How can we map masses orbiting the outer Solar System by determining the orbits of planets and smaller objects? Neptune was discovered 170 years ago with the help of observations of Uranus' orbit. A great demonstration of the scientific principle for its time. Recently, Earth and cis-Lunar based radio interferometry together with the Cassini orbiter have measured Saturn's orbit, 10 AU from the Sun, with very great precision. This has already helped determine some characteristics of a possible planet Nein 200-700 AU from here.

What would be the best way to gravitationally measure the yet unobserved mass concentrations in the Kuiper Belt and beyond, by observing orbits of objects among the known planets?

It takes alot to move Saturn to a measurable degree. What about a space probe with the sole purpose of being easy to locate at high precision, maybe in a highly inclined orbit? Or a family of them in different directions. Would its deplacement from the very weak gravity well of a planet Nine be detectable? If it is the history of Saturn that contributes crucially to the precision, what about an asteroid like Ceres which is known since 200+ years? Is there any special property of Saturn that makes it especially useful? Would a second object as well measured as Saturn be extra helpful by some kind of triangulation of distant gravitational influences?

Iorio (Preliminary constraints on the location of Telisto/Planet Nine from planetary orbital dynamics) has suggested transmissions from the New Horizons probe (currently beyond Pluto) could be used to test for the influence of the proposed new planet, and provide constraints to its possible position.

It is estimated that the position of New Horizons could be affected by the proposed planet, of the order of 100 metres or more by 2019. It is stressed that this is a preliminary proposal:

  Moreover, the present investigation should be considered just as a preliminary sensitivity analysis which, by no means, is meant to replace a full covariance study based on extensive simulations of the ranging data from New Horizons and their reduction with accurate models including also PX itself, which is outside the scopes of this paper. Nonetheless, the possibility of using the telemetry of New Horizons seems appealing, and would deserve further, more detailed and dedicated studies which will, hopefully, prompted by the present one.


World line coordinate finiteness If you trace two particles' world line backwards in time, according to current theory, both objects should converge at big bang.

Would both objects arrive there simultaneously?

Another way of asking the question:

Is there any evidence that supports the idea that new matter is "introduced" into the universe as a continuous stream, as opposed to everything coming from a singularity both in time and space?

The analogy would be from the inside of a black hole. For each particle in that singularity, they would all have a world line that starts at the same spacial coordinate, but the particles would not be "introduced" into the black hole simultaneously. You can argue that time stops having meaning in a black hole, but that's only true for an "outer observer" in my reasoning.

Particles are generated and destroyed all the time. This is obvious for photons, but also holds for massive particles. Today most particles are stable and long-living, but shortly after the big bang the universe was so hot that particle generation and destruction were in some equlibrium and all particles participated in this process (for photons this continued until the epoch of re-ionisation at redshift $z\sim1000$. There is definitely no particle around today that existed all the time since $t=0$.


Do Noctilucent clouds cause problems with telescopes? Noctilucent clouds are at around 85km altitude and are formed of ice crystals. The ISS recently posted a number of photographs of these clouds and pointed out the key feature, which is the bright blue glow.

Are these known to have an impact on ground-based or orbital telescopes?

Noctilucent clouds are not a problem for space telescopes because their orbits are always more than 85 km. The Hubble Telescope orbits at about 570 km.  Noctilucent clouds are a problem for ground-based telescopes (although only at high latitude sites,> 50$^{\circ}$), especially if you are trying to get accurate photometric brightnesses.  However, typically you can see them reflecting city lights or moonlight, or you can tell from the large variance in your photometry counts that they are there and you should try again another night.


Which rotational frame does the universe prefer? Which rotational frame does the universe prefer?
We say that the earth rotates once per 24 hours.
But is this the master frame?
What is the solution to Newtons Bucket Problem?

I would like if the answer is a detailed explanation.
I know about high school math and science and participated in a astronomy olympiad.

There's a lot of loosely connected questions here, so let's go through them one by one.

  Which rotational frame does the universe prefer?

The universe doesn't prefer any frame. There is no "universal" reference frame. This is one of the principle axioms of Special Relativity. You cannot pick a frame and say this is the "correct" frame. The reason being that you have no means of establishing what "correct" even means. One person may claim their chosen frame is correct, but another person might choose something different and claim theirs is correct. Neither person would be able to show their frame is better or more "correct" than the other's.

That being said, I think its important to point out that while no true universal reference frame exists, scientists sometimes use a de facto "psuedo-universal" reference frame. That frame is the one which is at rest with respect to the Cosmic Microwave Background (CMB) radiation. It is not a true universal reference frame, but because of the importance of the CMB, the frame at rest with respect to the CMB has been given more (arbitrary) importance by scientists, hence the term psuedo-universal. You can read more about this at this great physics Stack Exchange question.

  We say that the earth rotates once per 24 hours. But is this the master frame?

This is a good question. Anytime someone mentions motion, you should always ask, motion with respect to what? If I'm inside a car, I would say the car isn't moving at all because my motion, with respect to the car is zero. A person standing on the side of the road would say the car is traveling at 25 mph. So you can see that defining motion requires defining a frame to measure that motion in.

When it comes to talking about the rotation of the Earth, there are actually a few different frames that scientists use. They're all very similar and produce similar answers (all about 24 hours). I'll restrict myself to talking about the frame which defines the Solar Day (another example would be the Sidereal Day).

The Solar Day is the time it takes the Sun to return to the same position in the sky as from the previous day. As you can hopefully see, this measure of a day is fixed to a frame with respect to the Sun. It would be the same regardless of the motion of our solar system through the universe and thus wouldn't be considered a "master" frame (not that there is one).

  What is the solution to Newtons Bucket Problem?

This, I would say, is really outside the realm of the Astronomy Stack Exchange and really should be its own question, preferably on the Physics Stack Exchange. For that reason, I won't answer this question here, but I will point you to this great source which talks about the Bucket Problem is detail.


Is there a non-trivial limit to a number of nested orbits? In his popular youtube channel, Scott Manely addressed a question of whether an astronaut can orbit a space station. The provided answer is negative since such an orbit would require the space station to be impractically dense.

This got me wondering if similar arguments can limit the number of nested orbits in general. For example, say you are free to choose any solid bodies A, B, C,... s.t. A orbits B, B orbits C, etc. Is there a maximal number of such bodies and do we know it?

There is no theoretical maximum. It would be possible for the solar system to orbit a sufficiently large mass, (while the planets' orbits remain stable). And it would be possible for that large mass to orbit another, even larger mass. The process could, in theory, be repeated without limit. The masses of the central bodies would have to be exponentially larger with each embedding, and so could not be conventional stars, beyond a certain point. (They could be black holes)

This is the absence of theoretical limit imposed by the theory of gravitation. In practice, multiple levels of orbits don't exist, since there does not exist a mechanism that can create the massive central bodies in the right orbits.


Integral Field Spectroscopy vs Fabry-Perot Interferometry Why are Fabry-Perot Interferometers (FPIs) and Fourier Transform Spectrographs (FTSs less common than Integral Field Spectrographs (IFS) nowadays? My understanding is that:

With FPIs and FTSs you have to scan your spectral range over time. This (A) Takes more telescope time, and (B) Is affected by seeing variations over time (i.e., different wavelengths end up having different PSFs).
Limited spectral range. As a result, you have limited information about the observed object.
I was also under the impression that reducing FPI or FTS data is more complex. However I do not have a reference to support this claim. It this true? If it is, can someone explain why?
For some reason FPIs use Image Photon Counting System (IPCS) instead of CCDs. Why is that? Does this somehow make them "worse" than IFSs?

What other advantages does IFS have over FPI and FTS?

I think you've got it pretty well summarized. In particular, your first three points are dead-on.

(I'm going to use "IFU" = integral field unit instead of "IFS" = integral field spectrograph, but that's just because I'm more used to the former term; they're basically synonyms.)

Let me combine the "data reduction" question with one other key advantage of IFUs that you may have missed, which is that they are (almost) plug-in substitutions for traditional long-slit spectrographs, which means they can take direct advantage of existing spectrograph-design expertise and data-reduction software.

As an example, one common form of IFU is a closely packed array of small input apertures attached to fiber-optic cables. Away from the IFU, the cables can be re-arranged so that their output ends form a 1D row, very much like the output of a long slit in a traditional spectrograph. (And even more like a multi-object spectrograph design, where multiple individual input apertures are scattered across the field of view rather than being organized into a contiguous 2D array.) This means that the rest of the spectrograph design -- gratings or prisms, collimation optics, imaging onto a 2D CCD detector, etc. -- is pretty much the same as traditional spectrograph. There are instruments on some telescopes which allow you to switch between long-slit, multi-object, and IFU modes, with the same set of gratings, filters, and detector being used for all modes (for example, the GMOS instruments on the Gemini North and South Telescopes).

This carries over into ease of data reduction: the spatial+spectral image on the CCD detector (or near-IR detector array) produced by an IFU is very similar to that produced by a traditional long-slit spectrograph, so the same data-reduction techniques can (with a little modification) be used. In essence, you just chop up the big 2D data image you get from the CCD into a bunch of little sub-images, one for each fiber that came from the IFU, and you can treat each one as though it were a standard single-aperture or single-slit dataset. This contrasts quite strongly with the very different (unique) form of data and data reduction that F-P requires.


Why does Hellas Basin look white on some maps? One map of Mars I see a lot is this colourized map: the Viking colourized global mosaic (Source):

But one thing which strikes me as odd is just how white the Hellas Basin is (the white patch which is not on the poles, just in the middle of the lower right half of the above image)
On the older version of the Viking mosaic, the basin doesn't look that white. Here the Mars Viking Colorized Global Mosaic v1
(Source, link to image):

It certainly still looks at least paler than the surrounding areas, but nowhere near as much as on the first map (on the first map it is about as white as the poles, on the second it is not).
So: Is the Hellas basin really as white as the first map suggests? And if yes why?

That map of Mars is a composite of many different images that were taken at different times during Mars' year.  Mars' arctic circles are very similar to Earth's, and during that hemisphere's winter, that terrain is in permanent shadow, allowing seasonal frost to form.
Even though Hellas is equatorward of the antarctic circle, it gets cold enough, days are short enough, and it is low enough in elevation so the atmosphere is thick enough to allow seasonal frost to form on Hellas.  It is highly variable year-to-year, but it typically covers much of the floor of Hellas.
So, what you are seeing in the first map is southern summer and northern summer for the poles (or probably late spring given the extent of the south polar ice), and then very early spring near Hellas.  I would guess that those were the best images they could find from Viking to make that particular mosaic.
That is in contrast with the 2nd picture, which probably better matched seasons for a more smooth mosaic and one that better represents late spring / early summer at most pixels.


Amount of energy of the Big Bang What is the currently accepted estimated range of the amount of energy of the Big Bang event?

In joules at some estimated size, so a temperature may be calculated.

For context, I wonder if the temperature of the Big Bang would make its heat radiation wavelength shorter than the Planck length.

Let's start by making some points clear:
1. We don't know what the Big Bang was.
Rather, we know that the Universe is expanding. If you extrapolate backwards, you'd expect the Universe to be denser and denser. More specifically, we talk about this as a change in the scale factor $a$, and this gets smaller and smaller as we look further back in time. According to general relativity (our modern theory of gravity), 13.8 billion years ago, $a$ should have been $0$; however, you can't have a metric with $a = 0$.
Thus, we know that general relativity is necessarily incomplete. It breaks down at the conditions of the early universe, so we currently have no physical model to explain that time. Rather, we know that the early universe expanded, and the Big Bang is the time that perplexes cosmologists. Some theories, like quantum gravity, have emerged in an effort to explain the Big Bang; however, we currently have little understanding of what it actually was.
So no, we can't tell you what the energy output of the event was, since we don't know what actually happened.
2. The temperature of the early Universe was high
Our theories break down at the Planck epoch of the Universe. The Planck epoch was the earliest epoch of the Universe and lasted until $10^{-42}$ seconds after the Big Bang — that's 200 Planck times, which are the shortest meaningful measurement of time.
During this epoch, the entire Universe was at $1.417×10^{32} \; \mathrm{K}$, which is the Planck temperature. This is the hottest possible temperature; an object at this temperature will emit photons with wavelengths of a Planck length (you can read more about this in my answer here). The point is that there is no meaningful distance smaller than a Planck length, so the Universe couldn't be hotter than the Planck temperature.


What should I look for in an astronomy tripod? I have a Celstron SkyMaster 20x80 binoculars, and a tripod — Vanguard 263AT with a SBH-100 head.

The binoculars weighs 2.2kg, while the head is rated to a loading capacity of 10kg, and the legs to a tilted loading capacity of 5kg. This means that the tripod should work well, but it doesn't — it vibrates a lot even when my face gently touches the eyepiece, even without the center column extended, so much so that it becomes a pain to use the binoculars.

So I'm considering upgrading my tripod legs. I'd like this tripod to work well not just with the 20x80 binoculars I have, but also with a 25x100 pair of binoculars I'm considering buying (weight: 4.6kg), and maybe a small telescope (say with a 6-inch aperture).

Am I correct in understanding that:

I need to upgrade my tripod legs.
I don't need to upgrade the head, since it's rated to twice the capacity of the legs, and to twice the weight of the biggest binoculars I'll use.
I should look for legs that are rated to twice the weight of the biggest binoculars I'll use.
I should look for a set of wooden legs, since these have greater stability than aluminium legs while being affordable, unlike carbon fiber.
What else should I keep in mind when I look for an astronomy tripod?

I read that:

  [...] the most popular materials are aluminum, carbon fiber, and,
  somewhat surprisingly, wood. It’s one of those common compromises—pick
  two of the following three: weight, price, and/or vibration dampening.

  Wood is extremely good at absorbing vibrations, and is pretty
  affordable, but weighs a ton. Carbon fiber is light and stable, but
  you’ll pay for it. Aluminum is affordable, and fairly light, but is
  prone to channeling vibrations.

So I'll prioritise stability and affordability over weight.

I have the exact same binocular, and in fact I also have a very similar tripod and head. I'll take a photograph this evening to demonstrate my setup.

I am very happy with the setup, and I find it very stable. I usually sit the tripod on grass, but I do not recall ever having a problem with it on concrete or stone. I do try to keep the neck as short as possible, and use the legs for height. Also, the binocular can have it's tripod mount moved fore or aft, I think that mine is in the full aft position.

You might be able to hang a large mass (bag, perhaps) from the underside of the tripod; you should find a hook already there made specifically for this purpose.


What if you 'placed' the Chicxulub asteroid onto the Earth? We all know what sort of devastating impact an asteroid could have if it collided with the Earth.
But I'm curious to know what the effects would be if you removed the speed element from it, and were somehow able to casually 'place' a giant asteroid onto our planet?
I think I read somewhere that the Chicxulub asteroid was most likely around 9 miles long, and that the end of it would have extended beyond our atmosphere if it hit the Earth lengthways.
But would this have any effect if it was travelling at, say, 2mph? There would be nothing burning in the atmosphere or shockwaves etc.
Would it have any affect at all apart from squashing whatever was underneath?
If the answer is no... how about placing a moon-sized object onto the Earth?

An asteroid resting on Earth would be a mountain. Or, for smaller asteroids, a pile of gravel.
Mountains are limited in altitude by the strength of stone to resist compression: a too tall mountain would sink down as the base crumbled and spread out. The limit on Earth is about 10 km. Besides the strength issue mountains are also floating ("isostasy") on the fluid but denser interior of Earth, so even if it doesn't fall apart it may sink down partially.
So the impactor would not hold together that well. A big ball of rock 10 km in diameter would first definitely sink down several kilometres into the crust, and almost definitely break. That means that something like $10^{15}$ kg will descend several kilometres, giving an energy $mgh \approx 10^{19}$ J (and there are estimates 100 times larger for the mass). That is "just" like a big hurricane, but will mostly be released as localized seismic energy and heating. The end result would be a massive pile that likely destabilizes the crust in the vicinity a lot: far less than an actual high velocity impact, but still a major disaster.
If you do the same with the moon you will break up the crust, boil off the oceans and wipe out all life on the planet.


Why were the Magellanic Clouds named that way? From the Wikipedia article, the Large and Small Magellanic Clouds were observed prior to Magellan's expedition, and the name only stuck long after Magellan's expedition (at first, they were called nubecula major and nubecula minor, or Large Cloud and Small Cloud). So why were they finally named after Magellan?

Michel Dennefeld has written an arXiv paper, "A history of the Magellanic Clouds and the European exploration of the Southern Hemisphere" which goes into the history of the European discovery of the Magellanic Clouds and the history of the name. The term Magellanic clouds appears to have originated with sailors, as noted by Edmond Halley in his 1679 "Catalogus Stellarum Australium":

His work is known because of
the inclusion of a new (fleeting) constellation, Robur Carolinum, to please King Charles II. He
mentions in an appendix that:

Proxima, ex iis quas observavi, est in Cauda Apodis, in distancia paulo ultra 8 graduum, Duae
Nubeculae, quae a Nautis Nebulae Magellanicae appellantur, exacte referent Galaxiae
albedinem & Telescopio inspectae, hinc inde Nebulas Parvas & exiguas Stellas ostendunt…

It appears it was clear to him that the name Magellanic Clouds was used only by sailors, and
that those Clouds resemble the Milky Way.

Dennefeld goes on to describe the gradual spread of the term "Magellanic Clouds" beyond the nautical community, e.g. in Nicolas de La Caille's 1755 catalogue:

But his words,
“que nous appelons nuées de Magellan”, seem to indicate that around that time the
association of the Clouds with Magellan was already spreading beyond the nautical
community, even if the scientific term was still simply Nubeculae or les Nuages.

The astronomical usage switched from "Nubeculae" around the time that Latin was abandoned as the language of science, and the first usage in the scientific literature seems to have been by John Herschel:

John Herschel (1847), in a section however entitled "On the two nubeculae or
Magellanic Clouds", talks only about Nubeculae. Only in the accompanying figure (Fig. 8),
where he gives his visual observations (figure where it is however hard to recognise the
Clouds), does he mention again, "The two Magellanic Clouds as seen with the naked Eye".
So he uses both the scientific denomination and the more public name, and Herschel seems
to have been the first one to use the name Magellanic Clouds in a scientific publication.

So it appears to be that the name became Magellanic Clouds because this is the name that was originally used by European sailors, that then gradually became the common name as Latin was abandoned.


Clarification of radio spectrum terminology For a radio spectrum, when is it called thermal emission, synchrotron emission, self-absorbed synchrotron emission and inverted spectrum? They are all power-law and their difference is power-law index?
For a power-law $\nu^{\alpha}$, as long as $\alpha$ is positive, the spectrum is an inverted spectrum, right?
Please recommend reference books which introduce these models and their corresponding astrophysical phenomena.

Terminology
The terminology of thermal and non-thermal emission is somewhat unfortunate. Formally, non-thermal emission is continuum radiation from particles with non-Maxwellian energy spectra. Practically, we generally consider it to be emission which is not described by Bremsstrahlung or black body radiation (source).1 Note that this is true for the entire electromagnetic spectrum, not just the radio region.
Synchrotron emission is a form of non-thermal emission which occurs when particles are accelerated perpendicular to their translational motion, specifically when the particles are relativistic (otherwise it is called cyclotron emission).
Self-absorbed synchrotron emission is described well in the introduction of this paper, and occurs when the synchrotron emission becomes a non-negligible source of heating. It occurs only when the self-absorption frequency is larger than the cooling frequency, i.e.
$$
\nu_{c} < \nu_{a}
$$
The spectral index, $\alpha$, describes the dependence of the radiative flux density on the frequency of the emission. With frequency $\nu$ and radiative flux density $S_{\nu}$,
$$
S_{\nu} \propto \nu^{\alpha}
$$
An inverted spectrum results from $0 < \alpha < 2.5$,2 and this is generally indicative of thermal emission. However, the observed emission can be depressed by absorption processes, so a positive spectral index (i.e. an inverted spectrum) is not robustly indicative of thermal emission on its own.

 Radio Astronomy
Most radio astronomy uses radio interferometry instead of direct observation, there are a number of reasons for this: most emitters are very weak in the radio portion of the electromagnetic spectrum, radio telescopes for direct imaging have to be much larger than their shorter-wavelength counterparts, and radio interferometry allows the highest angular resolution of any technique (it is possible to create a radio interferometery array the size of the Earth, which we have done).
That being said, there are a number of phenomena and objects whose activity in the radio region lends them to direct imaging:

Masers - stimulated spectral line emission, usually in the range of $3$mm to $0.3$m. This phenomena can occur in the atmospheres of gas giants, the expanded fuel sheath of late-type stars, protosolar disks and nebulae, supernovae remnants, and black holes. The produced spectrum is characterized by high brightness, extraordinarily high black-body temperature equivalent typically on the magnitude of $10^9$ K, but as high as $10^{14}$ K, and very high polarization that is predominantly circular. A spectra which contains a sharp peak that would correspond to a black-body temperature above $10^9$ K would likely be the product of a maser.

Quasars - The supermassive black holes in the centers of galaxies sometimes have accretion disks or clouds which produce immensely luminous emissions as the material falls into the black hole. The sheer scale of the energy involved (often orders of magnitude larger than the total output of the rest of the galaxy) leads to a relatively uniform emission across much of the electromagnetic spectrum, from X-rays to far infrared.

Pulsars - The mechanisms for emission here are poorly understood, but in the formation of a neutron star (or even a white dwarf occasionally), the conservation of angular momentum results in a very rapid rotation. A strong magnetic field causes protons and electrons to accelerate on the star's surface which produces cyclotron and synchrotron emission. The emission exits the pulsar in a beam.

Radio galaxies - Active galaxy nuclei which are exceptionally luminous in the $30$ m - $300$ km region. The radio emission in these nuclei is generated by synchrotron action characterized by a smooth, broad-band radio spectra with strong polarization.

1 Not a published source, but does a pretty good job of explaining this in the introduction.
2 $2.5$ is the rough upper limit for $\alpha$ in the power law description.


Will a black hole swallow the universe? If black holes swallow stuff, will a big black hole eventually swallow the universe? The universe will eventually go dark, because there will not be enough matter to make stars, and all that will be left is a bunch of black holes and they can merge. So will the universe get swallowed by a big black hole someday?

If black holes swallow stuff, will a big black hole eventually swallow the universe?

As far as I know, no. Because the universe is expanding, and we have good evidence that the expansion is accelerating. Yes,  some people talk about the Big Crunch, but there's no evidence for it. I don't think there ever will be. I say that because it's space that's expanding, and a gravitational field isn't collapsing space. The universe didn't collapse when it was smaller and denser, so I see no way it can collapse in future.

  The universe will eventually go dark, because there will not be enough matter to make stars, and all that will be left is a bunch of black holes and they can merge. So will the universe get swallowed by a big black hole someday?

As above. Even if star formation ceased and a whole lot of matter ended up in black holes, the universe would still be expanding. As far as we know. The expansion would have to reverse for the black holes to coalesce. Mind you, it might be a good thing if it did. As Stephen Hawking said, the universe is like a black hole in reverse. If the universe ended up as one big black hole, it would be ending up the way (we think) it started out. Then you'd have a cyclical universe.


What weather is present in Jupiter's great red spot? What weather is present in Jupiter's great red spot?  Is it raining?  Is it snowing?  There is definitely a hurricane a blowing :-).  I know it is very cold, extremely high winds, has clouds of an orange / red color, and there is probably lightning.  I also know the gravity on Jupiter is strong, about 2.5x that of Earth.  Here is a similar question / answer that answers some of what I am looking for, but I still want to know if it is raining / snowing, etc...
What Would it Look Like Inside Jupiter's Great Red Spot?

I am making a level for Planetary Settlers that takes place in an air base trapped in the cloud tops of the Great Red Spot, and here is what I envision the red spot looking like so far (I am still working on adding the lightning): YouTube video of Jupiter Level Development.  I would like to know if I should add rain, snow, or any other unusual phenomenon present there.

You seem to know most of what there is to know about the Great Red Spot, except that there is not enough moisture in it for rain. It is cold enough for snow, but again there is a water deficiency. Water vapour is a comparatively rare gas in Jupiter's upper atmosphere. I think lightning would be admissible, and high winds too, and just possibly some frost. After all, there is frost on Mars despite the scarcity of water, though frost doesn't seem to be commonplace there.

The Great Red Spot would not be a good place for settlers, they'd be much better off on Mars. I presume this air base is carried around by a huge balloon, but what would they do for food and water? Normal aircraft would be no use because of the lack of oxygen, which is a pity, because there is hydrogen fuel in abundance.


Is the James Webb telescope going to orbit around the earth? If it’s going to orbit around the earth, what kind of orbit would it be (what distance)?
Are there benefits to sending a telescope further away from the earth if it’s just to photograph galaxies and not objects that are in our solar system?

Your first question - is JWST going to orbit Earth - is a little complicated. It will follow a mission profile that will send it to the Sun-Earth $L_2$ Lagrangian point. It will take the telescope about three months to achieve its orbit in $L_2$. Now, $L_2$ is unstable, and so some station-keeping - essentially, course corrections by thrusters - will be needed. It won't be orbiting Earth, but orbiting the Sun with Earth, in a nearby location.

You're not quite right that JWST will only be observing objects outside the Solar System. Some of the most exciting targets - young, evolving galaxies and exoplanetary atmospheres - lie far beyond. However, the telescope's capabilities for atmospheric analysis (thank you, NIRSpec!), useful for determining the composition of exoplanetary atmospheres, will also be used for observations of the atmospheres of Mars and the giant planets, as well as compositional data for other bodies.

Now, it's not really helpful to send JWST elsewhere in the Solar System, given that most of its observations will be of objects outside the Solar System. Okay, maybe you could get higher-quality data on Mars if you sent it to Mars orbit, but that would be quite costly, and not effective for a spacecraft designed to be used by many scientists for a number of purposes. Keeping it at $L_2$ simplifies the orbital trajectories immensely, although it's still much too far away to repair, unlike Hubble. You also lessen the communication time.

Well, you might say, why not just launch JWST into Earth orbit? We have scientific instruments there; why do we go to the trouble of putting so many things at $L_2$? It turns out that debris high above Low Earth Orbit (LEO) would be too dangerous for the sensitive optics on the telescope. The best instruments are often also the most delicate, and JWST would be damaged if it was truly orbiting Earth, at LEO or elsewhere.


How can I determine line luminosities from equivalent width measurements? I have a set of equivalent width measurements for different emission lines. How can I use them in order to determine the corresponding line luminosities?

If you also have photometry, then there is a reasonably accurate conversion between the apparent magnitude and the continuum flux (per unit wavelength interval) at the wavelength of the photometry.

Once you have this conversion factor, you multiply your equivalent width by it to get a flux.


What is the boundary in the atmosphere seen in this video? This video, which shows time-lapse views of Earth from the ISS, consistently shows some sort of hard boundary in the atmosphere, below which there seems to be significant reflection from sunlight and/or Earth's cities and above which there seems to be nothing impeding the view of black space.  What is this boundary?  I don't think its the edge of the atmosphere, partly because the atmosphere has no hard edge and partly because the aurorae (an atmospheric phenomenon) seen in the video exist, at least in part, above this boundary.

Yes this is neither aurora, nor reflection of any light.
This is airglow, that comes from upper layers of atmosphere.
The brightest emission is green 558nm light from excited atomic oxygen in a layer 90-100 km high. At that height most of the oxygen is in atomic state, not molecular, and is constantly being exposed to solar UV radiation.

Here you can find more details.
 airglow formation


Could vacuum energy dim standard candles? It is assumed that particles and anti-particles pop out in vacuum (vacuum energy). Could this have an effect on the paths of photons when they interact with these particles? The amount of photons going straight from the source would then gradually diminish. The supernovae would then appear dimmer than without this effect.

Locally, energy and momentum must be conserved. So no, a photon cannot lose energy or change momentum by interacting with virtual particle pairs that have a net energy and momentum of zero.


Total solar eclipse, supermoon, and spring equinox all happening at the same time: anything special about this? Today (March 20, 2015) is seeing a rare combination of the spring equinox, a total solar eclipse, and a supermoon.  I am wondering if there is anything special astronomically about all three of these happening at the same time, like is the equinox a particularly more or less likely time for solar eclipses to happen (my guess is that it has no impact eclipses).

The supermoon/total solar eclipse combination doesn't seem surprising to me, since if the opposite were to happen (moon far away and solar eclipse) we would end up with an annular eclipse instead of a total eclipse.

There wouldn't be a direct correlation between the equinox and the moon's orbit, so there really isn't any special - it's just a coincidence.


Can we measure stellar accelerations? Many people concern themselves with understanding the Galactic potential. To me, the most obvious way is to measure the acceleration of a star, as this is directly related to the potential. So, one could measure a LOS velocity (or proper motion) at different years and measure its change. I understand that orbital timescales of stars are huge however, so can this be done? If not, how far (order of magnitude) away are we from having the required uncertainties for achieving this?

A commenter points out that accelerations in the galactic gravitational potential are extremely small:

  The Sun's galactic orbital speed is roughly 200 km/s and its distance from the galactic centre is around 8 kpc. Using $a=v^2/r$ that gives an acceleration of roughly $1.6×10^{−10}\rm\,m/s^2$.

A century is only $\pi×10^9$ seconds, so in a hundred years the $\Delta\vec v$ for the Sun and its neighbors is only half a meter per second --- and that's mostly a difference of direction, not magnitude.  We can't measure three-dimensional stellar motions to a precision of a half-meter per second, and the career of a professional astronomer is usually shorter than a century.


Why is WISE J0521+1025's distance so uncertain? If you look at the Wikipedia page List of nearest stars and brown dwarfs
, you'll notice that all stars mentioned in the list have a very precise value; all are $\pm0.02$ ly (except for GJ 1005, but it is still within 1 ly)
Even though WISE J0521+1025 is within 5 parsecs of Earth, its uncertainty is 1.3 parsecs (4.3 ly)! What is going on here, and why is the distance so inaccurate?

The distance is that reported by Bihain et al. (2013), which is based on a mean relationship between absolute magnitude and spectral type that has a lot of scatter.
i.e. in contrast to most (all?) the other objects in that list, there is no reported trigonometric parallax measurement for this very faint T7.5 brown dwarf.
In fact, if it turned out to be a binary brown dwarf, the distance could be incorrect!


How does V838 Monocerotis look today? Any subsequent light echo images, or is the party over? This answer to the question V838 Monocerotis “light-echo” images morphed into nice video, but why so few original images? contains the following information below. There are other answers and more images there, so it's recommended to enjoy there as well.

Question: What does V838 Monocerotis look like today? Have there been subsequent light echo image sequences, or is the party over?

See also Why weren't the Hubble light-echo images of V838 Monocerotis supplemented by ground telescopes?

The V838 Monocerotis expansion (not a supernova) and the observation of the subsequent "spectacular" light echo was quite a notable event! From Nature 422, 405-408 (27 March 2003)

From Astronom. J. 135, 2, 2008 or ArXiv

  ".Galactic light echoes are extremely rare. The only other known example of extent similar to that of V838 Mon was the echo produced by Nova GK Persei 1901 (Kapteyn 1902; Perrine 1902; Ritchey 1902). Following early misunderstandings, light-echo geometry was properly described by Couderc (1939), and more recent discussions are given by many authors, including Chevalier (1986), Felten (1991), Sparks (1994), Sugerman (2003), and references therein".

It was the sole topic of an international conference photo from here:

Figure 2 of the Nature paper describes the preservation of the actual light curve (history) within the structure of the light-echo shell:

  "FIGURE 2. HST images of the light echoes
  The apparently superluminal expansion of the echoes as light from the outburst propagates outward into surrounding dust is shown dramatically. Images were taken in 2002 on 30 April (a), 20 May (b), 2 September (c) and 28 October (d). Each frame is 83" times 83"; north is up and east to the left. Imaging on 30 April was obtained only in the B filter, but B, V and I were used on the other three dates, allowing us to make full-colour renditions. The time evolution of the stellar outburst (Fig. 1) is reflected by structures visible in these colour images. In b, for example, note the series of rings and filamentary structures, especially in the upper right quadrant. Close examination shows that each set of rings has a sharp, blue outer edge, a dip in intensity nearer the star, and then a rebrightening to a redder plateau. Similar replicas of the outburst light curve are seen propagating outwards throughout all of the colour images."

also see: This is an artificially constructed animation, morphing eight existing images to suggest what a higher cadence imaging campaign might have seen: https://www.spacetelescope.org/videos/heic0617a/

I'm not an observational expert, but here's what I found.

I found this object (V838 Monocerotis) on the HST data archive.

https://archive.stsci.edu/hst/search.php

I input the 'Target Name' as "V838 Monocerotis", clicked all 'Imagers', 'Spectrographs', and 'Other,' and searched it.

(You can also do the exactly the same thing I did because this site is opened for everyone.)

And there are 199 images of this object. And the last observation was made in 2012. Unfortunately, I couldn't see any echoes in the picture taken in 2012. So I looked from the beginning. And I found this!

This picture was taken from 2006-09-09 20:04:57 to 2006-09-10 01:14:57. Dataset number is J9SF01020. You can see that the shape of the echo. There are other pictures in the archive, but I couldn't see significant echo shape anymore after that.

I'm sorry to say that the party is over, the echo was faded away.

I also love this picture, it means a lot to me because this picture made me study Astronomy. Thanks for making me remind this memory.

P.S.: If you want other pictures, especially color pictures, check out this link too.

http://hla.stsci.edu/hlaview.html#Images|filterText%3D%24filterTypes%3D|query_string=V838%20Monocerotis&posfilename=&poslocalname=&posfilecount=&listdelimiter=whitespace&listformat=degrees&RA=106.020093&Dec=-3.847399&Radius=0.200000&inst-control=all&inst=ACS&inst=ACSGrism&inst=WFC3&inst=WFPC2&inst=NICMOS&inst=NICGRISM&inst=COS&inst=WFPC2-PC&inst=STIS&inst=FOS&inst=GHRS&imagetype=best&prop_id=&spectral_elt=&proprietary=both&preview=1&output_size=256&cutout_size=12.8|ra=&dec=&sr=&level=&image=&inst=ACS%2CACSGrism%2CWFC3%2CWFPC2%2CNICMOS%2CNICGRISM%2CCOS%2CWFPC2-PC%2CSTIS%2CFOS%2CGHRS&ds=


Are there any unmanned large telescopes located on Earth? I just read the answers to the question "Why are all space observatories in Chile?".  In addition to the reasons provided, one of the answers also mentions that

  ...if the telescope should be manned by humans permanently, it cannot
  be located to high due to the difficulty for humans to function at
  extremely high altitudes, ...

Which leads to my question:

  Are there any unmanned large telescopes located on Earth?

I realize that, even if large telescopes are unmanned, they probably need to be serviced by humans on a regular basis. But are humans needed permanently on site?

I also realize that, even though there are humans monitoring the measurements from space telescopes, space telescopes are unmanned in the sense that there are no people at the space telescopes manning them.

The largest "robotic" (i.e., unmanned) telescope I'm aware of is the 2.4-meter Automated Planet Finder. Other large robotic telescopes include the 2.0-meter Liverpool Telescope and its copies (Faulkes Telescope North and Faulkes Telescope South). You can read about the automated control system of the Liverpool telescope here.

These are all located at existing observatories (e.g., APF at Lick Observatory, Liverpool Telescope at Observatorio del Roque de los Muchachos, La Palma, Canary Islands, Spain), which means there are people on site for potential maintenance work, etc., though said staff are really devoted to working on other telescopes.

The closest thing to an isolated, automatic telescope in an uninhabitable setting might be the small (0.5-m) Antarctic Survey Telescope at Dome A in Antarctica, which I believe is serviced annually by Chinese expeditions. It can apparently be remotely controlled via Iridium satellite communications, though actually retrieving the full data sets is part of what the annual expeditions are for.


Those are data from "roving observers", essentially any ground-based observation that isn't from an observatory blessed with a three character observatory code. The observatory code is columns 78 to 80 of the entries. Note that the observatory code for those roving observer entries is always 247".

The longitude, latitude, and elevation of the known observatories are known. This isn't the case with those "roving observers", which is why they take two lines. The first line of a "roving observer" entry contains a "V" in column 15 and "247" as the observatory code. The second line of a "roving observer" entry contains a "v" in column 15. The bulk of the second line contains the longitude, latitude, and altitude of the observer.


Should the synodic period or the orbital period be used to determine the diameter of the Moon? Aristarchus famously used a lunar eclipse to determine the diameter and the distance of the Moon, given that a lunar eclipse can last up to 3 hours and the angular diameter of the Moon is 0.5°.
Using this method, the diameter of the Moon is
$
\frac{(\textrm{earth diameter}) \times (\textrm{moon angular diameter})}{\frac{\textrm{duration eclipse}}{\textrm{orbital period of the Moon}} \times 360}.
$
Is it correct to use the orbital period here, rather than the synodic period? Should the orbital period or the synodic period of the Moon or some other period be used in this expression, and why?

What you need, as Aristarchus, is a frame of reference in which the Earth's shadow is stationary. Otherwise you are trying to deal with a moving thing (the Moon) relative to another moving thing (the shadow cast by the light of the Sun as it orbits the Earth).
In a stationary-shadow frame of reference, the period you need is the time it takes the Moon to orbit from the middle of the Earth's shadow to the middle of the Earth's shadow. That is to say, from new moon to new moon: which is to say, the synodic month.


Why does the face of the moon 'sync' with the earth? We never see the dark side of the moon. It rotates so that the same side always faces us.

I heard two (dinner party) theories on this:
1. The moon was made by smashing out a chunk of the earth, and so somehow preserved some orbital rotation from the earth
2. The gravitational drag from the earth to the moon is lowest when the moon is in 'sync' to only show one face to the earth

I don't trust either of these.

My question is: "Why does the face of the moon 'sync' with the earth?"

It is called "Tidal locking", or "gravitational locking" or "captured rotation". According the the Wikipedia page on "tidal locking" (check the references for sources), it is due to Earth's gravity causing a small tidal bulge on the Moon, which affects its rotation. Over time, the Moon's rotation, affect by Earth's gravity, makes it's orbital rotation synchronise with its lunar rotation.


When exactly does a sub giant become a red giant? A simple, but interesting question I think.
I understand that after the hydrogen exhaustion on the main sequence the post main sequence starts with the SGB where they burn hydrogen in a shell around the Helium core.
We also know that stars under $ 2 M_{\odot }$ experience a Helium flash and start Helium core fusion after the Red giant tip. Whereas stars greater than $2 M_{\odot }$ do not experience this Helium flash and starting the He-core burning immediately after the SGB without a helium flash (very massive stars burn Helium on the main sequence).

My question is now what specific physical process happens in a sub giant, which makes it a red giant afterwards?

Thank you in advance (I am very open for mathematical/physics explanations as well, since I study physics myself).

The straightforward definition is in terms of where a star lies on its evolutionary track in the HR diagram (see below). The subgiant branch stars are those which have exhausted their hydrogen core and which are burning hydrogen in a shell but their He cores have not begun to contract significantly. The distinct upturn in luminosity marks the beginning of the red giant branch. This occurs when the core grows significantly more massive, cannot support itself hydrostatically and begins to contract. At the same time the envelope expands and becomes convective and the H-burning shell moves inwards and increases in temperature and luminosity.
The tip of the red giant branch is where He is ignited. This takes place "explosively" in a degenerate core if the star is below about $2M_{\odot}$ (not $5M_{\odot}$), but begins smoothly in higher mass stars. This causes the core to expand, pushes the H-burning shell out and reduces the luminosity.

HR diagram showing the location of the subgiant and red giant branches on the evolutionary tracks for stars of different mass. The tracks begin at the main sequence and evolve to the termination of the main sequence, across the subgiant branch, turn up the red giant branch, which terminates when core He burning commences and the stars become less luminous and hotter.


How come the Sun does not make the Moon crash on the Earth? As the Moon loops around the Earth, on one side of the Earth it moves toward the Sun (let's call it downward side) and on the other side it moves away from the Sun (let's call it the upward side).

Now, I imagine the Sun's gravity attraction accelerate the Moon on the downward side and slows it on the upward side. This should raise the Moon's apoapsis on the upward side and lowers its periapsis on the downward side, gradually putting the Moon on a collision course with Earth.

Yet, this does not happen. I am curious to understand why. I suppose even a very very slim effect should build up upon billions of years, so is there a fault in my reasoning or some tidal effect correcting the Moon's trajectory?

The Sun's gravity does perturb the Moon's orbit but more subtly than you imagined.
The Moon's perigee and apogee migrate eastward in an 8.9 year cycle called apsidal precession.
Also the plane of the lunar orbit, inclined about 5.1 degrees to Earth's orbit, shifts westward in an 18.6 year cycle called nodal precession.
Lunar theory also addresses various effects on shorter timescales.

If the Moon were 4 times farther from the Earth (outside the Earth's Hill sphere), then the Sun would destabilize the lunar orbit.
Even so, the Moon would most likely drift into an independent orbit around the Sun, with little risk of collision with the Earth.


Events I cannot explain In the last few days I experienced two events I cannot explain. I watch the sky since I was a child so I'm aware of most events like satellites, ISS, Iridium flares, etc...
First event
It was about 22:30 local time and I was located in the central Italy. I was looking in the nearby of Hercules and Corona Borealis when some degrees at left of Arturo (say near M5) a white light flashed shortly. The flash was very short, I think below 150 ms without any fade. Because it was so short I cannot say if it was moving or not, but the other strange thing is it's apparent diameter. It was larger than any other star or planet and without any difference of luminosity between the center and the border. I mean, it was like a small disk with a matte surface and a uniform white light. I can estimate it's magnitude about -2 (slightly less Jupiter in these days). It was not like an Iridium flare at all.
Second event
Yesteday, norther Italy, 1:30 AM local time. I saw what looked like a very bright satellite in SW/NE orbit. But there were few strange things:

as far as I know, it was too late for any satellite to be illuminated by the sun for such a bright magnitude. It was as bright as Venus at its maximum brightness (I would say at least -3.5).

it was composed of two distinct lights, very close together. At first I thought it was a plane from the nearby airport with landing lights still on after takeoff, but I know them and this was different. Furthermore there were no red/green lights nor strobo. And the lights were too close for a a plane below 10000 feet. Then they didn't change their brightness across the sky, like landing lights do.

it moved across the sky like the ISS does, and just few degrees from my zenith it disappeared instantly without any fading.

Any idea of what they might be?

Both of these sound like satellites reflecting the Sun, even the first one which didn't fade.  There are so many satellites now with unusual orbits that being close to midnight in time doesn't mean you won't see them anymore like it used to.  Years ago, if the local time was midnight you could almost be assured no satellites would be visible but that's not the case now.
I think your second event was also a satellite, or more specifically a pair of satellites.  Since they vanished near your zenith at that time of the night, my guess would be they entered the Earth's shadow.  Polar orbiting satellites can catch the Sun at unusual times due to the geometry as can a multitude of the new sats being launched constantly.
I forgot to add the the first one you saw, somewhat near M5, would be in the right position for that one to have been a geostationary sat (perhaps tumbling) but it also could have been another type, as well.


A quick (elementary) check on arsecond conversion to sky distance in parsecs I just wanted to double check I had something correct. If an object, say a nebula, is measured to roughly have a diameter or extent of, say, 250 arcseconds, would its diameter in parsec simply therefore be 250 parsecs?

The pedagogical answer is this.  You're confusing the
width
of something, with it's
parallax!
Here's the fixed background

Here's something that happens to have width ...

Here's something that has parallax ...

That's the deal!
Note that width can just be measured, using one printout of the photo and a wooden ruler.
But parallax means two photos taken 180 days apart, and you measure how much the thing has moved.
(The mind-boggling GAIA does not actually use wooden rulers!)

One huge confusion with "parsecs" is this:
Parallax means the distance an object moves in two photos take six months apart. That technique can be used with objects from 1 to about 100 parsecs away from us.
However, astronomers (for convenience) still use the measure "parsecs" even when the distance is far more than 100 parsecs.  And, they even use the measure "parsecs" for things like "size", which just have no connection at all to parallax photo measurements.
If you hear that something is "5000 parsecs away", it is inconceivable the distance was determined using parallax photo measurements!!  And, if you hear that something is "1000 parsecs wide", that just has no connection at all to parallax photo measurements (it's a width - nothing to do with the distance from Earth!!)
In the question at hand here, the "width" of an object was quoted in parsecs. The OP was originally assuming that, since parsecs were mentioned in relation to the width, you could use something related to the technique of parallax photo measurement, in relation to the width.  In fact, parallax photo measurement just has no connection, whatsoever to width. You literally might as well use smell or taste - heh!  The confusion is that parsecs happen to be also used (very confusingly) to measure things like width: but remembering that the technique of parsecs (parallax photo measurement) has absolutely no connection at all to width, and no connection to anything more than 100 ps from us.


Why is there no hurricane on the Sun? I was reading about gigantic storms on giant gas planets, the great red spot on Jupiter and hexagon on Saturn to name a few, how about our Sun which is consist of plasma(hot gas which have some of their electrons stripped) why don't we see any big hurricane?

We see sunspots, which are giant, planet-sized storms on the surface of the Sun. There are however many differences between sunspots, the Great Red Spot on Jupiter, and tropical cyclones (e.g., hurricanes) on the Earth. Tropical cyclones are low pressure systems fueled by evaporation of warm ocean water and sustained by the Earth's somewhat rapid rotation rate. The Great Red Spot is a high pressure system that is sustained by Jupiter's quite rapid rotation rate. Sunspots are low temperature systems fueled by the Sun's magnetic field and carried along by the Sun's rather low rotation rate.


Gas halo of our Milky Way Galaxy This question relates to a diffuse hot gas halo of our Milky Galaxy. I've read that there is a hot diffuse halo of gas surrounding our Galaxy (NED, Caltech)

I was wondering why such a halo can exist? Why doesn't it collapse to a disk shape? Is it because the gas itself is still hot and so remains largely unaffected by the Galaxy potential?

The scale height of gas in a disk (if it were in equilibrium) is roughly $kT/mg$, where $T$ is the temperature, $g$ is the gravitational field, $m$ the mean mass of agas particle, and $k$ the Boltzmann constant.

If we assume most of the mass is in a thin disk, then Gauss's law for gravitation tells us that that $g = 2\pi G \sigma$, where $\sigma$ is the mass per unit area in the disk. According to Rix & Bovy, $\sigma \simeq 70 M_{\odot}$ pc$^{-2}$ at the location of the Sun (http://arxiv.org/abs/1309.0809).

If we assume hydrogen gas, then the effective particle mass is that of a proton, and this means the gas scale height is
$$ H = 4300 \left(\frac{T}{10^6\ K}\right)\ pc$$

Thus gas hotter than a million degrees will have a very substantial scale height and is not expected to be confined to the Milky Way disk.


How far does the Earth travel around the Sun during each month of the year? After reading about the Earth's orbit and how it moves faster in January than it does in July (based on its distance from the Sun), I started to wonder...

How far does the Earth travel around the Sun during each month of the year?

What is the math behind this calculation?

Note: I'm not asking about the average monthly distance, but the unique distance traveled for each month.

By using the Approximate Solar Coordinates formulas on the U.S. Naval Observatory website, you can calculate the ecliptic longitude of the Sun (L) and the distance between the Earth and Sun (R) for any date. Then, The distance travelled between two dates is approximately s=R*(180/pi)*(L2-L1), where R is the average distance during the interval, and L2 and L1 are the ecliptic longitudes. Since the Earth's orbit is almost circular, this approach is reasonably accurate even when performing the calculation over an entire month. These are the results that I get, where the distance s is in Astronomical Units (AU, the average distance between the Earth and Sun).

Month     Distance Travelled,  days ave. daily  ave. change  accl.
Jan 2019      0.542 AU         31   .00175 AU       0 miles 0 f/ss
Feb           0.488            28   .00174 AU   - 475 miles 0.000672
Mar           0.536            31   .00173 AU  - 1300 miles 0.001840
Apr           0.514            30   .00171 AU  - 1490 miles 0.002104
May           0.527            31   .00170 AU  - 1210 miles 0.001709
Jun           0.508            30   .00169 AU   - 651 miles 0.000921
Jul           0.525            31   .00169 AU       0 miles 0
Aug           0.526            31   .00170 AU     279 miles 0.000395
Sep           0.513            30   .00171 AU    1300 miles 0.001840
Oct           0.535            31   .00173 AU    1490 miles 0.002104
Nov           0.522            30   .00174 AU    1300 miles 0.001840
Dec           0.542            31   .00175 AU     744 miles 0.001053

Of course, the variation from month to month is heavily influenced by the number of days in each month, where as the variation from January to July is due to the distance from the Sun.

As a check, I calculated the distance travelled for each day in January 2019 and totaled the distance. That result was s=0.5419 AU compared to s=0.5422 AU when using the calculation for the first and end of the month.


Is there any practical use for astronomy? Although astronomy is very cool and the things we are learning are awesome, is there really any practical use to knowing the things we know about the universe?

Do other fields of science draw from the current tome of astronomical knowledge?

This question begs the question, does everything need a practical use? The answer is a resounding no. What's the practical use of the Louvre, or of your local neighborhood public park where you enjoy weekend barbecues?

There are some things that are very worthwhile that have little or no economical gain. Your local neighborhood public park in fact has negative economic gain. Admission is free, but maintenance is not. Think of how much money your city would make if they sold it to a condominium developer, and how much money it would save by not having to pay to have the park maintained.

Despite having no obvious economic gain, some things are nonetheless worth quite a bit. Many of the sciences fall in this category. For example, what is the practical use of archeology? (There are some, but that's not the point.)

Astronomy, like archeology, the Louvre, and your local public park, doesn't need a practical economical purpose. The purpose of the science is good enough.

That said, there are practical applications of astronomy. The key application has been and still is navigation. Knowing the location of a ship at sea or the orientation of a vehicle in space requires astronomy.

A less direct but still very important application of astronomy is in how it informs physics. Kepler was an astronomer, not a physicist. (Those two disciplines were very, very distinct in Kepler's day). Yet Kepler's work informed Newton on how to describe gravitation. More recently, astronomy has informed physics that its standard model was not quite correct. The observed neutrino flux from the Sun (see http://en.wikipedia.org/wiki/Solar_neutrino_problem) was a third of what physics at the time said it should be. This resulted in a change to the standard model. Neutrinos have a small but non-zero mass, and they oscillate from one form to another.

Astronomy continues to inform physics to this day. Physicists (and astronomers) remain clueless with regard to what constitutes dark matter and dark energy. But whatever they are, they certainly do exist.


How did we differentiate lights from stars? I got this question while watching the COSMOS TV series. I couldn't find the right answer in google.

I read about Stellar classification and it states that we can classify the stars based on their spectral characteristics. My question is how can we be sure of which particular star the light is coming from. Since Earth is getting light from so many stars How the the light is classified?  Even if we can get different spectra then how can find their exact sources?

If two lights are shining in your room, one red and one blue, how do you know which one is red and which is blue?

If you look at the two lights, the lens in your eye will focus the red light at a different place from the blue light, so you can see two lights in different places. Even though the light that reaches your eye is mixed, the lens can separate the light.

Telescopes have lenses (or curved mirrors that do the same thing). The light from different stars is focused by the telescope, to form an image, and so the light from different stars is separated. The light from individual stars can then be split into colours. This is the star's spectrum.

The job of the telescope is two-fold: collect a lot of light and then focus it so that individual objects can be seen and analysed individually.

There are many ways of classifying a spectrum, Stars have continuous spectra crossed by dark lines. The temperature of the star changes the spectrum, the stars are classified in order O B A F G K M from hottest to coolest. Nebula sometimes have emmission spectra which looks mostly dark, crossed by bright lines.


Does tidal locking also slowly reduce the orbiting body's axial angle? The Moon is tidally locked to the Earth, and it has an axial angle of 6.687 degrees relative to its orbital plane. I'd like to know: did the Moon start out with a higher axial angle?
In other words, does the process of tidal locking also slowly reduce the orbiting body's angle of rotation? If so, how does this work? And will it someday cause the angle to be reduced to zero?
A secondary question would be: suppose an orbiting body had an extreme axial angle, say 80 degrees (perhaps due to an impact during its formation), how would this affect the process of tidal locking?

In general, tidal forces will cause the obliquity to decrease for an initially prograde rotation, so unless something is acting to disrupt the situation you would expect a tidally-locked satellite to end up with a very small obliquity. This is the observed rotation state of most of the major satellites in the Solar System.
The Moon is something of a tricky case thanks to its location and its history of outward migration. The Moon likely formed close to Earth in a near-equatorial orbit. As it moved outwards it crossed the Laplace radius, which marks the region in which the Laplace plane (the plane about which the satellite's orbital plane precesses) transitions from being aligned with the planet's equator to being aligned with the planet's orbit. If the Earth was originally in a relatively high-obliquity state, this transition would have resulted in rapid changes in the Moon's orientation, and the Earth's obliquity becoming more similar to its present-day value.
The upshot of this is, even though tidal forces tend to damp the inclination with respect to the Laplace plane, the Moon's orbit is still inclined by a few degrees to the Laplace plane (likely as a result of picking up a fairly high inclination during the transition across the Laplace radius), and the nodal precession has a timescale of about 18.6 years. This timescale is more rapid than tidal forces can decrease the Moon's obliquity, so the Moon cannot evolve into the zero-obliquity state.


Why is it that we can"t feel how fast the Earth is moving? We are moving like 100,000 mph. And why do we not fall off going that fast

Because the rotation of the earth is very smooth and doesn't change, the centripetal acceleration we feel is very nearly constant. This means that the (small) centrifugal force from the rotation gets added to gravity to make up the "background force" we don't notice.

Taken from Physics SE.


Why can't I see Mars clearly? I just bought my first telescope - a Celestron CDC 1100.  It came with a 40mm lens so with a primary focal length of 2800mm the power should be 70x.

My problem is that I've looked at Sirius and Mars and they look clearer to my naked eye than with my telescope.

I am able to focus from way out so I can see the 'blind spot' of the secondary mirror, go through 'focused' and then continue moving the mirror in the same direction until I again see the blind spot.  So I know I'm focusing through when it should be 'good'.  But, like I said, what I see is a relatively unfocused spot of light that is clearer and steadier with my naked eye.

Admittedly, my 'seeing conditions' aren't very good since I am looking through my window in a suburban neighborhood where light pollution is pretty bad.  That being said, I still see Mars and Sirius quite brightly - it's just that they never come into clear focus - nor are they any larger than when I look using my eyes.  And when I move my eye around the eyepiece I see streaks of light in different directions originating from the target.

There are many stars to be seen, though they are just pinpricks as I would expect given the seeing conditions.  What I mean to say is that I cannot see all those stars with my naked eye but can see tens or hundreds more looking through the scope so I know it is gathering plenty of light as an 11" scope should.

I tried both with and without the diagonal with no discernable difference.

I have not collimated but supposedly it was collimated before leaving the factory and I would hope that even if it were off a bit from shipping I would still be able to see a sizable though fuzzy image of Mars instead of just a small speck of orangish light.

Any thoughts as to what I may have done wrong or what may be causing this lack of clarity?

You're experiencing a few issues all of which have one solution. Take the Scope outside. Let it acclimate to the ambient air tempurature, and then try looking.

The 1st issue is the mirror needs to be the same temp as the air you're observing in. Since you are indoors you have a very warm optical path this causes distortion and trouble focusing.

2: Because you're looking through a piece of window glass (or more if it's a multipane window) you're introducing distortions that way. It's no wonder you can't reach focus.

Light pollution won't really affect Mars or the other planets too much, so that's not your worry.

Why are you observing from inside anyway?


Why haven't more captured small moons been found? Shouldn't captured moons have the same size distribution as asteroids? And asteroids are more common the smaller size they are. Moons are likely captured if they are in highly inclined orbits, and those moons should've originated as asteroids or Kuiper Belt objects. But there are only two moons, out of 194 known, smaller than 500 meters in radius. Aegaeon and S/2009 S1 both of Saturn. Although it is expected to exist over a million asteroids smaller than that. And all of Pluto's moons were large enough to be found before New Horizons arrived (sorry Alan Stern, no moon for you!)

Is this purely an observational bias?
Is it expected that the planets are surrounded by a myriad of moons too small to have been detected yet, from hundreds of meters down to dust grains?
Do they all somehow gather into planar rings below a certain size? (I think the main line is that a ring system forms as a result of a single collision or tidal crunching event).
Or what mechanism makes them lack small to tiny moons and dust?

How would this asteroid size frequency plot compare to a moon size frequency plot?

Encounters between small planetesimals and the larger core of planets (especially Jupiter) transfer momentum between the two. But momentum is conserved. Momentum is:

$p = m v$, where $p$ is momentum

So, smaller planetesimals gain more velocity than a larger planet core, for the same momentum exchange. Thus, smaller planetesimals are preferentially boosted to escape the planet, either to be scattered outward to the outer system, or inward to the sun. Some will have the right velocity and aggregate with to core.

Since smaller planetesimals get a larger boost than larger ones, say 1000x rather than 20x (just an example, not real numbers), then they are much more sensitive to initial velocities. Thus a much smaller range of initial velocities will have that exactly right velocity to be captured by the planet, as either a satellite or aggregating with the planet. In the case I just made up, 1000/20 i.e. 50 times narrower. Thus, for an equivalent distribution of velocities, the smaller planetesimals in this example will be 50 times less likely to be captured.

As for dust, beyond it will eventually clump together into larger rocks due to electrostatic attraction, a lot will be sucked into the growing core, or scattered outwards. But it will exert a net drag on the core (see the Grand Tack hypothesis) and gain momentum, and through velocity, energy (a.k.a. heat). And this energy will be averaged out through elastic collisions. Eventually, the dust will either aggregate into larger bodies, sucked into the planet, or scattered into the sun or outwards from the solar system.


How do I calculate Allan Variance (maybe for pulsars)? I am studying pulsar timing this summer and have come across many Allan Deviation plots that include several popular clocks alongside pulsar measurements, like this one by N. Ashby and D.A. Howe:

I am unclear as to where to start. What equation would be appropriate? I have pulsar timing data from the NANOGrav Public Data Set, which I am able to view through Tempo2 (aka will be able to view it as soon as I figure out how Tempo2 works). If there is other software anyone would recommend for this task that would also be helpful.

I am also unsure where the data for the other clocks would be publicly available. I am hoping a scientist who has done a similar calculation can point me in the right direction.

So, after some discussion and paper reading I learned of the sigma_z calculation in this paper by Matsakis, Taylor, and Eubanks. Several astronomers recommended this to me. I was also told to try out Stable32, although I'm not sure if that will work for pulsars. The code for a sigma_z calculation can be found here: sigma_z calculation. I believe this is the best method for doing what I asked.


How can heavy elements exist when the universe is young compared to the lifetime of a star? Our sun is said to be incapable of producing heavier elements, and these are thought to originate from older stars dying and going supernova. If our sun is 4-5 billion years old, and has a lifetime of 10 billion years, how could an earlier sun have lived and died in the short time between the creation of the universe and the birth of our fine solar system?

First, stars evolve at radically different rates. The higher the star's mass, the faster it evolves.  The more massive stars had plenty of time to go through multiple generations before the Sun started to coalesce.

It went even faster than it does today, because the collapse of interstellar clouds to form stars is halted mainly by radiation pressure from the new star blowing the rest of its birth cloud away and ending accretion. Since hydrogen and helium are transparent to most of the new star's radiation, the very first stars could grow larger than stars do today. (They were pretty much made of nothing but H and He (with a bit of Li), since very little of the heavy elements had yet formed.) They went through their lifetimes and turned into supernovae and neutron stars very quickly. The largest had lifetimes on the order of just a million years.

So by a very short time after the first stars lit, they were already spewing heavy elements back into space and later stars already had some heavier elements in them at formation.  The fraction of heavy elements has increased steadily since then, but the Sun formed many billions of years after heavy elements became common.


Determining North-South Line Via Watch Method: Theory & Reason I recently read that if you're in the northern hemisphere and have an analog watch, then you can point the hour hand at the sun and know that a south line lies between (bisection) the hour hand and the 12 o'clock position:

Apparently the trick works in the southern hemisphere as well and needs to be adjusted for day-light saving (image taken from Wikipedia):

I would like to know the reasoning behind this trick, and would like to know if the trick can be applied to tell the approximate time provided I only have a Brunton compass.

I made an illustration if it helps:

When the Sun is on the meridian and is due South (local noon in Northern Hemisphere) your hour hand will point to (or close to 12) - assuming your local noon and civil noon are similar! (Daylight saving means 1pm fills this role).

The Sun crosses the sky in 24 hours, the hour hand travels round the watch in 12 hours. Hence the watch hour hand is twice as fast, so when the hour hand has reached 2pm, the Sun has only gone half as far (ie "to 1pm"), so if you point the hour hand towards the Sun in this case, the meridian is an hour further behind rather than 2 hours.

By the time we get to 4pm the Sun will have travelled "two watch hours" away from the meridian and so on.


What produces gravitational waves with "periods between about 100 - 8000 seconds"? The Ulysses mission has a compelling story. It was sent to Jupiter to perform a gravitational assist shooting it out of the plane of the ecliptic in order to fly over the Sun's north and south poles to perform "fast latitude scans". Because of its design it was used for several important lines of scientific study.
Ulysses contained a pair of coherent transponders which received signals from Earth, shifted them in frequency in a coherent way using phase-locked loops and beamed them immediately back to Earth at two different frequencies.
From ESA's write up of the Ulysses Gravitational Wave Experiment:

In the spacecraft Doppler tracking method, the Earth and spacecraft constitute the two objects whose time-varying separation is monitored to detect a passing gravitational wave. The monitoring is accomplished with high-precision Doppler tracking in which a constant frequency microwave radio signal (S-band) is transmitted from the Earth to the spacecraft (uplink); the signal is transponded (received and coherently amplified) at the spacecraft; and then transmitted back to Earth (downlink) in both S- and X-band signals. This Dual frequency downlink is required in order to calibrate the interplanetary media which affects the two frequency bands differently. The downlink signal is recorded at Earth and its frequency is compared to the constant uplink frequency f0 to extract the Doppler signal, δf / f0.

The article goes on to say:

Since the optimum size of a gravitational wave detector is the wave length, interplanetary dimensions are needed for detecting gravitational waves in the mHz range. Doppler tracking of Ulysses provides sensitive detections of gravitational waves in this low frequency band. The driving noise source is the fluctuations in the refractive index of interplanetary plasma. This dictates the timing of the experiment to be near solar opposition and sets the target accuracy for the fractional frequency change at 3.0 × 10-14 for integration times of the order of 1000 seconds.
SUMMARY OF OBJECTIVES
The objective of the gravitational wave investigation on Ulysses is to search for low frequency gravitational waves crossing the Solar System. Because of the great distance to the spacecraft, this method is most sensitive to wave periods between about 100 - 8000 seconds, a band which is not accessible to ground-based experiments which are superior for periods below 1 second.

You can read more about Ulysses in eoPortal's Ulysses where I found both the link above and the following:
B. Bertotti, R. Ambrosini, S. W. Asmar, J. P. Brenkle, G. Comoretto, G. Giampieri, L. Iess, A. Messeri, H. D. Wahlquist, “The gravitational wave experiment,” Astronomy and Astrophysics Supplement Series, Ulysses Instruments Special Issue, Vol. 92, No. 2, pp. 431-440, Jan. 1992

Question: What produces gravitational waves with "periods between about 100 - 8000 seconds"?

Any binary system produces gravitational waves at twice it's orbital frequency, i.e. with periods of half it's orbital period. So binary systems with periods between 200s and 16000s will produce such waves.
We can use Kepler's third law to say something about these:
$$ a = \left(\frac{GM}{4\pi}\right)^{1/3} P^{2/3},$$
where $P$ is the orbital period, $M$ is the total mass of the binary system and $a$ is the orbital separation.
For a binary with $M\sim 1M_{\odot}$ and $200<P<16000$s, then $0.11 < a < 2.00\ R_{\odot}$. Since normal stars of mass $\sim 0.5M_{\odot}$ have radii that are similar to this, then the stars would probably need to be stellar remnants (white dwarfs, neutron stars or black holes) except right at the longest period end, where it might be possible to observe W Uma binaries. More massive binaries have separations that increase as $M^{1/3}$, but the radii of normal stars increases more like $M$, so this conclusion is even firmer at larger masses.
It could be possible to have a compact binary involving a low mass star plus a compact object - perhaps a Roche lobe filling one, so as well as "double degenerates", the long period end of this range would include Cataclysmic Variables and Low Mass X-ray binary counterparts, with orbital periods of a few hours. Here is a prime example Time domain astronomy and fastest eclipsing binary ZTF J1539+5027 (+20 mag, 6.91 minutes): How to measure its minimum brightness?
Of course gravitational wave strain goes as something like $M P^{-4/3} d^{-1}$, where $d$ is the distance. These binaries are much longer period than the (presumably rare) massive, merging black holes seen so far and so probably need to be close, in our own Galaxy, to be detected.
e.g. LIGO was capable of detecting $M \sim 30 M_{\odot}$ merging black holes, with $P \sim 0.02$ s at distances of a billion light years. A similar strain amplitude would be produced by a $M\sim 2M_{\odot}$ binary with $P= 200$ s at a distance of 300 light years.


How to discover Neptune from the Uranus orbit (by computer simulation) I would like to demonstrate the existence of another planet (Neptune) by studying the discrepancy between the observation of the Uranus orbit and the mathematical prediction, this work was made from Le Verrier and I would like to understand his method.

I have read the Chapter 2, "The Discovery of Neptune (1845-1846)," in the biography Le Verrier -- Magnificent and Detestable Astronomer,
but it is too much in-depth and I did not understand very well his work.

I am studying the three-body problem (Sun, Uranus, Neptune) via Matlab and the two body problem (Sun, Uranus) taking the initial condition from here:

http://nssdc.gsfc.nasa.gov/planetary/factsheet/uranusfact.html

I have tried this method:
I put Uranus in the Perihelion with the Max. orbital velocity and I calculate the semi-major axis, and it is more accurate than the one obtained from putting Uranus and Neptune in the Perihelion with their respective Max. orbital velocity.

Here a cool pic made with Matlab:

Can anybody help me? what I have to do and with what data I have to compare my prediction? Even a simple link could be helpful.

Here is what I did:

Based on their masses, it is safest to initially consider Jupiter and Saturn as well as Uranus. It might also be fruitful to include the Earth in the analysis, to get relative positions, observation angles, etc. So, I will be considering:

Sun
Earth
Jupiter
Saturn
Uranus
Neptune

Get the standard gravitational parameters (μ) for all of them
Get initial positions and velocities via JPL/HORIZONS for all these planets. I had some data lying around from J2000.5, so I just used the state vectors from the 1st of January, 2000, at noon.
Write an N-body integrator with built-in MATLAB tools. Integrate this incomplete Solar system once without Neptune, and once with Neptune included.
Analyze and compare!

So, here's my data, and N-body integrator:

function [t, yout_noNeptune, yout_withNeptune] = discover_Neptune()

    % Time of integration (in years)
    tspan = [0 97] * 365.25 * 86400;

    % std. gravitational parameters [km/s²/kg]
    mus_noNeptune = [1.32712439940e11; % Sun
                     398600.4415       % Earth
                     1.26686534e8      % Jupiter
                     3.7931187e7       % Saturn
                     5.793939e6];      % Uranus

    mus_withNeptune = [mus_noNeptune
                       6.836529e6]; % Neptune

    % Initial positions [km] and velocities [km/s] on 2000/Jan/1, 00:00
    % These positions describe the barycenter of the associated system,
    % e.g., sJupiter equals the statevector of the Jovian system barycenter.
    % Coordinates are expressed in ICRF, Solar system barycenter
    sSun     = [0 0 0 0 0 0].';
    sEarth   = [-2.519628815461580E+07  1.449304809540383E+08 -6.175201582312584E+02,...
                -2.984033716426881E+01 -5.204660244783900E+00  6.043671763866776E-05].';
    sJupiter = [ 5.989286428194381E+08  4.390950273441353E+08 -1.523283183395675E+07,...
                -7.900977458946710E+00  1.116263478937066E+01  1.306377465321731E-01].';
    sSaturn  = [ 9.587405702749230E+08  9.825345942920649E+08 -5.522129405702555E+07,...
                -7.429660072417541E+00  6.738335806405299E+00  1.781138895399632E-01].';
    sUranus  = [ 2.158728913593440E+09 -2.054869688179662E+09 -3.562250313222718E+07,...
                 4.637622471852293E+00  4.627114800383241E+00 -4.290473194118749E-02].';
    sNeptune = [ 2.514787652167830E+09 -3.738894534538290E+09  1.904284739289832E+07,...
                 4.466005624145428E+00  3.075618250100339E+00 -1.666451179600835E-01].';

    y0_noNeptune   = [sSun; sEarth; sJupiter; sSaturn; sUranus];
    y0_withNeptune = [y0_noNeptune; sNeptune];

    % Integrate the partial Solar system
    % once with Neptune, and once without
    options = odeset('AbsTol', 1e-8,...
                     'RelTol', 1e-10);

    [t, yout_noNeptune]   = ode113(@(t,y) odefcn(t,y,mus_noNeptune)  , tspan, y0_noNeptune  , options);
    [~, yout_withNeptune] = ode113(@(t,y) odefcn(t,y,mus_withNeptune),     t, y0_withNeptune, options);

end

% The differential equation
%
%    dy/dt = d/dt [r₀ v₀ r₁ v₁ r₂ v₂ ... rₙ vₙ]
%          = [v₀ a₀ v₁ a₁ v₂ a₂ ... vₙ aₙ]
%
%  with
%
%    aₓ = Σₘ -G·mₘ/|rₘ-rₓ|² · (rₘ-rₓ) / |rₘ-rₓ|
%       = Σₘ -μₘ·(rₘ-rₓ)/|rₘ-rₓ|³
%
function dydt = odefcn(~, y, mus)

    % Split up position and velocity
    rs = y([1:6:end; 2:6:end; 3:6:end]);
    vs = y([4:6:end; 5:6:end; 6:6:end]);

     % Number of celestial bodies
    N = size(rs,2);

    % Compute interplanetary distances to the power -3/2
    df  = bsxfun(@minus, permute(rs, [1 3 2]), rs);
    D32 = permute(sum(df.^2), [3 2 1]).^(-3/2);
    D32(1:N+1:end) = 0; % (remove infs)

    % Compute all accelerations
    as = -bsxfun(@times, mus.', D32);              % (magnitudes)
    as = bsxfun(@times, df, permute(as, [3 2 1])); % (directions)
    as = reshape(sum(as,2), [],1);                 % (total)

    % Output derivatives of the state vectors
    dydt = y;
    dydt([1:6:end; 2:6:end; 3:6:end]) = vs;
    dydt([4:6:end; 5:6:end; 6:6:end]) = as;

end

Here is the driver script I used to get some nice plots out:

clc
close all

% Get coordinates from N-body simulation
[t, yout_noNeptune, yout_withNeptune] = discover_Neptune();

% For plot titles etc.
bodies = {'Sun'
          'Earth'
          'Jupiter'
          'Saturn'
          'Uranus'
          'Neptune'};

% Extract positions
rs_noNeptune   = yout_noNeptune  (:, [1:6:end; 2:6:end; 3:6:end]);
rs_withNeptune = yout_withNeptune(:, [1:6:end; 2:6:end; 3:6:end]);

% Figure of the whole Solar sysetm, just to check
% whether everything went OK
figure, clf, hold on
for ii = 1:numel(bodies)
    plot3(rs_withNeptune(:,3*(ii-1)+1),...
          rs_withNeptune(:,3*(ii-1)+2),...
          rs_withNeptune(:,3*(ii-1)+3),...
          'color', rand(1,3));
end

axis equal
legend(bodies);
xlabel('X [km]');
ylabel('Y [km]');
title('Just the Solar system, nothing to see here');

% Compare positions of Uranus with and without Neptune
rs_Uranus_noNeptune   = rs_noNeptune  (:, 13:15);
rs_Uranus_withNeptune = rs_withNeptune(:, 13:15);

figure, clf, hold on

plot3(rs_Uranus_noNeptune(:,1),...
      rs_Uranus_noNeptune(:,2),...
      rs_Uranus_noNeptune(:,3),...
      'b.');

plot3(rs_Uranus_withNeptune(:,1),...
      rs_Uranus_withNeptune(:,2),...
      rs_Uranus_withNeptune(:,3),...
      'r.');

axis equal
xlabel('X [km]');
ylabel('Y [km]');
legend('Uranus, no Neptune',...
       'Uranus, with Neptune');

% Norm of the difference over time
figure, clf, hold on

rescaled_t = t/365.25/86400;

dx = sqrt(sum((rs_Uranus_noNeptune - rs_Uranus_withNeptune).^2,2));
plot(rescaled_t,dx);
xlabel('Time [years]');
ylabel('Absolute offset [km]');
title({'Euclidian distance between'
       'the two Uranuses'});

% Angles from Earth
figure, clf, hold on

rs_Earth_noNeptune   = rs_noNeptune  (:, 4:6);
rs_Earth_withNeptune = rs_withNeptune(:, 4:6);

v0 = rs_Uranus_noNeptune   - rs_Earth_noNeptune;
v1 = rs_Uranus_withNeptune - rs_Earth_withNeptune;

nv0 = sqrt(sum(v0.^2,2));
nv1 = sqrt(sum(v1.^2,2));

dPhi = 180/pi * 3600 * acos(min(1,max(0, sum(v0.*v1,2) ./ (nv0.*nv1) )));
plot(rescaled_t, dPhi);

xlabel('Time [years]');
ylabel('Separation [arcsec]')
title({'Angular separation between the two'
       'Uranuses when observed from Earth'});

which I'll describe here step by step.

First, a plot of the Solar system to check that the N-body integrator works as it should:

Nice! Next, I wanted to see the difference between the positions of Uranus with and without the influence of Neptune. So, I extracted just the positions of those two Uranuses, and plotted them:

...that's hardly useful. Even when zooming in greatly and rotating the heck out of it, this is just not a useful plot. So I looked at the evolution of the absolute Euclidian distance between the two Uranuses:

That's starting to look more like it! Roughly 80 years after the start of our analysis, the two Uranuses are almost 6 million km apart!

Large as that may sound, in the grander scale of things this might drown in the noise when we take measurements here on Earth. Plus, it still does not tell the whole story, as we'll see in a moment. So next, let's look at the angular difference between the observation vectors, from Earth towards the two Uranuses to see how large that angle is, and if it can stand out above the observational error thresholds:

...whoa! Well over 300 arcseconds difference, plus all sorts of wobbly bobbley timey wimey rippling going on. That seems well within the observational capabilities of the time (although I can't find a reliable source on this so quickly; anyone?)

Just for good measure, I also produced that last plot leaving Jupiter and Saturn out of the picture. Although some perturbation theory had been developed in the 17th and 18th centuries, it was not very well developed and I doubt even Le Verrier took Jupiter into consideration (but again, I could be wrong; please correct me if you know more).

So, here is the last plot without Jupiter and Saturn:

Although there are differences, they are minute, and most importantly irrelevant for discovering Neptune.


What's the mass of the interstellar neighborhood of the Sun? What's approximatively the mass of the interstellar neighborhood of the Sun ?
(for example the neighborhood of radius 50 light-years).

Remark: The mass of the observable matter should be completed by the mass of the "dark matter".

You don't have to guesstimate to come up with the answer.

What you do is look at the dynamics of stars with respect to the Galactic plane - in particular, the velocity dispersions of stars with known distances from the plane, combined with a reasonable assessment of where the Sun is with respect to the plane (close), yields an almost model-independent assessment of the mass density in the disk of our Galaxy in the vicinity of the Sun.

This work was done using the Hipparcos cataogue by Creze et al. (1998). They found that the local mass-density (all forms of matter) in the Galactic disk was $0.076 \pm 0.015 M_{\odot}/pc^3$.

Thus within 50 light years (15.3 pc), because the Galactic disk has an exponential scale height much bigger than this (about 100-200 pc), we can assume a constant density and derive a total mass in the solar neighborhood of $1100\pm 200\ M_{\odot}$.

This mass is almost compatible with the total mass estimated to be in stars, white dwarfs, neutron stars, brown dwarfs and the interstellar medium gas. For instance, according to Chabrier (2001), the total mass density is almost accounted for in the form of 60% main sequence stars plus white dwarfs and neutrons stars, about 5% brown dwarfs and 30% gas, and so the contribution of dark matter to the local disk density is probably very small. Dark matter is not distributed in the same way as normal matter; it did not collapse to a disk, because it is dissipationless and probably exists in a spherical halo. So this consistency between the local baryonic matter density and the total mass density is not an argument against the dark matter hypothesis.

EDIT: I found a few estimates of the local dark matter density that use the kinematics of local stars as a constraint. The range of densities quoted is from 0.008 to 0.02 $M_{\odot}$/pc$^3$ (Bovey & Tremaine 2012; Garbari et al. 2012) i.e. small compared with the contribution from stars and gas.


Is Sun a part of a binary system? Does our Sun have a counterpart, i.e., is it a part of a binary-system? If so, how does the other star look like and where is it?

There is no observational evidence that the sun is a member of a binary (trinary, or more) star system, where "star" means an object that is at least ~80 times the mass of jupiter and emits energy/light via standard hydrogen fusion.

Some evidence that people point to is that the majority of stars in the Galaxy (perhaps 60% or so) are binary.  However, that does not mean that we are a binary system, just that we are among those which are not.

There is the Nemesis star idea which was a hypothetical binary companion to the sun to account for periodic mass extinctions roughy every 26 million years (as in, the two stars would therefore be on a 26 million year period, so the companion is a few light-years away).  There are two problems with that:  First, that a 2010 reanalysis of the data showed that it was too periodic over the past 500 million years, during which time we've gone around the galaxy twice, and so our hypothetical binary orbits should have been perturbed and not be perfect (though others say that the precision of the geologic timescale isn't good enough to say this).  But, second, we can see stars out to a few light years -- we can see small, faint stars out to thousands of light years.  We have all-sky infrared surveys that should be able to pick up the faintest even star-like objects to at least a few light-years, and yet ... nothing.

The other problem is what LDC3 pointed out in their answer: We should see some systematic motion of our own star in orbit around the common center of mass of the hypothetical binary.  This would not be a yearly wobble, but rather it would be a slow motion of the entire sky on top of the other >1 year motions that we see.  We now have very accurate astronomical records dating back at least a century of star positions, especially close stars.  Even if we were on a 26 million year orbit - and especially one much shorter as some people claim - we should see effectively that our star is making a small arc, a part of a circle as it orbits the center of mass of it and the binary.  We don't.  There does not seem to be any systematic signal in the motions of stars that require the binary star model.

So, to summarize: A binary companion simply lacks any observational data; if it existed, we should not only have been able to see it by now, but we should also observe not only the stars in our sky show a systematic motion due to our orbit around it, but we should again see THAT binary star also move very quickly, relative to other stars, as it orbits the common center of mass.


How is an exoplanet characterised as "Earth-Like"? Which features or characteristics are taken into account to characterize an exoplanet as "Earth-Like"?

There is no unique definition of Earth-like and it depends on what characteristics are important in the context of discussion.

The most simple one is to just compare radii and masses. That is observationally sensible as radius is directly deducable from transits (assuming stellar size known) and mass is often an accessible properties for transiting planets when you have access to complementary radial-velocity data. It's also a good starting point if you want to add  further restrictions. The habitable planet catalogue uses this definition (considering only planets in the habitable zones around their stars, using a simple equilibrium temperature definition)

If your quest is actually for a 2nd Earth not only in surface gravity but other characteristics, one approach is outlined by the Earth similarity index (ESI). While it is not directly a measure for habitability either, it is a good proxy by comparing more essential parameters with Earth like mass, gravity, surface temperature. For a selection of parameters you define this ESI as

$$
\mathrm{ESI} = \prod_{i=1}^n \left(1 - \frac{x_i - x_{i,o}}{x_i + x_{i,o} }\right)^{w_i/n}
$$

where $ x_{i,o} $ is the Earth's value of a property and $n$ the total number of properties. This is basically a weighted geometric mean of the property differences. It is a good idea to consider 'internal values' and 'external' ones. The Planetary Habitability Lab uses surface temperature and escape velocity as external and radius and density as internal and use this to create a nice 2D scatter plot for known planets.