Dark Energy Expansion The new Cosmos extra features mentions that at about 6,771,500,000 years ago the universe began an accelerated expansion. How do we know this? What evidence do we have for this renewed and accelerated expansion?

From the interactive cosmic calendar on the new Cosmos bluray, dates above taken from the heading

  Until [now], gravity has been slowing the expansion of the universe,
  but suddenly a mysterious, unknown force that we call dark energy
  takes over, the expansion of the universe begins to accelerate

How do we know it was dark energy? Given we can't define or detect dark energy, why do we believe this is the cause? How do we even know this new acceleration happened? Something in the red shift? Is there a gap in the red shift in things we can see that would imply a slow expansion and then a fast expansion? If there is a gap, why is dark energy the only explanation? Or are there other theories?

Yes that is a lot of questions, but the main question is how do we know an accelerated expansion happened at that time? the rest of the questions stem from that, mainly in asking for data to back up claims, for or against

Is Cosmos seriously using that exact number? Egads... if they are, don't take it too seriously, but otherwise they're probably conceptually correct.

  How do we know it was dark energy?

In cosmology, the ΛCDM model fits numerous observations, most notably those observed by the WMAP satellite, but also others, to the Friedmann-Robertson-Walker family of solutions of general relativity, which describe a homogeneous and isotropic universe. The $\Lambda$ refers to the cosmological constant, which is the simplest model of dark energy, and is equivalent to an intrinsic energy density of the vacuum--hence, it must be Lorentz-invariant, and since time and space mix in a Lorentz transformation, the only way for that to be so is for its pressure to be equal to exactly the negative of its energy density. With $\Lambda>0$, the pressure is negative, and this is what's responsible for the outward gravitational acceleration of the universe.

  Given we can't define or detect dark energy, why do we believe this is the cause?

The cosmological constant was defined eight decades before the discovery of cosmic acceleration, and we believe it to be the cause because it's theoretically very simple and it fits observations. There are other models for dark energy, though.

  How do we even know this new acceleration happened?

Cosmological evolution is described by the Friedmann equations, which has the principal contribution of radiation, matter, and dark energy, in addition to spatial curvature of the universe. As the universe expands, both radiation and matter get diluted, but dark energy does not. Therefore, if the universe keeps expanding, the proportion of dark energy becomes dominant over both radiation and matter.

Most likely, the article is simply an exaggerated reference to the transition between the matter-dominated era and the dark-energy-dominated era. The particular number $6,771,500,000$ years seems comically precise for much of anything in cosmology, but since the nine-year best-fit WMAP+eCMB+BAO+H₀ (see above WMAP link) for the age of the universe is $13.772\pm0.059\,\mathrm{Ga}$, I propose the following explanation:

The author hears that the age of the universe is $13.772$ billion years, and completely ignores any error bounds on that estimate.
The author hears from somewhere else that the dark-energy-dominated era began when the universe was around $7$ billion years old, and again treats this number as exact.
The author then subtracts the two.

This accounts for all but the last $0.5\,\mathrm{Ma}$, which about one-ten-thousandth. I conclude that my theory of the origin of the number agrees rather well with the empirical evidence of the author's words. Perhaps the $0.5$ as the next digit was rationalized as some sort of midway-point in the author's mind.

As to where $7$ billion years comes from, I have two guesses.

In cosmology, the Hubble parameter is sometimes taken to be unitless, by scaling to $100\,\mathrm{km/s/Mpc}$. For example, we sometimes say that $h = 0.704$ when we mean that $H_0 = 70.4\,\mathrm{km/s/Mpc}$. Thus, $100\,\mathrm{km/s/Mpc}$ is kind of a de facto standard of comparison. It might be a coincidence, but it's rather suspicious that if one ignores the contribution of radiation (which is justifiable, since it's negligible except in the very early universe) and pretends that $H_0 = 100\,\mathrm{km/s/Mpc}$, then the age of the universe when the matter and dark energy densities are equal would be
$$\frac{2\ln(1+\sqrt{2})}{3\sqrt{1-\Omega_\text{M}}}\frac{1}{H_0}\bigg\vert_{\Omega_\text{M} = 0.2865} \approx 0.70H_0^{-1}\bigg\vert_{h = 1.0} \approx 6.8\,\mathrm{Ga}$$
I'm using the Ryden reference given in wiki for matter-dominated era above. Thus, I think someone did a guesstimate using a nice, round number for the Hubble parameter and rounded to $7$ billion years.
The transition between matter-dominated and dark-energy-dominated is a somewhat arbitrary matter of convention. We could take it mean when matter and dark energy densities are equal, as above, but on the cosmological scale, pressure is three times as important as energy density (cf. Friedmann equations). Thus, we could take it instead as the time when the matter density was twice the dark energy density. This would be earlier than the $9.8\,\mathrm{Ga}$ given in the Ryden reference, but arguably more appropriate because the Friedmann equations would predict zero acceleration, i.e., it would be the transition between inward and outward acceleration. I haven't calculated when that would be, but this sort of conventional fuzziness is might be responsible for the discrepancy between the wiki's matter-dominated era and the dark-energy-dominated era pages. Although the latter claims $5\,\mathrm{Ga}$ instead, rather than $7\,\mathrm{Ga}$ (but it doesn't explain the source).

So Cosmos may or may not be right about dark-energy-dominated era beginning when the universe about $7$ billion years old, but concluding that it happened $6.7715\,\mathrm{Ga}$ ago is certainly taking some numbers as too exact to be justifiable.

  Something in the red shift?

I think at this point you're better off making a separate question on specific issues with the ΛCDM model, if you have any. For the purposes of this question, I'm taking ΛCDM as a given, as almost certainly Cosmos is doing as well.

  Yes that is a lot of questions, but the main question is how do we know an accelerated expansion happened at that time?

The standard cosmological model is that dark energy is a cosmological constant. Thus, its energy density does not change. It then becomes straightforward that at some point, cosmological expansion makes matter too dilute to compete with dark energy, which is when dark energy starts dominating. It was there all along, but before then, matter was dense enough to be dominant instead.


Is there a connection between the solar wind and the cosmic radiation? Cosmic radiation consists of high-energetic particles, mostly protons, that come amongst others from supernovae and our sun, right?
The solar wind also consists of charged particles, like protons and electrons, and comes, of course, from the sun.
So I wonder whether there is a connection between these two? Is the solar wind a part of the cosmic radiation?

Cosmic radiation (which consists of high-energy protons and atomic nuclei) comes from the sun, other stars, even other galaxies.
Solar wind is a part of that - the part coming from the sun, which mostly consists of electrons, protons and alpha particles with kinetic energy between 0.5 and 10 keV


How did Kepler determine the orbital period of Mars? As I understand it, Kepler used the orbital period of Mars, along with observational data of Mars' and the sun's position in the sky to derive the orbits of Earth and Mars.  (As described, here: https://faculty.uca.edu/njaustin/PHYS1401/Laboratory/kepler.pdf)

But how did he determine the orbital period of Mars without already knowing the orbit of the Earth?

He knew the orbital period of Earth. You can do this by observing the sun's motion in the sky over a year. (And then you know that the orbital period of the earth is a year.)

Now from earth, you can observe Mars from one Opposition (https://en.wikipedia.org/wiki/Opposition_(planets)) to another. You can find out the time ($T_\text{relative}$ it takes for Mars to do this, using which, you find the relative angular velocity between the orbital motions of earth and Mars ($\omega_\text{relative}=2\pi/T_\text{relative}$). This angular velocity is the difference between the angular velocities of earth and mars. Using this and knowing the orbital period of earth, you can find the orbital period of mars.

$2\pi/T_\text{relative}=2\pi/T_\text{earth}-2\pi/T_\text{Mars}$

Hope this is clear.


Does coronal mass ejection include radiation at optical wavelengths? It is known that during Solar/Stellar Flare, plasma from the corona of the star is released into the space, and travels comparatively slowly.

In an hypothetical situation, does the coronal mass also consist of electromagnetic radiation within the optical wavelengths? And if so, is it possible to observe the incoming coronal mass ejection from the ground/surface of the Earth way before it approaches?

Coronal mass ejections consist of a very hot, but thin, plasma. Their very weak intrinsic emission would be dominated by ultraviolet and X-ray lines and bremsstrahlung continuum. There is very little optical radiation.

However, CMEs can and are monitored at optical wavelengths using the light that they scatter from the Sun. The process is Thomson scattering from the free electrons in the plasma. This is done by spacecraft because the Earth's atmosphere makes it very difficult to achieve the contrast necessary to observe the scattered light close to the Sun (except during total solar eclipses).

A CME takes a couple of days to travel from the Sun to the Earth and such monitoring can give "space weather forecasts" associated with magnetospheric phenomena triggered by the impact of CMEs.

An example picture taken by the LASCO white light imager on board the SOHO spacecraft is shown below. Everything here is viewed in scattered light. The Sun is obscured by an occulting disk.I have picked an example showing the eruption of a CME.


Succinct explanation of black hole mass, diameter, shape? I don't have any formal education in physics and I have found it hard to find a succinct and somewhat simple to comprehend explanation of what we think is the actual reality of a black hole. I'm asking after seeing a statement saying that the mass of the central supermassive black hole in the Phoenix Cluster was 20 billion solar masses and the diameter we think to be approximately 118 billion kilometres across.

Is someone able to point me to a resource or give me a simple explanation as to what we think a black hole actually is in shape? Is it a solid sphere with a central point and a radius of approximately 59 billion kilometres? A spinning disc with a face that is 118 billion kilometres across and if so what are the other measurements?

Am I thinking about this all wrong with regards to trying to understand it as a shape at all? If I understand correctly, mass doesn't give any indication of size as it is dependent on density. Is a black hole just basically a somewhat spherical body of mass like any other that just has an absolutely massive density and spin?

I should start off by saying that black holes are enigmas. In describing our universe, we have the General Theory of Relativity (GR) to help us describe gravity and objects which are very massive. We also have quantum mechanics (QM) and quantum field theory to help us describe small objects such as particles. For the most part, these two realms are completely separate and these two important theories don't mix. After all, you don't need QM to describe the physics of orbiting the Sun, and you don't need GR to describe electrons orbiting an atomic nucleus.

A problem arises though, when you begin talking about objects which require both QM and GR, and black holes fall very nicely into this regime. Often we find that when we mix the two theories, we wind up with nonsensical answers. It is highly probable that our understanding of the mechanics of a black hole is incomplete. What we really need is a quantum gravity theory to talk about black holes and as yet we don't have one.

That being said, let me lay out the basic structure of a black hole based on what current theory suggests to be true. Fortunately, black holes are relatively simple in structure, as dictated by the No-Hair Theorem (which basically says you only need a few parameters to completely describe the state of a black hole, as opposed to what you'd need to fully describe the physics of a person or even an atom). The (somewhat simplified) anatomy of a black hole is shown below.

In this diagram there are a few important parts. The first being the singularity. When you put a bunch of mass together, gravity acts to squeeze that mass into a ball. That is what drives the fusion in our Sun and makes our planet round. Normally there is some resistive force that stops the matter from collapsing completely, such as the structural integrity of rocks in the Earth or fusion in the Sun. At some point though, if you pack on enough mass, that resistive force can't hold out and gravity wins. In this case, all the matter collapses down as far as it can go. It becomes a point of matter known as a singularity. This singularity has no size, it is literally (and mathematically) a point. All the mass of the black hole exists here. So when they say a black hole is 20 billion solar masses, they mean that the singularity contains the matter equivalent to 20 billion Suns, all confined to that single point.

The other important concept for a black hole is the event horizon. I should stress that the event horizon is not a physical object in space. It is a mathematical boundary. Without getting into the nitty-gritty and in basic terms, the event horizon is where our current physics "stops working". We cannot describe inside the event horizon with our current physics because we are in that region where both QM and GR are important and they don't play well together. The event horizon is also the boundary where, if crossed, you're trapped forever in the black hole. There's a lot more physics involved with the event horizon, but that's the gist. Usually when someone refers to the size of a black hole, they're talking about the event horizon. Thus if they say the black hole is 118 billion kilometers across, they mean that the event horizon has a diameter of 118 billion kilometers. Note, from the image above, that the radius of the event horizon is easily calculated from only the mass of the singularity (and a few physical constants).

There are further components to a black hole such as the ergosphere or photon sphere, and often black holes have accretion disks of material around them. I'll leave those concepts to another question though.


The light of which star is next to reach the earth? I heard that the sky wouldn't be dark at night if the light of every star had already reached the earth.

Of which (currently invisible) star will the light reach the earth next and when will that be?
What was the last star that we can now see at night to reach the earth and when was that?
How can we know (1) if the light hasn't even reached us yet? Or can't we know and will a star suddenly appear without anyone being able to predict that?

I think you may be thinking of Olbers' paradox
. This supposes that if the universe were infinite in time and extent, and stars were more or less randomly placed, then every line of sight would end at a star, and so the sky would be as bright as the surface of a star.

The paradox can be resolved by supposing that the universe is not infinite in time, but had a beginning.

We don't see stars suddenly appearing as their light reaches us. What we see is stars in all stages of their development. When we look the nebulae where stars are born we see clumps of gas in the process of collapsing to form stars. These are called protostars. Now the process of stellar formation takes a long time: about a million years for a medium sized star. The start of a star is not a sudden "turning on". Instead, as the core of the protostar shrinks and heats up, nuclear fusion starts; slowly at first, then increasing until it is able to prevent further collapse and heating.

Protostars usually are dim in visible light, they can be found with infra-red telescopes.

Since the process is slow and there isn't a sudden turning on, your questions cannot be directly answered. However, there is an object HOPS383 which is 1400 light years distant, which between 2004 and 2008 had a "growth spurt" and became visible in infra-red for the first time. It can't be seen in visible light yet - so in a sense this protostar answers your question 1 and 2, in different wavelengths.


Can I use the parallax coefficients for observatories as a proxy for latitude using arctan2(sin, cos)? In How are the three character IAU Minor Planet Center's Observatory Codes assigned? Why are some letters so popular? I looked at the distribution of the first alphanumeric character versus the value of the second two (36 x 100).

The table lists longitudes but no latitudes for these observatories, but using the cos and sin parallax coefficients in $\arctan2(\sin, \cos)$ and plotting as a function of tabular longitude produces a map that looks like it could be the distribution of telescopes on Earth.

Question: What exactly are the cos and sin parallax coefficients, and how closely can $\arctan2(\sin, \cos)$ be expected to serve as a proxy for latitude? Within a degree or so?

import numpy as np
import matplotlib.pyplot as plt
# blob is the unformatted lines from https://www.minorplanetcenter.net/iau/lists/ObsCodes.html
threes = [line[:3] for line in blob.splitlines()]
key = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ'
a = [[x] for x in key]
for t in threes:
    a[key.find(t[0])].append(t)
b = np.zeros((36, 100))
for i, thing in enumerate(a):
    c = [int(x[1:]) for x in thing[1:]]
    for d in c:
        b[i, d] = 1
plt.imshow(-b, interpolation='nearest', cmap='gray')
plt.gca().set_aspect(2)
plt.show()
pairs = []
todegs = 180/np.pi
for line in blob.splitlines():
    try:
        cos, sin = [float(x) for x in (line[13:21], line[21:30])]
        lat = todegs * np.arctan2(sin, cos)
        lon = float(line[4:13])
        pairs.append([lon, lat])
    except:
        pass
print(len(pairs), len(blob.splitlines()), float(len(pairs))/len(blob.splitlines()) )
lon, lat = np.array(list(zip(*pairs)))
lon = np.mod(lon+180, 360) - 180
plt.figure()
plt.plot(lon, lat, 'ok', markersize=1)
plt.ylim(-90, 90)
plt.xlim(-180, 180)
plt.gca().set_aspect('equal')
plt.show()

The title of the question reads:

Can I use the parallax coefficients for observatories as a proxy for latitude using arctan2(sin, cos)?

and the body expands:

Question: What exactly are the cos and sin parallax coefficients, and how closely can $\arctan2(\sin, \cos)$ be expected to serve as a proxy for latitude? Within a degree or so?

First the "How closely can (they) serve as a proxy for latitude?" part
tl;dr: to within +/-0.2 degrees; the range in the differences between geocentric and geodetic latitude.
The table provides the sin and cos of the same angle $\phi'$ multiplied by $r/R_E$ and the 2nd half of the answer will explain that $\phi'$ is the angle that a radius drawn from the observatory to the Geocenter makes with respect to the Earth's equatorial plane. Everyday latitude or GPS latitude which is a kind of geodetic latitude (what I want to use for my map) is a little bit different because it's drawn on an ellipsoid.
How different?
Well it turns out that I wrote an answer about geodetic latitude about 18 months ago. (@DavidHammen wrote a nicer answer elsewhere as well) Here is an excerpt from my answer:

The 3D cartesian coordinates $X, Y, Z$ in Earth-centered, Earth-fixed coordinates assuming an ellipsoidal shape is given by:
$$X =  \left(N(\phi) + h \right) \cos\phi \cos\lambda $$
$$Y =  \left(N(\phi) + h \right) \cos\phi \sin\lambda $$
$$Z = \left(\frac{b^2}{a^2} N(\phi) + h \right) \sin\phi $$
where $\phi, \lambda, h$ are latitude, longitude, and altitude, and $a, b$ are the equatorial and polar radii of the ellipsoid used, and
$$N(\phi) = \frac{a^2}{\sqrt{a^2\cos^2\phi + b^2 \sin^2\phi}}. $$
Lat, lon, alt in "GPS coordinates} is based on WGS 84 with $a, b$ of 6378.1370 and 6356.7523 kilometers, respectively.

The parallax coefficients are $\rho \cos(\phi')$ and $\rho \sin(\phi')$ and $\rho$ is the distance from the observatory to the Geocenter divided by "the Earth radius". That makes them:
$$\rho \cos(\phi') = \frac{\sqrt{X^2+Y^2}}{R_E}$$
$$\rho \sin(\phi') = \frac{Z}{R_E}$$
So if we chose a geodetic latitude $\phi$, choose a longitude of zero so that $Y=0$ and calculate $X, Z$ and plug them into this:
$$\phi' = \arctan2\left(Z, X \right) = \arctan2\left(\frac{Z}{R_E}, \frac{X}{R_E} \right) = \arctan2\left( \rho \sin(\phi'), \rho \cos(\phi') \right)$$
we can get the geocentric latitude, and then calculate the error by subtracting the two.
It turns out to be about +/- 0.2 degrees with extrema at about +/- 45 degrees. It is of course zero at the equator and poles.
So to an error of 0.2 degrees, yes that can indeed be used as a proxy!

Now the "What exactly are..." part
As pointed out in comments, the link explains how the cos and sin parallax coefficients are defined:

The following list gives the observatory code, longitude (in degrees east of Greenwich) and the parallax constants (rho cos phi' and rho sin phi', where phi' is the geocentric latitude and rho is the geocentric distance in earth radii) for each observatory. It is updated nightly.

What is geocentric latitude? It's the "easy one" to understand; it's just the angle measured at the Geocenter between the radius to the point and the closest point on Earth's equator. So while the definition *does not specify a specific Earth radius, it doesn't have to because the table only gives the unit-less parts cos phi' and sin phi'.
But why both sin and cos? Can't you just use sin between -90 and +90 degrees?
To just define an angle (without the radius), mathematically you might think so. However sin is insensitive near the poles and in the world of real computing numerical errors can crop up. Near the poles sin will still provide the appropriate sign and cos will provide the sensitivity. This way you will have low numerical errors (for this particular step) everywhere.
But this is the latitude and distance together, so you have to have two numbers. They could be a latitude angle and a radius explicitly, but this also works. (revised per @DH's comment)
But what are the parallax coefficients for exaclty? What are they really?
I don't know, but hopefully someone with a valid "astronomer's ID card" will come along and post an answer. I'm to scared to ask a new question on this topic given all the comments under the question! :-)

Script for the plot
import numpy as np
import matplotlib.pyplot as plt

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
a, b = 6378.1370, 6356.7523 # only their ratio matters in this case
todegs = 180/pi

phi = np.linspace(-halfpi, halfpi, 101)
h = 0.
lam = 0.
N = a**2 / np.sqrt(a**2 * np.cos(phi)**2 + b**2 * np.sin(phi)**2)
X = (N + h) * np.cos(phi) * np.cos(lam)
Y = (N + h) * np.cos(phi) * np.sin(lam)
Z = ((b/a)**2 * N + h) * np.sin(phi)

phi_prime = np.arctan(Z/X) # two quandrants so no need for arctan2

plt.figure()
plt.plot(todegs*phi, todegs*(phi_prime-phi))
plt.xlabel('geodetic ("GPS") latitude (degs)', fontsize=12)
plt.ylabel('geocentric - geodetic latitude (degs)', fontsize=12)
plt.xlim(-90, 90)
plt.show()


Sidereal time and right ascension I'm new to astronomy and have to make a star map that displays the stars above a current location. From the information I've pieced together I understand that declination is equal to the current latitude of the observer, however I am slightly confused about right ascension. My two quesions are;

If I work out GMST using the equations here; http://www2.arnes.si/~gljsentvid10/sidereal.htm will the answer be the current right ascension above Greenwich or are there more steps I've missed?
If I want to find the sidereal time of, for example, Sydney, Australia, do I need to account for daylight savings at either location or do I simply add 11 hours to GMST?

Thank you for your help.

I can't vouch for the formula, but you can try it out and check with some internet source, eg USNO, to see if it works.  If the formula is correct, that's it.

As for the sidereal time in Sydney, you need the difference in longitude between Greenwich and Sydney, and use the conversion rate of 15 degrees longitude is equivalent to 1 hour.


Does the accelerating expansion of the Universe contradict Hubble's law? Hubble's law gives a linear relationship between the distance to a galaxy and it's recessional speed.

Observations of distant type 1a supernovae showed that their red shift (and therefore their recessional velocities) were lower than expect, implying that the rate of expansion of the Universe was lower in the past.

However, doesn't this mean that Hubble's law only applies for short distances, as at great distances a plot of recessional velocities against distance would not be linear (as I have crudely tried to show on the graph below)?

The Hubble law gives the velocity of a distant galaxy right now. A galaxy at a distance $d$ recedes at a velocity $v = H_0\,d$ right now$^\dagger$.

However, the relation between $d$ and the redshift — which is the quantity that we observe — is a non-trivial function of the expansion history of the Universe, obtained by integrating the (inverse) scale factor over time, which in turn depends on the densities of matter, radiation, and dark energy.

Analogy

Paint dots on a rubber band, hold onto one end, and let your friend walk away with the other end. It doesn't matter whether your friend changes pace, maybe even stops and walks toward you for a while, then starts sprinting away from you. At any given time, the velocity of a given dot with respect to you is simply proportional to its distance from you.

$^\dagger$Strictly speaking, this is only true if our assumption is true that the expansion rate of the Universe is the same at every location, which is a corollary of the Cosmological Principle (i.e. that the Universe is homogeneous and isotropic).


Does a kilonova leave a high mass remnant? My understanding of Type Ia supernovae is that they are expected, in most cases, to destroy the white dwarf(s) that went in to them, leaving behind no high density remnants (i.e. no white dwarf, neutron star, or black hole). Black hole/black hole collisions are expected to leave behind a black hole, of course, with less mass than the sum of the masses of the black holes that went in to the collision. Do we expect a kilonova to leave behind nothing but gas and radiation or some kind of stellar remnant? If it leaves behind a remnant, what class to we expect it to be (white dwarf, neutron star, or black hole) and what mass? The addition of mass seems redundant, but the dividing lines between the masses of these objects are based on upper limits on the mass of the less dense class (Chandrasekhar limit for white dwarfs, Tolman–Oppenheimer–Volkoff limit for neutron stars), and don't actually apply as lower limits for the mass of the high density class. For example, neutron stars are sometimes referred to as giant nuclei, which would put the lower limit of their mass at $1$ or $2$ atomic mass units, depending on whether the presence of a neutron and stability to radioactive decay are requirements. Yes, those times people are being poetic due to the difference in stabilization mechanisms (gravity vs nuclear forces), but the point remains that it may be possible for a neutron star to be theoretically stable at less than $1.4M_\odot$. The only lower limit to black hole mass I know of would be the lifetime limit from black hole evaporation.

Put another way, do we think the density at any point in either of the inspiraling neutron stars gets high enough to form an event horizon? If that happens, this seems like a plausible way for producing black holes that are smaller than those produce by core collapse supernovae, kind of like how critical mass for atom bombs can be achieved by either bringing together enough fissile material or compressing the available material enough (e.g. the fat man detonation mechanism vs the little boy one).

I think it is generally expected that the merger of two neutron stars will lead to the formation of a black hole. What is more uncertain is whether that black hole forms straight away, or an intermediate state of a hyper(/supra)massive neutron star forms (see, e.g. Sec IIC of Hotokezaka et al, PRD, 044026, 2013 for a definition of hypermassive and supermassive neutron stars), which then collapses to a black hole after a short time. A few papers looking at the stability of such a hypermassive neutron stars formed from a binary merger are - 1, 2, 3 and 4 (check out the references and citations to them for more info). In the first three reference I think the collapse to a black hole is expected within a second (see Table 2 in Hotokezaka et al, PRD, 044026, 2013 for expected lifetimes before collapse times for various neutron star equations of state and masses), whilst in the final one a range of collapse times from fractions of a second up to several tens of thousands of seconds are found.
A search for a post-merger remnant from the binary neutron star merger GW170817 is now available here. In the introduction of this paper it describes in more detail some of the post-merger remnant possibilities that I mentioned above. The searches don't find any evidence for a signal, but given their sensitivities this isn't surprising (they'd require more than the total mass of the system to have been converted to gravitational waves to see anything!)


Could an ultra-hot Jupiter flare? Some of the ultra-hot Jupiters (WASP-33b, KELT-9b) have dayside surface temperatures similar to low-mass stars. Would they be capable of producing flares? To be specific, I am asking about the possibility of flaring on the planet itself, rather than the planet inducing flares on the star.

Flares in space are known for coming from stars. Sunlike stars, red dwarfs, etc, almost any kind of star has flares.

Hot Jupiters can be the size of Jupiter,(hence the name). They can be bigger; however, if they get too big, they become brown dwarfs. This means that hot Jupiters are too small to have flares.

In addition, many hot Jupiters have unstable atmospheres. They are so close to their parent stars, that they lose their atmospheres. No atmosphere, no flares.


How are the atomic clocks synchronised between worldwide VLBI telescopes? I know that VLBI needs precise synchronised atomic clocks to calculate the baseline distance between the radio telescopes at cm level. But i cant´t find any information how the atomic clocks are kept synchronised over long distances (several thousand kilometres around the world). As far as i know GPS is not accurate enough.... How are atomic clocks kept synchronised on all VLBI telescopes ?

Thanks a lot :)

They don’t have to synchronize the atomic clocks at each telescope.  Instead, they synchronize the collections using GPS (so the telescopes observe the same target in the same frequency range at the same time), as shown in the diagram from this paper.

When the data is brought together, they use differences between the arrival times of quasar data to calculate relative antenna locations.  From this paper from NASA:

During geodetic VLBI observations, signals emitted by distant sources
of radio frequency energy (quasars) are received and recorded at
several antennas. At each antenna (VLBI station) a very stable
frequency standard (hydrogen maser) provides a reference signal that
enables time tagging of the quasar signals as they are being recorded.
For each VLBI experiment, correlation of the time-tagged, recorded
information from the participating antennas yields the differences
among the arrival times of any specific quasar radio wave at the
antennas. These time differences are used to calculate the locations
of the antennas with respect to each other.


Stratospheric Balloon with an Xray camera Some colleagues and I were planning to launch a stratospheric balloon in December. A thought occurred to me: what if we were to put some sort of xray camera in the payload? We are expecting a maximum altitude of about 80,000 feet. Would it work any better at that altitude?

Long ago I was part of a group that flew high altitude balloons to study x-rays from pulsars etc. How much you detect and at what energy depends upon how much atmosphere is above your detector. Lower energy x-rays are absorbed more than higher energy x-rays. Realistically you won't detect anything except maybe the Sun at energies below 20 keV and at altitude below 25km. We aimed for altitudes near 40km. To detect high energy x-rays the cheapest method is to use a proportional counter. Not something you can buy off-the-shelf.

Also the Sun only emits a lot of high energy x-rays during solar flares.

An interesting experiment would be to fly a geiger counter see http://earthtosky.net/


Which direction do the constellation revolve in? If I am standing looking towards the East from the Northern Hemisphere, do the constellations appear to rotate towards the West via the North or the West via the South or am I completely off target all together?

Why do we say Capricorn is in December at the Solstice? Where do we see it in the sky in the Northern Hemisphere in December and likewise would we see Aquarius in that same position in February etc..?

The sky appears to rotate around the Poles. Stars that are in the northern Hemisphere will move in circles around the Pole. Some Southern hemisphere stars are also visible from the North, the move in circle around the southern pole (Although the star is sometimes visible the southern pole is always below the horizon from the Northern hemisphere)

A star that rises due East, will reach its highest position in the sky when it it is due South. It won't enter the Northern half of the sky.

Astronomers don't tend to talk much of Capricorn in December. (That leads to astrological nonsense) But it means that the sun in in the same region of the sky as Capricorn at that time (which means that you certainly can't see it) In fact the best time to see Capricorn is in Early Summer, when it is as far from the sun as possible.


Can we know where (South, East, West...) the moon will be at a specific time? Apparently, a few nights from now (more specifically on the night between the 27th and 28th of September this year, nights being Belgian nights), we will have a lunar eclipse.

That's great, and I wish I could find a nice spot to go see that with friends!

Sadly I never took a lot of time watching the moon with a compass in my hands, so I'm wondering:

Can we know at what orientation (as in North, South etc) will the moon show up from Brussels between 12PM and 4AM GMT in the night of 27th to 28th September?

What are the calculations and/or data source involved?

A planetarium programme will tell you exactly where to look, and can be invaluable when looking for the location of planets, but the Moon is very easy to find in the sky, even after several glasses of Flanders Red Ale.

So what you really want to know is a) is the eclipse visible from Belgium and b) roughly where in the sky will the moon be.

This map shows that the Eclipse will be fully visible from Belgium. So that's good.

The the Moon will be full (a lunar eclipse always happens at full moon) so the moon will rise in the East as the sun sets, and travel into the West. At Midnight (12 AM, but many people make that mistake) the Moon will be in the South, and at 4AM it will be South West, and quite low in the sky. But even eclipsed the moon hard to miss!

Don't go out too early. Nothing much will be visible until the "Umbral eclipse" starts at 1:07 UTC (=3:07 am in Belgium)

I wish you clear skies.


A question related to black holes Not a physicist just a simple guy, don't know much about physics as well. Just like to read things. I was reading about black holes and a question popped in my mind.

If I understand it correctly then the black hole is created by collapse of a super massive star at the end of its life. This star will have a finite mass. Then what leads to the black hole to have such massive gravity so that nothing can escape it (because if it was a star before with a finite mass then it would also have some finite gravity, right)?

And then they start absorbing more and more matter. So they grow bigger and bigger. Do black holes die?

Do all galaxies have a black hole at their center?

Thanks & Regards

Update-1

Found answer to the question, do black holes die?

Update-2

Found answer to the question, Do all galaxies have black hole in the center?

Posting the answer link here so anyone having this questions can find all answers in one place.

Answering your first question:

  If I understand it correctly then the black hole is created by
  collapse of a super massive star at the end of its life. This star
  will have a finite mass. Then what leads to the black hole to have
  such massive gravity so that nothing can escape it (because if it was
  a star before with a finite mass then it would also have some finite
  gravity, right)?

You are right that a black hole will have a finite mass that is smaller than that of the massive star that collapsed (some matter is carried off by the supernova). At large distances the gravitational field of this black hole will be no different than that of a star with the same mass.

The difference arises because in a black hole this mass is compressed into a much smaller volume. Consequently, you can get much closer to the concentration of mass, than you would be able to for a star (where the mass is scattered over a relatively large volume). It is here where the gravitational fields can become really "strong" and prevent light from escaping.


Astronomical data convertion from Jy/pixel to MJy/sr? I'm in the middle of collecting infrared data of various wavelengths. My problem is the following: SPIRE data (250µm) are in MJy/sr and PACS data (100µm) are in Jy/pixel. I would like to combine these data, but I'm a bit lost with these units... So I would have several questions:

How can we switch from Jy/pixel to MJy/sr?
Is "Jy/beam" equivalent to "Jy/deg^2" or "Jy/sr"?

For both conversions, you need a bit more information about the data.

How can we switch from Jy/pixel to MJy/sr?   The Jy -> MJy part is just a factor of $10^{6}$ Jy/MJy.  The pixel -> sr (steradian) part requires that you know the angular size of a pixel on the sky.  For PACS it looks like the pixel size is different for different wavelengths; see Table 3.1.  Once you know the pixel size in arcseconds, then the conversion is 206265 arcseconds / radian (or more precisely, 3600*180/pi).  Square the size in radians to get the area in steradians.   You also may be able to find the pixel size for your data by looking in the FITS header, if you have FITS images.

Is "Jy/beam" equivalent to "Jy/deg^2" or "Jy/sr"?  No, but you can convert between the two.  Both are flux per angular area on the sky.  You need to know the angular area of the telescope beam (e.g. in square degrees, or steradians) and then you can convert between the two.

The README for the ATLAS survey says this about the conversion in their SPIRE maps:
"To convert the maps from Jy/beam to Jy/pixel, in order to carry out aperture photometry,
the values in the maps should be divided by the ratio between the beam area and the pixel
area in arcsec^2 (469/36, 831/64 and 1804/144 at 250, 350 and 500 microns respectively).
The mean of the maps should also be subtracted. Users interested in the most accurate
aperture photometry of extended sources should consider scaling the SPIRE fluxes using the
K4 corrections for extended sources. They should also consider making an aperture
correction."


Generalised planets? There is somewhat of an abstract way of generalising the notion of planets.

Standard definition of planets is, obviously: "planets are the objects formed from the residual material surrounding a newly formed main sequence star through a complex formation process".

Let me introduce a bit more general, by no means commonly known, definition of "generalised planets as bound macroscopic objects formed from some former accretion process taking place near some gravitating body over many orbital periods".

Clearly, the latter definition is more general, as it doesn't specify the type of accretion event and the type of the central body. Now, for example, the accretion mechanism can be very different from that in protoplanetary discs: consider, for example, accretion on a compact object (a neutron star or a black hole), accrection on supermassive black hole, accretion from remnants of binary neutron stars merger, or any other possible accretion process.

As noted in some answers, some stars also form in accretion-involving processes in the vicinity of other stars. To avoid including stars into 'generalised planet' category, the definition specifies that the objects should be formed over many orbital periods, implying that they should be sufficiently bound with respect to the primary.

The question: Are there generalised planets other than planets/exoplanets?

The main flaw with your definition of a "generalised planet" would be the boundary between stars and planets. And, in particular, the status of brown dwarfs, that are kind of "failed stars", that could not start to burn their hydrodgen.

With this regard, the definition of a planet as it is now is probably the best one (or the less worst one), because it is then possible to draw a line between planets and stars (and in particular brown dwarfs), because a planet is then an object formed by accretion in the debris material of a forming/newly formed star, whereas a star is, in general, an independent object formed by collapse and accretion (there are now observations of forming brown dwarfs; see for example the review by Luhman (2102) or the recent paper by André et al. (2012)). So you have two types of celestrial objects, well defined by two types of formation mechanisms (core accretion for the planets vs. gravo-turbulent collapse for the stars (brown dwarfs included). Then you can also clearly distinguish between "real" free-floating planets (that are then planets expelled from their orbit and floating around in the galaxy) and brown dwarfs (that form independently, where they lay).

Bottom line: if your object, in any environment (including any type of accretion disk) is form by core accretion: it is a planet. If your object is form by gravitational collapse (with a later, unavoidable phase of accretion onto the central object): it is a star.


Do novas create a nebula? Could a planet survive if its star exploded in a nova? Does a nova, caused by runaway fusion in hydrogen accreted onto a white dwarf from a companion star, create a visible nebula?

If a planet were in a circumbinary orbit around a binary pair that produced a nova, would it survive?

A nova may leave behind a nova remnant: a shell of gas, expanding at about 1000km/s. In recurrent novae, they may be lit by light echoes from subsequent eruptions. They are much less massive than planetary nebulae, and at much lower energy than supernova remnants.

Novae displaying nebula shells or remnants include: GK Per, RR Pic, DQ Her, FH Ser, V476 Cyg, V1974 Cyg, HR Del and V1500 Cyg.

Planets in such a system would be bathed in high energy gas, this would not be enough to vaporise the planet, but it would make life on such a world extremely difficult.


How does one calculate inclination from a graph of radial velocity? If a binary star system has a non-zero inclination, how would that show up in a graph of radial velocity?

How could I then extract the inclination from the graph?

From its RV graph alone, with no other information, you cannot calculate inclination.  This is why RV velocity measurements are typically reported as "$v \sin{i}$", because what you're actually measuring is the orbital velocity projected along the line of sight.  Without other information, you cannot disentangle the orbital velocity from the viewing angle.

What additional information could be had that would allow you to disentangle?  The following list is probably not exhaustive.

If you can observe the two stars as separate objects, then you can track their relative positions through an orbit and can measure inclination that way.
If you have "light curves" (brightness measurements over time) for the star(s) and you see periodic eclipses corresponding to the period and correct epoch from the radial velocity data, then you know the stars are aligned such that they periodically have their light blocked by one, the other, or both.  In such a case, you know that the inclination must be close to 90 degrees else the eclipses could not occur.
If you have reliable distance measurements to the stars and determine somehow that they are main sequence stars, you can determine the star's masses via theoretical stellar evolution models and calculations given a measurement of absolute luminosity (from a measurement of the apparent luminosity coupled with the distance measure).  Once you know the star's masses, you can calculate the expected orbital velocity, compare with your $v\sin{i}$ measure, and determine the inclination.
If the stars are not main sequence stars in the above scenario, then you'll need some additional information (their ages should be sufficient) to correctly extract a mass from their apparent luminosity and color.


How do we know that 2MASS J0523-1403 is a red dwarf? I was reading about red dwarfs, and stumbled upon this article about the least massive red dwarf, and noticed several peculiarities:

The spectral type is L2.5
The luminosity class is V
The mass is 67.54±12.79 Jupiter masses, or 0.0644±0.0122 solar masses.

The first bullet point states that the spectral type is L, a brown dwarf class. Sure, some red dwarfs have spectral type L0-L2, but L2.5 seems a bit too much. Also, the luminosity class is V (main sequence dwarf), not VI or sd (subdwarf). The mass is also way below the red dwarf limit (~85 Jupiter masses). Even if the uncertainty was at its maximum, the mass would fall short by about 3 Jupiter masses.
So, how do we determine/prove that a potential red/brown dwarf is truly a red or brown dwarf? Also, has this 2MASS object been proved to be one or the other?

The brown dwarf "limit" is about $0.072 M_{\odot}$ at solar metallicity (e.g. Chabrier et al. 2000) and is composition dependent. It gets a little higher in metal-poor gas and a little lower in metal-rich gas. $0.064 \pm 0.012 M_\odot$ (the third significant figure is superfluous) is within one error bar of that limit, which in itself is only a 68% confidence limit, not an absolute limit of what is possible. Therefore this object could be a star or more likely, it is a brown dwarf.
The relationship between spectral type and mass is not well established. It requires researchers to actually measure the mass of objects near the stellar/brown dwarf boundary. A spectral type of L2.5 would a priori indicate that this was probably a brown dwarf, and that is indeed what the mass measurement suggests.
The paper referred to by the wikipedia article you cite, in support of a L2.5V spectral type, is by Dieterich et al. (2013). Those authors say the spectral type is L2.5. The wikipedia authors made up the luminosity class.
To establish whether something really was a brown dwarf I think you would have to detect (or not) neutrino emission from its core. A sort-of proxy is to look for lithium. (Old) objects with lithium must have a mass lower than about $0.06 M_\odot$ and are almost certainly brown dwarfs. Objects that have depleted their lithium, which fuses at lower temperatures than hydrogen, are most likely stars, but could just be brown dwarfs. For example, I imagine that this object would have depleted its lithium, so it doesn't really tell you one way or another.
In the end, it isn't hugely important to know. There isn't much difference between a ball of very slowly contracting gas that is approaching its minimum degenerate radius and a slightly more massive ball of gas that has a little bit of fusion going on and a stable radius that is ever so slightly larger.
Postscript: I just checked and the mass determination comes from Filipazzo et al. (2015). This is not a dynamical mass estimate. They use evolutionary models and their estimate of the bolometric luminosity of the object to estimate the mass, assumimg that the age is somewhere between 500 Myr and 10 Gyr. The error bar here is just the range given by these age estimates (an older object would be higher mass). The error bar could be hugely underestimated. When dynamical masses for low-mass stars/brown dwarfs are measured, they are usually considerably larger (by 30% or more) than obtained from evolutionary models, for poorly understood reasons (e.g. Ireland et al. 2008). So from that point of view, maybe this object is a star (with a larger mass).


Why would discovering life on another planet be important/matter to us? I had a funny thought, and would like to pose it to you:

When I was a kid, reading any kind of astronomy or similar books would all say that there was no other life in the universe, and made it pretty clear that anyone thinking there could be was probably off his rocker.

Nowadays, it seems this has become a very important question for astronomers and other scientists, a new deep-seated belief that must now be proven. Some seem to think it would be the most important discovery of all time.

I'm wondering why?

My guess is, if we did 'discover' life on another planet, it'll make the news for a bit, and then the average person will go back to their every day life without much more thought about it.

For me, I would think this is great, only insofar as it means a possible new source of food (i.e., some delicious exotic cuisine assuming it's edible, and possibly even some place to go conquer and colonize if there was any profitability/money in doing so).

For some scientists, though, I get the sense it's more of an axe to grind with the long-dreary he-said she-said argument of science vs. religion.

Why do you think it would matter?

My question assumes any life discovered is probably not going to be very 'interesting' life, at least nothing we're going to have a conversation with.

It will be very important in several different ways. Not as in Messiah having descended to the World, and maybe not during one single human lifetime. But it will profoundly change the world in our minds, and thereby change our collective long term actions. The next generation would grow up taking alien life in space for granted, their minds will be more different than we old guys can change our stiff brain flesh. Like Newton's physics dominate our everyday life now, even for those of us who don't understand how all of physics has been deduced.

If life survives in other places than Earth, then maybe we (as in life from Earth) can do so too in the future.
If there's life in other places than Earth, then maybe we came from it? All life on Earth is family, all living things are closely related with each other. The question of our origin has always had high priority in human minds, see all religions.
If alien life is biologically similar to that on Earth (as in using DNA and proteins) then we could use it as a huge laboratory for medicine and environment control and agriculture and what not in life science. Astronomy today contributes to physics in our everyday life by studying a laboratory (the universe) with scale far beyond what humans could ever construct. If biology could do the same, we would soon notice the results in the shape of useful products.
If alien life does not use DNA and proteins but is very different from us, then I can't even guess what discoveries would come out of that. Zweistein.
If alien life is not found, even when we have searched long and hard, then we can gradually make the greatest discovery of all times: That we are unique and special. Searching for ET without finding it would be the most mind boggling thing imaginable. I don't think this hypothesis has been well addressed yet, I suppose no philosopher knows even where to start.


Schmidt-Cassegrain telescope What is a Schmidt-Cassegrain (SC) telescope?

We have Newtonian telescopes and also Galilean ones. How is SC telescope different from other optical telescopes?
What is its construction detail and what different mechanism does it work with?

There are many other variations on telescope designs others than the ones you have mentioned. They differ in the light path, the shape of the optics utilised, the use of mirrors and/or lenses, and the physical structure.

A Schmidt-Cassegrain is a particular type of Cassegrain reflector. A Cassegrain reflector uses two mirrors, a primary and a secondary. Both are curved, and the secondary is placed in front of the primary, which blocks some of the light to the primary. The primary directs light onto the secondary, and the secondary directs light through a hole in the centre of the primary. This 'folds' the lightpath, resulting in a short telescope, even at quite wide apertures.

A Schmidt-Cassegrain is a Cassegrain with particular design details. The primary mirror has a spherical curve, and the secondary usually has a spherical curve, though some manufacturers modify the design with a parabolic secondary to improve the performance. A key characteristic is the addition of a lens to correct for the optical aberrations produced by the spherical mirror(s). This lens is also used to mount the secondary mirror, so no vanes are required to hold it in place.

A Galilean telescope is a refractor, and so it uses lenses exclusively instead of mirrors. Lenses may be able to out-perform mirrors, but it is easier to manufacture high-quality large mirrors than large lenses.

Newtonian telescopes are also reflectors with two mirrors, but one is flat. It also blocks some of the light in the centre of the objective, but it is held in place by vanes instead of a lens. The flat mirror angles the light out the side of the telescope near the objective end of the telescope. While this may frequently place the eyepiece in a more convenient position to access it, especially when aiming near the zenith, it is also necessary to rotate the telescope in the mount to get the eyepiece in the right place.

There are pros and cons to all telescopes, and selecting a telescope requires knowing what you will be doing with it, and choosing something that complements that purpose well with its pros, and that has cons that are not relevant.


Software with simple representation of all stars Do you know any program similar to TheSky (astronomy software), which enables us to identify stars after clicking on them with a simple view (like in the figure)? Many thanks

Guide9 from projectpluto.com will do this.  You choose Chart mode (Display, Background, Chart mode).  Right clicking on a star or any other object will give you the information you want.
It runs on all versions of Windows.  It is not freeware, and is a bit dated in some aspects.


How far apart can two people watch the moon simultaneously? Obviously there's a limited part of earths surface where the moon is visible at any given time. I imagine the area of this surface would be a lot smaller than the area where the sun is visible at any given time, due to the fact that the moon is a lot closer to earth. But how big is that area exactly?

Or, to put it differently: Imagine you are standing outside your house and looking at the moon. At the exact same moment someone else does the same in another location. But how far apart can you (theoretically) be?

If we take 6371.0 km as the mean radius of Earth, an apogee of Moon of 405503 km, and a perigee of 363295 km, we get ratios of 6371.0 km / 405503 km = 0.01571 = sin 0.9002° resp. 6371.0 km / 363295 km = 0.01754 = sin 1.005°. So on both sides of Earth between $$40030\mbox{ km}\cdot 0.9002°/360°=100\mbox{ km}$$ and $$40030\mbox{ km}\cdot 1.005°/360° =112\mbox{ km},$$ with an average circumference of $2\pi\cdot 6371\mbox{ km}=40030\mbox{ km},~$ get lost of the 20015 km of the half circumference of Earth as zone of visibility.

The calculation is simplified to the simultaneous visibility of a point at the distance of the center of moon. More precisely one part of the moon can be visible from one observer, while an other part of the moon is visible from a second observer. This adds between 29.3 and 34.1 arc minutes or between 54.3 (apogee) and 63.2 km (perigee) to the diameter of the zone of visibility.

The visibility of the Moon is influenced by additional factors like air temperature, height of the observer above sea level or the geographic position by Earth's flattening. But this applies also for the visibility of the sun.

The corresponding calculations for the sun: With aphelion of Earth of 152098232 km and perihelion of 147098290 km, we get ratios of 6371.0 km / 152098232 km = 0.000041887 = sin 0.0024000° resp. 6371.0 km / 147098290 km = 0.000043311 = sin 0.0024815°.
So on both sides of Earth between $$40030\mbox{ km}\cdot 0.0024000°/360°=0.26686\mbox{ km}=266.86\mbox{ m}$$
and
$$40030\mbox{ km}\cdot 0.0024815/360°=0.27780\mbox{ km}=277.80\mbox{ m}$$
get lost of the 20015 km of the half circumference of Earth as zone of visibility.

(For a rough estimate you may take the 100 km loss of the moon visibility, and multiply it by the quotient of the distances Earth-Moon and Earth-Sun roughly equal to 400,000 km / 150,000,000 km = 0.002667, to get 266.7 m visibility loss of the sun.)

This is again simplified, and applies to the center of the sun. The appearent diameter of the sun varies between 31.6 and 32.7 arc minutes, adding between 58.6 (aphelion) and 60.6 km (perihelion) to the diameter of the zone of visibility for observers looking at different parts of the sun.

Depending on which scenarios to be compared, the lost area varies. As an example, a loss of a radius of the visible zone of the moon in comparison to the sun of 100 km corresponds to a roughly cylindrical area of $$100\mbox{ km}\cdot \pi\cdot 40030\mbox{ km}=12.6 \mbox{ million square kilometers}.$$

How far can two observers be apart at maximum, to see two differnt parts of the moon at the same time? $$20015 \mbox{ km} - 2\cdot 100 \mbox{ km} + 54.3 \mbox{ km} = 19869.3 \mbox{ km}$$ (apogee) for the average Earth radius. For the same part of the moon it's $$20015 \mbox{ km} - 2\cdot 100 \mbox{ km} = 19815\mbox{ km}.$$


What is the length of time for each fusion process in a massive star? In a massive star (i.e. an O5 star), first hydrogen is fused into helium, then helium is fused into carbon. This process continues to neon, oxygen, and finally silicon burning. So I am wondering, how long does it take for each process to commence?

My stellar astrophysics text, Francis LeBlanc's An Introduction to Stellar Astrophysics, gives the following quantities for the stages of burning in a $25M_{\odot}$ star (citing models by Arnould & Samyn 2001). This is somewhat less massive than a typical main sequence O5-type star, and with higher masses come higher core temperatures and shorter evolutionary timescales. Nonetheless, it should give you a reasonable order-of-magnitude idea of how long it takes for these processes to play out in a massive star.

Phase
Central temperature (K)
Timescale

Hydrogen
$6.0\times10^7$
$7\times10^6$ years

Helium
$2.0\times10^8$
$5\times10^5$ years

Carbon
$9.0\times10^8$
$600$ years

Neon
$1.7\times10^9$
$0.5$ years

Oxygen
$2.4\times10^9$
6 days

Silicon
$4.0\times10^9$
1 days

(One major caveat is that these models are two decades old, and stellar evolutionary codes are constantly being improved. However, I don't believe that these are old enough to be significantly out-of-date!)


How do you refer to comparative temperature? Earth's average surface temperature is about 288 K.  If we had a hypothetical exoplanet with a surface temperature of 300 K, this would be about 4% higher than Earth's.
Because we say Earth radii and Earth masses, would we refer to this as 1.04 Earth temperatures?  Or, how about 1.04 Earth's temperature?  $1.04 \cdot T_⊕$?  Is there a standard in astronomy to use?

Possible, but rare.  The reason is that many things are not linear with respect to temperature.
Volume of gas is in direct proportion to temperature (at given pressure) but we rarely need to consider this.
But, for example, the density of water is markedly non-linear with temperature.
Moreover our physiometric response to temperature is also non linear. We dont experience 300K as being 4% higher than 288K.
As such, planetary temperatures are usually described in Kelvin, Celcius or Fahrenheit (or even Rankin if you look hard enough) rather than being referred to Earth temperaturees.


Was the use of silver on 'optical' telescope mirrors more common in the past? Why? In the past, weren't many astronomical telescope mirrors periodically removed and 'resilvered', and wasn't this actually done using silver?

I'm guessing that this is much less common now - new telescopes are so large and complex that routine invasive procedures like that needed for maintenance of a silver coating would not be acceptable, and mirrors in many if not most older telescopes would have been 'updated' to aluminum.

Was silver used in the past because aluminum coating was too technically difficult, or because silver has slightly better reflectivity in the red, or some other reason?

note: I'm only asking about 'optical' telescopes meaning NIR & Vis, not the longer wavelength IR telescopes where IR emissivity is a concern.

Aluminium coating is a relatively recent process - it became available around the 1920s or 1930s. The Hale telescope arrived just in time to take advantage of this new technology.  (It requires a reasonably good vacuum to work, which probably explains why it took a while to come along.)

Before that - around the mid-1800s - various chemical "silvering" processes, usually involving real silver, were used. These weren't ideal as silver tarnished over time - the silver literally oxidised on exposure to air.  (Aluminium also reacts with air, but forms a thin oxide layer which doesn't harm the image and prevents further oxidation.)  Also some of these silvering processes were either toxic or explosive in application...

Before that - the first telescope mirrors were made of metal that was cast and ground to the required curve then highly polished. The material was "Speculum" a copper-tin alloy. They were poor in quality; far less reflective than later mirrors.  This account gives a nice description of casting mirrors for the Parsonstown telescope using over a ton of molten metal. Two mirrors were necessary for the larger telescopes because the mirror in use had to be taken out and re-polished so often...


Age of the universe vs. its contents A couple of decades ago I read an article (could have been either Astronomy, Space, or Science News) that suggested the contents of the universe may be older than the universe itself.  Of course, that result was so preposterous that it was summarily dismissed.  However, when I read that article, I thought it made sense.  Surely the universe is much more massive than its contents, so as General Relatively predicts, should experience time more slowly.  This conclusion seemed so obvious that a) a bunch of other folks much brighter than I would have thought of it, and b) therefore my conclusion is completely erroneous.  So my question is: why wouldn't the mass of the universe cause it to experience time at a rate slower than its contents as GR predicts?  (There are a couple of questions on SE that are similar, but I do not believe this is a duplicate.)

Models are uncertain

I'm guessing that you are referring to age estimates of very old stars. The age of a star can be determined from certain observables, among one is its absolute luminosity, which in turn requires a precise measurement of its distance. Even today, the uncertainties involved yield estimates with mean values longer than the age of the Universe, which of course is impossible. For instance, the age of the star HD 140283 was found as late as in 2013 to be $14.46\pm0.8$ billion years (Gyr). This is not really in conflict with the estimated age of the Universe, $13.798\pm0.037$ Gyr; all this tells us is that our methods are still not perfect, but that HD 140283 was formed shortly after the Big Bang.

Time dilation is small

General relativity predicts that time evolves more slowly in a graviational field. This is with respect to an observer outside the gravitational field, and since there is no external observer wrt. the Universe, the global time of the Universe can be said to be the "right" time. You're right that in principle the total mass of the Universe makes time run slower than in a hypothetical, empty universe, but almost anywhere in the Universe, the gravitational field is way too weak to cause any noticeable dilation. Close to massive objects, time does run slow. On the surface of a star, the time dilation factor is roughly $(1 - GM/rc^2)^{-1/2} = 1.000002$. Inside the star, the factor is even closer to unity, because only the mass between the center and you contributes, so the farther you dive into the star (kids, don't try this at home), the less the time dilation. Only if you compress the star, so as to squezze it into a neutron star of a black hole, will you get a larger time dilation. For instance, at the "surface" of a black hole, time stops (but again, only wrt. an external observer; for the person at the surface, time evolves as expected)


Why is the solar system often shown as a 2D plane? Whenever I have learned about the solar system I always see the orbits displayed as a virtually flat plane.

Are all of the orbits in the solar system really like this? If so, why? It seems like a rather large coincidence for all of our planets + the asteroid belt to be on virtually the same plane.

This is not a coincidence at all, but a direct consequence of the way the solar system was formed.

The generally accepted model is that solar systems (including our own) form out of a Protoplanetary disc.  Gravitation causes mass to collapse around a protostar, which always has some angular momentum (as does everything).  Wikipedia explains it better than I can:

  Protostars typically form from molecular clouds consisting primarily of molecular hydrogen. When a portion of a molecular cloud reaches a critical size, mass, or density, it begins to collapse under its own gravity. As this collapsing cloud, called a solar nebula, becomes denser, random gas motions originally present in the cloud average out in favor of the direction of the nebula's net angular momentum. Conservation of angular momentum causes the rotation to increase as the nebula radius decreases. This rotation causes the cloud to flatten out—much like forming a flat pizza out of dough—and take the form of a disk.

And then, from this protoplanetary disk, planets form.  Consequently, they're all in the same plane.

As such, the inclinations of each planet's orbit are pretty close to that of Earth's:

Planet  Orbital Inclination
Mercury 7°
Venus   3.39°
Earth   0°
Mars    1.85°
Jupiter 1.3°
Saturn  2.49°
Uranus  0.77°
Neptune 1.77°


Neutron stars and distribution of heavy elements In the news, of course, is the theory that much of the mass of the heaviest elements in Earth would have come from colliding neutron stars.  This graphic gives one such analysis:

Origin of the Elements in the Solar System by Jennifer Johnson is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License. (Larger version at http://www.astronomy.ohio-state.edu/~jaj/nucleo/)

I don't know how accurate this is, but the URL looks good.  Also, this seems to be a new theory, and AFAIK this is the first actual evidence for it.

If this is correct, what are the implications for the quantity of heavy elements that would be found on planets outside our solar system?  In particular, I guess, the very heavy ones: gold, platinum, thorium, uranium...

We've been watching the skies for signs of neutron star collisions for several years now (IIUC), and I guess we're starting to get an idea of the frequency of this event.  And I've seen estimates of how much gold etc. was produced by this particular event.  How many planets would one such collision supply with the quantity of heavy elements in Earth?
(I guess we only really know about gold in the Earth's crust, and to some extent in the mantle; are these elements concentrated in the core?  Any idea about heavy elements in other planets of our solar system?)

So: given the frequency of the event per galaxy, the estimated number of stars with planets in a galaxy, etc., is it likely that most planets are deficient in heavy elements, as compared with Earth?

(Side comment: if so, then nuclear wars might be rare or impossible on those planets.)

I may do some more detailed calculation, but I think there is a misconception that has provoked your question.

The content of (e.g) gold, with respect to other elements in the solar system can be readily measured in meteorites. The Earth is expected to contain a similar abundance, but is very depleted in hydrogen and helium and somewhat depleted of lighter elements like carbon and sodium. The surface abundance of gold in the Earth is then a bit lower, because it sinks during the Earth's formation.

We expect the relative abundances of chemical elements to be quite smoothly distributed in space and time. Material that is produced in supernovae and neutron star collisions is mixed quite quickly in the interstellar medium. Thus the material that makes up the solar system is contaminated by the excreta of as many as a billion stars that lived and died before it's birth. Even if a neutron star merger occurred right next to the molecular cloud that formed the Sun, that cloud may have contained a million solar masses of gas, that would have diluted the signature of the kilonova event. We expect this to be true for all stars born in our Galaxy, except those that were born right near the beginning, where individual "pollution" events might become apparent.

The rate of these events in the local universe is only known observationally to an order of magnitude. In the past, it is also rather poorly known, but can be indirectly deduced from the rate of short Gamma ray bursts (of which we see only a small, uncertain fraction), if we assume that all sGRBs are merging neutron stars and that all merging neutron stars produce sGRBs. The rate in our Galaxy is extrapolated from the properties of the meagre number of identified double neutron stars and their orbital decay rates and is about 1 every 50,000 years (so about 200,000 over the lifetime of the Galaxy).

These rates require that about 0.0001 to 0.001 solar masses of very heavy elements is produced in each of these explosions to reproduce the solar system abundances. In the recent event, the amount of heavy elements in the ejecta appear to be in that range.


Can radio-based telescopes produce visible light images? It's a fairly simple question (at least in my layman's mind). Can a radio telescope that has no mirrors or lenses to capture visible light (to human eyes) still produce imagery like an optical telescope such as the Hubble? I'm talking just ROYGBIV light (no infrared or UV).

Can a radio telescope that has no mirrors or lenses to capture visible light (to human eyes) still produce imagery like an optical telescope such as the Hubble?

Of course, either via false color or pseudocolor. A false color radio image maps various frequencies (typically three) in a multispectral radio image to visible colors. A pseudocolor radio image maps various intensities in a single spectrum gray scale radio image to visible colors. The same techniques are also used for gamma, X-ray, UV, IR, and microwave imagery. We can't see in those frequencies, so something needs to be done to the captured imagery to make them visible to us.

If the question is asking whether a radio telescope can be used to capture visible light, the answer is no. I'll start with the receiver. It's designed to capture long wavelength radiation. Photons in the visible spectrum will have minimal, if any, impact on a radio telescope receiver.

An even bigger problem with some radio telescopes is depicted below.

Source: https://en.wikipedia.org/wiki/Haystack_Observatory

The above is the Haystack Radio Telescope at the Haystack Observatory in Westford, Massachusetts. The radome protects the telescope against weather, but also hides the antenna from optical light. A couple of other radio telescopes at the Haystack Observatory show yet another problem:

Source: https://en.wikipedia.org/wiki/Haystack_Observatory

The fixed zenith telescope antenna in the foreground and the fully movable telescope antenna in the background are made of a wire mesh, making them see-thru in the visible portion of the spectrum. This wire mesh construction is quite common for long wavelength radio antennae because it allows the wind to blow through it. Shorter wavelength radio antennae are more likely to be solid, but those are oftentimes painted.

A wire mesh with wires separated by centimeters is very smooth to a meter long radio wave, as solid surface smooth at the millimeter level to a centimeter long radio wave. A surface with millimeter-scale roughness would make for an extremely lousy optical mirror. There is no reason to make a radio telescope antenna smooth to the visible portion of the spectrum, and there are many millions of reasons (i.e., many millions of dollars) not to do so.


photon gas pressure and Independence on refraction index Brief context: when studying Stellar Structure one of the main goals, aside of determining equations that describe quantities of interest of the system, is to also determine equations of state that allows us to approximate solutions for the coupled differential equations system we find.

An important step is to find equations of state, e.g., an expression for the Pressure $P$ of the system. Following common derivations on this subject (e.g., see Rose, Advanced Stellar Astrophysics or Clayton, Principles of Stellar Evolution and Nucleosynthesis) an integral for $P$ over a cone of particles with momentum $p$ would be
$$P =\int\limits_0^{\pi/2}\int\limits_0^\infty 2 \, v_p \,\cos^2\theta \, n(\theta,p)\,d\theta\,dp$$
where $n(p$) is the gas' distribution w.r.t $p$. An example could be an ideal monoatomic gas, in the non degenerate, non relativistic case. In particular:
$$n(p) \, dp = 4\pi p^2 n \frac{\exp\left(-\frac{p^2}{2mkT}\right)}{(2\pi m kT)^{3/2}}\,dp$$
and $p = m \, v_p$ so
$$P = n k T$$

The problem: let us find then this same $P$ for a Photon Gas (radiation pressure) considering that now $n(p) \,dp$ is given by a Bose-Einstein Distribution. In particular $$n(p)\,dp = \frac{8\pi \, p^2\,dp}{h^3}\frac{1}{e^{pc/kT}-1}$$ where $p = h\nu/c$ and $v_p \overset{!}{=} c$.

The procedure is very straightforward:

$n(p)\,dp$ can be written as $$n(\nu)\,d\nu = \frac{8\pi}{c^3}\frac{\nu^2 d\nu}{e^{h\nu/kT}-1}$$ so the integral becomes $$P = \frac{8\pi h}{3c^3}\left(\frac{kT}{h}\right)^4 \int\limits_0^\infty \frac{x^3\,dx}{e^x-1}$$ where $$x \equiv \frac{h\nu}{kT}$$. Here the $\int$ (e.g., related to this question) can be either solved by using Polylogarithms,  Riemann zeta and Gamma functions, complex analysis, etc.

It's easy to check that $$P = \frac{8\pi k^4}{3c^3h^3} T^4 \zeta(4)\Gamma(4)$$ then $$P = \frac{8\pi^5 k^4}{45c^3h^3}T^4 = \frac{4\sigma}{3c} T^4$$ which matches evenly.

The question: if we're dealing with Stellar structure, potentially dense environments (the interior of a star!) then EM radiation is not necessarily propagating in vacuum (in fact, I guess it's not the case!). Then would be $v_p \overset{!}{=} n \, c  \leq c$ but this is not usually considered on calculations. Why is this? Do you know any resource where anybody have examined this and see if star interiors are affected? (it might be negligible)

Thanks in advance,

Photons always travel at the speed of light, wherever they are. This is a consequence of them being massless particles.
Although "light" travels slower than $c$ in a refractive medium, the individual photons making up a light beam do not.
e.g. https://physics.stackexchange.com/questions/19203/what-is-the-mass-of-a-photon-in-non-empty-spaces


Photometric redshift I have decided to do undergraduate thesis on "estimating photometric redshift" or something related to this using machine learning. Reading previous papers, I have come to know that work has been done on this topic using support vector machine, artificial neural network, Bayesian approach and even deep learning. Can you suggest what are the  modifications or additions that I can possibly make?

The process of determining photometric redshift includes, e.g.,:
1. objects of interest
2. types of data (e.g., one epoch multiple filters, one filter multiple epochs)
3. method of analysis (including model, and algorithm)

From what you mentioned, those are just algorithms. So, you might take a look around on other aspects as listed.


Frequency of transit in planar and non-planar planetary systems Supposing that another planetary system is in one plane like our solar system, if let's say that the transit of one planet is visible then is it safe to also assume that all planets in the system must transit?

Logically I feel like that this makes sense because since they are all in the same plane then they therefore must all transit as well. But then I feel like the transit probability equation could make this false? Could someone please clarify my doubts?

Also suppose that we are in another system where each planet doesn't necessarily orbit in the same plane.  Is it easier to find transits if the planets are orbiting in the same or random planes?

I think that you have a better chance detecting a planet if they are in random orbits. but i am a bit lost in regards to detecting multiple transits.

This is unfortunately not true.
If you already look at our solar system, even this is not co-planar.
The few degrees inclination that the planets have to each other would already be enough to make some of them undetectable, when others are detectable by transits.

The answer to your question therefore is not so much about the transit probability, but more about what the mutual inclinations are of the planets. Those can be stirred up by close-planet encounters or resonances after the gas disc dissipates in which the planets form.


Can Eight Retrograde (seen from Earth) planets occur? First of all, for the purposes of this question Pluto is a planet. So the eight planets being considered are Mercury, Venus, Mars, Jupter, Saturn, Uranus, Neptune and Pluto.

My question is: can all eight of these planets go retrograde† at the same time? I know that seven can do so on rare instances; there were 12 days of that in the 20th century all in the 1940's and the 1980's. But to show how rare that is there was a block of 400 years in the middle ages with no days of seven retrogrades. Has eight ever happened? Or is it impossible?

†Retrograde motion as seen from Earth, not retrograde orbits around the Sun.

EDIT: I used the DE431 ephemeris to get the results below. However, if you visit HORIZONS (https://ssd.jpl.nasa.gov/horizons.cgi) and create an ephemeris for Pluto (actually Pluto's barycenter, object 999), the results include this warning:

  The Horizons output log may report using DE431 as the source of target
  or center data, but it is nevertheless using DE433 as an override if
  Pluto is involved. This is to maintain ephemeris consistency for the
  New Horizons  spacecraft encounter at Pluto.

Therefore, my calculations may be inaccurate, above and beyond the normal inaccuracy of predicting planetary data in the past and future.

There are 2 times in the past 15,000 years and 5 times in the next 15,000 years where all 8 planets are in retrograde as observed from Earth.

The table below lists the times when this happens:

The last column is the ephemeris time, the number of seconds since '2000-JAN-01 12:00:00 UTC', which NASA refers to as "the epoch".
The first line in each pair shows the start time of when all 8 planets are in retrograde, and the final planet to enter retrograde to make this true.
The second line in each pair shows the end time when at least one planet is no longer in retrograde, and the planet whose retrograde period has ended.

B.C. 12435-AUG-10 07:52 JUPITER STARTS RETROGRADE -455482152466.038696
B.C. 12435-AUG-24 17:12 MERCURY ENDS RETROGRADE -455480909251.797852

B.C.  4139-MAR-23 15:02 MARS STARTS RETROGRADE -193692373029.630005
B.C.  4139-MAR-28 21:57 MERCURY ENDS RETROGRADE -193691916114.978455

A.D.  7662-APR-27 19:33 MARS STARTS RETROGRADE 178685595220.195740
A.D.  7662-MAY-12 05:19 NEPTUNE ENDS RETROGRADE 178686840014.478882

A.D.  8807-JAN-07 02:02 MERCURY STARTS RETROGRADE 214808680981.726929
A.D.  8807-JAN-20 01:05 JUPITER ENDS RETROGRADE 214809800771.851685

A.D. 10575-FEB-14 21:34 VENUS STARTS RETROGRADE 270604748086.593933
A.D. 10575-FEB-24 18:53 NEPTUNE ENDS RETROGRADE 270605602468.037781

A.D. 13679-NOV-06 16:03 VENUS STARTS RETROGRADE 368580427477.379639
A.D. 13679-NOV-19 01:05 MERCURY ENDS RETROGRADE 368581496795.683594

A.D. 16364-MAR-11 19:25 MERCURY STARTS RETROGRADE 453290109945.561157
A.D. 16364-MAR-13 08:53 MARS ENDS RETROGRADE 453290244824.895996

Notes and caveats:

I wrote https://github.com/barrycarter/bcapps/tree/master/ASTRO/bc-retrograde.c to print out retrograde times for all planets.
The file https://github.com/barrycarter/bcapps/tree/master/ASTRO/all-retrogrades.txt.bz2 shows start and end retrograde times for all 8 planets. NOTE: the lines indicating that planets leave retrograde at "17191-MAR-01 00:00" is incorrect-- that's simply the maximum time for which I could make calculations.
I wrote https://github.com/barrycarter/bcapps/tree/master/ASTRO/bc-moving-backwards.pl to determine which planets were in retrograde at any given time, and used the results to find the times above where all 8 planets were in retrograde.
I spot checked my results against https://www.calsky.com/ -- note that I am computing "ecliptic retrograde", when a planet's ecliptic longitude is decreasing, not "equatorial retrograde", when a planet's right ascension is decreasing.
As always, I invite people to double check my results, and remind everyone not to use my results for anything critical.


What are detached black holes? I understand what black holes are, and how they form, but I was recently introduced to the topic of detached black holes. What are they and how are they different from standard black holes?

A detached black hole is simply a black hole that is not part of a close binary pair.
There are two ways that we can detect stellar mass black holes. Firstly when they are close to a star, and matter is flowing from the star into a disc around the black hole.  Matter in the disc will orbit at substantial fraction of the speed of light, and friction will cause it to heat up to extreme temperatures, hot enough to emit X-rays.  If we can detect these X-rays and we can work out the mass of the object, we can deduce the existance of a black hole.
The second way is through gravitational wave astronomy, detecting the gravitational waves emitted when two black holes merge.
Detached black holes, that don't have a partner, are much harder to detect.  As there would be no disc of hot matter they would not emit x-rays.  In fact they would not emit anything. A detached black hole could be part of a binary system, but far enough from the star that no matter transfers from the star to the black-hole.  Such a black hole could be found by observing the radial velocity of the star. The gravity of the black hole will cause the star to orbit around it, the forwards and back motion of the star can be detected and the mass of the object can be deduced.  A large but invisible mass is a likely black hole.  See A detached stellar-mass black hole candidate in the globular cluster NGC 3201


What's the origin of Clyde Tombaugh's 9-inch Newtonian telescope? Astronomer Clyde Tombaugh is shown in a now-famous photograph at the age of 24 with his "homemade" 9-inch Newtonian telescope in this image also shown in Elva R. O'Hara's Clyde W. Tombaugh: Farm Boy Reached for the Stars.
Shown in the photo, the telescope tube is marked with something like

9 INCH AP
NEWTONIAN TELESCOPE
79 INCH FOCUS

along the length of the tube, and illegibly with

OHIO-...

and some other markings in a different, light-toned printing around the circumference of the tube.
According to the linked article:

Tombaugh was born on Feb. 4, 1906, in Streator, Ill. He might have been a typical farm boy but for his one fascination: astronomy. Through the influence of his father and an uncle, the 12-year-old Tombaugh began observing the night sky. He spent hundreds of hours using a 9-inch telescope bought at Sears and Roebuck, learning to methodically identify stars and constellations as well as to observe the movements of the planets.

But on page 243 of Gary Leonard Cameron
Iowa State University 2010 thesis Public skies: telescopes and the popularization of astronomy in the twentieth century it says:

Sears Roebuck had only a few cheap hand-held telescopes for sale in its catalogues from the 1920s to the 1940s. However, by 1961, Sears had a total of nine models of astronomical telescope for sale, all Japanese imports under the Tower label.

Question: So exactly what is the origin of Clyde Tombaugh's 8-inch Newtonian telescope?

Cropped from Elva R. O'Hara's Clyde W. Tombaugh: Farm Boy Reached for the Stars:

The description there says:

Image caption:  Clyde W. Tombaugh shared a homemade telescope with his father and uncle.  Photo courtesy of NMSU Library Archives and Special Collections

Description here says:

The astronomer Clyde Tombaugh, discoverer of Pluto here shown with his homemade 9-inch telescope. Date: circa 1930

In the first chapter of Out of the Darkness: The Planet Pluto, Tombaugh gives a colorful, personal account of his youth in Illinois and Kansas, including five telescopes:

his uncle's uncorrected 3-inch refractor
a 2.25-inch achromatic Sears-Roebuck refractor his father and uncle shared
a mediocre 8-inch reflector he made in 1926
a better 7-inch reflector he made for his uncle
an excellent 9-inch reflector he made in 1928

The last two benefited from optical testing in a storm cellar built extra long for that purpose.
Tombaugh's planetary sketches using the 9-inch reflector got him the job at Lowell Observatory.

The same photo appears on page 23 with this caption:

  Clyde W. Tombaugh, at the age of twenty-two, with his homemade 9-inch (22.9-centimeter) reflecting telescope on the family farm near Burdett, Kansas.
  This picture was taken in the summer of 1928, eighteen months before the discovery of Pluto.

APOD 1997-02-04 (© 1996 K. Beatty) shows him at age 89 with the same telescope in New Mexico.


Spectra: What is that Supposed to Mean? I've found this website (click here for the link), the link leads you to a problem about the velocity Andromeda is going towards the Milky way galaxy, and you can try three approaches. I like to challenge myself, so I click on the second one, about spectra. I already know what the Electromagnetic Spectrum is, but this one is a tad different. It tells you about the redshift equation, which I already know about, and I've used it a few times in examples I made for myself, but I am confused about the next page (to get there, click the button at the bottom that says "Use This Technique to Solve for M31's velocity"), and the next page is worded in a way that I can't really understand it, and using the clues I get from the Spectrum Graph, I can't figure out what the answer is. What is Relative intensity of light, and what does the Calcium have to do with it? Also, isn't their value for the speed of light inaccurate, as far as I know, it's $3\times 10^8$, not $3\times 10^5$. Even further on, they want you to calculate the radial velocity from the graph, which I have no idea how to do.
If an answer about this here is acceptable, please give me a feasibly worded explanation of all this madness, and thanks in advance!

Okay, well, calcium is just an easily recognizable spectral feature, so I guess that's why they used it. The y-axis shows the relative intensity. I am assuming this is a spectrum that is normalized or scaled down so that the highest value of the intensity is 1, but this is not so important. The laboratory wavelengths of the absorption lines are shown in red in the graph. If you look carefully, you can see that the minimum of each spectral line does not exactly line up with these red tick marks. That is because the lines are Doppler shifted. M31 is moving and thus the light that we see has a different color (or frequency) from what we would observe in the lab. Let's focus on the left absorption line. The minimum is a bit to the left of the red mark. I conceive this to be approximately 4 angstrom off. The wavelength that you measure here on Earth will thus be 3630 angstrom and not 3934. If I take the formula from the previous page and rewrite I get
$$\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}$$
where $\Delta \lambda$ is the difference between the observed and the lab wavelength of the spectral line. In this case $\Delta \lambda = -4$ angstrom and $\lambda_0 = 3934$ angstrom. The speed of light is given in km/s, not in m/s, so the value is indeed correct. Now you can get the speed of M31 from
$$v = c\frac{\Delta \lambda}{\lambda_0} = 3 \times 10^5 \frac{4}{3934} = -305km/s$$
I entered this value and it was accepted.


How is magnetic field is generated over the sun? Dynamo theory suggest how magnetic field generated on Earth. Due to flowing magma which contains molten iron generates magnetic field on earth. But how magnetic field is generated over sun?

The Sun's interior is partly or completely ionised and therefore supports the flow of electric currents. The solar magnetic field is explained by a solar dynamo that converts the kinetic energy of electrically charged fluid flows in the Sun (due to differential rotation and convection) into electromagnetic energy.


What observation can be expected on LIGO if any when Betelgeuse goes supernova? What observation can be expected from the LIGO†  gravitational wave observatory if any when Betelgeuse goes supernova?

Could we know that Betelgeuse has gone supernova before we see it light up our night sky?

†or other current or soon-to-be completed gravitational wave obervatories

Potentially a short (less than a second) burst of gravitational waves (GWs) would be detected. Much depends on asymmetries in the core collapse, since a spherically (or even axially) symmetric collapse would not produce GWs (e.g. Morozova et al. 2019). However, theoretical models suggest that the GWs start at low frequency (tens of Hz) and are associated with the core bounce at the end of core collapse. After about half a second the frequencies rise to kHz and are associated with the oscillations of the proto-neutron star at the centre. Since LIGO is sensitive from about 10Hz to 1kHz, then only the initial phase that lasts a few tenths of a second would be detectable, but snce GWs can travel through the star without much hindrance they will indicate the moment of core collapse quite accurately.

The theoretical calculations referenced above suggest GW amplitudes of $h_{+}d \sim 10$, where $h_{+}$ is the dimensionless GW strain and $d$ is the distance to the supernova in cm. Betelgeuse is at a distance of $\sim 6 \times 10^{20}$ cm (600 light years), so the strain at the detector would be $h_{+} \sim 2\times 10^{-20}$. LIGO is sensitive to strains of above $10^{-23}$ in this frequency range, so there should be no problem (according to these models) in detecting such an event.

If they were produced with detectable strength, then they should be received at the same time as ananticipated neutrino burst. The difference would be that the neutrino burst should extend over about 10 s. That is because the dense centre of a supernova is opaque even to neutrinos (but not gravitational waves) for that period and so they "leak out" over that period.

The electromagnetic signature of the supernova (i.e. a visible light increase) would be seen a few hours after that, since that is how long it takes for energy to be transferred to the envelope of the star.

NB: For all practical intents and purposes, over the distance to Betelgeuse, neutrinos, along with light and gravitational waves, travel at the same speed. i.e. The delay of the ($\sim  10$ MeV) neutrino burst due to their non-zero ($\
< 1$ eV) mass would be negligible. If the Lorentz factor is $10^7$, then the neutrino speed is the same as the speed of light to 1 part in $2\times 10^{14}$, which means they take an extra 0.1 ms to arrive from Betelgeuse compared to light speed.


How big does the moon have to be to have the Earth tidally locked to it I would like a earth-moon system like Pluto-Charon (but with earth instead of Pluto) where both will be tidally locked to each other within something like a billion years. As the moon would probably lock to earth before the opposite how big the moon has to be?

There's a lot of variables for the timing, but according to the formula from Wikipedia, the time ratio is inversely proportional to the square of the mass.

If our Moon will tidally lock the Earth to it in 50 billion years, a moon 7 times the mass should tidally lock the Earth, everything else being equal, about 7^2 or 49 times faster.  About 1 billion years.

There's numerous variables, such as how large the oceans are on your planet, the density, the radius, how long the surface stays molten after the impact, the initial rotation rate and initial distance.  (that one is proportional to the 6th power).

If somebody wants to give a better and more numerical explanation, please feel free, but this should get you started.


Does the Milky Way belong to a Galaxy cluster I know that the Milky Way is a member of the Local Group and the Virgo Supercluster / Laniakea Supercluster. However, I can't seem to find any information about the Galaxy cluster we belong to.

Does the Local Group / Milky Way belong to a galaxy cluster?

The Local Group contains 54 plus galaxies.
Don't know that that counts as much of a cluster.
Next up in scale, the Milky way is part of the Laniakea Supercluster
That contains about 100,000 galaxies, so it's a bit on the large size to call a simple cluster.
The well known Virgo Cluster contains about 1300 galaxies.
So it's clearly a thing of intermediate scale.


Does the mass of the observable universe ever change? First do we have anyway to even estimate the mass of the entire observable universe?  And then is there any data that shows mass being gained or lost?  Would we ever know if someone was playing with the til.

Also I want to be clear that I am not talking about small masses on the outskirts of the "universe" or small discrepancies in measurement or anything of the sort.

Note:  I would like to add that maybe we should define the observable universe as NOW (x-date) so that we aren't calculating a moving target.

Even if you're only referring the "ordinary" matter (such as stars, gas, and bicycles) and dark matter, the mass of the observable Universe does increase, not because mass is being created, but because the size of the observable Universe increases. In a billion years from now, we can see stuff that today is too far away for the light to have reached us, so its radius has increased. Since the mass $M$ equals density $\rho_\mathrm{M}$ times volume $V$, $M$ increases.

As called2voyage mentions, we have several ways of measuring the density, and we know it's close to $3\times10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$. The radius is $R = 4.6\times10^{28}\,\mathrm{cm}$, so the mass is
$$M = \rho_\mathrm{M} \frac{4\pi}{3}R^3 \simeq 10^{57}\,\mathrm{g},$$
or $5\times10^{23}M_\odot$ (Solar masses).

However, another factor contributes to the mass increase, namely the so-called dark energy, which is a form of energy attributed to empty space. Since the Universe expands, dark energy is being created all the time.