Determining latitude and longitude in bad weather In Sobel's famous book 'Longitude', she described the story of Harrison. I still have one question. What if it is cloudy on the sea? In this case, the sailors have nothing to compare with, right? This still means in many many cases, they have no means to determine their position, neither latitude nor longitude, right?

It seems that the problem is not so satisfactorily solved.

Sunstones are believed to have been used by vikings to determine the direction to the Sun on cloudy days. That helps you point the ship in the right direction, and by experience you can estimate the side ways drift in the water. But the longitude will still be very uncertain without precision time keeping. The news paper article I linked to says that a Sunstone was find at a wreckage in the English Channel from as late as 1592.


Is there a common metric to measure telescope power? It is rather hard for me to compare telescopes due to the various numbers used.
600mm focal length and a 50mm diameter and eyepiece is a 12mm.
VS.
Focal length of 500 mm and the objective diameter is about 90 mm, Using a 3x Barlow Lens.
Is there something like an APR for telescopes?  Is there any objective rating system?  Most of the review videos just sound like sales pitches.
GC_

Is there a common metric to measure telescope power?... Is there something like an APR for telescopes?... Is there any objective rating system?

If there was one single metric, it would be aperture, but there really isn't because there are several other aspects that can make one telescope much better suited for viewing than the other.

The diameter is most of the power of a telescope, dim stars always appear brighter, doubling the diameter of the objective results in 4x the area to collect light, so the apparent magnitude of a point object like a star goes up by 1.5 or so.  Also a greater diameter increases the detail visible to the point where you can use more magnification, or at least make the same view much sharper.
The ratio of the focal length to the diameter is the "focal ratio", for instance a short focus reflector with an 8" mirror and a 32" focal length has an f4 ratio, and a refractor with a 3" lens and a 36" focal length has a f12 ratio.
Shorter focal ratios are generally better for observing extended objects such as galaxies and nebulae since the available light is concentrated into a smaller angular size for the image, whereas long focal ratios are better for observing detail such as planetary observations and splitting close double stars because a given eyepiece will result in greater magnification. I take it you're a novice at observing astronomical objects, so I'd recommend dipping your toe in the water by buying a relatively small middle range focal length scope for general observing without breaking the bank.  Oh, and avoid those department store telescopes which are shaky and don't have good optics. Claims of 700x magnification don't do anything but make the wind blown blur more apparent.


How would Special relativity explain past-present-future if thinking about 13-billion-light-years distanced objects Halo.

I'm interested in how would Special relativity comment on this:

Suppose there are 2 objects, A and B. They are 13 billion light years away from each other.

When the light from the B object reaches the A, the observer at the A object sees what the (suppose planet, people, objects) were like 13 billion years ago on that object. Let's say it is also an Earth-like planet where the living creatures like exist.

And if the observer at the A object would like to imagine, what is now happening there, what would be the result:

Surely, he would have to wait another 13 billion years to actually
see
Would that mean that the inhabitants (if any) or just matter and
objects at object B exist in future relative to A ones. And vice
versa? If yes, then they both live in present, past and future
tenses at the same time?

Or have I completely lost my way of properly understanding the Special relativity?

First off, if Earth were point B, and you were an observer at point A looking at it with the most magnificent telescope ever imagined, you would still not see the Earth, because it didn't exist 13 billion years ago.  I assume you picked 13 billion years because it is roughly the age of the universe, so you'd see the universe as it existed then, but that doesn't get you what I think you want.  I will instead pick 4.5 billion years ago (or objects A and B 4.5 billion light years apart), because then you will have an actual Sun and Earth to look at but long before life evolved.

We'll also wave our arms over the tracking software of your telescope being able to locate Sun and Earth in their orbits over the course of 4.5 billion years as seen from point B.

Yes.  You would have to wait 4.5 billion years to see the observer at point A looking back at you through his telescope.  You would now be looking at a cooling hunk of molten rock with no moon or oceans.
No.  If you waited 4.5 billion years and saw Observer A, he would have been dead for 4.5 billion years by the time you see him.  In fact, his Sun would likely be expanding to the point of burning his planet to a cinder if not engulfing it entirely.  But of course, you would have to wait another 4.5 billion years to see that.  You would be observing the photons that bounced off Observer A all that time ago and have been traveling ever since.  It is important to note that this is not him, and does not mean he still exists.

Hopefully this answers your question, though it doesn't have much to do with Special Relativity, it's still good blow-your-mind stuff.  It's just speed of light stuff which we really knew and measured (albeit inaccurately) long before Einstein.


Do Satelites travel in train like formation? This morning approx 5.30am i seen roughly 12 travelling in perfect formation approx 15cm apart , im  200km south of sydney Australia travelling from a north west direction

Yes they can. It depends on how the satellites are deployed from the launch vehicle.


Why do astronomers like green laser pointers? When I've been to Astronomy-nights for the public at the local observatory, they always used green lasers to point things out.

Why is there such a preference for green lasers?

What's wrong with red lasers? (Which seem cheaper and more readily available.)

Mostly general purpose laser pointers are used for pointing things at smaller distances eg. Diagram or Equations in Powerpoint presentations, so the power of such laser pointers is quite limited/restricted to 5mw ( Class 3A or IIIa) or 10mW in some regions.
Because of the low power and small aperture of laser pointers if you point them through empty space, you can only see them impact spot where it hits a surface.

So to point stars/planets as in astronomy club during observation sessions these low power laser pointers are not useful as there is no surface where the beam can hit & you can see the spot.

Some higher-powered laser pointers project a visible beam via scattering from dust particles or water droplets along the beam path.
Higher-power ( Class 3B or IIIb lasers: greater than 5mW) and higher-frequency (Green or Blue) lasers may produce a beam visible even in clean air because of "Rayleigh Scattering" from air molecules, especially when viewed in moderately-to-dimly lit conditions. The strong wavelength dependence of the scattering (~λ^ −4 ) means that shorter (green & blue) wavelengths are scattered more strongly than longer (red) wavelengths.

[ Rayleigh scattering is the mainly the elastic scattering (in lay term collision) of light or other electromagnetic radiation by particles much smaller than the wavelength of the light.
Rayleigh scattering results from the electric polarizability of the particles (locally separating positive & negative charges by small distance apart inside the atom. i.e. temporarily separating centre of positive charge and centre of negative charge). The oscillating electric field of a light wave acts on the charges within a particle, causing them to move at the same frequency (hence the green colour). The particle therefore becomes a small radiating dipole (or small sources of light) whose radiation we see as scattered light (the beam passing from pointer to star pointed). ]

The intensity of such scattering increases when these beams are viewed from angles near the beam axis. (That's the reason, if you are standing close to the person pointing stars in astronomy club, you can see the beam bright and clearly whereas the people standing away can barely see it. So always stand near the presenter But avoid contact with your eye. :-) )

The apparent brightness of a spot from a laser beam depends on the optical power of the laser, the reflectivity of the surface, and the chromatic response of the human eye. For the same optical power, Green Laser light will seem brighter than other colours because the human eye is most sensitive at low light levels in the green region of the spectrum (wavelength 520–570 nm). The most sensitive pigment, rhodopsin, has a peak response at 500 nm. Sensitivity decreases for redder or bluer wavelengths.

Further, Green laser pointers are of moderate power (sufficient for Rayleigh  Scattering in clean air), they are compact and relatively cheaper than blue ones (even though human eye is less sensitive to blue colour, blue laser are a little bit more expensive than green one.)

Avoid their contact with your eyes and don't point them at aircraft.
People have been given up to five years in jail for aiming a green laser at an aircraft.


What is the density of stars in our region of the galaxy, up to say, 1000 light years distant? If we imagine a sphere of radius one light-year centered on Earth, it contains only one star - our Sun.

If we imagine a sphere of radius ten light-years centered on Earth, it contains about 10 stars, according to this wikipedia page.

If we imagine a sphere of radius one hundred light-years centered on Earth, about how many stars would it contain?

If we imagine a sphere of radius one thousand light-years centered on Earth, about how many stars would it contain?

Here's a crude way of estimating the answers:

The density of stars in the Milky Way disk decreases exponentially as you go out in radius and also exponentially as you go out in height above/below the disk midplane. The corresponding scale lengths are about 3 kpc (3 kiloparsecs = about 10,000 ly) radially and about 300 parsecs (about 1,000 ly) vertically ("scale height"); this means that if you go out in radius by about 10,000 ly, or up/down in height by about 1,000 ly, the density decreases by a factor of e (2.718...). Changing radius by about one tenth of the scale length or less, or changing height by about one tenth of the scale height or less, only changes the local density by about 10%, which isn't enough to worry about. That corresponds to about 300 parsecs (1000 ly) in radius, or about 100 ly in height.

So the for case of a 100-ly radius, you can assume the density everywhere is about the same as the local density suggested by the table you linked to. That density is 60 stars / ($(4/3) \, \pi \, 16.3^3$) = 0.0033 stars per cubic ly. For a sphere of radius 100 ly, you would then expect about 14,000 stars.

Naively, a sphere with a radius of 1000 ly would then have about 14 million stars. But a sphere that big extends above and below the disk midplane enough so that it samples about 1 scale height. So you can't just re-use the local density. Without getting into the proper approach of integrating the density functions in 3D, we can make a crude estimate as follows: If we used the density at one scale height (1/e times the local density), then the sphere would have only about 4.5 million stars. If we split the difference, that's about 10 million stars, which is probably not too terribly wrong. (I've been ignoring things like stars belonging to the "thick disk" and stars belonging to the halo, but they won't contribute a whole lot; also, the Sun isn't exactly at the disk midplane, but about 60 or 70 ly above it...)


Can gravity be amplified by other celestial bodies in its vicinity? I`m wondering if the gravitational pull from earth that should be constant around the planet, will get amplified by a celestial body like the moon orbiting the earth?

Gravity itself cannot be amplified but the forces are additive. The earth is held in orbit by the Sun's gravity but earth is also affected by the moon's gravity. Depending on their position the moon's gravity could be aligned with Sun's gravity, increasing the gravitational pull. Or the moon could reduce the effect of suns gravity if on opposite side of the earth relative to the sun.


How bright is Comet PANSTARRS C/2017 S3 now? The LiveScience article Incredible Hulk? Nah, This Glowing, Green Light in the Night Sky Is a Comet published just hours ago discusses Comet PANSTARRS C/2017 S3 and mentions:

  bright bursts exploding from its surface twice in close succession — first on June 30, and then again about two weeks later, Sky and Telescope reported. As Hulk himself might say, "Comet flash!"

The linked Sky and Telescope article PanSTARRS Comet, Rocked by Outburst, Goes Green is dated July 17, 2018 and offers the tantallizing information that

  Comet PanSTARRS (C/2017 S3) has erupted again! Now bright enough to see in binoculars, it might become a naked-eye object if it survives until perihelion.

Question: How bright is the comet now, and how is it predicted to behave in the next week or two (baring further outbursts)?

below: "On July 15, Comet PanSTARRS (C/2017 S3) appeared bright and well condensed with a narrow ion tail." Photo Credit: Michael Jäger. From here.

update: The Universe Today's article Catch Comet C/2017 S3 PANSTARRS In Outburst links to S Seiichi Yoshida's page for the comet, but there's no recent data!

The solar elongation (angle from the Sun to the object) is about 15 degrees right now and decreasing as the comet moves in towards perihelion (closest approach to the Sun) on August 15. As James K says this means that comet is lost in the Sun's glare. Anything below 30 degrees solar elongation is tricky to impossible to see.

Assuming it survives the close perihelion passage of 0.21 AU, it is predicted to enter the field of view of the SOHO LASCO Sun-observing coronographs on August 24:

  Assuming there is anything left of C/2017 S3, perihelion will occur this month on the 15th at a rather small perihelion distance of 0.21 AU. Being only 2 weeks from perihelion, it is already at a small elongation of 31 degrees on the 1st. Most observers will lose sight of it during the first week of August as its elongation drops below 25 degrees on the 4th and 20 degrees on the 6th. If the comet is still a going concern it will be visible in the SOHO LASCO C3 FOV between August 24 and September 13. The comet appear around the 7:30 position, passes to the east (left) of the Sun (halfway between the Sun and edge of FOV) and exits the FOV around the 12:30 position.

The above info is from the ALPO Comet Section. Latest LASCO image and movies are available here and many other places.


Why do these astronomers say that "Carbon chains with more than nine atoms are unstable"? The Phys.org article Interstellar iron isn't missing, it's just hiding in plain sight links to the paywalled article On the Structure, Magnetic Properties, and Infrared Spectra of Iron Pseudocarbynes in the Interstellar Medium (Tarakeshwar, Buseck and Timmes, Arizona State University), and says:

The new work may also solve another longstanding puzzle. Carbon chains with more than nine atoms are unstable, the team explains. Yet observations have detected more complex carbon molecules in interstellar space. How nature builds these complex carbon molecules from simpler carbon molecules has been a mystery for many years.
Buseck explained, "Longer carbon chains are stablized by the addition of iron clusters." This opens a new pathway for building more complex molecules in space, such as polyaromatic hydrocarbons, of which naphthalene is a familiar example, being the main ingredient in mothballs.
Said Timmes, "Our work provides new insights into bridging the yawning gap between molecules containing nine or fewer carbon atoms and complex molecules such as C60 buckminsterfullerene, better known as 'buckyballs.'"

Question: On Earth we don't see a big difference between hydrocarbon chains with lengths below and above 9 (think kerosene, wax...), why is there such a cutoff in stability in interstellar space? What is it about hydrocarbon chains longer than 9 atoms that makes them unstable there but not here?

Unlike the saturated hydrocarbons in kerosene, carbynes are unsaturated carbon chains with alternating single and triple bonds.
Molecules containing such chains are called polyynes, e.g. the short cyanopolyyne HC5N:

H−C≡C−C≡C−C≡N

Those carbon atoms readily interact, and long chains (if they form; see comments) are more likely to crosslink or form cycles than to remain linear.
Tarakeshwar et al. suggest that iron clusters in "pseudocarbynes" inhibit this by bonding to some of the carbons.

Loomis et al. 2016 looked for cyanopolyynes in radio spectra of Taurus molecular cloud 1.
They got a good HC9N signal but did not detect the HC11N lines Travers et al. 1996 observed in the lab.


How to calculate the statistically significant level for a guassian line profile? In some papers, they fit a line profile with gaussian model. For example, in
this paper, fig 1 draw three gauss line profile, O VII, O VIII, and Ne IX.
But the data seems noisy and the gaussian height (from bottom to the top of the gauss line, especially O VIII, and Ne IX) do not seems above 3*(the height of 1-sigma errorbar) of the residual data. But in the text they say (on the left bottom of page 3): "The centroid wavelengths of three lines are consistent with O VII, O VIII, and Ne IX at zero redshift,respectively. Those lines are statistically significant at the 6.5, 3.1, and 4.5 sigma levels.". So i'm puzzled with the calculation algorithom of the significance. In my memory that the sigma=sqrt(variance of the residual data) and the height of gauss Intensity=exp(a*(x-b)/c^2), the significance = a/sigma. Then any wrong with my calculation?

It is difficult to say, there isn't a description in the paper as to how they arrive at those numbers. Given that the equivalent widths of the lines have signal to noise levels that mean they are only greater than zero at significance levels of 2-3, then it isn't clear these are meaningful numbers at all.
However, here is how it might have been done.
You fit a continuum model to the spectrum, get a minimum chi-squared $\chi^2_1$ (or maximum [log] likelihood value); you then fit a continuum plus absorption line model and get a new, lower value, of chi-squared $\chi^2_2$ (or higher value of max log likelihood)
You then get the difference in chi-squared $\Delta \chi^2 = \chi^2_1 - \chi^2_2$ (or form the likelihood ratio) of the two models, which should be distributed as chi-squared with the number of degrees of freedom being equal to the number of new parameters in the more advanced model (i.e. 3 for a line centre, width and depth). You then check a chi-squared table for the corresponding p-value and significance.
In other words, if $\Delta \chi^2 = 6.25$, that corresponds to p=0.1, or if $\Delta \chi^2 = 11.35$ for p=0.01 etc.
In your comments you say:

First i use powerlaw to fit the continue get reduced chi-square1 = 1.00277 / dof=179. Then i fit with powerlaw+gauss, the reduce chi-square2 = 0.88091 / dof=176. The Δχ2 = 24.455. Is that right?

Yes. A chisquared test for 3 d.o.f. yields p=0.00002012
To convert to a number of Gaussian sigma, it is the stuff of standard statistics that
$$ n\sigma = \sqrt{2} {\rm erf}^{-1} (1 - p)$$
which in the case above suggests a significance of $4.26\sigma$.


Puzzle about proper distance So I tried deriving the radial co-moving coordinate formula to an observable galaxy for which the light observed at time  $t_o$ was emitted at time $t_{em}$. i.e  $r_0=\int_{t_{em}}^{to} c/a_{(t)} dt$

I tried deriving it without thinking about the Robertson-Walker metric, but with the relative velocity of light in space (the photon being like a boat on a water stream)

I think I might've made a conceptual error, and I hope that by posting my solution here someone will enlighten me.

In the above pictures $\tau$ is a chosen time with the property $t_o > \tau > t_{em}$; $v_{space}$ is the velocity of "space" (universe expansion) at the location of the photon (as you might've noticed, the relative velocity of the photon is given by the speed of light minus the speed of space); $r'$ is the "current" co-moving coordinate (at time $\tau$) while $r$ is the "old" (at emission) co-moving coordinate of the photon.

I'm sorry in advance if something is unclear. Feel free to ask for more clarification.

You can derive the comoving distance formula without the Robertson-Walker metric, but not using "the photon being like a boat on a water stream". You need to use that the speed of light is constant as peculiar velocity of the photon.
So let $R(t) = a(t) \, r(t)$ be the proper distance. The rate of change is $$\begin{aligned}R'(t) &= a'(t) \, r(t) + a(t) \, r'(t) \\ &= v_{\rm rec} + v_{\rm pec}\end{aligned}$$ where $v_{\rm rec} = a'(r) \, r(t)$ is the recession velocity, and $v_{\rm pec} = a(t) \, r'(t) = \pm c$ is the peculiar velocity. The formula follows.


Dark Flow: statistical limits on existence Dark Flow has been thrown around as a term a lot lately, especially with the latest Planck results.

New Scientist reported last year that it was now statistically improbable, but the paper by Atrio Barandela mentioned in the article places doubt on the statistical significance of the dipole measurement by the Planck team.

I'd like to know what observational limits lie on it's existence as a physical phenomena (giant attractor/multiverse/new region of spacetime/other weirdness) and not just a statistical anomaly in the overall matter distribution, I.e. what future observations could prove or disprove this theory?

The dipole in the microwave background indicates motion of the Milky Way and thus of the whole Local Group, at least, at about 600 km/s in a certain direction.  The straightforward explanation is that the density irregularities nearby from superclusters and voids result in a net gravitational acceleration that, over the age of the universe, resulted in this velocity.  A number of superclusters (Hydra-Centaurus, Shapley, and Norma) happen to line up along this direction and reasonably fit the picture.  Also there are some big voids in the opposite direction.
There are two basic ways to quantitatively understand what is the cause.  One can make measurement of local flows around these superclusters and voids to determine their masses and thus their net effect on the MW velocity.  Or one can try to measure the "dipole motion" on larger scales, that is at what size scale does this motion drop to 300 km/s and what scale does it drop to near 0?  A number of groups have tried to do this and the results are in disagreement.  In particular, one group found no drop off in motion out to fairly large distances.  Therefore, an alternate explanation was proposed that the dipole motion is primordial and due to objects beyond the horizon that interacted with us before inflation.

Will we ever settle this?  I am sure that we will eventually.  There is much work to do measuring the distances of galaxies out to 100 Mpc or so with greater and greater accuracy using new methods (SN Type 1a, surface brightness fluctuations, Tip of the Red Giant Branch) and larger telescopes with better instrumentation and then modeling the peculiar flow on all scales.  If the local distribution of matter cannot explain fully the dipole moment in the MWB, then perhaps Dark Flow from before inflation will be necessary to explain it.


How big is one arcsecond at various distances? I've been reading this article
https://en.wikipedia.org/wiki/Celestial_coordinate_system
and a question came to my mind.

The minimum units are seconds ( ″ ). I picture an angle of 1 second as being like a "cone" going out from Earth. It's really small at the beginning but as you go further its width increases.

How wide does this cone get as you get further into the universe? How big is an arcsecond on different astronomical bodies?

How big is one arcsecond at various distances?

An arcsecond is a small angle, 1/3600 of a degree or about 5 millionths of a radian ($4.85\times10^{-6}$). To estimate the size of something that appears 1 arcsecond across you can use the small angle approximation to trigonometry: Multiply the distance to the object by $4.85\times10^{-6}$

Examples:

One arcsecond on the moon is 1.87km
One arcsecond on the sun is 727km
One arcsecond on Mars (when it is closest to Earth) is 237km
One arcsecond at distance of one parsec is one astronomical unit (AU), by definition
One arcsecond on Alpha Centauri is 200 million km
One arcsecond on the Andromeda galaxy is 100 trillion km

The atmosphere limits how much detail you can see. Typically the smallest detail you can see is about 3 or 4 arcseconds, though professional telescopes do better by being built on the top of mountains and using various tricks, like adaptive optics.


Do we know the availability of any material suitable for solar cells on the moon? Disclaimer: I have no formal training in any scientific field.

Have we located any material suitable for solar panels already on the moon? Or does it appear that they would have be imported?

Welcome to Astronomy.SE!

The moon, like the earth, consists of 25% silicon by mass. Silicon is one of the main elements used in producing solar cells.

So sourcing the material is not the problem. The manufacturing is.

I assume your interest in this matter doesn't just stop at finding the materials, but also includes in-situ manufacture. I won't go for a detailed answer, but I will outline a few considerations.

You need your own silicon industry. It's the sort of thing that requires a large infrastructure behind it. For instance, monocrystalline silicon solar cells require high purity silicon (on the order of a few ppm) and pristine growing conditions for the crystal to grow. That means you need to have the infrastructure in place to mine moon rock, reduce the silica to silicon, and separate out as much impurity as you can. With the sorts of purity levels we're talking about, you'd probably need at the very least a medium-sized moon base already in place.
Making any sort of solar panel is hard. You need an industry to make the silicon. You need another industry to make the solar panels. This sort of stuff requires precision engineering, and complex (and heavy) tooling which you'll have to import from earth. You need a 'pyramid' of support structures and expertise to maintain such expensive equipment.
Amorphous SI is probably your best bet. Amorphous silicon seems to be your best bet. It can be applied through some form of vapour deposition, and from a brief bit of reading around, appears to be one of the easiest ways of making solar cells, though the efficiencies tend to be low, with the best reported efficiency being 14% and realistically achievable efficiencies being likely much lower than 10%. (see the bottom green line of this diagram)


Question about the Figure-eight periodic planar three-body motion If this is a posible periodic solution for a three-body problem, can anybody tell me where is the center of mass of the system? As we know, in a n-body problem the bodies orbit around their center of mass, but here I don't know where it is. Can anybody help me? Thanks.

There may be some three-body periodic solutions that orbit around a common point, but in general they get a little crazy-looking and may not always contain an immediately obvious center of mass from casual observation. But of course it will always exist.

If you watch closely, you'll see that whenever one body reaches the intersection point, all three are on a straight line with the other two equidistant (symmetric). If you look where this orbit was originally described mathematically, they probably defined the initial conditions with the green one in the middle - so that the y coordinates of all there were zero and the line at that point is horizontal.

At those moments (all six permutations are present) the center of mass is at the intersection point since the masses are equal.

Since there are no external forces, the center of mass doesn't change in between those moments, so the center of mass is always at the intersection point.


Why is the Earth's magnetic field stronger than its neighbors? [Edited] I've learned that the Earth's core is hot due to decay of radioactive elements, causing the liquid part of the core to stay liquid. This was stated as an explanation to why the Earth has a strong magnetic field.

Assuming the rocky planets were formed roughly the same way (planetoid collisions etc.), why was Earth the "lucky winner" of abundant radioactive elements, while Mars & Venus were left with much less, causing their cores to solidify, their magnetic fields to weaken, their water blown into space, and their fate sealed to become arid desert and boiling inferno, respectively?

I've learned that the Earth's core is hot due to decay of radioactive elements.

This is unproven, non-standard geophysics. There are several arguments against this. One is that all of the long-lived radioactive isotopes are isotopes of uranium (two isotopes, 235U and 238U), thorium (232Th), and potassium (40K). The problem: Uranium, thorium, and potassium are strongly lithophilic ("rock-loving") elements. These elements dissolve very nicely in molten rock, but not so much in molten metal. The presence of long-lived radioactive isotopes is enhanced in the Earth's crust, slightly depleted in the Earth's mantle, and by all rights should be strongly depleted in the Earth's core.

Another problem is that any significant amounts of uranium and thorium in the Earth's core have been ruled out due to neutrino detectors. Potassium-40 has not been ruled out because the neutrinos from 40K decay are too low in energy to detect, but that brings us back to problem #1.

The only hope for this conjecture is that potassium somehow becomes siderophilic at high pressure. There are some experimental results, most of which are highly controversial, that this might be the case.

Yet another problem is the conjecture of high radioactivity in the Earth's core was motivated by explaining the Earth's magnetic field. A number of recent papers say that there is zero reason for this motivation. The Earth's magnetic field is fully explainable without resorting to the chemically unsupported hypothesis that 40K somehow becomes a siderophile at high pressure.


Which worlds have unique words for a "day"? On Mars, we refer to a "day" as a "sol". NASA has developed a whole vocabulary around this term, like "yestersol" and "tosol", etc.

Do any other worlds have unique terms like this? How would one refer to a day on Venus, for example, or Mercury? Is "sol" used to avoid confusion with "day", like on Mars, or is "sol" an exclusively Martian term and "day" should be used for any other world until a unique term is agreed upon?

Word are created in response to need. There are only two planets on which we have any significant presence on the surface: Earth and Mars.

When running operations on Mars, the day/night cycle is significant, so a word was needed for the length of this cycle. Especially since the sol is so close to the day, there is greater potential for confusion. Other planets either have very long rotation periods or no solid surface.

On, for example, Mercury you can use "day" until there is a need for a new term.


Is there any hard evidence that rogue planets exist? A rogue planet (or a rogue anything, for that matter: a celestial body other than a star) is something that's drifting through space without being attached by gravity to any star. They're just out there in the big black void. Is there any evidence that they actually exist? Or are they purely hypothetical?

Rogue planets have been discovered by infra-red imaging because planets are hot when they form.  Here's a list of a few.

There's 2 types of Rogue planets.  One is failed stars.  condensing pockets of gas and dust that form similar to how our solar-system formed, but that are too small to form stars.   Source.

The 2nd type is planets that escape from a star's orbit.   This can happen by the star ejecting material and losing mass which causes the planets to expand in more distant orbits, some eventually escaping, or by gravitational assist either planet on planet or by two stars passing quite close to each other.    It's statistically impossible for Rogue Planets not to, from time to time, get ejected from solar systems, so they have to exist at least in reasonably large numbers throughout the galaxy, though I don't think it's well known how common there are.

Our solar-system might have ejected a planet - see here.   In general, larger planets can eject smaller ones but mostly not the other way around, but two stars that pass too close to each other can eject any planets, mostly ones with more distant orbits.  A star that loses a lot of it's mass or a star that goes nova can push planets out of it's orbit too.


Redshift quantization A full disclosure to begin with: I'm a PhD student in mathematics and while I understand most of standard cosmological-astronomical terms and I've followed a one semester course on cosmology, I don't really have a deep comprehension of the well known material neither on cosmology or astronomy.

I've recently heard about the possibility of a "redshift quantization". As I understood this is not a mainstream result, which means that as 2007 was still quite debated. Anyway I couldn't understand if it really was ruled out by some more recent observations or if it's a still ongoing idea. But to be more specific what I really don't understand are the implications of this result and if this result would invalidate some current cosmological theory and why. I've heard it might have some implication on Quasar formation and change the current timeline of the Big-bang but didn't get how nor why.

Can anybody give me some hints?

There's a pretty comprehensive wikipedia page on the subject. In summary....

Throughout the 2000's conflicting results were published using the same data. The conflict seems to center on the selection effects of the surveys, and the corrections applied to account for them. I did a quick search for 'redshift quantisation' on the Arxiv and found no papers since 2009, which suggests the supporting data has been comprehensibly discredited, or explained away by natural mechanisms.

There are two explanations for the quantization that do not require exotic cosmologies. The first is the large scale clustering of galaxies that traces the cosmic web. The walls and filaments formed in this web leads to groupings of galaxies around specific redshifts along the line of sight. The second explanation is the quantisation of emission from Hydrogen, known as the Lyman series. Since photons can only be released with certain energies, the measured redshifts are also quantised (this is only true in low pressure and temperature environments, otherwise pressure and thermal broadening occur).


What are the sources of antimatter particles in the magnetosphere? The PAMELA instrument detected antimatter particles in the Earth's magnetosphere.  What are the sources of these particles?

The answer to Source of high energy cosmic particles outlines the sources of cosmic rays in general, does it still hold true for antimatter particles?

Matter and antimatter particles are always produced as a pair. This particle physics' process is called pair production. If antimatter is detectable, it means that the
antiparticle of a matter-antimatter-particle-pair was able to survive long enough to be detected, for instance due to relativistic effects, as explained e.g. by the Hyperphysics page on muons:

The measurement of the flux of muons at the Earth's surface produced an early dilemma because many more are detected than would be expected, based on their short half-life of 1.56 microseconds. This is a good example of the application of relativistic time dilation to explain the increased particle range for high-speed particles.

You will now correctly say that a muon is not antimatter, but I just used the muon as example that some particles (or antiparticles) may be detected even if they have a short half-life.
So, what are the sources of antimatter in the magnetosphere? Strictly speaking, the source (in the sense of "location") of pair creation will probably be in the magnetosphere itself, as said by atmosphericprisonescape:

Anywhere where highly energetic reactions take place whose center-of-mass energy surpasses the rest-mass energy of 2*times some particle, this particle and its anti-particle counterpart can be created.

I suppose the real question is: How is their enough energy for frequent pair production in the magnetosphere?


When would Oumuamua have entered the solar system? So from what I've read, Oumuamua travels at around 26km/s when in interstellar space.

I also remember reading a while back that Voyager was travelling at around 18km/s when leaving the solar system.

Seeing as Voyager took around 35 years (I think) to exit the solar system, I'd guess that Oumuamua entered our solar system around 20–25 years or so ago?

One problem with answering this question is that it's not precisely clear where the solar-system ends or when (if) Voyager I crossed that border.

If we set the estimate at 19 billion km (per article) or 127 AU. (about 4 times the distance between Neptune and the Sun), then an estimate is possible.

Some conversions first.  Voyager left earth in 1977 and the announcement was in 2013, so it's usually listed as 36 years (it might be 35 and some months) but that's not all that relevant.

18 km/s x 31.5 million seconds in a year (close enough) is 567 million km per year.   It should have taken Voyager about 33.5 years at that velocity to reach 19 billion km.   However, Voyager's velocity wasn't consistent.  As it flies from the sun it slows down and it had several gravity assists along the way which sped it up and changed it's direction.   Mathematical calculations get a little tricky, but here's a chart of Voyager's velocity.

(Note: that's Voyager 2's velocity not Voyager 1.  I'm having trouble finding Voyager 1's velocity chart).

Here's it's direction, more of a fish-hook shaped path than a straight line, though it's been mostly straight since Saturn which is over 90% of it's journey.

Oumuamua's velocity at infinity, as it approached our solar-system was 26.33 km/s.   It speeds up as it approaches the sun and it will slow down as it leaves.   When it passed the 19 billion km "border", you have to add the escape velocity from our sun at that distance, which is about 3.7 km/s, so it entered the solar-system traveling at (close enough) 31 km/s or 976 million km/year.  At that velocity it would take 19.5 years to reach the sun (granted it will miss the sun), and in theory another 19.5 years to leave the solar system.   But that estimate doesn't take into account that it speeds up as it gets close to the sun, so it takes less time.   17-18 years (ballpark) might be a good estimate.  That's 17 (or so) years traveling in, another 17 (or so) to travel back out and reach the made-up border.

Unlike Voyager, nearly all of Oumuamua's journey is nearly straight, except for a quick curve around the sun.  That makes it's journey close to "V" shaped when viewed from a distance.  (I only mention that in relation to Voyager's more curvy fish-hook flight path, which was by design to pass by several planets).

There's probably someone here who can do the math (I'd prefer not to), and they can work out a more precise answer of when it entered the fuzzy 19 billion km border, when it will reach it's perihelion or closest to the sun, and when it will pass outside that border again.


Can someone calculate the age of the KOI-4878 star? The star’s low metallicity and fairly high space velocity suggest that KOI-4878 is older than the Sun.

But I don't know how to calculate an estimation for the age.

KOI-4878 data on Simbad

According to Wikipedia on KOI-4878.01 nobody did yet calculate the age of that object, but says exactly what you quoted:

its low metallicity and fairly high space velocity suggest that KOI-4878 is older than the Sun.

This supposidely relies on the following sources:

https://exoplanetarchive.ipac.caltech.edu/cgi-bin/DisplayOverview/nph-DisplayOverview?objname=KOI-4878&type=KEPLER_TARGET
https://www.drewexmachina.com/2019/09/05/koi-4878-01-keplers-most-earth-like-planet-candidate/

A search with the keyword KOI-4878.01 on different publication databases like arXiv or scholar.google did not help either.
Related

How do scientists determine the age of stars? focuses on the general question how to determine the age of a star.


Very near GAIA DR2 sources - has Proxima Centauri been dethroned? I made an extract of GAIA DR2 data for parallaxes greater than 500mas. Surprisingly I see a very significant number of sources with parallax greater that 1000, making them much closer than Proxima. This should lead to excited newspaper headlines, but i haven't seen those yet... Am I missing something?

You can see my extract of data here:

https://gist.github.com/agnes1/1492e94b5e9c0ef71f6ae7944964b666

Some GAIA source_id rows to look at include:

4039503676370417664

4042946384760661504

4042338732771057920

4046453826887613696

But there are many more. The closest source I find is 4062964299525805952 with  parallax of 1851mas. It has a phot_g_mean_mag of 19.63355... Would I be right in assuming this is a brown dwarf?

Another thing I note is that there is a surge of vmag 19/20 sources within 10 parsecs. These are obviously so faint GAIA doe not detect similar sources beyond 10 parsecs. To give some numbers I count 1722 sources (most mag 19/20) within 10 parsecs and only 606 sources in a shell of equal volume beyond that. Again, would these all be (a very large number of) brown dwarfs?

These are spurious data points. They are likely genuine objects, but the parallax value is incorrect. There are a small number of these in the Gaia data set.

There are 59 object with a parallax greater than Proxima Centuri. These don't represent genuine objects, instead they are when two sources are closely aligned (within about 0.2 arcsec). Most of these are found in the milky way, where there are lots of stars close together.

See
https://www.cosmos.esa.int/web/gaia/forum#!/thread/1688565/1688576

And section 9 of https://arxiv.org/pdf/1804.09365.pdf


Telescope on polluted environment this one bothering me,to buy a telescope.
I live in a city,unfortunately with dust and polluted air, is there any affect in viewing a nice capture of our galaxy? Or i need to buy a special telescope?

If you are going to buy a telescope in the city let me suggest a large aperture and a reflective telescope.  I live in a city as well and I use a reflector telescope.  I did have trouble with some dimmer galaxies to start with but if you buy a light pollution filter you can see them a lot better.  Also take a series of images of the same object and then stack them together!  There are many free astronomy image stacking programs on the web.

Have fun!


Ellipticals supported by random motions As opposed to spiral galaxies, which are rotationally supported, elliptical galaxies are said to be supported by random motions of their stars. I don't really have an intuitive understanding of this. How do random motions prevent a collapse?

The same way rotation does.  If you look at each star, it is following some kind of orbit, so it has a motion.  The only difference between a spiral and an elliptical is that the motions in a spiral are all in the same plane and the same direction, but in the elliptical, they are all over the place.  So "random velocity" just means you look at a bunch of stars-- each one is still in an orbit, so it's no surprise it stays in orbit.


Why haven't asteroid belts turned into new large bodies? If gravitation (attraction of mass) is the cause of the formation of all celestial bodies then how come the numerous small bodies found in asteroid belts spread over an orbit instead of clustering together ?!

From Wikipedia:
http://en.wikipedia.org/wiki/Asteroid_belt#Formation

Planetesimals within the region which would become the asteroid belt were too strongly perturbed by Jupiter's gravity to form a planet. Instead they continued to orbit the Sun as before, and occasionally colliding.[27] In regions where the average velocity of the collisions was too high, the shattering of planetesimals tended to dominate over accretion,[28] preventing the formation of planet-sized bodies. Orbital resonances occurred where the orbital period of an object in the belt formed an integer fraction of the orbital period of Jupiter, perturbing the object into a different orbit; the region lying between the orbits of Mars and Jupiter contains many such orbital resonances. As Jupiter migrated inward following its formation, these resonances would have swept across the asteroid belt, dynamically exciting the region's population and increasing their velocities relative to each other.[29]


Questions about terminology used for Mars and its moons I am having difficulty with some terminology:

sub-Martian hemisphere of Deimos ... what place on Deimos does it refer to?
leading/trailing apex of Phobos ... what place are those?
Apex ... in general it means a top of something, or an end of something pointed; but what does it mean on a relatively spherical body?
Anti-Mars meridian ... is this a meridian on the side of the moon facing away from Mars?

Thank you.

Deimos is, like our own moon, tidally locked to Mars, so one hemisphere of Deimos faces the planet, and the faces away. The Sub-Martian hemisphere faces Mars.

Since Deimos is locked, there is a point on its surface that always faces in the direction of orbital motion, that point is the leading apex. It is an apex in the sense of a "leading point"

A meridian is a line of longitude, running from the North to the South pole. One of Deimos's meridians goes through the centre of its hemisphere that faces away from Mars, this is the anti-Mars Meridian. As you say, the "meridian on the side of Deimos facing away from Mars.

The image here shows a map of Deimos:

The leading face is the left side of the image, and the trailing face is on the right. The central 50% is the sub-martian hemisphere and the anti mars meridian is the line on the left and right edge, where the 3D image of Deimos has been cut to make a 2D map


What consequences does a positive muon anomalous magnetic moment have for astronomy? On April 7th, 2021, the muon $(g-2)$ collaboration published Measurement of the Positive Muon Anomalous Magnetic Moment to 0.46 ppm, a result which made it to standard news, partly under headlines like Farewell to the Standard Model (German source: Abschied vom Standardmodell)
The following quote from phys.org nicely summarizes what it is in general about:

The experiment at Brookhaven indicated that g-2 differed from the theoretical prediction by a few parts per million. This miniscule difference hinted at the existence of unknown interactions between the muon and the magnetic field—interactions that could involve new particles or forces.

I am now wondering if there are theoretical studies which illuminate what concrete effects a $g-2 > 0$ would have on astronomy, as e.g. Dominik Stöckinger states (translation by me):

Some variants of supersymmetry, in which so-called superpartner particles also explain dark matter, are no longer tenable by this measurement.

Related

An answer to What is the closest candidate for Dark Matter currently? points out that WIMPS are currently the most appealing candidate for dark matter.

This is unlikely to have any implications for astronomy. They may have seen an extremely small deviation from the predictions of the standard model. The result may actually have to do with technical issues in theoretical methods, not with any new physics: https://arxiv.org/abs/2002.12347 . Even if it does have implications for new physics, there is no reason to believe that it would have implications for astronomy. It could, for example, be an indication of the existence of a new unstable particle with a high mass, whereas dark matter needs to be made of stable particles.


Could non-supernova carbon, oxygen, or silicon flashes be observed? I was reading about the helium flash, the short but sudden onset of helium fusion in certain red giant stars. As I understand, the upper (nondegenerate) layers of the star absorb the energy as they quickly expand, which leads to the stabilization of the core as thermal pressure takes over, leading to the period of helium burning.

I was curious as to whether this could be generalized to the onset of fusion of heavier elements later in a more massive star's life, and I came across a reference to a hypothetical silicon flash in Woosley & Heger (2015). More digging led me to a passage in Fundamental Astronomy (page 250), which states

  In stars with masses around 10 $M_\odot$ either carbon or oxygen may be ignited explosively just like helium in relatively low-mass stars; there is a carbon or oxygen flash. This is much more powerful than the helium flash, and may make the star explode as a supernova.

Let's say that a star undergoes a relatively low-power carbon, oxygen, or silicon flash, and enough energy is absorbed by the outer layers such that the star does not undergo a supernova. Would enough energy escape the outer layers of the star so that we be able to detect the flash using optical or neutrino telescopes? Obviously, this is distance-dependent, so I'll say that the hypothetical star is within five parsecs of Earth.

As another way to put the question, how luminous would the flash be?

There is also something called a "giant eruption", like eta Carinae, which is a type of "failed supernova" because it didn't destroy the star.  It is not known what causes giant eruptions, but perhaps one thing to think about is a flash of the type you mention.  Giant eruptions certainly are visible, eta Car got quite bright when it blew off much of its envelope.


Blue color of ion (plasma) comet tails What exactly is the cause of blue light of ion (plasma) tails of comets? Somewhere I have read that the source of blue light are CO+ ions which has just acquired the missing electron and became neutral molecule CO. Is this correct?

Close, but not quite right - the blue light is indeed emission from CO$^+$, but it's from the CO$^+$ ions themselves, with no need for recombination to CO; that (ionized) molecule has a strong set of energy transitions around 425 nm (4250 Angstroms), which is in the blue part of the visible spectrum:

Spectrum of Comet C/2016 R2 (Pan-STARRS), Figure 2 from Cochran and McKay (2018).
There's more on the physics of comet tails from Chris Mihos here.


Why do planets in red dwarf star systems tend to have a higher probability to get tidally locked? I have read about about the habitability of red dwarf systems on  Wikipedia, as well as some web articles on similar topics. The problem is, it does not explain why and how it happened. Google search comes up inconclusive. Since then, it became a riddle for me.

We all know that higher mass supergiant stars have stronger gravity than lower mass dwarf stars. Therefore, I can assume that stronger gravity results in stronger tidal forces as well. So, I can conclude:

High mass star systems should have a higher tidal lock probability than lower mass dwarf systems, right?

From that assumption that contradicts the reality, here comes another question:

Does it mean that the tidal lock probability has more to do with the size of the planet relative to the host star size than its mass or gravitational forces?

But here is my other assumption:
Probably a more concentrated (almost point source) of gravity from smaller red dwarf stars locks planets more effectively than a broad giant star.
(It is kind of analogous to a small speaker magnet that has stronger magnetic field compared to the Earth's magnetic field when measured).

A rough back-of-the-envelope way of seeing what's going on...

The tidal force is due to the difference in gravitational force, so follows an inverse cube law:

$$F_\mathrm{tide} \propto M_\ast R^{-3}$$

where $M_\ast$ is the stellar mass and $R$ is the distance. So at the same distance from a less massive star, the tidal force will indeed be weaker. But what's usually of interest is the effect on planets in the habitable zone. As a first approximation, define the habitable zone as the location where a planet receives the same amount of insolation from its star as the Earth does from the Sun. The inverse square law then gives:

$$R_\mathrm{HZ} \propto L_\ast^{1/2}$$

where $L_\ast$ is the luminosity of the star. For main sequence stars, we can approximate using a power-law mass–luminosity relationship:

$$L_\ast \propto M_\ast^{k}$$

where the exponent $k$ is traditionally given as 3.5, although it may vary between about 2 and 4 in the mass range of interest. Combining these gives:

$$F_\mathrm{tide,\ HZ} \propto M_\ast^{1-3k/2}$$

The exponent is negative for $k > \frac{2}{3}$, which is the case for the main-sequence mass–luminosity relationship. Therefore the tidal force in the habitable zone increases with decreasing stellar mass: the smaller distance to the habitable zone for these stars provides a stronger effect than the decreased mass.


Expansion ofthe Universe If the universe is expanding how come Andromeda is heading for us?
In isolated spatial areas does gravity overcome universal expansion?
What is the gravitational force between us and Andromeda?

What is the gravitational force between us and Andromeda?

The masses of the Milky Way and Andromeda (depending on exactly whay you consider "part of" each) are estimated as about 850 Billion solar masses and 1.7 Trillion solar masses respectively. Their mean distance is about 2.54 million light years, so it's easy to calculate the force using $$F = GMM'/R^2$$ We get a force of $$6.6\times 10^{29} N$$.


Determine distance of asteroid based on two observations from different locations Two astronomers A, B located in different regions A: 17° E, 46 ° 27 'N,
B: 17° E, 22 ° 34 'S observe an asteroid at the moment it is exactly above the meridian and is in position C which is a distance a from A and b from B.
Both observe from sea level and astronomer A sees the asteroid 25 ° south of his zenith, while astronomer B sees the asteroid 45 ° north of his zenith.
Consider the radius of Earth equal to 6400km.
Calculate the distance of this asteroid from the center of the Earth.
My work: We can create two triangles for example triangle AEC and BEC(Where E is the center of Earth) and from the data we have one angle for each (EAC = 180-25=155 and EBC=180-45=135) and one known shared side which is the radius of Earth. This is how far I have gone I suspect using law of sines and cosines to find some relations and do a system of equations but whenever I try and solve the system the numbers don't make sense.
Any ideas on how to continue with the soution will be appreciated.
Thanks in advance for your help.

Draw a diagram.
You get a quadrilateral EACB.  You should be able to determine the four angles in the quadrilateral.
Draw the diagonal EC, which is the length to be determined.
You know that ACE + BCE = ACB  (the known angle at the asteroid)
And you can use the law of sines twice, once in each of the triangles EAC and EBC
These should give you three equations, with three unknowns:  length EC,  and angles ACE and BCE.  Given three equations and three unknowns, a solution should be possible. You should find the distance is a little more than the Earth-moon distance.
Note that angles at the centre made by the diagonal: AEC and BEC are not 46° 27' and  22° 34', but the sum of those angles is the sum of 46° 27' + 22° 34'


What are the nearest galaxies I can observe? What are the nearest galaxies I can observe using my Telescope? Does it require to be out of the city lights?

The nearest one is the one you are on: Milky Way. It can be seen from any place, as it is all around us.

If you want to actually look at a whole galaxy in a simple view, then the nearest galaxies easy to be observed are the Small and Large Magellan Clouds, and the Galaxy of Andromeda.

There are galaxies nearer that those, but harder to be observed. A quite complete list is at https://en.wikipedia.org/wiki/List_of_nearest_galaxies


What measure should I use to help optimise the design of a telescope in a cubesat? I have started a project in Melbourne, Australia called nanosat eye in the sky to put a telescope in a cubesat and put it in LEO. The idea is then to add the telescope to itelescope.

I need to learn a heap about designing telescopes. I have limited space and want to understand how to measure the theoretical performance of our telescope.

So my question is- how do I measure the performance of the telescope?

I make telescopes and telescope mirrors. You're asking a hard question, since it depends on so many things. But it's an interesting project, so I'll try to help as much as I can.

The hard limit for performance is aperture, or diameter of the primary optics. Bigger aperture = higher hard limit for performance. On a nanosat you only have room for so much aperture, so performance will necessarily be limited.

Secondly, you'd need some kind of compact design. A reflector for sure, but not newtonian. Some kind of Cassegrain variant, you can make those pretty short and stubby. Unless you make it so that it "unfolds" in space, but that's going to be tricky.

Also, the whole optical stack has to be very rigid and must hold its shape through launch and while in space. The optical system must remain collimated at all times.

Finally, you probably have some weight limits. That will make it harder to create a rigid system.

One measure of performance is the resolving power, which indicates the angular size of the smallest details resolved by the telescope. The relationship is linear. An aperture of 100 mm provides a resolving power of 1 arcsec. An aperture of 200 mm provides a resolving power of 0.5 arcsec. And so on.

Telescopes operating at ground level are limited by atmosphere in several ways.

One is the so-called "seeing", or air turbulence. This blurs the image and reduces the effective resolving power. Since small telescopes have less resolving power to begin with, they are less affected. A 100 mm aperture is unaffected most of the time. A 200 mm aperture is affected most of the time.

Your telescope will be above the atmosphere, so the effects of seeing will be zero. It will operate at full resolving power all the time.

Another way the air influences telescope is via light pollution and other forms of luminescence and reflected light from the air. This reduces contrast and "erases" the faintest objects from the image; bright objects such as planets are not affected at all.

Your telescope will not be affected by light pollution, so its performance on faint objects (such as galaxies or nebulae) will always be 100% of the maximum for that aperture.

Bottom line: you are going to have a pretty small telescope, but operating completely free of the effects of the atmosphere.

Since it's operating in microgravity, the optics will not deform under their own weight, so you can make them quite thin and lightweight.

The instrument will operate under extremes of temperature. You will need some kind of dynamic focusing to compensate for expansion / contraction. Be careful, with some catoptric designs such as Cassegrain you also need to pay attention to the distance from primary mirror to secondary elements, etc. Using carbon fiber for the skeleton would reduce expansion / contraction a great deal, and would also provide excellent stiffness per unit of weight.

Quartz mirrors could deal very well with huge temperature variations, but quartz is expensive to process to optical precision. Borosilicate glass (the material used for most amateur telescopes) should also work well; Supremax 33 for example is cheap and fairly good. It depends on the optician who will make the mirrors. If you find someone who could give a discount on quartz, go for it; otherwise use borosilicate and don't worry about it.

You will also need some kind of inertial elements to change the orientation of the telescope in space. That's not a trivial task.

Bottom line: you begin with a performance budget limited by aperture. That's the ideal performance. Then real world issues will push real performance below that level: collimation errors, temperature shifts, optics quality, focusing errors, etc.

Performance of a real instrument is affected by everything.


How do you call it when two celestial bodies come as close to each other as they will in their current orbits? On Earth, we say Mars is in opposition when it is 180º from the Sun, which also marks the times when Mars and Earth come closest to each other.

Is there a specific name for such proximity events? For example, Saturn and Jupiter come to this maximal proximity every 20 years or so. Is there a name for the event itself, the "approaching" (previous) phase and the "departing" (next) phase?

This is a syzygy.  "A syzygy of the Sun, Jupiter and Saturn." In general a syzygy is when three bodies are in a line.  In your case the three bodies are the sun, and the two planets. You can also talk about "Saturn is at opposition as seen from Jupiter. Not every syzygy is "proximity event" but many are.  I know of no word or phrase for the "approaching phase" or the "departing phase".

I don't think that there is a specific term for "closest approach". Establishing absolute distance is generally difficult, compared with finding the direction. And the closest approach doesn't happen exactly at opposition (due to orbital tilt and eccentricity). Terms like "opposition" or "syzygy" are about the direction rather than distance.


Does the radio signal decay when it travels through the intergalactic space? When you emit the radio signal it starts moving at the speed of light. Radio beam is diffusing with each kilometer the signal has traveled. To the nearby receiver the signal is strong. But if the receiver is far away, the signal will become weaker and weaker until it becomes a noise. So the question is does the same happen in deep space and what distance radio signal (or any electromagnetic wave frequency) could travel until it becomes noise?

All electromagnetic radiation from a point source - which a normal radio transmitter is - propagates according to the inverse square law which means that the intensity of the signal is inversely proportional to the square of the distance. This happens on earth and in deep space equally.

So this will mean that for any signal there will be a distance at which it's becomes indistinguishable from the background noise of the universe. However, that distance will depend on the initial strength of the signal.


Do we make predictions in our time, or local time? I was scrolling idly through the Wikipedia article on Orion when I read:

  Orion will still be recognizable long after most of the other constellations—composed of relatively nearby stars—have distorted into new configurations, with the exception of a few of its stars eventually exploding as supernovae, for example Betelgeuse, which is predicted to explode sometime in the next million years.

Given that Betelgeuse is only 640 light-years away from Earth, the question of whether we're talking about Betelgeuse going supernova in the next million years, or about news of the supernova reaching us in the next million years, is irrelevant, since 640 years is hardly noticeable when you're talking about ten thousand centuries.

But what about a star further away? Say, something in the Andromeda Galaxy, 2.5 million light-years from us? If I read an article stating that a star in the Andromeda Galaxy was going nova in a million years, would that mean:

a) That we think the star will go nova one million years from now, and its light will reach us in 3.5 million years; or
b) That we believe the star went nova 1.5 million years ago, and its light will reach us in one million years

In other words, are we making estimates in our time, or in celestial-body-local time?

We express when events are going to happen (or have happened) in our observed time frame.

For example, we say SN 185 went supernova in the year AD 185; but it didn't. The star actually physically exploded about 9,100 years earlier.

When we say the sun will set tonight at 8:07 PM ET, that is when we observe the event. But our actual position relative to the tangent from the sun will have already passed about 8 minutes 20 seconds earlier (plus several minutes due to atmospheric refraction).

So when you say Betelgeuse is predicted to explode sometime in the next million years, that's how many more Earth years are expected to pass until we see it happen… from here. From Betelgeuse's point of view, it would have happened about 642 years earlier… or it may have happened already.


What makes the coming comet increase/decrease in brightness? I understand C/1979 Atlas Y4 is potentially going to rapidly increase in brightness and visibility to the point we can see it very well in May-ish, in the Northern hemisphere.

I am wondering why it's not predictable whether or not it will, or what processes affect the brightness.  If it's simply reflection, why would it dim or brighten? I really don't think I understand comets.

The nucleus of the comet is small and dark. We never see comet nuclei with the naked eye. The visible part of a comet is the coma and sometimes the tail which are made of dust and gas that have vaporised off the comet's nucleus by the heat of the sun.

Now comets are unpredictable.  They get more active, producing more dust as they get nearer the sun, because it gets hotter.  But sometimes there might be a sudden jet (as part of the surface of the comet gives way) and releases lots gas and dust.  Or the comet may develop a "crust" that prevents much dust from escaping. These events can cause a comet to become much brighter or remain dimmer than expected.

Comets are active and dynamic bodies. And so they are intrinsically hard to forecast.  Extreme examples of this are 17P Holmes, which went from magnitude 17 to magnitude 2.8 in two days, probably due to a crust layer cracking.


Does Pluto's atmosphere collapse? There were theories that Pluto's atmosphere collapses when it gets too far from the sun.

Then I've read papers that says that it doesn't.

But NASA says that the surface pressure went down recently (implying that it does).

Is there any clear data as to what's going on?

Found this link by Emily Lakdawalla, she wrote a detailed explanation about Pluto`s seasons and its movement thru the solar system. Here research seems to contradict the atmosphere collapse and and here 3 models and research on Pluto. Seems the most likely model gives the highest pressure in the atmosphere between 2020 ad 2040, but none a atmosphere collapse. source


Unit problem regards to proton In a textbook, we have an example:

The density of matter in the interstellar medium is of order one
proton per ${\rm cm^3}$, then the total mass of the interstellar
medium is of order (The $10^6$ term accounts for the number cubic
centimetres within one meter cubed). \begin{equation} m=1\times
V_\text{galaxy}\times m_\text{P}\times
10^6=3.3\times10^{40}{\rm kg}\approx 17 \cdot 10^9 M_{\odot}
> \end{equation}

So I was wondering, should the unit of the equation be:
$\frac{{1}}{{\rm cm^3}}\times{\rm m^3}\times{\rm kg}$?
And what exactly should be the unit of proton in the formula? Should it be unitless? Or ${\rm kg}$? Thank you.
Source: Beech, M. (2019). Introducing the stars. Cham: Springer Nature Switzerland AG.

The unit of the equation is "kg" because the unit of $10^6$ is $cm^3/m^3$
The unit of the mass of the proton is in kg. The  "1" is $cm^{-3}$ It represents 1 proton per cubic centimetre


Does a merging massive binary black hole ‘emits’ more than one gravitational wave? If we want to use gravitational waves (GW) to determine the Hubble constant, we need to find the source in the electromagnetic spectrum (EMS). However, we need to be lucky to ‘see’ it simultaneously in EMS and GW. This is a problem, but do BBH not ‘emit’ more than one GW? E.g. every time they spiral? Because than we have time to find it in the EMS, no? What makes it so difficult? Thanks in advance!

We can currently only detect gravitational radiation when it is extremely intense: in the last fraction of a second. For example the first gravitational wave detection lasted less 0.15 seconds. The black holes are releasing gravitational radiation with every orbit, but that radiation is too weak for us to detect. It takes a colossal amount of energy being released for it to be detected by current technology.  The inspiral of GW150914 released 3 solar masses of energy, nearly all in the final 0.15 seconds before merger.  Even so this only distorted space by less than 1/10000 of the diameter of a proton.  It is this that makes it so difficult.


Solid-body tides on the Moon
There are various good models for solid-body tides on the Earth. Are there any recommendations for tides on the Moon?

This answer starts with simplistic explanations and links with trivial calculations and progresses to links that, along with their references, provide exacting and lengthy approaches. @scb, you might want to skip halfway through the answer.
To make accurate calculations one needs to know about various parameters of the moon: composition of the crust and internal structure of the moon, Love numbers, precession, tidal acceleration, even time dilation - each results in a tiny error if omitted, a larger error if you skip them all.
An explanation of the anomalies of the Moon's crust is offered in Space.com's article: "Mystery of Moon's Lumpy Gravity Explained".
A particularly poor calculation is offered on this NASA webpage "SP-345 Evolution of the Solar System - 9. SPIN AND TIDES". It leaves out a lot (like Love numbers, precession, tidal acceleration, etc.) providing a quick and easy calculation along with a bit of unknown inaccuracies.
A short and quick explanation, less than you asked for, is provided by Space.com's article: "Moon Bumps: Earth's Gravity Creates Lunar Bulges" which explains:
"Earth's gravitational pull is so powerful that it creates a small bulge on the surface of the moon. ... The corresponding distorting effect on the moon, called the lunar body tide, is more difficult to see, because the moon is solid except for a molten core. But Earth's pull raises a small bulge about 20 inches (50 centimeters) from the surface on the near side of the moon and a matching bulge on the far side. [Rob's additional comment: they won't be equal.]
The same side of the moon always faces Earth, but the bulge does move around a few inches over time, wobbling and following Earth's pull like a magnet, as the moon shifts slightly during its orbit.
Scientists observed the lunar body tide using NASA's Lunar Reconnaissance Orbiter satellite, which is mapping the height of features on the moon's surface, and NASA's Gravity Recovery and Interior Laboratory satellite, which is mapping the moon's gravitational field. The satellites measured the height of 350,000 locations spread across areas of the moon closest to Earth and areas of the moon on the opposite side from Earth. The satellites passed over each location several times, so scientists could compare the height of each spot from one satellite pass to the next. By identifying which spots changed height, the researchers could track the lunar tide.".
A detailed calculation of the shape of the moon with plenty of math is covered in: "Evidence for lunar true polar wander and a past low‐eccentricity, synchronous lunar orbit". While covering a vast number of factors affecting the shape of the moon today body tides is amongst the points covered:
"Abstract
As first noted 200 years ago by Laplace, the Moon's rotational and tidal bulges are significantly larger than expected, given the Moon's present orbital and rotational state. This excess deformation has been ascribed to a fossil figure, frozen in when the Moon was closer to the Earth. However, the observed figure is only consistent with an eccentric and non‐synchronous orbit, contrary to our understanding of the Moon's formation and evolution. Here, we show that lunar mascons and impact basins have a significant contribution to the observed lunar figure. Removing their contribution reveals a misaligned fossil figure consistent with an early epoch of true polar wander (driven by the formation of the South Pole‐Aitken Basin) and an early low‐eccentricity, synchronous lunar orbit. This new self‐consistent model solves a long‐standing problem in planetary science and will inform future studies of the Moon's dynamical evolution and early dynamo.
1 Introduction
The long‐wavelength lunar figure is triaxial, due to the combination of rotational deformation (which creates a rotational bulge) and tidal deformation (which creates a tidal bulge along the Earth‐Moon axis). However, it has been known since Laplace [1878] that the observed deformation is significantly larger than the predicted deformation assuming hydrostatic equilibrium and the Moon's current orbital and rotational state (Figure S1). This difference has been ascribed to the presence of a fossil figure, frozen in when the Moon was closer to the Earth and the rotational and tidal potentials were larger [Sedgwick, 1898; Jeffreys, 1915; Lambeck and Pullan, 1980]. More recent work by Garrick‐Bethell et al. [2006] has shown that the observed deformation not only requires freezing the fossil figure while on an orbit with a smaller semimajor axis, but also with significant eccentricity. Matsuyama [2013] extended this work by taking into account the effect of finite lithospheric rigidity; however, the required initial orbital eccentricity remains large.
...".
That article includes many references which explain how the numerous equations were derived. As this illustration shows the article covers the development of the Moon's shape throughout history and attempts to calculate the minute bulging more exactly than the links referred to at the start of this answer.

"Figure 3. Our proposed model for the formation of the Moon's fossil figure. (a) The Moon coalesced from the debris of giant impact. (b) Under the action of tides, the Moon migrated outward and cooled from a magma ocean, perhaps experiencing periods of non‐synchronous rotation or high eccentricity. (c) As the moon cools, it forms an elastic lithosphere capable of supporting long‐term deformation, resulting in the fossil figure. (d–e) The South Pole‐Aitken (SPA) basin forming impact occurs, resulting in ~15° of reorientation, placing SPA closer to the south pole. (f) Subsequent large impacts and mare volcanism occur but do not significantly alter the lunar figure. (g) Ultimately, the Moon migrates to its current orbital configuration.".
Additional references:
"Lunar Laser Ranging Science" (18 Nov 2004), by Williams, Boggs, Turyshev, and Ratcliff.
"Lunar Geophysics, Geodesy, and Dynamics" (Homepage, direct .PDF link) (2003), by Williams and Dicky. That paper contains this useful text:
"The present lunar rotation model involves elastic solid body and fluid core. It will become more complicated in
the future. Numerically integrated physical librations (no fluid core) are available with DE403 at the JPL ephemeris
web site (http://ssd.jpl.nasa.gov/horizons.html).".
That website has moved, it's current address is: https://ssd.jpl.nasa.gov/?horizons - a bit of searching there turns up this URL: Lunar Constants and Models which says:
"Title:     Lunar Constants and Models Document
Author:     Ralph B. Roncoli
Date:   September 23, 2005
Reference:      JPL Technical Document D-32296
Summary: The primary purpose of this document is to provide a single source for the constants and models to be used in the trajectory and navigation design of missions whose objective is to orbit or land on the Moon. A secondary objective is to provide the mission analyst with some basic background information about the Moon, its orbit, and the previous missions that have explored the Moon. As a result, this document contains more information than the typical constants and models document. Some of the data are required for mission studies while other data are simply provided for "educational purposes". this document provides only brief descriptions of the constants and models. The user should consult the references if more detailed information is desired.
Format:     PDF
Link:   download here".
So that last list of references is "build your own rocket and land on the moon" accurate.


What is the nature of bright spots found on Uranus? The following text is from space.com which was written during the time of discovery of the first dark spot on Uranus:

During the past decade, many bright spots have been seen on Uranus in
both red and near-infrared filters. But this is the first dark spot
ever seen on the planet. A team led by Lawrence Sromovsky of the
University of Wisconsin and including Kathy Rages of the SETI Institute,
Heidi Hammel of the Space Science Institute (Boulder, CO), andPatrick
Fry of U. Wisconsin, observed the dark spot on Aug. 23 using the
Hubble Space Telescope (HST).

What was the nature of the bright spots? How were they formed? Was it some form of storm or vortex? I couldn't find any information anywhere regarding this.

Near-infrared pictures from the 8.1-meter Gemini North Telescope in Hawaii discovered a bright spot on Uranus. The spot is likely a tall methane cloud that reached high enough to become visible due to sunlight being scattered by its icy particles. The Uranian cloud is similar to an anvil cloud, a type of cumulonimbus cloud that's associated with severe thunderstorms. The cloud is also at a lower latitude on Uranus that could mean the spot is a storm that has migrated south.

Nobody knows how were the storms created. Some has speculated that the planet's tilt is the reason for such storms.
As Uranus is tilted on its axis is tilted, it virtually rolls around the Sun on its side. For half the orbit's period, the northern hemisphere is directed towards the Sun and the other half, the southern hemisphere is turned towards the sunlight. At two points in its orbit, known as the equinoxes, the Sun shines equally on both hemispheres. At these equinoxes, storms appear to become more frequent in Uranus’s disturbed atmosphere. Stormy weather has been observed a number of times in recent years, thus creating a notably bright spot appearing in 2014.
Reference:

https://www.nationalgeographic.com/news/2011/11/111021-uranus-planet-new-spot-storm-methane-gemini-space-science/
https://www.skymania.com/wp/mystery-bright-spot-appeared-on-uranus/
https://www.planetary.org/articles/08221504-new-spots-uranus
https://skyandtelescope.org/astronomy-news/bright-spot-uranus-atmosphere-11252014/


Day/Night Cycles and Seasons if the Earth's Rotational Axis and Orbit Changed I just enrolled in an introductory Astronomy course, and today in lecture the professor asked this question:

  Give 4 ways in which the day/night cycle and the seasons would behave on Earth if our rotational axis was in our orbital plane (Uranus is close to this). What about if the rotational axis was perpendicular to the orbital plane (so the ecliptic and Celestial Equator lay on top of each other), but the Earth's orbit was quite eccentric instead of almost circular?

For the first part, it means the Earth is rotating by North/South, right? Would this mean that the day/nights would be "flipped around," i.e. it would be night in a city when it would normally be day. I don't think the seasons would change, since the orbit remains the same.

I don't really know what the 2nd part is saying. Is it simply asking what would change if the orbit was eccentric rather than what it is now?

Any help or clarification on this would be greatly helpful. Thank you!

The Earth's rotational axis is essentially the line connecting the North and South poles.  This axis is slightly tilted, making the Earth's equator 23.5 degrees offset from the orbital plane (the ecliptic).

From Wikipedia

The first part of the question is asking what would happen if we were to make the rotational axis line up with the orbital plane.  That would essentially mean tilting it so that the Earth's equator would be 90 degrees offset from the ecliptic.  One way to do this would be to point the North or South pole directly at the sun, or to point it in the same direction as Earth's orbit.

The second part of the question is asking what would happen if the rotation axis was perfectly up and down.  This would mean that the equator would not be offset from the ecliptic at all.  However, for this part of the question, the Earth's orbit would also be highly eccentric, meaning that our distance to the sun would change a great amount.

I won't post the answer here, since this looks like it is probably a homework question, but to find the answer, first read up about what causes the seasons.  To talk about the day/night cycle, think about how the Earth's rotation causes day and night, by moving the sun across the sky.  If we changed how the Earth was tilted, how would that affect the seasons?  How would that affect where the sun appears in the sky?


This article says the moon may start moving closer to the Earth in the distant future; could this be true? Bruce Dorminey wrote an article for Forbes with the title of "Earth And Moon May Be On Long-Term Collision Course". The article notes:

"For now, our anomalously large Moon is spinning away from us at a
variable rate of 3.8 centimeters per year. But, in fact, the Earth and
Moon may be on a very long-term collision course --- one that
incredibly some 65 billion years from now, could result in a
catastrophic lunar inspiral."

It seems a little illogical that the Moon would suddenly start running toward earth at a far distance. Multiply 3.8 by 65 billion you get 2,470,000,000(May be innacurate). The article also notes this:

"As a result, the timing of the Moon’s ongoing recession is hard to
precisely predict. That’s because, as Barnes1 points out, the Earth
goes through glacial and interglacial cycles, causing the area of
shallow seas to change as sea level rises and falls."

Is it possible to understand how/why the author believes that the Moon could fall into a lunar inspiral? Please cite sources, thanks!

1Jason Barnes, a planetary scientist at the University of Idaho

Jason Barnes is certainly a highly credible source.  He is a planetary scientist and one of the leads on the Dragonfly drone now located on Mars.
The idea that the Moon will someday reconnect with the Earth is not new to him, however.  Here is a link to a Space.com article from 2007 which explains the same mechanism.
Earth's Moon Destined to Disintegrate
It is safe to accept that the article you link to reflects current thinking on the fate of the Moon-Earth system.


To what extent do the spin of smaller celestial bodies reflect the spin of the larger system’s of which they are a part? One can ask whether two objects are spinning in the same direction or not. For example, the Sun, Earth, Luna, most of our solar planets (except Venus and Uranus), and the planets’ satellites all spin in approximately the same direction. Likewise, the orbits of these bodies are in the same direction.

I understand that the rotational plane of our solar system is tilted about 60 degrees relative to our galaxy’s. If you project one rotation onto the plane of the other, I have the impression from maps that they may have opposite orientations — i.e. if we face north to south, our solar system’s rotation is counterclockwise, and our galaxy’s rotation is clockwise. Correct? What about the rotations at other scales of the larger systems we are in, such as our local star cluster, our local galactic cluster, and so forth? Thank you.

To what extent do the spin of smaller celestial bodies reflect the spin of the larger system’s of which they are a part?
What about the rotations at other scales of the larger systems we are in, such as our local star cluster, our local galactic cluster, and so forth? Thank you.

There's an amusing explanation here: Changing the rotational rate of a natural body, it's about this much (as far as most people are concerned):

Simple and incorrect answer: In Newtonian mechanics the gravitational field of a body depends only on its mass, not on it's rotation. If you go a bit further and assume that the spinning object has perfectly uniform density then it matters not whether it spins nor the direction.
More precise answer: The Lense–Thirring effect is very small—about one part in a few trillion. To detect it, it is necessary to examine a very massive object, or build an instrument that is very sensitive. The non-static stationary distributions of mass–energy causes frame dragging resulting in mass–energy currents and what is known as gravitomagnetism.

"This approximate reformulation of gravitation as described by general relativity in the weak field limit makes an apparent field appear in a frame of reference from that of a freely moving inertial body. This apparent field may be described by two components that act respectively like the electric and magnetic fields of electromagnetism, and by analogy these are called the gravitoelectric and gravitomagnetic fields, since these arise in the same way around a mass that a moving electric charge is the source of electric and magnetic fields. The main consequence of the gravitomagnetic field, or velocity-dependent acceleration, is that a moving object near a massive rotating object will experience acceleration not predicted by a purely Newtonian (gravitoelectric) gravity field. More subtle predictions, such as induced rotation of a falling object and precession of a spinning object are among the last basic predictions of general relativity to be directly tested.".

The equations for calculating the effect are straightforward but perhaps more than you wanted to know. Modelling this complex behaviour as a curved spacetime problem has yet to be done and is believed to be very difficult.
If you're trying to point a camera at a distant object it's a big deal:


Spotted a moving star-like object So a couple days ago, my husband and I spotted a moving object in the sky just like a star. I know it wasn't a plane or a drone, and it was way up in the sky just a star. It just moved smoothly across the skies and than disappeared. I love to observe the sky so that was priceless for me and my husband.

Based on your description, that might be the International Space Station:

  The ISS is visible to the naked eye as a slow-moving, bright white dot because of reflected sunlight, and can be seen in the hours after sunset and before sunrise, when the station remains sunlit but the ground and sky are dark.

So if what you saw was not long after sunset, and if the object was quite bright and moving slowly across the sky, that might be the answer.

If you want to check for yourself, you can input your location on this webpage and change the search period to see if the ISS passed above your heads on that day.

Other possibilities include other satellites (not very bright but also moving smoothly in the sky), or—more rarely—particularly large shooting stars.


What color is the Sky on other Planets? On Earth, the sky is blue.

What color would the sky be on other planets in our solar system? What about outside the solar system? Are there planets with purple skies, or green skies? Or are the all just blue?

The color you can see in the sky depends on two things, first the color (if you want to be more specific, the emission spectra of the light of the star) of the light shining, this is the Sun in our case (but may be another star in order to answer your question about outside the solar system), and second depends on the chemical components of the atmosphere of the planet in question, because this atmosphere is the reason for what is called the absorption lines.

Absorption light is something like that, if you have white light and a gas with green absorption lines, then after crossing the gas the light will turn the opposite of green, I don't know what color that is.

In conclusion if our atmosphere changes (for some unknown reason) here on planet Earth, then the color of the sky is going to change.


How do we know that an exoplanet is tidally locked to the parent star? Often in discussing exoplanets we hear that they are close to the parent star (usually a red dwarf) but then there's a caveat that the planet is tidally locked to that star and that that fact severely decreases prospects for that planet to harbor life.

But how do we know that an exoplanet is tidally locked to a star? I doubt that we can directly observe the rotation of that planet. Does that mean that we conclude that it's tidally locked because that's what our theoretical models tell us, and not that we observe it directly?

They are probably tidally locked, given some assumptions about age of the planetary system.

There's a formula to calculate how long it'll take for an orbiting object to tidally lock, and one of the outcomes of that formula is that large objects in close orbits don't take very long to tidally lock.

Since many of the exoplanets we've found are large objects in close orbits of old stars, tidal locking is a likely outcome.


Why isn't Earth a perfect sphere? In general, almost everyone describes Earth as an perfect sphere, but in reality it isn't a perfect sphere.

Is there any specific reason why Earth is not perfect sphere ?

There is no reason it should be.

There are bound to be irregularities due to local geological differences.

EDIT:

The theory as to why planets tend to be spherical is because any irregularity would cause a non-uniformity in the gravitational field at the surface, and the forces on the surface rocks would ultimately cause it to be spherical. Though, ignoring local irregularities, the Earth has a smaller radius (from its center of mass) at the poles, about 42 km less, than that at equator.

This can be attributed to Earth's rotation. The distance of a point on the Earth's surface located at the equator is larger from the axis of rotation than at any other latitude. As it rotates, this causes a bulge at the equator. The technical term given to this shape is the oblate spheroid.

You can read about the mathematical formula which describes the flattening and more about the formation here.


What came first: the Sun shining or the existence of helium? This seems to be a chicken and egg problem, the sun begins shining due to hydrogen becoming helium, but it's odd that there was no helium initially without the stars.

Is my logic flawed?

(Note: First year astronomy student)

The Big Bang theory predicts that (depending what assumptions you choose) the initial elements were formed from 10 seconds to 20 minutes after the Big Bang.

The initial elements were

mostly hydrogen
some deuterium
some helium-3 and helium-4
a little lithium-7
a couple of unstable isotopes that decayed to lithium-7 or helium-3.

As the linked Wikipedia article says

  Essentially all of the elements that are heavier than lithium and beryllium were created much later, by stellar nucleosynthesis in evolving and exploding stars.

So the answer to your question is: the Helium came first.


Are there expanding-universe cosmological models that do not include inflation? I'm happy that the universe is expanding, because we have good scientific evidence in the form of galactic redshift. The Wikipedia Hubble's law article gives what I think is a fair explanation of that. It redirects from cosmological redshift and maybe gives too much emphasis to Doppler shift, but no matter. Because like I said I'm happy that the universe is expanding. Even if there was no evidence, my understanding of general relativity would be telling me that the universe just has to expand. For the life of me, I still don't know why Einstein didn't predict it.

But what I'm not happy about is inflation. I take note of articles like Physicist Slams Cosmic Theory He Helped Conceive where Paul Steinhardt is critical of inflation. He says things like this:

  "The whole point of inflation was to get rid of fine-tuning – to explain features of the original big bang model that must be fine-tuned to match observations. The fact that we had to introduce one fine-tuning to remove another was worrisome. This problem has never been resolved".

I empathize. I don't empathize with his cyclic universe, but I share his view that we have not explained any feature of the universe by introducing inflation. In fact, I would dispute the claims that inflation explains why the universe is isotropic, homogeneous, flat, and bereft of magnetic monopoles. I'd even say that I think inflation is a solution to a problem that doesn't exist. But as the Wikipedia article says:

  "The basic inflationary paradigm is accepted by most scientists, who believe a number of predictions have been confirmed by observation;

I don't. I'm just some IT guy, but I'm like "the substantial minority of scientists who dissent from this position". Steinhardt is mentioned, as are John Earman and Jesus Mosterin. I can read their paper on SciHub. But what I'm struggling to find is any big-bang cosmological model or models that don't feature inflation. It seems to be included in all big bang literature and images, including this "wine flute" public domain image courtesy of NASA:

Can anybody point some out to me? In other words my question is this:

Are there expanding-universe cosmological models that do not feature inflation?

Variable speed of light (VSL) theories (for instance Petit (1988) and Moffat (1993)) have been introduced in the past to solve the horizon problem you mentioned, i.e. that the observable universe appears to be isotorpic and homogeneous. If certain universal "constants" (including $c$, the speed of light) are allowed to change, then it is possible for interactions between extremely distant sections of the universe to occur, solving the horizon problem. These modifications still fit in with general relativity and the idea of an expanding universe.

VSL theories, though, have not gained much traction, and inflation is still the prevailing theory in the scientific community today. It solves the horizon problem, the flatness problem (dealing with how fine-tuned the initial density of the universe appears to have been), and the magnetic monopole problem (i.e. why we don't see magnetic monopoles). Inflation isn't the only possible solution to these problems, but it works quite well.


How to calculate the true length of day on a celestial body? When I only know the sidereal rotation period, how do I determine the true length of day on a body?

We know the Earth's sidereal day doesn't have full 24 hrs unlike the true length of day. It's similar on Mars where the sidereal day isn't as long as a Sol (about 24 hrs 40 min) either. So how do we calculate the true length of day on a planet?

Let's do it on the example of the planetoid Ceres: its sidereal rotation period is 9 hrs 4 min and 27 sec. How do we calculate the true length of day on Ceres?

If you know the length of a sideral day and the length of a year, you can determine the length of a solar day.

From Wikipedia:

After one year, there will have been one more solar day than sideral day. Therefore :

solar day = sideral day / (1 - (sideral day/orbital period))


Relationship between the Casimir force and dark energy This is related to another question I asked. Might we actually be observing dark energy when we do experiments that verify the Casimir effect?

There are two problems that arise here. One obvious problem is that the Casimir effect is attractive1, while dark energy is repulsive. The other problem is one of scale.

Casimir (1948) shows that, between two particles (instead of the oft-cited case of two plates)
$$\delta E \propto R^{-7}$$
That's an enormous drop-off. On large scales, this should be negligible. Dark energy, however, is clearly non-negligible on these scales, while its effects are unobservable on small scales.

The Casimir effect is certainly one manifestation of vacuum energy, which may also be the cause of dark energy (in some theories), but that doesn't mean that dark energy is an example of the Casimir effect, or vice versa.

1 There are exceptions, as pointed out by Stan Liou, which arise under certain conditions, but they wouldn't play a part here.


Are there any stars that one day we could collide with? I'm not talking about tomorrow but in the far far future could we say our Sun could collide with another star or otherwise within our current detectable range?

Statistically speaking, the Sun will probably not crash into any other stars in the near future. Currently, we do not believe the Sun is on a collision course with any other star either, but we can only predict so far due to Chaos Theory and what I believe is called Lyapunov Time.

The reason for this is that solar systemic distances are minuscule compared to interstellar distances. The distance from the Earth to the Sun is 1 AU, or 150,000,000 km. Our sun is 0.004 AU in radius. The nearest star (which isn't on a collision course) is approximately 275,000 AU from the Sun. As you can see stars are generally far apart, and thus they rarely interact.

The locations of the stars do, in fact, change over hundreds of thousands of years, but again, stars generally stay far apart. That is not to say that stars never collide. However, it only happens in very dense globular clusters, young star systems (where they may form close), and in contact stars, as well as the occasional exception, which may be 1-2 in the entire galaxy when Andromeda collides with our Milky Way Galaxy in a few billion years (the Sun might already be dead though).