The star that died and became the Crab Nebula and Crab Pulsar? The supernova of this star was witnessed about 1 thousand years ago, and the star's remains are the Crab Nebula and Crab Pulsar. What is this no longer existent star called? How massive was this star? What was the star like before the supernova?

The crab nebula is more than 6000 light years away. Its progenitor was probably towards the lower end of stars that go supernova. And according to the luminoisity calculator had a magnitude of about 8.  This means that the progenitor was not visible to the naked eye.
As only the brighter naked eye stars have ever been named, and this star was not seen before it exploded, it had no name in any culture.
The star was probably about 8-10 solar masses. Prior to supernova it is assumed to have been a red giant.


If an asteroid twice the size of Earth passed super close would half of the Earth be pulled towards it? So let's say that an asteroid the twice the size of Earth passed by going relatively slow say in between the Moon and Earth.

Among so many other things (and perhaps you can touch on some of the biggest 'other' things), would anything on the asteroid's side of Earth be essentially 'sucked' towards the asteroid. That is if I am standing outside and it passes would I lift off the ground and float towards it?

Conversely, would things on the opposite side experience two times the standard weight? That is if I am standing on the other side, would I feel heavier as the asteroid passes?

I realise there would be many other devastating effects but I'm most interested in this one...

The possible answer I suspect is that objects on Earth will not 'feel' anything as the Earth as a whole will move towards the asteroid and not every little thing inside Earth...

In a close approach, Earth would feel tidal forces: the near side to the approach would be affected by the vistor's gravity more than the far side, because the force of gravity decreases with distance from a mass. If you were standing on the far side, though, your effective weight wouldn't increase -- it would decrease! This is because the visitor would be pulling the center of the Earth more strongly than it would be pulling on you, again because of the gravitational falloff with distance.

In a very close approach, closer than the Roche limit, the tidal forces would overcome the Earth's gravity completely, and Earth would begin to break apart. So it matters very much how close an approach we're talking about. Assuming the visitor has the same density as the Earth, the critical distance is about 10 Earth radii, or 64000km -- about a sixth of the distance from the Earth to the moon.


Can the Sun become a big ball of gold atoms? The atoms inside the core of sun are continuously fusing and forming other atoms. Can this continuous fusing make gold atoms?

Gold is heavier (further down the periodic table of elements, not necessarily more dense, although this is also the case) than iron. Iron's elemental number is 26. Gold is 79.

Any star that attempts to fuse iron will go into a supernova phase.

At that point, gold and all of the elements heavier than iron will be created, but only in the brief moments while the star is tearing itself apart.

No star can ever create gold in any circumstances other than during a supernova.  Our own sun is not nearly large enough to go supernova.

Any supernova will create gold, not just the type of supernova described above.

More reading: https://en.wikipedia.org/wiki/Type_II_supernova


Why isn't our local interstellar space colorised? Looking at answers to this question, people seem pretty convinced that human naked eye should be able to see the colour of ionised gas/dust cloud when they are inside an emission nebula.

And we know Solar system is surrounded by Interplanetary dust cloud, and possibly inside an Interstellar cloud. Why don't we see a colorised space backdrop?

Does it have to do with density of gas cloud? or how ionised they are? Does camera with long exposure see some background colour our naked eye can't?

(Excuse me for any wrong terminology.)

This is really more of a biology question, I think, so: The Mark-I Eyeball is far less sensitive to color than to intensity.  There are rods and cones in the retina. The neural signals from the rods are used to report intensity only. There are different spectral sensitivity curves for different cones (typically three types in humans), so the relative outputs from all cones are used to generate color information.  However, cones are less sensitive than rods, so for dim light such as that reaching us from nebulas, only the rods produce a neural output.


What exactly is a Hamiltonian telescope? Is this one? This comment on the current answer to Why is this telescope so short? How hard is it to make such a fast primary? says

In this forum topic Borisov appears to call it an f/1.5 Hamiltonian.

Wikipedia's Catadioptric system; Catadioptric dialytes) says:

Catadioptric dialytes are the earliest type of catadioptric telescope. They consist of a single-element refracting telescope objective combined with a silver-backed negative lens (similar to a Mangin mirror). The first of these was the Hamiltonian telescope patented by W. F. Hamilton in 1814. The Schupmann medial telescope designed by German optician Ludwig Schupmann near the end of the 19th century placed the catadioptric mirror beyond the focus of the refractor primary and added a third correcting/focusing lens to the system.

The image of the telescope in the linked question shows a very short and squat telescope with a 65 cm aperture that does not appear to have a "single-element refracting telescope objective", so I am confutes!
Questions:

What (the heck) is and is not a proper Hamiltonian telescope?
Is the Gennady Borisov telescope in the linked question (and shown here) one?

Gennady Borisov wrote on the astronomy.ru forum:

The first observations of the object were made in the morning of August 30 with a 650 mm telescope, F / 1.5 Hamilton

Первые наблюдения объекта были сделаны под утро 30 августа на телескопе 650 мм, F/1.5 Гамильтоне

Yes, this is Hamilton. All surfaces are spheres. Has a 3-lens corrector.
Good image quality, high aperture ratio. Large field of view: on FLI ML16803  FOV:  2.1 х 2.1 deg 1.87"/pixel

Да, это Гамильтон. Все поверхности сферы. Имеет 3-х линзовый корректор.
Хорошее качество изображения, высокая светосила. Большое поле зрения: на матрице FLI ML16803  FOV:  2.1 х 2.1 deg 1.87"/pixel


What could be the possible origin of boron found in the Martian soil if not from ancient groundwater? In a recent observation reported in the ScienceAlert article 'Boron has been detected on Mars for the first time" reporting that

  The Curiosity rover has found boron on the surface of Mars, indicating that, at some point, the Red Planet had long-term habitable groundwater.

Making the analogy that

  Boron here on Earth is associated with sites where there was once lots of water, but it's since evaporated away - like California's Death Valley.

However, they state that this may not necessarily be the case for Mars, so

What could be the possible origin of boron found in the Martian soil if not from ancient groundwater?

Research published in the PLoS One paper "Boron Enrichment in Martian Clay" (Stephenson et al. 2013) suggest a volcanic origin for the boron, in the chemical analysis of a Martian meteorite (found in Antarctica), they explain:

  A straightforward geochemical interpretation of our results is that boron, a relatively volatile and soluble element, was first concentrated in the fluid dregs of lava (4–7 ppm boron has been detected in the late stage mesostasis of other nakhlites) and then became further concentrated by any groundwater or hydrothermal fluids that came into contact with the rock.

Further modelling reported in the paper "Partitioning of light lithophile elements during basalt eruptions on Earth and application to Martian shergottites" (Edmonds, 2014), also point towards a volcanic origin. Their modelling is based on analyses of shergottite pyroxenes (Martian meteorites), stating that observed elevated boron alongside other light lithophile elements or LLE are to do with the earlier mantle chemical processes, extrapolated from those observed on Earth:

  The range in LLE and trace elements observed in shergottite pyroxenes are instead consistent with concurrent mixing and fractionation of heterogeneous melts from the mantle.

Also, a volcanic origin is suggested in the Goldschmidt Conference Abstract "Boron isotopes in the nakhlites: Implications for crustal fluids on Mars" (Spivak-Berndorf et al. 2008), also based on analyses of nakhlites:

  The lack of isotopic fractionation between the igneous and
  secondary alteration phases in the nakhlites indicates that (1)
  the B in the fluid that deposited the iddingsite was derived
  from the primary igneous minerals and (2) there was minimal
  B isotope fractionation between the fluid and the iddingsite


How are the days relating to right ascension around a polar star chart decided? If you look at this link they have days realting to right acension.

How did they calculate nov 20th == 0h, dec 6 == 1h?

I was trying to calculate this in astropy, but I end up with nov 20th, 8pm == 22h.

from astropy import units as u
from astropy.time import Time
from astropy.coordinates import SkyCoord, EarthLocation, AltAz

time_str =  "2017-11-20 20:00"

#Greenville SC, eclipse just went through there.
LON =  -82.3940
LAT = 34.8526
loc = EarthLocation(lat=LAT*u.deg, lon=LON*u.deg, height=0*u.m)

utcoffset = -4*u.hour  # Eastern Daylight Time
obstime = Time(time_str,scale='utc') - utcoffset

cord = SkyCoord(alt=90*u.degree,az=0*u.degree,obstime=obstime,frame = 'altaz',location = loc)
print obstime,cord.fk5.ra.hour, cord.fk5.dec.deg

>> 2017-11-21 00:00:00.000 22.505946538 34.7569396915

The date on the charts shows the meridian (north south line) at 8pm local time on that date.  Over a period of a month, the RA of the meridian will move 2hr (so that it gets back to where it started from after 12 months), and hence over 16 days, as you ask about, it will move 1.05 hours.  So you can either say that's close enough to 1 hr, or worry about the small discrepancy - that's assuming the 1hr you cite is actually 1.00hrs.


about H$_\alpha$ and H$_\beta$ lines in astronomical objects' spectra Generally there may be some line ratio between the two lines.

If  H$\alpha$ is strong, it is quite possible we can see H$_\beta$ too.

Is it possible we can see H$_\beta$, but can not see H$_\alpha$? In what kind of environments can it happen?

In a normal stellar atmosphere, where the temperature decreases with height, I think this is not possible.

H-alpha absorption arises from the $n=2$ to $n=3$ transition; H-beta from the $n=2$ to $n=4$ transition. Thus both transitions are governed by the number of atoms in the $n=2$ lower level.

The absorption coefficient for a transition can be written in proportionality terms by (in local thermodynamic equilibrium)
$$ \alpha_\nu \propto  \nu B_{lu} n_l ( 1 - \exp[-h\nu/kT]),$$
where $B_{ul}$ is the Einstein absorption coefficient, $n_l$ is the population of the lower energy level and $\nu$ is the photon frequency corresponding to the transition.

Using the well-known relationship between the Einstein emission and absorption coefficients and values for the emission coefficients found here, we can estimate a ratio of absorption coefficients for a given temperature.

$$ \frac{\alpha_{H\beta}}{\alpha_{H\alpha}} = \frac{g_{n=4}}{g_{n=3}} \frac{A_{H\beta}}{A_{H\alpha}} \left(\frac{\nu_{\alpha}}{\nu_{\beta}}\right)^2 \left(\frac{1 - \exp(-h\nu_{\beta}/kT)}{1-\exp(-h\nu_{\alpha}/kT}\right),$$
where $A_X$ are the Einstein emission coefficients and the statistical weights $g_n$ are given by $2n^2$.

Thus if we take $\nu_\alpha = 4.57\times10^{14}$ Hs, $\nu_{\beta}=6.17\times10^{14}$ Hz, $A_{H\alpha} \simeq 10^{8}$ s$^{-1}$, $A_{H\beta} \simeq 3\times 10^{7}$ s$^{-1}$, then
$$\frac{\alpha_{H\beta}}{\alpha_{H\alpha}} = 0.29 \left( \frac{1 - \exp(-29642/T)}{1 - \exp(-21956/T)}\right)$$

So when $T$ (in Kelvin) is small, the ratio is about 0.3. When $T$ becomes large then the ratio increases, but above $T \sim 12,000$ K, all the Hydrogen is ionised.

Thus in thermodynamic equilibrium is it very difficult to manufacture a circumstance where the optical depth in the H$\beta$ line is larger than that in the H$\alpha$ line.

[I'd be grateful if someone can tell me whether the ratio I calculated above is correct, since I could not find it in any reference and need to work it out from scratch].


What parameters cause sunburn? I am new to this site so if the question doesn't fit in this forum feel free to remove it.
I have read here https://en.wikipedia.org/wiki/Sunburn of all the causes of sunburn and I was mostly aware of all of them. I will list them here:

($1$) The time of day. In most locations, the sun's rays are strongest between approximately 10am and 4pm daylight saving time.
($2$) Cloud cover. UV is partially blocked by clouds; but even on an overcast day, a significant percentage of the sun's damaging UV radiation can pass through clouds.
($3$) Proximity to reflective surfaces, such as water, sand, concrete, snow, and ice. All of these reflect the sun's rays and can cause sunburns.
($4$) The season of the year. The position of the sun in late spring and early summer can cause a more-severe sunburn.
($5$) Altitude. At a higher altitude it is easier to become burnt, because there is less of the earth's atmosphere to block the sunlight. UV exposure increases about 4% for every 1000 ft (305 m) gain in elevation.
($6$) Proximity to the equator (latitude). Between the polar and tropical regions, the closer to the equator, the more direct sunlight passes through the atmosphere over the course of a year. For example, the southern United States gets fifty percent more sunlight than the northern United States.

My question is, can we compose ($1$), ($4$) and ($6$) into one statement: "Height of sun over the horizon"?
Wikipedia seems to hint that we can since it says:

Regardless of one's latitude (assuming no other variables), equal shadow lengths mean equal amounts of UV radiation.

However, the statement is kind of vague and I would like confirmation.
Another doubt I have regarding this is experience. I live in Europe where solar noon is at 1pm during summer and the day is longest on June 21 (or 22, whatever). But, people seem to get way more sunburn in, say, August 21 than in April 21, and also more at 4pm than at 10am (regardless of the month). It is also hotter in August than in April and at 4pm than at 10am. So it might be that people simply spend more time out because it's hotter and therefore get more sunburn, but it also might be that the temperature affects our skin somehow and we do get more sunburn when it's hotter.
So, taking clouds, reflective surfaces and altitude out of equation, is it true that sunburn is caused just by the height of the sun above horizon? If I spend 5 minutes at some place on a sunny day at noon in April, will I get the same amount of sunburn as I would at 3pm in June? (given that the sun is at the same height over the horizon at that times, and that 5 minute span is short enough to approximate that sun stays "fixed" in the sky)
My intuition says it should be true that height over horizon is the only parameter that should matter, but my experience shows different. Moreover, if I am correct, is there some graph/table showing how much intensity (or some other unit?) of sunburn I get with respect to height of sun? (say at zero altitude on a sunny day without reflective surfaces, or at some other, fixed, reference configuration).

Yes, all of those contribute to the total irradiance, which is the amount of sun power falling on a particular area, measured in Watts per square meter. You can imagine a 1m$^2$ "window" perpendicular to the sun's rays - excluding atmospheric and weather effects, the amount of sunlight passing through that window never changes. But when the sun is directly overhead, that window illuminates a 1m$^2$ patch of ground, while later in the day that same sun power is spread over a larger area. Only the angle of the sun affects how big that area is, and that is determined, as you point out, by time of day, season, and latitude.

Because of this, the length of your shadow is a good tool to estimate the angle of the sun and your risk of sunburn. I tend to be a bit sun-sensitive, but if my shadow is longer than I am tall (i.e. the sun is below 45 degrees), I'm usually fine - that's most of the day in winter, but excludes the hours around noon in the summer. That angle will of course have to be adjusted based on your personal sensitivity.

As for your experience, I would imagine that people tend to expose more skin in the hottest month of the year (August), increasing their sunburn risk over a period with higher irradiance but less exposure (June). Sunburns also will also often show up in the afternoon because they're a result of accumulated damage throughout the day. If an individual takes X hours of exposure to burn, they're more likely to hit that mark later in the day, rather than during the peak of irradiance.


Why can’t I see any detail on planets? I really want to know why I can’t see detail on planets.

This summer my Dad and I went on a trip to Utah I went outside to look at the planets and couldn’t see detail on Mars, Saturn, and Jupiter. Saturn's rings were visible but very faint and Jupiter was a white ball and then it’s four moons, this was all out in the middle of nowhere with 0 light pollution at 6000ft.

We then went into Bryce Canyon at 9100ft and still the same outcome the only difference was a very very faint strip on Jupiter, and it’s so dry in Utah and at an altitude of 9100ft where the atmosphere is not as heavy.

Is there still dew? Is my telescope not powerful enough I have an Astromaster 130 reflector and a 60az explorerscope refractor. I’m very confused

Planets are really small!  (when seen from Earth)

Deepskywatch has an example of what can been seen with a 150mm reflector and a magnification of x180 (ie slightly more than your Astromaster which has a 130mm main mirror, and comes with a x65 eyepiece)

You should be able to see the 4 moons, and one or two bands on Jupiter. This assumes good "seeing" (the amount of turbulence in the air, not the same as the clarity of the air), and a properly collimated and aligned telescope lens.

There are lots of other good deep-sky targets that benefit more from a dark sky than the planets.


Why is the time between equinoxes different? The path of the Earth around the sun is described as being an ellipse, with the Sun being at one of the foci of the ellipse.
Since an ellipse, by definition, is symmetrical with respect to the foci, I would expect the time for the earth to transit any two halves of the ellipse to be the same, so long as it moves at the same speed around the ellipse, or at least the speed characteristics are symmetrical as well.
However, this does not appear to be the case according to the following rounded figures for the time in days between equinoxes and solstices:
Summer Solstice to Winter Solstice    184.3
Winter Solstice to Summer Solstice    181.0
Vernal Equinox to Autumnal Equinox    186.4
Autumnal Equinox to Vernal Equinox    178.9

These values can be found, for example, in astronomer Robert Newton's book of 1976.
Why are the values different?

This is because the summer and winter solstices (approx. June 21st and December 21st) do not correspond to the aphelion and perihelion (approx. July 5th and January 4th). Therefore, the average distance from the Sun is longer in the period from the Summer Solstice to Winter Solstice than vice versa, so the Earth is moving slower (on average) and it takes longer.

I think the values you mentioned are incorrect (switched around); 181.0 is a more likely value for Winter to Summer (February is a few days shorter than August).


Did Edmund Scientific 4¼ inch Newtonians have 90° prisms as secondary mirrors in the 1970's? I mentioned my old telescope's secondary mirror in this answer (stiffness) but I'm not certain I am remembering correctly. My first telescope was a Edmund Scientific 4¼ inch Newtonian reflector. This was in the 1970's. I seem to remember that the secondary mirror was a right angle prism rather than an elliptical front-surface mirror.

Am I dreaming or mis-remembering, or did Edmund use prisms on at least some 4¼ inch Newtonians?

Either way, any discussion on the tradeoffs between front-surface elliptical mirrors and 90° prism secondaries would be appreciated.

Source: John M. Pierce's HobbyGraph Articles HobbyGraph #15; Diagonals and Diagonal Supports.

My Edmund F/10 4 1/4" was bought in 1975, and had a rectangular front-surface mirror as the diagonal, with a single arm support attached to the tube near the eyepiece holder.

I believe the Astroscan 2001 (or at least some models of it) used a prism for the diagonal, supported by the flat glass plate at the front of the tube; haven't found a reference though.

The advantage of a prism is that it is more maintenance-free then a mirror which will eventually need to be re-aluminized.  The disadvantages are that there are two optical surfaces instead of one in the light path that could reflect or scatter light, you can get color dispersion, and a rectangular prism will block more light than an elliptical mirror (especially important in a fast scope like the AstroScan).  The star diagonal Wikipedia page has a pretty good discussion of the trade-offs.


Are there actual asteroids in the Oort cloud? This answer to At what annual rate are new exoplanets being recognized? How does it compare to new asteroids? shows that the rate of new asteroid discovery is roughly two orders of magnitude higher than the rate of new exoplanet discovery, and once Vera C. Rubin comes online the disparity is likely to widen even further.
I quipped:

Thinking about the cross-over rate; at some point won't all the asteroids have been found?

there and then wondered if there is any distance limit to what is or isn't an asteroid, or are there new asteroids out there to discover all the way to the point where matter can no longer be gravitationally bound to the Sun because the pull of adjacent stars is just as strong.
Question: Are there asteroids per se in the Oort cloud, or for that matter, the Kuiper belt? Or are those basically considered comets? (trick question alert! see Do astronomers generally agree that the distinction between comets and asteroids is not so clear?)
Wikipedia sez the proposed Oort Cloud is full of icy planetesimals, and not being an astronomer I don't know how exactly to parse that.

Tl;Dr: It is a matter of consideration if you want to call Kuiper belt bodies/Oort cloud bodies "asteroids" or not.
Long answer:
The Oort cloud defines the cosmographic boundary of the Solar System and the extent of the Sun's Hill sphere and hence it is loosely bound to the Solar System, and thus is easily affected by the gravitational pull both of passing stars and of the Milky Way itself. These forces occasionally dislodge some bodies from their orbits within the cloud and send them toward the inner Solar System. This is how comets are originated. Based on their orbits, most of the short-period comets may come from the Kuiper Belt/scattered disc and the long period comets are thought to originate from the Oort cloud.
So, the general assumptions is that maximum comets(if not all) are Kuiper Belt/Oort cloud bodies but all Kuiper belt/Oort clouds bodies are not comets.
So, are they asteroids?
You "may" consider most of the KBOs and OCBs to be more like asteroids because most of the the them have fairly circular orbits around the Sun, and most of them don't come close to the Sun at all (only those which comes close to sun are considered comets). So if an object needs to have an icy composition and a highly elliptical orbit to be considered a comet, then KBOs and OCBs are more like an "icy asteroid belt and bodies" than a group of comets.
However, such classifications were done based on what we knew 60 years ago, and it's now clear that the Solar System is much more complicated. So, we need to have a better classification to avoid confusion. Now, we name celestial bodies in a very specific manner considering various factors like position, behavior, size, age, composition etc., for instance: "Main Belt Asteroids", "Kuiper Belt Objects", "Near-Earth Asteroids", "Long-Period Comets" etc., as opposed to just "asteroids" or "comets".


Supergalactic and Equatorial coordinates What's the difference between Supergalactic and Equatorial coordinates?

Equatorial coordinates have their equator and poles at the equator of the earth and the poles at the earth each projected onto the sky.

Supergalactic coordinates on the other hand have their equator in the same plane as the "Supergalactic plane", which comes from the distribution of nearby galaxy clusters. And the poles just perpendicular to the the plane.You can kind of see this preferred plane in the image below.

One is therefore just a rotated or tilted version of the other. To be precise the north poles of Supergalactic coordinates lies at approximately RA = 18.9h, Dec = +15.7° in Equatorial coordinates.

See also here here for a comparison of common celestial coordinate systems.


SDSS. Google Sky and DeCaLs (1) Please go through following image
Possible Super Red Giant?
However sdss shows it as galaxy! Am I missing something.

(2) Google Sky shows object in latitude and logitude format like this.
Is there any way to convert those numbers to Right Ascension (ra) and Declination (dec) numbers or opposite of that  so that I can spot google sky data on SDSS and SDSS object on Google Sky?

The Sloan Digital Sky Survey has saturation limits at around 13th or 14th magnitude. You are looking at $\phi$ 1 Cancri, which is a 6th magnitude star. It is hardly surprising that SDSS fails to morphologically classify it correctly.

No idea what your second question is about. You have given a link which has a latitude and longitude in the url. Longitude and latitude are not valid ways of expressing the coordinates of a celestial object. When I open this link, I see an image of your star and in the bottom left of the screen there is a display which gives the RA and Dec of where the cursor is.


Is Jupiter's rotation slowing? According to Wikipedia, re Europa's tidal flexing:

  the tidal flexing kneads Europa's interior and gives it a source of
  heat, possibly allowing its ocean to stay liquid while driving
  subsurface geological processes.The ultimate source of this energy is
  Jupiter's rotation, which is tapped by Io through the tides it raises
  on Jupiter and is transferred to Europa and Ganymede by the orbital
  resonance.

https://en.wikipedia.org/wiki/Europa_(moon)

Does this imply that Jupiter's rotation is slowing over time, since it is losing energy due to raising tides on Io?  (Or just generally, does raising tides cause rotation to slow?)

Yes

Tidal locking is a mutual process, and both the satellite and the parent body will eventually
permanently face each other. However, while tidal locking of a satellite can happen relatively quickly (that is, within the life span of the solar system), tidal locking of the parent body is extremely slow and will in a practical sense never happen (the current age of the universe is not enough in most cases).

The relative mass of the satellite and the parent body is one of the most important factors:

Charon and Pluto with a mass ratio of just 8.1 are already mutually locked.
The Moon and Earth has a relatively high mass ratio of 81. This causes a measurable effect on the Earth.

Compare that to the mass ratio of Juptier and Europa of 39,600 (or Io: 21300, Ganymede 12,800, Callisto: 17,600). While the time required for a locking of Jupiter itself is difficult to measure due to the fact that we don't have the necessary data about the moons, this is not a noticeable effect before the Sun dies.

Tidal locking of the parent body can most likely be justified by conservation of energy, but conservation of momentum gives a more intuitive view: As the satellite slows down and moves away from the parent body, its angular momentum increases (angular momentum is proportional to distance times velocity, and the orbital velocity is proportional to the square root of distance). This angular momentum must be supplied from somewhere. The only source for that is the rotation of the parent body. Even if all the moons of Jupiter were to completely escape, the angular momentum drain is not enough to stop the rotation of Jupiter.


How are mars years counted? In this diagram about the methane concentration in the martian atmosphere, there are data points labeled "Mars year 32", "Mars year 33", and "Mars year 34". How are Mars years counted? Is there a "Mars year 0"? What is the event that defines the start of the time scale?

Given that the methane variations are seasonal, the diagram is labeled in Martian years - the amount of time it take Mars to travel around the Sun once (about 1.88 Earth years). This makes it much easier to see the cycles rather than using Earth years. In particular, the x-axis has a length of one Martian year; the data point show seasonal methane measurements for that year. So, for instance, the red data point was take early in the spring of Mars year 34, the first blue point was taken at the start of summer of Mars year 32 (earlier!), and so on and so forth.

The specific year numbering is an arbitrary convention begun on April 11, 1955 (the northern spring equinox). The solar longitude system is applied to Mars; it's a geometric way of describing where in its orbit a planet is. The solar longitude is denoted $L_2$; on April 11, 1955, $L_s=0^\circ$; it increases to $L_s=360^\circ$, at which point a new Martian year begins.


Gravitational attraction of Sun on distant object A hypothetical situation which will help me with a real-life problem.
Several stationary objects are placed at great distances from our Sun -- say 0.1 light year, 0.5 ly, 1.0 ly. Assuming no other influences;
What velocity would each have when it reached the Sun?
How long would each take to reach the Sun?
Assuming the object-Sun path was perpendicular to the ecliptic & the Sun had its actual motion, would the objects hit the Sun, or (roughly) by how much would they miss it?

The full equation for the time for an object to drop is
$$t =  \frac{ \arccos \Big( \sqrt{ \frac{x}{r} }\Big)  + \sqrt{ \frac{x}{r} \ ( 1 - \frac{x}{r} ) } }{ \sqrt{ 2 \mu }  }  \, r^{3/2},$$
where $x$ is the radius of the sun, $r$ is distance of the object, and $\mu=GM=1.327\times10^{20}$. (in SI units, so you will need to convert your distance to metres, and get a time in seconds.)

As $x<<r$ you can assume $\frac{x}{r}=0$ so the formula simplifies to
$$\frac{\pi r^{3/2}}{2\sqrt{2\mu}}$$

For an object released at 0.1,0.5 or 1 light year that is 3000000 years, thirty million years or a hundred million years.

For the speed on impact, you can assume that the body will reach very close to the escape velocity of the sun, which is about 620 km/s. It matters little whether the body is released from 0.1, 0.5 or 1.0 lightyears, because the body will pick up speed only when it approaches the sun.

The last part of you question is unclear. If the body is at rest relative to the sun, and the various gravitational affects of other stars are ignored, then it will fall towards the sun. If on the otherhand the body is at rest relative to the centre of the galaxy, then it is moving extremely fast relative to the sun, much faster than the sun's escape velocity. It won't fall anywhere near to the sun, instead the sun will deflect it slightly as it falls towards the centre of the galaxy.


Which focus point does the sun occupy for each planet? I am aware that planets orbit in an elliptical fashion and that the sun occupies one of two focal points.

Let us say that the left focal point is f1 and the right f2.

Is the sun at the same focal point for each planet?

If not which ones are different?

In addition to the other answers, that are right, a different doubt seem to arise from the wording of the question:

Let us say that the left focal point is f1 and the right f2.
Is the sun at the same focal point for each planet?

If the question is if the Sun is in f1 or f2, the answer is that it doesn't matter. Ellipses are symmetrical, and therefore both foci are equivalent and any naming of them is just conventional. Just by saying that planets move in elliptical orbits and the Sun occupies one focus, Kepler's first law is unambiguously established.
If you need to mention a particular focus, one is just the Sun position and the other one is often referred as the empty focus.


How does radio bandwidth restrict the choice and use of science payload on interplanetary probes? I asked a question here about returning to Earth a physical memory capsule along with soil samples, as a complement to radio operations as today. This might return a much larger amount of data compared with radio today, but of course only at the end of the mission.

How useful would it be for a sample return mission like Mars2020 or Hayabusa2, to be able to return much more data? Would the choice of instruments change to allow for higher sampling rates, higher resolution, more filters et cetera? Is Big Data science instruments from an interplanetary mission even a possibility, or is today's bandwidth good enough for any conceivable payload?

Sample returns are expensive and risky.  Radio is simple and reliable.  Radio bandwidth on a particular mission is determined by several engineering tradeoffs.  Higher bandwidth requires either larger antennas, more output power, or more complicated processing.

Spacecraft are typically limited in mass, power and compute capabilities. What most missions have plenty of, is time.  Lower bitrate takes longer, but allows more mass and power to be used by other experiments.  This is often a highly desirable tradeoff.

Missions like New Horizons that have a fairly limited time to record lots of observations do so to a data storage device, and then spend months or years relaying the data back to Earth during the long coast to the next target.


How can I find the radius of a exoplanet using doppler spectroscopy I have the stellar and planet mass, planet velocity, radial velocity of star, stellar semimajor axis.
How should I go about to find the radius of this exoplanet?

In general, you can't. The spectroscopy tells you about the properties of the star and the properties of the orbit and gives some information about the mass of the planet ($M \sin i$, where $i$ is the orbital inclination).
If the spectroscopy were super-precise and with extremely high signal-to-noise and high time resolution, then you might be able to map the Rossiter-McLaughlin effect with such precision that you can estimate the planet radius. However, this would only work if the planet is transiting in front of the parent star - in which case you might as well have used transit photometry in the first place.
Another possibility is to search in the spectra for the Doppler-shifted scattered starlight from the planetary atmosphere (e.g. Martins et al. 2015). However the amplitude of this is tiny depends on both the radius of the planet and its albedo/atmospheric composition.


Can a pulsar hit us with both beams? Can a pulsar have an axial tilt close enough to $90^\circ$ to hit us with both beams for two pulses per rotation?

Yes. One of the features that helps this scenario is the gravitational bending of light close to the neutron star, which allows for a larger fraction of the surface to be visible at a given time.

Many pulsars have double-peaked light curves, for example here's the light curve of the Crab Pulsar. In the pencil beam model these peaks correspond to the beams from the two poles.

According to Annala & Poutanen (2010) "Constraining compactness and magnetic field geometry of X-ray pulsars from the statistics of their pulse profiles", the ratio of single-peaked to double-peaked pulsar light curves puts a most likely value for the maximum inclination of the magnetic axis relative to the rotational axis of 40°±4°.


Why is CMB an evidence of hotter and denser early Universe? In his book Gravitation and cosmology, Steven Weinberg says that CMB makes it  "difficult to doubt that the universe has evolved from a hotter denser early stage".
In my understanding, CMB is just a peculiar isotropic radiation representing a black body at ~ 2.7K.
How and why does CMB point to the early Universe being hotter and denser?
[The interested reader may refer to answers on Physics SE.]

By request:

Beyond the fact that the cosmic microwave background (CMB) is a direct prediction of the big bang model, there is the question of how you would produce it in any other way. It is remarkably close to being isotropic and remarkably close to being a blackbody spectrum - i.e. it is almost a perfect blackbody radiation field.

A blackbody radiation field is emitted by material in complete thermodynamic equilibrium (CTE). An example would be the interior of a star. A requirement for (CTE) is that the matter and radiation field are characterised by the same temperature and that the material is "optically thick" - meaning that it is opaque to that radiation at basically all wavelengths.

Given that the universe is mainly made up of hydrogen, helium and (presently) traces of heavier elements, we can ask how is it possible to produce a perfect blackbody radiation field? Cold hydrogen and helium are transparent to microwaves. To make them opaque they need to be ionised, so that the free electrons can be a source of opacity at all wavelengths via Thomson scattering. But this requires much higher temperatures - about 3000 K.

How do we uniformly raise the temperature of a gas (adiabatically)? By squeezing it. A smaller, denser universe would be hot enough to have ionised hydrogen and would be opaque to the radiation within it. As it expanded and cooled, the electrons combined with protons to form atoms and the universe becomes transparent, but filled with a perfect blackbody radiation spectrum. The light, originally at a temperature of 3000 K and mainly in the visible and infrared, has had its wavelengths stretched by a factor of 1100 by expansion of the universe, meaning we now see it mainly as microwaves.

Additional evidence for this model is that the radiation field is not absolutely isotropic. These small ripples encode information such as the expansion rate of the universe at the time of (re)combination and the density of matter. When inferred from measurements, these parameters agree very closely with other determinations that are independent of the CMB, such as  the Hubble redshift distance relationship and estimates of the primordial abundance of Deuterium and Helium.

There is now direct evidence that the CMB was hotter in the past and by exactly the amount predicted by an adiabatic expansion. The source of this evidence is measurements of the Sunyaev-Zel'dovich effect towards galaxy clusters (e.g. Luzzi et al. 2009); or more precisely by probing the excitation conditions in gas clouds at high redshift using even more distant quasars as probes (e.g. Srianand et al. 2008.


Are we really seeing the asteroids from the past? Is it actually relevant to study objects that are light years far? Because in theories we are actually seeing the past as the light takes years to reach us.

Lets say we are studying a asteroid that is going to hit as after several hundred years how can we confirm we are truly studying the actually position and everything of that object as we are seeing object from the past.

Most things in space move very predictably. For reasonably dense objects, like stars, planets or asteroids, gravity far outweighs all of the other forces acting on them, and it's not too hard to take account of gravity and predict their movements (if you have a powerful enough computer. For things like clouds of very thin gas it's a bit harder. So from the positions and movements of all the distant objects that we can see (at the time when the light we see them by left them), we can pretty accurately predict their movements a considerable way into the future, for instance up to possible collisions a few centuries out (for an asteroid) or a few billion years out (for Andromeda).

Of course there is a possibility of something unexpected happening -- between the light we see now being emitted and a possible future collision, a supermassive black hole travelling at close to the speed of light could pass through the Andromeda galaxy and scatter it, for instance, but we don't see things like that happening anywhere else in the universe, so it seems pretty unlikely to happen here.


How do I find the RA of sunset and sunrise in a specific location? I want to make some observations with a telescope in Hawaii on Mauna Kea but I am living somewhere else (not Hawaii) and I want to plan a night's viewing on the Hawaiian telescope. How do I find the RA range of the night? In other words; how do I find the RA range of the first object observable to the latest object observable for a particular time (of year) for a specific location (in this case Hawaii)?

Use Stellarium. You can set your location and date as needed and simulate all the conditions (ignoring weather...) :)

http://www.stellarium.org/


Properties of low-mass stellar remnants vs the Earth How does the Earth differ from a (low-mass) stellar remnant, which has stopped fusion and the outer layers of which have been blown away?

Could a stellar remnant end up with a similar relative abundance of elements as the Earth?

i.e. Is it possible that a star loses so much matter when it dies that the remnant doesn't turn into degenerate forms of matter?

Stellar remnants are completely different from planets.

The Earth was never a star and fusion has never occurred in the Earth's core at any time in its history. When a small to medium sized star dies, and the outer layers are lost, what remains is a white dwarf. It is still much more massive than the Earth, and is very hot. It is crushed by its own gravity, causing the matter to become "degenerate".

On Earth, atoms are held together by chemical bonds. Lots of interesting patterns can be formed by the atoms, producing minerals, rocks, seas and life. Nothing like this can happen in degenerate matter.

In degenerate matter, the atoms are pushed together by gravity. Degenerate matter is unlike regular matter. It is much much more dense, and chemical bonds are not a significant force between atoms.

White dwarfs are formed of carbon, oxygen, hydrogen and helium (with the lighter elements on the surface). Even after a white dwarf has cooled, it would still be degenerate, and quite unlike the Earth, in composition, properties and in density.


Is there online data on asteroid axial tilts? I am hoping to find axial tilts for asteroids and also their spring and fall equinox. Some of the asteroids I'm interested in are:
4 Vesta, 1 Ceres, 24 Themis, 65 Cybele, 153 Hilda, 624 Hektor

From this article, http://hubblesite.org/pubinfo/pdf/2010/33/pdf.pdf, here's the rotation axis for 4 Vesta: RA=307.5°±3.1°, Dec=43.1°±1.2°. Update, per TildalWave's comment: RA=305.8°±3.1°, Dec=41.4°±1.5°

Rotational motion of Ceres.

An old article (1986): Magnusson, P., Distribution of spin axes and senses of rotation for 20 large asteroids, Icarus 68:1 1-39 http://www.sciencedirect.com/science/article/pii/0019103586900722


Will comet 21P that just passed Earth on Sept 10 produce Draconid storm on Oct 9? Comet 21P/Giaccobini-Zinner just made its closest pass to Earth and also made its Solar perihelion on the same day (Sept 10) which is roughly one month ahead of the annual Draconids meteor shower (Oct 9).  In both 1933 and 1946 these same circumstances also occurred and there were reportedly 1000s of meteors per hour in those years - meteor storms.

Question - I know the expert forecasts are not calling for more than 20 or so per hour based on some reference data about debris streams, but is there any chance that they may be wrong this year in their forecasts? Could the favorable circumstances this year plus a new debris stream result in a meteor storm surprise next week?

Earthsky discusses this.

There isn't a storm every perihelion, and predicting meteors is tricky, as we can't observe the meteoroids in space, we don't know if the Earth will pass through a dense patch or not.

Key quote "How many Draconid meteors will you be able to catch? No one expects a Draconid storm this year, but one can always hope!"


What does the term alien planet actually refer to? This article left me wondering what do they mean by the term alien planet? Aren't all planets simply alien by default, or do they mean planets outside our solar system are alien? Do they mean there is alien life there? Is this a a new keyword the astronomy community is using to generate more buzz and get more attention to their articles?

Very interesting article!. The term alien planet is not a scientific one, so I think you already answered your own question by saying that it was used in the article to generate more buzz.

In the English language the word is used to describe something that's not familiar, so you could argue that this is in fact an alien planet, but because of the subject in question being an exoplanet I would bet they did this as a technique to get more attention.


Moon is a planet by "Clearing the neighborhood" condition? Whether the Moon (Earth's natural satellite) is a planet is an old debate in planetary science.
Though I do not agree that Moon is a planet, I found something difficult to clearly understand.

According to the 2006 IAU definition of planet, a planet must satisfy Clearing the neighbourhood condition.
And some criteria for the condition are introduced in the Wikipedia article about Clearing the neighbourhood.

Stern–Levison's $\Lambda$

$$\Lambda = \frac {m^2}{a^{\frac 3 2}} k$$

Margot's $\Pi$

$$\Pi=\frac {m}{m^{\frac 5 2}a^{\frac 9 8}} k $$

In the above equations, $m$ is the mass of the body, $a$ is the body's semi-major axis, and other factors are described in the Wikipedia article. Both criteria are said to be satisfied if the calculated value is greater than 1.

My calculations are simple. Let $m_{Moon} = 0.0123 \times m_{Earth}$ and Moon's other factors (such as $a$, $k$ and $M$) have the same values as Earth's. Then I can get...

$$\Lambda_{Moon} = 0.0123^2 \times \Lambda_{Earth} = 23.147$$

$$\Pi_{Moon} = 0.0123 \times \Pi_{Earth} = 9.963$$

Both $\Lambda_{Moon}$ and $\Pi_{Moon}$ thus satisfy the Clearing the neighbourhood condition. Though I do not calculate the Soter's $\mu$, it will not change my results.

What is wrong with my calculations and results? Do I miss something about the condition and the criteria? Or is there something wrong with the Clearing the neighbourhood condition?

Edit in response to comments:

This question can be related to the idea that the Clearing the neighborhood condition is inadequate and ambiguous for the planet definition.
More deeply, this question can be related to the idea that 2006 IAU definition of planet is logically ambiguous and clumsy.
For example, the 2006 definition of planet defines the Moon as a satellite and not a planet.
If one celestial body can be defined as a planet or not without rules and conditions, why other conditions are needed? All we need is just a planet list, not other planet conditions.

And...

Frankly speaking, the Clearing the neighborhood condition is like comedy.
The condition says "Planet must sweep out other bodies" ...
Then says "Except satellites" ...
Then again says "Except bodies having no significant size" ...
Then again "Except bodies having resonant orbit" ...
Then again "It may take billions of years" ...
Then again "Sweep-out process and time are influenced by other planets and bodies, actually" .......

It is obvious that sometimes the Clearing the neighborhood theory is useful. But it is inappropriate for the rule and definition requiring strictness.

According to the 2006 IAU definition of a planet, a celestial body must meet all 3 criteria to be considered a planet.  The Moon fails the first criteria "1. Is in orbit around the Sun".  The Moon orbits Earth, not the Sun, and thus it doesn't have a legitimate "neighbourhood" (solar orbit) to clear in that context.

Also to note, according to the Clearing the neighbourhood article, this criteria is contentious as it is vaguely written and Earth, Mars, Jupiter and Neptune have not "cleared the neighbourhood" of their respective solar orbits.


What would cause large jumps in brightness during a solar eclipse? This question on Physics SE about apparent brightness during a solar eclipse has a log-scale graph of light intensity versus time.  Parts of it are the shape I'd expect: a nearly-smooth curve going from "extremely bright" to "very dim".  However, there's a great deal of apparent noise in the graph, with jumps of nearly a full division in the graph (corresponding to about a two-fold or eight-fold change in brightness, depending on if it's a natural or base-10 log).  What causes these jumps?

Eric Pauer, who produced the chart, has an interesting web site that covers this eclipse.  It is evident that the weather was not great at the time of the eclipse, with intermittent cloud, and it will have been passing clouds that caused the variations in intensity.

You can see the evidence from this page
http://www.pauerhome.com/eclipse/eclipse99/1contact/1contact.htm


Horseshoe orbits and integration in C I'm studying a particular case of the restricted three-body problem. It's been found that some objects follow a horseshoe orbit pattern, and I'm trying to sort out something through an integration code in C. I'm following some advice in the article Families of periodic horseshoe orbits in the restricted three-body problem, which gives me ideal initial conditions and the equations in the centre of mass system. (m is the mass of Earth, and consequent position of the sun in the center of mass reference system, (x,y) are the coordinates of the third body, assumed massless (as the restricted problem requires).

$$ O=(x^2+y^2)/2+\dfrac{(1-m)}{r1}+\dfrac{m}{r2}+\dfrac{(1-m)m}{2} $$
$$r1^2=(x-m)^2+y^2$$
$$r2^2=(x-m+1)^2+y^2$$
$$ a(x)=\dfrac{\mathrm{d}{O}}{\mathrm{d}x} + 2v(y)$$
$$a(y)=\dfrac{\mathrm{d}{O}}{\mathrm{d}y}-2v(x)$$

The positions of the "sun" and "earth" are fixed to (m,0) and (m-1,0), in the same reference system. (rotating reference system, assuming the earth has a circular orbit.)

From all this I've calculated the equations to describe the system:

$$a(x)=x+\dfrac{(m-1)(x-m)}{((x-m)^2+y^2)^1.5}-\dfrac{2m(x-m+1)}{((x-m+1)^2+y^2)^1.5} + 2v(y)$$
$$a(y)=y-\dfrac{y(1-m)}{((x-m)^2+y^2)^1.5} -\dfrac{2ym}{((x-m+1)^2+y^2)^1.5} -2v(x)$$

I've used the algorithm of Runge-Kutta 4 to integrate those equations.
(I know the code is quite mind-twisting, but I just can't use pointers and I use structs everywhere).

#include<stdio.h>
#include<stdlib.h>
#include<math.h>

#define dt 0.0001
#define N 100

typedef struct{
    long double x,y;
}vec;

typedef struct{
    vec k1,k2,k3,k4;
}runge;

typedef struct{
    runge r,v;
}big;

double dS,dE,m;

double accx(double,double,double);
double accy(double,double,double);
void rad(vec);
big rungekutta(vec,vec);
vec moto(vec,runge);
double jacobi(vec);

int main(){
  vec r,v;
    big f;
    double J,t;
    int i,Num;
    FILE* s1;
    s1=fopen("HorseShoe.dat","w");

    Num=(int)N/dt;
    scanf("%Lf",&r.x);
    scanf("%Lf",&r.y);
    scanf("%Lf",&v.x);
    scanf("%Lf",&v.y);
    scanf("%lf",&m);

    for(i=0;i<Num;i++){
        t=(i+1)*dt;
        rad(r);
        f=rungekutta(r,v);
        r=moto(r,f.r);
        v=moto(v,f.v);
        J=jacobi(r);
        fprintf(s1,"%lf\t%Lf\t%Lf\t%Lf\t%Lf\t%lf\n",t,r.x,r.y,v.x,v.y,J);
    }
return 0;
}

void rad(vec r){
    dS=pow(r.x-m,2)+pow(r.y,2);
    dE=pow(r.x-m+1,2)+pow(r.y,2);
}

double jacobi(vec r){
    return pow(r.x,2)+pow(r.y,2)+2*(1-m)/dS+2*m/dE+m*(1-m);
}

double accx(double x,double y,double v){
    return x-(x-m)*(1-m)/pow(pow(x-m,2)+pow(y,2),1.5)-m*(x-m+1)/pow(pow(x-m+1,2)+pow(y,2),1.5)+2*v;
}

double accy(double x,double y,double v){
    return y-(1-m)*y/pow(pow(y,2)+pow(x-m,2),1.5)-m*y/pow(pow(y,2)+pow(x-m+1,2),1.5)-2*v;
}

big rungekutta(vec r,vec v){
    big f;
    f.r.k1.x=v.x;
    f.r.k1.y=v.y;
    f.v.k1.x=accx(r.x,r.y,v.y);
    f.v.k1.y=accy(r.x,r.y,v.x);
    f.r.k2.x=v.x+f.v.k1.x*dt/2;
    f.r.k2.y=v.y+f.v.k1.y*dt/2;
    f.v.k2.x=accx(r.x+f.r.k1.x*dt/2,r.y+f.r.k1.y*dt/2,v.y+f.v.k1.y*dt/2);
    f.v.k2.y=accy(r.x+f.r.k1.x*dt/2,r.y+f.r.k1.y*dt/2,v.x+f.v.k1.x*dt/2);
    f.r.k3.x=v.x+f.v.k2.x*dt/2;
    f.r.k3.y=v.y+f.v.k2.y*dt/2;
    f.v.k3.x=accx(r.x+f.r.k2.x*dt/2,r.y+f.r.k2.y*dt/2,v.y+f.v.k2.y*dt/2);
    f.v.k3.y=accy(r.x+f.r.k2.x*dt/2,r.y+f.r.k2.y*dt/2,v.x+f.v.k2.x*dt/2);
    f.r.k4.x=v.x+f.v.k3.x*dt;
    f.r.k4.y=v.y+f.v.k3.y*dt;
    f.v.k4.x=accx(r.x+f.r.k3.x*dt,r.y+f.r.k3.y*dt,v.y+f.v.k3.y*dt);
    f.v.k4.y=accy(r.x+f.r.k3.x*dt,r.y+f.r.k3.y*dt,v.x+f.v.k3.x*dt);
    return f;
}

vec moto(vec r,runge rk){
    r.x+=(rk.k1.x+2*rk.k2.x+2*rk.k3.x+rk.k4.x)*dt/6;
    r.y+=(rk.k1.y+2*rk.k2.y+2*rk.k3.y+rk.k4.y)*dt/6;
    return r;
}

Plotting the results I just get a spiral, while using the given inputs I should get a horseshoe orbit. I've tried many different inputs, (m=0.0001 and m=0.000003, the latter is in scale with the actual values of Earth and Sun masses (sun mass is 1-m)).

I just can't see what's wrong (probably everything :D), can anyone please help me?

I won't go through your equations carefully nor your code, but since you haven't mentioned the exact starting positions that you used, my guess is that you haven't first solved for a stable horseshoe orbit state vector first.

The positions for the Sun and Earth are fixed in the synodic (rotating) frame. But where did you start your astroid? For a given starting position, you have to also give the asteroid the correct starting velocity (speed and direction) otherwise it may not be stable and just wander off or spiral around as you mentioned.

Have a look at this answer where I show the correct equations and then solve for stable horseshoe orbits before plotting them.

I start with a position on the x-axis, opposite the Earth, and a little farther or closer from the Sun than the Earth is. Then I launch it with a low velocity (in the rotating frame) exactly in the direction of the y-axis. I watch it drift towards Earth and then boomerang back to the region of the starting area. When it crosses the x-axis again, I check to see how close the velocity is to y. I use an optimization zero solver brentq to adjust the initial velocity until the return velocity is in exactly the opposite direction.

If that's true, then the orbit will be periodic and repeating, and we can call it a stale orbit under the constraints of the CR3BP.

From this answer (there's a lot more to read there, including several references!):

  $$\ddot{x} = x + 2\dot{y} - \frac{(1-\mu)(x+\mu)}{r_1^3} - \frac{\mu(x-1+\mu)}{r_2^3}$$
  $$\ddot{y} = y - 2\dot{x} - \frac{(1-\mu)y}{r_1^3} - \frac{\mu y}{r_2^3}$$
  $$\ddot{z} =  -\frac{(1-\mu) z}{r_1^3} - \frac{\mu z}{r_2^3} $$

  above: half-cycles of some wobbly horseshoe orbits

  above: times to first x-axis crossings of the same wobbly horseshoe orbits, used to calculate half-cycle times.

  above: cycle times from this calculation (black dots) versus from the synodic period estimation method (red dots). Good qualitative agreement. Also the starting y velocities at each starting point in x.

  below: Python script for these plots.

def x_acc(x, ydot):
    r1    = np.abs(x-x1)
    r2    = np.abs(x-x2)
    xddot = x + 2*ydot  -  ((1-mu)/r1**3)*(x+mu) - (mu/r2**3)*(x-(1-mu))
    return xddot

def C_calc(x, y, z, xdot, ydot, zdot):
    r1 = np.sqrt((x-x1)**2 + y**2 + z**2)
    r2 = np.sqrt((x-x2)**2 + y**2 + z**2)
    C = (x**2 + y**2 + 2.*(1-mu)/r1 + 2.*mu/r2 - (xdot**2 + ydot**2 + zdot**2))
    return C

def deriv(X, t):
    x, y, z, xdot, ydot, zdot = X
    r1 = np.sqrt((x-x1)**2 + y**2 + z**2)
    r2 = np.sqrt((x-x2)**2 + y**2 + z**2)
    xddot = x + 2*ydot  -  ((1-mu)/r1**3)*(x+mu) - (mu/r2**3)*(x-(1-mu))
    yddot = y - 2*xdot  -  ((1-mu)/r1**3)*y      - (mu/r2**3)*y
    zddot =             -  ((1-mu)/r1**3)*z      - (mu/r2**3)*z
    return np.hstack((xdot, ydot, zdot, xddot, yddot, zddot))

# http://cosweb1.fau.edu/~jmirelesjames/hw4Notes.pdf

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint
from scipy.optimize import brentq

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]

mu = 0.001

x1 = -mu
x2 = 1. - mu

x = np.linspace(-1.4, 1.4, 1201)
y = np.linspace(-1.4, 1.4, 1201)

Y, X = np.meshgrid(y, x, indexing='ij')
Z    = np.zeros_like(X)

xdot, ydot, zdot = [np.zeros_like(X) for i in range(3)]

C = C_calc(X, Y, Z, xdot, ydot, zdot)
C[C>8] = np.nan

if True:
    plt.figure()
    plt.imshow(C)
    plt.colorbar()
    levels = np.arange(2.9, 3.2, 0.04)
    CS = plt.contour(C, levels,
                 origin='lower',
                 linewidths=2)
    plt.show()

ydot0s   = np.linspace(-0.08, 0.08, 20)
x0ydot0s = []
for ydot0 in ydot0s:
    x0, infob =  brentq(x_acc, -1.5, -0.5, args=(ydot0), xtol=1E-11, rtol=1E-11,
                           maxiter=100, full_output=True, disp=True)
    x0ydot0s.append((x0, ydot0))

states = [np.array([x0, 0, 0, 0, ydot0, 0]) for (x0, ydot0) in x0ydot0s]

times  = np.arange(0, 150, 0.01)

results = []
for X0 in states:
    answer, info = ODEint(deriv, X0, times, atol = 1E-11, full_output=True)
    results.append(answer.T.copy())

resultz = []
for x0ydot0, thing in zip(x0ydot0s, results):
    y     = thing[1]
    check = y[2:]*y[1:-1] < 0
    zc    = np.argmax(y[2:]*y[1:-1] < 0) + 1
    if zc > 10:
        resultz.append((thing, zc, x0ydot0))

if True:
    plt.figure()
    hw = 1.6
    for j, (thing, zc, x0ydot0) in enumerate(resultz):
        x, y = thing[:2,:zc]
        plt.plot(x, y)
    plt.xlim(-hw, hw)
    plt.ylim(-hw, hw)
    plt.plot([x1], [0], 'ok')
    plt.plot([x2], [0], 'ok')
    plt.show()

if True:
    plt.figure()
    for j, (thing, zc, x0ydot0) in enumerate(resultz):
        x, y = thing[:2]
        plt.plot(times[:zc], y[:zc])
    plt.show()

if True:
    plt.figure()
    for j, (thing, zc, x0ydot0) in enumerate(resultz):
        x0, ydot0 = x0ydot0
        cycle_time = 2. * times[zc] / twopi
        ratio = abs(x0/x2)
        T_simple_model = twopi * abs(x0/x2)**1.5
        T_synodic_simple_model = 1. / (1. - twopi/T_simple_model) # https://astronomy.stackexchange.com/a/25002/7982
        plt.subplot(2, 1, 1)
        plt.plot(x0, cycle_time, 'ok')
        plt.plot(x0, abs(T_synodic_simple_model), 'or')
        plt.subplot(2, 1, 2)
        plt.plot(x0, ydot0, 'ok')
    plt.subplot(2, 1, 1)
    plt.xlabel('x0', fontsize=16)
    plt.ylabel('cycle times (periods)', fontsize=16)
    plt.subplot(2, 1, 2)
    plt.xlabel('x0', fontsize=16)
    plt.ylabel('ydot0', fontsize=16)
    plt.show()


When it was discovered that the classical planet are not stars? The classical planets (Mercury, Venus, the Sun, Mars, Jupiter and Saturn) were considered to be "wandering stars" - and the word "planet" means just that.

Wikipedia says "By the 17th century, the idea of the stars being the same as the Sun was reaching a consensus among astronomers."

Since the planets were considered stars in those times, the question is when did the astronomers realize that they are something else?

The same question in put in another words: When did the scientists first hypothesize that the planets do not emit their own light, like stars, and they just reflect the star from the sun? And when was that proven for the first time?

On p.33 of the "Cambridge Concise History of Astronomy" it states that Hellenistic astronomers interpreted the variation in the brightness of the planets as a sign that they were closer or further away at various points of time. Unfortunately it doesn't have a deeper discussion here.

But on p.45 it discusses Ptolemy's thinking - he thought Mercury was, at its closest point, as close as the Moon and that the fixed stars were at the same distance as Saturn at its furthest point - the latter point suggesting that while the planets were seen as different, they might not have been thought of as being that different beyond the way they wandered.

To back that up further there is, for instance, a copy of an image from the Nuremberg Chronicle of 1493 on p.77 - here the planets are merely shown as wandering stars.


Is there an oxide layer on aluminum reflectors for telescopes? I was watching a Nottingham Science video linked here (periodic videos) and they mention modern reflectors are made of aluminum.

I can understand why mirrors are preferred over lenses, but what inhibits the aluminum oxide coating that would naturally form on the surface of the mirror? Why do they work so well?

The aluminum oxide layer forms as soon as oxygen comes in contact with the mirror surface.

Aluminum oxide is tough, firmly attached and transparent;  it serves to protect the aluminum surface.  Rubies and sapphires are slightly impure aluminum oxide


Minimum distance between planets In our solar system, MOIDs (minimum orbital intersection distance) of different planets reach a minimum of ~30 million miles (Mercury and Venus). However, other star systems have more compact planets. So I am wondering, what is the absolute minimum for planets of different sizes (list below)

1 Earth mass
5 Earth masses
10 Earth masses (Neptune-sized)
50 Earth masses (Super-Neptune)
100 Earth masses or 0.3 Jupiter mass (Sub-Jupiter)
1 Jupiter mass
5 Jupiter masses

P.S. Sorry if the list is too big, some can be omitted for sake of writing all of that.

I don't know how to calculate the minimum possible forbidden region for a planet of a specific mass orbiting a star of a specific mass at a specific semi-major axis. So I can't say what the absolute theoretical minimum separation between the orbits of two planets in the same planetary system can be.
A reductio ad absurdum calculation suggests that the minimum possible separation between planetary orbits could be about 2,500 to 5,000 kilometers.  Probably astrophysicists should be able to calculate a minimum possible separation of orbits which would be many, many times that.
There are also some theoretical configurations of planets which would result in two or more planets sharing the same orbit,  in which case one could be facetious and claim there is a zero separation between their orbits.
More Detailed Answer:
Part One of Six: Some "Dole ful" Calculations.
I do know that there have been formulas to calculate the minimum possible spacing between planets in a star system for decades.
Habitable Planets for Man, 1964. 2007, by Stephen H. Dole, is a scientific discussion of the parameters of planets which are habitable for humans.
https://www.rand.org/content/dam/rand/pubs/commercial_books/2007/RAND_CB179-1.pdf
Chapter Three: introduction to General Planetology, has a section "Spacing of Planets in the Solar System", on pages 49 to 52, discusses the spacing of planets in our solar system.
Dole bases his discussion of forbidden regions on his own paper:
"Limits for stable Near-circular Planetary or Satellite Orbits in the Restricted Three-Body Problem". ARS J, 31, no. 2 (February, 1961), pp. 214-219.
Dole calculated the limits of the "forbidden" regions of all the planets in the solar system, based on the masses of each planet and the Sun, the eccentricity of each planet's orbit, and the semi-major axis of each planet's orbit around the Sun.
The "forbidden" region is a ring around the Sun that a planet orbits within and which no other planet should be able to have a stable orbit within.
According to Dole's calculations, our solar system out to Neptune or Pluto is approximately half full of forbidden regions around planetary orbits:

This pattern of regularity should also be found in other planetary systems. Forbidden regions take up about 50 per cent of our solar system, and if this is true of other planetary systems (or multiple star systems); then it would be a simple matter to design any number of stable planetary systems by random mechanical processes.

So, according to Dole's calculations, a star like the Sun might have the same number and masses of planets as the Sun does, if its planetary system was about 50 per cent filled by planets and their forbidden regions.  If planets could be spaced close enough together that the limits of their forbidden regions touched, then our solar system could have perhaps twice as many planets with the same masses, about 16 to 18.
More importantly to me, if there are one, two, or three planets within the circumstellar habitable zone of the Sun, in the planetary system of a star like the Sun with the forbidden zones of the planets just touching, there could be only two to six planetary orbits within the circumstellar habitable zone of that star.  As a kid who liked science fiction stories where many planets in a single solar system were habitable, I found that idea rather "Dole ful".
Part Two: Some Early Ideas to Increase the Number of Habitable Planets in a system.
A number of science fiction stories have habitable planets in Trojan orbits, in which a planet orbits around a star in the L4 or L5 Lagrange point of another planet or another star, 60 degrees ahead of or behind the other astronomical object.
Such Trojan orbits are known to work when the primary, secondary, and tertiary objects differ in mass by hundreds or thousands of times.  In our solar system, the Sun is thousands of times as massive as planets which are thousands of times as massive as the asteroids in their Trojan positions.  And Saturn is thousands of times as massive as its moons Tethys and Dione, which in turn are thousands of times as massive as their Trojan moons.
But could two planets in the size range to be habitable share the same orbit, one planet being in the L4 point of the second one, which would be in the L5 point of the first one? I don't know.
But if that was possible, and if there can be two to six planetary orbits within the circumstellar habitable zone of a star like the Sun, then if there are two Trojan planets in each orbit that solar system could have four to twelve potentially habitable planets in its circumstellar habitable zone.
Another way to increase the number of habitable planets within the circumstellar habitable zone would be to replace each Earth mass habitable planet with a double planet having the same total mass as the Earth.  According to Dole, the minimum mass for a habitable planet would be about 0.4 Earth mass, so a double planet with twin planets of about 0.42 Earth mass would have a total mass of 0.84 Earth, and thus a slightly smaller forbidden region than Earth.
So theoretically about four to twelve habitable planets could orbit in the circumstellar habitable zone of a star identical to the Sun, if arranged as two to six sets of twin planets.
Part Three: Co-Orbital Planets?

In astronomy, a co-orbital configuration is a configuration of two or more astronomical objects (such as asteroids, moons, or planets) orbiting at the same, or very similar, distance from their primary, i.e. they are in a 1:1 mean-motion resonance. (or 1:−1 if orbiting in opposite directions).1
There are several classes of co-orbital objects, depending on their point of libration. The most common and best-known class is the trojan, which librates around one of the two stable Lagrangian points (Trojan points), L4 and L5, 60° ahead of and behind the larger body respectively. Another class is the horseshoe orbit, in which objects librate around 180° from the larger body. Objects librating around 0° are called quasi-satellites.
An exchange orbit occurs when two co-orbital objects are of similar masses and thus exert a non-negligible influence on each other. The objects can exchange semi-major axes or eccentricities when they approach each other.

https://en.wikipedia.org/wiki/Co-orbital_configuration
The space probes Pioneer 11, Voyager 1, and Voyager 2 discovered several new moons of Saturn when they passed it in 1979, 1980, and 1981.  This included the discovery of the tiny moons Epimetheus and Janus, which are co-orbital.

The Saturnian moons Janus and Epimetheus share their orbits, the difference in semi-major axes being less than either's mean diameter. This means the moon with the smaller semi-major axis will slowly catch up with the other. As it does this, the moons gravitationally tug at each other, increasing the semi-major axis of the moon that has caught up and decreasing that of the other. This reverses their relative positions proportionally to their masses and causes this process to begin anew with the moons' roles reversed. In other words, they effectively swap orbits, ultimately oscillating both about their mass-weighted mean orbit.

So it would potentially be possible for two habitable planets to share the same orbit in an exchange orbit like that of Epimetheus and Janus, thus having two habitable planets in the orbit and forbidden region.  That could be used to double the number of planets orbiting within the habitable zone of a star.
I can't help thinking that inhabitants of a planet in an exchange orbit with another planet would find the exchange process terrifying until and unless they could calculate that the two planets were not going to collide.
Part Four: Exoplanet Discoveries.
IN the last generation thousands of exoplanets have been discovered orbiting other stars.  And planetary systems with two or more planets orbiting the same star have also been discovered.  Because of the great difficulty in discovering exoplanets, it is reasonable to assume that in most cases there are more planets in a system than have been discovered yet, perhaps more planets than can be discovered until decades, centuries, or even millennia of future scientific progress is made.
Even though Dole wrote in 1964 that:

This pattern of regularity should also be found in other planetary systems.

The majority of planetary systems discovered around other stars have been significantly different in one or more major ways from our solar system.  Thus there must be a great variation in the processes which form and shape planetary systems.
According to Wikipedia's list of exoplanet extremes, the smallest distance between the semi-major axis of the orbits of two consecutive planets is between Kepler-70b and Kepler-70c, about 0.0016 AU, or about 240,000 kilometers, or about 149,129 miles, closer than the distance between the Earth and the Moon.
https://en.wikipedia.org/wiki/List_of_exoplanet_extremes
If each consecutive planetary orbit around the Sun was only 0.0016 AU farther out than the previous one, 262 planetary orbits could fit within Kasting's conservative habitable zone for the Sun, and 518 in his optimistic habitable zone.
However, the article on Kepler-70 indicates that it is now believed the planets do not exist and their detection was probably an error.
https://en.wikipedia.org/wiki/Kepler-70
I do not know what is the smallest known difference between the orbits of two confirmed exoplanets.  But the two exoplanets with the smallest ratio between their orbits are Kepler-37b & Kepler-37c.  Both planets are several times as massive as Earth but orbit very close to their star Kepler-37 and thus to each other.
The orbit of Kepler-37b has a semi-major axis of 0.1153 AU, or 17,248,634 kilometers, or 10,747,804.57 miles, and the orbit of Kepler-37c has a semi-major axis of 0.1283 AU, or 19,193,407 kilometers, or 11,926,230 miles, a difference of 0.013 AU, or 1,944,772.3 kilometers, or 1,208,425.49 miles.
If the planetary orbits around the Sun could have an average separation of 0.013 AU, 32 planetary orbits could fit within Kasting's conservative habitable zone and 64 planetary orbits could fit within Kasting's optimistic habitable zone.
0.1283 AU is 1.1127 times 0.1153 AU, and is the smallest known ratio between consecutive planetary orbits.  If I remember correctly I calculated that 4 planetary orbits with that ratio could fit within Kasting's conservative habitable zone and 6 planetary orbits could fit within Kastings optimistic habitable zone.
So should the distance of 0.013 AU between the orbits of Kepler-37b and Kepler-37b be considered the minimum possible spacing between consecutive planetary orbits?
Or should the ratio of 1.1127 between their semi-major axis be considered to the minimum possible ratio between consecutive planetary orbits?
If it is the ratio between orbits which determines the minimum spacing of planetary orbits, if a planet orbits very close to its star the minimum possible distance to the next planet's orbit could be less than the 0.013 AU between the Kepler-37 planets. For example, if the inner planet orbits at 0.01 AU, the next planet could orbit at a distance of only 0.011127 AU, a difference of only 0.001127 AU, a little less than the distance between the orbits of the alleged Kepler-70 planets.
If it is the ratio between orbits which determines the minimum spacing of planetary orbits, if a planet orbits very far from its star the minimum possible distance to the next planet's orbit could be many times the 0.013 AU between the Kepler-37 planets.  For example, if a planet orbits 100 AU from its star, and the minimum separation is 1.1127 times the inner orbit, the next planet out would have to orbit at a distance of at least 111.27 AU.
Par Five: Planets in Rings.
Astrophysicist Sean Raymond, in his PlanetPlanet blog, has a section called Ultimate Solar System designing planetary systems with as many planets in the habitable zone as he can fit in.
https://planetplanet.net/the-ultimate-solar-system/
In the Ultimate Engineered Solar system Raymond designs a planetary system with 416 planets in the habitable zone, using rings of planets sharing the same orbit.
That is based on this paper by Smith and Lissauer:
https://ui.adsabs.harvard.edu/abs/2010CeMDA.107..487S/abstract
Smith and Lissauer calculate that seven to forty two planets can share the same orbit around their star, if the planets have equal mass and are equally spaced around the star.
The planets would be separated by 8.57 degrees if there are forty two planets in the ring, increasing as the number of planets decreases, so that seven planets would have gaps of 54.42 degrees between them.  The size of those gaps in AU, kilometers, or miles would have to be calculated from the semi-major axis, and thus the circumference, of the orbit.
Raymond also states that planetary orbits should be separated by 5 to 10 times their Hill radius to be stable.  For example, a planet with the mass of Earth 1 AU from a star with the mass of the Sun would have a Hill radius of about 1,500,000 kilometers or 0.01 AU.
So planetary orbits of Earth-like planets orbiting Sun-like stars at distances of about 1 AU should be separated by at least 0.05 to 0.10 AU or 7,500,000 to 15,000,000 kilometers.
Part Six: Conclusion.
The minimum possible distance between the semi-major axis of the orbit of a planet and the semi-major axis of the the orbit of another planet orbiting the same star can be calculated from factors such as the mass of the planet and the mass of the star, and the semi-major axis of the planet's orbit.
Known examples of the separation of the semi-major axis of consecutive planets in a planetary system range from as much as about 662 AU between CVSO 30 b & CVSO 30 c down to as little as 0.013 AU between Kepler-37b & Kepler-37c.
Known examples of the ratios of the semi-major axis of consecutive planets in a planetary system range from as much as about 78,998 times the distance in the case  of CVSO 30 b & CVSO 30 c down to as little as 1.1127 times the distance in the case of Kepler-37b & Kepler-37c.
Of course if additional planets are discovered between the orbits of the known planets in those systems the records could change.
Astronomers are fairly certain that any planetary orbital separation within those ranges is possible.   I don't know the theoretical limits to how much larger or smaller the separation between the semi-major axis of two consecutive planetary orbits could be.
For minimum spacing I guess I can do a reductio ad absurdum calculation.
According to Sean Raymond here
https://planetplanet.net/2017/05/03/the-ultimate-engineered-solar-system/
The minimum possible separation between two planetary orbits should be five to ten times the Hill sphere radius of one of the planets - the planet with the larger Hill sphere radius.
The absolutely smallest the Hill radius of a planet orbiting very close to a very massive star could get would be at the surface of that planet.  If the surface of the planet extended above the Hill sphere radius the surface material would be ripped from the planet by the gravity of the more massive object until the planet was stripped down to to the radius of the Hill Sphere.
If a astronomical object has to be pulled by its gravity into a spheroidal shape to be considered a planet, that establishes a minimum size for a planet or a planemo (planetary mass object), and thus for its smallest possible Hill sphere.  The minimum radius necessary for astronomical object to be spheroidal is not known precisely.
More or less arbitarily making a radius of 500 kilometers the lower limit for a planetary mass object, and assuming that such an object could orbit close enough to its star to have it's hill sphere at its surface, the minimum possible separation between its orbit and that of the next planet in the system would be about 2,500 to 5,000 kilometers.
And of course if two planets shared the same orbit, such as Trojan planets, planets in an exchange orbit, or planets in a ring, someone could facetiously say there would be approximately zero difference between the semi-major axis of their orbits, since their orbits could be the same orbit.


Calculate the shadow on earth of a large orbital disk How to calculate the shadow on earth of a large orbital disk (low orbit)?

It might be easier to use similar triangles.

$$\frac{Rd}{b} = \frac{Rs}{a}$$

$Rs$ is 696,000 km and $Rd$ is 5 km. $a$ is about 150,000,000 km and c is 80 km.

The the smallest triangle is used next:

$$\frac{r}{b-c} = \frac{Rs}{a}$$

Solve for r and you get

$$r = Rd - Rs \frac{c}{a} = \text{4.63 km}.$$

Extra digits are not helpful because the exact diameter of the Sun depends on how you define it and the distance from the Sun to the Earth varies by almost +/-2 percent.

$Rs/a$ is about 0.00464 and that's also the half-angle of the Sun in radians. Convert it to degrees my multiplying by 180/pi and you get 0.266 degrees, or a quarter of a degree. The full diameter of the Sun is double that, or about a half degree.


How did Mars come to have a 24 hour 39 minute day? Mercury rotates three times for every two revolutions around the Sun, apparently due to a gravitational resonance with the Sun.Venus takes about 225 days to rotate, and rotates in the opposite direction of any of the inner planets. Maybe because its extreme nature makes it ornery.Earth rotates once every 24 hours, a condition caused by the tidal interaction between Earth and its Moon. It's believed that the Earth was rotating about once every 5 hours before the theorized collision with a Mars sized coorbiting object referred to as Theia.Mars shows no signs of a similar collision. Its two moons appear to be asteroids that were captured from the asteroid belt. So how did Mars come to have a day so close to the length of an Earth day?

"It's believed that the Earth was rotating about once every 5 hours
  before the theorized collision with a Mars sized coorbiting object
  referred to as Theia."

Almost. Theia did not have to be co-orbiting, just an intersecting orbit. We have no idea what the Earth's spin was before the collision, but it is theorized that the Earth rotation had a 5 hour period after the collision with Theia, at the time of the Moon's formation from the debris.

The fact that Mars and Earth have such a similar period is a coincidence, perhaps you are asking why Mars is spinning so fast? Well actually Mars is not the odd man out, Mercury and Venus are. Most planets spin fast. exactly which spin orientation is somewhat arbitrarily determined by the vagaries of the ways the planetesimals collided to form them. The fact that Venus and Uranus have unusual spin orientations is just the way things turned out.

Both Mercury and Venus used to spin much faster. Mercury's spin was tidally slowed down by the Sun and Mercury's orbit was (and still is being) driven further away by the Sun (just like the Moon and Earth: Why is the Moon receding from the Earth due to tides? Is this typical for other moons?). Eventually Mercury was held in that 2:3 resonance. Which, by the way had a certain amount of luck involved (see: Mercury’s capture into the 3:2 spin-orbit resonance as a result of its chaotic dynamics ). Venus, we are not so sure of.

The tidal force from the sun is much much less for Venus than for Mercury, but much more than for Earth. However Venus has a dense hot massive atmosphere, which can be forced into both gravitational bi-modal (two peaks) tides and thermal uni-modal (one peak) tides. The bulge lags behind the tidal forcing peak, which creates a torque by the sun to slow it down. This is fiendishly complex (See: Long term evolution of the spin of Venus - I. Theory )

P.S. Actually Phobos, and probably Deimos, are thought to be constructed fairly recently (millions of years) from debris from a collision of Mars with a large asteroid. There is no way to capture a whole asteroid into orbits that close.


What kind of triangle is formed by three unequal masses in a circular restricted three body orbit? This answer to Are the orbits of all triple star systems at least technically unstable? mentions:

There are known solutions to the gravitational three body problem that can be shown to be stable. Lagrange found a three body solution for general masses where all three orbit the common center of mass in an equilateral triangular formation. Gascheau proved in 1843 that this solution is stable if the component masses satisfy
$$ \frac{m_1 m_2+ m_1 m_3 + m_2 m_3}{(m_1+m_2+m_3)^2} < 1/27$$

When the smallest mass approaches zero the three masses are at the vertices of an equilateral triangle. In a realistic solar system this means Trojan asteroids are generally found in orbits of massive planets like Jupiter about 60 degrees ahead-of and behind it.
But if the smallest mass is large but the inequality above is still satisfied, what can we say about the triangle formed by the three bodies in a circular restricted three-body problem orbit?
Is it still known to be an equilateral triangle, but they spin around a point which is not the center of the triangle, but is weighted toward the heavier object?

If so, can this be shown by citing a math-based reference or shown here mathematically or computationally?
If not, is there an expression for two angles of the triangle as a function of to mass ratios?

 Source click for full size

These questions are answered by the same references listed in my previous answer.

In the Newtonian limit, an equilateral 3-body solution exists for any combination of masses. (However, it is stable only if the inequality in the previous answer is satisfied.). This equilateral configuration orbits the center of mass, which is generally not in the center of the equilateral triangle. Each of the bodies follows a circular orbit with radius (see e.g. 1212.0754

$$ r_1 = a\sqrt{\nu_2^2+\nu_2\nu_3+\nu_3^2}$$
$$ r_2 = a\sqrt{\nu_1^2+\nu_1\nu_3+\nu_3^2}$$
$$ r_3 = a\sqrt{\nu_1^2+\nu_1\nu_2+\nu_2^2}$$

and frequency

$$\omega = \sqrt{M/a^3}, $$

where $a$ is the length of the sides of the equilateral triangle, $M$ the total mass, and $\nu_i = m_i/M$.

The situation changes when one accounts for relativisitic effects. When one accounts for the leading (post-Newtonian) corrections, then  a circular restricted three body solution still exists for general masses (with a smaller region of stability than in the Newtonian case.). However, the triangular configuration is no-longer circular (unless all three mass are equal or two masses are 0). Keeping the distances to the center of mass $r_i$ as in the Newtonian case, the sides of the triangle are now given by (again see  1212.0754)

$$ r_{ij} = a(1+\frac{M}{a}\epsilon_{ij}+ O(\tfrac{M^2}{a^2}) )$$

with

$$\epsilon_{12} = \frac{1}{24}[(\nu_1-\nu_3)(5-3\nu_2)+(\nu_2-\nu_3)(5-3\nu_1)]$$
$$\epsilon_{23} = \frac{1}{24}[(\nu_2-\nu_1)(5-3\nu_3)+(\nu_3-\nu_1)(5-3\nu_2)]$$
$$\epsilon_{31} = \frac{1}{24}[(\nu_3-\nu_2)(5-3\nu_1)+(\nu_1-\nu_2)(5-3\nu_3)].$$

The angles of the triangle can be calculated from these lengths if one is so inclined.


What is the word for space that is in the solar system, but outside the heliopause? Wikipedia says the definition of interplanetary space is

  the region dominated by the interplanetary medium, which extends out
  to the heliopause where the influence of the galactic environment
  starts to dominate over the magnetic field and particle flux from the
  Sun

While it defines the solar system as the region in which the sun is the primary gravitational influence

  The Sun's gravitational field is estimated to dominate the
  gravitational forces of surrounding stars out to about two light years
  (125,000 AU).

So, if the region outside the heliopause but inside the solar system isn't interplanetary space, then what is it?

There doesn't seem to be such a word. Interstellar space within the Solar System is still just interstellar space.

There doesn't seem to be a demand for the word you're looking for either. We distinguish space by its contents; the space within the heliosphere is called the interplanetary medium (it contains solar plasma, dust, etc.), while the interstellar medium comprises sparse gaseous molecules from the Local Fluff and (if you go far enough) the rest of the Local Bubble.

Whereas there are about 10 oxygen molecules per cubic centimeter in the interplanetary medium, there is only 1 oxygen molecule per cubic centimeter in the interstellar medium. Seeing that there is no clear boundary for the Milky Way (nor is there any single definition for a boundary for galaxies), the intergalactic medium just vaguely describes the medium within galaxy filaments; here, the space is so rarified that there is only 1 oxygen molecule per cubic meter.


How to calculate the movement of the object passing near other object in space? Assume object A is moving through the space and is passing near the other object (B). Assume the gravitational influence of other objects can be ignored. How to find the equation describing the movement of the object B?

There are 2 cases, object A is moving straightforward or it's moving on orbit (around other object).

I think the problem is quite elementary, but I couldn't find anything that could help solve that problem using the physics on the level of basic university course (I've studied computer science, so I've got only 1 semester of physics, and basic mathematical knowledge - integrals, algebra etc.).

I know the problem can be solved numerically, but I'm interested in finding the equation describing the movement.

Since you're interested in the equations of motion, I would solve this problem by using Lagrangian mechanics. Essentially, find the kinetic and potential energies for these two bodies, A and B.

Construct the Lagrangian:

$$L = T - V$$

where T is the kinetic energy, and V is the potential energy. Then use the Euler-Lagrange Equation to achieve the equations of motion (I would add it here, but I'm not sure of the specifics of your problem).

Two-body motion can always be constrained to a plane, so you may have to throw a constraint in there through the use of a Lagrange multiplier.


Gravitational red shift vs Doppler redshift: Is the universe really expanding? Is it possible that the redshift observed by Edwin Hubble is really from a gravitational redshift and the universe isn't expanding as he has predicted?

What I think I know thus far is this:

Redshift due to the Doppler effect of a moving object is happening due to universe expansion. Galaxies further away are speeding away from us faster than those close to us, hence having a stronger redshift effect.
Gravitational redshift is happening due to light "escaping" the gravitational pull of the source (star) and it has nothing to do with the movement of such source.

Ultimately this is an application of Occam's razor

In order for the light to be gravitationally redshifted, it would have to be coming out of a deep gravitational well. For the red shift observed in galaxies to be gravitational, you would have to suppose several things.

First, that the stars in distant galaxies are somehow much denser: more than neutron star dense. (or possibly that entire galaxies are as dense as neutron stars.) A neutron star has a redshift of about z=0.35 distant galaxies have redshift of more than 7. No known object has a gravitational redshift like that.

Secondly that the density is proportional to distance from us. Placing us at a special position in the universe.

It is much simpler to interpret the redshift as a doppler redshift, and therefore an expanding universe


How many times brighter would full-earth-shine appear to people standing on the moon, than full-moon-shine does to people on earth? The definition of brightness, according to Wikipedia:

the term brightness in astronomy is generally used to refer to an object's apparent brightness: that is, how bright an object appears to an observer. Apparent brightness depends on both the luminosity of the object and the distance between the object and observer, and also on any absorption of light along the path from object to observer.

How many times brighter is full-earth-shine as seen from the moon's surface, than full-moon-shine as seen from the earth's surface?
Different internet sources give different figures. I have seen figures as low as  3x (here), and as high as 100x (here). What is the right figure?
1st reference to be found In the last paragraph of the article.
2nd reference to be found on page 26.
Ethan on this website claims that the figure is 43x.
Besides an original answer, it would be good to get explanations on where the references made wrong calculations to get to their different conclusions.

How much brighter is full-earth-shine on the moon, than full-moon-shine on earth?

On average, it's about 41 times brighter -- in terms of luminosity, or a bit over four times brighter ($2.5 \log_{10} 41 \approx 4.03$) in terms of the logarithmic response of the human eye.
There are two factors that come into play in the luminosity calculation: The much larger size of the Earth than the Moon (a factor of over 13, the square of the ratio of the Earth's radius to the Moon's radius), and the much larger albedo of the Earth than the Moon (a factor of a bit over 3).
This is an average value. If the full Earth view from the Moon happens to be centered on a cloudless Pacific Ocean, that factor of three due to albedo becomes a factor of less than one. On the other hand, if the full Earth view from the Moon happens to be centered on a very cloudy Earth, that factor of three due to albedo becomes a factor of five or more.
The Moon's albedo (0.12) is roughly the same as that of slightly aged asphalt. The Earth's albedo varies a lot. The Earth's oceans are rather dark when seen from space, as dark as freshly laid asphalt. Clouds can have an extremely high albedo, up to 0.9, or 18 times the albedo of water.


Why does this Lowell Observatory telescope have so many knobs? What do they all do? The Fox News article Arizona city played critical role in moon exploration history covers several interesting activities that took place in Arizona in preparation for the Apollo Moon landings.

It shows the telescope (shown below) and mentions that it was used to draw lunar maps.

There is more information about this telescope here and here as mentioned in this answer.

Question: Why does this Lowell Observatory telescope have so many knobs? What do they all do? Is it possible to identify the function of all of the annotated controls and devices on this complicated-looking telescope?

I've added some numbers to facilitate discussion. There is also a chain going between the locations of #4 and #5, though I don't know exactly what it couples.

click images for full size viewing

  above: Lowell Observatory telescope used by scientists who collaborated with artists to map out the moon for Apollo astronauts (Fox News) Source

below: Screenshot from July 9, 2019 Fox News video NASA lunar legacy in the Arizona desert

I contacted Dr. Danielle Adams, Deputy Director for Marketing and Communications at the Lowell Observatory. She was kind enough to reply, and generously provided the following (lightly edited for formatting):

  I spoke with one of our senior educators about the Clark knobs. The numbers below correspond to the numbers in the image posted in Why does this Lowell Observatory telescope have so many knobs? What do they all do?

  Slow motion control for adjusting Right Ascension (east-west movements) while the Right Ascension clutch (3) is locked down.
  Iris control that allows us to control the effective aperture of the telescope. There is an adjustable aperture that can be adjusted from 24" (wide open) to 6".
  Right Ascension clutch that when tightened prevents the telescope from being moved by pushing, allowing the telescope drive to operate.
  and 7. Slow motion controls for adjusting declination (north-south movements) while the Declination clutch (5) is locked down. The two controls are linked with bicycle chain so when one moves, so does the other. This allows operation of the control from either side of the viewing position.
  Declination clutch that when tightened prevents the telescope from being moved by pushing, allowing the telescope drive to operate.
  Sidereal clock that permits astronomers to point their telescopes to a given celestial object more easily.The sidereal day is about 4 minutes shorter than the 24 hour solar day. This is due to the fact that the Earth revolves around the Sun while it rotates about its axis.
  see (4)


Luminosity of Neowise So, I have been trying to figure out the bolometric luminosity of the comet C/2020 F3 (Neowise) on the 22nd of June, 2020. At that time, Neowise was $1.4$ AU away from Earth and had an apparent magnitude of $m=3$. By using the formula $$m=-2.5log\Bigl(\frac{F}{F_O}\Bigl)$$ and $$F=\frac{L}{4\pi D^2}$$ where $F_0$ is the flux of Vega, $F$ is the flux of the comet, $D$ is the distance from Earth to the comet. I found that the bolometric luminosity, $L$ ,of the comet Neowise was around $7.45\cdot10^{14}$ watts when it was $1.4$ AU away from Earth. However, on the 22nd of June, Neowise was also $0.4$ AU away from the Sun. I have tried to use the flux from the Sun at this distance to determine the luminosity of Neowise again. I have found the flux from the Sun at that distance to be $8488.26$ $Wm^{-2}$ using the above formula for flux, but now using $D=0.4 AU$. After that I thought of calculating the luminosity by using flux from the sun times the area of Neowise. Assuming the nucleus of Neowise has a diameter of $5 km$, and assuming that the comet is fully illuminated,
should I assume that the nucleus of Neowise is spherical or circular? And why should the 2 luminosities be different?

I think I get it - you are trying to calculate a luminosity based on some assumption about how much light is reflected/scattered from the Sun towards the Earth - hence the relevance of the flux from the Sun at the comet.
I think it's very hard to adopt this approach. There is uncertainty about what is the appropriate geometry to use for the comet and how the reflection/scattering should be treated. It is doubtful that assuming a comet diameter of 5 km will be appropriate because I would have thought most of the light we see is scattered from a coma around the comet, not from the surface of the comet itself.


What kind of process could create such stellar radiating structures, if real? https://www.nasa.gov/multimedia/imagegallery/image_feature_476.html
Please notice especially the bright star at the center of the image, and how the most massive stars have etched out a cavity in the HI and molecular cloud. It is noticeable that the star near the center that has a number of stars seemingly radiating from it.

What kind of process could create such stellar radiating structures, if real?
Is the spacing related to the Jeans radius?  Thank you.

When an embedded cluster forms and later blows away gas to become a young open cluster the stars inside interact gravitationally. They have essentially random velocities relative to each other, performing a chaotic dance. This typically makes the heavier (and hence brighter) stars end up near the core with low velocities and the lighter stars get into higher orbits. The phenomenon is called mass segregation and happens because all stars end up with roughly similar kinetic energy.

The effect is more extreme in globular clusters since they can persist for a long time, while open clusters tend to dissolve by interacting with the rest of the galaxy. In a way, as the cluster dissolves the lighter stars will be moving away ("radiating") from the core, although at this stage they are all mostly doing the kinetic energy exchange dance.