observational astronomy - SDSS ImgCutout webservice returns black image for dr12

The SDSS (Sloan Digital Sky Survey) ImgCutout webservice should return jpeg images for a given field that is defined by the parameters. The service can be tested here: http://skyservice.pha.jhu.edu/DR12/ImgCutout/ImgCutout.asmx?op=GetJpeg

A simple (and working) get query is given here:
http://skyservice.pha.jhu.edu/DR12/ImgCutout/getjpeg.aspx?ra=132.749&dec=11.656&scale=0.79224%20%20%20&width=400&height=400&opt=

If you enter the same information (the parameters) in the form of the first link and click on invoke you will get a black image only. You can test the base64 encoded result here: http://www.askapache.com/online-tools/base64-image-converter/

Now the question:

Do I require an additional parameter (non empty string) for either imgtype_ or imageField_ in order to not get the correct image? What are this parameters good for anyway?

(Is there a better place this question? It would be nice if there was a tag for SDSS)

light - Which planets have abundant amounts of photoluminiscent matter?

Sorry, but none of the planets have substantial amounts of phosphorescent chemicals.

Perhaps the closest to what you are thinking is the aurora, which can be seen on all the planets with atmospheres (to a greater or lesser degree) notably on the Gas Giants and on Earth. However, while the mechanism of aurora is similar to florescence, it is not phosphorescence.

No planets outside of the solar system have been directly imaged beyond a dim dot, so there is no information about exoplanets.

Milky Way stellar number density : is the stated equation in this paper incorrect?

The paper is :
http://www.astro.washington.edu/users/ivezic/Publications/tomographyI.pdf

The equation is equation #23 in the paper. It's a model for the density of stars in the Milky Way's disk. It has an exponential dependence on both $R$ and $Z$. $R$ is the distance from the center of the galaxy, and $Z$ is the distance above/below the plane of the disk.

The relation is roughly this:

$$varrho=text{constant}times e^{-frac{R}{L}-frac{Z+Z_0}{H}}$$

The problem I'm seeing is the $Z$-dependence of the formula. $R$ and $Z$ here are standard cylindrical coordinates. Thus, $Z$ can be positive or negative. The problem is that the formula blows up when $Z$ is negative. This is unphysical, since the number density must generally decrease going further away from the disk.

Am I missing something? Or should the equation really have an absolute magnitude of $|Z + Z_0|$?

[Added Later:]
I finished making a 3D demo of the stellar number density of the Milky Way. Note that your browser needs to support WebGL.

light - Can "dark matter" be the additional mass from the speed of the Galactic Cluster?

The first part, "could the galactic cluster orbit another at 10000000km/h". Is no. For simple observational reasons. If the local group were in orbit, we would be able to see and measure the supercluster which we were orbiting. We don't see it so it isn't there.

I take it that you hypothesise that the 28% of the mass of the universe that is not visible, is in fact the moving mass of the galaxies. That is not the case. Dark matter is not the moving mass of visible matter.

First remember that velocity is a relative concept, and when specifying it, you must give a frame of reference. The relativistic mass of an object is dependent on the frame of reference. For clarity lets agree to use the frame of reference co-moving with our galaxy (which is very nearly inertial)

Our own Milky way has an excess of mass. Since we are in the Milky Way, we are not moving relative to it, there is low relative velocity, low moving mass.

The first observation of dark matter was in the rotation rates of spiral galaxies. There is a missing mass in the galaxy in its own frame of reference, and in its frame of reference. Thus the missing mass is not relativistic mass.

There is a "real" missing rest mass in the universe, probably supplied by a weakly interacting massive particle.

Is there any planet/star bigger than VY Canis Majoris?

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.

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 frequently does a comet drastically change its orbit in the Solar System (like 67P did in 1959)?

How many known comets are known to have had their orbits substantially
changed by a single flyby?

Probably nearly all if not all of the (traditional) short period comets were deflected in closer to the sun by one of the gas giant planets. The reason for that is simple. Comets can't form in orbits with a near-sun fly-by so nearly all of them had to be deflected inwards by a gas giant planet. From the Wikipedia article quoted above.

Based on their orbital characteristics, short-period comets are
thought to originate from the centaurs and the Kuiper belt/scattered
disc

Short Period Comets also have a limited lifespan. Halley's for example, with a perihelion of 0.6 AU, which is closer to the sun than Venus, it loses a significant amount of ice every time it flies past the sun and Halley's may lose all of it's ice within a hundred or so more fly-bys. Source

Most Short Period comets are also planet crossers because they have highly elliptical orbits and planet crossing orbits aren't long term stable either. So of the 586 known short period comets, it's safe to say they have an lifespan of (probably) less than a million years when averaged out. Also, if you consider that Jupiter can work both ways, it can pull comets out of the inner solar system as well as toss them in, then you can work out that they would need to be replaced at least fairly often, perhaps on the scale of every 100 years or so, perhaps on the scale of decades. (there's probably more accurate estimates out there, my estimate is on the rough side)

There's a great deal of variation with individual comets life spans though. For example, a Halley-type orbit (see picture) doesn't cross Jupiter's orbit due to the 18 degree inclination, but Halley type comets are more rare (75 known per Wikipedia article above). Jupiter family short period comets are more common (511 known), and those are also more likely to get jostled by Jupiter since they cross Jupiter's orbit.

Source of picture here.

Wikipedia also mentions Main-belt comets, but those aren't what I think of as traditional comets as they orbit within the asteroid belt and don't have highly elliptical orbits. Because the Asteroid belt is mostly stable in a kind of gravitational zone between Jupiter and Mars, those "comets" may have been there for billions of years, and are only beginning to show comet like features as the sun has increased in luminosity.

The most common type of comet, long period comets, per the article, some 4,600 of them and about 10 new ones discovered every year and many many more exist but have not been discovered. The Oort cloud has perhaps more than a trillion "potential" comets, or cosmic icy objects 1 KM or greater in diameter. Source.

I don't know enough to say how Oort cloud objects are directed toward the inner solar system on average. With so many objects, it's possible that some Oort cloud objects naturally formed with high enough eccentricity to pass through the inner solar-system once an orbit. Some might be deflected by objects like Planet Nine (assuming it exists) or by repeated passes of dwarf planets or by the occasional pass of a rogue planet or the very rare close pass of another star, but it's way above my paygrade to try to estimate by which method Oort cloud comets are usually directed into the inner solar-system. A percentage of those probably pass close enough to Jupiter or one of the other gas giants for a measurable deflection but I couldn't even guess what percentage. Since these are the most common comets, it's very hard to give an estimate to your question in relation to these.

Loosely related, but Shoemaker-Levy was captured into Jupiter's orbit around 1960 or 1970. It's hard to know what it's orbit was before then because the calculations are difficult, but it had to be a Jupiter crosser of some kind. It's apogee (.33 AU) extended outside of Jupiter's Sphere of influence, which works out to about .22 AU, but it was a stable enough orbit to make a few cycles before crashing into Jupiter in 1994.

Jupiter's sphere of influence (.44 AU in diameter) is barely over 1% of it's orbital journey around the sun, but over hundreds or thousands of orbits, 1% is enough to clean out or deflect most of the objects that cross it's path.

Comet Ison, a long period comet was also (virtually) destroyed when it passed around the sun close enough to break apart in November 2013. Comets seem to die fairly frequently. Fortunately (or perhaps unfortunately), there's trillions more where they came from. :-)

Hope that wasn't too vague an answer, but I think the short answer, based on deduction is that events like the one you mention with Jupiter deflecting Comet 67P are reasonably common, maybe every few decades, maybe once a century. They're not rare events. Earth will measurably deflect a small asteroid in 2029. Asteroids are much more common than comets, but Earth is also much smaller than Jupiter. Icy bodes are likely to pass close enough to Jupiter to have a significant deflection every so often and there's about a 50/50 chance on whether that deflection will push them closer to the sun or further away with a small chance that they will be caught in Jupiter's orbit.

Hope that wasn't too general. Since this had gone unanswered, I thought I'd give it a shot.

planetary science - How can I find a daily record of the temperature on Mars?

There are a number of ways. Basically, you can either get a global record, or a record from a spacecraft. Curiosity provides us with the clearest cut version, available from its website.

Mars Climate Sounder, on MRO, shows daily maps of the temperature as recorded from MRO. That data isn't readily available, but can be found from the Planetary Data System.

radiation - Why Free-Free emission is regarded as Thermal emission?

Thermal radiation $neq$ blackbody radiation.

Thermal radiation is radiation that comes from a system where an equilibrium has been reached, where the various energy states are occupied according to the Boltzmann distribution and the particle velocity distributions are Maxwellian at some given temperature.

That does not necessarily imply that the radiation field is in equilibrium with the matter at the same temperature. It is this latter equilibrium that is required in order for the radiation field to approximate to a blackbody.

So the following statements are true:

Blackbody radiation is thermal radiation. Thermal radiation is not necessarily blackbody radiation.

An example of thermal, but non-blackbody, radiation would be thermal bremsstrahlung. The particle velocities in the gas follow a Maxwellian distribution at a characteristic temperature and this determines the spectrum of the emitted free-free radiation. However, the bremsstrahlung radiation is able to escape from the gas without further interaction - in other words the gas is "optically thin". The result is a thermal bremsstrahlung spectrum that is not at all like a blackbody spectrum, especially at frequencies $<kT/h$.

On the other hand, the bremsstrahlung process is purely caused by the interaction of free electrons and ions. If you allow there to be atoms and partially ionised atoms in the gas, then it is feasible for those to be in equilibrium at the same temperature. The population of the various energy states in the atoms and ions is determined by this temperature, and transitions between these levels produces spectral emission lines in a completely predictable, temperature-dependent, way.

the sun - Is there sufficient evidence to confirm additional Planet in solar system

This question already has muliple answers on this site...

However : Yes, astronomers DO have enough data to speculate about a possible Ninth Planet.

The orbits of six KBO are correlated, and a possible ninth planet could be the reason for those peculiar orbits. See image below for the computed results of the possible orbit of the ninth planet.:

However, since the Planet has not yet been SEEN or DETECTED, its existence is only supposed. We have nothing to prove it really exists. But if it exists, we know where it is :p

gravity - What do gravitational waves allow us to understand?

I think the wikipedia page gives a reasonable overview of why gravitational waves give us a window on the universe that is either not observable or is complementary to the view we get from electromagnetic waves (or neutrinos).

But let's have a specific example. The potential of LIGO, as we have seen today, is to detect the gravitational wave signatures of merging black hole binaries.

The power radiated in gravitational waves strongly depends on the orbital separation of the black holes, it also depends on their masses. The frequency of the radiation will change as the black holes spiral together. The "noise" that this makes in the final few seconds as the black holes merge is the "chirp" that is detected by LIGO. Have a look (listen!) to what a chirp might "sound" like at this web page from University of Birmingham.

The exact characteristics of the chirp (duration, frequency etc.) can for instance yield the masses of the black holes and of their final configuration and may tell us about spin rates of the black holes. Yet these same mergers may produce zero electromagnetic wave signatures that could be seen by conventional telescopes.

Furthermore, because the amplitude and the frequency of the waves depend differently on the masses and distance to the GW source, these binary chirps acts as "standard candles" in a similar way to type Ia supernovae. In other words, measurements of the chirp independently gives the masses of the merging black holes and the distance to them.

celestial mechanics - In a binary star system, what relation determines the eccentricity of the three orbits (for $m_1$, $m_2$, and the reduced mass)?

Isn't it just conservation of momentum? Without any external forces, the centre of mass of the binary system must stay in the same place.

For example, imagine that you had a binary system, where one star had a circular orbit and the other, with equal mass, was eccentric. Clearly, as the orbits proceeded, the centre of mass, which is half way between the stars, would have a position which oscillates with time. But with no external forces applied to the binary system, this is forbidden by the conservation of linear momentum.

temperature - Why is Rosetta's Comet so "warm"?

67P/Churyumov–Gerasimenko is actually not really warm when compared to these bodies.

You quoted figures for the minimum temperatures for Mercury and the Moon. Those are accurate but misleading. Both are for nighttime temperatures, taken from locations on the bodies that are facing away from the Sun. Daytime temperatures can skyrocket, to 427 degrees Celsius during the day on Mercury. The Moon can also get very hot during the day, although nowhere near as hot as on Mercury,

Looking at it like this, 67P/Churyumov–Gerasimenko's surface temperatures are pretty moderate, and nowhere near extreme. Rubin et al. (2015) suggest that it formed in extremely cold conditions - albeit ones not too extreme, for a comet.

The ESA has something related to say on the matter:

At these distances, the comet covered only a few pixels in the field of view and so it was not possible to determine the temperatures of individual features. But, using the sensor to collect infrared light emitted by the whole comet, scientists determined that its average surface temperature is about –70° Celsius.

The comet was roughly 555 million km from the Sun at the time — more than three times further away than Earth, meaning that sunlight is only about a tenth as bright.

Although –70° C may seem rather cold, it is some 20–30° C warmer than predicted for a comet at that distance covered exclusively in ice.

That said, evidence from the mission has made scientists think that an surface with a lot of dust may be the norm for comets.

67P/Churyumov–Gerasimenko isn't warmer or colder than the other bodies at this distance from the Sun. It's simply milder.

---

I would wager that the reason Mercury and the Moon can get so cold is because they don't really have atmospheres. Over the course of a day, Earth's atmosphere heats up a bit, and over night, it retains that heat, meaning that the surface can stay relatively warm. Mercury's atmosphere is extremely tenuous. The same goes for the Moon.

This might not be an issue, but one day on Mercury lasts 59 Earth days. One day on the Moon lasts for about one Earth month. This means that while there's time to trap some heat, there's also the same amount of time to lose it, and without an atmosphere, the heat gained is soon be lost.

67P/Churyumov–Gerasimenko also doesn't have an atmosphere to speak of, but it rotates quickly - rotating once every ~12.5 hours. This isn't necessarily perpendicular to the orbital plane, so not every part of it will face the Sun during a rotation, but it does mean that there's not as much time to lose heat.

luminosity - Conversion of magnitudes to Jansky and MAGPHYS?

I'm a bit puzzled and not an observer, so please bear with me if I'm being stupid here.

The code MAGPHYS specifies Jansky (Jy) as the input unit for flux through a filter (see Section 3.2.3 in the documentation). Meaning MAGPHYS wants me to use Jy as unit, when specify the total light of a whole galaxy in e.g. the SDSS-g band. But Jy is a unit of spectral density, i.e. W/m²/Hz. I would expect a flux as input unit when talking about the amont of light coming through a filter, i.e. in W/m² or in $L_odot$ (solar luminosity), so integrated over all frequencies that go through the filter and weighted by the filter response function.

If I have the absolute AB magnitude of a galaxy in the SDSS-g band, how do I convert it to Jy so that MAGPHYS will be happy?

I'm aware of:

$m_{text{AB}nu} = -2.5 log_{10} left(frac{f_{nu}}{3631text{Jy}}right) qquad text{or} qquad left(frac{f_{nu}}{text{Jy}}right) = 10^{-0.4 (m_{text{AB}nu} - 8.9)}$

But that is in one frequency, not a whole band.

Additionally, I have found the formula below in TOPCAT:

$left(frac{F}{text{Jy}}right) = 10^{left(23-0.4 (m_text{AB}+48.6)right)}$

Where does the 23 in the exponent come from?

What is driving the expansion of the Universe?

As far as I understood it, no one really knows. It is observable that the universe is expanding with increasing speed, therefore there has to be a form of energy driving it. We can meassure (or at least try) the amount of energy and find $Omega_lambda = 69.11%$ (see $lambda$CDM-Model)

To your questions:

1: I wouldn't imagine it as Big Bang momentum, for it is also changing. After the Big Bang, there was (or at least had to be, to make our comological model work) a period of inflation, meaning rapid expansion, which then slowed down again. I can't think of an explanation which includes some kind of finite momentum.

2: On small scales (speaking of galaxies), where densities are high, expansion doesn't act. Gravity is working against it.

3: It is responsible for the expansion. How exactly is not understood. (Little personal comment form my side. If a physicist calls something 'dark', it mostly likely means it is not understood)

environment - Are we so sure global warming is a result of humans burning fossil fuels?

We know a lot about the sun. It isn't the main cause of global warming, as we can monitor exactly how much energy it is producing. The sun's output can fluctuate though, and we have yet to measure how badly this affects the climate.

So what is the main cause of global warming?

The main cause of global warming is emissions of C02 (carbon dioxide) into the atmosphere. C02 is odorless, colourless and non-toxic, however, it is the main cause of global warming. C02 is a greenhouse gas and can stay in our atmosphere for about 20 years.

C02 causes something called the greenhouse effect. The greenhouse effect is the idea that long wave energy which is meant to go back to space, doesn't go back to space, and is instead captured by greenhouse gases such as C02. This causes lots of heat that should go off into space to be trapped inside the earth's atmosphere causing your temperature to rise.

Are there any other greenhouse gases?

Other greenhouse gases do exist, carbon dioxide is just the main one that causes the global warming. Other greenhouse gases include water vapor, methane, nitrous oxide and ozone.

Does global warming happen only on Earth?

Global warming happens all over the solar system, planets like venus have many greenhouses gases in their atmosphere too, infact, on venus greenhouse gases alone make the planet 465 degrees celsius hot, which is way over the boiling point of water - 100 degrees celsius.

Do I disagree?

Yes, as there is too little evidence to prove that the sun fluctuates enough energy to gradually increase our climate dramatically, and we have also proven the effects of greenhouse gases in a planets atmosphere.

the sun - Is the Sun slightly blue in the center? - Wavelength-dependent limb darkening of the Sun

I've slightly modified the title to try to attract some attention. If we call sunlight "white" and limb darkening is a result of seeing deeper at normal incidence and shallower at oblique incidence, then the center end edge of the solar disk viewed from earth should appear to have slightly different effective color temperatures, since the temperature is varying rapidly with depth.

I'd like to have an approximate expression for the wavelength-dependent limb darkening of the sun in the visible spectrum, either a relatively simple analytical expression that I can understand (with the appropriate coefficients for the sun) or just some linear images of the sun in various bands in the visible wavelengths so I can try to make one of my own.

This question received this helpful comment which links to here but honestly I can't make my way through that to a practical expression I can use. The Wikipedia article is not helping me much either, except for the image there. I plotted scans of RGB but by the time an image gets into the internet, things like sRGB and gamma mean it may not be linear.

update: At the Solar Dynamics Observatory (SDO) website, I just found the image sdo.gsfc.nasa.gov/assets/img/latest/latest_1024_HMIIC.jpg. The color gradient of the limb darkening seems very similar to the Wikimedia image below. I've discovered that it is called a "colorized intensitygram" and the color gradient is purely artificial - the data is single channel intensity. The limb darkening is certainly real (compare to the artificially "flattented" display!)

From http://www.solarham.net/latest_imagery/hmi1.htm

I appreciate (the existence of) the complexities and intricacies of photon transport, instrumental effects, and color perception, but I am just starting to do astronomically correct animations, so please for right now, something imperfect, or not absolutely correct is good enough for me.

20 pixel wide averages of horizontal (-) and vertical (--) "scans" across the center of this 600x600 pixel image from here

note: It seems this is a somewhat futile example, as the image is likely to be a monochrome continuum image with false color!

star - Is the sun too small to self-ignite?

Stan has essentially answered this in his comment, which I will attempt to spell out a little more laboriously.

The significant majority of our Sun's energy output comes from the proton-proton chain. This was advocated by Eddington back in the 1920's, but at that time your basic concern was a very real and major problem. Objects with like electrical charges repel each other. In particular, protons will repel other protons since all protons have a positive charge. What they knew of the sun then indicated that the core was far too cold for protons to overcome this repulsion—at least not even remotely close to the rate that was obviously necessary to produce a brightly shining sun.

With the development of quantum mechanics it was determined that a process known as quantum tunneling would give two protons a non-zero probability of "overcoming" this repulsion. But not in the sense of they just suddenly gain enough energy (hence why "overcoming" is in quotations). Instead, in the sense that the "fused together" state has the same energy as the "just about to repel each other before fusion can occur" state, and they just randomly switched from the latter to the former, despite every intermediate stage between the two of them requiring more energy than is available. That this is possible is one of the many non-intuitive features of quantum mechanics, and I think it would be beyond the scope of this question (and site) to try to get much more precise.

Still, even this failed to explain why our sun was very obviously fusing atoms to the extent that it was. If you fuse two protons together, you are left with an incredibly unstable state: the diproton. As soon as one of these forms it pretty much immediately breaks up into two distinct protons.

In the late 30's Hans Bethe (a man who eventually won the Nobel prize and was a part of the Los Alamos team that developed the atom bomb) proposed that yet another random quantum mechanical event would save the day: beta decay, a feature of the recently discovered fundamental force known as the weak force. In this situation one of the protons in the diproton undergoes a beta decay into a neutron before the diproton separates, at which point you have a stable deuterium nucleus: one proton, and one neutron.

That so many unlikely events have to occur for two protons to produce one deuterium is the reason why the sun's lifetime is approximately 10 billion years. That the sun shines as bright as it does is, as you suggest, a sheer numbers game: there are truly prodigious numbers of protons in the sun's core, so even though the proton-proton chain is really unlikely, we have so very many chances at it that there is a great many successful fusion events.

Basically every other stellar fusion reaction proceeds much more rapidly than those that produce deuterium. You may be familiar with the hydrogen bomb, which fuses its fuel very rapidly. One of the key differences (and there are many) between the hydrogen bomb and a star like our sun is that the bomb is fusing together deuterium (and tritium) into helium, and is not using the proton-proton chain. Stars have to produce large supplies of deuterium directly, and this is a lengthy process that consumes roughly 90% of a star's life—even for very massive stars, whose lifespans are orders of magnitude shorter than our Sun's.

(Note that very massive stars are rather more complicated beasts, in fact, but this is again beyond the scope of this question)

telescope - What are the advantages and disadvantages of a 2-inch eyepiece versus a 1.25-inch eyepiece?

When you look into the eyepiece, you've noticed that the image is round, framed by a black ring. That ring is called the field stop, and it's an actual round piece of metal or plastic inside the eyepiece. Its function is to limit the image to the size where the image quality in that eyepiece is acceptable.

If you were to remove the field stop, the image outside of it would start to look bad, and its edges would be fuzzy and barely usable. Also, most people prefer a neat, sharp edge to the field of view, so there's the esthetic element to consider also.

The size of the field stop is given by two parameters:

• the focal length of the eyepiece, measured in mm


• the apparent field of view of the eyepiece, or how wide the image appears when you look in the eyepiece, measured in degrees of angle

Longer focal length eyepieces will naturally have bigger field stops, keeping everything else in proportion. Eyepieces with a wider field of view will also require a bigger field stop.

Now, the field stop, obviously, must be smaller than the diameter of the barrel of the eyepiece, otherwise it would not fit in.

So, when you stick to short focal length eyepieces with a narrow field of view, the required field stop is pretty small, and it can easily fit into a 1.25" barrel.

But with longer focal length eyepieces, especially with modern eyepieces with a wide field of view, the field stop can grow pretty large, and eventually becomes bigger than the 1.25" barrel. That's when you need to move up to a 2" barrel. That's what drives the need for larger size barrels.

Long time ago, there used to be eyepieces with 0.965" barrel diameter, because back then opticians didn't know how to make wide field eyepieces. Then the field of view kept increasing, so opticians moved up to 1.25".

Eventually the newer 2" standard was adopted, as the field of view kept growing. At 32mm focal length, a Plossl eyepiece with a 52 degree field of view can fit in a 1.25" barrel, but a more recent 68 degree design requires 2".

Nowadays there's a new standard emerging, with a 3" barrel diameter, imposed by the very wide field eyepiece designs invented recently. At 30mm focal length, an eyepiece with an 82 degree field of view could use a 2" barrel, but if you demand 100 degrees FoV at 30mm, you need to move up to 3".

Of course, larger barrel diameters make heavier and more expensive eyepieces. So you need to mix and match. At smaller focal lengths, the 1.25" barrel works well. For longer focal length eyepieces, as you move up in terms of the size of the field of view, eventually you must switch to 2".

If you only use Plossl eyepieces, you can have all of them in the 1.25" format, because even at the longest focal lengths that make sense in practice, the Plossl field of view is narrow enough to not require a big field stop.

As a practical example, my collection of eyepieces currently is about evenly split between 1.25" and 2", with the 2" barrels being reserved for eyepieces over 18mm focal length. All my eyepieces have an 82 degree field of view.

planet - What causes celestial bodies to move like they do from Earth's surface?

The apparent motion of planets is complex but predictable.

The apparent motion is due to the combination of three different motions:

• The rotation of the Earth,


• the motion of the Earth around the Sun,


• and the motion of the planet around the Sun.

The Planets move in ellipses, slightly perturbed by other planets.

These motions are predictable, and their combination is straightforward trigonometry. The resultant motion is complex, the planets appear to loop relative to the stars, but entirely predictable. We know exactly where the planets will be in the sky for any time for many thousands of years in the future or the past.

solar system - Would I actually be able to see Ceres without shining a really bright torch onto it?

NASA's put out a semi-CGI video of a Ceres flyby using material from the Dawn mission.

Now I realise that visibility is relative, even on Earth: I can walk into my basement and not be able to see anything until I've stood there for a few moments letting my eyes adjust. But I'm wondering, since Ceres is so far from Earth, and despite the images we see beamed back from spacecraft, if I were on a spacecraft in orbit of Ceres, facing the side of Ceres that faces the Sun, would I be able to see its surface with the naked eye? Or is sunlight too dim at that distance to do the job on its own?

the sun - Should this photo of the sun's surface actually be white?

The term "color" is a label that humans have assigned to denote the ratio between the intensity at various wavelengths in the three different wavelength bands, or regions, that the human eye is able to perceive. These bands are centered roughly at 430, 545, and 570 nm, but are quite broad and even overlap:

^{Human cone response, normalized to the same height. In reality, the response of the blue cones is significantly smaller, and the green is somewhat larger (from Wikipedia).}

If an object emits light only at, say, 450 nm, the ratio is roughly 0.1:0.2:1 (in the order R:G:B); it then looks a special way to us, and we call it "blue", or maybe "violet". If it emits at 550 nm, or 650 nm, we call it "green" or "red". An object that emits light in a more continuous spectrum that covers the region 500–600 nm, we'd name something like orange-/brown-/olive-ish, depending on the exact spectrum.

The Sun emits photons at all wavelengths, but not in an equal amount at all wavelengths. The particular ratios between the three bands that we can see, we have labeled "white". However, when the Sun's light enters our atmosphere, some of the light is absorbed, especially at the blue wavelength. Filtering out the blue results in a spectrum that looks more orange to us. The figure below shows the Sun's "true" spectrum (in yellow), and the spectrum seen from the surface of Earth (in red):

^{The Sun's spectrum measured outside our atmosphere (yellow) and at sea level (red) (modified image from Wikipedia, with data from Global Warming Art).}

Sometimes we want to observe the Sun in a wavelength region that is invisible to humans, for instance in UV or X-rays. This can be done with a telescope and a detector that is sensitive to light in that particular region, but in order for us to see it, we represent the image with a color that we can see. The image in the top of the link you provide is taken with the European spacecraft SOHO's instrument EIT at 19.5 nm, which we call "extreme UV", bordering on soft X-rays. Since this is invisible to humans, they arbitrarily chose to represent it using green. They might as well have chosen pink or brown.

^{The Sun in Extreme UV, during a particularly violent solar flare (from the SOHO gallery).}

Several of the photos in your second link are images taken by the Japanese space telescope Hinode, which observes both in the optical (i.e. visible by humans), X-rays, and far UV. If these are shown in orange, again it's just to make them visible to us, and you may say that they have been "doctored to meet our expectation". In this way, I like better when they choose a color such as all green, so we know it's "false color".

distances - How to calculate position of an unknown star knowing positions of some other stars from an image?

First, let assume your image geometry is homogeneous, and has no peculiar distorsion in either direction.

Second, let assume you have the resolution of your image: the number of arcseconds / pixel.

Now, take one 'red-cross' star, call it A. It will be the origin of the triangle we will draw. Name your 'yellow-cross' star B. Now, take a new point, called 'C', that is at the same pixel-y coordinate of A, and the same pixel-x coordinates of B.

Drawing lines between A, B and C gives you a right triangle. You cannot simply apply flat geometry equations, since you are on the celestial sphere.

Hence, one must compute the 'Bearing' angle between A, B and C. See for instance here: (A bearing is an angle, measured clockwise from the north direction).

Here is a little piece of code you should be able to read:

double adjacent = pow(pow(B.x-C.x, 2.0) + pow(B.y-C.y, 2.0), 0.5); // dBC
double opposite = pow(pow(A.x-C.x, 2.0) + pow(A.y-C.y, 2.0), 0.5); // dAC

double theta = atan2(opposite, adjacent) * ONE_RAD_IN_DEGREES;

theta is the bearing angle, here expressed in degrees, thanks to the conversion constant ONE_RAD_IN_DEGREES. atan2 is the Arc-Tangent function that takes care of which quadrant you are in (it computes arctangent(opposite / adjacent), correcting for the quadrant, see the wikipedia article for instance).

Now, depending on whether you have East to the left or not (astro images have East to the left usually), you need to correct your angle. Again this little piece of code:

BOOL eastLeft = <true or false>

if (B.x < A.x && B.y > A.y) {
    theta = (eastLeft) ? theta : 360.0 - theta;
}
else if (B.x < A.x && B.y < A.y) {
    theta = (eastLeft) ? 180.0 - theta : theta + 180.;
}
else if (B.x > A.x && B.y < A.y) {
    theta = (eastLeft) ? theta + 180. : 180.0 - theta;
}
else if (B.x > A.x && B.y > A.y) {
    theta = (eastLeft) ? 360.0 - theta : theta;
}

Now, we have the correct theta value. Now, compute the distance (below, in degrees) between A and B, and call it delta.

Assuming the R.A. and Declination of A are called lambda1 and phi1, you can compute the R.A. and Declination of C, lambda2 and phi2, using the formulae given in here, under the section "Destination point given distance and bearing from start point".

In my code:

double phi1 = declination_A * ONE_DEG_IN_RADIANS;
double lambda1 = rightAscension_A * ONE_HOUR_IN_RADIANS;

double delta = degrees * ONE_DEG_IN_RADIANS;
double theta = bearing * ONE_DEG_IN_RADIANS;

double phi2 = asin(sin(phi1)*cos(delta) + cos(phi1)*sin(delta)*cos(theta));
double lambda2 = lambda1 + atan2(sin(theta) * sin(delta) * cos(phi1), cos(delta) - sin(phi1) * sin(phi2));

with the usual meaning of trigonometric functions (sin is sine, asin is arcsine, etc).

star - Betelgeuse and sun classification

If you take a look at the Hertzsprung-Russell Diagram that helps us to classify stars

(c) Wikimedia commons

(while this one uses data gathered from the Hipparcos satellice, roughly 100.000 stars) we see, that there are 2 main types of red stars: Giants and M-dwarves. The latter are stars on the main sqeuence (they have stable hydrogen-burning in their cores), while the giants are stars that have left the main sequence due to aging. They appear red as they bloat up to bigger radii compared to their main-sequence life which also makes them cooler, and thus redder.

Betelgeuse is now one of those giants, and he's actually more massive than the sun (somewhere between 7 and 20 solar masses, according to wikipedia), so there is no contradiction with M-Dwarves also being red.

Please let me know if you desire to know more details on this.

collimation - I'm having trouble achieving sharp telescope focus

Your telescope has a 2000 mm focal length, and 200 mm aperture. With the 20 mm eyepiece, you get 100x magnification. With the 10 mm lens, you get 200x. Several things could cause what you describe:

Seeing (turbulence) might be bad. It is common in many places that less than great seeing means everything at 200x and over might start looking blurry. But seeing changes with the seasons, or day to day, hour to hour, or indeed from one second to another. It's random. If you get the exact same results all the time, and the image never improves even after many weeks of trying, then perhaps it's not seeing that's the culprit.

Your telescope might not be collimated. Look in the user's manual. Is there a collimation procedure you need to follow? (EDIT: yes, there is - page 20) A miscollimated scope indeed behaves as you described - seems "sharp" at low-ish magnification, but quickly becomes blurry if you push magnification up.

Point the scope at a bright star and defocus. Are the rings perfectly circular? If not, it might be miscollimated.

http://www.astrophoto.fr/collim.html

Another factor is optics quality. Optics with lots of aberrations cause "mushy" images and a difficulty to find the true focus. High quality optics produce a scope that is very "snappy" when achieving perfect focus, and make very sharp images when seeing is good. There's nothing you can do about this, but the good news is - this problem is not very common nowadays; optics quality, while not always great, tends to be at least passable in many cases. Quality could be evaluated either on the test bench (which requires equipment), or via direct observation of stars (which requires a very experienced observer and takes time).

Finally, have you compared views with another instrument of similar aperture? The image at higher magnification will always seem a bit more soft, compared to lower magnification - even in perfect seeing with a collimated instrument. Also, the contrast always decreases as you increase magnification. This is why there is always an optimal magnification for each view - sometimes higher, sometimes lower, depending on many factors. The best scope in the world will look mushy and blurry if you push magnification up too much.

Can you easily see the Cassini division in Saturn's rings? If yes, then you're probably in okay shape (collimation and seeing), you're just not used to the softness at higher magnification. I have a very high quality reflector (self-made), similar aperture class to yours (a bit smaller), and Cassini is immediately visible at 180x and looks sharp; it's also visible at 255x but the whole thing starts to look mushy and faded. It looks super-sharp and high-contrast at 136x but it's kind of small. This is in perfect collimation, perfect thermal equilibrium, high quality optics throughout the stack (primary and secondary mirrors, eyepieces), and good seeing typical of N. California.

When seeing is bad, Cassini starts to fade no matter what I do.

Does Saturn look equally soft in the center of the image, as well as near the edge? Good optics (mirrors, eyepieces) make a good image everywhere, cheap optics make an okay image in the center and a blurry image at the edge.

My bet is that it's either bad collimation, or you haven't compared the view at 200x with another scope, or both. But it's hard to diagnose things over the Internet.

You should probably get a 15mm eyepiece as well. You'll use it a lot, probably more than the 10mm.

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Light pollution is never a factor for: planets, the Moon, the Sun, most double stars. These things are just too bright to care about it, so observing them from the city is fine. What does matter is seeing (turbulence). Also, it's important that the scope is in perfect collimation.

Light pollution only matters for the "faint fuzzies": nebulae and galaxies. These are low-brightness, low-contrast objects, that are difficult to see from the city. Being low contrast, your eye will not perceive a lot of detail anyway, so for these objects seeing doesn't matter.

Congratulations for being aware of, and following, the rule that says you need to acclimate the scope to ambient temperature. A lot of people are not aware of it. At least you're removing one unknown from the equation. It's always a good idea to let the scope "breathe" outside for an hour before you even begin to observe.

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One way to be 100% sure that your scope is focused perfectly is to use a Bahtinov mask. Just make (or buy) one to match the scope's diameter, put it over the scope's mouth, point scope at a bright star, and tweak the focuser until all spikes intersect exactly in the center of the star. Then remove the mask - the scope is in perfect focus. This will probably not help in your case, but it's just one way to remove another uncertainty. It also helps with regular observations - I use it almost every time I observe, it makes focusing so much easier.

The mask could be made from a piece of cardboard and a sharp knife in like 20 minutes. It doesn't need to be a marvel of engineering precision to work well.

http://www.deepskywatch.com/Articles/make-bahtinov-mask.html

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One thing that is certainly NOT a factor is the central obstruction. You'll often hear on the Internet how Cassegrain instruments, or other scopes with a large central obstruction (secondary mirror) are supposedly less "sharp" than scopes without a central obstruction, or ones with a very small obstruction.

Yes, the obstruction does matter a little, but not to the extent that Internet mythology would have you believe. A Cassegrain instrument, with good optics, in perfect collimation, can be a superb telescope even at a 40% obstruction. It simply performs like a somewhat smaller instrument, that's all (the physics of the central obstruction are very complex, and would require an entirely different discussion on this forum to fully elucidate).

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(Added in reply to a comment below)

Miscollimation is quite likely the #1 performance killer of all amateur telescopes. Some well-made refractors are more or less immune to it, but most reflectors need periodic collimation.

It's a good idea to google around for various collimation techniques, there's so many of them. Some require the observation of stars, some require special devices known as collimators. The technique is also different for different types of telescopes (newtonian, cass, etc).

If you own a reflector, it's a good idea to collimate it once in a while, just in case, like changing the oil on a car. Dobsonian telescopes probably require most frequent collimation.

Check Howie Glatter's site - he makes some of the most precise collimators currently available:

http://collimator.com/

Regardless, just learn any method you can and apply it. If the telescope is massively miscollimated, you'll get an improvement immediately.

Not being able to see Cassini is not a good sign. Maybe seeing is persistently bad (it can happen), or maybe the scope is off-collimation. Can anyone else in your area see Cassini these days in a similar aperture, in a scope known to be collimated well?

Surface detail on Saturn is not easy to see, but should be doable with apertures over 100mm. It's like zones of latitude similar to Jupiter's equatorial belts, but very, very pale and faded. You have to keep looking for a while to see it. Don't worry if you don't see that stuff for now.

Jupiter's Great Red Spot is harder to see nowadays. It's been shrinking and fading for decades now. Go on Wolfram Alpha and type in Great Red Spot. It will tell you if it's visible at the moment. Usually it looks more like a dent in the equatorial belts, rather than an actual visible spot.

http://www.wolframalpha.com/input/?i=great+red+spot

Jupiter works best at not so high magnification, otherwise the contrast gets too low. Around 140x ... 180x should give you decent contrast in your aperture.