What is the Fundamental plane for Elliptical Galaxies? An alternate explanation rather than from wikipedia is preferable.

I am not an expert, but it seems that the fundamental plane is a relation among characteristic quantities of the galaxy, showing a correlation that is the analogous of the Tully-Fisher relation for spiral galaxies.

You can start from the virial theorem:
$\frac{M}{R}\sim v^2$

($M$ mass contained in the radius $R$, $v$ velocity dispersion of the stars, also indicated with $\sigma$ in these cases)
meaning that the stars behave like an isothermal sphere.

Assuming that all galaxies have the same mass-to-light ratio
$L/M$, and that all galaxies have the same surface brightness

$\Sigma=L/R^2$

(with $L$ luminosity), you obtain the Tully-Fisher relation:

$L\sim v^4$

Of course, all galaxies do not have the same surface brightness.
So if you take $\Sigma=L/R^2$ and substitute that into the virial theorem (keeping the mass-to-light ratio assumption), then a relation between luminosity, surface brightness, and velocity dispersion is obtained:

$L \sim \sigma^4 \Sigma^{-1}$

You can plot the stars distributions in elliptical galaxy according to this relation, and will note that the stars do not distribute randomly, but are more concentrated along the fundamental plane (more or less the same happens for the HR diagram):

(Here $R_e$ is the so called effective radius).

For completeness, another illuminating picture from here:

References here and here.


Is it possible that quasars are not powered by black holes but some other kind of compact, massive object? AFAIK, objects in the universe thought to be black holes show evidence of radiatively inefficient accretion flow. But is this the case with Quasars? If not, is there a possibility that the core of a quasar is an object similar to a neutron star.

The solution to your question is surprisingly simple, I think:

A quasar that puts out energy around Eddington luminosity or higher, must accrete at a certain rate, corresponding to the energy output.

When consulting a lecture and a random paper from the archive on this topic, it is evident that those accretion rates correspond to 1-10 solar masses per year or higher. A neutron star accreting 1-10 solar masses in one year will become a very massive black hole quickly enough.


"Table of Astronomy's" depiction of the solar system models This is going to be a follow up question to my question about the "Table of Astronomy" from yesterday, What is this “Table of astronomy” about?.

I wanted to take a close look at the three depiction of the solar system models (left to right):

Fig. 43. Ptolemaic System; Fig. 45: Tychonic System; and Fig. 44: Copernican System; They are is this order in the original resource

I cant figure out what some of the symbols and circles are supposed to mean. I have marked everything im unsure about partially or completly in red.

(From left to right again)

Ptolemaic System:

The ptolemaic system is a geocentric system, meaning it has the Earth as its center. I have identified most of the symbols that mark the orbits, although I'm missing two. Furthermore, the 5 outer most orbits(?) are what I'm unsure about. As far as I understand it, Claudius Ptolemy believed not only that the Earth was the center of the universe (making this a model of the universe and not the solar system) but also, that the stars were laying on the inside of the universesphere. Therefore I can imagine that the innermost ring is just to close of the sphere. But there are still 4 rings left.

In addition to the rings, I cant identify the third (anchorlike) and the fifth symbol.

Tychonic System:

Here again I'm unsure about the two outer rings, and the third, not dotted, orbit around the Sun.
The two innermost orbits are Venus and ?

If the (red)dotted orbit is supposed to be the Earth's orbit around the Sun, then why does the earth have two orbits?

Also, what are the two stars under the Moon?

Copernican System:

This one is rather straight forward. I'd like to know which of Saturn's moon's orbits this illustration depicts and if my guesses on Jupiters moons are correct.

And at last, what are the 4 marks, A, B, C and D for?

I hope you can help me! Thanks

Ptolemaic system

Planets

In Ptolemy's system, the two missing symbols are Venus (♀) and Mars (♂), respectively. The arc in the bottom of the Venus symbol actually belongs to Mercury below (☿).

Stars

The two following spheres contain the stars, which were believed to be eternally fixed. The two next are, I believe, also containing stars, namely the zodiac constellations. The two last ones are the final boundary before the outside, which is the realm of God. The spheres are more discernible is this version (Petrus Apianus 1539):

Tychonic system

Planets and stars

In the Tychonic system, the two outermost rings again contain the fixed stars. The planets do not move between the spheres, but are fixed on the spheres, so the next sphere you ask about ("the third, not dotted, orbit") belongs to Mars (♂). The innermost one is Mercury, although its symbol (☿) isn't drawn in this version.

Earth's and Sun's orbits

The dotted orbit isn't Earth. In the Tychonic system, Earth is in the center, and the Sun and the Moon orbit Earth, but other planets orbit the Sun. This orbit isn't drawn in many other versions; I think it's the orbit that Earth takes around the Sun, if you considered the Sun in the center (which Tycho didn't). Although it's not exactly the same size as the Sun's sphere.

The Moon

The two stars underneath the Moon don't have anything to do with the Moon, I think, but belong to the spheres of Saturn and Jupiter. The reason I think this is that you see the same in some other versions, like this one from Tycho's De mvndi ætherei recentioribvs phænomenis (here taken from Graney (2010):

Copernican system

Saturn

Saturn's moons had not been discovered when Copernicus published his model in 1543. The drawing in your source is from 1728, at which time five moons had been discovered (Titan in 1655 by Christiaan Huygens; Tethys, Dione, Rhea and Iapetus between 1671 and 1684 by Giovanni Domenico Cassini). So since there are four rings, my guess is that they don't represent moons, but rather the rings, with the two midmost circles representing the Cassini Division, discovered in 1675.

Jupiter

Your listing of the moons of Jupiter is probably correct, since that's how they're ordered according to their distance from Jupiter.

Letters

I don't know about the  letters A, B, C, and D, but my guess is that they mark the direction in which the planets revolve around the Sun, which is correct if viewed from above the ecliptic plane, i.e. the plane with respect to which Earth has an axial tilt of some 23º.


Is our universe flat? I heard our universe is flat. Then one question is puzzling in my mind. If our universe is really flat, why we measure distance from any point across in a spherical way. In other words, why we say, the comoving distance (radius) is now about 46.6 billion light years. If this radius is only treated for observable universe, then how does science find that the universe is flat knowing nothing beyond the limit of the universe.

There are some questions in SO, however, they do not assimilate what I intended to know. In this question, for instance, OP asked about whats the shape of the universe. And also there are some terms related to black holes, 3D shape, ball etc. Apart from those complexity, my intention is simply to be clarified, why we call our universe flat? And by word flat, what I meant "having a level surface; without raised areas or indentations". Am i missing any alrernative meaning of flat here?

And also there are some terms related to black holes, 3D shape, ball
  etc. Apart from those complexity, my intention is simply to be
  clarified, why we call our universe flat? And by word flat, what I
  meant "having a level surface; without raised areas or indentations".

Imho in order to accept the answer, you have to realize the contradiction in the sentence above: You mean level surface, but we call the universe flat :) "Flat" in general relativity does not refer to "having a level surface; without raised areas or indentations". Please accept this first.

Now let's think of a ball (yes, a ball; it's not complexity). Would you call its surface flat or curved? I'd call it curved. Let's think of a pan. I'd call its surface flat.

Both the surface of a ball and a pan have two dimensions. The ball itself has three dimensions obviously, but its surface has two dimensions. I hope this is straightforward.

Now, everything described above about surfaces also applies to volumes. Volumes can be flat, and they can be curved. The issue is that humans cannot visualize things with three dimensions bending, so they cannot visualize a curved volume. But this is okay: We can still grasp the meaning of flat and curved, just by seeing it on surfaces, and understand that it can be extended to volumes.

However it's not just visualization and understanding: Flat and curved volumes (i.e. spaces) have very different physical and mathematical properties. A usual example is that in a curved space the sum of the angles in a triangle is not 180 degrees. There are also other properties and those enable astrophysicists to measure the actual curvature of the universe. I hope this was helpful.


How do we know Nemesis is not a black hole (or neutron star)? Nemesis, the hypothetical "death star", is supposed to be a massive body that orbits the Sun at long distances and periodically sends comets from the Oort Cloud into the inner solar system. These comets impact the Earth and cause extinction events. It hasn't been found, and the theoretical case for it is not compelling anyway.

My question here is about this curious line from Richard Muller's page at LBL.

  Fortunately, several all-sky surveys are underway that should find Nemesis in the next few years, if it is there, and rule out Nemesis if they don't. (Nemesis could hide if it were a black hole, but that is not very plausible.) These surveys include Pan-Starrs and the LSST.

How do we know it's not plausible for Nemesis to be a black hole? For that matter, how do we know it's not a neutron star?

If the Sun had been born in a relatively wide binary system with a star that was to become a black hole or neutron star via a supernova, then (a) it is quite likely that such a system would be disrupted by that supernova and we would not be in a binary system now; (b) there should be evidence of the supernova in the form of very high abundances of the daughters of certain short-lived radionuclides incorporated into solar system material. There is some evidence of the latter, but not I think enough for the Sun to have been in a binary system with such a star (though I might be checking this).

An alternative argument is that the Sun is captured in orbit by the stellar remnant at a later date. This avoids the supernova problems, but the capture process is inherently unlikely in our Galaxy once stars have left their birth environments, especially capture which is tuned rather precisely to yield just less than zero for the resultant system potential energy of a very wide binary. Capture by a "normal" star would in any case be much more likely than capture by a relatively rare compact object.


Why are the ALMA receivers' ADCs only 3-bits? above: Table 1 from Performance Highlights of the ALMA Correlators

The ALMA receivers use 3-bit ADCs for what would seem to be to be a high dynamic range application needing much finer quantization to get anything useful.

Then I found these sentences within the abstract of ADC bit number and input power needed, in new radio-astronomical applications:

  Abstract- For the most part, so far radio astronomy observations have been performed in protected frequency bands, reserved by ITU for scientific purposes. This means that, ideally, only the amplified equivalent system noise is present at the end of the receiver chain (i.e. the ADC input). So, typically, only a few bits are necessary to describe the signal (VLBI signals are digitised with only 2 bits), but today astronomers, in order to get more sensitivity and to boldly observe where no one has observed before, would like to study the radio sky even outside the protected bands...

And I even found a 1-bit ADC in Performance Measurements of 8-Gsps 1-bit ADCs Developed for Wideband Radio Astronomical Observations.

I think I am just missing something obvious, but I can't understand how a measurement requiring high dynamic range gets by using few-bit ADCs.

edit: Is it possible that the actual conversion of analog to digital is done to a far higher precision than suggested by the number of bits?

It is wasteful to sample with many bits because the signal to noise ratio at the ADC of a radio telescope is typically << 1, so using many bits would just be resolving noise. (An exception to this is when there is strong radio-frequency interference that needs to be resolved, but this is not a big problem for ALMA due to its location and observing frequencies).

High dynamic range measurements arise after averaging together many samples (or correlations of samples), which boosts the SNR to a meaningful level.

Using very few bits at the ADC does introduce quantization noise that reduces the efficiency of the instrument, but 3 bits is enough to achieve 96% efficiency [1].

[1] Convenient formulas for quantization efficiency


Cardinal directions in solar system? Does the solar system also have cardinal directions? Which one is North and in what direction does it point?

There are a bunch of different measurements for celestial coordinates, but in regards to the solar system - or specifically the Sun as its "master" - there's the ecliptic celestial coordinate system.

In that system, centred around the sun and our solar system by extension, the North pole is in the constellation Draco. According to Wikipedia, the coordinates from Earth are right ascension 18h 0m 0.0s (exact), declination +66° 33′ 38.55″.

Source


Right ascension and sidereal time difference I've read quite a few articles about this and watched visualisations but I still struggle to get a mental map of how this works

I understand that on March 21/22 RA 0 is on my meridian at local noon. Ignoring the sunlight, say there is a star also on that meridian at noon, in a month (April 21st) I understand that solar and sidereal time will have a difference of about two hours - does this mean the star will now be about 30 degrees west of the noon meridian?

Also I read that sidereal time means celestial objects are always appearing in the same place in the sky but the sidereal times can be different? I think my main confusion is how sidereal and RA relate.

To summarise I'm looking for more explanation regarding the two hour rule per month. Everything makes sense on the date of the vernal equinox but on say April 21st why is Sirius appearing two hours earlier? I know it's going to be obvious when it hits me :D

Thanks

Question 1: How do sidereal time and RA correlate together?

Think of the sky as a globe. The constellations and lines of right ascension (RA) are painted on the globe. Because of the Earth's rotation, it appears that the sky globe rotates east to west once every 23 hours 56 minutes 4 seconds.

The meridian is the line of RA that goes from due south to the zenith (the point overhead) to due north. The meridian splits the sky into an eastern half and a western half.
The sidereal time is just the RA that is on the meridian. Because of the rotation of the sky, the sidereal time increases by approximately 15 degrees (or 1 hour of RA) every hour of time.

Question 2: Why do the stars rise earlier each month (and each day!)?

The Sun (and Moon and planets) are not painted on the globe of the sky. They move on the globe at different speeds (depending on the distance from the Earth and Sun). The Sun moves west to east and goes through 360 degrees along the ecliptic in 1 year, or approximately 30 degrees per month.
Because we use the Sun to tell time, the Sun is on the meridian at local noon.
But one month after March 21, the Sun is at 2 hours RA (= 30 degrees east of 0 RA). Therefore, the star that is at 0 RA and was on the meridian at noon on March 21 is now 30 degrees west of the meridian at noon on April 21.

In other words, because the Sun moves west to east, at the "same time of day" the stars appear to move westward by 4 minutes per day, 32 minutes per week, approximately 120 minutes per month (2 hours per month). That's why Sirius appears 2 hours earlier each month.

Technically, the stars appear earlier because it only takes 23 hours 56 minutes for the Earth to rotate 360 degrees measured relative to the stars, but it takes 24 hours (on average) for the earth to rotate 360 degrees relative to the Sun. The 4 minute difference is due to the motion of the Earth around the Sun, or the apparent motion of the Sun on the globe of the sky.


Is it possible to see mercury transit "clearly" by the naked eye? I don't have a telescope but I'm interested in seeing events like eclipses and transits.

I'll use the atmosphere as my big natural lens. So I'll watch the upcoming mercury transit at the sunset time where the sun looks bigger than usual. Will this make the black dot of mercury bigger or big enough to be obviously seen by the naked eye? I've never seen it before.

Event: https://www.timeanddate.com/eclipse/in/egypt/cairo

I did see the Venus transit before the Sun in 2004.

I used no telescope, but of course I used proper solar eclipse glasses to protect my eyes. The black circle was small but clearly visible.

But Mercury is much smaller, the transit is not visible without a telescope. See Wikipedia.

Here an image of a Venus transit:

Here the much smaller Mercury:

The dark spots to the right and left are sunspots, Mercury is the very tiny black spot below the center of the Sun.

Images from Wikipedia.


Telescope in Sun's gravity lens focus - pointing, gain, distortions A telescope located in the gravitational focus of the Sun can use the Sun as a magnifying lens. The focus begins 550 AU away, but maybe a 700 or 1000 AU distance is needed to get rid of disturbances from the Corona, and the focus extends practically indefinately. Here are some slides by Dr. Maccone who has promoted this idea he calls FOCAL:
http://www.spaceroutes.com/astrocon/AstroconVTalks/Maccone-AstroconV.pdf

I intend to ask about the technical design and feasibility of such a project in the Space Exploration SE. Here I rather ask about the scientific value and challenges.

POINTING: The magnification would occur only in the exact direction of the Sun. But since the Sun moves and the background objects magnified move, I suppose that the observed targets would change continuously. Would it even be practically possible to give the telescope a trajectory which keeps it aiming at for example Alpha Centauri? Would there most of the time be nothing in the right direction as the line between the telescope and the Sun sweeps across space, or would there alway be some star or galaxy in sight? Like CMB if nothing else.

GAIN: In the slides linked above, Maccone has calculated the expected gain to 114 dB for infrared wavelengths. How many times "magnification" does this mean? I don't think I understand the units here, I get a ridiculously large number. Can it be explained somewhat intuitively? Would a FOCAL mission be a unique revolution in astronomy, or could similar results be achieved by building an interferometer with interplanetary sized baselines here nearer to the Sun? How does the science value of a gravity lens compare to that of a wide baseline? Are they good for different tasks?

DISTORTIONS: Could the lensed signals be reconstructed thanks to our knowledge of the Sun and measurements of corona activity? If pointed towards a central part of the Milky Way, wouldn't signals come from multiple objects at the same time, some much further away than others? Would the gravity lens of the Sun have bigger problems with distortion than the intergalactic gravity lenses we know of today?

And finally, can any natural strong lensing inside the Milky Way be used today, for example using a globular cluster as a lens?

The pointing is not a fundamental problem with the suggested design: The suggested trajectory is designed to include a Sun flyby as the last flyby. This ensures an asymptotically radial trajectory away from the sun after the flyby, hence maintaining the pointing relative to the sun. The proper motion of the observed object may be some challenge, but the trajectory could be adjusted appropriately by an additional burn.

114 dB are an amplification of a factor of about $2.51\cdot 10^{11}$.
It doesn't refer just to the magnification, but to the intensity of the signal.
Therefore interferometry with a long baseline isn't the same; the latter provides a high resolution.
Whether those numbers are achievable in practice is a different question; a factor of 1000 for naturally occuring gravitational lenses would mostly be regarded as excellent.
Theory allows arbitrary amplification for perfect alignment of observer, lense and observed object.

Most of the distortions can be deconvoluted, revealing the field of gravity of the sun, and the shape of the observed object.

Gravitational lensing is used today, here some lecture notes, and here quasars as an example.


How deep and shaped is the depth of a black hole? A common representation of a black hole is a black circle which if I'm correct, indicates its breadth.

This representation depicts it as a 'plate', so a really small depth compared to its breadth.

On this other representation, its depth has a funnel shape.

Question:

How deep and what shape has the depth of a black hole ?

The first representation isn't depicting the black hole; it's depicting its accretion disk, which is indeed plate-shaped. The accretion disk is around the black hole, and - if it's a rotating (Kerr) black hole - it's perpendicular to the black hole's axis of rotation. Accretion disks are disk-shaped, and so the representation is disk-shaped.

The other is a common representation of a black hole, but it's misleading. It's an extension of the "rubber sheet" analogy, which describes the bending of spacetime. A heavy object, placed on a semi-taut sheet, will curve it a little bit. This represents ordinary matter curving spacetime. A black hole is represented in that image as bending spacetime by an infinite amount. The issue is that this representation embeds the two-dimensional representation of spacetime in three dimensional space, making it appear as if the black hole has this extra depth. Black holes are not embedded in an extra spatial dimension.

Note also that the small object moving near the edge of the black hole is also represented by a tiny funnel. The creator seems to have tried to take the analogy to the extreme, portraying the object's funnel as finite, and the black hole's as either extremely large or infinite, in an attempt to explain that light cannot escape from a black hole. It's not too accurate, but it works to get the point across.

By the way, black holes do have size (the central point singularities of non-rotating black holes are infinitely tiny, while the ring singularities of rotating black holes are in the shapes of infinitely thin rings). You might want to read up on event horizons. The radius of the event horizon of a non-rotating black hole is a finite number, the Schwarzschild radius. It is calculated as
$$r_s=\frac{2GM}{c^2}$$

In this case, the event horizon is spherical. In rotating black holes, it may be in the shape of an oblate spheroid.


Why did the moon abruptly change positions in the sky? Three or four weeks ago I was outside showing my neighbors how cool the moon was through a 'backyard' telescope in broad daylight (~8PM PDT).  It was 3/4 of the way across the sky (setting behind the houses behind us.)  Two weeks later the moon was not even up yet at 8PM.  The planets I've been watching (Jupiter, Saturn, Mars) are all rising and setting a little earlier each day as I would expect.

Why did the moon abruptly change its position in the sky a couple of weeks ago?

Compared to the planets the moon changes it's rise and set times very quickly.  Using the calculator on this page it can be seen that if one were looking at the sky on May 1st, 2014 from Irvine, California the moon set at 10:08PM (which at 8PM would have had the moon most of the way across the sky as noted in the question.)

The moon rises and sets approximately one hour later each day.  So, by May 14th the moon did not even rise until nearly 8PM - which also correlates with the observations noted above.

Contrasting that with the other context point made in the OP, using this page we can see that on May 1st Saturn rose at 8:07PM and on May 14th rose at 7:11PM.

So while the moon changes its rise a set times by almost an hour each day, the planets change much more slowly by comparison, almost an hour over two weeks.

If one weren't continually watching the moon's progression it would appear to abruptly change positions when comparing it to the planets' movements.


What possible effects of life replicating on gas giant atmosphere? If bacteria managed to arrive then thrive somehow (if not already) on a gas giant could it cause the atmosphere to expand or contract in volume?

Would living matter potentially induce energy into the planets system to heat it up?

Yes, but probably not by a huge amount. There are two obvious ways this could happen:

Direct change in atmospheric composition. If our microbe does something like eat methane and excrete longer-chain hydrocarbons, that's going to eventually change the atmosphere enough to have some effect. (If the hydrocarbons weren't already forming, they probably aren't stable under local conditions, so there will be some equilibrium reached, the way the Earth's atmosphere isn't 100% oxygen despite hundreds of millions of years of photosynthesis.)
Change in solar heat balance. The change in atmosphere will probably also change the amount of greenhouse effect (one way or the other), and second-order effects or even just enough microbial biomass can change the albedo (again, either way). On Earth, these effects can combine to change the temperature by tens of Kelvins, which would be a few percent different in atmospheric volume.

In neither case is the added life going to change the total mass of the atmosphere.


Can I use an array of inexpensive cameras as an alternative to a telescope? The link below describes how an array of inexpensive cameras can create very detailed images. Would this work for astronomy?

https://graphics.stanford.edu/papers/CameraArray/

Yes it would.  However, it also depends on your definition of "Detail".  For some, it's a tight shot where the whole frame is only a few dozen arcseconds wide, and others think of detail as being wide field, but also "deep" - or having a very high signal to noise ratio.

If you want wide field, and a high signal to noise ratio, then yes, a multi-camera array will work well, and you'd be able to accumulate a large number of high integration time images in a relatively short amount of time.

So here's a breakdown of what you and others who may not know need to know about deep sky astrophotography:

The main goal for all photos is to produce as visually pleasing image as possible.  A single image taken of any target will accumulate light from many sources.  The first, and obvious source is the light from whatever target you're trying to photograph.  The second, less obvious source is light pollution from man-made sources, the moon, and natural sky glow.  The third source is the electronic and thermal noise of the camera itself.

Technically, all of these sources are "signal" - however two of the three are unwanted, so we will call the light pollution signal and the camera signal 'noise'.  Longer exposures produce more signal from what you want, but also more noise from what you don't want.

Luckilly, noise can be reduced by using several techniques in conjunction with each other.  First, you need to take long photos.  The signal will increase linearly with time.  A 60s photo with a signal of "x" will have a tenth of the signal as a 600s photo.  Noise, however, increases with the square root of the exposure.  Using the same example, the 600s photo will have a signal to noise ratio that is about 3.2 (sqrt of 10) times better than the single 60s photo.

Increasing the Signal to noise ratio by taking longer exposures is the first step to producing 'detailed' wide field images.  Camera noise can be further reduced by taking many images.  An array of cameras makes this trivial.  Two cameras will produce twice the images as one camera, and 10 cameras will produce 10 times the images as a single camera.  That's obvious.  Why does it matter?

When we stack images, the more we have, the better the stacking algorithms can preserve the signal we want to keep, and reduce the camera noise we don't.  Once again, the signal to noise ratio mathematics work out the same as above.  1 image will have a certain signal to noise ratio, but 10 images will have a signal to noise ratio that can be about 3.2 times higher - because the total integration time has increased ten fold.

Of course, if you're stacking photos, you should be calibrating them as well (dark frames, bias frames, etc) - which really reduce the amount of noise per sub-frame.  Because of this, your actual SNR boost by stacking should in theory be a lot higher than 3.2.

If photographed a target across 60 degrees of the sky (30 before meridian and 30 after) then you'd be able to acquire photos for 4 hours.  One camera taking 10 minute photos would get 24 photos in the 4 hour period of time.  If you had 5 cameras, you'd get 96 photos.  72 additional photos in the same amount of time is a lot of extra data to help reduce the noise, and really let the faint details stand out above the background - especially if all of your photos were well calibrated.


Can the Hubble Sphere ever extend over the cosmic event horizon? I am struggling to reconcile the Hubble volume with the idea of a cosmic event horizon.
As I understand, the Hubble volume is increasing over time because Hubble's constant is decreasing. This should allow photons emitted in superluminal regions to enter into the Hubble volume and become subluminal.
The cosmic event horizon, as I understand it, is the point where any light emitted beyond the cosmic event horizon will never reach Earth.
So, can a galaxy emit light that can reach us, but then pass over the cosmic event horizon, such that any light it emits after this point will never reach us? And, if the Hubble volume is increasing in diameter, why can't it overtake the cosmic event horizon, allowing the light in this region to now travel subluminally?

No, the Hubble sphere can never extend over the cosmic event horizon; it can only asymptotically approach it.
The Hubble sphere is the region within which galaxies (and other stuff) recede from us slower than light. The event horizon is the region outside which no light can ever reach us. If the Hubble sphere extended beyond the event horizon, it would mean that there exist a region in spacetime that recede subluminally, but which we could never see. But we have always been, and will always be, able to see light from subliminal regions.
However, a galaxy can cross the event horizon, such that it is no longer able to emit light that we can see. This is most easily seen in comoving coordinates, i.e. the coordinates that expand along with the Universe. In these coordinates, galaxies have fixed positions, but the event horizon is always shrinking.
In the spacetime diagram below you see time as a function of comoving coordinates. We are where the blue lines cross. The yellow region is the part of spacetime from which we may receive a signal, bounded by the event horizon (orange), and slowly approaching the Hubble sphere (purple). Galaxies at fixed coordinates follow lines parallel to the black dotted line, and may hence eventually cross the horizon.
For example, today the distance to the horizon is 16.5 Glyr. A galaxy at this distance may emit a photon now that we may detect in a almost infinitely distant future, redshifted to the extreme radio regime.
But tomorrow, it's too late.


Before Voyager 2, what were the highest quality photos of Uranus and Neptune? I'm in search of historical knowledge of the outer ice giants, which seems so hard to come by. Right now I would be happy with just a photo of each, the best photo we had of them before Voyager 2. Such a photo from the 60's to early 80's would be great. This will almost surely be from the Hale Telescope at the Palomar Observatory, because it's 5 meter primary mirror was the biggest for a long time.

Note: I'm not entirely sure that Neptune would even be a resolvable disk, but nevertheless I still want to see what our best photo of it (from that time) would be.

After googling this for so long, I've come to the conclusion that the best bet would be old astronomy books, which sadly, I do not have. Maybe someone here can help me out.

As per the books referred to in comments:

Both planets could be resolved as a disk, but no surface features could be observed on either. Spectrograms had been taken, so the general colour of the planets was known. The extreme axial tilt of Uranus was known about. But the relative warmth of Neptune was not.


Movement of the stars relative to my position I live in the Pacific North West (Washington State, US). I have watched the skies for years, looking for constellations that I recognize and keeping track of them. One of them is Orion. As I have watched Orion I noticed that it travels East to West as I look to the south. It traverses the line that is the Milkey Way...

My question is. If the stars that make up the constellation of Orion move East to West from my position, looking up and South. Why does the Milky Way seem to stay in the same position? How is it that the stars that make up Orion move but the stars that make up the milky way appear to be fixed?

Thank you in advance for any input.

The Milky Way consists of numerous faint stars near our galactic equator.
Many of these are visible in binoculars.
As time lapse videos show, they move across the sky along with the bright stars.
The light and dark patches in those digital images are easier to see visually near the galactic center in Sagittarius than in the opposite direction near Orion.

Due to its oblique angle, the galactic equator appears to go through a special cycle as everything revolves around the celestial pole.
From mid-northern terrestrial latitudes, it runs overhead NW-SE when Orion is highest, low in the north when Virgo and the Big Dipper are highest, and overhead NE-SW when Cygnus is highest.
We see no more than half of this cycle on any given night, which may explain the illusion you perceived.


How is the entropy of a blackhole stored at its horizon? I read that a blackhole has an entropy that is given by the area of its event horizon (measured in Planck units). Does that mean that this information is actually physically stored there? If yes, how can that be since there may be nothing there at all (no electrons for example that could carry this information in terms of spin up or spin down)? Is it perhaps stored in the vacuum somehow ?

Ultimately this is a highly theoretical question, and while appropriate here you may find perusing the Physics site helpful to get deeper answers.

But here's the basic idea for how I think about this.  It's based on asking "where else would the information be?"

Imagine the universe is a 2-dimensional sheet.  A piece of paper, say.  And that's all there is for things inside this universe: no desk it's sitting on, no air around it, etc.  Whatever inhabitants of this universe wish to talk about must be contained within this 2-dimensional sheet of paper.  Now imagine there is a hole in this paper.  What can in-universe inhabitants possibly use to learn about this hole and whatever properties it might have?  What to us, as outside observers is "the hole" is inaccessible to them: it's not part of the paper (it's an absence of paper).  But the border of that hole is accessible to them.  So whatever information this hole conveys to this universe must be accessible through that border, and the border only: anything else either doesn't exist, or is some distance away.

The hole in this paper is our analogue for the black hole in our universe.  "The inside" of a black hole doesn't make sense to an external observer, because the universe—the sheet of paper— ends at the border of the hole.  There literally is no inside to speak of.

The analogy gets complicated a bit by the fact that if you are not an external observer, but are infalling, then you can in fact pass through what the external observer calls the border. There's the whole information paradox as well when we bring in quantum mechanics.  But this is the basic gist of things: the black hole is a topological/geometric defect (as general relativity is a theory about the geometry and topology of space), and the only "part" of it which is contained in our universe is its border, the event horizon.

The actual physics at and near the event horizon is still a matter of significant debate and research.  This little exercise does not explain why the information of the No-hair theorem is accessible, but nothing else is, for example.  It just offers some intuition about why whatever information is accessible must, in some sense, be stored "on" the event horizon.  As far as external observers are concerned, that's the only place it ever could be!


What is period doubling in a variable star? I'm coming across the concept of "period doubling" in  RR Lyrae stars.  When I look up what period doubling is on the Internet, I find general-level explanations of bifurcations of dynamical systems and transitioning to chaos. However, I am unable to apply these things to stars in a way I understand.

What is period doubling, and what is physically happening in a star to lead to period doubling?

I think part of the confusion comes from the fact that "period doubling" can mean two rather different things, which are probably related but this is not completely known.  The first is a purely observational meaning, which happens when you have a clear periodic pulsation in a star, but the amplitude shows a higher-lower-higher-lower pattern that implies a "double" period to see the full cycle.  Then there's the theoretical meaning, which is a step in the approach to chaos.

The way the approach to deterministic chaos often works is, you have a simple system that shows a simple period.  But then you increase the degree of nonlinearity in the system, and you see bifurcations into something that requires double the number of iterations to be cyclic.  If each iteration represents a cycle time, like the time to go around an orbit or the time for a pulsation, then that means double the period.  Maybe you started with an interative mapping that converges to a fixed point, but then you increased the nonlinearity and it turned into a limit cycle where the iterations go back and forth between two points.

Now, it's not immediately obvious the connection between that and the amplitude modulation of a simple cycle, and I think many of the sources are purposefully ambiguous about that connection because nonlinear dynamics is still a mysterious field.  But the basic idea seems to be, if you have a linear system, it has uncoupled pulsation modes that are higher harmonics of the lowest frequency "fundamental" mode.  But if you have a nonlinear system, you can have coupling between those modes, which end up looking like amplitude modulation in a lower harmonic due to resonant coupling with a higher harmonic.  If the coupling has an odd-integer relationship, like a frequency of 2 coupling to a frequency of 3, then if the first oscillation in the frequency 2 mode couples resonantly with the frequency 3 harmonic, then the second oscillation will not couple well, it will hit the harmonic when it is 3/2 of the way through a cycle, so is out of phase.  Then the third cycle of the frequency 2 mode will hit the frequency 3 mode when it has gone through 3 full cycles and is back into resonance.  So that gives the "period doubling" where the amplitude of the frequency 2 harmonic gets a boost every second time around, owing to its ability to receive energy from the higher mode as well.

The connection between that and period doubling bifurcations is not entirely obvious either, but it seems that the idea is that if you have nonlinearity, the period doubling you find on the path to chaos might be related to this kind of resonant interaction between the simple behavior you see, and higher frequency behaviors that could exist but only make their presence known as a kind of residue from how they interact with the behavior you do see.  In other words, if the linearity is weak, some simple oscillation gets excited, but if you crank up the nonlinearity, that simple oscillation starts to overlap with more complicated behaviors that require more cycles to return to the original resonance relationship.  Crank up the nonlinearity even more, and all you see are the complicated behaviors, you have complete chaos.  The period doubling in variable stars represents only a small amount of nonlinearity, it's not chaotic and the relatively simple nature of the resonances allow for a large amplitude to build up and allow the stars to be classified as variable in the first place.

To do more research into this, you might look up the "Blazhko effect," or how period doubling also applies to other variable stars.  It sounds like an active area of research where there are still more questions than answers, but Kepler data has the promise of producing the necessary information to solve it.


White Dwarf and Degenerate Matter How do we know degenerate matter exists on white dwarfs? Is it purely hypothetical or have we observed it before? Have we ever created a form of degenerate matter on earth?

The density of white dwarfs is not hypothetical, it can be measured. The short answer is that the density is so high that a stable star can only be supported by electron degeneracy.

Sirius B is an example. The radius can be estimated by combining the luminosity of the white dwarf with its temperature estimated from spectroscopy. The mass can then be determined from the binary motion (with Sirius A), or gravitational redshift or from estimating the surface gravity from the width of absorption lines in its spectrum. These tell us that Sirius B has a radius of $0.0084 R_{\odot}$ and a mass of about $1 M_{\odot}$ (e.g. Holberg et al. 2012). This leads to an estimate of the average density of $2.4 \times 10^{9}$ kg/m$^3$.

Now, we know that for such an object to find any kind of equilibrium that the pressure must be greatest in the middle. Since for any plausible equation of state, the pressure increases with density, then the white dwarf must be even denser than the average at its centre.

From there it is a standard piece of bookwork to show why the white dwarf must be supported by electron degeneracy pressure and I will repeat it here.

An electron gas can only escape degeneracy if $E_F - m_e c^2 < k_{b}T$, where $E_F - m_e c^2$ is the kinetic energy associated with the Fermi energy of the electrons, $m_e$ is the electron mass and $T$ is the temperature.

The Fermi energy can be derived from the following formula (again, standard bookwork, this is not exotic physics, it is the same physics that describes how conduction takes place in metals).
$$ E_F = \left[ \left(\frac{3h^3n_e}{8\pi}\right)^{2/3}c^2 + m_e^2 c^4 \right]^{1/2},$$
where $n_e$ is the electron number density. Now it turns out that the temperatures we are going to need to avoid degeneracy (see below) are easily going to be sufficient to completely ionise the contents of the white dwarf. If we assume the white dwarf is made out of carbon or oxygen, then the relationship between mass density and $n_e$ is just $\rho = 2 n_e m_u$, where $m_u$ is an atomic mass unit (the 2 just comes from the fact that there are 2 mass units per free electron). We could mess about with the composition a little to change this relationship, but it doesn't matter very much.

Making the appropriate substitutions and using $\rho = 2.4 \times 10^{9}$ kg/m$^3$, we obtain $E_F = 0.75$ MeV (or $1.2\times10^{-13}$ J). If we now use the test
$$ T > \frac{E_F - m_e ,c^2}{k_b},$$
we find that to avoid electron degeneracy then the temperature inside the white dwarf must exceed $2.75 \times 10^{9}$ K (and be much higher in the centre of the white dwarf where the density and Fermi energy are larger). If temperatures were this high inside a white dwarf and it was made of something like carbon or oxygen then nuclear fusion would take place (rapidly) and the luminosity of the white dwarf would not be as low as it is.

But what if it were made of iron? There would be no nuclear fusion, even at these high temperatures. OK, leaving aside that if the interior really were that hot, it could not be hidden very easily by an atmosphere, then this would not be a stable situation. The star would be losing copious heat via neutrino losses. If the gas were non-degenerate, the star would rapidly contract and become even hotter and even denser, and if you are not going to allow for degeneracy (i.e. you postulate that quantum mechanics does not exist), then rather quickly the white dwarf ends up being a black hole.

A back of the envelope estimate would be to divide the gravitational potential energy by the current luminosity. Taking Sirius B as an example, the timescale is
$$ \tau \sim \frac{GM^2}{RL},$$
where $L$ is the luminosity. Taking the current luminosity from photons at the surface this timescale would be a few $10^{10}$ years (plausibly stable). But if the interior were at $\sim 10^{10}$ K, the neutrino losses would be many orders of magnitude greater and the timescale would be just millions of years.

We know that this isn't happening to white dwarfs because we know they are stable and long-lived. We see them in clusters of stars of known age and associated with stars that are much older than millions of years. For instance we know that stars like Sirius A are hundreds of millions of years old.
We can calculate using the theory of electron degenerate structures that the faintest white dwarfs we see have been cooling for 10 billion years without collapsing - precisely the age expected for the galactic disk. The cooling ages of white dwarfs in clusters match the ages of those clusters very well.

Degenerate matter is very common on earth. The conduction electrons in a metal form a degenerate gas. If you mean the kind of degenerate matter in white dwarfs, then no, at least not in any stable state. The pressures required to constrain such energy densities are too large to create in the lab.


Equations for coordinates of solar system objects I would like to write a program that visualizes the solar system. To do this, I would need equations for calculating the coordinates of each solar system object at a given time. Where could I obtain such equations? I hope they would be as accurate as possible. If there are such equations, it would be also nice to see the raw data they are based on and a description how they were computed. I would be grateful for all help.

It depends a bit on how precise you would want to be. A very good discussion on how to calculate the orbits of solar system objects is given in the book by Jean Meeus, Astronomical Algorithms (1999), which is at an advanced amateur level. At professional level you have the Explanatory Supplement to the Astronomical Almanac by Urban and Siedelmann.

For precise calculations Meeus uses the VSOP theory from Bretagnon (1987) (wikipedia link). You can download all files needed for these calculations from VizieR. These files contain a lot of numerical terms that are needed in the equations. The resulting positions are however very precise.

For the highest accuracy, you can download the predicted positions from the Jet Propulsion Laboratory (JPL) Horizon system.

There are also quite a few software library projects that implement the equations from Meeus. This might be the best option. Then you do not have to implement the equations yourself. For C/C++ you have for instance LibNova.

For highest precision you have professional software libraries such as for instance the NOVAS libraries from the Naval Observatory for Fortran, C, or Python. But to use that correctly you will have to have a good understanding of celestial mechanics.


If an object 1 billion light years away emits light, does it take more than 1 billion years to reach us because of the expansion of the universe? From page 7 of the recent (September 26, 2020) edition of Science News Magazine:

Detected May 21, 2019, the gravitational waves came from a source about 17 billion light-years from Earth, making this the most distant
confirmed detection.  Because of the universe's expansion, that
distance corresponds to a travel time of about 7 billion years,
meaning the gravitational waves were emitted when the universe was
about half its current age.

But that means the object(s) were only about 4 billion light-years or so away when the merger of two black holes occurred, right?  But it take 7 billion light-years, rather than approximately four, for the gravitational waves to reach us because of the expansion of the universe itself?
Is there an online calculator that can calculate this?  Does anybody know the equation(s) interconverting current distance, original distance and travel time of the radiation?

Yes, in the time it takes light — or, in this case, gravitational waves (GWs) from the black hole merger event GW190521 — to travel from a source to an observer, the Universe expands, thus increasing the distance further.
Various distance terms
In the following, "$\mathrm{Glyr}$" means a distance of a billion lightyears, while "$\mathrm{Gyr}$" means a time of a billion years.
There is a slight confusion, I think, in the quoted distance of $17\,\mathrm{Glyr}$ (Abbott et al. 2020): This is the so-called luminosity distance, which is the distance that satisfies the usual inverse-square law. This is not the same distance that you would measure if you were to freeze time and lay out meter sticks. This physical distance is smaller, only $9.5\,\mathrm{Glyr}$.
These values correspond to a redshift of $z=0.82$. That is, if GW190521 was luminous, its light would be redshifted by a factor $(1+z)=1.82$. In fact, in this case an electromagnetic counterpart was reported, albeit not with a measured redshift (Graham et al. 2020)
The time it took the GWs to travel to us is called the lookback time; it is the quoted $7\,\mathrm{Gyr}$. When GW190521 emitted the GWs we detect today, it was closer to us by a factor $(1+z)$. That is, its physical distance was only $5\,\mathrm{Glyr}$.
For a flat universe (which our Universe is to high precision) this is equal to the so-called angular diameter distance, called so because it is the distance that satisfies the usual relation between distance $d$, size $D$, and angle $\theta$, namely $\theta = D/d$.
Relation between lookback time and distance
In every-day life, all these distance measures are the same, and in the Universe, for small distances they also coincide. But because of the expansion of the Universe, and because the Universe's components (matter, radiation, and dark energy) influence its geometry, as the distance of an object increases they become increasingly different.
You can find the equations here, or use a cosmological calculator such as Ned Wright's (as commented by Alchemista). Alternatively, you can calculate them in Python using the module astropy like this:
>>> from astropy.cosmology import Planck15
>>> from astropy import units as u
>>> from astropy.cosmology import z_at_value

>>> dL = 5.3 * u.Gpc                                 # Lum. dist. in giga-parsec quoted in Abbott+ 20
>>> z  = z_at_value(Planck15.luminosity_distance,dL) # Corresponding redshift
>>> print(z)
0.8174368585313242
>>> print(Planck15.lookback_time(z))
<Quantity 7.11401487 Gyr>
>>> print(dL.to(u.Glyr))                             # Convert parsec to lightyears
<Quantity 17.28628801 Glyr>
>>> print(Planck15.comoving_distance(z).to(u.Glyr))  # Comoving dist. is equal to phys. dist. today
<Quantity 9.53452323 Glyr>

I used this to plot the current distance to GW190521 and other objects as a function of lookback time:

The answer to your title question
To answer the question in your title requires us to define exactly what we mean:

Light from an object which has a physical distance of $1\,\mathrm{Glyr}$ now, is redshifted by $z = 0.070$, its light has been traveling for $0.97\,\mathrm{Gyr}$, and it was $0.93\,\mathrm{Glyr}$ away from us when it emitted the light we see today.
Light from an object that was $1\,\mathrm{Glyr}$ away when it was emitted, traveled $1.03\,\mathrm{Gyr}$ before reaching us with a redshift of $z = 0.076$, and the object is now $1.076\,\mathrm{Glyr}$ away.

As you can see, the difference is not very large, but as you go to higher redshifts, it increases. The hitherto most distant observed galaxy, GN-z11, has a redshift of $z=11.09$. It was only $2.7\,\mathrm{Glyr}$ from us when it emitted the light we see today, but in the $13.4\,\mathrm{Gyr}$ it took the light to reach us (most of the age of the Universe), GN-z11 moved out to a current distance of $32.2\,\mathrm{Glyr}$!


Why are these telescope mirror makers not upside down? (Daniel K. Inouye Solar Telescope) The new open access Solar Physics article The Daniel K. Inouye Solar Telescope – Observatory Overview is the first of several papers about the new state-of-the-art solar observatory. For more info see Phys.org's Solar telescope releases first image of a sunspot
Figure 4 of the paper is shown below. It is the 4 meter primary mirror viewed at an oblique angle with several people in clean room garb standing behind it.
We can see their reflections in the mirror as well. The geometry is similar to people standing on the opposite side of a reflecting pool.
What struck me is that this differs in an important way; the people are not upside down!

Why not exactly?
Would the people see the camera upside down then?

It looks like the photo was taken from fairly close to the mirror based on the strong perspective scaling from foreground to background objects.

(Figure 4) The 4-meter primary mirror (M1) after coating at the AMOS coating facility on Haleakalā.


How do ASKAP's focal plane phased array feeds interact with the entire array phasing? Sky & Telescope's ASKAP Joins the Hunt for Mysterious Bursts says:

  A new telescope, the Australian Square Kilometre Array Pathfinder (ASKAP), has joined the search for energetic and elusive fast radio bursts. And in just a few days of looking, it’s already had success!

Later on, the article goes into some detail about ASKAP:

  A new player is now on the scene, however, and it’s already had huge success. ASKAP is a wide-field radio telescope made up of an array of 12-meter antennas. Using phased-array-feed technology, ASKAP is able to instantaneously observe an effective area of 160 square degrees — an enormous field compared to Parkes’ 0.6 square degrees! This capability significantly increases our chances of being able to detect fast radio bursts.

The article links to The Detection of an Extremely Bright Fast Radio Burst in a Phased Array Feed Survey which is not paywalled and contains some discussion of the array.

This sounds really interesting! It is a phased array of dishes, but at the focal plane of each dish there is also a phased array of receivers.

This is different than just adding an array of feed horns and independent receivers as I've discussed in my surprisingly robustly unanswered question What is the highest granularity focal-plane array on a dish radio telescope? Or is this the ONLY ONE? where receivers measure intensity as pixels, in a similar way to a camera (thus the counterintuitively placed astrophotography tag there).

Here, each focal plane is equipped with a true phased array, where if I understand correctly, phase relationships between elements in the focal plane are maintained and recorded.

According to Wikipedia and CSIRO's Australian Square Kilometre Array
Pathfinder – Fast Facts there are 36 separate 12-meter dishes in the entire array, and the focal plane of each dish is equipped with a 188 element focal-plane phased-array. 188 $\times$ 36 = 6768 channels, not counting polarizations because I don't know if they are separate channels in this case.

That would require quite a hefty correlater if the analysis was done flat, without any heierarchy in the calculation. This is a pathfinder for the SKA so pushing limits is important, but I'd like to know how this large number of signals is handled. Is there pre-processing?

Question: How do the phased array feeds on each dish of the ASKAP interact with the entire array phasing?

I'm not sure of the "focal plane array" is located precisely at the focal plane or not. If it were, then a given source in the sky would produce signals in only one, or a small group of receivers, and not the whole focal plane array! One normally uses a phased array instead of an imaging system so there must be something more interesting going on here, perhaps only nearest and next-nearest neighbor correlations within the focal plane array but full correlation between the dishes?

How does this thing really work?

below x2: Cropped from CSIRO ScienceImage 2161 Close up of a radio astronomy telescope with several more in the background.

The phased array feeds (PAFs) on ASKAP are indeed located at the focal plane. The ASKAP PAFs have 188 receivers, and from this digital beam-forming will create 36 independent beams on the sky. Some more information can be found from the CSIRO. Imaging is then done by scanning the dish (or letting the sky roll overhead). The scanning can then be done much faster than with a single receiver.

Its important not to confuse these independent beams with purpose of the array itself. The main reason for using an array is interferometry. We use an array for this purpose to have multiple baselines (the distance between the antennas). Having multiple baselines allows the telescope to 'see' structure various angular resolutions. It's true that you can use the array in a 'fly's eye' mode, and then every dish is providing 36 independent beams. In this mode even larger areas of the sky can be scanned quickly. This is useful for searching for transient events such as fast radio bursts.

So, the phasing within the PAF needs to be correlated and the beams formed digitally, and then the signals from each dish need to correlated. A huge task indeed! The computing breakdown for ASKAP is given here. A new supercomputer (Pawsey) has been built in WA for the purpose of processing ASKAP data.


How old is our Sun in Galactic years? A year is measured as the amount of time it takes Earth to orbit the Sun once, a Galactic year is the time it takes our sun to make one full orbit of the center of the Galaxy.

In Galactic years, how old is our Sun?

According to the blog How old is the Sun in Galactic years? (Siegel, 2008), a Galactic year is about 223 million Earth years.

The age of the Sun, according to the Stanford Solar Centre's page How old is the Sun? is approximately 4.57 billion years.

This would put the sun at approximately 20.5 galactic years old.


Would Pluto keep an orbit without its moon? I haven't studied space that much, but I know that Pluto has very little gravity, but just enough to keep a human on its surface. Would Pluto keep a regular orbit around the sun without some extra support gravity, such as its moon?

The orbit of a planet does not in any way depend in its moons, mass or own gravity. The orbit is the same for a grain of dust as for a giant planet. It only depends on the mass on the Sun and the distance to the Sun. The eccentricity of the orbit is then given by the initial angular momentum of the orbiting object. This might explain it a bit better.

This was a really BIG breakthrough in physics, actually the foundation of physics, about 400 years ago. Big because it is irrefutable although it is counter-intuitive. It is really difficult to believe unless one tries to accept the actual observations and logic.


Help with homwork - Astronomy We might expect to not find life around halo stars (“Population II” stars) because:

a)
they are out of the plane of the galaxy and thus exposed to more cosmic rays (energetic
nuclei)

b)
these stars are all too hot and thus will expose any planets to too much ultraviolet radiation

c)
these stars lack the heavy elements (“metals”) out of which life can form

d)
these stars have existed too short a time for life to have evolved

e)
the stars all pass too close to the central supermassive black hole in the Milky Way

I am stumped on this question, any ideas?

You will need to find out what population II
 stars are.

When you understand what a population II star is, ask yourself two questions about each statement:

For example a) "Are population II stars out of the plane of the galaxy, and do they receive more cosmic rays", and "Would receiving more cosmic rays actually prevent life from forming". If you can answer yes to both questions, you have a potential answer.

If you judge that several answers are possible, then re-read the definition of a population II star, and ask which statement is most pertinent to the definition of such a star. Good luck.


Are we moving ever closer to the center of our Galaxy due to a super massive black hole? I've seen in documentaries that at the center of each galaxy is a super massive black hole which holds the galaxy together.

Since black holes have such a strong pull, are we slowly being pulled in closer and closer to this black hole?

That's not really how gravity works.

When a galaxy forms, all the dust that falls into it retains angular momentum round the centre (and this pattern repeats locally round stars, and round planets round those stars etc) which leads to the relatively stable structures we see today with spiral arms etc.

Collisions and near collisions can change the direction of stuff orbiting a centre, but we are quite far out, so while we could change orbits we would be as likely to move wider than closer in the short term.

Over the longer term, it's a balance between the gradual loss of angular momentum of the Milky Way as a whole due to drag, gravity waves etc. and the slow growth of the black hole at the centre.


Can planets have orbits other than elliptical or circular orbits? Can planets have orbits other than elliptical or circular orbits? All of the planets in our solar system have either a circular orbit or an elliptical orbit (perhaps like Pluto).

Orbits are conic sections therefore can be either circular, elliptical, parabolic or hyperbolic.
Of these 4, only first two form a closed curve under 2 body hypothesis, while the later two extend to infinity.
If you talk about planet,  by definition it has to orbit a star which would require it to have a closed orbit hence circular or elliptical, with respect to the star. For any other kind of orbit the body will just fly away to infinity never to return back.  Infact comets are considered to be parabolic, but in theory they have highly elliptical orbits with their aphelion lying near the edge of the solar system.

But it is possible for a planet to have other kind of orbits if we  consider their motion from a different reference frame such as with respect to another planet.
So for an inertial frame of reference a planet will have a circular or elliptical orbit, even Pluto.


What does the "a" at the end of a spectral type mean? Looking at the chart for spectral type peculiarities (here), I noticed that the letter a was missing from the chart. This seems to be an error, as the star Vega has spectral type A0Va, with the a definitely being a peculiarity. What does the a in spectral types mean, and is this actually a spectral peculiarity?

If present, an a (or b or ab) do not refer to spectral peculiarities but are part of the luminosity class definition explained further up on the page.

Occasionally, letters a and b are applied to luminosity
classes other than supergiants; for example, a giant star
slightly less luminous than typical may be given a
luminosity class of IIIb, while a luminosity class IIIa
indicates a star slightly brighter than a typical giant.[27]


What might cause a planet to have a significant tilt in their orbit? Most of the planets in our solar system all orbit along the same plane, supposedly because of the almost flat dust clouds that formed the planets.

But some planets, such as the one featured below, have a significant tilt in their orbits that separates them from the normal plane of orbit for that star.

Image courtesy of Science Daily

What might cause a planet to have such a tilt? Is this the star capturing a wandering body into orbit or is this a phenomena caused by the formation of these systems?

Orbital tilts are almost certainly an effect of the formation and evolution of the protoplanetary disk (at least in the majority of cases). They could also be captured rogue planets.

Consider the eccentricity of Pluto. Like Pluto's eccentric orbit, it's probable that its current inclination is the result of orbital resonances with Neptune. This paper (unpublished grad student research) describes some such scenarios.

It is likely these resonances are between the Jupiter-induced precession frequencies of the ascending nodes of Pluto and Neptune. Pluto's inclination could either be relatively stable now, or it may still oscillate on the timescale of several million years.

I wish I could give a more definitive and detailed explanation - it's just one of these cases where no one knows for sure.


Could this three moons system be stable? In my fictional world I have selected all parameters to maximize the Hills Sphere of my planet. The planet has three times Earth's mass and it is located in 2 AU from its star that's 1.4 times more massive than our Sun. I assume that there are more planets on the system but in such way to have minimum interaction with my planet.

I want to have three  moons on a 1:2:4 resonance. I also want:

the first one to be as large as our moon to the sky
the second 1.5 times larger
and the third, half our moon's size.

To achieve this I have assumed:

the first moon having 1.1 lunar radius,
the second 2.3
and the third 1.1.

Their orbits are 18, 36, 72 days.

I don't really care about their density and I assumed that their masses are

1 lunar mass (probably rocky)
3 lunar mass (water or ice world)
0.3 lunar mass (probably ice world)

 I guess that to be more stable i should have eccentricities smaller than 0.05 and inclination equal to zero.
 I am not sure about the spin of my moons, is there a chance being tidal locked?
This is my effort to achieve maximum stability of the system, what parameters should I reconsider to make it more stable? I am willing to change almost everything even the resonance, but I want them to be visible from my planet in similar way moon is to Earth.
Have in mind this question is about a fictional world for a fantasy story so even the minimum possibilities for that senario to happen is enough for me.
So any ideas?

P.S I have asked a similar question on worldbuild.stackexchange but they directed me also here for a better answer.

I see no reason why this wouldn't work.  The innermost Galilean moons are in a 1:2:4 resonance so it's clearly a stable orbital configuration.  They could all be tidally locked if you want, but the planet itself can't be tidally locked with all of them.  If the planet were tidally locked it would likely be locked with the innermost moon.

Also of note, the Galilean moons all have eccentricity less than 0.01 and inclinations less than 0.5 degrees.


What is the temperature on surface of Pluto facing Sun? According to wiki page, the mean temperature of Pluto is 44 degree K (-229 degree C). However, given that Pluto is tilted greatly and there is daylight for long periods on part of Pluto facing the Sun and also given that atmosphere of Pluto scatters light to a high degree, the surface temperature on that part of Pluto may be higher. The atmosphere of Pluto contains methane which is a powerful greenhouse gas, and that also could push up the temperature.

What can be the maximum temperature on part of Pluto facing the Sun? Does available temperature estimate take into account findings of New Horizons probe which flyby Pluto earlier this year? Thanks for your insight.

According to this site, Pluto may experience temperatures of up to 55K, which I assume happens during the daytime. This low temperature is due to Pluto's distance from the sun, its high albedo (about 0.5), and its extremely thin atmosphere.


Can asteroids contain atmosphere? Gas is abundance in the universe, can a massive asteroid draws in these gas forming a thin atmosphere?

At a certain size, huge asteroids get classified as dwarf planets. Pluto has an atmosphere 100,000 times thinner than Earth, and Pluto is already one of the two largest dwarf planets known.

Asteroids (like everything) do have gravity, so nearby gas would be drawn to them. But it would take just very tiny distrubances for that gas to drift away, so what little there is would probably be close to undetectable.


Are the orbits of all triple star systems at least technically unstable? Background:

In the Circular Restricted Three Body Problem (CR3BP, CRTBP) some halo orbits are mathematically stable. That means that the orbit of the third body is closed, periodic, and stable against small perturbations as long as the two primary bodies are in circular orbits around their common center of mass. See answers to Are some Halo Orbits actually Stable?

Question:

But for three massive bodies such as stars, if they are interacting only with each other and only gravitationally, are all orbits still at least technically unstable?

Yes some systems may last longer than the lifetime of the individual stars, or even the age of the universe, but strictly mathematically are there some stable configurations or are they all mathematically unstable, i.e. one member can eventually be ejected like this?

(Ignoring that all orbits are technically unstable due to the emission of gravitational waves.)

There are known solutions to the gravitational three body problem that can be shown to be stable. Lagrange found a three body solution for general masses where all three orbit the common center of mass in an equilateral triangular formation. Gascheau proved in 1843 that this solution is stable if the component masses satisfy

$$ \frac{m_1 m_2+ m_1 m_3 + m_2 m_3}{(m_1+m_2+m_3)^2} < 1/27$$

More recently, it was shown by Kei Yamada and collaborators (1 2 3) that if one includes the first order Post-Newtonian correction, this solution is modified by a (small) correction to the  triangle legs based on the component masses, i.e. it is only an equilateral triangle if all masses are equal. The general effect of the 1PN interactions is to decrease the region of parameter space where this solution is stable, but for suitably far separated systems there will still be masses for which the system is stable. Moreover, they also proved that (4) these solutions are stable under the emission of gravitational radiation (i.e. a triangular system will adiabatically evolve into another triangular solution).

The effects of 2PN interactions (and beyond) on the stability of these triangular solutions is currently unknown (as far as I am aware).


Why is SPT0418-47 ("the most distant Milky Way look-alike") expected to evolve into an elliptical galaxy? Phys.org's ALMA sees most distant Milky Way look-alike describes the image reconstruction of a strongly and very nicely lensed z = 4.2 galaxy by a by a foreground galaxy at z = 0.263 and says:

"What we found was quite puzzling; despite forming stars at a high rate, and therefore being the site of highly energetic processes, SPT0418-47 is the most well-ordered galaxy disc ever observed in the early Universe," stated co-author Simona Vegetti, also from the Max Planck Institute for Astrophysics. "This result is quite unexpected and has important implications for how we think galaxies evolve." The astronomers note, however, that even though SPT0418-47 has a disc and other features similar to those of spiral galaxies we see today, they expect it to evolve into a galaxy very different from the Milky Way, and join the class of elliptical galaxies, another type of galaxies that, alongside the spirals, inhabit the Universe today.

and links to Rizzo et al. (2020) in Nature: A dynamically cold disk galaxy in the early Universe. Also see (YouTube and ESO)
Question: Why is SPT0418-47 ("the most distant Milky Way look-alike") expected to evolve into an elliptical galaxy? Is there something about this particular observation that indicates that, or is that just what galaxies "back then" did, even if they had a disk-like phase?

The full paper can be downloaded from the ESO site. This contains the following:

Dusty starburst galaxies are believed to be the progenitors of early-type galaxies (ETGs),
which are the most massive galaxies observed today, dominated by old stellar populations. The
most popular evolutionary track for this transformation predicts that the dusty-starburst phase
is followed by a quenching phase, during which AGN feedback leads to gas consumption and
heating with the consequent formation of a population of compact quiescent galaxies at $z \approx 2$. In
the final phase, dry minor mergers are expected to be responsible for a growth in galaxy size and
the transformation into present-day ETGs.

(AGN = active galactic nucleus)
They then compare the stellar/baryonic mass with the likely influence of dry mergers:

The comparison between the ETGs and the stellar/baryonic quantities for SPT0418–47 in the size–mass plane (Fig. 4a) indicates that this starburst galaxy should increase its stellar mass by a factor
of 6 (3 for the red diamond) and its effective radius by a factor of 11 (3 for the red diamond), in
order to evolve into an average ETG (yellow cross). This is in agreement with a simple toy model
for mergers, in which a single dry major merger event would be responsible for an increase in both
the size and stellar mass of SPT0418–47 by a factor of 3.

They also compare the distribution of dark matter with local ETGs:

Finally,
we derived the fraction of dark-matter mass within the effective radius and found that, with a value
of $f_{\rm DM} (< R_{\rm e}) = 0.095^{+0.004}_{-0.004}$ (red diamond), the central regions of SPT0418–47 are dominated by
baryons. As shown in Fig. 4c, such a low fraction of dark matter is compatible with observations
of local ETGs, implying that the physical mechanisms responsible for the mass and size growth
of this galaxy with cosmic time should preserve the dark-matter contribution within the innermost
∼1 kpc.

Note that the term "early-type galaxy" refers to elliptical and lenticular galaxies, which are near the start of the Hubble sequence. It does not refer to evolutionary stage, as noted in Hubble (1927) "The classification of spiral nebulae":

The nomenclature, it is emphasized,
refers to position in the sequence, and temporal connotations are
made at one's peril. The entire classification is purely empirical
and without prejudice to theories of evolution—comparison with
theories will be the more significant for this very reason.


Is Venus in some way tidally locked to... Earth? The video here demonstrates how Earth sees the same face of Venus every inferior conjunction (i.e., two planets are the closest to each other):

https://www.youtube.com/watch?v=4m_ouMC61-w (or https://twitter.com/physicsJ/status/1263402883275386880)

This might be caused by some kind of tidal locking, or it might be a coincidence. Does this type of gravitational resonance exist, either in theory or in the wild (e.g., between moons of Jupiter or Saturn)?

@BMF's comment links to (Gold & Soter 1969) Icarus 11, (3), November 1969, pp 356-366 Atmospheric tides and the resonant rotation of Venus. Since it is paywalled I'll add a short summary:

From the abstract:

  The observed spin-orbit resonance of Venus, whereby the same side of Venus faces the Earth at each inferior conjunction, cannot be explained adequately by gravitational interaction with the Earth alone. The expected solar tidal drag on the solid body of Venus would easily overwhelm the Earth's couple upon any reasonable permanent deformation of Venus. If there exists, however, a solar atmospheric tide, partly thermally induced and similar to that known on the Earth, its torque may counteract that due to the solar solid body tide at a particular rotation period. The small interaction with the Earth is then sufficient to lock the period to one of the resonances in the vicinity of that angular velocity.

The paper explains that if Venus' $J_{22}$ (transverse gravitational quadrupole moment) were about the same magnitude as Earth's it would be far too weak for this synchrony to be induced by gravitational interaction between these two planets because the tidal forces from the Sun would overwhelm the effect:

  The spin-orbit resonance between the two planets is undoubtedly due to the action of the Earth upon the net permanent de- formation (or transverse quadrupole moment) of Venus. But in order for this rather weak torque, which is significant only near inferior conjunction, to overcome the constant drag of solar tidal friction on the body of Venus, the permanent quadrupole moment required for Venus would have to be at least an order of magnitude larger than comparison with the Earth would suggest.

It then proposes that the tidal interaction with the Sun might be coincidentally cancelled by torque from Venus' atmospheric tidal effects. The paper discusses the history of understanding of transfer of angular momentum from a planets atmospheric tides to the planet itself:

  It was first shown by Lord Kelvin (Thompson, 1882) that the Earth's atmospheric tides transfer angular momentum and energy from the orbit to the rotation and tend to accelerate the latter. Holmberg (1952) suggested that the associated production of mechanical energy might be sufficient to balance the dissipation of oceanic and solid body tides, so that the Earth's present rotation rate would be one of stable equilibrium over geological time. Subsequent studies, however, appear to confirm the older view that the Earth is in fact slowing down; i.e., that the energy loss in conventional tides exceeds the input through atmospheric tides. Nevertheless, the notion that a planet's rotation rate might be stabilized in the balance between the opposing two kinds of tides is perhaps the key to the resonant rotation of Venus.

and concludes (in 1969) that this is probably what's happening. This is the extent of my answer, certainly in the fifty years since this paper it's been looked at in greater detail, but as far as I can tell there are no competing theories.

Update: The problem was also addressed by (Ingersoll and Dobrovolskis  1978) in Venus' rotation and atmospheric tides and by (Bills 2005) Variations in the rotation rate of Venus due to orbital eccentricity modulation of solar tidal torques J. Geophys. Res. 110, E11007

More recently Auclair-Desrotour, Laskar & Mathis (2017) Atmospheric tides in Earth-like planets A&A 603, A107 have addressed the general topic in great detail and Venus is examined in great detail, but I can't tell if they address this subject directly or not.

The paper also describes the rapidly evolving studies of Venus by Earth-based radar observation in the 1960's such that by 1969 Venus' apparent synchronicity became obvious at about the same time the Moon landings began. Radar studies of Venus started early, as discussed in

When did planetary scientists realize Venus' surface pressure was almost 100x that on Earth? How did they find out?
How was the radius of Venus measured so accurately (± 3 km) via radar in the mid 1960's, before Venera 4 and Mariner 5?

  FIG. 1. Radar determination of the retrograde sidereal period of Venus. With increasing accur- acy, the values have converged toward the reson- ant period of --243.16 days. The next nearest resonant periods are at -201 and -308 days. Figure adapted from Eshleman (1967); see also Shapiro (1967).


How could a supernova seed solar nebula? If the heavier elements such as Uranium came from a supernova, then this supernova had to sit somewhere right in the middle of the original solar nebula, because although the ejected matter traveled extremely fast by star or planet standard, it was still not fast enough to truly get across the empty space between clusters of stars. Does this mean the supernova was actually birthed and died in the same nebula or at least in the immediate vicinity in a region that is rich in hydrogen? Could the solar system be still in this nebula? What could endow the solar system as a whole with so much momentum that it acquired a different trajectory around the galaxy than its mother nebula, if the solar nebula is in effect no where to be found?

The Uranium on Earth did not come from a single supernova. The Earth formed from material that had been enriched by perhaps a billion stars that lived and died before the Sun and solar system were born. That material, enriched by heavier elements, is turbulently mixed within the galactic interstellar medium and is then gathered together to form new generations of stars. The mixing timescale is much shorter than the age of the Galaxy (see for example How can there be 1,000 stellar ancestors before our Sun?)

However, it is possible that one or two local supernovae did imprint a pattern of odd abundances for particularly rare or short-lived (radioactive) isotopes. It is widely believed that an explanation for the inferred high ratio of $^{26}$Al/$^{27}$Al (the former being radioactive with a half-life of less than a million years) in the early solar system, indicates that it was enriched by the ejecta from a nearby supernova. This may suggest that the Sun was born in a cluster of stars that was numerous enough to have contained very high mass stars capable of exploding as supernovae on short timescales.

Finding the solar systems birth place or its siblings is proving to be a difficult (and perhaps impossible) task, given the time that has elapsed and the multitude of effects that act to disperse stars (and their natal gas) on much shorter timescales (see Parent stars of our Sun - Where are its remains?).


Astrophotography: Is there a rule to the color mapping? As I understand it most astrophotography is colored using filters and multiple exposures. I would assume that blue filters would be used for higher frequency emissions and red filters for lower frequency emissions, as it at least keeps the "order" of information intact. But besides that, is there any rule or hints to go by when looking at colored astrophotography and deducing the depicted wavelengths? Because a fairly common problem would be, I assume, when something is only visible in either infrared or ultraviolet. How would I differentiate that? Do I just have to know what emission bands certain phenomena are visible in?

Rule in mapping color is just violet (high frequency) >>> red (low one). For example, it was just yesterday thay we had a speaker who who studied in three IR bands: W1 (high frequency = blue), W2 (green), W3 (red).

So, the image has to provide the description about the mapping to be informative.


What is the horizontal distance needed to observe an object just as badly as if it where in space at the zenith? When I see an object in space (let's say the ISS) above my head, my line of sight traverses ~$100\; km$ of atmosphere. The vast majority of the extinction, absorption and turbulence happens closer to the surface of Earth (since the density of air follows an exponential decay with altitude).

If I tried to observe another object, this time on the surface of Earth, the line of sight towards it would have to traverse huge amounts of air and the density would in principle be constant throughout the entire optical path.

My question is: How should I calculate the maximum distance of an object on the surface of Earth such that the image I can get from it is still better than the image I can get for any object outside the atmosphere in the zenith of an observer located on the surface of Earth? What is the distance for an object on the surface of Earth at which anything I photograph in space looks crispier than this object?

If the atmosphere could be redistributed so that it has sea level density $D$ up to height $H$ and $0$  density above that, then we can calculate
$$H = \frac{P}{gD} = \frac{100,000}{9.8 \times 1.27} \approxeq 8,000$$
(SI units), where $P$ is sea level pressure.

So it is the same as looking through $8$ km of air horizontally.


What to observe with a remote telescope close to sunrise time? I have an access to a remote telescope at 05.15-05.45 am at Hawaii time. The sun rises at 06.15 am. I will use this telescope for educational purposes for high school students. I wonder 30 mins is enough to make a variable star observation or a spectroscopic measurement.

The specs of the telescope could be found on https://lco.global/observatory/2m/

I have been an amateur astronomer for over 60 years. In all that time I have not had more that a passing interest in variable star observation or spectroscopic measurements. What makes you think a bunch of high school kids will?

As I recall the best teachers I had were the ones that could make a subject interesting. Kid want candy. Give it to them. Point that monster at the most visually interesting objects. If you want to start a fire you use use a bit of kindling not a pail of cold water.

Put yourself in their position. A once in a lifetime opportunity and you want to turn it into a lecture on page 347 of the Science textbook.


How variable does a star have to be, to be a variable star? Variable stars are stars whose apparent magnitude varies. But there are so many phenomena that can cause a star to be variable, that I would expect all stars to be variable. A rotating star has a starspot? It's variable! A planet is transiting? It's variable! A cloud of dust passes in front? It's variable!

This makes me wonder what exactly distinguishes a variable from a non-variable star? Must the variations in brightness be larger than a given magnitude? Must the variations in brightness be periodic? Do dips (or peaks) in brightness be observed more than once?

There is no lower limit, and as you say, all stars are somewhat variable.

However catalogues of variable stars exist, and they can record a wide range of levels of variability.  For example, the general catalogue of variable stars lists stars like Alpha Triangulum, with a variability of 0.01 magnitudes.

Ultimately a variable star is a star which has had its variability measured and studied and recorded in a catalogue.


Is it possible to detect gravitational lensing behind the Moon? Eddington waited for an eclipse to happen to be able to observe gravitational lensing of the stars behind the Sun. And nowadays, amateurs can do the same thing.

Of course, the Moon is much lighter than the Sun, so it doesn't bend light as much. But since Eddington's time, our observation capabilities have greatly increased. Is it possible to detect the gravitational lensing of stars that are behind the Moon?

The angular deflection caused by the lensing of a distant background object by the Moon is given by
$$\theta \simeq 4 \frac{GM}{Rc^2},$$
where $M$ is the mass of the lensing object and $R$ is the closest projected distance of the ray from the centre of the mass.

For the Moon, the maximum deflection would occur when the ray just grazes the limb of the Moon and would equal 26 micro-arcseconds.

At present the most accurate instruments for measuring precision positions are VLBI radio observations of point sources, where 10 micro-arcsecond relative positional precision can be "routinely achieved" (Reid & Honma 2014). To use this technique you need bright, radio point sources to be close in position to the Moon. This is certainly possible and has been done many times using the Sun as the lensing object, although the claimed accuracies of a short time-series of measurements on single sources is an order of magnitude lower and so probably wouldn't work for the Moon (e.g. Titov et al. 2018 ).

The Gaia astrometry satellite has likely end-of-mission positional precisions of about 5 micros-arcseconds for bright stars, but it is unclear what the precision is for a single scan (which would be needed because the Moon moves!). However, to counterbalance that, one could also average the lensing effect over many stars surrounding the Moon and of course average this for different stars with the Moon observed at different times over the entire Gaia mission.

Unfortutunately, although in principle this could be done, because of the way that Gaia scans the sky, it is always pointing away from the Moon and so there will be no observations towards the Moon that would be capable of revealing these small deflections (e.g. see here).

Conclusion: Measuring the gravitational deflection of light by the Moon is just out of reach of current observational techniques.


How far does the moon move relative to the sun during a solar eclipse Let's assume 'solar eclipse' occurred

Now, at what degree 'moon' will be in 'with reference to' earth (and)   at what degree 'moon' will be in 'with reference to' sun

In other words, by 'how many degrees' (out of 360 deg) the moon would have orbited 'earth' and 'sun' in the event of 'solar eclipse'?

Total and annular solar eclipses occur where the Sun-Earth-Moon angle is 0$^\circ$.
Where that angle is smaller than 0.5$^\circ$ but not 0$^\circ$, there is a partial solar eclipse.
The Sun-Moon-Earth angle at those times is 180$^\circ$.
Due to the Moon's orbital inclination, most new moons miss eclipse alignment, some by as much as 5$^\circ$.


Is this an actual nebula or CGI? I'm thinking of putting a large wall paper with an astronomical motive on a wall. I found quite a number of wall papers by various vendors, but most of them are obviously computer generated images, which is a no-go. It must be an image of an astronomical object, a nebula or a galaxy. (There's also a size requirement, not under 236 x 370 cm, so please do not suggest I use a printed image fitting a common frame format).

Now I found this wall paper. To my eyes it does not obviously look like CGI. But there are no diffraction artifacts visible for the bright stars; perfect circles it seems.

Can anyone identify the object?

This appears to be one of Sergii Tsololo's more convincing efforts in the synthetic "space" image genre.
It is listed on various stock photo sites as
"Blue and purple space galaxy,"
or with added 4-way spikes,
"Blue space starfield with flares."
A real astrophoto would identify the subject and not be tagged "abstract" or "fantasy."

To check it another way, I submitted the image to
Astrometry.net,
whose plate solver failed to match the field within their 10 minute time limit.
When I submitted an amateur photo of the
Trifid Nebula
as a control, Astrometry.net found a solution in less than 1 second.


How do we weigh a planet? How do we weigh a planet?
My friend asked me how do scientists weigh a planet. He is not from science background so I need to make him understand with simple analogy. How can I explain to him?

The other answer posted is correct, but let me try to explain it a bit more fully, and actually correct you on your terms.  I did an interview on this topic for a podcast literally just two days ago, so it's fresh in my mind.

Answer: The way we determine an astronomical object's mass is through Newton's form of Kepler's Third Law, which relates the distance another object orbits the object in question to the distance it orbits from that object, and some fundamental math and physics constants.  Doing this requires that there be a smaller, "test" object in orbit of the object whose mass you want to measure.
Conceptually, this means that we can only measure an object's mass via that mass's effect on another object.

What does that mean?  Kepler's Third Law states that the time it takes one object to orbit another, squared, is proportional to the semi major axis of its orbit around that object, cubed.  We often use that in our solar system in units of distance in A.U. (1 A.U. = Earth's average distance from the Sun) and Earth years.  So, before we knew how big an A.U. is in km, we still knew that Jupiter was about 5.2 A.U., and it took 11.9 Earth years to go around the sun.
Kepler's Third Law was formed in 1619, but it took Newton's Law of Gravitation in 1687 to provide the physics behind it.  That physics changed the "is proportional to" to "equals," and added 4π2/G*M into the formula.  We know what 4 means, we know what π means.  M is the mass of the big object that something is orbiting.  It's what we want to solve for.  And, assuming we can observe an orbiting object, we know its period and distance from the main object we want to solve for.
So then, all we have to do is figure out G, the gravitational constant.  Gravity is really weak, so it's really hard to measure, and it took nearly 100 years for it to be calculated kinda accurately, and the first real measurement that people like (done by Cavendish) didn't happen until 1798.
With a value for "big G" as it's often called in physics class, you can now calculate any object's mass, in space.  So long as it has another object orbiting it.  That smaller object is required so we can use Kepler's Third Law to observe that object's orbit around the big one we want to measure.  If we don't have that smaller object in orbit, we can't* measure the big one's mass.  So, our first reasonable measurements of Pluto's mass only came once its main satellite, Charon, was discovered.  Only asteroids with moons have reasonable mass estimates.
*There are slightly other ways to do this.  One is to put an artificial satellite in orbit, so once Dawn orbited Vesta, and then Ceres, we could get their masses.  Another way is a gravitational slingshot around an object, like sending a probe past Venus, we can watch how much Venus deflects the probe and get its mass.  We can also watch for gravitational perturbations in a system caused by smaller masses and figure out what those smaller masses must be to cause those perturbations, but this is much harder (but led to some early estimates of the mass of Venus).
The bottom-line is still that, conceptually, you can only do this by measuring the object's (mass's) effects on another object.

Terms:  Mass is a fundamental property of matter, it's a way to think of how much "stuff" is there.  It has the familiar units of grams (or kilograms) or pounds.  Weight is different.  Weight is how mass acts in the presence of an acceleration, like gravity.  It has the metric units of Newtons.
In the presence of a basically-the-same gravitational acceleration – Earth's surface – mass and weight are proportional because the gravitational acceleration is the same.  So my mass – which doesn't change (much) – is going to mean I weigh the same wherever I am on Earth's surface.  If my mass is 100 kg, then I will weigh about 1000 N on Earth.  If I go to the Moon, my mass is still 100 kg, but I will be in a different gravitational field and I will weigh about 160 N.
Because my weight and mass are proportional practically anywhere on Earth, we use the terms "mass" and "weight" completely interchangeably, but we still use them wrong.  In fact, when I did the podcast recording I mentioned above, we had to do up to seven takes of just one sentence because I kept using the terms wrong, even though I was explaining how to use them correctly!
This is important because your question is actually phrased wrong: The question is, "How do we determine the mass of a planet?"


Is the sun shrinking currently? The Sun loses about 5.5 million tonnes of mass every second, does this mean Sun's volume is also going down? If so can we tell by how much every year?
E:
"by how much" I mean can we tell if sun's radius is going down too

Is the sun shrinking currently?

It's the other way around: The Sun is slowly growing hotter and thus is slowly expanding. The Sun accumulates ever more helium in its core as it ages. This growing amount of fusion ash results in the core getting hotter, which in turn results in the Sun expanding in volume. This completely overwhelms what one would expect from looking only at the mass loss due to fusion and solar winds.
The solar mass loss rate currently is on the order of $10^{-13}$ solar masses per year. At that rate, it would take on the order of ten trillion years for the Sun to consume or expel all of its mass. That's a much longer time span than the Sun's much faster aging process. The Sun has five to seven billion years or so left on the main sequence. Shortly before the time the Sun leaves the main sequence, it will be about twice as luminous and considerably larger than it is now. Shortly after the the Sun leaves the main sequence, it will be much more luminous and even larger than it was just before it left the main sequence.