Why do planets tend to rotate in the same direction although they have formed from tumbling asteroids? The axial tilts of asteroids seem to vary randomly (let me know if this premise is wrong), while the planets have a strong tendency to rotate in the same way. If planets were formed by colliding asteroids, shouldn't the sum of random tilts result in random planetary rotation? Of course other factors are important, like angle and velocity of impacts, YORP-effect, centrifugal breakup and whatnot, but how could that taken together have any systematic effect on rotation?

Ceres does behave with 4° tilt, but the other of the first discovered asteroids have tilts like 84°, 50°, 42°. Dust particles (and gas molecules if applicable) surely rotate randomly. The Solar nebula had a net spin which gravity and friction has manifested in the orbits of the planets. But shouldn't the netting of rotation be individual for each planet, with uncorrelated tilts, as orbital orientation is for each star?

You are right that the tilt of the asteroids are distributed in very random way, and that the rotation of the Solar nebula is a minor contributor to that tilt, and only skews it a little.

However, you are not right that randomness simply adds up. The randomness does in fact cancel out more and more when you combine a large amount of asteroids, until the rotation of the nebula becomes the dominant factor. This is related to the Law of large numbers.

For instance, throw a dice. The outcome is random. Throw 10 dices, calculate their sum, and divide by 10. Not so far from the average any more? You can do the same thing with thousands of dices, or millions of asteroids. When the number of asteroids that form an object is really high, the tilt is not going to be far from the average value, determined by the nebula's rotation.

The same argument goes for inclination, and the fact that even though the planets' orbits are elliptical, they are not as far from circular that a random orbit would be.


How to view the surface of Mars with a 130 mm Newtonian? I bought a Celestron AstroMaster 130EQ telescope, but I am not able to see the surface of Mars through using the 10 mm eyepiece. Instead I only see a glowing dot.
What should I do to view the mars surface?

Keep in mind that you have a 650mm telescope and Mars is small.  While it is currently appearing larger because it is near opposition, it is still only 22 arc-seconds wide.
Here's a simulated view of what you should see using a 650mm f/5 telescope with a 10mm Plössl type eyepiece (51° Apparent Field of View - chosen because Celestron offers that so I suspect it may be the eyepiece included with your telescope):

It will be difficult to see much detail at that size even if seeing conditions were perfect.  Add in typical seeing conditions and it becomes even more challenging.
You could use a 2x barlow or a 5mm eyepiece and that will make it a little larger ... but not much.  Using a higher power barlow or a shorter focal length eyepiece (shorter than 5mm) will likely only result in a fuzzier image that doesn't resolve much detail.  My experience is that average seeing conditions will let you increase magnification to the 1x of your telescope aperture in millimeters (e.g. 130x) and it turns out that an eyepiece with a focal length that matches the telescope's focal ratio will do that (in your case, a 5mm eyepiece ... or a 10mm eyepiece with a 2x barlow).  Going beyond that (to the theoretical 2x aperture diameter limit or 260x in your case) is exceptionally rare and requires outstanding seeing conditions.
@AaronF offered suggestions to try to avoid factors that degrade seeing conditions (e.g. if you live in an area where it is currently warm during the daytime, viewing over rooftops or asphalt, etc. where a considerable amount of heat is still radiating from those surfaces will degrade the visual quality.  But this can be improved by relocating your telescope.


Apogalacticon and Perigalacticon What is the length of the apogalacticon and perigalacticon of the Sun and Milky Way?  The general terms seem to be apoapsis and periapsis.

My greatest efforts at Googling have failed miserably.  If you can provide references as well, please do!

According to this website we need another 15 million years to the perigalacticon.

  Recall the Sun's motion ... 30 degrees toward the galactic centre from circular motion, and 23 degrees upward out of the galactic plane. We will be at perigalacticon (closest to the nucleus) in 15 million years. At apogalacticon the Sun is seven percent farther out.

During the galactic year of about 250 million years our solar system isn't orbiting on a Kepler ellipse around the galactic center, but instead oscillating up and down the galactic plane.

  This vertical oscillation cycles at 3.5 times per galactic year.

We are now about 27,000 lightyears away from the galactic center, close to the perigalacticon. The difference between apogalacticon and perigalacticon is a little more than 4,000 lightyears (15% of 27,000 lightyears). The 15% number is calculated from the 7% eccentricity of the orbit by $1.07/0.93=1.15~~$ according to the definition of eccentricity, and Kepler's 1st law.


Magnification using a CMOS sensor I'm reworking my question regarding magnification of my 60mm, ⨍=700mm refractor. The video eyepiece is a CMOS sensor with a pixel size of 5.6μ x 5.6μ. The number of pixels is 720 X 480 with a 0.3 megapixel resolution. That said, I'm clueless regarding an algorithm from which you could decipher the magnification given these physical characteristics of the CMOS sensor. Any thoughts would be appreciated. Thank you!

Magnification using a CMOS sensor

The concept of magnification is most easily defined when we are comparing apples to apples, not apples to oranges.
When we use a telescope (and assuming normal or corrected vision) the situation is as follows.
Light entering a telescope from a point in space, it is essentially parallel, and when it is correctly focused light exits the eyepiece parallel as well. In this case the term magnification really applies to angles. If we have a 900 mm telescope and 7.5 mm eyepiece giving 120x magnification, then a 30 arc seconds wide Jupiter will appear  1 deg apart to our eye.
So here "apples to apples" means angles to angles.
There is no easy  way to talk about magnification when a recording device like photographic emulsion or a CMOS sensor is placed at the focus, because in this case angles have been converted to positions.
Here "apples to oranges" means angles to positions.
Instead we can talk about scale.
In [this comment] the OP correctly obtained 1.65 arc seconds per pixel using pixel size divided by focal length which gives the angle in radians and then the multiplier
$$\frac{180}{\pi} \frac{3600}{1000} = 206.2648$$
which converts radians to degrees, then to arc seconds, then divides by 1000 because that equation specified micrometers for pixels and millimeters for focal length.
If you wanted an apparent magnification based on viewing your image on a screen, you can use the following which is a ratio of angles
\frac{\text{camera pixel}}{\text{focal length}}
$$M_{apparent} = \frac{\text{angle}}{\text{angle}} = \frac{\frac{\text{screen pixel}}{\text{viewing distance}}}{\frac{\text{camera pixel}}{\text{focal length}}} = \frac{\text{screen pixel}}{\text{camera pixel}} \times \frac{\text{focal length}}{\text{viewing distance}}$$
Which means of course that the farther we hold the screen away from us, the smaller Jupiter looks, or the larger the pixel size on our screen, the bigger it looks.
But of course that only works if our device displays with a 1:1 correspondence between image pixels and screen pixels. If the device scales the pixels (most do) then you have a bigger challenge.
That's why recorded images through telescopes rarely if ever display "magnification". For microscope images, if they show a bacteria and say "magnified 5000 X" then they are comparing the size of the bacteria to the size of the printed image in the book.


If Sunlike stars become a red giant and eventually a white dwarf, what do red dwarfs become? The Sun is said to become a red giant at the end of its life (before that it will become an orange subgiant first and then an orange giant or so) and after ejecting its outer layers it should become a white dwarf. If yellow dwarfs like the Sun become red giants, what do red dwarfs become? Even redder giants? How do the red giants of former red dwarfs (say Proxima Centauri) differ from those of former yellow dwarfs (say the Sun) and how would Proxima's white dwarf be different from the Sun's white dwarf? I mean, a red dwarf's eventual white dwarf would be much less massive than that of the Sun, right?

A relevant paper here is Laughlin, Bodenheimer & Adams (1997) "The End of the Main Sequence". From the abstract:

We find that for masses $M_\ast < 0.25\  M_\odot$ stars remain fully
convective for a significant fraction of the duration of their evolution. The maintenance of full convection
precludes the development of large composition gradients and allows the entire star to build up a large
helium mass fraction. We find that stars with masses $M < 0.20\ M_\odot$ will never evolve through a red giant
stage. After becoming gradually brighter and bluer for trillions of years, these late M dwarfs of today
will develop radiative-conductive cores and mild nuclear shell sources; these stars then end their lives as
helium white dwarfs.

Section 3 of the paper provides a detailed description of the lifetime of a $0.1\ M_\odot$ star. A brief summary:

After approximately 2 Gyr of contraction, the star reaches the zero-age main sequence point with a temperature of 2228 K and a luminosity of $10^{-3.38}\ L_\odot$.

On the main sequence, the mass fraction of $^3 \rm He$ increases steadily over a trillion years. The completely convective nature of the star ensures that it is mixed throughout the structure of the star. The star slowly increases its temperature and luminosity.

The maximum mass fraction of 9.95% $^3 \rm He$ is reached at 1380 Gyr. After this, the mass fraction declines as the rate of consumption exceeds the rate of production.

Between 1500 and 4000 Gyr (the text appears to use values that are too small by a factor of 1000 judging by figure 1 and the statement of total lifetime at the start of §3.2) the star starts turning itself into $^4 \rm He$, with this isotope becoming the main component of the star (by mass) around 3050 Gyr.

By 5740 Gyr, the star develops a radiative core due to the helium mass fraction lowering the opacity. This causes a small amount of contraction of the star and a decrease in luminosity.

After the development of the radiative core, shell burning proceeds outward through the star, increasing the surface temperature to a maximum of 5807 K at 6144 Gyr. The luminosity at this point is about $10^{-2.3}\ L_\odot$. This star is called a "blue dwarf".

The star becomes cooler and less luminous. Shell burning continues during this time, eventually ending with the star having a hydrogen mass fraction of ~1%. The nuclear burning lifetime ends at 6281 Gyr, producing a helium white dwarf with temperature 1651 K and a luminosity of $10^{-5.287}\ L_\odot$.

A discussion of the appearance of the blue dwarf stage and how blue they actually are can be found in this question.
The $0.16 \le M_\ast / M_\odot \le 0.20$ range is transitional between the stars that become blue dwarfs and the stars that become red giants. From the paper:

In connection with their increased luminosity output, the
transitional stars in the mass range $0.16 \le M_\ast / M_\odot \le 0.20$
are able to produce increasingly larger expansions of the
overall stellar radius after the radiative hydrogen-exhausted
core has developed.

In the models calculated in the paper, the lowest mass object that unambiguously produced a red giant was $0.25\ M_\odot$, but as noted the transition region is not sharp. Nevertheless, this does mean that the higher-mass M dwarfs will eventually go through a red giant stage.


Perseid Meteor Shower Viewing events near Chicago I'm going to be in Chicago during the weekend of Aug 11th, 2018. Can you suggest any viewing parties that are not too far away?

I'm very very amateur. I don't really know too much about astronomy, but do like being around people that do.

Please post any events that you know of like this one that was one about an hour away at Indiana Dunes State Park. https://www.trippy.com/a/5b65a86ae4b0e99fd1765e1b/507c527fe4b0c31db879944f

thank you

The Astronomical League's list of astronomy clubs leads to these opportunities:

The Skokie Valley Astronomers have a meeting at the Ryerson Forest Preserve in Riverwoods on Friday the 10th.
Some Naperville Astronomical Association members will be at a stargazing picnic at Cantigny Park in Wheaton on Saturday the 11th.

Like most meteor showers, the Perseids produce more meteors in the early morning, but there should be some in the late evening.


Is there sufficient evidence to confirm additional Planet in solar system I have watched this scishow episode where Hank says that astronomers believe there is another planet in our solar system.

Link

He says that it's because how asteroids and comments are being pushed towards the sun (and us) from similar angles. Is this true? Do astronomers really have data that confirms the 'new ninth' planet?

EDIT: I did read the other related answers and articles, the way I understand this theory explained by tilt of 6 objects that could be explained by something massive enough pulling them. My question is do astronomers believe this is sufficient to confirm it that there is a planet.

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

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

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

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


Estimating star temperature I have an assignment, where I'm supposed to estimate the temperature of a star in at least two different ways, and determine which of the approaches gives the most reliable estimate. All that is given is the star's brightness at some random wavelengths. One way of estimating would be finding the peak brightness from the data, and use the corresponding wavelength as peak wavelength and solving

$$T = \frac{b}{\lambda_{max}}$$

So I've got 20 different wavelengths, ranging from 90nm to 1000nm. How would one go one about estimating T from a blackbody spectrum? Should I graph one with the given wavelengths and try different values for T until it fits the data?

However, I can't think of any other approaches with so little data.

Use Wien's displacement law - as you suggest.
Let's assume that the spectrum you have been given incorporates almost all the flux from the star. This might be ok, so long as the flux is heading towards small numbers at each end of the spectrum? If so, then you can use the temperature-independent shape of a blackbody function to argue that some fixed fraction of the flux is contained below a wavelength which depends only on the temperature. This is a more robust version of Wien's law.

For example 50% of the flux should lie below a wavelength of $4107/T$ $\mu$m.

Thus you sum up your 20 points, using Simpson's rule or Trapezium rule if they are unevenly spaced or have gaps. Then you work out how much flux there is up to the wavelength of each of the wavelength points in turn. When you reach the one where 50% of the total flux is below it, then the temperature is $4107/\lambda$, where $\lambda$ is in $\mu$m.

In principle method 2. is more robust, because you are always trying to find the effective temperature - which is the temperature  of a blackbody that gives the same observed integrated flux. Since method 2 is an integral method, it is less sensitive to the detailed shape of the spectrum. Which is good, because stars are not blackbodies.

In contrast, method 1 is really at the mercy of how close to a blackbody the spectrum is.

A possible problem with method 2. though, is that your spectrum does not contain all of the flux. In which case your temperature will be an overestimate (assuming the missing flux is longward of 1000 nm).


Why does the Hubble flow does not carry photons with it? I have this naive question. We know that the expansion of the universe "takes" or "carries" the galaxies with it, separating them, but this doesn't apply to photons. My guess is that it has to do with the fact that photons are massless (i.e. they always move with c), but I am not sure and would like a more physical explanation or insight.

Expansion isn't some mysterious force that "carries" galaxies. There's a very common misconception that it is, but it's not true. The expansion is just the recessional motion of galaxies and other matter.
Light does participate in the expansion. Even though it moves at a constant speed (and therefore can't move with the galaxies as you correctly observed), it still thins out with time. This is counterintuitive, but it happens even in a special relativistic cosmological model (the Milne model, which is a special case of FLRW cosmology). If you think of the big bang as an explosion of light bulbs, the bulbs that you see from a particular era in the universe's history have higher and higher relative speeds as time goes on (since the light from nearer, slower bulbs has already reached and passed you), and therefore are more redshifted, and therefore emit photons at a lower rate (since redshift affects the perceived rate of every process). So the relic photons from a particular era of the universe's history get more redshifted and less dense as time goes on.


Kepler space telescope undetected planets The Kepler space telescope detects planets based on the dip in brightness caused by planets moving past the star.

Wouldn’t that mean that there are an unknown amount of planets that have an orbit that wouldn’t be detected because their orbits don’t cross that path between the star and the telescope?

That's right. The inclination of the orbital plane around stars is considered to be random throughout the galaxy, thus the planets we can detect by the transit method is just a tiny fraction of the planets that we should expect in our stellar neighbourhood.

The transit method allows for planetary detection only when the line of sight from Earth to the system is contained, or almost contained, in the orbital plane of the planet. This means that only a tiny range of orbital inclinations on each star are good for detection.

Why did I say almost? Because there is some range of inclinations that still would yield a transit. This range is not fixed, and it depends on the distance of the planet to its host star. As you can see in this diagram:

Planet A is closer to the star and thus creates a wider shadow. If an observer is located in that shadowed region far away it can detect planet A. Planet B instead is farther from the star and thus its shadow is narrower. It is interesting to note, that even if both planets here share the exact same orbital plane there are places from where you would only detect planet A and never detect planet B (see the green arrows). This is the reason we have a bias towards planets orbiting closer to their star.

This effect is in fact quite strong: consider our Solar System from an exoplanetary perspective. If you were located in a random star in the sky, what are the chances you would spot an Earth transit? Well, it turns out that it is way more probable to detect a Mercury transit, even if Mercury is the smallest planet, just because of its vicinity to the Sun. A recent paper showed this diagram of the regions of the sky where some alien inhabitants would spot a transit for each of our planets:

As you can see Mercury has the wider strip. Also it's interesting to note that due to these differences in the size of the orbits (let's use the semi-major axis, $a$, as a reference) and due to small differences in orbital inclinations there is no place in the entire sky from which an alien could detect simultaneously more than four of our planets by the transit method. No place in the universe where all the Solar System's planets would be detectable.

The detection method also depends on the relative sizes of the star, $R_s$, and the planet $R_p$: A larger star has a larger disk (as viewed from Earth) that can be easily photobombed by a planet and a larger planet can photobomb more easily if it is larger.

The result is that the probability of detecting a planet increases as we increase both/either $R_p$ and $R_s$ and increases as we decrease the distance to the host star $a$. The relation is then of this form:

$P \sim (R_s+R_p)/a$

This relation imposes several observational biases. We can see exoplanets that are large and closer to their star, but we can't see planets that are small and farther. That is the reason the first detected exoplanets are the so-called hot Jupiters: giant planets much closer to their stars than Mercury is to the Sun. This diagram shows all the exoplanet detections plotted on size vs. orbital distance:

As you can see, small planets are only detectable if they have very small orbits around their stars. We have yet to find a planet the size of Earth (quite small) and with a 365 day orbital period (1 AU distance) using the transit method. There is no reason to think that this is representative of the overall population of planets. The black region of the plot is probably filled with dots, but our instruments can't scout that region yet.

The Kepler telescope had a camera with a field of view on which it could detect more than half a million stars, but the actual number of stars monitored during the mission was around 150,000 stars (these stars had good signals and were perfect targets for the mission). For these 150,000 stars Kepler found 2,345 exoplanets distributed in 1,205 stars. So we can say that for each star targeted by Kepler, the average probability of finding some planets there is around $0.8\;\%$. That should give you an estimate of the occurrence of orbital inclinations that result in transits.

The truth is that this number is too small, because Kepler has several more biases. For example, Kepler only confirmed planets after three transits were detected. Since the Kepler mission lasted for four years and four months we can say that in the best case scenario Kepler was able to detect a planet with an orbital period as long as two years and two months, but this is not even the case since for that to happen a transit should have been detected just at the beginning of the mission, halfway, and at the exact end of it, and this coincidence didn't happen. Thus Kepler had no chance to discover any planet with periods longer than two years (enough for Earth, but not enough for our Jupiter for example), even if the orbital inclination matched perfectly for the transit. So you might expect more possible transits than those actually portrayed by the Kepler telescope.

In fact, for planets close to their stars, it has been estimated that the probability of a random alignment to allow a transit gets up to $10\;\%$. For the case of stars as big as our Sun and planets at the same distance as Earth, the probability of this random occurrence drops to $0.47 \;\%$. So with all the diversity of planets (in terms of sizes and distances to their host star) it is reasonable to expect a $0.8\;\%$ detection rate for Kepler (if we also add the time restriction to observe three transits).

A $0.47\;\%$ is an amazing number! It means that for every Earth-like planet we detect by the transit method we should expect another 213 Earth-like planets orbiting other stars that are undetectable by the transit method.

This kind of reasoning has been expanded. We have many difficulties to detect them, but if you mathematically model that difficulty and the corresponding biases associated with the known instruments and you assume random configurations, you can see that each discovery yields statistical significance to the amount of possible planets that are really out there. There are so many detections now that we can finally establish with statistical confidence that there are more planets than stars in our galaxy (even if we have probed an infinitesimal fraction of the entire population), even if this was something that could be expected we have now strong evidence for that thanks to Kepler. This means that there could be around a trillion or more plants just in the Milky Way. Now we are also able to establish some statistical constraints on the occurrence of Earth-like planets (orbiting in the habitable zone of their sun-like star) thanks to Kepler. There are probably around 11 billion planets in our galaxy with these specifications.

TL;DR

There are many more planets than the ones we can detect by the transit method, between 10 and 100 times more depending on the size and orbital period of the planet you are searching for.


Could gravitational waves near merging black holes collapse to a black hole themselves? I have understood that merging black holes emit considerable amount of their mass as gravitational waves (up to 5-10%). Based on what I have understood from gravitational waves, the closer black holes are to merging, intensity (both frequency and amplitude) of these waves become stronger.

I am thus assuming much of the total GW output happens during the last "orbit" of the inspiral, and that the origin of these waves is located within "the final orbit", i.e. really close to event horizons of both black holes, my question is essentially, is it possible for such GW to collapse into a black hole themselves (something like a kugelblitz I believe), since there's just a lot of energy in GWs (say a few solar masses) in such small volume of space (say tens of kms cubed)?

I believe this might depend on the size of merging black holes, as both intensity of GWs as well as size of event horizons depend on the mass. I have a gut feeling that smaller black holes might be closer to this limit but maybe both the intensity and event horizon size depend linearly on the mass, and this does not matter at all.

You might like to look at some of the simulations from the SXS collaboration. Essentially the black holes and the gravitational waves are aspects of the same thing -- the curvature of space. In particular, the way black holes seem to merge in these simulations is that the two black holes come very close and space around them becomes more and more curved (and in more complex ways). Some of that curvature spreads out in the form of gravitational waves, becoming less intense as it gets further away, but some of it, very close to the two black holes reaches the "critical" level of curvature at which an event horizon appears (extending the event horizons of the two black holes) which then "rings down" becoming spherical and form the merged black hole.


Gaia: What is the difference between CCDs used for astrometry, photometry, and spectroscopy? My knowledge of CCDs is that these are sensors which collect photoelectrons. That's about it.

What is a difference between CCDs used for astrometry, spectroscopy, and photometry? As an example, each type of CCD is currently used by Gaia.

Recommendations of where else to read about this at a detailed engineering level is appreciated. Thanks.

CCDs are optimized for a certain wavelength range, and for a certain expected signal level. In astronomy, we tend to be short of light, so here we almost always want them to be as sensitive as possible (an exception may be observations of the Sun, which I don't know much about). But for instance, the Nordic Optical Telescope has a CCD which is optimized for blue wavelengths, but has quite a lot fringing in the near-infrared. And further out in the IR, CCDs aren't even used, instead using something which are just called "detectors".

However, whether the CCD is used for imaging (photometry and astrometry) or spectroscopy does not have anything to do with the CCD; it's just a matter of inserting a grism or not. I'm not really into the instruments of Gaia, but I assume that differences in the CCDs are due to different wavelength regions being probed. There may be a difference in how its sub-parts (it's actually an array of CCDs) are positioned (for instance, for spectroscopy in principle you don't need a large field of view, but can do with a long array rather than a more square one), but the design of the individual CCDs are the same.


Calculating the black hole evaporation time Black holes evaporate very, very slowly by emitting Hawking radiation, and eventually they disappear, maybe even during the lifetime of the universe ($\sim 1.5 \times 10^{18}{\rm s}$). That's what I recall from the lectures I attended.
I now stumbled upon a Black Hole Evaporation Time Calculator which uses this formula:
$$\tau_{\rm evaporation} \sim \left( \frac{M_{\rm black~hole}}{M_\odot}\right)^3 \times 10^{66} {\rm ~years}$$
As usual,  $M_\odot$ is the solar mass. My questions: How is that formula deduced? In which limiting case(s) does it hold?
Edit Just as a reminder: Right now, $4.3 \times 10^{17} {\rm s}$ passed since the big bang.

The derivation of life time formula for black holes:
It is not difficult to approximately estimate the life time of a black hole. Since Hawking radiation is indeed a black-body radiation, the rate at which the relativistic energy/mass of black hole ($M$) radiated away by its Hawking radiation can be computed by use of the Stefan-Boltzmann law as
$$- \frac{{dM}}{{dt}} = \frac{{dE}}{{dt}} = A\sigma {T^4},$$
where $\sigma  = 2.105 \times {10^{ - 33}}\,{\rm{kg}}{\rm{.}}{{\rm{m}}^{{\rm{ - 3}}}}{{\rm{K}}^{{\rm{ - 4}}}}$ is the Stefan–Boltzmann constant, $A$ is the object's surface (here black hole's horizon area, $A=4 \pi r_s^2$, where $r_s=2GM$ is the horizon's radius), and $T$ is the (Hawking) temperature. Note that this is the rate at which any blackbody radiates energy, no matter what is that object. Also, in this formula, $- \frac{{dM}}{{dt}}$ means that when black hole radiates, its mass decreases. Having this together with the Hawking temperature of the (Schwarzschild) black hole, i.e.,
$${T_{Hawking}} = \frac{{\hbar}}{{8\pi G{M_ \odot }{k_B}}}\left( {\frac{{{M_ \odot }}}{M}} \right) = 6.17 \times {10^{ - 8}}\left( {\frac{{{M_ \odot }}}{M}} \right)\,{{\rm{K}}^\circ },$$
we can evaluate the life time of the black hole ($M_\odot = 2×10^{30} \rm{kg}$ is the solar mass). For this purpose, we shall assume that the black hole initial mass ($M$) will eventually evaporate to nothing. So, integrating the Stefan-Boltzmann law, i.e. ${\tau _{{\rm{life}}}} =  - \int_M^0 {{{(A\sigma {T^4})}^{ - 1}}dM}$, yields
$${\tau _{{\rm{life}}}} = \frac{{256{\pi ^3}k_B^4}}{{3G\sigma {\hbar ^4}}}{(GM)^3} = (2.095 \times {10^{67}}\,{\rm{year}})\,{\left( {\frac{M}{{{M_ \odot }}}} \right)^3}.$$
This is much greater than the age of the universe!
In addition, I should emphasize that these computations have been performed in units in which the speed of light is set to unity ($c=1$). For example, in SI units, the Stefan-Boltzmann constant is given by $σ=5.67 \times 10^{-8}\,\mathrm {W\,m^{-2}\,K^{-4}}$), but in this answer we had $\sigma  = 2.105 \times {10^{ - 33}}\,{\rm{kg}}{\rm{.}}{{\rm{m}}^{{\rm{ - 3}}}}{{\rm{K}}^{{\rm{ - 4}}}}$.
And your final question:

In which limiting case(s) does it hold?

This estimation has been derived for nonrotating black holes. For rotating (Kerr) black holes, it is expected the order of magnitude of this estimation is approximately valid, at least in the case of slowly rotating black holes. I guess, the computation in such cases would be more difficult.
Edit: In comments, it has been discussed that there are other approximations. Of course, for example, the Hawking temperature for astrophysical black holes would be extremely small, even much smaller that the CMB radiation (see more details in my answer here, please), so black hole can absorb more matter and then they grow more and more! In order to derive that formula, you should omit this challenging part. Another approximation comes from the number of particles (scalars, photons, vectors, fermions). However, the Hawking temperature (e.g., according to surface gravity definition) is same for every type of particle, and the number of species only adds a numerical factor to the Stefan-Boltzmann law, but still the order of magnitude of the final answer is approximately valid. In addition, the particles scaping from the black hole need to pass trough the potential and this change the resulting spectrum by a greybody factor ($\Gamma(\Omega)<1$). In this answer, which is very common in GR books, I utilized the Hawking radiation of Schwarzschild black hole with $\Gamma(\Omega)=1$. For more (another) details, also see the following related SE links/papers/books:

Is Hawking radiation detectable?

https://www.amazon.com/Introduction-Quantum-Effects-Viatcheslav-Mukhanov/dp/0521868343

https://arxiv.org/abs/2011.03486

https://arxiv.org/abs/1711.01865


Actual observable horizon for satellites on an orbit line map? On a casual star walk, (for comparison of identifying different objects) I witnessed some meteors, quite a few satellites gliding on their smooth and straight trajectories, and some airplanes with their much brighter (sometimes blinking) lights, lower altitude and the characteristic sense of moving towards and away from me. Suddenly one of the satellites seemed as if its motion had been a bit jittery, and its path started to curve inwards in an asymmetrical, gentle and large C-shape. The airplanes made noise, this one did not. Visually perfectly identical with all of the normal satellites. I had never seen anything like that before so I wondered if it could've been some debris colliding into things, but probably not. The best explanation I have for now is that it apparently might have been a satellite on a Tundra or Molniya orbit, as their ground tracks aren't the usual straight or sine wave shaped trajectories that we're used to. Most satellite tracking websites either didn't seem to have tracking options for those special types at all, or you could only see their locations for now, not a custom time.

I'm in the capital city area of Finland, and when I saw the object, the time was 23:10 on the UTC+3 time zone, 12th of August 2016. Of what I found, the only strange looking ground track that comes even close would be Molniya 1-62, drawing a loop above Finland. What I can't find is a website that would allow me to rewind time to check where the satellite was at 23:10. But whether or not I should keep searching for that info.

I first have this question: If I look at the satellite tracks going across a map, how do I determine how far a satellite would be visible when viewed from the ground? If I looked at similar tracks for airplanes, it's much easier to understand that the ground area from which an airplane can be spotted doesn't span vast distances, but how this compares to a satellite is difficult to imagine as we don't perceive them in a three dimensional space, but spots gliding across the flat surface of a dome.

It is not that satellite. As it passes over Finland, it has an altitude of 27000km and a magnitude of about 12. You won't see it. All Molniya orbits are similar.

The website Satflare to shows the region in which satellites are visible (it uses a blue ring around the satellite) Your satellite was so high that it was visible from nearly the whole hemisphere.

I won't try to identify your object, except to say that an object as described is not in orbit.


If the earth was not tilted, what would be the north star? What is earth tipped toward? The simple answer is Polaris.  The axis (one end of it) points at Polaris, so that's what it's tipped to point at.  But when we saw "toward" what do we mean? When it is tipped "toward" one thing, what is it being tipped "away from"?

If earth were not tipped then the equator would lie in the same plane as earth's orbit.  Earth being tipped 23 degrees means it is tipped 23 degrees "away from" alignment with its orbit.  So a 0 degree tip would be "not tipped" and a 23 degree tip is tipped 23 degrees in some direction.  What direction is it tipped in?  Another way to ask this would be, what would it be tipped at if it was tipped a full 90 degrees in the general direction in which it is currently tipped 23 degrees?

Imagine a coordinate system with two axes in the plane of earth's orbit and one axis perpendicular to that plane.  Earth's orbit stays in the plane, and the angle that earth is tipped points to, let's say, negative X.  So the line through the north pole and south pole lies in the XY plane and the earth's orbit lies in the XZ plane.  I want to look at stars and be able to see those axes, so I want to know what stars they roughly point at.

A cleaner way to ask this might be, what would be the north star if earth wasn't tipped? What would be the south star? What if it was tipped 90 degrees? That gives four of the six stars I'm looking for. I'm not sure how to phrase the question for the other two.  I want to call these stars "untipped north", "untipped south", "winter", "summer", "spring", and "fall" because the direction of the stars would be the same as the direction earth is from the sun during the extremes of those seasons (northern hemisphere).

[EDIT]

The plane of earth's orbit is aligned with a celestial (infinitely distant) great circle called the ecliptic.  It inversely happens to also be the path through the sky that the sun takes over the course of a year.  There is a latitude/longitude coordinate system where the equator is the ecliptic called "ecliptic coordinates".  So the question is actually:

Using ecliptic latitude and longitude, what 6 stars are at:

0, 0
0, 90
0, 180
0, -90
90, 0
-90, 0

It turns out that 0, 0 on ecliptic coordinates is not the "winter" direction, nor the "summer" direction.  And it turns out to be a good thing.  Zero longitude is defined using the "spring" direction, and this means that converting these points to another coordinate system is really easy.  The earth's axis is tipped 23 degrees, but imagine the axis on which it is tipped.  Thats the spring/fall axis.  This means that if we tip the whole coordinate system 23 degrees, two points stay unaffected (the spring/fall points).  The tipped coordinate system is called equatorial coordinates and its more useful so everything uses it.  The point 0, 0 is the same in both coordinate systems, as is the antipode of 0, 0 which is 0, 180.  So you can get the spring and fall stars right away by browsing something like http://www.sky-map.org/

Note that the longitude coordinates are in hours (0 to 23) instead of degrees (0 to 360), and they are shown in the order longitude, then latitude.  Also note that stars you can see without a telescope have an apparent magnitude of 6 or less.

Zero longitude is the direction the sun is in when it is spring, so the 0, 0 star is actually the fall star.

Now the other four stars: because the difference between the two coordinate systems is a single rotation around the spring-fall axis, the winter/summer points actually keep their longitude the same during the rotation.  And because the amount of rotation is 23 degrees:

0, 90 becomes 23, 90
and 0, -90 becomes -23, -90

You can check this here: https://lambda.gsfc.nasa.gov/toolbox/tb_coordconv.cfm Note that equatorial coordinates are called "celestial" on this page.  Note that you have to enter a year because the tip of the earth's axis slowly changes.  Just enter the current year in both epoch boxes.

It should make sense that the untipped north star would have an equatorial latitude of 90 minus 23 degrees, and the untipped south star would have an equatorial latitude of -90 plus 23 degrees.  Using the same tool to do the conversion:

90, 0 becomes 66, -90
-90, 0 becomes -66, 90

If you go back to sky map and poke around in those locations, you find:

The untipped north star is in Draco (18h 66d)
The untipped south star is in Dorado (6h -66d)
The fall star is in Pisces (0h 0d)
The spring star is in Virgo (12h 0d)
The winter/summer stars are in Gemini (6h -23d) and Sagittarius (18h 23d)

Note that the latter four stars are in zodiac constellations - this is because those constellations lie on the ecliptic.

For winter/summer which is which?
Well the sun comes up in the east, so earth rotates counter clockwise, so it travels around the sun counter clockwise.  We know that after the line from the sun to the earth points toward fall, it points toward winter.  So whichever is left of Pisces in the sky would be the winter one:

The winter star would be in Gemini
The summer star would be in Sagittarius

Note that the four zodiac constellations we got are evenly spaced throughout the year, and that the months associated with them are 6 months off from the seasons we mapped them to.  This is because the zodiac months are assigned based on where the sun is in the sky.  You never see a zodiac's constellation at night during the month associated with it.


What are IRAS sources? Do IRAS sources refer to star forming regions? I searched google and I came to know that it stands for IR astronomical satellite.But what do they search for? Please help me

The IRAS Point Source Catalog, Version 2.0, is a catalog of some 250,000 well-confirmed infrared point sources observed by the Infrared Astronomical Satellite (IRAS), i.e., sources with angular extents less than approximately 0.5, 0.5, 1.0, and 2.0 arcminutes in the in-scan direction at 12, 25, 60, and 100 microns (um), respectively.

This includes some comets, planetary disks, and even minor planets. Some sources are likely to be star forming, but an IR source meeting the spec above could be many things.


How exactly does Hawking radiation decrease the mass of black holes? From what I understand so far, when one of virtual particles crosses the event horizon and the other does not, they can not annihilate each other. The latter wanders off into the universe (btw, is it still virtual at this point, and what does 'virtual' mean at this point, if so?), while the other gets consumed by the black hole. I don't see how this event contributes to evaporation of the black hole (since the particles do not originate from the black hole). Shouldn't the consumed particle actually add-up to the black hole mass?

The closest question to mine is Does Hawking radiation in fact bring mass into the universe?, but I don't find the answers satisfactory.

I.e. "the escaped virtual particle is 'boosted' by black hole's gravitational field into becoming a real particle", rather adds to the question then answer it.

EDIT: I am humbled by the knowledge presented in the answers and feel incompetent to mark any as the most suitable. I hope this is okay.

I'm going to give you an intuitive answer. Keep in mind, this is not the "actual" answer, as the Hawking radiation is quite a bit more complex than the typical pop-sci explanation with virtual particles. But some intuitive justification is possible nevertheless.

  I don't see how this event contributes to evaporation of the black
  hole (, since the particles do not originate from the black hole).

You're missing a key point here.

When the pair was generated, those were virtual particles. After one side of the pair was absorbed by the black hole, and the other side was released, the released part is a real particle. Huge difference there - virtual vs real.

Virtual particles don't really exist the same way that you and me exist. They seem to exist for a very short time; the more energetic they are, the shorter the interval of their virtual "existence", per the Heisenberg equation. In many ways they are just a mathematical trick.

Think of the vacuum, where no real particles exist. Before, it's just vacuum. Right now, a virtual pair flickers briefly, then it's gone. In the future, it's vacuum again.

What was the energy before? Zero. What is the energy in the future? Zero. What's the energy during the flicker? Well, it basically zero, within the limits permitted by Heisenberg's equations. Bottom line is, virtual particles come and go, and they do not contribute to the energy balance of some empty chunk of space.

(I am ignoring here the concept of vacuum energy, for the sake of an intuitive explanation.)

But let's say one of the virtual particles gets trapped by the black hole, so it cannot annihilate with its counterpart. The other particle flies off in the opposite direction and escapes the black hole. What's worse, this is now a real particle - we've exceeded the duration permitted by the Heisenberg equations, so the one that escapes is not virtual anymore.

How did that particle become real?

This is a big issue, because virtual particles don't require an energy budget to briefly exist, while real particles do carry energy forever. Something prevented the virtual pair from annihilating itself, and boosted one of the components to the status of real particle. The virtual pair has zero energy. The real particle that gets away has non-zero energy. That energy has to come from somewhere.

It comes from the black hole. The black hole gives up some of its mass / energy (same thing) to boost one particle from virtual to real. The other particle is captured - but being virtual anyway, it doesn't really matter.

What this intuitive explanation doesn't say is how the boost actually happens. I dunno, magic. Somehow one of the virtual particles gets a chunk of energy from the black hole and becomes real.

Again, this is not the actual process. The actual process is more complex. This is just a pop-sci fairy tale.

EDIT: To hit closer to home, Hawking radiation is more like a close relative to the Unruh effect. Say an inertial observer sees empty space here in this chunk of volume. An accelerating observer would not see empty space in the same volume, but instead would see blackbody radiation. That's the Unruh effect.

Well, gravity and acceleration are the same thing, per general relativity. So the strong gravity near a black hole is equivalent to strong acceleration. Something similar to the Unruh effect must happen there. That's the Hawking radiation.

http://backreaction.blogspot.com/2015/12/hawking-radiation-is-not-produced-at.html

EDIT2: The other answers currently on this page provide useful alternative points, so check them out too.


Can a planet's gravity rip out its moon's atmosphere? We all know that solar flares can rip a planet's atmosphere if the magnetic field is weak.Solar flares travel at tremendous speed so when they hit the gas molecules of a atmosphere the molecules are ejected from the gravitational pull and thrown away.

Can a giant planet's gravity pull out its moons atmosphere?Is gravity enough to pull its atmosphere or do we need an instantaneous force/impact to blow away an atmosphere?

It's probably rare that moons have atmospheres to begin with. Planets don't generate solar flares. Planets emit thermal radiation as a result of their heat and during formation and collision planets can get significantly hot.   Earth is believed to have glowed red hot following the theoretical giant impact and that heat actually may have helped give the Moon its lopsided crust.  See the Earthshine part of this article.  The Moon, which also likely glowed red hot after it formed, was unlikely to have ever had an atmosphere, but in theory, a hot planet could thermally strip a nearby moon if that moon had an atmosphere.

A planet with a surface temperature of a couple thousand degrees could heat the near side of its moon more than the star that they both orbit, perhaps giving off enough heat to cause the loss of the moon's atmosphere through thermal or Jeans escape which is basically a product of upper atmospheric temperature and escape velocity.  It might not even be that uncommon for moons to receive more heat from the planets they orbits than the star they both orbit when the planet is undergoing sufficient bomboardment.

A second way that a planet might strip a moon's atmosphere is if the moon was close to the planet's Roche limit.  The atmosphere is less gravitationally bound and even if the moon comfortably holds itself together outside the Roche limit, its atmosphere would extend miles above its surface and be less gravitationally bound where gradual atmospheric stripping would be possible.

Moons rarely have atmospheres to begin with, but ice moons at the right temperature can generate atmospheres through outgassing.   Pluto, for example, has a tenuous atmosphere from the thawing of frozen nitrogen and other ices on its surface and Pluto's low gravity means this atmosphere is very dispersed and it can extend quite far from the planet, far enough that Pluto and its moon Charon may exchange atmosphere (grain of salt warning with that article, to say they share an atmosphere is an interpretation that I'm not 100% comfortable with, but the article does make the point that an atmosphere can be tenuously held by a comparatively low gravity object like Pluto, and as a result, affected and even shared with a nearby object in a close orbit).

A third theoretical way a planet could strip a moon of its atmosphere would be through magnetism.   Jupiter's enormous magnetic field generates a radiation belt that extends beyond its inner moons, certainly beyond the inner 3 of its 4 Galilean moons.   The high temperature of the charged particles in a planet's radiation belt could strip a moon of its atmosphere in a similar way that the solar wind can strip planets of theirs.

You just asked about gravity and I gave you three possible scenarios; radiation belts and thermal radiation aren't gravity but they could theoretically strip a moon of its atmosphere.  The Roche limit scenario is essentially driven by gravity, so the answer is yes, it's possible.  It's not actually the gravity but the tidal forces, which is a product of the gravity that can strip the atmosphere of a moon close to its planet's roche limit.

Moons with atmospheres are probably pretty rare and moons that orbit close to their planet's Roche limits are also fairly rare.  So it might be a rare scenario where the two situations coincide.


Why does the Earth's orbital eccentricity oscillate with a period of about 100,000 years? This answer says:

  Earth's orbital eccentricity varies over time from being nearly circular (low eccentricity of 0.0034) and mildly elliptical (high eccentricity of 0.058). It takes roughly 100,000 years for Earth to undergo a full cycle.

Why does the Earth's orbital eccentricity oscillate with a period of about 100,000 years? What would be the simplest set of conditions and/or fewest body system that would lead to this particular type of periodic variation in eccentricity?

Saying why gets tricky beyond "because of Jupiter", but to clarify on the quote, the statement "Earth's eccentricity follows a 100,000 year cycle" is loosely true but it's also an oversimplification.    From Wikipedia.

  The Earth's orbit approximates an ellipse. Eccentricity measures the
  departure of this ellipse from circularity. The shape of the Earth's
  orbit varies between nearly circular (with the lowest eccentricity of
  0.000055) and mildly elliptical (highest eccentricity of 0.0679)2 Its geometric or logarithmic mean is 0.0019. The major component of
  these variations occurs with a period of 413,000 years (eccentricity
  variation of ±0.012). Other components have 95,000-year and
  125,000-year cycles (with a beat period of 400,000 years). They
  loosely combine into a 100,000-year cycle (variation of −0.03 to +0.02).

If the highest is 0.0679 and the variations are 0.012 and up to 0.03, that's over 50% variation peak to peak.  That's amplitude, not period, but if we look at the chart it's clearly not a neat and tidy cycle though it's somewhat close to a 100,000 year period (top line on the chart below).

Source of image.

The cause of Earth's eccentricity variation is the other planets, primarily Jupiter, with Venus and perhaps Mars (small but close) and Saturn also having effects.  Whether a particular planet is responsible for a particular period is a good question.   I'm not sure.

Ultimately this is a version of the 3 body problem or, more specifically, orbital perturbation, which is complicated mathematics.   Unfortunately, I can't give a good explanation of precisely why the gravitational perturbations work out to roughly 100,000 years or, more specifically, 413,000, 95,000 and 125,000.   Maybe someone else can.

  What would be the simplest set of conditions and/or fewest body system
  that would lead to this particular type of periodic variation in
  eccentricity?

The fewest bodies would be 3.   Sun-Earth-Moon is one example.  The Moon undergoes eccentricity variation on a much faster cycle than any of the planets.   Sun, Earth, Jupiter or any set of Sun and two planets would work too provided the planets were sufficiently large and/or sufficiently close to affect each other's orbits and (I think) you'd want the two planets to not be in orbital resonance.  That might create a different kind of pattern.


Why do transits of Earth across the Sun seen from Mars follow a pattern of occurring after 26, 79 then 100 years? According to this answer transits of Earth across the disk of the Sun as seen from Mars:

  The (Arthur C.) Clarke story is quite correct. Earth (and Moon) would transit the sun from the viewpoint of Mars on 8 May 1905, 11 May 1984 and again on 10 Nov 2084.

no surprises there, but they

  ...follow a pattern of occurring after 26, 79 then 100 years.

Why is that exactly? Why this pattern?

Such transit events occur when Earth and Mars are simultaneously at a node where their orbital planes intersect.

Between transits at the same node, both Earth and Mars must complete whole numbers of orbits, where 1 Mars year = 1.8808 Earth years.
Meeus and Goffin 1983 identify patterns of 79-year and 284-year intervals:

79.0 Earth years ≈ 42.0 Mars years
284.0 Earth years ≈ 151.0 Mars years

Their Table II arranges the May and November transit dates in separate "panoramas" with 284 years between rows and 79 years between columns.
These are somewhat analogous to a saros-inex panorama of solar eclipses.

Between transits at opposite nodes, both Earth and Mars must complete odd numbers of half-orbits:

Nov→May: 25.5 Earth years ≈ 13.56 Mars years
May→Nov: 100.5 Earth years ≈ 53.43 Mars years

These Mars year fractions differ from 0.5 because the half-orbit near perihelion takes significantly less time than the half near aphelion.
Meeus and Goffin Table I also shows some intervals of 21.5 and 53.5 years outside the present few centuries.


Star map - finding stars Assume you are given a star map that doesn't have the name of the stars on it. You know the latitude, longitude of the observer and the local time (year, month, day, local time). How could I proceed to find the stars, planets?

Notes: This requirement is given by a problem with the photo shown above. You cannot go outside, nor use other resources.
The stellar map represents the sky of a city ($44^\circ 26'$ N, $26^\circ 06'$ E, local time 00:47, date 04.03.2016).

Mark on the map the cardinal points
Identify on the map the planets from the solar system
Identify Polaris
Identy the circumpolarity zone. Identify at least 2 constellations outside the zone
Find the galactic equator and the ecliptic

NOTE: Information on question and comments has been changing a lot, so please beware that some parts of the answer may be related to previous versions of the question.

Short answer:

This is a question to check if you know the sky. Once you can identify constellations and planets on the sky, the answer on the map is quite obvious.

Long answer:

To answer this question you need to be familiar with the northern hemisphere sky. If you don't know it, you would need a properly labelled map to answer the question.

Suggested steps:

Assuming that the map just represents the visible portion of sky and the horizon is printed in it, first step would be to look for Polaris. The point of the horizon which is closest to Polaris is the North. The South is in the opposite edge of the map. To locate East and West, place the map with South lower and then East is at left and West is at right - just opposite as they would be in a map of Earth, but this is a map of Sky.
I think it would be easier to identify constellations before planets. The easier constellations are likely to be the circumpolar ones (those that we can always see in the northern hemisphere).
To find the planets, you should find the Ecliptic by looking at zodiacal constellations and then check for points that don't belong to those constellations. Of course, to do that you need to know the constellations. A helpful piece of information is that at 00:47 local time, the Sun, Mercury and Venus can't be seen. If the map shows the planets visible with naked eye, Mars, Jupiter and Saturn may be there in addition to the Moon.

Edit: addition about 1st step (finding north)

Your map is drawn inside of a circle. That circle represents the horizon, and Polaris is not in the centre of the map. Therefore, there is a point in the horizon that is the closest point to Polaris. That point is the north.

All this just comes from the fact that Polaris is always in the north. If you look at Polaris in the sky, you are facing north - just as if you look at the Sun at noon, you are facing south.

To better understand this, I suggest taking a sky map and a compass, going outside and trying to find Polaris tonight in the actual sky - tonight or the next available starry night.

Edit: finding Polaris

To find a constellation or a given star, you need to know the constellations or compare your map of the actual sky with a labelled star map. The question is not very different to (for example) "find Ireland in a blank world map": either you know the world and where Ireland is, or you have a labelled map to compare. Star maps aren't that different from geographical maps. In Earth maps you know where you are by looking at geographical features, and in sky you know where you are by looking at stars positions. To find Polaris, look for Ursa Minor; finding Ursa Major or Casiopea may help, too.

Addition after the map was included in the question:

North is in the right side of the map. Polaris is the dot halfway from the center and the right side.

The easiest clue to find Polaris is the conspicuous Ursa Major in the center of the map. As others have suggested, two stars of the Ursa Major point to Polaris. Furthermore, some stars from Ursa Minor are easy to spot over Polaris.


Can you create an orbit in a space station using balls? Would it be possible to spin a ball in a space station, and for it to create its own orbit for a smaller ball to rotate around it?

As PM 2Ring mentioned, you seem to have a misunderstanding that spin is involved in creating an orbit. What matters for gravitational attraction is the mass of the bodies and their distance, the effects of spin are more subtle and only an issue when bodies are non-spherical. Also, one body doesn't need to be smaller than the other, they can be the same size.

A lead sphere 10 cm across would mass 6 kg. If two such spheres were 10 cm apart (20 cm center-to-center), they would have a gravitational pull on each other of 0.06 micronewtons...not nearly enough force to feel. They would technically have an orbital period of about 5.5 hours, with an orbital velocity of a few millimeters per minute, but pretty much any random air current would blow them out of orbit and no human could place them precisely enough in the first place, and nearby masses (like the humans occupying the station) would throw them off.


What is the actual temperature of a quasar? My daughter asked me last night: What is the hottest thing in the universe?

The first thing that sprang to mind was the core of a fresh neutron star (this question agrees: What is the hottest thing in the universe?), so I told her that and then tried to verify it on the Internet. Unfortunately, most of the articles online claimed that the hottest things in the universe are quasars, since they have a "brightness temperature" of 10 trillion kelvin. That sounded fishy, and a review of the literature revealed that brightness temperature has no direct relationship to actual temperature.

Poor journalistic practice aside, what is our best estimate of the temperature of these objects?

A question/answer over at physics.se  discusses the issue of quasar spectral output.  Part of one answer says,

  The percentage of radiation that is thermal versus non-thermal varies
  from quasar to quasar, and for a single quasar is going to be
  different at different wavelengths, so there isn't one simple answer
  (and also it depends on what angle you are viewing the AGN from
  relative to the jet axis)

  And similarly, there isn't a single value you can quote for the
  accretion disk temperature. Different AGN have different black hole
  masses, different matter infall rates and generally different
  environments. And a single accretion disk doesn't have a single
  temperature either: the temperature is higher towards the center, but
  to the details of what the emission spectrum looks like depends a lot
  on what parameters you put in to the model and what assumptions you
  make about energy transfer within the disk etc.

So it's not really accurate to estimate quasar thermal temperature from the "Brightness temperature," because the latter only applies to objects closely fitting Planck's black body law.


What is the distance of Earth from the 7 planets, moons and the Sun at a point of time? I wanted to know the distance of the 7 planets (Mercury, Mars, Venus, Jupiter, Saturn, Uranus, Neptune ) each (or at least 5 planets) from the earth and the angle that it will make with a line connecting the centers of earth and the respective planet at a point of time. Also I need the same for the earth and moon system and sun and earth as they are big influences. At point of time because in the course of revolution of planets the distances with respect to earth will vary.

Being an engineering student I have the wish :) to calculate the mechanical forces acting on the earth at a point of time due to nearby heavenly bodies.

Basically I need the data or in case someone already attempted to calculate the force components I would be delighted to look at it.

There are plenty of sky simulators where you can calculate that information yourself. e.g. Celestia, KStars

Just install one of them and it will tell you position in sky and distance to most known bodies in Solar System, for any given time.


Where is the Solar system's barycenter located? Where is the Solar system's barycenter located?
The solar system as a whole, Where is the center of the mass for the combined mass of the Sun, inner planets, and gas giants, is it inside the Sun? Is there an AU measured distance of it?

The solar system barycenter (SSB) is sometimes inside the Sun and sometimes outside.
As an observer outside the solar system could detect with Doppler spectroscopy, the Sun is what's wobbling around.
The Sun's offset from the SSB is a vector sum of roughly:

0.00496 au ±5% away from Jupiter
0.00272 au ±6% away from Saturn
0.00083 au ±5% away from Uranus
0.00155 au ±1% away from Neptune

The other planets contribute much smaller amounts to the total.
Each planet's contribution is proportional to the product of its mass and its orbital distance.
When these components add constructively as in 2020-2023, the center of the Sun can be as far as 2 R☉ away from the SSB.
When they cancel as in 2029-2030, the center of the Sun is within 0.5 R☉ of the SSB.
The solar radius R☉ is 0.00465 au, shown here with a dashed line.


LIGO: How can laser interferometry (wavelength >$10^{-7}$m) detect length changes of arms <$10^{-18}$ m? I'm trying to understand the sensitivity of the LIGO interferometer. I've been reading around lots of discussion of how they manage noise cancellation between the two detectors, achieving a very pure laser signal, the many reflections to change the effective length of the interferometer arms to over 1000km, and other impressive tricks of engineering to achieve this remarkable feat of measurement. What I still can't get my head around is how, even in a perfect world with no noise at all, a phase shift of I guess $10^{-7}$ - $10^{-11}$ cycles or so (depending on the sensitivity claim you are taking) can show up as a signal.
From my fairly basic understanding of how an interferometer works. I am imagining that if both beams are initially in phase then a $10^{-7}$ phase shift would change combined amplitude by much less than this (1$0^{-14}$?) If they were at $\pi / 2$ then I guess the amplitude change would be approximately equivalent to the phase shift?
I appreciate that the computational details of how all this works are going to be a whole lot more complicated, but I would really appreciate some pointers on how best to understand the sensitivity challenge here:

Is it a case of measuring tiny amplitude variations in the combined laser signal?
Is the size of these amplitude variations (as a proportion of amplitude) equivalent to the size of the length variations (x number of reflections and as a proportion of laser wavelength)?
If yes to each of the above, am I right in thinking they are detecting laser amplitude oscillations down to $10^{-10}$ ish times typical amplitude?
If no to the first two points, please set me straight in any way you can!
Many thanks in advance.

The LIGO interferometer uses a homodyne detection technique. Basically, the light travelling in each arm of the interferometer is derived from the same laser source and is combined in the output channel and falls onto a photodiode.
The interferometer is operated so that when there is no gravitational wave (GW) passing through the instrument, the beams combine to produce a dark fringe (i.e. they are set to destructively interfere). There is a small offset from this, but basically the phase difference between the combiniing beams is close to $\pi$.
The phase difference caused by a GW, due to the changing length of one arm with respect to the other, can be derived as
$$
 \Delta \phi \simeq 2\pi \left(\frac{2hL}{c}\right) \left(\frac{c}{\lambda}\right) = \frac{4\pi}{\lambda} hL \ ,
$$
where $L$ is the length of the arms, $\lambda$ is the laser wavelength and $h$ is the strain amplitude of the fravitational wave signal. Actually, it is a little bit more complex than this, since the arms act as Fabry-Perot resonators which means the light effectively travels backwards and forwards many times in the arms (about 300 for LIGO, i.e. $L$ is effectively 1200 km).
For a typical dimensionless GW strain of $h \sim 10^{-21}$, $\lambda = 1064$ nm, then $\Delta \phi \sim 10^{-8}$ and is modulated at the frequency of the GW (typically 20-2000 Hz).
The problem then reduces to combining
$$ E_{\rm tot} = E_0\sin (\omega_l t) + E_0 \sin (\omega_l t + \alpha + \Delta \phi)\ ,$$
where $E$ is the electric field in each arm, $\omega_l$ is the angular frequency of the laser and $\alpha$ is the offset phase between the arms (close to $\pi$).
Using the identity $\sin a + \sin b = 2 \cos[(a-b)/2] \sin[(a+b)/2$ and squaring the total E-field to get an intensity:
$$I = 4E^2 \cos^2[(\alpha + \Delta \phi)/2]\, \sin^2[\omega_l t +(\alpha + \Delta \phi)/2] $$
Since $\omega_l$ is much greater than the GW frequency and much higher than can be sampled by any photo-sensitive detector, then the second term in the product above can be replaced by its time-average of $1/2$. If we now identify the total power $P_{\rm in}=E^2$  as the average input power to each arm of the interferometer and note that $\cos^2 (a/2) = (\cos(a)+1)/2$ and $\Delta \phi \lll 1$
$$ I = P_{\rm in} \left[1 + \cos(\alpha + \Delta \phi ) \right] \simeq P_{\rm in} \left[1 + \cos(\alpha) -\Delta \phi \sin(\alpha)\right] = 2P_{\rm in} \left[ \cos^2 (\alpha/2) - \frac{\Delta \phi}{2}\sin \alpha \right]\ .$$
It is the second term inside the bracket that contains the signal of the GW. That signal is proportional to the power in the interferometer and the phase difference between the arms. Note that although the signal-to-(shot) noise is mathematically maximised when $\alpha=\pi$, this would mean the SNR was 0/0 ! In practice there is always some other noise present so $\alpha$ is shifted a little bit away from $\pi$ - Fricke et al. (2012) suggests that $\alpha \sim \pi+ 6\times 10^{-5}$ is used.
The power input into each arm is about 600 W (the 100s of kW Steve Linton mentions in a comment is after accounting for the Fabry-Perot resonator, which I did above by talking about an "effective $L$").
In the absence of other forms of noise then photon counting (shot noise) becomes the limiting factor and is proportional to the square root of the power.
The output signal is the modulated GW signal discussed above which is recorded by detecting photons with the photodiodes. The response function that translates the photodiode signal into a strain is determined by acting on the test masses/mirrors with precisely calibrated lasers modulated at GW frequencies that can produce monochromatic phase shifts in the arm lengths.


Calculating RA/dec from JPL ephemeris data I'm using JPL planetary ephemerides to calculate the position of planets. Using DE405 ephemeris (using a parser provided by Project Pluto with all tests passing). I am fetching the position of Mars with the observer set as the Earth (geocentric).

My first assumption is that the result is a set of geocentric equatorial coordinates in rectangular form (x, y, z). This may be incorrect.

My goal is to calculate RA/Dec. My second assumption is that I should use a formula converting from Cartesian to Spherical:

$r =\sqrt(x^2+y^2+z^2)$
$ra =\arctan(y/x)$
$dec = \pi/2 - arccos(z/r)$

Using Julian day 2457134 (21st Apr 2015, noon UTC), and cross-checking the result with NASA's Horizons service which apparently uses the same ephemeris (DE405) gets me almost there but my results are a little off:

x 1.721427968227801 y 1.5802812301744067 z 0.6897426775298559

RA 02h50m12.51s Dec +16°26'41.37"

Horizons:

RA 02h50m11.52s Dec +16°26'36.5"

I'm not sure what I'm missing. Do I have to factor distance of the planet due to the time it takes for light to reach us? Am I not using the correct time? Are my assumptions wrong?

Read the fine print under the Horizons output when you select an OBSERVER table. It says (emphasis mine)

  R.A._(ICRF/J2000.0)_DEC =
     J2000.0 astrometric right ascension and declination of target center.
  Adjusted for light-time. Units: HMS (HH MM SS.ff) and DMS (DD MM SS.f)

You could do the same, but it will take some extra work. The Mars you see at some time $t$ was emitted from Mars several minutes earlier than that. You need to find the $\Delta t$ such that the distance between Mars at time $t-\Delta t$ and Earth at time $t$ is equal to the speed of light times $\Delta t$. One correction will get you very close, a second will get you even closer.

Another issue is time. JPL uses its own time scale, JPL ephemeris time (Teph). This is a relativistic time scale. It differs from Terrestrial Time (TT) by at most a few milliseconds, so for most purposes you can use TT in lieu of Teph in your homebrew ephemeris calculator. TT currently differs from UTC by 67.184 seconds, and beginning in July (there's a leap second at the end of June), it will differ from UTC by 68.184 seconds.

Or you could just live with it. If you are using your calculations for your backyard telescope, the handful of arc seconds of error that results from not accounting for the finite speed of light is small.


Is it just coincidence that the Moon follows the ecliptic? Or has it been gravitationally steered that way because of how it was formed?

No, this is not entirely a coincidence, though perhaps not in the way you expect (and other answers have indicated): it may have been coincidal at formation, but is required for long-term stability of the Moon-Earth system itself and, possibly, Earth climate.

The formation mechanism of the Moon (or the Moon-Earth system) is still debated. The favoured model involves a collision (between a pre-Earth and a Mars-sized object), when the resulting orientation of the Moon-Earth orbit depends critically on the impact parameters, which are essentially random, i.e. coincidal. Conversely, if the Moon-Earth system formed by some form of capture (possibly involving some surrounding gas to absorb the excess energy), a orientation in the eccliptic is more likely.

Howerver, the formation mechanism is not all there is to this. It is also required that the Moon-Earth system remains stable over 4.5 Gyr. This excludes any orbit initially circular and inclined by more than $i=\arccos\sqrt{3/5}\approx39.2^\circ$, when the Kozai mechanism would result in a collision of the Moon with Earth within a few years. The critical incliniation for a non-circular orbit is even smaller.

Even an inclination $i\approx30^\circ$ may not be viable, since it may imply that the Earth's spin axis (which is after all tidally coupled to the Moon's orbit) would be/have been tilted considerably stronger than by merely $\approx23^\circ$. A strong tilt of the Earth spin axis would have significant consequences for the Earth's climate and may well have prevented the formation of higher live forms. This is essentially a weak form of the antropic principle.


Will gravitational waves too far away ever reach us? Gravity is the curvature of spacetime, and its effects travel at lightspeed. However, space is expanding; eventually, light from distant galaxies will become more and more redshifted, and we will no longer be able to see them (source).

As such, there is a limit to how far we could ever possibly see, since light too far will never reach us due to the rapid expansion of space...or at least, if I'm understanding this correctly.

Now, gravitational waves travel at lightspeed. So, after enough time, when an object's light no longer reaches us, will its gravity no longer affect us either?

A better rephrasing is: at a certain point in time, will the gravity of any extremely distant object — even the most massive stars, black holes, or galaxies — simply not affect us whatsoever, in the slightest?

The answer here is very similar to if you were asking about light.

In principle gravitational waves might allow us to fractions of a second after the big bang. Electromagnetic waves can see back to where the cosmic background radiation formed, about 400,000 years after the big bang.

You are right, the universe has expanded. At the present epoch it is estimated that the observable universe, containing objects that have emitted light or GWs that may reach us now, is around 46 billion light years.

However, it seems quite likely that the universe continues well beyond this horizon, and sources beyond this horizon can never have emitted light or GWs that will reach us.

As the relevant section of wikipedia (https://en.m.wikipedia.org/wiki/Observable_universe) points out, the detection of GWs extends our view very slightly. We cannot "see" with electromagnetic waves beyond 45.7 billion light years because of the "fog" of the cosmic microwave background,  but GWs can penetrate this fog allowing us (in principle) to see signals from objects currently 46.6 billion light years away.


Simple question about the phases of the Moon If on March 20 the Moon were located at the position of the sun, how can I determine the phase in September 22, June 22, and December 20?

I drew the orbit and calculated by dividing sidereal days to get the position of moon. Then I moved the Earth to determine the position. But the answer was Full moon, First quarter, Third quarter respectively, which was different from mine.  I can’t figure out why I’m wrong.

The length of one lunar month is (on average) 29.53 days. You don't need to use sidereal days, since dates are based on a day of the sun, not of the stars. The calculation is just division

So you just need to count the number of days between March 20 and June 22 etc.

There are 94 days between those dates. And 94/29.53 = 3.18 lunar months.  So the moon will be crescent, on June 22nd (about 2 days before first quarter).  The other dates can be calculated similarly.

As a rough rule of thumb (that ignores things like the 30-31 pattern) the date of the new moon is earlier by about one day each month. By this reckoning, we would expect new moon on about June 17, Sept 14 and Dec 11, with first quarter one week later, etc.


How large part of the sky has Hubble observed? I'm curious for an estimate of how wide a field the Hubble space telescope has observed. Say, for example, how many times it has changed its pointing, and what fraction of that were revisiting the same field.

25 years of say 16 orbits/day is a (very over-estimated) maximum of 146,000 pointings.

The largest instrument is the various incarnations of the optical imager with a FOV of 6 (WFPC2) to 11 (ACS) square arcminutes. Thus the absolute maximum fraction of the sky (40,000 square degrees) that can have been covered is 1 per cent. And this is a big overestimate, since many parts of the sky have received many repeated pointings (e.g. the Hubble deep field).


Could a space radio telescope fill multiple purposes? Dream for a while and imagine that there will be a flagship+ mission within a decade to put a large radio telescope in space:

What kind of different tasks could the same radio equipment feasibly be used for?
How do the requirements for frequency sensitivity, electric power supply, dish size, and other radio technology (such as applicability of phased array?) and what else vary most importantly between such purposes?

For example, could one and the same big space radio telescope in the Asteroid Belt be used for any, some or all of the below, what would the synergies and contradictions be:

a) Radio astronomy on its own.

b) Interferometry together with a telescope on Earth to create an interplanetary baseline for huge resolution, like the spektr-R does cis-lunar today, AFAIK.

c) Radar characterization of the orbit, shape, size et cetera of any large enough asteroid in its current section of the asteroid belt.

d) Ground penetrating radar to characterize the interior composition of near flyby asteroids.

e) Communication with Earth so powerful that it need not engage the Deep Space Network (DSN), but can work with smaller Earth based radio antenna.

f) Extending the DSN by working as a relay between Earth and outer space probes, or between Earth and the backside of Mars or when Mars is in apposition behind the Sun from Earth.

See http://en.wikipedia.org/wiki/HALCA and http://en.wikipedia.org/wiki/Spektr-R.

HALCA was a Japanese 8 meter radio telescope satellite used for Very Long Baseline Interferometry (VLBI) from 1997 to 2003.  Spektr-R is a Russian 10 m dish launched into space in 2011 with apogee nearly as far away as the moon, and still going.
So, yes a radio telescope in space is quite useful for long baseline interferometry.  You probably do not want to also use it for deep space communications because that would interfere with the scientific use.  It is also not very useful for asteroid work because the odds of one going by is not greater than the odds of one going by the Earth and there are more powerful radio dishes on the ground.  Typically, one would not bother having a strong source for bouncing off asteroids on a radio telescope meant for extragalactic work.  And you sure would not want to risk sending it into the asteroid belt!


What are realistic and unrealistic values for the high energy gamma ray spectral indices of pulsar wind nebulae? This summer, I was working on a project fitting very high energy gamma ray spectra of the Crab Nebula, a pulsar wind nebula. At energies about $\sim$1 TeV, a simple power law suffices, i.e. $\phi(E)\propto E^{\alpha}$. Most power law models of the Crab find a spectral index of $\alpha\simeq-2.47$ (e.g. Hillas 1998, $\alpha=-2.49\pm0.06$). Using a normal analysis method on one dataset, I found $\alpha=-2.469\pm0.02248$, which matches the literature values quite well.

Using the modified analysis method I was testing, however, I found $\alpha=-2.316\pm0.02040$. We believe that method is a poor choice because I showed the calculated flux values were incompatible with previous observations. The spectral index also cast doubts on its viability. However, I was wondering if, for a random pulsar wind nebula, a slightly harder spectral index like $\alpha=-2.316$ is unphysical, or reasonable.

Therefore, I'm wondering: What is the distribution of spectral indices for power-law models of very high energy spectra of pulsar wind nebulae? Do we have a large enough population of pulsar wind nebulae that we can determine what a typical value for $\alpha$ is, and what range we would expect a spectral index to fall into? In other words, could if I had obtained a certain value for $\alpha$, what value would it need to have for me to reject the result as unphysical?

You might find these websites for your answers:

https://www.cv.nrao.edu/course/astr534/SynchrotronSpectrum.html
http://astronomy.swin.edu.au/cosmos/S/Synchrotron+Emission
https://ui.adsabs.harvard.edu/abs/2006MNRAS.371.1441S/abstract

We have a large amount of high energy emission these days, and the third reference shows the distribution from 394 GRBs.


How do we know Milky Way is a 'barred' spiral galaxy? In reference to the question, "How can we tell that the milky way is a spiral galaxy?"

The answers there clearly sum up the question asked. But Milky Way is not just a spiral galaxy. It is further classified as a barred spiral galaxy.

Question: Which particular feature in the distribution of stars, or in general a feature in observations led us to believe that it is a barred galaxy?

Note: The edge-on picture is not sufficient to establish it, since a dense distribution at smaller radii could also arise from a non-uniform density on the spiral disk if modeled to fit the observations. We don't have any data from other angles or orientations.

There are several different lines of evidence which together form a coherent picture: that of a barred galaxy. Moreover, as most disc galaxies are barred, we should expect the same from the Milky Way. The various evidences are:

The observed light distribution (2MASS) shows a left-right asymmetry in brightness and the vertical height. This is explained by the near end of the bar being located on that side.

The observed gas velocities show velocities which are "forbidden" in an axisymmetric or near-axisymmetric (spiral arms only) galaxy. These velocities occur naturally from the orbits of gas in a barred potential

The velocity distribution of stars in the Solar neighbourhood shows some asymmetries and clumping which is most naturally explained by orbital resonance with the bar rotation.

The extent, pattern speed, and orientation of the bar is consistent between all three of these.


Why are Shack-Hartmann sensors so expensive (4k+ USD)? Recently I am searching for a Shack-Hartmann wavefront sensor for university laboratory usage. I was expecting my target sensor to be cheap, which is:

Low spatial wavefront resolution (50 x 50) for beam measurement;
Low frame rate;
Portable design, and easy to use.

Most quotations are beyond my expectation. On ThorLabs it costs 4k USD (which is, "cheaper" compared to most competitors):
https://www.thorlabs.com/thorproduct.cfm?partnumber=WFS30-5C

Following reasons may to blame for this high price:

Expensive industrial sensors. But for https://www.flir.com/products/grasshopper3-usb3/?model=GS3-U3-14S5M-C it takes only 1.7k USD for a high performance sensor.
Microlens array (MLA) fabrication?
MLA-sensor alignment?
Software development? (SDK & GUI)

But I am not convinced by myself.

Any ideas why a Shack-Hartmann sensor is so expensive?

I have been deeply involved in both Shack-Hartmann and lateral-shear polarization interferometers. Now I want something simple and slow for hobby projects and had the same question. I don’t think such a project is as daunting as might be imagined.  One would like 12 bits in the camera, the Sony CMOS cameras are so good, the last bit is nearly noise free.
The camera need not be expensive: Any amateur telescope web store will have dozens for less than $400.  You want monochrome, preferably USB 3, with the largest diagonal focal plane size you can afford.  A Global shutter is nice if you have a dynamic situation. Rolling shutters are generally lower noise.
A lens array will be expensive.  Thorlabs, Seuss, and RPC Photonics, and others sell them for \$400-\$600.  The Thorlabs arrays are mounted.  In the end you want the focal spot to be at least 2 pixels across. Choose your micro lens F/# accordingly, usually quite slow, F/20 or more. Likely you want the closest pitch you can find as this sets the spatial resolution. A 4 x 5 mm focal plane will view 40 x 50 micro lens spots. Good enough.
You will need to craft a mount that mates the camera and the lenslet array at the micro lens focus distance.  They are slow, so the depth of field should be forgiving.
The real crux is the software.  There is public domain software that is close (git has AOtools) and there may be more. You will need to take a dark exposure for the shutter time used (I typically take 100 frames and average them).  Then the reaL GOTCHA issue: What to use as a reference?  You can use a shear plate to focus a collimated light source (a single mode VCSEL works well), but you are unlikely to have a shear plate about.  Place the VCSEL or other point source bright enough to see (an LED?) as far away in a dark room as you can manage.  Don’t use a mirror or lenses, turn off heating and cooling, etc. This gives you a known curvature. for 100 μm lenslets at F/20 the focal length is 2mm so a tilt of 1/2 a subaperture (to keep things easy) or 50 μm is only 25 milliradians (mr), something over a degree.  The edge of the camera must see the source as no larger than this angle.  So if you have 50 subapertures across your camera the edge is 2.5 mm from the middle.  Thus 2.5e-3 meters / d = 2.5e-3 radians so the source needs to be more than a meter away.  Easy!  With the distance from the source to the focal plane known you can now easily calculate the tilt distribution (should look spherical).
Use openCV or your favorite utility to get images into your computer and a) subtract background b) make sure the camera does not saturate (adjust exposure time, retake the dark background and try again) c) calculate the peak intensity of each spot and which pixel coordinates that is in d) Draw a box around each focal spot which is several spot diameters across each spot e) calculate the centroid of each spot using only pixels in the box (the rule is add no pixels that do not contribute to your signal, the centroid). You can play with the box size.  An F/20 blue-green 500 nm wavelength spot from a 100 μm lenslet will be about lambda F/# across (just 20 wavelengths, about 10 μm) which should be 3 to 4 pixels so a box 16 x 16 pixels should be a good place to start.  The centroids should be good to a small fraction of a pixel.  I have done better than 1/100 pixel with a really good focal plane array and 1/20 pixel for standard cameras such as the above. Now you can calculate the centroid location within the box, and reference that to the whole array.  A plot across a slice will give you a linear function which is somewhat less than the lenslet pitch.  Do this in both X and Y.  You may have a rotation in the spot coordinates. You can physically correct this, or use math.  The slope in X and Y should be the same.  If not, the lens array is tilted.  If you move the light further away, the spots should move to be more directly behind the lenslets.
In passing I would note that I found the accuracy of the lithography on the imager, and used to make the lenslet array was better than my best tilt calibration! Thus for a 3.5 μm pixel and 100 μm lenslet the spots should be 100/3.5 pixels apart.  This will be independent of focus etc.  I also found trying to uniformly illuminate the array to correct for the variation in responsivity pixel by pixel generated more noise, not less.  Silicon is awfully good.
The last calibration needed is to define where zero tilt is.  This should be close to normal to the focal plane array, so pick the middle lenslet and the pixel you declare is on boresight. Now you have the subaperture tilts calibrated.
The last messy task is to convert from wavefront subaperture tilt to wavefront.  There are books written on the subject. Notionally if you start at your zero tilt and add up the subaperture tilts as you move across the array, subtracting the zero tilt position of each subaperture, you have wavefront.  Note you can take any path from the center to the subaperture you want to know and add up the X and Y tilts and you should get the same answer.  Ideally you want to take the average of all possible paths (that are illuminated) as that would be the best average value you could get.  This is referred to as the minimum RMS error.  There are many ways to calculate this.  If you think of the tilts as the gradient in phase, and calculate the divergence of the gradients, you have a measure of the wavefront curvature.  It is also simply Laplace’s equation which can be solved by successive over-relaxation, or if you like fancy algorithms, use a multigrid method (it’s way faster). Greg Allen’s thesis has another approach, https://dspace.mit.edu/bitstream/handle/1721.1/120381/1084482108-MIT.pdf?sequence=1&isAllowed=y .   There are many more.
In summary, you can build a Shack-Hartmann wavefront sensor for about $1000 and a lot of software development.  The Thorlabs sensor seems cheap given the effort, at least until someone comes up with an open source sensor and software.
Jon


Why doesn't the black hole in the center of the Milky Way glow similarly to the famous M87 image? The M87 image made some astronomers famous recently as the first image of a black hole.
In the Milky Way, it has been concluded that there must be a black hole due to the movement of stars near the center. But why is there no such light effect in our black hole, as there is in M87? Or is it there, but we cannot see it for some reason?

M87 is an active galaxy with an accreting, rotating central black hole of mass $M\simeq6.5\times10^9M_{\odot}$ (Akiyama et al. 2019). It is the inner disk surrounding this black hole that produces, among radiation, the 1.3 mm synchrotron emission observed by the Event Horizon Telescope, as well as the relativistic jets emitted perpendicular to the disk's plane.
The Milky Way is not currently an active galaxy, and while there is some evidence for a cool accretion disk around Sgr A* (Murchikova et al. 2019), it is sparse. The strongest constraints on Sgr A*'s accretion rate are about five orders of magnitude lower than M87*'s $\sim0.1M_{\odot}\;\text{yr}^{-1}$ (Di Matteo et al. 2003). The very low accretion rate onto Sgr A*, a black hole three orders of magnitude less massive, combine to indicate that any emission of the same sort from the Galactic center should be much less intense than that seen by the EHT from M87*.


Exact measurement of a light year What is the exact measurement of a light year? I searched google for the answer in meters and came up with $9.461\cdot 10^{15}$ meters. When I calculated the answer considering $299\,792\,458 \;\text{m/s}$ as the speed of light, I came up with:

$$
299\,792\,458 \times 365 \times 24 \times 3\,600 = 9\,454\,254\,955\,508\,926 \;\text{m}
$$ Why is there such a gap? Did I miss something to add in the equation or is it just wrong?

By convention, astronomy uses the Julian Year for the computation of a light year:

  Although there are several different kinds of year, the IAU regards a year as a Julian year of 365.25 days (31.5576 million seconds) unless otherwise specified.

Wikipedia gives the length as

  $31 557 600 s \times 299 792 458 m/s = 9 460 730 472 580 800 m$ (exactly)

The reason that it's exact (and not subject to experimental error) is that the metre itself is defined in terms of the speed of light, so the quantities are fixed by definition.


Are there heavenly bodies between galaxies? I do understand that galaxies were formed in the famous Big Bang. However, I also understand that if such an explosion occurs in empty space it will never form distinct galaxies, it should more or less evenly fill the whole empty space.

Many of these might clot together to form galaxies but I cannot agree that between the void of galaxies there are no other bodies.

Am I missing something?

The Big-Bang was not an explosion in empty space.
Inter-galactic space is not empty, there is an inter-galactic medium, gas clouds and material ejected from galaxies, including stars and possibly globular clusters, by various mechanisms...


What makes scientists already so certain that Comet C/2019 Q4 (Borisov) is a comet and not an asteroid? US Naval Research Laboratory scientist Dr. Karl Battams has recently tweeted

  Unlike 'Oumuamua, whose asteroid-or-comet nature still gets debated, this one is definitely a comet. If it is unequivocally interstellar, it'll be fascinating to see how its composition (spectral properties) compare to the variety we see in comets from our own solar system.

Answers to Do astronomers generally agree that the distinction between comets and astroids is not so clear? suggest that the distinction can be a hard call sometimes, and some might instead call small solar system bodies formerly known as Prince comets and asteroids just that; SSSBs.

Question: What might cause astronomers like Dr. Battams be so confident that this object is a comet and not an asteroid, besides the fact that it's being called one? (Comet C/2019 Q4 (Borisov))

Is it just the orbit itself, or are there other observational data such as brightness in the infrared?

It's the appearance that is mainly used to distinguish comets from asteroids. As noted in the discovery Minor Planet Electronic Circular (MPEC), numerous observers observed extended diffuse emission (i.e it looks "fuzzy" with a Full Width Half Maximum (FWHM) greater than the field stars, or shows a tail) typical of a comet. In the case of 1I/'Oumuamua, deep searches with 8-m telescopes failed to record any evidence for any extended emission, indicating it was most likely a bare rock. The much brighter magnitude of C/2019 Q4 does play a role in this, making a search for extended emission easier.

The distinction has become more difficult in recent years with the discovery of activated asteroids such as (6478) Gault, Main Belt Comets but also objects like 1I/'Oumuamua on very long period, comet-like orbits which don't seem to show any sign of emission.


Can we observe changes in the CMB (surface of last scattering) over time? It is hard to determine the age of the Sun, because it won't change much over 100 million years or so. But what about the cosmic microwave background radiation, that is being mapped with ever better precision. It dates to only about 380 000 years after Big Bang. Is it possible to observe changes in it over a period of 10 years or 100 years, as it ages and cools, as the Hubble Bubble expands?

And how is the landscape of the CMB-map expected to change? Are relatively cool and hot areas permanent over time scale much larger than 380 000 years, or would our CMB-map be unrecognizable if we had lived, say, a billion years earlier or later?

It isn't all that hard to determine the age of the sun, provided you are content with an accuracy of 100 million years or so either way. The radiometric dating of zircon crystals, which are only a little younger than the sun, gives a date of about 4.6 billion years, and it is generally agreed that the sun is about 5 billion years old. This is compatible with its position in the main sequence of a Hertzsprung-Russell diagram. You certainly wouldn't notice any change on the CMB map in a mere ten years, but a billion years hence it will have a noticeably greater red shift, except for the fact that there will be nobody here to notice or record it. Our descendants, every last one of them, will all have gone to meet their maker in far less than a billion years.