Do we observe Mars Eclipse or Jupiter Eclipse? We frequently observe lunar eclipse throughout the year. Have we ever observed Mars Eclipse or Jupiter Eclipse or Saturn eclipse?

No.

There are two types of eclipse: A lunar eclipse is when the moon goes into the Earth's shadow. We can see a lunar eclipse because the moon is so close to the Earth. If you were standing on the moon, The Earth would look bigger in the sky than the sun. If the Sun, Earth and moon are aligned, the Earth will block out the light from the sun.

Mars and Jupiter are further away. If you were standing on Mars, The Earth would look like a bright star, it would be too far away to block all the light from the sun, even if the Sun, Earth and Mars were perfectly aligned.  So there can be no eclipse.

The second type of eclipse is a solar eclipse, when the moon blocks of light from the sun. Since Mars and Jupiter orbit further from the sun than the Earth, they can never block of its light. Venus and Mercury sometimes do cross the face of the sun. This is called a "Transit" of Venus. But all the planets are too far away to block of all the light, so there is no eclipse.


What distance has the moon been from earth at the closest? For every year, the moon is orbiting further away from earth.

What distance has the moon been from earth at the closest?

If we accept the Giant impact hypothesis, which says that the Moon was once a part of the Earth pushed away by a big impact, the answer is obviously 0.


Is the smaller companion of Pulsar J0453+1559 still the smallest known neutron star? Was it further confirmed to be a NS after Martinez et al. (2015)? @ProfRob's Physics SE answer to What is the theoretical lower mass limit for a gravitationally stable neutron star? was linked just now in their answer to Can a neutron star ever be less than about 1.44 solar masses (the Chandrasekhar limit)? Why not? which notes that a smaller neutron star companion to the J0453+1559 pulsar might (or might not) be the smallest currently known pulsar:

A number of neutron stars in binary systems have measured masses below $1.44 M_\odot$ (e.g. a pulsar of mass $1.251 \pm 0.021 M_\odot$, McKee et al. 2020). I think the current lowest mass contender is $1.174 \pm 0.004 M_\odot$ (Martinez et al. (2015) Pulsar J0453+1559: A Double Neutron Star System with a Large Mass Asymmetry.

The abstract of Martinez et al. (2015) includes the sentence:

If this companion is also a neutron star (NS), as indicated by the orbital eccentricity of the system (e=0.11), then its mass is the smallest precisely measured for any such object.

and Section 3, Results explains:

Given the orbital eccentricity (e = 0.11), the companion is very likely to be a neutron star – if it had evolved into a massive white dwarf star there would be no sudden mass loss associated with a supernova explosion and the system would have retained the circular orbit that is characteristic of compact accreting systems. This is consistent with the non-detection of an optical counterpart of the system in any of the optical catalogs. However, the high eccentricity does not entirely settle the matter: the recent discovery of a recycled pulsar with a massive (∼ 1M⊙) companion PSR J1727−2946 Lorimer et al. (2015) and an orbital eccentricity of 0.0456 bridges the previously observed eccentricity gap between systems with NS and massive WD companions. In the remainder of this paper we will assume, with caution, that the companion is a NS.

That the companion may be a neutron star is important for this analysis; the next sentence is

The pulsar’s orbital eccentricity allows a detection of the advance of periastron, ˙ω. If
both components are compact, as implied by the optical non-detection, this is given by:
$$\dot{\omega} = \dot{\omega}_{GR} + \dot{\omega}_{K}$$
[...]

And the conclusion section notes that this is a relatively small eccentricity for double neutron stars listed in Table 1 as of 2015.

It's now been six years, so I'd like to ask:
Question: Has the smaller companion to Pulsar J0453+155 been confirmed to be a neutron star by any further work beyond that of Martinez et al. (2015) and is it still the smallest known neutron star?
Has the smaller companion to Pulsar J0453+155 been confirmed to be a neutron star after Martinez et al. (2015) and is it still the smallest known neutron star?

Here is a picture from the most recent review I could find - Horvath et al. (2020).
The companion of J0453+1559 is still the least massive of the double neutron stars with very precisely measured masses. I cannot find any citations to the Martinez et al. paper (here is a list) that either update the mass or shed any further light on its true identity. It appears to have been accepted in the literature as a neutron star - though there is still a school of thought that it could yet be a massive white dwarf (Tauris & Janka 2019).
As you can see from the picture there are several other objects with only slightly higher masses, and some of these are pulsars and are unambiguously neutron stars (e.g. a pulsar of mass $1.251 \pm 0.021 M_\odot$, McKee et al. 2020). There are some neutron stars in X-ray binaries and with white dwarf companions that may have a lower mass; but the uncertainties could push them above J0453+1559.


Why temperature of dark side of moon is not 3 Kelvin The temperature of the space between Earth and the Moon is around 3 Kelvin ($\approx -270^\circ {\rm C}$). The moon has no atmosphere, therefore the heat convection is not possible from the bright side to the dark side on Moon.
Now, if the temperature of the space around the moon is $3 {\rm K}$, then how the dark side of the moon is having a temperature of $100 {\rm K}$?
References

https://sciencing.com/temperatures-outer-space-around-earth-20254.html
https://www.space.com/14725-moon-temperature-lunar-days-night.html

The "dark" side of the Moon is only truly dark during full Moon. Everywhere on the Moon there is day and night as well. The dark side of the Moon is called like that because we do not see it from Earth, since the Moon shows us always the same side due to tidal locking - not because it is always dark there.
In other words: During (lunar)  day, the Sun heats up the surface of the Moon and it gets pretty warm, even on the (not so) "dark" side of Moon. This explains the $100 {\rm K}$ you are citing.


Why Comet C/2020 F3 Neowise returns even it's orbit is near- parabolic? I have read in Wikipedia about  Comet C/2020 F3 Neowise that it's orbit is near-parabolic. But in another section, it says that it is a non-periodic comet whlith orbital period ranging from 4500 years to 6800 years.
My doubt is how a comet returns back if it is in a parabolic orbit and why it is non-periodic?
Here is the Wikipedia link: https://en.m.wikipedia.org/wiki/C/2020_F3_(NEOWISE)

Comet naming conventions limit "periodic comets" to those comets whose orbital period is less than 200 years. Comets in elliptical orbits with a period of greater than 200 years are classified as non-periodic. Comets in parabolic or hyperbolic trajectories are also classified as non-periodic (and they will never again approach the primary).


How strong would gravity need to be to prevent star formation This question got me thinking.

In another universe, with a different value for the gravitational constant G, how much bigger would G need to be for star formation to be prevented?

In other words in a hypothetical universe, the star would attempt to ignite, but the force of gravity would overwhelm the star's outward pressure and you would just get a black hole instead.

I'm not looking for a precise answer here... just ROM... 10 times bigger? 10,000 times bigger?

By way of starting a discussion, because I don't clearly see how to answer this question correctly or all the factors that may be involved.

If you increase $G$, then by the virial theorem, the central temperature of the ball of gas (that will contract quasti-statically, since radiation cannot escape unimpeded) will be proportional to $GM/R$. So for a fixed mass, the ignition temperature for nuclear fusion will be reached at a larger radius and stars would be physically larger and less dense.
$$T_c \simeq \frac{GMm_u}{10Rk_B},$$
assuming an ionised hydrogen gas and where $m_u$ is an atomic mass unit.

In order to avoid "starhood", you need to compress the gas either to within its Schwarzschild radius or to a density where it could be supported by electron degeneracy pressure without reaching a high enough temperature to ignite hydrogen fusion.

Considering the first of these. $R_s = 2GM/c^2$, so the central temperature reached at $R=R_s$ would be $\simeq m_u c^2/20k_B = 5\times 10^{11}$K, is independent of $G$ and is far higher than the $10^{7}$K required for H fusion (ignore deuterium). So direct collapse to a black hole cannot circumvent nuclear fusion.

The second possibility is likely to be more constraining.
There is a standard treatment given in Shapiro & Teukolsky (1983) thaat determines the radius $R$ at which electron degeneracy becomes important in supporting a star of a given mass $M$. The idea is that you work out the mass and radius for which the phase space occupied by an electron approaches $h^3$. The result is (for ionised hydrogen):
$$ R/R_{\odot} \simeq 0.07 (M/M_{\odot})^{-1/3}$$
However, digging into the maths a bit, this can be written as a proportionality
$$ R \propto G^{-1} M^{-1/3}$$
and thus the central temperature at which electron degeneracy pressure becomes important will be
$$T_c \propto GM/R  \propto G^2 M^{4/3}$$

Thus (for a fixed $G$) what happens is that a contracting ball of gas will reach a critical $T_c$ for fusion if the mass exceeds some critical threshold (above about $0.075M_{\odot}$ in our universe) then H fusion will begin before degeneracy sets in. If you increase $G$ then this threshold actually becomes smaller and smaller balls of gas will become stars.

But what about the Chandrasekhar mass? In our universe the threshold mass for fusion is much lower than the Chandrasekhar mass ($5.8M_{\odot}$ for pure hydrogen), so there is no problem in supporting an object just below the fusion threshold with electron degeneracy pressure. Dig a little deeper and we find that the Chandrasekhar mass is in fact $M_{Ch} \propto G^{-3/2}$. This means we can now express the critical central temperature in terms of mass as a fraction of the Chandrasekhar mass.
$$ T_c \propto G^2 \left(\frac{M}{M_{Ch}}\right)^{4/3} (G^{-3/2})^{4/3} \propto \left(\frac{M}{M_{Ch}}\right)^{4/3}$$

In other words, the threshold between a star that burns hydrogen and an object (brown dwarf) that will be supported by degeneracy pressure is a fixed fraction of the Chandrasekhar mass and does not depend on $G$.

My conclusion is therefore that although changing $G$ will do interesting things to stellar evolution (changing the rate at which it occurs for instance - see here where I show that the luminosity scales as $G^7$ and therefore lifetimes scale as $G^{-7}$ !), I do not think that changing $G$ allows direct collapse to a black hole without the central temperature becoming high enough to initiate hydrogen fusion.

Secondly, a failed star is possible in our universe - it just needs to be less than $0.075M_{\odot}$ and collapse is halted by electron degeneracy pressure. In a universe with an increased $G$ this threshold becomes smaller as $G^{-3/2}$.


What is the maximum possible separation between sun and moon in the earth sky for an eclipse to occur? How do we calculate the maximum angular separation between sun and moon during:

Solar Eclipse
Lunar Eclipse

We'll need the angular diameter of the Sun: 31.6-32.7 arcminutes; the Moon's is between 29.3 and 34.1 arcminutes (found on Wikipedia).

The solar eclipse case is 0 degrees; it can only be an eclipse if the Moon is touching the Sun. In that case, their centers are at most (32.7 + 34.1) / 2 = 33.4 arcminutes apart.

A lunar eclipse would occur if the Sun and Moon are diametrically opposite in the sky; the separation of their centers is then 180 degrees. The actual separation is at least (31.6 + 29.3) / 2 = 30.45 arcminutes less than 180 degrees.


What do you call it when a satellite goes once around its planet? When a planet or a satellite turns around itself, its called a rotation. When a planet goes once around the Sun, it's called a revolution. What do you call it when a satellite goes once around its planet? Is it also a revolution? But then how do you disambiguate with the satellite having gone once around the Sun?

Definitions:

Rotation can be defined as the circular motion of an object around its own axis, or upon itself.
Revolution can be defined as the circular movement of an object around an external axis, or around another body.

So a satellite moving around a planet once is called a revolution. Or an orbit. "The satellite stayed active for 3 orbits, and then the batteries ran out".
To disambiguate the case 'the satellite having gone once around the Sun': this depends. Is the satellite in orbit around a planet, or does it orbit the Sun directly?
If the satellite is in orbit around a planet, you have to describe what you mean. There's no separate word for this case, AFAIK.
If it's in orbit around the Sun, it's again one orbit or one revolution.


Convert coordinates between RA/dec and WGS 84 (SRID=4326) I've imported the HYG Database from http://www.astronexus.com/node/34 and now I need to convert all RA/dec stars positions to fit in a WGS 84 coordinates (SRID=4326) PostGIS map (-180 to 180, 90 to -90).

BayerFlamsteed  ProperName  RA            DEC
21Alp And       Alpheratz   0.13976888    29.09082805
11Bet Cas       Caph        0.15280269    59.15021814
88Gam Peg       Algenib     0.22059721    15.18361593
Alp Phe         Ankaa       0.43801871    -42.30512197
18Alp Cas       Shedir      0.67510756    56.53740928
16Bet Cet       Diphda      0.7264523     -17.98668410

Related to:

https://gis.stackexchange.com/questions/84389/spatial-database-containing-coordinates-for-stars-in-the-sky

https://gis.stackexchange.com/questions/2459/what-coordinate-system-should-be-used-to-store-geography-data-for-celestial-coor

EDIT: Ok, I think MerseyViking gives the answer in https://gis.stackexchange.com/questions/2459/what-coordinate-system-should-be-used-to-store-geography-data-for-celestial-coor telling to create a new coordinate system. Can someone give a look and some opinion?

EDIT 2 (After answered): Fantastic!! I need to show the results of whuber and Dieudonné's answer. The first picture I show the query in QGIS. In second picture you can see the real positions in a map. The scale is different - Find Shaula, Nunki and Antares.

This is the conversion method ('H' is for Hyparcus in my database):

insert into stars
    select
        id,
        proper_name,
        ST_SetSRID(ST_MakePoint( (-ra * 15), dec),4326),
        mag,
        'H'
    from hygn
    where (proper_name is not null)

Assuming negative longitude in WGS84 is west of Greenwich than you would get LONGITUDE = RA*15 and LATITUDE = DEC. The celestial longitude RA=0 through the vernal equinox would then be projected on the meridian of Greenwich.

If you subtract 180 from the LONGITUDE then you would just rotate the celestial sphere by 180°. The meridian of Greenwich would then equate to RA=12h.

If you want the stars to move with the rotation of the Earth, then you would use LONGITUDE = RA*15 - THETA*15, where THETA is the sidereal time at Greenwich in decimal hours.

You should be aware that for the celestial sphere you look from the centre towards the inside of the surface of the sphere, while for geographical purposes you look from the outside (above) towards the spheroid (down). So if you project the stars on a globe like this, you will notice that the constellations will look inverted from what you're used to in star maps. If you find one of those old celestial globes, you will also see the inverted constellations. For instance on this image, you'll see Leo to the left of Virgo and Hercules, while on most star maps, Leo will be to the right.

If you want the constellations to look 'right' then you would need to use LONGITUDE = -RA*15 (mind the minus sign). But then you would also need to invert the rotation of the Earth as well.


Surface conditions of lonely (rogue) planets The question How are rogue planets discovered? describes the difficulties in finding planets (or planet-sized objects) that are floating through space without being under the influence of any star or Galaxy system.

What would the possible surface conditions of these planets be? Could chemotrophic life arise on such planets?

Rogue planets have two formation mechanisms: Independent formation and ejection.

An independently formed rogue planet would have condensed out of the nebular material by itself and not formed from a young star's protoplanetary disk. We understand how individual stellar-mass objects condense and which have been tested by observation, but I'm not aware of any tested studies which predict rate at which Earth-massed objects independently form.  (We certainly have not observed a large population of them, and the gravitational micro-lensing studies which have been done would have detected them if they existed.)  But since Brown Dwarfs are fairly common, it seems safe to assume that there is a population of super-Jupiters that formed by direct condensation.

The planets formed by ejection should range in size from meteors right up to super-Jupiters and should be broadly similar to the size distribution of bound planets. (If small planets tend to be closer to their star, there may be some bias against them being ejected.)

Planets are mostly ejected early in a system's life, but it can happen at any point in a star's lifetime -- planetary system dynamics never becomes perfectly stable.  And if there's a near passage of another star or a large rogue planet, ejections can occur even after billions of years of stability.  (See Fritz Leiber's "A Pail of Air"!)  But the population of ejected planets is probably very heavily weighted towards planets ejected soon after formation.

The distinction between independently formed and ejected planets is important because when a planet forms it is very hot from the gravitational energy released in formation, and most likely initially inimical to life forming.  But it cools (the surface much faster than the interior) and eventually life can form.

If the planet remains in orbit around a star, the surface temperature drops asymptotically towards an equilibrium temperature where the sum of the star's radiational input and the heat leakage from the still-hot interior balances the IR radiation leaked by the planet into space. Frequently, that will be in a temperature range where life can form on the surface.

Once a planet is ejected, the equilibrium temperature will be much lower. For example, for Earth at 4.5 GY old, the Sun's radiation dumps 3000 times as much energy onto the Earth's surface than does leakage from the interior, (see https://en.wikipedia.org/wiki/Earth%27s_internal_heat_budget for details.)  The time it takes for an Earth-sized rogue planet to cool to Earth's current temperature is on the order of 10 MY and after that it would just keep on cooling down to a surface temperature on the order of 30K.

So the question of life forming on an Earth-like planet is one of whether or not there is time before the surface layers freeze solid, since that would appear to halt the formation of life.

All bets are off for larger planets which (a) cool more slowly and (b) probably have enough more water to have subsurface liquid oceans even after the surface freezes solid.

For planetary bodies where life does have time to form, the only source of energy would be the heat leaking from the interior either directly due to the heat gradient (a very diffuse source), or indirectly from the equivalent of Earth's deep sea vents.  The latter seems more likely since the steeper heat and chemical potential gradients are a lot easier to exploit.

Hal Clement wrote two scientifically excellent stories set on such planets.  One, Star Light, has high-gravity aliens working with humans to explore the surface of Dhrawn, a Brown Dwarf.  The other was short fiction and had humans meeting intelligent life living on an closer to Earth-sized rogue planet. (I think it was "Sortie" and sequels, but I'm not sure.)

In any event, it seems likely that there would be a great variety of different kinds of rogue planet that might potentially support life.


Difficulties in finding Jupiter I am new to astronomy. Jupiter is now visible to the naked eye but I have difficulties in locating it with my (new Newtonian reflector) telescope. The lowest magnification I have is 36X, which I guess is part of the problem(?). Any tips for the beginner..? I feel that if I knew any star constellations, it would help in locating objects.

Go out in the daytime and practice lining up on the leaves of trees on a distant hill or some such. It's easier to find targets when they aren't against a nearly featureless black backdrop.
If you have a finder scope, likewise align it to the scope's view using a distant daytime target.

At night, look first for the moon to get a feel for how aligning works.


Is it possible that in the Local Interstellar Cloud there are undetected stars? The Local Interstellar Cloud is supposed to be around 30 light years large. Is it possible that in this region there are stars which are undetected? I would have thought that it wasn't possible, but there are stars which are very small or are very close to another star. BTW, with star I mean an astronomical object which is performing fusion, because I've seen lists of smallest or coldest stars which includes brown dwarfs which don't perform fusion or white dwarfs.

It is unlikely, unless they are extremely faint. The reason is that the cloud obviously does not absorb enough light to keep us from detecting remote red and brown dwarfs, and hence a closer star would also be detectable.


Exactly how does Canis Minor look like a dog? Here's what Canis Minor looks like.

To me, all I see is two stars. Canis Minor is a constellation of the giant Orion's hunting dog. Two stars can give quite a meaning. Two stars can mean a stick, a telescope, et cetera. But how could that be interpreted as a dog?

Since this is a Greek constellation, are there any sources that explain why ancient Greeks considered those two stars to represent a dog?

Quoting from wikipedia:

  Canis Minor was one of the original 48 constellations formulated by Ptolemy in his second-century Almagest, in which it was defined as a specific pattern (asterism) of stars; Ptolemy identified only two stars and hence no depiction was possible. The Ancient Greeks called the constellation προκυων/Procyon, "coming before the dog", transliterated into Latin as Antecanis, Praecanis, or variations thereof, by Cicero and others. Roman writers also appended the descriptors parvus, minor or minusculus ("small" or "lesser", for its faintness), septentrionalis ("northerly", for its position in relation to Canis Major), primus (rising "first") or sinister (rising to the "left") to its name Canis.

It looks like the original (Greek) identification of the constellation by Ptolemy didn't attribute anything to it, let alone a dog. But over time, the constellation became known based on its proximity to Canis Major. Originally known as Procyon meaning "coming before the dog", i.e., it rises before Canis Major. This relation to Canis Major seems to have led it to being considered as a representation of a dog as well. The Minor attribute appears to be due to the fact that it is fainter overall than Canis Major.

Really though, you can't pick apart constellation patterns too much. Much of them don't look that much like their attributions and only evolved over thousands of years of stories and imagination. It's not like one guy sat down one day and said, what does this pattern of stars remind me of, and that was that.


How powerful a telescope would allow me to view the astronauts aboard ISS do a space-walk? This arose from a comment posted against a question elsewhere on the stackexchange

How powerful a telescope/binoculars would allow me to view the astronauts aboard the ISS do a space-walk? Aperture? Magnification?

This has been done before, so I don't have to go through all the heavy calculations using Rayleigh criterion accounting for atmospheric diffraction and visible light wavelength. Ralf Vandebergh, a Dutch astronomer, professional photographer and veteran satellite spotter has been busy trying to do exactly this since the 2007 and has indeed succeeded on several occasions by now using a 10 inch (25.4 cm) Newtonian reflecting telescope that has a resolving power (angular resolution on the CCD sensor) of roughly one pixel per meter at the distance to the International Space Station (ISS) that is currently in a 230 miles (370 km) orbit above the Earth:

   Ralf Vandebergh’s detail of an image he took on Mar. 21, 2009 showing astronauts working outside the ISS. Credit: R. Vandebergh

Vandebergh's personal page also hosts all kinds of other successful observations of the ISS through his telescope and recorded in both photographs as well as some short videos. Why short? Because targeting the ISS as it moves at the speed of 4.8 mi/s (7.7 km/s) is rather tricky, and the atmospheric conditions and times at which the ISS passes over some area on the surface of the Earth don't make it any easier either. But perseverance and hard work have paid well for this individual astronomer.

          Raw video of the ISS as seen through the air turbulence. Note the good visibility of the Lira antenna at the Russian          Zvezda Module in the lower part of the image. Credit: R. Vandebergh

So again, skipping the math to calculate the required angular resolution of a telescope and apply that to some arbitrarily selected image sensor size and resolution, we can see that using a well collimated 10 inch Newtonian or a Dobsonian telescope on a clear night can, with some near perfect targeting, produce a direct proof of a spacewalker doing his or her job 230 miles (370 km) high during an EVA. More powerful telescopes would of course produce better resolution images, but the atmospheric effects limit their use and are of course much harder to target an object moving fast over the skies with.


Why didn't we see those campfires on the Sun until now? What is it about its optics and instrumentation (aperture, sensors, filters), being in space and distance from the Sun on 30 May 2020 that allowed Solar Orbiter's HRIEUV telescope to see something that's never been seen before?
SOHO has been looking at the Sun in all kinds of wavelengths for a long time, and there's STEREO as well, and we've been looking at the Sun from high altitude balloons and sub-orbital rockets for over a half-century!
What could be the cause of solar "campfires"? provides links to a lot of information about the observed phenomenon, and eoPortal's Solar Orbiter Mission includes a lot of details on instrumentation .

tl;dr they're a bit too small for SOHO and Stereo and only visible in EUV
The "campfires" are described as a few hundred km across,

The smallest of those campfires are about the size of a European country, according to Berghmans.

which means that from Earth (or SOHO or Stereo) they will subtend from one to a few seconds of arc (a few micro-radians). They are observed in extreme UV which is blocked by the atmosphere, so only space-based instruments could see them.
SOHO's imaging instrument the EIT has about a 5 arcsecond resolution.
Stereo seems to be comparable to SOHO (2000 pixels across the sun is about 500km/pixel at the centre of the disk).
Solar orbiter has a resolution of about 1 arc-second and is at half the distance, so has effectively 10 times the resolution of the other probes.
I don't know about baloon and sub-orbital instruments, but I could speculate that achieving sub-arcsecond resolution in EUV while not having an aperture so large that the incoming light fries the sensors might have been hard and expensive and required recent developments in materials and sensors coming from the semiconductor industry.


Is it unlikely for a meteor to hit the day side of the earth? I have binge watched a lot of documentaries and movies about meteors hitting the earth. The one pattern I have seen, even in the science-fact documentaries like Nova, is that the animations show meteors hitting the earth on the day side. By that, I mean at a near 90° angle in the center of the day zone. As if it were to hit the earth at high noon.

What seems unlikely to me is where the meteor hits. It seems intuitive to me that statistics would favor a meteor strike on the night side of the planet rather than the day side. Or, possibly from the transition between day and night if the meteor came in from the side.

Is my suspicion correct? Is it unlikely for a meteor to hit the earth on the day side and far more likely to hit the night side?

To clarify, I'm thinking of significant meteor strike, like the Chicxulub impact  that killed off the dinosaurs. If something that massive came in from the day side, wouldn't it more likely to have been captured or redirected by the sun because of the sun's gravity?

Remember that the Earth, and the meteoroid, are both in orbit around the sun. As the meteoroid approaches Earth, it will also curve due to Earth's gravity. If you imagine the Earth to be stationary and the meteoroid to be moving in a straight line your intuition is likely to be wrong.

The result of the complex dynamics is that there is no time or direction from which it is impossible for a meteoroid to have come from. On average the direction of impact is 45 degrees, and the timing is somewhat more likely in the morning when the rotation of Earth combines with its movement through space. A direct impact at noon, from the apparent direction of the sun, is unlikely but possible.

When you look at lists of meteors, many are at night. Except the brightest fireballs, a meteor is much more noticeable at night.

Nasa hs a database of boloids: http://neo.jpl.nasa.gov/fireballs/ By combining the UT time with the latitude information it is possible to get an estimate of the local solar time of the fireball:

This histogram of time of impact of about 250 fireballs since 2007 shows no particular decrease around midday. The documentary graphics may show an apparently fully lit earth, and a perpendicular impact as it is easier to recognise the Day side of the Earth, and it looks more dramatic.


Is it possible to revive the Mars geodynamo? I’ve done a bit of reading and have found multiple theories suggesting how the Mars geodynamo stopped fully working.

This site suggests that one possibility could be volcanic activity ejecting water and other necessary elements from the mantle significantly slowed the natural convection process.

Where as this site says that it also could have been massive asteroids colliding with the surface, heating the planet up to temperatures that would disrupt convection.

No one is quite certain what happened, all that is known is Mars doesn’t have much of a magnetic field at this point.

Today I saw that there’s an idea to create an artificial field that’ll act in place of a geodynamo’s naturally generated one.

My main concern for this is that the units creating the artificial field could be destroyed and the entire planet would become unstable afterwards.

It would seem if the magnetic field is supported by the planet itself, it would offer a higher degree of stability and less involvement for humans in the long run.

What’s the possibility for ‘restarting’ the Mars geodynamo or perhaps ramping it up enough to support a strong enough field? And if so, how would we be able to do it?

Let's do a bit of math: magnetic energy density is at least $u = \frac{1}{2}\frac{B^2}{\mu_0}$. The magnetic field at the surface of Earth is of the order of 30µT which gives us a magnetic energy density in the order of $10^{-3} J/m^3$
Now we need to fill the entire volume of the planet and some surrounding space with this field in order to get some protection from the solar wind; let's assume a cube of $(20000km)^3 = 8\cdot 10^{21}m^3$. So that means that we need an energy in order of $10^{18}W$ to sustain a comparable field which equates to roughly 1 million state-of-the art GW nuclear power plants. We currently have significantly less than 1000 in the whole world combined and I have not considered losses of any kind (thus 100% conversion of power into magnetic field)

My rough calculation also does not include any additional power needed to sustain the field extent into space when subject to the magnetic pressure of the solar wind which will be needed in order to keep the atmosphere inside the magnetic field, especially in direction of the sun.

So please go and figure whether these scenarios are really realistic. Actually reviving any dynamo on Mars is an altogether different story and would need to understand how exactly it works/worked; it might entail to heat up the core which requires probably more energy and will pose even more severe technological challenges than a giant dipole magnet of planet size.


Does Dark Matter affect the motion of the Solar System? Would dark matter have a similar effect on the Solar System as it does on the Milky Way? I'd imagine it could help hold together some of the Solar System bodies or at least affect their motion, right? Maybe to a lesser extent, but would it be significant?

The dark matter model that is used to explain the "missing mass" problem relating to our Galactic rotation curve, consists of a pseudo-spherical distribution that is much more extended than the visible stars and gas. Even though this "halo" contains more than ten times the mass of the visible matter, when you work out what it's density should be in the solar neighbourhood, it turns out to be around 100 times less than the "normal" interplanetary medium.

This means, that inside the Earth's orbit, there could be as much as $10^{10}$ kg of dark matter. This sounds a lot and in principle, yes, this would speed up the orbits of planets in a way that depended on their distance from the Sun. However, the predicted effects of this are still about 6 orders of magnitude below the current precision of measurement.


What decided how the Kepler space telescope was pointed? I suppose that the Kepler field was well chosen. It is rich in stars and has minimal background light since it is pointing out from the disc.

But other things could have been considered. For example, a field which contains the most stars which are near enough to be followed up by ground based telescopes. And why wasn't Kepler designed to change its pointing so it could cover several fields, such as all of the above? Funny enough, it does so now as K2 after two of its reaction wheels have broken, so studying only one field doesn't seem to be a technical issue, but something which astronomers prefer. Why?

Changing fields breaks the timeline of the observations.  If you have variation with a period similar to when you switch, you might completely miss it.  Even if the periods are different, it degrades your observations and makes it harder to draw conclusions.  The more uninterrupted viewing you get, the better.

This mission document covers some of the reasons for the choice of the field.

Continuously viewable during the mission (you can't pick something where the earth, sun or moon constrains your viewing).
Rich in stars.  Observe as many as possible.
The size of the sunshade limits the field to be more than 55 degrees from the ecliptic.

Although it can image lots of stars, the actual field isn't all that large.  Attempting to pick a single field with some fraction nearby would be difficult, and was not Kepler's mission.  In fact the resource of follow-up observations from ground telescopes was mentioned as one reason a northern field was chosen over a southern one.


Why is the Sun's brightness and radius increasing, but not its temperature? On the Sun's article on Wikipedia, there is an image showing how the Sun's brightness, radius and temperature have changed over time:

For the past (and next) few billion years, I see the luminosity continuously rising, the temperature remaining about the same, and the radius also rising.

This seems bizarre to me. I would imagine that, for luminosity to increase, the temperature would have to increase. And for temperature to increase, the radius would have to decrease (as the core becomes denser, temperatures rise).

So why does it work this way?

PS: I would also appreciate a concrete explanation for why all three decreased at the beginning of the Sun's life, but suddenly began increasing.

The effective temperature $T_\mathrm{eff}$ of a star, which is presumably what's been plotted, is defined through its relationship with the star's radius $R$ and luminosity $L$ by

$$L=4\pi R^2\sigma T_\mathrm{eff}^4$$

This comes from the assumption that the star radiates like a black body at the photosphere. While this isn't strictly true, it's quite accurate, and regardless, that's how we define the effective temperature. The actual surface temperature will be slightly different but also behave roughly as plotted.

So, even if $T_\mathrm{eff}$ is constant, the star expands if it grows brighter.  Also, you can see that the sensitivity to temperature is steeper than radius, so a moderate change in luminosity can be absorbed by a relatively small change in effective temperature.

While the luminosity is basically determined by the simple behaviour of the nuclear reactions in the core (in terms of temperature and density), the surface properties depend on how energy is being transported near the surface. For radiation, you have to consider what the opacity of the material is, which itself depends on ionization states and whatnot. It's easy enough to see why the luminosity grows (the core gets denser and also hotter, producing energy faster) but the determination of the surface properties is more complicated. For the Sun, it turns out the way shown in the plot after you solve all the equations with the relevant opacities.

Also, as an extreme counterexample to "brighter means hotter and smaller", remember that red giants are much brighter but also much cooler!

PS: I'm not sure the source of the data but I would guess the wiggle at the start is because of the star finishing its contraction onto the main sequence.  That is, before the first minimum, energy is being released by gravitational contraction.  After it, the energy from nuclear reactions starts to dominate.


When will the Sun end all human life on Earth? If nothing else wipes out human existence prior to this, at what point will the Sun make Earth uninhabitable for humans?

The Sun is gradually getting larger and brighter. In fact, as called2voyage pointed out, its brightness is increasing by 1% every 100 million years. You can see how the Sun will change in the future from this graph: (Source)

According to this paper, within 1 billion (short scale) years from now, the ever-increasing luminosity will have made Earth nearly uninhabitable. The average temperature will have reached 47°C, compared to its current 15°C. Essentially no water will be left either, except at the poles. This may allow for simple life to survive for a while.

By 3.5 billion years, Earth will no longer resemble its current self. Its oceans, magnetic field and ozone layer and plate tectonics will be no more. Its surface temperature will skyrocket to roughly 1,330°C, hot enough to melt surface rock. No longer will our planet resemble a pale blue dot, and it will be more like Venus. Our planet is officially dead, along with all life on it. (Source)

~4.5 billion years from now, the Sun will become a red giant and possibly consume Earth. However, according to this paper, it may heat up potentially habitable bodies like Triton, to the point where they would support life. Unfortunately, the Sun won't remain in this stage for long enough — life usually takes billions of years to develop.


Plotting Galactic Longitude from 180 to -180 I made a scatter plot in Python that looks like this:

However, I want galactic longitude, l, to be plotted from 180 to -180 like these graphs from Iorio and Belokurov (2019):

Here is the code for my plot:

sns.set_style("ticks")
sns.set_context("poster")
%matplotlib inline
from matplotlib import pyplot as plt

fig = plt.figure(figsize=(8,5))

ax  = fig.add_subplot(1,1,1, aspect='equal')

ax.scatter(rrl_pm.l, rrl_pm.b, s=1, color='black', alpha=0.1)

ax.set_xlim(0, 360)
ax.set_ylim(-90., 90.)

ax.set_xlabel("l [deg]")
ax.set_ylabel("b [deg]")
textstr = 'N = ' + str(len(rrl_pm.l))
props = dict(boxstyle='round', facecolor='white', alpha=0.5)
ax.text(0.05, 0.95, textstr, transform=ax.transAxes, fontsize=16,
        verticalalignment='top', bbox=props)
plt.title("Scatter Plot of Cleaned Sample of RRLs in Sgr dSph with Spatial AND PM Cuts", fontsize = 16)
plt.savefig("SP-Clean-Spatial-PM.png", dpi = 1000)
plt.show()

If I simply change the xlim to run from 180 to -180 degrees the following graph is produced:

So essentially it only covers half the sky. Why exactly is this the case and what transformations do I need to apply to the galactic longitude for the 2nd plot to "wrap around" and display both sides?

Assuming

x, y = rrl_pm.l, rrl_pm.b

The transformation you want is

xx = [(q+180)%360 - 180 for q in x]

Add 180, do modulo 360, then subtract 180.

Then set your limits

ax.set_xlim(180, -180)

import matplotlib.pyplot as plt

x = (0, 10, 20, 40, 80, 160, 200, 280, 320, 340, 350)
y = (0, 10, 20, 30, 40, 50,  -50,  -40, -30, -20, -10)

fig = plt.figure(figsize=(8,5))
ax  = fig.add_subplot(1,1,1, aspect='equal')

ax.scatter(x, y)

ax.set_xlim(0, 360)
ax.set_ylim(-90., 90.)

plt.title('Old', fontsize=18)

plt.show()

xx = [(q+180)%360 - 180 for q in x]  # DO THIS!

fig = plt.figure(figsize=(8,5))
ax  = fig.add_subplot(1,1,1, aspect='equal')

ax.scatter(xx, y)

ax.set_xlim(180, -180)
ax.set_ylim(-90., 90.)

plt.title('New', fontsize=18)

plt.show()


Planets orbiting Alpha Centauri I read that a few years ago some planets were discovered orbiting Alpha Centauri. Has there been any more information since then concerning the possibility of life there? This is the closest star system to Earth, but it would still take us 63,000 years to reach it.

The detection of Alpha Centauri Bb unfortunately did not survive follow-up analysis and it is no longer thought to exist, see Rajpaul et al. (2016). This planet candidate was in any case too close to the star to support liquid water.

There has been a claim of a single transit of a longer-period planet orbiting Alpha Centauri B, although this has not been confirmed. The relevant paper is Demory et al. (2015). Once again, this planet candidate was inferred to be too close to the star for liquid water.

Recently it has been confirmed that Proxima Centauri is a bound member of the Alpha Centauri system (Kervella et al., 2017). The planet Proxima b has been found in the liquid water zone (Anglada-Escudé et al., 2016) and has a minimum mass of 1.27 Earth masses, so if the orbit is close to being edge-on it is likely to be a terrestrial planet.

Of course, just being in the liquid water zone is not sufficient for habitability (obvious counterexample: our Moon). A major issue is that Proxima Centauri is a flare star and the space weather environment experienced by Proxima b is extremely hostile. This brings into question whether or not Proxima b can maintain an atmosphere under such conditions, for example see Garcia-Sage et al. (2017).

There is also the issue that low-mass stars like Proxima Centauri spend billions of years in a luminous pre-main sequence stage of their evolution. This would likely trigger a runaway greenhouse on planets in the main sequence liquid water zone and result in the loss of significant amounts of water from the planet, as discussed in Ramirez & Kaltenegger (2014).

So while Proxima b appears to meet some requirements for being able to support life, in several other ways it appears to be far less than ideal and we should not be too surprised if it turns out to be non-habitable.


Considering that the moon’s gravitational pull on earth is so weak- we need special instruments to detect it- how does it cause the tides? That the moon’s gravitational pull is responsible for the tides is established. However, if we want to detect the moon’s gravitational pull on earth, this is very challenging. This seems to be a paradox considering the enormous force needed to drive the tide.

Gravity is an acceleration, and force = mass x acceleration.  Thus, if we want to detect the Moon's gravitational pull, we need either a giant mass or a really precise instrument.  Since the oceans are a giant mass, the force on the ocean is thus correspondingly large.  The fluidness of the ocean also allows it to move, even if the force per volume is small.
To expand on this, the tidal acceleration due to the Moon on the surface of the Earth is about 1 μm/s$^2$ (1), which means that the force due to the weight of a 1000 kg object will vary by about 0.001 N over the course of a day.  1000 kg of water is about 1 cubic meter, so every cubic meter of the ocean experiences that tiny variation in force as well.  Since the ocean has about 1.3 x 10$^{18}$ cubic meters (2) of water in it, you could say the tidal force is 1.4 x 10$^{15}$ N, which is an absolutely enormous force - translated into weight, it's more than the weight of all living things on the Earth (3).
There is also a corresponding acceleration from the Sun, about half as strong as the Moon, which also has some effect on the tides, resulting in a monthly very high and very low tide, known as the spring tide, and a somewhat averaged tide, known as the neap tide.  It might also help to realize that this 1/2 as strong force is what is also responsible for keeping the whole Earth in orbit around the Sun.


How much of the sun needs to be blocked in order to see a noticeable drop in brightness I live in NYC and just experienced the eclipse. What fascinated me the most was that despite the fact that the sun was about 70% blocked, the area was not noticeably dimmer. If the news wasn't saying there was a solar eclipse I wouldn't have realized it was happening.

How much of the sun needs to be blocked by the moon in order to noticeably dim an area?

Based on the CNN's reporting it appeared to get noticeably darker at near 95% coverage. That's just from watching the onset of darkness for locations experiencing totality.


Potential paradox with the Big Rip and the Multiverse? I came up with this confusing theory a few years ago, but did not ask it on this site. If the Big Rip and the Multiverse theories are true, then when the Big Rip occurs, the universe's size will expand to infinity. However, the size of the Multiverse is also infinite (supposedly), and there are an infinite number of universes (also supposedly), then there is a paradox of infinities. I am still rather confused about this paradox, and is there a work-around to it, or is this completely invalid?

I don't know whether this is a real paradox. Infinities are tricky :) For example, we don't have just one infinity. Also, our knowledge about the universe is too narrow too understand the system of our universe (or multiverse) and we can only imagine the universes as some balls which are expanding into some other space = multiverse. If multiverse exists, it is probably not some space with some balls in it. There are a lot of dimensions that are present in here.
So, this isn't a paradox: we shouldn't think about the universes as some physical objects expanding. But we also don't know what to imagine. We just haven't figured it out. But it is probably a lot different from our current physics, math and imagination.


Why does kerr black holes have a ring shaped singularity rather than point? I read about ring singularity on Wikipedia and didn't quite understand the flow of liquids inside a rotating body causing a ring shaped singularity with zero thickness. Moreover is it specifically ring shaped and not torus shaped in any case?

Firstly do not confuse the event horizon with the singularity. Wikipedia gives the formulae for the Kerr metric. There are a number of places where this formula appears to break down because you appear to be dividing by zero, essentially whenever $\Sigma = 0$ or $\Delta = 0$, in the notation of that page,  corresponding to the ergosphere and the event horizon respectively. There are two event horizons and two ergospheres but they all have the topology of spheres, one inside another (with some touching at the poles). However those are not singularities in general because you can choose other coordinates for space and time where the division by zero goes away. However you can't do this at points on the equator of the inner ergosphere, which makes those points a ring singularity. It is a ring with zero thickness (at least until quantum effects come in) .There is a helpful picture

Flow of liquids doesn't really enter the picture here. One can think of the properties of this metric as a "flow" of space-time, but that's an analogy at best.

There seems to be considerable doubt about what really happens inside a rotating black hole, or even, given that no information can escape from the black hole, about whether this is even a meaningful question.


What other definitions for a planet were proposed? This article from the IAU states

The first draft proposal for the definition of a planet was debated vigorously by astronomers at the 2006 IAU General Assembly in Prague and a new version slowly took shape. This new version was more acceptable to the majority

What were the other proposals for the definition of a planet? What was changed between the first draft and the final version that was voted on?

The IAU website has this draft definition of "planet" and "plutons" page, published on 16 August 2006 in Prague. It says (emphasis mine):

The world's astronomers, under the auspices of the International
Astronomical Union (IAU), have concluded two years of work defining
the difference between "planets" and the smaller "solar system bodies"
such as comets and asteroids. If the definition is approved by the
astronomers gathered 14-25 August 2006 at the IAU General Assembly in
Prague, our Solar System will include 12 planets, with more to come:
eight classical planets that dominate the system, three planets in a
new and growing category of "plutons" - Pluto-like objects - and
Ceres. Pluto remains a planet and is the prototype for the new
category of "plutons."

The proposed 12 planets Solar System looked like this (credit: The International Astronomical Union/Martin Kornmesser):

The final press release, published on 24 August 2006, says:

The IAU members gathered at the 2006 General Assembly agreed that a
"planet" is defined as a celestial body that (a) is in orbit around
the Sun, (b) has sufficient mass for its self-gravity to overcome
rigid body forces so that it assumes a hydrostatic equilibrium (nearly
round) shape, and (c) has cleared the neighbourhood around its orbit.
This means that the Solar System consists of eight "planets" Mercury,
Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune. A new
distinct class of objects called "dwarf planets" was also decided. It
was agreed that "planets" and "dwarf planets" are two distinct classes
of objects. The first members of the "dwarf planet" category are
Ceres, Pluto and 2003 UB313 (temporary name).

Which looked like this (credit: The International Astronomical Union/Martin Kornmesser):

So in 8 days, the IAU switched from a 12 planets system, with a planet subcategory called "plutons", to a 8 planets system, with an dwarf planet category distinct from the planet category. How did that happen? It's probably something only a General Assembly attendant who witnessed the discussion would know. It must have been quite a debate!


Is it true that open clusters have $E > 0$ and globular clusters have $E < 0$? I just need to confirm or refute something I'm believing since a very long time, maybe without real evidence.

Can we distinguish stellar clusters by their total mechanical energy ?

Is it true that all "open" star clusters have a positive energy (i.e. more kinetic energy that potential energy), while all globular clusters are "closed" systems with negative total energy (i.e less kinetic energy than total potential energy) ?

So for example, the Peiades M45 are "open" : they have $E > 0$, while M4, M53 and M80 (globulars) have $E < 0$ ?  Is this true ?

Stars in an open star cluster are - according to wikipedia - "loosly bound by mutual gravity". Whereas a globular cluster is more dense and will be "less loosly" bound - to use the same terminology.

Since the stars in the open star cluster still are gravitationally bound, the total energy $E$ must be less than zero. This criterium that $E<0$ is always correct for (gravitationally or otherwise) bound systems for exactly the reason you stated, the kinetic energy is 'less positive' than the potential is negative. Otherwise the particles (stars in this case) would break free from the gravitational interactions with the other particles - and that's the complete opposite of a bound system.

However, the fate of most open star clusters is that it will fall apart. This can be due to

mass loss of the cluster or,
internal processes like a slingshot due to close encounters between constituent stars.

From the get-go these open clusters are quite unstable, so $E$ will be close to zero, but still negative ofcourse because it's a bound system. But it is instable enough for the following to happen. Young, bright stars radiate a lot of energy, and this means that the radiation pressure on the surronding cloud from which these stars are born is high. The light from these young stars can blow large amounts of mass in the form of this gas and dust out of the system - which has an influence on the total energy of the system! Because the mass of the system drops, the total energy rises (read: becomes less negative) - wich makes it easier for the constituent stars to reach escape velocity. If you don't like thinking in terms of energy, the escape velocity is $v_e = \sqrt{2GM/r}$, so $v_e \propto M $. If $M$ drops, so does the escape velocity.

The second bullet speaks for itself, I think.

You can deduce that $E$ has to be negative from a conservation viewpoint. When the stars in the cluster are born from the same cloud, this cloud already is gravitationally bound, so $E<0$, and unless some energy is lost at some point, the cloud will stay bound. But like I said and as you said, most of these clusters don't stay bound for very long, and eventually will evaporate.


How to find the Orbital Period of two Binary Planets around a Single Star? This is kind of a weird one. Bare with me! :)

Background: (Skip to the math below if you just want to answer questions)

In my favorite game, Elite: Dangerous, I "own", or occupy I guess, a system named "59 Virginis", based on the real life counterpart inside the Virgo Constellation.

The game itself houses a unique, very close to life-like system called "Stellar Forge", which uses as close to real life accuracy with physics as possible.

In this system, my own group's "Home" is the Earth-Like planet "59 Virginis 4", which is in a binary (tidally locked) orbit with 59 Virginis 3 (a class III gas giant).

I am trying to create an "Alien Calendar" of sorts, and a key component of calendars is the year, or how long it takes to rotate the sun once. This is semi-important, seeing as the home planet itself is titled on it's axis, meaning the Earth-Like will have seasons! Meaning having a yearly cycle of seasons would be awesome, and making a calendar would be lots of fun!

This is where I'm incredibly stuck. The game doesn't outright tell us how long it takes for these two objects to circle the sun, but it does give a lot of extras on possibilities of finding it!

I've tried a few different methods, but I don't think I know enough math to be able to find such a thing.

Here is the information we have currently:

59 Virginis 3 (the Gas Giant)

"distanceToArrival": 964 light-seconds, (this value changes based on where the planets are in binary orbit at the time of information update),
"earthMasses": 2027.9021,
"radius": 72755.136 km,
"orbitalPeriod": 80.53994791666666 d,
"semiMajorAxis":  5.898593314670755e-5 au,
"orbitalEccentricity": 0.114867,
"orbitalInclination": 0.071319°,
"argOfPeriapsis": 246.659607°,
"rotationalPeriod": 1.7743948929398148 d,
"rotationalPeriodTidallyLocked": false,
"axialTilt": -1.236412°

59 Virginis 4

"distanceToArrival": 989 light-seconds,
"earthMasses": 1.795525,
"radius": 7228.5435 km,
"orbitalPeriod":80.53994791666666 d,
"semiMajorAxis": 0.0666149271869282 au,
"orbitalEccentricity":0.114867,
"orbitalInclination":0.071319°,
"argOfPeriapsis":66.659615°,
"rotationalPeriod":80.57560763888888 d,
"rotationalPeriodTidallyLocked":false, (it's pretty damn close tho)
"axialTilt":-0.293082°,

The above information can be taken from here which is the most accurate version that can be given. If you want it in an easier to read format, click on the bodies 3 and 4 here

I know that the orbital period of a binary object is this:

$T= 2\pi\sqrt{\frac{a^3}{G \left(M_1 + M_2\right)}}$

Which only describes the orbital period of the two objects around eachother, right?

Ideas that have been thrown around:

Using Kepler's Third law, which states "The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit."
Using the two object's "distance to arrival" to find the distance between the two objects, then finding where on that line the center of mass is, and then using that center of mass point to find the distance to the star again from that point, then using that distance to find the orbit. (This sounds like the most reasonable from my point of view, but possibly the hardest)

Here is a really bad paint drawing of what I think the one above could look like:

Possibly finding the sidereal period somehow, or synodic period in relation to the giant and the sun?

If you've gotten this far, you've already helped me more than most! Any advice in relation of where to go next would me MUCH appreciated! Thank you!!

The next step after finding how long it's sun's orbital period is to find how many years to reset the whole cycle, kind of like a leap year, but with the binary orbits. That should be slightly easier, but I'm stuck on this for now! :)

I think that at the moment, all you have available for the pair's orbit around their star is a snapshot of their two distances; 964 and 989 seconds, or about 289 and 297 million kilometers. Your snapshot doesn't have any velocity information, so I think that there's no way to calculate the mass of the star from that data. Since the first planet is 2000 times heavier than the second, lets say that the pair's center of mass is 289 million km from the star.
However your second link states explicitly that the star is 1.0195 solar masses. That means that the standard gravitational parameter or $GM$ is about 1.352E+20 m^3/s^2, so using

$$T= 2\pi\sqrt{\frac{a^3}{GM}}$$

I get a year of about 972 days for the center of mass of the pair. Since the lighter planet oscillates by +/- 10 million km (its semi-major axis) every 80 days, that's a 7% amplitude change in distance to the star or roughly a 14% amplitude modulation of the brightness of the sunlight every 80 days, so that may have some impact on the weather for the smaller planet.

Interestingly, the smaller planet goes "around the world in 80 days" if we name the larger planet "world".


What do falsifiability and verifiability actually mean as a precondition of a scientific hypothesis? What do falsifiability and verifiability actually mean as a precondition of a scientific hypothesis?
I have been hearing from science guys that unlike philosophical hypothesis or theory, to be a scientific hypothesis an educated guess must have to be falsifiable and verifiable. But I cannot comprehend falsifiable and verifiable conditions for some scientific hypotheses especially in astronomy field. For example,
Couple of scientific hypotheses exists about the shape of the universe,

Universe is positively curved
Universe is negatively curved
Universe is flat

These are the well known scientific hypotheses as far as my knowledge (correct me if I am wrong). It seems to me, though I may be wrong, none of the above hypotheses are anyway falsifiable and verifiable because it is never possible for us to measure the geometry as the universe is expanding beyond our scope of study. For example, we can may be able to test on larger amount space as time goes on and technology advances, but to measure the actual shape of the whole universe we need to know theoretically the full picture of the universe which is not possible for us according to current understandings of science.
Now, there might be many such hypotheses which we cannot take under testable and verifiable condition yet we call them scientific hypothesis. My question is,

Is a scientific hypothesis really required to be falsifiable and verifiable? If so, what are the falsifiable and verifiable conditions for above mentioned hypotheses?

[I am hesitant to post this answer, as I am unclear that this is a question about science rather than philosophy as I said in a comment.  But I've written it now, so.]
First of all, all scientific theories rest on a mass of assumptions not all of which are testable.  For example if I'm doing astronomy I'm assuming that I'm not just living in a simulation, or that, in fact, anything outside my own mind exists at all.  I can't test those ideas, but I'm going to assume them or I can't do any astronomy because there's no point.
Without assumptions like that all you can do is maths: Fermat's last theorem is true whether or not Fermat or Andrew Wiles, exist, say.
But there are assumptions which are made at a level above these kind of things which also matter in astronomy.  For measuring the spatial curvature of the universe one is very important: the cosmological principle.

The cosmological principle.  Viewed on a sufficiently large scale, the properties of the universe are the same for all observers.

(From William C. Keel (2007) The Road to Galaxy Formation (2nd ed.), Springer-Praxis, via Wikipedia.)
Here an 'observer' doesn't mean 'someone on Earth', it means 'someone anywhere you like'.
The cosmological principle implies two things: that on large enough scales the universe is homogeneous (it looks the same everywhere) and isotropic (it looks the same in all directions).
The cosmological principle is not fully testable I think.  We can check that the universe looks isotropic here, and people do this.  If it's not then the cosmological principle fails.  But if it does look isotropic then the principle still may be false: perhaps we are in some special location where the universe looks isotropic.  If we could get far away (very far away) and test that the universe looks isotropic from at least two other places as well, then we could conclude that the part of the universe we've observed is also homogeneous I think.
But it could still be fooling us somehow: the cosmological principle is really just something we assume is true.  It's a very reasonable assumption, I think, because what it's really saying is that where we are is not special: the universe is not, in fact, built around us.  The cosmological principle is the inverse of the Ptolemaic model, where everything goes around the Earth, really, and even more it is an inverse of the 'nothing but me exists' model which I said I was going to rule out at the start.
Well, if we assume that the cosmological principle is true, then the bit of the universe we can observe is just the same as all the other bits of it on large enough scales.  So, if we assume that general relativity is a good description of the universe on large scales (that's a lot more testable) and probably some other things which are testable then if we can measure the spatial curvature (or equivalently, the matter density) of the part of the universe we can see, we can conclude that this holds for all of it.
That is the sense, I think, that claims about the testability of measurements of spatial curvature are true.


Why are quasars so far away that they couldn't be optically resolved in the 1950's? Wikipedia's Quasar includes the following in its introduction:

The term quasar originated as a contraction of quasi-stellar (star-like) radio source, because quasars were first identified during the 1950s as sources of radio-wave emission of unknown physical origin, and when identified in photographic images at visible wavelengths they resembled faint star-like points of light. High-resolution images of quasars, particularly from the Hubble Space Telescope, have demonstrated that quasars occur in the centers of galaxies, and that some host-galaxies are strongly interacting or merging galaxies.

So today instead of "quasi-stellar" we might simply say "unresolved". The light is star-like because the light from galaxies has a strong stellar component. (this second sentence is argued against in comments)
Why are quasars so far away that they required high resolution techniques (i.e. space telescopes or adaptive optics) to resolve them? Is it just that they are fairly rare and a roll of the dice resulted in none close enough to be resolved in older large telescopes of the 1950's, or are there some cosmological effects in play as well?

Quasars have a very low density at the present epoch, but were much more common in the past. Their co-moving space density peaks at redshifts of 2-3.

The evolving space density of bright quasars. From https://ned.ipac.caltech.edu/level5/Madau/Madau7.html.
The fact that the quasar population density evolves was one of the key arguments against steady state cosmology.


Parts of the Universe Shows Fewer Named Stars I'm becoming interested in the currently identified stars in the universe. There is a wide swath where much fewer named stars are shown. Why is this?
The image below is a mollweide projection of the universe, with the earth at its center. It shows about 230,000 named stars located at their right ascension/declination points.

The database used for the above was for Deep Sky Objects, not stars.
When I changed the database the correct one, then the following image is distribution of 120,000 named stars. Thanks for flagging my error.


Difference in time between sunset and moonrise on a full moon So as a Sailor I have experienced some times where sunset and moonrise are very close together (this happens when it's full moon) which is particularly beautiful. I assumed that this was always the case but looking more closely at time difference between sunset and moonrise they can change quite dramatically.

2019 (Porto-Vecchio France)
Date        sunset   moonrise  diff
04/20       20:08    21:24     1:16
05/19       20:39    21:20     0:41
06/18       20:59    22:05     1:06
07/17       20:54    21:31     0:37
08/16       20:21    21:12     0:51

As you can see there is some degree of variation, can someone explain why this happens and how I can find out where and when they are closest together?

Technically, the Moon is Full at a specific time. Therefore you need to check the time of sunset/moonrise OR sunrise/moonset that is closest to the time of Full Moon. Then you can more accurately compare the variation. Here is the list for 2019 created using the U.S. Naval Observatory website.
Longitude E9° 8', Latitude N41° 35'
All times in Universal Time (UT)
Date of Full Moon   -------- Sun ----------     -------- Moon --------  Moon-Sun
     Date    Time              Date    Time              Date    Time     Diff
 1/21/2019    5:16   Rise  1/21/2019    6:45     Set 1/21/2019    6:54   +0:09
 2/19/2019   15:53    Set  2/19/2019   17:01    Rise 2/19/2019   16:55   -0:06
 3/21/2019    1:43   Rise  3/21/2019    5:26     Set 3/21/2019    5:58   +0:32
 4/19/2019   11:12    Set  4/19/2019   18:08    Rise 4/19/2019   18:14   +0:06
 5/18/2019   21:11    Set  5/18/2019   18:39    Rise 5/18/2019   18:14   -0:25
 6/17/2019    8:31   Rise  6/17/2019    3:49     Set 6/17/2019    3:54   +0:05
 7/16/2019   21:38    Set  7/16/2019   18:56    Rise 7/16/2019   18:49   -0:07
 8/15/2019   12:29   Rise  8/15/2019    4:32     Set 8/15/2019    4:09   -0:23
 9/14/2019    4:33   Rise  9/14/2019    5:02     Set 9/14/2019    4:59   -0:03
10/13/2019   21:08    Set 10/13/2019   16:46    Rise10/13/2019   17:02   +0:16
11/12/2019   13:34    Set 11/12/2019   16:07    Rise11/12/2019   16:25   +0:18
12/12/2019    5:12   Rise 12/12/2019    6:40     Set12/12/2019    6:42   +0:02

The difference is much smaller than your original list.
Why this happens
MCG and Carl Witthoft explained why it happens, namely because:

If the moonrise/set is not the same time as the exact Full Moon, then moonrise/set  occurs earlier or later than the sunset/rise.
Because the Moon passes above and below the ecliptic (the path of the Sun in the sky), the Full Moon is rarely directly opposite from the Sun. Therefore, it can rise earlier or later than the Sun even when exactly "Full".
Even if the Full Moon were exactly opposite from the Sun (180 degrees apart, and in a total lunar eclipse), there would still be a difference between the time of sunset/moonrise. Those times are when the limb of the sun/moon is visible, and the limbs are closer together than 180 degrees apart. Also, atmospheric refraction shrinks the separation a small amount.

Where and when are they closest?
This is more involved but can be reasoned using the following figure. The Sun is in the direction of 1, so the ring of sunset-sunrise is the circle A-A. The Full Moon is in the direction of 2, so the ring of moonrise-moonset is the circle B-B. Where these two circles cross at C is where the sunset and moonrise (or sunrise and moonset) occur at the same time. The highlighted area is where the Moon and Sun are both above the horizon at the same time. Note that this figure is highly exaggerated for illustrative purposes.


What orbital period would produce one New Moon (and one Full Moon) each year? What other effects would this produce? I am interested to know whether it's possible for the moon to orbit the earth at such a rate that we would see only one New Moon and one Full Moon a year. If so, what other effects would we experience?
I at first assumed this would be a one-year period, and technically I think that would produce one New Moon, but that single New Moon would last all year (i.e., you would never see the moon). So I'm interested to know if there is an orbital period that would give us one New Moon and one Full Moon, as seen from our earth, each year.
As I hold my hands up in front of my face and experimentally rotate them around each other, I want to say a six-month period would produce this effect, but I'm not sure that's correct, so I'd appreciate a word from anyone who actually knows what they're talking about. Thanks!
(Apologies if this has been asked before; I wasn't able to find it.)

This would naturally be achieved by a quasi-satellite. These are objects with a 1:1 orbital resonance with a different eccentricity to the planet. From the perspective of the planet, this leads to the quasi-satellite appearing to travel around the planet once a year in a retrograde direction. Starting with the quasi-satellite at its maximum distance from the Sun, this would be at full phase. Six months later, the quasi-satellite would be between Earth and the Sun, so would be at new phase. The diagrams below show an idealised version of the configuration based on Keplerian orbits in a non-rotating frame and in a frame where the Earth and the Sun are kept in a fixed position.

Real systems will evolve dynamically due to the gravitational interaction between the planet and the quasi-satellite, leading to precession and oscillations of the alignment of the orbits.
Earth currently has several known quasi-satellites, the closest of which is (469219) Kamoʻoalewa. The known quasi-satellites in our Solar System do not have long lifetimes: they are relatively recent captures into the quasi-satellite orbit that will eventually escape. Nevertheless, there are some conditions under which quasi-satellites can be stable over long periods, and perhaps some exoplanetary systems have more favourable conditions for quasi-satellites than ours does.


Birth and Evolution of Millisecond Pulsars How are millisecond pulsars formed, and what is the evolutionary track of a millisecond pulsar?

The standard model for the formation of millisecond pulsars is the recycling scenario.

The starting point of this model is a stellar binary system consisting of a neutron star and a secondary 'regular' star, where the neutron star is a normal pulsar with a rotational period of roughly $\sim7$ seconds.

Because a neutron star is such a compact object, its gravitational influence is very large ($F_g \propto M/R^2$). If the two stars are close enough, the outer layers of gas may be pulled of the companion star and flow towards the neutron star. All this matter will settle into an disk structure around the neutron star and start orbiting faster and faster as it draws closer to the surface ($v_{orbit} = \sqrt{GM/R}$).

By the time this matter touches down on the neutron star surface its orbital period can be as fast as a millisecond, certainly significantly faster than the neutron star. This bit of motion (or more precisely, the angular momentum) is then transferred to the neutron star (this process is called accretion), causing it to rotate slightly faster (spin-up).

After some $0.1\sim0.2 M_{\odot}$ (where $M_{\odot}= 2 \times 10^{30}$ kg gives the mass of the Sun) of material is transferred to the neutron star, its rotational period will have decreased from a few second down to milliseconds.

The transfer of this much matter takes a very long time, roughly $10^8$ to $10^9$ years. After this time, and with all the redistribution of mass, the binary system is expected to widen which causes the mass transfer to stop. As the region around the neutron star clear up, the radio emission mechanism turns back on (this mechanism does not work during the accretion stage) and we are left with a millisecond pulsar.

From this point on the neutron star will behave like a regular pulsar: over the course of a very long time it will convert its rotational energy in radio emission and slow down until it becomes to slow to act as a pulsar.

As for the evidence for this model; such binary systems containing an accreting neutron star emit massive amounts of X-ray emission (hence called X-ray binaries) and have been known to exist since the 60's, long before the first millisecond radio pulsar was discovered (in 1982). This entire scenario was proposed quickly after, but could not be confirmed for a very long time.

It was only in the late 90's that a neutron star was found whose X-ray emission was pulsed at millisecond periods (see this, and this), confirming that these neutron stars are indeed spinning up. The final piece of evidence; a millisecond pulsar which switches back and forth between X-ray and radio emission was only found very recently.


Is the diamond ring effect visible in a 40mm hydrogen alpha telescope? I'm going to Nebraska for the 2017 eclipse, and am weighing my options. I'm considering whether to pack my 40mm Coronado PST or to make a Baader AstroSolar filter for my 66mm William Optics apo. I've used the PST for transits and annular eclipses past, but have never seen a total solar eclipse before.

Can I see the diamond ring effect (aka Baily's beads) with my PST, or should I stick with white light?

Experimental answer, at least with a 40mm PST: No.

Don't worry, I didn't miss much of it :)


Sudden Effects of Sun Disappearing It takes 8 minutes for the sun's light to reach earth. If the sun were to instantly disappear, what would happen? I imagine that Earth would fly away on a path tangential to its orbit. What would happen to our heat and light during those first 8 minutes after the sun's disappearance?

Let's assume that every particle in the Sun quantum tunneled far away from us (there's a near 0% chance this will happen, but let's just assume). For the first eight minutes, nothing would change from our point of view. Gravitational waves propagate at lightspeed, so we would continue to orbit the Sun and see it there.

After eight minutes, however, the lights would go out. The Moon will no longer shine upon us (it receives its light from the Sun), so our only source of light will be that of the celestial objects. Mercury and Venus will have seemingly disappeared from our optical telescopes — they will no longer be illuminated by the Sun. It would take another 30 minutes to an hour for the reflected light of Jupiter to shut off, and soon all the objects in the Solar System will no longer be visible to optical telescopes.

According to Ahad (2004), the illumination by the rest of the Milky Way would be approximately 1/300 of that of a Full Moon. So we might be able to see around thanks to the night sky and our own man-made technologies. However, most plants will not be able to undergo photosynthesis and will die.

At this point, we would be traveling tangentially to our orbit around the Sun, as you mention. We would just keep going into the abyss; it's very unlikely the gravity of any object will pull us in. However, things will get quite cold. The Earth would radiate heat away at a moderate rate, and temperatures would be unsustainable in most parts of the world after a year.

Luckily, the Earth has a nice and toasty core that could provide life-sustaining heat in some places (like Yellowstone), so humans that migrate over there would likely be better off.


Are stars near the ecliptic the only stars from which Earth would appear to transit the Sun? The majority of exoplanets we have discovered have been found by the transit method, in which a planet passes in front of its parent star and by analyzing the light curve we can deduct its characteristics.
Say there was intelligent life around a nearby star, could they only see the Earth transit in front of the Sun if they were located around the ecliptic?

Yes. Heller and Pudritz 2016 define an Earth transit zone (ETZ) within 0.26° of the ecliptic and a restricted ETZ (rETZ) half as wide.
They also list 82 Sun-like stars in the rETZ and within 1 kiloparsec of the Sun.
Sheikh et al. 2020 report on a search for artificial radio signals from 20 of those.


Is it possible to image a meteor so the rock is visible? I was wondering what one would see if they were to look through a telescope tracking a meteor as it shot through the sky. Technology aside, would one be able to see the object tumbling or would it always be obscured by the light/dust/plasma?

Basically would there be any advantage to getting a high shutter-speed video with high magnification tracking the leading edge of a meteor or would it not offer any greater detail than just a bright streak in the sky?

Meteors are very small, typical ones are the size of a grain of sand, and bright fireballs are only a few grams. So they are nowhere near as big as "rocks" and there is no way to image them.

https://en.wikipedia.org/wiki/Meteoroid

http://curious.astro.cornell.edu/physics/73-our-solar-system/comets-meteors-and-asteroids/meteors/303-what-is-the-typical-size-of-a-visible-shooting-star-intermediate

https://www.amsmeteors.org/fireballs/faqf/

But let's say you could fly next to a shooting star in a UFO and image it that way. The next problem you would face is that the meteor is incandescent and and it would be difficult to get anything but an overexposed blob in a regular camera.


Given a star's ra, dec how can I compute one pair of lat, long and time where the star would be at zenith? As the questions says, I have the celestial coordinates of a star(ra, dec). My trigonometry is not so great so I haven't managed to figure out a solution to this. I want to find a place (lat, long) and a time when the star will be right above the computed point?

I.e. star coordinates: 7h 23m 23.97s -13° 25' 2.64", where would be the point(lat, long) and time where the star would be right above?

Latitude is the easiest; it's the same as the star's declination: -13° 25' 2.64".

Longitude and time depend on each other. A star with right ascension 7h 23m 23.97s will be at the zenith point at 7h 23m 23.97s local sidereal time (by definition). You need to convert sidereal time to local time (at a given day), e.g. with the script described here. On September 23rd (or whatever day the Fall equinox falls, thanks @MikeG) on the Greenwich meridian, this will happen at 7:23am UTC.


Are red giants stationary or do they move? A friend and I are having a friendly chat about this. They are stars so I'm claiming they still move in orbit.

Everything moves.  Entire galaxies move.  Our galaxy is moving towards Andromeda at over 100 km/s and our entire local group is moving towards the great attractor even faster than that.

When a less massive object orbits a more massive one, both move, but the less massive one moves more.  Often the calculation is simplified to make the central body unmoving, but that's a simplification and not accurate.

Our Milkyway has a center of mass with a giant black hole in that region.   If anything in our galaxy didn't move, it would slowly fall into the center.   Stars don't move like that, they move around each other while orbiting within the Milky-Way at high speeds.

Movement, I should add, is defined from a certain point of view, so it's always possible to pick one thing and say "this isn't moving" and that would be correct from that thing's point of view, but that object still sits in spacetime (relatively speaking), and space time defines a force, which (shockingly) moves things, so even if you say "Lets say this object isn't moving", the unevenness of space time will move it.

So, everything moves.   That's a universal law.


Is Jupiter made entirely out of gas? I heard that Jupiter is made out of gas. But in school I learned that Jupiter has gravity which is 2.5 times that of Earth (Gravity that can tear apart a comet) and gravity is proportional to mass.

So if Jupiter is made out of gas alone, how does it have such high gravity and maintain so many moons around it?

It doesn't matter if the body is made of gas, rocks, liquid or plasma, the four states of matter all have mass. So, as we know, mass create a gravitational field, and the more mass the stronger the gravity - and Jupiter has 317x Earth mass.


How has the average distance between asteroids in the asteroid belt changed over time? Just about every webpage related to asteroids states that, on average, asteroids are about 600,000 miles apart from one another.  Have planetary scientists hazarded a guess at how close they might have been 4 billion years ago?  2 billion years ago?  200 million years?  I'm not looking for a particular age range, just an understanding of the trend over time.  Unfortunately, I'm having trouble looking up an answer to this question on my own because the keywords I've thought of keep grabbing other unrelated things
Thank you!

The asteroids formed roughly where they are now, from matter in the sun's protoplanetary disc that didn't form into a larger body (at least partly under the influence of Jupiter's gravity)
Since early in the solar system's story, the asteroids have been stable. Most asteroids haven't changed significantly in 4.5 billion years.  And 4.5 billion years ago the separation between asteroids was comparable to how it is now.


Are there small voids of 10 - 20 megaparsecs also within, and not only between superclusters? I sincerely apologize if this question is stupid, however after all the articles I read so far, I could not confirm this assumption, which I think is actually a fact, but I'm not completely confident.

I'm writing on the topic of proportions of the universe, and in one of the statements I claim the following:

"Just one small void inside a supercluster of galaxies can span across 10 to 20 megaparsecs (32 to 65 million light-years ) in diameter."

Let me state that I understand that a void within a galaxy supercluster may not necessarily be completely surrounded by galaxies of that supercluster, and can extend far out from the supercluster it partially occupies, so to clarify what I mean by my question on a practical example:

If you were to draw a straight line across a map of a galaxy supercluster, can that line start inside a group of galaxies of that supercluster, go across a void of 10 - 20 megaparsecs, and end inside another group of galaxies of that same supercluster?

(I am basing my "small" classification of a cosmic void 10 - 20 megaparsecs in diameter on the Wikipedia article "Void (Astronomy)" https://en.wikipedia.org/wiki/Void_(astronomy) in which it says:

" ...Voids typically have a diameter of 10 to 100 megaparsecs; particularly large voids, defined by the absence of rich superclusters, are sometimes called supervoids. ..."

My only goal is basically to find out whether I would be wrong or not if I state that small voids of 10 - 20 megaparsecs, really actually exist within, and not only between superclusters.

For any eventual help on this topic, I would be sincerely grateful.

Kind regards

My only goal is basically to find out whether I would be wrong or not if I state that small voids of 10 - 20 megaparsecs, really actually exist within, and not only between superclusters.

The largest structure is called a filament.

We are in the supercluster Laniakea, in the smaller supercluster Virgo, in the Milky Way galaxy, in the Solar system, on Earth.

It looks like this:

$$\text{Click to zoom in - 23,624 × 2,953 pixels}$$
$$\text{Laniakea}$$

Here we are:

Wikipedia says:

Distribution: cosmic voids and sheets

Research has tried to understand the way superclusters are arranged in space. Maps are used to display the positions of 1.6 million galaxies. Three-dimensional maps are used to further understand the positions of these superclusters. To map them three-dimensionally, the position of the galaxy in the sky as well as the galaxy's redshift are used for calculation. The galaxy's redshift is used with Hubble's law to determine its position in three-dimensional space.

It was discovered from those maps that superclusters of galaxies are not spread uniformly across the universe but they seem to lie along filaments. Maps reveal huge voids where there are extremely few galaxies. Some dim galaxies or hydrogen clouds can be found in some voids, but most galaxies are found in sheets between the voids. The voids themselves are often spherical but the superclusters are not. They can range from being 100 million to 400 million light-years [that's 30.660139381 to 122.64055752 megaparsec] in diameter. The pattern of sheets and voids contains information about how galaxy clusters formed in the early universe.

There is a sponge analogy used often that compares a sponge to the pattern of clusters of galaxies in the universe – the holes are the voids and the other parts are the locations of the superclusters.".

So, yes - there are small and large voids.

See also this Wikipedia page:

Galaxy distribution

The number density of galaxies in the LS falls off with the square of the distance from its center near the Virgo Cluster, suggesting that this cluster is not randomly located. Overall, the vast majority of the luminous galaxies (less than absolute magnitude −13) are concentrated in a small number of clouds (groups of galaxy clusters). Ninety-eight percent can be found in the following 11 clouds (given in decreasing order of number of luminous galaxies): Canes Venatici, Virgo Cluster, Virgo II (southern extension), Leo II, Virgo III, Crater (NGC 3672), Leo I, Leo Minor (NGC 2841), Draco (NGC 5907), Antlia (NGC 2997) and NGC 5643. Of the luminous galaxies located in the disk, one third are in the Virgo Cluster, while the remainder are found in the Canes Venatici Cloud and Virgo II Cloud, plus the somewhat insignificant NGC 5643 Group. The luminous galaxies in the halo are also concentrated in a small number of clouds (94% in 7 clouds). This distribution indicates that "most of the volume of the supergalactic plane is a great void." A helpful analogy that matches the observed distribution is that of soap bubbles. Flattish clusters and superclusters are found at the intersection of bubbles, which are large, roughly spherical (on the order of 20–60 Mpc in diameter) voids in space. Long filamentary structures seem to predominate. An example of this is the Hydra-Centaurus Supercluster, the nearest supercluster to the LS, which starts at a distance of roughly 30 Mpc and extends to 60 Mpc.


A Sky Map of Gravity - what would it look like? PREAMBLE

Sky Maps exist which show the distribution and intensity of electromagnetic radiation received on Earth at various wavelengths.  An obvious form is a map of the stars (Star Chart).

Maps have also been made for the distribution/intensity of Cosmic Microwave Background Radiation and "to visualize the X-ray Universe".

I find it interesting to consider the appearance of a Sky Map which showed the variation in intensity of the Gravitational Attraction Force (as currently experienced on Earth) attributable to masses lying in different directions (looking radially outwards from the Earth).

Essentially (if we start with a Newtonian Model) this would be a map which indicated the magnitude of the sum of "mass/distance_squared" for all the masses falling inside each particular pixel of the map.  This then raises the questions of cosmological nature e.g. "how far do gravitational influences extend away from the Earth", along the lines of Olbers Paradox and "how was matter distributed through the Universe through time" - see The Cosmological Principle.

QUESTION

To bring this into the form of a question that can be answered succinctly, my question is: "What would a SkyMap of Radial Gravitational Force, looking outwards from Earth, look like (excluding that due to masses in the Solar System and Milky Way galaxy)."

APPENDIX

Calculation of Average Background Gravitational Force from Visible Universe

(Assuming, for simplicity, a static, spherical, universe of uniformly-distributed ordinary matter with Newtonian gravity and a test mass at the centre of the universe).

$\rho$ Average density of visible matter  $=3*10^{-28} Kg/m^{3}$

$R$ Radius of visible universe   $=1.7*10^{26} m$

Note:- values of $\rho$ and $R$ are obtained from here.

$G$ Universal Gravitational Constant  $=6.673*10^{-11} Nm^{2}Kg^{-2}$

$f$ Fraction of sky represented by $3.6 * 3.6$ degree rectangular pixel $=1/(50*100)$ $= 0.0002$.

The formula for Force per unit mass $F_m$ is obtained by integrating (over the interval $0,R$) the incremental force per unit mass $F_i/m = G.M_i/R_i^2)$.

Here $M_i$ is the incremental mass contained within the incremental volume $V_i$ of the spherical shell of radius $R_i=R_{i-1}+\text{d}R$ and given by $M_i = \rho.V_i \approx \rho.4\pi.R_i^2.\text{d}R$.

Thus the total force/unit mass is given by:-

$$ F_m = G\int_0^R \frac{\rho.4\pi.R_i^2}{R_i^2}\text{d}R  = 4\pi G \rho R $$.

And so the force per unit target mass per pixel is given by:-

$$F_{m,f} =4\pi G\rho R f  = 0.86*10^{-14} NKg^{-1}.$$

The rough answer is: just like the sun makes it hard or impossible to see planets and stars during the day, it dominates the gravity sky.

But there are interesting patterns there if we view the sky using a logarithmic scale.

The above plots are a combination of the gravitational force of the sun, moon and planets, the stars in the Hipparchos catalog, the gravitational field of the Milky Way, and the nearby galaxies. Each component have a somewhat different effect.

The solar system

The force from the sun is shockingly big: 0.0059 N (per kilogram - I will omit that henceforth). At first this might appear impossible since we never notice the force, but remember that the setup for the question is a fixed point in space rather than the free-falling orbit of Earth. Because of the orbit this force exactly cancels.

Note that for spherical objects we only need to calculate the force from the direction of their centre of mass, not for their full extent. Hence regardless of what resolution I use for my map planets and stars will only change one box.

The force from the moon is $3.3198\cdot 10^{-5}$ N, a factor of 179 less. (Why do we experience tides from the moon more than from the sun? Because the gradient for the gravitational field is steeper for the nearby moon than for the remote sun). The other planets have much less effect, from $10^{-6.8}$ N for Jupiter down to $10^{-9.5}$ N for Neptune (and Pluto at $10^{-13.4}$).

For positions I used the Matlab Aerospace toolbox planetEphemeris function. Incidentally, I might well have messed up some of the many different coordinate frames here, so don't use these maps to navigate. The important thing for this question is the magnitudes of the forces anyway.

Stars

I added the field of the stars in the Hipparchos catalog, but they do not show up even on my logarithmic map because they are overpowered by the galaxy at large. If we just plot their contributions to different boxes we get the following map:

Note that the peak force in each box is on the order of $10^{-14}$ N: even Pluto matters more than nearby stars. This map also shows an annoying problem: if we just sum the contributions in each 3.6 times 3.6 degree rectangle in the coordinate system the polar rectangles will be narrow and have smaller area; hence there are fewer stars included and less force. Dividing by $\cos(\delta)$ removes this bias and shows that the stars indeed pull on us roughly equally in all directions.

(I used a rough estimate of their masses: assuming them all to be on the main sequence I calculated their luminosity from their absolute magnitude, and then estimated the mass as $M\approx M_\odot (L/L_\odot)^{1/3.5}$.)

The Milky Way

I modelled the Milky Way using the approximate values and formulas in Carroll & Ostlie (p. 884), essentially a neutral gas disk, a thin disk, a thick disk, a central bulge, a stellar halo plus a dark matter halo. The disks for example have density scaling as $$\rho(r,z)=\rho_0 e^{-r/h_r}e^{-|z|/h_z}$$ while the dark matter halo has the Navarro-Frenck-White profile $$\rho(r)=\frac{\rho_0}{\frac{r}{R_s}\left(1+\frac{r}{R_s}\right)^2}.$$ I integrate these functions along a ray extending from the part of the sky I am interested in. Note that the integral of the force will be $$F=\int_0^\infty \frac{G\rho(\mathbf{x}_0 + \mathbf{d} r)}{r^2} r^2 dr = G \int_0^\infty \rho(\mathbf{x}_0 + \mathbf{d} r) dr$$ since it is really an integral of an ever widening pyramid of area $r^2$: the total force is only dependent on the total amount of mass there. Since these density fields change slowly I ignore lateral variations. Again there should be a correction for declination, but it does not make a great difference in the picture.

The end result is a fair amount of force, especially along the galactic plane. This causes "galactic tides" in stellar orbits and can sometimes rival the force from remote planets in the solar system. Again, we do not notice most of the gravitational field because the sun is free-falling in the galaxy.

Sag A* does not show up in this image because it is not very massive compared to the full galaxy. The force is just $9.3678\cdot 10^{-15}$ N.

Nearby galaxies

I used the NEARGALCAT catalogue to estimate the effect of nearby galaxies. This is in many ways the most dodgy estimate of all. While we know positions well we have weak data on actual masses. I used the estimated masses whenever they existed, and a linear fit between angular extent and mass derived from those to impute masses for galaxies lacking values. I am a bit uncertain if the mass estimates are just photometric or include dark matter; if not including DM, they should be an order of magnitude higher.

Typical forces are on the order of $10^{-19}$ N, although heavier galaxies might pull us with $10^{-15}$ N. They are hence on par with the nearby stars: totally overpowered by the galactic gravitational field and the solar system.

Technically, since galaxies are extended non-spherical objects, I should integrate over areas here. But there are limits to my energy.

In short, the gravity sky has nontrivial patterns if one has gravity-eyes with a logarithmic response.


What is the farthest we can see into the universe? Is there a limit in which the mass in space obscure the ability to detect anything farther even as technology progresses?

Related:

Can the range of the observable universe be extended through an intermediary?
Which galaxy is receding from the Milky Way the fastest? What is known of the mechanism behind its recession?

It depends. In some areas, far-away stuff is obscured by obstacles in the foreground. In other places, we can see almost all the way to the earliest (and furthest-away) galaxies, like in the Hubble Deep Field image:

The very earliest objects have so much redshift they disappear out of view for visible-light telescopes. That's where the JWST comes in: because it's an infrared telescope, it can capture objects that have too much redshift to be visible in a visible-light telescope.

Advances help us see more in other ways too. Gaia data revealed some small nearby galaxies which are mostly obscured by our own galaxy.


Would it be possible to detect a magnetic field around an exoplanet? Of course, we can't fly a magnetometer next to an exoplanet to measure the magnetic field, but might it be possible to find indirect evidence (e.g. polar auroras) of a magnetic field generated by an exoplanet? Has this already been done?

You usually look for electronic synchrotron emission in the radio, which corresponds to the magnetic field strength expected for giant planets. There has been a single detection claimed with LOFAR by Turner et al., (2020) around $\tau$ Bootes, but it's very noisy and hence disputed.
Other effects like Zeeman-splitting have been considered for detection, but require clean, high-resolution signals from planets. So far this is too difficult to perform due to stellar contamination.
Magnetic field searches around brown dwarves have been more successful, but those are not planets, of course.