Sundial (solar clock) in a binary star system What would a sundial look like in a binary star system? What is the path drawn by the shadows?

It probably depends on the system, the sizes of the stars, their distance... If you know a binary star system for which a sundial would look different from what we have on Earth, we can take it as an example!

Edit: thank you to barycarter for bringing to attention that the Kepler binary stars would look like one star from the surface of their planets. However, there must be, somewhere, a binary system where both stars are visible, such that the shadows are affected.

EDIT2 : A Google search shows that this question has been answered before, but all links to the answer are dead. - I searched "binary solar system clock", found a link by a "Charmaine Pauls" -
But after using "sundial" instead, I found this pintrest artwork, but it is not much help. It doesn't seem to be scientifically motivated.

The sun forms a useful basis for measuring time because to a reasonable approximation it stays put, and the rotation of the Earth causes it to traverse the sky once every 24 hours.

Provided the planet is spinning, a normal sundial can be used, but may not be quite as regular as on Earth.

A binary stellar system can have one of three possible configurations: Two stars in a close binary, with a planet orbiting at a much greater distance. This is the "Tattooine" configuration. Everything has a double shadow. A regular sundial can be used but it is not as regular, as the positions of the two stars changes as they orbit each other.

The second configuration is a planet orbiting a star with a second star at a much greater distance. If Jupiter had been a star this would be the second configuartion.  The secondary star would be much weaker, and a regular sundial could be used. The shadow from the secondary would just be ignored.

The last configuration is like the second, except the planet is in orbit around the smaller star. The more distant star is stronger, and compariable in brightness. In this situation it is quite likely that the planet is tidally locked to the star, so a sundial isn't useful for telling the time. The slow motion of the distant star might give a long day (which could last many years) to the side of the planet that faces away from the nearby star, followed by a long cold night. This would be a challenging planet on which to live.


About the reference for equation of state Well, I am a post graduate studying mathematics. Current I am working on the mathematical theory of star(not an expert thought). When I am making some assumption on the equation of state (for a star or sun of course), I simply sum up the radiation pressure, idea gas pressure and degenerate pressure together. In some sense I believe this choice is an appropriate approximate to some model. But when someone ask me why, I am kind of stuck, even thought some of the lecture note posted online supports my guess(for example, http://www.jb.man.ac.uk/~smao/starHtml/equationState.pdf). But I am still wondering, is there any standard text book introducing something like this? I know Chandrasekhar's book has already told something, but it looks like it did not really state that I can simply sum all the pressure together linearly. Thank you in advance.

You cannot just add the ideal gas pressure and degeneracy pressure together for any particular species of particle. (Well, you can, but it would be a poor approximation to the actual pressure).

The trouble is that they are not separate things. In general, the electrons in the gas are partially degenerate and always obey Fermi-Dirac statistics. Unfortunately there is no analytic solution that gives you pressure as a function of temperature and density; you either have to use Tables of pre-calculated results, or there are analytic approximations that work well in certain regimes.

Clayton's Principles of Stellar Evolution and Nucleosynthesis contains some of these tables and also some better analytical approximations that work in certain regimes of (partial) degeneracy.

What is true is that you are fine to add to this pressure, the radiation pressure due to photons and the ideal gas pressure of the ions.

EDIT: I realise that what may be troubling you is not the above, but why you can add the separate pressures together at all (eqn 122 in the notes you reference). The reason can be traced to eqn 31 in the same notes. This says that the pressure is an integral over the number density and momenta of individual particles. Note that this equation does not contain anything that identifies the particles (i.e. their mass). The integral can therefore be treated as a summation over all particles, which in turn can be separated into three integrals that cover the ions, electrons and photons separately.


How can one determine the angle between the line of nodes and the major axis of the orbit of Mars? Edit:

What I really was looking for, was the difference in longitude on the orbit of Mars between its aphelion and the point where the north pole axis of Mars would point towards the Sun (northern summer sostice).
I found out with this article of the Planetary Society that the solar longitude (Ls) is 90⁰ for northern summer solstice and Mars is at aphelion at Ls = 70⁰.

I know when northern summer is on Mars, the solar radiation on its surface has a minimum of 492 W/m², meaning that is also the time when Mars is furthest from the Sun.
So this would coincidence somewhat with the north pole of Mars pointing towards the Sun.
But what is exactly the angle between the line  of nodes and the major axis of its orbit ?

The name for this angle is the Argument of Periapsis.
The Argument of Periapsis (or in this case, Perihelion, since Mars orbits the Sun) is the angle, in the plane of the orbit, measured in the direction of travel around the Orbit, from the Ascending Node, through the body being orbited, to the periapsis of the orbiting object.
Princeton's Planetary Systems page lists the Argument of Perihelion of Mars as $286.5°$ in 2006.
This would make the interior angle between the major axis of Mars and the line of nodes to be $360°-286.5° = 73.5°$, as of 2006.


Can Liquid Water exist on Mars? Outside of a specially crafts environment, could liquid water exist on Mars? Not some time in the distant past, but within the last few years. Mars' atmosphere is too thin for water to exist, right?

Perhaps if the water was mixed with salts or other minerals, its sublimation point might be altered enough to exist? Mars can get warm during its summer days, up to about 20 degrees Celsius at the equator. Perhaps, in certain areas and time periods, we may get a kind of liquid water?

Sorry for the speculative nature of this question, but after NASA talked about the possibility of liquid water, it got me wondering if it was possible.

Liquid water can in principle exist at many locations on present-day Mars, but there are a few interesting twists to the story.

At low elevation, the atmospheric pressure is high enough. The triple point pressure of H2O is 611 Pascal, which corresponds to about mid-elevation. At low elevations, such as the northern lowlands, the atmospheric pressure will exceed the triple point pressure, so the liquid phase of H2O becomes possible.
There will be substantial "sublimation cooling" or evaporative cooling. When the partial pressure of H2O is almost as high as the total pressure, vapor rapidly escapes into the atmosphere, which takes a lot of energy. If you put an ice cube at the equator of Mars, it will not melt, because the sublimation cooling is larger than the incoming solar energy, even at noon.
However, if that ice cube is covered with a thin layer of dust, or contains dust that will accumulate on top as the ice sublimates, vapor loss will be reduced. In this circumstance, ice can melt to form liquid water.
That leaves the question of how the ice cube would have gotten there initially. I call it the "source problem". Places on Mars that get hot have already lost all their ice, and places on Mars that have ice do not get warm enough. So, it is difficult to get liquid water on Mars, but it is possible in principle.
With salts, which lower the melting point, everything becomes much easier, although supplies of ice/frost and salt are needed. When water with salt evaporates, the salt concentration increases, so the solution is driven toward the eutectic point, where the melting point suppression is the strongest.


Calculating Right Ascension and Declination from Latitude / Longitude / Time I have a latitude / longitude / time of a location on Earth, I need to calculate the Right Ascension / Declination that's overhead at the time provided.

So far I've found out that Right Ascension is the same as the sidereal time for the location but I haven't found any information about declination.

I'm trying to set my overhead viewing location in the web portal of World Wide Telescope.  It defaults to "now" but I want to show 10PM and there's no way (that I've found) to set time through the javascript interface.  I can however set Right Ascension and Declination, hence my question.

The declination of the point overhead (zenith) is the same as the observer's latitude. The RA of the point transiting zenith at any given time is the equivalent of the local sidereal time.

(Alternatively, the local sidereal time is RA of the observer's meridian.)


Explain to a 4-year old why people below the equator are not tilted or suspended upside down So I just bought a globe for my extremely inquisitive soon to be 4-year old son. He knows the very basics of the solar system and knows that we live on planet Earth and also knows the concept of gravity.

While i was explaining to him what the illuminated equator line is he asked me How do people stand below this line? (Look at the globe from a 4 year old's eyes and you will wonder the same)

I explained to him that earth is not as small of a sphere as this globe depicts and almost everywhere it will be relatively flat but then he said there are so many places on this globe at the bottom if you live there earth will be above you and gravity will hold you upside down.

How do I explain it to him?

Explain that "down" is more or less toward the center of the Earth, and that direction changes with location. (For a four year old, skip the "more or less" part. It's just toward the center of the Earth, period.) The direction of "down" doesn't change much on trips across town, but it does change with larger trips.

If you have taken your son on a long trip (just a few hundred miles or kilometers will do), show your son those two locations on the globe. The direction of "down" at that remote location and the direction where you live point in slightly different directions for a trip of a few hundred miles, vastly different directions for a longer trip.  Ask him if he felt tilted sideways at that remote location.


What is the precession of the star S62? What is the angle of precession of the star S62 orbiting the Milky Way supermassive black hole?

According to wikipedia it is 10 degrees.  Wikipedia cites Peißker et al,
which uses the formula:
$$\Delta\phi = \frac{6\pi G}{c^2}\frac{M}{a(1-e^2)}$$ to derive a relativistic periapse shift of 9.9 degrees per orbit.


Why are there no photos of exoplanets? There are a lot of photos of diffrent astronomial objects such as galaxies, nebulae and stars of which some are very far away.

However there are no real photos of exoplanets (by “real photos” I mean pictures that actually show something more than a blob of light). Why is that?

Distances of galaxies are by many orders of magnitude larger than distances to exoplanets and yet we have a lot of very detailed (and beautiful) pictures of them. Meanwhile, everything we got on exoplanets is limited to a few pictures with blurred blobs of light.

Galaxies are far away but very big. Planets are nearer but very small.

Even for planets that are not completely swamped by the much brighter light of their host star, their angular size is much, much smaller than even quite distant galaxies and certainly smaller than the best achievable angular resolution by any telescopes on Earth or in space.

A quick example will suffice.

The angular resolution of the Hubble Space Telescope (HST) is about 0.1 arcseconds. That is, unless something subtends an angle of much more than 0.1 arcseconds then you are never going to get a good resolved image of it.

A big spiral galaxy is about 50 kpc in diameter. Such a galaxy will still have an angular diameter of 0.1 arcseconds when it is at a distance of $10^{11}$ pc. i.e. We can get images of such things (if they exist and if they are bright enough) right across the observable universe.

A planet the size of Jupiter subtends an angle of 0.1 arcseconds at a distance of 0.01 pc. i.e. We could not resolve the surface of Jupiter even at only a hundredth of the distance to the nearest star.


What is the shape and size of a gamma ray burst? As I understand it, a gamma ray burst is created when a massive star collapses into a black hole.  The poles of the black hole eject a massive burst of energy which we know as gamma ray bursts.

So my question... Are gamma ray burst more or less cylindrical in shape?  Somewhat like a quasar, but in a spectrum that is invisible to our visible spectrum?  And if so, what would the diameter of it be?

When "shape" is used in the context of gamma-ray bursts, it is typically used to refer to the jets emitted on opposite ends of an axis through the source. These jets contain huge amounts of energy, most in the form of photons so high in energy that they are classified as gamma-rays, some of the highest-frequency photons known.

The jets are typically emitted along an axis through the progenitor object (I'm not going to talk about the object itself, as the causes of GRBs are not yet fully understood). A quick search engine search (I prefer duckduckgo, but you might like Google better) can give you hundreds of pictures (mostly artists' impressions) of GRBs. Notice how thin the jets are. They look like long cylinders, but closer inspection shows that they aren't. They are actually cones (truncated ones, but cones nonetheless).

The reason that these jets spread out is relativistic beaming, which also affect the beam's luminosity. The angle at which the beams diverge is so slim - in fact, this pre-print says that one observed GRB had a jet emitted at an angle of one tenth of a radian! That's only 18 degrees. (Note: On page 21, this calls an angle of 3 degrees "typical" for some jets ).

So the answer to your question is that gamma-ray bursts are typically conical, although the jets are often nearly cylindrical. The first paper I mentioned also did an analysis of spherical models, but concluded that they did not fit observations.


Does the astrology of constellations relate to its position during the year? For example, pisces birthdays are from Febuary 19th-March 20th. Would the sun be near or at the pisces constellation during Mid febuary to mid march? Basically, I was wondering whether the astrology we use now is related to the sun's position relative to the zodaic constellations.

Yes, but no.

The basic idea behind the zodiac is as you stated what constellation the Sun is located in during the year. But the time assigned to each constellation (1/12 of the year) is not accurate, for several reasons. Firstly, because of modern constellation borders, some constellations are larger than others. This has also caused the ecliptic to pass through Ophiuchus, not in the zodiac. Secondly, axial precession has rotated the zodiac by almost an entire constellation over the last 2000 years. That is not accounted for in astrological dates.


What are the curved lines in this X-ray image of the sky? This image was created using the NICER instrument on the ISS.

Source:

https://www.nasa.gov/feature/goddard/2019/nicer-s-night-moves-trace-the-x-ray-sky

What are the big curves between the X-ray sources?

GIFs too big to post here directly: 1, 2

NICER is all about time resolution and energy resolution and is used to collect the time and energy-resolved X-ray spectra from pulsars. It isn't an imaging telescope and so can't record "patches of the sky" like other devices can.

The curved lines are the paths that the narrow field of view (FOV) of the NICER telescope has traced out across the celestial sphere during this 22 month (intermittent) observation period.

According to this page the FOV is only about 30 square arc minutes:

Large effective area: ~1900 cm2 at 1.5 keV
Broad Bandpass: 0.2 < E < 12.0 keV
Absolute timing precision of < 300 ns
Moderate spectral Resolution: 6 < E/ΔE < 80 from 0.5 keV to 8 keV
Restricted field of view: 30 arcmin²

With limited observing time and effectively a narrow single pixel field, it can only trace out thread-like arcs across the sky.

NICER is mounted on the constantly moving International Space Station and has its own pointing mechanisms on top of that.

When you see several arcs meet at a certain point, this is an interesting X-ray source, usually a neutron star (the primary purpose of the device and the "N" in NICER), and the telescope passes over it several times in order to collect more data.

Watch the COOL YouTube video NASA'S NICER Does the Space Station Twist

Watch the COOL YouTube video ISS NICER - First Deployment & Range of Motion Checks

Watch more NICER videos here: https://heasarc.gsfc.nasa.gov/docs/nicer/nicer_gallery.html

Watch the YouTube video Unlocking Secrets of Neutron Stars with NICER

Read the presentation NICER Science Overview

You can see that it is a number of grazing-incidence curved crystals used as concentrators, with "single-pixel" X-ray detectors behind them in the photo below (from How will NavCube (actually) be important for the XCOM testing and demonstration?):

below: NICER X-ray timing telescope/concentrator array, from About NICER.


What will be the end of the universe? I am looking at this video about the timeline of the future.
It says that the universe will stop expanding in about 1 googol years and become a cold bubble but I am not sure that will be true. There are several theories that can be true, so I am not sure which one. Anyone have some proof that one will be true? Thanks!

There are several theories about the future of the universe, but I don't think we can be completely certain about any of them. The favourite at the moment, and probably the one with the most evidence to support it, is the dark energy universe which will go on expanding for ever, with the galaxies getting further and further apart and more isolated until they burn themselves out into total darkness. This raises problems about a unique, one-off beginning, with no cause, and vast quantities of matter and energy created magically out of nothing, thus violating the First Law of Thermodynamics.

My preferred option is the Big Bang-Big Crunch-Big Bounce universe in which the expanding universe is brought to a halt by gravity and collapses on itself to form a blue-shifted, shrinking universe which ends in a fireball and singularity similar to the one it was created from. Except it doesn't end. It erupts again (the Big Bounce) in a gigantic white hole which we call the Big Bang, and this cycle is repeated endlessly. It had no beginning, it will never end, and it doesn't violate the Fist Law of Thermodynamics.


What happened just before the Big bang? Does the CMB contain any clues to what caused the Big Bang? Does a mathematical model theorize it?

Caveat: I'm not a cosmologist, so this answer may not reflect the forefront of scientific knowledge/accuracy, but I have some knowledge so thought I'd share it and hopefully someone can correct me if I'm wrong.

The Big Bang theory states that everything is moving away from everything else, so it must have started closer together. We know from the Cosmic Microwave Background that everything was very dense and hot and small.

You may have heard that the universe started in a singularity. This means one point at which all matter was in the same place. In this state, all information in matter is lost. No particle can be labelled as having a position or size or spin, because every particle is in the same state as every other particle. This means we lose all possibility of having any information about what happened before this. There are many theories of what could have been before this, but no matter how simple/beautiful/mathematically rigorous the theory, it is physically impossible to know anything about it so they are impossible to test. In some definitions this makes these theories not actually science, as the scientific method involves making testable predictions.


What is Color Index? I found this Wikipedia page on this property called color index. Unfortunately, I could not find a source that explains this property well.

My interpretation is that color index is the brightness difference of an object through some 2 different wavelengths. Is my understanding correct? If not, could someone explain it to me?

Also, if my understanding is correct, why would there need to be a calibration table, as shown in the above link?

A colour index measures the ratio of brightnesses in two wavelength intervals, not the difference.

The reason an index is written as the difference between two numbers is due to the logarithmic astronomical magnitude system - the difference between two logarithms is the logarithm of the ratio.

The reason that a red colour is bigger than a blue colour is also because of the perverse way that objects with smaller magnitudes are brighter.

The calibration table you refer to appears to just show that stars of different spectral class and temperature exhibit different colours.

You do however need a calibration to relate real brightness ratios and real properties of stars to the colours. That is because each logarithmic magnitude scale has its own zeropoint (the flux associated with a zeroth magnitude star in that band). The baseline colours are defined using the star Vega as a standard, such that all the colours of an A0 main sequence star are approximately zero (as shown in that table).


Which planet is this (if any)? I live in Ahmedabad, in India, with an exact location of about 27.2048° N, 77.4975° E. Currently (as of this writing), it is 20:52 EST. I noticed a red object in the sky.

(Forgive the crude imaging. I took this with my phone camera.)
Now, the brighter object is the moon, and the less shiny object is the red one I saw. I read somewhere that Mars would be visible right now, as it is closer to the Earth.
I also read that Mars and Jupiter appear red, with Jupiter slightly dimmer.
I wanted to know whether the object I just saw was indeed a planet. If so, which one (and how can I identify them)? Or is it just another star that I spotted?

Stellarium shows the Moon and Mars very close together in the sky tonight (Saturday, 3rd October 2020), so yes, it was probably Mars that you saw.
Moon and Mars on 2020/10/03 (Stellarium)
Stellarium is a great tool for identifying astronomical objects (and satellites), and is absolutely free.
Very rough rules of thumb for identifying planets by eye:

Very bright, white, and low in the sky (early morning or evening): probably Venus (but might be Jupiter).
Bright, white, and high up in the sky: probably Jupiter (unlikely to be Venus, which stays close to the Sun).
Bright and red: probably Mars.
Moderately bright and pinkish: probably Saturn.
Faint, white, and low in the sky (early morning or evening): might be Mercury.
If it scintillates (twinkles), then it's a star.

Jupiter and Venus are easily confused, since they can both appear to be bright and white. You really need to consider the time of day and the elevation (how high it is in the sky). Since Venus is closer to the Sun than the Earth, you are more likely to see it in the morning or evening (before sunrise or after sunset). Venus can climb quite high in the sky, but it will never be more than 45° in elevation.


Is there a verification of the solar system barycenter as it is theoretically determined? The barycenter of our solar system is not the sun, but a changeable location outside of the center point of the sun. A often used diagram of that relation is shown below, originated from the Wikipedia article.

Does exist one or more practical quite observations that confirm the accordance of the real changing of position of the barycenter of the solar system in relation to the center point of the sun with the theoretical results?

Yes, measurements do exist. JPL takes very precise observations of many objects in the solar system and tries to obtain a location for the solar system barycenter.

The solar system barycenter is a hot topic especially in terms of PTA's, or pulsar timing arrays - we need precise a precise location for the solar system barycenter to do work with these pulsar timing arrays. I'll put some links at the bottom as well to help.

JPL does get its own calculations of the solar system barycenter, and AFAIK, they do this with use of what they call the Solar System Ephemeris - see more here and here, at their Solar System Dynamics page. They get precise locations of many objects in the solar system and then try to calculate a center of mass from there.

A cool walk-through of the effect of the eight planets and Pluto (and some other stuff) on the movement of the Sun.


Does the retrograde movement affect the true angular distance between 2 planets? I need to calculate the relative angular path between two planets in the geocentric view or in other words the number of degrees of a circle one planet travels away from another planet around the Earth. In the heliocentric system I simply subtract the longitude of one planet from the longitude of the other and add n*360 if they made full cycles in between. Would the same work for the geocentric coordinates? Does the loop a retrograding planet makes in the sky actually count as its path? Should I consider the retrograde movement as an optical illusion or include it in the formula and how?

In our moving Earth-based point of view,
retrograde apparent motion
means a planet's geocentric ecliptic longitude is temporarily decreasing
instead of increasing as usual.
The word "apparent" avoids confusion with a
retrograde orbit.
All major planets' orbits are prograde,
with heliocentric longitude increasing at all times1.
Most retrograde-orbiting Solar system objects are comets.

As seen from Earth, the apparent angular separation
between two planets near the ecliptic
is approximately2 the difference
between their geocentric longitudes at that point in time,
regardless of their rates of change in longitude.
Even if you plot this value versus time,
your calculation need not break apparent motion down
into positive and negative cases.

Except where longitude wraps around from 360$^\circ$ to 0$^\circ$.
When longitude difference is small (near conjunction), latitude difference is also significant.


what magnification is required to see jupiter, saturn, venus and mars with a 50mm refractor scope with f/12 ratio and 600mm focal lenth i have a beginner telescope of 50mm and want to see planets it is arefractor with f/12 ratio and two eyepieces of 5mm and 10mm and i want to see mars, Jupiter, Saturn and venus. my max zoom ix 120x and focal length is 600mm

Venus, Mars, Jupiter, and Saturn are visible with your eyes, so a telescope and magnification are not required to see them. Depending on the quality of the telescope, the magnification will show you some details.
Venus - you will be able to see the phase.
Mars - for several months every 26 months, you will be able to see some dark markings and a polar cap.
Jupiter - you will see the cloud bands and 4 of its moons.
Saturn - you will see its rings and moon Titan.

Enjoy the view!


Does Jupiter help to protect the Earth against comets/asteroids impacts? Does Jupiter help to protect the Earth against comets or asteroids impacts?

  Some astronomers believe that one reason Earth is habitable is that the gravity of Jupiter does help protect us from some comets. Long-period comets, in particular, enter the solar system from its outer reaches. Jupiter’s gravity is thought to sling most of these fast-moving ice balls out of the solar system before they can get close to Earth. So long-period comets are thought to strike Earth only on very long timescales of millions or tens of millions of years. Without Jupiter nearby, long-period comets would collide with our planet much more frequently.

From: http://earthsky.org/space/is-it-true-that-jupiter-protects-earth

Intuitively speaking, it would seem to me that while any massive body in the Solar System would deflect the trajectories of some comets/asteroids that would otherwise pose a threat to Earth, it also must be true that some few ones that were not in a collision course would then be directed to us after having their trajectories deflected. Why isn't that the case?

Wordy answer, mostly light on Math:

The key word here (and the article uses this word) is "long period comets".

First there's Jupiter impacts, but that's a relatively low percentage, cause even large Jupiter is quite small compared to it's orbit.  Even if you extend it out to it's Roche limit where a comet could break apart, it's still a very small target in the grand scheme of things.

But any comet that passes near Jupiter (but doesn't hit it, get captured by it or broken apart in it's Roche limit) gets it's orbit changed somewhat by gravity assist.  Jupiter's high mass and relatively high orbital velocity make it the gravity assist king of the solar-system.

The closer the pass to Jupiter, the bigger the change.  Gravity assists, change both direction and velocity.  They can't change only one.  It's safe to assume that half of those gravity assists add velocity to the comet and half reduce velocity.   Very loosely speaking, the direction of the comet affects it's Perihelion and it's velocity and any change in it's velocity affect's it's semi-major axis (explained a bit below).

The comet's orbital velocity is obviously dependent on how close the comet is to the sun.   (Kepler's law about equal areas over equal times), but any orbiting object also has an average velocity, which is closely tied to the semi-major axis and orbital period, and it's this average velocity and the velocity at Aphelion (it's slowest velocity) that matters for this discussion.

The orbital period of any object orbiting the sun is directly proportional to the 3/2 power of the semi-major axis.  The average velocity is very closely correlated to the square root of the distance between the semi-major axis and the Sun.

When you get comets with orbital periods of hundreds or thousands of years, their average orbital velocity, is relatively slow for a celestial object.  Pluto's average orbital velocity is about 4.67 km/s and it's minimum velocity at Aphelion is 3.71 km/s.  Source.

For longer orbital periods the orbital velocity drops, roughly by the square root of the relative distance, so a comet with an orbital period of 1,000 years (about 4 times Pluto's), would have an orbital velocity about 1/2 as much and a velocity at Aphelion of maybe 1.8 km/s or a bit less.

Escape Velocity for the comet requires just a 41.4% increase on Aphelion orbital velocity, so, with our 1.8 km/s orbital velocity long period comet at Aphelion, if Jupiter gives that comet a push of just about .75 km/s, that would be enough to push that comet out of orbit, where it escapes the solar system.

So, while it might seem logical that Jupiter could send a comet equally towards earth as away from Earth, the relatively small push that a long period comet needs to escape the solar system, makes that a far more likely scenario than a nearly perfect 100 foot put into the Earth's orbit.   The odds of a long period comet being thrown out of the solar system by Jupiter is relatively high, especially with multiple passes, where, hitting Earth is like hitting a 100 foot put blindfolded.   Earth is a small target.  Out of the solar-system is a large target.

For shorter period comets where the added velocity needed to escape the solar system is greater, this grows proportionately less true, but for long period comets which only need a small push to get thrown out of the solar-system, statistically, Jupiter tosses a lot more of those out than it sends towards earth.

The trick is to be very specific about what we're talking about.   If Jupiter was, for example, to migrate through the asteroid belt, or migrate outwards through the Kuiper belt, then it's size would send a lot of stuff in towards the inner solar-system.   It's not in and of itself, always going to protect Earth and reduce Earth-impacts.  It depends on where it is and what the orbits of the comets are, but specifically for long period comets, Jupiter is a-lot more likely to send them out of the solar-system than towards Earth.  Not all, just statistically more likely.

It's worth noting that there's a lot we don't know about the oort cloud and how many comets/icy objects are in it.   We don't know what percentage of long period comets are jupiter-crossers vs objects with more distant perihelions.   When a star or large rogue planet passes through Earth's Oort cloud, it can send some of the icy objects out there in towards the Earth.   (Planet 9 - if it exists, probably not so much, as that theoretical planet has probably already mostly cleared out it's orbit), so it's not a key player in sending stuff to the inner solar-systems, but a massive passing object that pass through the oort cloud can do that, at least, that's been theorized, though such events are quite uncommon because space is mostly empty.   When those rare events happens, then there's probably an increase in long period comets and when that happens, Jupiter, for the reasons described above, is pretty good at getting rid of them over time.

Scholz's star is thought to have passed within 0.8 light years of our sun some 70,000 years ago and such a pass might be close enough to send a lot of Oort cloud objects in towards the inner solar system (though 0.8 light years it's still a bit far, as the oort cloud is probably quite empty that far out.  For a really good number of new long-period inner solar-system comets you'd probably want a 0.1 lightyear pass, or closer . . . . but massive objects passing that close are extremely rare, and . . .  I digress).

But Scholz's star might have turned a number of Oort cloud objects into long-period comets that pass through the inner solar-system.  (We won't know for a long time, as it'll be perhaps a million years for those new comets to reach us), But assuming Scholz did that, any oort cloud objects entering the inner solar-system from that event would be extremely long period objects and Jupiter would be pretty good at (over time, it wouldn't happen quickly), but over multiple long period orbits which probably takes millions of years, Jupiter would be pretty good at removing most of those from the inner solar-system, but just by chance, it would probably send a few in Earth's direction too.   If there was no Jupiter, it would take a lot longer for those long period comets to get cleared out, so in that sense, it really does protect the Earth from Long period comets.   There would be a lot more of them if not for Jupiter (and to a lesser extent, Saturn) and more comets would mean more Earth impacts.


Open source alternative for RegiStar (align, rotate, and combine images into single frame) I'm looking for a free, open-source and Linux compatible alternative to the RegiStar software.

I need it to automatically combine several images into a single frame, where each individual image shows a portion of a large observed region.

Individual images also overlap by some amount, and need to be aligned and possibly rotated before superimposing them onto the final frame.

I believe RegiStar does all this but: a- it is not free, b- it is not open-source, and c- it seems to run on Windows only.

I've had success with Hugin. It's a panorama creator, but it can be used to rotate, resize, and relocate images into a coherent whole. It has a few different automated profiles, but sometimes they are a bit off with star stacks. You can manually select the guide stars though.

A really good program for after that is GIMP and the plugin G'MIC to stack and remove noise.

Unfortunately it's not a single package like RegiStar, but it's free and can do quite a lot!

Before

After


Is there a ceiling for stable L4 or L5 masses? L4 and L5, the Lagrange points 60 degrees leading and trailing an orbiting body, are famous for being stable.

A well known example are the Trojan aseroids at the Sun Jupiter L4 and L5. Nodding to these bodies I label the central mass S, orbiting mass J and the L4 mass T:

It's said S/J needs to be equal to or greater than 24.96 for the system to be stable. For a stable system there's a ceiling on J, it can't be more than 4% of S.

My question: Is there a ceiling on mass T? If T were as massive as J, could the system still be stable?

The Giant Impact Hypothesis, a theory of how the moon was created, says that once it exceeds 10% of the mass of your 'J', the orbit of an L4 or L5 (your 'T') destabilizes.

Possible origin of Theia

  In 2004, Princeton University mathematician Edward Belbruno and astrophysicist J. Richard Gott III proposed that Theia coalesced at the L4 or L5 Lagrangian point relative to Earth (in about the same orbit and about 60° ahead or behind), similar to a trojan asteroid. Two-dimensional computer models suggest that the stability of Theia's proposed trojan orbit would have been affected when its growing mass exceeded a threshold of approximately 10% of the Earth's mass (the mass of Mars). In this scenario, gravitational perturbations by planetesimals caused Theia to depart from its stable Lagrangian location, and subsequent interactions with proto-Earth led to a collision between the two bodies.


Under what situations can an aperture mask improve the resolution of a small/medium amateur telescope? Is this demonstrable mathematically? @antlersoft's answer describes some of the challenges to seeing any details in the small disk of Mars in small amateur telescopes. In the case of reflecting telescopes, it mentions the use of either an off-axis parabolic mirror, or an off-axis aperture mask on a standard Newtonian to produce a clear unobstructed view.

The answer links to a short article on the amateur telescope site BBAstroDesigns which is titled Off-Axis Masks: Light Reduction and Resolution

If I understand the explanation correctly, besides the light reduction function, the off-axis masking excludes the secondary mirror and support, thereby removing diffracted light.

I'm a bit surprised that for amateur viewing on Earth that diffraction from a Newtonian or Cassegrain's secondary mirror and supports would really have any impact on resolution of small extended objects like Mars. The example in the article is a 16 inch telescope. I would have thought that seeing would be the limitation, but maybe for bright objects a smaller diameter (1/3 to 1/6 of the full aperture) is actually better?

Is it possible to explain the physics/optic behind the use of off-axis apertures to improve resolution as discussed in these links?

If possible, can a bit of math be used to demonstrate this quantitatively and convincingly?

The vast majority of people - almost everyone, really - should not rely on this on a daily basis. The chances are overwhelming that they will reduce the performance of their instruments. There are so many variables involved, and it depends on so many factors, a likely result is worse performance.

The link is to a site maintained by Mel Bartels. Mel is a very experienced observer and telescope builder, with hundreds if not thousands of hours behind the eyepiece. He has enough practical experience so that, just by looking at the overall image, and the shape and size of the Airy disk and diffraction figure, he can tell exactly in what way the telescope's behavior is affected by changes.

The effects of the secondary mirror and spider are a topic which is much misunderstood. You will encounter many strong opinions on this topic with not much basis in reality.

The spider vanes, if straight, will each create two spikes, diametrically opposed, perpendicular to the length of the vane. So a 3-vane spider makes 6 spikes. A 4-vane spider makes 8 spikes, but coincidental two by two (look like 4 brighter spikes). If the vanes are curved, then the diffraction figure is more like a faint, diffuse cloud.

So using an aperture mask can remove the spikes due to the spider vanes. But will it make the image "better"? Spikes don't matter from a practical perspective. If a spike is blocking one of the components of a double star, just rotate the scope. And the theoretical resolving power is diminished. You are, after all, using a smaller aperture.

The secondary mirror will increase the size of the diffraction figure. The Airy disk and the first minimum become larger in diameter - and so do the rest of the rings. This is a matter of much confusion, with some claiming that obstructed telescopes (scopes with a secondary mirror) are essentially inferior to unobstructed designs - this is demonstrably false.

Simple rule of thumb: if D is the diameter of the primary, and d is the diameter of the secondary, then the performance of an obstructed scope is equivalent, from the p.o.v. of resolving power (high frequency components), to an unobstructed instrument of aperture D - d. This is not an exact formula but it's close enough. A 300mm reflector with a 60mm secondary has about the same fine detail performance on planetary observations like an unobstructed 240mm instrument.

There are cases when you're observing close double stars, and it would be great if one of the stars could fall into the diffraction minimum of the other, brighter star, just outside the Airy disk. Sometimes you can do that with an aperture mask.

The article by Mel Bartels makes an important point, and I wish he emphasized it more: the aperture mask should only be used at low or medium magnification. In other words, if you're planning anyway to use low or medium magnification, then in some rare cases the aperture mask might be useful. It NEVER makes sense to use an aperture mask at high magnification - it's like pushing a telescope above its maximum useful magnification, and the result is just a blurry mess.

The linked article also mentions the issue of brightness. Some observers claim that the Moon is too bright in a telescope. Mel Bartels sees this as an opportunity to use an aperture mask (again, NOT at high magnification). Others use Moon filters - neutral density filters that reduce the incoming flux of light.

There is a third school of thought here. Do not become dark-adapted while observing the Moon. Do not observe the Moon from a dark place. Observe it with lights turned on around you, like from the sidewalk of a well-illumined street. This is what I, and others like me, do. Suddenly the Moon is not too bright anymore. The telescope is not performance-limited by the aperture mask, and you can zoom in at high magnification to observe small details of craters and mountains. And your eyes are working in their optimal regime, when the pupil is not fully dilated.

Surface brightness of extended objects never increases in a telescope, it just seems to increase. In fact, it always decreases. The Moon seems too bright because you're dark-adapted - but in reality it's quite pale.

There is another case when the image is demonstrably improved by an aperture mask. This is when the primary mirror has strong defects near the perimeter, like TDE (turned down edge), which is a defect both quite common (for mass-manufactured instruments) and very damaging to performance. A scope with strong TDE has poor resolving power and poor high frequency contrast. Sometimes it's so bad that even an aperture mask smaller than D/2 will improve performance.

The solution here is not the aperture mask, but instead masking off the perimeter. Make a ring with an inner diameter somewhat smaller than the aperture and place it permanently on the mirror, so as to block off the defective edge. Aperture is somewhat diminished, but now the optics are good across the whole surface.

One more case: let's assume seeing is so bad, the image is shaking like a giant, blurry pile of jelly. There's no way you're going to use high magnification anyway. You operate at low-to-medium magnification. Now apply the aperture mask. Is the image better?

Objectively the answer is no. If you're doing lucky imaging with a camera, it's still better to leave aperture alone, and by luck you will capture a few frames where detail resolution is pretty good.

But the human visual system is weird. Some people prefer to let go of the higher performance ceiling, instead opting for an image that's perceived as more smooth and stable. Personally I prefer to let low magnification do its thing, and leave aperture alone, but not everyone is the same.

Even visually, if you don't mask the aperture, sometimes seeing gets better briefly, and you can see small details for a while. With an aperture mask, you will miss those.

But this is a subjective thing, it's not something that can be objectively measured, except as a study of the psychology of visual perception.

If you want to experiment with the aperture mask, by all means go ahead and do it. But be aware of the many, many pitfalls here. It's not a silver bullet. It's not the solution to all of humanity's problems. It's just a trick that sometimes works, in a fairly narrow slice of use cases.


During night on the Moon is there Earth light and Earth phases? I realise that night time on the Moon is two weeks, however I am curious if during the night there, whether or not the Earth reflects light in similar manner to the Moon.

On the one hand, I know it must (as seen in photographs from astronauts), but what I'm really wanting to know is:

Does the Earth have phases like the Moon and do they work in a similar manner for an observer on the Moon?

I assume the Earth isn't as reflective as the Moon since the Moon is much lighter in colour, however the Earth is also fairly larger, does this make any difference in the amount of light reflected?

Finally, if the Earth does appear in phases, does that mean that the two week long nights on the Moon are sometimes pitch black (new Earth) and sometimes less than pitch black (full Earth)?

From the perspective of the Moon, the Earth exhibits phases just as does the Moon when viewed from the perspective of the Earth. There are however some differences. One difference is that the phase of the Earth at a fixed local lunar time will be pretty much the same from one day to the next. This is a result of the fact that the Moon is tidally locked. Another difference is that earthlight is considerably brighter than moonlight. This is a consequence of the Earth's significantly larger albedo and significantly larger diameter compared to the Moon. A third difference is that except for locations close to ±90° lunar longitude, the Earth doesn't rise and set.

  I assume the Earth isn't as reflective as the Moon since the Moon is much lighter in colour, however the Earth is also fairly larger, does this make any difference in the amount of light reflected?

You have this backwards. The Moon is rather dark. It's reflectivity is about that of old asphalt. The Earth is significantly more reflective than is the Moon.

  Finally, if the Earth does appear in phases, does that mean that the two week long nights on the Moon are sometimes pitch black (new Earth) and sometimes less than pitch black (full Earth)?

No. This is the first difference between Moon phases and Earth phases. Earth phase on the Moon is a function of lunar longitude and time of lunar day. You won't see a full Earth one night and a new Earth some other night.


Do Electric Charges in the Van Allen Radiation belt move in Opposite Directions? The Wikipedia Article on Van Allen Radiation Belt gives the following explanation:

  A Van Allen radiation belt is a zone of energetic charged particles, most of which originate from the solar wind, that are captured by and held around a planet by that planet's magnetic field.

The term "energetic charged particles" hinted to me that these charges are in motion and in very large velocities.

Let us assume, a small portion of space in the Van Allen Radiation Belt, compared to the size of Earth. We can assume the magnetic field in this space to be uniform due to our previous assumption. From the formula for finding Magnetic force on moving charges, $$\vec{F}=q(\vec{v}\times\vec{B})$$ where $\vec{F}$ is the magnetic force on the charged particle, $q$ is the electric charge of the particle with an appropriate sign, $\vec{v}$ is the velocity vector and $\vec{B}$ is the magnetic field vector.

As per the above formula, the direction of force is dependant on the nature of charge as well as the direction given by the cross product. If we consider the charged particles to move with uniform speed, the net force acting on the charge must be equal to zero. I think we must neglect the gravitational force exerted by Earth on the charges since the mass of charges is negligible. The two forces, Centrifugal force and the magnetic force (exerted by Earth's magnetic field due to the motion of the charged particles) cancel each other when they have the same magnitude but opposite direction. From the formula, it is also evident that the direction of force changes with the sign of the charge. It must be opposite for positive and negative charges.

So, based on this is it correct to assume, if the positive charges orbit the earth in one direction the negative charges orbit the earth in the opposite direction? Do they really orbit Earth in opposite directions?

The following image (a view from the top of poles of Earth) depicts my idea on the sense (clockwise or anticlockwise) of rotation of charges in the Van Allen Radiation Belt:

Image Source : My own work :)

Yes and no. The question @JohnRennie links to in his comment  Analogy between magnetic bottle and Van Allen's radiation belt contains a lot of relevant information, but packaged as an answer to this question.

The question correctly concludes that, at least in the plane perpendicular to the magnetic field, particles will travel in circles, and those circles will be in opposite directions, but they do not, generally simply circle around the Earth as you suggest. There are two reasons for this: firstly the magnetic fields are too strong, or the particles too light, so the circles you get are much smaller and secondly because they are also moving in the third direction. So you should think of them travelling in fairly tight spirals around the magnetic field lines, which, roughly speaking run between the North and South magnetic poles.

As they approach the poles, the field gets stronger (the lines bunch up) and it turns out that this repels the particles, so they end up bouncing back and forth between the two poles (all in spirals around the magnetic field lines).  The protons and electrons do spiral in opposite directions.


Why is the difference between Apollo and Aten asteroids considered meaningful? I understand the difference between the two to be that they both cross Earth's orbit, but Atens have an orbit smaller than Earth's, and Apollo have an orbit larger, but I don't understand why they aren't simply lumped together in one category.

The first point to make is that categorization isn't an exact science.  It's done for convenience and practicality and sometimes, familiarity.  It's not going to be perfect.

The second point to make is that scientists like categorizing things.  Take the Taxonomic ranking of living creatures.  Everything from a paramecium  to an elephant has a species, a genus, a family, an order, a class, a phylum, a kingdom and a domain.   Categories are everywhere in science and they're useful.   Taxonomy classifications aren't perfect, but classifications don't need to be perfect to be useful and provide distinction and quick and familiar identification.   The same is true for asteroid classification.

Both Atens and Apollos are already lumped into one category called Earth-crossing asteroids, so, when needed, they can be discussed together.   It's more common to separate them by the two types, but it's not required.

Earth-crossing asteroids or Earth-Crossers are a sub-category of Near Earth Objects, and Near Earth Objects or NEOs are of particular interest because every so often, an undesirably large object hits the Earth.

NEOs can be divided into Atens, Apollos, Amors, Comets, and Atiras, sometimes called Internal Earth Objects or IEOs, but there are relatively few Atiras and they're sometimes included with the Atens, though technically, Interior Earth Objects aren't Atens if their aphelion is too far inside the Earth's orbit.

NEOs can be classified as potentially hazardous (PHOs) if they are over 140 meters across, per Wikipedia (NASA gives a different definition of PHOs below - so take that definition with a grain of salt).

Because of the infrequent but "could happen at any time" impact possibility, a well funded effort is underway to catalog, track and calculate the trajectory of every NEO larger than 100 meters, called the Near Earth Object Project.  Because so much attention is given to NEOs, the distinction and naming of Atens and Apollos makes perfect sense to me and doesn't seem unreasonable at all, but as pointed out above, there aren't any clear-cut rules when it comes to categorization.

What's most relevant to the NEO project, in addition to size is Minimum Orbit intersection distance or MOID.   PHO's by the NASA article, as opposed to Wikipedia, are categorized by Earth-MOID of less than 0.05 AU and an absolute magnitude (which is related to surface area, which is related to size) of 22 or brighter.

Not all crossing asteroids (determined by aphelion and perihelion) have orbits that come close.   Because orbits take place in 3 dimensions, orbits can be crossing, but never get close.  Pluto, for example, is a Neptune-crosser but the orbits never get close.

I added MOID cause I thought it was interesting, but back to categorization.

The other 7 planets have their own versions of Atens and Apollos, called inner and outer grazers.  See Jupiter crossing objects.

The 7 planets also potentially have libration asteroids, distinct from Crossers.   Libration asteroids are (captured, in horeshoe orbits, or Trojan).   The 4 inner planets have very few libration asteroids.   Jupiter has tens of thousands.  The outer 3 known planets have several known libration objects each.

There's also resonance objects (Jupiter's Hildas, Neptunes Plutinos).   Objects can also be defined by outer orbits (Trans) or inner orbits (Cis), and by prograde or retrograde, which isn't the same that planets appear to do, but orbits relative to the sun can move opposite to the 7 planets.   In the inner solar-system, there are very few retrograde orbiting objects but out beyond the 8 known planets, retrograde is more common.

The dwarf-planet Pluto, for example, is a Plutino, which is a subset of a Neptune-resonant-object, and a subset of a Trans-Neptunian object, which is a subset of a Kuiper-Belt object (KBO).  I mention that only to point out how many orbital categories there are.   Scientists like categories.

There's other categories include Centaurs, Scattered Disk Objects, Oort Cloud Objects and, of-course, Asteroid belt objects and lets not forget comets, which are distinct from Asteroids, though the distinction gets a little fuzzy for dead comets.

Another categorization example, Jupiter has so many trojans, that it's two camps are separated into Greeks and Trojans, which is odd, because Jupiter's Trojan camp is a subset of Jupiter's Trojan asteroids.  That's a naming system that stuck, but which I think is less than ideal.  You can also call them L4 Trojans (Greek camp) or leading the planet's orbit and L5 Trojans (Trojan camp), following the planet's orbit.

The overall point is that there's many of categories of asteroids.  Atens and Apollos seem to me, fine and relevant, not arbitrary.   (want an arbitrary definition, how about when a moon-planet system becomes a duel planet system - that's pretty darn arbitrary).  Many definitions are arbitrary.

(too long?)


What limits the usable focal length of telescopes currently? What barriers - of technology, physics and possibly economy (things that would be possible technologically but are simply too expensive) sets the upper bound on quality of telescopes for observation of the sky in visible spectrum - observing surface of distant objects at maximum resolution/detail?

To answer the question reworded as: "What limits the quality of telescopes currently?" The answer is mostly: money.

It used to be: atmosphere. But with advances in adaptive optics, ground-based telescopes are achieving what used to require a space telescope. Plus, we have the technology for space telescopes if we want, like the pending JWST.

So pretty much, it boils down to funding. Who is going to spend the money on the expensive technology to see better and further, when science is being squeezed for funding from all quarters, and astronomers can't put forward a business case for a return on investment like a biotech company can for developing a grass that will result in cows burping less methane.


How do I know what I'll be able to see? So I live in a suburb in Victoria, Australia. Less than an hour away from the city and I guess there is a bit of light pollution because from my backyard I can probably only see about 15 - 20 stars (probably less), I'm wondering what these stars are, and what I'll see when I get this telescope:

https://www.opticscentral.com.au/saxon-707az2-refractor-telescope.html?___SID=U#.WXQNMtN940r

This is going to sound stupid but how do I know when there are planets in the sky that I can see? I don't think I've ever actually seen a planet other than the moon.

Thank you.

Firstly, if you're planet spotting, don't worry too much about light pollution. The planets are some of the brightest objects in the sky and some (especially Jupiter) can easily be observed even with a full Moon - the full Moon (along with the Sun!) is the biggest contributor to light pollution!

Take a look at the list of brightest stars ( https://en.wikipedia.org/wiki/List_of_brightest_stars ), which also contains estimates for the brightness of the Sun, Moon and major planets. There aren't typically any stars brighter from Earth than Venus, Jupiter, Mars and Mercury and precious few brighter than Saturn. I'm going to suggest you probably have seen many of the planets - but just didn't recognise them.

+1 for Stellarium ( http://www.stellarium.org/en_GB/ ). It's free, intuitive and very visual to use. You can put in your local viewing location and it gives you a view for any time of the night, future or past. At the time of writing (23 July 2017), Saturn and Jupiter should be looking good for the Southern Hemisphere. This rotates throughout the year, and Stellarium will help with this.

Do a web search for "the sky at night in the southern hemisphere" and you'll find a number of examples of websites with highlights to look for when you get out.

Couple of final suggestions:

Do allow plenty of time to let your eyes adjust to the darkness - it takes me a good 20-30 minutes to get going. You'll see many more stars then
Do be aware that local conditions (houses / trees etc) might restrict visibility of things near the horizon. In any case, you find items over-head are much clearer
Refractors (yours included) come with an 'erecting prism'. Despite the name, the real benefit isn't that you see the image the right way up (that isn't really an issue in astronomy), the prism just allows you a more sensible and comfortable viewing position - you'll understand when you get going
Don't restrict yourself to planets. When you get going, learn about the deep sky objects too. For instance, the southern hemisphere has access to the famous Magellenic Clouds which we cannot see from the northern hemisphere!


How active was the Milky Way in its first 5 billion years? When I look into the history of the Milky Way, it seems like the Milky Way has been the pinnacle of stability. It seems highly likely that our galaxy has eaten several smaller galaxies/dwarf galaxies in the past. Also, our galaxy seems to be one of the first to form. I am wondering if our seeming stability is the result of our galaxy's age. So, was the Milky Way more active (as far as stellar formation, supernovae, etc.) in its early years? How do we/ could we know?

Although the question was asked a while ago, it still sounds interesting, so I did a bit re-search and I found a Nature news-feature by Adam Mann: Hidden history of the Milky Way revealed by extensive star maps

The team had spotted a set of 30,000 renegade stars. Unlike other objects in the main body of the Milky Way, which orbit in a relatively flat disk shape, these non-conformists were moving backwards, in orbits that were carrying them out of the Galactic plane.
Within weeks, the team had worked out that the luminous horde pointed to a long-hidden and especially tumultuous chapter in the Milky Way’s history: a smash-up between the young Galaxy and a colossal companion. That beast once circled the Milky Way like a planet around a star, but some 8 billion to 11 billion years ago, the two collided, massively altering the Galactic disk and scattering stars far and wide. It is the last-known major crash the Galaxy experienced before it assumed the familiar spiral shape seen today.

The highlighting in the quote is my attempt to answer your question - it is not talking about the Milky Way's first 5 billion years, but it somehow implies that its early history might have been indeed turbulent.
References
The original article is by Amina Helmi et al. is called The merger that led to the formation of the Milky Way’s inner stellar halo and thick disk


Why does the Solar Wind consist of charged particles? The Wikipedia article on Solar Wind gives the following explanation:

  The solar wind is a stream of charged particles released from the upper atmosphere of the Sun, called the corona. This plasma mostly consists of electrons, protons and alpha particles with kinetic energy between 0.5 and 10 keV.

If the solar wind consists of both electrons and protons, why don't they combine and give Hydrogen or any other elements? Then, why would Solar Wind be charged? Is that because of the high kinetic energy of the ejected particles? If yes, let us consider a proton and an electon. They must be of almost the same energy if so there will be no relative kinetic energy, so I think they must combine in the same way as if they were at rest or in motion at small velocities.

You are right about the kinetic energy (ie the fast motion) of the particles being the reason, but wrong when you say

  They must be of almost the same energy if so there will be no relative kinetic energy,

The individual particles are moving fast, but there is huge variation between them in just how fast and in what direction. In other words, the solar wind is very hot. The extent of disorderly motion in a gas is measured by its temperature.  The wikipedia article you link gives temperatures of 100 000 to 800 000 Kelvins. So if an electron and proton did happen to collide gently enough to form an atom of hydrogen, another particle would probably smash into them before very long and knock the electron off again.

You might think they are all moving away from the Sun, but you have to allow for the effects of magnetic fields. Charged particles moving in a magnetic field curve around in all sorts of ways, and, additionally, moving charged particles create a magnetic field which affects other particles. So  the movement is extremely turbulent.


How Wide is Earth's Roche Limit? My question is simple:

Given the gravitational relationship between Earth and the moon, how wide is Earth's roche limit?  How close does the moon have to orbit before getting disintegrated by Earth's gravitational pull, and how far does it have to orbit before being free from Earth's orbit entirely?

A little reading and effort is recommended and the Wikipedia page is straight forward enough, if you're not completely math-phobic.

But if you want an answer, Rob Jeffries 9,500 km is right, however, that's center point to center point.   Surface to surface, that would only be less than 2,000 km, which would be crazy close.   That close the Moon would take up about (guessing) close to 50% of the sky, maybe 70 degrees or so, when directly overhead.    But that 2000 km number doesn't take into account tidal stretching or rotation, as the Moon would almost certainly be tidally locked and rotating in sync with it's orbit like it does now and the Earth would be similarly tidally locked and have a tidal and gravitational bulge in the direction of the Moon.

Taking into account rotation, that pushes it out to the cube root of 3, not 2 or about 14% further, but that's center to center.  The surface to surface distance would be increased to about 3,000 KM.   Still insanely close.   Any orbital eccentricity would increase instability though, so for a safe distance, Earth to Moon where the Moon wouldn't break apart at all, but it would still stretch measurably, I wouldn't want it any closer than about 5,000 or 6,000 km from the Earth's surface.   That's a ballpark guess with tidal forces, tidal locking, solar perturbations and orbital eccentricity all in mind and even there, some loose moon dust and rocks might regularly blow off and shower onto the Earth.   6,000 KM (surface to surface) is less than twice the Moon's diameter, so it would still be enormous in the sky, but I think that's about the distance where the Moon would hold together in orbit.   The math says even closer, but that doesn't take into effect various issues like eccentricity, rigidity, etc.   It gets rather hard to calculate with rock solid accuracy taking into account all the details.  But, point is, the Moon would need to be stupid/gonzo/cray close before it begins to break apart.   Some 50-60 times closer (surface to surface) than it is now.

Now as far as escape, the math of the Hill Sphere or Sphere of Influence  is pretty straight forward, but the long-term stable zone is about 1/2 to 1/3rd of the Hill Sphere calculation, some 500,000 to 750,000 km.   Somewhere in there or perhaps just beyond that range the Moon probably becomes unstable and escapes the Earth's orbit becoming a very large and potentially very dangerous near earth object.   That doesn't mean that the Moon would just fly away past 750,000 km.  It would remain in orbit for multiple orbits, hundreds, maybe thousands even past that distance, but past the stable region it would probably destabilize over time and escape eventually.

The Moon is perhaps as much as 70% of the way towards where it could eventually escape the Earth, but that final 30% is a LOT.  The moon isn't going anywhere anytime soon.  It's still comfortably inside the long-term stable region and should remain orbiting the Earth for billions of years.

Hope that helps.


What are the most popular galaxies for which we have images? The only galaxies I can think of (not being an astronomer) are Andromeda and Milky Way. There are 51 near galaxies, but they all pretty much say "satellite of Milky way" or "satellite of Andromeda". There are 100k+ galaxies in the local supercluster, and that page seems to have a better list:

Corvus
Coma Berenices
Ursa Major
Virgo
Sculptor
etc.

If you had to rank them in order of prominence in the scientific community or in popular science, wondering what the top 10 or 20 galaxies would be (for which we have photos).

I am trying to come up with a list of images for educational purposes that are potentially somewhat familiar to laymen audiences, or which would be useful to introduce to laymen audiences.

Any such list is going to be terribly subjective. Since I'm an astronomer who studies galaxies, I'll go ahead and throw out a subjective list of the more famous, photogenic, and/or scientifically well-studied galaxies. The first six are in the Local Group (LMC and SMC are satellites of the Milky Way, M32 is a satellite of Andromeda).

Milky Way
Andromeda (M31)
Large Magellanic Cloud
Small Magellanic Cloud
Triangulum (M33)
M32
Sombrero (M104)
Pinwheel (M101)
Whirlpool (M51a)
M64 (Black Eye)
M74 (NGC 628)
M81
M82 (Cigar)
M87
M100
NGC 891
NGC 1068 (M77)
NGC 1300
NGC 1365
Centaurus A
Cygnus A


What is the lower limit of a coronal mass ejection? Reading an 2005 article by Fraser Cain called Smallest Ever Coronal Mass Ejection
 I became interesting in the exact definition of CME.
It seems like a simple question - but I did not manage to find the answer to the question myself. Is there a smallest possible coronal mass ejection (CME)? Is there a theoretical limit or is it a matter of observing capabilities?
Note: I left intentionally open whether "small" refers to mass, size, duration, or any other parameter.
The above mentioned article claims that the CME in the square in the following picture is actually the smallest CME:

Related

Solar Orbiter images first coronal mass ejections, Press release of the Max Planck Institute for Solar System Research, dated May 17, 2021
Wilson effect: How "deep" are sunspots?

No, as ‘coronal mass ejection’ is defined ambiguously and refers to an event that is not physically bound to some exact limit and therefore cannot have an undisputable lowest limit value, in the mathematical sense, in any parameter.
For example, the Oxford dictionary states:
“
coronal mass ejection
Noun - Astronomy
An event in which a large cloud of energetic and highly magnetized plasma erupts from the solar corona into space, causing radio and magnetic disturbances on the earth.
“
Clearly, any limit, if fixed to some exact value of mass, flux, duration, etc., would have to be fixed arbitrarily.
As a thought exercise, consider if someone came along and presented you such a ‘lower limit’.
You, or a third party, can respond with: “well what about the situation where an event occurs just one electron (or one Planck second, etc.) below such a limit? Would that be considered a CME or not?”


Visibility of earth from moon during day-time of moon Inspired by this question. I am curious whether earth, besides being nearly fixed on one place on the moon's sky, is it visible during the day-time on moon too?

My understanding is that earth should be visible as moon has no atmosphere. Also, if the NASA didn't edit the following photograph, it suggests that the day-time sky of moon is all black and earth should be visible in it.

I think, Stellarium don't take into account the atmosphere once you are on other planet.

Yes, the Earth would be visible even during day on the Moon. Without a (significant) atmosphere diffusing light, the day sky would be much like the night sky on Earth. If you were to look straight up such that you couldn't see the ground, you may not even notice that it was day at all (save for the Sun shining down at you).  This could actually be rather hazardous as the Sun would be more intense, but your eyes may not be adjusted to high light like they are on bright Earth days.


Fact Check: How bright will the Perseids be in 2017? A friend sent me an email which appears to be an extract from this Cambridge News article:

A meteor shower astronomers have labelled 'a once in a lifetime opportunity' will take over the sky in August. The upcoming Perseid meteor shower will be the brightest in human history so far, Physics Astronomy reported.
Anyone in the Northern Hemisphere should be able to view the shower on August 12. It is predicted to be so bright that stargazers should be able to catch a glimpse even during the day.
Astronomers believe that a shower this visible won't grace the sky for another 96 years. The Perseid meteor shower, one of the brighter light shows of the year, occurs annually between July 17 and August 24. The shower tends to peak around August 9-13.

However, this article sources its information from the website Physics Astronomy, a blog-type website which posts various articles on astronomical topics. Since Physics Astronomy doesn't seem to be a particularly reputable source, I will take this information with a pinch of salt.
So, is this article true? Will you be able to see the Perseids in the daytime with the naked eye this year?

There is variation in meteor showers from year to year, but for most showers, the big difference is how bright the moon is. If the moon is up, then it's hard to see many meteors.

The Perseids of 2017 will be rather ruined by the moon. There will be a waning gibbous moon, rising shortly after sunset and staying up for the rest of the night. It will make the sky too bright to see many meteors

2018, with a new moon on the 11th, will be much better.


Existence of a mass-volume astronomical chart? I've been wanting to make a mass-volume (or mass-radius) chart for astronomical objects ranging from asteroids to supermassive black holes on a log-log scale. A sort of 'Hertzsprung-Russel' type diagram but with mass and volume as the axes.

I wanted to check first to see if anyone else has done this before. Google doesn't seem to turn up anything of the sort, and so I'm wondering if anybody in the esteemed audience of Astronomy.SE would know of the existence of such a diagram.

Something like this?

https://arxiv.org/abs/astro-ph/9608168

But you can do literature search about the topic using NASA ADS for example, or astro-ph arxiv. They are pretty comprehensive databases of papers from astronomy/astrophysics field. If you dont find anything there you are pretty sure that no one did it. If you do find some results there (and I know for a fact that mass-size diagrams are explored/created for galaxies, clusters, etc) it might shed some light on your problem as well.

NASA ADS:
http://adsabs.harvard.edu/abstract_service.html

astro-ph arXiv:
https://arxiv.org/archive/astro-ph


How did the Milky Way and Andromeda Galaxy separate before they began their collision course? How did the Milky Way and Andromeda Galaxy separate before they began their collision course?

I understand that gravity is pulling together the two galaxies, but what separated them in the first place while gravity is pulling them together?

Thank you for your consideration.

I thought it was a fun question, I gave it some thought.  Layman's version, so corrections are welcome.

The very early universe, the first few seconds or so, it was quite dense.  This is fairly well publicized and the specifics and details don't really matter for this question, but the energy and stuff in the early universe that later formed the Milky-way and the energy and stuff that formed Andromeda were cosmic neighbors and in the first seconds of the universe, quite close to each other.  The basic Hubble rule is that things that are close (in cosmological terms) were always close, like the dots on a balloon that gets blown up.  Everything moves away from everything else, but things that were next to each other were always next to each other.  On smaller scales, like stars within galaxies or planets, this isn't true because of gravitational assists and different orbital paths, but its true for galaxies.   (there might be some small variation on this with "orbiting" dwarf galaxies, but it's still by and large, true).

Another important point to make about the early universe is that it was much more uniform, so, because of the uniformity, gravity was pulling nearly equally in all directions.   The nearby somewhat more massive clouds of matter and energy didn't dominate gravitationally because there was similar mass in all directions.    You might think that a denser universe means stronger gravity in the direction of nearby matter, but that doesn't happen if the matter is spread evenly enough in all directions.

So, generally speaking, in the very young universe, the stuff that became the Milky way and the stuff that became Andromeda started out much closer, but they didn't pull each other much early on because of uniformity.

Over time, local gravity overcame uniformity.  I couldn't tell you when that happened (maybe someone here can), but it took some time before the makings of galactic groups and clusters began to form.   We know from the background radiation map that the Universe was still quite uniform as late as 380,000 years after the big bang.   See picture:

(Edit due to correction).  And at 380,000 years, the known Universe is thought to be about 1/1100th it's current radius.   We know from the cosmic background radiation that at that time, the Universe was still largely uniform, though non-uniformity did lead to the formation of galaxies, clusters and super-clusters, so it was important, but it still took some time for local groups and galactic clusters to form.

See here.

  On this map, the hot regions, shown in red, are 0.0002 Kelvin hotter
  than the cold regions, shown in blue.

and

  These cosmic microwave temperature fluctuations are believed to trace
  fluctuations in the density of matter in the early universe, as they
  were imprinted shortly after the Big Bang. This being the case, they
  reveal a great deal about the early universe and the origin of
  galaxies and large scale structure in the universe.

But once local gravity overtook uniformity, then galaxies can start drawing each other together more efficiently.   The spiderweb pattern of galaxies shows that there is a tendency for this to happen, as galaxies draw other galaxies towards them, even as space expands, so you get lines of more dense regions with more galaxies and pockets of empty space, mostly free of galaxies.

Source of picture for further reading.

So, what basically happened with Andromeda and the Milky-way is that expansion combination with uniformity set them fairly far apart even as they were still forming, but they were still next door neighbors as far as galaxies go.

Both the Milky-way and Andromeda have absorbed smaller, closer dwarf galaxies, perhaps many times, so they weren't the closest 2, but they were the closest 2 non dwarf galaxies (at least, that we know of).

You might think that galaxies could orbit other galaxies, but the nature of Hubble expansion doesn't give galaxies much tangential velocity towards other nearby galaxies, so basically, 2 galaxies are either moving towards each other due to mutual gravity or away from each other due to expanding space.   Now in a 3 or more galaxy system with similar masses, they can get some tangential velocity, but generally speaking, the direction between any 2 large galaxies is mostly either towards or away, not an orbit.

Another thing to keep in mind, once the uniformity effect is overtaken by local gravitation of local groups or clusters, is that massive objects at these distances don't exactly pull each other quickly.   It takes a long time for gradual acceleration to turn into relevant velocity.

At our current distance from Andromeda (about 2.5 million light years), and it's current mass (about 1.5 trillion solar masses), it's G force on us, using the formula below (Math can be added if needed):

comes to about 1/2.8 trillionths of a meter per second squared.  That's less than 1/20,000th the gravitational pull Pluto has on Earth.   A pull that small is going to accelerate things towards each other extremely slowly, so (I think) for the first billion or few billion years, the expansion of space between the Milky way and Andromeda likely exceeded any gravitationally caused velocity between them.

The good news is, Andromeda and the Milky-way have had a long time to pull on each other.  About 13 billion years or so.   1/2.8 trillion m/s^2 over a billion years is works out to about 11 km/s.  Over 13 billion years, 143 km/s.   And, the actual velocity that the Milky-way is moving towards Andromeda isn't too far away from that, about 110 km/s.  That right there gives us a ballpark Netwonian estimate for how far apart they were when they started.   (Now, footnote, for objects of somewhat similar mass, you have to take the mass of both of them, not just the larger one, so it's closer to 15 or 16 km/s per billion years), but that doesn't change the overall numbers too much.

Dark Energy expansion (which currently, between the Milky-way and Andromeda is about 60 km/s), and that makes the calculations a bit more tricky, especially since dark energy expansion might not have been constant throughout the last 13 billion years.  Related article here.

At the current rate they are moving together, about 110 km/s or, 250,000 mph or 1/2,680 c, In one billion years at current velocity Andromeda will be 375,000 light years closer (and, in reality, a bit more as it will continue to accelerate by mutual gravity and dark energy expansion will be reduced as they get closer).

And if we work backwards, a billion years ago it was a bit less than 375,000 light-years further away than it's current 2.5 million light year distance, with diminishing distances further away over previous billion year intervals.   The farther back you go, and difficulties with knowing how close they were when local gravity became a key factor over uniformity makes estimates pretty difficult, but they were probably never too much further apart than they are now.  Perhaps double, as a very rough guess.   Because gravitational acceleration drops off at the square of the distance, I find it hard to believe that Andromeda was ever much more than twice as far as it is now.

I would imagine, based on my assumptions above, that by the time the galaxies were recognizable as galaxies, say, when the universe was a billion or 2 billion years old (give or take), Andromeda and the Milky way were probably moving away from each other initially, but over time, the gravitational attraction was able to overcome overcome the expansion of space.


How is Comet SWAN C/2020 F8 doing? Will it become easily visible by eye? The New York Times article How to See Comet SWAN in Night Skies says:

  Astronomers have their fingers crossed that the comet will keep brightening in the coming weeks as it heads north, passing 52 million miles from Earth on May 12 at its closest approach to our planet, and then rounding the sun on May 27.

and links to the comet's twitter account which is tweeting non-specific updates every few hours.

On May 12, 2020 https://minorplanetcenter.net/mpec/K20/K20G94.html puts the comet at

Date    TT    R. A. (2000) Decl.     Delta      r    Elong.  Phase   m1    m2
2020 05 12    01 17 42.4 +14 52 01   0.5572  0.5849    28.5   124.4   4.4

about R.A 1h 17m and Dec +14° 52' and an estimated magnitude of about +4.4 (if I'm understanding this correctly) and the JPL Horizons database puts it at

 Date__(UT)__HR:MN, Date_________JDUT, , , R.A._(ICRF), DEC_(ICRF),  Azi_(a-app), Elev_(a-app),   T-mag, N-mag,             delta,     deldot,     S-O-T,/r,     S-T-O,

2020-May-12 00:00, 2458981.500000000,*,m,    19.54137,   15.02829,   105.324535,    61.907026,      9.,  n.a.,  8.3196461568E+07, -3.4857438,   28.4079,/L,  124.6636,

But I'd like to ask if Comet SWAN C/2020 F8 is likely to become easily visible by eye at any point in time during this pass near the Sun.

The SWAN C/2020 F8 comet is passing near Earth, but its magnitude will be around 5. This means that you need to have a rather dark sky and good enough eyes to be able to see it.

From the comet observer's database

In addition to this, the comet will be very low on the horizon during the night. Here is its track from Wikipedia

Early June (when it will be rather bright), SWAN will pass near Capella which is roughly 15-20° above the North-West horizon at 22:30. This means that you need to have a pretty clear horizon to see it.

TL;DR : It will be visible by the naked eye if you have a clear horizon and a dark sky. So it won't become easily visible by eye at any point in time during this pass near the Sun.


What would Jupiter look like from a Galilean moon? If I were on a Galilean moon of Jupiter, like Europa or Ganymede, what would the planet look like? (Let's say I'm on a space station that provides atmosphere to make it slightly less unrealistic.)

Would it look like our moon, only a bit bigger and colored like Jupiter, with the stripes and the "eye", or would the moons be close enough for Jupiter to look gigantic and fill a significant portion of the sky?

The two moons that are closest to each other are Io and Europa. They seem to be about 250,000 kilometers away from each other. Would it be possible to see one from the other with the naked eye if they were aligned? For example, if Europa were in a position in its axis where it is lit by the sun, could I see it, looking at it from Io? Or would I need a telescope? A 3000 km sphere from 250,000 km away does not seem possible. The other way around, of course, Io is a bit bigger.

But the part I'm the most interested in is: what would Jupiter look like from one of these moons.

You can simply startup Stellarium and have a look yourself. Choose any locations you are interested in. The Galilean satellites of Jupiter are inside the default list of locations.

The attached image shows Jupiter as viewed from Io at the given time, Europa is the bright object to the right; with -9 mag it is considerably brighter than Venus when viewed from Earth, but slightly less so than Earth's Moon as seen by us.

Interesting to note is also the huge extend of Jupiter in the field of view, if you choose the right hemisphere (Io is in a tidally-locked state). With roughly 15° you need about your entire hand at arm's length to block it while the sun is a lot smaller (~30x larger than Sun or Moon when viewed from Earth). Also the daily eclipses of the Sun by Jupiter might be noteworthy.


How can I accurately find & apply the white donut to the center of my mirror? I have a Plettstone 18" Dobsonian telescope and use a Howie Glatter tuBlug, laser collimator, and Square Grid Projection Attachment to adjust the mirrors.  Proper collimation requires an accurate identification of the mirror's center. I would like to hear of good methods to very accurately place a white sticky donut concentric to the mirror's optical center.

The mirror already has a donut, but I wonder if it is accurately placed as to my eye it appears slightly off center, but perhaps the optical center is not the physical center?

I've seen a recommendation in Sky & Telescope to get two clear plastic rulers that bend and place them at exact right angles to each other with the donut upside down lightly stuck to the bottom center cross point and then positioning the rulers evenly and pressing down to bend them toward the mirror and affix the donut.  However, I don't think this method is well suited for an 18" mirror, perhaps I am mistaken?

I've also seen a recommendation by Gary Seronik to make a paper cutout that matches the mirror shape and to find its center and make a small hole and then place that over the mirror and use a pen to mark the center and center the donut around the marker, however, again, I don't think this is well suited for an 18" mirror.  This method is also mentioned here in some detail, but they specifically state not for use with a laser collimator.

Please suggest how I can first gauge the existing donut for accuracy and if it is found to deviate significantly how to remove it and the adhesive and place a new one more accurately.

I make my own mirrors and telescopes.

The method recommended by Gary Seronik, that you linked above, is very good, and it's widely used. A few comments:

Don't be afraid to touch the mirror with clean paper. The vast majority of mirrors these days are over-coated with SiO(x) and are quite sturdy. Maybe wash the mirror before you do anything else, this way you'll remove potentially abrasive dust particles. This is how: https://www.youtube.com/watch?v=9Y8xFnXFVGQ - it's a good idea to wash the mirror anyway, like once a year or every other year.

When making the big paper template, cut the disk from a piece of paper without wrinkles. Draw a good circle, and cut it carefully. Make sure it's flat and it doesn't want to roll up. When centering it, go over the edge again and again, and adjust it so that it's perfectly centered. You center it by working out the edge alignment with the mirror. Take all the time you need.

You could also cut the template out of transparent plastic, larger than the mirror, and just draw a circle on it, the same size like the mirror. This method is recommended by Jim Fly (of Catseye) even for quite large mirrors.

Both versions (paper or plastic) work about the same IMO.

It may take a bit of trial and error to put the central dot on the mirror without smearing it. The marker tends to smear up underneath the paper. Practice on a random piece of glass or polished metal.

The reason why Bob Fies (of Alcoat) says it's not for lasers is that, the way he shows it, the paper dot is solid. Of course it wouldn't work with a bare laser - you need a donut shape, with a hollow center, to allow the laser to reflect (when using a bare laser, not with a tuBlug). Just use a little donut instead of a solid paper dot, for the final step - like Gary shows on his website.

If you never use a bare laser to adjust the primary, a solid dot is fine. But a donut gives you more options.

I've done business with Bob directly, he's very experienced. Like Gary Seronik and Jim Fly, he should be considered an expert in this field. But take his advice in context and look at all the details - he says what he says for a good reason.

The way to gauge the existing donut is also by using Gary's method. Make a paper disk with a hole in the middle, center it on the mirror, and put a little dot with a marker on the glass. Compare that dot with your existing donut.

Wipe the marker dot off with alcohol and a q-tip when you're done. You can't damage the mirror unless you press very hard. The center is shadowed by the secondary mirror anyway, so it doesn't matter.

Do everything very slowly and be as precise as humanly possible. The precision of the central donut is critical for the precision of your collimation. Since you only put the donut on once, it makes sense to do a very slow but very high quality job.

If the existing donut is off-center even a little bit, don't hesitate to remove it. If alcohol soaking doesn't remove it, try acetone. You may have to clean the center afterwards, but that's okay.

After everything is finished, clean the inside of the donut with alcohol and a q-tip. You need that surface relatively clean, for the laser to reflect back into your collimator when not using the tuBlug (just bare laser). It doesn't have to be squeaky clean, just remove the marker smears, more or less, and make it nice and tidy.

Congratulations on taking the right steps. Many, many people don't care about collimation, and it's so important for overall performance. Keep that scope collimated.

The tuBlug is an excellent tool for collimation, it makes the process a snap. Great choice. Catseye is another great collimator, but if you have the tuBlug and a Glatter laser (one of the most precise lasers out there, BTW) you probably don't need anything else, unless you have some extremely low f/ratio on your scope (not your case).

I use the tuBlug every time I do an observation. It's taking maybe 2 minutes to do the adjustments. Totally worth it.


Why more matter still left after collision with anti-matter? Question implies all, why is that? Why we exist even after equally creation of matter and anti-matter, any possible exception?

I presume you're asking about the mystery of the missing antimatter. This features extensively in the popular and scientific media. For example the CERN webpage on the matter-antimatter asymmetry problem is fairly typical. It says the Big Bang should have created equal amounts of matter and antimatter, but everything we see today is made almost entirely of matter. It goes on to say “something must have happened to tip the balance. One of the greatest challenges in physics is to figure out what happened to the antimatter”.

I think it's something of a "clickbait" mystery I'm afraid, to try to curry favour with the paying public. Why do I think that? Because there’s baryon asymmetry, where we see lots of protons but hardly any antiprotons, and there’s also lepton asymmetry where we see lots of electrons but hardly any positrons. And they match. Take a look at positronium. It’s “a system consisting of an electron and its anti-particle, a positron, bound together into an exotic atom”. The electron is matter, and the positron is antimatter. Positronium however is neither matter nor antimatter. It’s both. It’s an exotic atom. Now see page 314 of photon and electron collisions with atoms and molecules edited by Philip Burke and Charles Joachain. It says “the positronium atom thus created is in essence a very light hydrogen atom”. So positronium is both matter and antimatter, and it’s like light hydrogen.

The clear inference from that is hydrogen is also a combination of matter and antimatter. We say the electron is matter, and the proton too. But why? Surely since the proton is more like the positron than the electron, we ought to say the proton is the antimatter. Like this:

That would mean antimatter is a lot more common than people say. The proton mass is 1836 times the electron mass. Which would mean that weight for weight, 99.95% of everything around you is made of antimatter. Where’s all the antimatter gone? Nowhere. Because it’s hiding in plain sight. Because truth be told and by and large, you are made of it.


Is the effect when two planets passing each other different, when they pass in the same or opposite direction? So is the effect between the passing planets different whether they encounter in the same or opposite direction in their orbits with the same relative speed difference?

From the any of the planet's point of view, the encounter is exactly the same regardless of direction if the relative speed and force is equal. This should be pretty clear for symmetry reasons. For any encounter, the law is $v_{in} = v_{out}$. However, if the encounter takes a long time, you would also have to consider the influence of the central body of the system. The problem then becomes a three-body problem and this case is not restricted enough for an analytical solutions. We can make a few observations though: As the velocity vectors of both objects before the encounter have no radial component, both orbits are going to get a higher eccentricity. Any other information depends on knowing the distance of the encounter and the relative masses.