How to convert a gravitational force to speed and direction I am doing some calculations, and I wanted to know that after using the gravity equation, $$F = G \frac{m_1m_2}{r^2}$$ if you can use $F$ and convert it into a speed, such as km/year, and also a direction or angle from the original direction. I am just looking into astronomy, and any help is appreciated.

You must apply Newton's law
$$ m\frac{\mathrm{d}\boldsymbol{v}}{\mathrm{d}t} = \boldsymbol{F}$$
which related the force $\boldsymbol{F}$ to the acceleration (=change of velocity). Note that positions, velocities, and forces are all vector quantities. Also note that as the objects move (change position), their mutual forces change (both direction and strength).

There is an enormous amound of literature on the resulting orbits for celestial bodies, going all the way back to Kepler and Newton. I suggest you look into elementary books or start with some Wikipedia articles.


Why the Milky Way appears to be arching across the sky? The Milky Way galaxy is disc-shaped and earth is in its plane:
 (source)

In the following image the milky way appears to be arcing across the sky:
 (source)

This looks like an out of plane view of the milky way... How is it possible? Why an arch and not a thick straight line?

The Earth is a sphere (or is nearly a sphere). So to make a map of the whole Earth you would need to project it onto a flat surface. When you do this you create distortions. For example, on many maps, the straight line between New York and Japan looks like a long curve. It is not actually a curve, but when you stretch the surface of Earth to make it flat, some straight lines become curved.

The sky is also a sphere. If you ignore distance and just think of the direction of each point in the sky it would be a huge sphere that surrounds you. When we take a picture of the sky we must project that sphere onto a flat surface. That causes distortions, and it means that what is a straight line on the sphere of the sky is distorted by the projection into a curve.

If you choose a projection in which the milky way goes straight across the middle, you can see that the milky way goes straight across the sky.

The appearance of an "arch" is a distortion caused by stretching the image to fit on your flat screen. It is not real.


Regarding the curvature of the Earth I have a doubt in Earth's curvature. This question may be dumb.

How much we need to travel on the Earth to feel the curvature of Earth? Hypothetical case: for example, we are building a vehicle. How much is the length of that vehicle to feel the curvature of the Earth?

A more question, what is about the speed of the Earth. Again a hypothetical case. If we are travelling in aeroplane from one country to other, whether we can take Earth's rotational speed (rotating around it's own axis), and aeroplane speeds as comparables and find the resultant speed?

About curvature, you can see it. Just look at a ship some hundred meters away: you'll notice if you are careful that you can not see where the ship touches water.

You can see at http://mathcentral.uregina.ca/qq/database/QQ.09.97/dyck2.html that the curvature is 8 inches per mile.

On the other hand, about rotation, planes do actually take advantage of earth rotation, but not directly. This was answered in https://physics.stackexchange.com/questions/16390/does-the-rotation-of-the-earth-dramatically-affect-airplane-flight-time


How does an Echelle spectrograph put such a long, high resolution spectrum into a square-ish format? In order to obtain a very high resolution spectrum to resolve closely spaced lines, or to measure doppler shits, spectrographs are operated with very high dispersion.

But photographic plates and CCDs usually have rectangular formats with aspect ratios not very far from square.

Echelle spectrographs can put a highly dispersed spectrum into this format, but how? This article explains to some extent, but a concise discussion here would also be helpful.

Echelle spectrometers operate in the higher spectral orders.  (CRIES+ uses orders 10-60 for IR light, I think.  Not sure about ESPRESSO, but you can count them...)

These high orders are highly overlapping, so, for example, if the input light is dispersed by the Echelle grating horizontally into a spectrum, then the light at a given point along the spectrum is has components contributed from several different orders.  But, since each contribution is a different wavelength from each order, the orders can be spread out vertically using a second grating, the "cross disperser", conveniently stacking them up for the camera.  You can see these components in this diagram of the ESPRESSO spectrograph.

TL;DR:  Each horizontal line you see is a portion of a single spectral order, and the orders are spread vertically by a cross disperser grating, oriented roughly perpendicularly to the first (Echelle) grating.

I found this presentation with more details & the equations.


why does venus flicker? I was watching Venus with the naked eye yesterday at about 7 pm and I noticed that it was flickering, almost like a star.

I have always learned that planets don't flicker to the naked eye, only stars (indeed, every other time I've seen a planet it wasn't flickering), so I was rather confused. I even checked Stellarium to make sure that what I was seeing was, in fact, Venus.

Does anybody have an explanation for that behaviour?

Thank you all.

Because planets actually do twinkle. Most people were told that the major difference between stars and planets is that only the former twinkle - but that's an oversimplification. Given the right conditions, planets will twinkle too, it just happens more rarely.

Several factors that contribute to it:

lots of air turbulence; or, as astronomers call it, "bad seeing"
closeness to horizon; if the planets are high in the sky, the air column is shorter so there's less chance they will twinkle; but when they are low, their light goes through more air and so it is perturbed to a larger degree

The observation you've made, Venus twinkling, is not very unusual. Many stargazers are used to seeing that once in a while. I've seen Venus scintillate several times in the past, always at sunset when it was about to drop below horizon; I would presume you could see the same behavior very early in the morning as Venus has just risen.


Regarding the age / size of the Universe I believe the current estimated age of the Universe is around 14 billion years.  I just read in another post on here that the diameter of the Universe is around 90 billion light years.  This doesn't make sense to me, shouldn't the radius of the Universe be equal to it's age?  Unless the explosion caused particles to spread out faster than the speed of light? Is that possible? I thought the speed of light was a speed limit that nothing can exceed?

Sorry if there is something I'm not understanding here or if I'm trying to compare apples to oranges.  I'm only an astronomy enthusiast.

The expansion rate of space is not itself the reason that the radius $R_\mathrm{Uni}$ of the observable Universe is larger than 14 billion lightyears (Gly). Just the fact that space expands is the reason. If space did not expand, then $R_\mathrm{Uni}$ would be the expected 14 Gly, as this is the distance that light can travel in the 14 billion years (Gyr) since the Big Bang. But since space has expanded in the meantime, then the distance to some particle that emitted a photon that we observe today, has been continuously increasing during the 14 Gyr, and hence $R_\mathrm{Uni}>14\,\mathrm{Gly}$.

An analogy to this is a worm on a rubber band that is stretched while it crawls along. The distance between its starting point and it's end point is not just a matter of how fast it crawls, but also of how much you stretch the rubber band.

The term "space expands faster than light" is a bit deceptive. Space expands *homogolously*$^1$, meaning that a given point in space recedes from you at a speed depending on its distance from you. If a galaxy is 1 Mpc (= 3.26 Mly) from you, it recedes at $70\,\mathrm{km}\,\mathrm{s}^{-1}$. If it's 2 Mpc from you, it recedes at $140\,\mathrm{km}\,\mathrm{s}^{-1}$. And so on. At some distance the recession velocity becomes larger than $c$, and in fact galaxies at a distance of $R_\mathrm{Uni}$ recede at more than $3c$. This doesn't violate special relativity which says that nothing can travel through space faster than $c$, because the galaxies do not travel through space. They lie approximately still in space, but space itself simply expands, i.e. the distance between everything increases.

Like three worms on a rubber band, with 1 cm between them. If you stretch the band to double length, the distance from worm #1 to #2 is 2 cm, while from #1 to #3 it's 4 cm.

$^1$Homogolous is not the same as homogeneous. Whereas the latter means "the same everywhere", referring to some physical quantity like density, homogolous means "proportional to distance".


Will Neptune be visible with the naked eye if I am standing on its satellite Assume that I am standing on one of the many satellites of Neptune. Will I be able to see the planet with my naked eyes (without any sort of visual aid).

If I was to stand on Moon I would be able to see the Earth because the light from Sun is reflected off Earth into my eyes and I am able to see Earth. In case of Neptune, the distance between Neptune and the Sun is so great that the sun would look like any other star (a bit brighter). In that case if I am looking at Neptune would I be able to see it ?

What about the other gas giants ?

NASA's Neptune fact sheet states that Neptune's irradiance is $1.51 W/m^2$. This is roughly three orders of magnitude less than at Earth (at $1367.6 W/m^2$). This sounds like a lot, and it is quite an attenuation. So much that for example solar panels for interplanetary probes are not worthwile at these distances from the sun.

However, if you compare this with the sensitivity of the human vision system, this seems still pretty bright. A sunny day outside will be approximately $10^5 cd/m^2$. So equally on Neptune a sunny day will probably be $10^2 cd/m^2$. This is comparable to an overcast sky, and still well within human photopic vision!

The perceived brightness of Neptune, as viewed from one of its moons, will not change, so yes, it will appear 1000 times less bright than earth does, but since you don't have any comparisons to brighter light sources there, you will perceive it as pretty bright!

As for the other gas giants, their surface irradiance will be equally higher, so they will in turn appear much brighter than Neptune.


What are the stages in the life of a universe? Geological periods describe varying phases in the last few hundred million years on earth.
Are there any names for the periods on a timescale 1-2 orders of magnitude greater, that is, for the stages of a/the universe?
I found some bits and pieces of information but nothing as clear as I did for geological periods, possibly because of the conjecture involved.

Yes there are. They are mainly based on what dominates the energy density of the universe at the time and they are known as epochs.
Thus we have the inflationary epoch in the first tiny fraction ($\sim 10^{-32}$) of a second, when the energy density was dominated by an inflationary field.
Then we are in the electroweak epoch, when the weak nuclear and electromagnetic forces were united.
This is followed by the quark epoch when the electromagnetic and weak nuclear forces separate at about $10^{-12}$s, which is followed by the hadron epoch between about a millionth of a second and 1 second, when quarks were able to be confined into multi-quark hadrons.
Then the lepton epoch between about 1 second and a minute when most of the hadrons have annihilated with each other and the energy density is dominated by electrons, positrons and neutrinos, although this also coincides with the "epoch of primordial nucleosynthesis" when the first multi-nucleon nuclei were built.
From there until about 300,000 years, we have the radiation epoch, when photons dominated the energy density of the universe.
After that, the universe entered the matter-dominated epoch for about 5 billion years and now we are in the epoch where the universe is dominated by dark energy.
There are other schemes and other nomenclatures depending on what physics is being focused on. For example, people interested in the formation of the first stars will talk about the "epoch of reionization", which occurred some 100 million years after the big-bang. Those studying the cosmic microwave background talk about the "epoch of recombination", which coincides roughly with the transition from a radiation to a matter-dominated universe.


How can a gas planet become tidally locked with a star? Supposing a planet is made entirely of gas, is it possible for the planet to become tidally locked with the star it orbits?

I know when a terrestrial planet orbits a star in a different amount of time than the length of its day, the planet warps continuously, resulting in tidal heating. This heating takes energy from the planet's rotation until it stabilizes in a tide-locked formation.

I believe a purely gaseous planet would also have some sort of internal heating for the same reason. But because the "surface" of a gas planet is fluid, it must also obey thermodynamic principles.

As a basic example, the side of the planet facing a star gets hot, and the side facing away gets cold. The hot gas expands and the cold gas contracts and you get convection, Hadley cycles, prevailing winds, and chaotic weather patterns.

So it seems like there's a battle between gravity, which wants the gas to stop moving and become tide-locked, and thermodynamics, which wants the surface to roil with convection.

Does one process dominate over the other? Does one process slowly take over as the other fades away? Can it go either way?

Gaseous planets would tidally lock, just as binary stars do.  That would occur in addition to convection-- tidal effects are weak and operate on very long timescales, convection is strong and operates on much shorter timescales.  It is true that a gaseous planet conforms more easily to the tidal potential even if it is rotating, so there is less of a turning of the tidal bulges, but the tidal bulges are larger because they fill the equipotential, whereas solid surfaces don't.  So the rapid response of gas could make the locking process either faster or slower, depending on which effect dominates.  But you are probably right to focus on energy dissipation as the key requirement, and gas is probably less dissipative, so I would guess the locking time is longer for gas.  Still, there is energy dissipation, there is a delay as the gas flows, and there is a tidal locking effect, all going on on timescales much slower than convection.


When and where are we exposed to most cosmic radiation? I understand the mechanical dynamics of being faced away from the Sun at night, having a lower altitude, not being in the weak points at the poles in the atmosphere, being in a valley and others decreasing cosmic radiation.

Are the particular areas on Earth where cosmic radiation is amplified or unprotected by the magnetosphere because of the  lensing effect on cosmic rays either by location or angle of the sun?

You seem to have a good understanding of cosmic rays, but not all of them come from the sun, so there will still be some coming in at night. These will tend to be the most energetic ones emitted from supernovae and other violent cosmic events. You are most at risk on a long distance airliner, which flies at abut 33,000 feet, but the risk is small and nothing to get worried about. For airliner crew it is more serious, but for the vast majority the time spent at 30,000 feet doesn't seriously affect their health, otherwise they'd be getting danger money. If you want to avoid dangerous radiation, the thing to worry about is radon gas. This is emitted by the decay of uranium in granite rocks. Radon is an alpha emitter, and is dangerous when inhaled or ingested. You can find a map on the internet giving radon levels in your area (they are very variable). If you are in a high radon area, put a gas proof covering on your ground floor and make sure your house is well ventilated. In some houses, radioactivity can reach levels which would be unacceptable in a nuclear power station or research establishment. I don't know of any place on Earth which is especially safe from cosmic radiation except, perhaps, a coal mine or a nuclear bunker.


When were the spiders first observed on Mars? On Mars, there are spider-looking features near the South pole that might be caused by CO2 sublimation. When were these features first observed on the surface of Mars?

I have the impression these were first observed in images taken by the Mars Orbiter Camera (MOC) of the Mars Global Surveyor, probably sometime between 1997 (arrival at Mars) and 1999 (when the main mapping program began) -- or possibly as late as 2000 (when they are first mentioned).
In Piqueux et al. (2003), a paper specifically focused on the spiders, they say

The term ‘‘spiders’’ was used by Kieffer [2000] (although the term may
have originally been coined by Andy Ingersoll [H. Kieffer, personal
communication, 2003]) and we use this term in this paper.

In Kieffer (2000) we find this statement

Dark radially converging dendritic patterns are visible in MOC images
of some portions of the spring polar cap [7]; these have been termed
"black spiders" by the MOC team [8].

Unfortunately, the published version of that short conference proceeding article is missing all of the references after number [4], so I don't know if there is more specific useful information available. The conference at which Kieffer (2000) was presented was in August, 2000, so I'd put mid-2000 as the latest possible date.


Is the moon only 60 pixels? In doing research on vision, I have learned that "20/20" vision corresponds to a visual acuity of being able to resolve details 1 arcminute in size, that most people have around 20/15 vision, and that due to the limits of physiology basically nobody has vision better than 20/10 vision. This is an upper limit of resolving details about 0.5 arcminutes in size.

According to Wikipedia the moon is around 30 arcminutes wide when seen by the naked eye.

Put these together, and it seems to say that when looking at the moon with the naked eye nobody can see more details than would be visible in a 60×60 image of the moon

and that the average person can't see any more details than in a 40×40 version

Those seem so small on my monitor. Can that really be all the detail that I can see on the moon with the naked eye?

Yes and no.

Yes, it's true that the apparent size of the Moon is 30 arcmin. It's true that the visual acuity of most people is 1 arcmin. So it's true that if you take the angular size of the smallest detail you can see on the Moon, and you put a bunch of those lined up straight in a row, you could span a Moon diameter with only a few dozen of them. In that sense, you are correct.

However, when you try to reproduce the situation on a computer screen, the comparison breaks down. First off, the eye doesn't see in "pixels". Like most optical systems, there's a point-spread function, that takes very tiny details and smears them up to a larger spot. The resolution of the eye is not the pixel size, but the size of the bell curve coming out of the point-spread function, and that has soft edges and is round, and it's everywhere and it's not fixed.

You assimilate the size of that larger spot with the size of a pixel on a digital screen, in your comparison. But that's not the same. The pixel grid in those thumbnails is fixed, so whatever falls between pixels is lost forever. Aliasing intervenes and creates artifacts that are not there in the original image. The dynamic range of the monitor is not the same like the dynamic range of the eye (the eye is much better). Color and brightness levels on the monitor are discrete, whereas the eye sees them as a continuum. Finally, the visual center in your brain is like a powerful computer that applies intelligent correction algorithms to the live image.

The list goes on and on. The bottom line is - all these effects combine and allow you to perceive a live image that is slightly more rich than those dead, frozen thumbnails that you posted. Not a whole lot better, but a little bit better. It's not like the eye can "work around" limitations, but it's more like you lose too much when you shrink a large image into a tiny fixed pixel grid on a computer screen.

It's very hard to reproduce reality on a computer screen. A much better way would be to take a 2000px by 2000px image of the Moon, put it on a big super HD monitor, and move it back to the point where the apparent size of that image is 30 arcmin. I know that doesn't sound satisfactory in the context of your original query, but it's a much better simulation.

Similar problems appear whenever you try to map the resolution of any continuous optical system (like a telescope) to a fixed digital grid (like a camera).

Let's say you're using a sensor with a pixel size of 4 microns. Let's say your telescope has a linear resolution in prime focus equal to 4 microns. You might be tempted to say - great, the sensor matches the telescope, right?

Well, not really. When that happens, you actually lose a bit of resolution. The image is good, but it's a little bit softer than it really should. See below an image of the Moon I took a while ago, with a system having exactly the parameters indicated above.

You can tell it's a bit soft, it's not really down to the pixel. Turbulence also plays a role, but part of the problem is that linear resolution is equal to pixel size.

Click the image below and open in new tab; if your browser shrinks it again to fit the window, left-click the big image to expand to full size - you must do this to see the full resolution image and notice the effects I'm talking about. The fuzziness is not visible on this small version here:

One way around that phenomenon, as an example, is to blow up the image in the telescope with a barlow until the linear resolution in prime focus is much greater than the camera pixel size, maybe 4x bigger. You do all your processing, and then you shrink it back, if you like, and you'll get a sharper image. Combine it with stacking multiple frames, and the overall quality can get pretty close to 100% the theoretical performance of the telescope.

TLDR: Continuous optical systems, and discrete grids of pixels, are very different things and cannot be easily compared.


How to calculate the current distance between two planets I'd like to add a feature to one of my apps that calculates the current distance between the Earth and Mars (or any other planet, for that matter). My google searches have led me to statistics of average distances (as a planets orbit is elliptical, and therefore the distance can vary greatly), as well as articles about Keplers laws of Planetary motion and more. I have also seen some mentions of HORIZONS, but looking over it I did not see anything to fulfill my questions.

I haven't found any readings explicitly describing how to calculate the current distance between two planets. So, how is this done?

Please explain your answer and the steps needed, so that I may understand better. Pseudocode is welcome.

Thanks!

There are many different ways of doing it.

One way is to use the Keplerian elements and their rates obtained from the JPL Solar System Dynamics web site to determine the Cartesian coordinates of the planets in the ecliptic plane at a given time. Information and data to allow computation of approximate positions for the planets is provided in the document "Keplerian Elements for Approximate Positions of the Major Planets" by E.M. Standish.

Then for each of the planets, $i$, we have the position coordinates in the form
$$
\mathbf{r}_i =
\left[\begin{array}{l}
x_i \\ y_i \\ z_i
\end{array}\right]
$$
Since these coordinates are all given in the same reference frame, the displacement vector from planet $i$ to planet $j$ can be obtained from
$$
\mathbf{r}_{ij} =
\left[\begin{array}{l}
x_{ij} \\ y_{ij} \\ z_{ij}
\end{array}\right] =
\left[\begin{array}{l}
x_j-x_i \\ y_j-y_i \\ z_j-z_i
\end{array}\right]
$$
and the distance between planet $i$ and $j$ can be calculated using
$$
r_{ij} = \|\mathbf{r}_{ij}\|
= \sqrt{x_{ij}^2+y_{ij}^2+z_{ij}^2}
$$
The method described in the document to obtain the ecliptic coordinates are very easy to implement in software.


How to convert the observed magnitude of a Quasar to it's bolometric luminosity? How can I convert the observed magnitude to Bolometric Luminosity? I know the relation for the star but should it be same for a Quasar? Basically I want to convert the X-axis of this plot to luminosity (or mass).

The bolometric magnitude of a quasar can be determined by the same method as a star (using total energy radiated)......but for the fact that quasars do not radiate isotropically and this distribution is dependant on wavelength
A further complication comes as many photons are absorbed and re-emitted , etc
Some standard methods of calculation can be found here......
Quasar bolometric corrections


Whose name is Fleming 1 nebula bearing? Fleming 1 is an unusual planetary nebula situated in the Centaurus constellation.

Yet it is not clear after who it was named

Annes Astronomy News says:

  Fleming 1 is a planetary nebula that lies about 10,000 light-years away in the constellation of Centaurus, while moving away from us at approximately 28.6 kilometers per second. It is named after the Scottish astronomer Williamina Fleming, who discovered the nebula in 1910.

Williamina Paton Stevens Fleming

  (May 15, 1857 – May 21, 1911) was a Scottish astronomer active in the United States. During her career, she helped develop a common designation system for stars and cataloged thousands of stars and other astronomical phenomena. Among several career achievements that advanced astronomy, Fleming is noted for her discovery of the Horsehead Nebula in 1888

That article's Notable discoveries section says

  During her career, Fleming discovered a total of 59 gaseous nebulae, over 310 variable stars, and 10 novae.

  Most notably, in 1888, Fleming discovered the Horsehead Nebula on a telescope-photogrammetry plate made by astronomer W. H. Pickering, brother of E.C. Pickering. She described the bright nebula (later known as IC 434) as having "a semicircular indentation 5 minutes in diameter 30 minutes south of Zeta Orionis". Subsequent professional publications neglected to give credit to Fleming for the discovery. The first Dreyer Index Catalogue omitted Fleming's name from the list of contributors having then discovered sky objects at Harvard, attributing the entire work merely to "Pickering". However, by the time the second Dreyer Index Catalogue was published in 1908, Fleming and her female colleagues at the HCO were sufficiently well-known and received proper credit for their discoveries.

  Fleming is also credited with the discovery of the first white dwarf:

    The first person who knew of the existence of white dwarfs was Mrs. Fleming; the next two, an hour or two later, Professor E. C. Pickering and I. With characteristic generosity, Pickering had volunteered to have the spectra of the stars which I had observed for parallax looked up on the Harvard plates. All those of faint absolute magnitude turned out to be of class G or later. Moved with curiosity I asked him about the companion of 40 Eridani. Characteristically, again, he telephoned to Mrs. Fleming who reported within an hour or so, that it was of Class A.

    — Henry Norris Russell13

  Fleming published her discovery of white dwarf stars in 1910. Her other notable publications include A Photographic Study of Variable Stars (1907), a list of 222 variable stars she had discovered; and Spectra and Photographic Magnitudes of Stars in Standard Regions (1911).

  She died of pneumonia in Boston on May 21, 1911

The quote by is from Notes on White Dwarfs and Small Companions Astronomical Journal, Vol. 51, p. 13 (1944)

Though effusing complements to Pickering, it seems that the first line "The first person who knew of the existence of white dwarfs was Mrs. Fleming" is meant to ensure some bit of historical record was set straight.

The article continues with substantial sections on Recognition and Legacy that I won't reproduce here, though other answers to this question can address these.

From this answer to Has great eyesight been necessary for astronomers?:

  The title of a book on this subject The Glass Universe is likely a play on words, the photographic plates were made of glass, and so was the ceiling. (Space.com review, The Atlantic, Magiscenter, BBC interview in YouTube, NPR)

  It is strongly recommended to see and/or listen to the author Dava Sobel talk in the YouTube video The Glass Ceiling and The Glass Universe and in Dava Sobel: The Glass Universe

Image source


Star formation analogy If I have understood correctly, stars form in big clouds of gas and dust that are pulled together by their gravity. And the stars are often ignited when something disturbs the cloud, such as a passing massive body, or a wave like the once which create spirals in galaxies.

Are there any phenomena on earth that behave like these clouds, which could be used to illustrate and explain star formation? Something people might be familiar with and which you could show a video clip of?

I'm thinking about some gas that is disturbed then takes solid form? Or something completely different which looks similar?

If I have understood correctly, stars form in big clouds of gas and dust that are pulled together by their gravity. And the stars are often ignited when something disturbs the cloud, such as a passing massive body, or a wave like the once which create spirals in galaxies.

You have the sequence backwards. Protostars ignite by themselves once their cores reach a critical temperature and pressure. The gravitational collapse that eventually results in protostars (and then a stars) oftentimes needs a little help from an external disturbance such as a passing star or a supernova shock.

  Are there any phenomena on earth that behave like these clouds, which could be used to illustrate and explain star formation?

Not really. The interstellar gas clouds that eventually collapse to form stars are very large, very massive, very cold, and very tenuous. The best way to understand them is to understand them directly rather than looking for a poor analogy. There's nothing on Earth like them.


Trisolar syzygy and it's effects on the planet Slightly spoilers here, but in the novel The Three-Body Problem, there's a scene in which Civilization #184 is destroyed by the stacked gravitational attraction of the planet's (aptly named Trisolaris) three suns.

This destruction is portrayed as things (like a sword) suddenly getting lighter, and people and buildings starting to float up in the air, as the atmosphere of the planet is being sucked into the three aligned suns. This is explained as being a "Trisolar Syzygy".

However, I don't understand how that is possible. As I understand it, a syzygy happens in any sort of gravitational alignment of some astronomical bodies, and it's apparently what generates the tidal variation on the earth from the moon's orbit. But would the tidal forces be enough to cause this effect?

In the novel, the planet is in the Alpha Centauri system, which may have a planet in the habitable zone, and is an actual trinary system. Could such an astronomical event actually cause some effect like that in this situation?

I'm asking this here because I thought it wouldn't fit in the Science Fiction Q&A, as I'm more interested in the astronomical side of it, but I'd appreciate a pointer if I don't need to ask this here.

Is it just for the spoiler?

My guess is that the 3 suns aligned create a sufficient gravity field for the outermost parts - less well attached - of the planet (atmosphere, people, buildings) to be in the Roche Limit of the system, causing them to be sucked out spiraling inwards.

Here is a funny vid I found to illustrate:  https://www.youtube.com/watch?v=6e5yyIcq40o .

Hope this helps.


What unit is used for region Surface brightness in DS9? The value of the surface brightness is in what unit? I hope it is in Rayleighs. Is that true?

ds9 is a common image analysis tool used in astronomy (http://ds9.si.edu/site/Home.html).
The units of surface brightness are simply detector counts per square arcsecond as that is all that can easily be deduced from an image. If you want to turn that into more physical units, you will need to know the gain of the CCD (how many photons produce each count) and you will need to make some assumptions about the wavelength of light that is producing the surface brightness of the emission this Note on Rayleighs may help; if all of this emission is coming from Halpha, then this is a reasonable assumption.


Could our Sun be the product of an ancient stellar collision? The canonical model for the formation of the Solar System involves the gravitational collapse of a nebula into (perhaps) several stars across several light years. Is it possible that two or more of the stars (or protostars) combined to form our Sun?  Or would any stellar collision release so much energy that our Sun could not have formed the way we see it today?  Have astronomers or astrophysicists created n-body orbital models that allow for a "soft" collision of two stars combined with the ejection of the other(s)?

The dynamics of the Solar System and the chemistry of the Solar System bodies don't support a hypothesis of a stellar merger later than formation of the protoplanetary disk which would have mixed-up things considerably and heavily disrupt any circumstellar disk. Thus this basically excludes any collision after the time one can start talking about a protostar, way before it even entered main sequence (as that's already the end of planetesimal formation).


What was the definition of a planet before August 24, 2006? In 2006, the IAU produced a definition of what it is to be a planet. This definition famously excludes Pluto, to the disarray of this small body's fans.
Before this decision, what was the definition of a planet?

What was the definition of a planet before August 24, 2006?

I think this deserves a bit more of an answer because it's not quite so simple.  Yes, it's true that there was no formal definition of what a planet was prior to the 2006 vote, but essentially everyone knew what was and wasn't a planet, just as we know what is and isn't a star, though the boundary of when it stops being a very heavy gas giant planet and stars being a brown dwarf star, or likewise, the boundary between hottest brown dwarf and coolest red dwarf is basically arbitrary.   Definitions are useful but at times, inevitably arbitrary.
I'm reminded of Plato when he said

"Man is a featherless biped"

His students accepted the definition until Diogenes plucked a chicken and brought it into Plato's Academy, saying,

"Behold! I've brought you a man."

After this incident, Plato added "with broad flat nails".
Little did Plato realize that there were some bipedal apes that still qualified as human with his new definition.
A more recent definition that Man is the only animal that fashions his own tools, goes back to Thomas Carlyle, and it worked for a while, until it was pointed out by Louis Leakey, based on observations by his protégé, Jane Goodall, that some chimpanzees fashioned and used tools of their own making.   So, back to the definition drawing board.
The irony to both these stories is that no person of any intelligence would have confused a plucked chicken or a chimpanzee using a tool with a human.  We didn't need either definition, though, to be fair to Thomas Carlyle, I think his quote reads as much about praise for the benefits of making and using tools more than it's meant as an infallible definition of man.   My point?  Definitions can be problematic.
Lets consider the case of the comet, I mean, planet, I mean asteroid, I mean, dwarf planet Ceres.  It's been all 4 of those things depending on time period.  Ceres was a planet from 1801 until the 1850s when it was demoted to asteroid.  The largest asteroid, but still, one of a few large asteroids in the asteroid belt.  It wasn't just Ceres smaller size that did it in, but also, the discovery of other good sized bodies in the same general region of space.  These larger asteroids were closer to Ceres in size than Ceres is to Mercury.

And what about Mercury?  Surely Mercury has more in common with Ceres than it does with Jupiter and Jupiter and Saturn have moons larger than Mercury.
That said, neither Jupiter nor Saturn's large moons are even half the mass of Mercury.   Mercury also has a large Iron core.  It's core alone is more massive than Ganymede.    Mercury has a lot going for it that Pluto does not.  There are at least dozens, possibly hundreds of bodies in the solar system similar to Pluto and Ceres, but there are just 4 known bodies similar to Mercury. Mercury checks a lot of boxes.
Before 2006, there may have been no formal definition of a planet, but there was a general agreement of what a planet and there was debate.  Scientists argued for the removal of Pluto from the list of planets prior to 2006.
One of the problems was the excitement that was generated when Pluto was discovered.  The initial calculation estimated Pluto's mass to be about 6 times that of Earth. (I'm curious why they were so wrong).
Initial observations of Pluto's distance and brightness still gave it an estimated size similar to Earth and I remember when I was first learning about the planets in School, Mercury, not Pluto, was the smallest planet.  Pluto was - from memory, about the size of Mars.  Pluto's companion moon and high reflectivity may have given the illusion of being larger than it was.   It wasn't until 1978 when Charon was discovered that an accurate estimate of Pluto's size could be made and it was tiny.  Smaller than the Moon.
Pluto's size relative to Earth here.

Despite all the outrage over Pluto's demotion, some of it genuine, some of it comical, such as jokes made in The Big Bang Theory TV show blaming Neil deGrass Tyson, it was, to my thinking, a pretty clear cut decision that to this day remains surprisingly controversial.  And it makes for some good reading.
Two years of committees, debates since the 1990s and finally a vote of just 424 people in attendance.  That's almost congress-like in terms of argumentiveness and lack of efficiency, but in the end, I think the official definition of planets to be a good one, though it's not hard to imagine that it may run into problems.
A binary planet probably isn't impossible, but it raises the question of whether that qualifies.  And it's not hard to imagine if Mars was a bit smaller and the asteroid belt a bit larger - at what point would Mars no longer make the cut?  What if Mercury was half it's current size?
Ultimately, as the article points out.

"For astronomers this doesn't matter one bit. We'll go out and do
exactly what we did,"

Caltech's Mike Brown
I'll put it in my own words.  The exciting stuff is in the details, not in the categorizing.   Astronomers knew what a planet was prior to 2006 and Pluto was on the border even before it got demoted.  The new definition still isn't air-tight.  It doesn't have to be.


Intercept a NEO trajectory I need help with an exercise in the book that I can't tackle. I need to calculate how much time we have to intercept and eliminate a NEO. I got the following orbital parameters:

a=-2791.44 km (semi-major axis),
i=22.3 deg /Hyperbolic),
e=2.77
v=247.96 deg,
r= 2,106,101.331 km (radius from Earth's center of mass)

My problem is that I don't know where to begin, but I did an attempt. I assume that the intercept with the target only can happen when the phase angle between Earth and the NEO are the same.
I know that one can calculate the velocity of the S/C with ($m_y$=gravitational param.):
$$v=\sqrt{(2m_y/r)-(m_y/a)}$$
If we calculate the escape velocity of the Earth with $V_e=\sqrt{2GM_e/r}$, we will get the difference in velocity. With this information, we can calculate the time needed.  Anyone that can help me?

Assuming that you have a phaser array sufficiently powerful to pulverize the NEO into a harmless meteor storm,
I read this as a question of when an object with those geocentric orbital elements arrives at perigee.
However, I get a different value for $r$ if I treat that $\nu$ value as a true anomaly; it could be a mean anomaly or hyperbolic anomaly instead.

I suppose your textbook covers this stuff in detail, but for other readers
these lecture notes
by M. Peet at ASU, using $f$ instead of $\nu$, may help.


Which kinds of astronomical observations most need to avoid the Moon being up? This comment to Did nobody in the Astronomy community think 12,000 new satellites in LEO might be a problem? links to Phys.org's New ESO study evaluates impact of satellite constellations on astronomical observations which includes ESO's Areas of the sky most affected by satellite constellations shown below.
It's a fisheye lens view (FOV wider than 180 degrees)looking straight up.
In Earth Science SE I've asked What causes this arc in the night sky where the background is brighter on one side than the other? but here I'd just like to ask about the effect of sky brightness associated with the Moon on observational astronomy.
Question: Which kinds of astronomical observations most need to avoid the Moon being up? Are some observations relatively insensitive to the Moon being in the sky but not necessarily nearby and others negatively impacted or impossible because of it?

This annotated image shows the night sky at ESO's Paranal Observatory around twilight, about 90 minutes before sunrise. The blue lines mark degrees of elevation above the horizon.
A new ESO study looking into the impact of satellite constellations on astronomical observations shows that up to about 100 satellites could be bright enough to be visible with the naked eye during twilight hours (magnitude 5–6 or brighter). The vast majority of these, their locations marked with small green circles in the image, would be low in the sky, below about 30 degrees elevation, and/or would be rather faint. Only a few satellites, their locations marked in red, would be above 30 degrees of the horizon — the part of the sky where most astronomical observations take place — and be relatively bright (magnitude of about 3–4). For comparison, Polaris, the North Star, has a magnitude of 2, which is 2.5 times brighter than an object of magnitude 3.
The number of visible satellites plummets towards the middle of the night when more satellites fall into the shadow of the Earth, represented by the dark area on the left of the image. Satellites within the Earth's shadow are invisible.
Crediti: ESO/Y. Beletsky/L. Calçada

Imaging and spectroscopy of very faint objects, especially towards the bluer end of the spectrum. The effects of trying to subtract the background sky brightness also become more demanding if the object is extended in size or if the angular extent of sources is broadened by atmospheric turbulence (aka "seeing").

The reason that the effects are worse at the blue end of the spectrum is that the presence of the Moon has a greater effect (because scattering is stronger) on the dark sky surface brightness, increasing it by the equivalent of about 4 magnitudes (per square arcsec). Thus the sky brightness is a factor of about 40 higher (at a good site, with little dust and at 90 degrees to the Moon) than when the Moon is well below the horizon.

The contrast is much lower in the red and near infrared. There is less scattered light and in the infrared, the sky background is dominated by airglow, and unaffected by the Moon.

Broadly speaking, observations of bright stars ($V<15$) or observations at near infrared wavelengths can be done at any time, including "bright time" when the full moon is up. Deep imaging, especially of extended objects, and spectroscopy of faint objects ($V>19$), especially at bluer wavelengths requires "dark time", with no moonlight. An exception might be observations of emission line objects (e.g. planetary nebulae), where all the flux is in very narrow wavelength intervals. In between, there is "grey time", when the Moon is up, but less than half illuminated.

For completeness - radio, mid-infrared and mm-wave observations are unaffected (unless the Moon is in the way!)


What is the effect of this expanding universe on the energy density of it? We know that our universe is expanding i.e.  volume of the universe is getting larger with time with total energy being conserved. My question is whether the energy density is getting smaller with time.
If this is true then what is the effect of this phenomenon on all the celestial bodies?

I am no expert and I am sure there will be a more rigorous answer later, but here my 2 cents:

It appears to me that the term "energy" and energy conservation becomes a bit more complicated in general relativity and cosmology, taking into account more abstract forms as the energy of the gravitational field or space curvature. As you seem to be more interested in the implications I just want to focus on the common forms of energy, i.e., radiation, matter and dark energy.

For a given expanding volume of space only the energy from matter (dark and normal) is dilutes as expected. You could say that with respect to expansion its energy is conserved. This means its energy density decreases with $\sim a(t)^{-3}$, where a is the scale factor (or just the diameter) of the universe.

Radiation additionally looses energy via cosmological red-shift, so its energy density decreases even faster with  $\sim a(t)^{-4}$.

Finally you have the dark energy density. What makes it special is, that its density does not depend on $a(t)$. One can think of it as the energy associated with space itself. So if the universe expands new space and dark energy "appears" and its density stays constant.

So over time the average (matter and radiation) energy density decrease and approaches the energy of empty space itself (or vacuum energy).

Note, however, that this is only true for the average densities. In regions of space, that are held together by gravity, as our local group, the matter density will not decrease. On the contrary, these region are still contraction due to gravity.

So to answer your question on the effect on all the celestial bodies:
Those which are gravitationally bound, won't notice much but others will drift apart never to see each other again.


What can a regular person use to view the stars? My favorite thing in this world is looking up at stars but I live in NYC so I can't ever see stars. I was wondering if anyone knew of a device I can attach to a smartphone that would let me look up at the stars and actually see some? I can't afford to buy an expensive telescope. I'm a student and I just want to see some stars. Thanks to anyone who can help.

Binoculars.

This isn't going to attach to a smartphone, not everything in life is best seen through a screen. The camera on a smartphone is remarkable, but your eyes are much more sensitive in low light. Binoculars will make a lot more stars visible, even in heavily light polluted places.

A pair of binoculars is cheap enough for students. They are categorised as, for example 10x50, which means 10x magnification, and 50mm lenses. You want the second number to be big, which means the binoculars gather more light. 10x50 is often thought to be a good compromise between light gathering and weight. You won't want "zoom" or "night vision" or other such features.


How does making a refracting telescope very long reduce the chromatic aberration of an uncorrected lens? Below are two cropped views of "Johannes Hevelius's 8 inch telescope with an open work wood and wire "tube" that had a focal length of 150 feet to limit chromatic aberration." from Harvard University, Houghton Library, pga_typ_620_73_451_fig_aa (found here) the first of which I've zoomed and enhanced the contrast.
According to Wikipedia's Aerial telescope; Very long "tubed" telescopes:

Very long "tubed" telescopes
Telescopes built in the 17th and early 18th century used single element non-achromatic objective lenses that suffered from interfering rainbow halos (chromatic aberration) introduced by the non-uniform refractive properties of single glass lenses. This degraded the quality of the images they produced. Telescope makers from that era found that very long focal length objectives had no appreciable chromatic aberration (the uncorrected chromatic aberration fell within the large diffraction pattern at focus). They also realized that when they doubled the diameter of their objectives they had to make the objective's focal length 4 times as long (focal length had to be squared) to achieve the same amount of minimal chromatic aberration. As the objective diameter of these refracting telescopes was increased to gather more light and resolve finer detail they began to have focal lengths as long as 150 feet. Besides having very long tubes, these telescopes needed scaffolding or long masts and cranes to hold them up. Their value as research tools was minimal since the telescope's support frame and tube flexed and vibrated in the slightest breeze and sometimes collapsed altogether.

A refractor's primary lens produces an image because the angle it deflects is proportional to the distance $x$ from the center of the lens of focal length $f$ as
$$\theta \approx x/f$$.
For a dispersive material the deflection at a given point on the lens varies with wavelength, and therefore the focal length of the lens varies with wavelength.
 Source
This is called chromatic aberration and its severity for a given type of glass is often expressed by its abbe number.
Question: Is there a simple mathematical way to explain how making the focal length longer could possibly improve a telescope's performance by decreasing the impact of chromatic aberration on its performance?
Or do I misunderstand what the benefits are to making these very long telescopes?

The actual math is a bit complicated, but there's a simple intuitive explanation.

Longitudinal chromatic aberration happens because, when you cut a convergent lens in two, and you look at the cross-section, the edge of the lens looks a bit like a prism, doesn't it? (look at the diagram that you've posted above, the top of the lens) And it does exactly what a prism does - it makes a "pretty" little rainbow. Not so pretty when you're trying to do astronomy, however.

Short, fat, triangular prisms make widespread rainbows. Thin, elongated, thin-wedge shaped prisms make tight, narrow rainbows. The spread of the colors increases as the two sides of the prism are separated by a wider angle. It decreases as the two sides get closer together at a narrow angle. At the extreme, if the two sides of a "prism" were parallel, there would be no rainbow - but then it's not a prism anymore, it's just a glass plate, completely flat.

All else being equal, a single-lens-objective refractor has a "fatter" lens when focal length gets shorter (the two sides of the lens are more strongly curved, and/or less parallel); it has a "thinner" lens when the focal length gets bigger (the two sides of the lens are less strongly curved, and/or closer to parallel). A "fat" lens at the edge is like a fat, blunt prism. A "thin" lens at the edge is like a sharp, slender prism. The former makes a big spread of color. The latter makes a much more tight rainbow.

So that's what the old telescope makers did, back in the day (Hevelius, late 1600s). They just pumped up the focal ratio, made these super-long refractors with weakly curved lenses to combat the rainbow edge images which are so annoying and ultimately can falsify your observations.

Now, if you study the properties of lenses, you notice that divergent lenses (the kind that make the image smaller) have a chromatic aberration opposite to convergent lenses (lenses that magnify, and that were used as primary, or objective lenses in telescopes back then). It could be argued that by combining a convergent and a divergent lens, chromatic aberration might be greatly reduced; as long as the convergent lens is stronger than the divergent companion, the combination should remain convergent.

Also, different glasses need to be used, so that the convergent lens is strongly convergent, but weakly dispersing, while the divergent lens is weakly divergent, but strongly dispersing. That's how you get near zero dispersion, but still pretty decent convergence.

In the early 1700s this was accomplished by British opticians. Chester Moore Hall is credited with the making of the first achromatic objective lens using the scheme shown above. But the actual work of grinding/polishing the glass was contracted to George Bass, who noticed that the two lenses he was making, when put together, had near-zero chroma. Hall failed to recognize the practical applications of the achromats, neglected his own invention, and years later Bass mentioned the achromat scheme to John Dollond, who understood its value and started to popularize the new design.

https://en.wikipedia.org/wiki/History_of_the_telescope#Achromatic_refracting_telescopes

https://en.wikipedia.org/wiki/Achromatic_lens

The achromatic combo guarantees zero chromatic aberration at two wavelengths. So you basically correct the aberration in the red and blue, or something like that, and the rest is "good enough". Things can be improved further by adding another lens - the apochromatic objective, or the "apo", or the "triplet". This was invented by Peter Dollond, John's son, in the mid-1700s.

The apochromat guarantees zero chromatic aberration at three wavelengths - usually in the middle and then closer to the two extremes of the visual spectrum. So that's overall better than the previous scheme.

https://en.wikipedia.org/wiki/Apochromat

The best apo refractors have extremely low chromatic aberration, basically invisible to the eye in most situations. But they are expensive.

Further improvements can be made. The most expensive amateur refractors are "quad" schemes, whereby using 4 lenses zeroes out chroma at 4 wavelengths. Takahashi is a brand nowadays known for excellent quality quad refractors, used in top-level astrophotography rigs.

BTW, not just chroma, but basically all aberrations can be reduced by increasing the focal ratios of single lens objectives. Pretty much every single aberration gets better if you put a piece of paper in front of it, with a hole smaller than the diameter of the lens. But the aperture gets smaller, resolving power gets worse, there's less light captured by the instrument, etc. Tradeoffs...

W.r.t detailed math of single lens chromatic aberrations, see chapter 5.3.1. Chromatic Aberration of a Singlet in 'Telescopes, Eyepieces, Astrographs' by G.H. Smith, R. Ceragioli, R. Berry. I believe that chapter is exactly the answer to your question, complete with working out the math for the aberration as a function of lens parameters.

http://www.willbell.com/tm/TelescopesEyepiecesAstrographs.html

Given your interest in the theory of optical systems, you should read that book.

Also worth mentioning:

Introduction to Lens Design, by J.M. Geary

http://www.willbell.com/tm/lens_design/index.htm

Practical Computer-Aided Lens Design, by G.H. Smith:

http://www.willbell.com/tm/lensdesn.htm

Some good theoretical info, as always, can be found on telescope-optics.net

https://www.telescope-optics.net/chromatic.htm


What's going on in this photo of Halley's comet's tail? When researching for this answer I was looking at Space.com's 2011 article Photos of Halley's Comet Through History

I saw the image below, and read the caption.

  Halley's Comet in 1910

  NASA/JPL

  Halley's Comet as photographed May 13, 1910, by a wide-angle camera at Lowell Observatory, Flagstaff, Ariz., during the comet's last appearance. A streak across the comet near the coma is a meteor trail, and not a scratch on the negative. Streaks at the bottom right are the city lights of Flagstaff Bright spot above the city lights is the planet Venus.

I know that 109 years ago astrophotography wasn't what is is today, but the combination of the overexposed venus, the coincident meteor trails, the non-uniform star trails blending in with the comet's tail breakup along with the unlikely looking "smoke trail" from the upper side of the comet's tail make this photo look quite unusual.

Question: What's going on with that second tail thing above the main tail? The shape looks so unnatural, it seems to have a parallel extent rather than radiate from the coma. What caused this?

"bonus points" for identifying the photo in some catalog, or in a publication.

Negatives taken with various intstruments at Lowell Observatory on the morning of 1910-05-13 appear in a 1981 conference paper by Henry L Giclas.
The wide-angle images were taken with cameras piggyback on a telescope tracking the comet nucleus.
These exposures are naturally deeper in the center than at the edges; modern astrophotographers take flat fields to correct for this.
The stars in the upper right of the image in question were low enough to the horizon to be obscured for part of the exposure, making their trails shorter.
To make this more intuitive, rotate the image so that the tail points to the right and upward.

Comet tails generally have two parts: a smooth, curved dust tail driven by steady radiation pressure; and a straight, blotchy plasma tail (also called an ion tail or gas tail) driven by the fluctuating solar wind.
A coronal mass ejection can severely disrupt the latter, as the STEREO spacecraft observed with Comet Encke in 2007.

The 1910 image in question may show a tail disconnection event in progress. If so, the highlighted feature is the old plasma tail drifting away from the nucleus, and the new plasma tail is not yet large enough to be distinct.
Barnard 1914 includes three photos of such an event on June 6-7, 1910.


Circular formation around the moon I have seen a large circular formation of cloud around the full moon sometimes.

Have anyone seen the same? What is the reason behind this formation?
Due to lunar attraction?

Most of the sky is empty and clear, and no clouds at all, only around the moon, far from the moon, not just very close, that cloud circle formed.
I saw this many times, like the below drawing:

That sounds very much like a 22° halo.

It's not an astrophysical phenomenon; it results from the refraction of light by ice crystals in Earth's atmosphere.

If this is what it is, it should be a fuzzy but regular circle centered on the Moon, with a radius of about 22 degrees. If you hold your fist out at arm's length with the thumb extended, the angular distance between the base of your fist and the tip of your thumb is typically close to 15 degrees; you can use that for visual comparison. Remember that 22° is the radius of the halo; the diameter is 44°, close to three time the width of your fist+thumb.


Why does the Sun track out a seemingly sinusoidal path on the celestial sphere? The question arose because I wanted to understand the path of the ecliptic after reading about it here: https://community.dur.ac.uk/john.lucey/users/solar_year.html

Is the sinusoidal path due to earth’s axial tilt? Or is it related to mapping the path on a 2D pane? Is there way to visualise this to aid understanding?

Both the equator and the ecliptic are great circles on the celestial sphere.
The appearance of each on a map depends on the map projection.
In an equirectangular projection centered on the equator, the equator (brown) is a straight line, and the ecliptic (blue) is approximately sinusoidal.

If the same projection is centered on the ecliptic instead, the ecliptic is a straight line and the equator is approximately sinusoidal.
As uhoh's answer illustrates, neither curve is exactly a sinusoid.

Lucey provides both views too, but his ecliptic-centered map doesn't show the equator.
The axes are labeled in ecliptic rather than equatorial coordinates.

He also provides a stereographic projection centered on the midday zenith at 55°N.
Here both equator (red) and ecliptic (green) are mapped as circular arcs.


50mm f1.8 for Astrophotography I am considering using a 50mm f1.8 maybe even try aN 85mm f1.8 on my Astro FI 102MM Telescope have you any thoughts or suggestions?
I have never taken images via a telescope so this is a whole new learning curve. Need to get my brain back in gear, stopped shooting over 5yrs ago due to the gear being too heavy and my back injury not coping so this will be a much easier way of holding the camera...I HOPE.

I'm not an astrophotographer but I'd like to make sure that you know that there are (at least) two ways to put a camera on a telescope from an optical perspective.

The first is to keep the eyepiece on which produces a virtual image at infinity, and keep the camera lens on, focused at infinity.

The second is to remove both the eyepiece and the camera's lens, which leaves the camera's CCD sensor right on the primary focal plane of the telescope.

The first method is the quickest, and it seems to be called Afocal photography (I didn't know that!). If you are careful and the image is very bright, say for example the moon, then you can just set the camera focus to infinity, hold it up to the eyepiece, moving around sideways until the inevitable vignetting is at least centered, and click! You can try to manage the vignetting by moving the camera closer or farther from the eyepiece. Ideally you want to place the entrance pupil of your camera's lens on top of the exit pupil of your telescope's eyepiece.
I predict you will quickly move to the second method which is called Prime Focus Photography. Find an eyepiece-to-camera body adapter and then you are a bona fide astrophotographer!
If you want to take advantage of your very nice f/1.8 lens then lose the telescope and just photograph the milky way or meteor showers. This works best if your camera can be programmed to take a sequence of regularly spaced images, because you can merge them later with a computer with offsets so that the motion of the stars doesn't produce trails.
For more on all of these, search questions tagged photography here and astrophotography in Photography SE.

For completenes' sake only, I'll mention a third way, called Eyepiece projection. While a properly focused eyepiece produces a virtual image at infinity, if you move it out, away from the telescope, it will produce a real image out past the eyepiece. If you are looking at the Moon or something very bright you could put a piece of paper out a ways away from the eyepiece, say 6 to 20 cm, and with some practice produce an image on the paper. You can then imagine having your camera's sensor at that plane. This gives you more magnification compared to prime focus photography, but it may not be useful magnification!
Further reading:

The Prime-Focus Method
Eyepiece Projection
What is eyepiece projection and how do I do it?
Sky & Telescope's Astrophotography With a Digital Camera
Afocal method
Afocal Deep Sky Astrophotography where you can clearly see what vignetting looks like!


Is there any planet/star bigger than VY Canis Majoris? Compared to VY Canis Majoris, our Sun is more like a speck of space dust. Is there any planet or star which is known to us that is bigger than the VY Canis Majoris?

Part 1: Assuming "larger" means greater in diameter.

Stars
Estimates on the size of stars are just that, estimates, and estimates based on rather fuzzy observations. VY Canis Majoris has been bumped down to size. The current thinking is that there are seven known stars larger than VY Canis Majoris, the largest of which is UY Scuti.

Current models indicate that the first generation of stars were much, much larger than anything we see now. It will be quite some time before we can resolve a first generation star. So far, they are just theoretical objects.

Planets
Jupiter-mass planets are about as large as a planet can get. There are some exoplanets that are larger than Jupiter, but that's because they orbit much closer to their parent star than does Jupiter. This makes them puff up a bit. The reason Jupiter-mass planets are deemed to be the largest possible is that planets of this mass are presently assumed to have a core of degenerate hydrogen. A funny thing happens to degenerate masses when mass is added to them: They shrink in diameter. (The shrinkage becomes catastrophic as the mass approaches the Chandrasekhar limit.)

This means that assuming all other things are equal (temperature, composition), a planet more massive than Jupiter will be small in diameter than Jupiter is. Even if all other things aren't equal, a Jupiter-diameter planet is (give or take) about as large as planet can get.

Part 2: Assuming "larger" means greater in mass.

Stars
In terms of mass, VY Canis Majoris doesn't even make the top ten, not even close! The most massive known star is R136a1. Again, these are estimates, but mass is a bit easier to pin down than is radius (or diameter).

As is the case with physical extent, the first generation stars are presently modeled as being much, much more massive than anything we see now.

Planets
There's not much difference between the largest planet and the smallest brown dwarf. I would argue there's very little difference. It's a spectrum with no distinguishing characteristic that lets one say "this is a planet" and "that is a brown dwarf". Ignoring the distinction between super-Jupiters and brown dwarfs the largest is about 80 Jupiter masses. V1581 Cygni C is 79 Jupiter masses. (More massive than that and they start burning hydrogen, thus making them a small red dwarf.)

The current factor that is used to distinguish between brown dwarfs and super Jupiters is mass. Anything larger than 13 Jupiter masses is a brown dwarf, anything smaller, a planet. That boundary is very arbitrary.


how accurate are the predictions for coronal mass ejections-(location, timing and magnitude)? I suppose this applies to all solar events.

I know centuries ago they knew the quantity of these things fluxuated in  solar cycles. Apart from that I know there has been advancement in solar weather forecasting but how accurate is it? Could, say ESA, have enough foreknowledge to launch a satellite or have one poised in solar orbit to 'go fishing' to investigate the emissions in someway? ?composition of particles- for example?

Old question; I hope you’re still active here…
Basically, we CAN’T predict coronal mass ejections in advance. See http://solar.physics.montana.edu/press/ssu_index.html
Other solar events, such as the number of sunspots, are somewhat predictable—not precisely, but at least have an idea about them. For example, sunspots follow a cycle of approximately 11 years. However, we don’t know before a cycle if it will be strong or not—our models are still too imprecise, though, like you wrote, there has been some advancement, but not much.


How much starlight do black holes absorb? Can it be calculated how much light from other stars a black hole of certain sizes would absorb? And how long will it take for the number of active stars to decrease to the point where Hawking radiation would surpass it?

tldr; extremely little. Space is big, even biggest black holes are ridiculously tiny in physical size

The amount of starlight absorbed by black holes is really minuscule. A black hole is not going to be sucking light in: it can be fairly well approximated to be an absolutely black object around the size of its event horizon (or photon sphere, does not really matter for the purpose for this thought experiment, since photon sphere is 1.5x EH).

Now, let's think of occlusions by more familiar objects: the Moon and the Sun. Now let's think our Moon would be a black hole. Retaining its current size, it would need to have mass in range of about 500 solar masses - pretty big for a BH. It would be also orbiting really close to our Sun; so close, it would probably get eaten by the Moon-BH relatively soon. Also, we would be observing this system really close.

But even in this case, we would still see the Sun most of the time unobstructed. Obstruction means absorption of light, thus there's really not much of this happening.

The amount of absorption of light from a star can be calculated by figuring out the ratio of black hole surface area to the surface area of a sphere with the radius equal to distance between the star and the black hole. In most cases you will be using square kilometers for the BH area, and square light years for the sphere. Which is almost exactly the same as comparing single atom mass to grams.

For the second question, star shine does not really contribute much to the temperature of space, so CMB will dominate this until very very far in the future, when all the stars are already gone


Why do certain moons have their rotational period equal to their orbital period? Our moon and Saturn's moon, Titan, have this feature.  Because of this we only observe one side (hemispere) of our moon. Why is this?  What is the Newtonian or astrophysics that would explain this?  It seems that it would be some kind of equilibrium that is arrived at over time, but I don't understand it.

The answer to this is certainly tidal forces, but that doesn't explain the exact mechanism for how tidal forces result in tidal locking, i.e., an orbiting body showing the same face to the central body as it orbits due to the rotation rate and revolution rate being equal. I'll describe this mechanism using the Earth-Moon system so I can be specific, but it equally applies for any system.

To start, tidal forces are a result of differential gravitational forces across the distributed mass of a body. The Moon is not a point mass, it has an extended size. The force of gravity on the Moon by the Earth is dependent upon distance (as is the force of gravity for anything). What this means is that on the Earth-facing side of the Moon, the force of gravity is stronger and as you progress through the Moon to the Earth-opposing side, the force of gravity gets weaker. This means the Earth-facing side of the Moon gets pulled stronger and closer to the Earth while the Earth-opposing side, while still getting pulled towards the Earth, doesn't get pulled as strongly or as close. Ultimately, this results in the Moon being deformed such that it becomes slightly oblate and stretched out in the direction of the Earth. This flexing of the surface is known as tides.

Now let's assume the Moon is not currently tidally locked with the Earth, and in fact spins slightly faster than it orbits. The Earth is causing tides on the Moon and the Moon is spinning on its axis. The tides, caused by the tidal forces, want to remain aligned with the Earth-Moon line since that's the direction the tidal forces are applied. However, it takes time and a lot of energy to deform the Moon. Once the Moon is deformed, it's going to spin along and try to pull that tidal deformity along with it, effectively moving the tidal bulge ahead of the Earth-Moon line. The Earth is still applying the tidal force along that Earth-Moon line to try and pull the tidal bulge back in line. This constant force trying to pull the tidal bulge back (or ahead if the Moon is spinning too slow) allows for a transfer of momentum to slow down the Moon (or again, speed it up if it is too slow). The main point here is that tidal locking is a equilibrium state because if the Moon is spinning too slow or too fast, the Earth trying to pull the tidal bulge into the Earth-Moon line will change the rotation rate of the Moon until it becomes tidally locked. Once it is tidally locked, that tidal bulge will always be along the Earth-Moon line and this force will disappear.

That covers about half of the answer though. When trying to tidal lock the Moon to the Earth, you have to consider two time periods. The first, discussed in the preceding paragraph, is the rotation time of the Moon around its axis. The other is the revolution time of the Moon around the Earth. Both must match. The previous paragraph described how the rotation time of the Moon around its axis could be affected, but there's also a means of affecting the revolution time of the Moon around the Earth. Fortunately, this is by a nearly identical mechanism as above. In effect, the Moon also causes tidal bulges on the Earth and since the Earth is spinning these tidal bulges won't be directly aligned with the Earth-Moon line. This unaligned tidal bulge on Earth acts to transfer energy to the Moon's orbital speed, causing it to speed up or slow down. Incidentally, through conservation of angular momentum, this necessarily causes the Moon to drift away from us at a small but persistent rate.

In summation, tidal forces cause tidal locking, but it happens through intricate and slow forces over a long time period affecting both the Moon's orbital velocity and rotation rate until a equilibrium is found. That equilibrium is tidal locking.


Star rotational axes and Solar system orientations How does the orientation of stars axes of rotation and any planetary systems relate to the plane of the galaxy? Is there any relationship that has been detected, are they grouped from their parental nebulae. Do galactic magnetic fields have coherent orientations that would affect nebula growth and consequent stellar growth.

No. there's in no relation as far as we know. Ecliptic plane orientations of other planetary systems with respect to the galactic plane are completely random as far as we know.

We know this also because we can find exoplanets by the transit method nearly in any direction of the sky; if we look at $N$ stars located at galactic latitude of (let's say) $20^\circ$ (thus their ecliptic plane has relative inclinations of $20^\circ$ with respect to the galactic plane) we find roughly the same number of transit events as if we look to $N$ stars at $0^\circ$ latitude (thus their ecliptic planes have relative inclinations of $0^\circ$ with respect to the galactic plane). Since we are probably not the center of a planetary conspiracy on which everything is aligned so that we see the same distribution in all the sky we can assume that inclinations are random and we just happen to detect a similar fraction of transits everywhere because we are looking at at the same fraction of the same random population in each direction.

Humanity has just started to observe other planetary systems so this might change a little but in principle if there is any "preferential orientation", for now, with our current knowledge it is completely camouflaged by the main random component of this distribution.


Planet with acid ice caps I remember reading an old astronomy book mentioning a certain planet and the main point of interest was that this planet has or may have polar ice caps but of acid. I can't remember the name though, what planet or planets fit this description?

Now there some very strange worlds out there though I'm unsure whether this one may or may not be all that strange. Is this something of an oddity or is a planet containing frozen acid not that uncommon?

Of the planets, two have ice caps: Earth and Mars.  The Earth's ice caps are water ice, and Mars has ice caps of Carbon dioxide, and some Water ice. The CO2 on Mars would give a weakly acidic solution if melted but it would not be a caustic acid.

Titan has hydrocarbon lakes at its poles, and other planets are too hot, or are uniformly frozen.


What would a terraformed Mars look like from Earth? Question in the title. Currently, Mars is a small orange dot in the sky. But what if green, blue, and some white were added? I wouldn't expect it to be too different- a large portion of it will remain pretty much the way it was, but with more snow (Tharsis bulge is REALLY high up).

To terraform Mars, broadly speaking would make it more like an analogy of Earth. As you say, it might be a little colder, but we would expect a terraformed Mars to have oceans, snow and maybe some imported plant life.

How would Mars look from Earth? A good starting point would be to see it the opposite way around. Here is how Earth looks from Mars.

Like Earth from a distance, a terraformed Mars would likely appear as a pale blue or white dot. If you want more snow or ice than water, it would surely lean towards white rather than a deep blue.


Why is the symbol of Vernal Equinox ♈? Since the direction of Vernal Equinox lies in the Pisces constellation, why is its symbol Aries? Shouldn't it be ♓ ?

Is this a historical anachronism, only invalidated by axial precession of Earth?

As Wikipedia explains in the article on the Zodiac, the Zodiac was originally developed with the assumption of fixed equinoxes. It is assumed this is because the Babylonians who developed the Zodiac were unaware of the precession of the equinoxes. At that time, about 2500 years ago, the vernal equinox did in fact lie in Aries. For this reason, in astrology, that time is referred to as "the Age of Aries".

Of course, in modern times these astrological concepts are scarcely important to astronomy, and in the few cases where the astronomical symbol for the vernal equinox is used, it is more practical to have a fixed symbol than a changing one.


Why has there been no moon for 2 nights straight I have been watching the sky all night for a week now I see flying stars everywhere well things that look like stars, what are they? And the moon has been gone for two nights now? Why?

The Moon is out of sight for a few nights every four weeks, more or less. It moves around the sky slowly. Sometimes it's far from the Sun (as seen from our planet), and that's when you see it at night, up in the sky. Other times it seems close to the Sun, and then of course you won't see it at night. Then the cycle repeats, over and over again. Takes a bit over 4 weeks to repeat.

The star-like things moving in the sky can be many things. Usually they are satellites in orbit around Earth, or airplanes flying at high altitude.


Perception of an objects history passing as an observer moves toward it If a star is 20 light-years away and I look at it through a telescope, what I'm observing is what the star was doing 20 years ago yes? So, if I fix my sight on the star and move toward it (for example's sake, let's say 0.5C or half the speed of light in a vacuum) in a linear path (which is not practical for real space travel) it would take me 40 years to reach it. By the time I reach it I would be observing in (almost) real-time what is happening. So to me that says that I've watched 60 years of the star's history in 40 years worth of time. So then, would it look to me as if I was watching a movie of the star's history in 1.5 times 'normal' speed? (60/40 = 3/2 or 1 1/2)

Yes, it will be 20 Julian years ago.

Yes, it will take 40 years to reach there with speed half the speed of light through vacuum without any obstacles and star must be stationary (though you can't travel with speed of light since physically it's not possible to achieve that speed).

And as you are travelling with speed half to that of the speed of light you will experience a phenomenon of time dilation. Check all the details about time dilation here:

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

You will see time dilated by $\gamma = 1/\sqrt{1-v^2/c^2}$, factor so if you substitute this values it will take you 60 years to reach that star. And it squares all the things.

I hope this explanation helps you. If it is not understandable do let me know; I will explain it in a more mathematical way.


Will we ever end up in the black hole? Since we are spiraling around the black whole will we spiral into the black hole?

It is very difficult to predict since the Milky Way galaxy and the Andromeda galaxy are going to collide in about 5 billion years. Since we cannot predict the gravity influences so far in advance, it is impossible to say.

There are 3 outcomes of this collision. The solar system falls into the black hole, the solar system goes into orbit in the new galaxy, or the solar system get propelled away from the new galaxy.


Rising & Setting time of a star on a specific date Given the local sidereal time (LST) of a star when it rises, how can I find out the time when it will rise (according to my watch) on a specific date?

You'll need to convert local sidereal time to local time on the specific date, using the longitude of your location. This can be done with the formula described here:

  $$
UT = \frac{LST - \frac{L}{15} - 6.697374558 - 2400.051336*T - 0.000025862*T^2)}{1.0027379093}
$$

  where $UT$ is the time in Universal Coordinated Time, $LST$ is the Local Sidereal Time, $L$ is the longitude, and $T$ is:

  $$
T = \frac{JD-2451545.0}{36525.0}
$$
  where JD is the Julian Date at noon UTC.  Once I've calculated the time in UT, I then add or subtract 24 until the result is within 0 - 24 hours.  Then I can convert from UT to local time.


With all the stars in the universe, why isn't the sky saturated with light? Space is dark. But if we direct telescopes to seemingly dark portions of sky, we see that it is filled with galaxies. If the universe is theoretically infinite, wouldn't an infinite number of photons be reaching the earth and illuminating the night sky brighter than it appears?

Logic1: If the universe is infinite AND static, the sky will be saturated with light.

Note "AND."

Logic2: If the sky is not saturated with light, the universe is not infinite or not static.

Note "OR."

Logic1 = Logic2

Observation: the sky is not saturated with light.
Therefore: the universe is not infinite or not static.

We know that the universe is dynamic, at least.


List of Moving stars on the northern hemisphere? [Stellar kinematics] Anyone know if there is a List of Moving stars on the northern Europe hemisphere I can search for? Like Barnard's

I wish to image them [when possible] and compare their yearly movement.

Thank you

What you're looking for is stars with "High Proper Motion".  A Google search of "High proper motion stars" will provide a number of such lists.  A quick look at Wikipedia gives this link:

https://en.wikipedia.org/wiki/Proper_motion#Stars_with_high_proper_motion

You can follow the links to individual stars if you wish to determine their latitudes.


Retrofitting a DSLR camera to a super-cheap Tasco reflector telescope I was given a 1995 vintage Tasco "302003" Newtonian telescope recently and have had some exciting success with it.  I have managed to view the Moon in beautiful detail and also had my first glimpse of Jupiter (with Enceladus Ganymede & Europa in the same shot!).

I was able (with very considerable difficulty) to photograph both Moon and Jupiter scenes by holding my camera up to the eyepiece.  Far from ideal but it has really fired my imagination about doing more astronomical photography.

Here's a picture of my telescope and it's accessories.  Despite spending the last 15 years in the back room of a charity shop, it's in brand new condition and all the parts are present including the tripod and all fixings!...

(Blue arrows indicate places where eyepiece lenses can be inserted).

My camera is a Nikon D40 (APS-C format, DX lens ring system).

Is it possible to modify this telescope's eyepiece (a rack-and-pinion tube-length adjustable type) so that I can attach my camera in order to take photographs of astronomical subjects?  Photographs of the Moon would be my primary goal.

I am happy to make severe modifications to the telescope if necessary.  But I realise that certain measurements would be absolutely critical to making good photographs possible.  I think I can make an adaptor that goes between the eyepiece socket and the camera.  But I don't know all of the parameters I will have to know to design such an adaptor.

Problems I believe I need to solve...

I think that holding the camera's central axis perfectly concentric with the eyepiece's axis is critical.
I need some sort of Relay Lens between the secondary mirror and the camera body.
Keeping the rack-and-pinion tube adjustment (actually a fairly sturdy mechanism) is important and would allow better focusing of the picture as the camera is not likely to have enough auto-focus range to cope with the 700mm F length.

I do have a small Milling machine and most of the basic machine tools.  I have a ready supply of aluminium stock and industrial plastic stock (RG1000) that I can build an adaptor from.

Is this project feasible?

I'm not looking for perfect photos, but rather something that is better and more convenient than holding my cellphone camera close to the eyepiece and holding my breath! :)

My astronomy/optics knowledge is practically zero but I'm a fairly savvy mechanical engineer and can build things to a specification.

Thank you for reading, I look forward to your responses.

The problem you have with many newtonian reflectors - especially smaller ones - is that to use them for prime focus photography (where you're effectively using the telescope like a big telephoto lens) , you need to set things up so that the camera sensor ends up at the focal plane of the telescope. And in a DSLR, the sensor is some way inside the camera body (behind the shutter, which is behind the mirror.

With a newtonian scope, the focal plane is usually somewhere inside the focuser tube, and many newts don't have enough focuser inwards travel to let you get the focal plane the 40mm or so beyond the end of the focuser that you need to get it to reach a DSLR sensor.

(With other types of scope that use a separate star diagonal, things are usually easier - you can leave the star diagonal off, which has the same effect as a big chunk of inward focus).

With a newtonian, if you can't reach prime focus, there are three basic options - either take photos through the eyepiece (afocal photography) as you already tried,

or move the main mirror further up the tube (which moves the focal plane further out, giving you more in focus). It also means the light cone is fatter when it reaches the secondary mirror, so you may lose some light as the outer part of the light cone misses the secondary.

or you can try using a barlow lens - that replaces the last bit of the light cone with a longer, narrower angle version (which increases the magnification and moves the focal plane further out). That MAY or MAY NOT let you reach infinity focus; it's not guarenteed, but may help. You can get barlows that have a T-Thread on the output end, so you can attach a T ring directly to the barlow (yours doesn't look like one of this type) and then hook the camera up that way. (as an added benefit, since you end up with a closed tube setup (the barlow lens closes off the access to the camera, it also keeps spiders and dust out of the camera :) ).

If you have a look at the bottom bit of the barlow (with the lens) you may find that you can unscrew it. You might be able to make a short top section with a T thread for the T ring at the back, and a suitable thread for the lens at the front; you'd probably need to make it the same diameter as the eyepiece barrel (to fit in the focuser) with a fatter bit at the back for the T thread and to stop it sliding all the way into the focuser.
I've used a commercial 1.25" barrel T threaded barlow (somewhat shorter than yours) for lunar and solar (WITH A SUITABLE FRONT OF SCOPE FILTER - DON'T TRY LOOKING AT THE SUN WITHOUT ONE) imaging with an APS sensor camera.

(You may need to check your scope - older ones often used a .965" barrel rather than the current 1.25" (or 2") barrel size used by most modern astro scopes.)


Martian Constellations I was wondering if the constellations visible from Earth would also be visible from Mars.  Would they look the same but just shifted slightly in the sky due to the larger circumference of the orbit?

The stars are so immensely far away that to the human eye there would be no noticeable difference. The nearest stars are moving roughly 1.5 arcsecond wrt. the background when viewed from Earth's position with half a year's interval (i.e. half a revolution around the Sun). Mars' orbit is only 50% larger, but the human eye cannot resolve better than roughly one arcminute.

However, the sky would look different for another reason: Although stars are point sources, on Earth stars are smeared out due to the atmosphere. On Mars, there's almost no atmosphere, but on the other hand there's lots of dust whirling around. Dust tends to absorb blue light more than red and thus redden objects. However, Martian dust happens to scatter red more than blue (Ockert-Bell et al. 1997), so the stars should appear more blue. Also, the dust will scatter the Sun's light so much that no stars are visible during the day (as opposed to e.g. the Moon).
Note also that the north pole of Mars points not toward Polaris, but toward Deneb in Cygnus.

Since a shift in position of 1 AU (the distance from Earth to the Sun) corresponds to a shift in position of 1 arcsec for a star which is 1 pc (3.26 lightyears) away, if you want to go to a place where the nearest star (Proxima Centari; 1.3 pc away) has moved a full moon diameter, i.e. ½ a degree — you would need to travel roughly 2340 AU away, i.e. much farther away than Pluto, but still inside the Solar system.

Most stars are much farther away, however. The brightest stars that give the constellations their appearance are tens to hundreds of pc away, so to really change the constellation, you should go 10 or 100 times farther away, i.e. roughly 1 pc. In other words, the constellation will look (slightly, but conspicuously) different from out nearest neighbor stars.


Calculate Luminosity of Vega I want to calculate the Luminosity of a star e.g. Vega but can be any star and came across the formula for Luminosity at the University of Oregon (http://pages.uoregon.edu/soper/Light/luminosity.html).

I'm not an expert on Maths, could anyone help me with the figures to put into the formula.

$$L = (4\pi d^2)b$$

$d$ is the distance but the page doesn't say Parsecs or Light Years.

$b$ is the apparent brightness.

I'm thinking

$$L = ((4\times3.141) \times (25.04^2) ) \times 0.026 = 204.81 \mathrm{(light\ years)}$$

$$L = ((4\times3.141) \times (7.68^2)) \times 0.026 = 19.26 \mathrm{(parsecs)}$$

Which gives me 204.81 which is different to Luminosity which is quoted (40.12) on the Wikipedia page.  Is this the right formula for what I'm after.  Please no scientific equations like what you see on the page linked above.  Any calculations would need to be the type you would put into a non-scientific calculator.

The $b$ in your equation, brightness, is what we call a "Flux".  That's the amount of energy per unit area at a certain distance.  Flux can be calculated using the Stefan-Boltzmann law.
$$
J=\sigma T^4
$$
where $\sigma=5.670373\times10^{-8}\space W\space m^{-2}\space K^{-4}$is the Stefan-Boltzmann constant, This is measured in Watts per square meter.  That means that we will need our distance to be in meters, and we will get a luminosity in Watts.  We can go a little further and divide our answer by $3.828\times10^{26} \space W$ to get our final value in solar luminosities (L$_\odot$).

That will make our final equation:
$$
L=\frac{4\pi R^2 \sigma T^4}{L_\odot}
$$
Because we're using the Stefan-Boltzmann equation, instead of the distance to the star, we have to use its radius.  Vega's radius is $2.362 \space R_\odot$, which is $1.64\times10^9$ meters. Its surface temperature is 9,600 K.  Plugging in those numbers yields a luminosity of:
$$
L=42.7L_\odot
$$
This value is slightly different from the value quoted by Wikipedia, due to rounding, but is well within the acceptable range.


Rule of Exception in All Nearest Objects I once heard of a strange phenomenon, but now cannot find anything about it on the Internet.  I might be looking for the wrong keywords or misremember some relevant aspect.  So maybe someone here can help me out and point me in the right direction.

The described phenomenon goes like this, as I remember it:

During the course of the scientific research, people found out more and more about the objects surrounding us.  Moon, Sun, planets, other stars, other galaxies, etc.  But each time (more or less) a little later (after a new discovery was getting an established fact) it showed that the closest object of this kind to Earth was rather an exception among its peers.  For instance, the closest other galaxy is blue-shifted.  Or that the closest Pulsar is somehow an exception among its peers.

It might be that the rule was not about the closest but about the first found object of a kind.

I think I remember Prof Harald Lesch (a German physicist lecturing on TV in a popular format called "Alpha Centauri") mentioning this peculiar phenomenon and fleshing it out with some examples, but now I'm not able to find the clip.

I understand the point about closing this question, but maybe leave it open for some weeks (or months), just in case somebody drops by who has something to contribute.  After a grace period the non-reaction can be counted as a vote for "isn't common knowledge".  Without such a grace period closing the question can only be based on "I don't know such a thing" (which should not be enough).

The concrete questions (because that was asked for, so it seemed to be unclear to some people) are:

Is such a phenomenon known?
Is there a source describing it?

Nice to have:  Which of the Alpha Centauri lectures was probably the one introducing me to the topic?

Is there a name to it?
What are examples of such objects?

I have not heard of any idea that the nearest object is exceptional in any significant way, and in the case of the nearest pulsar, nearest star, nearest (large) galaxy etc. the properties are usually typical of the underlying population, though perhaps not the observed population because of biases and selection effects.

Take Proxima Centauri as an example. It is unlike any star that can be observed with the naked eye. However, it is a very good example of a "modal" star - much less massive than the Sun, about half the age of the Galaxy and a member of a multiple system. The reason for the discrepancy is of course that stars have a wide range of luminosities and distances; in a "flux-limited" survey, we always pick out the brightest sources, which tend to be the most luminous sources at large distances, where our survey volume is largest.

So, it is true that nearby objects are different to the observed population.

The same may apply whenever some unusual chacteristic makes a type of object more likely to be detected. There is certainly a phenomenon in astronomy where the prototype of a class of objects (usually the first discovered) turns out to be atypical of the class as a whole.

As an example we could consider RS CVn binaries, named after the object RS CVn, which is a close binary system containing an evolved (subgiant) component, and exhibiting extreme levels of magnetic activity (star spots, coronal X-ray emission).

It was initially believed that the presence of an evolved component was a defining feature of the class. However, over the years it has become apparent that there are many more examples of close, magnetically active binaries with main sequence components. The prototype is atypical in having an evolved component - the explanation is that, although short-lived and rarer, evolved magnetically active stars are much more luminous and easier to find/study.

As a final word on this, there is a set of nearby/nearest objects, including the Sun and Moon, that are somewhat unusual. The Sun is more massive than 90+% of stars in the Galaxy; no other planets in our solar system have a moon that is so (relatively) large. An explanation for these things may lie with the anthropic principle, which argues that we would not observe any circumstance that would preclude the development of intelligent life on Earth or, to turn that around, we are more likely to find ourselves in a situation that is favourable for the development of intelligent life on Earth.