Telescopes for beginner astrophotography I'm a fairly experienced photographer and have enjoyed taking lots of photos of the night sky with my existing gear - wide angle starscapes over pretty hills, mostly.

I'd like to move on, having seen what's possible with telescopes and mounted cameras. I've read around, and am now confused....

The plan sees me getting a telescope, tripod and mount of some sort (equatorial I believe I need?), and periodically putting that carefully into my car, driving into a rural area with dark skies and clear horizons, and pointing it at interesting things - my urban gardenless flat really isn't a good location for this.

I have Nikon full and crop frame and Olympus m4/3 cameras available - I believe I'd need a T mount adaptor to fit them? Is one likely to be hugely better than the other for this in a way I've not foreseen? I know the Nikon has much better low light performance but the Olympus puts more pixels on the subject with its small chip and has the very useful live time mode. I'm used to raw processing, but I'm expecting there'll be more specialist software I'll need to learn.

Am I anywhere near right or is this all mad?

What sort of things do I need to watch for on telescopes?
Am I going to trash a telescope by moving it around like this? I'm used to moving big lenses which I'm hoping is tolerably similar (accepting the very different optical designs)?
What have I completely missed?

Ah, welcome to the world of AP... check your sanity at the door, and that loud sucking sound you hear is your bank account being emptied.

My first and best piece of advice for you is to NOT buy anything at all right off.  Find your local astronomy club(s) and join.  Most clubs have at least a few people doing imaging, and you would be well-advised to meet these people and learn from them.

AP is not like conventional photography at all, for the most part. It's closer to digital signal processing.  It is a blend of science, engineering, and art.  What you know from conventional photography is not useless information, but you have a whole slew of new things to learn.

When we talk about telescopes and equipment for AP, we don't talk about "tripods" all that often.  A tripod is one kind of support structure for a mount head, which does the actual work.  The mount's role is to carry the telescope, camera, and associated equipment and point it at the object you wish to capture.  While the concept is quite simple, the execution is extremely complex.

There are two fundamental problems that we must overcome to capture good images of astronomical bodies.  The first is the fact that the majority of these objects are very faint.  While people often think the purpose of a telescope is to magnify, this is a misnomer: an astronomical telescope's primary purpose is to collect more light and concentrate it into a smaller area (the eye or the image sensor).  Magnification, while not irrelevant, is a secondary concern (and comes with its own issues).  Because these objects are so dim, we must resort to long-exposure photography.  While typical daylight photography uses exposure times measured in tenths and hundredths (even thousandths) of a second, exposure times for astrophotography often are measured in whole minutes.  Actually, total exposure is often measured in hours, with individual sub-exposures done in minutes.

And this is where the second key problem emerges.  While our shutter is open to capture an exposure of 1 or more minutes, we have to contend with the fact that the objects we are capturing are moving relative to our instruments.  Due to the rotation of the earth, a given star, planet, galaxy, nebula, etc... in the sky above moves across the sky from east to west in an arc with a radius depending on the object's position relative to the celestial north or south pole.  In a full day, an object moves 365 degrees around the pole (well, we rotate, but for practical purposes, the object moves).  In a single hour, it moves 15 degrees.  In one minute, 15 arcminutes.  And in one second, 15 arcseconds.  This may seem like a small amount of motion, and when you look up at the stars in the sky, they do seem to be fixed in place.  But, when you start to reduce your field of view by using a telescope or camera lens, the motion starts to become more and more obvious.

As a conventional photographer, you are probably familiar with the standard alt-az style mount motion.  Alt-az is short for altitude-azimuth, and a typical camera tripod works this way.  The mount angles up/down (altitude) and left/right (azimuth) to point at an object.  In order to follow an object across the night sky, then, you have to move the mount in both axes.  The rates of motion between the axes will vary from one another as well as depending on the location of the object in the sky above.  To automate this process requires a great deal of computerization, and even then it is difficult to do so accurately enough to handle the demands of long exposure imaging.  And even if you do overcome those issues, you will find that the object you are following appears to rotate in the field of view as it crosses the night sky, a phenomenon we call frame-rotation.  This is because the frame of reference of an alt-az mount is the horizon, not the sky itself.

The solution to this issue is the Equatorial mount.  An equatorial mount has four axes.  The first two, altitude and azimuth, are only used for alignment purposes, not for motion.  They are used to align the third axis, known as Right Ascension (RA), with the celestial pole.  Once that is accomplished, motion in the RA axis shifts the telescope's aiming point east and west.  The fourth axis, Declination (or Dec), handles north and south.  When properly aligned, the mount then needs only to move in the RA axis and at a single rate of speed to follow an object across the sky.  This doesn't even require a computer, and thousands of mounts have been manufactured with clock-drive mechanisms to handle this motion.  Computerization only enhances the process and makes it more accurate.

But not all equatorial mounts are created equally.  Many just do not have the accuracy needed to handle long exposure imaging, and the level of accuracy needed increases as you increase focal length, focal ratio, and exposure time.  A mount that could handle a 300mm camera lens or 80 mm short-tube refractor and produce 60 to 120 second exposures might not be able to handle an 8" f/4 Newtonian telescope.

Weight is a common issue.  Mounts like this are all rated for a specific payload weight amount.  Payload here includes not only the telescope, but also any cameras, adapters, finders, guide scopes/cameras, and mounting hardware - everything attached to the top of the mount.  But these payload ratings are typically given for total payload maximums. When doing AP, the general wisdom is that you should keep your imaging payload to 1/2 or less the mount's rated maximum.  So if your mount can handle, say, 40 lbs. of weight, for imaging you shouldn't load it with more than about 20 lbs.  The rule isn't a hard limit, and not all manufacturers rate their mounts the same way, but it is definitely a strong suggestion.

As I said, the tolerances become tighter as you increase focal length and ratio.  First, let me discuss ratio. The focal ratio of a telescope is calculated by dividing the focal length by the aperture.  Thus, if you have a telescope with a 200 mm (roughly 8") aperture and a focal length of 1,000 mm, you have a focal ratio of f/5.  If you have a focal length of 2,000 mm with the same aperture, it's f/10.  In general, the lower the focal ratio, the better for imaging.  The reason is the concentration of light.  Imagine you have the two telescopes I just mentioned, both 200 mm apertures, one f/10 the other f/5.  Now imagine you have identical cameras in each and are tandem-mounted on a GEM that can handle the payload without issue.  Now, let's say you are imaging an object that appears roughly circular and evenly illuminated.  Now, let's say in the shorter scope, f/5, the object you are imaging covers a spot on your camera's image sensor about 100 pixels across.  The total area of pixels covered is then around 7,850 pixels.  Let's say each pixel receives about 20 photons of light per second and you run a 1 minute exposure.  Thus each pixel gets a total of 1,200 photons of light in the exposure, for a total of about 9.4 million photons collected.  Now, if we look at the f/10 telescope, we find the image is larger.  A longer focal length increases magnification.  Double the focal length, you double the magnification.  So now the object we are imaging has a diameter of about 200 pixels on the image sensor.  This means that it has an area of about 31,400 pixels - double the diameter means 4 times the area.  Now, because the aperture is the same, we do NOT get more light, we get the same number of photons, they are just spread out over the larger area.  Instead of 20 photons per second, each pixel gets only 5.  In one minute, each of them gets only 300 photons.  The total number is the same, about 9.4 million, but they are less concentrated.  This means the image is only 1/4 as bright as that in the f/5 scope.  To get the same level of exposure, you would need to increase your exposure time by a factor of 4 to 240 seconds.  Yes, you get a larger, more detailed image, but it takes you four times as long, and this can be a problem.

No mount is perfect and all will show drift over time, even with autoguiding technology.  Let's say that the mount you are using has an error that is equal to about 1 pixel per minute in the f/5 telescope.  In that 1 minute exposure, everything is stretched by 1 pixel.  This is hardly noticeable and not likely to be a problem.  Let's say you had a star near the center of the image that has a diameter of 5 pixels.  In the exposure, you would see it as 5 pixels wide by 6 long.  Hardly noticeable.  But now let's look at the f/10 telescope.  Because we double magnification, we now have 2 pixels of stretching.  Now the star is stretched 2 pixels.  Still, this isn't a huge problem, but we also need to increase exposure, so we quadruple the stretching to 8 pixels.  Now we have significant stretching in our image and blurring of our object (and that's assuming the mount is THAT accurate, which is not a given).  This is why astrophotographers refer to shorter focal ratios as "faster" telescopes.

Aperture is typically the critical measure for a telescope.  But for AP, this isn't always the case.  A lot of very high quality imaging is done using short-tube refractors with fairly small apertures.  This also gives you a much wider field of view.  And this, in turn, reduces the exacting tolerances necessary for the mount.

However, this isn't ideal for all targets.  And this is another key problem with AP: there is no one-size-fits-all solution.  What is ideal for planetary imaging is not ideal for imaging large nebulae, which is not ideal for imaging most galaxies... you need to consider multiple factors in equipment selection. Personally, I use an 8" f/10 SCT (with a Barlow lens giving me f/20) for planetary imaging, an 8" f/3.9 Newtonian for globular clusters, medium-sized nebulae, and most galaxies, and an 80mm f/3.75 refractor for wide-field nebulae and the Andromeda Galaxy (which is 3° across, so very large).

But the mount is the key to all of it.  The conventional wisdom in AP is to focus on the mount first.  You can have a million dollar top-of-the-line camera and a million dollar top-of-the-line telescope, but if you don't have them on a mount that can provide accurate motion, you won't get good images.  Alternatively, you could put a mediocre camera and telescope on a good mount and get pretty decent images - the mount is THAT critical to the process.

Prices on these kinds of mounts varies, and you can easily spend into the tens of thousands.  The cheapest mounts in this range are designed to handle DSLR Cameras and lenses, usually not more than 300 mm in length (perhaps a little shorter).   The price on a mount like this is around \$350 to \$600 USD, depending on quality and capability.  And, again, these are limited in what they can handle.

Stepping up from that are basic German Equatorial mounts with powered tracking ability.  While computerization (What we call GoTo) is not required, it is usually found on these kinds of mounts and allows for the addition of autoguiding technology.  Autoguiding is the use of a second camera and, usually, telescope that is harnessed to software to lock onto a guide star and make minor adjustments to the mount's motion to keep it on-target. Autoguiding will not make a bad mount into a good one, it will simply allow an already good mount to capture longer exposures.  The lowest-price mounts in this range usually have a payload weight maximum of 30 lbs. or less and a price tag between \$800 and \$1,200.  However, I wouldn't actually recommend ANY of them.  The two most popular here are the Celestron AVX, running about \$900, and the SkyWatcher HEQ5, at about \$1,100.  The SkyWatcher might actually be usable, but the 30lb limit means only 15lbs for imaging, and that's very low indeed.  You can still probably fit a short-tube refractor on here with a DSLR and lightweight guiding package.  The Celestron AVX promised to provide entry-level AP capability, but the reviews of the mount show that it hasn't lived up to its promises.  This mostly seems to be due to poor quality control in the production line, resulting in some that work just fine, and a lot that do not provide the accuracy needed for imaging.  They are still usually fine for visual observing, just not for imaging.

The next step up finds mounts capable of handling between 40 and 45 lbs. of payload with prices between \$1,500 and \$2,000 USD or so.  This is where I STRONGLY recommend people start.  While they are not cheap (though you can often find them used for about 30% or so less), they often have the quality and accuracy necessary.  There are four models in this range I recommend: The Celestron CGEM II, Orion Atlas EQ-G, SkyWatcher EQ6-R, and the iOptron iEQ45 Pro.  My personal preference is the iOptron (I have the early version of this mount and love it), but all should be competent.  My mount is capable of 45lbs and I run it at around 26 (yes, over the recommendation, but hot hugely so, and it handles it fairly well) with an 8" f/3.9 Imaging Newtonian and CCD Imaging camera.

You CAN go with a cheaper mount.  But you rapidly lose capability and accuracy.  The lower mounts aren't really intended for imaging use, the heavier ones are generally engineered to stricter tolerances needed for imaging.

One nice thing about these four mounts, and most others in this range, is that they all use a common interchangeable modular mounting system for attaching the payload.  I personally have three sets of payloads and can switch them out in a matter of minutes. It also means that if you buy the mount now with a low-end telescope option, you can later upgrade the telescope itself without having to get a totally different mount.  It also means you can have different payloads for imaging and for visual observing.  I've never heard of anyone having too much mount for imaging.

Now, as for cameras....

DSLRs are not the best tool for the job, but they are widely used all the same.  There's a few considerations here.

First, DSLRs tend to have a great deal of signal noise - noise which must be dealt with.  The preference for AP is generally thermoelectrically cooled CCD imaging cameras.  These use Pelletier cooling systems, heat sinks, and fans to cool the image sensor and dramatically reduce signal noise.  Remember, we're dealing with faint objects, and the noise can often drown out the image signal in the exposure.  These kinds of systems also typically better isolate noisy components and reduce their effect on the image. Also, the nature of CCD technology makes them inherently less noisy than CMOS sensor technology prevalent in DSLR cameras. However, CMOS cameras are improving and many manufacturers are now offering cooled CMOS cameras.

With or without a cooled system, you still will need to capture calibration frames. There are three common types: Dark frames, Bias frames, and Flat frames. Dark frames are images taken with the shutter closed.  They are done with the same exposure time/gain/ISO settings and, preferably, the same temperature range.  These are then digitally subtracted from the "light" frame image you capture.  Some cameras actually have built-in technology for this (my Nikon D5000 does).  But for AP, it is usually done through capturing separate dark frames and processing them in bulk after the fact.

The Bias frame is also done with a closed shutter, but for the shortest exposure time possible. This maps the noise created when the image data is read off the image sensor itself.  It is also digitally subtracted in processing.  I don't believe any DSLRs have technology to do this automatically, though I could be wrong.

The flat frames are a different beast altogether.  They are taken with the camera still attached to the telescope or camera lens.  To produce flat frames, you want even illumination entering the optical train (e.g. the front of the lens/scope) and an exposure length that produces an average of 50% illumination.  These images are used to map the optical system's light and dark spots and even out the illumination (compensating for problems like vignetting).

As I said, noise is a major problem in DSLR cameras. In well-lit images such as daylight, well lit interior, or flash photography, the noise is nearly irrelevant.  But in low-light long-exposure imaging, it can be dramatic.  This is especially true if you use higher ISO levels to increase sensitivity.  CCD imaging systems are not immune, but they are designed specifically to handle it.

CCD and CMOS systems designed for AP are also typically more sensitive.  The term Quantum Efficiency (Qe) describes the sensitivity level of an image sensor to various wavelengths of light.  Essentially, it indicates what percentage of photons that hit the image sensor are detected.  A typical DSLR has a peak Qe around 30-40%, with some newer ones moving up closer to 50% and a handful of newer ones in the 70% range.  CCD and CMOS systems for AP typically have peak Qe starting around 50% and some have peak Qe around 95%.  This means much more efficient imaging and shorter exposure times.  The actual Qe varies across the spectrum, with the peak being the highest sensitivity.

These issues, so far, are fairly simple to deal with, though often require longer exposure and/or more frames in the stack (I'll come back to stacking).  There's one problem with DSLR cameras, however, that isn't as easy to deal with.  With the exception of the Canon 20Da, Canon 60Da, and Nikon D810A, all DSLRs, and, in fact, nearly all digital cameras in general including cell phone cams, webcams, etc..., have a built-in filter for color-balance improvement.  In most cases, it's mounted directly over the imaging sensor "chip" and also provides some degree of protection from dust and damage.  This filter is designed to reduce the amount of deep red and near IR energy that reaches the sensor.  Human eyesight is not very sensitive to these longer wavelengths, and the filter helps produce a more "life-like" color balance to match human vision. Without the filter, images tend to be shifted toward the red.  Software color balance profiles can reduce this significantly, but for conventional imaging, the filter is the best option.

However, for astronomy, this is a problem.  There is a lot of astronomically interesting light in the deeper red part of the spectrum.  When an electron in a Hydrogen atom falls from its second-highest to third-highest energy state, it emits a photon at a wavelength of 656.28 nanometers.  We call this Hydrogen Alpha, and it's a major component of emission nebulae and star-forming regions in galaxies.  When you see red in a photo of a nebula, it's often H-Alpha.  Some nebulae, such as the Rosette Nebula, are extremely heavy in H-alpha.  Unfortunately, the filter in a DSLR blocks the majority of this light from reaching the sensor - something like 90% if I remember correctly.

The solution for this problem is to not have the filter in place.  To date, only three factory-issued DSLR models have been released with this filter removed (actually replaced with one that is transparent to these wavelengths).   For other DSLR cameras, you would need to have the camera modified. There are several services you can find on the internet that will do this for you, but there's a cost.  It is not something I recommend people do themselves unless they have experience with camera repair and/or can afford to run the (very large) risk of ruining the camera.  Doing so also makes the camera ill-suited for continued conventional imaging, unless you get a clip-in or screw-on replacement filter (more cost).  You can do AP without a modified camera, but you will be significantly handicapped.

Of course, for best results, a monochrome CCD and filter wheel is the best option of all, but this significantly increases the budget.

You mentioned one of your cameras has more/smaller pixels.  In most cases, this is not actually desirable.  Larger pixels tend to be more sensitive (more collection area to gather photons), and more pixels may provide higher resolution, but also means larger file sizes and MUCH slower processing.   The Hubble Space Telescope's primary cameras have a resolution around 2048 x 2048 pixels (about 4 megapixels). The issue is not resolution as much as sensitivity.  I've worked with some very high end astronomical imaging CCD's with resolutions of only around 1 megapixel: but they produce FAR better images than any DSLR on the market.

When it comes to brands, Canon and Nikon are far and away the leaders, with Sony gaining some ground.  But overall, Canon is really the brand of preference among astrophotographers using DSLRs.  Canon has been very welcoming and accommodating to the AP community. They produced the first two pre-modified cameras well before Nikon produced their first.  Canon has made a great deal of information available, including detailed specifications and even firmware source code. Nikon is slowly coming around, but has been a LOT slower to do so and still has not offered anywhere near as much support as Canon.  Because of this, you will find a lot more resources and, especially, software for camera control of Canon cameras than Nikon or anyone else.  For people just getting started in AP and on a budget, I generally recommend looking for a used pre-modified Canon DSLR as an imaging camera.  Sites like CloudyNights and Astromart frequently have these in their classified ad sections.

And now... software...

A major difference between conventional photography and AP is the processing.  I already mentioned calibration images.  But the big deal here is image stacking.

Most Digital AP, whether DSLR or CCD based, uses stacking processes to produce better images.  Stacking is, essentially, a signal processing method using statistical analysis.  You capture a set of images of the same subject and exposure time.  You then run each through a calibration routine which removes the dark and bias signals and then applies the flat frame to even out the illumination.  The images are then aligned (nearly always through lining up stars in each frame).  At the end of the alignment phase, any pixel all the way through the stack represents the same point in space.  Then, each pixel in the stack is statistically analyzed.  For example, if you have 100 images in a a stack, and a particular pixel in the stack has brightness values between, say, 100 and 105, you average them out to come to what the most likely "true" value is, say 103.   In many cases, a weighted average is used in which pixels outside of 1 or 2 standard deviations above or below the average are discarded entirely, and only values in a consistent range are finally averaged together to produce the most likely "true" value.  This has the effect of reducing noise and, using drizzle techniques, can actually increase resolution.  Even without drizzle, you end up with a much better overall image, with greater clarity and less noise.  The larger the sample size, the better.   My most recent image of M51, the Whirlpool Galaxy, uses 3.5 hours of 120 second (2 minute) exposures (so over 100 individual exposures). The more exposures I use, the better the final result and the more faint detail I can gather.

Of course, this processing takes time.  The larger your image files, the longer it takes.  My main imaging camera produces image files in FITS format (the most common format for AP) that are about 50mb in size each.  100 of those comes to about 5 gigabytes, not including the flat, dark, and bias files (usually 10-20 of each).  This ends up taking a lot of storage space and a lot of processing time.  There are other file format options, but it is critical that you use a lossless format (like FITS or TIFF), and the only way to reduce those files is through compression, which works against you in processing (you have to decompress each frame in processing, so you save space, but dramatically increase processing time).

There is good processing software available, some for free.  I do most of my calibrating and stacking currently using DeepSky Stacker, which is a free product.  I do most of my image capture in Nebulosity, which is not free, but works well for what I need and I got before the last price increase.  While live view on a camera is helpful, it's not necessary with a program like Nebulosity which will loop exposures and provide metrics to help achieve accurate focus.  You can get MUCH better focus this way than through the live-view on a camera.  It will then also allow you to automate the capture, setting the number of exposures, length of exposures, what to name the files, where to store them on your hard drive (typically these store directly to hard drive, not on the camera), and in what format (usually FITS or TIFF, not the camera's proprietary RAW/NEF/etc... or JPG - and you really do NOT want to use JPG, it's FAR too lossy).

After you've captured, calibrated, and stacked, you will then want to do post-processing.  Some people use GIMP, which is an open-source option.  I mostly use Photoshop, which has a lot of capability and there's a wealth of information and tutorial advice.  I am also trying to learn to use PixInsight, which is probably the best astrophotography processing software on the market, but is neither cheap nor easy to use.

At this point, you probably are starting to see how very difficult Astrophotography can be.  Doing "pretty decent" AP is neither cheap nor easy.  I'm about 10 years into it and have spent about \$10,000 total (that includes several bad choices in equipment and also includes a backyard roll-off-roof observatory), and I'm just now feeling relatively competent.  My main issue now is time and weather.

For someone just starting, my recommended starting budget is at least \$2,000 and preferably closer to \$4,000 or more.  At \$2,000 you can get a competent mount and low-end telescope to get you started (and, depending on deals and what used equipment is available, a pre-modified DSLR or lower-end CCD camera).  You CAN get started cheaper than that... but you rapidly start losing capability and flexibility.  A \$1,500 or so mount can handle a camera/lens combo for very wide field imaging, a smaller scope for normal wide field, a medium sized scope for narrower field deep-sky (e.g. galaxies and smaller nebulae) and longer focal length scope like an SCT for planetary imaging.  If you go with a lesser mount, you're not likely to have as much flexibility there.  And, again, most of these mounts allow for interchangeable payloads.  So you can start with something simple like a short-tube refractor, and add to your equipment options as you have the ability to do so (my annual tax refunds have often ended up going to the hobby).

In the end, I return to my original advice: joint a club.  Most clubs have people doing this you can learn from.  A lot of clubs have people regularly selling well-maintained used equipment to finance their next upgrade.  Some clubs have loaner equipment to borrow.  Most either have a dedicated observing site or have scouted out good locations away from light pollution. And many also have their own small observatories with higher-end equipment, and, sometimes, even imaging equipment.  Most clubs in the US have annual dues less than \$50 per year.  It becomes a very cost-effective way to pursue the hobby.

I know I've given you a lot here.  And I've honestly barely scratched the surface.  AP is not for the weak!

Good luck and clear skies!


What are the differences between TESS and PLATO exoplanet telescopes? Within a year now it has been decided to launch TESS in 2017 and PLATO in 2024, space telecopes of NASA and ESA. They seem to have very similar science goals, both searching for transiting Earth sized planets in the HZ of the brighest (nearest) stars across all or most of the sky.

Are these missions as overlapping as they seem on first sight, or are there important complementarities? Is it perhaps the intention for PLATO to continue the observations of the planets discovered by TESS in order to get a longer time series?

Here is what one commentator has said about TESS in comparison to PLATO:

The Transiting Exoplanet Survey Satellite (TESS), to launch in 2017, seems superficially to be a similar mission to Plato. It will potentially discover hundreds of planets before Plato even gets off the ground in 2024. However, the limited sensitivity of its cameras mean it is completely blind to Earth-like worlds around sun-like stars. Astroseismology is also off-limits for TESS, meaning the size of any worlds it does discover will be highly uncertain. Unlike Plato, it will also move between patches of sky every 30 days, allowing only hot, short-period planets to be found.
With all other new telescopes, both in space and on the ground, limited to finding super-Earths around small stars, Plato is the only mission on the table truly capable of discovering an Earth-like world around a star like our Sun. And by targeting bright stars that allow atmospheric follow-up, it is not impossible to think that, as well as the first truly habitable planet, Plato could find the first inhabited one too.

Credit: Hugh Osborn, PhD student, University of Warwick

http://lostintransits.wordpress.com/2014/01/30/what-can-plato-do-for-exoplanet-astronomy/


Going to Gliese 581 c Gliese 581 c is described as one of the most promising exoplanets to sustain life. In this New York Times article, Dimitar Sasselov says: "It’s 20 light-years. We can go there."
https://www.nytimes.com/2007/04/25/science/space/25planet.html
What does he mean by that?
It would take roughly 40000 years with the fastest thing we've got today if I'm not mistaken?
I do understand that he's not talking about a crewed space mission!

Well, you are not mistaken that travelling at low speeds you would never get there.
However, there are moves afoot to send (uncrewed) probes to the nearest stars at significant fractions of the speed of light and to send back pictures to the Earth - (e.g. the Breakthrough Starshot project). In which case, information might be returned to Earth on timescales of 100 years.


What leads an observer to choose a narrow slit over a wide one (and vice-versa) when observing? In spectroscopy, a wide slit will lead to lower spectral resolution while too narrow of a slit will reduce incoming flux from the object. From what I read, it appears as though there is always a balance between spectral resolution and throughput. I'm looking for examples when an observer may choose one "extreme" over the other. ie. a very narrow slit or a wide slit (let's leave slitless spectra out of the convo for now unless relevant). What types of scenarios would motivate an observer towards one of the tradeoffs?

You seem to have all the ingredients apart from the variables of what size your detector pixels are (either physically or binned in software/hardware) and the angular extent of the object you are taking a spectrum of.
The basic trade-off, as you say, is between flux and spectral resolution, but there are limits to that trade off.
You should not reduce your slit width such that its projected size at detection is less than 2 pixels. This would be undersampling and you will not get the gain in spectral resolution you hoped for.
There is no point widening your slit in the hope of more flux if it already encloses most of the light from the object. In other words, if observing a point source, you don't usually open up the slit much beyond the "seeing" width.
If you do, this can actually damage your data because the spectrum you get will start to vary depending on exactly where the star was positioned in the slit. Definitely bad news for accurate radial velocities.
The exception might be if trying spectrophotometry where you really want to get most of the flux and not be affected by seeing variability or guiding errors.
So your extremes would be: Use a narrow slit to achieve maximum (2-pixel) resolution on bright objects where radial velocity precision is important and the seeing is good enough so that your narrow slit still accumulates enough flux. Use a broad slit if the seeing is really bad but you don't care too much what your spectral resolution is. Or open the slit wide if you want to do flux-calibrated spectrophotometry and avoid flux variation due to seeing changes or guiding errors.


Is light emitted from a heavier body more redshifted than light from a lighter body? Is light emitted outwards from, say, a region very close to a black hole more shifted than light with the same wavelength that began next to a less massive object?

Redshift is not related to the time it takes to travel from the source to the viewer, not to the distance from the source to the viewer.
But light is red-shifted by travelling out of a gravitational well.  One way to see this is to consider the equivalence principle: A person at the front of an accelerating spacecraft would see light shone from the back to the front of the craft become redshifted, and would say this is due to the fact that they are moving faster when the light reaches them than when the light was emitted.  If the same craft were held stationary in a gravitational field the same experiment must have the same effect, ie the light is observed to be red-shifted.
Alternately you can consider energy:  If you fire a bullet vertically up, it will slow down, and lose kinetic energy as it moves out of a gravitational well.  If it is travelling fast enough it might be able to escape the pull of gravity, but it will be going slower by the time it reaches a great distance.
Light, on the other hand, can't lose energy by going slower, but it does lose energy and momentum as it rises out of a gravitational well.  This is seen by the photons becoming red-shifted as they rise.
Redshift can be caused by a Doppler effect (of when an object is moving away from you) or by a cosmological effect (as the space between you and the object is stretched with the expansion of the universe) or by gravitational effects as above.  Redshift is not caused by the time it takes to travel from the source to the viewer.  But as more distant sources are receding faster,  the red-shift of more distant galaxies (that has been travelling for longer) tends to be greater.


Do asteroids have a gravitational field? I know that asteroids are huge chunks of rock, orbiting a solar system. Do asteroids have a gravitational field and do they gravitationally attract each other to form planets?

By definition, gravity is a result of mass. Any body with a non-zero mass (even atoms) will have a gravitational field associated with it. The higher the mass the stronger will be the field. This is basic of classical mechanics. Until we reach quantum scale where the gravitational force is dominated by other 3 forces and the gravitational field becomes irrelevent.

When it comes to graitational field of asteroids, it exists, but is very weak. However over a course of few million years these small asteroids combine together to form large masses of bodies that we now call planets. That is one of the prominent theory of Solar system formation, where the gravity of small dust particles from first our generation disintegrated star over a course of time accumulated to give us what we now know as our Solar system.
Think of it like this, every planet that you see now would once have been an asteroid at some point during its evolution.

Another proof to support this is the presence of numerous binary asteroids that orbit each other around a common center of mass, which requires gravitational attraction.


How can the Kuiper belt coexist with the 9th planet? There are asteroid belts, and there are collections of bodies in resonant orbits, such as Trojans. Aren't these separate things?

The recent evidence for the 9th planet is an alignment in various parameters of known Kuiper belt objects. Conversely, few or no objects are known in the parts of the belt that would be swept away by the hypothetical planet.

If the Kuiper belt straddles the gravitational neighborhoods of Neptune and this other planet, and its populations reflect their influence, is it really a belt at all? A lopsided belt? Or just a region with a collection of distinct populations, each associated with a dominant body?

The recent evidence for the 9th planet is an alignment in various parameters of known Kuiper belt objects.

These are not Kuiper belt objects. The Kuiper belt is taken to refer to the neighborhood of Neptune resonances. The scattered disc is another population, further out, and the objects relating to Planet 9 are further still.

Legend: grey = Kuiper belt, green = integer Neptune resonances, blue = scattered disc. The objects in the recent announcement are all off the chart, although their perihelions are <100 AU.

Source: Wikipedia.

The shape of the grey splotch at the left of this chart shows that the Kuiper belt is indeed a collection of populations dominated by Neptune. This isn't news, and it's still taken to be a belt. It goes all the way around the solar system and the orbital eccentricities are low.

This postulated planet might help explain the "Kuiper cliff", the otherwise unexpected cutoff before the 2:1 resonance. But, that's another question. Failing that, there's simply no relation between the Kuiper belt and Planet 9 except that they're both very far away.


Are there observable changes in a star about to become supernova, minutes or hours before the explosion? I am writing a science fiction novel, where a ship is stranded in a single star system (a red supergiant). One of the plot points is the star becoming supernova in several hours, so the characters have to fix their ship before that happens.

I have basic knowledge of how it works: Iron generated from nuclear fusion gets accumulated in the core, until it reaches a point when iron fusion starts. As iron fusion is an endothermic reaction, the core is no longer able to generate enough energy to hold against its own gravity and external layers pressure, so it collapses, and explode.

I have read that once the iron fusion starts inside the core, the collapse occurs within minutes, that the collapse itself lasts a few seconds (even less than a second), and that the shockwave takes several hours to reach the surface. Is all that correct?

The thing is that I need the characters to bee able to predict the explosion in a short term. A few hours or even minutes. It would be great if they could be aware of the core collapse and start a countdown.

So, are there any external cue of these events, like changes in luminosity or color? Does the star spectrum change when iron fusion starts, or when the core collapses? I know that the core collapse generates a huge amount of neutrinos. Is this amount so intense that it can be easily detectable? (that is, without a huge detector in an underground facility). Can the amount of iron in the core be estimated from tha star spectrum and size, so the aproximate time of the collapse could be predicted?

I think your best bet would be detecting neutrinos generated by nuclear burning inside the star (as we do for the Sun). Once the star hits the carbon-burning stage, it's actually putting out more energy in neutrinos than in photons. During the silicon-burning phase, which lasts for a few days and is what creates the degenerate iron core (that collapses once it is massive enough), the neutrino flux increases to about 1047 erg/s a few seconds before core collapse. (The peak flux during core collapse is about 1052 to 1053 erg/s). This paper by Asakura et al. estimates that the Japanese KamLAND detector could detect the pre-supernova neutrino flux for stars at distances of several hundred parsecs, and provide advance warning of a core-collapse supernova several hours or even days in advance. Since your characters are in the same system as the star, they'd hardly need a large underground detector to pick up the neutrinos.

This plot shows an example of neutrino luminosity (for anti-electron neutrinos) versus time for a pre-supernova star (from Asakura et al. 2016, based on Odrzywolek & Heger 2010 and Nakazato et al. 2013); core collapse begins at t = 0s.

By measuring the spectrum of energies for different types of neutrinos and their time evolution, you could probably get a very good idea of how far along the star was, particularly as we can probably assume your characters have much better models for stellar evolution than we currently do. (They'd also want to get accurate measurements of the star's mass, rotation rate, maybe internal structure via astroseismology, etc., in order to fine-tune the stellar-evolution model; these are all things they could do pretty easily.)

The core collapse itself would be signalled by the enormous increase in neutrino flux.

This "What If" article by Randall Munroe estimates that the neutrino flux from a core-collapse supernova would be lethal to a human being at a distance of around 2 AU. Which, as he points out, could actually be inside of a supergiant star, so your characters would probably be a bit further away than that. But it does show that the neutrino flux would be easily detectable, and that your characters might well get radiation poisoning from it if they were closer than 10 AU. (Of course, you'd want to detect it more directly than just waiting around till you started to feel sick, since that might take longer than the shock wave takes to reach the surface of the star.) This is just to bring home the fact that they wouldn't have any problem detecting the neutrinos....


Material exchange between a planet and its host Of the exoplanets found so far, is there one that can exchange material with its host? How could we prove that?

Io might be a good example, but it is not a planet.

I believe WASP-12b, a hot Jupiter, is a good example of what you're looking for. Li et al. (2010) first predicted that there should be gas transfer - easy, because the planet has a semi-major axis of only about 3 times the radius of WASP-12. Aside from the mass loss, extreme temperatures and distortion from tidal heating make the planet stand out.

The gas should be transferred into a small disk around the planet, which could reach temperatures of 3,000-4,000 K. Eventually, there should be significant angular momentum transfer between the star/disk and the planet. However, the authors note that certain assumptions of their model mean that this angular momentum transfer may not be as significant. There is also the possibility that a super-Earth in the disk could influence the interaction.

Follow-up observations by Hubble showed that there is in fact mass transfer between WASP-12b and the star, with gas reaching peak velocities of around 100-200 km/s.

The mass loss rate $\dot{M}$ should be approximately
$$\dot{M}\sim\frac{\dot{E}_tR_p}{2GM_p}$$
where $R_p$ and $M_p$ are the radius and mass of the planet, and $\dot{E}_t$ is the tidal heating rate, given as
$$\dot{E}_t\sim\frac{e^2GM_pM_*}{a\tau_e}$$
for stellar mass $M_*$, semi-major axie $a$ and characteristic eccentricity damping timescale $\tau_e$.


Concerning the "Lithium test" for Brown Dwarfs Both low-mass PMS (pre-main-sequence) stars and young brown dwarfs can fuse lithium in their cores and the lithium can be depleted throughout the star/brown dwarf very quickly. Wiki. Then the Li I 6708A absorption line disappears from the object's spectrum.

Why and how can the Lithium test be used as an indicator to distinguish between young brown dwarfs and young low-mass stars?

Can somebody summarise in what cases we can tell whether an object is a brown dwarf or a low-mass star?

The "lithium test" for a brown dwarf involves measuring two things - the lithium content and the spectral type (a proxy for surface temperature) or luminosity. But you may also need to know (or assume) something about the age of the object. The method is useful because the tracks of low-mass stars and brown dwarfs run together very closely in the usual luminosity versus Teff plane of the Hertzsprung-Russell diagram. In addition to the considerable uncertainty in the placement of these tracks, one really needs to know the age of an object in order for its luminosity or Teff to lead to even a crude mass estimate and hence the Li-test is pursued.

Li is completely and rapidly burned in a fully convective, low-mass object, once the core temperature reaches about 3 million degrees. The time it takes for a contracting pre main sequence star (or some brown dwarfs) to reach this temperature is mass-dependent. It takes about 20 million years at 0.2 solar masses, about 120 million years at 0.075 solar masses (roughly the boundary between stars, and brown dwarfs that will never become hot enough to fuse hydrogen), and if the object is below 0.06 solar masses, it never gets rid of its Li. The physics of this process is well understood and hence it has become a very useful and trusted tool for mass and age estimation.

At the same time as this happens, the surface temperature and luminosity are also diagnostic of mass, but in an age dependent way. An M7 spectral type is on the brown dwarf limit at about 120 Myr, but a similar spectral type in an older and more luminous object would mean it was more massive.

Historically then, the Li test was applied to low-mass objects in the Pleiades cluster (see Rebolo et al. 1996), which has an age of about 120 Myr. Objects with spectral types cooler than M7 at that age should be brown dwarfs. This was confirmed by finding Li in their atmospheres.

The Li test is not a completely unambiguous indicator of whether an object of unknown age is a star or a brown dwarf. Both stars and brown dwarfs can have spectral types M6.5-L2, but a star would be older at the same spectral type. If you have measured the luminosity (or spectral type) of an object and know whether it has lithium or not, then this provides a constraint on both mass and age. The diagram below, from Basri et al. (1998), is I think as clear as it can be made.

The more-or-less diagonal curves are the tracks along which low-mass stars and brown dwarfs of the labelled mass evolve (getting cooler with age). The diagonal dashed region shows the area of the temperature versus age plot where Li is expected to be depleted. I have labelled the equivalent spectral types. The stippled (dotted) area marks where the presence of Li guarantees that you have a brown dwarf with mass $<$ 0.075 solar masses.

Thus an M6.5 object with Li could either be a brown dwarf or a star younger than about 140 Myr, but an M6.5 object without Li is a star. An M8 dwarf with Li is a brown dwarf with an age less than 200 Myr, but an M8 dwarf without Li could either be a more massive brown dwarf with age just over 200 Myr or an older low-mass star. At spectral type L0-L2 then an object with Li will be a brown dwarf if its age is less than 500-1000 Myr, but could still be a brown dwarf with mass above 0.06 solar masses with an age just a little older than this, or an even older low-mass star. Objects cooler than L2 are always brown dwarfs and always have Li - but this test may fail at much cooler temperatures (cooler than abut L8) because the Li atoms form LiCl and LiOH molecules, the atomic Li I line at 6708A disappears from the spectrum, and the molecular features are too weak/confused to be identified.

A similar diagram can be made with luminosity on the y-axis and actually, that is more secure, since Li depletion as a function of luminosity is much easier to
predict than the behaviour of effective temperature, which depends a lot on the details of the atmosphere. Of course to measure a luminosity you need the distance to the object though.

Nowadays, the presence of Li in a low mass objects atmosphere is more likely to be used as an age, rather than mass indicator. For example, if one sees Li in an M7 object, then it is likely to be younger than 120 Myr (and vice versa). The prediction of Li depletion, particularly as a function of luminosity or spectral type is thought to be more secure than predicting the position of such an object in the HR diagram. This forms the basis of the
lithium depletion boundary age determination method for clusters.

To sum up.

An object with a spectral type of M8 or cooler with Li is a substellar brown dwarf.
An object with Li but a warmer spectral type may be a brown dwarf if it is old enough.
An object younger than $\sim 150$ Myr without Li is a star.
An object cooler than about spectral type L2 is a brown dwarf.


Does this CHEOPS first light image imply bad astronomy? @KeithMcClary's comment under lousy mirror corrected by software links to Bad Astronomy's First Light for the Exoplanet Hunter Mission CHEOPS Goes Tetrahedral which shows the image below.
I understand that in order to do precision photometry with a pixelated sensor one wants to spread the light over several pixels in a controlled way, but is this particular PSF the result of some painstaking optimization, some problem with the figure of one of the telescope's elements (oh no I hope it's not 1990 all over again!) or just because it's preliminary and they'll tweak something to make it more symmetrical?
If it stays this way, will it degrade CHEOPS' performance?

CHEOPS launched on 18 December, 2019, and on 07 February, 2020, it took its very first image of a star! The target chosen was HD 70843, a star about 155 light years away that’s a little bit hotter and brighter than the Sun.
The image, traditionally called First Light by astronomers, is itself … surprising.

As the article you reference makes clear, the defocusing is deliberate. It spreads the light of bright stars (the main targets for CHEOPS) over more pixels and hence mitigates saturation and non-linearity problems in the detectors.
The first light images look very similar to simulated pre-flight images (e.g. Hoyer et al. 2020; Futyan et al. 2020). The first light images are in line with expectations. There is therefore no unexpected problem with the telescope or optics, as there was with HST, and therefore no unexpected compromise of the mission sensitivity or goals.
From Hoyer et al.

The focal plane is defocussed to deliver a large PSF with a 12 pixels radius encircling 90% of the
flux. As a result of the combination of the Ritchey-Chrétien design and other specific features of the building of the telescope,
the PSF exhibits sharp and peaky features at sub-pixel level.

The reasons for the weird defocused PSF shape are discussed in section 11.1 of Benz et al. (2020). The triangular pattern is caused by strain on the primary mirror from a three-point fixation mechanism. This fixing mechanism was added to ensure that it wasn't shaken out of alignment during launch. In fact, Benz et al. say that the measured PSF (in zero gravity) is a little bit more symmetric, wider and smoother than found in the laboratory calibration tests. As a result there are some positives for observing bright stars - reducing the possibility of image saturation, but the situation is slightly worse for fainter stars because the wider PSF introduces more background signal into the photometry aperture.


Find a planet with incomplete information of a star chart I am wondering why it was so hard to locate Luke in Star Was Episode 7, if you already have the small part of the map.

Suppose we have a sector of our galaxy that roughly equals the one missing in this image. (I found no image on the web of the part in question. In the movie version on netflix you can take good look at it at 52:56.)

Is it possible and how hard would it be to identify a planet within that part of the map, given our knowledge of our galaxy and

we only know the small part and have no information about its size.
there might be "landmarks" (planets, stars, nebulars, and the like) missing.

If it is not possible, what minimum requirements would have to be met (on the available equipment, the map, our knowledge, ..) to make it possible?

Please note: this is not a Star Wars question, that is just the motivation.

About which galaxies do we have enough knowledge to be able to do this?

http://ko.starwars.wikia.com/wiki/%EC%9D%80%ED%95%98%EA%B3%84?file=TFA_galaxy_map.jpg

I don't know if this will be closed as too general, but I think there's a fun question in here about galaxy mapping.

The trick is to have a good map of the galaxy.   We can't see large sections of the Milky way, so we might have a better map of Andromeda than the Milky way overall from our perspective.

And yes, there are landmarks.    Pioneer and Voyager used Pulsars as landmarks in a diagram on their plaques.

There are by this estimate, at least 200,000 pulsars in the milky-way.  A map of 14 of them won't do someone on the other side of the galaxy if our map has some errors (which Voyager's might) or if they don't have a good chart on the other side of the galaxy.  Pulsars are very precise and they can make  very landmarks if you have enough of them and properly spaced.

Also, stars move, so maps age and lose accuracy over time, but in your star-wars example, a human lifetime wouldn't change things much.

Other possible landmarks include X-ray binaries,  very large stars (less common than smaller ones), black holes, neutron stars, the age of stars could be used, contact binaries, nebulas (or is that nebulae), stellar nurseries.   Identifying an individual planet would be easy given good data and a complete map.   A typical modern computer could do it, but the trick is having a good map.   There's about 100 billion stars in the Milky way, so it comes down to how good the map is, but you're correct, if the small map is good, R2's larger galaxy map is irrelevant.   To be relevant, the smaller map should have been maybe 100-500 stars and R2's map nearly the entire galaxy, and there would be ways to make the map useless if it fell into enemy hands, by not including pulsars or star size, making it very very difficult to identify which part of the galaxy by just 100 stars, none of them distinct.   The principal of needing both maps is valid and could be done, but you're right that the size of the 2nd map was ridiculous, also, why would they need to track his journey like in the picture?.    Movies often ignore the finer points like that and just focus on what looks cool.


How does gravity have an effect from the inside the event horizon of a black hole with the rest of the universe? According to general relativity, as I understand it, the space around a black hole's event horizon is distorted by gravity, such that the distance to a point approaching the event horizon from an observational point further out from the EH approaches infinity.

So if the distance to any mass inside the EH appears to be infinite, how does this mass assert gravitational effects outside the EH?

Perhaps a different question, what is the difference between the gravitational effects of a mass inside the EH of a black hole and a similar mass that is outside the cosmic event horizon of an observational point?

(edit as follows)
My original intent for this question is rethink how we think about black holes.  The distance to a point on the event horizon of Sag A is as far to the point on an observer's cosmic event horizon.  You can travel no farther than such a point on the CEH than you could to a point on Sag A's event horizon, at least using our 4D coordinate system.  BTW, I think all talk about a black hole having a singularity at its center is bunk, first of all, we can never get there to confirm or deny. Secondly, the density of Sag A if it were uniform, would mathematically would be around the density of liquid water as it exists on earth, but we would never be able to know, just as we would we would ever be able to know the density of the universe outside of our cosmic event horizons.  Any answers to those two questions would be meaningless anyway, as long as there is no away around the speed of causality rule enforced by our universe.

Now I believe I have read that Hawking Radiation has been shown to exist.  However, does that mean the part of whatever is outside our cosmic event horizon is evaporating as well?

As is the case even in Newtonian gravity, the gravitational attraction from a spherical body exterted on a far-away object is identical the that exterted by a point mass located at the the center of the body. This is the so-called shell theorem.

To an outside observer, due to the infinite time dilation at the event horizon of a black hole (BH), all mass that has ever fallen into the BH has actually never reached the event horizon, but is instead placed in a shell around the BH, still falling ever-so-slowly toward the BH.

Thus, what we feel is in fact the gravitation not from a point mass, but from a shell of matter, but there is no way for us to tell the difference.

Now you may ask, "What if we drop a massive object, e.g. a planet, into a symmetric black hole? Since it never reaches the BH, won't the effective gravitational potential be asymmetric?" What happens is that the way that space is "warped" around the black, the gravitational field is bent along with it in such a way that sufficiently far from the BH+planet, the gravitational field looks symmetric. Sort of like the following drawing shows close to (left) and far from (right) the BH.


Is Sofia a radio telescope proper? I usually think of SOFIA, the Stratospheric Observatory for Infrared Astronomy as an infrared optical telescope:

SOFIA uses a 2.5 m (8.2 ft) reflector telescope, which has an oversized, 2.7 m (8.9 ft) diameter primary mirror, as is common with most large infrared telescopes. The optical system uses a Cassegrain reflector design with a parabolic primary mirror and a remotely configurable hyperbolic secondary. In order to fit the telescope into the fuselage, the primary is shaped to an f-number as low as 1.3, while the resulting optical layout has an f-number of 19.7. A flat, tertiary, dichroic mirror is used to deflect the infrared part of the beam to the Nasmyth focus where it can be analyzed. An optical mirror located behind the tertiary mirror is used for a camera guidance system.

and one thing that makes that happen is that it looks just like a "normal" telescope with shiny polished and metal coated mirrors.
I've just seen Phys.org's Stellar feedback and an airborne observatory: Team determines a nebula to be much younger than previously believed (which links to Stellar feedback and triggered star formation in the prototypical bubble RCW 120) which talks about a "1.9-THz fine-structure line".
As far as I can tell this is done with little antennas as pixels, a heterodyne down-converter and radio receivers just like the Atacama Pathfinder Experiment data at 345-GHz it compares to.
In other words, it sounds just like a low granularity focal-plane array on a dish radio telescope.
I am not asking about the difference between the overlapping labels of radio and infrared as they apply to the electromagnetic spectrum, though that could well be part of an answer. These days radio astronomers call what they receive "light" as often as optical astronomers do. However the equipment used is different. Radio offers the possibility of recording phase for off-line interferometry, whereas optical detector arrays are generally intensity based only. So instead, am simply asking:
Question: Is Sofia a radio telescope proper?

Source

A) SOFIA [CII] integrated intensity, scaled from 0 to 260 K km/s. The red circles indicate the approximate inner and outer PDR boundaries defined from Spitzer GLIMPSE 8-μm emission (55), and the red star shows the location of the ionizing source, CD −38°11636. The yellow “+” indicates “Position 1” (see the Supplementary Materials). (B) Spitzer GLIMPSE 8-μm emission. The contours are of [CII] integrated intensity, scaled from 40 to 160 K km/s in 40 K km/s increments. (C and D) APEX 12CO(3-2) and 13CO(3-2) integrated intensity, scaled from 0 to 260 K and 0 to 90 K km/s, respectively. The contours are the same as in (B). The areas enclosed by the dashed yellow lines in (A), (C), and (D) were used to extract the position-velocity diagrams shown in Fig. 2.

The GREAT receiver attached to SOFIA to detect far-infrared radiation incorporates elements reminiscent of both radio and optical detectors, but since it works by mixing a signal from a tunable oscillator with the oscillating field from the incoming radiation rather than by the photoelectric effect as in most optical detectors, you might consider it more akin to a radio telescope receiver.


What is the role of the mesh on which dipole elements of the MWA antennas are placed? I was reading about the radio telescope - Murchison Widefield Array (MWA) situated in Western Australia. Antennas of this telescope are quite unique and different from the usual dish radio telescope. In MWA a four by four regular grid of dual-polarisation dipole elements are arranged on a 4m x 4m steel mesh.

So, I was wondering why these dipole elements would have been mounted over the mesh? On searching over internet I found out that this mesh is called a counterpoise (though I am not sure if I am correct). I read more about it and I partially understand that the mesh and ground (Earth) act as two plates of a capacitor but I didn't get why it was done and what was the use of it? In this image mesh can be seen on which dipole are mounted - Image courtesy: SKA

 Source  click for full size

The mesh acts as a ground plane.  An antenna design project can be divided into two theoretical regimes: free space, and ground plane.  A free space antenna is an antenna that is located far enough away from anything, that nothing will alter its electrical properties.  This is the simplest way of doing a design, but not applicable to this antenna
This antenna is mounted on the Earth.  The soil under the antenna has electrical properties that can not be precisely known.  The properties may change with the amount of vegetation or rainfall through the year.  To avoid having to deal with the exact properties of the soil, we place a metal shield between the antenna and the soil.  The shield is call the ground plane.  Its purpose is to present known electrical properties to the antenna.  Ideally the ground plane is always at zero Volts relative to the antenna.
The reason for using a mesh as opposed to a solid sheet of metal due to economy of materials facilitated by a neat property of electromagnetic waves: a wave cannot pass through an opening that is smaller than its wavelength.  Therefore instead of spending all the money, and gaining all the weight of a solid sheet, we can adjust the width of the mesh to block (that is, reflect) the frequencies we are interested in.


Is there light on the surface of Venus? A simple question: is there light on the surface of Venus?

The best way to answer this question is from images taken from probes that landed and took pictures from the surface of the planet.  According to the Lunar and Planetary Institute, images from the Russian Venera 13 lander revealed

  The effect of sunlight filtering through the dense atmosphere appears to give the surface an orange tint.

Despite some uncertainty of the actual colour due to possible chromatic distortions because of the atmospheric composition etc. It is clear in the image from Venera 13 (below) that there is indeed light on the surface f Venus.

Image attribution (from the link above): Images courtesy of James Head (Brown University), in collaboration with USSR (now Russian) Academy of Sciences.

Observations, reported in the NASA article NASA Scientist Confirms Light Show on Venus reveal another potential source of light - lightning, and theorised to be

  maybe even more activity than there is here on Earth


Could the Earth survive a red giant Sun? So my question is that, is not a lie that we are flying away from the sun slowly, that is increasing by time because the sun loss mass, so when the scientist usually talk that we will get swallow by Sun when it get red giant, do they count on this? I mean, we can get far enough to survive or they just calculate with the actual orbit?

The answer is yes, there are many authors who do take mass loss from stars into account when trying to work out the fate of their planetary systems. Examples include Schroder & Smith 2008; Adams et al. (2013); Adams & Bloch (2013).

It is thought that (e.g. Kalirai et al. 2008) that the Sun will lose half its mass by the time it ends up as a white dwarf. Almost all of this mass loss occurs in the final hundred million years or so (in just over 7.5 billion years time) whilst the Sun is on the red giant branch and asymptotic giant branch. The mass loss that occurs before then is small in comparison - about $10^{-13}$ of a solar mass every year (see Noerdlinger 2008), so losing less than 1% of its mass prior to the RGB.

Consideration of conservation of angular momentum alone leads to the idea that $M(t)R$ is constant, where $M(t)$ is the mass of the Sun as a function of time and $R$ is the semi-major axis of a planetary orbit. Thus prior to the RGB phase, the Earth's orbit expands by maybe 1.4 cm per year, or by a total of $7\times 10^{-4}$ au prior to the RGB.

However, during and after the RGB we see that the Earth ends up with an orbital radius of $\sim 1.5$ au (or in solar units, about 290 $R_{\odot}$) after the RGB and maybe 430 $R_{\odot}$ after the AGB. So, whilst the Sun is in the RGB/AGB phase and losing mass, the Earth moves outwards from about 215 to 430$ R_{\odot}$. Meanwhile, as it ascends the RGB, the Sun will likely increase in size to about $200 R_{\odot}$. Therefore it is a close-run thing as to whether the Earth will be engulfed within the outer envelope of the evolved Sun.
Major uncertainties include just exactly how big the Sun will get, how much mass it will lose on the RGB and then on the AGB, and it turns out that the time-profile of the mass loss (which will happen in pulses, rather than continuously) may also play a role.

The plot below (from Schroder & Smith 2008) may help to make some sense of this. It shows the (log) of the radius of the Sun versus time in the RGB and AGB phases as it loses mass. Also shown with a dashed line is the orbital radius of the Earth. Fractions written on the plot show what fraction of the Sun's present mass remains.

However, even if the Earth escapes immediate engulfment, the Sun would be several thousand times more luminous than it is today, so the temperatures at the surface of the Earth could reach $>1000$ Celsius even if it stays an astronomical unit away from the Sun's surface.

A further issue is the effects of tidal dissipation. The Earth will be dragged back by tidal bulges induced in the Sun's extended envelope. The rate at which angular momentum is lost is however exquisitely sensitive to the exact radius of the Sun and the radius of the Earth's orbit, to the extent that their orders of magnitude uncertainty in whether the Earth will be dragged into the Sun's envelope even before it reaches the tip of the RGB (this is what Schroder & Smith 2008 claim) or whether the loss rates will be too low to have much effect. It all depends on how much mass loss there is (Schroder & Smith use a slightly lower estimate than above) and therefore what radius the Earth reaches and what the final radius of the Sun becomes.

At the moment I would say there is sufficient uncertainty that the fate of the Earth is undecided, but it will unarguably be uninhabitable.


Where can I find a visible star dataset? I'm trying to do a project for a databases class where I have a user enter their current location and I tell them what constellations/planets/etc are visible in their area. I've seen this data compiled into images, PDFs, and website entries, but I can't seem to find a csv for it. I get that this data is bound to be massive or extremely complex, but I feel like there should be something out there. Any clues?

You can probably extract the data from Stellarium. Also, does it need to be all sky, or not? What Wavelength? If you want an all sky visible catalogue, the United States Navy produced the NOMAD catalog by combining 2MASS, UCAC, and the USNO-B catalog. You could also use the APASS all sky catalog, meant for helping astronomers with photometric calibration.

If you don't care about whether the wavelength is in the visible, CalTech's IRSA has a bunch of infrared source catalogs (the smallest/shallowest one being the IRAS catalog). The deepest all sky infrared catalogs are the WISE sets.

Other good databases to dig in to include the Hipparcos set of parallaxes, or some part of the Gaia catalog.


Are all the photos of the universe really so colorful or is it just visual effects that are added later? Do the colorful pictures of the universe show how it looks like that in reality or is it just a visual effect that is added later.

There is no straight yes or no answer.

Some objects do have significant radiation in visible part of the spectrum which would make them colorful if we would be close enough while observing them (our eye needs large number of photons to distinguish color so that means, astronomy wise, that object needs to be either very bright or very close or both).

On the other hand lot of them are being adjusted for color later. Usually most beautiful images are mosaics of several images from different filters (even from parts of spectrum our eye cant see) where each image is given a specific color based on filter used and then later all of them get combined.

If you want to learn more check this very helpful paper, which has some tutorials as well:

http://www.stsci.edu/stsci/resources/imaging/papers/rector2007/files/Rector2005.pdf

Btw, as a curiosity - images from rovers on Mars are mainly black and white and are colorized later (thats why the rovers usually have a color wheel on them for calibration purposes as seen here). This lead to uncertainties of color of Mars sky which turned into a long lasting debate.


The effect of gravitational wave is like tidal forces? What would happen if a strong gravitational wave passed through a human body or any other kind of material?

The shape of the human body would change?

Would the human body be distorted apart?

If so, would the person survive?

It depends what you mean by "very strong".

The announced first detection (14th September 2015) was considered a very strong signal and that still only stretched the 4 km detector arms by less than the width of a proton (a tiny particle inside the nucleus of an atom). This wave would have no effect on a person.

You could hypothesis a stronger wave that would stretch and squish a person, I'm not sure what effect this would have but there are no objects in nature capable of producing waves this strong, even something as close as our sun becoming a supernova wouldn't be that strong, so it will never happen in real life.


What percentage of galaxies rotate with trailing arms? I am looking for the percentage of galaxies that rotate with trailing arms. I know that only a small percentage of galaxies rotate with leading arms and would like to know whether there are any results that determine the percentage of spiral galaxies that rotate with tailing arms.

I am interested in publications in this topic.

Note:
So far I have found the following:

Land et. al (2008): "Determining the rotation
direction in this way is accurate for 96% of galaxies
(Pasha & Smirnov 1982)"
Pasha & Smirnov (1982)
Buta et al. (2003)

Do you know of any newer publication on this question? More recent than Pasha & Smirnov (1982) where the sample is quite small?

What percentage of galaxies rotate with trailing arms?

  I am looking for the percentage of galaxies that rotate with trailing arms.

  ...

  Do you know of any newer publication on this question? More recent than Pasha & Smirnov (1982) where the sample is quite small?

I think the best answer is that exact numbers are very difficult to determine.

Take for example page 5 of "Arm classifications for spiral galaxies" (1987) by D. M. Elmegreen and B. G. Elmegreen, cited over 200 times.

Affiliation: IBM Thomas J. Watson Research Center, Yorktown Heights, NY

Publication: Astrophysical Journal, Part 1 (ISSN 0004-637X), vol. 314, March 1, 1987, p. 3-9.

  Page 5:

  "II. THE CATALOGS

  a) Selection Criteria

  Table 2 includes all 708 spiral galaxies listed in the Second
  Reference Catalogue of Bright Galaxies - [3rd] Edition link (A. de Vaucouleurs, G. de Vaucouleurs, and H. G. Corwin 1976), with declinations $\delta \gt -35°$, inclination-corrected radii at 25 mag arcsec$^{-2}$, R$_{25}$, greater than $1^\prime$, and inclinations less than 60°; 654 galaxies could be classified, 54 galaxies which met the criteria but were too faint, had no details (were overexposed on the POSS or were too early in Hubble type to distinguish structure) or were too inclined for reliable classification are also included in Table 2, but with designations f n, or i, respectively, instead of arm classes.

  ...

  b) Self-Consistency Checks

  Most galaxies in Table 2 were classified at least 3 times, separated by intervals of several months, as a consistency check. They were classified from both paper and glass copies of the POSS. In the de Vaucouleurs sample, ~10% of the galaxies were given different arm classes during independent examinations on the paper prints. An 8% variation occurred when the glass images were subsequently examined.

  ...

  For the high-resolution galaxies listed in Table 3, again ~10% of the arm classes changed upon reclassification. In these cases the large amount of detail presented confusion; irregular features were given more or less weight at different times of classification. Arm classifications become increasingly difficult for very small, faint galaxies. The cutoff of $1 ^\prime$ radius for the galaxies in Table 2 represents a minimum practical size for classification from the POSS.

As you can see, even with expert examination multiple times of the same data, determination of exact numbers with any accuracy is impossible at the present time. A guess is "most of them", an exception example is: NGC 4622 (also called a backward galaxy).


Why don't we detect planets around OB stars and no terrestrial planets around A or early F stars? Looking at an exoplanet database, I noticed that there are very few planets detected around main-sequence OBA stars, and most of them are gas giants/brown dwarfs. Why can't we detect low-mass planets around these stars?

There are a number of reasons and you will see that my answer is subtly different to HDE226868's.
OBA stars are less common than FGK stars, but they are much brighter and any magnitude limited sample would contain plenty. However they have not been observed as intensively and planets are much more difficult to find around them. And the latter has actually led to the former!
Transits: OBA stars are bigger than the Sun by factors of 2-10. This increases the probability of seeing a transit by a similar factor for a planet of a given size and orbital radius. However, the amplitude of the transit is reduced by the square of that factor. This makes transits harder to detect - more or less impossible in OB stars - and they are harder to validate by observing radial velocity variations (see below).
Radial velocities: OBA stars are 2-30 times the mass of the Sun. For a given planetary mass and orbital radius this decreases the radial velocity amplitude observed in the star as $M_*^{-2/3}$. However, given the precision of current spectrographs this would not prevent the detection of hot Jupiters/Neptune's, or even cool Jupiters. No, the big problem with hot stars is the lack of spectral lines and the broad width of these lines due to the intrinsic rapid rotation of these stars. This prevents achieving the precision necessary to detect exoplanets.
More success has been obtained in identifying planets around the giant stars that evolve from A-stars. That is because as they expand, they cool, spin down and they have many sharp spectral lines. Many exoplanets have been confirmed around giant stars that would have been A stars on the main sequence.


Would you recommend this telescope to a beginner? Celestron Astro FI-5 125 mm f/10 Schmidt-Cassegrain with a computerized altitude-azimuth single fork mount

https://www.celestron.com/products/astro-fi-5-schmidt-cassegrain-telescope
https://www.amazon.it/dp/B06WD5JVJ8/ref=cm_sw_r_other_apa_i_kxXnEb3XD5AR4

A friend of mine wants to buy a telescope and begins observing in person. The area won't be among the most polluted in Italy, say the Milky Way is often visible by eyes and she already gets some traces of it while playing with modern smartphone and the apps they are equipped with.

She has spotted the above telescope.

Now, both me and my friend like amateur astronomy but we totally lack of knowledge on observational backyard astronomy and the related optics.

Would the linked telescope be a good starting point for a total beginner? Does it allow for the observation of nebulae and Andromeda while being OK for planetary seeing as well?

Alternatively feel free to signal if this can be a sort of overshooting. Though the budget is not an issue, considering that it just to begin is somehow mandatory.

What I began to consider is avoiding to invest some money in something that requires a very very dark environment in order to be used to see diffused objects. The telescope will stay, at least most of the times, in a dark extraurban area nevertheless close to several towns.

Would you suggest this scope or is there something making it not a good or reasonable choice for beginners?

PS beginning astrophotography is also considered, it can be as simple as having a smartphone mount. But playing with smartphones is how we became enthusiastic about astronomy and photography.

The Celestron SCTs at all apertures have been honed by 30 years of critical amateurs and a market-attentive manufacturer, so you will almost certainly get high-quality images from the OTA, the telescope itself. The Kellner-design eyepieces are another matter. Their apparent field diameter (AFOV, the size of the circle as you look through the eyepiece) is only about 55°, or roughly the diameter of the circle if you close your index finger to your thumb and hold it about 12.5 cm (5 in.) from your eye as you look at the sky. Moreover, these are mass-manufacture products sold as original equipment supplied to many telescope makers. Their quality assurance is not guaranteed. Most people find the star images are generally not as crisp as more modern designs. You would be wise to budget another €200 to €300 for two wider 70°–82° AFOV eyepieces to your budget. A good focal-length suggestion would be 25-30mm and 8-10mm. Quality wide-field eyepieces vastly improve the viewing experience. Objects stay in view longer, and generally are crisper and "deeper" (more pinpointy stars in the field than with mediocre optics). There are numerous telescope equipment retailers which you can locate with search terms such as "1.25 telescope eyepiece 82° 25mm". It is best to patronise a recognized astronomy-specialist than a drop-ship outfits like Amazon. Professional dealers will know most of the problems you encounter and have a vested interest in answering your queries promptly. Since you represent possible future business, they will go out of their way to give you info and service you don't get from sell-&-forget outfits.

The scope's 127mm aperture is a good starter choice. With the exception of M31 the Andromeda Galaxy, galaxies will not be impressive; "faint fuzzies" aptly summarises it. However, nearby star clusters and nebulae more than make up for it; you will find plenty of eye candy just roaming the Milky Way band. If your observing location is Italy, you may find that "rough air" (atmospheric turbulence) degrades image quality much more than any optical problems in the telescope or eyepieces. If you just can't get a good focus on any of the stars in a given field, it means the "seeing" is not great. In bad seeing, the moon will look like the seething shimmer of a distant village on a hot day. Except for really bright inner-city or moonlit skies, light pollution will be less of a problem in telescope fields compared with naked-eye impressions. You are effectively looking at a star cluster through a soda straw with a magnifying glass, so the inherent brightness of the stars will easily override the slight grayness of the background field. Yes, you do see a lot more under truly dark skies, but if you can see the MW visually where you live right now, your telescope will give you years of pleasure before you have to move on to that cabin in the woods. Finally, have a look at the amateur astronomy community www.cloudynights.com. Just about any telescope or astronomy info you need can be answered by the enthusiasts on that website. Moreover, there are four amateur astronomy forum/discussion groups in Italy. Check them out, you have a lot of friends out there you never knew you had.


How are galactic rotation curves determined from radial velocity measurements? I have been reading about the deviation of galactic rotation curves from the predicted Keplerian curve to a flat curve at large galactocentric distances.

They were the first indicators of the existence of dark matter but I do not understand how the curve is obtained from Doppler shift data (as mentioned by some papers as the method of determination for the curve.)

Doppler shift data would allow measurement of radial velocity but how can that alone be enough to extrapolate the rotation curve?

Imagine a spiral galaxy where the rotation of gas and stars within the disk is close to circular (generally true for gas clouds, though less true for stars) in a single plane (the disk), and imagine that it is tilted at some intermediate angle with respect to your line of sight. A circular disk galaxy will then project to form an apparent ellipse, with a major axis and a minor axis (see figure).

If you look at points which lie along the minor axis (say, points b1 or b2), you will see stars and clouds which are moving perpendicular to your line of sight, so they have zero radial velocity relative to you, and there will be no Doppler shift.

If, instead, you look at points which lie along the major axis (say, points a1 or a2), your line of sight will intersect gas clouds and stars which are moving (mostly) along your line of sight (either towards you or away from you), so you will observe the maximum possible radial velocity and thus the maximum possible Doppler shift (either positive = velocities away from you, or negative = velocities toward you).

There's one further factor, which is the effect of inclination. If the galaxy is genuinely face-on (inclination $i = 0$), then you won't seem any Doppler shifts, because all the motion is in the plane of the sky, perpendicular to your line of sight. (In practice there will really be some motions up and down with respect to the galaxy plane, but this won't tell you about the rotation curve.) If the galaxy is edge-on, then you get the maximum possible Doppler shift, which gives you the rotation velocity $V$ directly. If the galaxy has an in-between inclination, then the radial velocity is the part of the cloud/star velocity in the direction of your line of sight, which is $V \sin i$. So if you can determine the galaxy's inclination, you can divide the observed velocity along the major axis by $\sin i$ to get $V$.

Up until relatively recently, the practice (in optical astronomy) was to use a long-slit spectrograph and orient it so that the slit lies along the galaxy's major axis. Then at each point along the major axis, from the center outward, you measure the Doppler shift, apply the $\sin i$ inclination correction, and then you have the rotation curve: $V$ as a function of radius $r$ along the major axis.

If you have fully two-dimensional spectroscopy (e.g., from a radio telescope interferometric array observing emission from atomic hydrogen or molecular gas, or an optical or infrared integral field spectrograph), then you can use the data at intermediate locations within the disk (not just along the major axis), where the observed velocity is $= V \sin i \cos \alpha$, where $\alpha$ is the angle between the major axis and the location you're looking at (so for locations on the major axis, the observed velocity is just $V \sin i$). Then you can fit a model to the set of observed velocities (which you get from the Doppler shifts) and work out $V$ as a function of radius. (The standard approach is called a "tilted ring" model, where you assume the velocity is a simple function of radius, but allow for the possibility that the disk might be warped, so that the orientation -- including the inclination -- might change with radius.)


How open is China in sharing data from the Chang'e 4 mission? It's really good to see that man (well, machine) has gone back to the moon again.

Chang'e 4 is studying potential mantle material from the lunar depths and will be conducting radio astronomy in frequencies that simply aren't possible from Earth.

There's a huge amount to be learned that will fill some gaps in our knowledge of the early Moon, Earth, and the Solar system as well as the wider galaxy and beyond.

Will this information be open for the greater scientific community, or will we have to rely on the Chinese scientists to collate and publish the data as and when they can do?

Different space agencies and different governments have different policies about archiving their data.  One commonality that I have found, though, is that their systems are often difficult to find, they are often difficult to navigate or search and get the data, and then just as difficult sometimes to process.  Sometimes, they also require an account.
Doing an internet search for simply "Chang'e 4 data archive" linked me to this conference abstract as the first link, which was presented at the mid-2019 Data Users' Workshop in Flagstaff, AZ.  The abstract explains the different data types (such as level 0A, level 1, level 2C, etc.) and the different instruments, and it also explains their data archiving plan.
Perhaps more important for this question, it provides a link to China's "Data Release and Information Service System" within China's Lunar Exploration Program.
It has a lovely English translation button in the upper-left, so clicking on that (because I do not speak nor read any type of Chinese), and the news to the right shows they have Chang'e 4's 12th data release as their most recent posting.  You can follow through the links and find the data archived there, or, this is a direct link to the search engine.
Of course, as with any mission, agency, or government, you have to assume that they are publishing everything, but I see no specific reason to doubt that in this case.  Keep in mind, though, that there is always a waiting period between the time the data are taken and the time they are posted on public archives.  With NASA, that is usually 6 months, though at the start of mission data returns it can be more like 12 months.  So, just because data might not go up through today in these archives, that does not mean that they won't in 6–12 months.


Relation between phase and magnitude Given the phase of a planet or satellite, I can find the area illuminated visible to us, if we consider a 2D surface. But how do I find the percentage of illuminated area visible considering it a 3D surface and thus finding the magnitude of the partially illuminated planet/satellite.

In general, you'd have to project the light from the sun (Assuming we're talking about objects in our solar system) onto the 3D surface and project it onto a 2D surface. This can be accomplished via a ray tracing algorithm.

This procedure will give you the fraction of the surface illuminated.

Now if you are concerned about the fraction of the illuminated surface that is reflected in a specific directly -- so that it is "viewable" from that direction -- you have to do the ray tracing in 3D from the source (Sun) to the object and on to the viewer. There packages you can google that will do this type of ray tracing if you can model the objects in 3D...


Could a near collision with small black hole account for the effects of the 9th planet? Much interest has been raised about the existence of a 9th planet. One reason for this is the strange orbits of bodies in the outer solar system. Could the orbits be explained by a near collision of a passing black hole?
Recent work is looking to see if the 9th planet could be a small black hole. But my question is different. Could a passing black hole, say a million years ago, have thrown the objects in the outer solar system into the strange orbits they exhibit?

To quote from Batygin and Brown (2016)

Trujillo & Sheppard (2014) point out that the Kozai mechanism allows libration about both ω = 0 as well as ω = 180, and the lack of ω ~ 180 objects suggests that some additional process originally caused the objects to obtain ω ~ 0. To this end, they invoke a strong stellar encounter to generate the desired configuration. Recent work (Jílková et al. 2015) shows how such an encounter could, in principle, lead to initial conditions that would be compatible with this narrative. Perhaps a greater difficulty lies in that the dynamical effects of such a massive perturber might have already been visible in the inner solar system.

So the short answer is that it is unlikely a single encounter with a massive object could cause the perturbations without leaving other traces in the dynamics of the solar system. Furthermore, stellar encounters are rare enough as it is - encounters with black holes should be even rarer since stellar evolution and the critical mass requirement of black hole formation dictate there should be far fewer black holes than stars in existence.
To me that means favoring an encounter with a black hole to be responsible for the "unexpected clustering in their respective arguments of perihelion" of Kuiper belt objects over other theories would be a violation of Ockham's Razor.
All that being said, due to the chaotic nature of orbital dynamics and the relatively short Lyapunov time of Mercury in particular, it is very difficult to rule something like that out. It may be the case that an encounter occurred long enough ago that the only remnant is in the perturbations of long-period Kuiper Belt objects. And that all other traces of its influence on the inner solar system have been obscured by its characteristic emptiness - it could in fact have even been the cause of that unusual emptiness.
It should be noted of course that this would not have been a 'small black hole' (also sometimes referred to as micro black holes) as has recently been theorized in place of a 9th planet, but a stellar black hole (roughly 5 or more times as massive as our sun).


Life forms in brown dwarf atmosphere Some of brown dwarfs generate the enought amount of heat to warm their atmosphere up to 300 K. This is a condition to liquid water and, possibly, organic life forms.

Is where are a some studies about life forms in BD atmosphere?

Although you might think that Y-dwarfs (brown dwarfs with temperatures $<500$ K) might become cool enough to support liquid water. In fact that does not happen. As discussed by Burrows et al. (2003) and Morley et al. (2014), in Y-dwarf atmospheres, when the temperature becomes low enough for water to condense out of the gas phase ($<375$K), it goes straight to water ice particles.

Thus, although water vapour will condense  in clouds in a cold brown dwarf, it will be in the form of ice particles, and there is not expected to be liquid water in a brown dwarf atmosphere.


LIGO gravitational wave chirp signal frequency This question is about the LIGO gravitational wave signals' frequency. The signals start from about 35 Hz to peak at about 250 Hz, giving evidence of gravitational waves. The question is about the exact frequencies the waves' detection started to the exact peak of the wave.

It seems the LIGO graphs show the peak is a little
above  250 Hz. Are there tables published by LIGO that put the start and peak frequencies as 35 to 250 Hz? The graph in the paper by Abbott, et al shows the start and peak (the start is around 36-37  Hz, the peak seems to be at around 256 Hz) - are there exact numbers (tables) for the start and peak, that one can reference?

There's a detailed graph of it here:

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

What is the most precise reference for your query is the Time-Frequency graphs in black and green, also called waterfall graphs, spectrographs, sonograms and so forth.

On the Y axis, the width of the green smudge is about 30-50Hz, That's the frequency-uncertainty for the TF graph for every wave of the roughly 10 waves that make up the signal.

That's the actual real measurement that LIGO returned, and there are idealized versions here with an uncertainty range of about 5Hz.

If you are interested in sound theory, you can research about Wigner-Ville TF graph, which is the most precise sound analysis mathematics known, and which uses quantum uncertainty estimates to minimize the quantum uncertainty of placing single audio waves in time at a given known frequency. If you have java JVM you can compute a more deeply processed TF graph using sonogram 4 visible speech demo version.

As an synth-reverse-engineering sound scientist, I have to say that the precision of the wave is incredibly clear, it's the same as recording an individual bird chirp from 20 meters away on a clear microphone. Gravity actually has sound, like a bird call, except it's not carried by air compression it's carried by space time at c.

https://losc.ligo.org/s/images/omega_sample.png


As the Sun dies, does the Earth's orbit change? As the Sun nears the end of its life and expands, would the Earths orbit get bigger as it moves out or would it stay in the same orbital plane and not move out?  Therefore in the far off future, we'd have 512 days instead of 365 days?

As the Moon is moving out at a distance of 3.8 cms a year (ref: http://curious.astro.cornell.edu/about-us/37-our-solar-system/the-moon/the-moon-and-the-earth/111-is-the-moon-moving-away-from-the-earth-when-was-this-discovered-intermediate), are we moving out or in towards the Sun or is this not possible to work out because there is nothing to measure against?

The answer is yes. As the Sun ages, it will become a red giant and the mass loss rate from its surface will increase. This effect will increase (dramatically) further when the Sun enters the asymptotic giant branch phase, where thermal pulsations drives a cool wind that may carry away a millionth of a solar mass per year, eventually leaving a burned-out core in the form of a white dwarf with about half a solar mass.

At any point in this evolution we can model the evolution of the Earth's orbit using some simple approximations - that the wind from the Sun escapes to infinity, that a negligible proportion is actually accreted by the Earth and nor does it exert a torque,  that the mass loss takes place on a timescale much longer than the Earth's orbit and that the mass of the Earth $m$ is always much less than the time-dependent mass of the Sun $M(t)$.

In which case we consider the orbital angular momentum of the Earth:
$$ m a \omega^2 \simeq G\frac{M m}{a^2},$$
where $a$ is the semi major axis. So the angular momentum $J = m a^2 \omega$ is given by
$$ J^2 = m^2 a^4 \frac{G M m}{m a^3} \propto M a$$

As the angular momentum of the Earth's orbit is conserved, the $M(t) a(t)$ is constant and as the Sun loses mass, the semi major axis increases by the same factor.

Coming to the specifics - when the Sun is a half solar mass white dwarf, the semi-major axis will be 2 au (assuming the giant Sun did not quite engulf it - it will be a close-run thing) and Kepler's third law $(P^2 \propto a^3/M)$ can be used to estimate an orbital period of 4 years.

The tidal effects of the Sun on the Earth's orbit are quite negligible compared with these mass loss effects.


When black holes forms out of a supernova do they have a very high velocity? When black holes forms out of a supernova do they have a very high velocity ?? Or do they remains at the position of parent star ??

There is no specific answer to this -- anything from "just sits there" to flys away at high speed is possible.

It all depends on the symmetry of the supernova (SN) explosion. Extensive modelling shows that the explosions can be quite asymmetrical, and if they are the gravitational waves created can give the new black hole (BH) quite a kick.  If the explosion is relatively symmetric, the new BH can more or less stay right where it was formed.

Note that if the exploding star is a member of a close binary, the SN explosion will usually give the companion star a high velocity. Before the explosion, a close pair may be rotating around their joint center of mass at 1000 kps. When one star goes SN, it blows most of its mass right out of the system and high velocity. This doesn't change the velocity of the companion star which suddenly is no longer bound and has escape velocity from the star system -- so it more-or-less goes flying off in a straight line at its orbital velocity.

Some of the runaway stars discovered by Gaia are probably companion stars to SNs.