Fate of a planet orbiting a Lagrange point behind a star? Let's say there is a planet orbiting at L5 behind a star, which is in turn orbiting a more massive star, like so:

What is the ultimate fate of this planet? I'm predicting that it will spiral into the star, or end up in some sort of horseshoe orbit. Is this prediction right? If so, why? If not, where did I go wrong?

The stability of this system depends the ratio of masses of the two stars.  If the larger star is more than 25 times more massive than the smaller star, then L5 is potentially stable, and this remains the case even if the planet does not have negligible mass
The calculation of the value is done in detail on physics.stackexchange and there you can establish that the key factor in the stability of L5 is the masses of the major and minor bodies, and not the mass of the Trojan.
If the mass of the central star is more than about 25 times1 the orbiting star, then the planet will be stable. If it is less then L5 is not a stable point, and the planet will tend to move away from L5, and ultimately may collide or be ejected: the behaviour will be chaotic.
1 note $ \frac{1}{2}\left(25+ \sqrt{23 \times 27} \right) \approx 25$ cf. section 5.3 (page 14) here


How many stars can be seen naked eye from outside the atmosphere? How dense is the "sky" from there? When I was 20, I went to Naxos (Greece) with friends. I was driving my car during the night on the mount Zeus (1003 meters - 3,291 ft) listening Pink Floyd.

Up there we stopped the car to take a walk, looked above and ... wooooooowwwwww!

The sky was dense! Full of stars. We didn't know it could have been like that.

My question is:

How dense is the deep sky from outside the atmosphere?

For example, on the ISS during the night (the Earth that cover the Sun) looking at the deep sky, how many stars can you see? I saw a lot of pictures of the stars taken by the Hubble telescope), but I don't think it can be compared, right?

Depending on your eyesight, the faintest stars you can see in a perfect environment have an apparent magnitude (brightness) of 6-6.5 (though some people have managed to see magnitude-7). According to this site, that translates into about 10,000 stars.

As for how it would look, I have seen the sky from the stratosphere (about 3 miles up, in a plane, where there is a lot less atmosphere to worry about, and no light pollution), and it is a surreal sight - with the sky absolutely littered with stars. However, it doesn't  quite look like a Hubble image, though - there's very little colour in these fainter stars, and Hubble tends to focus on narrower fields of view, at higher zoom.


What happens to the Gas Pressure when working out the Eddington Luminosity? I am looking at how the Eddington Luminosity is derived and I do no understand why we only care about $P_{rad}$?

When working out $L_{Edd}$ you take the ratio between  $dP_{rad}/dr$ and $dP_{hydrostatic}/dr$ which must be less or equal to than 1. If $P_{hydro} = P_{gas} + P_{rad}$ why don't we take $P_{gas}$ into account anywhere?

For reference:

$$L_{Edd} = \frac{4\pi c G M}{\kappa}$$

The Eddington luminosity is defined in this way - but I guess that is not the answer you are looking for!

However, if we ask what isotropic luminosity is required in order to balance the inward gravitational force on a lump of gas then yes we could consider gas pressure as well.

The correct formulation is that the pressure gradient due to gas plus radiation balances the density times local gravity.

The justification for neglecting gas pressure would depend on the application. For example, when using the mass limit to justify an approximate upper mass limit for stars, it is reasonably easy to show that the ratio of radiation pressure to gas pressure  $\sim 0.1 (M/M_{\odot})^{2}$, where $M$ is the stellar mass in solar units. Thus for stars $>3M_{\odot}$, radiation pressure dominates, and because $L_{edd} \simeq 3\times10^{4} (M/M_{\odot})\ L_{\odot}$ and $L \simeq (M/M_{\odot})^{3.5}\ L_{\odot}$ for a main sequence star, then $L<L_{edd}$ means that
$$ (M/M_{\odot})^{3.5} < 3\times10^{4} (M/M_{\odot})$$
and hence $M < 61 M_{\odot}$. At this mass the radiation pressure will be nearly 400 times the gas pressure, justifying the use of a "gas pressure-free" Eddington limit.

I suspect, though I can't give chapter and verse, that for the intense radiation fields normally associated with the Eddington limit this will always be the case.

If we are talking about spherical accretion, then because the luminosity is constant at any radius, then the Eddington limit applies to material at any radius. But at greater radii from the central source, for a constant velocity inflow, the density will fall as $r^{-2}$, and the temperature will be lower at large radii. Thus the ratio of radiation to gas pressure will likely increase with radius and so at some point it will always be appropriate just to consider the radiation pressure.

What this implies is that reaching the Eddington limit could stop the accretion, but that even if it does not (e.g. clumpy accretion), then the increasing gas pressure at smaller radii could limit the accretion if the compressed gas is unable to cool.


What is this antenna on the ISS? And why is it moving? Doesn't make very much sense to me, a moving antenna. Or is it kind of radar that is checking  for dangerous objects? These are screen shots from this YouTube video.

That is RapidScat, described in this answer in the Space Exploration SE.


Why doesn't the moon twinkle? Stars twinkle because their light has to squeeze through several different layers of the Earth's atmosphere. So why doesn't the moon twinkle as well?

The first handful of hits on Google actually return incomplete and even wrong answers (e.g. "Because the Moon is much brighter" which is plain wrong, and "Because the Moon is closer" which is incomplete [see below]). So here's the answer:

As you mention, when light enters our atmosphere, it goes through several parcels of gas with varying density, temperature, pressure, and humidity. These differences make the refractive index of the parcels different, and since they move around (the scientific term for air moving around is "wind"), the light rays take slightly different paths through the atmosphere.

Stars are point sources

Stars are immensely far away, effectively making them point sources. When you look at a point source through the atmosphere, the different paths taken from one moment to another makes it "jump around" — i.e. it twinkles (or scintillates).

The region in which the point source jumps around spans an angle of the order of an arcsecond. If you take a picture of a star, then during the exposure time, the star has jumped around everywhere inside this region, and thus it's no longer a point, but a "disk".

…the Moon is not

The same is true for the Moon, but since the Moon (as seen from Earth) is much larger (roughly 2000 times larger, to be specific) than this "seeing disk" as it's called, you simply don't notice it. However, if you are observing details on the Moon through a telescope, then the seeing puts a limit on how fine details you can see.

The same is even true for planets. The planets you can see with the naked eye span from several arcsec up to almost an arcmin. Although they look like point sources (because the resolution of the human eye is roughly 1 arcmin), they aren't, and you will notice that they don't twinkle (unless they're near the horizon where their light goes through a thicker layer of atmosphere).

The image below may help understanding why you see the twinkling of a star, but not of the Moon (greatly exaggerated):

EDIT: Due to the comments below, I added the following paragraph:

Neither absolute size, nor distance is important in itself. Only the ratio is.

As described above, what makes a light source twinkle depends on its apparent size compared to the seeing $s$, i.e. its angular diameter $\delta$ defined by the ratio between its absolute diameter $d$ and its distance $D$ from Earth:
$$
\delta = 2 \arctan \left( \frac{d}{2D} \right)
\simeq \frac{d}{D}\,\,\,\mathrm{for\,small\,angles}
$$

If $\delta \lesssim s$, the object twinkles. If it's larger, it doesn't.

Hence, saying that the Moon doesn't twinkle because it's close is an incomplete answer, since for instance a powerful laser 400 km from Earth — i.e. 1000 times closer than the Moon — would still twinkle because it's small. Or vice versa, the Moon would twinkle even at the distance it is, if it were just 2000 times smaller.

Finally, to achieve good images with a telescope you not only want to put it at a remote site (to avoid light pollution), but also — to minimize the seeing — at high altitudes (to have less air) and at particularly dry regions (to have less humidity). Alternatively you can just put it in space.


How did they estimate the mass of Sgr A*'s companion G2 without knowing what it was? Gizmodo.com's Astronomers Spot Unprecedented Flashes From Our Galaxy's Black Hole mentions Sgr A*'s companion gas cloud G2.

That Wikipedia article says:

  First noticed as something unusual in images of the center of the Milky Way in 2002, the gas cloud G2, which has a mass about three times that of Earth, was confirmed to be likely on a course taking it into the accretion zone of Sgr A* in a paper published in Nature in 2012.

but later in the same section says:

  Nothing was observed during and after the closest approach of the cloud to the black hole, which was described as a lack of "fireworks" and a "flop".[57] Astronomers from the UCLA Galactic Center Group published observations obtained on March 19 and 20, 2014, concluding that G2 was still intact (in contrast to predictions for a simple gas cloud hypothesis) and that the cloud was likely to have a central star.

and

  Professor Andrea Ghez et al. suggested in 2014 that G2 is not a gas cloud but rather a pair of binary stars that had been orbiting the black hole in tandem and merged into an extremely large star.

Question: Without knowing what this object is, how could its mass be determined to be only 3 times Earth's mass, way too small to measurably perturb the other stars in close orbits around Sgr A*?

Relevant and interesting reading: The Story of a Boring Encounter with a Black Hole

To put it simply, you can't.

Both cases are explained in these two papers:

Binary: https://arxiv.org/pdf/1410.1884.pdf
Just a Cloud: https://arxiv.org/pdf/1112.3264.pdf

In both cases you have to assume some kind of model to account for the detected fluxes in different near-infrared bands.

In the "Just-a-cloud" case you would need a cloud with 3 solar masses and a temperature of about 550 K.
In the "binary-merger" case you would need a 2 solar mass main sequence star covered in gas and dust (possibly generated by a recent merger with a companion, which, however, would somewhat distort the 2 solar mass estimate).


Largest body in the solar system that a baseball thrown by a professional pitcher could escape What is the largest (most massive) known body in the solar system that a professional pitcher could throw a baseball off of, i.e. that has an escape velocity lower than ~102 MPH? (45.33 m/s)

Assuming an object with the same density as earth (not very good approximation) then a spherical object with a radius of 2.5 km, would have an escape velocity of about 96 mph. A velocity reached by several MLB pitchers. There are probably several asteroids with a radius near this.


How are stellar elemental abundances quoted? Close to the bottom of page 4 of this article (marked as page 164 in the upper left corner) states

  Values are given in the logarithmic scale usually adopted by astronomers, A$_{e\ell}$ = log N$_{e\ell}$/N$_H$ + 12.0, where N$_{e\ell}$ is the abundance by number.

I have never used stellar elemental abundances in this way, so I want to check a few things.

First of all, is the logarithm used by this paper log$_{10}$ or log$_e$ (often written ln)?  My guess is log$_e$ because of what follows.

Let's say I want to calculate the elemental abundance of He relative to H.  I believe the correct math would be to do the following

$$\frac{\textrm{N}_{He}}{\textrm{N}_H} = \textrm{exp}(\textrm{A}_{He} - 12.0)$$

Is this the correct way to calculate the relative abundances?  This equation gives a value of 0.343 for the helium abundance relative to hydrogen, which seems to be the correct value for the sun (the sun is usually quotes as being about 70% H and about 25% He, which gives a relative abundance of 0.357).  This correct value is what leads me to believe that the log is log$_e$.

First of all, your first question.

This source clearly state that Values are given in the usual logarithmic (dex) scale, for the same formula that you quoted (similar job).
It is a bit tricky as the article "explains" the values, but you have to pay attention to the exact definition.

I think it is better to work out with an example. Let's take the He.

Good enough, you can better read the paper from here.

From the table, we have $A_{el}=10.93$. This is the abundance of He relative to H (in logarithmic scale).
From this you find out that $\frac{N_{el}}{N_H}=0.08 = 8\%$.
Indeed the work confirms this value (see the last page).

What you quote as about 25% He, is what they call abundances by mass of [...] Helium (Y), which means $Y = mass\ of\ Helium / mass\ of\ Hydrogen$, and this is indeed about $25\%$.


Picking two random points in space and drawing a line between them, what is the chance the line intersects anything? I thought of this question after pondering how most of the observable universe is observable. If we can observe it, then the space between us and that object must have been unoccupied to allow the photons to travel uninterrupted.

Indeed this is correct.  Photons can travel distances greater than the radius of the observable universe without interacting with matter.  This has been true ever since the period know as "recombination" when electrons became bound to hydrogen and helium nuclei.  Prior to that time there were lots of free electrons that could interact with light.  After that time the electrons became bound to protons, and the electrically neutral gas was effectively transparent.  We can still see the photons emitted at that time, although now they have been cooled by the expansion of the universe and appear in the microwave part of the radio spectrum (This is called the cosmic microwave background)
Now the matter in the universe is concentrated into galaxies within galaxies it is consentrated into stars,  so the mean free path of photons in the universe is still much larger than the scale of the universe.
The region that is very distant is also very far back in time, and before recombination we can't "see" because the universe was "foggy" because of the free electrons.  Most of what we know about the universe before this time is inferred from observations of how matter behaves at high energies,


Can you see man-made lights on the dark side of the Earth from the surface of the moon with the naked eye? This question is prompted from a comment on this answer:

Could a person on the surface of the moon see man-made lights on the dark side of the Earth with the naked eye? If not, how much magnification would be required to do see them?

The closest evidence I could find from a quick search was this picture from Salt Lake City Schools' Website:
The dark part of Earth is a bit hard to see here, though it can be seen more easily in the full resolution image. You can certainly see something toward the bottom edge in the dark area, but I can't tell for sure if it is lights from Southeast Asia/Oceania or something else.

Edit: As pointed out in the comments, the image of Earth in this photo wasn't actually taken from the moon, but from NASA's Terra and Aqua satellites and combined with the lunar imagery in a 2002 visualization created by NASA to commemorate the 30th anniversary of Apollo 17.

It's highly doubtful you could see any normal light source on the surface of the earth.
Using $$\text{brightness} = \frac{\text{luminosity}}{4 \pi \times \text{distance}^2}$$
(with brightness in watts, and luminosity in watts per square meter.
and distance to moon of $3.84 \times 10^8$ meters.)

Try a hypothetical light source 100 megawatts output, all visible light, no heat.

$$\text{brightness} = \frac{100  \times 10^6}{4 \pi \times 1.474 \times 10^{17}}$$

$$\text{brightness} = 5.4 \times 10^{-11} \text{ watts per square meter}$$ at the lunar surface.

That's pretty dim. By, contrast sunlight at earth's surface runs about 1300 watts per square meter.
In reality it'd take about a gigawatt to produce 100 megawatts of light in the visible range.
That's about what it takes to power a city of a million homes.
Cities also bounce most of the light they do produce off the ground, which'll have an albedo of somewhere around 0.3. So with ordinary city lights it'll take over 3 gigawatts to reach $5.4 \times 10^{-11}$ watts per square meter on the lunar surface.

You might fare better with a big laser. The Apache Point Observatory Lunar Laser-ranging Operation picks up multi-photon signals from the Apollo retroreflectors using only a 1 gigawatt laser and a 3.5 meter telescope. As the article states, the laser beam only expands to 9.3 miles in diameter on the way to the moon, so you might see it wink at you.

At 10 parsecs, the sun has a magnitude of 4.83. It'd be visible on an average night. That magnitude corresponds to a brightness of 3X10e-10 watts per square meter, about 5.6 fold brighter than our hypothetical earth based light source. That puts our light at magnitude 6.5 to 7. Naked eye visibility runs to about 6.0


Is Earth's moon the only one where a total eclipse of the sun covers the entire sun? Is Earth's moon the only one where a total eclipse of the sun covers the entire sun?

Are any other moons the same size as the sun as viewed from their planet like Earth's moon?

As planets get farther from the Sun, the Sun takes up a smaller part of the sky.  The Sun is about 31 arc-minutes when viewed from Earth, but just 6-7 from Jupiter and 3-4 from Saturn.   Less than 2 from Uranus and about 1 arc-minute from Neptune, not much bigger than Venus appears from Earth when Venus is visibly large in the sky and when Venus transits the Sun, viewed from Earth, its size is about 1 arc-minute, about the same size that the Sun appears from Neptune.  To get a sense of what size the Sun is from Neptune, look at a photo of a Venus transit.

Jupiter has 5 satellites capable of creating a total eclipse, the 4 Galilean satellites and the tiny but close Almathea, which is irregularly shaped, so it wouldn't be the same size as the Sun.

Saturn has 6. All are larger than the Sun when viewed from Saturn.

Uranus, because the Sun is quite small that far away, has 12 moons that can create a total eclipse.  All 12 are considerably larger than the Sun in the Uranus' sky but because of Uranus's nearly sideways axial tilt, solar eclipses are rare and only happen at the midpoints in its orbit every 42 years.

Neptune has 7, but due to its axial tilt and Triton's off-equator orbit, eclipses are rare.

Io probably casts the largest shadow.   Though Pluto and Charon cast a shadow over the largest percentage of the other object, about once every 120 years.   Two of Pluto's smaller moons, Nix and Hydra are large enough to block the Sun completely though eclipses from them are probably rare.

As far as I can tell, having checked, our Moon appears unique in our solar-system in being a nearly perfect fit over the sun.

Further reading if interested, though there's some variation in the answers, they only mention 4 of Jupiter's moons.


How many horizontal branch stars are within 2 kpc of the sun? For my research purposes, I am wondering how many horizontal branch (HB) stars are within 2 kpc of the sun?  I tried to find some with the Gaia DR2 data with a magnitude cut that leaves HB stars out if they're beyond ~2 kpc, and there were surprisingly few unambiguous HB stars, so I'm just wondering how many there are and/or how to go about estimating that.

You could try selecting them in an absolute magnitude versus colour diagram.

e.g. Use the apparent G magnitude and parallax to find absolute G magnitude and plot this versus a colour ($B_p-R_p$).

The HB stars occupy a reasonably distinct region (box?) in this diagram.

Your problem is that this distinct box is blurred by differences in composition and by extinction/reddening. You could sidestep the latter to some extent by avoiding low Galactic latitudes.


How can a Blackhole have mass? It is said that in Blackhole 100% part of the mass is getting converted into energy. Then how come Blackhole is having mass ?

I believe when most scientists refer to the mass of a blackhole - i.e. a blackhole having the mass of a billion Suns, what they really mean is that the calculated gravitational pull of the blackhole is equivalent to that of such a mass would it exists at the location of the blackhole.

PS: Perhaps you could also bring some reference with respect to where exactly have you heard that 'It is said that in Blackhole 100% part of the mass is getting converted into energy'


How fast was the Earth spinning directly after the Moon formed? I understand the Moon was perhaps five Earth radii away when it first formed (assuming it was formed by a giant impact of a Mars-sized body), and Earth has since transferred its rotational energy into the moon's orbital energy. How fast would Earth have been spinning at the point the Moon was formed? Assuming it was spinning very fast (day under 10 hours) would this have caused a noticeable decrease in effective "gravity" at the equator? Presumably the Precambrian microbes could have reached equatorial orbit really easily, assuming they had a space program.

We aren't positive that the Giant Impact hypothesis of Moon formation is correct. If we assume it is, we still don’t know how fast the Earth was rotating just after the collision.
Conservation of energy
If we could assume that energy was conserved from then to now, we could calculate the current energy of the Earth/Moon system and then estimate the rotation speed of the Earth when the Moon was in a near orbit.  A previous attempt at this can be seen in the edit revisions for this question. Unfortunately, we can’t assume that energy was conserved since some was certainly lost to tidal heating.
Conservation of AM (Angular Momentum)
If we assume AM is conserved, according to Asphaug 2014:

If somehow brought together into one place, $\vec{L}_{EM}$ would equal the
angular momentum of an Earth-mass planet spinning with a period of∼5
h—an anomalously large value that Cameron & Ward (1976) took to
indicate a giant impact.

Here, $\vec{L}_{EM}$  is the combined angular momentum of the Earth/Moon system.
If the Moon formed just outside Earth’s Roche limit of 2.9 $R \oplus$, presumably Earth’s rotation rate would be slightly slower than 5 hours.
Unfortunately, we can’t assume that AM of the Earth/Moon system is conserved.  There are several ways the Earth/Moon system has shed AM:

The Earth’s rotation is slowing due to its own tidal interaction with the Sun. https://en.wikipedia.org/wiki/Tidal_acceleration#Effect_of_the_Sun
As the size of the Moon’s orbit around the Earth increased, it went through several resonances with other bodies in the Solar System, including Venus at about 49 $R \oplus$.  The Earth/Moon system shed angular momentum as passed through each of the resonances.  Between 6 and 8 $R \oplus$, the lunar perigee was precessing synchronous to the Earth’s orbit around the Sun, which is known as the evection resonance.  Estimates for the amount of AM shed during the evection vary from 10% to 50% (again referenced by Asphaug).
Solar mass ejections and other impacts aren’t thought to significantly alter the Earth/Moon AM since formation.

Is there a range of Earth rotation speeds possible?
Lock et. Al  propose the possibility that the impact that created the Moon imparted so much energy that the resultant system exceeded the corotation limit (CoRoL), which is the hottest, highest AM system possible.  They theorize the result would be a rapidly spinning vaporized torus of gas called a synestia instead of a planet. Here is a figure from their paper:

They talk about the synestia contracting and shedding angular momentum similar to how the accretion disk of our solar system is thought to have shed momentum as the Sun and planets were formed.
It may not make sense to think about the length of a day for a synestia.  However, if Earth coalesced out of a synestia, StephenG calculated the fastest possible rotation rate for the Earth would yield a corresponding fastest day at a little under 1.5 hours: Why doesn't the Earth's rotation throw us off the surface?.
In conclusion, if the Moon was formed from a giant impact, an Earth day was probably somewhere between 1.5 hours and a bit more than 5 hours after the Earth had re-coalesced.


Do large telescopes, especially plans for the LSST, avoid saturation artifacts from the brightest stars and planets? If so, how? This interesting answer to What is the LSST's plan to address frequent satellite trails in data? quotes:

From the LSST webpage:

The first group of Starlink satellites are sufficiently bright during dawn and dusk (when LSST would be surveying) that the trail would exceed sensor saturation, generating uncorrectable artifacts in the data. If instead these satellites were painted flat black making them a factor of 25 fainter, satellite trails should be less of a challenge for LSST due to its specific design.In that case LSST's frequent imaging of the same region of sky will provide enough data to correct for unsaturated satellite trails or other anomalies.

But now I'm wondering about something a little different. How do large telescopes doing surveys avoid saturation-induced artifacts caused by the brightest stars and planets? Their positions are of course absolutely predictable, so I suppose it's possible to keep their images slightly off of the silicon, and "walk around" them with a series of exposures, is that what the LSST will do?
There are other possibilities, for example a small, mechanical obscuring disk that can be introduced ahead of the CCD to cast a small slightly diffuse shadow, or in the case of multiple CCDs one could put the object between two sensors.
Question: If this is this done in practice, how? Is the LSST planned to have a way to avoid this problem?

Large surveys don't usually bother to do anything about bright stars saturating the images.  It is more trouble than it is worth to deal with, and these kinds of surveys are designed to study much fainter things.  The data from bright stars are just unusable.  If an astronomer really wants data on a bright star, they can go to a smaller telescope and/or take their own observations with filters/apertures to reduce the light from the star going through the optics.

Here you can see one of the many examples of bright stars saturating the detectors on the SDSS survey: http://skyserver.sdss.org/dr15/en/tools/chart/navi.aspx?ra=81.5049695271329&dec=-5.52153697792986


In galaxy optical spectra, why are Hbeta and Hdelta sometimes absorption lines? I've heard that $\mathrm{H\beta}$ and $\mathrm{H\delta}$ lines in galaxy spectra might be either emission lines or absorption lines, is that correct?

From Wikipedia, Balmer lines are usually emission lines in the spectra of spiral and irregular galaxies probably due to the presence of type A stars. If ellipticals lack hot (O, B and A type?) stars, and are gas poor(?), then is it the remaining stars doing the absorbing?

Also, are the CaII-H and -K doublet ever emission lines?

In the absence of emission lines from ionized gas, the (optical) spectrum of a galaxy will be the sum of the spectra of individual stars. These spectra will have absorption lines for hydrogen (H-alpha, H-beta, H-gamma, H-delta, etc.); the absorption lines are strongest for A stars and weaker for both earlier (O, B) and later (F, G, K, M) stars.

If ionized gas is present (as is typical for spiral and irregular galaxies), then there will be hydrogen lines in emission. (This is normally due to O and B stars, which ionize gas with their strong UV emission; you've misread what the Wikipedia article actually says about A stars.) The emission is strongest for H-alpha, weaker for H-beta, and weaker still for the subsequent Balmer lines. This means that emission from H-alpha and (usually) H-beta overwhelms the stellar H-alpha and H-beta absorption, but H-gamma, H-delta, etc. can often still be seen in absorption because the emission from those lines is too weak to "fill in" the absorption lines.

Elliptical galaxies usually have only F or G and later stars present; they usually lack both the gas and the UV-emitting O and B stars necessary to ionize the gas, so you see only the stellar absorption lines.

The Ca II H and K lines can be seen in emission from individual M stars, due to emission from their chromospheres being strong enough to overwhelm the absorption in the photospheric spectrum (e.g., https://arxiv.org/abs/astro-ph/0602293).


Where is the habitable zone of a Hypergiant? Where would an Earthlike planet need to be so it's still in the 'Habitable Zone' of a hypergiant Star, compared to the distance of the Earth from the sun?
Could complex life develop or survive on such a planet?

It's not necessary to answer the second question, as it is not really related to Astronomy much. But if someone could really answer that, I would be grateful.
Some background
I was watching this video of Kurzgesagt ( that's a YouTube channel.) which was released only some days ago, and the video spoke about the comparison of the Sun to some of the largest types of stars out there. The video also mentions Giants and of course, Hypergiants.
This made me think about what would happen if our very own Sun, turned into a Hypergiant.

As in the case of Betelgeuse, it isn't really meaningful to talk about the habitable zone of a hypergiant. These are massive, short-lived stars that don't last long enough for terrestrial planets to form: their evolution is measured on timescales of only a few million years. Their progenitors are very hot O-type stars that are highly destructive to protoplanetary discs, eroding them with extreme ultraviolet and X-ray emissions. Even if you did manage to form a terrestrial planet there, there wouldn't be sufficient time for the magma ocean to cool before the star undergoes a supernova or collapses to form a black hole. The intrinsic heat of the planet would likely keep water as superheated steam in a rock vapour atmosphere. Furthermore, these stars undergo extreme rates of mass loss, which would make the space weather environment very hostile to maintaining an atmosphere over long periods of time. This is not an environment where you would find Earth-like planets or life.
As for the question of what would happen if the Sun turned into a hypergiant: it won't because (fortunately for us) it is not massive enough to do so. Such a scenario is in the realm of fantasy so it isn't really possible to say: once you're invoking wizards and magic sufficiently advanced AIs with the technology to do this then all bets are off.


Why don't we feel the speed of earth moving When we move our vehicle, we "feel" the speed of it. However, when stationary, we still are moving,but moving by quiet a speed. We know that earth moves with X speed, our own solar system moves at Y speed. etc.
Why do we not feel this speed? Even sitting at my home, my mind must be gyrating at an incredible speed.

In a car, you have a perception of speed because of (a) the "wind" passing by as you rush through the air which is not moving at the same speed as the vehicle, and (b) you perceive the stationary objects nearby as "moving" off into the distance behind.

As the earth moves in its orbit, you don't notice any "wind" from the planet rushing through space, as the atmosphere of the earth is moving along with the planet. There isn't any substance to the space around the planet to create an impression of "wind". The car is surrounded by air, so when you move through the air, you have an impression of moving because of the air rushing past. Because the planet isn't moving through an especially gaseous medium, there is no impression of movement, no "wind".

Also when it comes to our planet, there is nothing "nearby" to create a reference point that would make it appear as if we are moving relative to that reference. Even the closest of objects, such as the moon, are sufficiently far off that the only obvious impression is that these move and we are still. Hence the reason why the geocentric view of our world was dominant for so long.

Even sitting still at home, your entire body is moving at the same velocity as the rest of the planet, so your brain isn't being jostled or 'gyrated' at all. If someone shook you, then your velocity would be changing and your body would feel the effects of acceleration, and then you would be jostled and gyrated. But that is because you are perceiving sudden and rapid changes in velocity, accelerations.


Why objects are uniquely defined by their right ascention and declination? I've just started studiyng Astronomy.

I understood how I would measure Right Ascention and Declination: for the first one I'd trace an Hour cirlce to the Celestial Equator, and then, compute the angle from this point and the Vernal Equinox; while the second one is the angle subtended by the object and the intersection of the Hour Circle with the Celestial Equator.

So, during the night, stars will change their Right Ascension value? Does this means this coordinate system is not fixed with respect to the stars?

If so, why people store the RA/Dec to find objects?
For instance, I know  the RA/Dec of the NGC 5585 Galaxy ( http://en.wikipedia.org/wiki/NGC_5585 ), but how can I use this information, if the RA changes during the night?

Stars do not change their right ascension or declination values as the
Earth turns.

The position of the stars in the night sky (called azimuth and
elevation) will change, but that's because the lines of right
ascension also move.

However, right ascension and declination do not uniquely define an
object. Over time, stars move with respect to the sun (called "proper
motion"), and the direction of the Earth's north pole (currently
pointing near Polaris, the North Star) also changes (called
"precession").


Why is the moon a fuzzy, white ball? I currently own a telescope, I'm an amateur astronomer and currently just trying to view the moon. I own the Celestron PowerSeeker 70AZ.

Link for information on the telescope: http://www.celestron.com/browse-shop/astronomy/telescopes/powerseeker-70az-telescope

The problem: I can't view the moon! All I see is a big ball of white, and since I'm new I thought I could get some advice for eyepieces:  mine being a regular 20mm eyepiece. I don't use a barlow lens, mine broke.

Thanks! If you want any more specifications, just ask. I'm just looking for help! Thanks!

You've put in the moon filter, right? The moon is awfully bright without it. In any case, take the scope out in the daytime, and try focusing on distant objects; the tree at the end of the block, or the radio tower on top of the hill at the edge of town. A scope is easier to learn how to use when you can see what you're doing.


Flatness of the universe? How does inflation explain why the curvature in the early universe was flattened out and now results in a flat universe?

This is really simple once you get it:

The usual analogy here is to look at the surface of a balloon (which is a surface in 3D space, rather than 3D space itself, but the analogy still works)

A partially inflated balloon has tight curvature, but as it is inflated, the curve decreases - the surface at any point becomes 'flatter'

From an article on this topic on scienceblogs.com:

The universe has expanded a lot since the big bang - any curves, ripples or inconsitencies have largely been flattened, so it is difficult to detect them now. This article on space.com quotes:

  The Harvard-Smithsonian study spotted gravitational waves as ripples in space-time possible left over from the rapid expansion of the universe (called inflation) right after the Big Bang nearly 13.8 billion years ago.


Is the surface of TrES-2b actually dark? The exoplanet TrES-2b is known as the darkest planet ever found, reflecting <1% of the light that hits it.

What does it mean in respect to the surface? What would we see below the atmosphere of this planet? Is it a "shadow planet" with an atmosphere so thick and dense that it blocks all light that would reach the surface?

I don't know, how it works there, but perhaps it is easier to imagine, how it could look.

TrES-2b is a gas giant, so it doesn't have a well defined surface. When you are somewhere in the atmosphere then:

-When looking up, you will see the light, as deep sea creatures do. The amount of light and its color will depend on how deep you are.

-When looking down, it should be dark (black), comparing to how it looks up. Otherwise, the atmosphere would be reflecting light.

-When looking to the sides, it should also be dark, for if the light were scattered at some angles, it would also be scattered upwards.

So, one simplest possible model is: like in deep sea.

Note, that at some other wavelengths, particularly in infrared, it all can look very differently.


At what size do objects burn up in the atmosphere when falling from orbit? As the title suggests, I am wondering the maximum size an object can be so that it will completely burn up when entering and falling through Earth's atmosphere. I'm focusing on man made objects, not asteroids or meteors. Also, how much does the make-up of the object affect its burn rate? I.e. metal vs plastic vs glass, and so on.
Thanks!

The Earth atmosphere protects us from small impacts from both asteroids and man made objects. This is well known from meteoroids, where meteoroids as large as a few tens of meters in diameter usually fail to penetrate into the lower atmosphere because they get fragmented and dispersed at high altitude. Fragmentation height depends mainly on the meteoroid strength, only strongest irons reach the surface in one piece.

We could extrapolate from meteoroids to man made objects that even large do not have the same physical strength than a massive meteoroid body. This means than nearly all man machine objects will disintegrate before reaching earth. Heavy metallic ones (iron) will disintegrate at lower altitudes than lighter ones such glass and plastic.

Two comments:

This is not true for objects not having reached orbit, e.g. rocket 1st or 2nd stage engines will fall without disintegrating.
Even if they disintegrate, nuclear powered satellites will cause some level of radioactive pollution, however widely scattered in the atmosphere.

Although the risk to human beings walking on earth is next to nil, there is a real risk of impact from man made objects littering Earth orbits to other spatial vehicles.

Reference: "The Impact Hazard", David Morrison, Clark R. Chapman and Paul Slovic, in "Hazards due to Comets and Asteroids", T. Gehrels, Ed., 1994, The University of Arizona Press


When does a star rise with the sun? Problem: Suppose an observer's latitude is 45 S and a star's RA/DEC is
3h15min/41S. On what date will the sun set with the star.

I have been stuck on this problem for a while.
I found out that the time of setting of a star is found using the formula:

$$\alpha+\arccos(-\tan(\delta).\tan(\phi))$$
$\implies3\frac{1}{4}^h+\arccos(-\tan(-41)\tan(-45)) = \alpha_{sun}+\arccos(-\tan(\delta_{sun})\tan(-45))$
$\implies 47.8 = \alpha_{sun}+arccos(tan(\delta_{sun}))$

I am not able to solve this any further.
Is there something wrong in my approach? Something I am missing?
If not how do I solve the above equation further?
Thanks in anticipation!

The formula you gave is to find the hour angle of the star while setting, not the setting time itself. Suppose the hour angle of star is $HA\star$, and $RA=\alpha$,then the Local Sidereel Time is given by $LST=HA\star+\alpha$. At the time of setting, $HA\star=10^h1^m30.2^s$. Thus, $LST=13^h16^m30.2^s$at the time of setting.

Now, $RA\odot=6^h$ on $\text{June} 21^{st}$ and $HA_\odot =\arccos(\tan(23.43^{\circ}))=4^h17^m17^s$ at sunset. Thus the star sets with the sun after $\text{ June} 21^{st}$. Since $RA_\odot$ increases approximately by $2^h$ every month, we can assume that the star sets with the sen near $\text{July }21^{st}$. Let it set after $n$ days. At $\text{July }21^{st}$, $\delta_\odot = 20.09^{\circ}$ which gives us $HA_\odot=4^h34^m11^s$, and $RA_{\odot}=8^h9^m44^s$. The $HA_\odot$ at sunset is approximately the same for nearby dates. Since $\delta_\odot$ is also near the maximum, we can approximate $\Delta RA_\odot= \omega_\odot n$. Thus,

$$13^h16^m30.2^s=8^h9^m44^s + \omega_\odot n +  4^h34^m11^s$$
$$ \omega_\odot n =0.5433^h$$
$$n=8$$

At  $\text{July } 29^{th}$, $\delta_\odot=18.23^{\circ}$, $HA_\odot= 4^h43^m6^s$, $RA_\odot=8^h42^m12^s$ which gives us $LST=13^h25^m18^s$, which is a bit more than expected. Trial and error for few days before gives us that on $\text{July } 26^{th}$, the sun sets at LST $13^h15^m9^s$ which would be the most appropriate answer.

Thus, the sun sets along with the star at $\text{July } 26^{th}$


Can smaller habitable planets (or moons) orbit a larger habitable planet? I am doing some research for my fantasy novel, but I'd like to really understand how my planet system would cause life to grow if at all possible.

My idea would be, there is a large core planet that is orbiting a sun. Two smaller planets (or moons) are in an orbit around it on polar opposite sides of the core planet. Is this even possible? Would the necessary means to spark life occur?

First off, let me point out that if you have a central planet which is orbited by other bodies, those other bodies must necessarily be considered moons and cannot be planets. This is due to the official definition of a planet stating that it must orbit the central star. The objects orbiting your central planet are not primarily orbiting the central star and thus would be considered moons or satellites.

Orbital Dynamics

What you're proposing is that your central planet is orbited by two moons that are on opposite sides of the planet. This condition provides some strong constraints on your system as dictated by the laws of physics.

If we are to consider that the moons are orbiting around the central planet, then the planet must be fairly massive. If all three of your bodies are of a similar or comparable mass, you'd have a much different (and not necessarily stable) type of orbit than what you're describing (see this link for example). The fact that you have two moons orbiting your planet, without your planet having a noticeable orbit about the moons, implies the planet has most of the mass of the system. This results in one of two scenarios, either

your central planet is very large (e.g, the size of Jupiter or Saturn) with moons the size of Mercury or Ganymede, or
your central planet is relatively small (e.g., the size of Earth or Venus) with moons the size of asteroids (e.g., Phobos and Deimos, the moons of Mars)

Of course you could have any gradation between the two possibilities I just listed. The main point is, your planet must be much more massive than your moons in order to be the dominant gravitational player in its system.
You also state your moons are orbiting such that they're on opposite sides of the planet. Such an orbital configuration is known as a 1:1 resonance. It necessarily implies that your moons are orbiting at the exact same distance from the planet and take the same amount of time to complete a single orbit. While such a system is technically stable, I can't say with certainty that it is stable for a long period of time. More than likely, small perturbations will occur that will build up over time and cause your moons to leave their 1:1 orbital resonance. This will result in unpredictable outcomes, but my best guess is that the moons may start gravitationally interacting if they get close enough after no longer being on the same side of the planet. Eventually this may result in them exchanging energy and one getting kicked to a closer orbit while another getting kicked to a further orbit. But that's just a guess.

So short answer, yes your system is possible but it comes with some constraints and is likely not going to be stable for a long time period.

Habitability

You also ask about the habitability of such a system. This is a bit tricky and there's no single answer. There are so many conditions that affect habitability. To make matters worse, we only have a single example of a habitable planet. However, there are general conditions we believe must be true for a planet to be habitable (for humans at least).

The big condition is that your system must be at a good temperature such that water can exist as a liquid. If your central planet is a Jupiter/Saturn analog, this is not going to happen since such planets have large, mostly hydrogen gaseous envelopes. There's not really any pools of water for humans. However, your moons could still host water. It really depends on how far out your planet is from the star. These types of planets form very far from the star, where water doesn't easily exist as a liquid so either your planet must migrate in to the inner stellar system early on (and stop at a respectable distance), or else some sort of tidal effect has to keep your moons much warmer than they would be other wise (see Enceladus, for example). The former means your chosen moon orbits are likely to be less stable and the latter means you're going to need other moons to help provide the heating, again making your orbits less stable. If your central planet is Earth-like, it can certainly host water and be as Earth-like as you want, but the moons are unlikely to be big enough to be habitable.

There are various other conditions for habitability such as nice atmospheres or magnetospheric protection from the star but one could write a whole book on this subject. I'll leave off by pointing you to this question on habitability.


Could we parallax measure stars just based on the Earth's size? Imagine two major surface observatories, perhaps the "northermost" and "southernmost" such (ideally on a similar longitude).

For nearby stars, could they each take a photo at the same time, and achieve a distance measurement to that star, based on about the size of the Earth as the baseline?

(I appreciate that either of the scopes need only wait a few days, for their viewpoint to have moved a much longer baseline distance!)

Or is that distance far too short?

What then is about the minimum baseline we could parallax-measure the nearest stars, with our currently best telescopes?  100,000km, a million? Far more?

In principle, it's not impossible.

The Gaia spacecraft, designed primarily for measuring stellar positions, is able to measure parallaxes up to 10 kpc away with 20% uncertainty. Its baseline is 2 AU; $2.3\times10^{4}$ times larger than the diameter of Earth. Thus, placing two Gaias on each side of Earth would be able to measure parallaxes of stars up to a distance of $10\,\mathrm{kpc} \,/\, 2.3\times10^{4} \simeq 0.4\,\mathrm{pc}$, meaning that you would almost be able to measure the distance of our nearest neighboring star, $\alpha$ Centauri, which lies at 1.3 pc. So you would just need improve your Gaias a little bit.

This ignores small complications such as the atmosphere, but if you're willing to put them outside the atmosphere, you could do it. Of course, it would sort of be a waste of time, since we already know the distances to the nearest stars, but hey, go ahead.


How do I calculate the sun's azimuth based on zenith, hour angle, declination, and latitude? Wikipedia cites this formula for calculating solar azimuth:

Where:

$\phi_s$ is the solar azimuth angle
$\theta_s$ is the solar zenith angle
$h$ is the hour angle, in the local solar time
$\delta$ is the current sun declination
$\Phi$ is the local latitude

I'm implementing this equation in JavaScript, but it doesn't seem correct. Can someone here help me confirm whether or not this equation is correct?

Here's my JS implementation. The only strange thing is at the end: limiting the acos domain to [-1, 1] to prevent NaN values.

const getSolarAzimuth = function(
    zenithAngle, hourAngle, declination, latitude
) {
    const cos = Math.cos;
    const sin = Math.sin;

    const cos_phi = (
        sin(declination) * cos(latitude) - cos(hourAngle) * cos(declination) *
            sin(latitude)) /
          sin(zenithAngle);

    return Math.acos(Math.min(Math.max(cos_phi, -1), 1));
};

I've confirmed that this formula is in fact correct. I did have the wrong units in the code, which @Glorfindel suggested: sun's hour angle was in hours instead of radians.

I did need to adjust the azimuth's angle offset based on the hour angle:

     // The angle offset needs to be adjusted based on whether the hour angle
     // is in the morning or the evening.
     //   https://en.wikipedia.org/wiki/Solar_azimuth_angle#Formulas
     if (hourAngle < 0 || hourAngle > Math.PI) {
         return az;
     }

     return (Math.PI * 2) - az;


How do we know that Singularity is not an object with reasonable radius? So for example: a Neutron star has a radius of 10km and 2 solar masses.

A singularity can have a radius of 1m and 10 solar masses, with a 1km event horizon radius. Basically it will just be a more compact object than a neutron star. How come we are so sure that singularity is an infinitely small point instead?

We don't know, but general relativity implies that no force can stop the object from shrinking to infinitesimally small.  Quantum Mechanics is more uncertain.

I just want to point out that 10 solar masses would have a radius 29.53 km at the event horizon, not 1 KM.   A Neutron Star is pretty close to its event horizon already.

The problem with "some distinct size inside the event horizon" is that things change when the acceleration is greater than the speed of light.   If the acceleration is strong enough that the gluons are drawn towards the center faster than they an move (speed of light), then it's entirely reasonable that the structure wouldn't hold.   Another way of saying this is that the gravity becomes stronger than the Neutron Degeneracy Pressure and there's no known force that would prevent continued shrinkage once the Neutron Degeneracy Pressure is overcome.

The Pauli exclusion principal says that two neutrons can't occupy the same space, but this can be overcome somewhat by the weight and pressure approaching infinity.   The neutrons can get closer and closer as their momentum approaches infinity and that happens at the singularity.

So, both relativity and QM (the two best theories for dealing with this question) say that a ridiculously small object may exist at the singuarity, not meters, but maybe the size of an atomic nucleus or size of a quark or even, possibly zero size, but without a quantum field theory of gravity it's impossible to say with much confidence.

Other possibilities exist like space might get stretched into some very strange properties, where virtual particles become real but fall back towards the singularity, similar to hawking radiation but without the escape, or, the angular momentum could form into a ring or band, not a point but all that is just guesses.    The mathematics of the two main theories both say that a very tiny point is entirely reasonable and the argument for what you suggest, an object with a distinct radius of say 1 meter, is far less supported by current theories.

That's my layman's answer anyway.   I can delete this if I got too much of the physics wrong, but I thought I'd give you a quick answer.


Is the centre of axial precession perfectly perpendicular to the ecliptic? It is my understanding that the Earth's axial precession is such that the direction of the North pole pointing to the celestial sphere traces out a circle with a period of roughly 26,000 years. My first question on reading this was, where is the centre of that circle? In other words, what axis is the Earth precessing about? I had assumed this would be somewhat arbitrary but have since learned that it points towards the north ecliptic pole. Am I then correct to assume that planets naturally tend to rotate in the same plane as their orbits, but the effects of precession caused by other massive bodies cause the axis of rotation to oscillate around the north ecliptic instead?

If so, then what is going on with Uranus and it's unusually high axial tilt? Is this another phenomenon or does Uranus' rotation axis also precess about it's own ecliptic north pole? Albeit with a much larger amplitude.

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).


Why don't we build an interferometer between Earth and the Moon? The Event Horizon Telescope (EHT) does interferometry from an array of telescopes spread all across Earth. The data is locally stored on a hard drive and shipped to a central location, so the telescopes in the array don't need to be connected to each other in real time. Would it be possible to make an EHT-like interferometer with telescopes on the Earth and on the Moon? It seems like such a telescope would have a much higher resolution than the EHT, and would make it possible to observe in wavelengths shorter than radio.

Of course, money is likely an issue here, but is there any technical reason why this wouldn't be feasible?

Put simply: controlling the phasing would be impossible (not just difficult).  Not only do you have to deal with a constantly changing physical distance but also with the dynamics of the atmosphere.  It's difficult enough to maintain transverse phasing across an earth-based receiver (think adaptive optics);  I can't imagine a reliable way to track the radial variation.  Without that, unless you are sticking with multi-meter wavelength radio systems, you cannot hope to phase up the receivers.


How to correctly create planetary orbits from kepler elements I would like to build a gravity model (with 3D graphics) that would build the orbit of the planets based on orbital elements input and the gravity calculation for each element.

My initial code is live here (Click the 0.005 to turn it into 0.05 and then click on restart time to run the orbits).

The example above has all planets on Solar System (up to Neptune), the Sun and the asteroid Apophis. For clarity, the Sun and planets other than earth have been hidden, but are still considered for gravity calculations.

However, the orbits do not match the correct estimate intersection date. In this article, it says that Earth-Apophis close-encounters will happen on 13 April (meaning that the earth should be in the orbit's intersection point every 13 April, or very close to it).

The data from the program is taken from Nasa JPL.

Apophis Data
Planets Data
The sun is positioned at the origin

Start date is 2454441.5, as given in JPL Apophis data. Planets orbital elements have been corrected to this date, with the calculations in the above PDF.

Why the orbits intersection date is wrong? Should I be doing something else? Am I doing something wrong?

The above Apophis link also cites a "Standard Dynamical Model". If anyone knows how to correctly create one, I believe this would answer my question as well.

TLDR: How can I correctly create planetary orbits from kepler elements?

Thank you in advance if anyone can help me!

If you're running your own N-body dynamics,
you can get initial Cartesian positions and velocities for each body in your system (including the Sun) by asking
JPL HORIZONS
for Ephemeris Type: Vectors.
I would choose Coordinate Origin: Solar System Barycenter so that the Sun does not drift away from the origin,
and Target Body: [Planet] Barycenter (or Sol) unless you are modeling the Earth and Moon as separate bodies.

If you still want to compute initial Cartesian positions and velocities from Keplerian elements, first do a sanity check: Earth should cross the X axis at the equinoxes and the Y axis at the solstices.
Second, those approximate elements for the major planets may not be precise enough for your purpose.
Instead you might ask HORIZONS for Ephemeris Type: Elements, and Time Span beginning with the epoch of your elements for Apophis.
You could also get elements for Apophis in a more convenient epoch this way.

Note that Keplerian elements are subject to dynamic evolution and are only valid for a limited time near their epoch.
If you ask HORIZONS for Apophis's orbital elements one week before and after its close approach to Earth, you can see that they are quite different. Of course the latter are also highly uncertain.

For the gory post-Newtonian details of the dynamical model the Apophis prediction article discussed extending, see JPL's
notes on DE430 and DE431, part III.


Can we detect the wobble of neighbouring stars around their barycenters? The wobble of a star around its barycenter is interesting (the wobble caused by the pull of the rest of the solar-system on the sun).

This question explains why we don't see this wobble from within our own solar-system (because we are mostly orbiting the sun, and not our solar-system's barycenter):
What point does Earth actually orbit?

But is it possible to detect from here on Earth, the wobble of neighbouring solar-systems relative to the wobble of our solar-system?

Yes it is. The Gaia astrometric satellite is and will be doing exactly that. The subtle movement of the photocentre of a target star due to any orbiting planets can be separated from the proper motion and the annual parallax due to our motion around the Sun.

The technique is most sensitive to massive planets that are widely separated from  parent stars that are close to the Sun and have low mass themselves, since these provide the biggest angular shift in the star's position with respect to the barycentre. However, the separations cannot be too wide or the orbital period of the star/planet system will be too long for the motion to be detectable on the timescales of the mission. In practice therefore, Gaia will be most sensitive to planets with orbital periods of 1-10 years (or separations of 1-5 au). Simulations suggest that tens of thousands of exoplanets may be revealed in this way (Perryman et al. 2014).

ESA have even provided a little animation to show how it works.


Are red dwarfs really 30-100 times our Sun's density? This article by K. E. Edgeworth claims red dwarfs tend to be around 30-100 times our Sun's density.  However, this seems a bit high.
With smaller stars, mass tends to be proportional to radius but have a notably smaller volume, leading to higher densities.  This Wikipedia article labels red dwarfs as massing between 0.08 and 0.45 solar masses.
However, following mass being proportional to radius, a 0.25 solar radii star with 0.25 solar masses has a density of 22.5003 g/cm³, or about 16 times our Sun's density.
A star 1/4 the size and mass of ours is smack in the middle of the red dwarf range, but is only about halfway to the 30 times our Sun's density minimum the above article outlines, while based on the above article I would expect a red dwarf 1/4 the size and mass as our Sun to have a density more like 60 times our Sun's.
What is the cause of this discrepancy?

Red dwarfs, depending on your definition, can range from 2.5 to 150 times more dense than the Sun.

What is the cause of this discrepancy?

They give no calculations, so I can only guess.

The article is from 1946 and we've gotten a lot better at science.
It's 1946 and information exchange is limited. No internet, no TV, and long distance calls are expensive.
Red dwarfs are tricky to observe and we've gotten a lot better at that.
Their hypothesis is that red dwarfs do not fuse hydrogen which is incorrect.

Edgeworth's other contemporary works might provide insight.
"Red dwarf"
There is no strict definition of a red dwarf. Some use class M stars below a certain temperature and mass. Others include some class K stars. Fortunately they give a definition.

low luminosity (say, not more than one tenth that of the sun), small mass (say, not more than three quarters that of the sun) and high density (perhaps 30–100 times the density of the sun).

Luminosity and mass are basically correct for the modern definition of a red dwarf, M-class, but red dwarf density runs a considerably larger range from 2.5 to 150 times more dense. Still, they're inside the range.
They give no rationale for their density calculations. Being 1946 its probable their mass and radius calculations were considerably off. Red dwarfs, being so faint, are very difficult to observe.
Red dwarfs fuse hydrogen

In the case of stars belonging to the main sequence, it seems to be necessary to assume that the general conditions in the interior of the star are such that convection currents occur which are of sufficient importance to convey hydrogen in adequate amounts from the outer layers to the central core...

Main sequence stars are convective, but it's complicated.

...although this does not imply that the convection currents are of
sufficient magnitude to be the dominating factor in determining the
distribution of temperature.

Incorrect. The lower the mass the more convection is the "dominating factor" over radiation.

...red dwarfs probably contain a normal proportion of hydrogen...

Correct.

...but that convection currents in these stars have never been of
sufficient magnitude to convey the hydrogen in adequate amounts from
the outer layers to the central core...

Incorrect. The lower the mass the more convection. Low-mass red dwarfs are fully convective with no helium accumulating in the core.

When the mass of the star is less than [0.75 M☉], the hydrogen which
the star contains never reaches the core in sufficient quantity to
initiate and sustain the process of transforming hydrogen into helium,
and dynamical equilibrium is secured only by contraction involving
comparatively high densities.

Incorrect. Red dwarfs fuse hydrogen. What they're describing is more like a brown dwarf.

On this theory the red dwarfs are not stars of great age which have consumed their substance in riotous living; they are simply feeble things which have never been able to make use of the supplies of fuel which they actually possess.

Red dwarfs are of great age, but they aren't bigger stars which have run through their fuel. They also are not "feeble things" which can't fuse hydrogen. They are low-mass stars which use their fuel very efficiently and so live to a great age.
With an incorrect model they will get incorrect answers.


Help converting ECI to RA DEC Using the library OrbitTools I can input a TLE and get back ECI coordinates. I want to find the RA DEC though, so I use

right_ascension = atan(y / x) * 180.0 / PI;

declination = acos(z/ sqrt(x* x + y * y + z * z)) * 180.0 / PI;

I figure there is likely some constant I need to add (PI/2, PI, etc.) but comparing the calculated RA to that in the TLE, there is no constant that would fix it.

I also noticed x,y,z can be very low (on the order of ~100 km in some cases) which would suggest its measured from earth's surface, so I would need to add the earth's radius to x,y,z, but this also doesn't fix my problem.

Alternatively, I tried using the built in look angle function from OrbitTools, assuming that (0,0,0) (latitude, longitude, height) would be equivalent to RA DEC but... its not. Thanks!

Assuming you are calling cEci and have the units and sign conventions correct (longitudes are notoriously inconsistent as to which way is positive) then you should be able to calculate it almost as you have as:

distance = sqrt(x*x + y*y + z*z)
RA = atan2(y/x)
# Normalize right ascension into the 0..2*PI range
if ( RA < 0. )
  RA += PI + PI
Dec = asin(z/distance)
# Convert radians to hms and or degrees to taste

This code was taken from Bill Gray's sat_code, specifically observe.cpp. Other bits of that code handle the conversion from a latitude, longitude, height (geodetic coordinates) to a X,Y,Z vector from the Earth center to the observing site (geocentric co-ordinates) and then the rotation by the Local Mean Sidereal Time (site longitude + GMST, where GMST is the Greenwich Mean Sidereal Time) and then the 2 vectors (Geocenter->Site and Geocenter->Sattelite) are added. However it looks cEci already has this as I see references to the flattening factor and LMST.

One thing to note with TLEs is they are not a very good representation of the satellite motion/orbit over long periods of times. You should make sure your TLEs are up-to-date or use something like SGP4 to apply the perturbations and update the TLEs.


Is Enceladus the moon with salty water? May anyone tell me does it known if Enceladus has water and is that water salty?

Like many bodies in the outer solar system, Enceladus is covered in water ice. It is also strongly believed to have liquid water underneath the ice, as we have observed (with the Cassini probe) plumes of water from its polar regions.

The water is salty and alkali due to the dissolved salts and the reactions of water with the underlying rocks. The plumes contain sodium chloride.

So there is a salty ocean under the ice of Enceladus.  There is also likely to be water under the ice of other moons, such as Europa.


How to measure the altitude and azimuth of a star? Given that the star is crossing the local meridian line in a certain location, I've tried calculating the altitude of a star by finding the difference between the declination of the star and the latitude of the position from where it is observed (which I believe is not correct). I'm so confused about that and it would be of great help if someone could explain the altitude calculation. Also, I'm wondering how to calculate the position of a place (latitude and longitude) while the declination and the right ascension of a star at the zenith is known.

For the second question: If you know the coordinates of the Zenith, your latitude is exactly Zenith's declination. For your longitude you can not rely on the Zenith: the same star will be at the Zenith at the same sidereous time for every place in the same latitude, so you need a clock besides the telescope. cf. http://en.wikipedia.org/wiki/History_of_longitude#Problem_of_longitude

For the first question: The altitude of a star that is crossing the local meridian is also quite easy.

For Northern Hemisphere, a star that is crossing the North local meridian (that is, between the Zenith and the North Horizon) the latitude is
$90-alt=dec-lat$

For Northern Hemisphere, a star that is crossing the South local meridian (that is, between the Zenith and the South Horizon) the latitude is
$90-alt=lat-dec$

Same goes for Southern Hemisphere if you change also North Horizon and South Horizon.

You did not asked specifically for the Azimuth, but again you need a clock for that.


What is the schedule for science runs of aLIGO (and VIRGO)? Advanced LIGO seems to operate only intermittently. Is there a schedule for at what times it will be able to register new gravity waves? Even if no formal schedule is available, what main factors determine the scheduling?

I suppose VIRGO in Italy, to get started in 2016, will operate mostly simultaneously with aLIGO. Or would its design require another type of schedule?

There is no set date due to the complexity of the machines and their upgrades but this is the closest you will get to an answer:

  O1 is planned to last for four months, closing mid-January 2016. Then work will start on upgrading the instruments for our second observing run, which is called O2; those upgrades will be informed by what we have learned about the instruments during O1. O2 will start in 2016 and last around six months. Hopefully, around this time Advanced LIGO will be joined by Advanced Virgo.

They have no fixed date for the second observing run (O2) but it should be up an running later this year (my guess would be June-July time).


Has the most luminous object in the universe been found? It's hard to search through the entire (observable) universe, but the most luminous thing there is must on the other hand be easy to find. Can we now say that the most luminous quasar we know, is the (intrinsically) most luminous one that exists, or will we soon?

As it stands I think the answer is no. The most luminous object in the (observable) universe is probably hidden behind many magnitudes of dust extinction. There are lots of faint, infrared-detected objects that may be extremely luminous galaxies in the early universe (and indeed some have turned out to be) but it is difficult to get their redshifts, so it is difficult to be sure we have a complete luminosity function. We now know that objects with central black holes $>10^{9} M_{\odot}$ already exist in the universe after a billion years or so. To get to that size there must have been an awful lot of accretion and an awful lot of accretion luminosity.

In addition, the most luminous galaxy/quasar is going to be a rare object and only ridiculously small areas of the sky are covered by things like the HST ultra-deep field. i.e. There is almost bound to be a more luminous one in some other patch of sky that has not (yet) been surveyed. The logistical problems of getting a UDF over the whole sky (or even a significant fraction of the sky) means that isn't going to happen any time soon.

Finally, the brightest quasar is certainly not the most luminous. 3C273 is quite a modest object, in terms of luminosity, but it is very close to us as quasars go. But perhaps you didn't mean bright in that sense? Brightness is normally reserved to refer to the apparent brightness. Luminosity is the extinction-free, distance-corrected power.


How Did Early Astronomers Measure Distances? Prior to the era of radar and other forms of radio/RF/EM ranging, what approaches, methods, and techniques did early astronomers (e.g., Kepler, Cassini, Copernicus) use to measure the distance(s) from Earth to various objects (such as: Mars, Venus, Alpha Centauri, etc.)?

Is there an especially informative article (say, Scientific American, Sky) or book that explains their efforts and results (e.g., how accurate were their computations?)?

References and citations are warmly welcomed!

They were not very successful, but not for a want of trying.

The ancient greeks tried to measure the angle between the sun and the moon when the moon is half full. They got a value of 87 degrees, (when really it is 89.85). From this they could estimate the relative distances of the Moon and the sun.

Further observations of the moon from different locations on Earth could, in principle find the relative distance to the moon in terms of the radius of the Earth. And the radius of the Earth could be determined by noting the different lengths of shadows at the same time in different locations.

Several Greeks attempted to put these together to find the distance of the sun, and while their methods were geometrically correct, their observations were not accurate enough to achieve an accurate value for the distance of the sun. Typically they estimates were about 5% of the true distance. Attempts based on similar methods were made in other centres of astronomy, in China, India and the Middle East. They all tended to dramatically underestimate the distance.

Even after the invention of the telescope allowed the observations to be done more accurately the distance was usually under-estimated. only when careful measurements of the transit of Venus were made, was Jerome Lalande able to calculate the distance of Venus, and use Kepler's laws to derive the distance of the sun and other planets.

Kepler had created a model of the solar system that predicted the relative distances of each planet very successfully, but not the absolute distance

Nearly all other distances are based on the Earth-sun distance. The distances of the planets can be found using Kepler's laws. The actual distance of stars was unknown we were able to measure the slight change in position of a star as the Earth orbits the sun. If you know the distance to the sun you can calculate the distance to a star. Prior to the observation of this "stellar parallax" estimates of the distanct to stars was essentially guesses, and usually significant underestimates.

So prior to radar ranging, the methods were geometric, and based on a ladder of measurements. The size of the Earth was used to determine the distance to the moon and sun. Then the distance to the sun was used to find the distance to the stars.

For more distant stars there can be considerable uncertainty about the distance. If a star is too far to have parallax, then it's true distance will not be known to accuracy.


Focusing light through fiber optic cable to extend telescope viewer I want to build a system to view my telescope without being uncomfortable. Currently I have to stand up and bend over to view the eye piece.

I want to build a system that allows me to 'pipe' the output of the view piece in a non-digital format along a flexible fiber optic cable to my eye and possible split the view into two for a sort of VR headset.

I know the cable will transfer the light, but will it be just burly light or break down too much (one meter or less) or will I be able to focus it into view-able picture?

I could just try it, but the cable itself is really expensive to get, and I want to know if this is worth trying before I spend.

I'm thinking of using this 14 mm 'solid core' cable

In principle a fiberscope eyepiece could work but not very well.
By putting one end of the cable flush with the field stop of an eyepiece (few allow this) and the other end exactly in the telescope's focal plane, you'd get a fiber-pixelated image.
A binoviewer would require two cables, difficult to align.


Could the Universe really be 12.5-13 billion years old? My initial reaction is that "this must be wrong" and apparently that's a lot of people's initial reaction according to the article.

https://www.nbcnews.com/mach/science/universe-may-be-billion-years-younger-we-thought-scientists-are-ncna1005541?cid=sm_npd_nn_tw_ma

My question is, is this study actually making scientists scratch their heads, and, 2nd part, if accurate, how could the 13.78 (or whatever it is) estimate have been wrong, as it seemed to be the result of more and more accurate estimates over time.

One of the two estimates has to be wrong.

Because some measurements are difficult to make with absolute precision, it often happens that there is some uncertainty about them. The half life of the neutron was once thought to be 12 minutes, but in the 1980s this was amended to 10.5 minutes. In the early days, C14 dating was often substantially out by comparison with dendrochronology and other methods. It depends on the amount of C14 in the atmosphere at any particular time, and when they looked into the matter, scientists found that this varies from one century to another. When they took that factor into account, C14 dates became much more accurate and reliable. Back in the 1970s the age of the universe was thought to be about 15 billion years, but that has been revised to 13.8, and as new data comes in it may need to be revised again. This is why gravitational collapse of the universe has not been entirely ruled out, despite the advent of dark energy, which is thought to be accelerating the expansion when it should be slowing down. It is possible that like some other measurements in science, the dark energy measurement may also need to be revised at a later date.


Transit Depth Color Versus Transmission Spectroscopy One of the tools being developed for the study of exoplanets is called transmission spectroscopy - basically looking at how passing through the exoplanet's atmosphere changes the spectrum of the star. The problem, though, is that spectroscopy is hard, and requires a big telescope, probably space-based. This makes the process slow. It seems that it would be very useful if we would triage the data using broad-band photometry to more quickly limit the presence of compounds like water, $\mathrm{CO}_2$, and $\mathrm{O}_3$. Looking at a plot of Earth's atmospheric transmission, with a breakdown by chemical compound, it seems like looking at the color of the transit depths in and between the windows in the wavelength range from 1 to 20 microns, might provide a plausible alternative to quickly get information on the atmospheres of planets and brown dwarf stars.

For the exoplanets, specifically, the wavelengths blocked by an atmosphere make the planet, effectively, bigger at some wavelengths than at others, depending on the opacity at that wavelength, the thickness (density) of the atmosphere, and vertical extent of it.

The question is: is this practical? Specifically, with a 1 to 2 meter class space telescope that simultaneously images in 2 to 8 filters?

But this is already possible. Spitzer has performed "low-resolution" transit spectrophotometry of several exoplanets in its near infrared bands - see for example Tinetti et al. (2007);  Knutson et al. (2011); Desert et al. (2011).

I dare say you could get better data on more planets with a dedicated telescope with a larger mirror and the ability to image in several bands at once. All that is required is money.


Understanding the angular peaks in the CMB power spectrum I have read up quite a bit on the CMB power spectrum, but one feature evades my understanding. The first peak is often referred to as the "first compression peak", meaning that the plasma had time to compress exactly once. The way I imagine this is that we have a central overdensity, and then a ring of plasma around it which moves closer, i.e. the ring is being compressed. In the same vein I think of the second peak as corresponding to the ring being compressed and then rarefacted. But I don't see how this fits in with the power spectrum: the first peak corresponds to a larger separation between the center and the ring, while the second peak corresponds to a smaller separation. With the way I have described it, I believe it should be the other way around: compression should correspond to smaller scales, while rarefaction should correspond to larger scales.
What am I misunderstanding about this?

My attempts at a short answer failed, so here is my best shot at an explanation.
When you look at what the photon-baryon plasma does in the dark matter background, you will bump into an equation that is almost the equation for a harmonic oscillator, but it contains a time dependent quantity $\chi = \frac{3\rho_B}{4\rho_\gamma}$ that spoils the fun. It is, however, possible to approximate the solution:
$$\Theta_k(\tau) + \Phi_k \approx A_k \cos\Big(k\int_0^\tau c_s(\tau')d\tau'\Big) - \chi\Phi_k$$
where $\Theta$ is the temperature fluctuation $\delta T/T$ (in Fourier space), $\tau$ is the conformal time, $\Phi$ is the potential caused by dark matter and $c_s$ is the sound speed in the plasma. The observable quantity is $\Theta_{0k} = \Theta_k + \Phi_k$. These are the intrinsic temperature fluctuations plus the background potential.
The prefactor $A_k$ is determined by evaluating this expression at $\tau = \tau_{LS}$, the moment of last scattering. This then yields:
$$\Theta_{0k} = \frac{1}{3}(1 + 3\chi)\Phi_k\cos(k\lambda_s) - \chi\Phi_k $$
with $\lambda_s = \int_0^{\tau_{LS}} c_s d\tau$ the sound horizon (i.e. the distance acoustic waves can travel in a certain amount of time). The quantity $\Theta_{0k}$ is called the monopole contribution to the CMB. The dipole contribution, that is due to the Doppler effect, is much smaller and I will not consider it here.
Let's take a look at an overdensity ($\Phi_k < 0$) and place our coordinate axes right in the middle of it. At the "time of the Big Bang", $\tau = 0$ the monopole is $\Theta_{0k} = \frac{1}{3}\Phi_k <0$. The argument of the cosine is 0, so the cosine itself reaches its maximum value.
Half a period of the cosine later: $\Theta_{0k} = -\frac{1}{3}(1 + 6\chi)\Phi_k$, which is a lot larger! In the standard cosmological model, $\Phi_k$ remains constant, so the only thing that could have gotten bigger is the intrinsic temperature fluctuation $\Theta$. This temperature fluctuation is directly linked to the density fluctuations of the photons and baryons.
What happens intuitively is the following: the baryons love falling into dark matter potentials. Baryons are approximately pressureless, so nothing would impede them to do so if it weren't for the photons to which they are coupled before $\tau_{LS}$. The photons are dragged down into the potential well by the baryons, which makes them hotter (photons do have pressure), so thermodynamics tells them to cool again by escaping the well. It's much harder to get out of a potential well than it is to fall into one, so the amplitudes of compression ($B$ and $\gamma$ falling into a well) and expansion ($\gamma$ pushing $B$ out of the well) are not equal.
Look again at the cosine at $\tau_{LS}$. It will reach maximal compression when $k\lambda_s = (2n + 1)\pi$ for $n = 0, 1, 2...$, which defines the wavenumbers of the fluctuations that reach maximal compression. It will reach maximal expansion when $k\lambda_s = 2n\pi$ for $n = 1, 2, 3...$.
The first peak reaches maximal compression at last scattering. It is huge because it is least affected by Silk diffusion. It corresponds to a large angular scale (compared to the other peaks) because its wavelength is $\lambda = 2\lambda_s$ and is therefore the largest fluctuation.
And finally, I think you already found Wayne Hu's website, but here is the link to a gif that shows you the oscillation of the baryon-photon plasma in a dark matter potential well.


If rotation has nothing to do with gravity, why do astronomers always link swirling dust and gas to stellar evolution? All my life I was taught in school that we have gravity because the earth spins.  Now I know that is not true.  Online academics were baffled why anyone thought so. I can only say that's what they said in school.
But if gravity is more mass and everything in the universe has some gravitational "pull", then why do astronomers always use spinning and whirling dust and gas as a way to explain stellar evolution?  They make it sound like the spinning around created the gravity to form the mass and the bigger it got the more it spun and made more gravity.

Any time you have a large cloud, different parts of it will randomly be moving in different directions. If you pick any line through the middle to consider, slightly more particles might be moving (say) clockwise than counterclockwise around that line, just by chance. And there has to be some axis where that random difference is the largest.
Over time, the gas and dust may start to contract under its own gravity. Like a spinning skater pulling in their arms, the rotation will speed up, and there will be collisions so that over time there will be a consistent rotation in a certain direction, just arising from the original random motions.
Instead of producing the gravity, it kind of does the opposite thing. The objects really want to move in straight lines, and gravity is the only thing that keeps pulling their paths around the middle. If there were no motion, the objects would just fall straight down towards the middle, so in that sense, the rotation is acting against the gravity. Explanations of the formation of the solar system mention the rotation of the initial cloud because it was the precursor to the revolution of the planets about the sun, along with the rotations of the planets and of the sun itself. It is not because the rotation somehow creates the gravity.
When you spin an object, things inside it swing out towards the edge. That's really just their momentum carrying them in a straight line until (say) the wall of the can gets in the way. But if you're in a spinning room and the wall is constantly keeping you from flying off in a straight line, it kind of feels like gravity is pushing you against the wall. But it's not gravity, just the wall pushing inward on you and steering you in a circle. The inward force of the wall is called centripetal, and the outward force that you perceive is technically not a true force, but it is a meaningful concept called centrifugal force. But, again, it is a very different thing from gravity.


Survival on a rogue planet Are there any planets not orbiting a celestial body which can support life despite the temperature? Also what is the absolute minimum temperature that life can survive in?

We have not yet discovered any planets that are certainly capable of supporting life, (except Earth) and the few rogue planets that we do believe that we have found have been gas giants, no good for life.

It is unlikely that life could exist on a rogue planet. The surface of a terrestrial planet that was not in orbit around a star would be exceedingly cold, as cold as Pluto.

The only place for life to start would be in a layer of liquid water, similar to that on Enceladus. If the core were sufficiently warm, it could maintain a liquid water layer, under a thick layer of ices.

No such rogue planet is known to exist.


Is there matter forming in the inflationary space? Following the answers and the line of thinking from here Is there enough hydrogen left after a star dies so another star will have enough to light up?

I was wondering if is there any possibility to know if, where new space is created between the galaxies, is there also created any kind of new matter ( i.e. hydrogen, or subatomic particles that could eventually form atoms, which will lead to forming new stars and galaxies ) ?

A note: don't talk about inflation for the current expansion. Inflation is the brief period immediately after the BB when the universe expanded at an outlandish rate.

Matter was formed when the universe was so compact that its temperature was billions and even trillions of Kelvin hot. The energy of the vacuum in any given volume was so high that matter-antimatter pairs were created, which usually quickly annihilated. Sometimes, however those short-lived particles collided, releasing large amounts of energy. For instance, to create an electron you need an energy of 0.5 MeV. The graph below shows that at a temperature of $10^{16}$ K collision energy could be as high as 1 TeV.

Because of its expansion the universe has cooled so much that its temperature is less than 3 K, which isn't nearly enough to create any new particles.

The graph shows how temperature and energy drop as the universe expands in the first seconds after the BB. Note that at 1000 seconds (about 20 minutes) the energy is too low to create electrons (0.5 MeV), let alone heavier elements.

What is created in the expansion, though, is dark energy. Its density is constant, so when the universe expands to double its size the amount of dark energy will also be doubled. (That's the reason why the expansion goes ever faster and faster, with nothing to stop it.)


Solar visible light spectrum So the visible light from stars can be used to identify the elements in that star by looking at the spectral emission lines and comparing those lines to emission lines of various elements (and their isotopes) here on earth.  These well-known line wavelengths can also be used to determine the relative motion of that star to the earth by looking at the red (or blue) shift of wavelengths of those spectral lines.  It would then be reasonable to assume that our own sun emits those same well-defined spectral lines, which should show up at specific wavelengths (based on the elements in the sun).

However, sunlight appears (mostly) white, and using a prism shows a rather consistent rainbow of color across the visible spectrum, rather than what one would expect to see (based on analysis of emission spectrum from other stars) of spikes at specific wavelengths.  So, what is causing the full-spectrum light from the sun to appear at the surface of the earth if what I'm assuming is specific spectral lines being generated from fusion on the sun itself?

I'm sure I'm missing something in my understanding, but I'm not sure what it is.  I understand (from this question: How deep is the "A" Fraunhofer line in the solar spectrum? Is it from the Sun's or Earth's atmosphere?) that the earth's atmosphere absorbs specific wavelengths of light, but why would the sun's light be consistent across its visible spectrum if the typical star light is generated at specific spectrum wavelengths?

When we look at spectral lines in a star's spectrum, we're actually looking for absorption lines, not emission lines. A star's spectrum usually resembles that of a black body, continuous and smooth. However, there are elements in the star's atmosphere that absorb some of the emitted light; these create characteristic absorption lines in the spectrum we observe.

There's a reason that a star's spectrum closely resembles that of a black body. H- opacity means that some of the solar photosphere is optically thick for a wide range of frequencies. This in turn means that the emission no longer comes from just a few frequencies corresponding to a few elements. This spectrum appears continuous, and therefore like a black body.


What is the light that we see from a neutron star come from? What hold the electrons in orbit? Is it gravity or the atomic attraction between the protons and neutrons?

I asked this question on Physics SE I hope I improved on the quality enough to do better here.

Is the light we see from a neutron star from the neutrons or the electrons?

The visible "surface" of the neutron star contains no free neutrons. It consists of completely ionised atomic nuclei and free electrons. The exact composition is uncertain - either at or just below the surface will consist of iron-peak elements, becoming increasingly neutron-rich with depth, however the outer few cm (which is where any electromagnetic radiation arises) could consist of ionised hydrogen and helium that has been accreted.

Neutron stars also have magnetospheres and these are also a source of radiation. Charged particles are accelerated away from the neutron star magnetic poles, spiral in the magnetic fields and emit (mainly the electrons, which undergo more extreme acceleration) synchrotron and curvature radiation.

The high density "neutron fluid", containing mostly free neutrons, with a small proportion of protons and electrons, is not reached until about 1-2 km below the neutron star surface.


Planetary phase angle of Hohmann transfer This is Question 23 from USAAAO First Round 2015. The answer is D. Can someone show how to get the answer?

Calculate the planetary phase angle (counterclockwise from Earth, a =
1.0 AU) that a probe may correctly complete a Hohmann transfer orbit to Venus (a = 0.7 AU)

a. 141 degrees
b. 17.5 degrees
c. 121 degrees
d. 241 degrees
e. 343 degrees

What I tried:

Semimajor axis of Hohmann orbit: 0.85 AU
Period of Hohmann orbit: 0.7836 yr
Time of flight: 0.3918 yr
Period of Venus: 0.5856 yr
Angular distance travelled by Venus during flight: 241°

The last one looks suspiciously similar to the answer, but using that angular distance the phase angle should be 61°. Feel free to correct me anywhere that I’m wrong

Let's assume that Earth and Venus have circular, coplanar orbits.
The spacecraft advances 180° in heliocentric ecliptic longitude in 143 days; meanwhile, Venus advances 241° and Earth advances 141°.
When the spacecraft departs from Earth (left), Venus should be 61° behind Earth as seen from the Sun.
When the spacecraft arrives at Venus (right), Venus should be 39° ahead of Earth.

The Sun-Earth-Venus angle marked in green in your second figure is Venus's elongation, 43° east of the Sun as seen from Earth at departure time.

In another context, "phase angle" could refer to the Sun-Venus-Earth angle.
At departure time, an Earth-based observer would see Venus in a 76° phase angle or 62% illuminated.


Is the local group bound to the Virgo cluster? The virgo cluster is currently red-shifted by 0.003 according to Wikipedia; this indicates that the cluster is expanding. Will the cluster's enormous gravity eventually "win" and pull the milky way/local group back toward itself, or will our current recessional velocity combined with dark energy cause us to recede off to infinity?

It would seem so. The local group is part of the Virgo cluster and as such is considered to be gravitationally bound. Although the Virgo cluster and the local group are currently moving apart, the mass if the Virgo cluster will likely slow and reverse the recession over time, with the local group ultimately merging with the cluster.

References:
http://astronomy.swin.edu.au/cosmos/V/Virgo+Cluster
http://heasarc.nasa.gov/docs/cosmic/local_supercluster_info.html


How to calculate a meteor's entrance altitude to Earth's atmosphere? Also, in which altitude did fireball went thermal explosion?

There is a example here of the process required. But basically you need observations of the track from two locations and then just perform some relatively simple trigonometric triangulation. The Nemetode network regularly does this using images from a network of cctv cameras.


What's the actual speed of electromagnetic radiation in space? The speed of EM radiation is very slightly less than $c$, because space is not quite a vacuum.  Say EM travels at $(1-\varepsilon)c$.

For example, this results in a slight delay between receiving a gravitational wave and detecting an associated EM emission from the body that caused it.  For an object about 130 million light years away (GW170817) I've seen this delay quoted as everything from $1.74$ seconds (so $\varepsilon\approx 4\times 10^{-16}$) to 27 minutes (so $\varepsilon\approx 4\times 10^{-13}$).  I'm sure these figures depend on all sorts of interesting astronomy to do with black hole mergers and gamma ray bursts, but the actual value of $\varepsilon$ doesn't need to know any of this.

So I'm asking here because anyone working on LIGO must surely know our best estimate for the value of $\varepsilon$.

I believe this question was addressed in the paper , Abbot, et. al., Gravitational Waves and Gamma-Rays from a Binary Neutron Star Merger: GW170817 and GRB 170817A , ApJ. Lett., 848:L13, 2017 October 16.

  "We use the observed time delay of ($+1.74 \pm 0.05$) s between GRB 170817A and GW170817 to: (i) constrain the difference between the speed of gravity and the speed of light to be between $-3 \times  10^{-15}$ and $+7 \times 10^{-16}$ times the speed of light, ..."

In Section 2.2 of the paper they discuss the uncertainty of delay between GW and GRB emission, predicted to be a few seconds. In Section 4 of the paper they compute the range of speed uncertainty quoted above. They conservatively use 26 MPc, their lower bound on the 90% credible distance interval, and for the upper bound on speed difference they assume the GW and GRB were emitted simultanously, and use the 1.74 s $\Delta t$. For a lower bound on speed of gravity, they assume the GRB were emitted 10 s after the GW, and a faster EM signal made up some of the difference.

I believe that X-rays were also detected several (~8-9) days later. This delay was attributed to time needed for the shock wave from the merger to interact with surrounding matter. I was unable to find any reference to signal arrival 27 minutes after GW detection in the LIGO papers.


What's the bleeding-edge model for how Population III stars are born and evolve? I'm trying to do a brief literature review to find out what the best current models for population III stellar evolution are. I was hoping that someone with more expertise in the area could perhaps summarise or point me in the right direction.

This post on physics.SE is a great intro, but it's now over 4 years old.

Recent observations (see here and here) suggest hints of these kinds of stars at relatively low redshift (z ~ 6) - does this provide significant tension on current models? Have there been other observational hints?

This is a more recent review http://arxiv.org/abs/1209.4688 from someone hoping to understand population III by observing their supernovae

Since CR7 was discovered, there's been a bunch of papers on whether it could be pop III - http://arxiv.org/abs/1506.07173 for example. Try searching for "CR7 population III".