Is it possible to get a glimpse of the Big Bang through gravitation waves?

Gravitational waves from the big bang may be "heard" but not by LIGO. The waves emitted at or around the inflationary epoch of the big bang are expected to be at much lower frequencies (milli-Hz or lower) than those announced today by LIGO. There are various sources of noise that make LIGO insensitive to GWs at frequencies below about 10 Hz.

It will take space-based interferometers like the proposed LISA, with longer interferometer arms and well away from terrestrial sources of noise to stand a chance of detecting such GWs.

If they are detected - they might "sound" something like this (if upshifted into the audible range) - from the LIGO website. It sounds like white(ish) noise because of the broad continuum of frequencies expected.


Ice chunks would be melted off during closer approaches, potentially blasting non-ice bits off of the exocomet as it goes. This could affect the trajectory of the exocomet both by the inertial force of the separation and because the mass would be reduced.

Eventually it would run out of water, which means it would no longer be a comet--the streaming tail of gas wouldn't exist anymore, even on closer approaches to the sun. What would remain--assuming anything remained at all--would be a hard rocky core.

From there, the orbit would either stabilize or decay to a point that it crashed into the sun. Even a fairly stable orbit could eventually bring it close to other gravitational bodies in the solar system which could throw it into an unstable orbit, or alternatively, it could collide with a planet or asteroid.


Why does e.g. Hubble's secondary mirror not block part of the picture?

A particular point in the image is formed by the convergence of all the rays of light that leave from that point in the object, and bounce off the mirror.  As long as there are some rays that can do that, then the only effect of the secondary is to block some of the rays, dimming those parts of the image slightly.


Gravitational waves are affected by gravitational lensing in the same way as light. Most lensing events are not caused by black holes, but by more common objects with similar mass.

There largest deflection examples of the light from distant galaxies, quasars etc being diverted by foreground sources (clusters of galaxies) is about an arcminute (e.g. Zitrin et al. 2014). Current gravitational wave detectors are not capable of locating sources even to a few degrees. So source location isn't an issue at present.

Source strength might be. A lensed source can be magnified or even split into multiple images that arrive at different times. This will lead to a luminosity distance underestimate from a gravitational wave source and an overestimate of its mass.

For those that are interested, Oguri et al. (2018) have done an extensive set of simulations, showing the effects of gravitational lensing on estimates of mass and redshift distributions for black hole binary gravitational wave sources.

The conclusion is that for detectors like LIGO there should be a (very) small tail of highly magnified events with underestimated redshifts and overestimated masses. These are predominantly at (chirp) masses $>40 M_{\odot}$ (which is also roughly the minimum mass of the primary component in a binary) and more massive than any of the LIGO events seen so far.

The "optical depth" to lensing as a function of source redshift is shown in Fig. 3 of that paper. What this shows is that the chance of any individual source being strongly ($>$ factor of 10 or split into multiple images) lensed is less than one in a million at typical LIGO source distances of a billion light years (redshifts of less than 0.1). But this probability can increase by several orders of magnitude at the high redshifts that might be probed by detectors in the future.


Is the eccentricity of a planets orbit relevant to finding whether the distance between said planet and its host star determines seasons?

Yes it would. If the sun was much further away the season on earth would certainly be more winter-like. The reason distance is not the deciding factor for earth is the variation in distance is fairly small.


Do objects that are invisible to the naked eye occlude much of the stars to Earth based observers?

From your description, an object that blocks out light from a star in only one location cannot be an astronomical object.

Occultations by asteroids are fairly common. You can see a list at  http://www.asteroidoccultation.com/. When a star is being occluded by an asteroid there will be a fairly narrow strip from which the star will be hidden. But it is still about 50km wide. The star appears to rapidly fade, and it remains hidden for at most a few seconds, then reappears. Your description of "only hidden to one of my eyes" doesn't fit an asteroid occultation.

The object hiding the star must, therefore, have been local, and not moving. It would be speculation to suggest what it could be.

There are may objects that are hidden in visible light, not by asteroids, but by interstellar dust clouds. We can see them by using other wavelengths.


The faint green smudge looks to me like a flare (that is, an optical defect called a flare).

However I'd expect a flare to be symmetric with the bright object it relates to about the center of the field of view.  This does imply either the image is off center (e.g. the camera is off center or the image is a crop off center) or there's a misalignment of your optics.

This question on Physics SE is about correcting flare in a telescope and may be relevant.

The similar question Does this smartphone photo show Mars just below the Sun? and its answers contain some other examples.


Do changes in the orbit of the Earth contribute to global warming

They found in 1976 that the axial tilt and precession changes matched the geologically known ice age time periods. (see this pic for precise details)

The graphs say the world will get a tiny bit warmer for a few hundred years and then colder, if it weren't for CO2 being at the highest level for many million years:

Figure 1:  Orbital Parameters:  Eccentricity, Precession and Obliquity- click for larger image

How is it measured? It's mostly straight mathematics, say if the earth is 10% nearer to the sun it casts a larger shadow into outer space and collects more solar radiation.

This figure is developed from A.L.Berger, 1978, Long Term Variations of Daily Insolation and Quaternary Climatic Changes, Journal of the Atmospheric Sciences, volume 35 (12), 2362-2367.

M.F.Loutre and A.Berger, 2000, Future Climate Changes: Are we entering an exceptionally long interglacial?, Climatic Change 46, 61-90

The effects of galactic cosmic rays on the atmosphere (via cloud nucleation) and those due to shifts in the solar spectrum towards the ultraviolet (UV) range, at times of high solar activity, are largely unknown. The latter may produce changes in tropospheric circulation via changes in static stability resulting from the interaction of the increased UV radiation with stratospheric ozone. More research to investigate the effects of solar behaviour on climate is needed before the magnitude of solar effects on climate can be stated with certainty.


Radioactive dating works based on the assumption that once a particular rock (in this case, a tiny zircon crystal) has cooled and solidified, the material within it will not be exchanged with material outside it. Since the rock contains trace (known) amounts of radioactive isotopes at its formation, measuring the amount of radioactivity today gives the amount of time since that rock solidified and cooled.

The USGS claims that the age of the zircon crystals is roughly the age of the Earth. They are doing so based on the assumption that the zircon crystals solidified shortly after the Earth was formed. Considering that Earth formed when a diffuse cloud of tiny particles collapsed into a comparatively tiny ball, releasing a bunch of potential energy in doing so, it is reasonable to suspect that Earth, just after it formed, was extremely hot, certainly hot enough to melt zircon at 1900 degrees C. As discussed above, melting something "resets the clock" with respect to isotope composition, as new radioactive isotopes readily diffuse in to replace the old ones. Therefore, their assumption seems to be well grounded.

As to the last question, it depends on what you mean by "know". Obviously we don't have planet-sized clouds of dust to play around with in laboratories, and we also don't really have enough observation time to see these things happen (even though I said they occur "quickly", not many processes on the scale of planets take less than a few thousand years to happen). So we will probably never see more than a snapshot of this evolution process directly. Then again, if you trust the simulations that people are doing nowadays (considering how much thought is generally put into those things, it's reasonable to do so), we seem to have a general idea of how long the collapse should take, which is on the order of a million years, starting from the phase where the dust has already collapsed into 1 km-wide bits: https://www.cfa.harvard.edu/~kenyon/pf/terra/t0/index.html.


What is the reason the Chinese want to measure the <30Mhz radiation behind the moon?

Frequencies below 30 MHz have a hard time penetrating the so-called ionosphere - the series of layers of ionized atmosphere that exist at high altitudes.

This is good news for folks like me, amateur radio operators, because it allows us to speak into a transmitter at home, and let the ionosphere bounce the radio waves back to Earth, then again up into the ionosphere, and so on, eventually reaching another radio operator on the opposite side of Earth.

But it's bad news if you're a radio astronomer trying to receive such frequencies from distant objects - the ionosphere simply bounces those waves back into space.

So, apparently, you could raise a receiver above the ionosphere, in a satellite, and have it receive signals from outer space. But there's a catch. At 30 MHz and below, the wavelengths are huge. 27 MHz - the wavelength is 10 meters. 14 MHz - wavelength is 20 meters. 3.5 MHz - 80 meters. And so on, the lower the frequency, the bigger the wavelength.

The size of the antenna needs to be comparable to the wavelength if you want to receive signals from one direction only, so as to be able to tell where the signal is coming from. This also helps with efficiency. A very good antenna needs to be much bigger than the wavelength for it to be directional and efficient.

The problem is, even with the ionosphere, there's a lot of signal "leaking" out of the Earth, through ionosphere, into space; some of it is bounced back down, but some of it penetrates the ionosphere and escapes into space. A satellite in low orbit would be swamped by transmitters from Earth, completely erasing any signals from deep space. You'd need a humongous antenna to ignore the signals from the nearby Earth. This is not feasible.

It would be a lot better if you could just put the Moon in between Earth and your antenna. Even these low frequency transmissions cannot penetrate the Moon, and can't go around the Moon either. So the Moon acts like a screen, casting a shadow on the antenna and eliminating parasite transmissions from Earth.

Now your satellite can receive low frequency transmissions from deep space.

TLDR: Too much noise near Earth at these frequencies, need to hide the antenna behind the Moon.


Temperature of Earth if tidal locked with the Sun?

That's very, very difficult to answer with a lot of if's. If the earth had no atmosphere it would be 255K-277K (-18-3°C) on the day side and a lot colder on the nightside. But if we include the atmosphere into our calculations we can get widely differing results.

For the sake of simplicity let's assume that earth tidal locks and nothing else happens. No change of atmosphere because many plants die or human population and therefore CO2 emmissions go down or something like this.

In this case I guess that the temperature would stay about the same. While one side of the earth enjoys permanent sunlight, the air on this side heats up and therefore rise to higher layers of the atmosphere. Cooler air from the other side of the earth streams in and the hot air in exchange streams to the cooler side where it cools down and returns to the bright side. So while one side of the earth is permanently sunny it is also permanently cooled down by air from the other side.

The dark side would not cool out because it is permanently heated by the hot air from the sunny side.

On the whole we would live on a very windy planet. But thats it.


The title of the question reads:

Can I use the parallax coefficients for observatories as a proxy for latitude using arctan2(sin, cos)?

and the body expands:

Question: What exactly are the cos and sin parallax coefficients, and how closely can $\arctan2(\sin, \cos)$ be expected to serve as a proxy for latitude? Within a degree or so?

First the "How closely can (they) serve as a proxy for latitude?" part
tl;dr: to within +/-0.2 degrees; the range in the differences between geocentric and geodetic latitude.
The table provides the sin and cos of the same angle $\phi'$ multiplied by $r/R_E$ and the 2nd half of the answer will explain that $\phi'$ is the angle that a radius drawn from the observatory to the Geocenter makes with respect to the Earth's equatorial plane. Everyday latitude or GPS latitude which is a kind of geodetic latitude (what I want to use for my map) is a little bit different because it's drawn on an ellipsoid.
How different?
Well it turns out that I wrote an answer about geodetic latitude about 18 months ago. (@DavidHammen wrote a nicer answer elsewhere as well) Here is an excerpt from my answer:

The 3D cartesian coordinates $X, Y, Z$ in Earth-centered, Earth-fixed coordinates assuming an ellipsoidal shape is given by:
$$X =  \left(N(\phi) + h \right) \cos\phi \cos\lambda $$
$$Y =  \left(N(\phi) + h \right) \cos\phi \sin\lambda $$
$$Z = \left(\frac{b^2}{a^2} N(\phi) + h \right) \sin\phi $$
where $\phi, \lambda, h$ are latitude, longitude, and altitude, and $a, b$ are the equatorial and polar radii of the ellipsoid used, and
$$N(\phi) = \frac{a^2}{\sqrt{a^2\cos^2\phi + b^2 \sin^2\phi}}. $$
Lat, lon, alt in "GPS coordinates} is based on WGS 84 with $a, b$ of 6378.1370 and 6356.7523 kilometers, respectively.

The parallax coefficients are $\rho \cos(\phi')$ and $\rho \sin(\phi')$ and $\rho$ is the distance from the observatory to the Geocenter divided by "the Earth radius". That makes them:
$$\rho \cos(\phi') = \frac{\sqrt{X^2+Y^2}}{R_E}$$
$$\rho \sin(\phi') = \frac{Z}{R_E}$$
So if we chose a geodetic latitude $\phi$, choose a longitude of zero so that $Y=0$ and calculate $X, Z$ and plug them into this:
$$\phi' = \arctan2\left(Z, X \right) = \arctan2\left(\frac{Z}{R_E}, \frac{X}{R_E} \right) = \arctan2\left( \rho \sin(\phi'), \rho \cos(\phi') \right)$$
we can get the geocentric latitude, and then calculate the error by subtracting the two.
It turns out to be about +/- 0.2 degrees with extrema at about +/- 45 degrees. It is of course zero at the equator and poles.
So to an error of 0.2 degrees, yes that can indeed be used as a proxy!

Now the "What exactly are..." part
As pointed out in comments, the link explains how the cos and sin parallax coefficients are defined:

The following list gives the observatory code, longitude (in degrees east of Greenwich) and the parallax constants (rho cos phi' and rho sin phi', where phi' is the geocentric latitude and rho is the geocentric distance in earth radii) for each observatory. It is updated nightly.

What is geocentric latitude? It's the "easy one" to understand; it's just the angle measured at the Geocenter between the radius to the point and the closest point on Earth's equator. So while the definition *does not specify a specific Earth radius, it doesn't have to because the table only gives the unit-less parts cos phi' and sin phi'.
But why both sin and cos? Can't you just use sin between -90 and +90 degrees?
To just define an angle (without the radius), mathematically you might think so. However sin is insensitive near the poles and in the world of real computing numerical errors can crop up. Near the poles sin will still provide the appropriate sign and cos will provide the sensitivity. This way you will have low numerical errors (for this particular step) everywhere.
But this is the latitude and distance together, so you have to have two numbers. They could be a latitude angle and a radius explicitly, but this also works. (revised per @DH's comment)
But what are the parallax coefficients for exaclty? What are they really?
I don't know, but hopefully someone with a valid "astronomer's ID card" will come along and post an answer. I'm to scared to ask a new question on this topic given all the comments under the question! :-)

Script for the plot
import numpy as np
import matplotlib.pyplot as plt

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
a, b = 6378.1370, 6356.7523 # only their ratio matters in this case
todegs = 180/pi

phi = np.linspace(-halfpi, halfpi, 101)
h = 0.
lam = 0.
N = a**2 / np.sqrt(a**2 * np.cos(phi)**2 + b**2 * np.sin(phi)**2)
X = (N + h) * np.cos(phi) * np.cos(lam)
Y = (N + h) * np.cos(phi) * np.sin(lam)
Z = ((b/a)**2 * N + h) * np.sin(phi)

phi_prime = np.arctan(Z/X) # two quandrants so no need for arctan2

plt.figure()
plt.plot(todegs*phi, todegs*(phi_prime-phi))
plt.xlabel('geodetic ("GPS") latitude (degs)', fontsize=12)
plt.ylabel('geocentric - geodetic latitude (degs)', fontsize=12)
plt.xlim(-90, 90)
plt.show()


The proton-proton chain ultimately converts four protons into one helium nucleus. The charge of the 4 protons was balanced by 4 electrons, but helium contains 2 protons (and 2 neutrons), so it only needs 2 electrons to be balanced.

As you point out, the process of converting a proton to a neutron releases a positron (and an electron neutrino), and that positron quickly annihilates with an electron.

Here's the diagram from that Wikipedia page of the main p-p chain.

So the process actually consumes 6 protons, and emits 2 protons, a helium nucleus, and 2 positrons (plus a couple of neutrinos), and a couple of gamma photons. The positrons annihilate with 2 electrons, releasing more gamma photons (usually 2 or 3 apiece, depending on the spin alignments of the positron & electron).

If you add everything up, you'll see that the electromagnetic charge balance is unchanged.

We started with 4 protons, which are balanced by 4 electrons nearby in the stellar core plasma. (We can ignore the intermediate pair of hydrogens that are eventually re-emitted). We end up with a helium nucleus that only needs 2 electrons to be electrically balanced, so if those other 2 electrons weren't annihilated then the star would build up an excess of negative charge.


How many moons in the solar system have water?

It's probably easier to consider the reverse question: which ones don't have water.

Talking about the major satellites of the planets, there are only two which are not ice moons: our own Moon, which has ice deposits at the poles, and Io. As far as I am aware, the evidence for water on Io is tentative at best (e.g. Douté et al. 2004), so of the major planetary satellites, the answer is all of them have water (usually in the form of ice, but some of them also have liquid subsurface oceans) with the possible exception of Io.

The small satellites of the giant planets are probably icy but I'm not sure the possibility that one or two of them are dry rocky asteroids that have been captured has been fully excluded. Phobos and Deimos have rocky surfaces but the possibility of ice in the interior of the Martian moons has not been ruled out.

There are probably lots of ice-free asteroid moons but finding out which ones they are would require going there and checking whether there's any ice in the subsurface, so confirming this is not currently possible. Small solar system bodies in the outer system are probably going to be icy, so dwarf planet satellites (Charon, Vanth, etc.) all likely contain ice.


How can they tell no asteriods will hit earth in the next hundreds years?

A couple of points based on some basic orbital mechanics

They don't need to get a "good" view, like the clear, crisp photos of Pluto to see one coming.   They only need to get a picture over time to calculate trajectory.   The unclear snapshots work just fine to calculate if it'll hit us or miss us.

Also, an object as far as Mars at it's closest pass to earth, a bit over 1/2 AU, would still take a few months to reach earth if it's in solar orbit.   Mostly we don't need to track anything as far out as Pluto, they can look much closer to the Earth and still have sufficient warming time.   The hard part, is tracking things that approach from the Sun side, cause those are harder to see.  That's why the Chelyabinsk meteor wasn't spotted.   it was also on the small side, smaller than NASA is currently looking for.

The good news is that, we don't get struck by things that size very often.   The Solar system is pretty enormous and pretty empty and pretty big strikes like that one are rare, like, maybe once a century.

Also, virtually all of the injuries from the Chelyabinsk meteor were from people who didn't know what to do.   If you see a big fireball in the sky, it's human nature to watch it, but use some common sense.  A space rock of that size will make a shock-wave that travels at roughly the speed of sound and the shock-wave can break windows, even knock over trees and buildings if it's big enough.   You don't want to be standing in-front of a window when the shock wave its.  Lay down next to a couch or under a table in case your building gets shaken and cover your ears.   If everyone had done that, there would have been very few injuries.   You only need to wait maybe 2 minutes or so to be on the safe side.

If you're in a car, stop, cause the shock-wave could knock down trees or debris in-front of you and stay in the car, cause that's safer than being outside.     All told, the damage to buildings was tiny compared to natural disasters like Earthquakes, floods or volcanoes which happen to us several times a year.    It's good that NASA is watching for this kind of thing, but it's also a pretty rare event.


How to plot orbit of binary star and calculate its orbital elements?

If it's a binary, it's fairly simple (in comparison with a ternary system), because the stars of the binary orbit about their common barycenter in Kepler ellipses.

An orbit simulator, see here; replace the central star by the barycenter of the binaries.

Calculation of an Orbit from Three Observations. Kepler problem on Wikipedia.

If you don't find a ready-to-use software, try to solve it numerically by simulating the orbit of the binaries according to Kepler's law, and vary mass, distance, excentricity assumptions until they match to the observations. Use optimization methods, e.g. hill climbing algorithms, or gradient methods.


Would Venus have any significant effects on Earth if its orbit were entirely within the habitable zone?

Changes on Earth:

Except for possible tides and possible eclipses, if it were near enough, no.

Changes on Earth's orbit:

Venus would establish some resonance with Earth's orbit, if it were near enough or in one of the exact possible spots. This will not have a major impact except on year's lenght and Sun illumination on Earth.


Stratospheric Balloon with an Xray camera

Long ago I was part of a group that flew high altitude balloons to study x-rays from pulsars etc. How much you detect and at what energy depends upon how much atmosphere is above your detector. Lower energy x-rays are absorbed more than higher energy x-rays. Realistically you won't detect anything except maybe the Sun at energies below 20 keV and at altitude below 25km. We aimed for altitudes near 40km. To detect high energy x-rays the cheapest method is to use a proportional counter. Not something you can buy off-the-shelf.

Also the Sun only emits a lot of high energy x-rays during solar flares.

An interesting experiment would be to fly a geiger counter see http://earthtosky.net/


What does "negligible mass" mean in the formulation of geodesics equation?

Yes its means that you ignore the effect that an object such as a spacecraft has on the central mass that it is orbiting, that the orbiting body has infinitely less mass.

Unfortunately, I think there is no exact solution found of the geodesics equations when that approximation is no longer valid or when you try to devolop geodesics equation for a spacecraft under the influence of two (or more) spherically symmetric mass distributions such as the combination of planets and the Sun in our solar system.

In the solar system JPL who calculates orbits of planets and other celestial objects use what is called the "post Newtonian expansion". You can look at their official documentation, Formulation for Observed and Computed Values of Deep Space Network Data Types for Navigation, expression 4-26 on page 4-19. Some of the the terms may be zero if you only have two non-neglibly massive objects.

Note that this is not really a "geodesics equation" per se. Nasa/JPL uses what I believe is a first order expansion of the Schwarzshild solution in isotropic coordinate and then uses a scheme I do not quite follow to get to the relativistic acceleration terms of the post-Newtonian that they add to the classical Newtonian gravitational acceleration term to calculate the orbits.

I do not know if there is a simpler way to do it.


Shouldn't the estimate of the universe's age be higher, not lower, after the attractive strength of gravity is taken into account?

Attractive gravity means a shorter time since the Big Bang, for any given Hubble value. Here's a picture (from lecture notes by Sean Carroll):

The averaged distances and speeds of nearby galaxies give us the slope of the curve at the horizontal position marked "now". If you assume they've always moved at that speed, you get the scale factor indicated by the dotted line, while if you take into account that they're slowing down, you get the solid curve, which reaches $0$ (intersects the $t$ axis) at a more recent time.
Wilczek is wrong when he says that linear extrapolation gives an age of 20 billion years. It actually gives about 14 billion years, close to the correct value. This is more or less an accident: it's only true during the transition from matter-dominated decelerating expansion to dark-energy-dominated accelerating expansion, which happens to be happening now. In old matter-dominated cosmological models (predating the discovery of dark energy), linear extrapolation gives you about $1.5$ times the correct age, so it's possible that Wilczek got 20 billion by multiplying a modern estimate of the age by a ratio from an obsolete model.


The standard picture of vacuum decay events is very much like nucleation in a phase transition: it starts somewhere, and then expands at a constant speed - lightspeed in this case. So the future lightcone of the nucleation event will be true vacuum.

In an accelerating expanding universe like our current $\Lambda$CDM cosmology future lightcones actually have a finite radius as measured by co-moving coordinates (e.g. from the perspective of matter moving passively with the expansion of the universe). Beyond a certain (current) distance there are galaxies that will never be reached by a light signal we send now - when it gets to the distance they currently have they will have moved much further away, and it never manages to close the gap. Same thing for vacuum decay: the front will always expand outward at lightspeed, but large parts of the observable universe will expand away so fast that it will never catch up with them.

In about 100-200 billion years galactic clusters will become separated by exponentially growing gulfs and quickly lose contact (they stay the same size because they are gravitationally bound). They become separate "island universes". Beyond this point vacuum decay would just be a local issue to one island universe.

Extra question: if a galaxy has too much energy it will implode as a black hole - mass and energy are equivalent in relativity. Whether that counts as dislocation is unclear. But overall regions with more matter density tend to contract while less dense regions expand more, which is why we get the cosmic web of sheets and filiaments of dark matter plus galaxies, and expanding voids. Galaxies in the voids (they do exist) could be described as energy poor (even though their stars are doing just fine) and they would experience everything moving away from them not just because of universal expansion but also gravitational contraction towards remote masses.


Is this star group a constellation recognized by the IAU?

The IAU recognizes 88 constellations. The charts for each constellation can be found at the bottom of this page from the IAU website. You can go through them yourself, but I haven't found any that really resembles your image. So it probably isn't a constellation recognized by the IAU.

But it could still be part of the sky, without being a constellation. So I also ran the image through astrometry.net, but it didn't find what field the image could belong to.

This suggests that it is just a random group of stars, rather than an actual image of the sky.


I can't vouch for the formula, but you can try it out and check with some internet source, eg USNO, to see if it works.  If the formula is correct, that's it.

As for the sidereal time in Sydney, you need the difference in longitude between Greenwich and Sydney, and use the conversion rate of 15 degrees longitude is equivalent to 1 hour.


tidaly locked celestial body in three planet system

Tidal locking just means that one side of the Moon will face the planet all the time. The period of rotation around its own axis is the same as the period of rotation around the planet. It is not absolutely "locked" exactly in that configuration, but on average the two periods will be the same. Another moon or two could possibly also be in the same condition, and there wouldn't be strong interference between them. There could be some slow processes that could work against both remaining tidally locked, or some slow processes that kept them that way, but it's a complicated problem


What causes these asteroid trails to be so strangely shaped and repeated?

The linked image is a composite of several long duration exposures captured by the Hubble Advanced Camera for Surveys of the same region of the distant sky. Each individual exposure was about ~2200 seconds long. Nearby objects such as asteroids would have been subject to the proper motion of the asteroid with respect to the Earth and parallax due to the Hubble being in orbit about the Earth during those ~2200 second long exposures.

Each individual streak comprises both proper motion and parallax effects. The parallax effects are what make the streaks have a nonlinear shape. One orbit later (~5725 seconds), another ~2200 second long exposure showed a similar path for the same asteroid, but shifted a bit because of the asteroid's proper motion. This is why the composite image shows twenty streaks (per the linked article; I see sixteen) for only seven asteroids.


As months are not of fixed length it is only possible to calculate an average.

From Wikipedia:

The average frequency of a blue moon can be calculated as follows. It is the period of time it would take for an extra synodic orbit of the moon to occur in a year. Given that a year is approximately 365.2425 days and a synodic orbit is 29.5309 days, then there are about 12.368 synodic months in a year. For this to add up to another full month would take 1/0.368 years. Thus it would take about 2.716 years, or 2 years, 8 months and 18 days for another blue moon to occur.

So for a 70 year life span the average number of blue Moons would be 70/2.716 or 25.77.


Why don't we see evidence of glaciation on Mars, similar to Earth's Ice Ages?

Obliquity anything from 11 - 28 degree swings which would obviously
  affect seasonal changes. Is this a fact (where's the evidence) or
  "scientific fable".

I don't like to use the word fact in a situation like this.  What that variation is, is a mathematical model based on well understood but pretty complicated n-body gravitational perturbations between planets, so, it's likely accurate.  Whether it's a fact is perhaps a matter of semantics, but I don't like to use the word fact in this situation.

That said, it's very likely true that Mars undergoes both significant axial tilt changes and eccentricity changes over thousands or tens of thousand year cycles, and if variations like that happened on Earth, they would cause havoc to our climate.
Everything else being equal, that kind of variation should generate considerable glacial movement on Mars' poles, but everything isn't equal between Mars and Earth.

Ice ages on Earth require large land masses on or near one or both poles, where ice can build.  Glaciers can't form in open deep ocean water, they require land, where the summers are cold enough that ice can build up, which, over thousands of years of buildup, becomes a glacier.

Mars lacks the water to expand it's glaciers.   Nearly all Martian water is either underground or already frozen on it's North pole, so there's no water to add.   Literally about 99.99999% of Mars' surface water is already part of one of it's two glaciers on it's North & South poles.

Unlike Earth - where there's always some water in the atmosphere and there's always the possibility of snowfall, so Earth's glaciers can always grow bigger if the seasonal temperature allows it.   Mars has no ice to add to it's glaciers.

That limits Mars' glacial expansion to CO2, and that does happen to a degree.  Mars glaciers do add CO2 during the winter and return it during the summer, but that's seasonal and Mars has only so much CO2, so it's glaciers could only expand as far as it's atmosphere would allow.

Mars' atmosphere is about 2.5 x 10^16 kg.   Mars' surface area is about 1.4 x 10^14th square meters, which means it's entire atmosphere works out to about 180 kg per square meter, which would be about 7 inches high if frozen into dry ice.   7 inches means that Mars can't form glaciers like Earth can, no matter what it's axial tilt or eccentricity variation.  It lacks the material.

If things were very different on Mars, like, lets say it had ammonia-water oceans and a thicker atmosphere and regular snow and rivers, then it might undergo glaciation periods and ice ages comparable or perhaps exceeding Earth, but it can't do that because it lacks sufficient atmospheric material.

Mars can create a thin ice cover and it can do that seasonally:

Mars' seasonal ice, winter to summer, is prone to considerable adjustment.

And, I don't want to say 100% that Mars undergoes no glacial changes.   It still might, as it goes through it's Milankovich cycles.  I'm not sure anyone knows for sure what Mars' ice caps looked like 10,000 or 100,000 years ago, so it's possible there were noticeable changes, but Mars' still lacks the material to form new enormous glaciers the way Earth can.

As a sidebar, and perhaps this could be looked into in more detail in a separate question, but Mars may undergo some kind of below the surface glaciation.  Posting article as a curious sidebar only.


The complexity of the formula depends on the precision you need.
We can make a crude approximation with two sine waves: one for the Moon's travel eastward around the ecliptic over a
tropical month

$$\delta_1 = 23.4^\circ \sin {2 \pi (u - 10.75) \over 27.32158}$$

and one for the Moon's travel north or south of the ecliptic over a
draconic month

$$\delta_2 = 5.1^\circ \sin {2 \pi (u - 20.15) \over 27.21222}$$

both modified as the Moon goes faster or slower over an anomalistic month

$$u = t - 0.48 \sin {2 \pi (t - 3.45) \over 27.55455}$$

where t is the number of days since 2000-01-01 12:00 TT (J2000.0).
Then

$$\delta = \delta_1 + \delta_2$$

is within 1° of the Moon's geocentric declination most of the time.
For a topocentric declination, subtract
1° sin (observer's geographic latitude).

If you need higher precision, see
van Flandern and Pulkkinen 1979
or Meeus 1999
for more complicated formulas.

Note: the u and t offsets above are near a 0° longitude crossing, an ascending node, and an apogee in January 2000 but are tweaked to minimize RMS differences from JPL HORIZONS lunar positions for 2000-2018.


How frequently do Jupiter & Regulus have triple conjunctions?

With any given star, Jupiter has 3 or 4 triple conjunctions per century, with either 12 or 71 years in between.
These occur with Regulus in 1873, 1885, 1956, 1968, 2039, 2051, 2063, 2134, and 2146.

Jupiter takes 11.9 years to orbit the Sun, advancing at a fairly steady ~30° per year.
From Earth, Jupiter appears to move ~40° forward and ~10° backward in a 13-month cycle.
The retrograde loop happens when Earth and Jupiter are on the same side of the Sun.

We see a triple conjunction with a star when the retrograde loop encompasses it.
Around the February 2051 opposition, Jupiter has conjunctions with Regulus in October, February, and June:

If Jupiter passes a star while Earth is on the other side of the Sun, we see only a single conjunction, as in August 2003:

Jupiter-Venus conjunctions occur at intervals of 10 or 14.5 months.
The Jupiter-Venus conjunctions of October 1884 and July 2146 occur reasonably close to a Jupiter-Regulus conjunction in a triplet listed above and far enough from the Sun to observe easily.


Why do we continue to find a discrepancy in the Hubble Constant?

To make a long story short, the measurements from Planck and the Hubble Space Telescope disagree, and the reason behind this isn't known.

First, let's look at the values with the uncertainties. We then have three different results that are, perhaps, not as inconsistent as they originally seemed:

$70.0^{+12.0}_{-8.0}\text{ km s}^{-1}\text{ Mpc}^{-1}$ from LIGO.
$67.8\pm0.9\text{ km s}^{-1}\text{ Mpc}^{-1}$ from Planck.
$73.48\pm1.66\text{ km s}^{-1}\text{ Mpc}^{-1}$ from Hubble Space Telescope (HST) observations.

The LIGO measurement is consistent with the others, as the authors happily note in their conclusions. However, there is a discrepancy between the results from Planck and the results from the HST. This isn't a new problem; the Hubble group previously determined a value for the Hubble constant of $73.24\pm1.24\text{ km s}^{-1}\text{ Mpc}^{-1}$ in 2016. The uncertainties in the measurements describe one standard deviation; it doesn't mean that all values beyond the interval of the measurement plus or minus one standard deviation are impossible.

Now, the difference between Planck results and the 2016 HST data is $3.4\sigma$, which is a lot. The group came up with several explanations for the discrepancy:

Systematic errors in the data collection, including in the use of the Cepheid variables used for the measurements.
Small-scale variations in the Hubble constant, which appear unlikely to have propagated through the data.
Systematic errors in the Planck results.
Corrections needed to the $\Lambda$CDM model of cosmology, used by Planck.

In the 2018 paper, the authors note that the conflict with the Planck data is even stronger - $3.7\sigma$ (and we're now getting results that say it could be $4.4\sigma$; I should check those out). They haven't been able to point a figure at any single problem that could be behind the discrepancy; they do optimistically say that it could be the sign of new physics.


How to deal with the celestial meridian?

I think you are overlooking these factors:

Iowa State University is at 42 degrees North latitude, so the Sun and celestial equator that they are showing in their figures are always to the south at the meridian. Naturally, observers in the southern hemisphere would see the sun and celestial equator to their north at the meridian.
The lines of right ascension go from pole-to-pole.
So the hour angle that the website is describing is the angle between the plane of the meridian and the plane of right ascension. This angle applies regardless of where you look in the sky: north, south, or overhead.


Why does iron consume more energy in the fusion process than it produces?

It would be good if you referenced your sources, because you may be misunderstanding them. We'd be able to see what they actually say, and help you understand them.

Nucleosynthesis of iron does not use more energy than it produces.

It is, however often referred to as the heaviest element created in fusion that results in more energy produced than consumed.

However, that isn't quite true. Heavier elements can be produced by fusion that produce more energy than is used, except these fusion reactions don't occur in stars. (Eg 40Ca + 40Ca)

Also, it is possible for heavier nuclei to be fused in stars that result in more energy being produced than is used, but these are unstable isotopes and they decay quickly.

So, more accurately, iron is the heaviest element produced in stellar nucleosynthesis in any significant quantity that produces more energy in fusion than the fusion consumes.

This is called the alpha process ladder. Keep adding alpha particles to the newly generated nuclei, until you stop getting more energy out than you put in.

The last step in the alpha process that does produce energy is 52Fe + 4He => 56Ni(excuse the rubbish notation; if this answer is considered helpful at all I will try to tidy up the notation)

56Ni + 4He => 60Zn uses more energy than it produces.

56Ni has a very short half-life of just 6 days, decaying to 56Co which has a half life of 77 days, which decays to 56Fe, which is stable. So when a star is on the limit of collapsing, it will be producing a lot of iron in it's final stages - some of it from decaying heavier radioactive isotopes.

Why does the alpha process stop producing energy at this point? It is because that is where the peak binding energy is. More.


Calculate mass of exoplanet from transit method

For the case of a single planet, you can't. The transit method is sensitive to the radius of the planet, not the mass.
For a multi-planet system, you can use variations in the timing of the transits caused by the gravitational interaction between the planets to infer the masses, for example as done in this paper by Nesvorný & Morbidelli (2008). These timing variations can be very significant if the planets are located close to mean-motion resonances (i.e. the ratio between the orbital periods being close to a fraction with small denominator, e.g. 1:2 or 2:3).


The Jupiter Trojans are not at the Sun-Jupiter L4 and L5 Lagrange points. They are instead in pseudo orbits about those points. These pseudo orbits appear from the perspective of a frame of reference that rotates at Jupiter's orbital rate (i.e., a frame in which the Sun and Jupiter are more or less fixed). Unlike central body motion, which results in planar orbits, these pseudo orbits in general are not planar, and they librate between in front and behind the L4 or L5 point over a period of about 150 years.
The very large volume means interactions are rare, and the crazy pseudo orbits means collisions are more likely to be violent than not. That said, at least one Trojan asteroid, 624 Hektor, is most likely to be a peanut-shaped contact binary.


What are the raisons d'être for the Large Binocular Telescope "binocularity"?

I asked myself the same question and back then, I found the reasoning on the German Wikipedia page pretty comprehensive. I am translating, summarizing and expanding the corresponding section in the following:
By using two mirrors instead of a single one has a couple of advantages:

The double telescope collects the same amount of light as an 11.8-metre mirror, but is from an engineering point of view easier to built than a single large mirror, because the center of mass is outside the telescope(s). An example structural problems of large moving telescope was e.g. the failure of the Green Bank Telescope in November 1988. In radio astronomy, the approach of having multi-antennas on a single moveable structure is used for long, e.g. in the Pluton Complex
The LBT's optical resolution can be increased by interferometry to that of a 22.8-metre mirror.
Interferometry will also be used to virtually block out the light from central stars in order to make their possible planets directly visible, sometimes called nulling interferometer.

Last but not least, I remember hearing that LBT's mirror size is the largest feasible non-segmented mirror. This hypothesis is somewhat supported if you look at the list of largest optical telescopes.


This is a difficult question to answer. Physics really starts after the big bang. Scientist don't know about the laws, if any, before the first instance after the big bang. The time before and during the big bang is really the province of philosophy.


If you lived on the far side of the Moon, how could you infer the existence of Earth?

A body tide seismometer on the lunar far side would pick up both the solar tide, and the 20 inch body distortion produced by the earth.
While "tidally locked", the moon is not in a perfectly circular orbit, and also wobbles a bit; libration. Your seismometer should pick up both effects.

Watching the parallax of mars cycle every 28 days, as suggested in comments above, might be a simpler way to go.


Does science need support from religion or philosophy to explain the creation?

The question seems to be based on the notion "before". This notion relies of the concept of the arrow of time, meaning time "flowing" in some "direction". The arrow of time itself is based, among others, on the notion of "time".

The Planck epoch - the first epoch of the current Big Bang model - lasted a little less than $10^{-43}\mbox{ s}$. That's the shortest unit of time. To ask within this short time interval for an arrow of time doesn't make any sense. Hence a before/after didn't exist within  the Planck epoch, and hence asking for a "before" isn't meaningful within the Planck epoch.

The Planck epoch to be able to exist needs the mere existence of space and time, not yet an arrow of time.

The cause for the existence of space and time at all is a matter of the pre-Planck epoch. In this phase using any notion which is based on space or time isn't meaningful. Cause and effect in the temporal sense, which underly most "why?"-questions, are based on time, and are meaningless in this epoch.

It's possible to construct other "pre-Big-Bang" models. But it's hard to find observational evidence for any of them.

If you suggest a creator of whatever kind, you just push the ultimate question a little bit further to the "past", and making it much more complicated. How could a creator be created, able to create such a complex thing like the universe? You need a chain of ever more complex meta-creators without end.

If you argue, that the first creator doesn't need a cause to exist, this applies also for the universe, formed from a pre-Planck medium. Hence nothing would be won.

Therefore it's rather unlikely, that religion or philosophy could be able to provide a more satisfying answer than science.

(Here some metaphysical multiverse theories.)


I like to classify solutions of the problem of the time evolution of the
complete initial state of a set of objects at some epoch time, where the
objects are subject to Newtonian gravitation into two main groups.
One approach is to use orbital elements of some sort.
The other is to use a numerical initial value problem solver,
aka a numerical integrator. The latter is the primary subject of this answer.

Note well: This classification isn't quite perfect as hybrid approaches are
also possible, wherein one uses a numerical integrator to integrate
time-varying orbital elements.

Keplerian orbital elements work quite nicely in the case of two point
masses, or more generally, two objects with a spherical mass distribution.
The anomaly is the only Keplerian orbital element that changes over time.
Keplerian elements can be used in situations where the underlying
assumptions are approximately correct by developing a model of how
those supposedly unchanging elements vary with time. One way of doing
this is to use Lagrange's planetary equations. (There are other related
approaches such as Gauss' planetary equations, Delaunay's planetary
equations, etc.)
Lagrange's planetary equations yield expressions for how Keplerian orbital
elements vary over time given a set of perturbing forces.

Another approach is to use something akin to those Keplerian elements
such as Delaunay elements), coupled with planetary equations for those
alternative elements. Yet another approach is to use orbital elements
(e.g., Brouwer-Lyddane elements, SGP4 elements) in which the planetary
equations are embedded in the orbital element to Cartesian state
transformation algorithm. This final approach is used to this day to
describe vehicles in Earth orbit.

The other approach is to use numerical integration.

I'll start with a discussion of how to solve for the value of a scalar
function $x(t)$ at some time $t_1$ given an initial value $x(t_0) = x_0$
and some well-behaved (continuous and bounded) derivative function
$f(x(t),t) = dx(t)/dt$ that describes the time evolution of $x$.
This falls in the very broad category of initial value problems.

Suppose the ordinary differential equation cannot be solved analytically
and cannot be expressed terms of a useful power series. This doesn't mean
nothing can be done. There are a number of techniques for solving this
problem numerically.

Note the dependence of the derivative function on the dependent variable $x$.
This becomes the much simpler problem of numerical quadrature if the
derivative function can be expressed independent of $x$.
The discussion that follows assumes that the derivative function $f$ does
indeed depend on $x$.
Note well: Newtonian gravitation falls in this category.

It also assumes the derivative function is well-behaved.
Numerically integrating across a discontinuity is a bad idea.

The foundation of the integration-based techniques for a scalar function
is the mean value theorem, which says that at some time $t_c$ between
$t_0$ and $t_1$, the value at $t_1$ is exactly
$x(t_1) = x(t_0) + (t_1-t_0)\,f(x(t_c),t_c)$. If only we could find that
magical $t_c$ and the derivative at that point. There's a chicken and
egg problem here: that magical point in time is not known. Even if it was,
the derivative function depends on state, and that too isn't known.

A very simple approach around this problem is to assume that this magical
point is the initial point:
$$x(t_1) = x(t_0) + (t_1-t_0)f(x(t_0),t0)$$
This works quite nicely for values of $t_1$ that are very close to $t_0$.
It doesn't work very well at all where $|t_1 - t_0|$ is not small.
This suggests splitting the interval $(t_0, t_1)$ into a number of smaller
intervals. This results in Euler's method: Apply the above to advance
state to time $t_0+\Delta t$, then to $t_0+2\Delta t$, and so on, eventually
reaching the desired time. Euler's method is rather lousy, even for a simple
first order scalar ODE. We can do much better than this. The key reason for
discussing Euler's method is that it is the basis for many other integration
techniques. Learn how it works, then toss it.

One approach to approving on Euler's method is to somehow correct the
result from Euler's method. For example, take an Euler step and compute
the derivative at the end point. Then use the average of those two
derivative values (the original value used to make the Euler step, and
the other from the end of the Euler step) to recompute the step from
$t$ to $t+\Delta t$. THis is Heun's method.

Another approach is to guess that the magical point $t_c$ lies
somewhere between $t$ and $t+\Delta t$. Perhaps the middle? We can use
Euler's method to advance state to the midpoint, and then use the
derivative at that point to advance state from $t$ to $t+\Delta t$.
This is the midpoint method.

Both Heun's method and the midpoint method appear to be steps backwards,
computationally. While Euler's method requires but one evaluation of the
derivative function per time step, these improved methods require two.
However, the error growth is in general so much smaller with either
Heun's method or the midpoint method compared to that from Euler's method.
This means that those "improvements" most definitely are improvements.
The expense of calling the derivative function twice per step is more than
offset by the fact that imprpovements enables take steps that are orders
of magnitude larger than one can make with Euler's method.

Both Heun's method and the midpoint method are simple improvements.
This problem has been studied in many guises. There are many more advanced
techniques. One is the class of Runge-Kutta integration techniques.
Both Heun's method and the midpoint method fall into this class.
The most popular of these, classical Runge-Kutta 4, is a significant
improvement on those two methods. There are even higher order Runge-Kutta
integrators than RK4.
Heun's method also falls into the broad class of predictor-correctors,
wherein one method (the predictor) advances state to the end of the interval
and another method (the corrector) uses the derivative at this approximate
endpoint to correct the guess made by the predictor.

The above focused solely on first order ODEs involving a scalar function.
What if the problem is multidimensional or involves higher order derivatives?
The mathematics of the techniques described above can easily handle
multidimensional data: Simply use the vector-valued time derivative.
Since a higher order ODE can be converted to a first order ODE via an
augmented vector-valued state, the same approaches used to address
multidimensional data can also be employed to address higher order ODEs.

There's a problem with doing this: It throws out geometry. For example,
consider the rather simple first order ODE $\dot x = -y, \dot y = x$.
The solution to this multivariate ODE is uniform circular motion.
Applying Euler's method to this results in
$$\begin{aligned}
x(t+\Delta t) &= x(t) - \Delta t\,y(t) \\
y(t+\Delta t) &= y(t) + \Delta t\,x(t)
\end{aligned}$$
The square magnitude of this new vector is
$(x(t)^2+y(t)^2)(1+\Delta t^2)$, which is always greater than magnitude
of the vector at the start of the step. This is not uniform circular motion.
The solution obtained via Euler's method instead spirals out. Other techniques
spiral inward. A geometric integrator on the other hand will somehow
maintain the constraint that $x^2+y^2$ is a constant of motion.

The above example showed why we don't want to toss geometry in a very
simple problem. The geometry of Newtonian gravitation, along with much
classical mechanics in general, is symplectic geometry.
This is why symplectic integrators are of great concern. A simple example
again, with Euler's method: Suppose the second derivative of position
is given by some function $\ddot {\vec x}(t) = \vec f(x(t),t)$.
Applying the basic Euler method against 3+3 dimensional phase space
dictates that
$$\begin{aligned}
\vec x(t+\Delta t) &= \vec x(t) + \Delta t \vec v(t) \\
\vec v(t+\Delta t) &= \vec v(t) + \Delta t \vec f(\vec x(t),t)
\end{aligned}$$
A simple change makes this symplectic:
$$\begin{aligned}
\vec v(t+\Delta t) &= \vec v(t) + \Delta t \vec f(\vec x(t),t) \\
\vec x(t+\Delta t) &= \vec x(t) + \Delta t \vec v(t+\Delta t)
\end{aligned}$$
As is the case with the scalar techniques discussed at the start,
Euler's method is a starting point rather than the end with
regard to symplectic integration techniques. Symplectic Euler's method
is rather lousy. But at least orbits don't spiral outward.

In making an N-body gravitation simulation, the size of N (the number of
bodies) is a key concern. Simulating a galaxy is a very different concern
from simulating a star system. The techniques used in simulating the
formation of a galaxy are very different from those used to develop a
solar system ephemeris. Galactic scale simulations cannot afford to
calculate all of the N2 gravitational interactions amongst all
the particles, and because N is so large, it cannot afford anything
more complex than very simple integrators. A star system model that does
not calculate all N2 of the gravitational interactions
or that uses very a simple integrator will by viewed in disdain.


This was not Mars.  Mars is close to opposition now, so it almost opposite the sun, and visible as an obviously red "star" in the Southern sky at midnight.

Your photo shows that the sun has just set, and you note that other stars were not visible.  This is the planet Venus. Currently shining at magnitude -5 it is 100 times brighter than the most bright stars. It is common, shortly after sunset for it to be the only visible "star", and as it orbits closer to the sun than the Earth, it is always seen in the sky close to the sun, as your photo shows. The when the Planet Venus is an "evening star" it is seen in the west. This is consistent with your plane travelling in a roughly northerly direction.

It is hard to be sure what time zone, you are in, or exactly where you are, but Venus will be visible shortly after sunset in most parts of the world.


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

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

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

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

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

  maybe even more activity than there is here on Earth


Information on the star at the centre of the Helix "Eye Of God" Nebula

The central star is a hot DAO white dwarf. As far as I can tell, the numbers that are the most reliable are those from Napiwotzki et al. (1999). They use a direct spectral fitting technique to estimate the surface gravity and effective temperature of the central white dwarf star. These are $g = 10^{(7.00 \pm 0.22)}$ cm/s$^2$ and $T_{\rm eff} = 103600 \pm 5500$ K.

They use these numbers and theoretical white dwarf evolutionary models (there is little alternative for a star that isn't part of a binary system) to estimate a mass of $0.57 \pm 0.02 M_{\odot}$.

At a distance now known to be around  $219 \pm 24$ pc (Harris et al. 2007), this makes the luminosity of the white dwarf around 100 times that of the Sun. A bit of calculation shows that for this gravity and mass, then although the star is small (about $0.04 R_{\odot}$), it is much bigger than most white dwarfs. That is because it is recently formed (planetary nebulae are recently formed objects) and hot; the white dwarf is still contracting towards its final equilibrium size of around $0.015 R_{\odot}$.

The CDS SIMBAD entry for the star is http://simbad.u-strasbg.fr/simbad/sim-id?Ident=WD+2226-210&NbIdent=1&Radius=2&Radius.unit=arcmin&submit=submit+id

It is listed in several catalogues, so goes by various names e.g. 2MASS J22293854-2050136, WD 2226-210.

The most recent (and probably most accurate) distance measurement is from the Gaia DR2 parallax survey, giving $201 \pm 4$ pc ($655 \pm 13$ light years).


Where is the flaw in this hypothetical concept of escaping event horizon using another's blackhole flyby?

In short, it doesn't work.

the event horizon is defined as the region of space which you can't escape from, If you are between two black holes, and you can escape then by definition you are not inside the event horizon. In the case when there are two black holes, the black hole is not a Schwazchild black hole (which is a solution of Einstein's equations for a single point, uncharged non-spinning mass) and the event horizon is not spherical. To determine the shape of the event horizon for two masses you need to use numerical solutions of the equations, you can do this yourself with the Einstein Toolkit, but the computing requirements are very high. A supercomputer is required to get very far.

In the exact situation you draw, it is likely that the release of graviational radiation would lead to a merger rather than a pass but you would need to run the simulation to confirm this.

Now you can't just remain stationary inside the event horizon. If you started inside it you can't stay in the same place and wait for a the pass of the other black hole to pull you out. Inside the event horizon spacetime is flowing towards the singularity and even if the shape of the event horizon is changed by the passing of another black hole, nothing passes from inside to out.


Is there any scientific correlation between cold nights and star-lit skies?

Yes. Clear skies allow the Earth to cool more effectively at night. If there are clouds, these re-radiate some portion of the infrared flux from the Earth back towards the ground, keeping it warmer.


There is no 1:1 mapping between density and composition/structure. You have to look at detailed planetary models. For example, some hot Jupiters are extremely dense ($\geq 10$ g/cm$^3$) but they are undoubtedly gas giants. The origins of this diversity are the source of much speculation and theory, but are certainly within the realms of known physics.

An example of the difficulties can be gleaned from this plot from Lissauer et al. (2014) that shows the mass-radius plane for small planets together with the loci of models for various compositions. Lines of constant density are drawn on the plot too, but notice that a planet of a given composition can have a range of densities depending on its mass.

The situation appears even more strange with giant planets. Generally speaking, more massive giant planets are more dense, which is as expected from basic theory of degenerate gases, but objects of a given mass can have almost an order of magnitude spread in density. Below is a plot I made from exoplanets.org. Almost certainly, the change in the nature of the relationship at around 0.1 Jupiter masses marks the transition from gas giants to ice giants and other rocky-type planets, but notice that gas giants can be just as dense as smaller planets. i.e. A classification based on just density wouldn't work, but mass and density gives more of a clue.


The sun isn't on fire.   Fire is actually very rare in the solar system.   It requires chemical potential energy, which happens on earth because of life.  Photosynthesis uses solar energy to build things out of carbon, oxygen, nitrogen and hydrogen (and some other elements).   It's these carbon chain structures that are ultimately flammable, and pretty much, only in an Oxygen atmosphere.   Oxygen atmosphere is produced by the same photosynthesis.

The sun's heat comes from nuclear fusion which doesn't make smoke.    Also, the nuclear fusion only happens deep the inside of the sun, the outside is just a thick blanket of molecules - mostly hydrogen.   The sun also isn't transparent.  The light we see from the sun is like the light we see from a red hot piece of metal or lava.   Red hot metal doesn't smoke (unless you drop water on it - which is actually steam.)   The sun, like red hot metal or lava, glows bright because of temperature.

In a certain sense, the tail of a comet is kind of like steam being made by the heat of the sun.


Gravity works exactly the same around a black hole as it does everywhere else in the universe:

The mass of the black hole distorts spacetime, and then objects follow the shortest route in the curved spacetime. The only difference between the gravity of a black-hole and the gravity of Earth is that the Black hole has a much larger mass in a much smaller volume, so the gravitational field is much stronger.

The strength of gravity has several consequences I'll mention three:

First close to the black hole spacetime is so curved that light can't escape. Now all gravity bends light a bit, but you usually don't notice it. A black hole bends light a lot.  The edge of the black hole is called an event horizon and it is the closest distance that light can go to a black hole and not fall in. Nothing can ever pass from within the event horizon.

Next there are tidal forces. All gravity can cause tides: the force of the earth gravity is a bit more on your feet than on your head. Usually you don't notice. However a solar mass black hole has such extreme tides that you would be pulled apart by the difference in force (this is the spaghettification you mention)

Finally, inside the black hole, space works like time. You can't change your direction in time (unless you have a Delorean) Inside a black hole you can't change direction in space.

But remember the only difference between the gravity around a black hole and the Earth is a matter of degree.


From where does the energy for gravitational waves come from?

Radiating gravitational waves makes an inspralling binary orbit closer and faster. (Rob Jefferies)

The source of the energy for both increased kinetic energy, and the gravitational radiation is the same: gravitational potential energy. (PM 2Ring)

Two black holes at a distance of 1 light year have a huge amount of potential energy, about 10^48 Joules of potential energy. As they spiral, a significant amount of that energy is radiated as gravitational waves

This is real mass lost. The mass of resulting black hole is smaller than the sum of the two merging black holes, though at no point does any black hole itself become smaller.


Do more massive stars become larger or smaller white dwarfs?

More massive stars have a more massive core and produce more massive white dwarfs. The relationship between the initial mass of the main sequence star and the final mass of the white dwarf is monotonic, but not linear. The Sun is expected to produce a white dwarf with a mass of around $0.5 M_\odot$, whilst a $8M_\odot$ star is expected to produce a carbon/oxygen white dwarf of mass $\sim 1.1 M_\odot$. It is possible that some stars with $8-9M_\odot$ could result in a slightly more massive oxygen/neon/magnesium white dwarf.
The difference between the initial and final masses is accounted for by mass loss in the red giant and asymptotic giant branch phases. Note that white dwarfs close to the Chandrasekhar limit are probably not produced by the normal evolution of a single star and that stars with initial mass $>9M_\odot$ probably end their lives with a supernova of some kind, that either destroys the star or leaves behind a neutron star or black hole.
White dwarf stars are supported by electron degeneracy pressure. This form of pressure, does not depend on temperature (to first order), but does depend on density. To work out how big (in terms of radius) a white dwarf of a given mass is, one uses the density-dependent pressure to solve the hydrostatic equilibrium equation and one finds that for lower densities (which means $<10^{10}$ kg/m$^3$ in this context!), the pressure $P$ is proportional to density $\rho^{5/3}$, and this results in a radius $R$ that is proportional to mass $M^{-1/3}$. That is, a more massive star has a smaller radius. The increased weight of a more massive star requires a bigger pressure gradient to support it and this is provided by a bigger central pressure and smaller radius.
The mathematics of this is a very standard piece of bookwork involving "polytropes" (where the pressure is proportional to density to some power) and I won't produce it. A simple argument is the following. The hydrostatic equilibrium equation is
$$ \frac{dP}{dr} = - \rho g,$$
where $g$ is the gravity due to the mass within radius $r$.
If we work in proportionality, then $P \propto \rho^{5/3} \propto M^{5/3} R^{-5}$. Thus $dP/dr \propto M^{5/3} R^{-6}$.
On the right hand side, $\rho \propto MR^{-3}$  and $g \propto MR^{-2}$.
Thus the hydrostatic equilibrium equation tells us
$$ M^{5/3}R^{-6}\propto M^2 R^{-5}$$
and so
$$ R \propto M^{-1/3}.$$
As a more massive white dwarf gets smaller, the density goes up as $\rho \propto M^2$ and the degenerate electrons become relativistic. This drives the pressure equation towards $P \propto \rho^{4/3}$. This leads the radius to decrease more steeply as a function of mass and it asymptotes to zero radius and infinite density at the Chandrasekhar mass.


Is Earth's 1g solid surface gravity unusually high for exoplanets?

According to this paper, fig. 4, solid planets of arbitrary mass (up to 3,000 times Earth's mass) don't grow much about three- or four-times the Earth in diameter. That's because interior parts of the planet become compressed by the high pressure. Planets heavier than 3,000 times Earth's mass are in the transition zone to stars. In the case of rocky planets they would become substellar objects, similar in their physical properties (electron degeneracy in the core) to white or black dwarfs, not capable to start nuclear fusion due to lack of light elements in their core.

Surface gravity is proportioal to the mean density $\rho$, and to the radius $r$ of the planet: $g=\frac{4\pi}{3}G\rho r$, with $G$ the gravitational constant.

A planet with roughly the composition of Earth, and 3,000 times the mass of Earth would have about 3-times the Earth's radius, hence the 27-fold volume, and the 3000/27 = 111-fold density. Its surface gravity would hence be 333 g. That's close to the upper bound of surface gravity of what would be called a planet. With a pure iron planet we could go a little beyond. Any surface gravity below this upper bound is possible for a solid planet, at least in theory.

According to page 1284 of the paper:

  Massive solid exoplanets of
  hundreds to thousands of Earth masses may be able to form around
  massive stars (B and O stars; 5-120 solar masses) where the protoplanetary
  disk would contain enough heavy elements.

B and O type stars are rare (0.13% of main sequence stars) and short-lived (less than 100 million years; more accurately less than $10^{10}\cdot 18^{-2.5}$ years, for mass > 18 solar masses). Hence large solid planets of the described type will also be rare, although a priori not impossible.

The planet with the highest estimated surface gravity discovered thus far (March 8, 2014) is
CoRoT-Exo-3b (add "surface gravity" column to the exoplanets table, and sort by that column):

  CoRoT-Exo-3b has a radius of 1.01 ± 0.07 R_Jup and transits around its F3-type primary every 4.26 days in a synchronous orbit. Its mass of 21.66 ± 1.0 M_Jup, density of 26.4 ± 5.6 g cm-3, and surface gravity of logg = 4.72 clearly distinguish it from the regular close-in planet population, making it the most intriguing transiting substellar object discovered so far.

A surface gravity of logg = 4.72 means surface gravity equals $10^{4.72}\mbox{ cm}/\mbox{s}^2=52,480 \mbox{ cm}/\mbox{s}^2 = 53.5~g.$
This research has made use of the Exoplanet Orbit Database
and the Exoplanet Data Explorer at exoplanets.org.

CoRoT-3b on Wikipedia.


Is it possible to use Photolithography for telescope image sensor?

The pixel size is not a major constraint on the imaging capabilities of an astronomical telescope. The angular resolution of the image is limited by diffraction at the aperture of the telescope. For a reasonable pixel size, the telescope geometry can then be designed so that each pixel corresponds to a region of sky smaller than that angular resolution, so that making the pixels smaller would gain you very little.
Responding to a comment. A diffraction limited telescope with an aperture of $d$  observing at a wavelength of $\lambda$ has (at best) an angular resolution of $\lambda/d$ radians. So if it has an effective focal length of $l$, the smallest useful pixel size will be around $l\lambda/d$. The effective focal length depends on the whole optical setup and will usually be designed to make this size big enough.


Yes. There is no reason why one wouldn't. But we don't know any exo-dwarf planets (yet). See the answer to this question why.


How to make a 65 cm lens with a 20 cm hole in it for a Hamiltonian telescope?

Drill or core the blank
(anneal if needed ... most likely not if annealed to begin with)
use pitch to glue a plug of the same material in the hole
grind, polish and figure as normal then remove the plug.


How to get ellipticities of galaxies from SDSS using casjobs?

Just found out this method in casjobs.Previously I came across one SQL code in Stackexchange in which you have to input one ObjID to get its stokes parameters and isoA and isoB values.In sample codes in DR7, you can edit the line which gives Feild values and coordinates to get Stokes parameters(type q_r,u_r in their place).

From stokes parametres you get the ellipiticity of the galaxy from this equation (which can be done in python.)

$e  = 1 - \frac{b}{a} = 1 - \frac{1 - \sqrt{Q^2 + U^2}}{1 + \sqrt{Q^2 + U^2}}.
$

Note that if you are downloading the Casjobs query data in CSV or excel format,You need to write python code to read the file and apply the formula
(to learn how to read csv files in python click here)


Did I see the dragon capsule on Friday night?

I have certainly watched the ISS and an undocked Shuttle Orbiter chase each other across the sky. Since they appeared to be of equal brightness, I didn't know which was which. Their angular separation from my POV was approximately 20°, so their actual separation would have been about 150km. It was one of the last Shuttle flights, so I was very lucky to see it.

It is possible that you saw the Dragon capsule, and it would certainly have appeared to be be fainter than the ISS. The easiest way to establish the fact is by using a program like Stellarium. This will show you if the ISS was visible at your location, given the date and time of day.


Are astronomers waiting to see something in an image from a gravitational lens that they've already seen in an adjacent image?

What you do is cross-correlate the observational datasets for the multiple sources and look for the "lag" that maximises the cross-correlation function. Generally speaking, the "events" are not really individual flares or dips, but the summation of all the time variability that is seen.

The variability in question usually comes about from the central portions of the "central engine" of a quasar or active galactic nucleus. For a supermassive black hole at the centre of a quasar, the innermost stable circular orbit is at 3 times the Schwarzschild radius ($= 6GM_{\rm BH}/c^2$). This basically defines the inner edge of any accretion disk and if we divide this by $c$ then we get a a timescale for the most rapid variations in luminosity output. So this is very nearly the same formula as presented in the linked question
$$\tau \sim 3\times 10^{-5} \left(\frac{M_{\rm BH}}{M_{\odot}}\right)\ {\rm sec}\, ,$$
except that the supermassive black holes are much less massive than entire foreground lensing galaxies (usually). This the timescale of variation is much shorter than the potential delay time due to gravitational lensing. It is this difference in timescales that means there is plenty of "structure" within the light curves that can be locked onto by the cross-correlation.

There is a notable example however of a type Ia supernova being seen in a multiply lensed image (Goobar et al. (2017), but the predicted delay in the light curves was $<35$ hours and the light curves are not good enough to measure this. This technique is an active area of research and a major bit of science that is exprected to be achieved by the Large Synoptic Survey Telescope (Huber et al. 2019).

Finally, the thing you are really looking for has happened in terms of SN "Refsdal". This was a type II supernova seen to "go off" in a multiply imaged galaxy, seen through/around a galaxy cluster. A prediction was made, based on a model for the cluster gravitational potential, that another image ought to appear within a year or two. This further image was then detected by Kelly et al. (2016) in a paper entitled "Deja vu all over again".

From Kelly et al. (2016) ("Deja vu all over again"). See "SX" in the third panel:

  Figure 1. Coadded WFC3-IR F125W and F160W exposures of the MACS J1149.5+2223 galaxy-cluster field taken with HST. The top panel shows images acquired in 2011 before the SN appeared in S1–S4 or SX. The middle panel displays images taken on 2015 April 20 when the four images forming the Einstein cross are close to maximum brightness, but no flux is evident at the position of SX. The bottom panel shows images taken on 2015 December 11 which reveal the new image SX of SN Refsdal. Images S1–S3 in the Einstein cross configuration remain visible in the 2015 December 11 coadded image (see Kelly et al. 2015a and Rodney et al. 2015b for analysis of the SN light curve).

  Kelly, P. L., Brammer, G., Selsing, J., et al. 2015a, ApJ, submitted
  (arXiv:1512.09093)

  Rodney, S. A., Strolger, L.-G., Kelly, P. L., et al. 2015b, ApJ, in press
  (arXiv:1512.05734)


Calculate Dec and RA of a star from Euler angles and GPS data

Obtain from the publisher Willmann-Bell the book ''Astronomical Algorithms'' by Jean Meeus. If obtaining elsewhere, be sure to obtain the 2nd ed. with corrections as of August 10, 2009. The equations you want are in Chapter 13, "Transformations of Coordinates".

Some variables must be defined:

$\alpha$ = right ascension, if obtained from formula it is in radians
$\delta$ = declination, positive north, negative south
$h$ = altitude, positive above the horizon, negative below horizon
$A$ = azimuth, measured westward from the South, other sources often measure from the North
$\psi$ = observer's latitude
$H$ = local hour angle
$\theta$ = local sidereal time

The first step is to transform horizon coordinates (azimuth and altitude) to equatorial coordinates (local hour angle and declination).

$$
\tan H  = \frac{\sin A}{\cos A \sin \psi + \tan h  \cos \psi}\\
\sin \delta  = \sin \psi  \sin h  - \cos \psi  \cos h  \cos A
$$

Then the local hour angle H is transformed to right ascension $\alpha$:

$$
\alpha = \theta - H
$$


calculation of characteristic expansion time of the universe

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.


This paper, Quasi-Periodic Behavior of Mini-Disks in Binary Black Holes Approaching Merger, which shares authors with the paper linked in the question, sheds some light on the terminology.

  We present the first magnetohydrodynamic simulation in which a
  circumbinary disk around a relativistic binary black hole feeds mass
  to individual accretion disks ("mini-disks") around each black hole.

They are talking about active binary black hole systems. The system as a whole has an accretion disk, the circumbinary disk, and each black hole has its own accretion disk, which they call the mini-disk.


Is there a term for stars without any planets?

A star is described as such whether or not it has planets.

There doesn't appear to have ever been a need, and to be honest, I can't think of any need for one. When it was assumed (in historical times) that we were on the only planet, stars were not thought to have planets. The title worked then. Now, as we know most stars are likely to have planets as part of normal formation, we can still call them stars.

Typically words are invented as there is a need for them.


Why Helium-burning Stars are found in a Horizontal Branch?

This is explained in the Wikipedia article

  Stars on the horizontal branch all have very similar core masses, following the helium flash. This means that they have very similar luminosities, and on a Hertzsprung–Russell diagram plotted by visual magnitude the branch is horizontal.

To expand a little. In stars of a certain mass range, helium builds up in the core until it reaches a specific mass, at which point the "Helium flash" occurs and burning of helium to carbon and oxygen starts throughout the core. When things settle down, helium burning is going on in the core, which is more or less the same size, independently of the original mass of the star. Since this is the main power source of these stars, they all have about the same luminosity.

The variation across the branch comes from how much remaining gas there is outside the helium-burning shell -- more gas means a larger cooler star radiating the same total amount of energy


A basic orbital rule is Kepler's third law, which states that the closer the moon is to the planet, the quicker is its orbit. A body that is close to the Earth will move in orbit faster than one which is distant.

The fundamental weirdness: If a body is in orbit, and you give it a push in the direction it is moving, it will gain energy, move away from the planet, and slow down. Contrariwise, If you give it a pull back, it will drop to a lower orbit and speed up. This is fundamentally weird because we are used to the idea that if you push something it will speed up, for bodies in orbit the opposite happens. Make sure you understand why this is.

Now suppose we have two moons, a big one and a little one. The big one is in a circular orbit (and is big enough that the orbit won't change) The little one is on the opposite side of the planet, and slightly closer to the planet than the big moon (only slightly closer). The little one is also in a circular orbit, in the same direction as the big moon.

Since the little one is closer to the planet, it orbits faster than the big moon, so will slowly catch up the big moon.

As it catches up the big moon, the pull of the big moon's gravity will pull the little moon forward. Now remember the fundamental weirdness: if the little moon is pulled forward, it will move away from the planet and slow down. The big moon is also pulled, but we'll suppose the big moon is big enough for the gravity of the little moon to be negligible.

Now the little moon has slowed down. It is still going in the same direction as the big moon, but now it is going slower, so it falls behind the big moon. Eventually it falls so far behind that the big moon begins to catch it up.

Now the gravity of the big moon is pulling the little moon back, and because of the fundamental weirdness, this makes the little moon drop towards the planet, and speed up.

The process can now repeat itself, with the little moon repeatedly catching up, then falling back from the big moon. An observer on the Big moon would see the little moon slowly approach then fall back, go behind the planet, approach on the other side, then fall back again. It would appear to move in a horseshoe shape, relative to the big moon. Relative to the planet, both moons move in circles or near circles.

How stable it is depends on all sorts of things: tides, other planets, the effect of the little moon on the big. However such orbit can be stable over long periods of time. But perhaps not over the 5 billion years of the solar system.


Can a high enough density ignite a gas giant into a brown dwarf?

Or did I miss something about hydrogen bombs?

You missed something about hydrogen bombs.
The center of our Sun is the equivalent of a warm compost pile in terms of energy produced per unit volume. A multistage thermonuclear weapon briefly (very briefly) compresses and heats the fusible material in the bomb to conditions far beyond those found in our Sun, let alone Jupiter.


Note: I am self-answering my own question in hope that someone post another answer that beats this one.

Earth is near the inner edge of the Sun's habitable zone. And since the Sun is expected to grow and increase it luminance, Earth might be unhihabitable for any life somewhere between 1 or 3 billion years in the future. So, a planet that have longer time to develop before its parent star moves it beyond the inner edge of the habitable zone is more favorable than Earth.

Since Earth itself always was inside the habitable zone since it formed 4.6 billion years ago and will still be for lets say more 1.4 billion years, with a large uncertanity factor, this gives roughly 6 billion years of time for complex life to develop. Given that it is unlikely to form early due to the time needed for evolution to take place and due to an elevated level of large bollides collisions, we could discount the first 2 billion years from any life bearing planet, including Earth, as unlikely to develop complex life. Further, it is unlikely that complex life would finally evolve out from simpler forms when the planet is already overheated and already crossing the inner edge of the habitability zone, so lets take out the finishing 10% of that period for any planet (probably something more than 10%, but lets keep this as a conservative estimative). So, for Earth, this gives a window of a size of 3.4 billions years to complex life evolve. Similar planets with larger windows have better probabilities.

Stars larger and more luminous than the Sun tends to be more unstable and live shorter. As a result, it is expected that planets around stars larger than the Sun has less time to develop complex life, and thus a shorter time-window. On the other hand, this means that stars smaller and less luminous than the Sun gives a larger time-window to the planets to develop life.

For stars smaller than the Sun (a G-type yellow star), we could consider the K-types (aka, orange dwarf) and the M-types (aka, red dwarf) as specially favorable.
An orange dwarf star may live for 10 to 30 billions years in the main sequence. A red dwarf star may live in the main sequence for trillions of years.

However, planets in the habitable zone of red dwarfs are likely to become tidally lock, and we don't know if this is really that bad or not for life biodiversity. Lets assume that this is really bad, so a planet orbiting an orange dwarf in the habitable zone is likely to have a better habitability than Earth.

Accordingly to this, a planet with two times the mass of the Earth, will have stronger gravity, and thus it is likely to be flatter. Further, it is likely to have a ticker atmosphere that would protect the surface from UV radiation better than Earth. It would be geologically active for a longer time, resulting in more carbon cycling. With the right quantity of water (not a desert nor a global very deep ocean), it might be an archipelago world, since its flatness would not allow the ocean to be very deep nor the continents to be very large. As a result, life would flourish in a number of rich biologically favourable environments significantly larger than Earth. Further it magnetic field is likely to be stronger than Earth's one, protecting the surface from cosmic rays.

As a result, a planet with two Earth masses orbiting an orange dwarf star in the habitable zone has a good chance to be more habitable to life than Earth itself.

Needless to say, near-circular orbits are more favourable than excentric ones, since excentric orbits may make the planet enter in periods of freezing or boiling. However, a reasonably excentricity that periodically changes the environment in a significant manner, but not as too much that it would extinguish non-extremophile life, might give to the planet life a selective pressure needed for developing rapid evolution to face the always changing climate.


Firstly, there is no centre to the universe and the universe seems to continue indefinitely in all directions. It is best to imagine the universe as infinite in size.

Now we need to explain why the CMB appears to be a sphere surrounding us.

Let's imagine, for a moment that I have filled the entire infinite universe with (magic) mud. This mud glows, and because its magic mud, it doesn't change the orbits of the Moon, Earth or anything else in the universe.  But if I hold up my my hand in front of my face I can't see it, because of the mud.

Now then, at one moment all the mud vanishes, everywhere, throughout the whole universe at the same time. One second later, if I hold my hand up I can see it. But if I look in the sky I can't see the moon. The moon is still hidden in mud.  In fact if I look at the sky, all I see is mud at 1 light-second (300,000km) distant. And because it is glowing mud, the sky appears to glow.  It looks like I am in a middle of sphere with radius 300000, surrounded by glowing mud. But this is an effect of the fact that light travels at 300000km/s

After 1.2 seconds, the moon becomes visible, after about 8 minutes, the sun becomes visible too. But it takes 4½ years for the first stars to be visible. At this point it would look as if I'm in the middle of sphere with radius 4.5 light years surrounded by mud.

Now what really happened is that 13.5 billion years ago the whole infinite universe was filled with hot gas that blocked light. As this cooled this became transparent. The whole universe became transparent at the same time, and while it wasn't instantaneous, it happened pretty quickly. So what we see now is that we seem to be in the middle of a sphere of hot gas 13.5 billion light-years across.

Except that in those 13.5 billion years, space has stretched. This has stretched the sphere from 13.5 billion light-years to 45 billion light years, and it has stretched the light from that hot gas, so it appears to be very cold.

The photons that we see in the CMB now did not come from the hot gas that once filled this part of the universe, but from hot gas that was in part of the universe that is now 45 billion light years away. That being so, The photons that we saw last year came from gas that was in part of the universe that is now 44,999,999,999 light years away. Since we can't detect structures as small as few light years at that distance, the CMB appears as a constant even pattern in the sky, changing very little if at all.


It all comes down to the brightness of objects (not their size). For all intents and purposes we can assume that the most distant galaxies and the small, but much closer, objects in the Oort clouds are unresolved point sources.The Oort cloud objects are too faint to see, with JWST, but it should be able to see bright galaxies and quasars even at 13 billion light years.

JWST will have a magnitude limit of something like 30th AB magnitude at red and near-infrared wavelengths. It turns out that quasars at Z~10 can have brightnesses that enable them to be detected at these magnitudes and brighter.

OK, but now work out how bright an Oort cloud object is seen in reflected light from the Sun. Taking a solar luminosity and calculating the flux at say the nearest hypothesised Oort cloud objects at about 2000 au from the Sun, we get  $1.4\times 10^3/(2000^2) = 3.5\times10^{-4}$ W/m$^2$. Now let's be generous and say that as much as 10 per cent of this is reflected (a very high albedo) and that it is a big Oort object with a radius of 20 km. If this acts as a Lambertian reflector, then it will receive $3.5\times 10^{-4} \times \pi \times 20000^2 = 4.4\times 10^{5}$ W and it will radiate $4.4\times 10^{4}$ W isotropically back into a hemisphere towards the inner solar system.

Now reversing the calculation. After this light has travelled back 2000 au to the JWST, the flux received will be $4.4 \times 10^4 /(2\pi [2000 \times 1.5\times 10^{11}]^2) = 4.9 \times 10^{-25}$ W/m$^2$.

We can compare this with the original flux of the Sun at the Earth. This is $1.4\times 10^{3}$ W/m$^2$. Now assuming that the reflection doesn't change the solar spectrum, we can take the ratio of the solar flux at earth to the reflected light from the Oort cloud object to calculate that the Oort cloud object is 68.6 magnitudes fainter than the Sun as seen from the Earth (or close to the Earth - JWST will not be in orbit around the Earth) in any magnitude band you care to choose.

So in the near infrared, the Sun has an apparent magnitude of around -27.5. This means that the Oort cloud object would have a near infrared apparent magnitude of 41 and would thus be 11 magnitudes too faint to be seen by JWST. (And remember we chose a big object that was as close as an Oort cloud object is expected it to be and gave it a high albedo),

Now consider that 30th mag detection threshold. For a solar-type spectrum this equates to a flux threshold of $\sim 10^{-20}$ W/m$^2$. Ignoring the cosmological nuances of proper distance etc., then something at 13 billion light years would have to have a luminosity of a mere $10^7$ solar luminosities to produce this kind of flux - the brightness of a modest galaxy. Of course there is also the extreme redshift to consider, so this calculation is way too optimistic, but even allowing for an order of magnitude or two, one should easily be able to see the brightest galaxies and quasars at 30th magnitude.


Here is 13 hours 48 minutes.
That's a mighty early sighying:
https://www.universetoday.com/107700/ultra-thin-young-crescent-moon-sighted-from-u-s-southwest/
11 hours 40 minutes was the world record as of 2014.


Basically, if two particles are placed with no other interaction between them, the distance between them will increase.

Imagine living on the surface of a balloon which is being blown up. Your size stays fixed, because you're more or less rigid, but items not attached to you will move further away. Your ruler, another rigid body, stays fixed in size (though it may bend to accommodate the new curvature — this isn't so important). But two rulers (which are not attached to each other) move further away.

  ...also, if everything, including all our measurement devices expands at the same speed, how can we determine the fact it's expanding? :D

On the outset this seems true, however there are other forces at play here. Our measurement devices are held together by electromagnetic interactions, and the strength of these will not change. So the measurement device will hold itself together.

Imagine two faraway atoms. When space expands, the distance between the two atoms increases. However, the size1 of the atom does not — this is determined by electrostatic equilibrium (and quantum mechanical considerations), and this remains unaffected. Even if the atom was stretched, it would rebound.

This scales up to measurement devices, so they don't get distorted either. Indeed, the expansion of space only really makes sense when you look at galaxies — these are pretty far away (when not in the same supercluster) and they don't have any interactions maintaining an equilibrium distance between them.

1. Whatever closest analog we have to "size" for atoms; eg the area which contains 99% of the charge density; or the nth Bohr radius.


How do we know a star's age based on its spectrum?

There are two separate issues when you talk about a star's age.  Are you talking about its absolute age, in years, or its age in terms of how far along is it within its own evolution?  The difference is, massive stars go through their evolutionary phases very quickly, and low-mass stars very slowly, so essentially all very low-mass stars are older in years than essentially every very high-mass star.   Also, you can know more than the "spectral type" from a spectrum, you can also know the "luminosity class"-- that's the roman numeral that tells you if the star is, for example, a main-sequence dwarf or not.  So if you pick an M spectral type main-sequence star, and an O spectral type main-sequence star, as two stars at random, it is extremely likely that the M star is much older than the O star, and you can know that purely from the spectral type and luminosity class.  But if you want to know the age of the M star, or the O star, in years within their evolutionary stages, then it gets more difficult because there is rather little main-sequence evolution in the spectrum.  There is some evolution though, especially late in the main sequence, so you could tell its age if there is evidence of main-sequence evolution in the spectrum (for one thing, the luminosity rises with age, within a spectral type, but that's pretty hard to tell from the spectrum).  If the star is both low-mass and early in its main-sequence evolution, the best way to tell its age is to look at the rotational broadening of its lines, as low-mass stars spin down rapidly as they age (due to strong magnetic fields and weak winds).

Another good way to tell the age of stars is if they are in clusters that all formed together, you can look at what is called the "main sequence turnoff", which basically means you look for the lowest mass star that has gone supernova, and you know how long that takes for that mass of star.  You tell this from the spectrum because the turnoff is in the "Hertsprung-Russell" diagram, which you make from the spectra, but it requires a lot of stars in the cluster, so it's not really a property of a single spectrum.

So the bottom line is, you need a suite of different techniques for assessing the ages of stars, it is not always easy or straightforward but you can take each case individually and look for ways to infer age.  Much easier is determining which evolutionary class a star is in, because stars change dramatically in one phase or another, and the spectrum alone can tell you if you have a dwarf or a giant, etc.  (Basically, the lines get much narrower as a star puffs out in radius, because the pressure drops.)