text:
Is there any habitable exoplanet close to Earth?
I would like to know please if there is any habitable exoplanet close to Earth.

Habitable, in the context of exoplanets, means "terrestrial and water could exist in liquid form". It does not mean we could actually live there. It doesn't mean that there is actual water, or oxygen, or an atmosphere. There are three "habitable" planets in the solar system, But you couldn't live on Venus

There's no such thing as "close" when you're talking other star systems.    Even the closest star system, 4.2 light years away is ridiculously far.

Yes, but potentially habitable. The closest potentially habitable exoplanets are Proxima b, Ross 128 b, Gliese 273 b, Gliese 3323 b and Gliese 667 C c. Here you can find a video about them: https://www.youtube.com/watch?list=PL3RiFKfZj3pv1ZqpFxuZinoGtUGEOankw&v=Pd6smyz2tUk

I think this is a perfectly reasonable question, it doesn't deserve down-votes. An exoplanet is, by definition, outside our own solar system, so "close" is easily understandable as meaning something like "nearby", i.e. in the stellar "neighbourhood". I'm sure we all realise that interstellar distances are big. Anyway, I've posted a comprehensive answer which SE readers might find useful as a summary.

Habitability is related to size or mass (i.e. likely to have a rocky composition and maintain surface water), and distance from the host star (the “habitable zone” – not too hot, not too cold – which depends on stellar temperature and stellar flux). The Planetary Habitability Laboratory (University of Puerto Rico at Arecibo) publishes a catalog of potentially habitable exoplanets.
The PHL catalogue is divided into two lists. The first list covers the “conservative” habitable zone, where the combination of stellar temperature and flux is optimal (equivalent to an orbit between Earth and Mars around a Sun-like star). Exoplanets are also given an ESI (Earth Similarity Index) rating, a measure of how similar the planet is to the stellar flux, mass, and radius of Earth (Earth = 1.0).
As at March 2018 there are only 10 confirmed exoplanets within the conservative habitable zone, the closest (in order) being:

Proxima Centauri
b (4.2 ly = 40
trillion km), ESI 0.85
Gliese 667 C
c (23.6 ly), ESI 0.84
TRAPPIST-1 e (39.6 ly),
ESI 0.85
TRAPPIST-1 f
(39.6 ly), ESI 0.68
TRAPPIST-1
g (39.6 ly), ESI 0.58
LHS 1140 b (40.7 ly), ESI
0.68

After these 6, the next nearest in the list is Kepler-186 f at a distance of over 561 ly.
Note that while Proxima b might be “in the zone”, it’s unlikely to be habitable. At only 7 million km from its host star (compare with Mercury, 46 million km from the Sun at perihelion), it’s likely to be tidally locked, restricting any habitable area to a thin border wedge between the two extreme regions. It also receives 400 times more X-ray radiation than the Earth; and the stellar wind pressure is 2000 times that of Earth, sufficient to blow away any atmosphere. Not an ideal candidate to look for life!
Kapteyn b (12.8 ly) is in PHL’s conservative list and at 12.8 ly would be the 4th closest habitable planet, but PHL lists it as “unconfirmed”, as recent research has cast doubt on its existence: Robertson, Mahadevan & Roy (2015) concluded that "Kapteyn b is not a planet in the Habitable Zone, but an artifact of stellar activity."
PHL’s second list covers exoplanets that are a bit less likely to be habitable, either because they are less able to maintain surface liquid water due to their composition (based on size/mass), or because they’re in a less hospitable orbit (the “optimistic habitable zone”). There are 33 confirmed exoplanets in this list, the closest being:

Ross 128 b (11.0 ly), ESI 0.86
Gliese 273 b (12.4 ly), ESI 0.86
Gliese 3323 b (17.4 ly), ESI 0.89
Wolf 1061 c (14 ly), ESI 0.82
Gliese 832 c (16 ly), ESI 0.81
TRAPPIST-1 d (39.6 ly), ESI 0.90

There are some possible planets in the Trappist-1 solar system.  Particularly planet 1e looks to be in the habitable zone but we won't know until the Webb telescope is launched next year.
Close to Earth is a relative term, Trappist 1 is about 40 light years away.


text:
Could Hot Jupiter systems sustain Trojans?
My understanding is that there is a gravitationally stable region between the Sun and Mercury where asteroids could - in theory - orbit for long periods without being perturbed. However these hypothetical vulcanoids are seemingly non-existent; it being likely that their orbits decay through other means besides the influence of a planet, such as the Yarkovsky effect. We also have trojan asteroids, which share the orbits of planets such as Jupiter but spend their time in the leading and trailing Lagrange Points, keeping their orbits stable.
If a Hot Jupiter such as 51 Pegasi b were to host trojan asteroids, would their proximity to the planet allow their orbits to remain stable for longer, or would their orbits decay regardless? If a Hot Jupiter migrated inwards towards its star, is it more likely to have swept away any possible trojans?

Small point, and more of a pet peeve of mine than a correction, but "stable for longer" are you talking years or orbital periods.   The orbital periods obviously get considerably shorter as the orbiting objects get closer to the star.   It seems to me that orbital stability should be counted in number of orbits, not in amount of time.  (Maybe I should make that a separate question).

I'd say 'long period' meaning as long as the planet has been around, or at least from the time the hot jupiter settles into its current orbit to the present.

Yes, hot Jupiters can have stable Trojans.
A great analysis of this very problem is given by Kislyakova et al. (2015). They suggest that super-hot outgassing Trojans may be an explanation for the ultraviolet anomalies seen in the WASP-12 and HD 189733 systems. As part of their analysis, they explore the photogravitational stability of Trojans of various size, and while they show that meter-sized bodies can be ejected on timescales of around a million years, Io-sized objects can be stable over the lifetime of the system. It may also be worth noting that although they found many stable tadpole orbits, they did not find any stable horse-shoe orbits (although they do not show that such an orbit cannot exist).
As for the actual formation and capture of such a Trojan, Beauge' et al. (2007) modeled the inward migration of a massive planet, with its semi-major axis decaying over a time scale of $10^5 - 10^7~\mathrm{yr}$. Their results suggest that a Trojan satellite with a tadpole orbit close to $L_4$ will be maintained during the migration.

A nice find! Will definitely have to go over the linked papers in more depth. I like possibility that a migrating planet can drag its trojans along with it!

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Mercury mapping attempts before Mariner 10?
Mercury's surface was not mapped up-close until Mariner 10 passed by in 1974. This was followed by MESSENGER in 2008. I have seen several attempts of mapping Mars that came before space exploration, such as Schiaparelli's maps from 1877.
With regards to Mercury, were there any documented attempts to map its surface with ground-based observations prior to unmanned space exploration? If so, however unlikely, did any observed features from Earth match up with Mariner or MESSENGER close-ups?

It's very hard to see anything on the surface of Mercury. It is always seen against the glare of the sun. It is always low in the sky, so we see it through a lot of atmospheres, the seeing is worse than it would be for an object higher in the sky. With CCD imaging technology you can see some details, by taking very short exposures and combining them(source). You can't do that with conventional photographic plates.
Lowell produced a map at the start of the century showing supposed linear markings that he thought were the result of cooling.
The best that could be done in the middle of the twentieth century, direct observation images from the Pic du Midi telescope

It wasn't until radar-based observations in the 60s that the rotational period was established. There were some mapping attempts at mapping in the 70s https://www.sciencedirect.com/science/article/pii/0019103572900231 (behind paywall)
It would be hard to identify any features on these drawings with know regions or features on the Mariner images.

These are some great finds! I'd suspected due to its size, distance and proximity to the Sun, there would be few or no mapping efforts. I like how Lowell's theory of surface features resulting from the planet's cooling is similar to what we actually see (even if his own observations don't quite match up!).

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
In a planetary system close to the galactic core, would it be possible to see the supermassive black hole?
I understand that radiation on planets near the galaxy’s center make life on these planets nearly impossible, and that one cannot truly “see” a black hole. However, if you could stand on the surface of a planet that orbits a star close to or within the galactic core, could you theoretically look up at the sky and see an absence of light/stars indicating the location of the central supermassive black hole?
Would it be too far away to see, obstructed by debris, or too small to notice?

I don't think radiation excludes life, even remotely

In the movie Interstellar they posit an idea that an accretion ring around a black hole might be hot enough to give off light. I don't get exactly how that works since a black hole is supposed to suck in even light, but maybe its because the disc is just outside its event horizon? Anyway, that might give you some ideas. Oh, L Dutch gives you that answer below.

If the black hole is active, meaning it is still capturing matter from its surrounding, it will have a large accretion disk all around, which is the only way to dissipate angular momentum for the matter falling into it.
As a result of this dissipation all the matter will warm up and emit radiation. This disc will be fairly large and thus clearly visible as a bright object in the sky.
This page shows an image captured by Hubble of such a disk:

Strictly speaking then, one cannot see directly the black hole, as its view will be covered by the bright emission of the disk.
However, the presence of the disk will allow to observe the black hole thanks to its gravitational effect.

You couldn't see it as a black patch in the sky, because it's far too small. It's only 17 times the radius of our sun, which of course you can't see as a disc even from the outer reaches of our own solar system. What you could easily see is the much larger area of light and other radiation from matter falling into it.

There's a thing called gravitational lensing, which means that light coming from behind the black hole would bend towards it, and since the galactic core has a lot of stars, it might be that instead of a black spot in the sky, you'd see a great accumulation of light in and around the black hole's position.
I'm not sure how literal "lensing" is, so I don't know whether there is a focal point based on gravity and the energy of the light and whether it would matter where the planet is from this focal point.
http://www.cfhtlens.org/public/what-gravitational-lensing (google will provide more links if you search for it)

It works like actual spherical lenses. And you can see it in Nolan's Interstellar. The black hole in the movie was designed and rendered according to the most up to date knowledge on astrophysics, with help from Kipp Thorne himself.

@Renan is that what it was? I was wondering why there was a halo around it. That question is answered, than you. :)

@RealSubtle They actually "discovered" that halo thanks to their advanced rendering efforts. Nobody was really expecting it, but when they plugged all the numbers into the simulation and told it to render, there it was, and after checking the math over and over they realized it must actually appear that way in real life. We just don't have a clear enough view of one to observe it directly.

Rainbow cloud.
A source of radiation from black holes is stuff spiralling into the black hole, heating up as it "fell" and released its gravitational potential energy.  Again this is black body radiation but this time the regular kind: the hotter the emitters are is the shorter the wavelength.  This radiation comes from next to the black hole, not out of the hole itself.
https://physics.stackexchange.com/questions/24958/how-can-a-black-hole-emit-x-rays

The X-rays come from hot gas orbiting around the black hole in an
  accretion disk. As the gas orbits, magnetic stresses cause it to lose
  energy and angular momentum, thus spiralling slowly in towards the
  black hole. The orbital energy is transformed into thermal energy,
  heating up the gas to millions of degrees, so it then emits blackbody
  radiation in the X-ray band.
Once the gas gets closer than a few times the horizon radius, it
  plunges into the black hole, so while some X-rays can still escape
  just before the horizon, most are emitted a fair bit outside.

Telescopes to detect black holes look for the most energetic rays, which are emitted from the hottest areas of gas nearest the hole.  We cannot stand on a planet and look up and see xrays.  But consider: if there is very hot gas, next to it there is less hot gas, and next to that gas that is less less hot.  The cooler a blackbody is, the longer wavelength the emitted radiation is.  Somewhere in that progressively cooler cloud is gas which emits radiation in the visible wavelength.
I here assert that the gradual change in temperature of this cloud as it is progressively farther from the hottest innermost gas should produce a gradual change in the frequencies emitted. The first visible light would be in the far violet, nearest the hole.  This will grade through blues and greens farther away and then to red at the farthest coolest part of the cloud.
This prediction should be true not just for black holes, but for any cloud of gas heated from within.  Now let me look... here we go.

https://www.space.com/12051-bright-nebula-photo-supergiant-star-betelgeuse.html
The black hole rainbow cloud will be more symmetric than this one.  The star is spewing this stuff out willy-nilly but the hole is sucking gas in, so it will be a symmetric spiral.

The question is about supermassive black holes, which emit no detectable Hawking radiation, so none of this is relevant.

@Mike Scott:  Take a look at the bold heading "You could see high energy xrays / gamma rays from black hole associated gas clouds.".  Maybe you missed that the first time through.

@Willik The question’s not about that either; it asks if you could see an absence of light and stars. You could edit this to actually answer the question that is asked by removing the first section and turning the second part into a negative answer, pointing out that there will be light and not an absence of light.

@Mike Scott: you are right; he was asking specifically about the lack of light and  not about detection more generally.  Well, I learned something reading up.  I will tweak this answer.

Also note that your photo of Betelgeuse was taken in the infrared band, so the colours are fake.

@PM 2Ring noooooooo!


text:
What is the canonical method to calculate the surface brightness limit of a 2D image, e.g., in a 1 arcsec2 aperture?
What is the canonical method to calculate the surface brightness limit of a 2D image, e.g., in a 1 arcsec2 aperture?
...if image has erg/s/cm2 units and each pixel has some arcsec2 dimension.

If the units are as you say then the image is already in surface brightness units. Do you mean how do you calculate the average surface brightness over a square arcsecond? How big are your pixels - presumably much smaller than 1 arcsecond? Add up the surface brightness over $n$ pixels and divide by $n$.

@RobJeffries thank you!
I need to calculate the "surface brightness limit" (how deep the image is) in 1 square arcesec. Following what you said, shall I sum the surface brightness over a region of 1 arcsec (far from any source, so that I take only noise) and divide by the number of pixels in the region?
Other question..what about the surface brightness radial profile?
Is it the sum of pixels in any annulus divided by the area in the annulus?

I don't think you can use ergs, unless you're just asking for energy output rather than photopic brightness, or lumens, etc.


text:
Will the TESS telescope be much better than the Kepler telescope?
I would like to know please whether or not the TESS telescope will be much better than the Kepler one.
Thanks.

Rather than one being better than the other, I would say that they have different targets.
Whereas Kepler focused on a specific area of the sky, TESS will cover an area 350 times bigger.
Also, most of the exoplanets discovered by Kepler are far away; TESS will focus on stars within 300 light years away.
For more info, feel free to watch this video I recently made: https://www.youtube.com/watch?list=PL3RiFKfZj3pv1ZqpFxuZinoGtUGEOankw&v=7bsxlG-rh1E

Looking up NASA fact sheets for TESS and Kepler, the literature ( https://www.jpl.nasa.gov/news/press_kits/Kepler-presskit-2-19-smfile.pdf) says that Kepler could detect brightness changes down to 20 parts per million. For perspective they say that an Earth-like planet orbiting a sun-like star (I assume at an Earth-like orbit) creates a brightness change of 84 ppm. By contrast, the one I found for TESS (https://tess.gsfc.nasa.gov/documents/TESS_FactSheet_Oct2014.pdf) claims 200 ppm with a system noise ratio of 60 ppm. I'm not calling this an answer since I may be misinterpreting.

Rather than one being better than the other, I would say that they have different targets.
Whereas Kepler focused on a specific area of the sky, TESS will cover an area 350 times bigger.
Also, most of the exoplanets discovered by Kepler are far away; TESS will focus on stars within 300 light years away.
For more info, feel free to watch this video I recently made: https://www.youtube.com/watch?list=PL3RiFKfZj3pv1ZqpFxuZinoGtUGEOankw&v=7bsxlG-rh1E


text:
Are harmonic relationships among the rotation periods of the planets well known? Are they factors in other planetary phenomena?
Every 39.88 minutes 10 degrees ($\frac{360^\circ}{36}$) of arc passes over the zenith. It can also be stated that the 36th harmonic of the rotation period of the earth (1435.8 min) is equal to 39.88 min. Note the similar relationships with the rapidly rotating planets:

Mars, rotation period of $\frac{1477.2 min}{\text{harmonic of }37} = 39.92$
Jupiter: $\frac{595.8}{15} = 39.72$
Saturn: $\frac{639.6}{16} = 39.98$
Uranus: $\frac{1035}{26} = 39.81$
Neptune: $\frac{966.6}{24} = 40.28$

For the slowly rotating planets, and the Moon, the numbers are larger, but still very close to 39.88:

Mercury: $\frac{3520.2}{88} = 40.00 \text{     } (88 = 22 \cdot 4)$
Venus: $\frac{14582.4}{368} = 39.63 \text{     } (368 = 23 \cdot 16)$
Pluto: $\frac{9223.2}{232} = 39.76 \text{     } (232 = 29 \cdot 8)$
Moon: $\frac{39343.68}{992} = 39.66 \text{     } (992 = 31 \cdot 32)$

The questions are: Are these relationships well known? Are they factors in other planetary phenomena?

Voting to close: it seems like all you've done is divide planetary rotation period in days by 39 or so and then used that as your divisor on the left side.

True. The odd thing is that all the results are integers within a few seconds. Coincidence?

See, how well this site supports Latex! :-) Just type $5\cdot5$ and you get $5\cdot5$.

I'm voting to close this question as off-topic because it's about coincidental numerical relationships, not astronomy.

I'm very sorry for you, but this is just (mathematical) probability at work and no (astronomical) pattern. I'm on mobile right now so I can't do a full analysis, but note that if you divide Mercury's period (3520.2) by 89 instead of 88, the result is 39.55. Still pretty close to 39.88. In other words, the larger the numbers get, the easier is to find a 'divisor' close to 39.88.

So you divide the rotation periods of the planets by any number you need it to be for the end result to be 40, and wonder why the result is 40?
Circular argumentation works because circular argumentation works. It's just not meaningful in any way.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Icy moons: why hasn't the ice sublimated away?
Let's take Europa for example:
The predominant model suggests that heat from tidal flexing causes the ocean to remain liquid and drives ice movement similar to plate tectonics.
The heat generated from the tidal flexing has to go somewhere. Is it all carried away by the water vapor plumes we recently detected? Is more water from the ice cover melting away? Is it just radiated away?
I would expect it to cause sublimation as well: There is hardly an atmosphere (molecular oxygen at 0.1 micro Pascals, or 10-12 times that of the Earth), and a tenuous ionosphere.
And that same article suggest that the surface ice is also broken down by radiation:

the oxygen [...] is formed through the process of radiolysis, where ultraviolet radiation from the Jovian magnetosphere collides with the icy surface, splitting water into oxygen and hydrogen. The same radiation also creates collisional ejections of these products from the surface, and the balance of these two processes forms an atmosphere.
The hydrogen, meanwhile, lack the mass needed to be retained as part of the atmosphere and most is lost to space. This escaped hydrogen, along with portions of atomic and molecular oxygen that are ejected, forms a gas torus in the vicinity of Europa's orbit around Jupiter.

Or do we have reason to believe that any losses are compensated by ice being deposited from external sources?

"Europa's surface temperature at the equator never rises above minus 260 degrees Fahrenheit (minus 160 degrees Celsius)." https://www.space.com/15498-europa-sdcmp.html   That's mighty cold for rapid sublimation.

There are some interesting answers to If there is actual ice on the moon, why hasn't it sublimated? (in a different site) that have some bearing here.

I'll pick up on your example as it can be used very well for illustrating the main points:

First: The solar intensity in Jupiters orbit is less than 4 percent compared to Earths. So much less energy comes from the Sun which limits possible sublimation significantly
Second: Sure tidal flexing leads to fountains. These fountains blow material into space. But this does not mean that all the material will leave Europas surface forever. It can fall back or it stays in Europas orbit and returns to the surface due to the effects of Jupiters magnetic field or gravitational forces of the other moons.
Third: Sure the heat of the tidal flexing has to go somewhere. And it is largely irratiated away. As Europas surface is quite bright it reflects most sunlight away instead of absorbing it. This makes Europas surface quite cool (50-125K). Compare it with the darker neighboring moon Ganymede (70-152K). As you can see the albedo (which means the reflectivity of the surface) is far more important than the tidal heating.
Fourth: Jep some of the water is split up by the radiation. This leads to an oxygen 'atmosphere' on Europa with 0.1 µPa surface pressure. While this is detectable it still is practically a vacuum. There is very little oxygen there, probably not more than a few hundred or thousand tons on the whole moon. Not enough to threaten the existence of moon with a mass of 4.8 * 10^22 kg

In summary: Yes, icy moons lose some of their material. But for exactly this reason they only appear in the outer solar system where the solar radiation is low enough to make this effect negligble.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Do the cometary gas ions which make up the plasma tail of a comet reach the velocity of the Sun's solar wind?
Do the cometary gas ions which make up the plasma tail of a comet reach the velocity of the Sun's solar wind?

Yes, the ions are said to be "swept out" by the solar wind, such that they reach the speed of the solar wind.  That speed varies a fair amount, however, because there is both a "fast" (about 600 km/s) and "slow" (about 300 km/s) solar wind.  It is really the magnetic fields entrained in the wind that sweep out the charged ions, since each ion has such a low mass it is easily affected by those fields.  Note the dust particles from a comet, which is the tail that we see, are much more massive and are not affected by the solar wind, so they are moving out much more slowly and often make a noticeably curved shape.

Is the curvature due to the path the comet follows, or to the fact that ions in a magnetic field follow curved paths?

It's more like the path of the comet, like the curvature of the smoke coming from the smokestack of a ship in motion.  It is the pattern that is curved, the particle motion is a bit different-- the dust particles continue to orbit the Sun in ways that depend on the size of the dust grains.  The largest grains, those that make meteors in Earth's atmosphere, continue to follow an orbit not so different from the comet itself.  Smaller grains get pushed more by sunlight, so their orbits are more eccentric, but still cross the comet path.

@KenG you’re saying the radiation pressure is stronger than the solar wind for small grains? What about for gas molecules released from the comet?

It's a matter of cross section per gram, which peaks somewhere around the wavelength of the light.


text:
What's the profound reason(s) why the star formation rate drives the scatter of the mass-metallicity relation in galaxies?
Since decades it's well known that a correlation between stellar mass and metallicity (Z; both stellar and gas-phase Z, but here I focus on the gas-phase Z) exists (e.g., Tremonti+04; analysis of >53000 galaxies from SDSS).
More recent results have shown that the scatter of this "mass-Z" relation is mainly driven by the star formation rate (SFR; e.g., Mannucci+10), i.e., at a fixed stellar mass, galaxies with higher SFRs have lower Z. This happen systematically over a wide range of stellar masses.
What's the physical mechanism(s) responsible for that?
Also, even more recently (e.g., Bothwell+16) it has been shown that the "real" driver of the "mass-Z" relation's scatter is the molecular gas mass (the previously found dependence on SFR may be only a by-product of the Schmidt-Kennicutt relation between gas and SFR).
How this would change the interpretation of the first problem?

This is an open research problem! There are still many papers being published on the topic. You could write a paper on the topic if you came up with a plausible explanation.

I would wonder here is if we could assume a correlation between Z and the "age" of a galaxy. I would think that larger galaxies would tend to be older (larger due to mergers which take time). Also, when comparing similar size galaxies I would think that those with higher SFR would be "younger" and thus have a lower Z. I could be way off track here and I am sure there are many anomalies that contradict this.


text:
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?

This is not really about astronomy, but about space exploration, for which there is a separate forum. I'm pretty sure the answer is "this won't work", or at least "work work nearly as well as you think. But on space exploration.se there are people who can tell you for sure.

Check this: https://space.stackexchange.com

You should ask this in space.stackexchange.   That said, you can't increase the force that the solar wind imparts on a surface and your tube, a relatively small surface area, that appears to be a problem to me.  The idea of solar sails is that they're large and they catch a tennis court sized area of the solar wind.   The solar wind is very disburse, so you need to catch a lot of it.   I don't think the sail model can be improved upon if you're using the solar wind as a means of propulsion.

I wonder at the reason for asking this question...

I did post a question about this solar wind-container concept in space.stackexchange yesterday and I suppose that this question in astronomy.stackexchange can now be closed. However, I would like to share one more perspective on this idea and get some feedback on it before it is closed. Since the negatively charged oxygen ions repel each other, when the solar wind charged particles slam into them, won't the oxygen ions bounce around like billiard balls on a pool table and pass that kinetic energy to the closed end of the tube, and if so, won't this propel the tube away from the Sun?

Solar wind runs about 200-600km/sec, and about 1-14 protons per cubic cm. Oxygen ions might hold the things in your tube a little longer, but I'm not sure you'd get any more xfer of kinetic energy with them. Wind: http://www.spaceweather.com/

Come to think of it, an empty tube is probably better. Elastic and Inelastic Collisions: http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html

maybe have a solar sail and then have a rocket follow it and blast it with a laser?

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 you say is true and I have studied many of the solar sail ideas that are currently being considered, yet I have not seen any solar sail idea that would use water as it’s primary means of propulsion. So I feel my design is a unique one and worth consideration.

@FanofComets  Water is heavy.   Sails are light.  Have you calculated the minimum mass of your approach and the maximum kinetic energy transfer per square meter?

@userLTK, no, I have not done those calculations. I posted this as a conceptual/theoretical question and at this point I'm just looking to get feedback on whether the working principle of this type of rocket is viable or not.


text:
Do we know how old the matter that makes us is?
Last weekend, I had the pleasure of visiting an observatory where there was a series of lectures to cover up the fact that there was too much cloud cover to actually observe much at all.
One of these lectures involved the life-cycles of stars, which mainly involves Hydrogen and Helium.
Out of curiosity, I raised the question

How old is the matter that make us and where does it come from?

The "where" is covered by Nucleosynthesis, but the "how old" was shrugged away with "maybe 2-3 sun lifetimes".
Is it possible to more accurately estimate the age of the matter that made up our solar system's accretion disk?  Or is the likely source of the supernova/collisions involved unlikely to be known?

Interesting but, other than radioactive items, unlikely to be determinable.

The age would be the age of the universe if you refer to all the nucleons involved. Undeterminable if you want to know when and where the nucleons were fused to nuclei.

The material (heavier than helium) that makes up the solar system was made in millions, if not hundreds of millions of stars that lived and died  in the ~7 billion years between the formation of the Galaxy and the birth of the Sun. This material has been thoroughly mixed in the interstellar medium and so the heavier elements arise from countless individual stars of different ages. You can say something probabilistically, based on what we understand about the build up of chemical enrichment in our Galaxy.
Starting with hydrogen - the vast majority of hydrogen is primordial. That is, it hasn't been inside another star.
The carbon in the solar system is made in a variety of stars. Some will come from high mass stars and dispersed either when they exploded as supernovae or via winds from their surfaces during the Wolf-Rayet phase; some will have been manufactured in lower mass asymptotic giant branch stars and expelled in their winds. The distinction is important - massive stars enrich the Galaxy very promptly (see below, when discussing oxygen), near the beginning of the life of the Galaxy, long before the Sun was born. The enrichment rate by supernovae and massive stars tailed off considerably at later times. On the other hand, enrichment by lower mass stars takes longer because of their longer lifetimes. Thus the carbon in the solar system will have a spread of ages from about 13 billion (the oldest stars in the Galaxy), right up to the age of the Sun.
Elements like oxygen, phosphorus, magnesium, sulphur and silicon are primarily made and dispersed by massive stars and supernovae. These are likely to be at the older end of the range discussed above. On the other hand, nitrogen is also made in a mixture of stars, both high and low masses, with a consequent range of ages.
Iron and nickel are mostly dispersed into the interstellar medium by type Ia supernovae. These are exploding white dwarfs, which in turn are the remnants of comparatively long-lived stars of a couple of solar masses or so. Most likely, these elements were made not long before the Sun was born.
See also Parent stars of our Sun - Where are its remains? and How can there be 1,000 stellar ancestors before our Sun?
A more technical description of this differential enrichment history can be found in Pippino & Matteucci (2008).

Thank you for such a great answer. It beautifully demonstrates the incredibly random fragility of our world and how very small we are in the universe. As my dog sleeps at my feet, I wonder how many stars died so that he might live.


text:
Will the Sun's solar wind push a permanent magnet out of our solar system?
If you were place a strong permanent magnet into the solar wind, like a 1 inch cube neodymium magnet, would the charged particles of the solar wind colliding with the magnet's magnetic field push the magnet out of our solar system, that being if it does not encounter any planets or objects in its path?

Magnets tend to align themselves with magnetic field lines.   More likely the magnet would turn in the direction of whatever magnetic field it crossed.   The solar wind is created by magnetic storms but in and of itself, I don't believe the wind is strongly magnetic.  It has both positive and negative charged particles, mostly protons, electrons and alpha particles.   Think of a compass needle, the needle turns, the needle isn't pushed away from magnetic north.

I think if the magnet is placed in line with the Sun's equator then the magnet should align itself in a north-south orientation in relationship to the Sun. This would mean the the magnet's magnetic field lines would be perpendicular to the flow of charged particles flowing out from the Sun. If so, then the charged particles should be deflected around the magnet's magnetic field and this should produce a drag on the magnet and push it away from the Sun.

How about we make it easy and magnetize Russell's Teapot?  :-)

@Carl Witthoft, ouch that hurts, but its funny at the same time  :)

@FanofComets - One would be skeptical about the existence of Russell's teapot but the lid of the coroner's jar is a real thing - also an appropriate shape.

No. The magnetic field has two poles, the force on the two pole of the magnet is equal and opposite. This is why a compass needle will align in the North South direction, but is not pulled towards the North or the South.
Two magnets will attract or each other because the North pole of one is nearer the South pole of the other. You can get a net force on a magnet when the field-lines are getting bunched together. But the magnetic field in the plasma is not like this. To a first approximation they are parallel to each other and perpendicular to the equator of the sun. In fact the magnetic field is complex as it is carried with the flow of solar wind.
There would be no net force on the magnet, but it may tend to align with the magnetic field. This might cause a turning force, but no acceleration.
There would be some force due to solar wind particles impinging on the object, but since there is only something like 5 atoms per cm3, the force isn't very big, and it has nothing to do with the object being a magnet.

Yesterday I was thinking about the solar wind density of 5 atoms per cm3 at 1 AU and something occurred to me. If the magnet was put in motion, such as going around in a circle over a certain distance, then it would come in contact with many solar wind atoms. For example, if the magnet is placed at the end of a 10 ft pole and the pole is then spun around, it will travel 754 inches/1915 centimeters in one second. 1915 x 5 atoms = 9575 atoms. So, this means that the 9575 atoms in the path of this magnet will be deflected away and thus produce a force on the magnet, pushing it away from the Sun.

Actually the way magnetic fields sweep out material is generally not the Lorentz force anyway, because the Lorentz force cannot impart kinetic energy.  It is electromagnetic forces owing to the fact that the fields are moving, and those forces work in the same direction on both signs of charge, which is how the solar wind can push out ions of either charge.  For a magnet, what would be most relevant is Faraday's law, the way an EMF is induced when you try to change the magnetic flux.  I suspect moving magnetic variations would push out a permanent magnet, as an application of Lens' law.

I realized that I miscalculated in my last post. Since the magnet is a 1 inch cube magnet, that means that the  surface area of each side of the magnet = 2.54 cm squared = 6.45 sq cm. So, the number of solar atoms that are in the path of the magnet would be 9515 x 6.45 = 61,372. Also, this magnet would be a N52 strength neodymium magnet for maximum effect.

I just redid the math and the answer is 61,759.

on the other hand, the magnet is composed of about 10000000000000000000000000 atoms.  (I may have miscounted the zeros) . You are trying to move a planet with a peashooter. It doesn't really matter if the planet has a magnetic field, or if the peas are electrically charged.

If you were place a strong permanent magnet into the solar wind, like a 1 inch cube neodymium magnet, would the charged particles of the solar wind colliding with the magnet's magnetic field push the magnet out of our solar system ...

Technically speaking, yes. Practically speaking, it would take an extremely long time. The solar wind consists of 5 atoms per cubic centimeter per second. Solar sails are usually huge and work by reflection, simply pushing on a tiny object is inefficient, being magnetic helps very little.

I'm not sure that's true, though I could be wrong myself.   The solar wind pressure is very slight and past a certain distance, virtually nil.  (exiting the solar system would take a very long time).  There's also the Poynting-Roberton effect for an object that size that .   1 inch is a little large for that, so that effect would perhaps be minimal too.   http://planetfacts.org/poynting-robertson-effect/

At .5 AU, the density of the solar wind is 20 atoms per cubic centimeter per second. So, it the magnet was placed at this distance from the Sun, it would be pushed along with a higher dynamic pressure and its acceleration would be faster. The farther out it goes, the atom density will decrease, but I believe it will maintain the speed it gained between .5 to 1 AU.  I'm not sure about the Poynting-Robertson Effect, but I suppose it would have some effect on the magnet's overall trajectory as it travels out of the solar system.

As Ken G points out, you need to know whether the magnet started out in an orbit, or stationary and about to fall directly into the sun, etc.

My idea is that it would begin its journey as a stationary object placed at a distance of .5 AU. I agree with Ken G that the Sun's gravity would pull the magnet in despite moving against the solar wind.

@FanofComets - A metal cup or parabola would likely work better but a 1" cube neodymium magnet is only $10. I thought your question presumed that the journey would start sufficiently far away from the sun because you wrote "... that being if it does not encounter any planets or objects in its path?" - the sun behind it, and sufficiently close, puts the sun's gravitational field in the cube's path. I didn't assume things you didn't write and answered your question as-is - a solar sail is thought to be practical.

@Rob, I think what is all comes down to is that I have overestimated the potential power of the Sun's solar wind. So, it is thus very unlikely that the solar wind can push a permanent magnet out of our solar system.

Solar sails have a very high ratio of surface area to mass, and don't work on the solar wind, they work on sunlight.  The momentum flux in sunlight is vastly higher than the momentum flux in the solar wind.

The Sun's gravity would pull in any large object not already in orbit, that would be dominant over magnetic forces from the very weak wind.  If the object was already in orbit, the situation would be akin to dust particles leaving a comet-- their path is more affected by light pressure than magnetic pressure, but for a large object, neither of those could rival gravity.

Exactly - the total force diagram is what matters, and the solar wind - be it mag fields or particle momenta impinging on the magnet - are small factors.

This is a good point that Ken G brought up and one that I did not consider. I was thinking that the magnet could start off as a stationary object and then the solar wind would begin pushing it away from the Sun, yet its logical that the Sun's gravity would win out and draw the magnet in especially if the stationary magnet is placed at a distance of .5 AU.


text:
Fundamental axioms in LCDM
What are the axioms (if any) behind the LCDM model of cosmology?
NB: axioms, not postulates (e.g., inflation)

What do you mean by "axiom" and "postulate" -- normally they pretty much mean the same thing. What distinction are you trying to get at?

I think if you [edit]ed to indicate prior research, you might get some more activity.  Do you mean something like the Wrightman axioms?  There is no satisfactory realisation for Wightman axioms in the case of Yang–Mills fields. (wikipedia)

@MarkOlson I think it’s commonly understood that axioms are accepted as true (eg universal speed limit is c) whereas postulates are equivalent to hypotheses.

@Chappo That distinction was shown to be meaningless during the investigations into the foundations of mathematics in the 1800s, so I'm not assuming that that's what to OP meant. (Maybe he did. In that case, I'd like to hear it.)

It's also worth noting the LCDM model isn't a mathematical theory, but a physical theory. It's certainly wrong to call inflation a postulate -- Inflation itself is a theory that explains certain observations better than any other theory we presently have. But it's still just a fairly poorly validated theory.

Sorry for the wording. What I mean is
"What are the fundamental assumptions (if any) on top of which one can build LCDM?"

The fundamental assumptions of LCDM cosmology are:

General relativity is valid on cosmological scales.
Universe is dominated by "cold" dark matter (origin, composition unknown).
The metric of the universe is given by the  Friedmann–Lemaître–Robertson–Walker metric.

Reference: https://en.wikipedia.org/wiki/Lambda-CDM_model
 A combination of 1 and 2 helped explain large scaled structure formation, which is smaller masses merging to become bigger systems while 3 helped explain the expansion of space depending on the matter density.
The lambda parameter comes from invoking general relativity. Although Einstein added it to support his own view of a static universe (disproved a decade later by Hubble's observations), it was later re-interpreted as a negative energy density of space (see Harvey, Alex 2012 https://arxiv.org/pdf/1211.6338.pdf and/or Carroll, Sean 2001 https://arxiv.org/abs/astro-ph/0004075).
Hope this helped.


text:
What explains the existence of energy/matter if it cannot be created or destroyed?
If the energy/matter that was contained in the singularity before the Big Bang had always existed, and if it’s true that energy/matter cannot be created or destroyed, then what explains its existence?

I'm voting to close this question as off-topic because it is a philosophical question

@Carl Witthoft, how can the creation of energy/matter before the Big Bang be a philosophical question? I think it’s a fair question concerning both astrology and cosmology. Explain to me how it is off-topic?

Sorry, I meant to say astronomy

@CarlWitthoft I'm not keen on questions involving "before the Big Bang" as we have no evidence for a "before" in any human common-sense meaning of before, but I think the general thrust of the equation ("where did the energy come from") is a valid one.

@CarlWitthoft OP is asking for a physical explanation for a physical phenomenon. Although the premise is somewhat flawed, I don't see any philosophy here.

@StephenG Why do you have a problem with people discussing the biggest mysteries in the universe? I presume it's because questions such as these are invasive and unwelcome to the 'mainstream physics' bubble, and to 'mainstream' theories.

@WhitePrime This is not a discussion forum.  It's a straight Q&A website and that's the reason many question which can't be reasonably answered in a brief and reasonably definitive way are going to be off-topic.

@StephenG Okay, mate. My apologies.

That's a very complicated question!
First, let's remember that Moses didn't bring the Law of Conservation of Energy down from Sinai on stone tablets -- it's something that we've observed to be true of the universe today, and which our theories use to make very accurate predictions. But this doesn't mean that it was always true, and particularly that it was true early enough in the Big Bang that our current theories don't apply.  We know that our current theories fail during the Planck Era, for example.
Second, Noether's Theorem says that energy conservation exists because the Lagrangians of mass-energy are time-symmetric. Time symmetry is a very good symmetry today in regionals of space small compared with the entire universe, but there may well have been eras in which it wasn't.
In fact, there probably were such eras, just as time-symmetry is not a good symmetry even today at very large scales.  (The universe is expanding, and on the face of it, this causes a non-conservation of energy.)
Thirdly, if you try to save the law of conservation of energy by calculating the total energy of the entire universe including the gravitational energy, one reasonable set of arguments (energy calculations in General Relativity can be very difficult, since gravitational energy tends to be non-local) yields a total energy of zero! The mass-energy in the universe is exactly balanced by the gravitational potential energy.  If this is correct, the total energy has always been zero.
But this is all speculative!  Remember, that real physics is an experimental science and if you can't test a theory it may be interesting, but it isn't physics.

Well, maybe... You could make an equal argument (shudder) that us physicists invented the FudgeFactors of potential energy and "energy in a field" to make the equations balance :-)

@Carl Witthoft Not really. The energy of the gravitational field in GR is significantly different from the energy of, say, the electromagnetic field, and this is due to the significantly different underlying nature of the two fields. (Oversimplifying, the energy in the EM field is local, that in GR gravitational field isn't.)

Sure, but that's just another FudgeFactor - creating a different kind of "field" thing that nobody can see but energy sneaks into and out of.  (Devil's Advocate)

It's possible you may not have a full understanding of the differences, here.

Mark, that's not my point.  I fully understand that the theory of fields holds up because experiment shows consistent predictable results (as does the theory of potential energy).

Splitting of virtual particle pairs at event horizons also violates the energy/matter not changing thing.  It's an approximation, useful for most of the stuff we encounter on earth.

The simplest answer might be that mass/energy have always existed.  For field-based inflation, their past eternality is prohibited by the Borde-Guth-Vilenkin theorem, but, as that theorem only prohibits it in a universe that is expanding "on average", they appear compatible with some of the bouncing cosmologies developed more recently, of which the simplest that's compatible with CMB data may be Poplawski's entirely relativistic version (using the fermions of the Einstein-Cartan version of GR, which have spatial extent) of the "baby universe" scenario, in which the gravitational collapse of any large rotating star includes tidal effects which initially separate the particles and antiparticles of virtual pairs, that are subsequently blown outward by energy until then tied up in particulate spin, when the collapsing star's fermions lock up against each other at extremely high densities.  The "parenting" universe (which is on vastly larger scales of space and time) is analogous to the 3-D skin of a basketball, and is causally separated by the inertia of its own continuing expansion, as it would've begun in an earlier (and larger-scaled) version of the same process.  Poplawski feels this cosmology to require a prevalent direction of rotation, so it is falsifiable and consequently scientific, however marginal that prevalence may be.  (My reading of physics notation is very limited, so I've relied on his English verbiage for some of the inferences I've made here.)

Although they don't reference Poplawski, the larger scale of the "parenting" universe is shown diagrammatically in the 2018 paper by Iljas and Steinhardt, "Bouncing cosmology made simple".

A pop-sci description of a very recent CMB study, which substantiates Poplawski's description of the local universe as closed while remaining consistent with its indefinitely-prolonged expansion, can be seen at https://www.youtube.com/watch?v=F2s7vyKucis .


text:
Have we sent radio signals to exoplanets that may be habitable?
I would like to know please whether or not we have sent radio signals to exoplanets that may be habitable.
Thanks in advance !

Short answer : Yes
Long answer : Yes
We have actually sent radio signals to distant exoplanets that are thought to be capable of hosting (hopefully intelligent) life.
Some examples?

The radio signal directed at a world 12 light-years away included music and math lessons from Earth
Are There Any Habitable Exoplanets That Can Hear Us?

The second link only 'mentions' a planet that had been sent radio planet :

Those waves, which are all electromagnetic like light, would have since traveled about 70 light years away from Earth. The newly confirmed Kepler 452b, at 1,400 light years away, will not be seeing 'I Love Lucy' for more than a millennium.

Hope this answers your question.


text:
Why don't all perseid meteors follow the same trajectory?
Since these bits of debris crossing Earth's orbit have the same relative speed, why aren't they drawing parallel paths in the upper atmosphere? (consider one hour of observation to minimise Earth's rotation influence)

Meteoroids belonging to the same stream (so, this is valid not only for the Perseids stream) actually do follow the same trajectory (orbit) in space but meteors caused by them apparently appear to originate from a particular spot in the sky (the so called radiant) trailing in all possible directions away from it only due to our perspective of view. In fact, you could even spot a so called stationary meteor which would appear point-like because it appears exactly in the radiant and is directed along your line of sight.
Usually, to explain this it is compared to the same effect when you'd drive a car at a high velocity through a snow-fall: the snowflakes would apparently appear only to you to come from some frontal area radiating then in all directions away from it, although they all have the same trajectory in space for an external non-moving observer.

They are all following parallel paths. They are not on exactly the same trajectory but they are all travelling parallel to each to other in their orbit around the sun, in a stream of particles.
What you see is then multiple meteors on parallel paths. When you look in the direction that the meteors are coming from you see them appearing to radiate from a point
These meteors are all travelling parallel to each other. It is only perspective that makes them appear to be radiating from a point.

This is similar to the "flying through space" screensaver that used to be popular on windows.

Why don't all perseid meteors follow the same trajectory?
Since these debris crossing Earth's orbit have the same relative speed, why aren't they drawing parallel paths in the upper atmosphere ?

Every 133 years the comet Swift–Tuttle orbits the Sun, it last passed by Earth during November/December 1992 (26 years ago) and is due for another pass on August 14, 2126 (IAUC 5636: 1992t).
When a comet travels through space it disintegrates leaving a shower of debris, every year the Earth passes through the tail's remnants (between July 14 - August 24 this year). The comet nucleus is 26 km in diameter, thus it produces a wide tail of debris that can fall to Earth from various angles, appearing in the direction of the constellation Perseus (but not actually coming from that constellation).
The meteor shower appears to radiate from a single point.

That is a result of passing through the trail of debris.

See NASA's webpage: "Perseids":

"Where Do Meteors Come From?
Meteors come from leftover comet particles and bits from broken asteroids. When comets come around the sun, they leave a dusty trail behind them. Every year Earth passes through these debris trails, which allows the bits to collide with our atmosphere and disintegrate to create fiery and colorful streaks in the sky.
The Radiant
Their radiant — the point in the sky from which the Perseids appear to come from — is the constellation Perseus. This is also where we get the name for the shower: Perseids. However, the constellation for which a meteor shower is named only serves to aid viewers in determining which shower they are viewing on a given night. The constellation is not the source of the meteors.

The answer has been given, it is the effect called "perspective."  I would simply add an example that should make the situation quite clear: sunbeams.  When you see sunbeams come through clouds, it is natural to think they are not parallel rays, because they look very far from parallel.  Your eye tells you that the Sun is sending out those rays in all directions, from its apparent location in the center of the clouds.  But your eye cannot tell that the Sun is a trillion times farther than the clouds, so it cannot tell the divergence of the rays is a complete illusion.  Actually those rays are almost perfectly parallel, because they all point to that incredibly distant Sun.  But look at them carefully the next time you see sunbeams through the clouds-- they seem as far from parallel as they could possibly be.  So it is with the streaks of meteors, it's because your eye cannot pick out that third dimension so tends to imagine it is not important.


text:
Which part of the milky way does the image show?

As a general question and more in specific regarding the image above: Which part of the milky way, planets, stars, clusters are included in the image?
How can I know which entities are included in different pictures?

I wholeheartedly recommend astrometry.net for this sort of thing. It Just Works(tm); running on your image produced this output with absolutely no hints or guidance from me:

For avoidance of doubt, I have no association with astrometry.net.

I didn't even try astometry.net since some times it returns nothing back. I guess I got lazy, anyhow I appreciate the effort I' ll come back tommorow.

That's remarkable!

I joined just now just to +1 this answer. I've never heard of this site and now I'm going to use it all the time. I'm in love. Thank you for this.

Continuing on Philip Kendall's answer, Saturn is the bright object to the lower left of center (on the line from 14 Sgr to Kaus Borealis in the astrometry.net version of the image).

@nick is the photo from mid-July?

Then Saturn it is!

In addition to Saturn and the stars astrometry.net circled, the frame includes several Messier objects:

open clusters M16, M18, M21, M23, M24, M25
globular clusters M22, M28
bright nebulae M8, M17, M20

This chart from freestarcharts.com indicates which is which.

Very useful info indeed! Yes the image was taken 14th of July from Drimonas mountain.


text:
Does Celestia respect 'setframe' coordinate systems?
Through trial and error (see script below that summarizes what I've found), I've learned Celestia's coordinate system appears to be:

The x axis points to the vernal equinox
The y axis points to the ecliptic north pole
The z axis points to 18 hours right ascension, -23.4393 declination (ie, the obliquity of the ecliptic)

regardless of whether you choose the "equatorial", "ecliptical", or "universal" frames.
I wrote the script below to confirm this: regardless of which setframe command you uncomment (or if you leave them all commented), the results are the same.

{

# using Celestia 1.6.1

# set to equatorial frame
# setframe {ref "Sol/Earth" coordsys "equatorial"}
# setframe {ref "Sol/Earth" coordsys "ecliptical"}
# setframe {ref "Sol/Earth" coordsys "universal"}

# 27 Psc is near the vernal equinox at a distance of 234.48 ly
# this brings me within 14 ly
setposition {base [234.48 0 0]}
select {object "27 Psc"}
center {time 0.01}
wait {duration 3.0}

# 42 Dra is near the north ecliptic pole at 18h, 66.56 declination
# this brings me within 15 ly
setposition {base [0 314.82 0]}
select {object "42 Dra"}
center {time 0.01}
wait {duration 3.0}

# 7 Sgr is near the 18h, -23.4393 declination
# this brings me within 20 ly
setposition {base [0 0 1079.99]}
select {object "7 Sgr"}
center {time 0.01}
wait {duration 3.0}

}

I am aware that https://en.wikibooks.org/wiki/Celestia/Binary_Star_File suggests the coordinates are stored internally as some gyration of ecliptical coordinates, but assumed "setframe" would change this.

I'm voting to close this question as off-topic because it seems like a question suitable for e.g. the Celestia forum rather than Astronomy SE.

Maybe setposition always operates in "universal" coordinates. What effect does setframe have on other movement commands?

@StephenG I went ahead and added it there, but I think it's appropriate here too. There are plenty of questions about stellarium here, and I think it's reasonable to discuss popular astronomy software here, just as it's reasonable to discuss qgis, grass, and other GIS software on gis.SE

My message was rejected on the Celestia Forums: ": The reported message has the only purpose to advertise for a website or another product..". I guess there ARE sites with worse moderators!


text:
Does time dilation increase within overlapping gravitational fields?
If a spaceship was to travel to the barycenter of a binary star (the stars having identical size and mass) and come to a stop there, would the people on board this spaceship find that time is moving slower there (relative to the passage of time experienced outside of the gravitational fields of this binary star) due to the two stars overlapping gravitational fields at this point in space?
Does time dilation increase within overlapping gravitational fields?

Gravitational time dilation depends on the observers' relative positions in the gravitational potential.
For relatively weak gravitational fields the rate at which a clock ticks for observer A compared to the clock rate for observer B is given by
$$\frac{d \tau_{\rm A}}{d\tau_{\rm B}} \simeq \sqrt{1 + \frac{2\Delta \Phi_{\rm AB}}{c^2}} \sim 1 + \frac{\Delta \Phi_{\rm AB}}{c^2},$$
where $\Delta \Phi_{\rm AB}$ is the difference in gravitational potential between observers A and B.
Gravitational potential is a scalar quantity, so the basic answer to your question is yes - the effects of the two neutron stars would be summative. However, I would hesitate to use the approximation above when dealing with the strong gravity near neutron stars.


text:
Is there a limit to how fast spacetime can bend/warp around an accelerating object?
If in the future a spaceship can accelerate to nearly the speed of light, will it encounter any kind of resistance/drag from the spacetime in front of it not being able to bend/warp fast enough around it?
It is my understanding that as an object accelerates, its mass will increase and this increase in mass should result in the object having more gravity, and this gravity will continue to increase as the object's speed increases. The creation of more and more gravity should result in an increased area of spacetime surrounding the object being bent/warped.
So, if there is a limit to how fast spacetime can bend/warp around an accelerating object, along with an increased area of spacetime being affected by an accelerating object's increased gravity, will this spaceship start to heat up and begin to melt the faster it accelerates?

Photons emit from atoms in picoseconds to nanoseconds.

Realize that the object's rest mass won't increase-- just the mass as perceived by objects traveling at a different speed. It's not really "spacetime in front of it", but rather actual particles that will now hit the ship at tremendous speeds.

Changes in the bending of spacetime propagate through spacetime at the speed of light. If I wave my hand thus the (very, very, very tiny) changes to spacetime will propagate outwards as gravitational waves at the speed of light and won't reach the Sun for 8 minutes.
No matter how massive an object or how great the acceleration, spacetime elsewhere only "feels" that acceleration after it has passed into the acceleration's lightcone.

I have nothing to add to Mark Olson's answer except to point out that there are far more severe problems to contend with if you want to have a spaceship travelling at highly relativistic speeds
The cosmic microwave background will be blue shifted, boosted and beamed such that it might become a hazardous source of X-rays/gamma rays. Of course these problems would be much, much more severe should you fly towards anything brighter - like another star!
Secondly, cosmic dust and debris is going to become a serious problem. Any relativistic mass increase your ship manages will be mirrored by an increase in the kinetic energy of anything that runs into you (from the point of view of the rest frame of the ship).


text:
Is it possible that the Universe is expanding due to additional Big Bang events?
Is it theoretically possible that the Universe is expanding due to additional Big Bang events that have occurred at the same location that the original Big Bang occurred at?
Perhaps there has been an ongoing series of smaller Big Bang events over the past 13.7 billion years, which although were not as massive in scale as the original Big Bang, have created a large amount of new space-time and new matter that has resulted in the Universe expanding.

One problem with this argument is that the expansion appears highly uniform.   Different "big bangs" would suggest a lack of uniformity, where as one, that happened everywhere, implies high uniformity and that's what we observe.

Suggest you read Did the Big Bang happen at a point ? on the Physics SE site.

@userLTK There were (are) ideas that maybe the Higgs field may have a different value in the for us unreachable regions of the Universe.

You may need to revisit the theory behind the Big Bang. The Big Bang is more or less a description of what the early universe looks like if you view expansion in reverse. The single "Big Bang" at the beginning of our universe is the only "Big Bang" event predicted by expansion. Also, the Big Bang did not happen in a single place, it happened everywhere in the universe. Finally, the Big Bang did not create space-time, and the expansion of the universe is not the creation of new matter. The Big Bang describes a time when the metric of space-time was extremely dense, and expansion is the metric of space-time increasing as below:

The distance between points in space increases, but the same mass-energy is there as before.


text:
Number Density of Dark Matter Halos
Is there a way to calculate the expected number density of Dark Matter Halos above a given mass, in a certain redshift range, and in a certain area?

The function you're requesting — i.e. the number density $N$ of DM halos above a given mass $M_\mathrm{h}$ — is called the cumulative halo mass function (cHMF). It is obtained by integrating the halo mass function (HMF) from a given mass to infinity. The HMF, in turn, is thus the function that describes the (differential) number density of DM halos of a given mass.
In other words,
$$
\boxed{N(>\!\!M_\mathrm{h})
= \int_{M_\mathrm{h}}^\infty \!\!\!dM_\mathrm{h}'\, \frac{dN}{dM_\mathrm{h}'}.}
$$
Halo mass function
So, the problem is to determine the HMF, i.e. $dN/dM_\mathrm{h}$. This was first calculated, analytically, by Press & Schechter (1974) assuming spherical collapse of structures from an initial, smoothed density field. It may be written as
$$
\frac{dN}{dM_\mathrm{h}} = \frac{\rho_{\mathrm{m,0}}}{M_\mathrm{h}}
                    \left| \frac{d\ln\sigma}{dM_\mathrm{h}} \right|
                    f(\sigma),
$$
where
$\rho_{\mathrm{m,0}}$ is the present-day average mass density of the Universe,
$\sigma = \sigma(M_\mathrm{h},z)$ is the rms fluctuations of the (smoothed) density field, and
$f(\sigma)$ is the "multiplicity function" (note that because the number density decreases so fast with halo mass, for computational purposes often the HMF is expressed as $dN/d \ln M_\mathrm{h}$ rather than $dN/dM_\mathrm{h}$).
If you want, I can give you more details about how to calculate $\sigma(M_\mathrm{h},z)$ and $f(\sigma)$. You can also find details on this in sec. 2.1 of Laursen et al. (2018), but note that there's an error in Eq. 1. If you're happy with using a "black box", you can get both the HMF and cHMF for your favorite cosmological parameters using this online HMF calculator.
Only in the Press-Schechter formalism can an analytical form of $f(\sigma)$ be derived, and it was subsequently found that it over(under)predicts the collapsed fraction at the low-(high-)mass end (see e.g. Governato et al. 1999); in general one must obtain it by fitting halo abundances in cosmological $N$-body simulations.
Defining a halo
This is what N. Steinle describes in his answer, but actually you don't need to assume a density profile of the halos. So, how do you count halos in a simulation? There are several ways, arguably the two most popular ones being the spherical overdensity (SO) and the friends-of-friends (FoF) method.
Spherical overdensity
In the SO method, you first calculate the center of mass (CoM) of a clump of particles in the simulation, and then you calculate the average density $\bar{\rho}$ of particles in successively larger spheres centered on the CoM. As you increase the radius of the sphere, $\bar{\rho}$ drops because you include less and less dense regions. When the density has reached a certain factor $\Delta$ times the average density $\rho_\mathrm{m}(z)$ in the Universe you stop, and the total mass of all particles inside that sphere is then the halo mass. The overdensity factor $\Delta$ is usually chosen to be around 200, but other choices exist, e.g. 500 which then gives slightly smaller masses (because you stop counting earlier).
A variant of this method is using ellipsoids rather than spheres, allowing for more realistic results for elongated structures.
Friends-of-friends
In the FoF method, you start at an overdensity and count all particles that are "linked" together, meaning they are within some chosen distance (the "linking length") of each other. This method may give more realistic results for very non-spherical structures, but also tends to include particles in filaments streaming into the galaxies, which perhaps shouldn't be considered a part of a galaxy.
The figure below (from Klypin et al. 2011) shows halos from the Bolshoi simulation identified with the two methods; SO (red points) and FoF (blue points), compared to an extension of the P-S HMF (solid black line) to allow for ellipsoidal structures (Sheth & Tormen (1999, 2002). At all masses, FoF halos are seen to be more massive than SO halos.

Redshift range and area
Re-reading your question, I think maybe you're interested not in the number density, but in the absolute number in a volume spanned by a redshift range $dz$ and area $dA$. If that is the case, then you simply multiply your $N(>\!\!M_\mathrm{h})$ by the cosmological volume given by $dz$ and $dA$. Let me know if you also want to know how to calculate that.

Is there a way to calculate the expected number density of Dark Matter Halos above a given mass, in a certain redshift range, and in a certain area?

There are ways of counting mass distributions of dark matter halos. I'm not sure what you mean by "expected number density," perhaps you could elaborate if my answer is not what you wanted? Do you mean the number density of the background galaxies if one is using lensing data? That is a very different question.
As with all astrophyiscs, limitations on computational power are limits on "experimental" power, since we use simulations as our laboratories, and the modeling of Dark Matter Halos is no exception.
So, a standard technique is to run N-body simulations of a dark matter universe and try to see how everything arranges itself by fitting functional forms to local regions (hence the density of a halo out to a certain mass and radius). Thus, the density you get depends on the functional form you choose! The standard profile is the NFW profile which treats the dark matter halo spherically, but this profile has many limitations that one must be careful of, for instance it is only valid approximately out to the virial radius of the galaxy. But, nowadays there are numerous available profiles (each with its own peculiarities) that try to include ellipticity of the dark matter halo. It is an area of active research to compare these profiles and improve them further, for instance see this recent paper, where the Deimer and NFW profiles are compared - the Deimer profile is capable of going beyond the virial radius because it's composed of an "inner density" that is similar to the NFW, and transitions at approximately the virial radius to an "outer" density profile.


text:
Why does the sun appear 13x bigger through diffraction in Voyager 1 image from 1990, but not from Earth?
The sun in the Voyager 1 photo here from 1990 appears 13x bigger than its diameter, diffraction of course. On Earth, photographs of the sun do not make it appear 13x bigger. Why?
A description of the photo from jpl.nasa.gov,

"The wide-angle was taken with the camera's darkest filter (a methane
  absorption band), and the shortest possible exposure (5 thousandths of
  a second) to avoid saturating the camera's vidicon tube with scattered
  sunlight. "

Where do you get the value "13" times bigger from?  Why do you say "diffraction of course"

distance to Venus, 77 solar diameters, see position of planets and Voyager 1 here,
https://en.wikipedia.org/wiki/Pale_Blue_Dot#/media/File:Voyager_1_-_14_February_1990.png

Can you show your working?

its known data, the distance to Venus from the sun is on average 77 solar diameters, Venus is in the composite closest to the sun, and Earth the box next to it

Apparently the Sun also has 8 spikes coming out of it and an off-centre corona with 2 distinct zones with different brightness.

yes sure, glare rays and lens flares

the hypothesis that the suns corona, which is luminous as you see in a solar eclipse for example, appears from distance with surrounding space adding contrast as luminous, does not reject optics in any way

Along the ecliptic there are dust particles that orbit the sun on the ecliptic plane just like every planet in our solar system does. These reflect light, and can be seen for example with the "zodiacal lights". As you know, light pollution makes luminous objects less visible in contrast to the surrounding space, that is why you cannot see the stars as well in cities, or during day time. The Voyager 1 was 32 degrees above the plane of the ecliptic.

Don't base the size of the Sun on what you see in that image because the Sun is over-exposed.  The big spot in the center of the diffraction spikes are saturated pixels because they all appear to have the same value (the spot looks uniform) so you don't know what their real values are; you would need to take a shorter-exposure image (or put filters in front of the lens) if you wanted to take a proper picture of the Sun.  Since Venus is a lot dimmer than the Sun, they set the exposure long enough to see it and accepted the fact that the Sun would be over-exposed.  You see the same thing in some of the Hubble images where to bring out the background stars, the brighter ones are over-exposed.
Another way of looking at it, if you see an image where you can very clearly see the diffraction spikes, that part of the image is over-exposed.

the image where Venus is visible was a separate photo, it is composited into the photograph, same with the image of Earth

Either way, you can't resolve the Sun in that picture because it is washed out.  For instance, look at these streetlight images.  Clearly the streetlights aren't as physically large as the images suggest they are.

what do you mean either way, the images are composited. there is no either way, there is one way since the image is what it is. most images of the sun, if you Google, are not 13x larger, they're less than 2x larger, why would the Voyager 1 image capture it as 13x larger, if it was making a composite why overexpose the image for the sun?

By either way, I meant that it doesn't matter whether it was a composite image or not because the image of the Sun is overexposed just like the image of some of the stars in the Hubble image are, and the images of the streetlamp are.  I don't know the source of the image so I can't tell you what the Voyager image team was trying to look, but I would be surprised if they were trying to image the Sun.  Does the image not look overexposed to you?  Here is another example of overexposed stars.

The composite is of the sun, earth and venus, would guess it was planned, the planets were in nice positions, and have been photographed one by one,
https://en.wikipedia.org/wiki/Pale_Blue_Dot#/media/File:Voyager_1_-_14_February_1990.png

isn't the most likely possibility that they were imaging the sun ("trying to image the sun"), and also imaged the planets with higher exposure, and then combined the images into a composite?
here is from jpl,
"The wide-angle was taken with the camera's darkest filter (a methane absorption band), and the shortest possible exposure (5 thousandths of a second) to avoid saturating the camera's vidicon tube with scattered sunlight. "
https://www.jpl.nasa.gov/spaceimages/details.php?id=PIA00450

so the exposure time of the image the sun is in was 0.005s, and the one of earth and venus 0.5

From what you said, it seems they were trying to get the "family picture", which in itself is really cool.  But also from what you said, they used their darkest filter and their shortest exposure, so they did all they could do for the Sun.  Even with those settings the Sun was just too bright (but not so bright as to wash out the whole picture).  Voyager wasn't meant to image the Sun.  If it was, they would have included an appropriate filter, like an H-$\alpha$ filter.  Check out SOHO for a mission designed for that.

are there, like the street lights you used as an example, any other images of the sun at 0.005s exposure that show it 13x larger than it is?

I was thinking, and that may sound absurd, that at that distance, 40 AU, the furthest photo ever taken of the sun, the suns corona was visible as a luminescent disk, for whatever reasons. The corona is 13x the suns diameter, and if you measure the disc in that image, it's 13x the suns diameter.


text:
Light pollution and apparent magnitude of objects in solar system
Dust along the ecliptic scatters light, causing “light pollution” throughout the orbital plane of the solar system. If an object like, say, the sun's corona, had its brightness measured from above the ecliptic, would the measurement reflect that?

Just to be clear, are you assuming the dust lies primarily only in the ecliptic?

no assumptions, data, the orbital plane has lots of dust, see zodiacal lights etc

Unclear what you mean by "would the measurement reflect that".

had its brightness measured from above the ecliptic - in contrast to what?

just in contrast to ambient light, like always when measuring brightness. re: would measurement reflect that, similar to measuring brightness of stars in day time or night time

I have to say that I also can't understand what "reflect that" means. Can you go back and edit the wording of your question (not add more comments) and rewrite it to express more clearly what it is you'd like to know? Perhaps you could add an equation, or ask in terms of "would a measurement X be brighter than a measurement Y, or would they be equal?" or "how are measurements of X compensated for the effects of scattered light from the ecliptic?"

The answer is yes: dust is highly constrained to the ecliptic plane and so viewing from above the plane would result in less light that needs to be subtracted from observations of the solar corona.  The zodiacal light is quite strong in the direction of the sun because the dust strongly scatters in the forward direction. Near the Sun, its surface brightness is somewhere around 21 - 23 mag/arcsec$^2$ from our viewpoint in the ecliptic, and this would be an additive contamination to any measurement of the Sun's corona.  From the ecliptic pole, this light foreground would be much less.  Of course, a careful observer would know to also measure the zodiacal light contribution and to subtract it off, so both observers would get the same brightness for the corona roughly.  The observer at the pole would have lower noise for a comparable instrument and observing time.

Zodiacal light can also be subtracted because it has different polarisation properties to the Thomson scattering in the Sun's k-corona. It is also not time-varying like the Sun's corona.

@RobJeffries your proposition contradicts the answer. since collaborative Q&As are for selecting the accurate answer, if you have propositions that falsify this one, could you contribute an answer, so that the selection process can do its magic?

@juli I have upvoted this answer. It seems a perfectly reasonable and clear set of statements. I haven't a clue whether it answers you question, but the answerer is blameless in that regard


text:
Why has the orbit of Venus such a low eccentricity?
The eccentricity of Venus is almost 14 times smaller than that of Mars, and 2.5 times smaller than that of Earth, and 30 times smaller than that of Mercury. Is it just a pure fluke? Or is there some “physical reasons”? I realize those are the ratios as of now, and that they change over time: will the evolution of those eccentricities dramatically change those ratios? If not, then the question of a “physical reason” emerges again.

Possibly related:  https://astronomy.stackexchange.com/questions/28368/can-orbits-change

Of course there is a physical reason, there is a physical reason for everything. It's more about the type of answer you're expecting: The answer, as illustrated by JamesK, is of the type "multi-planet interactions", and not "this combination of natural constants gives the eccentricity of Venus", which people sometimes assume is behind everything.

@AtmosphericPrisonEscape I had the first kind in mind definitively but I hoped a narrowed answer akin to the resonances which explains that some periods are nearly commensurate.

Ok I see, just wanted to clarify that.

It has a low eccentricity, but there may not be a particular reason.

Image by Kheider on wikipeda using Gravity Simulator by Trevor Dunn
In a simulation of the solar system, both Earth and Venus had orbital eccentricities that were much below those of Mars and Mercury (note the two axes. Mars and Mercury are on a scale that is 10 times bigger). But in 25000 years, Earth will actually have a smaller eccentricity, and 20000 years ago Venus's eccentricity was much larger.
So there may not need to be a special explanation for this. Planetary eccentricities do change over 1000s of years due mostly to perturbation of other planets.

So the hierarchy endures then, Mercury and Mars on one hand, and Earth and Venus on the other hand. More striking, Mars and Mercury have a remarkably stable eccentricity over this long time span. It feels that it needs an explanation but this might just be the classic mental bias to find causes!

@frapadingue  Keep in mind, the variation on that chart for Venus and Earth (eyeballing it) is about 0.02 to 0.0025.   The variation on Mars is about 0.08 to about 0.105, so Mars actually varies more than Venus and Earth in the chart above.    It only looks smaller because the scale is shrunk by a factor of 10.   But a question on Mars might be fun.  It should be a separate question though.  Stack guidelines.   In a nutshell, as I understand it, Mars is more influenced by Jupiter, so it's orbital eccentricity swings are larger.  They appear to be longer in time/period as well.


text:
Can orbits change?
I was reading about how the planet Mercury changes its orbit. I was wondering if farther away planets, particularly ones with elongated orbits, could do that as well? Planets and objects such Halley's comet, Pluto, Sedna, the dwarf planets, or even planet Nine (if there is one). Could the same happen to those objects, and if so how great are those changes? Can other things change as well like the speed of the orbit, or the size?

After reading this answer https://physics.stackexchange.com/a/137399/37754 I'd ask them if the outer solar system is chaotic.  Given a long-enough time span, and something like if Mercury gets tossed into a highly elliptical orbit), it seems quite reasonable that one could get pulled out of it's orbit.

Venus and Jupiter elongate and deelongate the orbit of the Earth in cycles of 700,000 years.

Halley's comet must change its orbit due to its outgassing near the Sun.

Precession
I assume you're talking about the precession of Mercury, one of the famous tests of general relativity. One reason Mercury is a good planet to test the relativistic predictions of precession is that its precession is more dramatic than the other planets. We can calculate the shift in the longitude of perihelion by
$$\frac{\mathrm{d}\varpi}{\mathrm{d}t} \propto n^3a^2$$
with some other (important!) factors multiplied in. $n$ is the mean motion (where $n\propto 1/T$) and $a$ is the semi-major axis. Over one orbit, the net change is approximately
$$\Delta\varpi=\frac{\Delta\varpi}{\Delta t}\Delta t\approx\frac{\mathrm{d}\varpi}{\mathrm{d}t}T\propto n^3a^2T=\frac{a^2}{T^2}\propto a^{-1}$$
Orbits with larger semi-major axes should have smaller $\Delta\varpi$, meaning that Mercury's shift is the largest in the Solar System; therefore, the other planets' orbits do precess, but by smaller amounts.
Planetary migration
theRiley mentioned this in their answer. Planets' orbits can change over time through a process called planetary migration. This often occurs by one of two mechanisms:

Disk-driven migration, where interactions with a protoplanetary disk change a planet's semi-major axis. This is the main theory for the formation of hot Jupiters.
Scattering, where interactions with other planets or planetesimals causes drastic orbital changes. This is thought to have happened in the Solar System several billion years ago (see the Nice model, the Nice II model, and variations).

The relativistic precession of Mercury, 42 arcseconds/century,  pales in comparison to the ~500 arc seconds/century Newtonian precession of Mercury caused primarily by Venus, Jupiter, and the other planets.

Yes, especially early in a system's life-cycle, when things are less sorted.  Resonances between neighboring planets' orbits can arise, inducing eccentricities which can enhance their mutual interactions, resulting in altered orbits, sometimes catastrophically of course, as with Theia and the proto-earth.
There is a theory that the Late Heavy Bombardment of the earth (4.1 - 3.8Ga ago) was caused by disruptions in the orbits of Saturn and Jupiter, which further disrupted asteroids in the asteroid and/or Kuiper belts.  This is called the giant-planet migration hypothesis.


text:
Could an ultra-hot Jupiter flare?
Some of the ultra-hot Jupiters (WASP-33b, KELT-9b) have dayside surface temperatures similar to low-mass stars. Would they be capable of producing flares? To be specific, I am asking about the possibility of flaring on the planet itself, rather than the planet inducing flares on the star.

To get flares, you need  nonhomogeneity somewhere.  EIther a large atmospheric storm - analogous to tornado -- or strong magnetic fields, as happens in the sun.

Flares in space are known for coming from stars. Sunlike stars, red dwarfs, etc, almost any kind of star has flares.
Hot Jupiters can be the size of Jupiter,(hence the name). They can be bigger; however, if they get too big, they become brown dwarfs. This means that hot Jupiters are too small to have flares.
In addition, many hot Jupiters have unstable atmospheres. They are so close to their parent stars, that they lose their atmospheres. No atmosphere, no flares.

Just because hot Jupiters are smaller than brown dwarfs it does not necessarily follow that they are too small to have flares, unless you can show that the mechanism is mass-dependent. Incidentally flares have been observed from brown dwarfs (e.g. LP 944-20) so being smaller than a star is apparently not a blocker. Furthermore the estimated lifetimes of hot Jupiters in the face of evaporation tend to be in the 10^10–10^11 yr scale, which means that for G-type and earlier host stars they will likely last the entire main sequence lifetime. Hot Neptunes would be another matter obviously.

Yes, but hot Jupiters are planets, while brown dwarfs aren't. Can gas giants produce flares?

that's a substantial part of the original question. You assert they cannot because they are not stars, but semantics probably doesn't have much bearing on the physical mechanisms involved.

Hot Jupiters are often larger than brown dwarfs. Possibly you mean less massive.


text:
Total number density of galaxies and problematic expression in practise
This post comes from physics exchange forum, I have transfered it here, maybe I will be luckier since I have not received answers on the other forum.
Here is the issue :
I am asked to give the formal expression of the total number density of galaxies and explain why is this expression problematic in practice.
From what I saw from my research and into my lectures, I have found the follwing relation which gives the number of galaxies $N$ with magnitude $(m < M)$ (number counts) :
$\text{log}\,N(m<M)\,\propto\,0.6 M + \text{constant}\quad(1)$
$N(m<M)\,\propto\,\text{exp}(0.6 M)\,\text{exp(constant)}\quad(2)$
I don't know what the teacher wants to highlight, i.e by saying that the formal expression of the total number density of galaxies is problematic in practice, given that I even haven't the kind of expression or equation which gives the estimation of galaxy density.
1) Concerning this distribution to use, do you advise me to use a distribution as a function of magnitude, mass or a distribution as function of luminosity (like Schechter) ?
2) I have seen on web the HMF (Halo Mass function) but it seems to be about the dark matter halos : can we count one dark matter halo per galaxy ?
3) But the problem is here is that a dark matter halo is more massive than the galaxy hosted in this halo : how can we deal with this ?
If someone could help me, this would be kind.
PS: I have also to explain why the estimation of galaxies density is difficult and problematic.
4) What are the issues when we want to estimate density of galaxies ?
Concerning the distribution as a function of luminosity, I have found the Schechter luminosity function :
$$N(L)\ \mathrm {d} L=\phi^{*}\left({\frac{L}{L^{*}}}\right)^{\alpha}\mathrm {e}^{-L/L^{*}}{\frac{\mathrm {d} L}{L^{*}}}$$
UPDATE 1: my apologies, $m$ and $M$ represent apparent magnitude and not masses.
But the questions remain about the counting of galaxies and the computation of total density at large scale.
In practise, what are the major problems when we want to estimate this density ?
UPDATE 2: people here not really motivated by this bounty ... feel free to ask for precisions about this post .
For the moment, I am faced to know at which scale I can compute a mean density : have I got to take into account this scale since it seems to have a dependance between density and volume considered, doesn't it ?
Is there a general formula to express this dependance ?
Regards

The problem with the form "N(m<M)" is that you have to define when a clump is so small that you no longer consider it a galaxy. Usually one considers N(m>M), since the number density decreases rapidly with halo mass / galaxy mass. I don't know where you got Eqs. 1 & 2 from, but usually one is interested in a average number density, not an absolute number, unless you're talking specifically about some particular cluster or whatever. I think you'll find the answers in the links I give below to other astro.SE posts.

Possible duplicate of Number Density of Dark Matter Halos, Number of galaxies per redshift, and How many galaxies are there in the Hercules–Corona Borealis Great Wall?. See also Boötes Void numbers, 2016

-@pela . Thanks for your quick answer. For the moment, I have found the Schechter Luminosity function. I still try to find out what are the main problems where one wants to estimate the density of galaxies : I have proposed different hypothesis into my UPDATE 1 : feel free to tell me if you think these issues prevent a good estimation for this density of galaxies in universe, or if you know relevant criteria which make things harder to calculate. regards


text:
number density of Dark Matter Halos Versus number density of Galaxies
Can we count only one dark matter halo for each Galaxy ?
Indeed, I am trying to estimate the total density of galaxies in given volume and I have often seen on physics exchange and other forums the estimation of dark matter halos distribution (as a function of the mass of halo or as a function of Luminosity).
If I could count one galaxy per halo, this would make the things simplier but I think that things are more complicated, especially since a halo and the galaxy associated to this halo have not the same mass, so distribution mass is different (mass of halo is not linear towards the mass of galaxy, isn't it ?).
If someone could help me on this issue, regards

No. You need to specify structure formation and galaxy evolution for details, but in general, you will expect many galaxies in a dark matter halo. If you need a relation, the halo occupation distribution might be a start. The Wikipedia page references three good texts on the matter.


text:
From original burst, fraction of stellar mass still surviving on Main sequence
Suppose that all stars in this galaxy were born in a single major-merger burst event about 10
Gyr ago. From this original burst, I want to compute the fraction of stellar mass still surviving as stars in the
main sequence ? For this, I have got to use a Salpeter IMF, and a star formation range between 0.1 and 120 solar
masses.
What I have done is starting from Salpeter IMF : $$\Phi(m)\text{d}m=\Phi_{0}\,m^{-2.35}$$
with $$\Phi_{0}$$ a constant normalization.
From this, I integrate from $$m_{1}=0.1\,\text{M}_{\odot}$$ to $$m_{2}=120\,\text{M}_{\odot}$$
$$N(0.1<m<120) = \int_{0.1}^{120}\,\Phi(m)\,\text{d}m = \Phi_{0}\,\bigg[\dfrac{0.1^{-1.35}-120^{-1.35}}{1.35}\bigg]$$
This result depends on the valeur of $$\Phi_{0}$$ and I don't know how to deal with it in order to get $$N(0.1<m<120)$$ ?
Moreover, it seems that I have to take into account of the age of the major-merger burst event (10 Gyr).
From these 2 principles, how could I calculate the fraction of stars surviving in the main sequence ?
Any help is wlecome, Regards

I would like to start a bounty on this post but unfortunately, the link doesn't appear, could anyone tell me why ?

When you are calculating fractions, rather than absolute numbers, the value of $\Phi_0$ does not matter, since it will be a multiplying factor in both the numerator and denominator.
You have (almost, see below)  successfully got an expression for the denominator of your fraction.
The numerator is found by calculating an equivalent integral from your lower limit to an upper limit that is instead defined by the most long-lived main sequence stars that are still "alive" - i.e. those with a lifetime equal to the age of our Galaxy.
Finally, you were asked to find the fraction of stellar mass surviving, not the fraction of stars. The stellar mass existing between two mass intervals is
$$M_* = \int_{m_1}^{m_2} m\Phi(m)\ dm$$

-@Rob Jeffries. Thanks for your quick answer. So the fraction of stellar mass $\eta$ still surviving as stars in the main sequence is equal to : \begin{equation} \eta = \dfrac{\int_{m_{min,\text{10$,$Gyr}}}^{m_{max,\text{10$,$Gyr}}} m\Phi(m), dm}{\int_{0.1}^{120} m\Phi(m), dm}\end{equation} ? Is this the definition of mean mass towards the distribution $\Phi(m)$ ? What values can I take for $m_{min,\text{10$,$Gyr}}$ and $m_{max,\text{10$,$Gyr}}$ ? regards

@youpilat13 To first order, yes. I think you should look at your notes on stellar evolution to see how lifetime depends on mass, but all the stars with 0.1 solar masses that were ever born are still alive.

-@Rob Jeffries. For the moment, I have taken for the lower limit integral : $m_{min,\text{10$,$Gyr}}=0.1,\text{M}{\odot}$ and for upper limit : $m{max,\text{10$,$Gyr}}=1,\text{M}_{\odot}$, Do you think  that's a good  approximation ? With these values, I get 60% of mass surviving into Main Sequence.


text:
Compute the total stellar mass of galaxy with major influence of Red Giants luminosity
Suppose that all stars in this galaxy were born in a single major-merger burst event about 10 Gyr ago
If the luminosity in the B band (absolute magnitude in B-band is equal to -21.22) is dominated by stars of in the RG branch, with masses $m \sim 1\,\text{L}_{\odot}$ (within $\sim 10\%$) and average luminosities $\sim 1000\,\text{L}_{\odot}$.
How can I estimate the total stellar mass of this galaxy using the Schechter relation ?
I think that I have to use the definition of Schechter law:
$$N(L)\ \mathrm {d} L=\phi^{*}\left({\frac{L}{L^{*}}}\right)^{\alpha}\mathrm {e}^{-L/L^{*}}{\frac{\mathrm {d} L}{L^{*}}}$$
or maybe Salpeter relation : $$\text{d}N=0.06\,\times\,M^{-2.35}\,\text{d}M$$
But how to introduce the parameters of Red-Giants of $1\,\text{M}_{\odot}$ with $L=1000\,\text{L}_{\odot}$.
Initially, I calculate the fraction of masse between $m_{1}=0.9$ and $m_{2}=1.1\,\text{M}_{\odot}$ :
$$\text{d}N(m_{1}<m<m_{2})=\int_{m_{1}}^{m_{2}}\,\Phi(m)\,\text{d}m=0.06\,\dfrac{(0.9^{-1.35}-1.1^{-1.35})}{1.35} = 1.22 \%$$
Anyone could see the trick to compute total stellar mass from these parameters with above laws ?
Regards

No one has an idea about the trick to compute total mass stellar from the parameters that I have used in my little calculation ?

The Schechter function for galaxy luminosity is not relevant; it describes the relative numbers of different galaxies as a function of their luminosity, and has nothing to do with luminosities of individual stars within a single galaxy.

-@Peter Erwin. ok, let's forget Schechter. Have you got an idea to compute the mass of this galaxy, from Salpeter IMF, the fact that we suppose a burst $10^9$ years ago and that the luminosity in B-band is dominated by red-giants ?