Two luminous points disappearing in the sky I saw a few days ago two luminous points in the sky, which were quite close (about 1-2 degree of separation between them) and not apparently moving (or to slow to be noticed with the naked eye). The apparent luminosity of these points was close to the one of Jupiter (a bit less maybe, but far more luminous than Saturn).
The luminosities of these points increased a bit and after a few seconds they faded. I guess these were not satellites entering in the shadow of the Earth as they were not moving. And I think that the probability for this to be two meteors falling exactly towards me is quite low...
Any suggestion this might be ?

Edit 1: @JohnHoltz correctly pointed out two errors - geosats are on the equator, not ecliptic. My brain was thinking "equator" but my fingers had other plans. I read about geosat flares quite some time ago and thought it was solstices when they could be seen but it's around equinoxes, not solstices.  I've changed my mind and think planes are the most likely answer as to what was seen.  However, I do stand by the statement that geosat flares end due to the Earth's shadow. They are unlike other sat flares in this regard according to CalSky which I've always trusted above all other sources for info like this. The link is in the first sentence of my original post below. End of edit.
Edit 2: (by @uhoh) answers to How bright are geostationary satellites due to reflected sunlight? explain that flares from GEO can be bright enough to be visible, and when you look at how big those "mirrors" are it's not hard to imagine!
The heat radiators are dark in thermal IR but mirror-like in the visible to prevent absorbing heat from sunlight. The solar panels are larger but have fairly low reflectivity.

I'm 90% sure that you saw geostationary satellites catching the Sun just right causing a flare.  It's the right time of the year for them to be seen, too. Jupiter is sitting on the ecliptic at the moment which is where geosats are parked in clusters (the reason there were two dots visible at 1-2 degrees in separation). Their brightness increased because they were moving towards opposition (reflecting more sunlight), then faded because of the Earth's shadow exactly as you thought.  Geosats move slowly against the background stars since their job is to remain over the same point on the Earth; this is how people can get satellite TV with satellite dishes on their rooftops pointing at the same spot all the time. It'd be hard to notice their motion without keeping a steady eye on them for 15 mins or so.
Your description matches perfectly with a geosat flare and I'm a bit jealous because I haven't caught one yet!  Congratulations, you witnessed an event that isn't often visible to the naked eye!  They show up in my astropics all the time, but I have yet to see one with only my eyes.
A comment on the video in the link:  Geosats stay in the same position over the Earth but this video makes them appear to be moving left as the stars set behind them.  I was puzzled by this until I realized the ground objects visible in the bottom of the video are slowly moving, so the time lapse video was made with a camera slowly panning the to the right.  The video shows the clusters well but with the wide angle the individual geosats in each cluster aren't visible.  I think the ones you saw were in the same grouping.


How can a Grade-11 Student avail TESS data and learn to process? As mentioned, I am a Grade-11 student. I have always been interested in Data Processing activities. This sums up my experience of working on Asteroid hunt Campaign under IASC (International Asteroid Search Campaign). I yielded more than 22 Preliminary discoveries at once. Now I wish, to hunt for KBO (Kuiper belt objects) and Exoplanets. Definitely, hunting exoplanets is not a kids' task, but I possess the ability to learn and process and render astronomical data under guidance! I am ready to do everything. I am more inclined towards searching Outer-Worlds than the inner objects! Mentor-ship in this regard will be gratefully appreciated. I have searched quite a few things but none helped. I'm a Zooniverse user too. And over there, project slots were filled by other Citizen-Scientists!

I'll use the TESS satellite as an example, which you might use to look at transiting exoplanets for example.

Go to https://heasarc.gsfc.nasa.gov/docs/tess/data-access.html this gives you a list of links to other useful TESS websites, data access and analysis tools. There are links here to get raw and processed light curves or even to get light curves for stars with known exoplanets. This page also explains what data are available.
What I want to do is look at some lightcurves. So you go to the MAST portal. First thing is to decide what target you want to find. There's a box at the top where you can type something in. Not every star in the sky will have data. There are limitations to the coverage and the types of stars for which data are acquired and you probably need to familiarise yourself with that if you are doing something systematic.
So I try HD 36705 (AB Dor, this is a star I know - about 6th magnitude in the constellation of Doradus. Up pops an image of the sky (on the right of the screen). HD 36705 is the bright star in the middle. In the centre panel you see a list of observations by whatever satellites are in the archive (not just TESS, but IUE, Kepler etc.) This star has lots of TESS observations listed. Each row is a dataset. You want the TESS lightcurves, which are some way down. You can identify them because they are labelled as "TESS", they have a little icon next to them which indicates you can view the data (see below), they are labelled with a target name of "149248196" and you want the ones which have an Observation-ID that ends in "-s".
These are lighcurves. You can quickly view them using the internal viewer, by clicking on the little lightcurve icon.
 The light curve viewer allows you have a basic look at the data, but you need to download it to do any analysis.
Click on the download icon  and put it on your own computer. This is a zip file. Unzip it. The good stuff is here:
inflating: MAST_2020-02-05T1007/TESS/tess2018292075959-s0004-0000000149248196-0124-s/tess2018292075959-s0004-0000000149248196-0124-s_lc.fits
This is the light curve in fits format. So you need a fits viewer to do anything with it or you need to import it into your own (python) code using something that reads fits files. To get started I would download a java program called topcat. This is free and works on windows, linux or mac.
Load the light curve file into topcat. Plot "SAP_FLUX" against "TIME" and that is your basic lightcurve (see below). You can zoom in, manipulate the data, fold the data, subtract or add things to the data, lots of cool things. Plus make plots, anotate them, save them in various formats etc.

It took me 15 mins to write this answer while downloading the data and doing those plots. You can do it!


How can the observable universe shrink in a Big Rip? As far as I know, the Big Rip occurs when the scale factor reaches infinity in a finite time. This will only happen in a universe dominated by phantom energy (i.e. a universe with an equation of state of $< -1$). However, according to Wikipedia,

  A universe dominated by phantom energy is an accelerating universe, expanding at an ever-increasing rate. However, this implies that the size of the observable universe is continually shrinking; the distance to the edge of the observable universe which is moving away at the speed of light from any point moves ever closer. When the size of the observable universe becomes smaller than any particular structure, no interaction by any of the fundamental forces can occur between the most remote parts of the structure. When these interactions become impossible, the structure is "ripped apart".

On Wikipedia, there is a "[why]" tag, indicating that this wasn't explained very well.

This raises two questions. In a universe dominated by phantom energy, why would the observable universe not continue growing and approaching the cosmic event horizon? After all, its comoving distance is just $\int_0^t \frac{\mathrm{d}t}{a(t)}$, so shouldn't it just monotonically increase?

More confusingly, Wikipedia implied that the Big Rip occurs when the observable universe's radius becomes incredibly small and the fundamental forces can no longer hold any structure together, rather than when the scale factor reaches infinity. Am I right in assuming these are separate events, or do they occur at the same time?

I've found the answer to this is actually very interesting and (to me, at least) surprising — it is not just the light cone that shrinks, but the cosmic event horizon as well! To explain this, let us first look at a de Sitter universe. The scale factor $a$ of a de Sitter universe can be expressed as a function of time with:

$$a(t) = e^{kt}$$

where $k$ is some constant, depending on the specifics of our universe (in a de Sitter universe, $k = H$, the Hubble parameter). A universe dominated by phantom energy is actually pretty similar to a de Sitter universe, but the expansion rate gets exponentially faster. Thus, in a Big Rip universe $k$ is time dependent ($\dot{k} > 0$), so the scale factor can be written as:

$$a(t) = e^{k(t)t}$$

Why is this relevant? Well, the comoving distance to the event horizon is actually inversely proportional to $a$ and $k$:

$$d_{eh}(t)\propto\frac{1}{a(t)k(t)}$$

In a de Sitter universe (and our own universe, assuming $w = -1$), $k$ is equal to the Hubble constant, so the event horizon’s distance will remain constant (in a de Sitter universe, the $\dot{H} = 0$ because the deceleration parameter $q=-1$). In a Rip universe, $k$ is time dependent; as $k$ increases, the distance to the horizon decreases.

From this formula, we can see that the event horizon’s distance is inversely proportional to the scale factor; thus, if $a(t)$ reaches infinity, the event horizon’s distance will reach zero. This means that the Big Rip can be defined as either the scale factor reaching infinity, or the observable universe’s diameter reaching zero — these two events happen at the same time.


Conversion of magnitudes to Jansky and MAGPHYS? I'm a bit puzzled and not an observer, so please bear with me if I'm being stupid here.

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

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

I'm aware of:

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

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

Additionally, I have found the formula below in TOPCAT:

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

Where does the 23 in the exponent come from?

Caveat this with the fact that I am not familiar with this code.

The input asks you for the flux through the filter, but a filter has a bandpass, so if you have a 10 nm bandpass filter and you measure $1 W/m^2$, then you measured $0.1 W/m^2/nm$ (sorry, I'm a physicist so I never got used to measuring wavelengths in Hz).

As for your second question, the factor of 23 is because in cgs units a Jansky is $10^{23} erg/s/Hz/cm^{-1}$ and that TOPCAT equation is in cgs (the equation is $m_\text{AB} = -\frac{5}{2}\,log_{10}\,f_v - 48.6$) and if you are normalizing it to Janskys, you need to divide by $10^{-23}$.


Could an ejected "extra ice-giant" still be lurking in distant solar orbit? BACKGROUND

Hot Jupiters are thought to have migrated inwards, implying that another giant planet has been ejected in order to conserve the orbital momentum of those planetary systems. The number of interstellar "vagabond planets" has been estimated to be as many as the number of stars. And simulations show that if the Solar System started out with an extra gas giant next to Jupiter which was ejected, then one can end up with a very good fit with today's Solar System.

QUESTIONS

Where would such a hypothetical ejected ice giant be today? Should it be closer or further away than hot Jupiter-ejected planets because it is much smaller? Could it stay in an Oort cloud type orbit during several billion years (maybe ~15 orbits around the Milky Way) without being pulled off by passing stars? At what distance could we today observe a Uranus sized planet? Should it stay in the ecliptical plane or could its orbit have become greatly inclined when it was ejected?

Since I'm a hobbyist, I usually wait to see if someone a bit smarter wants to answer first, but I can give a couple thoughts on this.

  Hot Jupiters are thought to have migrated inwards, implying that
  another giant planet has been ejected in order to conserve the orbital
  momentum of those planetary systems.

In the article you posted (I'll pull the quote)

  The conservation of orbital momentum dictates, as planetesimals move
  from the outer disk inward, that Saturn, Uranus and Neptune must move
  outward. This process is known as the planetesimal driven migration

The conservation of orbital momentum he's talking about happens when objects cross each others orbits and interact, one gains orbital energy, the other loses energy.   That's not the only way planetary orbits can change.   Over time, there's tidal effects, orbital resonance and a sun can lose mass leading to gradual drifting or perhaps a sun can gain mass if the solar system passes through a dense dust/gas cloud.

But mostly what we're talking about here is a form of gravity assist as 2 planets get close enough that they kind of swing past each other, one moves in the other out.

More on that here:   http://science.howstuffworks.com/nomad-planet2.htm

Now, were gas giants ejected from our solar system, according to this question, somebody said no:

Is there any evidence that the Gas Giant planets in our solar system are experiencing orbital migration?

But, this article, and the article you posted that suggests otherwise, that it probobly did happen based on models, but when the solar system was young, like about 4 billion years ago, this model suggests the Solar system could have lost 1 or 2 ice giants.

http://www.space.com/13584-extra-giant-planet-solar-system.html

and we could still lose Mercury (though we probobly won't).   http://www.universetoday.com/14032/could-jupiter-wreck-the-solar-system/

  Where would such a hypothetical ejected ice giant be today? Should it
  be closer or further away than hot Jupiter-ejected planets because it
  is much smaller?

Tough call.  Using basic gravity assist mathematics, the closer to the sun the faster an object could be ejected from the solar system so a hot Jupiter probobly eject inner planets at a faster velocity on average than more distant Jupiter, but the bigger factor is the size of the star system, not the size of the planet.   The more massive the star, the faster the orbital velocities, the faster the Ejection is likely to be, even after accounting for escape velocity.

it's also important that the object being ejected is smaller than the object doing the ejecting, for example, Jupiter could pull Mercury away from the sun but Mercury couldn't pull Jupiter into the sun.   More massive wins the tug of war, but how much smaller the ejected object needs to be isn't that relevant.  Jupiter could throw a Neptune sized planet out of a solar system at almost the same speed as it could throw a Mercury sized planet out of a solar system.

Now, as to where it would be, planets aren't likely to escape a solar system at too fast a speed.   Jupiter orbits the sun at 47,000 KM/Hour and an optimal gravity assist could eject something way from Jupiter at 94,000 KM/Hour, (plus any velocity of the object's own, which would depend largely on eccentricity, but since it would cross Jupiter's path, probably wouldn't be hugely different than Jupiter's), and if you subtract escape velocity from the sun at that distance (47,000 * square root of 2), or 66.4 KM/Hour, your left with a maximum exit the solar system velocity of, roughly 28,000 KM/Hour, plus a bit - and that's maximum.  Average escape velocity would likely be less, but lets say 20,000 KM/Hour, which works out to be 1 light year every 50,000 years.  considering it left the solar system an estimated 4 billion years ago, that's 80,000 light years, which from where we are, is pretty much the far end of the Milky way and who can say how it's orbital velocity might have changed over that time as it passed by other stars, but basically, it could be anywhere in the Milky way, and, if you consider, our sun might have come from the Sagittarius dwarf galaxy, not the Milky way, it could be practically anywhere, even ejected from the Milky way.  4 billion years is a long time, even for a slow ejection.

or, it could be orbiting another sun - article on that here:  http://www.universetoday.com/94656/rogue-planets-can-find-homes-around-other-stars/

  Could it stay in an Oort cloud type orbit during several billion years
  (maybe ~15 orbits around the Milky Way) without being pulled off by passing stars?

Entirely possible though billions of years is a long time.   Passing stars usually don't pass very close, but I'm sure it's happened at least a few times in the past 4 billion years.

close enough to disrupt the inner planets - that's different and more rare.   Close enough to disrupt the Oort cloud, probobly happens every so often - but I wouldn't want to guess how often.

  At what distance could we today observe a Uranus sized planet? Should
  it stay in the ecliptical plane or could its orbit have become greatly
  inclined when it was ejected?

it depends if we know where to look.    If we saw it pass in-front of a star and we could plot it's path, then a number of light-years.   Looking into the blackness of space, much less distance, and it would depend on the planet's reflectivity too.   I wouldn't even know what to guess, but Eris, for example is 1/20th the diameter of Uranus and it's one of the larger Kuiper belt objects.   With relative ease, they could likely spot a Uranus sized object at well over 20 times the distance of Eris.   Now, if the question is, what's the furthest a Uranus sized object could be and not have been seen yet - that's hard for me to say but an object that size and at that distance would likely have it's own Trojans and I find it hard to believe it would have gone unnoticed.

as to "would it stay in an elliptical plane" - not necessarily.   A gravitational assist around a planet can alter the elliptical plane depending on the angle of approach.

hope that's not too wordy.  I can try to tidy up later.


Is there any formulae to calculate whether the sun will depress 18 degrees or more below the horizon on a certain day / location? I'm trying to code a piece of software which will calculate some times for me that would greatly help me in my day to day but none of my formulae work if the sun doesn't go at least 18 degrees below the horizon.

In this case, I have different formulae to use but I need to program the software to be able to use this other formula on these days. I have no way of determining the maximum the sun will depress below the horizon on any specific day or location.

I am not an expert in any sense of the word in astronomy and literally only know what is relevant to what I need, but I think what I am trying to ask is if it is possible to determine the if the minimum solar elevation / altitude angle is less than -18 degrees on a day / location.

I have attempted to search but was unsuccessful and am even unsure if what I am asking for is even possible.

Thanks for any help!

An approximation of the Sun’s declination for a certain date is:
sin δ = 0.39795 ⋅ cos [ 0.98563 ⋅ ( N – 173 ) ]
where N is the number of days since January 1.
That, combined with the formula given by @user21, viz.:
$ a_{min} = \lvert \phi + \delta \rvert - 90° $
will give you the dates where the Sun is at least 18° under the horizon.
(Oddly enough, a simple instrument, invented by the Greeks almost 2,000 years ago and perfected by the Muslims over the next few centuries, the astrolabe, would give you the answer almost instantly, without having to do any calculations. Below is a photo of one I made in brass.)


Is it plausible to use other stars for the proposed FOCAL mission instead of the Sun? For some time, the far-reaching and speculative idea of using the Sun as a gravitational lens has been floating around. See this and this. This would require sending a spacecraft about ~550 AU of a distance away from Earth, so the idea is not currently realistic.

But I am wondering, is there any plausibility to using other stars (like the Alpha Centauri stars)?

Is there there enough parallax for this to be useful? How far away must a satellite go to be able to manipulate parallax to its advantage? Are there any shortcomings to this idea that make the FOCAL proposition superior? And of course, are there any shortcomings to both propositions?

I can imagine there's not much freedom in where the so-called lens can be used, but maybe the limited field of view can still be useful if the satellite could roam the solar system?

I'm curious if anyone can give useful feedback or calculations.

Gravitational lensing works from anywhere beyond the focus, so in that sense, we could use any star as a gravitational lens. The problem is that the field of view is tiny. We only get any useful information from alpha centauri as a gravitational lens if the target object is almost exactly behind alpha centauri from our point of view. To look in a slightly different direction, we need to move a huge distance sideways.

In fact we do use gravitational lensing from other bodies for astronomy. We use very large clusters of galaxies to amplify the images of other galaxies which lie behind them, and we watch for brief flashes as planets and stars line up for a moment as a way to detect planets (called "microlensing").


Very bright star in the east at northern hemisphere. What is it? From some time now (few weeks as far as I can remember), there is a very bright star in the eastern sky.

I first thought it was Venus, but according to this link, Venus is in the western sky and sets 2 hours after sunset. But this (very) bright star is still up even though it's midnight (6 hours after sunset).

Does anyone know what is this star (or planet) ?

Please excuse my silly question as I know nothing about astronomy.

Thanks.

It will be very difficult to tell you without knowing where you are located.  Particularly, if you are in the northern or southern hemisphere.  At this time of year, most of the brightest stars are in the sky.  Also where in the sky, the particular star that you are wondering about is.  And any other constellations that you can identify.

For stars rising this time of year, particularly bright ones you might be thinking of Sirius.  Which would be near the constellation Orion below the hourglass shape of the it (http://starchart.polaris.net/chart.php?constel=ori&layoutType=Key&layoutOri=Portrait&project=SANSONS&type=PNGC).

Another possibility would be Aldebaran, it is located in a "V" shape of stars that make up the constellation Taurus.
http://starchart.polaris.net/chart.php?constel=tau&layoutType=Key&layoutOri=Portrait&project=SANSONS&type=PNGC

To find out I would take a look at some of the online star charts that are available for your time and location.  That will give you a good idea of what you are seeing.  Here is a list of some online:

http://www.skyandtelescope.com/resources/internet/3305011.html?page=1&c=y


How will they know when to start taking the picture of the black hole at the center of the Milky Way? @csm's answer to Why not take a picture of a closer black hole? points out that it's necessary for the supermassive black hole at the center of a galaxy to be actively feeding for it to generate a radio-bright accretion disk that we can image. M87 is always feeding, but our own black hole only nibbles occasionally as a dust cloud passes by.
Question: How will they know when to start taking the picture of the black hole at the center of the Milky Way? Are there bits of food being tracked and they'll have all the EHT's telescopes ready when accretion begins? Or is it long enough that once it starts they can shuffle observing time around and still collect sufficient data?
For background see

What is it exactly about these flares of infrared light from Sgr A* that "confirms" it is a supermassive black hole?
How did they make a video of the center of the galaxy, and what is it exactly that's flashing there?

The ESA video ESOcast 173: First Successful Test of Einstein’s General Relativity Near Supermassive Black Hole includes a clip of images of stars at the center of our galaxy orbiting around SgrA*, a presumed supermassive black hole.
GIF made from video at around 02:50:

Six annotated frames from GIF highlighting the flashing that I'm seeing.

The Milky Way's central supermassive black hole (SMBH) is feeding, albeit at a very low level. Radio emission from the accretion disk (and/or weak jets) is responsible for the long-lived "Sgr A*" radio source.
Here is a paper from 2000 (Falcke et al.) arguing that VLBI (as used by the Event Horizon Telescope) should be able to image the "black hole shadow", based on the known sub-mm and mm-wave emission. And in fact the EHT has been observing the Milky Way's SMBH.
As I understand it, the real reason we haven't seen a formal, published detection of the Milky Way's SMBH by the EHT is that its emission is highly variable on short time scales (e.g., minutes to hours). In the case of M87's SMBH, the variability of the (sub-mm and mm-wave) emission is slow (days to weeks), so they could combine observations taken over several hours and two nights in April of 2017 under the assumption that it was all of the same static configuration. Figuring out how to properly account for the short-term variability of the Milky Way's SMBH emission is much more difficult, which is why the (relatively) easier case of M87 was solved and published first.
See also Rob Jeffries' answer to this physics.stackexchange question.
Edited to add: Unfortunately, I don't think there's any validity to the idea that we can track incoming "food" and predict future accretion flares for the Sgr A* SMBH with any useful accuracy. There was some excitement a few years ago when a group reported detection of an apparent gas cloud ("G2") on an orbit that would take it in to about 2000 Schwarzschild radii from the SMBH at pericenter (in 2014), possibly allowing it to be tidally shredded and increasing the accretion rate. But as a review article published in 2013 pointed out, "The actual free-fall time scale from ∼2000 $R_s$ is roughly one month and the viscous time scale could be anywhere between months up to hundred years depending on the viscosity parameter $\alpha$."
And in fact the actual pericenter passage produced... nothing much at all. There's a discussion of the "fizzle" here: "with the majority of the simulation parameters used, only 3–21% of the material Sgr A* accreted from 0–5 years after periapsis is from the cloud".
So in the one case where potential "food" was identified and tracked, one couldn't be sure in advance whether the possible increased accretion would happen on timescales of months to years, and so far nothing significant has happened. I very much doubt the EHT team is basing their observing schedule on this kind of thing.


Eccentricity and Inclination of a Star I understand Planetary Eccentricity and Inclination.  However when it applies to a star, does the centre or ecliptic plane refer to the Sun or the Galactic Centre?

The star Sirius for example, 136.336 ± 0.040° inclination, the angle at which it is inclined to Sun, Earth or Galactic Centre?

The star Sirius has an Eccentricity of  0.59142 ± 0.00037.  The only thing the star orbits is the Galactic Centre (Sagittarius A*) so I guess that is the centre?

Sirius is a binary system, composed of a main sequence star (Sirius A) and a white dwarf (Sirius B). The two orbit each other with a period of about 50 years, and the eccentricity given on Wikipedia is the eccentricity of the stars' orbits around one another. The inclination, according to Bond et al. 2017 - the paper Wikipedia references for both numbers - is described as the inclination of the "relative visual orbit", leading me to believe that the inclination is defined based on a plane relative to the point of view of an observer on Earth.

Something it's important to realize is that stars don't follow Keplerian orbits, because of the non-point-like distribution of matter in the galaxy; the motion of a star around a galaxy is therefore somewhat more complicated than the motion of a star in a binary system.


Why aren't new stars in Earth's relative proximity constantly discovered? Okay, bear with me. In layman's terms:

The estimates vary, but on a clear night, away from big cities, one can see up to 5000 stars in the sky. All those are in the Milky Way, but they're hardly equidistant from Earth, which would mean that they revolve around the galactic center at different orbits. According to some sources, some are positioned very closely to us while others hang out as far as 1000 light year away.

Now by astronomical standards, 5000 stars is not a great number. It's still pretty impressive, though.

Some stars are larger, and others are smaller. It stands to reason that some of them might be "hiding" behind others. Because they all orbit the galactic center at different speeds, using different orbits, wouldn't it follow that from time to time, over the centuries, decades, or even years (months?) a "new" star would emerge from behind another one? Shouldn't this be happening often; or are they so sparsely distributed around the galaxy that no "new" stars have appeared out there over the past three thousand years or so?

To the naked eye, the answer is almost certainly no because of the enormously slow movement of stars across the sky and because 5,000 stars may be a lot but only a tiny percentage of the sky is covered by visible stars.

To Hubble, which can see perhaps tens of millions of stars, maybe more, the link here, has a picture of two stars that are approaching crossing each other's path from our point of view.  With a big enough telescope it probably happens from time to time, though I wouldn't want to try to calculate how often, but to the naked eye, I'm comfortable saying no, in fact, it was often assumed that stars didn't move and were fixed in the sky (contrary to what Macrobius said).  That was the popular point of view prior to Halley's observation.

There was also Tycho Brahe's "De Nova Stella" or "new star" which we now know to be a super-nova, and that was quite the surprise at the time.   Nobody thought a new star could appear because they thought the stars were fixed and permanent, but that appearance wasn't by the method you suggest.

Consider how small stars are from our point of view.    Alpha Centauri A, the larger one, it's about 1.7 million KM across and it's about 4.3 light years away, or, 41 trillion KM.   It's diameter is 23 million times smaller than it's distance from us.  That's the equivalent of looking at a golf ball from nearly 200 miles away.    Now if you scatter 5000 golf balls each 200 miles away across the sky and you let them move around very very slowly, how often do you think one golf ball passes in-front of another?   Not very often.   Granted, that's not quite right as the atmosphere spreads stars out a bit so each golf ball is smudged to maybe the size of a basketball, but they almost never pass infront of one another, at least, not if we only take into account 5,000 visible stars.

Now, binary stars, it happens more often if they are lined up right, then they can pass infront of each other and this has certainly been observed by telescope but not to the Naked Eye, we can't visibly tell that Alpha Centauri is 2 stars (3 with the more distant Proxima but that can't be seen by the eye).  They are on average about a billion miles apart but that can't be seen by the naked eye.   It was observed by telescope in 1689.

There simply aren't enough visible stars (and taking HDE's point that most of the 5,000 visible stars weren't cataloged until recently), there's essentially zero chance that it was ever observed that a star appeared "new" by passing from behind another star.

Using Hubble, it can happen, but not to human sight.


More recent data and simulations of "Milkomeda", the collision of the Milky Way and Andromeda galaxies? The Space.com headline Hubble Telescope Spots Two Galaxies in a Doomed (but Dazzling) Dance; The galaxies will ultimately crash into each other was probably overstated as seems to be policy in some popular press sites. The galaxies are not "doomed". However the paragraph below is interesting:

  Our Milky Way, for example, is on an inevitable collision course with the neighboring behemoth galaxy — Andromeda. Individual star systems like ours will likely be largely undisrupted, but distant observers will see the two galaxies gradually become one in some four billion years. ESA nicknames this new merged galaxy "Milkomeda."

Question: How much more is predicted now about the "upcoming" collision and possible merger of the Milky Way and Andromeda galaxies than has already been described in this answer's citation of a 2012 Astrobites review The Fate of the Milky Way? What measurements (if any) have contributed to this additional level of prediction?

  The current data could be improved by future additional HST observations. Also, soon it will be possible to compare them with independent water maser measurements for individual sources in M31 (see this Letter), which might allow measurement of other cool effects, such as the M31 proper motion rotation, and the increase in Andromeda’s apparent size due to its motion towards us.

Since the second data release (DR2) of the European Space Agency's Gaia mission there has been a revolution in astrometry, including measuring the motion of the Andromeda Galaxy.

On February this year van der Marel et al (also in ArXiv) published interesting results on that matter by using Gaia's DR2 measurements. The results reveal that the collision is going to happen 600 million years later than the previous estimate (in 4.5 Gyr instead of 3.9 Gyr). Also Andromeda appears to have more tangential motion than previously thought and thus "the galaxy is likely to deliver more of a glancing blow to the Milky Way than a head-on collision".

Also interesting, M33 (Triangulum galaxy) is going to make its first infall towards Andromeda and might interact gravitationally quite a lot with it. The absence of stellar streams between Triangulum and Andromeda show that this is the first time they are going to meet each other.

Thanks to Gaia the measurements are now so precise that for the first time we are able to notice even the minute rotation of both Triangulum and Andromeda galaxies astrometrically (and not only by using doppler shifts).


Are we still going to have rainbows if Sun is replaced by another star? Rainbow is a property of light splitting due to the water molecule present in the clouds and is related to the light emitted by our Sun. Our sun is a yellow star. Now if we replace sun by a blue giant star or red giant star, or any other star of different color, are we still going to have rainbows? If yes is it still going to compose of seven colors? What kind of possible change will occurs, if any?

Rainbows would lack most blue, and some green for red stars.
For a blue star, the blue part of the rainbow would be more intense.

For more complex colors, the rainbow may show some gaps. A rainbow is essentially a spectrum of that star light portion, which is visible to our eyes, and to which the atmosphere is transparent.

Stars vary in brightness. A blue giant would be large and glaring, a red dwarf faint.

Colors in the rainbow would be blurred, hence closer to white, for large stars, and sharper, more distinct, for small stars, according to the angular size of the respective light source.

... This all assumes, that there is still rain. With small, red stars, it would get too cold for rain. With large blue stars, Earth would heat up too much.
To adjust for these effects, the distance to the star would need to be modified.
And of course, the length of a year, and the orbital velocity may change.
This could then cause different tides, changes in volcanism, etc.

Exchanging the star could cause various other effects, too, other polar lights, effects to the ionosphere, the ozone layer, atmospheric erosion, more...


Why does gravity make everything round? I am studying ways to make an anti-gravity machine (!!). I believe it can be done, or theoretically be done using gravity itself as a "force" that pushes outwards instead of inwards.
A similar question was asked here, which speaks about why objects in the universe are round, which as stated, is due to Gravity.
I understand Gravity is the strongest, most consistent "force" in the universe and to traverse it we have to push against it - jets, propulsion, etc.
Never the less, I am perplexed why gravity's forces make objects round - eventually or over time. Why can't it turn a planet into a cube for example? Why can't Gravity make objects flat, and perhaps round at the perimeter?
If I can answer this question, I am sure I can find a way to "bend" Gravity for my invention.

I am studying ways to make an anti-gravity machine (!!).  I believe it can be done, or theoretically be done using gravity itself as a "force" that pushes outwards instead of inwards.

It doesn't work that way.  No one has found a repulsive version of gravity and there is no evidence this is even possible.

And in science, we go by the evidence.

  A similar question was asked here, which speaks about why objects in the universe are round, which as stated, is due to Gravity.

Which the question's answer explains by the attractive nature of gravity.  No evidence (that word again) of repulsion.

  I understand Gravity is the strongest, most consistent "force" in the universe

You understand wrong.

It is considered the weakest of the fundamental forces.  This Physics SE answer explains why.

It appears to be strong only because on average, over a large scale, matter is generally neutrally charged, whereas mass (and energy) just gets bigger the more space you have (a little simplistic, but that's the general idea).

As for "consistent", well all the forces behave consistently or we would not be able to model them with mathematics.

  and to traverse it we have to push against it - jets, propulsion, etc.

More evidence for no repulsion !

  Never the less, I am perplexed why gravity's forces make objects round - eventually or over time. Why can't it turn a planet into a cube for example? Why can't Gravity make objects flat, and perhaps round at the perimeter ?

Because these are unstable configurations.  Even if they could form from natural processes (which is unlikely as I'll explain) the resulting forces and stresses (for large objects) would be too great to prevent collapse into a more stable shape.

Now smaller objects can form less rounded shapes - small asteroids can be all sorts of shapes.  Only their relatively small mass (and hence small gravity) allows them to resist compression into "nicer" round shapes.

But larger objects have lots of gravity simply because the have lots of mass.  The materials they are made of will not withstand compressive forces on them and they will form roughly round shapes, because that's the most resistive configuration.

So why more-or-less spherical ?

Because these larger objects tend to form in a way that builds up material equally all over the surface, so there is some tendency to be spherically symmetrical just from the way material accretes as it grows.

However once they get really big (planets, larger moons and so on) the forces on the materials will be so great that they will as a general rule, behave a little like fluids.  And fluids will end up in that nice spherical shape as they lack the rigidity to resist compression.  Again a little simplistic, but this gets really complicated quite quickly.  They key here is to stop thinking of e.g. rocky planets as rigid objects.  Maybe on a timescale of seconds they are, but on astronomical timescales they're not so rigid.

So for larger objects, more-or-less spherical is inevitable simply because they're large.

  If I can answer this question, I am sure I can find a way to "bend" Gravity for my invention.

No anti-gravity to be seen here.

Really, learn the basics of physics if your interested and hopefully you'll find that not only is it interesting, but it can explain things without the need to invent other explanations.  Usually. :-)


How do we detect if a planet has a liquid core? In case of Earth, we have many hints about the internal structure of our planet, from what I know, the most important of which is analyzing seismic waves.

Do we have any instruments to detect if other solid planets, such as Mars or Venus, have solid or liquid core? Is it possible using remote sensing?

A good way to show how this is achieved is to show you a case study, below I will detail how we came to make the theory that Mercury has a liquid core.

It is first important to note that we are never definite of the properties but we are able to use the data to infer and create a theory that we believe to be an accurate representation of the truth.

The MESSENGER spacecraft was launched for the purpose of analyzing MErcuries Surface, Space ENvironment, GEochemistry and Ranging (hence MESSENGER).

Messenger had a number of pieces of equipment on board to allow it to conduct it's experiments, as detailed in the below image:

Mercury Laser Altimeter

So how does the MLA aide us in determining if Mercury has a liquid core? Well the purpose of the MLA is to:

  track the planet’s slight, forced libration – a wobble about its spin axis – which will tell researchers about the state of Mercury’s core.

How does a Laser Altimeter work?

  A laser altimeter is operated from a plane, a helicopter or a satellite. It determines the distance to the [Planet's] surface by measuring the time-of-flight of a short flash of infrared laser radiation. The instrument emits laser pulses which travel to the surface, where they are reflected. Part of the reflected radiation returns to the laser altimeter, is detected, and stops a time counter which was started when the pulse was sent out. The distance is then easily calculated by taking the speed of light into consideration.

Using this technology one can build up a 3D visualization of a planets surface

What scientists discovered from this mapping is that:

  The planet's crust is thicker in low latitudes and thinner at the poles, a distribution that suggests the planet could have a liquid outer core. Its core is also large relative to the planet, comprising 85 percent of the planet's radius, much more than Earth.

How did they come to this conclusion? This paper explains in depth. (The full version is free, but requires signing up to the site, signing up is also free and has no costs and has the ability for you to turn off receiving emails and newsletters from them)

The most notable part is this excerpt and image:

  Resulting moment of inertia parameters for more than 1 million Monte Carlo models that include a silicate crust and mantle, as well as an Fe-rich core that may contain solid and liquid layers are shown in

(A) Outer radius of Mercury’s liquid core. (B) Average density of the solid shell that overlies the liquid core

Conclusion

Essentially, specific readings taken by a Laser Altimeter allow us to map out the three dimensional shape of a planets surface. Analysis of the shape shows us that the crust is thinner at the poles which is likely due to a liquid core.

On top of this, Mercury is believed to have a small magnetic field (1% of our own) and this might be due to a liquid outer core circulating a solid inner core, acting like a dynamo. Though there is still much speculation given over what causes the magnetic field so this is unreliable to use for a solid footing.

Aside from that there is the logical thought process that larger planets are capable of creating huge amounts of heat and pressure at their cores which create a molten core, or at least a molten layer. This is why there is much skepticism over Mercury because it is such a small planet that its core should have solidified long ago.


Is T Tauri a compulsory stage for stellar evolution? Do all stars go from Protostar stage to T Tauri and then to Main Sequence or is T-Tauri stage only for certain stars?  If only certain stars, is that those that are less than 3 solar masses which i see mentioned?  (https://en.wikipedia.org/wiki/T_Tauri_star)

Yes, only certain stars that are less than about 3 solar masses were T Tauri stars. Here are some reference and quotes that I saw to answer this question. There are all review papers.

  Our planetary system was formed in early stages of the suns evolution. What the sun was like at that time can be learned by observing young solar-mass stars.

https://link.springer.com/article/10.1023/B:ASYS.0000003267.35552.f7

  T Tauri stars are newly formed low-mass stars that have recently become visible in the optical range,

https://www.annualreviews.org/doi/abs/10.1146/annurev.aa.27.090189.002031

  This review presents the main aspects of the body of knowledge on T Tauri variable stars, which are presumed to be young, gravitationally contracting low-mass stars.

http://iopscience.iop.org/article/10.1088/0034-4885/47/2/001


What does this "web on the surface of the Sun" image reflect? What does 789 nm show us? What is this web on the surface of the Sun? has got me thinking.

This is probably not a normal color photo.
The cooler areas are really dark!

update: Comments point out

The NSO press release says the passband is 789 nm.
"Images have been processed to remove noise and enhance the visibility (contrast) of small-scale (magnetic) features while maintaining their shape. The movie frames have been smoothed to remove noise."

Question: What does this image at 789 nm show us? Are we looking at changes in blackbody radiation or is there a spectral feature that tracks something more specific? Does intensity in the image actually track temperature directly, or reflect it in a more subtle way?

There are two videos linked in Phil Plait's Bad Astronomy article

DKIST first light high-resolution video of solar granules
DKIST First light video of solar granulation (wide angle).

From The Universe Today's This is the Highest Resolution Image Ever Taken of the Surface of the Sun

Well let me take a stab at it. The line in question is said to be a probe of an Fe XI line, that is iron atoms with 10 electrons removed.

You do not get such ions in the solar photosphere, it is far too cool; the radiation from the photosphere is probably a pseudo-continuum at that wavelength.

However, much hotter material in the chromosphere and corona may contain Fe XI ions. Plasma in these structures could absorb light from the underlying photosphere, if it were dense enough, or more likely, if you point your camera above the limb of the photosphere, you could see optically thin coronal structures emitting light at this wavelength.

Why is this important? Normally you would need to use EUV or X-ray emission to probe the coronal plasma, but the imaging quality is not so good. You can do much better at optical wavelengths, but there are precious few diagnostic lines that can be used.

Edit: Indeed this presentation on DKIST coronal diagnostics appears to confirm this hunch and also mentions the additional possibilities that polarimetry of optical/IR spectral lines offer in investigating coronal magnetic structures. The Fe XI line in question is sensitive to Zeeman splitting, offering the opportunity to prove the strength and direction of magnetic fields where the line is formed. The Zeeman effect is proportional to the square of the wavelength, so the more usual X-ray and EUV diagnostics just can't do that.

On p.2 of that presentation it clearly states that these kinds of measurements are limited to coronagraphic images taken off the limb of the Sun. Pointing at the photosphere will not yield useful information for these diagnostics, because the weak light from the chromosphere and corona is swamped by the normal photospheric emission.

Sextus Empiricus has pointed to a press release, which although unclear in its exact meaning, implies that the dark features around the granulation may offer sufficient contrast to see the much thinner and hotter chromospheric gas directly above, producing bright points in the dark lanes. Observing these through different polarising filters could then reveal details about the structure and strength of the magnetic field.

However, on further research, these photospheric bright points are nothing of the sort. They are concentrated magnetic flux tubes that allow a deeper (and therefore hotter and brighter) view of the Sun. The typical temperatures of the deeper material is still  only around $10^4$ K (e.g. Shelyag et al. 2010) and nowhere near enough to excite Fe XI.

The picture below, which comes from this site, referred to by Sextus Empiricus, shows the situation. A bundle of magnetic flux "hollows out" a passage further into the solar interior and the light we see comes from deeper, hotter, brighter regions. Nothing to do with coronal emission.

I arrive at the conclusion that this image (which was taken for science verification purposes) was just using the Fe XI filter as a narrowband filter. Everything we are seeing in the image is essentially continuum from the photosphere at temperatures between 4000K and $\sim 10^4$K. The contrast therefore just arises from the differing monochromatic intensity of material at different temperatures.


The dominance of Gravity vs Strength and its dependency on crater size So I am trying to write a research paper (about 3700 words) for high school about the topic "The dominance of Gravity vs Strength and its dependency on crater size" or something similar.

I have found the following information which I am struggling to understand and has been holding me back from continuing the research : (Can someone explain what dominance is and how can I approach reading more about this topic?)

Making a crater requires both material strength of the target and gravity to be overcome.
which of these dominates depends on the size of the crater

$$ \rho g d = \frac{1}{5} \rho g D = Y $$

Where Y is the yield strength, ρ is density of the target material, d is the crater's depth and D is its diameter.

A related graph to this topic is this (Holsapple 1993) :

So please, can someone explain what all this means in simple terms ?

I guess the point is that there are two ways to calculate how big a crater a given impact will create. One is based on calculating how much rock the impact energy can shatter so the main constraint is the strength of the rock) the other on calculating how much it can lift (so the main constraint is the weight of the rock). For small impacts the first is the main consideration ("dominates") for large impact's its the second and the formula identifies the point around which it changes over.


South or North Celestial Pole I'm really confused with this one. Almost all documentation says you must start with your telescope pointing to the Celestial Pole and for the lucky ones who live in the north you can use Polaris.

But I live in the Southern hemisphere (Chile), so besides no Polaris for me, the real question is: Should I start with the telescope pointing to the South or North Celestial Pole?

I have had a month of frustration trying to align.

I have looked into other answers here and other sites, but no one explicitly says where to point it and how to find it.

If your telescope has an equatorial mount, its polar axis should point toward the south celestial pole (SCP).
The bad news is that all the stars within 5 degrees of the SCP are magnitude 5 or fainter.
The good news is that visual observing does not require precise polar alignment.
Even if you are off by 5 degrees, the mount's drive should keep most targets in view at least 10 times as long as if the drive were not running.

You don't have to aim the telescope's optical tube at the pole, only the polar axis of its mount.
Find your geographic latitude, set the polar axis to the same angle above the horizontal, and point it south.
The skymaps.com Evening Sky Map,
Southern Edition, can help you get your bearings.
The SCP makes a right triangle with Crux and Canopus,
both of which are fairly high in the southern evening sky in March, April, and May.
If you really need a precise alignment, this
Australian article
has a finder chart for
$\sigma$ Oct
and describes the drift method.
This article in Spanish may also be useful.


How does a Bahtinov mask work? For focusing the image of a telescope, one can use a Bahtinov mask. How does this mask work, and how did Mr Bahtinov get the idea to cut such a peculiar shape? Is it possible for me to design my own focusing masks?

The idea is to create a specific diffraction pattern so that you know if you image is in focus.  The pattern is so that you end up with 3 lines crossing the source.

The mask consists of three separate grids, positioned in such a way that the grids produce three angled diffraction spikes at the focal plane of the instrument for each bright image element (star). As the instrument's focus is changed the central spike appears to move from one side of the star to the other. In reality, all three spikes move but the central spike moves in the opposite direction to the two spikes forming the 'X'. Optimum focus is achieved when the middle spike is centered on the star and symmetrically positioned between the other two spikes. Small deviations from optimal focus are easily visible.

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

You can construct a mask with some materials:

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

There is also a Hartmann Mask that splits the object into multiple when not in focus.  This seems a little easier to construct.

http://www.cloudynights.com/item.php?item_id=518


What natural mechanisms could lead to the unlikely case of the same rogue asteroid or planet passing through our solar system twice? I find he following answer to the question What's the soonest Oumuamua could return? unsatisfying and oversimplifying and the last five words "Therefore it will never return." unsupported at a minimum.

  Oumuamua as an object is remarkable, because it has a positive net energy, which means it is not bound to the gravitational well of our sun. Therefore it will never return.

So I'd like to ask: *What natural mechanisms could lead to the unlikely case of a rogue asteroid or planet passing through our solar system twice?"

Stick to accepted scientific facts and theories
Let's give an ample time limit of 10 billion years between visits.
For the purposes of this question, the solar system includes the Kuiper belt (50AU) but not the Oort cloud
Don't include intervention by artificial means or intervention (divine or otherwise)

Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is (Douglas Adams)

The wikipedia page notes that 'Oumuamua "has circulated the Milky Way several times" so it was unlikely to have formed around any nearby star. While it would take 'Oumuamua  600,000 years to travel here from a local star, the time before it can be expected to have another stellar encounter would be much greater, hundreds of millions of years. And the next star that it meets is unlikely to be a currently local star, it could be almost any star in the galaxy.

Once an interstellar object has left the solar system it will travel in a roughly straight line, (slowly curving with the general gravity of the galaxy). For it to return to the Earth it would have to have a gravitational encounter with another star (or more likely a series of encounters) that send it back towards the solar system. Even given a vast mind-boggling long period of time like 10 billion years the chance of this happening is minute.

Such an object will probably orbit the galaxy several times between stellar encounters, perhaps it has an encounter every 250 million years. And it is equally likely to encounter any of the disc stars (conservatively 50 billion stars). This means that the mean time between encounters of the sun is on the order of ten quintillion years.

It isn't impossible.  It is quite easy to design a free-return flight to the moon.   There is no reason why a free-return flight to another star can't be done. However, space is big "therefore ʻOumuamua will never return".


Any live moon stream? I searched Google for a while and of so many available cameras, I couldn't find a moon live stream video camera. Does anyone know any free website that offers this? Where can you see live the moon right now?

Since the moon isn't visible all the time from any single camera location on Earth, such a website would have to be an aggregate of many different cameras around the world, displaying a single camera based on current time and weather conditions. Even in this scenario, there's a chance all the cameras will be obscured by clouds and there will be no view of the moon.

Such a website doesn't exist that I know of. Perhaps it would be an interesting project if you're interested in such a thing.


Currently proposed ways to distinguish between a really big planet and a really small star The Next Big Future article Rogue Exoplanet 12.7 times bigger than Jupiter is 20 light years away

Astronomers using the National Science Foundation’s Karl G. Jansky Very Large Array (VLA) have made the first radio-telescope detection of a planetary-mass object beyond our Solar System. They found a rogue planet 12.7 times the mass of Jupiter twenty light years from earth.
It is at the boundary of Jovian objects and Brown dwarf stars.
[...]
Last year, an independent team of scientists discovered that SIMP J01365663+0933473 was part of a very young group of stars. Its young age meant that it was in fact so much less massive that it could be a free-floating planet — only 12.7 times more massive than Jupiter, with a radius 1.22 times that of Jupiter. At 200 million years old and 20 light-years from Earth, the object has a surface temperature of about 825 degrees Celsius, or more than 1500 degrees Fahrenheit. By comparison, the Sun’s surface temperature is about 5,500 degrees Celsius.
The difference between a gas giant planet and a brown dwarf remains hotly debated among astronomers, but one rule of thumb that astronomers use is the mass below which deuterium fusion ceases, known as the “deuterium-burning limit”, around 13 Jupiter masses.

In the spirit of

Do astronomers generally agree that the distinction between comets and astroids is not so clear?
Is this object an asteroid or comet, and how can it produce so many tails?

and the relatively recent reclassification of Pluto as a dwarf planet and not a planet proper, I'd like to ask what are the ways that have are currently proposed and given consideration to distinguish between a really big planet and a really small star?
Is the "rule of thumb" proposed in the block quote the leading contender or are there other definitions that receive a similar amount of support

The dividing line between star and brown dwarf is the mass at which hydrogen fusion via the pp-chain occurs at its core. A brown dwarf below this mass limit will never reach a sustainable equilibrium due to this reaction and will continue to cool as it gets older. A star will reach the long-lived main sequence.

The dividing line between planet and brown dwarf is much less well defined. Some want the definition to be mass-based, and there is a mass-based dividing line at about 13 Jupiter masses between brown dwarfs that fuse deuterium in their interiors (which provides only a brief hiatus in their cooling) and "planets" that do not.

Others argue that any definition of a planets should be based on whether they formed around a star, but this is problematic, because such objects may then be disrupted or escape and also it is not clear where a binary brown dwarf becomes a star plus planet.

The reason that some campaign for a different definition is that the composition and structure of a planet formed around a star could be different to that of a planet that collapses from a small gas cloud. According to the "core accretion model" for gas giant formation, the core of such a planet would consist of solid ice or rocky material, whereas it would be gas all the way down in a less massive version of a brown dwarf. However, the waters are further muddied by ideas that large planets can also form by direct collapse  in the disc of a forming planetary system.


Convert a Decimal into RA or Dec http://exoplanets.eu/ displays R.A. on screen in hh:mm:ss format but when you export the table, it gives the R.A. as a decimal.  Likewise for Declination.

How can I turn for example 24.35417 (WASP-18) into 01:37:25.0 ?  And likewise for Declination -45.67778 to -45:40:40?  I'm looking for a method, not a website that will do it for me.

Either a way to convert between formats or how do I export in non-decimal format?

Thanks

DecRA is the decimal right ascension
RAh, RAm, RAs are the hms form

$${\rm DecRA} = {\rm RAh}\times 15.0 + {\rm RAm}/4.0 + {\rm RAs}/240.0$$

$$ {\rm RAh} = {\rm INT}({\rm DecRA}/15.0) $$

$${\rm RAm} = {\rm INT}(({\rm DecRA}-{\rm RAh}\times 15.0)\times 4.0)$$

$${\rm RAs} = ({\rm DecRA}-{\rm Rah}\times 15.0 - {\rm RAm}/4.0)\times 240.0$$

where INT is the operation that truncates to an integer

e.g. DecRA=24.35417

$${\rm RAh} = {\rm INT}(24.35417/15.0) = 1$$

$${\rm RAm} = {\rm INT}((24.35417 - 1\times 15.0)\times 4.0) = 37$$

$${\rm RAs} = (24.35417-1\times 15.0 - 37/4.0)\times 240.0 = 25.00$$

DecDE is the decimal declination

DEd, DEm, DEs are the dms form

posneg is -1.0 for a position below the celestial equator (negative declination) and +1.0 for above the equator

$${\rm DecDE} = {\rm DEd} + {\rm posneg}\times {\rm DEm}/60.0 + {\rm posneg}\times {\rm DEs}/3600.0$$

$${\rm DEd} = {\rm INT}({\rm DecDE})$$

$${\rm DEm} = {\rm INT}(({\rm DecDE} - {\rm posneg}\times {\rm DEd})\times 60.0\times {\rm posneg})$$

$${\rm DEs} = ({\rm DecDE} - {\rm posneg}\times {\rm DEd} - {\rm posneg}\times {\rm DEm}/60.0)\times 3600.0\times {\rm posneg}$$

e.g. DecDE = -45.67778   posneg=-1.0

$${\rm DEd} = {\rm INT}(-45.67778) = -45$$

$${\rm DEm} = {\rm INT}((-45.67778 - (-1.0)\times 45)\times 60.0\times (-1.0)) = 40$$

$${\rm DEs} = (-45.67778 - (-1.0)\times 45 - (-1.0)\times 40/60.0)\times 3600\times (-1.0) = 40.0$$


Do solstices and equinoxes shift over time? Right now, I think the winter solstice is around December 21. Has it always been around this date constantly?

Since the earth's axis's 23.45° tilt will be on the exact opposite side in 13000 years (or was 13000 years ago) due to precession, it'd seem like the seasons would flip every 13000 years. So, +/- 13000 years from now, it would've turned summer around December 21 in the Northern Hemisphere, right?

The Gregorian Calendar was created so that annual astronomical events, specifically the vernal equinox (used to determine when Easter is), would on average keep their places in the calendar year over time.  It is the best official approximation to the definition of the tropical year, which is defined as "the length of time that the Sun takes to return to the same position in the cycle of seasons".  Because this calendar describes 97 leap years out of every 400 years, it defines the average year as exactly 365.2425 solar days, or exactly 365 days, 5 hours, 49 minutes, and 12 seconds.

However, the mean tropical year is in reality about 365 days, 5 hours, 48 minutes, and 45 seconds, or 27 seconds shorter.

Because the Gregorian Calendar is based on the tropical year, the calendar dates of the year will keep up with the solstices and equinoxes, and thus the seasons.  If this calendar were exactly the length of the tropical year, then the calendar would keep the vernal (northward) equinox around March 20th for all time.

But because of the slight inaccuracy, it will take about 3,200 years (60 s/min * 60 min/hr * 24 hr/day / 27 s/year) for these 27 seconds to add up every year to be 1 full day, and that will result in the solstices and equinoxes marching backwards in the calendar by 1 day every 3,200 years or so, depending on the accuracy of the 27 seconds difference.  This very slow shift is due to the slight inaccuracy in the Gregorian calendar in, on average, matching the tropical year, not because of the precession of the equinoxes.

3,200 years from now, if the Gregorian Calendar is still used, the date of the vernal (northward) equinox will be on average one day earlier in March.  The precession of the equinoxes will still occur, so the Earth's axis tilt will be significantly different from today.  The Earth will be at a noticeably different position with respect to the Sun on the vernal (northward) equinox from where it is today, in 2014, on the vernal (northward) equinox, but it will still be in March.

This inaccuracy may very slowly increase over time, because according to the same Wikipedia page for the tropical year, the tropical year is very slowly getting shorter, and the mean solar day is even more slowly getting longer.  But for 10,000 years to come, the Gregorian Calendar will keep the vernal (northward) equinox in March, even if it slowly shifts earlier in the month.

This is in contrast to the scenario that you imply, where the calendar date would correspond to the relative position of the Earth in its orbit around the Sun.  That describes the sidereal year, the time taken for the Sun to reach the same spot in the sky relative to the stars, which is 365 days, 6 hours, 9 minutes, and 10 seconds.  A sidereal calendar would explain why you might think that precession would cause the dates of equinoxes and solstices to change in the calendar year.  That would result in a shift in the calendar of one full month in 1/12th the cycle length of the precession of the equinoxes, or about 1 full month about every 2,000 years.


Do lower mass stars consume more hydrogen than higher mass stars? If I understand correctly, red dwarfs consume all of their hydrogen and will eventually shrink into helium white dwarfs, whereas the higher mass stars will consume a fair amount of their hydrogen fuel, not necessarily all of it. Correct me if I'm wrong.

Stars less than about 0.3 solar masses are fully convective, so hydrogen throughout the entire star can be transported to the core and consumed.  At higher masses, the core is radiative, so only hydrogen that starts in the core is consumed.


What kind of halo is this? My friend sent me a video of a kind of a halo phenomenon and asked me what kind of halo it is.

There is a 22 degree circular halo around the sun.

But there is also a ring which seems to go horizontally around the entire sky. The ring is the same height as the sun. It seems to start and end where the 22 halo is at the same height as the sun (3 and 9 o'clock on the halo) and goes horizontally around the sky.

So it looks something like this? https://cloud.planetmaker.de/index.php/s/RkmkBNfb8YyDHSE

That sounds like the circum-horizontal ring (yes a name as simple as that). It (probably) is caused by hexagonal ice crystals floating horizontally. Where it touches the 22° halo you might see sun dogs which can be slightly colourful brighter patches.

See also https://www.meteoros.de/themen/halos/haloarten/ (German) and http://www.atoptics.co.uk/halosim.htm


Converting parsecs to light years in the dumbest way possible I have to convert 132 parsecs to light years and I'm struggling to get the expected result. Here is my method:
$132 \text{ pc} = 132'' = 2,2' = 0,03(6)°$
I have an equation for stellar parallax which looks like this: $d = \Big( \dfrac{1 \text{ AU}}{\tan(0,03(6)°)}\Big)$
This should give me the distance to the object in Astronomical Units right? Well It doesn't and I have absolutely no idea why, I'd be very glad if someone could help me out

You asked for a dumb, tedious way of doing it. So here it is, in all its glory
One parsec is defined as $$\dfrac{1}{\tan(1'')} \text{ AU}=\dfrac{648000}{\pi} \text{ AU}$$
As $1 \text{ AU} = 149597870700 \text{ m}$, one parsec is equal to $$\dfrac{648000}{\pi} \cdot 149\,597\,870\,700 = \dfrac{96\,939\,420\,213\,600\,000}{\pi} \text{ m}$$
One light year can be calculated by multiplying $c$ by the number of seconds in a year: $$299792458 \cdot 86400 \cdot 365.25 = 9\,460\,730\,472\,580\,800 \text{ m}$$
Dividing the former by the latter, we get $\dfrac{1 \text{ pc}}{1 \text{ ly}} = 3.261564$


Can a human land on Mercury? Can a human in a normal astronaut suit designed for the moon, land on Mercury and walk on it's surface. If not what would happen to him/her?

No sir/madam. An astronaut or a cosmonaut cannot walk on Mercury's surface. The reason is plainly simple. During daytime, its temperature surges up to 430 degree C making it 2nd hottest planet after Venus. Whereas, night time temperatures are as low as -180 degree C.

Despite of having ice caps at its polar regions which might seem attractive, its totally not feasible to land there and then mine the frozen ice to make up a base.

For a much more descriptive read, kindly refer this link : What it would be to live on Mercury


Using setting circle for Right Ascension I'm a little bit confused about the setting circle of my EQ5 mount. Basically, I'm trying to find M42 nebula using Sirius star as reference. So, considere that I did the following steps:

A) Polar aligned for south pole
B) Pointed the telescope to Sirius star
C) Adjusted the RA wheel to 6h45m and DEC wheel to -16º42'

Now, I want to view the M42 nebula basing only in the RA and DEC wheels (without using the finderscope or whatever).

Then, I did:

D) Changed the coordinates in the EQ5 mount to match RA 5h35m and DEC -5º23
E) Spent 5 minutes performing the step above (D)
F) Looked in the eyepiece

So my questions are:

1) After doing all these steps above, will the M42 nebula be 5 minutes away from the center of my eyepiece? So instead of using RA 5h35m I will need to use RA 5h40m?(??)

2) If yes, will I need to perform all the above steps every time that I want to see a new object? I mean, as fast as I can, to minimize the RA error...

3) And if the answer is yes again to Question (2), how useful is the setting circle under these conditions for you?

I will appreciate any kind of help.

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I think I mis-read your question before. Let me rephrase it to make sure I understand. In step D (moving the scope from Sirius to M42), that process took 5 minutes to complete. Or maybe it took 30 seconds to complete, but you then took a 4.5 minute coffee break :-). So during that 5 minutes of time since you calibrated the setting circle, M42 has drifted another 5 minutes to the west.

Yes, the object will be 5 minutes west of the eyepiece. Since your setting circle does not update during that time, you need to position the scope 5m to the west, or to 5h 30m right ascension (not 5h 40m R.A.).
Yes. Before moving to the next object, you should reset the right ascension setting circle to the coordinate of the current object. Then move to the next object. Then compensate for the time it took to move the scope. Of course, switch to a low power eyepiece just before starting the move to increase the chance that the next object will be in the field of view.
Mechanical setting circles are useful for small jumps. I have used them on my scope to locate Venus near the Sun. I have also used the circles on the scope at my club's observatory to locate planets in the day. (That is slightly different that your setup since the club's scope is permanently mounted, professionally made, and has really large circles -- all of which help to reduce any errors.) If I remember correctly, the "rule of thumb" is that mechanical circles should be about the same diameter as the optics in order to be "accurate". A 200 mm reflector (the diameter of the mirror = 200 mm) should have circles that are about 200 mm in diameter. But anything should definitely get you close to the target.


The consequences and the mechanisms of a shift of the Earth away from the sun First premise: I have already made this question in the worldbuilding section, but since it is a very specific question many have recommended me to write in the astronomy section so here we are.

Second premise: I'm Italian so please excuse any mistake

So, I'm writing a story about a fast and cataclysmic climate change. Obviously there are some artistic licenses but I would like nothing in my story to be completely impossible, only (at most) very unlikely.

The climate change that I would like to include in my story is very extreme, so I had to discard any terrestrial forcing to cause it (no change of ocean currents, supervolcanoes and methane eruptions) because they could not act fast enough and would not be so catastrophic (compared to what I imagine). So I focused on astronomical forcing. I excluded the impossibly forcing (for example, a magical and radical reduction in solar luminosity) and those that would be more catastrophic than the climate change itself (i.e. a planetary impact).

The last possible (though very very unlikely) option I have left is a dramatic change in the Earth's orbit. I have some knowledge in physics and astronomy so I did a personal search to figure out what could cause such an orbital change.The most plausible option would be to pass a rogue planet (or star) into the solar system, but since I want to complicate my life, I would like the orbital change to happen without any notice, which is quite impossible if it is the passage of a planet or a star in the solar system to cause it.

The only plausible explanations I have found to fully satisfy my scenario are either the passage of a primordial black hole in the solar system (a black hole with Jupiter's mass would have a schwarzschild radius of 2.8 meters and, if it did not come in direct contact with some celestial body, it may take in the solar system without notice) or passing a high-density cluster of dark matter (assuming an self-interacting form of dark matter) near or inside the solar system. Both are immensely unlikely options and very speculative (primordial black holes are hypothetical objects and we still have no idea what the dark matter really is and how it behaves) but I do not think they are impossible in the beginning, or is it wrong?

Assuming I want to move the Earth 2x10^7 km (if I do not mistake at this distance the average temperature of the planet would be between 15 and 18 ° C lower than now) farther from the sun, could this change take place in less than a decade and the earth's orbit would remain stable or simply the planet would be expelled from the solar system?

I also wondered what effects this sudden and drastic lowering of the Earth's temperature would have on the weather (especially considering that the poles would cool down much faster than the equator) but I think it's a pretty OT question so I leave it as optional

It sounds more like world-building than astronomy to me, but since they punted it here, I'll give a crack.

There's a few ways you could approach this and stay within good science but making change relatively quick is difficult.  This might be possible if the planet is in a situation like mars, with no moon and between two more massive planets, so it's more vulnerable to orbital perturbations.   Mars axial tilt is believed to have changed quite a bit and fairly quickly.   While the total solar energy is constant, a variation in axial tilt can lead to a variation in ice buildup where, generally speaking a lower tilt is better for ice buildup, as the higher tilt leads to warmer summers and that melts ice more easily.   Mars axial tilt varies from 13 to 40 degrees, but that variation can take hundreds of thousands or millions of years, which is probably longer than you're looking for.

Nearly any orbital change, even a rapid one would probably take tens of thousands of years minimum.

Geological changes are probably a little faster.  While it's still in the direction of theoretical, and I think you're right to avoid a large volcano or something like that because the effect from the volcanic ash would be greater than any long term effect from climate change.

About 3 million years ago, the formation of the Isthmus of Panama is thought to have significantly changed ocean currents that cooled the Earth by a few degrees that lead to the Pleistocene epoch and ice age cycle.   You'd need to get creative to figure out a way for a land bridge like that to make a radical change, but it's possible.

Another method, if the oceans are shallow on your planet, you could have a valley like the Mediterranean valley about 5.3-5.6 million years ago, with a natural land bridge keeping the ocean water out.   That would be a hot, dry, sandy valley.   Then when the land bridge gives way the ocean quickly flows in.   If your valley is big enough the oceans could drop several feet and if you have land or small islands near the poles, the falling ocean level could increase polar land which could trigger ice formation.   Glaciers don't form over water, only land.   This is a bit of a stretch cause you'd need two geographical oddities, very shallow oceans over the poles and a very large dry valley ready to be flooded, but it could turn a planet towards a major ice age.   Ice tends to keep growing once it starts.   It's a natural feedback mechanism.

Probably the best method would be a passing rogue planet giving your planet a gravity assist.  You'd want a larger rogue planet and it would need to pass fairly close to your planet to tug your planet into a longer orbit.  You'd probably want a gas giant mass and as it passed, it would probably need to be visibly larger than our moon in the night sky.

You mentioned 20 million km, or a 13% increase in distance and about a 22% drop in solar energy received.  You wouldn't need that much because of feedback mechanisms.   Ice ages can be triggered by as small as a 1% difference and for radical, rapid climate change, pushing your planet just 4-6% further away should be enough.

You'd need to make your planet's year longer, using Kepler's 3rd law, you'd use the 3/2 power rule, so 4% further would make the planet's year 6% longer or an additional 22 days to a 365 day year.

As to how fast that would make the planet cold.  That's hard to say.   You'd probably notice longer colder winters and frosting pretty much immediately, but for ice buildup and enormous glaciers you're still probably looking at at least a few hundred years, but that's kind of a ballpark guess.

You could do the same thing with a theoretical primordial black hole but they've kind of fallen out of fashion since it's believed they don't exist.

If you wanted to get creative you could have the rogue planet pass in-between say your earth like planet and a mars like planet, pulling the Earth orbit more eccentric and longer and the mars planet shorter, possibly leading to the two planets having crossing or nearly crossing orbits - which could lead to all kinds of scenarios and predictions like a possible planetary collision in 6,000 years or something or a near pass and enormous tides.

It would also stir up the asteroids and comets in our system likely leading to an increase in impacts following the pass for the next 100 or more years.

The rogue planet would create tides too if it appeared larger than the moon in the sky and maybe an earthquake or two, but wouldn't need to be a global disaster.    Rogue planets passing that close would be very rare, but not impossible.

rftwqa


The moon: how to determine its size, speed and distance from earth using a simple telescope? Assuming I only have a simple telescope, how can I determine the size of the moon, its angular velocity and distance from earth?

Finding the distance is tricky. To find the distance you need to carefully note the position of the moon (relative to the stars ) from two different places, *at the same time *.  The moon will appear to be in a slightly different position from the two viewpoints (an effect called paralax). That done, finding the distance, and size is a simple exercise in trigonometry, and Khan academy has a page which shows hoebto find the distance to the moon.

Greater accuracy could be obtained by measuring the time that a star is hidden by the moon, from two locations, and using that to determine the position of the moon to greater accuracy than possible by direct measurements.  Hipparchus apparently used a solar eclipse to obtain two positions atbthe same time needed for calculation of paralax.

Knowing the distance makes finding the diameter and speed simple to do, by observing the angular size, and using a little trig.  Working with photographs takenby the telescope is convenient but not strictly necessary.


How do astronomers find interesting events? I always wondered how those tiny dots representing moving stars or whatever forming an interesting event (supernova explosions, stars being sucked into black holes etc.) get caught in the huge solid angle of $4\pi$? It seems absolutely improbable to find any such event in that big solid angle.

Do astronomers just appear lucky looking at the right place at the right time to find an interesting event, or do they use some systematic methods of finding them? Do they maybe use special software which analyzes tons of data and identifies such interesting events?

The answer to your question is yes. Many times discoveries in astronomy are serendipitous in nature - some of the most well-known discoveries fall into this category: the discovery of the cosmic microwave background, the discovery of some of the planets in our solar system, are but two examples of this.

However, many times they are not serendipitous. Astronomers plan to discover things - look at the Kepler satellite which finds nearby exosolar planets by detecting planetary transits. The Sloan Digital Sky Survey was a planned galactic survey which gave us new insights about the properties of galaxies as well as structure formation.

One also has to consider time scales. Many things astronomers care about have timescales which are millions or billions of years (merging clusters, the evolution of main sequence stars, etc..), however, some phenomena are on very short time scales, namely supernovae. Microlensing events can also be somewhat serendipitous in their discovery, though researchers still devise experiments to look for them (see OGLE).

It is certainly a combination of luck and planning. This answer was intended to be a bit vague, since the question doesn't specifically ask about one phenomenon. Software does play a big role in identifying and classifying objects and events in the universe. I can talk a little bit more about this in the context of gravitational lensing if people care to know about it.


Could population III stars develop directly into population I stars? Hundreds of millions of years after the Big Bang, the very first stars began to form consisting of mostly hydrogen, a bit of helium, and maybe some lithium. These stars lacked any "metals" (elements heavier than helium) and are categorized as Population III stars. We have yet to observe any of these Pop III stars, and observation remains unlikely. The reason for this is because Pop III stars are thought to have been incredibly massive, so they died out so long ago that we can no longer see their light. Pop I stars are metal-rich; their metallicity is 1/10 to 3 times that of our Sun's. In my research, everyone seems to agree that that the first generation of stars had no metal, second generation stars had very little, and third generation stars have some metal.But, no one ever addresses the possibilities. So, is it possible that some Pop III stars were so massive that they were able to skip Pop II and go straight to Pop I? Maybe if several are close by when they go supernova?

No they couldn't. Population I stars contain elements like strontium, barium, gold, lead etc. that are not formed (much) in type II (core collapse) supernovae. They are also much richer in iron, nickel, manganese etc. than are stars formed from gas enriched only by supernovae.

Population I stars (like the Sun) can be termed third generation stars because they contain material that has been through at least two stars - but not just two high mass stars that exploded as supernovae.

Iron-peak elements are mainly formed and disseminated by type Ia supernova that are exploding white dwarfs. White dwarfs are the remnants of low-mass stars with long lives.

Many heavy elements (strontium, barium, lead etc) are formed by the s-process in intermediate mass stars that also have (relatively) long lives and do not explode as supernovae. Furthermore, this neutron capture requires iron-peak nuclei to act as seeds, so these stars must in turn have formed out of material already enriched with iron-peak nuclei.

It is also now thought that other elements (silver, gold, osmium etc) are formed by the merger of neutron stars, where there must be a significant delay between the initial supernovae that produced the neutron stars and the subsequent merger by orbital decay.


Why is Neptune true anomaly decreasing? So I've been having fun with trying to determine where the perihelia of different planets are (using JPL Horizons).
And by the time I got to Saturn, something weird started happening. I've noticed that at some moments true anomaly (that I've used to track perihelion time) started falling rather than rising. I've also noticed that it seems to be only a slight effect, which also demonstrated ~ 17 days period. So I though: perhaps that's the effect of Titan, right? Problem solved.
Yet now I look at Neptune's true anomaly (see the last column on the picture) and have absolutely no idea what's going on.
Can somebody explain, please? How is something like that even possible? It's like the planet decided to go backwards for a change.
My settings are:
Ephemeris Type [change] :   OBSERVER
Target Body [change] :  Neptune [899]
Observer Location [change] :    Sun (body center) [500@10]
Time Span [change] :    Start=1900-01-01, Stop=2100-12-01, Step=1 Y
Table Settings [change] :   QUANTITIES=18,41
Display/Output [change] :   default (formatted HTML)

Short answer: because Neptune's argument of the perifocus, the reference point for the true and mean anomalies jumps around a lot because Neptune's orbit is nearly circular.

If you run HORIZONS w/ the following parameters:

you'll notice Neptune's mean anomaly also decreases at times.

Both of these occur because Neptune's argument of the perifocus (from which both anomalies are measured) is itself jumping around quite a bit.

Except when Neptune is actually at perihelion, the argument of the perifocus is a theoretically computed point, not representing any actual physical quantity.

The function NASA uses to calculate the argument of the perifocus is oscelt, and specifically notes:

   6) If the eccentricy of the orbit is very close to but not
      equal to zero, the argument of periapse may not be accurately
      determined.

To summarize, using the true anomaly to determine's Neptune's position is a bad idea, unless you also account for the changing argument of the perifocus.

As you note, Neptune's ecliptic longitude behaves much better, because the zero point of the ecliptic, which, while based on the Earth's precession and thus not completely fixed, nonetheless moves very slowly, much slower than Neptune's orbit.

Of course, the "best" coordinate system to use is something like ICRF, which is unchanging with respect to the fixed stars.


Why does the Moon seem larger when it is close to the horizon? I've noticed that when I look into the sky at night, sometimes the moon appears very large near the horizon. Sometimes it also looks like it has a yellow tint.

Is this the light rays bouncing off the moon differently, or the atmosphere of the earth, or some type of sunspot blocking part of the light to the moon?

Interesting fact. It Isn't bigger!

This is a well known optical illusion that dates back hundreds of years. It's all in your head, it seems bigger because your brain perceives size as something that is relative.

This is an effect called perspective distortion and is will understood and used in photography. Phil Plait's Blog (A.K.A - The Bad Astronomer) also writes on this matter and would be worth reading.

Consider the below picture:

As the focal length of a lens gets longer the difference in relative size of foreground and background objects becomes more pronounced. Your brain then interprets this and respectively the moon seems huge!

But why does it look yellowish?

It appears yellow sometimes because our atmosphere blocks and stretches certain wavelengths. So when the moon is low in the sky, it has to pass through the most atmosphere to get to you, stretching and distorting the light through the yellows and oranges.


What is the composition of an asteroid in percentages? I am developing a RPG game (or so I like to tell myself) within outer space. Within the game, players would be able to mine small asteroids and collect resources, in order to make money. I'm having some difficulty determining exactly how much of the resources there would be in relation to everything else gotten from the asteroid.

This image has provided me with a good idea of what asteroids generally contain. It does not, however, show me the quantities. For example, I'm fairly certain that an asteroid would have more oxygen than palladium, but I don't know by how much.

Question: Can someone tell me (or better yet show me) the percentage of materials within an asteroid? I realize not all asteroids are the same, so categorizing or averaging is fine. The more detailed the better. The ideal information would be a list of values that I can scale with asteroid size.

You might find the information you seek in this article by Nature on the compositional mass distribution of asteroids.

From Wikipedia Asteroid Spectral Types:

  The majority of bodies fall again into the three broad C, S, and X
  categories, with a few unusual bodies categorized into several smaller
  types:

C-group of carbonaceous objects including:[citation needed]
    B-type largely overlapping with the Tholen B and F types.
    C-type the most 'standard' of the non-B carbonaceous objects
    Cg Ch Cgh somewhat related to the Tholen G type
    Cb transition objects between plain C and B types.
S-group of silicaceous (stony) objects including:
    A-type
    Q-type
    R-type
    K-type a new category (181 Eucharis, 221 Eos)
    L-type a new category (83 Beatrix)
    S-type the most 'standard' of the S group
    Sa, Sq, Sr, Sk, and Sl transition objects between plain S and the other types in the group.
X-group of mostly metallic objects including:
    X-type the most 'standard' of the X group including objects classified by Tholen as M, E, or P-type.
    Xe, Xc, and Xk transition types between plain X and the appropriately lettered types.
T-type
D-type
Ld-type: a new type with more extreme spectral features than the L-type
O-type a small category (3628 Božněmcová)
V-type (4 Vesta)

You will have to go to the Wikipedia article to drill down for more detail on the probable minerals contained in these asteroid classes.  We really only have good evidence on those classes that we think we have meteor samples of.  The rest would be guesses.


How to detect emission lines in optical spectra? Is there any handy module to detect emission lines in a spectrum like one we get from the Sloan Digital Sky Survey (SDSS)?

You can see there are many emission lines like Ha,OI in the spectrum below. In fact the spectrum is a two column data set, wavelength and flux. It is a combination of lots of scatter points. The typical interval between two points nearyby is 1.5 angstroms. I need to find all the Gaussian bumps and their line center.

So,first I need to find the the bumps and then I need to fit them to get their centers.

The programs I am aware of that you could use do require some programming expertise to operate. I would recommend looking at either IRAF (or PyRAF, which uses Python to interface with IRAF), or SPECTRE. Unfortunately, I'm not aware of a "black box" solution where you can just press a button and go.

SPECTRE is written in FORTRAN77 and is quite easy to use once you get the hang of it, though it has a somewhat limited feature set and the documentation is brief. It will help you to manually fit Gaussians to spectral lines, giving their central wavelengths and equivalent widths. A caveat for SPECTRE is that it requires the plotting package SuperMongo, which is not free software. If you are affiliated with a university, most Physics and Astronomy departments should have a license.

IRAF is a quite fully-featured, free, cross-platform application purposefully built for astronomical image analysis. It can help you measure the spectral lines, and can analyse the spectrum to determine the object's radial velocity.


Does astronomical observing "count" as remote sensing? I have long considered astronomical observing as a form of remote sensing, though quite different than typical remote sensing in that many objects observed are not resolved.

I am wondering, though, if the  formal definitions of remote sensing include or exclude astronomical observing (and specifically, observing things outside the solar system)?  And/or are there any resources that can be pointed to that either do or don't call astronomical observing remote sensing?

As suggested by, e.g., the Wikipedia page (or this page at the US Geological Survey), "remote sensing" is used pretty much exclusively for airborne or satellite-based observations of the surface of the Earth (including the ocean floor in some cases).
This makes sense if you consider that "remote" is used in contrast to observations made locally (or "on-site"). E.g., if you walk around on the ground and make observations, those are "local" observations; if you fly a plane and make observations of the same place, those are "remote" observations.

Since astronomical observations are overwhelming of places you simply can't make local observations of, it's unnecessary to specify them as "remote" observations, just as no one call looking up at high-altitude clouds or distant mountain peaks "remote sensing". (And it seems more than a little silly to claim that humans have been doing "remote sensing" for millions of years, even though humans and proto-humans have indeed been looking at the Sun, the Moon, planets, and stars "remotely" for about that long.)

(I have occasionally seen satellite observations of planetary surfaces -- e.g., imaging by probes orbiting Mars -- referred to as a type of remote sensing; this is clearly based on the direct analogy with Earth remote-sensing satellites. But that's about as far as you can push the term.)

You're free to consider astronomy to involve "remote sensing", based on an overly literal and overly logical interpretation of the term, if you like; in practice, no one working in either astronomy or remote sensing does.