What ever happened to the "Small Comet Theory" of water influx on present-day Earth?

The "small comets" buzz was temporary and has been completely superceded by subsequent evidence that it was all noise in the detectors.  A careful investigation that produced a null result looking for the comets is published in http://adsabs.harvard.edu/abs/2000JGR...10524907M.  The moral is, beware of the "1 sigma effect," where a researcher sees a hint of a signal and is beguiled into seeing it as something real.


How to know the orbital and physical characteristics of the solar planets with only a non-professional telescope?

This is a fascinating goal, and very difficult. It would make a book on recreating the history of science.

It took many people spending lifetimes collecting data to get anywhere. However, you don't need to rediscover that the planets go around the Sun I would think.

There are some key milestones that need to be reached. For example, you can distances to planet orbits in terms of ratios of a reference distance, like the Earth Sun distance, WITHOUT knowing the actual Earth-Sun distance. You only know that Mars' orbit is x times bigger than Earth-Sun orbit.

The first calculation of Earth-Moon distance was done by the Greeks. Aristarchus or Eratosthenes perhaps. One of them found the diameter of the Earth using shadows  of monuments and in wells at noon, and the other estimated the arc of the Earth by Its shadow on the Moon, and thus the size of the Moon.

I think then it goes something like Newton measure the mass of the Earth by measuring g for a dropped object. Then using estimates of the Moon's distance and its orbital period he could determine the constant G in his law of universal gravitation. Then you can calculate the Earth-Sun distance and the rest of the planets fall into place based on their periods.

Now the other challenge is collecting useful data for determining orbits of planets. There are a couple of ways to "determine an orbit from observations". And this is a particularly useful project. Getting a mental picture of what to do can be difficult. But basically, you record the time and the position of a planet in some Earth centered coordinate system, like Right Ascension and Declination. Or relative to the know position of nearby stars.

Now, the orbits are ellipses with the Sun at one focus, so the Earth based system is not much good. You convert all your observations to Sun centered coordinates in the form of lines or vectors that point to the planet. But they point to the Planet from where the Earth was when you took the measurements. These are a set of lines in space and they don't go through the Sun or anything like that. They are just from some point in space (where the Earth was) pointing to the Planet.

C.F. Gauss (I think Laplace improved on this - not sure) showed that you only need three observations to find an orbit because only one ellipse will have points on the three or more lines and have a focus at the Sun. Two points are enough if you have other information like is it getting closer or further and which way is it going (it might be in retrograde so you can't tell).

More measurements taken further apart will increase accuracy and refine the orbit.

Also, the distance to Jupiter or Saturn could be found by watching their moon's orbits and noticing how they stray from prediction with distance due to the speed of light at opposition versus being further away.

So, you can see that the steps are not too daunting if you don't need to wait for a lunar eclipse and take a few other things as known.

I think you could do the temperatures with a small reflecting telescope. It is by detecting the infra-red radiation from the planet. A reflector is used so as not to absorb the IR. The instruments used were a bolometer that responds to heat and a galvanometer. The galvanometer is an incredible precise and sensitive instrument with no electronics.

Search on "calculate orbits from observations" and there are bunch of PDFs from good sources.


Why do stars seem not to move relative to each other?

They do move - just far too slowly for you to detect by eye even over several human lifetimes.

  Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is. I mean, you may think it's a long way down the road to the chemist, but that's just peanuts to space

Douglas Adams, "The Hitch-hikers Guide to the Galaxy"

Even the closest stars are a very, very, very long way away so their apparent movement relative to each other is going to be very small.

You can see the same effect when looking out of a moving vehicle through the side windows. You see the objects closest to you rushing past, but objects on the horizon appear to move much more slowly. Now, scale that up by many orders of magnitude to interstellar distances and you'll see why the stars don't appear to move relative to each other.

There's software that simulates the night sky and you can run time backwards and forwards - if you wind it far enough you'll see the constellations change.


Need help understanding the recent observation of extra-galactic planets!

Instead of reading half-comprehensible newspaper articles, I found it more helpful to go to the original source.

There, the authors explain their method very clearly:

There is a background galaxy, the Quasar galaxy. This object, being a Quasar, has an actively accreting SMBH. The SMBH, being an actively accreting one, will emit strongly in X-Rays. Those X-Rays, being emitted from the rotating accretion disc, deep inside the SMBH's gravitational well, will suffer from strong red- and blue-shifts.
The paper works with only one line here, the Iron $\rm K\alpha$ line, observed by Chandra in 2009, 2012 and 2017.
The red- and blue-shifted light of the SMBH, the 'image' will then travel through space and time.
Eventually this light will be lensed by a foreground galaxy, and in lucky cases the foreground galaxy can create Einstein rings, or at least multiple paths so that the same quasar image crosses the lensed galaxy at multiple positions through the foreground galaxy.
The object of interest now, (RXJ1131-1231) consists of a quasar at $z=0.658$ and a foreground galaxy at $z=0.295$ that lenses the background quasar four times, giving four images in X-rays that can be spectrally resolved, to measure the distributions of red- and blueshifted $\rm K\alpha$-lines in each pixel.

And where the 'micro' in microlensing comes in is

  Light from these Quasar images crosses different locations of the foreground galaxy and is further lensed by nearby stars in the region in the lens galaxy.

So what is important here is:

No individual stars or planets are actually resolved
The amount of microlenses must therefore be inferred by spectroscopy (that's why all the talk about red and blueshift).

Now comes physics into play. The authors argue that they find anomalously high red- and blue-shifted $\rm K\alpha$-line distributions, that must be explained by planetary-mass objects.

To understand how, let's consider fig. 1 from the paper:

This shows the redshift distribution (The factor $G = \frac{E_{observed} }{ E_{rest-frame} }$) around the background Quasar accretion disc. Also shown is an example trace of a foreground object that could lens the emitted $K \alpha$ light from the disc.

** The rest of the problem can now be understood in a very visual way:

If the passing foreground object can actually lens the light in a way that would be seen in the spectrum is now determined by the ratio of the SMBH Gravitational radius and the foreground's object lens radius. This works as follows:

The gravitational radius is just the Schwarzschild radius $r_g = GM/c^2$ and sets the scale of the above image. The lens radius $r_l$ is a more involved function of cosmological distance and lens mass. One can imagine it however, being a circle traveling on the green trajectory in fig 1. that will lens all the background light that it hits into one observational data point.
The latter point must be emphasized, as the situation in fig 1. will be vastly under-resolved in the actual observation.

So if $r_l \gg r_g$ the lens will lens red- as well as blue-shifted $\rm K\alpha$-light into the image pixel and no net energyshift will be detected. If $r_l \le r_g$, then the lensed light will probably be very red or blue-shifted, according to the G-values it hits on it's trajectory.

** We're nearly done:

Now the lensing configuration of RXJ1131-1231 and their cosmological distances conspire to make the case $r_g = r_l$ correspond to Jupiter mass lens masses. This means objects of Jupiter masses and below will be able to contribute to a measurable Fe $\rm K\alpha$ redshift, but only if they are not buried in the larger lensing radius of a star (which would have $r_l \le r_g$)!

But, as we already said, the authors only can spectroscopic information about the lensed images. So they try to reproduce the redshift-distribution of the spectrae of their images with a star-only distribution, which fails.

Subsequent tests they make, indicate that they have to inject a large number of objects with $r_g=r_l$ into their test-lensing distribution to reproduce their data.

Summarizing:
This is a purely statistical argument for microlensing by a certain number of planetary-mass bodies in the X-Ray spectrae of 3 of their 4 macrolensed images. It's not a discovery and as another Q&A pointed out, there are no planets to name or point to.
It is even not the only explanation for the X-Ray data seen, as there are other competing ones.
But the presented planetary explanation is the only one that can reach sufficient redshifts seen in the data.

(P.S. I hope I didn't make it more complicated than the original paper, I've actually tried to summarize it...)


What unit is used for region Surface brightness in DS9?

ds9 is a common image analysis tool used in astronomy (http://ds9.si.edu/site/Home.html).
The units of surface brightness are simply detector counts per square arcsecond as that is all that can easily be deduced from an image. If you want to turn that into more physical units, you will need to know the gain of the CCD (how many photons produce each count) and you will need to make some assumptions about the wavelength of light that is producing the surface brightness of the emission this Note on Rayleighs may help; if all of this emission is coming from Halpha, then this is a reasonable assumption.


Why does the Moon seem larger when it is close to the horizon?

Interesting fact. It Isn't bigger!

This is a well known optical illusion that dates back hundreds of years. It's all in your head, it seems bigger because your brain perceives size as something that is relative.

This is an effect called perspective distortion and is will understood and used in photography. Phil Plait's Blog (A.K.A - The Bad Astronomer) also writes on this matter and would be worth reading.

Consider the below picture:

As the focal length of a lens gets longer the difference in relative size of foreground and background objects becomes more pronounced. Your brain then interprets this and respectively the moon seems huge!

But why does it look yellowish?

It appears yellow sometimes because our atmosphere blocks and stretches certain wavelengths. So when the moon is low in the sky, it has to pass through the most atmosphere to get to you, stretching and distorting the light through the yellows and oranges.


What is Charon's ascending node in reference to?

I'm assuming this is Pluto's equatorial plane.

That is incorrect. Charon's right ascension of ascending node with respect to Pluto's equator is undefined. Seemingly paradoxically, it is well defined with respect to Earth's equatorial plane, and since that has become the universal plane of reference, that is what is used.

Charon and Pluto are tidally locked to one another. Their mean axes of rotation and orbital axes are identical. This means that Charon's orbit has zero mean inclination with respect to Pluto's mean equator, which in turn means that Charon's mean right ascension of ascending node with respect to Pluto's mean equator is undefined.

Charon's orbit about Pluto is however inclined with respect to the Earth's mean equator. In particular, it's inclined with respect to the Earth's mean equator at the J2000 epoch (noon Terrestrial Time on 1 Jan 2000). The Earth's mean equator at the J2000 epoch is the ICRF equator, to within a fraction of a milliarcsecond. The linked data clearly states that Charon's right ascension of ascending node is measured with respect to the ICRF equator.

Reference:

Brozović, Marina, et al. "The orbits and masses of satellites of Pluto." Icarus 246 (2015): 317-329.


Could the merger of two rotating black holes create another that breaks the rotational speed limit?

The simple answer is that no, black holes cannot merge into a new black hole with a spin greater than the maximum allowed spin. The reason being that such an act would cause the black hole to show us a "naked singularity". Black holes have an event horizon which shields their singularity from observation by the outside universe. As they spin faster, that event horizon shrinks. At some point, it completely disappears, causing the singularity to become "naked" and observable to the universe. Some believe that naked singularities cannot possibly form, mainly because if they could, causality and physics would break down. This concept is known as the cosmic censorship hypothesis.

I found a statement concerning this matter in Numerical Relativity: Solving Einstein's Equations on the Computer, which I believe is relevant (emphasis mine).

  Black hole spins that are aligned with the orbital angular momentum
  increase the binary's total angular momentum. If this total angular
  momentum exceeds the maximum angular momentum of a Kerr black hole,
  then the binary cannot merge until a sufficient amount of angular
  momentum has been radiated away. Quite generally, we expect binaries
  with black hole spins aligned with the orbital angular momentum to
  merge more slowly than binaries with spins that are anti-aligned.
  This effect, sometimes referred to as "orbital hang-up", has been
  explored with numerical simulations.

One such numerical simulation looked at aligned and anti-aligned mergers and found that the aligned black holes took much longer to merge and radiated away more energy, in the form of gravitational waves, before they merged.

There is another way to burn off some of that spin angular momentum too. When spinning black holes merge, they experience a "kick" in their linear momentum. In other words, they suddenly speed up in their motion through space. This kick is a result of converting some of the two distinct black holes' orbital and spin momentums into the merged black hole's linear momentum. Quoting the same book as above concerning these kicks:

  Most of the initial calculations focused on black hole spins that are
  aligned or anti-aligned with the orbital angular momentum. The
  resulting kicks are several hundred km/s in magnitude, easily
  exceeding the maximum kick of approximately 175 km/s found for
  non-spinning black holes.

In summation, two spinning black holes cannot merge into a black hole which is spinning faster than the maximal rotation. Gravitational waves and the linear momentum "kick" play an important part in helping binary black holes to lose spin energy, such that the merged black hole doesn't exceed a maximal rotation.

I think the really interesting question that should be asked is, how do two black holes know, before merging, that their merger cannot happen as it would violate the cosmic censorship hypothesis? How do they know they need to radiate away extra energy before they can merge? What is the mechanism that prevents them from merging? Only time can tell for now.


How would you relate the respective solar gravitational pulls of two planets?

For the most part, you can assume Newton's law of universal gravitation, (see link for more information). Basically, this law can be stated as the following.

For the gravitational force between two celestial objects (or really, any two objects with mass), the force is given by the equation,

$$ F_G = G\frac{m_1 m_2}{r^2} $$

Given the mass of the two objects, $m_1$, $m_2$, and the distance between them, $r$. The constant of proportionally given by $G$, the gravitational constant (not to be confused with $g$, the acceleration of an object with respect to the conditions on the surface on the Earth).

For the specific problem(s) that you are given, you can say that, using the information from Wikipedia, for the mass of the Sun, Earth, and Neptune,

$$ m_\text{Sun} = 1.988 \times 10^{30} \text{ kg} $$
$$ m_\text{Earth} = 5.972 \times 10^{24} \text{ kg} $$
$$ m_\text{Neptune} = 1.024 \times 10^{26} \text{ kg} $$

Your question gives us a preliminary distance. The $r$ for the distance between the Sun and the Earth is given as $1 \text{ AU}$, the $r$ for the distance between the Sun and Neptune given as $30 \text{ AU}$. Note that $1 \text{ AU} = 1.496 \times 10^{11} \text{ m}$.

Therefore, for the gravitational pull (really another way of saying the gravitational force) between the Sun and Neptune,

$$F_{N \leftrightarrow S} = \left( 6.674×10^{-11} \text{ N m$^2$ kg$^{-2}$} \right) \frac{\left( 1.024 \times 10^{26} \text{ kg} \right) \left( 1.988 \times 10^{30} \text{ kg} \right)}{\left( 5.983 \times 10^{12} \text{ m} \right)^2} $$

Therefore, for the gravitational pull between the Sun and the Earth,

$$F_{E \leftrightarrow S} = \left( 6.674×10^{-11} \text{ N m$^2$ kg$^{-2}$} \right) \frac{\left( 5.972 \times 10^{24} \text{ kg} \right) \left( 1.988 \times 10^{30} \text{ kg} \right)}{\left(1.496 \times 10^{11} \text{ m}\right)^2} $$

Solving for these quantities (note that the answer that one might get may be numerically different depending on the sources of the planatary constants),

$$F_{N \leftrightarrow S} = 3.796 \times 10^{20} \text{ N}$$

$$F_{E \leftrightarrow S} = 3.541 \times 10^{22} \text{ N}$$

A bit of extra information, the gravitational force felt by the Earth or Neptune from the Sun is exactly the same as the force that the Earth or Neptune acts upon the Sun, as per Newton's third law of motion. But, by now, I suspect that the problem is answered to where taking the ratio of the two is a trivial manner.

Suggested by uhoh:
The force exerted by the Sun on Neptune (and vice versa) is about 93.3 times weaker than that of the force exerted by the Sun on the Earth(and vice versa).


Finding Asteroid Geocentric Longitude and Latitude

This can be accomplished through the following steps :

compute the position of the Earth in the same heliocentric frame,
express both positions (Earth and asteroid) in heliocentric rectangular coordinates,
change the origin of the frame to get the location of the asteroid in geocentric rectangular coordinates,
convert the position of the asteroid in geocentric latitude and longitude.


Effect of Hot/Warm Jupiter atmosphere loss on other planets

The primary effects wouldn't be the things you ask about.   First, an object the size of Jupiter would change the day and night sky drastically.    The sky now is basically, Sun, Moon and a bunch of yellow dots (and one reddish dot) mostly fixed, 4 move around visibly to the naked eye (Venus, Mars, Jupiter, Saturn).

Venus stands out a little as larger and brighter, but it looks pretty much like an abnormally bright star.   All that changes if you put a Jupiter sized planet in the orbits around where Mercury or Venus are.   Jupiter is about 1/10th the diameter of the Sun.   If you put it about midway between Earth and the Sun it would be about 1/5th the diameter and transits over the sun would be obvious, unlike Venus transits which are observable (not a good idea without protective lenses), but easily missed.

Jupiter in an orbit between the Earth and the Sun would have apparent phases like the Moon and visible size variation.  It's likely that in such a scenario, the geocentric model would never have been adapted, cause Jupiter would raise too many questions in this hypothetical.   Aristotle and Ptolomy were smart.   This situation would have been enough to perhaps look at things differently.

There's also the 3 body gravitation problem.   Jupiter inside the Earth's orbit might wreck havoc with the Earth's orbit.  Earth's eccentricity might vary a lot more perhaps leading to more extreme climate change periods.   There's a good chance an Earth like planet's relatively stable orbit would be subject too a lot of undesirable changes with a hot Jupiter between it and the sun.  Comet impacts could be a problem too.   A distant Jupiter changing the orbit of passing comets helps the Earth avoid comet impacts.  A Jupiter that close to the sun could do the opposite.    There's also the migration problem.   If Jupiter forms out far from the sun and migrates inwards, that would wreck havoc with any planet orbits it passes.    So there's a huge gravity problem with your scenario.   If you're looking for Earth like planets, solar-systems with hot Jupiters are probably not the best place to look.

that said, onto your questions.

  What effect will this bled atmosphere from HJ have on EL, if any?

  If there is an effect, when does it happen? On all of EL's orbit,
  despite the different inclinations? Only when EL crosses a transit
  node? Only when there's a transit (or rather, when EL passes through
  the transit node a certain amount of time after HJ does, to account
  for the velocity of the atmospheric particles through space)?

  Will this bled atmosphere be (naked-eye) visible from EL? What would
  it look like?

It's difficult to say with precision.   Certainly if the bleeding was significant enough it might be visible, but on average I think probably not.   As a bad comparison, we can't see Coronal Mass Ejections.   Most of the "bleeding" from a hot Jupiter would be surface gas molecules, primarily hydrogen and helium, most of it uncharged.  Those gas molecules, spread out in space would be essentially transparent, unless it was an absolutely gonzo insane amount.    Very sensitive cameras might pick them up, but they should be largely invisible from Earth.

Comet tails are visible, even occasionally visible during the day, but comet tails are different.  They're full of dust that breaks off the comet and the escape velocity off a comet is low, so the dust that breaks off from the comet tends to follow in a similar orbit, hence the relative density and apparent cohesion of a comet tail.   With a hot Jupiter and gas escaping in a variety of directions presumably due to heat escape or Jeans escape primarily, there would be very little cohesion.   Escaped gas molecules would quickly spread out in space.   The charged particles that escape from this theoretical Jupiter might be enough to create a faint Aurora, but should be significantly smaller than the charged particles escaping the sun.   The uncharged particles, a small percentage would collide with Earth, most would miss.   There's no scenario I can imagine that would be visibly impressive from such bleeding unless the hot Jupiter was super-close to the sun, say closer than Mercury and the bleeding was extreme.   Even then, the primary visual would be the stream of gas falling into the sun, not the spread of gas moving towards and past the Earth's orbit.

Something not too different happened in the Earth's distant past.  Venus used to have oceans and it bled probably at least an Atlantic ocean of water, perhaps a Pacific ocean of water into space that drifted mostly past the Earth.  Likely there was very little effect to Earth drifting through a very disburse amount of water-vapor in it's orbit.

That's a ballpark answer anyway.  Corrections are welcome.


When the sun explodes, will some of the planets in the Solar System survive and become rogue planets?

The Sun will not become a supernova, it will never explode. A star must have about 8 times more mass than the Sun in order to cause a supernova explosion. When it starts to fuse helium, the Sun will become a red giant and expand out to about Earth's orbit and throw out gas to form a beautiful so called planetary nebula. Mercury, Venus and maybe Earth will be inside of its photosphere when it is at the largest. I don't know if they will evaporate completely inside the thin hot gas of the red giant Sun.

The remaining planets will keep orbiting the white dwarf remnant, which is not a star (no fusion) but a compact planet sized object. It will shine like a star because of its stored heat. Orbits will change because the Sun will have only 40% of its mass left, the rest having been thrown out to form a nebula, over very long time. Orbits should become larger, and planets maybe will be pushed out a bit by the ejections from the red giant too. Gas planets maybe get their atmospheres a bit stripped off, icy dwarf planets like Ceres and Pluto could melt and evaporate a bit of their mass since the helium burning red giant is hotter than the Sun today. But I'm kind of guessing here, I don't know if those are significant effects.


Annes Astronomy News says:

  Fleming 1 is a planetary nebula that lies about 10,000 light-years away in the constellation of Centaurus, while moving away from us at approximately 28.6 kilometers per second. It is named after the Scottish astronomer Williamina Fleming, who discovered the nebula in 1910.

Williamina Paton Stevens Fleming

  (May 15, 1857 – May 21, 1911) was a Scottish astronomer active in the United States. During her career, she helped develop a common designation system for stars and cataloged thousands of stars and other astronomical phenomena. Among several career achievements that advanced astronomy, Fleming is noted for her discovery of the Horsehead Nebula in 1888

That article's Notable discoveries section says

  During her career, Fleming discovered a total of 59 gaseous nebulae, over 310 variable stars, and 10 novae.

  Most notably, in 1888, Fleming discovered the Horsehead Nebula on a telescope-photogrammetry plate made by astronomer W. H. Pickering, brother of E.C. Pickering. She described the bright nebula (later known as IC 434) as having "a semicircular indentation 5 minutes in diameter 30 minutes south of Zeta Orionis". Subsequent professional publications neglected to give credit to Fleming for the discovery. The first Dreyer Index Catalogue omitted Fleming's name from the list of contributors having then discovered sky objects at Harvard, attributing the entire work merely to "Pickering". However, by the time the second Dreyer Index Catalogue was published in 1908, Fleming and her female colleagues at the HCO were sufficiently well-known and received proper credit for their discoveries.

  Fleming is also credited with the discovery of the first white dwarf:

    The first person who knew of the existence of white dwarfs was Mrs. Fleming; the next two, an hour or two later, Professor E. C. Pickering and I. With characteristic generosity, Pickering had volunteered to have the spectra of the stars which I had observed for parallax looked up on the Harvard plates. All those of faint absolute magnitude turned out to be of class G or later. Moved with curiosity I asked him about the companion of 40 Eridani. Characteristically, again, he telephoned to Mrs. Fleming who reported within an hour or so, that it was of Class A.

    — Henry Norris Russell13

  Fleming published her discovery of white dwarf stars in 1910. Her other notable publications include A Photographic Study of Variable Stars (1907), a list of 222 variable stars she had discovered; and Spectra and Photographic Magnitudes of Stars in Standard Regions (1911).

  She died of pneumonia in Boston on May 21, 1911

The quote by is from Notes on White Dwarfs and Small Companions Astronomical Journal, Vol. 51, p. 13 (1944)

Though effusing complements to Pickering, it seems that the first line "The first person who knew of the existence of white dwarfs was Mrs. Fleming" is meant to ensure some bit of historical record was set straight.

The article continues with substantial sections on Recognition and Legacy that I won't reproduce here, though other answers to this question can address these.

From this answer to Has great eyesight been necessary for astronomers?:

  The title of a book on this subject The Glass Universe is likely a play on words, the photographic plates were made of glass, and so was the ceiling. (Space.com review, The Atlantic, Magiscenter, BBC interview in YouTube, NPR)

  It is strongly recommended to see and/or listen to the author Dava Sobel talk in the YouTube video The Glass Ceiling and The Glass Universe and in Dava Sobel: The Glass Universe

Image source


Are there upper or lower mass limits for black holes?

In classical General Relativity, black holes can exist at any size (mass) without any problem. The upper limit is given by the available mass of the universe and there is no theoretical lower limit.

As already noted in the question, quantum effects like Hawking radiation set up lower limits on stable black holes; the ones with too low mass will decay quickly into radiation.


Absolutely.  You are looking for a binocular eyepiece, or a binocular viewer.  Most of them require you to insert two matching ocular eyepieces, so it can be fairly expensive.

Most major telescope eyepiece manufacturers sell them.  I'm not sure what the rules are for mentioning brand names,  but a quick google for 'telescope binocular viewer' will turn up many examples.

You will require a decent telescope that can take standard 1.25" or 2" eyepieces.  Also, the telescope needs a lot of 'back focus', which many Newtonian telescopes do not have. You may also need a barlow lens for focus, which will increase magnification but cut down the field of view and light gathering.

It would be best to check the specifications of the specific bino viewer and your telescope to see if they are compatible.  If you have a Schmidt-Cassegrain scope, you should be fine and don't need the barlow.


Are there any electrons inside the sun / a star?

Plasma consists of ionised particles. The electrons are not bound to their nuclei and are free to move around within the star.

However, they are not free to leave the star. Without the electrons to maintain a neutral charge, there would be massive electrical repulsion from all the positive protons and helium nuclei, the result would probably be something like the formation of a black hole larger than the observable universe... The "What if", blog considered something like this in Proton Earth, electron moon and concluded that it would be one of very few things that could actually destroy the universe.

In conclusion, the sun contains about $10^{57}$ electrons. One for every proton.


Spectral type is determined by (and is an indicator of) photospheric temperature and surface gravity. The classification of G2V applies to a star with a photospheric temperature of about 5800K and a surface gravity of about 100-1000 m/s$^{2}$.

The Sun's temperature and gravity have not changed greatly since it first appeared on the "zero age main sequence" - it has got a little bigger and more luminous and a little bit hotter (by about 170 K- Schroder & Smith 2008) - so maybe it was a G3V-G4V star at the zero age main sequence (Mamajek 2018).

However, there were much bigger changes during the 20 million years or so before the Sun reached the ZAMS. It was cooler and briefly much larger. This would be reflected in a spectral classification that was something like K5IV- M0IV; indicating a subgiant-like gravity and a temperature of about 4000K. As it aged, it would then shrink and become hotter, passing through the G5V-K5V spectral types before reaching the main sequence.

If you can synthesise these two diagrams in your head, you can see what I mean.

Pre main sequence evolution in the Hertzsprung-Russell diagram

Spectral types on the Hertzsprung-Russell diagram


What constellation would the Sun be in from an exoplanet?

The sun would appear roughly half-way between Capella and the W of Cassiopeia. The constellations are a human invention, and don't correspond to actual groups of stars. Since all the nearby stars would be in somewhat different positions the constellations would be mixed up, for example, Sirius would be close to Betelgeuse. You can't really talk about constellations except from the perspective of Earth, but you could say that the sun would be near some of the stars that form Perseus and Camelopardalis.

From more distant exoplanets the constellations would be completely mixed up, so there would be no sense in describing the sun's location in terms of a constellation. The sun would also be too dim to be seen with the naked eye from many exoplanets.


What kind of collision of celestial bodies would be visible to the naked eye?

OK, so... you said the collision with Shoemaker-Levy 9 and Jupiter happened on the non-visible side. What if we'd seen it? The energy released by the largest fragment collision was estimated at 6 million megatons of TNT, which is 2.52e22 joules. The Sun's luminosity is 3.846e26 watts, by comparison. So obviously the collision won't rival the Sun. Using the luminosity-distance relation and taking the 6 million megatons as pure luminosity (an overestimate, but we're not going for precision here) gives the result of 0.0033 watts/square metre.

Relating flux to apparent magnitude, we get an apparent magnitude of the collision of... approximately -12. That's about the brightness of a full moon. Bear in mind this is a very rough calculation and it makes the assumption that all the energy was released over a second and all of it was light, so probably the real value would be a few magnitudes down But, assuming I've made no terrible calculation errors, it would definitely have been visible.

Given that was a relatively tiny comet, your example of two of Jupiter's moons colliding would release FAR more energy, therefore would be clearly visible, I'd imagine. All depends on how much energy gets emitted as light.


What equipment and techniques were used to study Betelgeuse's diameter in 1920?

These measurements were made by Michelson & Pease (1921) using the 100-inch Cassegrain telescope at the Mount Wilson Observatory. Modern-day interferometry typically uses multiple telescopes (e.g. IOTA and the Keck Observatory interferometer), as this is typically needed to achieve long baselines (with long baselines admittedly being less important for short-wavelength observing than for, say, radio waves). However, the interferometer the two astronomers used was slightly cruder: a set of four mirrors mounted above the 100-inch mirror. These mirrors were movable, reaching separations of up to 20 feet. Light would reach the outer two mirrors (spaced 20 feet apart) and then be reflected to the inner two mirrors, which would then send the two beams to the main mirror, turning the telescope into an interferometer.

This was indeed the first time the diameter of a star (excluding the Sun) was measured; Michelson had built the instrument three decades earlier, but tested it by observing the Galilean moons. Betelgeuse is an obvious choice, as its large envelope and proximity to the Solar System mean it has one of the largest stellar angular diameters at any wavelength as seen from Earth.

I believe the interferometer is still in existence, in the observatory's museum. Here, however, is the diagram presented by Michelson and Pease in their 1921 paper:

With regards to the star's physical diameter: The pair measured an angular diameter of 47 milliarcseconds, which is certainly in line with the accepted values today. There is variation in Betelgeuse's angular diameter, of course, due in part to the star's variability, but 47 milliarcseconds falls right in the middle of the range expected at optical wavelengths. Had they had the correct parallax, they would likely have arrived at a very reasonable radius.


Was the Earth-Sun distance ever calculated based on the time difference between night and day?

You have a good idea. You mentioned 4 effects already (clock accuracy, latitude, time of year, and clear horizon), but there is another effect that is larger than those: atmospheric refraction. Refraction causes a rising object to appear to be half a degree higher than reality. Refraction depends on atmospheric pressure and temperature, so it may be harder to compensate for it.
This article on Sky & Telescope indicates that the amount of refraction is not well understood: We Don’t Really Know When the Sun Rises

But Michigan Tech dissertation work by Teresa Wilson (now at the U.S. Naval Observatory) suggests that our estimates are often off by 1 to 5 minutes.
She found that, overall, predicted times varied in accuracy by location and season, with sunrise times over land generally being early in the summer months and late in the winter. Summer showed the largest discrepancies, probably due to the pronounced refractive effect that the large temperature difference in the atmosphere has during those months. Mirage effects due to cold air topped by warm over water horizons also exacerbated lags in sunset times throughout the year, sometimes by up to 5 minutes.


Why would there be no day/night cycle even though the firmament would still be rotating, according to Copernicus

Find a spherical object and imagine it's a celestial sphere with the world at the middle... consider the daily rotation of the sphere around as a spinning top, and a yearly tilt through the constellations which is the same as an unstable spinning top which is about to tip over.

Now, rotate it and dont spin the days, and only spin the sphere for the yearly seasons.

How fast are all points on the sphere rotating as it tilts, same as a spinning top is not straight, and omitting the individual rotations?

All points on the sphere are going at the same speed, the top of an orange describes a 1cm axial tilt, and the sides of the sphere also describe a 1cm axial tilt... put the orange sideways and tilt it again, the rind is at the same speed as the side whichever way the orange is viewer. How to combine the slower star speed of rotation in 24 hours with the equal star speed of rotation through the seasons? The previous theory didn't give two speeds and axes of rotation.

So, the 7 heavens other than the stars can be described as an epicycle on a 2D, as a turntable and epicycle.

For the stars, the 2D epicycle is more complicated, because it has to account for two speeds of rotation, the constant one around the seasons and the one that is slower in the middle. The epicycle geometry breaks down in 3D as a sphere, and only a sphere can explain the two types of rotations in the stars, and the 2100 year one.

Copernicus said: A sphere cannot move as an epicycle. Take an orange and try to move it as an epicycle.


How many planets does Omicron Persei have?

Currently http://exoplanet.eu/ (which lists more exoplanets than the official NASA archive) lists no known planets of Omicron Persei. So currently we know of none, it is outside of the Kepler field so has no Kepler candidates either.

Though just because none are known that does not mean it has none, it may be that their mass and or orbital plane is such that it is very difficult to identify planets or no one has looked with a suitable instrument.


What suggests the cyclic nature of the universe? Proposed by Stephen Hawking

There's nothing "cyclic" in the universe Hawking is describing. Rather, Hawking's thought experiment applies to a particular shape of the universe.

The observable universe is, for practical purposes, a sphere 93 billion light years in diameter, with us at the centre. Observations suggest that this observable universe is pretty much isotropic and homogenous. If anything exists beyond this sphere, it's so far away that the light from it hasn't had time (i.e. the age of the universe) to reach us.

It's rational to assume that there is a universe beyond what we can see. This is called the global universe. Since we can't observe it, we can only theorise what its properties might be. There are three broad possibilities for the shape of an isotropic global universe:

  flat (no curvature)
  closed (positively curved)
  open (negativelycurved)

Current observations indicate that the universe is flat, with only a 0.4% margin of error - in other words, if it's a simply-connected universe (e.g. not a torus) then it extends infinitely in all directions. Hawking's thought experiment, on the other hand, takes place in a positively curved universe. This can be thought of as a three-dimensional hypersphere. One way to picture this is as though we're an atom in the surface fabric of a balloon: from where we are, everything looks flat and infinite, but in fact the flatness curves away to form a spherical shape. Importantly, this shape is "closed" - in other words, finite, as is the surface area of a sphere.

What Hawking is suggesting is that if you set out in a straight line on the surface of this balloon, and so long as your movement away from the starting point is faster than the rate the balloon is expanding, then eventually you'll circumnavigate the balloon and end up where you began.

It's purely meant as a demonstration of what a closed (positively curved) universe "looks like". Given that our local, observable universe appears very close to flat, the curvature would have to be so small that the "balloon" would be truly immense; circumnavigating it would therefore take so long (say, a billion billion years) that our Milky Way would be unrecognisable, if it existed at all, by the time you returned to your starting point.

To be fair to Hawking, in 1988 when his ground-breaking A Brief History of Time was published, there was good reason to think that the universe was positively curved, since a truly flat universe (curvature exactly zero) seemed an improbable "special case". However, the discovery in 1998 of the accelerating expansion of the universe, which provided direct evidence of dark energy and firmly installed Lambda-CDM as the preferred model, rewrote the science and was probably one of the reasons Hawking decided to rewrite his book.

Even so, when A Briefer History was published in 2005, only the earliest data from the Wilkinson Microwave Anisotropy Probe (WMAP) was available, putting the total density of the universe at 1.02 (allowing a realistic possibility of a slight positive curvature), whereas by 2012 the WMAP data had reduced this to 1.0027 (very close to flat).


How "fast" do astronomical events occur?

A supernova (and many other interesting events) starts abruptly -- a flash of neutrinos lasting just a fraction of a second, but then different aspects of the event play out over minutes, days, hours or years.  The astronomy community has for some years been developing increasingly sophisticates ways to deal with such events early on. Instruments with a very wide field of view (gravitational observatories, neutrino observatories, gamma ray detectors on satellites) detect the events initially and bring first networks of robotic telescopes and then humans and still larger telescopes into the loop to try and get as much information as possible about the violent and fast developing early stages. See for example http://growth.caltech.edu/

Other things are slower. The "something coming out of a black hole" (not really that, just something coming from very close to a black hole) event takes a long time, this was just the first occasion when they'd observed that black hole with the right instrument.


The star that died and became the Crab Nebula and Crab Pulsar?

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


There is an accurate depiction of the moon phase and shape in Stellarium (see below) that you might find useful. You can select your observing location and time anywhere on Earth (or another planet if you so wish). Then you can just search for the moon in the search window (if you cant spot it in the sky), and you have a real representation of how the moon looks at that specific time and location. However if you need to repeat this calculation many times then this isn't the best solution.


Roche Zone of a black hole vs Roche Zone of Neutron Star

The Roche limit is defined in classical physics, dealing with materials that behave classically. That said, of course people have tried to estimate the counterpart in relativistic conditions. It turns out that it behaves roughly like the classical case when considering a liquid body orbiting a black hole. For the neutron star case and a more massive stellar mass black hole the limit is not too far away from the ISCO.

Would a black hole be affected by tidal forces? A tidal force is after all just the effect of the radially varying curvature of spacetime, and a black hole embedded in such a curved spacetime does show distortion. Not only that, but one can apparently treat the black hole as a fluid sphere that can even dissipate! (paper, presentation2) As the plot in the second presentation shows, yes, the horizon can get distorted by a nearby massive orbiting object.

However, the laws of black hole thermodynamics do not allow the horizon area to go down. Splitting a mass 2M hole into two holes of mass M would change the area from $4 \times (4\pi G^2 M^2/c^4)$ to $2\times (4\pi G^2 M^2/c^4)$. So that is not going to happen (unless there was a process that could release a black hole's worth of entropy in the outside universe - a very tall order).

In the end, what would happen is that the deformations would dissipate orbital energy (by heating the neutron star and gravitational radiation) until the objects spiralled together, the neutron star would cross the relativistic Roche limit and shortly after the remnants would plunge inside the ISCO and be absorbed.

The magnetic field case would complicate things in various ways, but remember that even magnetar magnetic fields only have the density of $10^4$ times lead - compared to the neutron star density this is nothing. There would presumably be a lot of electromagnetic dissipation, jets, and complex torques, but I suspect the endgame is the same.


How will microshutter arrays be used in the James Webb and future space telescopes?

The role of the microshutters is NOT to act as pinholes or coded aperture and to produce a focused image, similar to a mirror or lens.
Its job is just to separate the the light from different sources. This is achieved simply by opening only the shutters corresponding to the sources of interest. See the following image:

This is important for spectroscopy as here the light is split in its wavelengths and dispersed in one direction by a grating or prism. That is why Webb's micro-shutter is only used for spectroscopy.  In order to not have a bunch of overlapping spectra you need some shutters to block everything you are not interested in.
The spread out image of multiple sources on the detector than looks like this:

In other words: As you can not detect a 3D image (2 spacial + 1 spectral) on a 2D detector, you have to make some sacrifice by blocking much of your field of view.

This sketch makes the process quite clear:

Further sources: Wikipedia Closed-source micro-shutter decription


This is a common question that comes up - one I've (at least tried to) answer a few times.

Let's try to understand what the Big Bang is. The Big Bang Theory is a model, and scientists came up with it when they observed the universe expanding - and thought that since the universe is now expanding it must have been smaller in the past. Running the clock backwards eventually means all of the universe was squeezed into a "singularity". Our physical models of the universe break down at this point - so we are not too sure of what the universe was like at the Big Bang.

Notice that I said the universe was squeezed into a singularity - not the stuff inside the universe. This means space-time was also compressed into a single point. The universe is not expanding out into empty space - it is space-time itself that is expanding. Hence, there was no "before" the Big Bang - time (as well as space) was created when the Big Bang happened. There is no outside the universe! This means there is no outside the Big Bang!

You can't go back in time and watch the Big Bang - since there is no universe or space (or time) outside the singularity. This can be very confusing, but I hope this helps!

For a few places to look for more info:

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

https://en.wikipedia.org/wiki/Big_Bang#Singularity


What determines if a comet's debris will produce periodic meteor showers?

Meteor showers tend to be formed from dust from short period comets.  As they swing repeatedly around the sun, they build up a stream of dust in orbits close to their own orbit. Some will be pushed slightly off the comet's orbit and end up orbiting faster or slower than the comet, and eventually the stream of dust will go all around the sun.
The other thing that there needs to be to get a meteor shower is a comet whose orbit is close to crossing the Earth's orbit. Comet Swift-Tuttle is one such body. Its orbit passes only 130,000 km from the Earths orbit (close enough that there was at one point some concern that it could collide with the Earth in the future. This has now been ruled out for the next two thousand years).
Neowise will release dust, but most of that dust will still be near the comet now and it will continue to orbit with the comet. So that dust will be leaving the inner solar system and will become too spread out to cause a significant meteor shower by the time it returns in several thousand years. Neowise doesn't approach the sun often enough to build up a significant stream of dust particles.
So to summarize it's not really the comet's tail intersecting the Earth's orbit that leads to showers but the debris stream built up along the comet's orbit as it repeatedly goes around the Sun. In long-period comets the debris stream is spread out too much along the orbit to produce a significant stream but short-period comets can produce a significant stream leading to showers.


What was the age of the universe when the average density was one atmosphere?

The critical density of the universe is
$$\rho_c=\frac{3H_0^2}{8\pi G},$$
with $H_0=67.8 \frac{\mbox{km}/\mbox{s}}{\mbox{Mpc}}$ the Hubble constant and $G=6.673848\cdot 10^{-11}\frac{\mbox{m}^3}{\mbox{kg }\mbox{s}^2}$ Newton's gravitational constant.
Hence with one parsec $1\mbox{pc}=3.0857×10^{16} \mbox{m}$,
$$\rho_c=\frac{3\cdot (67.8\cdot 10^3\cdot \frac{\mbox{m}/\mbox{s}}{3.0857×10^{22} \mbox{m}})^2}{8\pi\cdot 6.673848\cdot 10^{-11}\frac{\mbox{m}^3}{\mbox{kg}\mbox{s}^2}}=8.635 \cdot 10^{-27}\frac{\mbox{kg}}{\mbox{m}^3}.$$

Ordinary matter content is $4.9\%$, the universe has total density close to critical density, hence ordinary matter density is
$$0.049\cdot 8.635 \cdot 10^{-27}\frac{\mbox{kg}}{\mbox{m}^3}=4.23 \cdot 10^{-28}\frac{\mbox{kg}}{\mbox{m}^3}.$$
(The Hubble constant is not precisely known, so the calculated density depends a bit on the precise value.)

Density of water:
The density of water is about $10^3 \mbox{kg}/\mbox{m}^3$.
That's a factor $4.23 \cdot 10^{31}$ denser than the average universe.
The space needs to have been reduced by a factor $\sqrt[3]{4.23 \cdot 10^{31}}=3.48\cdot 10^{10}$ in each if the three spatial dimensions. That's the inverse of the scale factor, hence corresponds to a redshift of $z=3.48\cdot 10^{10}$.

Using the Cosmology Calculator on this website, the cosmological parameters $H_0 = 67.11$ km/s/Mpc (slightly different to the above value),
$\Omega_{\Lambda} = 0.6825$ provided by the Planck project, and setting $\Omega_M = 1- \Omega_{\Lambda} = 0.3175$ the age of the universe was 0.022 seconds at the redshift $z=3.48\cdot 10^{10}$, hence when it was of the average density of water.

Density of air at sea level:
The density of air at sea level is $1.225 \frac{\mbox{kg}}{\mbox{m}^3}$. The corresponding redshift is about $z=1.423\cdot 10^9$ yielding an age of the universe of 12.68 seconds.


How can I calculate moon phase and height?

The first question is quite simple if you imagine how the Earth is looked from above. To help you with that I've added a little drawing below.

Probably you know that the Sun rises on the East and sets on the West.  Imagine now how the sun illuminates the Earth: Europe is in the afternoon when the East cost of US is waking up. Imaging this you can see that the Earth is turning anti-clockwise if looked from the north.

Now, imagine you are somewhere in US (so we can use the drawing from below), the sun is rising (on the East) and the moon is above you (something close to the zenith).  This means the moon the configuration of the Sun-Earth-Moon system is something like in the drawing.  As you can see in the drawing I've kept in white what's illuminated from the Sun.  In Spanish we use to say that the moon is a bit of a liar, this comes because it shows a C shape when is waxing (decreciente) and a D shape when is crescent (creciente), so just the opposite letter of its action (don't know if there's a similar expression in other languages, and notice that the moon is not a liar for the Spanish speakers in the Souther hemisphere).  So, coming back to the drawing, you can imagine that from the red-star position you will see the moon with a C shape, therefore is waxing... which means in 7 days or so would be new moon so is going to move between the Sun and Earth.. and we can see how the moon rotates around the Earth in the same direction than the earth rotates around itself.

I don't really understand your second question, so I cannot help much with that.  But, if you mean to get a roughly calculation as on where the moon is at a certain time knowing its phase you can calculate it by knowing that the Earth spins 15°/hour and that the moon positions for each phase.  Therefore, new moon the moon will be closer to the zenith at noon (local solar time), 6am for waxing moon, 12 am for full moon and 6pm for crescent moon. However, this is just a roughly approximation as the moon movement happens on a plane which is 5.14° inclined from the ecliptic plane which corresponds to positions between 18.29° and 28.58° latitude on Earth.  So the same phase moon can be at different heights at different times of the year at the same hour.


Is the centre of axial precession perfectly perpendicular to the ecliptic?

The answer lies in the cause of the precession. Spinning objects don't simply precess of their own accord, it is the effect of other bodies acting on the rotation object to induce a torque.

In the case of the precession of the equinoxes, the precession is caused by the other bodies in the solar system, most significantly, the sun and the moon. Because the Earth is slightly oblique (the equatorial radius is greater than the polar radius), there is an equatorial bulge. Further, because of the tilt of the Earth's axis, this bulge is tilted with respect to the ecliptic. The gravitational pull of the sun acts more strongly on the side of the bulge closer to it and creates a small torque. Similar effects take place due to the presence of the moon, the combined effect is known as lunisolar precession. Other near by massive bodies also play a role.

A complete understanding of how/why this torque produces the effects of precession requires an understanding of angular momentum in general and is relatively trivial given that.

To answer the question explicitly, the centre of axial precession points roughly along the ecliptic pole because the effect is caused by the sun (which is always on the ecliptic) and the moon (which is very close to the ecliptic).


Can we create neutron star by pressing earth? Or other objects?

It depends on what you mean by "can".

If you want to make the Earth into a black hole, you would have to compress it down to about the size of a marble.  That is impossible.  That is not just impossible with the technology of today, that is impossible with anything less than magic.

However, if we did make an Earth-mass black hole, it would be stable. (Except to Hawking radiation)

Now if you could somehow compress the Earth into a slightly smaller space, you would need to overcome electron-degeneracy and at some point before making your black hole you would form neutron matter. This is impossible. Nothing can do this, not now and not ever.

An Earth-mass piece of neutron matter, however, would not be stable. Its own gravity is not sufficient to keep it from expanding, so it would very very rapidly explode. The same with Jupiter. A body needs to be larger than about 15% of the mass of the sun to have enough gravity as a neutron star to be stable. Jupiter has about 0.1% of the sun's mass

We can't create a small neutron star by compressing the Earth, unless we use magic, and if we stop squeezing it would explode.


Is it possible to see mercury transit "clearly" by the naked eye?

I did see the Venus transit before the Sun in 2004.

I used no telescope, but of course I used proper solar eclipse glasses to protect my eyes. The black circle was small but clearly visible.

But Mercury is much smaller, the transit is not visible without a telescope. See Wikipedia.

Here an image of a Venus transit:

Here the much smaller Mercury:

The dark spots to the right and left are sunspots, Mercury is the very tiny black spot below the center of the Sun.

Images from Wikipedia.


Problems with computing X,Y,Z positions of planets at a certain time in Unity3D (C#)

I think E and M are in degrees in your code, but the trig functions in the math library in C# expect an argument in radians.

One way to handle this is to define constants  radians = Pi/180 and degrees = 180/pi then always do Mathf.Sin(M*radians). You will also need to do this for the lines float xx=.

Alternatively, you could work entirely in radians. Convert all angles to radians at the start and only use radians in the code. You would need to convert n, i, w and M to radians and remember that e and v will also be in radians.


An 8 or 10 inch Newtonian telescope is principally a fine thing. Much telescope for little money.
A Dobson is a Newtonian telescope mounted on a rockerbox (see Alt/Az mount for a discussion). Consequences are: cheap, lightweight, trivial setup, your're either on your knees or on a chair, too seldom in between, it needs constant adjustment because the sky doesn't move that way.
If you want to track stars you need an equatorial mount, a cross of 2 axes of which one points to the celestial north (or south) pole. The same tube from the Dobson can also be mounted there. They can be motorized on one or both axes or/and have a computer built in to automagically find an object out of a database, then they are decorated with the "goto" attribute.
They are heavy, expensive (multiple times the price of a newton tube) and need a setup ritual each time they are moved. But once oriented, a motorized mount, connected to some software and computer, can be left alone for hours and will track the stars it points to.

So, the question is: do I want it light and cheap and don't I care about gymnastics, then the Dob. Do I want to go further and track things, do photography, than the other way. Perhaps with a sexy apochromatic refractor. But that's easily more than 20 times the price of a dob without camera stuff.
I personally don't do goto. Finding things is half of the fun for me. I go with equatorial mounts, a small portable one and one for photography setup I can drag outside. How about, you start with the Dob. If you bite, you're going to invest more anyway :-)
Edit: there is a third way (and potentially others) that might be worth considering, that would be a telescope like the one in this thread, a small reflector mounted on a computerized alt/az mount.


Does the edge of the Universe travel faster than the speed of light?

The universe has no edge, as far as we can tell.

Now, the observable universe does have an "edge" - it's the most distant places we can see. But that's just us and our limited instruments.

Due to the expansion of the universe, it is true that two objects, far enough from each other, could actually move faster than light relative to each other. That means light sent from one of them would never reach the other.

However, the speed of light, no matter how you measure it, is always the same. It does not matter how fast or slow its source was moving - when you measure the speed of that light, it's always, ALWAYS the same, for you, locally.

So how can that be?

Well, it's because the universe is expanding because space itself is growing. There's more space all the time in between distant objects. And the further apart two objects are, the faster the space between them is growing.

It's like a canvas with galaxies painted on it, but the canvas itself is growing. However, when light is traveling across the canvas, its speed is always the same relative to whatever galaxies it may find along the way.


How is the bar formed in a Barred-Spiral galaxy?

"Galactic bars develop when stellar orbits in a spiral galaxy become unstable and deviate from a circular path. The tiny elongations
in the stars’ orbits grow and get locked into place, forming a bar. The bar becomes even more pronounced as it collects more and
more stars in elliptical orbits. Eventually, a high fraction of the stars in the galaxy’s inner region join the bar. This process has been
demonstrated repeatedly with computer-based simulations."

Source:
http://hubblesite.org/hubble_discoveries/science_year_in_review/pdf/2008/barred_spiral_galaxies_and_galactic_evolution.pdf


Mars is currently experiencing a global dust storm. There are virtually no features visible.

Otherwise, some dark markings and the polar cap would be visible when conditions are favorable (meaning when it is close to us as it is now.)


Why does Saturn have a more prominent equatorial bulge and higher flattening ratio than Jupiter despite rotating slower?

Consider the dimensionless parameter $k$ for some planet, given by
$$k \equiv \frac{\Omega^2 a^3}{G M}$$
where

$\Omega$ is the planet's equatorial sidereal rotation rate,
$a$ is the planet's equatorial radius,
$G$ is the universal gravitational constant, and
$M$ is the planet's mass.

The values for this dimensionless parameter are

0.003461 for the Earth,
0.08919 for Jupiter, and
0.1576 for Saturn.

The factor $k$ is essentially the ratio of centrifugal effects to gravitational effects at the "surface" at the equator of the planet, surface in scare quotes because giant planets don't quite have a surface. Ignoring higher order terms, the flattening of a uniform density planet in hydrostatic equilibrium can be shown to be $\frac54k$. The mathematics that go into this derivation are a bit hairy, even for a planet of uniform density, so for now I'll just state this as a fact. (But see Rotational Flattening, for example. Note well: The derivation at this site involves a bit of hand-waving. Overcoming the hand-waving involves even more mathematics.)

The assumption of uniform density is not that good. Even if one does allow for density variations as a function of radial distance from the center of the planet, the parameter $k$ defined above plays a significant role in the computation of what the flattening should be. The observed flattening of Saturn versus that of Jupiter is consistent with the fact that the value of $k$ for Saturn is 1.76 times that of the value for Jupiter.


Why do astronauts move so slowly in zero gravity?

It's more for safety than anything else. Space is a very dangerous place for so many reasons. And making mistakes can very easily cause death.

Being weightless does not mean you lose mass, so momentum is just as difficult as ever. But whereas on the ground you can easily use friction to stop, in space if you try to stop against the floor you will just move off it. You can only stop by holding something, or pressing against something close to perpendicular to your movement.

As an example, imagine you jumped with all your force from one wall in the ISS. You will notice as you approach the other end that you are travelling at speed, head first, with no safe way to stop. Even reaching out to a handhold on a side wall will whip you round and into that wall, possibly injuring yourself or damaging instruments on the wall.

Look at any video from the ISS to see how carefully they move.

Similarly, outside the ISS, you want to do everything slowly so you don't damage your suit, miss a handhold or otherwise cause death.


The Earth and the Moon do interact, but because they have different sizes and masses the effect on each is different.

The Moon rotates once in its orbit around the Earth, causing it to constantly show the same face. This was the Earth's effect on the Moon due to tidal dragging by gravity.

It's pretty stable, the Moon will continue to show the same face to the Earth even as the tidal forces cause it to orbit farther and farther away.

We actually see slightly more than 50% of the Moon. The Moon's orbit causes its aspect to nod because its orbit is slightly elliptical and shimmy because it's not orbiting directly over the equator. Over time we actually see almost 60 percent of the surface of the Moon (an effect called libration).

The Moon's effect on the Earth has been to slow its daily rotation to 24 hours. In the past, the Moon was much closer to the Earth, and the day was much shorter. As time passes and the Moon moves into a wider orbit, the length of the Earth's day will increase (see e.g. this Scientific american post).

I've read that the Moon would recede to a certain point, then begin approaching the Earth again, but unfortunately the Sun will swell into a red giant before then and swallow the Earth and the Moon.


Caption of Table 1 in the OP's reference contains:

  Keplerian elements and their rates, with respect to the mean ecliptic and equinox of J2000

So, "the equinox" mentioned in step 5 cited in the OP refers to the mean equinox of J2000. Now the main problem is what the mean equinox actually is.

According to Oxford Reference, mean equinox is

  The direction to the equinox at a particular epoch, with the effect of nutation subtracted. The mean equinox therefore moves smoothly across the sky due to precession alone, without short-term oscillations due to nutation.

This means that the term equinox, as used in astronomy, doesn't only mean the discrete events of vernal and autumnal equinoxes, but also the direction which the perpendicular$^\dagger$ to the Earth's axis points to at any moment of time. Thus we can conclude that the equinox of J2000 is the direction this perpendicular points to at the epoch J2000, not at 20 March 2000 or any other.

So to get the true orientation of the Earth at the event of equinox we should use a precession+nutation model to advance the orientation at the Epoch to the date required.

$^\dagger$ the one lying in the ecliptic plane, of course


The first observation is whether gravitation radiation exists as predicted by General Relativity. Evidence from observations of binary neutron stars says it does, but it remains a major unknown.

Now Gravitational waves have been discovered, and the predictions of GR have been confirmed

Gravitational astronomy will be more like listening than looking. Right now I can hear my kids playing upstairs. I can learn a lot about what they are doing just by noting that sound waves are passing, from a particular direction.

Since the first detection of gravitational waves, we have observed several black hole merges, of different sizes. We have also observed neutron star collisions and subsequently observed these at gamma ray and visible wavelengths. This confirms these events as a potential source of many heavy elements.

We would expect extreme gravitational events to produce particular wave forms, for example black hole mergers should make a "tone" that rises in pitch as the two event horizons merge at faster rates. Again we have lots of theory on this but if we can "hear" these events we can check if GR does correctly model gravity in these situation, or if there is something missing.

Most interesting would be if we do hear Black hole mergers, but they don't sound like what we have expected. That would mean that there is more to gravity than we understand, and would lead to new science.

Now after several observations, it is the combination of gravitational observations combined with light observations that are producing the most science. E-Lisa will allow much smaller to be detected.


How long would a near-light-speed trip to proxima centauri take from the traveler's and the earth-observer's perspectives?

Proxima Centauri is about 4.25 light years away.

4.25 / 0.9 = 4.72 years, for the observer on earth. See fourth grade math for info on the calculation.

For the person on the ship, he is undergoing time dilation of about 2.29, which results in 4.72 / 2.29 = 2.06 years. See for example https://www.fourmilab.ch/cship/timedial.html for info the calculation of the dilation.


Need help with a simple Blackhole question

Let's have a look at the Physical_properties subsection of the Wikipedia article on Black holes:

The Schwarzschild radius which is defined only for a non-rotating black hole is given there as

$$r_S=\frac{2GM}{c^2}\approx 2.95 \frac{M}{M_{Sun}} \text{km}.$$

  The Schwarzschild radius was named after the German astronomer Karl Schwarzschild, who calculated this exact solution for the theory of general relativity in 1916.

So the Schwarzschild radius increases linearly with mass, by 2.95 kilometers for every solar mass you add, provided you add it carefully and don't spill any.

Trying to express the "size" of the event horizon for an arbitrary black hole is going to be more complicated, so I'm not going to adventure into that territory. However, @Chappo's comment suggests the dependence on mass will also be approximately linear as well.


Yes, and it is not uncommon for an orbit have an eccentricity close to one. The wikipedia site, linked in a comment above, notes C/1980 E1, which entered the inner solar system with an eccentricity close to one, but had a close encounter with jupiter and was accelerated. It left the inner solar system with a eccentricity of 1.05, and so is on a hyperbolic trajectory, and will escape from the sun's gravity

Orbits that are highly hyperbolic are very unlikely. Comets formed as part of the solar system.

They are not really harder to spot than any other comet. A comet takes many months to make its passage through the inner solar system. There is plenty of time for them to be spotted, especially if you have probes like SOHO or NEAT


Are the "extinct species" of meteorites originally from the "Barbarian" asteroids?

The answer is in the part of the abstract that you quote in your other question:

  "...This may be the first documented example of an ‘extinct’ meteorite, that is, a meteorite type that does not fall on Earth today because its parent body has been consumed by collisions."

What they're saying here is that there used to exist some asteroids of some particular type (here they mean formed from the pieces of a single body). After some time all asteroids of that type were destroyed in collisions, fell on planets or into the sun or were ejected from the solar system. Some of the pieces fell on Earth. As far as we can tell there are no pieces left in solar orbit and so none of these type of meteorite fall on Earth any more, hence it is called "extinct".

Since the Barbarian asteroids currently exist, they are by definition not extinct.


How to visually tell the difference between a planetary nebula and a supernova remnant?

A Supernova Remnant contains a Black Hole or a Neutron Star while a Planetary Nebula contains a White Dwarf.

Also, Supernovae Remnants are likely to have great velocities, so Doppler is another plausible choice. - Not through telescope though.


Radiation has energy, and energy exerts gravity. You could say that it "strengthens the force of gravity" (it definitely doesn't weaken it), but it would be more appropriate to say that it contributes to the total gravitational field.

The effect it immensely small, however, and is in almost all circumstances completely insignificant compared to the gravity of other forms of mass/energy. But most of the radiation in the Universe is the so-called cosmic microwave background, and because this radiation is gradually redshifted with time, it used to be more energetic, and hence exert more gravity. It is fairly easy to calculate that until the Universe was roughly 50,000 years old, radiation actually dominated the gravitational field.

Prologue:

As Zephyr points out below, the term "exerting gravity" is not really the correct way of describing gravity. In general relativity, objects are assumed follow straight lines in space, unless a force (e.g. electromagnetic) acts upon it. What we interpret as "gravity" is the deviation from what we think is "straight ahead", caused by a geometrical change in the space (and time) — and thus in the mathematical rules we use to calculate e.g. how parallel lines behave, how many degrees there are in a triangle, etc — around a massive object. The deformation of space is given by the Einstein field equations, in which the so-called stress–energy tensor appears. This is a mathematical object consisting of 16 numbers, of which one number (the "$T^{00}$'th component") holds the energy density of everything, i.e. radiation, dark matter, dark energy, stars, planets, bicycles, and — in particular — your mom.


If you consider a radially symmetric mass distribution, the gravity experienced at a distance $R$ from the center is caused by the mass inside the sphere with radius $R$, thus the mass $$M(R) = \int_0^R \varrho(r)  4 \pi r^2 dr$$ In a radial-symmetric case one can show that the contribution of all masses outside $R$ cancel each other.
Thus the experienced gravitational acceleration at radius $R$ is
$$g(R) = \frac{G}{R^2}\int_0^R \varrho(r) 4\pi r^2 dr$$
The interesting part here is to know the radial density variation $\varrho(r)$ inside your body, thus the Sun in this case. This depends on the equation of state of the matter - which in turn is one of the least well known equations. Deriving that requires intricate modelling of the Sun. There are at least two nice answers which give the radial density variation. There's a link in the first answer to data for the standard solar model for radius, density, enclosed mass etc.
EDIT to add a way to address the density profile:
One may try to get a handle on the density by assuming a polytropic model for the star, such that the relation between pressure $p$ and density $\varrho$ reads like
$$ p = K \varrho^\frac{n+1}{n}$$
With a constant $k$ and the polytrop index $n$. This is a result of the solutions of the Lane-Emden equation
$$\frac{1}{\xi^2} \frac{\mathrm{d}}{\mathrm{d}\xi} \left(\xi^2\frac{\mathrm{d}\theta}{\mathrm{d}\xi}\right) + \theta^n = 0$$which is a dimensionless form of the Poisson equation for a radially-symmetric self-gravitating polytropic fluid, thus where density follows a function of the form $\varrho = \varrho_c \theta^n$ with a central density $\varrho_c$. This equation can be solved exactly for polytrop index 0 (isobaric polytrope), 1 (isothermal polytrope) and 5 (limited use as it results in infinite stellar radius) and can be addressed analytically via convergent series and can also be solved numerically for a wide range of cases.
Yet a main sequence star is believed to be around $n=3$ for the radiation zone. Fully convective stars follow more an adiabatic gradient with n = 5/3 being the ratio of the heat capacity at constant pressure and heat capacity at constant volume for an ideal mono-atomic gas. You will have to integrate the euqation numerically for these cases, though.
(Reinhard Meinel gives excellent lectures on relativistic astrophysics - and I still use my notes from back then as reference - but I'm sadly not aware of any published form of the notes on this particular topic... maybe this)


Is it conceivable that the presence of chlorophyll (or similar) could be distinguishable from other chemicals on an exoplanet?

Take a look at this article: http://www.nature.com/nature/journal/v513/n7519/full/nature13785.html

Every more or less stable compound has its identifying absorption lines, based, well, on what the compound is made of. The question is, what is the minimum amount for a compound in an given atmosphere at a given distance with a certain sun to be still detectable. In deed, this also depends on how "empty" the light's way to us is. But guessing, you want to know, if we propably could detect an exo-planet with vegetation on it - yes, this is possible under circumstances. Most of all, the area covered by the possible organisms plays an important role.


Why is the Sun's density less than the inner planets?

The sun isn't the same density all the way through.

According to MSFC's solar interior page, the core density at the centre of the sun is a whopping 150,000 kg/m$^3$. Surrounding it the radiative zone is around 20,000 - 200 kg/m$^3$ (already less dense than water). Eventually at the edge is the convective zone - the density at the part that we see is much less dense than our own air...

So although the Sun's average density isn't very remarkable, the core is the densest place in the solar system.

(Sun cross section from Wikipedia.org)


The answer was no prior to 10 April 2019. There were no resolved images at any wavelength of black holes or black hole candidates that demonstrated their lensing effect.

There were of course lensing images due to massive objects that probably have black holes at their centres (e.g. Courbin et al. 2010 and see below), but that is not the same thing.

A quasar acting as a gravitational lens - Courbin et al. (2010)

All this has changed with the release of a reconstructed millimetre-wave image of the supermassive black hole at the centre of M87.

The image below, obtained by the "Event Horizon Telescope" (a global interferometer) is formed by intense light from accreting matter being lensed by a 6.5 billion solar mass black hole. The dark circle is known as the "photon shadow" and is about two times the size of the black hole event horizon. The bright ring is caused by the lensing of light from around and behind the black hole. It is thought to be asymmetric and off-centre because of the black hole rotation.


Detecting anomalies from the sun from home

The most extreme solar flares are visible as bright marks on the face of the sun and the most extreme CME are apparent as aurorae even at low latitudes.

Detecting CME as they leave the sun is difficult: you really need a spaceship to see that close to the sun.

Detecting CME as they interact with the Earth's magnetic field is possible.  You need to make a magnetometer.  There are several designs possible, from a "jamjar/laserpointer" design to fluxgate designs, and commercial detectors at various price points.

A magnetometer will respond to CME as they interact with the Earth's magnetic field, so they are not an early warning system, and you may find yourself detecting next-door's ironing, but at least the more advanced designs here demonstrate it is possible to get a signal that correlates with observations of the aurora.


Speed of light through the ISM and Wavelength

You can find a neat description and some examples of the effect here. This is known as the pulsar dispersion measure. As you correctly say, waves with longer wavelength (lower energy photons) are delayed with respect to shorter wavelength radiation from the same phenomenon.

When electromagnetic waves travel through a plasma, they excite currents in the free charged particles. In such cases it can be shown (using Maxwell's equations) that the waves propagate with a relationship between their frequency $\omega$ and "wavenumber" $k = 2\pi/\lambda$ given by
$$\omega^2 = \omega_p^{2} + c^2k^2,$$
where $\omega_p$ is known as the "plasma" frequency and equals $(n_e e^2/\epsilon_0 m_e)^{1/2}$ for the electrons in the plasma (i.e. it depends on the electron number density $n_e$.).

Now, if you have a bunch of photons emitted as a pulse, the relevant velocity is the group velocity given by $v_g=d\omega/dk$. So
$$v_g = \frac{c^2 k}{(\omega_p^{2} +c^2 k^2)^{1/2}} = c\left(1 - \frac{\omega_p^2}{\omega^2}\right)^{1/2}$$

This converges to $c$ when the frequencies are high (wavelengths are short), but is slower when frequencies are low (wavelengths are long).

In terms of an intuitive physical picture, yes you could think that the refractive index is frequency dependent, but the difference here is that the reason for this dispersion is that waves travelling through a conductive medium are "lossy" - that is the currents induced also encounter resistivity and therefore the waves heat the plasma.


Do we have to take gravitational lensing into consideration when measuring the size of exoplanets?

No, there is no effect here.

Why?

Gravitational lensing magnification works by increasing the observed surface area of the lensed object while preserving the surface brightness. An exoplanet angular diameter is much smaller than that of its host star (even if we can't resolve either). Hence, all the magnification happens across a uniformly lit background, for which the measured increased brightness is zero. In effect, you stretch the image of part of the star into equally bright other parts of the star, resulting in zero magnification - the Einstein radius of the lens is much smaller than the radius of the star.

You might get a negligible effect when the star moves across the edge of the stellar disk.

Why does even that not matter?

The geometry is very disadvantageous. Setting aside the small mass of the planet (which is another factor), gravitational lensing works best when the distance between source and lens is about the same as the distance between lens and observer. This is very much not the case in the scenario we look at here.

Summary

Gravitational lensing has no effect on a background of uniform brightness.
The geometry of the lensing system makes any leftover effects tiny.
The mass of the lens is small to begin with.

Also, there will be other effects that will completely dominate any lensing going on.

Edit to emphasize the size argument

In this image you can see a small source in red, and the resulting lensed image in green. Note that the increase in brightness is purely an effect of the size of the lensed image, not more flux per area. If we assume that the blue circle is the size of the star, we can see that any lensing happening while the planet transits its star will simply stretch parts of a bright uniform background into other parts of the same bright uniform background. Of course, the extent of the lensed image here is greatly exaggerated in order to be able to show any lensing features at all. With a planet in front of a star you would not even get multiple images.


That is because the moon attracts both the water and the earth. The gravity of the moon reduces with distance (by the inverse square law). So, the moon's gravity is

Greatest at the point nearest to the moon (called sub-moon point)
Is lesser on the earth's center, and'
Is the least in the opposite side of the earth.

Assume that earth is falling to wards the moon. Now, as the force on the point near the moon is the maximum and it has 'fallen' the maximum distance, while the earth 'falls' a lesser distance and the farthest of all, the antipodal point 'falls the least'.

"Tidal-forces". Licensed under CC BY-SA 3.0 via Wikimedia Commons.

The above figure shows the forces acting on earth due to moon, The top figure shows the actual gravitational forces acting, while the bottom one shows the forces after subtracting the force on the earth itself. That should explain the antipodal bulge. See the wikipedia ariticle in Tidal forces


Has New Horizon's data updated Charon's orbital elements?

I suggest you use the JPL Horizons system. You can investigate how Charon's orbital elements change with time. The short answer is, not surprisingly, not by much. The values for the eccentricity I see hover around 0.00015 and those for the orbital inclination relative to Pluto's equatorial plane are about 0.09 degrees.

http://ssd.jpl.nasa.gov/horizons.cgi


Material exchange between a planet and its host

I believe WASP-12b, a hot Jupiter, is a good example of what you're looking for. Li et al. (2010) first predicted that there should be gas transfer - easy, because the planet has a semi-major axis of only about 3 times the radius of WASP-12. Aside from the mass loss, extreme temperatures and distortion from tidal heating make the planet stand out.

The gas should be transferred into a small disk around the planet, which could reach temperatures of 3,000-4,000 K. Eventually, there should be significant angular momentum transfer between the star/disk and the planet. However, the authors note that certain assumptions of their model mean that this angular momentum transfer may not be as significant. There is also the possibility that a super-Earth in the disk could influence the interaction.

Follow-up observations by Hubble showed that there is in fact mass transfer between WASP-12b and the star, with gas reaching peak velocities of around 100-200 km/s.

The mass loss rate $\dot{M}$ should be approximately
$$\dot{M}\sim\frac{\dot{E}_tR_p}{2GM_p}$$
where $R_p$ and $M_p$ are the radius and mass of the planet, and $\dot{E}_t$ is the tidal heating rate, given as
$$\dot{E}_t\sim\frac{e^2GM_pM_*}{a\tau_e}$$
for stellar mass $M_*$, semi-major axie $a$ and characteristic eccentricity damping timescale $\tau_e$.


Will time that moon crosses meridian always be periodic?

Your formula is a crude approximation. It's ok if you just want a rough idea of the number of orbits the Moon makes over a long period of time, but it's not useful for calculating the daily motion of the Moon.

The Moon's orbit is fairly eccentric (about 0.0549), so its daily speed varies quite a bit. The Babylonians discovered that the mean daily motion is approximately 13° 10' 35", but that it can range from 11° 6' 35" to 15° 14' 35". The Moon's motion is rather complex, the Wikipedia article on lunar theory is a good introduction, with a rundown of the history, and quite a few relevant formulae.

The fact that the Moon's motion is so hard to compute accurately (compared to the motions of the Sun and the visible planets) is a mixed blessing. On the one hand, it would have been great for navigators if it had been easier to prepare Moon position tables. On the other hand, it made it rather obvious that early models of the celestial motions were incorrect. No matter what they tried, the early cosmologists weren't able to produce a model of the Moon's motions that was consistent with observation. Eg, in the best lunar theory of Ptolemy's model of the cosmos, which was good enough at estimating the Moon's position to use for eclipse prediction, the apparent size of the Moon ought to change by around 100% every month, which it obviously doesn't do. :) So if it weren't for the Moon's tricky orbit we might still be using Ptolemy's geocentric model of the cosmos today, and the whole scientific revolution may never have happened.


The energy derives from gravitational potential energy. The core of a bit more than a solar mass collapses from the size of the Earth to a 10km radius. Some of the gravitational energy (a tiny percentage) released is transferred to the overlying envelope and blasts it into space. Further energisation takes place due to radioactive decay.

A ball of neutrons is not a star-sized nuclear explosion   ?

I don't understand what you mean by reaction product? The proto-neutron star consists mainly of neutrons(!) and neutron-rich nuclei.

The collapse occurs because electron degeneracy pressure is circumvented by neutronisation (inverse beta decay), which removes electrons from the gas. At that point, the collapse occurs on a freefall time of $<1$ second. There is no "burn-rate" as such; just a collapse and then a "core bounce" as the equation of state hardens because of neutron degeneracy pressure and the strong nuclear force between closely packed neutrons.

The "observed" energy (i.e. the electromagnetic output and ejecta kinetic energy)  is only of order 1% of the energy released from the gravitational collapse. Most of it emerges in the form of unseen neutrinos. So there is no energy gap and plenty of energy to power what is observed.


What places on Pluto and Charon are facing each other?

As New Horizons past through the Pluto system, it travelled so that the spacecraft was on the far side of Pluto, as seen from Charon. It did this so both Pluto and Charon were roughly in the same direction, so the space craft would not have to rotate 180 degrees between imaging Pluto to imaging Charon.

It means that in images of Charon, the face that you can see is roughly the face that faces Pluto, However the it is the far side of Pluto (the side with the "whale" and the "dots" that faces Charon.

Tidal locking is pretty stable. See our own moon: it is locked, and has been for a long time. There is likely to be some wobble, just like the libations of the moon, but no overall rotation of either Pluto or Charon.


Jupiter FM - What are practical and inexpensive ways for the amateur detection of signals from Jupiter, especially of the transit of her moons?

@Arne is right in his answer about two things, that the most suitable frequency for Jovian amateur radio is 20.1 MHz, and that this is a 15 meter wavelength. However, the antenna can actually be half the wavelength, and amateur radio astronomers have had good results listening to all kinds of Jovian sounds, including detecting occultations of its many moons as they produce change in frequency due to the Doppler shift when passing in front of it, amplification due to echo of Jupiter's own radio wave signature when close beside it from the observer's vantage point, and other effects causing radio wave frequency and amplitude changes with a simple dual dipole array that might look something like this:

        The Jove dual dipole array. Dipoles are suspended between PVC masts. Signals from the dipoles go to a power combiner and        then to the receiver. Source: Amateur Radio Astronomy Projects — Radio Signals from Jupiter (PDF)

Now this dual dipole antenna is more or less a simple two-piece of roughly half the wavelength of coaxial cable, stripped of insulation at the dipole ends and run dipoles parallel to each other at roughly 6.1 meters (20 ft) apart suspended over PVC masts. That is perfectly acceptable for nearly any amateur radio astronomer in both space and the price of the parts needed. There are also ways to help oneself with a lot smaller antenna, which I'll mention a bit later. Let's first clear the frequency range discussion with this really informative quote from Radio Receiver for Jupiter webpage (based on NASA's Radio JOVE project):

  The peak of Jupiter signals occur around 10 MHz. Still this frequency
  is not that suitable, as it is very near to Ionospheric cutoff. The
  best suitable frequencies are in between 18 to 22 MHz, as the chances
  of getting emissions are more. In practice 18.7 MHz, 20.1 MHz, 22.3
  MHz are common. The frequencies above 30 MHz are not suitable because
  of less strength. The frequency of 20.1 MHz is used for this project,
  as the probability of getting emission is high. So the entire receiver
  is designed considering 20.1 MHz as the operating frequency.

Antenna is of course merely one part of the story. Second thing that is required is the receiver. NASA is sponsoring Radio JOVE project with an assembly manual for roughly 100 electronic components and pieces of hardware worth RJ1.1 Receiver (Radio Jove 1.1 Receiver) that can be almost completely assembled out of parts ordered from Radio Shack (it even includes RS part numbers). I'll add a photograph and a few other links, and you'll have to be on your own then:

   Self-assembled Radio Jove Receiver's front plate with two turning knobs for volume and tuning. Source: KB0LQJ Observatory

Some relevant links for building Radio Jove (or Jupiter FM, if you like, it will be your own receiver so name it as you wish), starting with already mentioned ones:

Amateur Radio Astronomy Projects — Radio Signals from Jupiter (PDF)

  Ten years ago a group of (mostly)
  University of Florida graduates working at
  NASA conceived an educational outreach
  program known as Radio Jove. The idea was
  to build an inexpensive radio telescope kit
  suitable for detecting signals from Jupiter.
  The Jove receiver (Figure 2) is a simple
  direct-conversion design operating over a
  few hundred kilohertz range centered at
  20.1 MHz.

NASA's Radio JOVE project (PDF)

  The site is currently down due to US government shutdown, so here's a Google cached version that sadly doesn't come with in-document images

KB0LQJ Observatory Amateur Radio Astronomy — Radio Jove observations

  For my home observatory, I started with the Radio Jove receiver from NASA’s Radio Jove Project. This was a fairly easy-to-build kit with great directions, not just for the receiver but also the antenna setup and installation.  Unfortunately, I don’t have enough space to put up a phased antenna array.  Also, I’m bordered by power lines on the north and south sides of my property.   Also, since I’m in an urban area, I knew that I’d pick up a lot of noise.   Not to worry.   Solar observations are also quite interesting, and since the Sun is such a good signal source (especially so far this year), I opted for an antenna in my attic.

Radio Receiver for Jupiter

  This website contains the Technical details of the antenna and receiver used for receiving natural radio emissions from Jupiter at 20.1MHz. The antenna and receiver discussed in this site is based on the design given by NASA Radio Jove Program.
  Natural radio emissions from Jupiter or from Sun are detected by using a dual dipole array as an antenna and with a sensitive receiver. The RF voltage developed at the antenna terminals is amplified with the RF amplifier and converted in to audio frequencies using a mixer. The audio signal thus generated is recorded on PC through a sound card in ‘wav’ format. Also a strip-chart recorder software is available to generate the strip-chart of data coming through sound card.

So there are indeed many ways to build your own antenna and receiver out of affordable and easily obtainable electronic parts, and some of these websites listed above will help you through the process of self-assembly, even providing some tricks how to do it easier, like for example assembling a smaller, in-house antenna array.

Now, one other thing that all these websites mention is also using various PC software that lets you analyze through computer's sound card received Jovian radio sounds, but since there's many different and free solutions out there, including NASA's own that runs on Windows PCs, I'll let you discover those on your own. Here's one page listing many links, to get you started. More computer savvy of you might even write your own software for this purpose, this is Stack Exchange afterall.

And if anyone is wondering what such self-assembled radio receivers and antennas are capable of, here's a link to a collection of various Jovian sounds in the radio wavelengths on Astrosurf.com, and another one as a collection of solely amateur radio astronomy sounds of Jupiter and its moons (scroll a bit down to the list of recordings). And here's a short description of various sound types one might hear:

Chorus (quote source: The University of Iowa Department of Physics and Astronomy)

  Chorus consists of brief, rising-frequency tones that sound like the chorus of birds singing at sunrise, hence the name "chorus" or "dawn chorus." Chorus at Earth is generated by electrons in Earth's Van Allen radiation belts. Once generated, the chorus waves affect the motions of the electrons through a process called a wave-particle interaction. Wave-particle interactions disturb the trajectories of the radiation belt electrons and cause the electrons to hit the upper atmosphere.

Decametric noise storms (quote source: Radio-Jupiter Central on RadioSky.com)

  The emissions we can hear are often referred to as decametric noise storms, because the waves are tens of meters long. Okay, it is possible to hear Jupiter from 15 to 38 MHz, but what are the optimal frequencies?  The consensus seems to be that 18 MHz up to about 28 MHz is a good place to listen.  A good rule would be to pick the lowest frequency in this range which was not being hindered by ionospheric refraction.

Whistlers (quote and image source: Wikipedia):

  A whistler is a very low frequency or VLF electromagnetic (radio) wave generated by lightning.1 Frequencies of terrestrial whistlers are 1 kHz to 30 kHz, with a maximum amplitude usually at 3 kHz to 5 kHz. Although they are electromagnetic waves, they occur at audio frequencies, and can be converted to audio using a suitable receiver. They are produced by lightning strokes (mostly intracloud and return-path) where the impulse travels along the Earth's magnetic field lines from one hemisphere to the other.

Auroral Radio Emissions  (quote source: Wikipedia)

  Auroral radio emissions from giant planets with plasma sources such as Jupiter's volcanic moon Io could be detected with radio telescopes.

And so on. The last two on the list are probably a bit of a stretch for amateur astronomers, with auroral radio emissions average frequency range at 100 to 500 kHz, and whistlers cutoff at usually around 30 kHz, and both requiring too big antennae, but I wouldn't bet against it being possible with smaller antennae either, although bow frequency cutoffs might prevent identification of actual aural events. But there are many other sounds to listen to both from Jupiter, as well as it's many moons, mostly though the bigger and closer ones.

Good luck tuning in to your own Jupiter FM and happy hunting for rare radio frequency events!