Are the stars distributed in uniform distribution, on the celestial dome, with respect to brightness? Has there ever been a statistical analysis of the distribution of stars in the sky (on the surface of the celstial dome), by brightness? I want to know if they are uniformly distributed.

For example, all stars with Apparent Magnitude from 1.0 to 2.0. Are these uniformly distributed? Then all stars from 2.0 to 3.0, etc.

EDIT: To avoid bias for the Milky Way, there should be some cutoff magnitude. I don't know exactly what's the best cutoff, but I would guess somewhere around 4.0 to 5.0 apparent magnitude. If the cutoff is 4.0, then that gives us 3 tiers to look at (1 to 2, 2 to 3, 3 to 4). Are each of those tiers uniformly distributed?

Btw, the "obvious" distribution of theta from 0 to 2pi and phi from 0 to pi will NOT produce a uniform distribution. That will produce clustering near the poles. Wolfram Alpha explains it.)

I did google around for this. The results are all about star clusters or binaries, not the actual starfield of the night sky, or sometimes about how to program a random set of stars on the sphere.

That's not what I want. I'm looking for a statistical analysis on our own starfield. Has it ever been done before? What were the results?

Naked eye stars are not distributed uniformly in the sky. That is because the median naked eye star is at a distance of 440 light years, and this is far enough away that some of the details of Galactic structure start to become apparent. Most importantly, the density of stars increase towards the Galactic midplane and has a scale height of a few hundred light years and thus there is an overdensity of stars towards the Galactic plane.

I did an analysis of this in https://astronomy.stackexchange.com/a/10260/2531 , but the important plot is repeated below. It shows the normalised density of stars (per square degree) as a function of Galactic latitude. You can see the peak at low Galactic latitude and even work out that our viewing point is probably slightly above the Galactic plane, given that the peak is at about $-10^{\circ}$. The peak is still there in 3rd magnitude stars. You could repeat for a sample limited to very bright stars, but you run into number statistics trouble.

The distribution of bright stars with Galactic latitude.

EDIT: Extra mile time. Below I show the Hipparcos data, divided into 4 apparent magnitude bins. Points are plotted on a RA vs Dec Aitoff projection. Superimposed in red is a Galactic coordinate grid. There are too few stars with $V<2$ to form much of an opinion, but for fainter stars, a ring-like structure, coincident with the Galactic coordinates equator (i.e. Galactic latitude of zero, or just below, as I showed in the 1D projection above) is clearly seen.


Habitable Planet around Red Dwarf Is there the possibility for a habitable Planet around a red dwarf? It is not required that the whole surface is habitable, but it shouldn't be a tiny spot.

After reading a bit I assume the planet would be tidally locked to the star, with the side that faces the sun looking like an archery-target (from center to border: scorching, hot, warm, cold, freezing).

Also from space it would look like a Rugby Ball (exaggerated) pointed at the star.

But would such a planet be stable enough to develop and support life more complex than the simplest forms?

I do not need a specific star, but the general answer if this is possible, however unlikely (though a value how unlikely would be nice).

Well, it's a relatively new field of study so I don't think there's any exact answers, but I've read up on this a bit out of interest.  Red Dwarfs have 2 main disadvantages.   One, which you've mentioned.  They're relatively cold compared to our star so the planet would need to be comparatively close and therefor, probobly tidally locked.   The other problem is red dwarfs tend to have very active sunspots compared to our sun, so the planet would need a crazy strong magnetic field to protect the planet.   Neither of those are "deal breakers" but perhaps less likely than a sun-like star.

A tidally locked planet wouldn't necessarily have a scorching hot side and a freezing cold side.  Modeling suggests that temperature would circulate.   If it had extensive oceans on the tidally locked side, for example, that could create a cloud cover and not cause intense heat.   You could also have the sun side be far enough away to be temperate and the star-side icy, or a livable section in a ring around the planet.

There's also the possibility that a tidally locked planet could have winds that we would find very difficult to adapt to.

I think you'd ideally have the sun side be the habitable side for photosynthesis.

On the flip side, if the star is too large, the star's life would probobly be too short for the planet to cool down enough to be habitable for us.  The life of a star is the inverse cube of the size, so a star twice the size of our sun would only have a main sequence of 1.2-1.3 billion years.  When you get a bit bigger than that, there's not enough time for a planet to cool off and a solar system to get past it's heavy bombardment period.

So I'd give a red dwarf a better chance of having a habitable planet than a sun 2.5 times or larger than our sun.   That, and red-dwarfs are the most common stars in the galaxy, so I think there's a probability that some red dwarfs could have habitable planets.


What is the closest galaxy? I think that the galaxy closest to the Earth, that is, to the Milky Way, is the nebula called Andromeda, which is 2.5 million light years away. Despite this great distance, M31 is visible, and very pretty. Are there any galaxies closer to the Milky Way than the Andromeda galaxy?

At last 80 nearby galaxies are believed to be part of a small cluster called the Local Group.
The three largest members of the Local Group, have their own system of satellite galaxies.
Over 50 galaxies are satellite galaxies of the Milky Way Galaxy and are closer to it than the Andromeda Galaxy is.
https://en.wikipedia.org/wiki/List_of_nearest_galaxies
The Andromeda Galaxy is actually number 86 in that list of nearest galaxies.
The Small Magellanic Cloud is number 23 and the large Magellanic Cloud is number 19.
The nearest galaxy on the list is the Canis Major Dwarf Galaxy, altough its status as a galaxy is soemewhat disputed.
If the Canis Major Dwarf Galaxy is not actually a galaxy, the next galaxy on the list would become the nearest galaxy. That is Draco II.


Is it possible to determine an observer's position on Earth from a photo of the night sky and the time of the shot? To my knowledge, this question can be related to celestial navigation on one hand and to attitude (or orientation) determination from stars direction on the other hand. However the combination of this two methods does not solve the problem, so I am wondering if there is a different approach.

In celestial navigation, the altitude of the stars (elevation wrt to the horizon) at a given time can be used to determine the observer's position. The limit of this method is that the horizon must be visible, which is not necessarly the case in a photo of the night sky.

An image of the night sky can also provide information about the attitude (orientation) of the camera wrt to an inertial frame. The measured direction of a few stars (from their position on the image) and their known direction in an inertial frame give an estimate of the attitude. This method is used in devices called star trackers onboard satellites for instance.

The combination of this 2 approaches fails to solve the problem because the information about the horizon is still missing. Indeed, the orientation is found with respect to the inertial frame, not the local frame, and thus the horizon is still unknown.

What can you tell from a picture of some stars?

At the very least, you need a recognisable asterism

If the exposure is too low or the light pollution too high to identify unambiguously an asterism of three or more stars, you're simply not going to be able to tell much from the photo.

An asterism on its own can only tell you where the picture wasn't shot.

If a star in a picture has a declination of $60^o$, you can rule out all shooting locations south of $-30^o$ latitude. A star like Polaris can only really be seen in the northern hemisphere, while Sirius can be seen everywhere south of Svalbard.

An asterism plus the horizon is sufficient for determining latitude

If the edge of land in the picture is sufficiently close to being the actual horizon (and not a range of mountains) and it is present alongside a recognisable asterism in the photo, you can calculate the latitude. This is true for any magnification, as you can use the angular distance between stars (a known quantity) to measure the angle to the horizon. I can't think of any special locations where you wouldn't be able to determine the latitude from the stars + horizon.

I haven't really done the math for how accurately one would be able to determine the latitude, but here are some considerations: There are two sources of error, the position of the stars and the position of the horizon. The star image may spread out over several pixels, and if the picture was shot through a standard camera and not a telescope, there would be an error, at a wild guess, of around ten arcminutes. The horizon can appear lower than $0^o$ depending on your distance above ground, but the effect is small. Looking out from Mt Everest, the horizon will appear a twentieth of a degree lower down than it ought to. I believe the horizon's exact position can also be obscured through atmospheric refraction.

All in all, the error in the angle between an identified star and the horizon equals the error in latitude of your identified position. For a 10 arcminute error, that corresponds to ten arcminutes of latitude, which is about 18 kilometres.

For longitude, you need global, not local time

If you only have local time, the photo could have been taken anywhere along that latitude. If you have some form of global time (UTC, GMT, whatever, it doesn't matter which as long as you know what it is) you can figure out longitude from there.


Do the actual false colours in the M87 black hole picture convey information? As for the title. The picture could have been plotted in grey scale, B&W, whatever couple of colours, or perhaps using a window of wavelength in order to show shift of the Doppler/relativistic Doppler kind.

Discarding the obvious dark, shall the picture be watch at merely in term of brightness, or considering yellow and red as well?

And independently, why those colours have been choose?  Just an aesthetic and suggestive choice, or that should be the apparent image as seen by eyes if we could go there?

I resume: is the picture a radio waves intensity or brightness map, or conversely a kind of "as it would be" picture in the Vis (modelled from the behaviour of the collected radio waves and theories)?

See this article recently published by the EHT collaboration describing how they generated the image. Specifically, chapter 5 of that article describes the origin of the image. To quote the article (specifically the caption of figure 3):

  The image is shown in units of brightness temperature, ${T}_{{\rm{b}}}=S{\lambda }^{2}/2{k}_{{\rm{B}}}{\rm{\Omega }}$, where S is the flux density, λ is the observing wavelength, $k_{B}$ is the Boltzmann constant, and Ω is the solid angle of the resolution element.

The false colors in the image convey the surface T brightness (a sort of measurement of the intensity or flux from that area) of the material surrounding the black hole's event horizon. Brighter color means higher brightness temperature. There is not information about the physical T of the in falling materials as the collected radiation is of synchrotron type and not that of a black body. As you say, the image could have been shown in grayscale or any other color scale. The fact that it is shown in an orange-ish color is merely the convention chosen by the scientists who generated the image.

BUT there's a lot more information in the image than just how bright the material is. For example, the fact that the material is brighter on one side than on the other can tell us about the geometry of the material surrounding the black hole, or about how that material is rotating around it. If you have the time, check out this article and read up on the actual journal articles published by the Event Horizon Telescope collaboration linked at the bottom.


Orbiting around a black hole Is it possible (for either a satellite or a planet) to orbit around a black hole? Do they attract everything around themselves into the center? Or they just affect gravitational force just like stars?

Absolutely possible. There's nothing magical about a black hole. The gravitational pull of a black hole reaches as far as gravity would for another object of the same mass.

If you replace the Sun with a black hole of the same mass, everything would continue to orbit it just as it currently does.

Anything with mass has a gravitational force itself, and a black hole will attract anything with mass. Again, it's the same as our star having an effect on the Earth, and the Earth having an effect on the Moon.


Status of new telescopes on Mauna Kea? I know that there has been conflicts between scientists and the native population. Astronomers want to build new telescopes on Mauna Kea, however, several sacred sites would have to be built on.

My question is, are there any current plans to move forward with the observatories, or have they reached a gridlock?

You're talking about the construction of the 'Thirty Meter Telescope'.
Currently the construction has been halted by the State Supreme Court of Hawaii as a reaction to the violation of native sacred land.

Negotiations are going on about moving the construction site to the canary Islands, which has been deemed an acceptable alternative in the northern hemisphere, even if only barely.


Is there any way a meteor can hit at less than escape velocity? Note:  I have figured out one case:  The meteor is splash from an impact on a close-in moon.  I'm looking for any path from deep space.

My understanding is that anything from deep space will inherently pick up the escape velocity in it's fall and thus there is no possibility in the simple case.  I'm not so sure about what could happen after an encounter with one or more moons, though.

Edited.
No if you talk about the escape velocity form Earth. This follows simply from the fact that energy $E$ is conserved. An object that is not gravitationally bound to Earth has $E>0$ and hence $v>v_{\mathrm{esc}}$ when hitting ground.

Yes, if you meant the escape velocity from the Solar system, because the Earth moves with $v_{\rm Earth}=v_{\rm escape}/\sqrt{2}$ relative to (but not towards) the Sun. Here $v_{\rm escape}=\sqrt{2GM_{\odot}/1{\rm AU}}$ is the local escape speed from the Sun, while $v_{\rm Earth}=\sqrt{GM_{\odot}/1{\rm AU}}$ is the speed of the local circular orbit. An object at 1AU form the Sun and bound to the Sun cannot have speed greater than $v_{\rm escape}$.

Now, the impact speed of an object that moves at $v_{\rm escape}$ can be as low $v_{\rm escape}-v_{\rm Earth}=v_{\rm escape}(1-1/\sqrt{2})$ if it hits Earth "from behind", i.e. moving in the same direction as Earth at the time of impact.

Note also that meterors typically move not faster than $v_{\rm escape}$, for they don't come from outer space, but from the Solar system.


How does angular resolution of a telescope translate to its parallax precision? We can often read in the scientific and also more casual reader literature and articles about the angular resolution of various telescopes and other optical equipment, be it ground based or onboard space probes. They would often list their angular resolution, or in other words, their ability to resolve or distinguish small, distant objects with today's digital era mostly on a per sensor pixel basis.

                Finding a star's distance from its parallax. Trigonometric Parallax method determines distance to star by measuring                its slight shift in apparent position as seen from opposite ends of Earth's orbit. (Source: Measuring the Universe)

What I'm interested in is, is the precision in measurements of parallax and with it our ability to determine distance of observed objects directly analogous to mentioned radial resolution and how could it be calculated using data on angular resolution of a telescope alone, if we assume both ground based and in space observatories have more or less the same perihelion to aphelion distance (i.e. the space observatory is in Earth's orbit).

Parallax measurement in practice is not as is explained above using the popular diagram you have used. The parallax causes the star to prescribe an ellipse in the sky, the semi-major axis of whose is equal to the parallactic angle.

The telescopes generally measure the shift in co-ordinates of star(RA and Dec) and then translate the information to that of the ellipse prescribed and determine the parallactic angle. Now, the ability of telescopes to note this change in RA and Dec is based on their least count, obviously which is dependent on their angular resolution.

Hence, distance measurements using parallax method have their range based on the angular resolution of the telescope being used.

Image source: http://documents.stsci.edu/hst/fgs/documents/handbooks/ihb_cycle16/c01_intro4.html


Leonid meteor showers and the Tempel-Tuttle comet I've been reading about the Leonid meteor showers and am struggling to understand one piece of information. On Wikipedia (Tempel-Tuttle), it states:

  The orbit of 55P/Tempel–Tuttle intersects that of Earth near exactly,
  hence streams of material ejected from the comet during perihelion
  passes do not have to spread out over time to encounter Earth. This
  coincidence means that streams from the comet at perihelion are still
  dense when they encounter Earth, resulting in the 33 year cycle of
  Leonid meteor storms. For example, in November 2009, the earth passed
  through meteors left behind mainly from the 1466 and 1533 orbit.

Why are we going through the meteors left behind by the orbits from 1466 and 1533? What happened to the orbit of 1500 and every subsequent orbit?

My guess is that the perihelion moves slightly from orbit to orbit so the earth moves through a different one each time but I'd love to get an expert explanation!

When material is shed from a comet, it mostly continues to follow the same orbit as the comet, but drifts out ahead or behind the comet in a complex way due to interactions with solar wind and the gravity of other objects.

If you imagine the elliptical orbit of 55P/Tempel-Tuttle, it will have patches of denser dust in clumps spaced out around the ellipse corresponding to different perihelion approaches.  The 1466 clump will be in a different place than the 1500 clump, and they will both be travelling more or less around the comet's orbit.

Since the comet's orbit only crosses Earth's at one place, and we only cross that spot once a year, we will only encounter whatever clumps of dust happen to be at that point at the same time the Earth is.

Calculating exactly where these clumps of comet dust are is non-trivial.  And that is a big reason why predicting how good a meteor shower will be is a rather tricky business.


What are the Mathematical Algorithms/Formulas behind hindu Sunrise and Sunset Calculations? There are differences between Astronomical Sunrise/Sunset calculation and Hindu calendar Sunrise/Sunset calculations. The astronomical algorithms to calculate the sunrise/sunset can be found over the internet. Here is a calculation that I followed to correctly calculate the astronomical sunrise/sunset time.
Sunrise and Sunset algorithm
In the Hindu calendar, sunrise/sunset is considered when the center of the Sun disk is at the horizon, also the refraction effect must be considered.
Sunrise on the Hindu calendar = Astronomical Sunrise + Time taken by the Sun to rise half of its diameter + Time taken by the Sun to rise further to neutralize the refraction effect.
Similarly, Sunset = Astronomical Sunset - Time taken by the Sun to set half of its diameter - Time taken by the Sun to set further to neutralize the refraction effect.
I am able to derive the Astronomical Sunrise/Sunset, but how to do the calculation for the following?:

Time taken by the Sun to rise half of its diameter.
Time taken by the Sun to rise further to neutralize the refraction effect.
Time taken by the Sun to set half of its diameter.
Time taken by the Sun to set further to neutralize the refraction effect.

In the linked algorithm,  the "zenith" is the position of the sun at sunrise.  This is the number you need to change.
If you want to calculate the time when the top of the sun reaches the horizon you use the "official" zenith of 90⁰50' This is intended to account for refraction and should estimate the time when the sun first appears over the horizon.
Since the sun in 30' across, if you want the time when the centre of the sun is on the horizon, just subtract 15'  ie use a zenith of 90⁰35' (90.583 degrees)
It isn't possible to be more accurate as the amount of refraction varies with the weather. But this should give values that are correct to the minute.


Season cycle < Year? Is it possible for a planet's axis to precess or wobble in such a way that its season cycle is of lesser duration than its year? The gyroscopic precession will thus give seasons to one or the other hemisphere during a possibly hundreds-year-long revolution around its star. Earth's precession is very much longer than it's revolution period, but is it possible that these could be reversed and a planet still be habitable?

Is it possible for a planet's axis to precess or wobble in such a way that its season cycle is of lesser duration than its year?

For this, definitely yes, if you tweak some parameters. On the Wikipedia page for Axial Precision, you'll find the set of equations and parameters that allow you to calculate the precessional rate for Earth's axis. Mainly, the Earth's precession is caused by a variety of factors but primarily due to the Moon and Sun (and thus related orbital/body parameters). If you look at those equations, and start tweaking values in them, you'll find you can force the axial precession period to be less than a "year". I'll go through and construct just such a system.

First though, I need to figure out the orbital parameters of the planet such that it has a 100-year long revolution period. This is acheived by using Kepler's Third Law, relating orbital period, orbital distance, and mass of the Sun (and Earth, but that's negligible). The relevant form of the equation, for our purposes is:

$$P^2 = \frac{a^3}{M_{\mathrm{Sun}}}$$

where, $P$ is the orbital period in years, $a$ is the orbital distance in AU (1 AU is the distance between Earth and the Sun), and $M_{\mathrm{Sun}}$ is the mass of the Sun in units of solar masses. If you're dictating that $P=100\ \mathrm{years}$, then I find I can set $a = 10\ \mathrm{AU}$ and $M_{\mathrm{star}} = 0.1\ M_{\odot}$ and make this equation true. What that means is that this new planet orbits a star which is 10 times less massive than our own star (making it a red dwarf) and orbits 10 times farther than Earth. My choice of parameters is all serendipitous and you could certainly find other parameters that work as well.

Now, given those parameters above, we can try to calculate the precessional rate for this new planet. I'll assume this new planet has an orbiting moon, just like Earth does. The two important equations for calculating the Earth's precessional rate are to account for the contributions from the Sun and the Moon. The equations are given by:

$$P_{\mathrm{Sun}} = \frac{3}{2}\left(\frac{GM_{\mathrm{Sun}}}{a_{\mathrm{Earth}}(1-e_{\mathrm{Earth}})^{3/2}}\right)\left(\frac{(C-A)}{C}\frac{\cos \epsilon}{\omega}\right)$$

$$P_{\mathrm{Moon}} = \frac{3}{2}\left(\frac{GM_{\mathrm{Moon}}(1-1.5\sin^2 i_{\mathrm{Moon}})}{a_{\mathrm{Moon}}(1-e_{\mathrm{Moon}})^{3/2}}\right)\left(\frac{(C-A)}{C}\frac{\cos \epsilon}{\omega}\right)$$

In these equations, $G$ is the Gravitational Constant, $M$ is mass in kg, $a$ is the orbital distance in meters, $e$ is the orbital eccentricity, $i$ is the inclination of the Moon's orbital plane wrt to the ecliptic, $(C-A)/C = 0.003273763$ is a measurement of the Earth's shape, $\epsilon$ is the Earth's axis' angle, nominally $23.43928^\mathrm{o}$, and finally $\omega$ is the rotation rate in rad/s.

Plugging in all the values for the Earth-Moon-Sun system (which you can find on the wiki page), you find that $P_{\mathrm{Sun}} + P_{\mathrm{Moon}} = 7.789\times 10^{-12}\ \mathrm{rad/s}$ and to convert to a time scale, you can multiply by  $2\times10^{-7}$. Doing so gives you the nominal Earth precessional period of about 26,000 years.

Given all of that, we can start playing with parameters to get it such that the precessional period for our new system is less than 100 years. I already said that $a_{\mathrm{Earth}}$ is 10 times larger and that $M_{\mathrm{Sun}}$ is 10 times smaller so we could get the period of 100 years. I found that I could make the following additional changes.

The moon is now 5.8 times more massive than our Moon.
The moon is now a quarter of the distance of our Moon.
The planet is spinning twice as fast.

With those changes, I find that the precessional period falls down to about 50 years, or half an orbit. You can continue to play with the numbers to get a different result.

An important consideration here is whether or not such a system is likely to even form. I think, given the system I've created, it is not outside the realm of possibilities. To have such a large, almost binary planetary system be so far away from a small star is unlikely, but I don't think impossible. The real difficulty here is that you've set the orbital period to such a long timescale. If you were to relax that a bit to something much shorter (say 10 years) you'd get a more realistic (and likely habitable) system.

  Earth's precession is very much longer than it's revolution period, but is it possible that these could be reversed and a planet still be habitable?

Certainly there's nothing to say a planet with a precessional period shorter than the revolution period is uninhabitable. What will make it uninhabitable is being too far from the star. The extreme orbital time you've provided necessitated having a planet that is likely too far from the star to be in the "habitable zone". You just have to play with all the parameters and find something that fits all your needs.

Another thing to consider is that if a planet has such a large and close moon as I've constructed, it will likely cause the planet to slow its rotation rate over short (astronomical) timescales. The planets rotation rate has a significant affect on the precessional rate and slowing it down over short timescales may mean your system doesn't keep the precessional rate under the rotation period for too long.


If the solar system is nearly flat, then why don't all the planets appear to lie on the same axis when viewed from earth? For instance, consider the image below. The location of the planets seem to be so random relative to Earth. If the solar system was indeed flat, then I would expect the planets to lie on the same 'plane', and relative to us, it would look something like the second image. I mean, maybe Jupiter and Mars and Saturn could be on the same plane, but I fail to see how Venus would be.
What is wrong with my understanding here?

The first image is from a July article by TV meteorologist Brent Watts.
Approximating that view in Stellarium:

The yellow arc is the ecliptic, the plane of the Earth's orbit around the Sun.
It looks curved due to the map projection.
If the view is centered on it, the ecliptic looks straight:

Unlike the Sun, which appears on the ecliptic at all times, the planets can appear a few degrees north or south of it because their orbital inclinations are all slightly different.
The more another planet's orbit is inclined relative to ours, and the closer it approaches us, the more its ecliptic latitude varies.


Is extra-planetary phosphine actually a new discovery? For all the buzz on Venus’s phosphine, I’m curious as to what sets it apart from the findings of phosphine on Jupiter,  Titan, etc.  There are several papers on NASA.gov dating to 1979 and more recent on the subject.   Per request in comments:  https://astrobiology.nasa.gov/nai/annual-reports/2010/jpl-titan/the-commonality-of-life-in-the-universe/index.html
Why is this latest finding of Venusian phosphine a special case?

You're correct; extraterrestrial phosphine is not actually a new discovery. As you said, we've known for decades (see e.g. Ridgway et al. 1976) that phosphine can and does exist in some gas giant atmospheres, including Jupiter's. However, detailed modeling has been done of the associated photochemistry beginning at around the same time, and as far as I'm aware, the important thing is that the observations aren't in conflict with the models.
The recent claim of phosphine detection (Greaves et al. 2020) went beyond all of that because as it indicated that phosphine appeared to exist in higher concentrations in the Venusian atmosphere than could be produced by known processes. From what I can tell, the group behind the detection was fairly exhaustive in their attempts to account for the concentrations by other means - lightning, known photochemical processes, meteorites, etc. All of them fall short by quite a lot, by orders of magnitude in most cases.
That's the key: It's not that phosphine was claimed to be detected, but that too much phosphine appeared to be detected. All that said, doubt has been cast on the results. Recent work (Snellen et al. 2020) found that the method used by Greaves et al. involving a 12th-order polynomial can also lead to numerous false positives when applied to other spectral features near 267 GHz, the frequency of the claimed detection. The reanalysis of the data implies a feature detected at a signal-to-noise ratio of 2, which is certainly too low to imply a statistically significant result.

(As an aside: the paper you cite (Pasek et al. 2011) isn't an observational paper - that is, it doesn't report an actual discovery of phosphine on Titan. Instead, it describes theoretical modeling and some laboratory work involving phosphorus chemistry. It's one thing to say that "Hey, here's how phosphine might participate in chemical cycles in this environment" and another to actually confirm the existence of that compound.)


Under Earth's gravitational pull, how dense would an object have to be to reach its Schwarzschild Radius? The Schwarzschild Radius is a theoretical radius, that if the mass of the object was compressed into that certain radius, the escape speed from the surface of the compressed object would equal the speed of light, therefore creating a black hole. But I wanted to know, under Earth's gravitational pull, how dense would an object have to be to reach its Schwarzschild Radius?

Earth's gravity is quite weak. It manages to compress air (luckily for us), but that's about it. Since the Earth is not a black hole it can never compress something else to be a black hole.

That being said, lets consider some object and assume something compresses it to a black hole.

The Schwarzschild radius is defined as
$$
R_s = \frac{2GM}{c^2}.
$$
Now lets say the object is a sphere of radius $R_s$, then its uniform density is
$$
 \rho = \frac{M}{ \frac{4\pi}{3} R_s^3 }.
$$
Using this to get rid of the mass term leaves us with
$$
R_s = \frac{2G \rho \frac{4\pi}{3} R_s^3}{c^2}
$$
Inverting this expression gives the black hole density for a given radius
$$
\rho = \frac{3 c^2}{8 \pi G R_s^2}.
$$

Notice that if you want a really small black hole, say $R=1$ m, the density has to be about $1.6\times10^{26}$ kg/m$^3$. That's about 20 times the mass of Earth stuffed in a cubic meter, so quite a lot.

But if the black hole can be as large as we please, then the density can get extremely low. A great example of this are supermassive black holes, which have about the same density as water (that's $10^3$ kg/m$^3$).


Is the duration of a sidereal year stable? According to this essay by Phil Plait, the duration of Earth's sidereal year is 31,558,149 seconds.

Is this figure stable? If I were to meaure Earth's orbit several million (or billions) of years in the past or future, should I expect to see roughly the same duration?

If not, what would be the cause of any change observed?

It changes some.   The solar energy pressing against the Earth from the sun, the kinetic energy from coronal mass ejections hitting the Earth, tidal forces from the Earth-Sun relation and even gravitational tugs form other planets and meteors and space dust that hit the earth all change the Earth's orbit a teeny tiny bit, also, changes in the Sun's mass either by the sun accumulating mass from space dust or comets or losing mass, which most suns do over time.   Each tiny orbital change changes the sidereal year (and all the other years) a tiny little bit.    Perhaps this only amounts to millionths of a second per year but some of these forces are consistent and can lead to predictable changes over time.  Others variable, leading to uncertainty.   The long term changes to the Earth's orbital distance are currently unpredictable.

It's generally believed that the Earth is slowly moving away from the sun, but I don't believe there's much agreement on how fast this is currently happening.   Articles here and here.   The 2nd one estimates 1.5 CM per year or 15 KM per million years which even over a billion years (15,000 KM) is quite a small distance compared to 150 million KM.   But estimates are difficult for a variety of reasons.

I asked a related question here, with a bounty and didn't get any definite answers.   It's possible to mathematically model the movement of some planets (Jupiter to an extent and Neptune/Uranus and perhaps Saturn based on the Jupiter model).  It's thought Jupiter moved both towards then away from the sun over the first few billion years of the solar system.  The Moon's distance from the Earth is also mostly predictible and can be modeled back into the past based on the Oceanic tidal bulge, but modeling the Earth;s distance from the sun over the past is much harder.  It's a tiny object compared to the Sun and Jupiter and N body systems have a lot of chaos and unpredictability to them.   4 billion years ago it's possible that the Earth was quite a bit closer to the sun, but I don't think anyone has a good model for how much closer.   The faint young sun paradox and the theory that young suns eject much more matter and have more active coronal mass ejections suggests to me that the Earth was likely closer to the Sun billions of years ago and not further away.

The earth's orbital eccentricity also changes with time, on a cycle of roughly every 100,000 years but according to Wikipedia, that doesn't change the length of the year, only the shape of the orbit.

Finally, and just for fun, it makes logical sense to me that the Moon orbiting the Earth creates small apparent fluctuations at least from the point of view of the Earth, because the Earth in effect orbits the Earth-Moon barycenter in an ellipse roughly 1/81st the size of the distance to the moon, or about 4,480 KM in radius every 29 days.     Depending on whether the Moon is ahead of the earth or behind it, the Earth year can change by as much as 5 minutes.    I suspect that this is largely ignored when measuring Earth years though and the Earth-Moon barycenter is used.   If we measured the Earth's center we should get much larger variations year to year.  (see picture).

Source.

(if anyone can improve on this, feel free, I have the feeling it's only a partial answer, cause logically it makes sense that the Earth-Jupiter positions or Earth-Venus should change, if only slightly, the Sidereal year).*


How much gold is there in our sun? XKCD 1944 claims that there is "more gold in the sun than water in the oceans". Is this really true?

The mass of the sun is 1.989 × 1030 kg.

Abundance in the Sun of the elements gives a percentage 1 × 10-7 % for gold *, so that leaves you with a mass of 1.989 × 1021 kg of gold.

HowStuffWorks states that there is 1.26 × 1021 kg water on Earth, of which 98% is in the oceans, i.e. 1.235 × 1021 kg.

This would mean the XKCD statement is true: there is 1.6 times as much gold in the sun as there is water in the oceans.

* They cite WolframAlpha as their source. Executing SolarAbundance "Gold" there confirms this (mass) percentage.


How far into intergalactic space would you have to go to see the Milky Way? If in the distant future a ship were somehow able to move beyond the galaxy and into intergalactic space, how far would it have to go before we could look out a window and see the Milky Way in its entirety, rather than just edge-on as we see it from Earth?

We can calculate the distance we need to be from an object in order for it to be any arbitrary angular size using the formula $D=\frac{d}{2\tan(\frac12\delta)}$.  In this case, $D$ is the distance from the object, $d$ is its diameter, and $\delta$ is the angular size.  The Milky Way's diameter is 100,000 light years.  If we want the Milky Way to appear as large as the Sun appears from the Earth (0.5$^\circ$), we would need to be 5,700,000 light years away.  If we want it to appear much larger, we could be closer.  For the Milky Way to be the size of a 1 meter circle held at arm's length (90$^\circ$), we would need to be 50,000 light years away.


Distinction between signal from terrestrial and signals from celestial sources How is it confirmed that a radiation source (not in visible light) is either terrestrial or celestial?
If in-depth explanation and/or references are possible, I would be grateful.

There's a few indicators where a source is located:
Direction: Radio waves are just like light waves only with longer wavelength: you have to look into the direction of the source in order to see it. Terretrial sources usually are not high up in the sky.
Apparent movement: Another important aspect is Earth's rotation: the sky rotates with respect to Earth's surface. Thus if the source moves with the rotation speed of Earth and remains stationary in a celestial reference frame, it is very likely a source outside of Earth - and vice versa.
Location / Parallax: If you have several receivers, they will disagree on the direction a signal comes from, but will agree much better the further away a source is. Celestial sources are far away enough that the direction will be (nearly) identical. The amount of disagreement, the so-called parallax gives you a geometric indicator about the distance to the source.


Is there a public record of planetary disks apart from ours? Our planetary disk is not aligned with the rest of the galaxy.  Are there publicly available records of the orientations of any other planetary disks?
Especially interested in nearby stars - but I'd expect that to be easier for us to measure?

A useful resource for protoplanetary discs is the recent DSHARP survey using the ALMA telescope. With the ALMA telescope the spatial resolution is good enough that orientation (i.e. position angle and inclination) can be derived from the images.
See here.


Black Hole - energy and matter Can someone explain what is happening with energy (e.g. light) and matter which was collected by a black hole?
I suppose that matter builds a black hole's mass and size, but what about light? Does it just disappear? But to generate light, the star consumed matter (mass).

No, the light doesn't just disappear, it adds to the black hole's mass in accordance with $E=mc^2$. So if a star produced light from $m$ kilograms of mass and all that light fell into a black hole, the black hole's mass would increase by $m$ kilograms.


How to determine the eccentricity, knowing the orbital speed and distance How to determine the orbital eccentricity of a planet, knowing the instantaneous orbital speed of it and the distance from the Sun to the planet at that moment?

How to determine the orbital eccentricity of a planet, knowing the instantaneous orbital speed of it and the distance from the Sun to the planet at that moment?

We are given three things:

$v$ "instantaneous orbital speed" (a scalar)
$r$ "distance from the Sun to the planet at that moment" (a scalar)
$GM$ since the OP specifies "the Sun"

Let's see how far we can get with the vis-viva equation and look at the problem purely from an energy standpoint.

$$v^2 = GM(2/r-1/a)$$

becomes

$$a = 1/(2/r-v^2/GM).$$

Let's look at a numerical example. Say $r = 1AU \approx 1.5 \times 10^{11}$ meters, and $v = 25,000$ m/s.

We know then that the semimajor axis of the orbit must be 0.773 AU. If the orbit reaches 1AU, the velocity will be 25,000 m/s. If it reaches 1.1 AU (which it might or might not, depending on eccentricity) it's velocity will be 21,500 m/s.

We're almost there but not quite. We need one more piece of information.

The periapsis and apoapsis $r_p$ and $r_a$ are related to the semimajor axis and eccentricity by

$$r_p = (1-e)a$$
$$r_a = (1+e)a$$

So for example if you had mentioned that the apoapsis was 1.1 AU, then we could calculate eccentricity from

$$e = r_a/a -1$$

and get $e = 0.4228$ AU and a periapsis $r_p = 0.387$ AU.


What explains the distribution of new impact craters on Mars? 24 minutes into this van Karman lecture by Dr. Tamppari, the slide below is shown with symbols for transient events on Mars. Red dots represent meteoroid impacts which have occurred while MRO has been in Mars' orbit.

New impacts are almost only found at lower latitudes. And longitudinally there are two equally sized gaps at the equator. Is this because of some observational bias or is there a natural explanation?

I was on the targeting team for one of the cameras that discovered most of these new impact craters. The reason for the distribution is simply because it's easiest to find "new" impact craters in the dusty regions of Mars. Often what we see is the dark-toned blast zone created by the impact in lower-resolution data, which has a large areal footprint. Then, we target those dark-toned splotches with the higher resolution cameras---which have MUCH smaller footprints---to confirm whether or not the splotches are craters. See the attached image as an example. The dark splotches really stand out in dusty regions, but in the non-dusty areas there's nothing obvious to pick out to say "hey, there's a new crater" as a human without tediously comparing images.


statistical techniques for estimating distribution of mass I occasionally hear claims along the line of ' the mass of the universe is not uniformly distributed but actually is concentrated along a '  line, or some other lower  dimensional manifold & I would like like to know what statistical techniques, if any,  are used to make support these claims.  I'm not interested  in whether the claims have held, just in the what was used for the original analysis.  I have data  whose dimension I would like to estimate, and am familiar with correlation dimension.

Matter is indeed distributed in knots (clusters), filaments, and sheets (e.g. Bond et al. 1996). This is predicted from theories and numerical simulations of structure formation under gravitational collapse, and is directly observable in large galaxy surveys by eye, such as The 2dF Galaxy Redshift Survey:

Two-point correlation function

To quantify this clustering, usually a correlation function is used, either in 2D ($x$ and y position on the sky), or 3D (if redshifts are available for the galaxies, giving the $z$ dimension). The (two-point) correlation function $\xi(r)$ gives the excess probability of finding two galaxies at a distance $r$ from each other, relative to a random (Poisson) distribution.
If $\bar{n}$ is the mean number density of galaxies, then the probability $dP$ of finding a pair of galaxies at a separation $r_{12}$ in volume elements $dV_1$ and $dV_2$ is
$$
dP = \bar{n}^2 \big(1 + \xi(r_{12})\big) dV_1 dV_2.
$$
The power spectrum $P(k)$ of the galaxies is related to the two-point correlation function by
$$
\xi(r) = \frac{1}{2\pi^2} \int dk\,k^2\,P(k)\,\frac{\sin kr}{kr}
$$
Observationally it is found that the two-point correlation function is given by a power-law:
$$
\xi(r) = \left( \frac{r}{r_0} \right)^{-\gamma},
$$
where $r_0$ is the correlation length (of the order $5\,h^{-1}\mathrm{Mpc}$) and $\gamma=1.7$–$1.8$.

Note that the value of $r_0$ depends on galaxy type; brighter and redder galaxies have a longer correlation length, i.e. they are more strongly clustered (e.g. Wang et al. 2008).

Excursion set theory

Using a formalism know as excursion set theory, the mass fraction in knots, filaments, sheets, respectively, can be predicted analytically. The idea of this theory is to model the collapse of structure in an expanding Universe as a stochastic process, where an overdensity fluctuates and forms a structure if it crosses a given threshold. See e.g. Shen et al. (2006).

Dynamical classification

Forero-Romero et al. (2009) proposed a dynamical classification of the cosmic web, based on evalution of the deformation tensor
$$
T_{\alpha\beta} = \frac{\partial^2 \phi}{\partial r_\alpha \partial r_\beta}
$$
of the gravitational potential $\phi$, counting the number of eigenvalues above a certain threshold at each grid point in a cosmological N-body simulation, where the case of zero, one, two or three such eigenvalues corresponds to void, sheet, filament or a knot grid point.

Other methods

I quick googling leads me to other methods I don't really know anything about, e.g. "the Shapefinder diagonistic" (Prakash 2016), and large-scale halo ellipticity–ellipticity and ellipticity–direction correlations (Lee et al. 2008).


What coordinate system is at rest relative to the center of the galaxy? I am given equatorial coordinates and I need to transform them into a coordinate system that is at rest relative to the center of the galaxy. The closest thing I could find is galactic coordinates which seems like they are fixed with the sun at the center. Wouldn't this meant that they are moving with the same velocity as the sun and they are rotating?

Galactic coordinates are sun centred with a zero longitude chosen to align with the centre of the galaxy when the measurements were made in 1958.

As defined, the galactic coordinates don't rotate, and so the position of Sag A* is very slowly drifting. Currently Sag A* is about 4 arc-minutes from the defined zero longitude, but decreasing by 5.7 milli-arcseconds per year (a tiny amount)

The task is ill defined, You can convert between different coordinate systems of the sky, but they all must be centred on the solar system.


Are there are any photographs online that approximate what the Milky Way looks like to the unaided eye? I love seeing all the long-exposure photographs of the Milky Way, but I'm going to Cherry Springs State Park in Pennsylvania next month, and I'm wondering if there are any photos that approximate what our galaxy looks like to the unaided eye. I'm really just trying to set my expectations, because I know it's not going to look like what you generally see in long-exposure photos. (Naturally, people are generally showing off their carefully crafted and enhanced images, rather than trying to show what we actually see with our eyes.)

Edit: I did find this post, which, according to the author, comes close to approximating the view with the unaided eye. For those that have seen the Milky Way from a Dark Sky site, though, I'd be interested to hear your opinion on the relative accuracy of this image.

It may be difficult to find proper photos that emulate this, both because given the tools and the subject, most people would opt for longer exposures, and also because of the general difficulty in emulating how our brain processes visual information into an image, especially in low-light or high contrast situations like this.  I think the photo at the bottom of this article may be a decent recreation of it, though it's a bit fuzzy and not so wide. Note: There are star trails forming in the image, which is indicative of it being a longer exposure. The image (unless cropped) doesn't appear to be from a wide-angle lens - I would guess somewhere in the 35-50mm range.  Perhaps a 10 secondish exposure, but my judgement is more based on result than technique.

What you can see will depend a lot upon the conditions of the night. You're probably not expecting a lot of light pollution, but you also have to keep in mind that your eyes will adjust to light as well.  A cell phone or a full moon can be enough to prevent you from being able to see well.  If conditions are perfect - little to no light pollution, no light from cars, flashlights, etc, a new moon, low humidity/no clouds, etc, and the milky way is in a good position, you will be able to pick out the galaxy quite well. You should be able to notice both the high concentration of stars as well as the 'milky' lighter shade caused by the many stars too dim or distant to appear bright.


Why is today the shortest day but the solstice is actually tomorrow? Here in Ashland, Oregon, today (Dec 20, 2020) is the shortest day length of the year at 9:05:45.  But the winter solstice isn't until 2:02 AM PST tomorrow (Dec 21, 2020). Tomorrow's day length is a whopping three seconds longer at 9:05:48.  Why is tomorrow's day length greater than today? It would seem to me that tomorrow's sunrise and sunset are slightly closer to the magic moment of winter solstice and should therefore account for a slightly shorter day.

The other possibility is https://sunrise-sunset.org is inaccurate.

An app that I use, LunaSolCal, show the length of day on Dec 20, 21, and 22 in Ashland OR are 9:05:45, 9:05:44, 9:05:49.
The website https://www.timeanddate.com does not have Ashland OR in its database, so I used nearby Medford OR. The length of day on Dec 20, 21, and 22 are 9:04:50, 9:04:49, and 9:04:53.

Although those two sources have a difference of 1 minute in the length of the day, the pattern is correct and expected: the shortest length of daylight is on the day closest to the winter solstice. Therefore, I think sunrise-sunset.org is off by one day with its calculations.


Determining distance from semi-major axis and eccentricity I am trying to obtain the distance covered by an object in orbit around the earth within a specified amount of time past its perigee passage. The object is in an elliptical orbit with eccentricity 0.2 and has a semi-major axis of 9600km. Where would I look to find the objects position 90 mins after it passes its perigee? Thanks for any advice

Supposing that the mass of the object is negligible compared with the mass of the Earth, you can derive the orbital period $T$ from the 3rd Keplero's law:

$\frac{T^2}{a^3} = \frac{4\pi^2}{G(m_E + m_b)} \approx \frac{4\pi^2}{Gm_E},$

where $a$ is the semi-major. With $T$, for each time istant you also know the mean anomaly $M$, given by (suppose $t = 0$ at perigee):

$M(t) = \frac{2\pi}{T}t$.

Solving numerically the Keplero's equation for the eccentric anomaly $E$ (where $e$ is the eccentricity)

$M = E - e\sin E$

and then use the following equation to derive the true anonaly $\nu$, which is the angle between the direction of periapsis and the current position of the body, as seen from the Earth:

$ \cos \nu = \frac{\cos E - e}{1 - e\cos E}$
and
$ \sin \nu = \frac{\sqrt{1-e^2}\sin E}{1 - e\cos E}$.

The distance from the Earth is just given by the orbit equation

$ r = \frac{a(1-e^2)}{1 + e\cos \nu}$.

If I'm not wrong with calculation, it should be:

$ T = 9364 s = 2.6 h.$

$ M(90min) = 207.60°$

$ E(90min) = 203.11°$

$ \nu(90min) = 198.95°$

$ r = 11'366 km$

To get the covered distance you should calculate the line integral of the orbit equation.


Planet-timer "device" from 1970s or 1980s possibly by Edmund Scientific Many years ago (early 80s or late 70s), I bought a cardboard "device" which I think was called "Planet Timer" and I think was sold by Edmund Scientific. Does anyone have any information on this device? Google was unhelpful.

The device had concentric revolving circles around a fixed Sun at the center. By revolving the circles to the current time (via tables provided with the device), you could presumably see what constellation each planet was in. You could also use it to approximate planetary conjunctions.

The circles were equal distance from each other, so I'm pretty sure the device couldn't actually have worked, but I'm curious to see if I can find out more about it.

Note: this is not a "planisphere" device, which did work, and for which I can still find plenty of information.

I ended up finding this on page 14 of the Edmund Scientific Holiday Sale catalog, with sale ending on December 31, 1983:

except that it's "Planetimer" (and not "Planet-timer"), catalog number H9454.

A 1200dpi image somewhat shows how it's supposed to work:

My original question was going to be: could this device have worked?

I would also appreciate any other information on this product (including a better scan, ideally of the product itself, if anyone still has one).

For reference, here is the page on which the ad appears and the cover of the catalog in which I found it:


What is the significance of using baseline pairs in radio interferometry? Radio interferometry utilizes arrays of smaller telescopes that are linked together to synthesize a larger aperture telescope. Astronomical radio observatories, such as the Very Large Array in New Mexico, consist of multiple radio antennas in various configurations (ie. 27 antennas for the VLA). When I read about the basics of radio interferometry, a lot of time is spent on the delay and baseline between a pair of telescopes. In other words, the light from the object will enter one telescope first before the other (unless pointed directly up) and a pair of telescopes act as 1 baseline. I don't understand the significance of these telescope pairs. Do the pairs act as a single pixel? I read that a pair of antennas measure one point in the u,v plane. Why are pairs of telescopes needed to do this and what is the significance to radio astronomy?

The basic idea behind interferometry is that of interference, the combination of two waves (in this case the electromagnetic waves from distant sources).  Interference inherently implies two signals to interfere with each other, and that is why the pairs of telescopes (also referred to as baselines) are important.  The interference pattern between the signals received at two different telescopes carries information about the spatial structure of the source.  Widely-separated pairs are sensitive to small-scale structure, and close pairs are sensitive to larger-scale structure (which is why you need both to reconstruct a good image).  North-south pairs are only sensitive to structure in the north-south direction, and likewise for other orientations of pairs, so again you need pairs/baselines oriented in different directions.  (The rotation of the Earth helps somewhat by changing the orientation of baselines over time, though modern interferometers with many telescopes, such as ALMA, have pretty good coverage in a single snapshot without waiting for the Earth to rotate.)
The signal from a source is recorded at each telescope, and then those signals are combined (or interfered) with each other for each pair of telescopes, and all of those per-pair interference signals constitute the overall data collected by the interferometer.  In a radio interferometer this interfering/combining of signals is done in a computer called a correlator. For radio, the signal at each telescope is converted to a voltage and these voltages are correlated; for an optical or infrared interferometer, the light from each individual telescope is split into multiple paths so that each pair of light beams can  be interfered directory in a beam combiner before the interference pattern is recorded.


Why does hydrogen ionization happen in HII regions? Why does hydrogen ionization happen in HII regions? Why is the hydrogen there ionized?

Stars are responsible.

HII regions$^\dagger$ can refer to several things, but usually I guess one thinks of the volumes around star-forming regions. The more massive a star is, the faster it burns its fuel, and at a higher temperature, meaning that the peak of their spectra are more toward the high frequencies. The most massive stars of a stellar population — the so-called O and B stars — produce enough photons with wavelengths below the hydrogen ionization threshold of $\lambda = 912$ Å that they carve out bubbles in their surrounding HI clouds, giving rise to HII regions.

Right: The HII region NGC 604 (from Wikipedia). Left: The spectra of three different stars. Only the B star has a significant portion of its spectrum above the hydrogen ionization threshold. Note the logarithmic scale on the intensity (from here).

Because of the high densities, the HII quickly recombines to HI. If the recombination goes directly to the ground state, a new ionizing photon is emitted, which again is absorbed by a hydrogen atom, but if it goes to one of the higher states, the emitted radiation is no longer capable of ionizing the gas. In this way the ionizing radiation is converted into photons of specific wavelengths, corresponding the energy differences between the excited level of hydrogen, most notably the Lyman $\alpha$ emission line with $\lambda = 1216$ Å.

Because hydrogen is the most abundant element in the Universe, and because Lyman $\alpha$ is the most common transition, the Lyman $\alpha$ line is an excellent probe of the most distant galaxies where other wavelengths are not observable. This is especially so because the most distant galaxies are also the earliest and hence still in the process of forming, meaning a high star formation rate which, in turn, means that the shortlived OB stars are still present.

In addition to this distinct regions of HII, ionized hydrogen also exist in a more diffuse component between the stars of a galaxy, in huge bubbles caused by stellar feedback and supernovae, and in the intergalactic medium.

$^\dagger$The terms HI and HII refers to neutral and ionized hydrogen, respectively.


Existence of gravitons? For a lot of my uninformed life, I have doubted the existence of gravitons or even that gravity is an actual "force" (like electromagnetism). This is because my vision of general relativity was that mass curves space such that objects are still traveling in a "straight line" when being acted upon by "gravity", so that no "force" is needed. I know now that this is a naive view, but I'm not 100% sure why. I was thinking the other day that just the fact that gravity follows an inverse square law implies that it is a force carried by particles (falling off in flux intensity due to the geometry of 3D space).

My question would be: Does the fact that gravity follows an inverse square law naturally fall out of general relativity equations or is it an assumption used when developing the equations?

And, just now, I had the thought that other forces may curve space also (just in higher dimensions).

Gravity is a fictitious force, actually, much like centrifugal force.  In a free falling frame of reference it disappears.  In general relativity (GR) gravity is just a result of (differential) geometry: space-time curvature.  The inverse square law is just the low energy approximation, but the actual equation for gravity derived from GR is more complex than that.  The massive success of Newtonian gravity tells us that any model of gravity must be approximated by the classical inverse square law at low energies.

Whether GR does that by (Einstein's) design or something else is a matter of personal opinion.  Einstein definitely knew he had to get approximately Newtonian gravity at low energies, so he would have discarded or modified any ideas that failed this criterion.  However, there are standard arguments for why gravity must obey an inverse square law, at least in low energy situations.

Now in GR a lot more than just mass contributes to the gravitational field.  Spin and electric charge, for example, and quite importantly: energy (the famous $E=mc^2$ tells us how to express a mass as energy, so we can treat these things on a common footing).  So, yes, all forces and particles contribute to gravity.  Even photons.

GR itself makes no predictions (or requirements) for the existence of any new particles outside of the standard model, such as gravitons.  GR and quantum mechanics (QM) are famously incompatible: in extreme situations where both GR and QM are relevant (neutron stars and black hole formation, for example), they stop making sense rather quickly.  Especially GR.  "Gravitons" and assorted variations are hypothetical particles that are proposed to resolve this issue by creating a quantum theory of gravity.  The only "evidence" we have for them at this stage is that our two most massively successful theories about the workings of the universe, GR and QM, are so painfully incompatible.  So we know these theories are flawed (aka wrong) and that some other theory is necessary that can handle these situations, while also incorporating all of the successes of QM and GR—they're amazingly accurate when only one of them is particularly relevant, after all.

Exactly what that theory is is an ongoing and substantial research area.


Do the heliospheres of stars at the center of globular clusters interact? I read that the stars in the center of globular clusters are incredibly close to one another and that these stars can be spaced 100 to 1000 times closer than the stars near the sun. Because these stars are so closely spaced, would their heliospheres interact with each other? There are binary stars whose coronas interact and create powerful x-rays. Would something like that happen with those central globular cluster stars, but on a much larger scale, what with the larger amount of stars involved?

Similar to what @Barrycenter said in his comment, There's a wide range of distances between 100 or 1,000 times closer on average and as close as binary stars.

Alpha Centauri A and B get as close as 11.2 AU.   The Sun and Alpha Centauri are roughly 270,000 AU apart so at 1,000 times closer, the close stars in a globular closer would still be (ballpark), 250-300 AU from their nearest neighbors.  In a crowded ball of stars you're not likely to get binary systems that far apart, so most of the interaction that you're looking for would be in close binary systems.   Neighboring stars in a globular cluster would mostly be too far apart to have much interaction.

There's also a world of difference between the corona and the heliosphere.  The Sun's corona extends about 12 solar radii or about .05 AU.   The heliosphere, while, it's one, not a sphere and two, probably has some variation in it's boundary, extends roughly 100 AU (121 by this estimate), over 2,000 times further than the corona.

Binary systems like Alpha Centauri A & B probably have a shared heliosphere and/or some form of heliopause between them where their two solar winds meet.   It would be interesting to study that in more detail.  I'm not sure how much is known about the heliopause between two binary stars, especially given that they orbit each other.   I'm not sure what form it would take - perhaps that would make a good question.

Binary stars need to be much closer than 11.2 AU to have corona to corona interaction.  Stars that close are probably quite rare but they do happen.   Here's one example of a binary star system so close they're actually thought to be touching.


Time domain astronomy and fastest eclipsing binary ZTF J1539+5027 (+20 mag, 6.91 minutes): How to measure its minimum brightness? Per Wikipedia's ZTF J153932.16+502738.8

ZTF J153932.16+502738.8 is a double binary white dwarf with an orbital period of just 6.91 minutes. [...] The light curve shows eclipses. One dip in the light curve is 15%, and the other is close to 100%.

Below is a light curve plotted on a linear scale in Fig. 1a of Burdge et al, (2019) General Relativistic Orbital Decay in a 7 Minute Orbital Period Eclipsing Binary System. The dip at 0 phase is deep.
A logarithmic magnitude scale might tend to show just how deep but noting that the minimum is dimmer than +21.5 mag and probably lasts only a tiny fraction of the orbital period (i.e. a few seconds) makes a good measurement of the minimum quite a challenge.
It seems to me like this might be a situation where a photomultiplier tube and photon counting might be competitive with photometry based on CCD imaging and clever readout schemes.
Questions:

What are the technical challenges to measuring the minimum brightness of this eclipsing pair?
If it could be measured, would such a result even be particularly useful in this case?

Figure 1: Lightcurve of ZTF J1539+5027 a) The binned CHIMERA g' lightcurve of ZTF
J1539+5027, phase-folded on the 6.91 minute orbital period. At phase 0, the lightcurve exhibits
a deep primary eclipse, indicating that the hot primary star is producing most of the observed
light. Outside of eclipse, there is a quasi-sinusoidal modulation because the primary star heavily
irradiates one side of its companion. At phases ±0.5, the secondary eclipse occurs as the hot
primary transits the irradiated face of its companion. b) The phase-folded ZTF g-band lightcurve
of the object. We were able to discover the object because of its periodic behavior. c) A binned g' lightcurve obtained with KPED, phase-folded on the orbital period. Error bars are 1σ intervals.

Unless I've done my maths wrong, the period of total eclipse is about 18 seconds.
The CHIMERA camera at Mt Palomar, the instrument which followed up the discovery of this system, can take exposures at up to 8 full 1k$\times$1k frames/second and considerably higher if windowed on an object. There is no need for photon-counting equipment for such a slowly varying source. Indeed, the CHIMERA observations used 3 second exposures.
The problem in time resolution I suspect is just how faint the object is. So moving to photomultiplier technology, which would mean a hit in terms of efficiency, is not going to help here. The hot primary has $g = 20.38 \pm 0.05$. Given the the ratios of radii and temperatures in the discovery paper - $R_1/R_2 = 0.5$ and $T_1/T_2 > 4.9$ - then just a simple bolometric scaling suggests the flux from the secondary is $>144$ times smaller (or 5.4 mag).
It is quite challenging to accumulate much in the way of signal-to-noise on a magnitude $\sim 26$ object in $\sim 10$ seconds, even on the biggest telescopes on Earth. Probably the best bet would be to use a standard CCD camera on say the Gemini-N telescope (or Subaru) and just take a ten second exposure exactly at the predicted time of the minimum and then exposures either side (to confirm you were in the right place!). This can be repeated every 6.91 minutes for a whole night if necessary.
It doesn't look hopeful though. I used the Gemini-N integration time calculator with a $g=26$ source of spectral-type $\sim$A0V (ok for a white dwarf) observed in the darkest skies. It would take $215\times 10$s exposures to get a signal-to-noise ratio of 5, and the the noise is dominated by sky background at that magnitude. So basically unfeasible.
Determining the effective temperature of the unheated side of the secondary would be interesting because it might give you a constraints on its intrinsic luminosity, its cooling age, and hence on the evolutionary pathway that led to the less massive, but larger secondary helium white dwarf forming before the hotter primary C/O white dwarf.


Is the lower mass limit of a neutron star the same as the upper mass limit of a white dwarf? If not, when can a white dwarf be more massive than a neutron star?

No, the two limits are not the same - there is some range of masses that both white dwarfs (WDs) and neutron stars (NSs) can have.

The Chandrasekhar mass limit suggests that WDs cannot be more massive than about $1.4\,M_\odot$.  However, this is true for non-rotating WDs.  Rapidly rotating WDs might be as massive as $2\,M_\odot$.  Accretion in a binary system could both spin up a WD and increase its mass.

NSs that have not undergone accretion are regularly observed with masses around $1.3\,M_\odot$, while NSs who have 'recycled pulsars' have an observed mean mass of $\sim 1.5\,M_\odot$ according to the same paper.


How did Ptolemy prove in the following excerpt that the Earth is not cylindrical? I have trouble interpreting the following excerpt:

  “Nor could it [the Earth] be cylindrical, with the curved surface in the east-west direction, and the flat sides towards the poles of the universe, which some might suppose more plausible. This is clear from the following: for those living on the curved surface none of the stars would be ever-visible, but either all stars would rise and set for all observers, or the same stars, for an equal [celestial] distance from each of the poles, would always be invisible for all observers.”

I understand that if the Sun rotated around the curved side of the cylinder, no one living on the curved side would see any ever-visible stars because their view of the celestial views is blocked. However Ptolemy states that “all stars would rise and set for all observers,” which I believe would be wrong because the two poles and the curved surface would be oriented in different directions on a cylinder. He also states that the other possibility is that “the same stars, for an equal [celestial] distance from each of the poles, would always be invisible for all observers.” However, I don’t understand how his explanation in the excerpt directly above would prove that “the same stars… would always be invisible for all observers.”

Here’s the link to my source (page 41, chapter 5:) https://isidore.co/calibre/get/pdf/Ptolemy%26%2339%3Bs%20Almagest%20-%20Ptolemy%2C%20Claudius%20%26amp%3B%20Toomer%2C%20G.%20J__5114.pdf

If the Earth were a cylinder, and stationary in the centre of the universe, then the apparent daily motion of the stars must be due to the stars' actual motion in the heavens. The stars he thought to be on a sphere that encircles the Earth, and there is a pole to this sphere, at the Northern pole there is a bright star.

If the Earth were a cylinder, and the stars moved in circles around the axis of the cylinder; the axis of the cylinder pointing directly at the poles of the celestial sphere; then for any observer on the curved surface the pole star would be on the Northern horizon. The Southern pole would be on the southern horizon an all the stars would rise and set. That is clearly not the case.

Ptolemy then considers the possibility of a cylindrical Earth with the axis of the cylinder not pointing towards the pole of the sphere. Then some stars would rise and set and others would remain always below the horizon. Ptolemy (in Egypt) would see the Pole Star at about 30 degrees above the horizon. But he knew that the position of the pole star was not the same in the Northern part of the Empire. There were stars that were sometimes visible in Alexandria that remained always below the Horizon in Rome. In Rome the Pole Star was 40 degrees above the Horizon. This is not consistent with a cylindrical Earth.


If I want to learn about astrophysics, should I study up on mathematics? I am currently reading A Brief History of Time and while the book itself does not contain much mathematics in it, I want to learn more about the concepts Prof. Hawkings talks about more in-depth.

Now I am aware that I will never get the expansive knowledge that a university tuition in the field would give me, but I have already branched off into something else that does not allow me to pursue that field even if I really wanted to so I want to make do. (I also understand that any knowledge I do get will be basically useless to me, but I love space that much that, in spirit, the knowledge would feel useful to me.)

My question is, before I read books more technical, should I brush up on my math? I am aware that a more applicable subject like physics has more correlation with what I am interested in, but I don't want to read about physics if I have little to no knowledge what the hell each equation means. (Also, how I will learn is through online courses/websites/books. I know it is not the ideal way to go about it, but again, I will do what I can.)

Of course, if you believe I should somehow go back a few years and try my hardest at physics and follow the most tried path, I am all ears. It is just that I am now going to a technical college for IT and if I want to leave, I would have to spend a lot more time in education than most - meaning (as I live in the UK) will probably have to pay extra when I pass a certain age.

I know this question is not exactly linked to astrophysics, but I do not know where else to ask it. If I had read A Brief History of Time earlier in my life, I would not be in this sticky situation. Thanks in advance.

(ADDITION: I have always had a love for space and I was interested in physics, but my teacher was uninspiring and I was going through a tough period in my life. I am willing to work towards it now, even if my teacher is the worse thing to ever exist.)

Yes, absolutely. At bare minimum, you'll need to study some geometry to understand how distances are calculated and how the shapes of orbits are related. Next, you'll want to study some algebra, so you can understand the likes of the inverse square law for how the intensity of light drops off as you move away from an object emitting light equally in all directions. Algebra also helps in understanding the magnitude system astronomers use to describe how bright objects are in the night sky.

After geometry and algebra comes trigonometry, especially spherical trigonometry, because of its use in dealing with how we describe the location of objects in the sky, and how to relate positions in one coordinate system to positions in another.

Finally, if you want the really deep understanding, you'll want to go for integral and differential calculus. As my high school physics teacher used to say, the difference between an ordinary formula and one that uses calculus is like the difference between prose and poetry - the calculus encodes deeper meaning and is applicable in more situations than any single formula that can be derived from it.


Why did the edge of Olympus Mons collapse? On Earth, the maximum height for a mountain is ~10 km. Beyond that, the mountain will start to spread or collapse. On the surface of Mars, the gravity is not as strong, so the maximum height for a mountain is higher.
The edges of Olympus Mons are very steep cliffs, rather than being a shallow slope all the way down. Surrounding the mountain, there is a large aureole that looks like it might be left over from landslides.
Did the edges of Olympus Mons collapse and cause a landslide because the mountain reached the maximum height for Mars? Or was some other process involved?

This source from NASA says:

Most scientists think the the cliffs formed by landslides. This collapse is driven by the weight of the huge volcano exceeding the strength of the rocks it is built of.

In other words, one theory is as you have surmised: The outer 7km shell of the volcano appears to have slid down, all around the mountain, forming the cliffs.  This paper has more information.
Another theory says:

The part of the Olympus Mons edifice surrounded by cliffs is morphologically similar, albeit at a much greater scale, to the volcanic seamounts on Earth like
the Canaries and the Hawaii.

This says that the Hawaiian volcanoes have a similar cliff around them, so perhaps Olympus Mons was formed in an ancient sea.  This paper by the same author gives the reason for why the cliffs don't go all the way around as more evidence for a sea on the portion where the cliffs did form.  A third paper by this author states that this also explains the aureole extending out so far.
A third theory is that as the lava and ash spread out from the volcano, eventually they reached a point where rather than forming rock they more formed sand dunes - which eventually started to blow away.
So, in summation, it could be because the mountain is too tall; because the bottom 7km or so was underwater as the volcano was forming the mountain; or because the outer portion of the mountain was only formed as ash.


What star was the asteroid 2014 MU69 predicted to occult on 2018-08-04? From a search, I found a list of some preparatory HST observations made for the August 4, 2018 occultation of a star by the asteroid 2014 MU69

  We propose observations in support of the upcoming New Horizons encounter for 2014MU69. One visit is for support observations of a star to be occulted by 2014MU69 on 2018 August 4 from which ultra-precise astrometry and shape information about the target can be derived. The second visit is for a backup epoch of astrometry on 2014MU69 itself at the end of the 2018 observing window at a time when New Horizons will also be observing the target. (minor typographical edits made)

Five lines are shown, for measurements between 2018-04-30 01:03 and 01:48. The target is listed as MU20180804 but that name seems to be the identification of an occultation, not the name of a star.

So far I have not figured out the actual identity of the star that was predicted to be occulted by 2014 MU69 on 2018-08-04.

Here is a very nice prediction for three occultations by 2014 MU69 last year (2017). However when I go to the main page for that asteroid occultation site and look at 2018-08-04 I don't see any events with MU69 as the asteroid!

See also Will the upcoming observations of occultation by “Ultima Thule” (2014 MU69) be of a single object, or two?

A VizieR query within 15 arcseconds of Buie's coordinates returns a 13th magnitude star, variously identified as:

2MASS 19042146-2035362
UCAC4 348-170642
Gaia DR2 4082342058110504064


How precisely we know the Gaia spacecraft baseline? [Question related to Planet 9 and the cosmic distance ladder] The European Space Agency mission Gaia is performing an outstanding job by measuring the parallaxes of almost 2 billion stars, many of which are thousands of light years away from the Sun. For this amazing archivement, two main things are needed: 1) An accuracy of less than a milliarcsecond when measuring angles with respect to distant quasars (which it uses as a reference frame) and 2) An accurate measurement of the parallax baseline (which is not exatly 2AU since Gaia is located in L2 and the distance to the Solar System barycenter is not always the same).
My question is about the second of these measurements. How precise is the distance between Gaia and the Solar System barycenter currently known?
In a recent paper it is said (if I read it correctly) that the position of the planets with respect to the Solar System's barycenter is known within uncertainties of the order of $\sim 100\;m$. I guess this was archived with lots of observations coming from radar ranging echoes bouncing over the planets and radio links from planetary probes (correct me if I'm wrong, I would like to know more on this).
In another paper I read that:

The location of the barycenter relative to the Sun, Moon, and planets
depends on the set of bodies modeled. In particular, inclusion of
trans-Neptunian objects such as Sedna and Eris in the ephemerides from
the Institute of Applied Astronomy causes a difference of the location
of the barycenter with respect to the Sun of about 100 km.

Which made me wonder if the addition of large unknown Solar System bodies, like the hypothetical Planet 9, would change significantly the parallax baseline of Gaia as to make the distances to stars wrong by much (this in turn would make the Cepheid distance estimator more unreliable and that in turn galactic distances up to the entire Cosmic distance ladder, possibly affecting the value of the Hubble constant in the end). Am I mixing things (models and actual data, direct and indirect measurements)? Our distance measurements inside the Solar System are accurate enough to avoid any of these issues really?

This page of the documentation sort of answers the question.

An important part of Gaia data processing is the Gaia ephemeris allowing one to compute Gaia position and velocity in the BCRS for any moment of time covered by observations. Clearly, the accuracy of Gaia ephemeris is crucial for the project. The required accuracy of Gaia velocity is driven by the aberration of light: 1 μas in direction corresponds to about 1 mm s−1 in velocity. The requirements for the accuracy of Gaia position comes from the paralactic effect for the near-Earth objects (NEOs) and was assumed to be 150 m. The latter requirement is only important for data releases that include solar-system data.

So the position is known to 150m and the velocity to a few millimeters per second. This is probably relative to the Earth, but the Earth's orbit is known to comparable precision. In fact a far less precise determination of the position would do for observing the parallax of stars. The largest such parallax is about 1 second of arc and the Gaia's observations of the position of stars is accurate to a few micro arc seconds, so the errors there are at least a few parts per million. On the other hand an error of (say) 1km in the diameter of the orbit is a few parts per billion. The precision location data is only needed for observing objects within the solar system.


How large does refraction become in radioastronomy? For atmospheric refraction of visible light, Wikipedia gives the order of 1 arc minute at 45° altitude above the horizon, and 5.3 arc minutes at 10°. This is caused by the dielectric polarizability of all of the bound electrons in all the atoms of the atmosphere.

At the much lower HF frequencies of radio, the free electrons and ions will contribute, and some forms of radio communication have relied on refraction at large incident angles to deflect terrestrial signals back to the Earth at a distant ground station.

So I expect that at the lower frequencies used in radio astronomy, corrections to the observed location of radio sources due to ionospheric refraction could be much larger than those at visible wavelengths, but I am not sure.

How large does can this effect ever get? At what frequency? Are there ever corrections as large as 1 degree?

I started thinking about this after asking How many stations could one hear with an AM/FM radio in front of the ISS' cupola window? which includes the image below.

below: from the Radio Jove Project's exercise The Effects of Earth's Upper Atmosphere on Radio Signals.

The refractive deviations in position are very similar for both radio and optical astronomy, until you consider very low frequency radio waves ($<200$ MHz) when the effect becomes rapidly larger.
For plane parallel refraction an approximation for the deviation you are talking about is
$$\Delta \theta \simeq (n-1) \cot \theta,$$
where $\theta$ is the observed elevation, $\Delta \theta$ is the change in elevation from its true value due to refraction and $n$ is the refractive index averaged over airmass.
According to this source from the Green bank radio telescope, they use something like this, with an added model for how $n$ varies with height, scaled by the atmospheric pressure. The largest value of $n$ quoted is 1.00031 at ground level. This is basically the same as the refractive index of air at visible wavelengths and amounts to about 60 arcseconds at elevations of 45 degrees.
So, to my surprise, the effects of refraction on radio telescope pointing are similar to those for optical telescopes. It simply turns out that the real part of the refractive index (that controls the phase velocity of light and hence refraction) is just as close to 1 for radiowaves as it is for visible light.
Here is another source that gives some algorithms to calculate the effective (small) real refractive index for radio waves, with similar results.
This source claims the calculations are valid for mm-wavelengths and upwards. Of course they cannot be valid as one approaches the ionospheric cut-off at around 40 MHz (wavelengths of metres), where the refractive index will depart sharply from unity and the deflections must get commensurately larger.
I did manage to find something about positional refraction at these low frequencies. The LOFAR radio array can apparently do work down to frequencies as low as 10MHz, but the practical limit may be a little higher. Anyway, they do need to account for refraction in the ionosphere and I found this presentation, which contains a section on refraction and in particular, the figure below.

Thus for low frequency radio astronomy (<200 MHz), refraction is certainly a bigger effect than in the optical. e.g. At elevations of 45 degrees, the refracted positions are shifted by about 0.1 and 0.4 degrees at 50MHz and 30MHz respectively.


How to determine what directly imaged objects are? A directly-imaged object found close to a certain star could be another star in the background or a  companion planet?

How would one possibly discern between the two?
Also would we know it works?

Simple really. You observe again and see if the pair are moving together. If the star plus "planet" are quite widely separated (e.g. tens of au or more), then you would expect negligible orbital motion in the course of say a year and their relative position should remain quite constant.

On the other hand, if the "planet" is a much more distant star then it is quite likely to have a very different proper motion on the sky to the foreground star and so they will show an appreciable difference in their relative positions second time around.

More than two observations might be needed to be sure. The bigger the proper motion of the foreground star, the more decisive the test.


Rotational speeds of the material forming planets should increase, not decrease We know that planets form as a result of gaseous material revolving around a center of mass that froze to form planets. However shouldn't the angular speed of that matter increase as they go closer to the center to maintain the angular momentum?

Why then the average rotational period of the external planets are faster than that for the terrestrial planets?

You need to consider which material the planets are formed of.

Terrestrial planets are formed from less material (way less), and with less momentum (since they were orbiting near from the Sun). Gaseous planets (which is another name for external planets, due to their material), also known as gaseous giants, are formed with a lot more of material with a lot more of angular momentum.

Angular momentum formula is $\vec L=\vec r\times m\vec v$.

Orbital speed formula is $\vec v=\vec\omega\times\vec r$.

Thus: $\vec L=\vec r\times m\vec\omega\times\vec r$

Since all vectors are perpendicular:

$L=r^2 m \omega$

Thus a higher mass $m$ makes a higher momentum $L$, and a higher distance to the Sun $r$ makes a quadratically higher momentum $L$.


Observing a point 13.82b ly away, 1b years ago I understand that we can observe far away galaxies to get an idea how the universe looked like in the past.

Assuming the universe is 13.82b years old, would it be correct to say that if we looked at a galaxy that was 13.82b ly away, we would see whatever happened just after the big bang? If this is correct, what would we have observed had we looked at the same point 1b years ago?

Is this an application for special relativity?

Time is kind of funny when you look at such distances. Let's imagine that you are running away from a person throwing a ball at you. The ball will travel further to hit you, based on your speed, than the distance that was present when you started to run.

The universe is expanding. The early Universe expanded very quickly, giving large deviations from the expected position today than what we see through a telescope. Something that we see that is 12 billion years old is actually quite a bit further away today, although we can't know exactly where it is. As I understand, this hypothetical object would be around 40 billion light years away today (See this question), given current theories.

But yes, if you looked at a galaxy that appears in today's distance to be 13 billion light years away, then the universe was 800 million year old, roughly speaking. Galaxies took about a billion years to form, however, so you probably wouldn't see such a galaxy.


Orbital Period of an Asteroid What's the orbital period of an asteroid which has synodic period of 429 days?

I know that the formula of orbital period for inner planets is $$P_{orb} = \frac{P_{syn}}{P_{syn} + 1}$$ and for outer planets is $$P_{orb} = \frac{P_{syn}}{P_{syn} - 1}$$ where $P_{syn}$ is the synodic period of the planet measured in years. However, I don't know whether the asteroid should be counted as an inner planet or outer planet.

Just to clarify, inner planets usually refers to Mercury, Venus, Earth, Mars.    But in relation to your question, the "inner" planets are just Mercury and Venus and the outer planets are Mars and out, because the +1 / -1 formula is in relation to whether the Earth is passing the planet as it orbits the sun or being passed by the planet.

Someone else can do the math if they like, the gist is explained here.  Because it's the motion relative to the stars in the background that you're tracking, anything that orbits closer to the sun and faster than Earth would move in a different direction and require a different formula than orbits further out and slower.

If your asteroid remains outside the Earth's orbit, then it would obey the outer planet formula.

If your asteroid is an Earth crosser, which isn't uncommon, then it would have a different type of orbital period that would follow neither formula, and I'm not sure what formula would apply for an earth-crossing asteroid.


Approximate distance from Earth to Mars at a given moment with reasonable (+/-100km) accuracy? I've been scouring the internet for a straightforward way to approximate the distance between Earth and Mars at a given time. I hope to achieve an accuracy of about 100km.

Do not underestimate the astronomers! 100 km? That's absurd. Saturn('s center of mass) is located to within one mile's distance from the Sun. Mars is of course even much better located.

Here's a site I googled that seems to give current distance to Mars to within single kilometers. I don't know how that site makes up its numbers, but it is certainly possible to measure it like that. With mutual radio communication, there are really good opportunities for precis distance measuring. 135,834,832 km just now! Pretty close and that's why the ExoMars space probe was launched towards Mars a few days ago. And here is a table with the conunction distances just past and soon coming up.


Would exoplanets like those of Kepler-70 possess magnetic fields? The B-type subdwarf, Kepler-70  (aka KIC 5807616) is believed to have two planets orbiting it, at average distances 0.006 AU and 0.0076 AU (source). Due to their proximity to their parent star, they are almost certainly tidally locked to it (although I don't know if tidal forces between the planets themselves might counter this, since they apparently pass within a few hundred thousand miles of each other and have short orbital periods.)

Now, the "day side" of each of these planets would have surface temperatures exceeding that of the Sun! (i.e. > 6000 K.) If, indeed, they are tidally locked, we might also expect a significant temperature difference between the "night" and "day" sides. The planets are believed to be the rocky/iron cores of gas giants that were engulfed when their parent star expanded. If I have understood the paper by Nordhaus et al. correctly, the gas giants would have had to be significantly larger than Jupiter. Certainly, however, the Chthonian planets that they turned into are now hot enough for metal/rock to be evaporating on at least one side, and possibly both. It would seem unlikely that they would be mostly solid.

If this is correct, then:

We have two planets, each mostly liquid/gaseous, with significant temperature differences between different regions. So it looks to me as though there would be convection.
If the planets have significant metallic content, they would also be conductive.
They are certainly also rotating, at the speed required to keep one side continuously facing the star.

Am I correct, therefore, that they probably have magnetic fields generated by the dynamo effect? Or is there some factor, possibly related to the extraordinarily high temperatures involved (or my own lack of knowledge :-) ) that would imply otherwise?

Sources:

J. Nordhaus, D. S. Spiegel, L. Ibgui, J. Goodman, A. Burrows, Tides and tidal engulfment in post-main-sequence binaries: period gaps for planets and brown dwarfs around white dwarfs, Monthly Notices of the Royal Astronomical Society, Volume 408, Issue 1, October 2010, Pages 631–641, https://doi.org/10.1111/j.1365-2966.2010.17155.x
Charpinet S. et al. A compact system of small planets around a former red giant star. Nature vol. 480 (2011)
Bear, E., & Soker, N. (2012). A tidally destructed massive planet as the progenitor of the two light planets around the SDB star KIC 05807616. The Astrophysical Journal Letters, 749(1), L14.

Although it's not proven, I've just read a research paper which provides evidence for the planets having magnetic fields. Assuming that they exist, of course.

The paper is "A tidally destructed massive planet as the progenitor of the two light planets around the SDB star KIC 05807616."

The authors state:

  Considering that the planets have suffered a recent strong tidal deformation with continuous strong tidal force from the central EHB star, and that they are heated by the central star radiation, they are expected to be hot and liquid, and possess a differential rotation. It is very probable that a strong dynamo is operating in each planet, leading to the required magnetic field.

Bear and Soker also calculate the expected time of evaporation for the closer of the two planets, resulting from the star's intense UV radiation. They arrive at a figure of $\approx 10,000,000$ years. However, Kepler-70 has been a B-subdwarf for 18.4 million years (source: Charpinet et al.) and its exoplanets would therefore be at least that old.

They state:

  the expression is highly uncertain, and we might overestimate the evaporation rate.

However, their calculations show that a planetary magnetic field about 10 times as strong as that of Earth is plausible, and would suffice to protect the exoplanet from evaporation. They find this more likely than uncertainty in the aforementioned calculation.

  Our preferred explanation for a low evaporation rate is the existence of a magnetic field ...  More likely we find the possibility that a planetary magnetic field, about 10 times as strong as that of Earth ... will substantially reduce the evaporation rate by holding the ionized gas.


Photosphere is relatively transparent. Is that right? This book, Astronomy in Minutes: 200 Key Concepts Explained in an Instant, by Giles Sparrow, says the photosphere is a visible surface at first, but after that, it says the Sun becomes cool and then transparent, so I am confused.

Our local star is a fiery ball of gas 1.39 million kilometres (863,000 miles) across. Powered by nuclear fusion, it has a three-layered internal structure (see pages 272 and 286). Its visible surface or 'photosphere' marks the zone where the Sun's hot gases grow sparse and cool enough to become transparent, at a temperature of around 5,500°C (9,900°F), but the Sun's sparse atmosphere or corona extends much further outwards.

I think the sentence highlighted in bold (not highlighted in the original) represents the definition of chromosphere... Is that right?

See the other answer by @RobJeffries for a more thorough explanation. Here I just address the the question of how something transparent can also be visible.
Transparent things can still emit bright light. Under some circumstances it can even approximate blackbody radiation. Think about red hot glass, and then imagine it is much bigger and the glass is now gas.

above: from Glowing Glass

above: from Wikipedia


If the expansion of the universe is speeding up, how long until everything nears the speed of light, and what would happen? Since dark energy is causing everything to speed up infinitely, wouldn't everything near the speed of light? And since it's nearing the speed of light, wouldn't all kinds of strange things start happening such as time dilation?

The answer to your question is that if we assume the Universe is infinite then there will always be objects receding from us at the speed of light, simply because recession velocity depends linearly on distance. Indeed there are objects that we can see (i.e. they are within our observable Universe) that recede from us at the speed of light (this is a very misunderstood issue, so I'd advise anyone taking issue with this statement to make sure they've read this paper first).

If we assume dark energy takes the form of a cosmological constant, then the Universe asymptotically becomes like De Sitter space over time. This means that the sphere of objects that are receding from us at c (the Hubble sphere) settles down to a constant radius which is inversely proportional to the square root of the cosmological constant.

Edited to add: recession velocities though are purely a function of cosmological coordinates so don't confuse them with local velocities which are always constrained by c.