What is the shape (along the plane, not up-down) of stellar orbits in flat spiral galaxies What I mean is, with a central mass orbits are relatively simple, but orbits around the galaxy are different, in essence as the star orbits through the dark matter halo, the further it moves away from the center of the galaxy the greater the gravitational field, so while orbits tend to slow down as kinetic energy is turned into potential energy, stars are effectively orbiting around more mass the further they are from the center of the galaxy, which might speed them up. The equal areas over equal time rule would seem to no longer apply.

So, ignoring up and down motion relative to the plane of the galaxy, is there a common shape for galactic orbits? Is there a formula? I think eccentricity would still apply, but I don't think the orbits would be ellipses, so it would be a different kind of eccentricity. Would these orbits tend to circularize (in an ideal dark matter cloud, ignoring perturbations. Obviously perturbations would lead to, in a practical sense, unpredictable orbits, not neat and tidy formulas. I gather from a show I caught on the science channel that the Milky Way's magnetic field also plays a role in forming its spiral arms, but I'm asking in an ideal, mathematical sense, what shape would orbits be through a galaxy, from a 2D point of view, ignoring up and down motion.

Update

It occurred to me that the question might be more simply asked by a black hole in effect "orbiting" through a planet or even a low-velocity neutrino (if there are any low-velocity neutrinos). The inside a planet orbit is quite different than the Kepler orbit, because gravity increases as the object moves further away from the center. The overall orbit might still be somewhat like an ellipse but the laws for such an orbit would be different, you would see the highest acceleration at the furthest points from the center though you'd probably still see the highest velocity at the closest point to the center.

Anyway, I'm mostly just curious if those shape has been worked out and what it looks like.

This is an interesting question and so often interesting questions cannot easily be answered with any current knowledge, but this one can be answered to some degree. I'm going to run through the basics of orbital theory and describe how they can apply to galaxies and how it differs from Keplerian systems. You should have a reasonable understanding of Newtonian physics (after all orbits derive precisely from Newton's laws) and a strong knowledge of math. If you don't have these things, just jump to the end of each section where I'll try to summarize the important points behind the math.
A quick note on math notation I'll be using. A dot over a symbol indicates a time derivative (e.g., $\dot{a}$) and un-italicised, bold symbols are vector quantities (e.g., $\bf F$). Let's get down to business.
The Orbital Equation of Motion
Consider a mass $m$ as some position $\bf r$ and moving with a motion described by $\bf \dot{r}$. This mass experiences a force $\textbf{F}(r)$ which is only a function of radial distance, $r$, from the center of the coordinate system. The goal here is to determine the equation of motion that can describe the orbit of the mass due to this force. This equation can then be used to solve for $r(\theta)$. By Newton's law the equation of motion can be initially defined to be
$$\textbf{F}(r) = m\textbf{a} = m\Big(\ddot{r} - r\dot{\theta}^2\Big)$$
Note that in this case $r$ is simply the radial component of $\bf r$ and $\theta$ is the azimuthal angle of the body in a spherical coordinate system. I'll leave it to you to determine how to break down the acceleration into the two components above, under the appropriate coordinate system. Let's try to remove our $\theta$ dependence so that we have only a function of $r$. This can be achieved by using angular momentum conservation. The angular momentum per unit mass is given by $\ell = r^2\dot{\theta}$ so that $\dot{\theta} = \ell/r^2$. This gives
$$\textbf{F}(r) =  m\big(\ddot{r} - \ell^2/r^3\big)$$
This is now a differential equation that allows us to solve for $r(t)$, but we want $r(\theta)$ so we need to do some conversion. Let's re-parameterize by defining $u \equiv 1/r$ (the reason will become clear in a bit) and determining $\ddot{r}$ in terms of $u$ and $\theta$.
$$\frac{\mathrm{d}}{\mathrm{d}t}(r) = \frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{1}{u}\bigg)=\frac{1}{u^2}\frac{\mathrm{d}u}{\mathrm{d}t}=\frac{1}{u^2}\frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}\theta}{\mathrm{d}t}=-\frac{\dot{\theta}}{u^2}\frac{\mathrm{d}u}{\mathrm{d}\theta}=-\ell\frac{\mathrm{d}u}{\mathrm{d}\theta}$$
Note the substitution of $\ell=r^2\dot{\theta}=\dot{\theta}/u^2$. Now differentiate again to determine $\ddot{r}$.
$$\frac{\mathrm{d}^2}{\mathrm{d}t^2}(r)=-\ell\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\mathrm{d}u}{\mathrm{d}\theta}\bigg)=-\ell\frac{\mathrm{d}\theta}{\mathrm{d}t}\frac{\mathrm{d}}{\mathrm{d}\theta}\bigg(\frac{\mathrm{d}u}{\mathrm{d}\theta}\bigg) = \ell\dot{\theta}\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2}=-\ell^2 u^2 \frac{\mathrm{d}^2u}{\mathrm{d}\theta^2}$$
Putting this into our expression for the equation of motion and making the transformation that $r=1/u$ finally gives
$$\textbf{F}(1/u) =  m\Big(-\ell^2 u^2 \frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} - \ell^2 u^3\Big)$$
Writing in a more convenient form we finally arrive at
$$\boxed{\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} + u = -\frac{\textbf{F}(1/u)}{m\ell^2u^2}}$$
Remember that, $m$ is the mass of the body, $u(\theta) \equiv 1/r(\theta)$, $\ell$ is the angular momentum per unit mass, $\bf F$ is a purely radial force acting on the body, and $r$ and $\theta$ are the radial and azimuthal coordinate positions of the mass.
Punchline: The final result here is a general equation of motion for a body orbiting according to some arbitrary force. This could be gravity, electromagnetic, a spring force, or anything else we decide. It is purposefully derived under general and nonconstrictive assumptions and hopefully you can see that it can be used to understand the orbital motion of a star orbiting in a disk galaxy. The goal with this equation should be to plug in your force (whatever that may be) and solve for $u(\theta)$. From there it's easy to determine $r(\theta)$.
Keplerian Motion
Before we go about looking at orbital motion in a galaxy, let's look at standard Keplerian motion so we have something to compare to. Keplerian motion derives from assuming our mass $m$ is orbiting a single, point-like mass $M$ and under the influence of simple gravity. In that case, we can write our force as $F(r) = kr^{-2}$ and thus $F(1/u) = ku^2$, where $k\equiv GMm$ is a constant, defined here for mathematical simplicity. Note that $G$ is the gravitational constant. The general orbital equation, under this force, now becomes
$$\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} + u = -\frac{k}{m\ell^2}$$
This is a standard second-order, nonhomogeneous differential equation with a constant forcing function. If you know your Diff EQ, you should know the solution almost immediately.
$$u(\theta) = \frac{k}{m\ell^2} + A\cos (\theta - \theta_0)$$
In this equation, $A$ is some unknown constant, and $\theta_0$ represents the starting angle of the orbit, which we can arbitrarily choose to be zero. Our ultimate goal is to get $r(\theta)$ so let's do that. I'm going to do a few steps in one though and leave you to work out the intervening math. I'll fill in for $k = GMm$ and define the angular momentum as $L = \ell\mu$ where $\mu$ is the reduced mass of our system. I will also state, without proof, that $e = A(m\ell^2/k)$ where $e$ is the eccentricity of the orbit.
$$\boxed{r(\theta) = \frac{L^2/GM\mu^2}{1+e\cos (\theta)}}$$
Punchline: We've found a final equation which represents the orbital motion of a mass under the influence of a gravity due to a point-like mass $M$. If you know your stuff, you'll see that this equation precisely describes the conic sections, depending on the value of $e$. If $e = 0$, you get circular motion (since $r(\theta)$ becomes a constant). If $0 < e <1$, you get elliptical motion, $e=1$ is parabolic motion and $e > 1$ is hyperbolic.
As expected, Keplerian motion (i.e., having a central force such that $F\propto r^{-2}$) resulted in conics, which is precisely Kepler's first law. Kepler's second and third laws more or less derive from the same assumptions. It stands to reason then, that any system where $F \not \propto r^{-2}$ does not follow any of Kepler's laws. The orbits are not perfect conics (e.g., ellipses, circles, etc.), they do not sweep out equal areas in equal times, and the standard $P^2 \propto a^3$ certainly does not apply.
Orbital Motion in a Galaxy
Your question correctly describes the situation for stars (or anything really) orbiting in a galaxy. Stars are not orbiting central, point-like masses. They are embedded within both the baryonic and dark matter that comprises the galaxy and are orbiting through it. It is a well known concept in physics that spherically symmetric mass distributions do not have a net gravitational pull on objects interior to that distribution, which means that for stars in a galaxy, the mass affecting its orbit is the mass interior to its radius. If that radius changes, the mass changes!
The central force on our star is still going to be gravity, but the mass acting on it will be all the mass interior to some radius, denoted by $M_r$. We can see that $F(r) = GM_r(r)m/r^2$. If we want to determine the force acting on our star (and thus the exact orbit, via the differential equation above), we first need to figure out what the mass interior to some radius is. This can be achieved by using the mass continuity equation.
$$\frac{\mathrm{d}M_r}{\mathrm{d}r} = 4\pi r^2 \rho(r)$$
You can essentially figure out all the mass interior to $r$ by integrating up all the mass density as a function of $r$. Here of course, you need a good equation for $\rho(r)$. A simple, but physically unrealistic density profile is the Singular Isothermal Sphere (SIS), whereas a more realistic, but mathematically complex equation might be the NFW profile or the Einasto profile.
I've now laid out all the steps you need to figure out the orbital motion in a galaxy, but I have to say, it's not pretty. We can look at part of the simplest case though, that of the SIS.
Single Isothermal Sphere
For the singlar isothermal sphere, you have that $\rho(r) = v^2/(4\pi Gr^2)$, where $v$ is the rotational velocity of your star. This profile relies on a crucial fact of disk galaxies $-$ their rotational profile is flat! This was well established, for example by Ruben et al. 1978. I've reproduced a figure from this paper below which shows the rotation curve for several galaxies. The important point here is that this shows $v$ is constant and does not depend on radius! (Assuming we're not near the galactic bulge or center. That's a whole different beast.)

With this crucial piece of information, we can solve for $M_r$ by integrating $\rho(r)$ (which I'll leave up to you). The result is that
$$M_r = \frac{v^2r}{G}$$
This means your force is given by
$$F(r) = \frac{v^2rm}{r^2} = \frac{v^2m}{r}\Rightarrow F(1/u) = v^2mu\propto ku$$
You can see here that unlike the Keplerian case, our force is proportional to $r^{-1}$ rather than $r^{-2}$. You can do this process with other density profiles (such as the NFW or Einasto I listed above), but you'll wind up with the same result.
If you're so inclined, you could choose to plug this into the orbital equation of motion above and solve it out, but you're now working with a non-linear differential equation and things can get messy fast.
Punchline: I'm not sure if this actually answers your question or not. I've led you partially down the rabbit hole, but hopefully you can appreciate how complex it quickly gets. All the above work was using broad assumptions and simplifications. I suppose the short answer to all of this is that stars orbit galaxies in a complex, yet closed orbit which is not easily described precisely (even for our own galaxy) via calculable equations. We can approximate and do our best to work through the math, but in the end its an approximation. In the crudest approximations though, you may as well consider an orbit such as our star to be circular and be done with it.


Are there any astronomical phenomena that could emit strong radio waves with multiples of a discrete frequency? In the New Scientist article Is this ET? Mystery of strange radio bursts from space, it is reported that several times since 2001, astronomers have detected fast radio bursts that seem to have a frequency of multiples of 187.5.  In the article, the theory that it could be a pulsar is discounted.

Are there any astronomical phenomena that could emit strong radio waves with multiples of a discrete frequency?

I think you've misunderstood the article - the quantity that seemed to be occurring at integer multiples of some number isn't the frequency of the radio emission but rather the dispersion measure (DM) of the source. As photons travel through the interstellar medium, interactions with free electrons mean that lower frequency photons take longer to reach the observer than higher frequency photons. The time delay of between photons with frequency $\nu_1$ and $\nu_2$ is
$$t_1-t_2=4.15\left(\frac{\text{DM}}{\text{pc cm}^{-3}}\right)\left[\left(\frac{\nu_1}{\text{GHz}}\right)^{-2}-\left(\frac{\nu_2}{\text{GHz}}\right)^{-2}\right]\;\text{ms}$$
with the dispersion measure defined as the integral of the free electron number density $n_e$ over the path between the source and the observer:
$$\text{DM}\equiv\int n_e\;d\text{l}$$
In a sense, the dispersion measure is a distance proxy. If we had some uniform $n_e$, then the DM would scale perfectly linearly with the distance to a source. That said, the Galactic electron density is not uniform in all directions; it's obviously much higher in the disk, meaning that a source inside the disk some distance from us will have a lower DM than a source in the halo the same distance away. Note that the DM isn't an intrinsic property of the source - it depends on the observer's location.
The point of the paper cited in the article (Hippke et al. 2015) is that the dispersion measures of the 11 fast radio bursts (FRBs) appeared to fall near integer multiples of 187.5 pc cm$^{-3}$. There's certainly variation, although I don't know how the deviations compare to the measurement uncertainties. The authors argue that this could be explained if FRBs are actually artificial - in other words, if there's some sort of human-made source emitting at different frequencies with each frequency emitted at a different time mimicking the effects of interstellar dispersion.
Our knowledge of FRBs has since grown as the sample size has expanded dramatically, and most known DMs vary quite a bit from the trend. I don't know if anyone has looked at this small sample again, but at this point, the extremely broad consensus is that FRBs are astrophysical (mostly extragalactic) sources. A broad range of mechanisms have been proposed over the years, with some since ruled out and others supported by further observations.
A note on pulsars: The article isn't stating the pulsars aren't FRB sources - the sample size back then was far too small to rule them out - but rather that fast radio bursts differ significantly from typical pulsar emission, which tends to be uniform. There are still viable FRB mechanisms involving pulsars, such as giant pulses of young pulsars, and neutron stars in general play a role in some currently-favored theories, such as magnetar flares.
All that said, I think it's worth answering the title question. It is indeed true that some radio sources emit at integer multiples of some frequency - namely, spectral lines from rotational transitions in molecules. In a molecule with moment of inertia $I$, the energy of the state with the angular momentum number $J$ is
$$E_{\text{rot}}=\left(\frac{\hbar^2}{2I}\right)J(J+1),\quad J=0,1,2,\cdots$$
A transition from $J$ to $J-1$ then releases an energy
$$\Delta E_{\text{rot}}=\frac{\hbar^2J}{I}$$
and leads to the emission of a photon with frequency
$$\nu=\frac{\hbar J}{2\pi I}$$
As a notable example, the $\text{CO}$ molecule$^{\dagger}$ emits a photon at $\nu\approx115\;\text{GHz}$ during the $J=1\to0$ transition, at $\nu\approx230\;\text{GHz}$ during the $J=2\to1$ transition, and so on. Keep in mind, though, that fast radio bursts are broadband sources, and while spectral radio lines will exhibit some broadening through certain mechanisms, it's simply not comparable - the emission profiles are completely different.

$^{\dagger}$Specifically, the common isotope $^{12}\text{C}^{16}\text{O}$. Other isotopes like $^{13}\text{C}^{16}\text{O}$ have corresponding lines at slightly different frequencies.


Is there a general term for epicycles, deferents, and eccentrics in Ptolemaic astronomy? According to Ptolemy's (c. 150 CE) account of the motions of planets, planets moved in circular paths ("epicycles") around center points that in turn moved around the center of the earth along a path called a "deferent", or around point near the center of the earth, in a path called an "eccentric".  (I think I have this right.)

Is there a general term for these three kinds of paths?  "Orbit" doesn't seem correct, since none of the three are orbits in the modern sense.

I'd also be interested in knowing of a general term of the centers of these circles.

EDIT: I originally posted this on History.SE, because I think of it as a question for historians of science, and the History of Science SE site is in Area 51.  Someone at History.SE suggested it would be better to ask this question on English.SE.  I did, but I left the original question up at History.SE.  The question got downvoted at English.SE, and was migrated here.  There is now an answer at History.SE that I've accepted. Perhaps this question should be deleted from Astronomy.SE?  Or I should not have posted it in two places?  I'm reluctant to delete a question for which someone has posted an answer.

Orbit can also mean "the usual course of one's life or range of one's activities" so I would find Ptolemic orbit quite acceptable if you hesitate to use just orbit when the path isn't elliptical.


Challenges when trying to build a telescope using two lenses I have some lenses - a 10 cm and 7 cm convex lense and a 4 cm and 1 cm concave lenses. I am trying to make a telescope by combining one of convex and concave lenses, but I am struggeling to do so. The picture that is created using these two lenses doesn't show anything other than the thing that can be seen by eyes. Is it possible to make simple telescope using these lenses?
PS: All of numbers are lense diameters.

Diameter is important, but if you're trying to get any magnification out of that system, you need to look at the focal length.

If you have a very distant light source, like the Sun, focused through a convex lens, until the image of the source is as small and as clear as possible, then the focal length is the distance between the lens and the image you're creating. Like when you burn paper with the lens, focal length is the lens-paper distance.

For concave lenses focal length is a little more tricky to define, but they do have a focal length anyway.

If F is the focal length of the forward lens (the objective), and f is the focal length of the lens near your eye (the ocular, or the eyepiece), then the magnification of the instrument is:

M = F / f

So you need an objective with a long focal length, and an eyepiece with a short focal length.

Both the objective and the eyepiece could be convex actually. It's just that the image will be reversed. If the eyepiece is concave, the image will be straight-up, but the field of view of the telescope will be narrow (like looking through a peephole).

If you use both convex lenses, the distance between them need to be close to the sum of their focal lengths, F + f. Start there and adjust it slightly for best results.

If you use a concave lens for the eyepiece, then the distance between lenses needs to be the difference of their focal lengths, F - f.

Both lenses need to be perpendicular on their common axis, to minimize aberrations.

It helps if they are mounted in a tube, to reduce glare, but it's not absolutely necessary. They could simply be attached to a long rod or something - but you'll get a lot of glare into the eyepiece that way.

Do not necessarily try to get very high magnification. If the diameter of the objective is D, measured in mm, then the maximum usable magnification is 2D. So an objective 100 mm in diameter could never provide more than 200x magnification - I mean it could, but the image would be blurry. Even then, this is a theoretical limit, and in practice the image will get pretty bad long before you reach the limit, unless you use high quality lenses designed specifically for telescopes, etc. Also, at high magnification it's difficult to hold the scope steady, and the image is jumping all over the place.

Try and make a 5x ... 8x instrument first, see how that goes. A lot of binoculars are in this range.

I've built several refractor telescopes from lenses when I was in high school. It's totally doable. It just takes some patience to figure out how to solve all problems.


Neutrino interaction with dense molecular structures Would dense molecular structures on heavier planets (crystalline or other compounds which are generally unknown to us) allow neutrinos to pass through as easily as on Earth?

Neutrinos wouldn't pass through white dwarfs or neutron stars as easily as through Earth, measured by the scattering rate per distance, since these types of stars are very dense: white dwarfs about one million times denser than water, neutron stars about as dense as atomic nuclei.

To scatter a neutrino with a chance of > 50%, you "just" need the inner of a neutron star along 10 times the diameter of Earth's orbit around the sun:

The neutrino-nucleon scattering cross section is about $10^{-38}cm^2=10^{-42}m^2$ for neutrinos with energies above about 1 GeV, according to this paper. A proton has a diameter of about $1.75\cdot 10^{-15}m$.
With $$1.75\cdot 10^{-15}m/10^{-42}m^2=1.75 \cdot 10^{27}$$ we need about $1.75\cdot 10^{27}$ nucleons in a row to scatter a neutrino. That's about $$1.75\cdot 10^{27}\cdot 1.75\cdot 10^{-15}m=3\cdot 10^{12}m=20\mbox{ a.u.}$$ as a rough estimate. (For simplicity I've neglected the exponentially decreasing intensity of the neutrino beam in the scattering medium.) Neutrinos will probably scatter a little more easily in a neutron star than in the same number of single traversed neutrons, but that's a guess based on the scattering cross section for heavier nuclei.


How to measure distances to stars by means of spectroscopic parallaxes? How to measure distances to stars by means of spectroscopic parallaxes on practice?
What is the accuracy of measuring distances using this method compared with distances based on HIPPARCOS trigonometric parallaxes? My sample of objects consists of A-B type stars.

Actually the method described on Wikipedia is not the method that is meant by Spectroscopic Parallax. To determine the spectroscopic parallax, you'll need a spectrum of the star and measure the widths of the spectral lines. Compact stars have higher surface gravity, which means that their spectral lines are broader. This means that the radius of the star can be calculated from line widening. Using the observed effective temperature $T_{eff}$ and the radius R, you can get the Luminosity from Stefan's Law, which gives an absolute magnitude and together with the apparent magnitude, a distance modulus and therefore a distance.
This gives an accuracy of 10% in luminosity and 5% in distance for main sequence stars and 25-30% in distance for giant stars. (see de Grijs, 2011)

Wilson and Bappu (1957) proposed to use the width (W_0) of the emission core of the CaIII absorption line at 3933 Angstrom. Using data from Hipparcos the relation between the absolute magnitude in V and the width of the absorption line was calibrated to:

$M_V = 33.2-18.0 \log(W_0)$

by Pace er al. (2003).
So to use this method you would need to be able to measure the width of the CaIII absorption line, but other calibrations using the MgIIk line also exist.

As these distance measurements are calibrated using Hipparcos, they will never be able to be more accurate until better calibration data is available (i.e. with Gaia).

The much more simpler method as described on Wikipedia might also be used but that is even less accurate. NB. The relation between the colour (colour index B-V) and the spectral class is not linear. Many Hertzsprung-Russel diagrams show both colour and spectral type on the X axis but this is not really correct. A K0Ia supergiant has a different colour index than a K0V star on the main sequence.
On page 50 in de Grijs, 2011, this relation is shown in Fig 2.11. It should also be somewhere on Wikimedia (to which it is attributed) but I cannot find it.
EDIT: Here it is: http://en.wikipedia.org/wiki/File:H-R_diagram.svg


How distances are measured in the Solar System, in the Galaxy, and in the Universe? Distances in our universe can be calculated in several ways:

Stellar Parallax
Standard Candles
Redshift

How are these methods linked, and how they are actually used to calculate   distances are measured in the Solar System, in the Galaxy, and in the Universe?

Stellar Parallax
Stellar parallax uses differences in perspective to determine the distance from an object. When the earth goes around the sun, our perspective of the star, galaxy etc. changes and so the angle from us to the object changes. Because we know how the earth moves around the sun, we know the distance between the points that we take the measurements. From this we can use the Sine Rule to determine the distance from the object.
For Example, if the distance between the two points of measurement were 2 units, and the two inside angles (A and B) were (A)80° and (B)75° (and therefore the other angle (C) is 25°) we can say that:
$a = \frac{2\sin80}{\sin25}\ $
where "a" is the opposite side to angle "A". This means that the distance between the object and the second (B) point of measurement is approximately 4.66 units.
Diagram here.

Standard Candles
 A standard candle is an astronomical object that has a known absolute magnitude. (source)
By using the known absolute and the measured apparent magnitude (what is seen by instruments etc.) it is possible to determine the distance of the object by using the formula:

m-M = 5 log d – 5
where m is the apparent magnitude of the object, M is the absolute magnitude of the object, and d is the distance to the object in parsecs.(same source)

One of the most widely used examples of these standard candles is the supernovae type Ia, which was used to discover that the universe is accelerating in its expansion. This led to the awarding of the 2011 Nobel Prize in Physics to Saul Perlmutter, Brian Schmidt , and Adam Riess.
Additionally, RR Lyrae are another form of standard candle used to measure galactic distances (SN Ia are much more luminous and therefore can be used to much higher distances). RR Lyrae are a type of variable star with a relationship between pulsation period and absolute magnitude, which is what makes them good for use as standard candles.
Other information here.

Redshift
Redshift is the effect where light from an object is "stretched" out so that the wavelength is moved towards the "red" end of the spectrum. Distances are calculated by using Hubble's Law, but I'll let others explain it: here, here and here.


Longest and shortest wavelength What is the longest and what is the shortest wavelength of light? How fast would a light-emitting object have to recede in order to shift the frequency of light to increase the wavelength from the shortest to the longest? (If there are limits)

Regarding visible light, the nominal range is usually given as 400 nm (violet) to 700 nm (red). For light of the former wavelength to get Doppler-shifted to the latter means a z value of (700–400)/400 = 0.75, corresponding to a recession of 152,000 km/s, just over half the speed of light.


Did big bang happen only one time or it might happen again somewhere million time away from our observable universe? We know, everything is in existence due to big bang from singularity. Time also started from that big bang point.

So, did big bang happen only once or it might happen again somewhere million time away from our observable universe ?

We don't know it, because the question isn't feasible thus far to experiments or observation.

But it's possible to develop theories far beyond our observable universe, even far beyond the multiverse theories described here in Wikipedia. There exist theories about a tranfinite hierachy of meta-levels of theories in mathematical physics, each level comprising all possible theories on the level below (beyond any notion of infinity) and all their interconnections (including hierarchies of interconnections between interconnections). To explain this in reasonable detail would need hundreds of mind-boggling pages, at least. So I won't even try to explain more details. At the end it's not even clear, whether there is a difference between mathematics (including a transfinite number of meta-levels) and physics (including a transfinite number meta-levels).

The anthropic principle tries to answer the follow-up question "Why do we live in exactly this universe with exactly these laws of nature?". But this kind of answer isn't satisfying from a scientific point of view, when looking for cause and effect.

If you're looking for a final answer comprising everything: There is no, because the notion "everything" leads to paradoxa, when used in an absolute sense. (See Gödel's incompleteness theorems, if you need more details.)


Is there really a blood wolf moon this month? Just wondering so I can see it because I don't know if  it is happening this month or not?

On January 20th, 2019, there will be a total eclipse of the moon. Since Rayleigh scattering of sunlight passing through the earth's atmosphere and then hitting the moon during a total lunar eclipse makes the moon look reddish, total lunar eclipses are sometimes given the (portentous and slightly silly) name of "blood moon". (This seems to be the result of vague New Testament references to lunar eclipses being promoted by a couple of apocalyptic fundamentalists in the last decade; see here for more.)

In addition, there are various folk traditions which give special names to the (first) full moon in each month; it's apparently an old Anglo-Saxon tradition to call the first full moon in January the "Wolf Moon". Since total lunar eclipses necessarily happen during full moons, you could thus (somewhat) legitimately call this a "blood wolf moon". (Some discussion of other "full moon names", including the "harvest moon", can be found here.)

Finally, there is a recent trend of referring to a full moon that happens near to lunar perigee (when the moon makes its closest approach to the earth and appears somewhat larger in the sky) as a "super moon"; this term was apparently coined by the astrologer Richard Nolle in 1979. This sort of thing happens three or four times a year.

So you could call the upcoming lunar eclipse on January 20th a "blood wolf super moon" (or "super blood wolf moon", or whatever), if you really wanted to....

(Personally, I don't think that's nearly enough in the way of unnecessary and portentous adjectives. I'm holding out for "blood wolf demon vampire super extreme power moon".)


What is the Terra phase? Does the Moon have a night and day like Earth? When sitting on the Moon and viewing the Earth what does it look like when the Earth goes through phases like the Lunar Phase? How does the sun travel through the Moon's sky facing the Earth like this gif?

Part of my question was a duplicate so I deleted it and posted the other link below.
How bright is the full Earth during the lunar midnight?

The Earth-Moon System is a system where the moon is tidally locked to the planet, but the planet is not tidally locked to the moon. Earth has a 24 hour day, whereas the Moon has a 29.5 Earth day long day. It takes the Moon 29.5 Earth days to orbit around the Earth once as well, which means the Moon is tidally locked in a 1:1 ratio. This means that if you were to stand on a particular point on the Moon, the Earth would not appear to move because the Moon is spinning on its axis at a rate such that it matches how much the Earth should appear to move from orbiting around it. Basically, the same side of the moon is always facing the Earth, which is why there is a concept of a "dark" side of the moon. The "dark" side is not always actually dark, it's just that no one had ever seen it before until we sent some satellites up there to look at it.

Here is a gif showing tidal locking. The tidal locked moon is on the left.

Now imagine you are back on Earth. Throughout the course of the day you can see the Moon move across the sky. That's because 24 hours is drastically shorter than 29.5 days. However, if you were standing on the Moon, the Earth would appear to stand still. Now, it would actually appear to wobble a bit because of the Moon's orbit's eccentricity, but it more or less stays in the same place.

Here is an artistic yet scientifically accurate rendition of what it would look like. It's from Space Engine. Notice how the gas giant (the planet) wobbles a bit but stays in roughly the same place. It would look similar if you were on the Moon, but without the atmosphere, and the Earth would obviously be smaller and further away.

Yes, the Earth would go through the same phases as the Moon. It would complete a cycle every 29.5 days because that is how long it takes for the Moon to orbit around the Earth. However, something different is that you would be able to see the Earth rotate, since it takes only 24 hours to rotate. When we look up at the Moon from Earth, as I said before, we cannot see it rotate because it is tidally locked, and thus only one side gets shown.

Brightness depends on the phase, but generally the Earth is brighter from the Moon than the Moon is from the Earth because the Earth has a higher albedo (reflects more light) than the Moon, and the Earth appears larger from the Moon than the Moon appears from the Earth. The Moon takes up roughly 30 arcminutes of the sky, whereas the Earth takes up roughly 1 degree and 50 arcminutes.

At full moon, the Moon has an apparent magnitude of roughly -12.75. At full earth, the Earth has an apparent magnitude of -16.75. That's about 40x brighter! If we were to qualify phases as a % from 0% being a new moon to 100% being a full moon, then every 10% you go below 100%, the apparent magnitude of each phase decreases by about the same amount. For example: 90% Earth would be roughly -16.5 (+.25 magnitude), and 90% Moon would be roughly 15.5 (also +.25 magnitude). The differences in magnitude only get larger when you go below 10% phase, where each little bit of the celestial body that shows adds a lot more to the magnitude.

Here is a rough chart:

If you have more questions feel free to ask.


Can a dying, swelling, star be rejuvinated by enveloping orbiting gas giants? Many hot jupiter type exoplanets have been found, orbiting near their parent stars.

At the end of a stars life, they swell up and sometimes envelop closely orbiting bodies.

Combine these effects and what could/would happen? Could the unfused hydrogen of a gas giant be dragged into the stars core and end up rejuvinating the star and extend its lifetime? Have these conditions been contemplated before? What might be expected to happen? Or is the environment of a stars outer layers just not dense enough to keep a gas giant from just continuing in its orbit?

Stars turn into Red Giants not because they're running out of fuel, but because they're accumulating material they can't use for fusion (yet) in the core. The star isn't so much dying of starvation as it is wallowing in its own muck.

Red giants form when the fusion is no longer taking place in the centre of the star, but instead in a shell around the centre. The centre is filled up with the products of fusion, which themselves cannot fuse until temperatures and pressures get high enough. The exact reason why this results in red giant formation eludes me still, but you'll find plenty of questions asking about it on this SE.

Furthermore, most stars that are larger than red dwarfs never fully use up their fuel. There is often plenty of hydrogen in the outer reaches of the star, way more than you'll find in Jupiter (which, as zeta-band mentions, is only about 1/1000th the mass of a sun-mass star).


Are features in the CMB correlated with features in the SDSS map of the most distant galaxies? This is a follow up to an earlier question to which @pela gave an excellent answer. Apparently the statistics of the length-scales of the CMB fluctuations are similar to those obtained from other measurements.
This question is more specific in asking whether any of the structures observed in the CMB are also observed in the distribution of the most ancient galaxies such as from the Sloan Sky Survey SDSS.  So this is specifically asking whether there is a significant positive cross-correlation between between the different measurements (rather than whether their autocorrelations are similar).

There shouldn't be any correlation. The CMB light that we see is from a spherical region in the early universe. Its homogeneity strongly suggests that the interior of the sphere was just as homogeneous, but we can't actually see CMB light from the interior. The galaxies that we can see formed from matter inside the sphere, and quite far from the edge. Between the CMB redshift ($z\approx 1090$) and the redshift of the most distant observed galaxy ($z\approx 11$), light traveled a comoving distance of roughly 14 Gly, around 30% of the radius of the sphere. Therefore only the very largest features of the CMB should be reflected in the distribution of observable galaxies.


More than 67 Jovian moons? Wikipedia says that there are now 67 known moons of Jupiter.

It got me wondering: is it really possible in our day and age that there are more, but we haven't observed them yet?

On the other hand, there is a Jovian moon discovered as recently as in 2011, so I wouldn't be amazed if the answer was 'yes'.

Really though, choosing Jupiter for this question was a bit arbitrary on my part, as the same question can be asked for other more distant and bigger planets of our solar system, like Saturn.

Yes, there probably are more.

There is a minor problem of definition. Unlike the case for "planet", the IAU has not produced a definition of "moon". There is no requirement to be round, or have cleared an orbit. So any chunk of rock or ice that we can see orbiting a planet can reasonably be called a moon, and Jupiter has, over the years, picked up a lot of stuff: Small asteroids that it has captured from solar orbit.

It is likely that there are other, sub-1km sized objects in orbit around Jupiter, that we haven't noticed yet.

Juno is, as of 2016, in orbit around Jupiter. However her main focus of study is the planet itself, not hunting for moons. The recent moon discoveries have been made by terrestrial observations.


Would a low albedo substance warm up Mars? If Mars were coated in a dark substance that absorbs sunlight (including infrared light), would this heat up Mars?  I am envisioning a fine dust that is dispersed into the atmosphere, but possibly settles on Mars's surface all around.  If we were able to change the albedo, possibly down to as low as 1%, would that make Mars habitable (temperature wise)?  If the albedo were changed all the way to 1%, could this even make Mars too hot?

Of course! Actually this is one of the possible ways we have in order to make Mars friendlier
to humans. This technique is called Terraformation and it inludes the altering of the Mars surface's albedo. However, the feasibility of this plan is yet questionable and there are maybe better ways to achieve this increase in the temperature like CO$_2$ emissions. Here is a nice article discussing exactly this subject.


Searches for dipole form of anisotropies (on luminosity distances) in the universe with type Ia supernovae There are many articles like this one Testing the isotropy of the Universe by using the JLA compilation of type-Ia supernovae (PDF, arxiv.org) trying to search for a dipole effect in cosmology with type Ia supernovae (used as standard candles). The idea of this kind of search is to do a test of the cosmological principle on a phenomenological approach independantly of some given models (e.g. Tolman-Bondi anistropic model).

They consider a dipole effect on the distance modulus $ \mu $: $$ \mu \leftarrow \mu \times \left(1+A_D (\hat{\textbf{n}}\cdot \hat{\textbf{p}})\right)$$ where $\hat{\textbf{n}}$ is unitary vector pointing in the direction $(l,b)$ of the dipole with an amplitude of $A_D$ and $\hat{\textbf{p}}$ points the direction of each Type Ia supernova.
The distance modulus is related to luminosity distance $d_l$ as $ \mu =5 \log \left(\frac{d_l}{10 pc}\right)$.

There is something that bothers me a bit:

Our peculiar velocity to the universe is usually estimated with the CMB dipole measuments, right? And so the reshifts of SNIa are computed in the CMB frame using this correction.

So, supposing we have that kind of cosmological ansisotropy, how it is not mistakenly corrected with the dipole anisotropy of the CMB (and could also be why all searches of this kind give us an amplitude of anisotropy compatible to 0)?

I mean, the CMB dipole implies a velocity of $369.5\pm3.0$ km/s in the direction of $l=264.4^{\circ}\pm0.3^{\circ}$ and $b= 48.4^{\circ}\pm0,5^{\circ}$ according to COBE measurments. But what if the CMB dipole is not just due to our peculiar velocity but also to an unknown "cosmological effect"? This effect would be corrected and nothing will be see when try to make a dipole fit on SNIa measurement, right?

Isn't the point of using type Ia supernovae that they have an estimated distance that is independent of redshift?

${\bf \hat{n}}$ here is not the direction of the CMB dipole, it is a free parameter in their model.

If distances were being estimated using redshift I guess you would have a point.


What could be past the particle horizon? Forgive any kind of "dumb" question I may ask as this is a new interest of mine and I know it's purely hypothetical.

If one were able to surpass light speed and the expansion of space to go beyond the particle horizon, what could be out there? Could it just be an empty void? Could it loop around to the other side of the particle horizon? I'd love to read about any kind of theories or research done on this subject.

It sounds like your question is about the single most important principle of cosmology, called the cosmological principle.  This asserts that as long as all our observations are consistent with the idea that the universe is the same everywhere at a given age, we will continue to use a model with that attribute.  There are lots of good reasons for adopting that principle, the least of which is not that it allows us to have a model in the first place.  So what this means is, our answer to what is beyond what we can see is that we don't know, but we are going to model it in a way that makes sense to us, so we will use the cosmological principle.   You have to be able to distinguish what we know in the sense of having direct information about, versus what we know about our own best models.


Why can we detect gravitational waves? Now that LIGO has finally measured gravitational waves using a huge laser interferometer, to me, the question remains, why was it possible? As it is explained in many news articles, gravitational waves are similar to water waves or electromagnetic waves, they just do not exist in a medium like water or space, but space-time itself is the transport medium. If space-time itself gets contracted and expanded by the gravitational waves, so does any means of measurement. The ruler you use for measurement (the laser beam) gets deformed while the wave travels through the measuring device. Otherwise the "ruler" had to live outside of space-time, but there is no outside. If space-time was a cup filled with pudding, on which we had painted a straight line with 10 marks, pushing into the pudding slightly with our thumb does bend the line, but for us, there remain 10 marks on the line, because to measure the extension, we had to use a ruler, outside of our space-time (pudding) to measure, let's say, 11 marks. But, well, there is no outside. I assume the same happens not only to the 3 spacial dimensions but also to the time dimension. Because they "did it", what am I missing?

The short answer is that waves that are "in the apparatus" are indeed stretched. However the "fresh waves" being produced by the laser are not. So long as the "new" waves spend much less time in the interferometer than it takes to expand them (which takes roughly 1/gravitational wave frequency), then the effect you are talking about can be neglected.

Details:

There is an apparent paradox: you can think about the detection in two ways. On the one hand you can imagine that the lengths of the detector arms change and that the round-trip travel time of a light beam is subsequently changed and so the difference in the time-of-arrival of wavecrests translates into a phase difference that is detected in the interferometer. On the other hand you have the analogy to the expansion of the universe - if the arm length is changed, then isn't the wavelength of the light changed by exactly the same factor and so there can be no change in the phase difference? I guess this latter is your question.

Well clearly, the detector works so there must be a problem with the second interpretation. There is an excellent discussion of this by Saulson 1997, from which I give a summary.

Interpretation 1:

If the two arms are in the $x$ and $y$ directions and the incoming wave the $z$ direction, then the metric due to the wave can be written
$$ds^2 = -c^2 dt^2 + (1+ h(t))dx^2 + (1-h(t))dy^2,$$
where $h(t)$ is the strain of the gravitational wave.

For light travelling on geodesic paths the metric interval $ds^2=0$, this means that (considering only the arm aligned along the x-axis for a moment)
$$c dt = \sqrt{(1 + h(t))}dx \simeq (1 + \frac{1}{2}h(t))dx$$
The time taken to travel the path is therefore increased to
$$\tau_+ = \int dt = \frac{1}{c}\int (1 + \frac{1}{2}h(t))dx$$

If the original arm is of length $L$ and the perturbed arm length is $L(1+h/2)$, then the time difference for a photon to make the round trip along each arm is
$$ \Delta \tau = \tau_+ - \tau_- \simeq  \frac{2L}{c}h$$
leading to a phase difference in the signals of
$$\Delta \phi = \frac{4\pi L}{\lambda} h$$
This assumes that $h(t)$ is treated as a constant for the time that the laser light is in the apparatus.

Interpretation 2:

In analogy with the expansion of the universe, the gravitational wave does change the wavelength of light in each arm of the experiment. However, only the waves that are in the apparatus as the gravitational wave passes through can be affected.

Suppose that $h(t)$ is a step function so that the arm changes length from $L$ to $L+h(0)/2$ instantaneously. The waves that are just arriving back at the detector will be unaffected by this change, but subsequent wavecrests will have had successively further to travel and so there is a phase lag that builds up gradually to the value defined above in interpretation 1. The time taken for the phase lag to build up will be $2L/c$.

But then what about the waves that enter the apparatus later? For those, the laser frequency is unchanged and as the speed of light is constant, then the wavelength is unchanged. These waves travel in a lengthened arm and therefore experience a phase lag exactly equivalent to interpretation 1.

In practice, the "buildup time" for the phase lag is short compared with the reciprocal of the frequency of the gravitational waves. For example the LIGO path length is about 1,000 km, so the "build up time" would be 0.003 s compared with the reciprocal of the $\sim 100$ Hz signal of 0.01 s and so is relatively unimportant when interpreting the signal (the detection sensitivity of the interferometer is indeed compromised at higher frequencies because of this effect).


Maximum spin rate of a black hole? I have just been watching a podcast called "Deep Astronomy" and the discussion was about a super fast spinning black hole discovered with the NuSTAR space observatory. This black hole was modeled with high confidence to be spinning at about 99% of the maximum spin rate. They stopped short of saying that the tangential velocity of this spin rate is "c" (and how can a singularity have a "tangential velocity"?) They did say that the event horizon at maximum spin of a stellar black hole is about 1-1/2 km. and that if a black hole were to spin faster then the result would be a "naked black hole" that would defy the laws of physics (GR).

Also, shouldn't all black holes spin extremely fast (conservation of angular momentum) or would a retrograde accretion disk slow it down. Could somebody clarify this whole "black hole spin thing" without getting too complicated??

Since I like math, let's throw some math into this. I'll try to keep it as simple as possible though.
Kerr Black Holes
A rotating black hole is known as a Kerr Black Hole (named after Roy Kerr who found the numerical solution to GR equations for rotating black holes). In the case of a rotating black hole, there are two important parameters used to describe the black hole. The first is of course the mass of the black hole $M$. The second is the spin $a$. Really $a$ isn't the spin itself $-$ it's defined by $a=J/M$ (see footnote) where $J$ is the angular momentum of the black hole $-$ but it's a good proxy for spin so often you'll see scientists get lazy and just call it the spin of the black hole. The mathematics will tell you that Kerr black holes have the limitation that
$$0 \le a/M \le 1$$
Black Hole Event Horizon
The important parameter we want to calculate is the black hole's radius. If you run through the math, you find that this radius is given by
$$r_e = M+(M^2-a^2)^{1/2}$$
In the case when $a/M=0$ (and thus $a=0$), this reduces to just $r_e=2M$, or in regular units (instead of geometrized units) $r_e=2GM/c^2$. Hopefully you can see that this just reduces to the normal Schwarzchild radius for a non-rotating black hole and thus the equation above is a generalization to account for spin. Let's look at the other limit when $a/M=1$ (and thus $a=M$). In this case, you find that the radius is $r_e=M$. When $a/M=1$, you have a maximally rotating black hole, and your radius is half the normal Schwarzchild radius of a non-rotating black hole. This equation defines the radius of the Event Horizon, the point after which there is no returning from the black hole.
Ergosphere
As it turns out, when you define your equation to calculate the radius of the black hole, there are actually multiple solutions! The section above shows one such solution, but there's another important solution as well. This radius, sometimes called the static limit is given by the equation
$$r_{s} = M+\left(M-a^2\cos^2(\theta)\right)^{1/2}$$
Notice that this is almost exactly the same as above, except for that extra $\cos^2(\theta)$. This defines a different, slightly larger and somewhat "pumpkin-shaped" horizon which encompasses the inner Event Horizon defined above. The region between this outer horizon and the inner horizon is known as the Ergosphere. Without getting into the nitty gritty details, I'll just say that one important point about the Ergosphere is that anything inside it (that is, $r_e<r<r_s$) must rotate exactly with the black hole - it is physically impossible to stay still here!
Answers

They stopped short of saying that the tangential velocity of this spin rate is "c" (and how can a singularity have a "tangential velocity"?)

When you talk about the tangential velocity, there are multiple components of this black hole you/they may be talking about. One such tangential velocity is the tangential velocity of the event horizon (defined by $r_e$ above). We can take a look at the case of a maximally rotating black hole and say that the angular momentum, based on the equations above, of such a black hole is given by
$$J_{max} = a_{max}Mc = M^2c$$
Note that I've dropped the geometrized units just to be completely explicit. This has introduced an extra $c$ now. Remember that $a_{max}$ is achieved when $a/M=1$.
We can also define the angular momentum using the standard equation from physics 101, $J=rMv_\perp$, where of course $r$ is the radius of your object, and $v_\perp$ is the perpendicular, or else tangential, velocity of your spinning object. Recall from above that for a maximally rotating black hole, $r_e=M$ so we also have that
$$J_{max} = r_eMv_{\perp} = M^2v_\perp$$
You can see that these two equations for $J_{max}$ only equal each other if the tangential velocity $v_\perp$ is equal to the speed of light $c$. So yes, you are correct to presume that at the fastest possible rotations, the black hole's event horizon is rotating at the speed of light!
I said though that there are multiple components you could talk about when discussing rotating black holes. The other, as you allude to, is the rotating singularity. You correctly point out - "how can a singularity have a tangential velocity"? As it turns out, Kerr black holes don't have point singularities, they have ring singularities. These are "rings" of mass with zero width but some finite radius. Almost like a disk of no height. These rings of course can then have a tangential velocity. You were correct to be suspicious of a point singularity having tangential velocity though. That is not possible.

They did say that the event horizon at maximum spin of a stellar black hole is about 1-1/2 km. and that if a black hole were to spin faster then the result would be a "naked black hole" that would defy the laws of physics (GR).

We know the equation exactly, since I defined it above. The radius of a stellar black hole (that is a black hole with mass exactly equal to the mass of the Sun, $M_\odot$) is given by
$$r=\frac{GM_{\odot}}{c}=1.48\:km$$
So yes, they were correct on their radius. They also state that spinning any faster results in a naked singularity. This is entirely true. To see this, go back to the equation for the event horizon. Remember that our upper spin limit is that $a=M$. What happens to our event horizon radius when $a>M$ (and thus $a/M>1$)? For arguments say, lets say $a=2M$. Then our event horizon radius becomes
$$r_e=M-(M^2-a^2)^{1/2}=M-(M^2-4M^2)^{1/2}=M-(-3M^2)^{1/2}=M-i\sqrt{3}M$$
Suddenly our radius is complex and has an imaginary component! That means it is not physical and thus cannot exist. Now that we don't have an event horizon, our singularity can't hide behind it and is "naked", exposed to the universe for anyone to see. GR tells us such an event shouldn't be allowed to happen because it results in all sorts of violations of physics. So somehow, something has to prevent black holes from spinning faster than a maximal black hole.

Shouldn't all black holes spin extremely fast (conservation of angular momentum) or would a retrograde accretion disk slow it down.

Yes, that is true in general. All black holes should spin extremely fast, simply because of conservation of angular momentum. In fact, I don't think I can come up with a case where a black hole was found to be not spinning. Shown below is a plot from this Nature paper which shows the measured spin of 19 supermassive black holes. They're all spinning pretty fast with some of them almost at the speed of light. None of them are even close to not spinning.

Footnote: In GR, to make the math easier, often scientists adopt special units known as geometrized units. These are units chosen in just such a way that the gravitational constant, $G$, and the speed of light, $c$, are equal to one. There's infinitely many units which allow this. Essentially this means no GR equations have $G$ or $c$ in them, but they're implicitly there, they're just equal to one and so not shown.


If the Milky Way galaxy is as big as planet earth, how big is Earth? If the Milky Way galaxy is as big as planet earth, how big is Earth?

Well, they are different shapes, but we can get a rough answer by doing a ratio calculation based on their radii.

Let $g$ be the radius of the galaxy, about $5\times 10^{20}\ \mathrm m$.
Let $e$ be the radius of the Earth, about $6371\ \mathrm{km}$.
Let $s$ be the radius of the scaled-down Earth.

Then,
$$\frac e g = \frac s e$$
So $$s = \frac{e^2} g = \frac {(6371\ \mathrm{km})^2} {5\times 10^{20}\ \mathrm m}$$
which is approximately 80 nanometres. So it's about 1000 times the size of an atom.


Did I see a supernova explosion? I think I just saw a supernova explode with my own eyes through my GSO 12 inch Dobsonian. Please tell me what it was! I am still trying to find what it was it was mindblowing!
I went out on my roof around 18:20 (New Delhi, India) and was looking up thinking we are going to have clear skies tonight.
Suddenly I could see the bright star without the telescope and I was surprised to see such a bright object near zenith so I ran in and got my telescope out.
When I looked through the scope It was looking like a bright bubble very shiny and the outer surface seemed brighter. I thought my collimation was out so I collimated the scope and it was not too off.
After collimation, I could see a small object in orbit around the star/object. As I continued to observe it, it suddenly exploded like fireworks and tiny particles shining very brightly started to swirl in waves and dimmed quickly. This was around 18:38 I took pictures with my phone through the eyepiece and the sky was still lit and blue in colour.
The below pics are edited in snap seed on my phone.

Supernovae increase in brightness over several days and decrease over months.  Thus, whatever you saw, was not a supernova, sorry.


Why don't multiple eclipses happen over the course of a few days? From my understanding over the course of the seasons the moon and sun rise at varying locations. Though the day to day differences are no noticeable, to me at least. For instance, in the summer, the sun takes a longer path overhead as a result of the earths tilt towards the sun in the summer months -- in the winter, short paths.

With the seemingly negligible difference in path lengths of the sun from day to day why then do solar eclipses only happen on a single day and not over the course of many days? It seems that if the moon and sun paths during the day were to intersect they should do it multiple days in a row as their path lengths across the sky during the day change.

I assume the answer to this lies particularly with the moons orbit around the earth and its shape/inclination. But I am far from my last astronomy courses and have lost all the jargon and knowledge needed to determine this.

The reason is because the moon moves much faster (in terms of angle per day) than the sun. In fact it moves on average 12.5 degrees per day, and all in the same direction.  Given that the sun is only about 0.5 degrees in diameter, it doesn't take long for the moon to overtake it, and once it's past  it doesn't go backwards.


3 years later, how strong is the evidence for planet 9? I came across post in quora where somebody said that "most astronomers no longer think Planet 9 exists" and being somewhat shocked by that, I did a google search and found a fairly recent article that claims exactly that.

It can take several years to prove a negative within pretty good certainty, as opposed to a discovery which can be checked pretty quickly.   Because I don't want to wait several years, I wanted to ask if the general consensus is that Planet 9 probably doesn't exist?  Or is there still a respectable amount of belief among astronomers in the original evidence and any new evidence that's followed?

If it ends up not existing, I'll be sorry to hear it.   It was a fun story when it lasted.

It's worth noting that Siegel's article seems to extrapolate from one paper (Shankman et al., 2017) to "most scientists" without much justification, so it's not particularly clear what most scientists think anyway (and that's before we get into the question of whether the "most scientists" referred to includes scientists whose discipline is not astronomy-related). Usually in these kind of situations there's a lot of back-and-forth as new objects are found in the outer solar system and the analyses of systematic errors and biases in the various surveys are refined. If there is an actual detection of Planet 9, that would resolve things fairly definitively, the reverse situation is going to be rather more difficult to prove.

In response to the OSSOS paper (Shankman et al., 2017) Michael Brown and Konstantin Batygin released a new analysis of the clustering in the outer solar system, taking into account the OSSOS results that are the subject of Ethan Siegel's article. The paper can be downloaded from the arXiv. Brown also has a Twitter thread summarising the result. From the paper, their conclusion on the apparent absence of clustering in OSSOS:

  That is,
  the uncertainties in the measurement of clustering from
  the OSSOS data are so large that OSSOS would not be
  capable of confidently detecting the clustering seen in
  the larger data set even if it were real and present in the
  OSSOS data. Because of the limited survey region and
  small number of detected objects, OSSOS observations
  are equally consistent with being drawn from a uniform
  distribution of longitudes of perihelion and with being
  clustered in longitude of perihelion as strongly as seen
  in the ensemble data. No conclusions on clustering of
  longitude of perihelion observed in the complete dataset
  can be drawn from the OSSOS data.

They do note that this does not necessarily prove the existence of Planet 9, but in the absence of Planet 9 there would still need to be an explanation for the clustering.

Batygin et al. subsequently posted a review of the Planet Nine hypothesis on the arXiv, favouring a lower planetary mass and a less eccentric orbit than originally proposed. The new parameters do not favour Planet 9 as an explanation for the solar obliquity.

On the anti-Planet 9 side of things, it's worth looking into Kavelaars et al. 2019, a subsequent paper from the same authors as the study based on the OSSOS data, who note that there seems to be an absence of objects with perihelia ($q$) between 50 and 75 AU that would be expected if a planet was causing the alignment, which so far doesn't seem to be explainable by detection bias.

  A survey
  that might have detected a TNO at $a ∼ 500\ \mathrm{au}$
  and $q ∼ 75\ \mathrm{au}$ is actually more likely to have detected objects with similar $a$ but smaller values
  of $q$. The same is true of the other $q > 75\mathrm{au}$
  detections: the lower-$q$ but similar $a$ detections
  are always more likely. Thus, the lack of detections in the $50 < q < 75\ \mathrm{au}$ range may be indicating that there really is an absence of TNOs
  on orbits in this range. This strongly contradicts models of orbital evolution that include
  an additional planet, as the gravitational action of such an object would cause TNOs to
  be distributed across a range of $q$ values at any
  given moment (Figure 2; Shankman et al. 2017a;
  Lawler et al. 2017). Thus, if the lack of objects
  in the $50\ \mathrm{au} < q < 70\ \mathrm{au}$ range is real, the
  hypothesized external planet can be excluded.

It still is an active area of research, so it remains to be seen how the various studies will hold up.


Can a comet orbit a planet? Given that moons commonly orbit planets, why do we never encounter a comet orbiting a planet?

What would happen (exactly, in detail) if a large one did settle into a stable orbit around the Earth?

Sorry... just making the question a bit clearer!  The accepted answer will describe where all the water ends up.

Its very unlikely for a comet to become a satellite of an inner solar system planet. Much less likely than it is for an asteroid. Most asteroids are on fairly circular orbits, and so the relative velocity between asteroids and planets is quite low. In comparison comets have very elliptical orbits, and their relative velocities to the planets are much larger.

For an asteroid to be captured it must lose momentum. This is possible, though rare. For example, a binary asteroid can be captured if it is separated by tidal forces. For an comet with much more momentum, the chance of being captured is much much lower. Asteroids are captured by the Earth moon system, but not into stable orbits, they don't stay long.

If it did occur, the comet would still be active, with a coma of gas, which would be visible just like a very nearby comet. It wouldn't be particularly bright, since the surface brightness of a comet doesn't depend on distance from the Earth.

Over time the comet would run out of volatiles and become more or less indistinguishable from a captured asteroid. If it were in the Earth's orbit it probably wouldn't last that long, as there are not many orbits that are stable in the long term around the Earth, due to perturbations from the moon.

The dust and gas, including water vapour, will initally remain in orbit, forming a faint ring. It will, over time, be disrupted, and either end up in the atmosphere, on the moon, or ejected from the system. A comet doesn't contain enough water to make a difference to the Earth's ecosystem.


How long was the hyper inflation phase? How long was the hyper inflation phase? Haw can we know that it lasted for a period of time and not more or les time?

Googling "hyperinflation" mostly returns articles about Zimbabwe and the Weimar Republic, so I'm going to assume that you are referring to what is usually just called (cosmological) inflation.

We are still pretty uncertain about when and for how long inflation occured, but we do however have some sort of constraints. In general, inflation is thought to be caused by some sort of scalar field, usually written as $\phi(\mathbf{r},t)$, and its associated potential energy $V(\phi)$. Various forms of how this field changes as a function of time exist, but a popular one is, or has been at least, the slow-roll field, which makes it behave like a cosmological constant, enabling it to drive an exponential inflation.

Inflation can be initiated by a symmetry break when the unified forces of electromagnetism and the weak and strong forces are separated in the strong and the electroweak force (the "GUT" era, for Grand Unified Theory), at temperatures of $T\sim10^{28}$ K. It will continue until the field has "rolled" from its metastable false vacuum state to a true vacuum. During this time, the Universe expands something like $e^{100}$ times, and its temperature drops by the same factor, i.e. to a freezing temperature of $T\sim10^{-15}$ K. This doesn't last long, though, because the energy of the field is converted to photons and other relativistic particles. This is called reheating.

The factor by which the Universe expands during inflation is essentially set by the minimum factor needed to explain in particular three cosmological "problems", viz. the flatness of the Universe, the causality of the horizon, and the missing monopoles, and the maximum factor allowed in order not to blow stuff too far apart that no structure would have had the time to form before gravitational attraction was too small, so that we wouldn't be here to ask the question. This requires at least 60 e-foldings (i.e. expanding by a factor of $e^{60}$), but could be 100 or maybe more. The factor is discussed e.g. in Galaxy Formation And Evolution by Mo, Bosch & White, Sec. 3.6.2.

Since the size evolution of the Universe during inflation is exponential, the scale factor can be written as
$$a(t) \propto e^{Ht},$$
where $H$ is the Hubble constant during inflation, so for an expansion of, say, $e^{100}$ we have
$$e^{100} = \frac{a(t_{\mathrm{end}})}{a(t_{\mathrm{begin}})}
          = \frac{e^{H t_{\mathrm{end}}}}{e^{H t_{\mathrm{begin}}}}
          = e^{H(t_{\mathrm{end}} - t_{\mathrm{begin}})},$$
or
$$\Delta t \equiv t_{\mathrm{end}} - t_{\mathrm{begin}} = 100/H,$$
where $t_{\mathrm{begin}}$ and $t_{\mathrm{end}}$ marks the time of beginning and end of inflation, respectively.

During inflation, the Hubble constant should not evolve, but have a constant value of $H=(\Lambda/3)^{1/2}$, where $\Lambda$ is a cosmological constant. If inflation started when the GUT era ended, which was around $t_{\mathrm{GUT}} \sim 10^{-36}$ s, the Hubble constant was of the order $H \sim t_{\mathrm{GUT}}^{-1} \sim 10^{36}$ s. Then a hundred e-foldings will take you to $t_{\mathrm{end}} \sim 10^{-34}$ s. Wikipedia states that inflation should last somewhat longer, something like $10^{-33}$ to $10^{-32}$ s. This might be if you do a more careful analysis, or maybe a different model. For instance, H is not exactly constant during inflation, but does evolve a little.

Various observations can constrain the physical parameters that enter the equations, e.g. cosmic microwave background and baryon acoustic oscillations. The latest results from the Planck satellite can be found here. Wikipedia also has a summary of the observational status of inflation.

So, I guess the answer to your question is, "We don't, really, but we have several theories that together with several observational constraints tell us that the duration of inflation should be of this order of magnitude. If it were much shorter, it wouldn't solve the cosmological problems that gave rise to the idea of inflation, and if it were much longer, the Universe would be blown apart".


In regards to the holographic principle, what exactly is information? I'm having trouble understanding what exactly "information" is in the context of the holographic principle suggested by string theory. Can it be equated to a matrix of ones and zeros? Does this information have its own laws of physics or are all laws of physics in our universe a result of this information, as in, does information supersede everything?

The answer here is deceptively simple, and has very little to do with gravity directly. You only really need to know a bit of quantum mechanics, and the answer comes almost for free.

In any quantum mechanical theory, we have a Hilbert space $\mathcal H$. (Actually, we really want a rigged Hilbert space, but this distinction isn't particularly relevant here.) This can be as simple as the space of a single q-bit, or it can be as complicated as you need it to be (e.g. the Fock spaces which arise in quantum field theory). The Hilbert space describes all the possible configurations of any system described by this theory; an individual vector is a specific configuration. We also have a linear operator $H$ on $\mathcal H$, called the Hamiltonian, which describes the dynamics of such a system via the Schrödinger equation.

The "information" is just knowing what state $| \psi \rangle \in \mathcal H$ your system is in. This state vector tells you the result of every possible measurement, and thus contains all the information of your system. If you really wanted to, it's possible to express any such state uniquely as a sequence of zeros and ones, but that's a very classical way of thinking, and we're dealing with quantum information, which means that the fundamental objects aren't zeros and ones, but state vectors.

So, when we talk about holography, what we're really saying is that we can determine the state $| \psi \rangle$ that our system (e.g. the universe) is in simply from knowing the results of experiments we perform on the boundary of the spacetime (e.g. infinitely far away). The holographic principle alone doesn't say how this reconstruction works, only that it is possible.

Since this is a bit broad, it may be helpful to see how it works in the only really well-understood example, the AdS/CFT correspondence. In this case, we have a theory of quantum gravity  in $d+1$ dimensions with Hilbert space $\mathcal H$ and Hamiltonian $H$, constrained to have spatially-asymptotically AdS$_{d+1}$ metric (AdS is just a special, maximally symmetric solution to the vacuum Einstein field equations which has nice properties that make this work). It turns out that, at least morally speaking (there is ongoing work to understand the full extent of this), when we collect all the observables in this theory and shuffle them around in well-defined ways, we can construct out of them vectors in a different Hilbert space $\mathcal H'$. That is, we have a linear map $T: \mathcal H \rightarrow \mathcal H'$. This map is 1-to-1, meaning that we don't lose information. Strictly speaking, it doesn't need to be onto, but this distinction isn't crucial at first pass, so you can think of $T$ as a 1-to-1 correspondence (i.e. a linear isomorphism) if you like. In addition, the Hamiltonian $H$ can be mapped to $H'$, which describes compatible dynamics to $H$ on $\mathcal H'$ (in the sense that one can first evolve in the original theory and then map to the new one, or first map and then evolve, and the results will be the same). When we look at the pair $\mathcal H', H'$, we recognize this not as a $d+1$-dimensional theory of quantum gravity, but as a $d$-dimensional theory without gravity, but with extra symmetries that turn it into a so-called "conformal field theory" (which come from the extra symmetries of the AdS spacetime). In some sense, this new theory can be thought of as living on the boundary of AdS. So when we say that all the information is contained on the boundary, we mean that given a state $| \psi' \rangle \in \mathcal H'$ which describes the boundary theory observables, we can reconstruct the full state in the quantum gravity theory simply by $T^{-1}(|\psi'\rangle)$. What I've given here is just the 2-minute summary; AdS/CFT is still an active area of research and everything I said above is only approximately and/or morally true.

You might wonder why we need to rely on quantum mechanics. It turns out that (at least in the AdS/CFT correspondence), we can't do it classically. Highly quantum mechanical behavior in one theory is recovered by the classical limit of the other, and vice-versa. This is both a blessing and a curse, but at any rate there's no real classical equivalent. This is suspected to be true in any nontrivial example of the holographic principle, so we're pretty stuck describing things in terms of quantum information.


Water Content of Space Bodies In general, are carbonaceous chondrites (CC) or comets more water rich?

I know that evidence has suggested that both CC and comets are partly composed of water, but which body is generally more water rich?

I compared C type asteroids and comets for this question. C type asteroids have around 30% of empty space, meaning that that much space can be for water.

Comets have around 40% of empty space.

Obviously, just because they have more empty space does not mean they have more water. However, evidence for water on Earth originally comes from comets. Not only that, but comets are occasionally balls of ice with some rock. Asteroids are known for being just rocks, and only have a limited amount of water.

In conclusion, comets are more water rich.

Hope that helps...?


Can a black hole with the density of water exist? During an outreach event, the speaker mentioned the fact that black holes weren't necessarily dense objects and that black holes of any density could exist, even black holes with the density of water.

Now indeed, the schwarzschild radius of such a black hole would be
$R = c\sqrt{\frac{3}{8 \pi G \rho}}$
with $\rho$ the density of water.

But could such a non-dense black hole exist in practice? If not, what would prevent it from forming or from being stable?

Applying a numerical density to a black hole isn't possible. The material inside the event horizon will fall to a "singularity" (or some other ultrahigh density state that we currently have no adequate theory to describe) on a relatively short timescale.

What you can do, is exactly what you have done, which is divide the gravitational mass of a black hole by the volume defined by a simple Euclidean estimate using the Schwarzschild radius$^{*}$.

If instead of substituting out the mass as you have done, you instead substitute out the Schwarzschild radius, then
$$ M = \left(\frac{3}{4\pi \rho}\right)^{1/2} \left(\frac{c^2}{2G}\right)^{3/2} = 1.4\times10^{8} \left(\frac{\rho}{\rho_w}\right)^{-1/2} \ M_{\odot},$$
where $\rho_w = 1000$ kg/m$^3$.

So, if you have a black hole of mass $1.4 \times 10^{8} M_{\odot}$, then it has the "density" of water if you calculate it like this.

There are lots of black holes with this mass or even higher that are situated in the centres of galaxies. The black hole at the centre of the Andromeda galaxy has a mass of about $10^{8} M_{\odot}$, so fits the bill perfectly.

Note that it is a vast simplification (and incorrect) to use $4\pi r_s^3/3$ as the volume inside the event horizon. There is in fact no uniquely defined volume; it depends on choice of coordinate system (see DiNunno & Matzner 2008).


Can we conclude with reasonable certainty that there is/are no Type 3 Civilizations in our galaxy? As per the Kardashev Scale, type-3 civilizations are defined as "civilization(s) in possession of energy on the scale of its own galaxy, with energy consumption at ~ 4 x 10^44 erg/sec". Considering that we have detected none, despite projects like SETI - can we conclude with reasonable certainty that there is "currently" no type-3 civilization in the milky way? Or Is it perhaps possible that there could be such civilization(s) primarily harnessing dark matter / energy and hence gone undetected?

I am assuming there could yet be type-2 civilizations that extract fusion energy ("normal/known" sources of energy), information, and raw-materials from multiple solar systems - within our own galaxy - yet undetected.

Short Answer: We won't have to worry about the Galactic Empire any time soon.

Long Answer: First we have to consider some of the properties of a Type III civilization - or, to maybe put it better, what kinds of things it would do. Some are shared properties of Type II civilizations, but on a larger scale. You can probably list some off the top of your head:

Advanced interstellar travel
Large-scale engineering projects
The ability to capture energy from stars, black holes, and pretty much any other source you can think of.
Lots and lots and lots of signals
Colonies on most star systems
The intelligence to not contact us

You can extrapolate #1 from just about every science fiction movie ever made. But I don't mean anything like the starship Enterprise, or the Death Star. To become a Type III civilization, a civilization first has to reach Type II status. What does that entail? Well, among a lot of other things, shunning war. I know that might be hard to believe, but think about this: Could a single nation get off this planet on a large scale all by itself?

A good example is the International Space Station. To get the project going, it needed the combined help of the U.S., Russia, Japan, China, and the E.S.A. Admittedly, we're entering a new age of space travel. SpaceX, Blue Origin, Virgin Galactic, Reaction Engines, and other companies are showing that they can get people and things to orbit or suborbital trajectories. Maybe an entire country isn't needed to do something grand; maybe a single company can do it.

So I can conclude that any spacecraft we find from such a civilization would most likely be peaceful - a passenger ship, perhaps, or a freighter. Such a craft would be hard to detect - admittedly, we have trouble finding anything the size of a planet or smaller, once we get outside our Solar System! But that brings us to #2: Large-scale engineering.

Picture the Death Star. "That's no moon!" Yep, and it looks - from afar - a lot like one. Type III civilizations would have the capacity to build such structures - for peace, likely (see #1) - and these would be a lot easier to detect. Wait. Didn't I just say that anything the size of a planet or smaller would be hard to detect? Yep. But a Type III civilization wouldn't be limited to making small faux moons. Planet-sized objects, or bigger (we'll get to that in #3). How could we detect such objects? Well, if they're big enough, we could find them the same way we find (if we're lucky) exoplanets - transits of the primary star, wobbles in the star's orbit, etc. There is another way, though - for specialized creations.

Ah, 3#. Energy-collecting devices. A quick Internet search of "Dyson sphere" should give you a good idea of what I mean (and if that's not enough, you can see Dyson's original paper or see this question). Dyson spheres/rings/any-other-configuration-you-can-think-of would affect the star whose light they were collecting, and we could figure out if the object in question was a Dyson sphere (see the "Dyson sphere" question). But a Type III civilization could transcend that (Dyson spheres are often cited as the trademark of a Type II civilization). Imagine collecting the energy of jets emitted from the accretion disk of a black hole - not an easy job, but still possible.

Up next: signals. Humans have been emitting radio waves since Marconi (okay, he wasn't the only one) turned on his transmitter. Over 100 years later, we're still going at a pretty good rate. A civilization would need millennia to reach Type III status, and should have emitted a bunch of signals along the way. Also, such a civilization would have spread out, and should be emitting them today. Now, the galaxy's a pretty big place. It is tens and tens of thousands of light years in diameter (although there is a debate about where the galaxy ends), so our radio waves won't have reached much of the galaxy. But a Type III civilization should, theoretically, be almost everywhere, and we should have gotten messages (even inadvertent ones) long ago. Unless you think the "Wow" signal was actually a message, there's not a lot of hope. You hit the nail on the head with your SETI remark.

For the next (#5) property, we turn to . . . well, just about any sci-fi show ever made (again). What would Star Wars or Star Trek be without colonies all over the galaxy? Technically, the definition of a Type III civilization is one that merely uses all the galaxy's energy; it doesn't have to be all over. But it's not much of a conceptual leap to realize that such a civilization would actually have to be pretty much everywhere in the galaxy in order to collect all that energy. So we should, in theory, have a decent chance of stumbling upon a colony if we headed to any star. Is that going to happen any time soon? Nope. So there goes that method.

The last property is something of a joke, but also something of a parting message. As Calvin (from the comic strip Calvin and Hobbes) once said,

  Sometimes I think the surest sign that intelligent life exists elsewhere in the universe is that none of it has tried to contact us.

But think about it. We're a pretty violent race. Just turn on the evening news. Remember what I said before: advanced civilizations may not want to contact us. So even if we don't receive any messages, or find any Dyson spheres, of fail to find evidence of any kind, doesn't mean that we're alone on the Milky Way. Unlikely? Yep. But plausible? Yep.

So, to summarize: There are a bunch of different properties that would scream "Advanced civilization" at you. We could detect them by looking at their effects on stars, by searching the sky for radio waves, or by simply hopping over to a few nearby star systems. But being as we really haven't found anything yet: We can conclude, with reasonable certainty, that there is/are no Type III civilizations in our galaxy.


Do neutron stars have an ergosphere? Ergospheres are a property of black holes, and do neutron stars have it? If yes, why seldom article mentions that; if no, why can't neutron stars have this ergosphere component?

Neutron stars are generally not compact enough to have an ergosphere. Typically, the radius of a neutron star is several times the Schwarzschild radius corresponding to its mass. The ergosphere however starts at the Schwarzschild radius.


Does a planet's mass or gravity affect the height of it's mountains? According to this Wikipedia page, the top five highest mountains on Mars (and the highest on Venus) are all taller than Mount Everest (and Mauna Kea as measured from the ocean floor).

Does a planet's mass or gravity affect the height of it's mountains?  Is there an upper limit that a mountain can reach given the mass or gravity of a planet?

Yes, gravity definitely affects the maximum heigh of mountains.

Think on a solid bar of steel. It sticks straight because of the electronic forces. But when you make it larger and larger gravity makes it bend: gravity starts being considerable, but still smaller than electronic forces.

If you make the bar larger, there will be a moment in which the weight of the whole bar will be larger than the short-range electronic force: your bar will break purely due to gravity.

Exactly the same happens to mountains made of solid rock (as opposed to sedimentary ones quoted by Hobbes). There is a point,k depending on the strenght of planetary gravity, where it takes over short-range electronic forces, making the mountain collapse.

This is exactly the force that "rounds up" the planets, as opposed to the non-spherical asteroids.


Is it certain that dark matter consists of particles? (And not just curved space) The main idea about dark matter today seems to be supersymmetry. That there exists a set of yet undiscovered fundamental particles which (in a mathematically most convenient way) mirror the particles already known.

But has it been established that dark matter consists of particles at all? Couldn't it for example be a continuous property of space itself in some regions? That some parts of space are curved without the influence of any particles or waves. Is anything but particles ruled out?

Also asked at https://physics.stackexchange.com/questions/174080/how-do-we-know-dark-matter-isnt-curved-spacetime

Basically no. Or at least you can't have this idea and General Relativity. GR demands that you have something (matter/energy density) to cause the curvature. Curvature without cause is not part of the model. That's not to say that what you propose couldn't be true, but one would then have to ditch GR, which is a reasonably successful model (though some would argue that the dark matter problem is one of its biggest failures!).

I think a lot of parameter space has been ruled out for dark matter. We know it can't be baryonic from our understanding of primordial element abundances and also from the way it interacts (or doesn't interact) with normal matter in forming the basic structures in our universe on the right timescales. The microlensing surveys that have been conducted rule out large (planetary sized and above) dark matter bodies (and black holes, brown dwarfs, cold white dwarfs etc.), since to account for all the mass, these would produce lots of microlensing events that simply aren't observed.

An excellent primer on these topics has been produced by Garrett & Duda (2011).


What is the orbital period of our sun around its barycenter? In years, what is the time it takes our sun to orbit around our solar system's barycenter?

As you can see from the diagrams below of the motion of the Solar System's barycenter relative to the Sun (courtesy of Wikipedia), the Sun's motion with respect to the system's barycenter is not a simple closed curve, and it doesn't have a simple period.

(Click the above image for a SVG version).

It is primarily dominated by Jupiter's period (11.862 years) because Jupiter has most of the non-solar mass of the Solar System. (On a related note, Jupiter has most of the angular momentum of the solar system). Jupiter's mass is approximately 0.0009543 solar masses, so the Sun is almost 1048 times more massive than Jupiter.
According to Wikipedia

To calculate the actual motion of the Sun, you only need to consider
the motions of the four giant planets (Jupiter, Saturn, Uranus,
Neptune). The contributions of all other planets, dwarf planets, etc.
are negligible.

Here is a Sage / Python script which displays an interactive 3D plot of the above diagrams, using data from
the JPL Horizons on-line solar system data and ephemeris computation service. Horizons can produce data covering a wide range of dates, please see the Horizons documentation for details.


What was the largest telescope ever fitted with an eyepiece Serious modern telescopes use CCDs to capture the results. This makes it convenient to process or display the results halfway around the world. However, there is something visceral about looking through an eyepiece and seeing the results live.

My telescopic experience has been with smaller (<=8") devices in not-great-seeing conditions, and I've generally found the results dim and fuzzy. I've imagined what it would be like to look through an eyepiece on a serious telescope, showing a nebula bright and clear. Imagine stereo eyepieces on the Large Binocular Telescope!

Hence my question: what is the largest aperture telescope that was ever fitted with an eyepiece so that someone could look through it in real-time? This should include situations where a normally-digital telescope was temporarily fitted with an eyepiece.

From this answer, it seems that the Harlan J Smith telescope located at the McDonald Observatory, in Texas, with a 107-inch (2.7 m) aperture is the largest aperture telescope that was ever fitted with an eyepiece.

Some independent evidence is there to indicate that this indeed had an eyepiece.


Planetary system orientation & the search for exoplanets A quick review of photographs of galaxies will reveal that not all galaxies are orientated in the same plane as the field of view from Earth. Some galaxies are parallel to our field of view and we see them side on. Others are perpendicular to our field of view and we see them as if we are looking either from above or below them.
Given this fact, it would be fair to assume that the orientation of planetary systems is similarly varied. Not all of them are parallel to our field of view.
All the exo-planetary systems I am aware of have been seen in a side view perspective  and any exoplanets have been detected by either a wobble of the star at the center of the planetary system or via dimming of the star's light as the exoplanet transits the star.
Have any exo-planetary systems been discovered that are perpendicular to our field view, looking from above or below the system and if so, how have any exoplanets in these systems been detected? If no such systems have been observed, how would astronomers detect exoplanets in such systems?

The multi-planet system HR 8799 is pretty close to face-on. Here's a video made by interpolating images taken over 7 years showing the orbital motion.
The planets of HR 8799 were detected by imaging. The majority of exoplanets have (at the time of writing) been detected by the radial velocity method, which requires radial motion and thus is most sensitive to planets close to edge-on, and the transit method which requires that the planet passes in front of the star as seen from Earth (thus will only spot planets extremely close to edge-on). The imaging method does not have this issue, so can find both face-on planets (like HR 8799) and edge-on ones, like Beta Pictoris b (video).
Astrometry is another exoplanet detection method that would allow detection of face-on systems. It has been plagued by a number of false starts (e.g. the infamous "detection" of planets around Barnard's Star by Peter van de Kamp) but hopefully once the full Gaia dataset is released there will be numerous astrometric exoplanet detections, which may include near-face-on systems.


Can black holes collide with the Sun? In that case what will happen? Can a black hole collide with the Sun?

What will happen in that case? Will the Sun survive?

Yes, in theory. But;

No, this is very unlikely to happen during the life of the sun. If it did happen it would be Bad.

Many black holes are much more massive than the sun, even though they are much smaller. If a black hole were to enter the sun, some the matter that forms the sun would form into a accretion disc around the black hole and eventually fall into the black hole. The rest of the sun would be utterly disrupted by the massive release of energy as matter falls towards the black hole.

The solar system would be wrecked. The Earth, even if it survived intact, would not be able to support life.

Fortunately black holes are rare, and space is BIG. The chances of a black hole coming anywhere near the sun in the next billion years or so is very very small. There are more pressing things to worry about.


Stellar mass limits for Neutron Star and Black Holes Don't hate on me if I am asking a very basic and straightforward thing. I have a few questions about black holes and neutron stars.

What is the mass range (in terms of solar masses) for a main sequence star to end its life as . . .

A) a neutron star?
B) a black hole?

Is it there a (practically observed or proven) possible method for a main sequence star to form a neutron star (or black hole) at the end of its life cycle without undergoing through the process of supernova?

If yes, please explain or guide me some article on this matter.
What is the mass range of a main sequence star to end up as a pair instability supernova?

If the range of pair instability supernova overlaps with that of neutron star or black hole forming supernovae, how do we determine what type of end a star would have?

A succinct summary of supernova types is given in the following image based on Heger et al. (2003):

Image courtesy of Wikipedia user Fulvio 314 under the Creative Commons Attribution-Share Alike 3.0 Unported license. The graph is based on the graph in Fig. 1 of the linked paper.

The pair instability realm is upwards of ~100 solar masses, though it is metallicity-dependent (Question 3). As Figure 1 (below) shows, neutron stars form in the mass range of >9 solar masses - again, this is metallicity-dependent (Question 1a). Starting at around 25 solar masses, black holes will form (Question 1b).

It is thought to be possible for a black hole to form without a supernova (see the section of the graph marked "direct black hole"). This has not been confirmed observationally, although there are some possibilities. I've written about this in more detail here.


Are Dwarf Galaxies a Subset of Galaxies? The question is a rather obvious one.
In normal English a word like "dwarf" used as an adjective is assumed to modify the following word.  Thus a "dwarf galaxy" would be assumed to be a galaxy of a type which is noticably smaller than galaxies of at least one other type.
And in science there might possibly be an established dividing line between dwarf galaxies and ordinary sized galaxies, just as there might possibly be a diving line between ordinary sized galaxies and giant galaxies.
But in 2006 the International Astronomical Union created new definitions, in which a dwarf planet is not a subcatebory of planet, but a member of a totally separate class of astronomical solar system object.  Dwarf planets are not included in planets, and planets do not include dwarf planets.
That is ungrammatical in English, but it follows a pattern.
A minor planet is mostly equivalent to what is usually called an asteroid.

A minor planet is an astronomical object in direct orbit around the Sun (or more broadly, any star with a planetary system) that is neither a planet nor exclusively classified as a comet.[a] Before 2006, the International Astronomical Union (IAU) officially used the term minor planet, but during that year's meeting it reclassified minor planets and comets into dwarf planets and small Solar System bodies (SSSBs).1

https://en.wikipedia.org/wiki/Minor_planet[1]
So in astronomy a minor planet is not a planet, and a dwarf planet is not a planet. The category of planets does not include minor planets or dwarf planets.
And thus it seems possible to me that perhaps in astronomy a dwarf galaxy might not be a galaxy.
If dwarf galaxies are a subcategory of galaxies, it would correct to call a dwarf galaxy a "galaxy" for short.  But if dwarf galaxies are a category of astronomical object separate from galaxies, it would not be correct to call a dwarf galaxy a "galaxy" for short.
So are dwarf galaxies a subcategory of galaxies, or are they considered to be a separate category from galaxies?

The IAU has not defined "galaxy", therefore you should not use the word  (Joke)
In all seriousness, language is a tool.  You are trying to dance around definitions.  There are many types of galaxies: spiral and elliptical, barred and unbarred, flocculent and unflocculent, small, medium and large.  They are all galaxies, of one kind or another.  The IAU has chosen to define "Planet" and "Dwarf planet" in a particular way, so that children don't have to learn the names of 46 solars system bodies.  They haven't chosen to give a definition of "galaxy", nor "cluster" nor "nebula".  But it doesn't matter.  You use language to communicate. You are not in thrall to the IAU. So you can use the word "galaxy" as they do to mean "galaxy".
Yes: dwarf galaxies are galaxies.