A moon in eccentric orbit dipping below Roche limit Let's imagine (or do we know one?) a moon in a very eccentric orbit around a planet, with periapsis below the Roche Limit. How would it behave? Would it break apart when speeding past the planet, then reform back as it spends a long time near the apogee, as the debris and parts are pulled back together by gravity over and over until tidal forces circularize its orbit enough that it stabilizes, or is one of these states considerably dominant, and it would either never get to break apart in the short periods near periapsis, or maybe gradually break into a ring and never get to reform?

Bonus question: what considerations / problems would exploration of such a body (artificial satellites, landers, rovers, maybe even a base) face comparing to similar endeavors involving a typical moon?

Comet Shoemaker-Levy 9 (SL9 for short) is a great example of a moon in a highly eccentric orbit that passes through the Roche limit at periapsis.
SL9 was discovered in 1993, but is thought to have been orbiting Jupiter for 20-30 years prior to discovery. It passed through Roche limit of the Jupiter/Comet pair and is thought to have broken apart in July of 1992. The pieces then collided with Jupiter in July of 1994.
SL9 had an extremely eccentric orbit ($e \approx 0.9986$), with apojove of almost a 1/3 AU!  This "loose" orbit allowed massive orbital perturbations by the Sun at apojove, which eventually led to its demise. Here is a nice figure of its last passes:

Note:

In this case, the tidal forces were dominant over the gravitational forces, so the SL9 fragments never came back together once they were broken apart.  But SL9 passed deep into the Roche boundary of Jupiter prior to its collision. This may only be a partial answer to the above question. If SL9 had grazed the Roche boundary just enough to be barely broken apart, I don't know if it would have come back together prior to hitting Jupiter.

After SL9 was split apart by tidal forces, as is shown in the above image, the pieces got further and further apart. This is exactly what we expect to happen since the pieces in higher orbit have a slower orbital speed and the pieces in lower orbit have a faster orbital speed. If SL9 had been in a stable eccentric orbit, the fragments would have continued to break up until they were small enough to withstand the tidal forces. Over many orbits, they would spread out and intermingle into an eccentric ring.

This may not be the only time an object is first broken up and then later collides with a larger body.  Check out this set of craters on Ganymede!


Is stacking welder's glasses a safe way to watch at the eclipse? You can find in many place on the Internet that welder's glass #14 is good for looking at an eclipse. Tomorrow (March, 20th 2015 at 10:45 CET) there's a solar eclipse and yesterday I could only find glasses #11 and #9. I tried briefly this morning, and combining a #11 and a #9 was giving me a good view of the sun. Using 2 #11 gave a too dark image. Now, I read different opinions on the internet.

Against (Perkins Observatory):
http://perkins.owu.edu/solar_viewing_safety.htm

  Be careful that you use the right kind of glass! Welder's glass is
  numbered from 1 to 14 with 14 being the darkest. It is only number 14
  glass that is dark enough for solar viewing! And NO STACKING! A pair
  of number 7's or a 10 and a 4 together DO NOT have the same protection
  as a single piece of number 14 (see unsafe methods for more details).

Favorable (Royal Astronomical Society of Canada):
https://www.rasc.ca/tov/safety

  If SN14 filter is not available, it is possible to combine lower shade
  numbers to get roughly the same level of eye protection from solar
  radiation, e.g. combining SN 6 and SN 8 filters. However the image
  quality may be considerably poorer than that seen through the single
  SN14 filter

I could not find a table or something explaining which kind of protection gives each number; according to the Canadian website, the only concern is about how much infrared light goes through, ultraviolet does not seem to be a problem in almost any case (I was surprised to read that).

Note: It's not my intention to open here the discussion on what could be other safe methods to watch the eclipse, this is well explained everywhere around. I read too late about the eclipse to order specific glasses.

Edit:
One of the answers here report the following formula: (more insight at this link)

  13 or darker is safe enough. Also, you CAN add up welding glass, using
  the formula S(sum) = S1 + S2 -1. S(sum) should be greater than or
  equal to 13

YES

You have summarized pretty much everything available on the subject. There are loads of people saying you cannot, but fortunately for us, the Royal Astronomical Society of Canada, that you quote, includes evidence to back up their claims. They are a respected institution and the author of this article appears immensely qualified. The article itself does not include evidence that welding goggles can be stacked, but the inclusion of such for all the major points adds credibility to the entire article.

It does appear like the glasses are not exactly additive, this Reddit post goes through some calculations, but they are pretty close such that aiming for 1 tint higher will more than offset it and if anything is overkill.


Origins of meteorites? I am aware that the vast majority of meteorites come from the asteroid belt (including 4 Vesta), with some from the Moon and Mars. Are there any other locations from which meteorites have originated (the Kuiper belt? Ceres?). Here it mentions that carbonaceous chondrites formed farthest from the Sun. Is that still within the asteroid belt?

Kuiper belt (and beyond) objects do sometimes come into the inner solar system: we call them "comets".  However comets have a large proportion of volatiles (water etc) and the non-volatile matter tends to be in the form of small dust particles, neither of which form useful meteorites  Ceres is also covered mostly in ice, so even if some was ejected into space and reach the Earth, it wouldn't survive the descent.
An older paper wonders if very old dead comets might be a source of meteorites, but notes that none have been found. A much more recent paper describes a small (100 µm) particle of cometary dust found embedded in another meteorite.  This dust particle would have come from the outer solar system.


At what depth on Mars would the atmosphere have equal pressure of that on Earth? I know the atmospheric pressure on Mars is less than that of Earth, and so is its gravity.  However, I know that the deeper you go (e.g. in a cave or a hole that is dug), the more atmospheric pressure you would experience.  How deep would you have to go on Mars (e.g. below the surface) for the atmospheric pressure to be equal to the pressure we experience on the surface of the earth?

Nasa has a atmospheric model of mars:
$$0.699 *e^{-0.00009 h} $$

A naive application of this model, solving for a pressure of 101 kPa, gives a depth of -55 km.

The Armstrong limit depth (at which water boils at body temperature) is -24km

The model assumes constant temperature, and gravity (it doesn't correct for the fact that at 55 km below the surface you would be well into the martian mantle and the temperature would be very much higher, and deep enough for a measurable difference in gravity). There's no "goldilocks depth" at which you would only need an oxygen supply.

These depths are not achievable with current technology. The deepest mines on Earth are about 4km deep, and even the Kola superdeep borehole only managed 12km


Navigation using moon phases How to navigate with possible maximal precision using moon phases?

The Moon is the brightest celestial body seen in the night sky, and it is possible to find even through moderate clouds. So it's a good natural object to use in navigation when a compass is not available.

A full Moon is quite an easy case - it's directly on the opposite site from the Sun. However, in the other phases, are there tables or mnemotechnical methods allowing to determine the north based on moon shape, assuming the time is known? Are they different depending on geographical height? Is it possible to estimate both north and the time of the night, when only the geographical height is known?

In other phases, the angular distance of Sun from Moon is equal to the phase of the moon, in the direction of the lit side. One of the methods could be to estimate the position of Sun based on this, and then you can, assuming time is known, estimate North.

If you don't know time, use constellations (which can form a separate question) to estimate North and the reverse of above method for time.

Now, if there are clouds and you cannot see any constellations, well, you need to be able to calculate cosines and sines and their inverse in your heads. This method, which I will not describe depends on the rate of change of moon's position, which you'll have to estimate using your hands.

I will have to search for my notes for actual calculation of time, let's see if someone else answers it. But let me tell you how to find direction just using Moon, though this is less accurate than if you already know the time.

From the center of the Moon, draw a line in the sky and follow it towards the lit side of the Moon, the spot where it cuts on the Horizon is either East or West. It is East if you are observing this in Waning Phase i.e. after a Full Moon, else it is West (if after the days of a New Moon). You can get the rest of the directions this way.

Here's how it works. The line you just drew was approximately the ecliptic, the apparent path of Sun in sky. It is at an angle of 23.5° from the equator, so that is the maximum inaccuracy that you might have for your cardinal directions. This would actually lead you off in a wrong direction initially, but the ecliptic keeps on moving, so keep on doing this for the whole night, the positions of your North will be varying but the averaged out value will be better than your initial value.


What qualifies as a local star? My professor used the term 'local star'. What range of distance is generally considered 'local' for a star? And what would be considered distant?

This is too vague a term to be useful, because the Sun does not have any kind of sphere of influence (beyond the solar system) and nor do the nearby stars affect us with their light or gravity (Oort cloud perturbations aside).

Instead it would normally refer to the stellar population in the solar vicinity. This is a mixed bag of objects, with few shared characteristics. For example, objects within 10 light years of the Sun now, will not be so in 10 million years time.

A working definition might be those stars that are close enough to the Sun that we think the census is complete to that distance. Unfortunately, incompleteness remains in all such samples at low masses.

The RECONS programme has been working hard for the last decade to complete the census of objects that are within 10pc (33 light years).

The definition of a distant star is even more vague - all stars are distant. The Sun is 5 orders of magnitude closer than the next stars, so if anything, that is the only natural division between local and distant!


Do the gases in the Earth's atmosphere affect the color of a lunar eclipse? As the sun's rays pass through the Earth's atmosphere only the red light gets through. Is this the result of specific gases in the Earth's atmosphere filtering the red light?  Does all of the nitrogen and oxygen affect the sunlight shining through it?

Let's say the Earth (or just about any earth-like planet) has an atmosphere made mostly of green chlorine. When the sun shines through the chlorine atmosphere and hits the moon during a lunar eclipse, would the moon appear green? Would different gases result in the moon becoming different colors?

From the way you have formulated your question, it seems to me that you are aware of the light diffusive process named Rayleigh Scattering and why it explains the various tints and colours of the sky depending on the Sun position. Other readers can refer to https://en.m.wikipedia.org/wiki/Rayleigh_scattering.

We are used to our atmosphere and as such often discuss the RS merely on geometrical bases, as I shortly did just above.

However, the intensity of RS does not only depends on the inverse of the fourth power of wavelength, but also on the size of the scattering particles and specifically on the square of the molecular polarisability.

As such, changing the nature and composition  of the atmosphere does effect the intensity of RS. An example is SO2, which tends to intensify the yellow tint of sunsets in polluted area.

This should answer the first part of your questions. Let us move to the second part, specifically using the example atmosphere that came to your mind, chlorine as a green gas, i.e. molecular chlorine Cl2.

In this case, the gas is not colourless. Different respect to nitrogen and  oxygen, chlorine does absorb in the visible part of the electromagnetic spectrum.
It absorbs the colours opposite to green in the colours wheel, to keep things simple as possible.

In an inherently coloured atmosphere the relative importance of scattering will be less or at least variable depending on parameters such as atmosphere thickness, density etc. Let me ignore RS in such a toxic atmosphere :)

What is important now is that when you point your eyes up to the sky, the green colour you see is not that one whose propagation is hindered, but rather the one that can propagate (given that we are ignoring RS). In other words, the light will be always richer of green colour.

To make the final answer simple let us look at the moon when the only light that hit it is that coming throughout the planet atmosphere.

With Earth, the Moon is bloody as for RS has removed the blue violet part of the solar spectrum.

A hypothetical change on Earth atmosphere composition will also have a minor impact, as we have seen that the polarisability of the molecules counts.

About Chloro, the planet having chlorine atmosphere. During the eclipse,  its moon will have a greenish tint, or likely appear green. In this case the moon is lit by sunlight depleted of the wavelengths absorbed by chlorine.

To resume, the colour of the moon is always that less affected by the planet atmosphere but the mechanisms and the final result differ.

When the atmosphere is intrinsically colourless, the colour of the moon is somehow "complementary or so" respect to that of the daily planet skydome. When the atmosphere absorbs, the moon and the sky dome will tend to the same colour.

Note: I did assume that sunlight can pass the intrinsically coloured atmosphere.  Again depending on various parameters, this is of course not given. Most likely we won't see the Moon. We shall really care of our atmosphere.

Also the intrinsic properties of the moon do not change. But this is even more trivial.


Calculating obital velocity from radial velocity I'm not really a physicist so I don't know if it's possible to get the real orbital velocity of a star from its radial velocity. I'm working on a paper for school and I want to get the mean velocity of selected open clusters to graph the galaxy's rotational curve, but in the databases I only encounter radial velocities. How do I get the real orbital velocity from those parameters? Thanks!

You can't without assuming something about the overall velocity.

The radial velocity is one component of a velocity vector; you are missing the other two components, which could in principle be anything. However, you could assume that as open clusters are mostly quite young and members of the Galactic disc population, that they are moving in the disc on roughly circular orbits. NB. This cannot possibly work for globular clusters, which are an entirely different population, many with non-circular orbits and orbits out of the galactic plane.

Having made that assumption, your radial velocity is the orbital velocity resolved through an angle that is formed between your line of sight to the cluster and a tangent to the Galactic centre at the cluster's Galactic position.

This method will not work well for clusters with a Galactic latitude much different to zero, or clusters with small Galactic longitude (which should have zero radial velocity in this model).

For all this to work you need the radial velocity, the Galactic latitude and longitude of the cluster, the distance to the cluster and an assumed distance to the Galactic centre.

I've had a go at sketching the situation. The cluster is on a circular orbit around the Galactic centre with a true orbital velocity $v$ (shown as a vector). We measure the component of this in our line of sight as the radial velocity, where
$$v_r = v \cos \alpha.$$
The trick is to find $\alpha$ using the triangle formed by: us, the Galactic centre and the cluster. If the distance to the galactic centre is $d_{gc}$ and the distance to the cluster is $D$ and the galactic longitude of the cluster (angle measured from the direction of the Galactic centre) is $l$, then I think that the distance of the cluster from the Galactic centre $d_{Cl}$ is given by the cosine rule as
$$ d_{Cl} = \left( d_{gc}^{2} + D^2 - 2d_{gc}D \cos l \right)^{1/2}$$

Then the angle $\beta$ at the apex defined by the cluster id given by
$$\cos \beta = \frac{d_{Cl}^2 + D^2 - d_{gc}^2}{2d_{Cl}D}$$
which leads to
$$ \cos \beta = \frac{D - d_{gc} \cos l}{d_{Cl}}$$

You then look at the diagram and note that $\beta = 90 - \alpha$, so that
$\cos \beta = \sin \alpha$. Hence you can calculate $\alpha$ and hence calculate $v$.


Should BD in this sector means Brown dwarfs or Black dwarfs? The one in the sector as BD/NS/BH. And please explain why Brown/Black dwarf should be here among neutron stars and black holes. Thank you for your help.

Source: Beech, M. (2019). Introducing the stars. Cham: Springer Nature Switzerland AG.

Looking at the diagram as a whole, the bottom line seems to be the type of object that a star in that category will evolve into.  So a large star that is variable due to mass loss will explode in a type II SN and evolve to either a Black Hole, or a Neutron star (which may become variable as a pulsar but ultimately evolves to a non variable NS.
Looking down the right. We see "mass exchange variables (ie novae) and dwarf novae such as U Gem or Z And will ultimately evolve into one of three types of stellar remnant: Black dwarfs, neutron stars or black holes. Black dwarfs are the final stage of evolution of white dwarfs.
Therefore I surmise that BD means Black Dwarf.


How to calculate the position of the sun in long/lat? I have a world map, and I want to know, given a timestamp, where is the sun located right now, in terms of long/lat.

Here is a website that does that:
https://www.timeanddate.com/worldclock/sunearth.html

But they do not share the calculations.

How can I calculate the sun's position in earth gps terms given a timestamp?

Edit: some context

I have a plane that is determined by 2 points on earth, and that plain is perpendicular to the earth in those points. I want to know where is the sun relative to that plane, so I can know if the sun is to the right of it, or to the left of it.

Instead of working in 3 dimensions, I decided to work in 2. I already have that plane in 2D (just a line on the map), now I need to know where the sun is

Edit2: Using the wiki page here, this is what I got so far

Note: The interesting part is getCurrentPosition

class Sun {
    private julianTime(timestamp: number) {
        // Source: http://stackoverflow.com/questions/466321/convert-unix-timestamp-to-julian
        return timestamp / 86400 + 2440587.5;
    }

    private convertEclipticEquatorial(lambda: number, beta: number, R: number, n) {
        const epsilon = 23.439 - 0.0000004 * n; // Obliquity of the ecliptic
        console.log("epsilon", epsilon);
        const alpha = Math.atan2(Degrees.cos(epsilon) * Degrees.sin(lambda), Degrees.cos(lambda));
        console.log("alpha", alpha);
        const delta = Math.asin(Degrees.sin(epsilon) * Degrees.sin(lambda)); // declination
        console.log("delta", delta);
    }

    getCurrentPosition(timestamp: number) {
        const JD = this.julianTime(timestamp);
        console.log("JD", JD);

        // Source: https://en.wikipedia.org/wiki/Position_of_the_Sun
        const n = JD - 2451545;
        console.log("n", n);
        const L = (280.460 + 0.9856474 * n) % 360;
        console.log("L", L);
        const g = (357.528 + 0.9856003 * n) % 360;
        console.log("g", g);
        const lambda = (L + 1.915 * Degrees.sin(g) + 0.020 * Degrees.sin(2 * g)) % 360;
        console.log("lambda", lambda);
        const R = 1.00014 - 0.01671 * Degrees.cos(g) - 0.00014 * Degrees.cos(2 * g); // Distance
        console.log("R", R);

        this.convertEclipticEquatorial(lambda, 0, R, n);
    }
}

class Degrees {
    static sin(angle: number) {
        return Math.sin(angle / 180 * Math.PI);
    }
    static cos(angle: number) {
        return Math.cos(angle / 180 * Math.PI);
    }
}

console.log(new Sun().getCurrentPosition(new Date().getTime() / 1000));

Running this now, gives:

  JD 2457844.130512928 (can confirm, this is correct)

  n 6299.1305129281245 (simple math)

  L 9.181612328272422 (simple math)

  g 85.95292328111373 (simple math)

  lambda 11.09465308568091

  R 0.9990992788207762

  epsilon 23.436480347794827 (can confirm, that is roughly correct)

  alpha 0.17801325468993906

  delta 0.0766106706512887

By latitude and longitude of the sun, I think you mean latitude and longitude of a point on the earth's surface where the rays of the sun fall perpendicular to the surface.

In astronavigation, this point is called the Geographical Position (GP) of the sun and terminology to describe this point is slightly different:-

Latitude of the GP is called Declination, and it is numerically equal to what one might call "Latitude". It is measured in degrees North or South the equator.

Longitude of the GP is called Greenwich Hour Angle (GHA). Just like longitude, GHA is measured East or West of Greenwich Meridian.

Describing the sun's position in this way is called the Equatorial Coordinate System.

The Wikipedia page you have used gives the additional formulae below to convert from Ecliptic Coordinates (the steps you followed) to Equatorial Coordinates.

Follow these steps to determine Declination $\delta$ (Equivalent to Latitude).

To get the sun's GHA, you need to first calculate $RA$ (Right Ascension) as shown above and then convert this to GHA by these= rules...

$SHA° = 360 – 15 · RA[h] $

$GHA_{Aries}° = 15 · GST[h] $

$GHA = SHA + GHA_{Aries}$

For the second step, you will need $GST$ (Greenwich Sidereal Time) which is you can calculate if you know UT (UT can be obtained from the time shown by a clock). Use this link to convert directly from UT to GST

If you want a simple, concise way to be able to calculate the positions of all manner of celestial bodies, the sun, moon, planets, stars, then I highly recommended the book Astronomical Formulae for Calculators by Jean Meeus. It is the simplest, most comprehensive collection of astronomical formulae you are likely find anywhere on earth.


Can there be elliptical rings around a planet? The rings of Saturn and Jupiter are circular. Orbits can be elliptical, so can rings be elliptical?

Short answer: no.

Long answer:

There are many collisions within ring systems, and collisions always work, over time, to push orbits to a circular shape (or destroy the rings). Any deviations tend to be quickly corrected.

For the same reason, rings tend to be extremely flat and extremely thin. E.g. Saturn's rings are only dozens of meters thick; given their diameter of 300 thousand kilometers, if you cut a model of the rings from ordinary size printer paper, the paper model would be too thick. (would not represent the thickness of the rings to the correct scale)

Another characteristic of rings is that they are equatorial. If they are not in plane with the equator, precession will drive them towards that configuration.

You could apply a perturbation that makes a ring non-circular, or non-flat, or non-equatorial, but then the system will quickly evolve towards circular, flat, and equatorial (or will be destroyed).

https://www.uwgb.edu/dutchs/AstronPowerPoint/Rings.ppt


Dark Matter Curvature Does Dark Matter affect Space Time Curvature? If so, then what is the formula to calculate space time curvature due to dark matter?

Yes, the same equation as is used for other matter.
Marc Postman, Ray Villard and Donna Weaver of the Space Telescope Science Institute posted the results of a CLASH study: "News Release 2011-25":

"This image of galaxy cluster MACS J1206.2-0847 (or MACS 1206 for short) is part of a broad survey with NASA's Hubble Space Telescope.
The distorted shapes in the cluster are distant galaxies from which the light is bent by the gravitational pull of an invisible material called dark matter within the cluster of galaxies. This cluster is an early target in a survey that will allow astronomers to construct the most detailed dark matter maps of more galaxy clusters than ever before.

On the NASA website: "A Clash of Clusters Provides New Clue to Dark Matter":

"... this composite image, made using data from the Hubble Space Telescope Chandra X-Ray Observatory, researchers have discovered clues to increase our knowledge of dark matter. This image is a powerful collision of clusters. These clusters show formations of dark matter, where it separated from the ordinary matter. The different color disparities, along with the interesting evidence for dark matter ...".

Another article with the same photo: "heic0818 — Science Release -- Clash of clusters provides new dark matter clue":

"New Hubble and Chandra observations of the cluster known as MACSJ0025.4-1222 indicate that a titanic collision has separated dark from ordinary matter. This provides independent confirmation of a similar effect detected previously in a target dubbed the Bullet Cluster, showing that the Bullet Cluster is not an anomalous case.".

Wikipedia webpage on CLASH - Cluster Lensing and Supernova survey with Hubble.


Calculating area of visible sky Can we calculate the area of sky visible to us from the point we are standing?I mean is there any idea or experiment to calculate it?

Approximating the Earth as a sphere with radius $R$, then when viewing from a height $h$ above the surface, the Earth blocks out a cone of some opening angle $2\vartheta$, where $\csc\vartheta = 1+\frac{h}{R}$. Thus, the visible portion has a solid angle of
$$\Omega = 2\pi\left(1+\cos\vartheta\right) = 2\pi\left(1+\frac{\sqrt{h^2+2Rh}}{R+h}\right)$$
steradians. Divide this by $4\pi$ to obtain the fractional area of the the visible sky, compared to what you could have if the Earth wasn't blocking your view, since a full sphere subtends a solid angle of $4\pi$ steradians. That is probably a more natural measure of the visible sky than a literal area.

For an actual area, you need some sort of reference distance $r$ to measure from, with the visible sky a distance $r$ away having area $A = \Omega r^2$.


Formula for sun's atmospheric refraction below the horizon I have searched in Google and found interesting articles for atmospheric refraction influence on sun's elevation above the horizon (from 0° till 90°).
eg. Bennett G.G. etc.
I would like to know if there is a formula that approximates the refraction when the sun's center is below the horizon (from 0° to ca. -3°) and taking in account for temperature and air pressure.

Without taking temperature and pressure into account, from my old code I found this approximate formula: (angles in degrees)

$\Delta \alpha_{reflect} = tan[(90°-\alpha_{technical}) \times 0.9679] \times 0.023626$

If this calculated reflection angle exceeds $\frac{34.9}{60}$ degrees, use that.

Observed elevation would be $\alpha_{observed}=\alpha_{technical}+\Delta\alpha_{reflect}$

Example

a point technically (without atmosphere) 34.9' (minutes of arc) below horizon, appears on the horizon.
a point technically on the horizon, appears 28.1' above the horizon.


Will 1036 Ganymed's orbit be changed by its close encounter with Mars in 2176? In 2176, the asteroid Ganymed will come as close as 0.02868 AU (4,290,000 km / 2,670,000 mi) from Mars. That's pretty close to Mars imho. Can Mars' gravity alter Ganymed's orbit significantly in 2176 and if so, how much would it change? Just curious. Ganymed is a near-Earth object after all, and one at global killer size.

Ganymed's Mars approach in 2176 is only moderately close: 4.4 times Mars's Hill sphere radius or 11 times the average Earth-Moon distance.
JPL HORIZONS predicts these heliocentric orbital elements (J2000) before and after the encounter:

Date
e
a
i
Ω
ω

2176-12-06
0.525078
2.65922 au
27.7778°
212.085°
136.031°

2176-12-26
0.525093
2.65922 au
27.7772°
212.085°
136.025°

The change is slight.
With respect to Potentially Hazardous Asteroid criteria,
Ganymed is large enough (H = 9.4 < 22)
but can't get close enough to Earth (MOID = 0.34 au > 0.05 au).


How can I obtain raw observation data of stars surrounding Sgr A*? Are there any publicly available telescope that offers the sensitivity to observe the stars surrounding Sgr A*? If not, is there any raw data available that I could download?

Observations of stars near Sgr A* are done in the near-infrared, usually with adaptive optics. Most of these are done by two groups: Andrea Ghez's group at UCLA, using the Keck Telescopes in Hawaii, and Reinhard Genzel's group at MPE (Max-Planck-Institute for Extraterrestrial Physics), using the European Southern Observatory's Very Large Telescope (VLT) in Chile.

Both telescopes have public data archives, so in principle you can retrieve the raw data for these observations, as long as the observations are at least a year old (which most of them are). You need to know which are the relevant instruments; NACO has been used for most of the images at the VLT, and I think NIRC2 has been used at Keck. (There are also spectroscopic observations using both telescopes.)

So you could, for example, search using the NACO-specific search form in the ESO archive. I would suggest entering "Sgr A*" in the "Target name" field, restricting the "Search Box" field to something like 30 arc seconds (change the default "00 10 00" to "00 00 30"). Note that these observations involve lots of short exposures in different filters, and with no other restrictions on the search, you will get more than 10,000 hits.

(In order to actually retrieve the data, you will need to register as a user of the archive, which is free.)

Something similar might work for the Keck Observatory
Archive, though I've never really used that.

You'll need to be familiar with the reduction and analysis of near-IR imaging to get anything from the data, of course. ESO does provide "pipeline" software that can do some of the reduction (search on their site).

You can, in principle, apply to use the VLT yourself, but unless you really know what you're doing, it's very unlikely you'd get time in competition with, say, Genzel's group. Also, NACO has been taken off the VLT; its replacement instrument, ERIS, won't be available until sometime in 2020. (You can't generally apply for Keck time unless you're at one of its member institutions.)


How do you determine the apparent magnitude of the stars leaving main sequence with CMD diagrams? So I have a CMD diagram and on the left, I have apparent magnitude and at the bottom, I have the color index. I would just estimate the magnitude of stars leaving the main sequence and continue from there, however, I was wondering if there's a way to do it more accurately?

I would take stars between certain magnitudes and calculate the average magnitude, but again, I'm estimating from where to where the MSTO point is...

There is no clearly-defined "turn-off" magnitude. Stars move from the zero age main sequence (roughly speaking, where core hydrogen burning begins to dominate the luminosity) to the terminal age main sequence (where the core hydrogen runs out) relatively slowly, increasing in luminosity as they do so. Their evolution then accelerates redward in the colour magnitude diagram. The turn-off is roughly the most luminous star for which this acceleration in evolution is yet to occur, but this is confused/blurred by observational uncertainties, binarity, differential reddening and spreads in age and rotation rate.

In practice what you do if you really want to do a proper job of estimating the age, with uncertainties, is you do a population synthesis (including binaries and observational uncertainties) for clusters of stars with a variety of ages. You then compare these with the observations using maximum likelihood or Bayesian techniques (e.g. Naylor & Jeffries 2006).


Orbital Elements Mars near June 2014 do you know any website in which I could obtain orbital elements of planets at some date?

For instance, I'm now needing the orbital elements of Mars around June, 2014.

Thanks!

As @barrycarter mentioned you can use the Horizons web interface from JPL: http://ssd.jpl.nasa.gov/horizons.cgi

You can change the setting in the web form by clicking on the [change] links for the different settings (for instance the time span or the planet). There are several different settings that can be changed to get different outputs (table columns) and such.

If you really want the orbital elements (and not just the Right Ascension and Declination) you will have to select ELEMENTS in the Ephemeris type settings.

You then get a table with the following elements:

JDCT     Epoch Julian Date, Coordinate Time
EC     Eccentricity, e
QR     Periapsis distance, q (AU)
IN     Inclination w.r.t xy-plane, i (degrees)
OM     Longitude of Ascending Node, OMEGA, (degrees)
W      Argument of Perifocus, w (degrees)
Tp     Time of periapsis (Julian day number)
N      Mean motion, n (degrees/day)
MA     Mean anomaly, M (degrees)
TA     True anomaly, nu (degrees)
A      Semi-major axis, a (AU)
AD     Apoapsis distance (AU)
PR     Sidereal orbit period (day)


Can we get the gas compositions from a gas planets? Is it possible to get some gases like hydrogen from gas planets (Saturn, Jupiter,  Uranus or Neptune)?

Yes.  It's certainly feasible to obtain a sample from Jupiter or Saturn, effectively getting gas from the planet, but it's not easy.

Massive planets, as a byproduct of their high mass, create large gravity wells which require a lot of fuel to fly close enough to get a sample from and return with it to where it's needed.  It would take perhaps a ton of conventional fuel to return from Jupiter with a few pounds of gas, which isn't a good return on investment.

Collecting hydrogen from Jupiter would be expensive almost certainly not worth the cost.   It would be much easier to obtain hydrogen in space from frozen water ice, frozen methane or frozen ammonia, where hydrogen is abundant.     This can be obtained in icy comets and there's even some water-ice on the moon in craters, which has a much smaller gravity well.   Also on Mars' ice caps and inside Ceres, it's believed there's about an entire oceans worth of water, perhaps more water inside Ceres than on Earth.

That's a lot of hydrogen that would be far easier to harvest than hydrogen from Jupiter.   And while it would need to be separated from Oxygen, chemical separation would require many times less energy than collecting hydrogen from a gas planet's gravity well.   You'd also have the advantage of not just harvesting hydrogen but oxygen as well.

It's not impossible to harvest hydrogen from Jupiter or Saturn, it's just difficult to imagine it being worth the trouble.


Can a star emit a larger fraction of its total light in the visual range than the Sun does If one sets both the bolometric and visual luminosity of the Sun to 1 (and use a linear scale), can a star then have a higher visual than bolometric luminosity?

It makes sense that this will not be the case for most stars, as smaller stars will tend to radiate more of their energy as infrared, while larger go more into the ultraviolet. That is, this is inherently linked to what wavelengths the human eye has evolved sensitivity for. But are there exceptions? If there are, I would think they have parameters very close to the sun.

Surely this is exactly what the V-band bolometric correction assesses. The numerically larger the bolometric correction, the more of a star's flux (as a fraction) is in the V-band.

The Sun's bolometric correction is about -0.06 to -0.11 mag, depending on which sources you look at. Cool stars and hot stars have numerically smaller BCs and emit a bigger fraction of their flux outside the V band.

Some sources put the BCs of A5-G0 stars as very slightly larger than the Sun's, by up to 0.1 mag (e.g. Pecaut & Mamajek 2013).


If the Universe is Flat, why does it look 3D from our View? So I know that the universe is flat in a 4D point of view, but what about a 3D point of view? The universe does not SEEM flat from our view, so how can it really be so? Please answer more simple, im not an astronomy pro.

Sorry if this is a dumb question, I don't have a degree in astrophysics or anything, i'm just a kid who is really interested in the universe, so i'm sorry (in advance) if this is a simply answered bad question.

So I know that the universe is flat in a 4D point of view, but what about a 3D point of view?

You seem to be misunderstanding the term "flatness". In this context, it does not mean the universe looks like a piece of paper in any sense. It instead refers to whether space is warped.

  i'm just a kid who is really interested in the universe

I'll try to simplify as much as possible. I'm assuming you've learned that triangles have 180°. For the mathematics you'll be dealing with in middle school, high school and even most of college, you'll only have to deal with that.

The more accurate statement is that triangles have 180° when our geometry is not warped. When triangles have 180°, we say we are dealing with Euclidean geometry (named after Euclid, an ancient Greek mathematician). There are in fact cases in higher mathematics when this does not hold, which we call Non-Euclidean geometries. Just as a side note: Euclidean geometry was developed thousands of years ago. On the other hand, Non-Euclidean geometries only began to develop in the 1800s, so they're pretty recent.

Anyway, Euclidean geometry deals with flat spaces. For example, if you draw a triangle on a piece of paper, it will indeed have 180°. However, Non-Euclidean geometry deals with curved (warped) spaces. For example, if you draw a triangle on the surface of a ball (which is not flat), the triangle will actually have more than 180°. If you instead draw the triangle on a potato chip or saddle-shaped surface (also not flat), it will have less than 180°.

So with regards to the Universe: let's say we draw a triangle connecting three stars in outer space (we can actually use any object or simply points, but I'm using stars as an example). Will that triangle be less than, equal to or more than 180°?

According to our measurements, it will actually be equal to 180°; the Universe is indeed flat. This has major implications for physics and astronomy.

  i'm sorry (in advance) if this is a simply answered bad question.

Future advice: don't be afraid to ask questions. Yours wasn't bad and is pretty common.


Where and how to look up the coordinates and magnitude of the star in this occultation prediction? The motivation for this question comes from this question. A star is named in the occultation prediction shown below, from AsteroidOccultations.com's News & Announcements for 2014 MU69 on 2017-Jun-03 03:11.

I believe that the star is in the Omega Centauri cluster, and I believe it is between +12 and +14 visual magnitude roughly. The name given in the prediction is NH -08194514G The location shown in the prediction is R.A: 19 03 34.493 DEC -20 34 39.34 but according to Wikipedia the coordinates of the cluster are roughly R.A. 13h 27m, DEC. -47deg, 29m.

I need help understanding why the two coordinates seem so different (am I misunderstanding the prediction?) and finding out how to look up this particular star’s visual magnitude by using its name. I’ve tried a few quick searches but realized that I was over my head and not likely to get the right information.

From the linked question:

People have fanned out over South America and South Africa with telescopes with cameras and GPS clocks and have attempted to watch the shadow of a roughly magnitude +13 star cast by a rock in the KUIPER BELT pass over the Earth!

Early Twitter reports reported clear skies and images of the Omega Centauri star cluster. There is a Twitter hash tag for these events, #mu69occ. Here is a 500 millisecond exposure from a few days before the occultation. The short exposure time is necessary to resolve small scale debris from the main occultation.

above: Omega Centari Star Cluster, 500 millisecond exposure. Tweeted by Alex Partker.

The goal is actually to look for shadows cast by even smaller debris, orbiting around the rock in the Kuiper belt. This is for New Horizons mission planning.

above: From AsteroidOccultations.com's News & Announcements for 2014 MU69

I believe this is a misinterpretation of this tweet: https://twitter.com/Alex_Parker/status/870169900970639360

If you look at the reply chain, it starts with https://twitter.com/Alex_Parker/status/870168057032921088 where Alex Parker states "Also, last night while we were packing up I took a very quick image of the Carina Nebula with one of our #MU69occ cameras."

The tweet in question states "This is what Omega Centauri looks like after just 500 milliseconds of exposure with one of the #MU69occ cameras."

In other words, Alex Parker was taking images of various clusters with the camera he plans to use for the MU69occ event. The image targets have nothing to do with the occultation itself.

I also went to https://ssd.jpl.nasa.gov/horizons.cgi and took a geocentric view of "2014MU69" for the month of June and it confirms this object is nowhere near Centaurus.

EDIT: I emailed Amanda M. Zangari (azangari@boulder.swri.edu) and her response was:

Hi Barry,

I do not know where the star numbers come from.

Gaia IDs for each of the stars are here:

http://www.boulder.swri.edu/MU69_occ/gaiainfo.html


Coordinate system for space I am high school student, I  have doubt if I have to locate a planet or any star in space, how do I locate it, because we have no perfect coordinate system, I have read in some books, that time is define as the one of the coordinate, in coordinate system, can we define something only by time as coordinate.

For locating objects in the sky, the horizontal and equatorial coordinate systems are commonly used. These systems describe the position of some object in the sky very well, but do not explain the position of the object in space (if you know the distance you know "where" the object is, but this is relative to the equatorial/horizontal plane and is tricky to wrap your head around).
If you want to precisely locate the position of some object in space relative to the ecliptic, the ecliptic coordinate system exists for that, and for locating deep sky objects, the galactic coordinate system is normally used, as it nicely describes positions of objects relative to our galactic disc.
The space with the fourth coordinate is used for relativistic models (it fits well into special relativity), but not for describing positions of objects (to see more about that, look up Minkowski space).


Could a close passing star be captured by the Sun's gravity? Earlier I read an article here stating that a binary pair of stars passed within 0.8 light years from the sun. That made we wonder why the stars weren't captured by the Sun's gravity since the sun's gravitational influence spans over 1 light year.

Could a passing star be captured by the Sun's gravity? if so how close would it have to be before it was captured?

Unless the stars comes so close that they actually collide, two stars will not be able to catch each other gravitationally. The reason is energy conservation: As they approach each other, their potential energy is converted into kinetic energy, increasing their velocities. When they are closest, their velocities are at their highest, but since there's nothing to take out energy from the system, their kinetic energy will be converted back to potential energy, propelling them far apart again.

If a third star is present, however, this may be able to extract energy from the system, so that one star is slung out while the other two start orbiting each other.

Note however that the probability of two stars encountering each other extremely small, and that of three stars is close to zero.

In the above, I have ignored the presence of planets. In principle, a large planet could help, but since their masses are much smaller than the stars', their effect is quite small.


Can I get help getting image of Celestial Body at RA 2.220 and Declination +15.59? My mom had a star named for my dad who died a few years back.   I told her that I'd try to find an image of that star and send it to her.  My dad's Birthday was 3/31/1931 if that helps.  Would love some help with this.  Thanks so much, in advance.  :-)

I think your star is an 8th magnitude star in Aries: properly catalogued as HD 14610. It is an F5 white dwarf, a little larger and brighter than the sun, but about 300ly away, and too dim to see without a telescope.

It is located at RA 02h 22m 01s, dec +15* 59' 30''

An image from the sky surveys: http://simbad.u-strasbg.fr/simbad/sim-id?Ident=HD++14610 (click on "Interactive Aladin-lite to get a wider view)

Others have touched on the unofficial nature of these "star naming" services, which some feel are fraudulent.  However if it makes your mum happy.


Comparing Star Formation rate in different galaxies I was looking for some scale to compare my result of star formation rate in order to find out whether it is moderate or high or low, but I couldn't find any scale. Is there any paper related to such scale where Star formation rate form minimum to maximum of each galaxy is given and once can compare their result with scale.
I found that one of dwarf galaxies has 0.2 star formation rate, but since I don't have scale to compare, thus I have no clue whether it is high or low or moderate? What is the maximum star formation rate?

Not being an expert in star formation, I found a well-written paper summary from which I conclude that typical star formation rates range between $6 \ldots 24 M_\odot / yr$.
The blog quotes the following graph
 by M. Boquien, V. Buat, and V. Perret, see https://arxiv.org/abs/1409.5792

In this paper we investigate in isolation the impact of a variable star formation history on the measurement of the SFR.
We combine 23 state-of-the-art hydrodynamical simulations of 1<z<2 galaxies on the main sequence with the cigale spectral energy distribution modelling code. This allows us to generate synthetic spectra every 1 Myr for each simulation, taking the stellar populations and the nebular emission into account. Using these spectra, we estimate the SFR from classical estimators which we compare with the true SFR we know from the simulations.

In summary: To me, $0.2 M_\odot / yr$ looks rather small, but I cannot judge whether it is unrealistic for a dwarf galaxy.


Can two comets travel together as one? If two comets come together in space, will they bounce apart, merge into a single body, or could they travel together through space, either touching or orbiting a shared center of gravity?

Yes, binary comets do exist; 8P/Tuttle is an example of one. This is an area that is still being heavily researched, but it is suspected that binary comets can "form from collision, mutual capture, or fission."

In recent news, the comet 67P/Churyumov-Gerasimenko has also been revealed to be a contact-binary by the European Space Agency's Rosetta spacecraft.

Sources:

Radar observations of 8P/Tuttle: A contact-binary comet - Harmon, Nolan, Giorgini, Howell
Rosetta's target is 'double' comet - BBC


What are the craters on the moon with peaks in the middle of them? Was taking some extremely amatuer pictures of the moon from my cheap telescope with my camera phone...

You see the crater Im talking about? Top right.The one with a little nub in the middle- which Im sure is actually massive. How was this formed differently than the other craters? What couldve caused the hill?

Nice picture! Those peaks are actually a very common feature in craters formed by large impacts. Craters with these features are known as "complex craters". Basically what happens is that when the meteor hits, it pushes in the surface. Then the underlying mantle produces an "elastic rebound" and pushes outwards. This is a similar effect to what happens when you hit a drum. Your drumstick rebounds off the drum, and the drum begins to vibrate. Since rock (even molten rock) is much more dense than a drumskin, it does not vibrate very much. The vibrations solidify, resulting in a crater with one central peak.

How exactly this happens is still being investigated. This paper provides a pretty good explanation of what happens, but it's quite technical.

Even larger impacts will produce ring structures. These are relatively rare on the Earth, since such structures are quickly weathered away. A good example of this is the Chicxulub crater off the Yucatan peninsula (the impact that killed the dinosaurs).


Isn't the date of the Big Bang a bit bogus? If I understand correctly, the date of the Big Bang is an extrapolation of acceleration of the universe's expansion through time based on the erroneous assumption that the universe is approximately 13 billion years old. The age of the universe, as far as I can tell is based on the speed of light and the fact that the furthest we could possibly see into the depths of the universe is approximately 13 billion light-years. Isn't that sort of a cat chasing his tail?  It seems that there is no valid reason for believing that the universe is really of a radius of ~13 billion years. That is just as far as we can see. If this is wrong and there is a valid reason for believing that the universe is ~ 13 billion years old, then that fact makes for a very interesting possibility.  That the expansion rate of the universe is equal to the speed of light, which leads to a very interesting suggestion that I pursued for quite awhile (until it became apparent that the suggested size of the universe is not what it seems), that the expansion rate of the universe is equal to the speed of light. Or, put another way, is it possible that the speed of light is based on the expansion rate of the universe?

I'd like to correct a couple misonceptions that might be confusing you.

  If I understand correctly, the date of the Big Bang is an extrapolation of acceleration of the universe's expansion through time based on the erroneous assumption that the universe is approximately 13 billion years old.

This is a false assumption. The observable universe is much, much larger than a sphere with a radius of 13 billion light-years. The radius is actually ~46 billion light-years (see, for example, Gott et al. (2005)).

  The age of the universe, as far as I can tell is based on the speed of light and the fact that the furthest we could possibly see into the depths of the universe is approximately 13 billion light-years.

Look at pela's answer here, if you don't want to go off Stack Exchange (or see Wikipedia for another starting point). See how the age of the universe is calculated. As an approximation,$$t_\mathrm{Uni} \sim \frac{1}{H_0} = 14\,\mathrm{billion\,years}.$$This only relies upon observations of the movements of stars.

  That the expansion rate of the universe is equal to the speed of light.

The universe is not expanding at the same rate at all distances from us. Objects further from us recede at greater speeds.


Why do stars twinkle but the Sun doesn't (I'm asking this because the Sun is also a star) My friend asked me

  "Sun is also a star then why doesn't it twinkle like other stars"

I brought this question here so that I have (and have more) infomation on this topic.

Stars twinkle because they are effectively a point of light. This point of light can be distorted and magnified by movement of patches of varying density in the atmosphere. These act as lenses causing the twinkle effect.
If you get to space, the stars don't twinkle. Stars appear as a point only because they are so very very distant.

Planets twinkle less because they appear as little discs of light. If one point is being made brighter by the atmosphere perhaps another is being made less bright. By being a disc, the twinkling is smoothed out by the average amount of light making it to your eye.

The sun is a very large disc because it is closer than the other stars. The sun is so large in the sky that the distortion of the light is not very noticeable. However when the sun is setting and its light has to travel through a lot more air to reach the ground, you will notice a shimmering around the edge. This is due to the same kind of atmospheric distortion.


How can gravitational waves escape from the gravity of black hole? I know that even light can't escape from a black hole's gravity and the velocity of light and gravitational waves are the same. How can only gravitational waves escape from its gravity?

I see this phrase all the time, and I have to say I've come to greatly dislike it because it's a very bad misnomer. Nine times out of ten, when someone is talking about a black hole, they describe it as an object with such strong gravity that "not even light can escape".

However, this unqualified statement presents a strong misconception as to what black holes actually are and how they work and accomplishes nothing but confusing innocent bystanders such as yourself. The gravity of a black hole is no more or less strong than any other object in the universe. Black holes are not cosmic vacuums which use their powerful gravitational forces to suck up all nearby matter, light, etc. In fact, if you replaced our Sun with a black hole of exactly the same mass, all the planets in our system would go about orbiting exactly the same way and wouldn't notice a difference at all (aside from the mass extinction on Earth due to no longer receiving any energy from the Sun).

That being said, let's paint a better picture of what a black hole is and how it works. A black hole is a clump of mass which has become so massive that the gravitational force of that mass on itself, trying to pull it together, actually collapses the mass into a singularity. The singularity is a point-like region of space where all the mass becomes contained. Slightly outside this singularity, the physics gets weird. For example, if you're right next to this singularity, and you calculate the speed needed to get away from that singularity (e.g., you need to travel ~11 km/s to get away from Earth) you find a speed which is much greater than the speed of light. That's the origin of the phrase "not even light can escape". But, if you start farther away from the singularity, you need less speed to escape it because you feel less gravitational pull from it (gravity decreases with distance). This means, at some distance from the singularity, the speed of light actually is fast enough to escape the black hole. This distance is so important that scientists have given it a special name, the event horizon. It can get a lot more complicated than the simple picture I've painted above, but that's the general idea.

If you put all that together, then that tells you that any light which is outside the event horizon has no trouble escape the black hole. It is only the light inside this event horizon which cannot escape. Likewise, any gravitational waves outside the event horizon can escape just as easily. This is what the answer by StephenG meant by saying they were "outside" the black hole. By outside, he meant outside the event horizon. And it is true that as long as the creation of the gravitational wave happens outside the event horizon it will escape the black hole.

And just for reference of size, the supermassive black hole in the center of our galaxy, which is 4,000,000 times more massive than our Sun, has an event horizon which only extends ~10,000,000 km. That's barely out to the orbit of Mercury if it were in the position of our Sun. So you can see, it's not very hard to be outside the event horizon as the event horizon isn't that big in astronomical terms.


Is it proper to refer to objects beyond Neptune as "Kuiper Belt Objects?" It seems to me that nowadays the more proper term is "Trans-Neptunian Object".  Is there any difference between Kuiper Belt Objects and Trans-Neptunian Objects?  If not, is one term preferred over the other?

Yes there is a difference, Kuiper Belt Objects (KBOs) are a subset of Trans Neptunian Objects (TNOs). Other subsets are Oort-cloud objects (OCOs) and scattered disk objects (SDOs). These are not KBOs but they are TNOs. See for instance this wikipedia page.


.mt file convert to .fits/.ascii/.dat How to convert .mt file with a spectrum to a more convenient format, please? I am able to display that in fv: Fits Viewer, but I am lost in converting. Thank you

.mt files are the output files (or possibly intermediate ones as the final products in the ESO archive have .fits extensions) produced by the ESO-MIDAS data reduction system for the (in this case) FEROS echelle spectrograph. The individual echelle orders have been extracted from the original 2D CCD images, flattened, wavelength calibrated and joined together into a 1D spectrum.
The .mt files are FITS files however the fact that they are 1D can cause some issues for normal image viewers; the ESO archive has some help on how to setup various tools (including fv) for displaying 1D spectra. In this case, since the file is a standard FITS file, regular FITS tools e.g. AstroPy, IRAF, MIDAS etc will work to convert the file and the ESO archive link above has info on how to do this in Python, IDL and IRAF.
One relatively pain-free option could to use Splat-VO to do the plotting, investigation and conversion. (When loading the .mt spectrum you will need to either rename the extension to .fits or tell Splat-VO the file format is FITS as it won't be auto-recognized). Once it's loaded in, you can convert to a variey of formats, including ASCII/text.


What is the theoretical ages for Pop I stars How old could population I stars theoretically be? What's the earliest they could have formed?

Since Pop III stars are the oldest stars, then they could have theoretically began to form in the Cosmic Dark Age (CDA), see section 5, since the oldest galaxy we've observed so far (in infrared) is about half way through the CDA (~$480$ million years). As the article states,

  Stars and galaxies are formed when dense regions of gas form due to the action of gravity, and this takes a long time within a near-uniform density of gas and on the scale required, so it is estimated that stars did not exist for perhaps hundreds of millions of years after recombination.

So not only must the temperature of the universe be sufficiently cool, gravity must have acted for a long enough amount of time for baryonic structure to begin to form.


What is the format of the data from the JPL's HORIZONS system? I'm looking for accurate positions and velocities of the planets in the solar system over several decades. I want to simulate their trajectories using Newtonian laws of motion, and compare the precession of Mercury's perihelion with observed data. I used the JPL's HORIZONS system but I can't manage to find these positions and velocities in the output data.

Could you explain how the data is formatted? (The HORIZONS web-interface Tutorial didn't help.)

WebGeocalc (http://wgc.jpl.nasa.gov:8080/webgeocalc/) is actually a
fairly useful tool, and the next best thing to downloading the CSPICE
libraries, so let me provide a slightly more detailed answer to your
question.

When you first visit WebGeocalc, you will see something like this:

Because you are looking for perihelions, which involve distance,
scroll down and choose "distance finder":

Since you want to find the perihelion of Mercury, fill out the top
half of the form like this:

For this example, I'll find all perihelions from 1900 to 2201. A
perihelion is a minimum of distance, and since you're looking for all
perihelions, choose "is local minimum" on the lower half of the form,
so it looks as follows:

Scroll down once again (you can leave the remaining form values as is)
and click the "Calculate" button:

When you get your results, click "Save All Intervals" to save the
times of Mercury's perihelions. Note that each result is an interval,
but the interval is 0 seconds long, so it's actually a point in time:

Now click on "Calculation Menu" to get back to the main menu:

We'll now find Mercury's position at all of these perihelions so you
can see how the perihelion position changes.

To find Mercury's position at these times, first click on "State
Vector", which will take you to a form. Fill out the top half of the
form as follows:

Fill out the lower half of the form as follows, dragging the results
window (for the times of Mercury's perihelions) into the input times
window:

Click "Calculate" to receive your results:

See also http://naif.jpl.nasa.gov/pub/naif/toolkit_docs/Tutorials/pdf/individual_docs/36_webgeocalc.pdf

NOTE: if you're going to do extensive calculations, you'll probably
want to download the C SPICE library and kernels:
http://naif.jpl.nasa.gov/naif/tutorials.html

EDIT: I emailed Jon D. Giorgini (jdg@tycho.jpl.nasa.gov) re the open-sourceness of HORIZONS, and got this answer (which I'm not sure is helpful, but...):

Horizons output is copyrighted in the sense one can't take a
table output by Horizons and re-publish it as one's own.
JPL is a contractor organization and so the output is not prepared
by an officer or employee of the US government, and it is not
transferred to the US government, so copyright remains relevant.

The data values themselves, say the RA/DEC of Pluto at
some instant, are not legally copyrighted, though it would
be unethical to publish it without citation.

Legal copyright protection is for fixed representations
analogous to a book; the book has copyright protection, but
an individual word extracted from the book (i.e., into a
different representation) is not itself protected.  Up to
a point

Note that Horizons reads SPICE files for planets, natural
satellites, and spacecraft, then derives results it outputs
from scratch from that basis. But for comets and asteroids,
it creates all information on demand from scratch.

Further, we (JPL Solar System Dynamics) put the planet/satellite/
spacecraft information into the SPICE format to begin with. So
a SPICE file might be like the blank pages of a book; not
copyrighted until the original work is written into it, but not
itself copyrighted (beyond the source code, of course).

But if you need a more trustworthy legal opinion for a specific
situation, I can put you in touch with legal staff. If this is an
actual issue and not just a point of curiosity, that's the way to
since perhaps other rules would apply to your actual situation.


Why didn't the Big Bang produce heavier elements? Shortly after the Big Bang, temperatures cooled from the Planck temperature. Once temperatures lowered to 116 gigakelvins, nucleosynthesis took place and helium, lithium and trace amounts of other elements were created.

However, if the temperatures were so high shortly after the Big Bang, why weren't much heavier elements produced? 116 gigakelvins is obviously far above the temperature required for elements like carbon and oxygen to fuse. In addition, shouldn't most of the protons at those temperatures have fused, leaving the Universe with mostly heavier elements?

I think that your thought process is flawed in that you assume that by drastically increasing the temperature you are guaranteed to get heavy elements. As odd as this may sound, this isn't the case (especially during the Big Bang Nucleosynthesis (BBN)) for a few reasons. In fact, if you took a hydrogen-only star and made it go supernova, you wouldn't get heavy elements like you see in current stars going supernova.

BBN Timescale

One major point to consider is that the BBN era is calculated to be only ~20 min long. That isn't really much time to form elements. Sure, supernovae happen in an instantaneous flash, but there are other things going on there, which I'll get to in a second. The main point here is that fusing takes time and 20 min isn't that much time to form heavy elements.

Deuterium

To get heavy elements, you need to build up to them. You can't just smash together 50 protons and 50 neutrons and get tin. So the first step is to smash together a proton and neutron to get deuterium, but here you already run into a problem known as the deuterium bottleneck. As it turns out, the huge temperatures actually (and somewhat counterintuitively) impede the creation of deuterium. This is mainly because the deuteron will end up having so much energy that it will be able to overcome the binding energy (and deuterium has pretty low binding energy being that its only two nucleons) and will likely break apart again. Of course, given the density and temperature you can still get a good amount of deuterium simply by force of will, but not as much and not at the rate you'd expect otherwise. Another point that makes deuterium form less frequently that you'd naïvely expect is that the proton to neutron ratio before BBN was about 7:1 due to the proton being more favorable to be created since it has a slightly lower mass. So 6 out of 7 protons didn't have a corresponding neutron to combine with and had to wait for deuterium to form first before it could combine with anything.

Tritium, Helium, Lithium, Oh My!

Deuterium is then the catalyst for forming all the next stages of particles in your soup. From here you can throw them together with various other things to get $^3\mathrm{He}$, $^3\mathrm{H}$, and $^4\mathrm{He}$. Once you've got a good amount of deuterium, tritium, and helium isotopes floating around, you can start making lithium and if you're lucky a bit of beryllium.

To Boron and Beyond

But now, once again you run into a bottleneck, and one more severe than the deuterium bottleneck. You can't easily jump to heavier elements with what you have on hand. The next fusion chain, and the way stars do it, is the triple-alpha process which helps to form carbon but to perform this chain and build up enough carbon you need a lot of time. And we only have 20 minutes! There just isn't time to form the carbon we need to progress along the fusion cycle. As I hinted at the beginning, pure hydrogen stars also wouldn't produce heavy elements upon supernova for this reason. They're able to produce heavy elements now because they've had billions of years before their SN event to build up a base amount of carbon, nitrogen, oxygen, etc. that can aid in the heavy element fusion processes.

So you don't have the time to follow the triple alpha process and make carbon — what about other processes? Surely the temperatures are high enough that you can do different fusion methods not seen in stars. Well... no. You can't even smash together lots of $\mathrm{He}$ or $\mathrm{Li}$ to get really heavy elements because of the fact that heavy nuclei are only stable if they have way more neutrons than protons. And we already said there was a big neutron deficiency early on so the chance you have enough neutrons hanging around to smash together to get, say $^{112}\mathrm{Sn}$ (that's tin with 62 neutrons), is pretty small. What's more, you can't even try to either skip carbon by making something slightly heavier or form something intermediary between lithium and carbon. Again, this is because of stability issues. So with no other options, you have to shoot for carbon after lithium, and as stated above, you just don't have time for that.

TL;DR

Overall, BBN is limited to getting only to lithium because of limited time, proton to neutron abundance ratios, and fusion bottlenecks that slow things down. All these come together to produce ~75% $^1\mathrm{H}$, ~25% $^4\mathrm{He}$, ~0.01% $^2\mathrm{H}$ and $^3\mathrm{He}$, and trace amounts of $\mathrm{Li}$.


What do we know about the characteristics of stellar objects, 13,3 billion light years away from us? To receive light from objects like MACS0647-JD, I suppose, it has to has extraordinary dimensions and intensity of electromagnetic radiation. What are the mass, the dimensions and other characteristics for this objects?

Holger,

I highly recommend you check out the NASA story on the discovery MACS0647-JD, which includes many dimensional facts (mass, distance, age, etc.) and comparisons:

http://hubblesite.org/newscenter/archive/releases/2012/36/full/

From the article:

  The estimated total mass of the stars in this baby galaxy is roughly
  equal to 100 million or a billion suns, or about 0.1 percent to 1
  percent the mass of our Milky Way's stars.

Measuring the mass of a galaxy is very tricky indeed (especially one as far as 13.3 billion light-years from earth!), but there are a couple different methods that can one can use, see here:

PSU Astro Measuring Mass of Clusters

And this Physics Stackexchange thread as well:

https://physics.stackexchange.com/questions/123/how-does-one-measure-the-mass-of-a-galaxy-and-other-such-large-quantities


Why are there no ISM clouds with temperatures between 100 and 6,000K? In the interstellar medium, there are several different diffuse phases of gas, distinguished by their density and temperature. Specifically, the cold neutral medium has temperatures from ~50-100 K and the warm neutral medium ranges from 6,000-10,000 K. According to my professor, there have been few gas clouds found with temperatures between these two.

Does anyone know why such a large gap exists?

Let $n$, $T$, and $x_i$ be the number density of hydrogen, the temperature of the gas, and $n_i/n$, where $n_i$ is the number density of the $i$th component of the interstellar medium. We can then write the criteria for thermal equilibrium as
$$n^2\Lambda(n,T,x_i)-n\Gamma(n,T,x_i)\equiv n^2\mathcal{L}=0$$
where $\Lambda$ and $\Gamma$ and the heating and cooling functions, respectively, and $\mathcal{L}$ is defined through those functions and $n$. If the equilibrium is unstable,
$$\left(\frac{\partial\mathcal{L}}{\partial S}\right)<0$$
for entropy $S$. This leads to different instability conditions, termed the isochoric and isobaric instabilities (Field (1965), $\text{Eq } 4a,4b$). These can be determined from the temperature, pressure, and density of the gas (also assuming that the gas can be approximated as an ideal gas).

In general, $\Lambda$ and $\Gamma$ are complicated to determine, although combinations of power-law and exponentially decaying factors may often be sufficient. An example curve that seems to pop up a lot as an example was calculated by Dalgarno & McCray (1972), Figure 2:

More accurate measurements have been made since then, but the general shape is still applicable. The warm neutral medium occupies the area near the sharp change at around $\sim10,000\text{ K}$, and the cold neutral medium occupies the area in the left of the diagram. Another way to visualize this is on a $\log P/\log n$ diagram like this one (from these slides, annotated from Wolfire et al. (1995)):

In reality, the two-phase model is an oversimplification, and the ISM has more distinct components. However, the isochoric/isobaric instabilities still limit the range in which clouds can exist in stable equilibria, and explain the dearth of gas in the relevant temperature range.

Let me elaborate on the terms isobaric and isochoric. In thermodynamics, it is sometimes convenient to assume that some thermodynamic variables remain constant in a certain situation. Isothermal processes occur at constant temperature; likewise, isobaric processes occur at constant pressure and isochoric processes occur at constant volume.

The equations for the two instabilities are
$$\left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\rho}<0\tag{Isochoric}$$
$$\left(\frac{\partial\mathcal{L}}{\partial T}\right)_{p}=\left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\rho}-\frac{\rho_0}{T_0}\left(\frac{\partial\mathcal{L}}{\partial\rho}\right)_T<0\tag{Isobaric}$$
In the first, we assume that the cloud is at constant volume, and since the total amount of matter in the system is constant, the (mean) density must also be constant. In the second case, we assume that the cloud is at constant pressure. Perturbations that lead to instabilities thus arise from perturbations of other thermodynamic variables.

As a final note on notation,
$$\left(\frac{\partial\mathcal{L}}{\partial A}\right)_B$$
means that we take the partial derivative of $\mathcal{L}$ with respect to $A$ while keeping $B$ constant. It's a common thermodynamical convention.


What does the word "shock cooling" mean in supernova physics? I cannot understand the meaning of "shock cooling" which occurred in supernova.
And I saw the sentence like "Early photometry and spectroscopy reveal evidence of shock cooling of an extended helium-rich envelope". Could you explain the mechanism of shock cooling in supernova?

Shock cooling in a SN is the scenario where kinetic energy carried by the shock transforms into other energy forms: adiabatic expansion, radiation, acceleration. A shock is actually a term meaning discontinuity. For example, a density profile of steady-wind medium is $r^{-2}$ while that of external part of ejecta is $r^{-n}$ where $7 \le n \le 12$ as typical values in core-collapse supernovae, but at the interface between the medium and the external ejecta, where they interacts, its density profile spikes before asymptotically decaying to the characteristic profile of each region. The interface is the shock.

The mechanism of shock cooling in SN involves processes which are not completely understood. What we know is that the shock is formed as a double-shock structure: forward and reverse. The forward shock propagates outward, while the reverse one goes inward in mass coordinate. (Note that in spatial coordinate both shocks propagate outward). Kinetic energy in each shock is transformed by interactions. At a forward shock, typically the energy is transformed into acceleration of relativistic particles, which in turn radiate via synchrotron or inverse Compton processes, and collisionally interacts with the medium. At a reverse shock, the shock energy thermalizes ejecta, which in turn thermally radiates.


Stars located outside of a galaxy How common or rare is it to find stars outside of a galaxy? Can such a star still have planets that contain life? Recently, I think that I heard of a star with a planet that entered our Galaxy. Did this star come here from another galaxy or was it originally from the Milky Way?

I don't think we have a good estimate of the number of stars not in galaxies -- they're actually pretty hard to find unless they're really bright, since even if you spot it, how do you know that it's extragalactic?

But galaxy mergers (which are very common) tend to throw out large numbers of stars from the merging galaxies, so it seems likely there are many stars in space between the galaxies.

There is no reason why such stars can't have planets, but it's likely that the planets formed along with the star inside a galaxy and got ejected from the galaxy along with the star.

Stars (and planets) form when Giant Molecular Clouds collapse under their own gravity, the collapse frequently being triggered by tidal forces from a passing star or a nearby supernova.  It seems unlikely that there are many GMCs in intergalactic space, and there are certainly fewer encounters which might trigger collapse. So it seems likely that relatively few stars (or planets) actually form in intergalactic space.

Life does not seem to have evolved because of the existence of stars in the sky, so once a planet exists in a habitable zone, life should form regardless of where it is.

The Gaia mission is providing us with superb position and velocity data for billions of stars, and it has identified a handful (maybe more by now) of stars which seem clearly to have come from intergalactic space.  Here's an article which covers the report last year.


Where can I still find solar film? I, like many others, am excited to see the eclipse. I want to view the eclipse through my telescope, however, I don't have any solar safety equipment for solar viewing and I cannot find any online that I can get before the eclipse. Could anyone lead me to a reputable site where I could purchase solar (preferably mylar) film before the eclipse?

Amazon has solar film sheets if that's what you're looking for. It is sold by Amazon for Thousand Oaks Optical. Amazon sometimes has next day or 2 days delivery.


Did Hubble see a pair of UFOs? (Identify this artifact) The image below is this one, found in this group of Hubble space telescope images of the interstellar comet 2I/Borisov (C/2019 Q4).

Can anyone recognize this pair of artifacts in the upper right corner of this HST diagnostic image?

I'm going to guess that these are reflection artifacts of one or two nearby bright stars. The fact that they have about the same length as the stellar streaks, suggests they are smeared by the telescope's tracking of the comet. The fact that they are oriented with a slightly different angle than the star streaks may be due to the reflection shifting slightly as the telescope is moved.

In fact, Figure 6.9 on this page from the WFC3 Handbook (the instrument used for the comet images) shows a clear examples of "figure-eight shaped ghosts" due to bright stars in the lower-right quadrant, described thus:

  When a point source is positioned in the lower right quadrant of the
  UVIS detector, out-of-focus reflections between the CCD and windows
  appear along a diagonal from the source towards the upper left, well
  removed from the source. These figure-eight shaped ghosts gradually
  move outside the field of view as the target moves out of the lower
  right corner. They contain a few percent of the flux of the target.
  The ghosts of several bright stars are visible in the exposure of 47
  Tuc shown in Figure 6.9.

Figure 6.9 -- note the clear example paired elliptical rings ("figure-of-eight shaped ghosts") on the left part of the upper half of the image.

(Note that the comet images were taken with a sub-array of the WFC3 UVIS detector, which is probably why you don't see the star that produced the ghosts.)

You see ring-like ghost images in ground-based images, too; I've been examining an image taken by some collaborators with a telescope in the Bavarian Alps, and it's got several examples (fortunately not on top of the galaxy we're interested in!).


Why are Delta Aquariids "for the southern hemisphere" while the Perseids are "for the north"? In the video How to spot Comet NEOWISE, plus the largest 3D map of the Universe | Night Sky News July 20 after 05:28 Dr. Becky (Smethurst, Oxford astrophysicist) says:

…and it just so happens that we’ll be passing through the debris trail of two of the comets whilst comet NEOWISE is going to be visible in the sky, and so that’s going to cause to very popular meteor showers. There’s the Delta Aquariids meteor shower at the end of July, and then the Perseids meteor shower mid-August time.
And the Perseids meteor shower is just hands down the BEST meteor shower of the year. There’s something like sixty meteors per hour during its peak which this year is going to be the 12th of August.
Now the Delta Aquariidss are probably more like fifteen to twenty meteors per hour, so not quite as spectacular as the Perseids, but the Delta Aquariids really are more for the southern hemisphere where the Perseids is really going to be for the Northern hemisphere.

Question: Why are Delta Aquariids for the southern hemisphere while the Perseids are for the north? The Earth is moving along its orbit at 29 km/sec in both cases, why the big difference?
note: there are notes below the video that you might find helpful if looking for comet NEOWISE.

I suspect that that is simply a statement about where you would have the best visibility of each meteor shower.  The name of each shower tells you where the radiant is, the point in the sky from which the trails of the meteors appear to originate.   Perseus in the northern sky (with the specific radiant point of the Perseids at +58 degrees declination), while Aquarius is farther south (with the radiant of the delta Aquariids at about -16 degrees declination), so a northern observer would have the Perseids radiant higher in the sky and would see more of those meteors, and vice versa for a southern observer and the delta Aquariids.
Since the meteors go in all directions from the radiant, at least some from either shower would be visible for observers over a large range of latitudes, but you’d see the most if the radiant is high in your sky.
Edit: this nice image from spaceweather.com shows clearly why the radiants have their respective locations:

(From spaceweather.com, linked from Sky & Telescope.)
These are reconstructed orbits of fireballs seen as part of these showers.  The light green are the delta Aquariids, and the darker green / teal are the Perseids.  You can see that they have very different inclinations where they intersect Earth, so they appear to originate from different directions in our sky.


Lunar illumination (percentage) data on an eclipse day needed I am looking for the information on the percentage illumination of the moon during the different stages of a total lunar eclipse. Is there a database from where I can acquire information of lunar illumination, distance and angle for the July 27th 2018 lunar eclipse for a given location?

I found some chart regarding this, which is here:
http://www.mkrgeo-blog.com/remote-observation-of-the-total-lunar-eclipse/
The light level drops from about 50-60% prenumbral phase. It becomes noticeable.

If you want to get the percentage illumination of the Moon during the eclipse, install the Cartes du Ciel and turn on the Earth's shadow. Then you will be able to gauge the percentage value.
https://xjwfriends.com/2018/02/18/astronomical-diary-vat-4956/


Objects entering or leaving the observable universe Have there been any observations of objects entering or leaving the observable universe?

When looking at the physical limit of observing something like this I will to assume that the "object" would be an galaxy. I think it would be a good assumption to choose the size of such an galaxy, $G$, to be in the order of $10^{5}$ light years and the radius of the observable universe, $OU$, is $46 \times 10^{9}$ light years. Using the limit of angular resolution and the fact that the current biggest telescope has a diameter of 22.8 meters, then the longest wave length which would still yield would be about $40 \mu m$. When reversing the redshift, assuming this would be around $z=9$, then the longest emitted wave length which would be detected would be $4 \mu m$.

So observing such an event would technically be possible, however relative to the astronomical timescale we have only been able to observe with such a high resolution for a very short time, so the probability will not be very high.

And if such an event would be observed, would such an object suddenly appear/disappear. It will take along time for an entire galaxy due it size, so this would translate to more of a fade. Or will something else happen?

I don't know if there have been any observations of objects leaving the observable universe, but I'll admit that I have a hard time keeping up with the latest discoveries in observational astronomy. I'll try to approach your question from a theoretical viewpoint.

As you pointed out, the radius of the observable universe is $4.6 \times 10^{49}$ light-years. To figure out how fast an object is moving away from Earth (or any observer), we can use Hubble's law:

$$v = H_0D$$

where $v$ is the recessional velocity, $H_0$ is Hubble's constant, and $D$ is the proper distance from the observer to the object (see Wikipedia for the difference between proper distance and commoving distance). The trouble here is that the exact value of Hubble's constant isn't exactly well-known. There have been many different observations since Hubble proposed his law that attempted to find values for the constant, but they range widely. Here I'll use $82$ $km$ $s^{-1} /Mpc$ (where $Mpc$ is a megaparsec - 1 million parsecs).

Let's say that the object in question is $13$ gigaparsecs away (so $13,000$ megaparsecs away). The light we see from the galaxy would be from when it was very young. One of the most distant observed galaxies is this far away, and only $2,000$ light-years across (note: do not confuse the "light-travel distance" with the "proper distance"), so let's say that this galaxy is smaller in size - negligibly small, in fact. We can then use Hubble's law to calculate a recessional velocity $v$ of

$$(82)(13,000) = 1.066 \times 10^6$$ kilometers per second, or $3.36 \times 10^{13}$ kilometers per year. That's pretty fast! An object $13$ gigaparsecs away would be $42.38$ billion light-years away, not far from the edge of the observable universe. Yet if the edge of the observable universe is $46$ billion light-years away, the galaxy would still be $3.62 \times 10^{21}$ kilometers away, and it would take a long time to get there. A galaxy this young would be maybe $1,000$ light-years across, so, moving at its current speed, it would take it many years to cross a distance its own length - that is, if the furthest edge of it were at the edge of the observable universe, it would still take many years to fully disappear. And that's even factoring in that it would be moving away faster!

The reason I don't have the object being further away is that any objects at the edge of the observable universe would be so young when they released the light we see today. That means that galaxies would be small, and wouldn't emit a lot of light. One of the furthest objects we have detected is only $30$ billion light-years away, and $2,000$ light-years across. So it would be terribly hard to observe an object at the edge of the observable universe (ironic, isn't it?). But here I have shown that it would take years for the object to completely disappear.

I hope this helps.

My sources for the data and conversion factors (i.e. light-years to kilometers):

Hubble's Law

Wolfram Science World

Most distant objects

Spacetelescope.org

Light-year


Why do we we even exist? It came to my mind one day that why does this universe even exist and why do we even exist and like we are enclosed in the so called atmosphere and a floating rock in space called the Earth, so is there any possibility that the universe is also enclosed in something and if it is, then what is outside that enclosure?

That's an interesting question, but as the comments say might be a better place to ask this question in the Philosophy SE site, but the question about the universe being englobed maybe by another much bigger 'space' is something we might never be able to prove right or wrong. This 'existential' question might refer to the holograph theory, the multiverse theory. The thing is, at the moment the only possible explanation of whats beyond the 'observable universe' boundary is nothing, neon, an infinite multi-dimensional plane of nothingness.


Conventional matter to dark matter ratio, outside of galaxies Consider say this outstanding diagram from another question,

To begin with galaxies have a typical density D. Intergalactic space has a much lower density d.

As we know: in galaxies (with density D) most of the stuff is dark matter.  Let's say for convenience it is 10:1.

So, the ratio of dark matter in galaxies is (we'll say) 10:1.

But here's what I can not find explained anywhere:

(1) In intergalactic space. First of all simply is there any dark matter?  Or, do we believe that dark matter exists, simply, only in galaxies? Not between?

(2) If there is dark matter between galaxies. Do we know or believe: what is the ratio of dark matter out there?  In fact, is it just the same ratio (say, 10:1) as inside galaxies? Or, is it perhaps even more dominant, or perhaps much less dominant?  What is the ratio out there?

(3) If there is dark matter everywhere. Galaxies exist in filaments and walls.  Does dark matter do this? or is it evenly-spread?

Let me see if I can answer at least some of this.

Yes, there is dark matter between galaxies. This is demonstrated by the fact that in galaxy groups and clusters, you need more dark matter than is found in the galaxies themselves to explain what's going on: in terms of why the groups or clusters are gravitationally bound in spite of the extreme velocities of the galaxies within the groups/clusters; why the X-ray-emitting gas in clusters is so hot and high-pressure and yet still confined to the cluster; and in order to explain the gravitational lensing of background galaxies by clusters. (Also supported by observations of things like the Sunyaev-Zeldovich effect in clusters.)

(In fact, the oldest evidence for dark matter was the extremely high velocities of galaxies within clusters as measured by Fritz Zwicky back in the 1930s. This is also one of the failure modes of Modified Newtonian Dynamics as an alternative to dark matter: even if you think you can get rid of dark matter within galaxies, you still end up needing dark matter within galaxy clusters, in addition to the galaxies themselves.)

And, of course, cosmological models indicate that dark matter accounts for about 25% of the critical density of the universe. In order to have that much dark matter, you have to have more dark matter than what's inside individual galaxies.
"If there is dark matter between galaxies. Do we know or believe: what is the ratio of dark matter out there?" I think the simplest way to answer this would be to take the dark-matter/baryonic-matter ratio for the universe as a whole, as determined by current cosmological observations and models. This gives a ratio of about 5:1.
"Galaxies exist in filaments and walls. Does dark matter do this? or is it evenly-spread?" Yes, dark matter does this. The reason we see galaxies in filaments, clusters, and walls is because the dark matter has gravitationally clumped into those structures; on this scale, the galaxies are basically along for the ride, following the gravitational pull of what the dark matter is doing. The reason for galaxies existing in filaments is because of the uneven distribution of the density of both regular and dark matter throughout the universe. An overdensity leads to the matter collapsing to form (for instance) a galaxy, but the overdensity can be asymmetric. This leads to the collapse happening faster in one particular direction, causing galaxies to form in filaments (see e.g. this summary paper for an overview).

This page has images from a large-scale cosmological simulation (the Millennium Simulation). Most of the images showing filamentary structure on different scales are showing the dark matter only.


Moon looks too small in 3D model I'm not sure if this is more a question for a 3D graphics forum, but I'm trying here first in case the answer has more to do with astronomy than 3D.

I am creating a model of the Earth and Moon in a piece of 3D software. To avoid the enormous scale of using real-world sizes, I am using a 1000:1 scale ratio.

These are the dimensions I'm using:

Earth radius = 6,371 m
Moon radius = 1,737 m
Distance from Earth to
Moon = 384,400 m

Since I have scaled everything equally, I would assume that the proportions of things would look accurate, but the moon looks very small indeed.

Here is a quick render from a camera placed just above the surface of the Earth, looking out at the moon:

The camera's focal length is 24mm, which I understand to be relatively similar to the human eye. I was expecting the moon to look similar in size to how it looks when you look up at it in real life but, as you can see, it's a tiny little dot.

Is there something obvious I'm missing here, or some explanation for this discrepancy?

I decided to answer because the role of perception is not yet taken into account.

The way we see is not only dictated by the optics of our eyes as cameras. It depends on how the brain elaborate or interprete the signals.

It is well known but still surprising that a finger at arm length can accommodate something like seven to ten moons (if not ten to thirteen, I can't look right now). Still, it quite possible that asking a person (not aware of that) something like  "how many moon can you cover with a finger?" we get an answer such as "may be half" or " one? two? Tell me...".

Specifically to Moon:

https://en.m.wikipedia.org/wiki/Moon_illusion

Photographing the Moon with a smartphone or a point and shoot camera we get things like this picture (3.5 mm, aperture 2.6).

The moon looks much smaller in the photograph than as seen in reality. The result does not look too dissimilar from the pic posted by OP.
(Let me add that my photo was cropped a bit, so the native size of the moon is even smaller).

All this is to say that whatever is aimed at giving a feeling of realism should consider perception of reality rather then size of reality.

This is done in Stellarium and many others sky viewing apps where a "realistic" option is used to alter the angular size of the moon paradoxically away from reality. I.e. the moon is arbitrarily scaled up to fit our common perception.

I think this is what the OP is dealing with.

PS I am pretty sure that my pic has been uploaded. For some reason it does not appear to be the case. If necessary let me know. By the way it was a house with sky and a small than commonly perceived Moon ;)

Per this comment here is a cell phone image with a horizontal FOV of about 60°. However in this pic the moon is not that small. I suspect the brain still enlarges it as it is down at horizon and trees are close :)


Would the dynamic pressure of the Sun’s solar wind be strong enough to trap plasma ions within an open container? If you were to place a long cylinder, open at one end and closed at the other end, and position this cylinder so that the open end is facing the Sun, and then you were to inject plasma ions, such as oxygen ions, into the cylinder near the closed end, would the solar wind entering the open end of the cylinder be strong enough to keep the oxygen ions pinned against the closed end of the cylinder, effectively trapping them there? How close to the Sun would this cylinder need to be for the solar wind to be strong enough to 'entrap' these oxygen ions?

So you basically said that we could use the Sun to propel a rocket. We already have rockets like this equipped with solar sails. These solar sails have a "sail" that catches the stuff thrown out of the Sun. I'm no expert on this way of travel, but that is the basic description of how it works.


What is projected separation and how can I make sense of its unit (h^-1 kpc)? I have been reading this paper on the separation of a binary quasar in which it states:

  At a projected separation of 8.3/h kpc (Omega_matter = 1), CTQ 839 would be the smallest projected separation binary quasar currently known.

What is meant by projected separation and where does this unit come from?

It means the separation between the two objects if they were both at the same distance away. This separation is found by multiplying the angular separation (in radians) by the distance to the objects.

Often, the distance to extragalactic objects depends on the assumed value of the Hubble parameter. In this case, $h$ is a dimensionless number corresponding to $h=H_0/100$, where $H_0$ is the Hubble parameter in km/s per Mpc.

The current best value of $H_0$ is about 70 km/s per Mpc, so $h\simeq 0.7$.


How would astronomical seeing on Mars differ from that on Earth? Astronomical seeing is the limiting factor for the resolution of all but the smallest Earthbound telescopes.

 Source

Stunning advances in adaptive optics (along with it's predecessor speckle interferometry and it's budget-minded cousin lucky imaging) get around this but only with substantial compromises in (some combination of) throughput, cost, complexity and wavelength range (see answers to Why aren't ground-based observatories using adaptive optics for visible wavelengths?)

If identical telescopes sat on the surface of Earth and Mars and looked at a distant and equi-distant  body, how would the following differ between the Mars telescope and the Earth telescope?:

As far as I know, "seeing" (or rather the effects influencing optical wave propagation) is caused by turbulence in the atmosphere.

Using the Reynolds number Number $ Re = \dfrac{\rho L v}{\mu}$ as a measure for turbulence:

density $\rho$ drops due to the reduced pressure (about 1/100 earth pressure), additionally the gravity is smaller than on earth
characteristic length $L$ will stay similar
average wind speed $v$ is about 2 times higher than on earth
the dynamic viscosity $\mu$ of an (ideal) gas is independent on the pressure, and the temperature dependency can be approximated with $\sqrt{\dfrac{T_{mars}}{T_{earth}}} \approx 0.9$

So for an average martian day an atmospheric Reynolds number would be much smaller than on earth and I would expect a far better seeing due to less turbulence

Additonally, I would guess that the effects of a thinner and dryer atmosphere would help as well, since the index of refraction is dependent on the gas pressure as well.