We're fairly sure that no such black holes have been created at CERN.  Particle collions of much higher energies occur when cosmic particles hit the atmosphere, and again there is no evidence of black holes being created but we are looking: https://arxiv.org/abs/hep-ph/0112247


Broadband spectrum of Sun Broadband spectrum is one which is inclusive of all the observed wavelength ranges. In the case of the BB spectra of our Sun, there is a sharp (non-differentiable) kink in the curve in the UV-range (as shown above). What is the significance of this kink?

The point here is the "the solar spectrum" received from "the Sun" actually comes from different places depending on the frequency. The (stylised) smooth curve at optical and IR frequencies is pseudo-blackbody radiation from the photosphere. Pseudo-blackbody, because the radiation comes from different depths and temperatures in the photosphere, which gives rise to absorption features.
The photosphere is not hot enough to produce significant amounts of blackbody UV radiation, which accounts for the sharp "Wien tail" dying away into the UV.
The dashed lines in the UV and X-ray frequency range are emission that comes from the chromosphere and corona. i.e. They arise from a physically distinct region and thus there is no reason why they should form a continuous curve with the radiation from the photosphere.
The UV and X-ray emission is a combination of thermal radiation from hot plasma in the corona (at $\sim 10^6$ K), which produces continuum bremsstrahlung and recombination lines from highly ionised metal ions like Fe X or O VI, and plasma at $10^4$-$10^5$ K in the chromosphere and transition region that is dominated by discrete recombination lines produced by ions like Mg II and C IV.
The reason that a discontinuity can persist is that the photosphere, chromosphere and corona are not radiatively coupled. i.e. The underlying photosphere is not the source of heat for the chromosphere and corona. The heating agent is the solar magnetic field.


How bright will Sirius be when it's closest to the solar system? Preferably in terms of percentage. As a layman, visual apparent magnitude means little to me.

From Sirius: Brightest Diamond in the Night Sky by Jay B. Holberg we have current distance of Sirius ~8.5ly, closest approach will be ~6.7ly. The intensity of of light from Sirius assuming constant luminosity scales as distance squared, so the intensity (power per unit area at the eye) at closest approach will be: $\left(8.5/6.7 \right)^2 \approx 1.6$ times that today, which is a change of magnitude from about $-1.4$ to about $ -1.6$


Convert MJD date range to number of days elapsed in IDL Hopefully I'm in the appropriate section as my question is mainly about coding, but it's for astronomy.

Right now I'm trying to figure out how I can take a range of MJD values and convert them into number of days elapsed so I can normalize a plot of two supernova and their absolute magnitudes over time in IDL. Is there a simple conversion for this I'm not seeing or do I need to tediously convert every value to determine the number of days to use for my x-axis?

Note: I do not wish to convert to the date, but the actually number of days elapsed as to normalize my two plots for comparison.

Thanks!

Like @James K said, the MJD is simply a count, so you can set some arbitrary date to be 0 and go from there.

For example, take today's MJD date and say there was one event today:

Today = 57657

Add three months and 12 days to the next event:

Today + 103 = 57760

So if you called ten days before the first event 0, then the "dates" of the first two events would simply be the following:

Start = Event_1 - 10 = 0
Event_1 = Start + 10 = 10 <-- Semantics right!
Event_2 = Event_1 + 103 = 113

It really is that simple. Just set a variable to equal whatever starting MJD you want to be zero and count earth days between events.


Can a planet exist with the same mass as the earth, but have a different diameter? Can there be a planet with the same mass, and a similar composition, as the earth, but with a larger or smaller diameter, and how could that theoretically occur? I've tried to see if there is any research or sites that tackle this idea, but I, understandably, haven't found anything.

Yes, a planet can exist with the same mass as Earth, but have a different diameter - if it has a different density. The lower the density, the larger will be the diameter. In terms of an exact formula, we have:

$$d(M, \rho) := 2 \cdot \sqrt[3]{\frac{3M}{4\pi \rho}}$$

The trick is that density is specifically a function of composition: hence, the extent to which the diameter can vary depends essentially completely on what value you have in mind for when you ask for a composition as being "similar". Dramatically different densities, and hence diameters, will only be possible with a dramatically different composition - e.g. a heavy gas envelope will provide a significantly reduced density, as would a significant mass percentage of water (think of this hypothetical Earth-mass planet as being like a big version of certain moons of Jupiter and Saturn, such as Europa and Enceladus. In fact, some percentage of discovered extrasolar planets are believed to be of this type: though you should not think of it as having liquid ocean all the way down, because after about 100 km of depth or so on comparable-to-Earth gravity, water freezes due to compression.).

If by "similar" composition we mean that, say, the planet is still to have a solid, hospitable surface, and to have all the same materials as on Earth but just perhaps in somewhat different abundances, one "easy" way might be a reduction in the mass of iron. A lot of the Earth's density is supplied thanks to its relatively large central iron core: compare it with the Moon, which has little to no iron core - the density for Earth is 5.5 g/cm^3, while that for the Moon is about 3.3 g/cm^3, althuogh one should also be mindful of compression effects due to gravitational crushing of the interior. A small iron core would give one a density in the intermediate range between these two.

With such a smaller iron core, to give one Earth mass (about 6000 Yg), it must be exchanged for a greater quantity of less-dense rock, hence the planet's diameter will be larger. Conversely, we could imagine having more of an iron core, similar to Mercury, and then the diameter would be smaller (and the surface gravity larger) as we should have less rock to avoid going over the one Earth-mass prescription.


Why is twilight longer in summer than winter and shortest at the equinox I recently decided to set my alarm clock to wake me up when it is "dark" out.  In the end, I decided to set my clock to the earliest time that nautical sunrise is in my state (Illinois) and stick with that all year.

While doing some research for this, I noticed something that surprised me.  Check out this disparity between astronomical sunrise to civil sunrise for the solstices and equinox (the latter two adjusted for daylight saving time):

Date/Astronomical/Civil/Disparity

Dec 20: 0533 - 0640 (67 minutes)
Mar 20: 0526 - 0630 (64 minutes)
Jun 20: 0320 - 0451 (91 minutes)

To be honest, these sets of ranges surprise me for multiple reasons.  I clearly don't know what I don't know, but here are some questions I can formulate:

Why would twilight be longer in summer than winter?  Before seeing this data, I had assumed that since the sun makes a more perpendicular path through the horizon in summer that twilight would be shorter in summer than in winter.  After all, in winter the sun takes a "slanted" path across the horizon.  Wouldn't the summer's path be more direct and therefore quicker?
OK: seeing this empirical evidence I conclude that something is wrong with my premise that forms my first question.  Summer twilight is longer in summer than winter.  However, I still would have assumed that March 20 would have had a twilight length in between the two solstices.  But it's not!  Why does the equinox have the shortest twilight?

Appended 5/13/2014:

I didn't want to leave my original incorrect statement in here without flagging it. As Cheekhu points out below, the sun does not follow a more perpendicular path in summer than winter, as I had erroneously assumed and stated above.  See his post for more details.

See this diagram,

Now, the slant appears to be the same for June and December Solstices, doesn't it? First thing, it is a wrong conception that Sun travels more "perpendicular" to horizon in the summers. It just reaches a higher point as you can see, not that the slant is more.

Where does the disparity come in?

Civil Sunrise begins when the Sun is 6° below the horizon, Nautical Sunrise begins when it is 12° below the horizon and 18° for Astronomical Sunrise.

So, the Civil/Astronomical Disparity is due to the extra 12° of altitude that the Sun has to cover. Now, the diagram I have posted is better to explain the sunset, so let it be that way. It is just the same for our purpose.

The motion of the sun is uniform along the circles that you see. But these angles which decide our sunrise are measured perpendicular to the horizon, from the zenith. Are you aware of what happens when you project circular motion along a straight line?

That's right. You get a sinusoidal curve. The special thing about this curve is that it changes fast at some points and slow at others.
Transit is the position when the Sun is highest in the sky. In the curve for equinox (as in Fig 1), if you assign 0° to that point, the Sun is at 90° at sunset and -90° at sunrise. Both positions give you the peak of a sine curve and the change is lowest at that point. That might explain why you get the least difference at Equinox.

Now, if you consider the curves for June/December solstice (as in Fig 1), the position for sunset/sunrise is a little off 90°, hence rate of change is faster there as compared to the equinox.

That should explain why the disparity is larger at solstices as compared to equinox. But what about the difference in June and Decemeber solstice?

Now, my dear, you are entering dangerous territories. Let me remind you that the line/plane along which we are making our degree measurements is perpendicular to horizon. But the plane of circular motion is still at an angle to it. So, the sinusoidal curve has to go through another set of projections before you get the exact thing.
On Equator, the disparity between the solstices vanish, yes, as the aforementioned two planes align.

Anyway, I was hoping to give you a qualitative description just so you get the gist of it. Here are the equations if you need a mathematical idea

$$\cos a = \sin \delta \sin \phi + \cos \delta \cos \phi \cos H$$,

$a$ is (90+6)° for civil sunrise/set, and (90+18)° for astronomical sunrise/set,

$\delta$ is declination of Sun on that date. Google it.

$\phi$ is latitude of the place

To get the disparity, you need to find $H$ from the formula for two values of $a$ and get their difference. Remember to divide by 15 to convert from degrees to hours.

Here is a graph showing disparity along declination(-23.5° to +23.5°)


How to know when the moon is visible on a particular night? I am looking for beginners information on how I can know when the Moon is going to be visible on a particular night ?

Google-fu says to look for moonrise information about my city from websites such as https://www.timeanddate.com/moon/australia/darwin
So if I followed the above guide for today and started walking on a plane field West-NorthWest direction, at 10.01 PM I should be able to see the Moon. Is my understanding correct? Or is there another way to know when Moon will be visible on a particular (clear) night.

Google-fu is correct... It really depends upon where you are, for an exact time. However, if you understand that there are 8 named phases and they typically have a "standard" rise time, you can know approximately what time and what phase will be visible. The one caveat is that you need to look at the moon to know where to start.


Is it fruitful to search for a correlation between rates of in-falling matter in black holes and the expansion rate of our universe? To my understanding, white holes are hypothetical objects that are time-reversed black holes; black holes gravitationally "suck in" matter whereas white holes "spit" matter back out. As such, I think it's a nice idea that the Big Bang could be the "spitting of matter" via a white hole (assuming such an object exists).

This got me wondering how one could experimentally test for such an object. Since the expansion rate of the universe is increasing, could one possible explanation be that a white hole is continually "spitting out" matter such that an increase of flux of in-falling matter/space could be observed as an increasing rate of expansion of out-pouring matter/space? While I don't know of a way to detect this, I was wondering whether the following idea was fruitful or not.

Given that a supermassive black hole is believed to be at the center of every galaxy, given that most galaxies (sans Andromeda) are receding away at a speed proportional to their distance relative to us observers, and given that the spatial distribution of matter in our universe is isotropic and homogeneous, could one build a simulation of such a concept? To emulate isotropy and homogeneity, one could use a random number generator corresponding to a uniform distribution to place "galaxies" along a grid; accounting for gravitational forces, one can simulate the time-evolution of these galaxies and compare rates of in-falling matter to find a correlation between the rate of in-falling matter and the expansion rate of the universe (likely as a function of time-elapsed, and age and size per black hole). While the simulation does not definitely prove anything, providing a correlation could be interesting in my opinion.

Is this a stupid idea, or is there a potential to build upon this idea? If it is not fruitful, I am curious as to why.

As a layman, I can tell you that there are some issues with your proposal.

The biggest one is that White holes probably don't exist.  They're purely theoretical and, even worse than that, they're mostly not even well defined theoretical objects.  For a physical hypothesis to work, every element of the theory needs to be precisely defined.  White holes aren't well defined.  (now, let me add that, perhaps White holes are well defined by Einstein's field equations, I'm not sure, but the specifics go above my paygrade).

And running time in reverse, a black hole really doesn't look anything like a big bang, and, granted, we don't know what the/a big bang looked like, but we know what a black hole swallowing matter looks like.   It absorbs matter by accretion disk and it shoots out jets along the accretion disks axis.   Running time backwards you can see streams of matter and the accretion disk reforming into stars and dust that flies away from the black hole but that gravitational consumption in reverse That looks NOTHING like the big bang is thought to have looked.

The moments after the big bang are believed to have been a very hot, dense and rapidly expanding quark/photon/fundamental particle soup kind of thing.   Nothing in today's universe, backwards or forwards in time, looks anything close to that, so running time backwards doesn't really help.

The very early universe and the ratio of hydrogen to helium and the presence and near perfect uniformity of the background radiation all imply a mostly uniform, very hot, very dense, and only the slightest bit lumpy rapidly expanding origin, that only works under modern physical laws if space expands faster than the speed of light.

We can re-create those conditions, but only for the tinyest fractions of a second, and with only a handful of particles, in particle accelerators. You might think that inside a black hole, similar conditions exist, but that's not entirely true.  Inside a black hole, gravity is dominant.   Due to faster than light expansion and uniformity, it's not believed that gravity was a dominant force in the moments after the big bang.  It was a force, but not a dominant force.   The space of space-time, for example, is very different inside a black hole than in the moments after the big bang.

A final issue with your question is whether the big bang was an explosion, or, more specifically that it originated from a source-point-origin at all, (as it's often presented), or whether it was an expansion that took place everywhere (everywhere stretch, so to speak).

A theoretical white hole ejecting matter would be a source point and a source point is subject to the inverse square rule and we'd expect to see a variation in density and evidence of a direction towards the origin.   We don't see that.  The observable universe appears remarkably homogeneous, which is only possible, from a source-point, if the size is enormously large.   The rate that distant galaxies move away from us is also very difficult to explain without some kind of dark energy expansion.

The variations in the image are exaggerated.  The cosmic background radiation is very homogeneous, implying we were inside the expansion, not part of some material that was ejected.

Source.

I realize that you didn't say explosion specifically, but asked about a source object ejecting material, but both have a similar effect.  Ejected material spreads out from it's source.   The universe doesn't appear to have been formed in that way.   Nice summary of the expansion theory here.

That's my layman's understanding anyway and I'm not sure I should have gone there.  Origin of the universe questions are kind of out there.


What is the official classification and naming system of objects in the universe(, or what would my URI look like)? Poetic human centric approach:

Universe/Laniakea/Virgo/MilkyWay/Sol/Earth/Humans/AndreasT

Little bit more technical:

Universe/0/SuperCluster/0/GalaxyCluster/0/Galaxy/0/StarCluster/0/Star/0/Planet/3/../HomoSapiensSapiens/2834929849289

More ... multiversal?

Or is there something in the spirit of the rebasing of SI units to universal concepts that makes us not be the origin(hah!) of the system?

Objects aren't named like that.

Either objects are given "names" that have no structure. For example, an asteroid may be named "Vesta".

Or they are given essentially meaningless catalogue numbers. For example, a star might be catalogued as HD 138987. It is number 138987 in the Henry Draper catalogue. Some major stars do have their constellation as part of the name. Alpha Centauri is the brightest star in the Centaurus constellation.

Comets are often just named according to the year of discovery

Other objects are named by their position in the sky, using polar coordinates.

There is no URI-like system in use.

The reason for this is that there is no natural hierarchy. When naming species of animal, they naturally group into genus, family, order, class as a result of evolution. No such structure is found among astronomical objects, so a hierarchical approach is not appropriate.


Given sunrise, noon, sunset, longitude and latitude can I calculate the "hight" or "ascension" of the sun? I'm trying to make a simple app for my own use to be able to check which of my favorit restaurants have outdoor seating sun and for how long etc.

Given the data in the title (sunrise, noon, sunset, longitude and latitude) I can draw a path where the sun should be at any given time, but I want to display the hight as well.

I live in sweden where the sun changes "hight angle" (sorry I don't know the exact term for it) which make the angle relevant when deciding on seatings amongst high buildings.
My end goal is to be able to use google maps 3D data to actually cast shadows on the map in a realistic way!

I have no Idea if I'm in the right place but I figure I'll give it a shot :)

To the precision that you need, you can calculate the azimuth A and altitude h of the Sun from the following equations:
$$
\tan(A) = \frac{\sin(H)}{\cos(H)\sin(\phi)-\tan(\delta)\cos(\phi)}\\
\sin(h)=\sin(\phi)\sin(\delta)+\cos(\phi)\cos(\delta)\cos(H)
$$
where

A is the azimuth measured from due south (positive towards the west)
H is the hour angle which is the number of degrees before (negative angle) or after (positive) solar noon. Calculate this as 15*1.00274*(time of day - time of noon).
$\phi$ is the latitude (positive for northern hemisphere)
$\delta$ is the declination of the Sun. This can be calculated from the second equation by assuming the altitude of the Sun h=0 when it is rising or setting, and H=15*1.00274*time from sunrise to solar noon. The assumption here is that the time from solar noon to sunset is the same.
h is the altitude of the Sun above the horizon.

All of the above quantities are in degrees. You may need to convert those to radians to perform the trigonometry calculations.

For a more accurate calculation, you would calculate the position of the Sun (its right ascension $\alpha$ and declination $\delta$) and Greenwich Mean Sidereal Time (GMST) based on the date and time throughout the day. Then using the latitude and longitude, you would calculate H, A, and h for various times of the day.


Can a natural satellite exist in a geostationary orbit? While browsing through Physics SE, I noticed a question about satellites in geostationary orbit (unrelated to the one I'm asking here), and for a moment I interpreted it as referring to natural satellites (e.g. a moon). So I wondered: Could a natural satellite exist in geostationary orbit?

Then I stopped and thought. For large gas giants, such as Jupiter, having moons too close to the planet can be fatal (for the moon). If it ventures inside the planet's Roche limit, it's toast. But there is good news: the Roche limit depends on both the masses and densities of the primary body and the satellite. So perhaps this reason is non-applicable, as a high-mass natural satellite might be able to survive. So the question changes:

Could a sufficiently high-mass, high-density natural satellite occupy geostationary orbit over its primary body?

Yes. Charon is in Pluto synchronous orbit. Pluto and Charon are mutually tide locked.


Do main sequence stars in binaries transfer mass by Roche lobe overflow? It is known that binary star components containing red giant can under go mass transfer when their radii exceeds the Roche limt. Do main sequence binary stars also undergo mass transfer by this mechanism?

Yes. An example would be the Cataclysmic Variable stars (CVs), where the donor star (often in the main sequence), loses material via Roche Love overflow onto a compact white dwarf companion.
The other obvious example (though less common) are the Low Mass X-ray Binaries (LMXBs), where the Roche Love overflow is onto a neutron star or black hole.
Nelson & Eggleton (2000) did a theoretical study of Case A mass transfer (where the donor is hydrogen burning star) and found several subcategories where all the mass transfer takes place whilst both stars are on the main sequence (labelled AD, AR and AS). In all three cases, the endpoint is a contact binary of the W UMa type, where both stars fill their respective Roche lobes.
W UMa stars are very common, but that is because their lives are long, whereas the period over which mass is transferred between the progenitors is very short by comparison, so objects "caught in the act" will be rare. The basic picture is that they start out as short-period, deteached binary systems. But as they evolve, two things happen. (i) If they are of low-mass then dynamo generated magnetic activity can result in angular momentum loss which brings them closer. (ii) Normal stellar evolution on the main sequence leads the stars to get bigger. The two combine and lead to the primary star filling its Roche lobe, transfering mass and accelerating the process towards a W UMa status.
The exception is the AS case, where the transfer occurs on the "nuclear timescale" (e.g. the Sun's nuclear timescale is 10 billion years) and the mass transfer is very slow.
Thus semi-detached main sequence systems are either rare or not doing much, which makes them hard to spot. Nevertheless, they are found if you look in the literature. e.g. Deb & Singh (2011) identify several in a sample of contact or close-to-contact eclipsing binaries, from their distinctive light curve shapes. Other authors refer to these as Pre-Contact W UMa Binaries (PCWBs) and have catalogued some tens of objects (Samec et al. 2013).


Meteorites bring water The latest information suggests that Earth's water came from Meteorites . The Moon was also bombarded by Meteorites and yet it has no water. Is this because it lacks an atmosphere and its water was lost in space ?

On further thought on this matter, it occurred to me that Mars has an atmosphere and yet it has no oceans either, just like the Moon . So Mars I assume was bombarded by meteorites, just like the Earth was and it has a stronger gravity than the Moon, yet it doesn't have oceans either ? So what happened ?

It would have to be a big meteorite to supply the quantaties we have here and therefore support the nuclear break down theory.ie- hydrogen to helium etc.

In respect of delivery by cosmic objects, the atmospheric pressure on the moon would result in a low boiling point and the evaporation velocity(due to low gravitational attraction) would exceed the escape velocity of the moon and meet with no atmospheric resistance.
This is the way I see it.
The body of water, although spread over the surface, would consider itself as an independent body and have its own "existence parameters" with regards to its environment.

The ability of Mars to retain water vapour is much reduced by its lower gravitational strength and atmospheric density. Can water exist at these parameters continuously, the evidence shows not.


Comet 289P/Blanpain sudden outburst? So Sky Safari has shown that comet 289P/Blainpain has had a sudden increase in brightness, and is currently sitting at mag -1.4!!! Last week it was sitting at +15. It is due to reach Perihelion on 20th December (2 days time) so is this due to some outburst? Or is there a glitch in Sky Safari? Looking at THIS site, it states:

  When the perihelion distance is larger than 1.0 A.U., it becomes very
  bright if it passes the perihelion on a special day in early December.
  However, only if the perihelion passage shifts by a several days.

But the Perihelion distance is 0.089AU for this year according to Google

This comet could be prone to outbursts as it was recently rediscovered in July 2013 due to a sudden brightness increase from an outburst (although nowhere near this much!)

However, currently TheSkyLive is showing that it is currently at mag +18.

I have had a quick search to try and verify this, unfortunately, I live in England and our skies are very cloudy at the moment so I can't go out and see for myself. Does anyone have any information as to whether this is a Sky Safari glitch or if not, what has happened to cause such a huge brightness increase?

When I add 289P/Blanpain in the solar system editor of Stellarium,
it too shows the the comet blazing away at apparent magnitude -1.4.
The problem must be upstream.
Planetarium software typically gets comet and asteroid data from the Minor Planet Center,
whose own ephemeris service gives a similar result under m1:

0289P
Date       UT      R.A. (J2000) Decl.    Delta     r     El.    Ph.   m1     Sky Motion
            h m s                                                            "/min    P.A.
2019 11 30 000000 22 29 37.8 -16 25 42   0.280   1.006   85.9  78.0  -0.2    1.25    023.6
2019 12 15 000000 22 46 29.4 -06 51 49   0.201   0.963   78.0  90.2  -1.1    2.39    021.8
2019 12 30 000000 23 17 53.0 +14 42 59   0.123   0.968   79.2  93.7  -2.2    6.04    019.2

However, MPEC 2019-W179 from the same organization lists some actual observations from late November and gives this ephemeris:

Date    TT    R. A. (2000) Decl.     Delta      r    Elong.  Phase   m1    m2
2019 11 30    22 29 37.7 -16 25 45   0.2802  1.0058    85.9    78.0  17.3
2019 12 15    22 46 29.3 -06 51 54   0.2013  0.9628    78.0    90.2  16.4
2019 12 30    23 17 52.7 +14 42 50   0.1232  0.9678    79.2    93.7  15.3

MPC has collected other observations since then.
The JPL Small Body Database record, based on the same observations, estimates a total absolute magnitude M1 = 20.8±1.
HORIZONS gives this ephemeris:

Date__(UT)__HR:MN     R.A._(ICRF/J2000.0)_DEC  T-mag  N-mag
2019-Nov-30 00:00     22 29 37.44 -16 25 42.6  18.05   n.a.
2019-Dec-15 00:00     22 46 28.88 -06 51 47.8  17.25   n.a.
2019-Dec-30 00:00     23 17 52.05 +14 43 16.0  16.19   n.a.

If I edit my .stellarium/data/ssystem_minor.ini

[289pblanpain]
absolute_magnitude = 2.5

and replace the faulty 2.5 with 20.8, Stellarium predicts a more realistic magnitude 16.9 for the comet today.
The MPEC ephemeris is consistent with M1 ≈ 20.0.

The MPC data file CometEls.txt lists a few other comets at absolute magnitude 2.5.
JPL estimates for those comets range from 4.8±0.8 to 10.7±1.


How many galaxies are there in the Hercules–Corona Borealis Great Wall? The Hercules–Corona Borealis Great Wall is "to date (February 16, 2015), it is the largest and the most massive structure known in the observable universe" (source). How many galaxies is the Hercules–Corona Borealis Great Wall believed to contain?

I'm sorry for the long answer. The result is in the bottom. :)
The number of galaxies in the "wall", as in any volume of space, is the number density $n_{\mathrm{gal}}$ times its volume $V_{\mathrm{wall}}$. However, the number density of galaxies depends on their masses/sizes: Small galaxies are much more common than large galaxies. Usually we describe this by the halo mass function (HMF), which gives the (differential) number density dN/dM of dark matter halos hosting galaxies as a function of halo mass M. Depending on your favorite cosmology, it looks something like this (created with this tool):

The solid line shows the HMF today in the local Universe, i.e. at redshift zero, while the dashed and dotted line shows how it was at redshift 1.6 and 2.1, respectively, roughly 10 billion years ago, which mark the beginning and end of the wall (according to Horváth et al. 2014, A&A, 561, L12). Notice how the number of large halos grow with time, at the expense of small halos. This is due to the smaller ones accreting onto larger ones.
The HMFs are characterized by a power law at low masses with an exponential cut-off at high masses. To get the total number density we need to integrate over the entire range of masses, and here comes the problem: Whereas the upper limit is set by the maximum size a galaxy can have (roughly $3\times10^{13}M_\odot$, but the exact number isn't very important, as these galaxies are so rare), the lower limit depends on your definition of a galaxy. Large galaxies have smaller galaxies orbiting as satellites, and there isn't a well-defined threshold for when a clump of gas, stars, and dark matter is called a galaxy. These guys report on the detection of a $\sim10^3M_\odot$ clump which they call a galaxy. That's probably not what you think of as a galaxy, so for the sake of this exercise, let's use Small Magellanic Cloud-sized galaxies as our lower minimum, i.e $\sim 6.5\times10^9 M_\odot$.
The total number density of galaxies is thus found by integrating the HMF. Luckily, the online HMF tool can do that as well. The result, i.e. the number density of halos above a given mass, looks like this:

Note that the $x$ axis is given in terms of $h^{-1}M_\odot$. Since $h \simeq 0.7$, the mass of the SMC is $M_\mathrm{SMC} \simeq 5\times10^9 h^{-1}M_\odot$. From the graph, the number density of these objects is $\sim0.5\,h^3\mathrm{Mpc}^{-3} = 0.17\,\mathrm{Mpc}^{-3}$ (1 Mpc, or megaparsec, equals 3.26 million lightyears). In principle we should subtract the number density of halos above $3\times10^{13}M_\odot$, but since these are four orders of magnitude smaller, they can be neglected.
The wall starts at redshift ~1.6 and ends at redshift ~2.1. This translates into starting at a distance of 4.65 Gpc (billion parsec) and ending at 5.47 Gpc, i.e. a depth of ~800 Mpc. These numbers are in comoving coordinates, i.e. what it has expanded to today, as opposed to physical coordinates, i.e. what it was at the time it emitted the light that we see today. But the HMF is given in terms of comoving volume, so that's cool.
To get the volume, we also need the area. The finders report that it's roughly 2 Gpc times 3 Gpc, but they don't seem to state whether this is in physical or comoving coordinates (which differ by a factor (1+z) ~ 3). I suspect it's comoving, though.
Thus, the total volume is $V_{\mathrm{wall}} = 0.8 \times 2 \times 3\,\mathrm{Gpc}^3 = 4.8\,\mathrm{Gpc}^3 = 4.8\times10^9\,\mathrm{Mpc}^3$, or 166 cubic gigalightyears. Here, I have ignored the fact that the physical area spanned by the angle of observation is a bit different at the front and the back of the wall, but since these numbers are already quite uncertain, it's of lesser importance.
The considerations above are all for the average Universe. But the wall represents an overdensity in space, so the number will be larger. I can't find any estimates of the overdensity of this particular supercluster, but in general galaxies in superclusters are more numerous by a factor of $\delta \sim 5$–$10$. Let's use 5.
That is, the total number of galaxies the size of the Small Magellanic Cloud or above is
$$N_\mathrm{gal,tot} = n_\mathrm{gal} \times V_\mathrm{wall} \times \delta$$
$$= 0.17\,\mathrm{Mpc}^{-3} \times 4.8\times10^9\,\mathrm{Mpc}^3 \times 5$$
$$\sim 4 \mathrm{\,billion\,galaxies}.$$
But if you go through the same exercise for halos of masses above, say, $10^3\,M_\odot$, you'll find a number which is a million times larger.
In reality, the number density of luminous galaxies is smaller. At the high-mass end, the gas has trouble cooling sufficiently to condense and create stars, and active galactic nuclei in massive galaxies heat the gas. At the low-mass end, gas is more easily blown out of the potential well by stellar feedback from supernovae.


Our sun becoming a red giant If our sun was already in the process of becoming a red giant, would the gradual rise in luminosity be noticeable to our eyes at some point in human existence?

If our sun was already in the process of becoming a red giant, would the gradual rise in luminosity be noticeable to our eyes at some point in human existence?

I'd say no for a couple of reasons.

@PM2Ring's comment:

FWIW, in about 1.1 billion years it will be too hot for most lifeforms currently on Earth, long before the Sun starts moving off the main sequence and becoming a red giant. See Timeline of the far future.

Our eyes and vision processes accommodate (adjust for) changes in brightness, we walk indoors from a sunny day to a well lit room and the brightness of things has dropped by 99% or nearly a factor of 100 within a few seconds and yet though we are somewhat aware of the change of tens of percent per second we hardly notice it.

Our vision system is constantly adjusting its white point to accommodate changes in the color of available light.† A white sheet of paper looks white to us even as the color of ambient light changes because our vision is always "color balancing" (not exactly the right term but it's something like that).

†From @Luaan's comment

It's called "white balancing" - calibrating vision for ambient light to make white surfaces (hopefully) appear white. Sometimes it doesn't quite work properly, and you get fun things like The Dress (since different people's vision chooses different arbitrary ambient conditions, and end up disagreeing wildly about the colors involved)


Why do not we use optical telescopes to study cosmic rays directly? HESS, CTA, and VERITAS are used to study cosmic rays.

What they receive are optical (maybe very blue) photons.
So my question is: Why don't we use optical telescopes to study cosmic rays directly?

The telescopes in your picture are optical telescopes. They are used to detect cosmic rays using the   Cherenkov radiation caused by charged particles moving faster than the speed of light (in air) in the atmosphere. These are normally secondary particles that have been produced in collisions between cosmic rays and nuclei in the upper atmosphere.

Radiation produced by particles moving at close to light speed is beamed in the direction of motion. An array of optical telescopes picks up the beam, and the orientation of the image is a projection of the original track of the particle. By combining images taken with telescopes covering a wide area, and taken at almost the same time, one can reconstruct the paths of the secondary particles to work out which direction the original cosmic ray came from. The intensity of the images tells you something about the original energy of the cosmic ray.

Possibly this is the answer to your question - a single optical telescope might detect the Cherenkov radiation from a single secondary particle, but it would be difficult to reconstruct the path or energy of the original cosmic ray on its own - at least a pair of telescopes, or better still, a network is preferable. The properties of these "cosmic ray" telescopes are also quite different from "ordinary" astronomical telescopes. Cosmic ray telescopes need as much collecting area as possible, because the Cherenkov radiation is faint - but they can do this at the expense of image quality. They use very wide angle cameras (5 degrees or more) and use huge faceted/sgemented mirrors that give massive collecting area, but image quality (maybe 3 arminutes full width half maximum for a point source) that would be completely unacceptable for conventional optical light astronomy.

Or (thanks Conrad Turner) you are asking why we cannot use the mirrors of an optical telescope to focus the cosmic rays onto a detector and image their source? The reason for that is simply that cosmic rays are not reflected from glass/silvered surfaces in the same way that light is. They are extremely energetic particles that either pass straight through the mirrors or are absorbed within them. i.e. They do not reflect or refract in the same way as light.

In other circumstances, cosmic rays are a nuisance when conducting observations with optical telescopes. They are a source of background noise in CCD detetor images. Cosmic rays (or the secondary particles) are capable of liberating electrons in the silicon and therefore simulating small intense light sources in the sky. Often these are seen as small bright pixels, groups of pixels or trails on the CCD image, confusing what you were originally trying to look at. However, these cosmic rays are not focused by the telescope, they are essentially being picked up by the detector itself and would have been so if the detector was not even attached to a telescope at all. They are nearly impossible to shield against because they have very high energies and your detector needs to have a hole to let the optical light in!


Is the Earth Really Spinning? (honest question) Most people consider it common knowledge that the Sun's "movement" in the sky is only perceived due to the Earth's "spin and movement".

Based on this, you'd think that the stars in the night sky (when viewed with time lapse photography) would take a path in the sky similar to the sun. After all, both the sun and stars are practically stationary over the course of one day, given their accepted distances from the earth.

Yet, time lapse photography shows this is not the case.

Compare these photos:

1) Time Lapse of Sun

2) Time Lapse of Stars

3) Time Lapse of Moon

If the Earth's spin is the primary reason for "movement" of things (farther than clouds in the sky), then why do the sun and moon move through the sky in a similar fashion that is totally different from the sky-path of the stars? The Earth is said to spin around 360 degrees every 24 hours. It seems to me that everything in the sky should move in one direction where I live (Texas).

To me (in Texas), it seems the stars should sweep from horizon to horizon over the course of the Earth's night-spin (like the sun and moon do). Yet, the time lapse photos show circular paths in the sky for the stars. That would make sense at the poles, but not in Texas!?

-Lonnie Lee Best

You are already starting to get it.

  That would make sense at the poles

What about one meter from the poles?
Or a kilometre?

As long as you can see the celestial pole in the sky, you can see the stars revolve around it at night.

Let us see if you are able to see the celestial pole.

Texas was about 30 degrees north last time I checked:

That explains the circular movement of the stars, the Sun and the Moon.

This is true for all locations on the Earth, except for the equator:

Is the Earth spinning? That depends, you can always choose a frame of reference that suits you. However, only one of them are non-rotating, the Inertial frame. In all the others we have fictitious forces acting, like centrifugal or Coriolis forces.

We can test if the Earth rotates by watching a pendulum throughout a day. The pendulum would then seem to slowly rotate during this period of time, meaning some fictitious "force" is acting on it. That means that we are located in a rotating frame of reference, and thus the Earth rotates.


Daily magnitude of Betelgeuse I tried searching on the internet and could not find anything, so does anyone know if there's a website which tracks and shows Betelgeuse's magnitude daily?  I just went out today and saw that it's now about the same brightness as Bellatrix.

The American Association of Variable Star Observers tracks observations of a number of variable stars, including Betelgeuse. There are typically many such observations each day, primarily at visible wavelengths. It's quite easy to generate a light curve in the band(s) and date range of your choice.

I'm also aware of a Twitter bot that tweets light curves from AAVSO data several times a day, including long-term projections, but I assume you'd be more interested in looking at the individual measurements yourself.


What explains the existence of energy/matter if it cannot be created or destroyed? If the energy/matter that was contained in the singularity before the Big Bang had always existed, and if it’s true that energy/matter cannot be created or destroyed, then what explains its existence?

That's a very complicated question!

First, let's remember that Moses didn't bring the Law of Conservation of Energy down from Sinai on stone tablets -- it's something that we've observed to be true of the universe today, and which our theories use to make very accurate predictions. But this doesn't mean that it was always true, and particularly that it was true early enough in the Big Bang that our current theories don't apply.  We know that our current theories fail during the Planck Era, for example.

Second, Noether's Theorem says that energy conservation exists because the Lagrangians of mass-energy are time-symmetric. Time symmetry is a very good symmetry today in regionals of space small compared with the entire universe, but there may well have been eras in which it wasn't.

In fact, there probably were such eras, just as time-symmetry is not a good symmetry even today at very large scales.  (The universe is expanding, and on the face of it, this causes a non-conservation of energy.)

Thirdly, if you try to save the law of conservation of energy by calculating the total energy of the entire universe including the gravitational energy, one reasonable set of arguments (energy calculations in General Relativity can be very difficult, since gravitational energy tends to be non-local) yields a total energy of zero! The mass-energy in the universe is exactly balanced by the gravitational potential energy.  If this is correct, the total energy has always been zero.

But this is all speculative!  Remember, that real physics is an experimental science and if you can't test a theory it may be interesting, but it isn't physics.


How does 'centre of mass' concept work? We know that two celestial bodies rotate around a center of mass. How does this concept work in case of solar systems where there are several planets versus one/two stars? (Even if the center of mass lies within a star, the center of mass for every planet would be different.) How is it that the rotation of celestial bodies does not become unsteady?

We know that two celestial bodies rotate around a center of mass.

"We" do not know that. Rhetorical question: Pick up an apple and drop it at high noon. Does it fall towards the solar system barycenter? The answer is of course "No." It falls away from the barycenter, towards the Earth.

I'll steal a graph from an answer I wrote to a question at the sister site, physics.stackexhange.com, Do the planets really orbit the Sun?:

Given the above graph, it appears that it's much better to say that Venus orbits the Sun (red curve) rather than the solar system barycenter (black curve). So why do people consistently use a barycentric frame rather than a heliocentric frame? The answer is simple: The barycentric frame is the frame in which the equations of motion undertake their simplest form. This is particularly so when one wants to incorporate some of general relativity into the equations of motion.

Our solar system is well-behaved. Other star systems aren't quite as well-behaved. For an example of a hypothetical system that is extremely far from well-behaved, I suggest you read Saari and Xia, "Off to infinity in finite time," Notices of the AMS.


Can Fermi Bubbles be produced in non-spiral galaxies? In 2010 observations with the Fermi Gamma-ray Space Telescope discovered giant bubbles which have been named Fermi Bubbles. Similar structures have also been discovered by Pshirkov et al. (2016) surrounding M31 (the Andromeda Galaxy).

Our galaxy and M31 are barred spirals. Could Fermi Bubbles also form around other galaxy types, for example lenticular or elliptical galaxies?

The discovery paper (Su et al. 2010) mentions that bipolar shells on the order of $0.5\text{ kpc}$ were found on both sides of the nucleus of Centaurus A (Quillen et al. 2006). Centaurus A is a galaxy of contested morphology, likely either elliptical or lenticular - certainly not a spiral. This would seem to be a strong indicator that similar bubbles can indeed be found in non-spiral galaxies.

Now, the nuclear shells Quillen et al. detected were seen at mid-infrared wavelengths (5.8, 8.0 and 24 $\mu\text{m}$), rather than via the gamma rays with which Su et al. discovered the Fermi bubbles. However, the authors believe they may have been caused by a burst of star formation a few million years ago, driven by winds. This is one of two main scenarios considered by Su et al. for the formation of the Milky Way's bubbles (the other being AGN activity), meaning that the same processes could indeed have been at work in both galaxies. I don't know if corresponding gamma-ray emission from the nuclear bubbles in Centaurus A has since been detected, but it certainly seems plausible.

I find it a little strange that Pshirkov et al. were so careful to focus on spiral galaxies. It's clear that starburst galaxies and the host galaxies of AGN can obviously have non-spiral morphologies. Perhaps the group was just interested in searching for other nuclear bubbles on galaxies similar to the Milky Way which might have similar conditions and evolutionary histories.


Any book on `How do they know it (in Astronomy)`? I am reading my college level undergraduate astronomy textbook. It has lot of ínfromation but every page rises a similar question in my mind: "How do they know it?". It starts from simple things like radius of earth, other planets, masses, composition, distances of stars, temperatures, luminosities etc. How did scientists know these is a question that pops up in my mind about every fact I read. Do you have any book recommendations which mostly (if not exclusively) focuses on the question "How do scientists know it?"

There are books that treat the general history of astronomy, such as those mentioned in the previous post.

The History of Astronomy: A Very Short Introduction by Michael Hoskin.
History of Astronomy by George Forbes

Some books focus on discoveries from a certain period of history

The Cosmic Century
by Malcolm S. Longair
details discoveries that were made in the twentieth century.

Other books retrace the "how we know it" of specific phenomena

Flash
by Govert Schilling
traces the story of how we discovered what we know about gamma ray bursts.
Cosmic noise : a history of early radio astronomy,
by Sullivan, Woodruff Turner III
focuses on what we learned from radio astronomy.


Why do we believe that the super massive black holes at the centers of two merging galaxies would themselves merge? When listening to podcasts or watching youtube videos of astronomers discussing galaxy mergers, I often hear talk about how the super massive black holes at their centers will themselves merge during or shortly after the collision. Why do we believe this to be the case?

A priori, I would expect SMBH's to behave the same as all other galactic objects. They might be extraordinarily massive, but physically they are still minuscule compared to the vast empty space between stars in a galaxy. Collisions between objects (not counting gigantic clouds of gas and dust) would be extraordinarily rare, so why do we make an exception for the SMBH's?

I could see a case for them merging in the (rare?) instance where the host galaxies hit each other in such a way that the mutual center of mass happens to coincide with their individual centers of mass. In that case, the SMBH's could be close enough to orbit about one another, losing energy to gravitational waves and eventually merging. I would think that this scenario is rather rare, however. I find it more plausible that an average galaxy merger would leave the SMBH's independently orbiting the center of the combined galaxy, too far apart to lose any significant kinetic energy to gravitational waves.

The astronomers who talk about galactic mergers know a heck of a lot more than I do about the subject, so I presume there are flaws in my assumptions or my understanding of the physics. What am I missing?

The SMBHs reside in the bottom of the galactic potentials, which are dominated by the galaxies' dark matter halos. But although dark matter dominates gravity, collisions between gas and dust particles in the interstellar medium causes enough friction that the baryonic component of the galaxies is decelerated. This will cause the other components of the galaxies to decelerate as well, through gravitational attraction.

Moreover, despite dark matter (and, in practice, stars and black holes since they're so small) being collisionless, there are several ways of "relaxing", i.e. to evolve towards an equilibrium. In the context of galaxy merging, the most important mechanism (I think) is "violent relaxation", where the rapid change of the gravitational potential causes particles to relax, e.g. more massive particles tend to transfer more energy to their lighter neighbors and so become more tightly bound, sinking towards the center of the gravitational potential.

Although SMBHs are…, well, supermassive, the potential will (usually) be dominated by dark matter, gas, and stars, so the new gravitational potential will also cause the SMBHs to seek towards the bottom in the same fashion, and eventually merge.


Does the earth spiral around the sun's movement/motion path? I have watched the following video (How Earth Moves by Vsauce) regarding how earth moves:

Here are some screenshots:

I have some questions:

Does the earth spiral around the sun's path as it is shown in the video (exact time is 19:49) and screenshots? Is the spiraling movement caused by sun movement (on its path shown above) where the sun is dragging the earth (and other planets).
Does the earth go in front of the sun then back, then fourth and back like it is shown on the video? I see both the sun and earth racing with each overtaking the other over and over. Is this assumption true? In this manner, would earth, at one point in time, be closer to the destination where the solar system is heading (I think they call the destination Vega) than the sun?
I am working on a poetry piece. If I mention that earth and sun are locked in a never ending race with each taking the lead (earth moving in front of the sun's path / winning) and earth actually dropping out of the race (spiraling/moving out the race and falling behind), that they are destined to never meet (collide), would I be correct?

P.S. My knowledge of astronomy/physics is very limited. I have tried my best to be as clear as possible. Sorry for any confusion. Thanks.

Does the earth spiral around the sun's path as it is shown in the
video (exact time is 19:49) and screenshots?

The Earth does spiral around the Sun's path, but not quite as it's shown in the video.

It isn't clear if the video accounts for the 60 degree tilt of the Solar System with respect to the galactic plane.

The sizes and distances are not to scale.  The Sun and the Earth are depicted far too large with respect to their distance from one another.

The relative motion is not to scale. The distance from one side of the spiral to the other side should be about 2 AU, or 3x10^8km.  But the Sun is moving at about 200km/s around the galaxy, or 6.3x10^9km per year.  So the distance between two consecutive spirals should be more than 20 times the distance from one side of the spiral to the other!  This can be visualized as an overextended slinky.

Is the spiraling movement caused by sun movement (on its path shown
above) where the sun is dragging the earth (and other planets)?

No, the Sun's and the Earth's mutual gravity cause the Earth to rotate around the Sun.  The Milky Way's gravity causes our Solar System to revolve around the galactic center.  The Sun and the Earth are on the same plane and moving through the galaxy together.  The Earth would still rotate around the center of the Milky Way if the Sun magically disappeared.

Does the earth go in front of the sun then back, then fourth and back
like it is shown on the video? I see both the sun and earth racing
with each overtaking the other over and over. Is this assumption true?
In this manner, would earth, at one point in time, be closer to the
destination where the solar system is heading (I think they call the
destination Vega) than the sun?

If you consider a plane tangent to the Sun's velocity vector, then the Earth will pass across that plane twice per year, appearing "in front of" and then "behind" the Sun.  This is simply a geometric result since the Earth and Sun share another plane. I wouldn't call any future position of the Solar System a "destination" without evidence the Solar System was sentient, but I am not a philosopher.

I am working on a poetry piece. If I mention that earth and sun are
locked in a never ending race with each taking the lead (earth moving
in front of the sun's path / winning) and earth actually dropping out
of the race (spiraling/moving out the race and falling behind), that
they are destined to never meet (collide), would I be correct?

We haven't ruled out the possibility that the Sun may someday envelope the Earth: What will "the Earth being consumed by the Sun" look like?. Also, the Earth isn't likely to leave its orbit before that time, but the Solar System is chaotic on long time scales. However, your poem would be consistent with current models, and we (humans) don't conclusively know how the Solar System will end.  So my two cents is that your poem doesn't contradict modern astronomy.  Besides, what's the use of a poetic license if you don't use it?


Earth Orbit Simulator I'm trying to simulate Earth's orbit using 2nd Newton's Law and cartesian coordinates. The program seems ok, but when plotting graph, it's a parabola.

I know the shape of orbit changes according velocity. I'm using vy0 = sqrt(G*M/R) and it supposed to be an circle.

(Sorry the bad english)

Here's the code:

'-------------------Bibliotecas Importadas-------------------'

import math
import matplotlib.pyplot as plt

'-------------------Listas Criadas-------------------'

pos_x = []
pos_y = []
vel_x = []
vel_y = []
teo_virial = []
tempo = []

'-------------------------- Definição de Funções -----------------------------'

def velocidade_atualizador(velocidade_inicial, aceleracao):
    velocidade = velocidade_inicial + (aceleracao * dt)
    return velocidade

def tabela(nome, lista):
    arquivo = open(nome + '.txt', 'a')
    for i in range(len(lista)):
        arquivo.write(str(lista[i]) + '\n')
    arquivo.close

def graph(x, y, eixo_x, eixo_y):
    plt.plot(x, y)
    plt.xlabel(eixo_x)
    plt.ylabel(eixo_y)
    plt.show()

def aceleracao_equacao(pos_1, pos_2):
    r_cubo = (pos_1**2 + pos_2**2)**1.5
    k = constgrav*massasol
    aceleracao = (k*pos_1)/r_cubo
    return aceleracao

def posicao_equacao(posicao_inicial, velocidade_inicial, aceleracao):
    pos = posicao_inicial + (velocidade_inicial*dt) + ((aceleracao*(dt**2))/2)
    return pos

'-------------------Condições Iniciais-------------------'

constgrav = math.pow(6.67, -11)             #constante gravitacional
massasol = math.pow(1.98, 30)               #massa do sol
massaterra = math.pow(5.97, 24)             #massa da terra
xo = math.pow(1.5, 11)                      #posição inicial no eixo x
yo = 0                                    #posição inicial no eixo y
vxo = 0                                   #velocidade inicial no eixo x
vyo = math.sqrt((constgrav*massasol)/xo)                               #velocidade inicial no eixo y
npasso = 10000                                #numero de passos
t = math.pow(3.1536, 7)                               #tempo total = tempo de um ano
conttempo = 0                             #contador de tempo
dt = t/npasso                               #delta t

'-------------------Laço de Repetição-------------------'

while (t > conttempo):

    ax = aceleracao_equacao(xo, yo)
    ay = aceleracao_equacao(yo, xo)
    a_modulo = math.sqrt(ax**2 + ay**2)

    vx = velocidade_atualizador(vxo, ax)
    vy = velocidade_atualizador(vyo, ay)
    v_modulo = math.sqrt(vx**2 + vy**2)

    x = posicao_equacao(xo, vxo, ax)
    y = posicao_equacao(yo, vyo, ay)
    r = math.sqrt(x**2 + y**2)

    energia_pot = (- constgrav * massasol * massaterra)/math.sqrt(x**2 + y**2)
    energia_cin = (massaterra * (v_modulo**2))/2
    virial = energia_pot / energia_cin

    print('x =', x)
    print('y =', y)
    print('virial =', virial)

    pos_x.append(x)
    pos_y.append(y)
    vel_x.append(vx)
    vel_y.append(vy)
    teo_virial.append(virial)
    tempo.append(conttempo)

    xo = x
    yo = y
    vxo = vx
    vyo = vy

    conttempo = conttempo + dt

tabela('Lista x', pos_x)
tabela('Lista y', pos_y)
tabela('Lista vx', vel_x)
tabela('Lista vy', vel_y)
tabela('Lista tempo', tempo)

'-------------------Gráfico-------------------'

graph(pos_x, pos_y, 'x (m)', 'y (m)')
graph(vel_x, tempo, 'vx (m/s)', 't (s)')
graph(vel_y, tempo, 'vy (m/s)', 't (s)')

Thank you very much, guys!! I was thinking but don't know what to do anymore.

After changing the code, it works. I can rest in peace now. Hahaha.

There is no obvious astronomy error, but your use of math.pow is wrong

You write constgrav = math.pow(6.67,-11).  That means $G = 6.67^{-11}$.  You mean constgrav = 6.67e-11 or $G=6.67\times 10^{-11}$.

With completely different initial conditions, it is not surprising that the Earth flies off into space.

As a suggestion, try not using SI units for this, as it makes for some very small and very large intermediate values, the computers have difficulty in rounding.  Instead measure distance in Astronomical Units, Mass in "solar masses" and time in years. You can convert back to SI units when printing or plotting.


What proportion of planetary systems have been found with 'Hot Jupiters' According to the NASA website "The strange attraction of Hot Jupiters", one of the main types of exoplanets that have been detected are Hot Jupiters, which are

  These are behemoth worlds that orbit close to their parent stars, blocking a fraction of the star’s light when it transits in front.

Planets that are from as large as Saturn through to far larger than Jupiter, orbiting in a matter of days very close to their parent star.  Something our solar system does not have.

What proportion of star systems have Hot Jupiters within their systems?  Is it a case that Hot-Jupiter-less systems such as our own are unusual?

In 2011, about 20% of the exoplanets found were hot Jupiters. That is a lot, but it is strongly biased simply because there are the easiest planets to detect. You can detect exoplanets by transit (that means that the planet passes in front of its host star which decreases the luminosity of the observed star during the transit), hot Jupiters being closer to their host star and bigger than smaller planets will therefore be easier to detect. You can detect exoplanets by radial velocity, which means that you detect variation of the position of the host star due to planet's gravity. The closer and the heavier the planet is, the larger the variation will be, and therefore the easier to detect.

That's why it is important to say "In 2011"; as our detection techniques (and telescopes) improve, the proportion of hot Jupiters should decreases.

Edit:
To put that in perspective, you can read this article from Jason Wright in which he tries to estimate the "real" ratio of hot Jupiters orbiting around "normal" stars (F, G and K type stars). This ratio is actually much smaller than the currently observed ratio, about 1.2%.


Why do comets come from our local Oort cloud instead of from interstellar sources? Comets form where stars form. And they are then ejected by the orbital changes of giant planets. But why are they ejected to form specifically a local Oort cloud instead of going interstellar and becoming vagabonds? The fact that comets come from all directions doesn't discriminate between them coming from an Oort cloud or from interstellar sources.

The main argument against the idea of one star capturing a comet from another star, seems to be that the relative velocity is too high for that. But consider the huge age of the Sun which is now on its 18th or so orbit through this crowded galactic star field disc. And which by chance or co-formation might have now, or at some earlier time, roughly the same velocity as some other millions of other stars. And what if there are many (many) more comets formed and/or ejected from young stars than currently supposed?

Is it really unlikely that the Oort "cloud of comets" instead largely represents an interstellar source?

Since most comets are on a predictable orbit that has them circling the sun (several times), then they cannot be from outside the solar system.

Any comet that originates outside the solar system, will pick up enough velocity approaching the sun to be able to leave the solar system. Therefore, we would only see the comet once. It is unlikely that the comet approaches Jupiter (Saturn, Neptune or Uranus) where the trajectory shifts enough so that it will remain in the solar system.

Since the solar system is believe to be formed from the gravitational attraction of interstellar gas and debris, it is very unlikely that comets will be ejected unless their orbits are perturbed.

So, there is a low probability of a comet being ejected from one system, another low probability that the comet will encounter another system (after millions and billions of years) and yet another low probability that the comet gets captured in the system.

Added:
If the object was pushed out to 1 LY and it then returned as a comet, it was still under the influence the sun's gravity and would not be considered as extra-solar. The object would need to come from another system to be called extra-solar. Since the ejected objects are in a nearby orbit to the solar system, they will either return, or travel outward. Think about a satellite with a period of 12 hours. A slight push will put it in orbit 2 meters higher, and will increase the period significantly.

There is a lot of space at Neptune's orbital distance, and the intersection with Neptune's gravity is quite small (ratio of area of influence (0.1g) / surface area = ${8.6\times 10^{-11}}$). The reason the Jupiter captures asteroids is because they are in the same plane and are moving in the same direction. This increases the probability of changing the orbit significantly. It could also take several hundred passes before the orbit changes significantly.

The area of influence is the cross section area where the gravity is greater than 0.1g. The surface area is the spherical surface area of the mean distance to the planet.

Jupiter - ${5.325\times 10^{-8}}$ ~= 0.0000053%
Saturn  - ${4.703\times 10^{-9}}$ ~= 0.00000047%
Uranus  - ${1.791\times 10^{-10}}$ ~= 0.000000018%
Neptune - ${8.605\times 10^{-11}}$ ~= 0.0000000086%


Confused about the significance of sidereal time My understanding of sidereal time is that the position of a star will be in the same place 4 minutes earlier the next day. So a 20 days later the star will be 1hr 20 mins earlier. So if the star is in the same place it must be a multiple of 4 minutes earlier (and the seconds that I have ignored). So it is not possible for the star to be in the same place and not be that time multiple apart.
Its just that I have some data points where the locations are identical and they are all multiples of 4 mins apart but I am worried that that is not significant because it is inevitable. Is that correct?

Let's start with a description of sidereal time. From my knowledge, sidereal time at any location can be described as:
the right ascension currently at the zenith
Here are the definitions of zenith and right ascension, if you need them.
So, if the sidereal time at some location is 12 hours, 0 minutes, and 0 seconds, the right ascension directly above that location at that time will be 12:00:00. So, one sidereal day is the time it takes for the sidereal time at any location to "repeat itself" - i.e., the time between one 00:00:00 and the next 00:00:00. This means that one sidereal day is the time it takes for Earth to rotate 360 degrees.
The time that full rotation takes - i.e., one sidereal day - has been calculated to be roughly 23 hours and 56 minutes, or 1436 minutes. Since you're measuring your time in solar time, you would expect to see this 4-minute discrepancy in a lot of places, because, well, 23 hours and 56 minutes is 4 minutes less than a full solar day of 24 hours. From one day to the next, most objects (the Moon is a notable exception) will rise 4 minutes earlier.
So, yes, your analysis is correct - this daily 4 minute change is not really "significant" since it is a consequence of the Earth's rotation around the Sun, and you should expect to see that.

A side note:
Additionally, we can see that the time it takes Earth to rotate one full degree is:
$\dfrac{1436\: \rm{minutes}}{360 \: \rm{degrees}} = 3.99 \dfrac{\rm{minutes}}{\rm{degree}}$
So, this 4-minute daily change is also a nearly 1-degree daily change.


Supermassive black holes at the center of galaxies The fact is, many, if not most large galaxies have a supermassive black hole at their center. My question is why? Is it because when these galaxies were first formed supermassive black holes were created out of super-dense clouds of matter, which then attracted the rest of the galaxy? Or is the black hole attracted to the center of the galaxy in such a way that due to its large mass and ability to perturb everything around it, the interactions with the galaxy tend to move the center to where the black hole is located?

It seems interesting that Sagittarius A* is at the precise center of the galaxy, when it has so much less mass than the rest of the galaxy put together. The sun being at the center of the solar system makes a lot of sense - it accounts for 99.8% of the mass of the solar system. But the mass of Sagittarius A* is on the order of one millionth the mass of the galaxy.

So, to phrase it like a chicken and egg problem – what came first, the super-massive black holes, or the galaxies surrounding them?

This is a very well posed question, thanks!
The problem is, we still don't know.
What we suppose, is that dark matter came first.

Primordial fluctuations of dark matter made the right conditions to accrete enough mass to build SMBHs, and then the surrounding galaxy.

But still we don't know if, the SMBH comes from merging, of other SMBHs or galaxies, or from high massive BH that are called seed (of the order of $100\,M_{Sun}$).
Perhaps, there are very recent developments, which could be known by the experts, but until few years ago, this was the scenario.

About the mass issue, we could have a better reasoning if we think to the influence spheres. We discovered a relation between the SMBH mass and many bulge properties (i.e., the luminosity, the velocity dispersion, and others). This means that an object of the order of $\sim1$ pc, influences an object of the order of $\sim1$ kpc (maybe $\sim3$). Plus, the bulge structure is essential to the formation and characterization and classification of the host galaxy.


Could a binary system of two planets with oceans reflect each other? If there were two earth-like planets in a binary system and one of them had landmasses, and the other one had oceans covering a significant amount of its surface, could you see the continents of the first planet reflected onto the oceans of the second one?

If so, could this also work with oceanic moons?

The surface of the ocean isn't smooth, so there would not an image formed.

Look at how the Earth's oceans reflect the sun.

(credit: NASA/ESA)

Obviously the sun is reflected by the ocean, but the rough surface of the ocean scatters the light enough for no image to be formed. It would be the same if there were a planet being reflected: No detail would be visible.


If two black hole event horizons overlap (touch) can they ever separate again? Hypothetical question based on my understanding that two event horizons that overlap (touch) can't ever separate again:

Imagine a 1 billion solar mass black hole (so the event horizon is massive and very gravitationally weak) is travelling at a velocity of 0.9c through empty flat intergalactic space; now imagine an identical 1 billion solar mass black hole travelling at 0.9c but in exactly the opposite direction so the two are heading roughly towards each other. The black holes' paths, once all the space time warping is taken into account, aren't on a direct collision but the outermost edges of the event horizons will just 'clip' each other, ordinarily only overlap for a fraction of a nanosecond as these two bodies are travelling at such incredibly fast velocities and in opposite directions to each other.

So firstly, am I right in thinking that if two event horizons overlap they can never 'unlap'?

Secondly, what would happen to this incredible amount of momentum of each other the black holes? Would it just get instantly turned into gravitational energy? Bearing in mind when black holes normally merge, it happens very slowly as black holes slowly move closer and closer together over millions of years giving off gravitational energy as that happens, so not in a fraction of a nanosecond as in this case.

And thirdly, what would this look like? Would the event horizons remain fairly spherical and the radiated energy just insane or would they stretch and warp into a kind of long thin elastic event horizon as they shoot past each other and then over time slow down and snap back to each other?

You have already got some good answers, but I'll just try to provide one more intuitive solution on why the event horizons will never separate again if overlapping each other:
First, imagine a speck of dust that comes inside the EH of a black hole. I believe we'll agree this speck can never escape the black hole, because nothing can come back from behind the event horizon.
Now, imagine the same speck of dust, but inside the overlapping parts of the EH of two black holes passing each other. This speck of dust will never escape any of those two black holes, because it is inside the EH of them both. If these black holes would be able to separate again, the speck caught between them would obviously escape at least one of the black holes, after being behind it's event horizon.
Since this can not happen, the two black holes will be united from the point their event horizons are overlapping, no matter their speed.


Are there any sounds in space? Is our universe noisy or silent? Can people hear them? When astronauts are out of the Earth and near an explosion, can they hear its noise.

There is sound, but you cannot hear it.

1) There is no sound in space:
This shows how sound gets fainter in an evacuated glass jar. The same happens with an explosion (or a shout) in space - there is no air to transmit the pressure waves, so no sound.

2) There is sound in space:
The interstellar and even intergalactic medium is not completely empty, even though it is much more empty than the best vacuum we can create on earth. Nonetheless, there can be pressure waves in those media.

 (The bullet cluster, taken from NASA's APOD)

This image shows the mass distribution of two colliding galaxy clusters in blue, and the X-ray gas in red. The galaxies in the clusters have passed through each other without much disturbance, but the intra-cluster gas has collided. The red cone to the right of the centre looks a lot like the sonic boom familiar from aircraft.

For more regular sound waves (repeating pattern as opposed to the one-off of an explosion), you might find those in the interplanetary gas / solar wind of oscillating stars (RR Lyrae or Delta Cephei type stars). I am not sure if anyone has ever measured this.

At any rate, while human ears can pick up sound in the range of ca. $10^2$ to $10^4$ Hz, these sound waves would have a much much lower frequency. Also, the force exerted by these pressure waves on a human ear would be much too small to register.


Why is the peak value(s) of the spectral radiance of the CMB different if you calculate with freq. instead of wavelength? On Wikipedia, the calculated value of the peak frequency of the cosmic microwave background is 160.23 GHz, but it says if you do the calculating with wavelengths, then convert to frequency, you get a value of 282 GHz.
Why the difference?
There used to be an answer on Quora, I believe, but I can't find it anymore...
Something to do with the summation using integral calculus, I believe...

Because $B_\lambda$ is not just $B_\nu$, with $\nu$ replaced by $c/\lambda$.
The relationship between the two functions is that
$$ B_\lambda\ d\lambda = B_\nu\ d\nu$$
since one is defined in terms of flux per unit wavelength, the other as flux per unit frequency.
Thus
$$B_\lambda = B_\nu\ \left|\frac{d\nu}{d\lambda}\right| = \frac{c}{\lambda^2} B_\nu, \tag{1}$$
where you can then substitute $\nu =c/\lambda$ into the expression for $B_\nu$.
To find where the maximum of a function, you differentiate and set to zero. Because of the above, the wavelength where $B_\lambda$ peaks is not $c/\nu_{\rm max}$. We can see that by differentiating both sides of eq (1).
$$\frac{dB_\lambda}{d\lambda} = -\frac{2c}{\lambda^3}B_\nu + \frac{c}{\lambda^2}\frac{dB_\nu}{d\nu}\frac{d\nu}{d\lambda}$$
$$\frac{dB_\lambda}{d\lambda} = -\frac{\nu^3}{c^2}\left(2B_\nu + \nu \frac{dB_\nu}{d\nu}\right).$$
The usefulness of this expression is that you can easily see that if $dB_\nu /d\nu$ is zero at some frequency, then $dB_\lambda/d\lambda$ cannot be zero at that same frequency.


Why is there no Proxima 'a' planet? I'm confused about the naming convention Why is there no exoplanet around Proxima Centauri designated with the letter 'a'?  There is, famously, a 'b' and now a newly confirmed 'c'....

I have read that exoplanets around a star are designated from 'a' to z or whatever in alphabetical order, starting from the closest to the star OR beginning with the first one discovered, then the second....

Usually, the primary host star reserves the right to have the 'a' designation. Therefore 'a' never gets assigned to a planet.

Please check the following post for a detailed explanation and some examples.
Where did TRAPPIST-1a go?


Can you identify this planet seen in the night sky? I saw a very bright star in the night sky on Monday (Jan 30, 2017). The star was in the Western sky and was directly above the crescent moon.

My location is Toronto, Ontario. Canada. It was about 8pm at night, but by 10pm the star had disappeared over the horizon.

I'm sure this was a planet as it shined brighter then any of the other stars, but I couldn't guess which one.

Can you identify this planet for me?

This Community Wiki answer was created to preserve knowledge while other answer(s) are under review.

From Dr Chuck:

  It was Venus. It should have been 10 - 15 degrees above the moon.

From Wayfaring Stranger:

  Venus is whiter than Jupiter, Saturn, or Mars, which have yellowish, yellow, and reddish tints instead. It's also usually much brighter than the other planets, including Mercury. Mercury always hangs closer to the sun than did Venus in early Feb 2017.


The expansion of the universe and how stars and galaxies still collide How can it be that all galaxies and stars are moving away from each other yet they still collide?  From what I have researched it as an open universe (or many open universes) yet the idea of stars and galaxies colliding boggles me.

The expansion is overridden on a small scale by gravity. Our galaxy is not expanding, and the stars are bound together.  In fact, a whole group of galaxies don't notice the overall expansion.