Closest path of solar eclipse central line I'm trying to calculate the closest path of the center line of a given eclipse at a given time by using its besselian elements; However i'm getting an error range up to 30 km (18.6 mile) in the worst case and 5 km (3.1 mile) in the best case (both cases was compared with F. Espenak prediction)
Lets use eclipse of August 12, 2026 at 18:00 UTC.
F. Espenak prediction: 58°16.1'N   021°34.4'W (58.2683, 21.5733)
My prediction: 58°16.8'N 21°47.4'W (58.2795, 21.7901)
The difference is close to 13 km (8 mile) which is too big for my application so I need to refine/review the procedure/algorithm I'm using.
Calculus
Given the following besselian elements:
e20260812 = {
    t0: 18.00, At: 72.3,
    x0: 0.475510, x1: 0.518923, x2: -7.727e-05, x3: -8.039e-06,
    y0: 0.771185, y1: -0.230167, y2: -1.246e-04, y3: 3.767e-06,
    d0: 14.796683, d1: -0.012065, d2: -3.098e-06, d3: 0.0,
    l10: 0.537974, l11: 0.000094, l12: -1.212e-05, l13: 0.0,
    l20: -0.008142, l21: 0.000093, l22: -1.206e-05, l23: 0.0,
    u0: 88.747810, u1: 15.003090, u2: 1.764e-06, u3: 0.0,
    tanf1: 0.004614,  tanf2: 0.004591
}

(At represents $\Delta T$)
0 - Reference timestamp.
Where t1 = 18.00
t = t1 - t0 + At/3600.0

1 - Time dependent calculus
x = x0 + x1 * t + x2 * t^2 + x3 * t^3
y = y0 + y1 * t + y2 * t^2 + y3 * t^3
d = d0 + d1 * t + d2 * t^2 + d3 * t^3
u = u0 + u1 * t + u2 * t^2 + u3 * t^3

2 - Location dependent calculus
Initially I'm using ro=1.0 and following the below sequence to obtain an approximation of phi, which is the Latitude.
ro = 1.0
e = 0.0818192 // earth eccentricity
sin_d = sin(d * pi / 180.0)
cos_d = cos(d * pi / 180.0)
zeta = sqrt(ro^2 + x^2 + y^2)
cos_theta = (-y * sin_d + zeta * cos_d) / ro
sin_phi_1 = (y * cos_d + zeta * sen_d) / ro
cos_phi_1 = sqrt( (x/ro)^2 + cos_theta^2 )
phi_1 = asin(sin_phi_1) / pi * 180.0
phi = atan(sin_phi_1 / cos_phi_1 / sqrt(1 - e^2)) / pi * 180.0

At this point phi = 58.1224 for 18:00 hs UTC.
Once phi was calculated, we recalculate ro as follow to get a better approximation of phi and varpi (not yet calculated)
sin_phi = sin(phi * pi / 180.0)
ro1 = sqrt( (1 - 2e^2 * sin_phi^2 + e^4 * sin_phi^2) / (1 - e^2 * sin_phi^2) )

Applying the updated ro (now ro1), we update sin_d and cos_d as follow:
sin_d1 = sin_d / ro1;
cos_d1 = sqrt(1-e^2) * (cos_d / ro1)

And update zeta and cos_theta
zeta1 = sqrt(ro1^2 + x^2 + y^2)
cos_theta1 = (-y * sin_d1 + zeta1 * cos_d1) / ro1

Finally we calculate vapri (the Longitud) as follows:
tan_theta = x / cos_theta1
theta = atan( tan_theta ) / Math.PI * 180.0
vapri = u - theta * ro1

and recalculate phi by doing:
cos_phi2 =  cos_theta1 / cos(theta / 180.0 * pi)
phi1 = acos(cos_phi2) / pi * 180.0
phi = atan(tan(phi1 / 180.0 * pi) / sqrt(1 - e^2)) / pi * 180.0

At this point phi = 58.2795 , vapri = 21.7901 for 18:00 hs UTC (both values MUST be properly signed but I omitted that trivial part) a significant difference compared with the NASA prediction: 58.2683, 21.5733 (13 km or 8 mile)
My feeling is that ro1 in not good enough precise, so the final results get shifted, but I have no idea at all about how to getting unblocked.
I'm using 64 bit floating point calculator; It is ok for this kind of calculus?
Is this method good enough to get an error range near to 500m? if not, could you provide a method/fix to achieve that aim.
Any help would be appreciated.

Your methodology seems to be based on an algorithm that uses successive approximations to obtain the desired values. I was not able to arrive quite at the same values as you mentioned, but I noticed an error in the posted script, zeta = sqrt(ro^2 + x^2 + y^2) clearly should be zeta = sqrt(ro^2 - x^2 - y^2), so perhaps slightly different values or operations were used somewhere else too, but I couldn't find out where. I also noticed that you did not take into account the difference between Ephemeris Longitude and True Longitude in your calculation, True Longitude shifts towards the East by 1.002738 * At (Delta t) / 240 degrees. Moreover, executing more iterations could possibly converge to a better result.

However, to compute the intersection of the Moon's shadow axis from Besselian Elements, I suggest using the following closed-form solution, which is derived from solving the intersection of a parametric line (Moon's shadow axis) and the Earth ellipsoid.

f = 1 / 298.257223563
b = 1 - f
sin_d = Sin(d * PI / 180)
cos_d = Cos(d * PI / 180)
y0_r = y * cos_d
z0_r = -y * sin_d
y1_r = y0_r + sin_d
z1_r = z0_r + cos_d
v = y1_r - y0_r
w = z1_r - z0_r
a_quadratic = v ^ 2 + b ^ 2 * w ^ 2
b_quadratic = 2 * z0_r * w * b ^ 2 + 2 * y0_r * v
c_quadratic = z0_r ^ 2 * b ^ 2 + y0_r ^ 2 + b ^ 2 * (x ^ 2 - 1)
p = (-b_quadratic + Sqrt(b_quadratic ^ 2 - 4 * a_quadratic * c_quadratic)) / (2 * a_quadratic)
y_i = y0_r + v * p
z_i = z0_r + w * p
dist_axis = Sqrt(x ^ 2 + z_i ^ 2)
lat = Atan(y_i / (dist_axis * (1 - f) ^ 2)) * 180 / PI
lon = Atan2(x, z_i) * 180 / PI - u + 1.002738 * At / 240

Here is the explanation of the formulas:

We use the flattening value to define the WGS84 ellipsoid, and calculate the semiminor-axis:

f = 1 / 298.257223563
b = 1 - f

Convert declination angle d to radians and execute sine and cosine functions:

sin_d = Sin(d * PI / 180)
cos_d = Cos(d * PI / 180)

Select 2 points (points 0 and 1 in the code) on the shadow axis, at z = 0 and z = 1, and apply the declination rotation (_r) on them:

y0_r = y * cos_d
z0_r = -y * sin_d
y1_r = y0_r + sin_d
z1_r = z0_r + cos_d

The x value always remains the same, so we don't need to recalculate it.
Next, evaluate the parametric line's vectors v,w (delta y, z) from point 0 as a base:

v = y1_r - y0_r
w = z1_r - z0_r

By substituting the parametric line's equations in the ellipsoid equation, and rearranging to quadratic form and solve:

a_quadratic = v ^ 2 + b ^ 2 * w ^ 2
b_quadratic = 2 * z0_r * w * b ^ 2 + 2 * y0_r * v
c_quadratic = z0_r ^ 2 * b ^ 2 + y0_r ^ 2 + b ^ 2 * (x ^ 2 - 1)
p = (-b_quadratic + Sqrt(b_quadratic ^ 2 - 4 * a_quadratic * c_quadratic)) / (2 * a_quadratic)

'p' here represents the distance from point 0 along the parametric line, expressed in Earth's equatorial radius.
The contents of the square root must ideally be tested beforehand for negative values to avoid exceptions if the line does not intersect the ellipsoid.
With this, we find the coordinates of our points of intersection:

y_i = y0_r + v * p
z_i = z0_r + w * p

Then we convert this to geographic latitude and longitude:

dist_axis = Sqrt(x ^ 2 + z_i ^ 2)
lat = Atan(y_i / (dist_axis * (1 - f) ^ 2)) * 180 / PI
lon = Atan2(x, z_i) * 180 / PI - u + 1.002738 * At / 240

Direct conversion for the Latitude is possible here because our point is on the surface of the ellipsoid.

At the end, you can use the following logic to capture out-of-bounds longitude values:

If lon <= -180 Then
    lon += 360
ElseIf lon > 180 Then
    lon -= 360
End If

In terms of accuracy, you can expect pretty good results with this method, but it does not take into account some factors, like local topography, geoid height, refraction, etc. Topography can lead to differences of a few kilometers, the geoid, +-100 meters, and refraction can bend the shadow by a few meters to a few hundred meters depending on the incident angle. Also, there is no correction for the moon's center of figure, that can shift the umbral center by several hundred meters as well.

With the above formulas for UT1 = 18h and delta t = 72.3s, the computed intersection point is at 58.26828°N, 21.57331°W, which matches F. Espenak's values.


What if we are looking for the wrong signs of life on other planets? I am by no means an astronomy expert, having only taken 2 semesters in college. But I am an avid follower of news reports that look for signs of life on other planets. Today it was announced that 20 new planets "with the potential for life" have been found. Along with the usual signs (temperature and water) these planets were also cited because one of them has a year of 395 Earth days, a very similar year to our own.

However, I'm always bothered by these reports because they are all predicated on the premise that life on other planets will closely resemble our own. What if life on other planets doesn't need water? What if they are not carbon-based life forms? What if they can live in extremely hot (or extremely cold) temperatures?

It just seems like a massive mistake to (effectively) say, "this planet doesn't look anything like ours, there must be no life there" or to imply that only Earth qualities (or similar) can support life. Are there any fields of astronomy that look beyond the "near-Earth" life comparisons and look for life in other (drastically different) forms?

What if we are looking for the wrong signs of life on other planets?
Are there any fields of astronomy that look beyond the "near-Earth"
life comparisons and look for life in other (drastically different)
forms?

and

I'm always bothered by these reports because they are all predicated
on the premise that life on other planets will closely resemble our
own.

Short answer:
It doesn't matter what people think.   What matters is what we can observe.   Just because they will probably look for life on Earthlike planets first, doesn't mean they won't look elsewhere.  The answer will depend on what they find, not what anyone expects to find.
Long answer:
I think there are a lot of good comments below your question, but I want to try to address a few points.
On looking for life.
The scientists that look into space through our biggest telescopes aren't exactly "looking" for life on other planets because life is too hard to see.   Mostly, they look at stars.   Our best telescopes today can't see exoplanets at all.  Exoplanets are detected indirectly by variations in the stars they orbit.    If exoplanets are too far away most of the time to even be seen, looking for signs of life on them is obviously out of reach.   We're not actively looking for life because we can't.   We need much bigger telescopes for that.
So, basically "20 Earthlike planets discovered" is the upper limit of our technology right now.  That's today's cutting edge discovery.   I understand that it might be boring.  I know that they're just adding more planets to the directory, but that's the limit that our best equipment can deliver right now.
Almost nothing is known about all those planets, just the very basics, like how close they are to their star, and about how large they are and sometimes, about how dense they are giving us a limited idea of their composition and if they are hot, cold or perhaps in the Goldilocks zone.   If there's only 5 lines of data per exoplanet, the tag-line "Earth-like" often makes the best headline, not because NASA is focused on only Earthlike planets, but because there's not a lot of detail.
It's kind of like "I see a bird, but it's far away and I don't have my binoculars.  It's flying south", but I can't see what kind of bird, I can make out the size, to some extent, I can see that there's just one, but I can't make out the shape.   So, I write in my bird journal "One largeish bird flying south - too far away to see what kind".
That's basically where we stand with exoplanet observation right now.
This will change some when the James Webb Space Telescope is operating.   The JWST should be able to see star-light filtered through some exoplanet atmospheres, which will tell us a whole lot more than we currently know about some of those exoplanets.  Information such as how thick an atmosphere and what it's made of will tell us a lot, and give us some information on the climate of the planet.  JWST will be a HUGE (all caps for emphasis) step forward in the search for alien life, though it still won't observe life directly, but it will hopefully get a good read on some exoplanet's atmospheres, and from that, deductions can be made, such as abundant oxygen isn't expected without some form of photosynthesis.   Oxygen is too reactive to be likely to exist on it's own.
A combination of CH4 and O2 could be especially interesting and one of the first things to look for, but they also have to be careful.  There's always the chance of more than one explanation for a planet's atmosphere.  We shouldn't expect any kind of guarantee of discovery, but JWST should be an enormous step forward in what we can observe about exoplanets.
On reporting
Scientists in this field, their job is essentially to analyze and report the data.   Their job is not, for the most part, to speculate on life but to simply report what can be verified (and today, relatively little other than "I found a planet" can be verified).
Theoretical biology is great.  Don't get me wrong.  It's fun and it's real science, but the majority of telescope observation focuses on observe, analyze and report.  That might sound dry, but that's mostly how science works.  "20 Earthlike planets found" was impossible 10-15 years ago.  It really is cutting edge.
On Theoretical Alien Life:
Scientists, today (Neil deGrass Tyson among them) who would love to get a closer look at some of the Moons of Jupiter and Saturn, because they are thought to hold the best chance of having life beyond Earth in our solar-system.   The desire to find alien life, if it's out there, is one of the highest held goals of NASA and Scientists in this field.   Nobody's ignoring or dismissing anything.  They will likely point JWST towards Earthlike planets first, but that's only natural.
I'd like to also point a basic rule of astronomical observation.   "Weird" is actually good.   Weird leads to discoveries like dark matter and dark energy.  Like the Isaac Asimov "That's funny" quote, finding things they don't expect is often the best find.   Scientists love that.   If weird atmospheres are found by JWST, it won't matter if they are on Earthlike planets or not.  Anything weird or unexpected will be of great interest and closely looked at.
Like those theoretical Alien Mega-structure or Dyson sphere stars (which probably aren't Dyson spheres).  I'm highly skeptical that it's aliens, but I'm glad they're looking into it.  Anything unexpected is worth studying.
When JWST starts to get better data on atmospheres, then you'll see more interesting headlines than "Earthlike".  For now, and for the next year or so, "More Earthlike planets found", is as far as our technology can take us.
Finally, I want to point out that on theoretical biology, there are issues.  (The other answer covers this better than I did so I'll abbreviate), but there are reasons to suspect that life might need to be carbon and liquid water based, so planets thought to have liquid water should be the first places to look.
And ultimately, it doesn't matter what people think is out there.   What matters is what JWST (and telescopes that follow) will discover.   What we know about the Universe is based on what has been observed, not on the various prejudices and assumptions people made.   Sometimes what is observed goes against the scientific belief of the day, and that's what's great about science.  Evidence trumps belief.


Do stars have exactly sphere shape? Planets (at least, some of them like Earth) aren't exactly spherical - but what about stars?

Is Sun perfectly spherical, for example? What may be the reasons if it isn't? What about other stars?

No, no stars have an exactly spherical shape. The reason for this is that the centrifugal force of the star's rotation is much greater at the equator of the star than it is at the poles, for the simple reason that the rotational velocity is greater. This greater centrifugal force pushes the equator outwards, stretching the star into an oblate shape. This is called gravity darkening.

Because we have only visually resolved the surfaces of a few other stars, it's not something that is commonly directly observed (though the effect can also be observed from stellar spectra). Regulus is one star that has been observed as an oblate spheroid spectrographically. The image below shows the star Altair, directly imaged using the CHANDRA space telescope. Go here for an animation!

Some stars are also non-spherical due to the effects of a nearby star in a close binary orbit. Much like how the Moon causes tides on Earth, two stars can stretch each other's surfaces. If they're very close to one another, as in the picture below, there can even be mass transfer between them (ref common envelope).


calculation of characteristic expansion time of the universe Hubble's constant is estimated at 15km/s at a distance of million light years, without considering effects of mutual gravity, relativity, etc.
I am trying to calculate the characteristic expansion time of the universe.
Its formula is given as reciprocal of the Hubble constant, because time is distance by velocity. But:

$$v=kx$$

This implies that velocity and distance are exponential functions of time and so time cannot simply be calculated by dividing 1 million light years by 15km/s.
So what am I missing? Why is the characteristic expansion time simply the reciprocal of Hubble's constant?

Here is some remarks on the issue, straight from Ryden:

  If galaxies are currently moving away from each other, then it implies
  they were closer together in the past. Consider a pair of galaxies
  currently separated by a distance $r$, with a velocity $v = H_0r$
  relative to each other. If there are no forces acting to accelerate or
  decelerate their relative motion, then their velocity is constant, adn
  the time that has elapsed since they were in contact is

  $t_0 = \frac{r}{v} = \frac{r}{H_0r} = H_0^{-1}$,

  independent of their current separation $r$.

Which part of this are you confused about? I can edit my answer once I better understand your question.


Orange star like ight Tonight high up in the night sky, it was 8.20 precisely, I saw a bright orange light floating slowly across the sky ,occasionally slowing as if looking for something and flickering, it was too high up and far away to see an exact shape, but it looked like a star, so was it a star, do we have bright orange stars

I saw a bright orange light floating slowly across the sky ,occasionally slowing as if looking for something
it looked like a star, so was it a star,

For the sake of teaching some scientific reasoning skills, I'll point out that just because you think it looks like a star does not mean it is a star.
In fact the motion of the object suggests it was not a star.  Stars do not move except as part of a general motion of the night sky dues to Earth's rotation.  What you describe is not that.
When we try and work out what something is, we use all the evidence.
What would it be like ?  Perhaps a helicopter with lights.


What cluster of stars is this with a "dark donut" to one side? In the 2011 Sixty Symbols video Spy Satellites (from Deep Sky Videos) after about 01:44 amateur astrophotographer Nik Szymanek shows an image of a cluster of stars and I think I see what looks like a dark donut shape (dust?) to the right of the center of brightness, as shown in the screenshots below.

The vertical streak is a satellite trail, the topic of the video but not related to the present question.

Question: Can someone recognize what object this is and explain what is thought to cause the dark donut?

Screenshots from video, click for full size:

Cropped and shamelessly processed to enhance the dark donut's visibility:

OK, having (finally) actually looked at the video, it's clear that Szymanek is looking at the center of M33. There is in fact a nuclear star cluster in the center of that galaxy; not knowing the field of view or the resolution, I can't tell how much of the central condensation is simply the unresolved, smeared-out nuclear star cluster (which is small -- you really need something like HST to resolve it) and how much is that plus the surrounding central region of the disk.

The "dark donut" is undoubtedly the diffraction shadow due to a dust grain on the filter or other optical element in front of the detector; see the "Dust Rings" subsection of this page for an example and brief discussion.

  Dust on either the filter, the window protecting the CCD, or any of the corrective optics will leave little donut shapes on an image like the one below. They appear as rings because the dust grains lie on optical surfaces above the focal plane so when they cast a shadow on the CCD, it is out of focus. Astronomers can measure the size of a dust ring and tell exactly where in the optics the grain of dust lies. Dust grains on the CCD itself leaves little dark spots. Calibration frames completely remove dust spots and rings from images.


Does the mass of a collapsing star change? Does a star, when collapsing into a neutron star, or even a black hole, pick up any significant relativistic mass due to its rapid collapse rate?

This is an old question, but I'd like to answer it anyway.

  Does a star, when collapsing into a neutron star, or even a black hole, pick up any significant relativistic mass due to its rapid collapse rate?

No. Conservation of energy applies. When you drop a 100000kg asteroid from some high elevation above the Earth, some of the asteroid's gravitational potential energy, which is mass-energy, is converted into kinetic energy. The mass-equivalence of the system does not increase as the asteroid falls. Instead the asteroid's rest mass reduces as the relativistic mass associated with its kinetic energy increases*.

When the asteroid hits the Earth (BOOM!) some of the kinetic energy will get dissipated and radiated out into space, and then you're left with a mass deficit. The mass of the system is then less than it was. See Wikipedia: "This missing mass may be lost during the process of binding as energy in the form of heat or light, with the removed energy corresponding to the removed mass through Einstein's equation E = mc²".

The same applies to a collapsing star. There is no "pick up" of relativistic mass. Conservation of energy applies. But some of this energy is typically radiated away, so you're left with a reduced mass.

The same also applies to a brick. When you drop a brick, gravity converts some of its rest mass into kinetic energy. This gets dissipated, and you're left with a mass deficit. Then when you lift the brick you do work on it. You add energy to it, thereby increasing its mass. Then you're back where you started.

The same also applies to an electron.

* this also applies to the Earth, but to a much lesser extent, so we tend to forget about it.


Why is iron responsible for causing a supernova? Why is the element iron responsible for supernova? Can any star create more element than iron within the span of its life?

I understand that when star dies due to supernova, other elements are created (Gold and other 92 elements) due to more heat generated than required not only for fusion of Iron element but also for the fusion of higher element with each other.

Your question is a bit oversimplified because there are many types of supernovae based on the size and configuration of the star. But I can answer your question about "why iron" by considering what keeps a star from exploding in the first place.

In the simplest terms of star formation, when material from an interstellar nebula starts to collapse under its own gravity, the pressure and temperatures involved will become great enough to eventually start fusing hydrogen into helium (it's a bit more complicated than that, but I'm speaking in generalities). If you were to consider the helium atoms created by that process, you'll notice that each helium atom weighs just a bit less than the two hydrogen atoms that formed it. That bit of extra mass is given off as energy which is produced in great quantities as the hydrogen continues to be fused into helium.

During the star's "main sequence", the release of energy by the hydrogen-helium fusion helps counteract the weight of the star's gasses pushing inward. Material presses in; energy pushes out in perfect balance. This balance of gravity and energy output continues until the star uses up most of its hydrogen.

It's at this this point (when there is no hydrogen left in the core of the star to fuse into helium) that the fusion reaction stops and gravity will resume to collapse the star further. As this star collapses, it will quickly become denser and hotter until the temperature and pressures of the interior are great enough to start fusing helium into heavier elements… and the process continues.

That is, until the star starts fusing elements into iron…

The fusion into iron is the first element that does not create more energy than it takes to produce. The effect is that there is no net energy being produced to counteract the gravity pushing inward. So the outer layers will quickly collapse into a much denser and smaller ball causing the remaining star material to undergo fusion all at once, causing the supernova.

So, in that sense, iron is not the cause of the supernova, but its presence marks the inevitable end of this star's life cycle… in this particular scenario.

But understand that this is an oversimplification to illustrate the process you asked about. There are many other sequences of a star's life cycle. Our sun, for example, does not have sufficient mass to keep collapsing down with sufficient pressures to fuse heavier elements into iron. Without getting into other pathways for the production of heavy elements (even in smaller stars like our sun) — once our sun starts creating carbon and oxygen, the fuel starts to run out and the core will simply collapse and rebound as it swells up into a red giant, before losing its outer layers as a planetary nebula while the core shrinks to become a white dwarf (and eventually cooling into a black dwarf).


Colossus telescope, trying to outsmart aliens? I was listening to Jeff Kuhn's talk
on SETI's Colossus telescope project.

Background:

He explains his theory that a civilization living somewhere the galaxy would want to hide, and thus would shortly learn to avoid broadcasting radio signals all over the galaxy, so as to avoid attracting attention (presumably to avoid inter-stellar predators).

Then he comes up with a great idea; due to the second law of thermodynamics, they would have to dissipate heat-energy, and thus, one can detect them anyway, because they have no way to hide this heat-energy. This, particularly when they are a Type I civilization, and nearing Type II; in this case, to avoid global warming, they would capture and use almost all solar energy, and emit an equivalent amount heat-energy.

Thus he proposes a telescope to focus on single stars, which would be capable of detecting this heat-energy (by the distinctive infrared light emmited).

My question:

If a civilization was smart enough to hide all of its radio signals, wouldn't they know that they cannot avoid emitting heat-energy? Thus their predators would be able to find them anyway, and thus, it is pointless to hide the radio signals?

Therefore, we should already be seeing their radio signals, and this telescope is being built for a theory with some holes.

Note: I am not suggesting that the telescope shouldn't be built; I think it could have other very valuable uses, and it would also be interesting to eliminate the theory if it is indeed incorrect (or prove it correct ... somehow). I am just asking if there might be justifications for the theory, in light of my point-in-question, which I am missing.

The real reasoning has nothing to do with some civilization "deliberately" hiding its radio emissions. Rather, the problem is that we can not expect some other civilization to do something we would not do ourselves.

It makes no sense whatsoever to radiate large amounts of energy into space when there exist other, more economical alternatives. Radio transmitters circa 1911 required whole dedicated dynamo machines to run at abysmally low efficiency - these days we are using tiny battery powered devices with directed radio beam steering and adaptive power minimization as to not transmit a single joule more than absolutely necessary to get the data through to intended destination.

Another problem arises from the fact that most of the "radio SETI" effort to date was concentrated around the "cosmic waterhole" (the frequency of interstellar hydrogen emission). This is a reserved frequency here on Earth, so nobody broadcasts on it. Common sense suggests that aliens would probably do the same - after all, we expect them to have developed radio astronomy as well. Unfortunately, listening on any other frequency will probably pick up locally generated broadcasts first and completely mask the weak alien signal.

Given the above, what choices are we left with? One would be to construct a large radio telescope somewhere Earth radio emissions can not reach - the far side of the moon or even a trans-solar location on the Earth orbit come to mind. Unfortunately, such project, while doable, is hugely expensive.

The other approach is to develop "distributed" telescopes which can be built gradually on low budget and to develop techniques to cancel the local noise. For low frequency radio telescopy such project is already ongoing:

http://en.wikipedia.org/wiki/LOFAR

For infrared, we've got this "Collosus" proposal, but place our hopes on JWST (http://www.jwst.nasa.gov/).

And we don't expect to see much in visible light for all the obvious reasons.


If the universe keeps expanding, can we travel to the edge of the universe? If the universe keeps expanding, is it possible to travel to the edge of the universe?

When we talk about the universe, we are really talking about one of two things:

The observable universe, which is everything we can possibly see.
The Universe, which is everything that has ever existed, currently exists, and will exist.

The observable universe has its own center, usually the Earth. It is a spherical region of everything that we can see, essentially anything whose light has reached us. We usually refer to this when we say things like "there are $10^{86}$ atoms in the universe."

In reality, everyone has their own observable universe, and it can change depending on where you are. An exoplanet far away has its own observable universe, and can receive light from different places. Essentially, you are the center of your own observable universe. As more light reaches us, our observable universe will grow (or expand, if you will). If this is what you are referring to, then your answer is here.

If you are referring to the latter, then there is a totally different answer and "expansion" refers to something completely different. The Universe (notice the capital "U") is all of space and time and its contents. Anything that has existed, will exist, and currently exists is part of it.

The Universe is thought to be infinitely large, so it can't have a center. The center of something is the point equidistance from the edges, but if something spans infinitely long, it would just keep going. It wouldn't have an edge, and thus it wouldn't have a center. You couldn't find the point equidistant from the edges if it just spans infinitely.

The Universe is not like a ball. Rather, you can think of it like a flat grid, and its "expansion" just means that the distances between objects on the grid are getting larger. In essence, more space is being created between the objects. That's what we mean by expansion — that objects are moving away from each other, since more space is being created between them.

Here's an easy analogy: imagine you are walking your dog. Suddenly, the ground begins expanding between you. You and your dog will separated and continue receding away from each other.

That's essentially happening everywhere: space is expanding between everything, so we are drifting away from other galaxies. The Universe is infinite, and we can constantly drift apart from other objects because space is being created in between us. Here's a GIF I made that might help you get it:

You can see how the galaxies drift apart as the space between them increases. And this happens everywhere in the Universe. The Universe is infinite, but more and more space is being created between matter.

Fun fact: These objects can actually drift away from each other faster than the speed of light. That is, light from them eventually won't make it to us, since they'll be drifting away too quickly.

Now, this doesn't actually go against Einstein's theory that the speed of light is the fastest thing in the Universe. Einstein said that nothing can travel through space faster than light — but here, space itself is actually being created between the objects. Distances are increasing because space itself is dilating, and thus we can drift apart from other objects faster than light.


By how much does Haumea's fast rotation affect its surface gravity? The equation for surface gravity is $\frac{GM}{r^2}$ but I'm not sure how to include the effects from its rotation.

The numbers are going to be approximate because Haumea's size is uncertain.  It may have a ring.  If it does, then it's smaller, if it doesn't, then it's larger.
The formula is simple enough.   For force:    $F= \frac {m v^2} {r} $ where v is the rotational velocity and r is the same r in the gravitational formula.   or $\frac {v^2} {r}$ if you want the reverse acceleration.
Haumea has a rotational period of 3.915 hours or 14,090 seconds (I'm rounding) and an average radius of 780,000 meters (assuming it has a ring, I'm going to go with that assumption), but an estimated equatorial radius of about 1,050,000 meters and that's the number we need.  Divide the larger radius by number of seconds, multiply by 2 Pi and the velocity is 468.2 m/s.   A tiny bit faster than Earth's equatorial velocity.
Some Maths, the lifting force on the equator is $ \frac {468.2^2} {1,050,000} $ = $ 0.208 \frac {m}{s^2} $, slightly over half it's listed equatorial surface gravity of $ 0.401 \frac {m}{s^2} $.
That's a pretty high ratio.   Possibly the highest in our solarsystem for objects that are massive enough to be spheroid shaped by their mass and gravity.   There are asteroids that are thought to effectively have negative gravity at their equator, held together by the stickiness of the material, but those are a lot smaller.


Could the Sun's bending of light be measured on photographic plates before Einstein's prediction? It was predicted from Newtonian physics already in the 18th century that gravity should bend light, but not by as much as general relativity predicts. This was first confirmed during a solar eclipse in 1919. Surely, solar eclipses were imaged long before that. Didn't anyone notice that stars near the Sun were displaced? Hadn't anyone thought of testing the newtonian prediction of bent light?

I suggest reading the paper on the 1919 expedition to get a clearer picture of why they did it at that time and why it hadn't been done before. From reading chapter 2 I think the main reason was the astrophotography equipment required for the experiment and the alignment of bright enough stars close to the sun to observe the effect.

Of course before Einsteins prediction in 1911 no one had any specific reason to observe the stars close to a solar eclipse before and during totality, it might seem obvious now but that's hindsight for you.


After Oumuamua passage, has an "urgency" telescope time request procedure been set up? During the rapid passing of ʻOumuamua through our solar system, it was difficult to get instant telescope time. Usually, formal requests and a review process are necessary before telescope time is provided to a particular project.

Now everybody is excited and eager to observe more interstellar objects. Did "they" set up an appropriate request procedure for immediate telescope usage for that matter? Or can one expect that similar hurdles and paperwork are required when the next such object is discovered?

Is there any kind of international standard, or are such requests managed individually by each observatory?

For observing "predictable" transient events (e.g. Gamma Ray burst or gravitational wave events), observers apply (i.e. write a proposal for peer review) for "target of opportunity" (ToO)time in advance. Typically, you might say we expect LIGO to give us 2 reasonably located, strong GW events dutring the next semester and we would like 3 hours of telescope time to follow each one of these up.

If the proposal is approved, then the ToO time can be "triggered" by the proposers at any time (up to their maximum of 2 events per semester, or whatever they got approved for), by contacting the observers who are currently at the telescope and asking them to execute the ToO observing that they asked for. The current observers programme is interrupted (and they may not be compensated for this, other than perhaps co-authorship on a paper arising from the ToO).

Totally unpredictable events like Oumuamua or a new Galactic Supernova have to rely on director's discretionary time (DDT). This is a fast track, way of getting some observing time. It is still peer-reviewed to some extent and only meritorious proposals will get any time allocated. Often a DDT proposal could be approved within 48 hours.

If something truly extraordinary happened (and I would not classify Oumuamua in this way) that required observing right now, such as for example a local supernova, then I imagine that any observer at a telescope would be happy to break off what they are doing and point whatever instruments they had at it. It should be noted though that there is a risk here. At many observatories (for example ESO) you are only supposed to observe what you said you were going to observe. Departures from this can get you into trouble. A wise observer would seek permission from observatory staff/astronomers before chasing up some object somebody phoned up and asked them to observe.

See https://www.eso.org/sci/facilities/paranal/sciops/ToO_policies.html for some more information on how these policies and procedures work at ESO.


Why is there "modulus" word in the "distance modulus" term? The difference of apparent (m) and absolute (M) magnitudes of an object is called "distance modulus", which is related to the distance $d$ to the object in parsecs with equation

$$m - M = 5 \log \Big(\frac{d}{10 \ \textrm{pc}}\Big) \tag{1}.$$

I'm trying to understand why $m - M$ is called "distance modulus". The word "distance" makes sense, because there is a distance d term in Eq. (1). But what is "modulus" word doing there? In math, modulus means the remainder after division of two numbers. For example, modulus of 5 and 3 is 2.

Another meaning of "modulus" is length of a vector. I don't see how either of these two meanings can be relevant to Eq. 1.

According to Oxford Dictionary of English, the word "modulus" is the diminutive of the Latin modus, meaning measure (modus, in turn, comes from Proto-Indo-European mod-os, Nocentini & Parenti 2010.

Hence, the distance modulus is a "measure of distance", just like the modulus of a vector is a way of measuring its size, and Young's modulus is a way of measuring the stiffness of a material.


Frequency of gravitational wave detection You may have heard in the news that the LIGO experiment recently detected a gravitational wave.

Though I'm not an astronomer, the paper is a good read and mostly accessible. The detection of the gravitational wave is one thing, but the black hole merger is quite new to me. From the data I could collect in the paper and this site, the source is estimated at 1.3 billion years, and the chirp lasted only a few milliseconds.

My question: what is the frequency of events of this size order?
Is there any estimate of the density of such events in the universe?

Harry (2009) cited several different sources stated that, as far we know, the rate of detectable events will be

40 neutron star mergers per year
30 10 M$_\odot$ black hole mergers per year
10 neutron star/black hole mergers per year

This is within a radius of about 200 Mpc. This cannot, however, be used to extrapolate the total rate of such events, because of detection bias - the more massive the objects, the more easy they are to detect. The same thing happens with exoplanets, but for different reasons (e.g. planets that are more massive or closer to their stars are easier to detect via transit or by radial velocity methods).


Can we visually perceive exoplanet transits with amateur telescopes? Is it possible for an amateur astronomer to visually perceive an exoplanet transit by the change of brightness of the star via observations spaced over time, or is the variation of the brightness too insignificant to be seen?
I have a 70mm scope, but no astrophotography equipment or software to compare data. My question is about really perceiving the brightness variation with our own eyes over time as we see it on variable stars.

Amateur equipment is good enough. But you cannot detect it with a naked eye.
The change in flux for a passing exoplanet in transit is roughly 1%...2% at most for the larger exoplanets - and it is a gradual change. That's a change you do not notice with the naked eye, but it needs photographic equipment to create a sequence of images which allow analysis of brightness of all seen stars in order to detect the variation in the one interesting one.
With an appropriate camera and stable mount your 70mm telescope will suffice to detect these light changes in a sequence of moderately long-exposure images over the course of a few hours (exposure time a few minutes each).
Many of the professional ground-based exoplanets surveys don't use or didn't use much larger telescopes (e.g. see the HATnet programme). The difference often is more in the grade of automation, the sensitivity of the photographic equipment and automation of data processing pipeline.


Why not us interferometry to take a picture of Pluto? Interferometry is among the best ways (if not, the best way!) to have an image of a very distant object.

Recently a picture of the black hole at the center of M87 was released. It is the result of several data collected by the Event Horizon Telescope, a series of arrays all across our world, working as one like a giant Earth sized telescope. The picture is not really cutting edge and high definition however it is still very surprising and in a way detailed enough, considering the fact that M87 is 53.49 million light years away... and this is where I arrive to Pluto which is just around 6 to 7 billion kilometers from us.

If we used an interferometer, perhaps the same size as the EHT (or just a smaller one, the size of an entire continent), and point all the arrays at Pluto, then we should have a picture with a resolution that is at least higher as the Hubble Space Telescope, but likely not as high as the pictures from the New Horizons spacecraft which directly made a fly by of Pluto... right?

If so then why don't we use interferometry to take pictures of Pluto from Earth?

Radio interferometry can combine observations over very large baselines.  But optical interferometry cannot.  According to a list of interferometry instruments on wikipedia, the largest baseline for optical measurements is less than a kilometer.  We can't take optical measurements with continent-sized instruments.

Then if you drop down to radio where the instruments do have that capability, I think you'll find Pluto is quite dim (it's not a radio source, and there's no strong radio emissions that it can reflect to us). There's no radio signal from Pluto that can be imaged.

From a page on optical interferometry:

  Interferometers are seen by most astronomers as very specialized
  instruments, as they are capable of a very limited range of
  observations. It is often said that an interferometer achieves the
  effect of a telescope the size of the distance between the apertures;
  this is only true in the limited sense of angular resolution. The
  combined effects of limited aperture area and atmospheric turbulence
  generally limit interferometers to observations of comparatively
  bright stars and active galactic nuclei.


Interstellar Dust properties While practicing for an upcoming examination, I ran into this problem:

Which of the following best describes dust grains in the interstellar medium:

They are a few hundred nanometers in size (size of optical light).
They are much larger than regular dust.
They are primarily composed of helium and hydrogen.
They emit large amounts of blue light.

I did some research and discovered that they scatter blue light, their size ranges form $1 \ldots 100 {\rm \mu m}$ while regular dust is about $1 \ldots 100 \mu{\rm m}$. Also, interstellar medium is composed of helium and hydrogen but that doesn't mean the same about dust. My guess: It is the first answer.
Is this correct? Any assistance much appreciated.

Yes, #1 is the correct answer.

✅ For small particles such as dust grains and molecules, light generally interacts most easily with particles roughly the same size as its wavelength. This is seen as the maximum in the figure below. That is, the particle’s cross section doesn’t depend on its exact shape, but rather on the ratio between the size and the wavelength of the incident light: As the particles become larger, you enter the regime of “geometric” scattering, where the cross section approaches the geometric cross section (e.g. $\pi r^2$ for a sphere etc.), whereas as the particles become smaller, you enter the “Rayleigh regime”, where the cross section drops proportionally to the wavelength to the power of $-4$.
Since observationally we know that interstellar dust affects optical
and, in particular, UV light significantly, we know that the dust
grains must be roughly the size of that light’s wavelength. More carefull modeling shows that a typical cosmic dust size distribution can be described by a power law with the probability of size $r$ decreasing roughly as $P(r) \propto r^{-3.5}$ (Mathis et al. 1977). In other words, for every dust grain of size $r$, there are 3,000 times as many dust grains of size $r/10$.

 "Normal" dust particles are in the range 1–100 $\mu\mathrm{m}$ (thanks for the sources @uhoh & @PeterErwin, unless they mean dust bunnies which are more like mm- to cm-sized. At any rate, cosmic dust is not much larger than normal dust, so #2 isn’t correct.

 #3 also isn’t correct. While it’s true that 98% of the interstellar gas is hydrogen and helium, dust is composed of the last 2%. That is, iron, carbon, silicon, oxygen, etc.

 Finally, while it’s true that blue light is absorbed or scattered more efficiently by dust than red light, dust doesn’t emit blue light. If it’s absorbed, the energy goes into heating the dust and is then later re-emitted as infrared light. So #4 also isn’t correct.

In the figure below you can see how the cross section of a(n idealized) particle depends on the ratio between its size $r$ and the wavelength $\lambda$ of the incident light.

Credit: Catslash, Public domain, via Wikimedia Commons with own annotations.


Computing orbit positions of Jovian Satellites / Moons using JPL data I've been trying to build a model of the solar system in a game.  Thus far I've succeeded in placing each of the planets in position using Keplerian elements and formulae from https://ssd.jpl.nasa.gov/?planet_pos  An example for Jupiter is:

               a              e               I                L            long.peri.      long.node.
           AU, AU/Cy     rad, rad/Cy     deg, deg/Cy      deg, deg/Cy      deg, deg/Cy     deg, deg/Cy
-----------------------------------------------------------------------------------------------------------
Jupiter   5.20288700      0.04838624      1.30439695       34.39644051     14.72847983    100.47390909
         -0.00011607     -0.00013253     -0.00183714     3034.74612775      0.21252668      0.20469106

At this point, I want to add several of the Jovian moons, however the same site doesn't seem to provide the same type of data for these moons, perhaps because the moons behave differently.  Unfortunately, my math is not great; I can usually implement something that's written down, but I don't have the wherewithal to work it out myself.

For the Jovian moons, the data is listed at: https://ssd.jpl.nasa.gov/?sat_elem#jupiter  Here is an example for Ganymede:

 Sat.           a       e          w           M        i            node      n             P       Pw      Pnode    R.A.        Dec.     Tilt
 (km)        (deg)     (deg)      (deg)       (deg)    (deg/day)    (days)    (yr)          (yr)    (deg)    (deg)    (deg)
 Ganymede  1070400.    0.0013    192.417    317.540    0.177        63.552    50.3176072    7.155   63.549  132.654   268.168    64.543    0.068

Here they state that the data is "Mean orbital elements referred to the local Laplace planes", as opposed to "Keplerian Elements for Approximate Positions of the Major Planets".

Whilst the first 6 parameters appear to be the same (a, e, w, M, i, node), the Keplerian elements also have "rates", or time derivatives (according to this post: https://space.stackexchange.com/questions/8911/determining-orbital-position-at-a-future-point-in-time).  The forumlae kindly provided by NASA for the Keplerian elements relies on these rates.

If I could understand how to determine the "rates" for the Jovian moons, then I might be able to use the same formula (is this the right thing to do?)  It looks to me like those rates are something I should be able to compute from the other data provided for the moons, however I haven't been able to determine how.

My suspicion is that I either need a different formula for the computation using the Laplace Planes, or I need to compute those 'rate' values and use the same formula I already have.

I'm trying to generate cartesian (x, y, z) coordinates for each Moon where (0,0,0) is the centre of Jupiter itself, for a given date/time (typically "now").

I'm beginning to wonder if I'm barking up the wrong tree on this.  Can anyone shed any light

An excellent book by Jean Meeus, "Astronomical Algorithms", provides the calculations for planetary positions and Jupiter's Galilean Moons (Io, Europa, Ganymede, and Callisto). Meeus does not really explain the math to a great degree, so it is relatively simple to follow it as a formulaic approach. He uses VSOP for planetary positions and Lieske's E5 theory for the moons (you can get an explanation of both on Wikipedia). As a bonus, Meeus shows how to make calculations for both parameters with a lesser degree of accuracy which might improve your calculation speed. Though this book can be a bit dense, you might want to read up on the difference between "ephemeris time" and "terrestrial time", as this will affect your accuracy as well.


Is there a term to describe all stars that are not neutron stars or white dwarfs? I am looking for a official term that describes all stars that are not neutron stars or white dwarfs.  "Still burning stuff stars" or "ordinary stars" are all I have been able to come up with.  Is there an official term for stars that are still evolving and haven't "died" yet?

Every so often someone will note that a "star" is a ball of plasma that is supported by nuclear reactions at its core, and so technically neutron stars and white dwarfs are stellar remnants, and not stars.

The technical term is therefore "star". And if you think that could be ambiguous then just spell it out: "Stars with supported by fusion reactions"/"Stars, excluding stellar remnants"


C-Type Asteroids and Carbonaceous Chondrite Meteorites I was just wondering, what is the relationship between C-type asteroids and carbonaceous chondrite (CC) meteorites? Do CC meteorites come from C-type asteroids?

The difference between asteroids, comets and meteors is discussed in What is the difference between asteroids, comets and meteors?

Some groupings of asteroids have been correlated with meteorite types:

  C-type – Carbonaceous chondrite meteorites
  S-type – Stony meteorites
  M-type – Iron meteorites
  V-type – HED meteorites

The S- and C-type asteroid classifications originate from the taxonomy proposed by David Tholen.

Meteors are generally considered to be asteroids which enter Earth's atmosphere.
Meteorites are meteor remnants which reach the ground. So essentially, carbonaceous chondrite meteor(ite)s start out as C-type asteroids; they're made of the same stuff.


How can I convolve a template spectrum with a photometric filter response spectrum? Suppose I have a template stellar population spectrum (say, from Bruzual & Charlot 2003) which runs from like 1000 Angstroms to 160,000 Angstroms and which has x-axis wavelength units of Angstroms and y-axis flux density units of $L_{\odot}$ per Angstrom. Now, let's say I want to get the integrated luminosity within a bandpass that runs from $\lambda_1$ to $\lambda_2$ Angstroms, and I have another "spectrum" which gives me the filter response function for this photometric bandpass.

I know that I'm supposed to "convolve the stellar template spectrum with the bandpass response function" in order to get the integrated stellar luminosity within my bandpass, but how do I do this exactly (preferably in python)? The stellar template and filter response spectra don't have the same number of data points and are not defined on the same wavelength grid (the stellar template spectrum obviously extends much further than the filter response function).

I feel like I could interpolate the filter response spectrum onto the wavelength array of the stellar template spectrum, and assign a value of 0 to the interpolated response spectrum for wavelengths outside $\lambda_1$ to $\lambda_2$. Then I would be able to do a simple element-wise multiplication, but this does not seem like the correct approach using traditional "convolution."

Your feeling is right: You shouldn't convolve the spectrum and the filter, you should only multiply so that flux outside the bandpass is suppressed. Subsequently you integrate the resulting function over wavelength, so that flux density (in energy/time/area/wavelength) becomes flux (in energy/time/area).

Simply setting the flux to 0 outside $\lambda_1$ and $\lambda_2$ (or, equivalently, just integrating from $\lambda_1$ to $\lambda_2$) corresponds to a "top-hat filter". Most realistic filters are more smooth.

So, something like (untested):

import numpy as np
from scipy.integrate import simps

lamS,spec = np.loadtxt('spectrum.dat',unpack=True) #Two columns with wavelength and flux density
lamF,filt = np.loadtxt('filter.dat'  ,unpack=True) #Two columns with wavelength and response in the range [0,1]
filt_int  = np.interp(lamS,lamF,filt)              #Interpolate to common wavelength axis
filtSpec  = filt_int * spec                        #Calculate throughput
flux      = simps(filtSpec,lamS)                   #Integrate over wavelength
print 'Total flux is {0:8.3e}'.format(flux)

Magnitude

I'm wondering, however, if "the total luminosity within the bandpass" is really what you're interested in. This quantity doesn't really bear any physical significance for the source, as it depends on the particular filter. Usually, one would normalize to the filter, thus getting the magnitude, which is independent of the filter shape.

In the AB magnitude system, the magnitude $m_\mathrm{AB}$ of a source is (Oke & Gunn 1983; note that they make a sign error in their own definition):
$$
m_\mathrm{AB} = -2.5 \log f_\nu -48.6,
$$
where the average $f_\nu$ in $\mathrm{erg} \,\mathrm{s}^{-1} \,\mathrm{cm}^{-2} \,\mathrm{Hz}^{-1}$ is given by
$$
f_\nu = \frac{1}{c}\frac{\int ST\lambda\,d\lambda}{\int T / \lambda\,d\lambda},
$$
where $S$ is the spectrum with units $\mathrm{erg} \,\mathrm{s}^{-1} \,\mathrm{cm}^{-2}$ Å$^{-1}$, $T$ is the filter curve (un-normalized as above), $\lambda$ is the wavelength in Ångström, and $c$ is the speed of light in Å $\mathrm{s}^{-1}$. That is, a flat spectrum (in $\nu$) of height $f_\nu$ would have the same integrated flux over $T$ as $S$ would.

So:

import numpy as np
from scipy.integrate import simps
c_AAs     = 2.99792458e18                          # Speed of light in Angstrom/s
lamS,spec = np.loadtxt('spectrum.dat',unpack=True) #Two columns with wavelength (in Angstrom) and flux density (in erg/s/cm2/AA)
lamF,filt = np.loadtxt('filter.dat'  ,unpack=True) #Two columns with wavelength and response in the range [0,1]
filt_int  = np.interp(lamS,lamF,filt)              #Interpolate to common wavelength axis
I1        = simps(S*T*lam,lam)                     #Denominator
I2        = simps(  T/lam,lam)                     #Numerator
fnu       = I1/I2 / c_AAs                          #Average flux density
mAB       = -2.5*np.log10(fnu) - 48.6              #AB magnitude


Laser Collimator For Dobsonian vs Newtonian Will a laser collimator like the Gosky 1.25 Metal Laser Collimator, https://www.amazon.com/gp/product/B01M4IVUYG/ref=crt_ewc_title_dp_1?ie=UTF8&psc=1&smid=AHQ6VCR020F8X, work for a dobsonian even though it says it's for a newtonian?

A Dob is a Newtonian, so yes it will work.  Strictly speaking the Dob bit is the mount, but it's nearly always a newtonian that's mounted.


What is this bright "glow" in the center of galaxies? It was always my belief that at the center of many galaxies, there are supermassive black holes. If this is the case, then we should not see a "light" coming out from the center since light get's sucked in black holes. Doing a quick search of galaxies on google image, I came across this:

These are famous galaxies, and real, and so I was wondering what this very white-yellow glow is in the center. If it is a collection of stars, then why are there so many stars in the center of a galaxy? If there is a black hole there, then why is there light?

If this is the case, then we should not see a "light" coming out from the center since light get's sucked in black holes.

You are overestimating the size and the capabilities of a supermassive black hole. Contrary to pop sci portrayals of black holes, black holes are not giant vacuum cleaners in space that suck up anything and everything close by. While the supermassive black hole at the center of the Milky Way is indeed very massive (about four million times the mass of our Sun), it isn't very large physically. It's less than a couple dozen solar diameters across. It also isn't that hungry, gobbling up perhaps the equivalent of four or so Earth masses over the course of a year.

On the other hand, the central bulge of a spiral galaxy contains several million stars in a fairly small volume. That central bulge is what you are seeing in those images. The supermassive black holes near the centers of those bulges gobbles only a tiny, tiny fraction of the light emitted by those millions of stars.


Questions related to the reflection of the moon What would happen if we painted the whole moon with black, will it still reflect light, or it will become invisible?
What would happen if we painted the whole moon with silver paint or make its surface shiny?

The moon is already a pretty dark grey. It only reflects about 12% of the light that hits it. It is about the colour of (worn) tarmac on the road. You can see how grey it is in NASA's gif from a million miles away.

"Black" paint reflects less light. Common household black paint reflects only about 5% of light. So the moon would be rather less than half as bright as it is now. But you could still see it.  There are some specialised black substances that reflect as little as 0.035%. A moon coated in these would actually be hard to see.

White paint and silver paint reflect about 80% of light, so if the moon were white it would be about 7 times brighter, but much less bright than the sun which is about 400000 times brighter than the full moon.


Does time dilate during night time when earth orbits around the sun? Number of the solar middays (noon) = Number of the solar midnights (ALWAYS). The diagram (Top view but not to the scale) in the following link depicts the path traced by the earth in its orbit (either circular or elliptical) around the sun.

https://i.postimg.cc/C5K2wX6V/IMG-3066.jpg
Both Midnight and Midday happen simultaneously and instantly. Any point on the outer circle represents solar midnight while on the inner circle solar noon. It's quite clear that the length (circumference) of the outer circle is greater than the length (circumference) of the inner circle. This means after the completion of one year, the occurrences of the Midnights are more the occurrences of Middays.
Earth not only rotates around its own axis but also revolves around the sun in its orbit however both axial and orbital motion of earth independent of each other. There is a possibility that the earth may glide w/o rotation for a short while and time dilate but such things never noticed either.

There's no gliding and no significant time dilation.
Your speed relative to the sun is not the same at midnight as it is at midday, since you must combine the rotation of the Earth with its orbital motion around the sun.
But for any point on Earth there will be one midnight and one midday each day. There aren't "more midnights".


Could black holes be creators of dark matter? At the tremendous forces of gravity inside a black hole, can "matter" be transformed into a more  degenerate form; "dark matter" (if that truly what dark matter is)?

I know the inside of a black hole isn't fully understood, but can this be an acceptable possibility? Why?

Ultra-high energetic cosmic rays are thought to be caused by black holes as one option. Those energies should be sufficient for the formation of (hypothetical) heavy supersymmetric particles, which should decay to stable (hypothetical) neutralinos, candidates for (hypothetical) WIMPS.

Assuming this theoretical framework, black holes (or similar dense objects) may cause the generation of dark matter. But the amount of dark matter formed that way won't explain the amount of dark matter needed for the Lambda-CDM model. Also dark matter forming this way should be hot (fast moving). Lambda-CDM needs cold dark matter (CDM).


Do all the stars visible to the naked eye belong to Milky Way? It is surprising that even late in 1920, people do not know there are galaxies besides the Milky Way.

Does this mean that all the stars visible to the naked eye are near to us?

All stars visible individually to the naked eye belong to the Milky Way. A handful of galaxies are visible to the naked eye, but here you only see the combined light of millions to billions of stars.

The only exception is when a massive star ends its life, exploding as a supernova. This can be visible to the naked eye for a brief perdiod of time. Most notably, the gamma-ray burst GRB 080319B was visible to the naked eye for roughly 30 seconds in 2008, despite its distance from us being $\sim10^5$ times the size of the Milky Way.


Old Telescopes versus New I am having trouble finding the quality of old telescopes. My question is how does a decent modern telescope that cost about a couple hundred USD compare to the top-notch telescopes in the nineteenth century. Can one make the same observations now with such a modern telescope as one could do in the nineteenth century with a top-notch one?

Modern telescopes, even those that can be purchased from stores for, say, under $10000 (which is already a lot), were produced with computer-controlled equipment to polish the mirror and make the lenses and mirrors, and the lenses usually have nano-coating to reduce reflections and increase light transmission.

Overall you have a century of progress & experience building them since then. Mounts are better as well, mechanically and electronically (you have auto-guiding whereas it was manual a century ago, with an astronomer watching a star through a reticle). In other words, modern telescopes, even amateur ones, are vastly improved.

You can compare photos (from various books such as https://www.amazon.com/Day-We-Found-Universe/dp/0307276600) to modern astrophotography done by amateurs (eg https://www.astrobin.com/explore/top-picks/). The modern ones (also benefiting from sensors more sensitive than film) are much better.


Is it possible to see a spacecraft orbiting the moon with a pair of binoculars? On this clear early morning over London, I was studying the moon with a pair of binoculars and noticed a tiny dot of light that appeared to trace a slow orbit around the moon. I reckon it moved roughly 30 degrees in as many minutes. I'm wondering whether I was seeing a star or if it's remotely possible that I was seeing sunlight reflecting off an orbiting craft? Although the sky was clear it certainly wasn't starry; it was about 100 minutes before sunrise.

No. There are currently 5 spacecraft orbit the Moon (Or the E-M L2 point). These are LRO, Artemis P1 and P2, Chang'e 5 T1, and the Queqiao relay satellite. Of these, the largest is probably LRO. The solar panels, which are the largest part of most spacecraft, "he three panels together are 168 inches wide and extend out from the spacecraft 126 inches". Converted to meters, that is 4.25 x 3.2m. Just to give an idea of how visible that would be, Elon's Tesla Roadster, according to JPL Horizons, when it was at the distance of the Moon, had an apparent brightness of about 15.27. Using this chart, an aperture of about 50 cm is required. LRO should be a bit brighter, but not enough to make a huge difference. And none of this takes in to account the glare of the Moon itself. Whatever you saw, it probably wasn't a satellite.

I suspect it was a star of some kind. If you can provide the date and time we might be able to figure out a candidate star.


Why oddly shaped constellation areas? Can somebody tell me why the constellations I've marked here have areas that resemble pieces of a puzzle, with very varying width and height and shapes, instead of being logically/evenly laid out on the sky?

Virgo is much bigger than Cancer, for example.

The image is a screenshot from Stellarium.

The constellations are artifacts of visual pattern recognition, many dating back to antiquity.
As the stars are distributed unevenly, naturally some constellations are larger than others.
Ptolemy catalogued 48 of them around 150 CE.
The other 40 were added between 1596 and 1763, either to cover the far southern sky or to fill gaps between the older figures.
Ian Ridpath gives a comprehensive history in Star Tales.

The modern constellation boundaries were standardized in the 1920s by an IAU committee led by Eugène Delporte.
Their goals were to assign each square degree of sky unambiguously to one constellation, to be consistent with scientific literature already published, and to preserve the traditional figures where possible.
Equal size wasn't on the agenda.
The boundaries align with the equatorial grid as of 1875:

The zodiac of 12 equal parts was invented by the Babylonians around 400 BCE and later refined by Ptolemy.
Each sign corresponds to a box 30 degrees wide and about 20 degrees high.
It's important to distinguish between a constellation (a visual group of bright stars) and a sign (a uniform box along the ecliptic).
For example, Virgo the sign spans ecliptic longitudes 150$^\circ$ to 180$^\circ$, whereas Virgo the constellation is somewhat wider and farther east.
In Ptolemy's time the signs and constellations were not far apart:

Ptolemy defined the tropical zodiac in reference to the equinoxes and solstices; for example, the boundary between the signs of Gemini and Cancer lies at the June solstice, wherever that may be among the stars.
Consequently, precession makes the signs drift westward about 1.4$^\circ$ per century relative to the constellations.
In the present day, the zodiac is roughly one sign ahead of its constellations:


What was the orbit of the meteor that impacted the Moon on 2013 September 11? On 2013 September 11th, a meteor impacted the Moon leaving a 34-meter crater. So far the only info I've found was from youtube (https://www.youtube.com/watch?v=zCFDkj2JtyA) and this site: http://lroc.sese.asu.edu/posts/810

I'm surprised this doesn't have its own wikipedia article (none that I could find, anyway), and google wasn't helpful either. Is the astronomical world more or less oblivious to this event?

Anyway, the data I'm most interested in is the meteor's orbital trajectory. And also, why would such a small meteor leave a "fireball" for 8 seconds according to the youtube video.

Out of curiosity I looked into this a bit.    North/South on the moon is pretty irrelevant for the objects orbit, but it struck at East 339 degrees (or West 21 degrees) near the Lubiniezky craters in the northwest part of Mare Nubium - which can be seen here - towards the center but a little bit in the south eastern part of the map.

and you can see the phase of the moon here. It changes slowly enough that a 24 hour range doesn't make a huge difference.

http://lunaf.com/lunar-calendar/2013/09/11/

but apart from the crater looking roughly straight on, it's probably impossible to say the orbit it came from because the angle of approach is too big an unknown.   It hit from kind of both the earth side and sun side direction, so it could have been a temporarily captured asteroid around the earth or a small comet or asteroid coming back from around the sun.  Either could have made that impact I would think.


What is a stellar component? I have a question where I have to calculate the orbital inclination of a planet, so that it isn't a exoplanet but a stellar component. The mass of the star it is orbiting is 1.1 $M_☉$. Does the exoplanet has a special mass if it is a stellar component (for example 0.5 $M_☉$) or is there another relation?

The question was made out of a few sub-questions. It is about 51 Peg b. In the first question, I had to calculate the half long ax (=0.05284 AU), the eccentricity (=0.013) and the $mass*sin(i)$ (=0.500 $M_J$)
In the second question, I had to calculate the inclination at which the mass of the planet is 4 $M_J$ (=7.18°) and the inclination at which it wasn't a planet, but a stellar component.

The term "stellar component" refers to a part ("component") of a system that is in the form of one or more stars.

For example, the Alpha Centauri system has three stellar components: Rigil Kentaurus, Toliman and Proxima Centauri. The planet Proxima Centauri b is also a component of the Alpha Centauri system but it is not a stellar component because it isn't a star.

Another example: the stellar component of a galaxy refers to the fraction of the galaxy that is in the form of stars, as opposed to dust, gas, dark matter, etc.

In the case of your question, it is asking what the inclination would be if the companion were a star (i.e. a stellar component of a binary star system).


How to connect my phone to a motorized telescope via a cable? Some motorized telescopes (e.g. Celestron NexStar Series) have no WiFi by default. Can I connect my phone to them to be used as a hand control via a cable? The telescopes have AUX port and phones USB C ports. Even when they have hand control or even WiFi, connecting the phone by cable can be better. I've heard that some telescopes have issues with WiFi connection and that the hand control does not provide the same level of convenience as an Android app (e.g. SkyPortal).

Group dedicated to all versions of the Celestron "C14" 14-inch Schmidt-Cassegrain Telescope.  This includes the original C14 and the newer "EdgeHD" series. All other Celestron SCT telescopes are welcomed to participate.
https://groups.io/g/C14EdgeHD


Has radio astronomy ever been done on objects that appear very close to the Moon? Is this avoided? This answer to Which kinds of astronomical observations most need to avoid the Moon being up? mentions

  For completeness - radio, mid-infrared and mm-wave observations are unaffected (unless the Moon is in the way!)

The Moon is of course opaque to all wavelengths used in astronomy, from very low frequency radio waves to gamma rays. So if the moon eclipses a target radio signals will be blocked.

But are there any more subtle or non-obvious effects? For example if a target is very close to the Moon but not covered by it, would that affect an observation in any way?

Question: Is this routinely avoided out of an abundance of caution? On the other hand, has a lunar occultation of a radio source ever be leveraged in some way for a specific measurement?

"Bonus points" for any anecdotes of strange results or confounding measurements in Radio Astronomy that turned out to be the result of not taking the position of the Moon in to consideration.

Yes, and lunar occultations have proved useful in several cases.

Hazard et al. 1963 used a lunar occultation to produce a high-resolution brightness profile of the now well-studied radio quasar 3C 273. Scheuer 1965 goes into a little bit of detail on general computations.

A slightly different tack was taken by Vedantham et al. 2015. They were attempting to create interferometric maps of the cosmic 21 cm line signal using LOFAR. Unfortunately, they needed a calibration source to be able to measure this "global signal". The Moon, occulting portions of the field of view, provided a calibration source as they observed that patch of sky.


What is the highest angle the moon ever makes above the horizon at the North Pole? The question is in the title. I'd like to find out the maximum angle the moon ever makes above the horizon at the North Pole. By "the horizon at the North Pole," I mean the tangent plane to the earth at that point.

Note that I'm not asking for the highest angle that the moon ever appears to make above the horizon. Appearances are distorted by refraction, etc., although I'm not sure by how many degrees. Though I'd be interested in the exact distortion: is there a data repository where I can look up the highest observed angle?

I would make the following prediction. The moon's orbital plane is inclined to the earth's equatorial plane at about $23.5+5\approx 28.5$ degrees. Let $R_e$ denote the radius of the earth and $R$ denote the radius of the moon's orbit around the moon. A little geometry for the highest point suggests

$$\theta_{\text{max}}=\tan^{-1}\left(\frac{R\sin(28.5^{\circ})-R_e}{R\cos(28.5^{\circ})}\right)$$

But because $R>>R_e$ (about 385,000 compared to 6300 km), $\theta_{\text{max}}$ should be just a little less than $28.5^{\circ}$.

Is this right?

${R_e}/{R}$ is about 0.017 radian or 1°.
Using your trigonometry and these values:

Lunar orbit inclination = 5.15° (varies between 5.0° and 5.3°)
Lunar perigee = 362600 km, apogee = 405400 km
Earth polar radius = 6357 km
Ecliptic obliquity = 23.44°

I get 27.7° at perigee and 27.8° at apogee.
At that altitude, atmospheric refraction is only about 0.03°.
The highest apparent altitude I found with JPL HORIZONS was 27.91° on 1987-09-15 around 17:15 UT.

Due to nodal precession,
the Moon doesn't reach ±28.6° geocentric declination every year, only near a major lunar standstill every 18 or 19 years, e.g. in 2006 or 2025.
Near a minor lunar standstill, e.g. in 2015, the Moon's declination is limited to ±18.3°.


How long do thunderstorms on Venus last? A while ago, I learned about field of exometeorology (the study of other planets ' atmospheres) which fascinates me. In particular, I read about thunderstorms on Venus, e.g. about the results from ESA's Venus Express Mission in 2006, or according to Universetoday.com: Lightning Storms on Venus Similar to Those On Earth:

“We have analyzed 3.5 Earth-years of Venus lightning data using the low-altitude Venus Express data, which is about 10 minutes per day,” Russell said. “By comparing the electromagnetic waves produced at the two planets, we found stronger magnetic signals on Venus, but when converted to energy flux we found very similar lightening strength,” said Russell. Also it seems that lightning is more prevalent on the dayside than at night, and happens more often at low Venusian latitudes where the solar input to the atmosphere is strongest.”

This is mainly a statement about the intensity and time distribution of the lightning on Venus. I am now wondering whether there is more known about the mesocale convective systems on Venus. My search gave me the impression that there is not much literature on that topic (yet), or am I mistaken?
The main question I am after: How long do storm cells on Venus last? Is any mean or median known? Can thunderstorm systems live for multiple (Venus) days?
References
Although I did not find any answer to my question, here some references I found:

C. C. Covey & G. Schubert: Mesoscale convection in the clouds of Venus, Nature 290, pages 17–20 (1981)
M. Nakamura, T. Imamura, et. al. Overview of Venus orbiter, Akatsuki, Earth, Planets and Space volume 63, pages 443–457 (2011)
List of all missions to Venus according to Wikipedia
NASA's In-Depth Page about Venus mentions the clouds and their lightnings briefly.
Akatsuki (あかつき, 暁, "Dawn") as Venus-orbiter, is currently (as of 2020) the only existing artifical Venus satellite, I suppose.

Independent observation led to the conclusion that lightning is prevalent on Venus. Electromagnetic signals detected by the 4 Venera landers explained the main cause of lightning on Venus. Venera 9 spectrometer detected a lightning stroke. The Pioneer plasma wave instrument detected waves which explained lightning might be at low altitude. VLF observation suggested that lightning is a intra-cloud  phenomenon which is more common in the afternoon and evening time and the occurrence rate is more as compared to Earth. Venera 11, 12, 13 14 were equipped to detected magnetic signature(acoustic data were also obtained from V11 and V12).
Lightning on Venus occur from strong convective cumulus cloud and occasionally from dust storms and volcanic clouds. The clouds may vary in altitude from few km to 30 km . Lightning on Venus occur due to two distinct forms: cloud-to-ground discharge and intra-cloud discharge. Such discharge occur when electric field exceeds from 106 V/m in wet air to 3 $\times$ 106 V/m in dry air. Cloud to ground discharge carry current from 200-300 A which increase every 50-100m, travelling at a velocity of 1.5 $\times$ 106 m/s which followed with a return stroke with velocity 0.6 $\times$ 108 m/s carrying 10-20 kA which decays in about 20-50 us. There are 3 or more stroke in one flash separated by 40 ms and lasts 0.2 s.
Discharges within clouds are 2 to 4 times more frequent than that of cloud-to-ground discharge and carry currents upto 4kA but they generate less energy(4.5 $\times$ 108 J in cloud-to-ground discharge). Number of discharges on Venus is more than that of Earth. Over 7000 discharges occurs on Venus per day  You can check the details and observation in ref.1 and ref.3

Fig. 1 Venusian atmosphere where lightning is most likely to occur

Fig.2 Magnetic field due to intra-cloud discharge (altitude 50 km)
References

Venus lightning by Russell, C. T., Space Science Reviews, vol. 55, Jan-Feb 1991, p. 317-356. (link and the author's paper here)
Russell, C.T. et.al. (2011). Venus lightning: Comparison with terrestrial lightning. Planetary and Space Science - PLANET SPACE SCI. 59. 965-973. 10.1016/j.pss.2010.02.010.
Lorenz, R.D. Lightning detection on Venus: a critical review. Prog Earth Planet Sci 5, 34 (2018). https://doi.org/10.1186/s40645-018-0181-x


Contradict the expansion of the universe? Lets say that we observe in a distant galaxy cluster, which in all the galaxies move far away from us.
and here we see in the same cluster a galaxy that move away from us in a much lower velocity than the other galaxies.
is this contradict the expansion of the universe?

No, it doesn't contradict the paradigm of the universal inflation. It would merely mean that the example galactic cluster that you mention has its proper (own) motion at a smaller velocity relative to the proper motion of the Milky Way than the speed of the universal inflation at the distance between them. I.e. it would move towards us, if the Universe wasn't expanding. But since it does, galactic cluster's proper motion only managed to decrease relative velocity at which it moves away from us.

To put it in analogy, say you're on a speeding train that moves at 120 km/h relative to the station it just passed by. You then lean out of its window and throw an apple towards the station that the train is now moving away from, but you can only manage to throw the apple at a velocity of 40 km/h. That apple, since it came from the train moving at 120 km/h and shared its momentum, is now travelling away from the station's platform at 80 km/h, even though you threw it towards the station, because you only managed to reduce its velocity relative to the station for 40 km/h. So it would still move away from it, but for 40 km/h slower than the train does.


What is the difference between asteroids, comets and meteors? Some celestial objects seem to be referred to as asteroids, some as comets, and some as meteors or meteorites.

What is the distinction between all of these different objects? Are any of them the same?

The objects you are refering to are actually two different objects: asteroids and comets. Meteor and meteorite are other names for an asteroid, at a given time of its interaction with our planet. We'll get to that.
So first, what is the difference between an asteroid and a comet?
A comet is a small solar system body that display a "coma" (an atmosphere of a sort) and sometimes a tail passing close to the Sun. They are mostly made of ice and dust, as well as some small rocky particles. We distinguish two kind of comets, with short or long orbital period. The short orbital period ones originated from the Kuiper Belt, a region composed of small bodies beyond the orbit of Neptune. The long orbital period ones originated from the Oort cloud, a scattered disk of icy planetesimals and small bodies laying around our solar system.
An asteroid is a small body, composed mostly of rocks and metals. In our solar system, they can be originated from the asteroid belt, laying between Mars and Jupiter, or from the orbit of Jupiter (the Jupiter Trojans), or from actually almost everywhere in the Solar System.
A small asteroid ('meteoroid') that enters the Earth's atmosphere becomes a meteor, what we also call a "shooting star".
Eventually, a meteor that was massive enough not to be completely distroyed entering the Earth's atmosphere and hitting the ground is a meteorite.


Can the order of planets change in a young planetary system? The Nice model and the Grand Tack model both show that the distance between a planet and its host star can change greatly over time. Is it possible in either of these models for the order of planets to change as well? Or is the order of planets around a star the same as when the planets formed?

Yes, the order of planets in a system can change over time. Indeed, in some variations of the Nice model, Uranus formed further out than Neptune and then switched places when Jupiter and Saturn underwent their resonant interactions. According to Tsiganis et al. 2005, one of the original three papers on the model, Uranus and Neptune switched positions in ~50% of all simulations (presumably meaning that in some where Uranus formed past Neptune, ended up further in, and in some where Neptune formed beyond Uranus, it traveled further inwards). Therefore, a number of models reproducing the parameters of the giant planets in the present-day Solar System involve Neptune forming closer to the Sun than Uranus. This seems to fit better with models of the Sun's protoplanetary disk.


How strong is the gravitational stretch we experience from the edge of the universe? How much gravitational effect do we experience (e.g. maybe -.00001 G or smaller) from the edge of the visible universe?  By edge of the visible universe, I am talking about the region of the cosmic microwave background and beyond.  In the past, the universe was very hot and dense, and we are seeing the light from this time period as the Cosmic Microwave Background.  It took slightly less than the age of the universe for that light to reach us, and it has become extremely redshifted with the expansion of space.  Since gravity apparently travels at the speed of light, we should be experiencing a gravitational effect from this shell around us as well.  As I understand, the universe was extremely dense back then, and it is around us in every direction, which would intensify the gravity greatly.  However, it is extremely distant, which would weaken the gravity greatly.  I don't know if redshift would affect gravity as it has the light from the cosmic microwave background, but if so, this would weaken the effect greatly as well.  With all this in mind, I am curious, what would be the effect of gravity from the edge of the universe on us?

Note: I would expect this to be a very small outward effect in all directions, something like -.00000001 g or something like that, but not 0.

In the past, the universe was very hot and dense, and we are seeing the light from this time period as the Cosmic Microwave Background.

This statement is (partially) correct.
However....

  Since gravity apparently travels at the speed of light, we should be experiencing a gravitational effect from this shell around us as well. As I understand, the universe was extremely dense back then, and it is around us in every direction, which would intensify the gravity greatly.

This statement perhaps indicates that there are some misconceptions about the present state of the universe and also about what Cosmic Microwave Background Radiation really is (emphases added to identify the problem areas). So, in order to give a reasonable answer to the question asked, it would require addressing these misconceptions too.

The Big Bang model suggests that everything we see in our Observable Universe was once all concentrated in an infinitesimally small, hot and dense region of space. However, the Big Bang happened everywhere, which was followed by a super-luminal expansion of the space by a process called Inflation. The universe remained too hot still for any photons to roam around freely. It wasn't until the next 380,000 years that it continued to expand and eventually got cooled enough for the first hydrogen atoms to form. This is when the light first became visible. The remnant of this light is what we now now call Cosmic Microwave Background Radiation.

Again, it's important to note that the space was expanding in all possible directions continuously when the first light was emitted everywhere. In other words, the "dense" region from which the spacetime had begun, had now become comparatively much less dense overall. All through these years, this expansion hasn't stopped and every region where the light was first emitted, has now receded to great distances (46 billion light years in any direction). All the regions have cooled down further still to evolve into stars and planets, galaxies, clusters and any other celestial objects imaginable.

So, to answer the questions asked (for simplicity, we'll stick to CMB as the "edge")....

There is no space outside our Observable Universe which is surrounding us as a concrete shell. To be precise, it is merely a limit to how far back into the past and how deep we can peek into the space with our current technologies. Also, this has got nothing to do with the present state of the universe there. The universe is homogeneous and isotropic — meaning, you could go to any other place in the universe and it would still appear the same as it does from here — stars and galaxies everywhere.  Your vantage point would determine which galaxy appears as a distant blob to you and which region of space would merely appear as a CMB radiation.
The regions which we now see as CMB, 46 billion light years away, are indeed one of the farthest reaches of space from where anything, including gravity, would have had any effects on us. By now, everything that we see as CMB would have coalesced to form celestial objects. But, there's a catch.... Remember, if the Sun was to disappear from the Solar System now, it would still take approximately 8 minutes for the Earth to be set free from its orbit. Meaning, the state of the Sun 8 minutes ago determines what gravity we feel because of it now.
So, to compute the gravitational tug from CMB (46 billion light years away), we need to look at the state of the matter from 13.8 billion years ago minus 380,000 years (when the CMB was emitted). We can roughly assume that, back then, it was just Hydrogen atoms all over. The overall mass of all those atoms combined would have been equivalent to the mass of the Observable Universe, i.e. $10^{53}kg$.
But again, the CMB was emitted everywhere and the isotropic nature of CMB tells us that departures from uniformity are only up to seven parts in a million! Meaning, when the CMB was emitted 13.8 billion years minus 380,000 years ago, the mass of the Observable Universe was distributed almost perfectly uniformly in every direction possible. From the distance of 46 billion light years then, this translates into almost equal gravitational pull from everywhere that light (and gravity) has reached us since then. Hence, the gravitational effects of the matter that is (i.e. was) spread uniformly in a radius of 46 billion light years from us, would be equivalent to what one would feel if they were to somehow reach the center of the earth — i.e. experiencing a nearly perfect equilibrium, with a net force of zero.

PS: Should there be any discrepancies above, I'd wholeheartedly welcome the suggestions on correcting or improving the answer.


What are the prerequisites for a meteorite to reach the ground? I'm sure it's a function of atmospheric entry angle, meteorite mass (and possibly composition/density), and its velocity, but what function is it?

Above all, how does it depend on velocity? Will the slower meteors impact more often (less kinetic energy to dissipate) or the fast ones (less time spent in the atmosphere = less time to melt and evaporate all the way through)?

Your first resource here can be the impact effects calculator, It can show you when it is likely that a high velocity impact will occur. Somewhat surprisingly, faster objects are less likely to hit the ground at hyper-velocity, as they are more likely to break up and explode at altitude. They will likely form multiple meteorites, as the Chelyabinsk meteor did

If you are interested in meteorite formation, rather than craters, then most impactors will in some part survive the fall, provided they are rocky rather than icy. Objects smaller than 5cm are mostly cometary in origin. Larger than that and they are more likely to be rocky.