What makes a really good observatory site, besides altitude? Let's limit this to optical telescopes.

I understand that the higher you are, the less atmosphere is above you to get in the way of observations. From reading around, there seem to be a few more but I'm not sure:

Not near an active or dormant volcano (but Mauna Kea seems to disprove this?). This is kinda a bummer because a lot of tall mountains seem to be volcanic.
Somewhere with clear and/or dry weather for as much of the year as possible
Not near major light pollution like cities
Cold weather is better than hot weather? Not sure I understand that; if it's uniformly hot or cold then I don't see how it makes a difference.

Am I missing any? Or are these wrong in some way?

As a concrete example, would the Tibetian plateau make a really good observatory site? (Not considering construction or transportation costs to a remote area, of course.)

• Not near an active or dormant volcano (but Mauna Kea seems to disprove this?). This is kinda a bummer because a lot of tall mountains seem to be volcanic.

There's nothing intrinsically wrong with dormant volcanos, as Mauna Kea and the observatories in the Canary Islands demonstrate. Not sure where you would have gotten that idea.

  • Somewhere with clear and/or dry weather for as much of the year as possible

Clear weather is essential -- you can't observe when it's cloudy!

Dry weather is very good, especially for infrared observations (water vapor blocks a lot of infrared light).

  • Not near major light pollution like cities

Yes.

  • Cold weather is better than hot weather? Not sure I understand that; if it's uniformly hot or cold then I dont see how it makes a difference.

Cold climates tend to have drier air. (Antarctica is an extreme case.) Of course, being on top of a mountain means colder air, which is why Mauna Kea is good despite it's being in the tropics.

  Am I missing any? Or are these wrong in some way?

You also want stable air with good "seeing", which rules out places with turbulent air and lots of wind (and is another reason mountaintops are good: they're usually above the most turbulent layers of the atmosphere).
Quoting from this page (which has a good discussion of the general topic):

  Seeing ... requires at minimum a lack of extra turbulence at all atmospheric levels, and seems to be best satisfied in the convergence zones just outside the tropics, at latitudes about ± 30º. Also, minimal local turbulence is often associated with mountain peaks that reach into the otherwise undisturbed oceanic airflow, as on islands or coastal ranges (and given the direction of the planet's rotation, this generally favors western coast ranges).


Finding the solar noon based on longitude and latitude? Please could someone advise the best method of finding the solar noon based on location?

http://pveducation.org/pvcdrom/properties-of-sunlight/suns-position

The stuff on here has been helpful but I can't figure out what I should be finding out!
LST sounds about right, but I'm not 100% sure.

Taking solar noon as being the time when the sun is at its highest elevation (altitude) in the sky for a given day, then your assumption of the LST (local solar time) is correct, as it is stated in the website you linked as being:

  Twelve noon local solar time (LST) is defined as when the sun is highest in the sky. Local time (LT) usually varies from LST because of the eccentricity of the Earth's orbit, and because of human adjustments such as time zones and daylight saving.

This is something I have, in part, programmed into an app for a separate project (quite unrelated to yours).

You would have to equate the precedent quantities first (Local Standard Time Meridian, Equation of Time and Time Correction factors) as they have done on the website you linked (as these are reliant on the observer's position and time of observation), but to determine at what time solar noon occurred, the final step needs to have local solar time (LST) set to 12:00 (solar noon).

A worked exampled is given within the first 7 pages/slides of Sunlight and its Properties II.


What is the light source in the black hole images? What is the light source in the black hole images?

I've searched the site before asking, as well as tried to google it. But perhaps the answer is buried due to the recent news.

My understanding, is that the ring in the images is an illuminated accretion disc.

What I could not find explained is the light source, and by extension, why is a bright accretion disc – which is round by definition – a proof for light bending? (I'm not disputing the fact.)

The radiation being detected in this case is mostly synchroton radiation, caused by energetic electrons spiralling around magnetic field line, rather than thermal radiation, but it comes from the same place (the disk). The actual evidence for light bending is, I believe the dark area in the middle, which is several times larger than the actual event horizon and is caused because the light that would have been coming to us from that direction has been bent away.


How much light would be reflected from Jupiter to Europa (in Europas night)? Suppose Europa is found habitable and humans start living on it. Europa is close to Jupiter and during the night Jupiter would be reflecting light to Europa, just like our full moon reflects light to earth. Our moon is small compared to Jupiter. Would humans still need artificial light during Europas nights?

It's a pretty straight forward calculation of 3 factors.   Distance from the sun, apparent size and albedo.   I'm going to compare Jupiter to our full moon, since we're all familiar with that.

Jupiter averages between 4.95 and 5.45 AU from the sun.   That puts it, in comparison with our full moon (about 1.02 AU on average), that means individual square meters on Jupiter would receive between 1/23.5 and 1/28.5 the light that hits our moon.  I'm going to round and say between 1/24 and 1/28.

Brightness is also important.  The Moon is very dark with an albedo of about 12% - about the color of worn or greyish asphalt.  Jupiter's bond albedo is about .34, so it would reflect nearly 3 times more light.   That means per square meter, Jupiter is up to about 1/8th to 1/10th as bright.

The big difference is size.   From Earth, the Moon takes up about 1/2 of 1 degree of arc.   Jupiter's diameter is about 37 times the Moon's diameter.    It's distance to Europa is about 1.73 times Earth's distance to the Moon, so it's apparent size would be about 21 times the diameter (about 10 degrees of arc), the reflective area is closer to 450 times.   If you divide that by the ratio of 8 to 10 depending on how close or far Jupiter is to the Sun in it's orbit, A full Jupiter would be the brightness of roughly between 45 and 55 average full moons.   It wouldn't glow as brightly, though some of it's lighter bands would be almost half way there.   But the entire package would be 45 to 55 times as bright (ignoring any atmospheric reflection, which would make it appear even brighter).

That's probably bright enough to read under, but not so bright that it would hurt your eyes if you stared at it.

This only answers your question for a full jupiter, but the ratio, 45 to 55 times brighter should be about right for all the phases.   And about 450 times larger than an average Moon.   The Moon's size varies fairly significantly from Earth, though we don't always notice.   Europa has a much more circular orbit around Jupiter, so it's size would stay about the same from the surface of Europa.


Eclipse Cycle Calculations Needed I am writing a work of fiction about a world where:

The length of a year is 335 days.
The world has two Moons.
Moon A is Larger and further away with a 78 day lunar orbit.
Moon B is smaller and closer with a 31 day orbit.

How often would both moons be full at the same time, overlapping each other like a bullseye?
How often would lunar eclipses occur for Moon A?
How often would lunar eclipses occur for Moon b?
How often would both eclipse at the same time?

This is a work of fiction, so liberties can be taken.  I just want it to be close.

Frankly, four bodies do not periodically line up, even if the orbits are planar, unless there exist a common multiplum. Because all your periods are integers, that holds, but be aware that your double alignments, like the double solar eclipse or double lunar eclipse is not periodic if one of the periods does not exactly follow that integer system. (so it does not hold for 31.01 days, for example. How you handle quazi-periods is talked about here).

  How often would both moons be full at the same time, overlapping each other like a bullseye?

Generally non-periodic, but can only happen at the synodic period of the two moons, that is:

.

$\frac{2418}{47}$ for the two moons. However, that must also happen when both of them is on the same line as the Sun, in other words when The relative synodic period between one of the moons and the sun, and the moons' synodic period both divides the requested period. For the inner moon and the Sun, the synodic period is $\frac{10385}{304}$. Since 47 does not have any factors in common with 304, the first common multiplum appears at.

$$\frac{10385}{304} \cdot \frac{2418}{47} = \frac{12555465}{7144}$$, or 1757.484 days. (@Jonathan gets 2418, because his model consider the Sun to stay static in the sky, but it does in fact revolve around one time a year.) The exact same period is also valid for your question 4.

  How often would lunar eclipses occur for Moon A?

Here, the answer is the synodic period of the Sun and the moon, already stated to be $\frac{10385}{304}$, or 34.161 days. That is a little bit longer than the Moon's orbital period, because the planet has moved a little in its orbit, slightly changing the Sun's position in the sky.

  How often would lunar eclipses occur for Moon B?

Synodic period here too, $\frac{26130}{257}$ = 101.673 days.


What is the cosmic mean density? I know cosmic mean density for baryons is around $(2\ \mathrm{to}\ 5)\times10^{-31}\ \mathrm{g/cm^3}$, assuming that Hubble's Constant is $100\ \mathrm{km/s/Mpc}$. But what is the current cosmic mean density including all particles in the universe (dark matter etc.)?

The simplest and shortest answer to this question is that since our universe is flat (to within our error bars), its density must be that of the critical density - that's what it means to be flat! In that case the overall density of the universe, accounting for matter, dark matter, dark energy, etc. is (assuming $H_0=100\ \mathrm{km/s/Mpc}$)

$$\rho_c = 1.8788\times 10^{-26}\ \mathrm{kg/m^3} = 1.8788\times 10^{-29}\ \mathrm{g/cm^3}$$

If you want to be more precise about it and actually use measurements of our universe, you can go about calculating it by using the density parameters. By definition, the density parameter is defined as

$$\Omega \equiv \frac{\rho}{\rho_c}$$

where $\rho_c = 1.8788\times 10^{-26}h\ \mathrm{kg/m^3}$ (notice that little $h$ now included). Numerous missions such as WMAP and Planck have measured these density parameters for the variety of mass-energies in our universe and you can look them up and put them all together to get the total mean density of our universe. Overall, you should find that

$$\rho = (\Omega_m + \Omega_c + \Omega_\Lambda + ...)\rho_c$$

where the various $\Omega$s are defined for the various mass-energies in our universe. You can look up values for the density parameters from recent results such as from WMAP or Planck. Just decide on your chosen model and pull out the $\Omega$ values $-$ or more properly the $\Omega h^2$ values if you want to use your own value of $H_0$. It's not too difficult to plug everything in. You'll pretty much always find though, that the sum of the density parameters very nearly equals one and to an order of magnitude you can say $\rho=\rho_c$.


Gravitational pull needed to keep a gas in atmosphere How can you determine the gravitational force needed to keep a particular gas in the atmosphere of a planet (for example, carbon dioxide (CO2))?

I came across the following formula $\left(\frac{8RT}{πM}\right)^{0.5}$

here: http://www.tau.ac.il/~roichman/CVI/hw1/hw1.pdf

$M$ = Molecular mass

$R$ = Gas constant

$T$ = Temperature

Would this formula apply?

Yes, the formula you quote applies to some extend.

It can also be written in terms of Boltzmann's constant $k_\mathrm{B}$ as
$$v_\mathrm{rms} = \sqrt{\frac{3k_\mathrm{B}T}{m_\mathrm{m}}},$$
with $m_\mathrm{m}$ the mass of the molecule in question,
gives the mean ("root-mean-square") velocity of the gas molecules as a function of temperature $T$. Comparing this to the escape velocity
$$v_\mathrm{esc}=\sqrt{\frac{2GM_\mathrm{p}}{R_\mathrm{p}}},$$
where $M_\mathrm{p}$ and $R_\mathrm{p}$ are the mass and radius of the planet, respectively, gives an order-of-magnitude estimate of whether or not the planet will be able to maintain its atmosphere.

However, if the two velocities are equal, it would still mean that half of the atmosphere evaporated instantly, and that over time, a large fraction would disappear, although the time scale for this to happen may be very long. In order for the planet to really sustain its atmosphere, $v_\mathrm{esc}$ needs to be roughly 6 times larger than $v_\mathrm{rms}$.

Gravity is not the only factor determining the stability of the atmosphere, though. Radiation pressure from the star ("Solar wind") can easily rip a planet of its atmosphere. On the other hand, if the planet has an efficient magnetic field, this will shield it against the wind.

Other factors are also involved. But you're right in that at the very least, the planet needs a certain gravitational field to keep its atmosphere.


How can one explain the apparent motion of the Sun from a heliocentric point of view? From a horizontal frame of reference, the path of the Sun looks like a part of a circle in a tilted plane, as shown in this picture:

If you go to a heliocentric system, one would say that this is because the Earth is rotating around it's axis. Is there any more detailed explanation of this change of reference?

So how can one explain in detail the following facts (seen from a geocentric point of view - horizontal reference system) starting from a heliocentric system and the fact that the Earth is rotating around it's axis:

Why is the path of the Sun a part of a circle?
Why is this path in a plane?
Why is this plane tilted?
How to get this "tilt-angle"?

Why does this motion seem to be uniform from an equatorial point of view, and can I neglect the ecliptic to understand the points above?

I tried to model it using a ball as Earth, attached a camera to it and used another ball as the Sun, but I didn't succeed in reproducing the effect of the apparent motion in this model. Perhaps there is any animation which makes the points above clear.

The picture shows 3 paths of the sun over a single day, plotted on three particular days: Midsummer, midwinter, and an equinox. The apparent motion of the sun is due to the rotation of the Earth, not the motion of the Earth around the sun.

Why is the path of the sun a part of a circle?

Because the Earth is spinning, so you are moving in a circle around an axis that passes through the poles of the Earth

Why is this path in a plane?

The local surface of the Earth is modelled as a plane tangent to the sphere of the Earth. As the Earth spins, this plane rotates, and so relative to this plane, the sun appears to move. Because the plane is rotating in a single axis and at a uniform rate, the sun appears to move in a circle, across the sky. As it is a circle, it is in a plane.

Why is this plane tilted?

It is tilted because you are not at the poles, the surface of the Earth on which you stand is tilted with respect to the axis of the Earth

How to get this "tilt-angle"?

It is your latitude, or more exactly 90 - latitude. If you are at the poles the sun would travel in a horizontal plane. If you are at the equator the sun travels in a vertical plane.

The different paths (red green and blue in the picture) are due to the tilt of the Earth at 23.4 degrees, and so in Summer (in the north) the sun is higher at midday and further North at rise and set.

Why does this motion seem to be uniformly from a equatorial point of view.

The image appears to show the path of the sun from a location about 45 degrees North, for example Northern USA. Again, I may be misunderstanding the question.


Why do astrologers use the “wrong” sign? I realize this may be the wrong site for this question... I apologize if you consider it inappropriate but hope this community knows the answer.
We just experienced the spectacular conjunction of Jupiter and Saturn (yay!). When I look at the sky, I see these two planets on the edge of Capricorn, almost in Sagittarius. Yet astrologers claim the conjunction is “in Aquarius”.

What is different about their (astrology’s) coordinate system? Any (historical) perspective would be appreciated!
PS I have no interest in astrology as a tool for predicting the future - just trying to understand where this discrepancy comes from.

Precession of the equinox, caused by the cyclical change in the direction of the Earth's tilt, causes the position of the vernal equinox (the point where the path of the sun crosses the celestial equator) to move over time.  When the constellations were first described, the vernal equinox was in Aries, and so it is still called "The First Point of Aries".  Now it is in Pisces and moving towards Aquarius. Astrologers use "signs" named according to the position of the vernal equinox when the constellations were first described.  So if a planet is at the vernal equinox (and therefore in Pisces, close to Aquarius) an Astrologer in the Western tradition will say that planet is "in Aries".
So for an astrologer, "Aries" is not a pattern of stars in the sky nor the region of the sky surrounding those stars, but the strip in the sky between 0 and 30 degrees of longitude, even though that region no longer contains the stars of the constellation Aries.


How do you define the diameter of the Sun That is, it's twice the radius where the radius is from the centre of the sun to some edge. But what is that edge?

Most literature will define the diameter of the Sun up to the photosphere, the layer of the solar atmosphere you would see if you were to observe the Sun in white light.

The base of the Photosphere is defined as the region where the optical depth is around 2/3, or the region where the plasma becomes transparent to most optical light wavelengths.

Of course the true edge of the solar atmosphere could be considered as the heliopause, where the direct influence of the Sun's magnetic field and solar wind end and interstellar space begins.


Why can't this be the simple (and obvious) explanation for the dimming of KIC 8462852? Couldn't it be just a relatively large object, anywhere in the interstellar space between the star and Earth? I mean, look out the window at a far away car or a tree, then stretch out your hand and you can block it entirely with your thumb. What if the thumb is that object (a planet, a star, a comet, some space debris, etc.) that's between Earth and the star, and it's just the right size to only block that star, and no other stars. Why has this very simple explanation been ruled out (if it has been)?
I know that the debris next to the star explanation has been ruled out because of the lack of IR that would come from that debris being heated up if it was close to the star. But this is not a problem here, if the blocking object is far away from the star.

The simplest reason why your proposed explanation is not really that obvious, and thus was not considered is that the dimming of KIC 8462852 was not an isolated event. The dimming is a pattern that has been observed over years. There is no known phenomenon in interstellar space that would cause a repeated obstruction of that star, so it doesn't make any sense to consider that possibility when there are other phenomena (such as those captured in Boyajian's paper) that are more likely to be the source of the dimming.


Scenarios: Abusing a black hole So a black hole is this large, sphere-blob shaped fluid thing that distorts space and time, making a weird hole in space(even after 5 years, I still don't exactly get what a black hole is exactly and where the mass goes when it eats xD).

So, let's say, we make a huge antimatter(opposite charge, same mass as matter) star and send it into the black hole. What would happen? The antimatter would annihilate with the matter made black hole, or is it wrong to assume a black hole is made of matter?(I assume not much would happen if the antimatter can't react with the matter, because the black hole would probably get its original charge messed up.)

What about if we created a huge exotic matter(same charge, opposite mass) star and sent it into a black hole? Would the repulsive force successfully destroy a black hole? According to Kurzgesagt and several articles cited by them, exotic matter is able to keep a wormhole stable, so wouldn't it be able to annihilate the black hole or something? Or have I made a misunderstanding?

Black holes are not "made of matter". They are better described as structures of gravity/warped spacetime. However, they do grow when absorbing things with positive mass-energy.

Antimatter still has positive mass-energy, so the black hole cheerfully gobbles it up.

Now, exotic matter with negative energy density might indeed decrease the size of the black hole, akin to some form of enforced evporation process. (paper1, paper2) This is highly speculative since there are various theoretical reasons why exotic matter does not exist, and many kinds of hypothetical exotic matter that might work differently.


Is there a luminosity class for brown dwarfs? I've been reading about brown dwarfs, and checking their spectral types, and I noticed that they do not have a luminosity class (V, VI, etc). So I am wondering, do brown dwarfs have a luminosity class, or is it intentionally absent, or is it defaultly V or VI?

The SIMBAD page for the binary brown dwarf Epsilon Indi B lists the components as T1V and T6V, indicating they use the normal dwarf luminosity class V.
There are also T subdwarfs, e.g. Burningham et al. (2014) give the spectral type of ULAS J131610.28+075553.0 as sdT6.5. This follows the usual convention I've seen where subdwarfs are written with a prefixed "sd" rather than using the Roman numeral luminosity class VI.
Schneider et al. (2020) also indicate that WISEA J041451.67-585456.7 and WISEA J181006.18-101000.5 are extreme subdwarfs, of spectral classes esdT0±1. Again this is indicated by a prefix ("esd") rather than a Roman numeral luminosity class. They also mention 2MASS 05325346+8246465 as being of spectral type esdL7.
Zhang et al. (2018) list a number of L subdwarfs (sdL), extreme subdwarfs (esdL) and ultra-subdwarfs (usdL).


Is Limb Darkening because at centre we are directly looking core, whereas at limbs we are seeing only photospheric part? Is limb darkening of the Sun because at centre we are directly looking the core and different zones, whereas at limbs we are seeing only the photospheric part (and less beneath photospheric part)? Or there is some other physics behind this phenomenon?

We can't see the core, but we can see deeper into the photosphere of the sun. The deeper layers of the photosphere are hotter and brighter, and this is the cause of limb darkening.


Information speed and our view of the cosmos It is common knowledge that when we look at the stars we look back in time. And most stars are thousands of lightyears away from each other. The information that photons cary about the position of a stellar object can be really deceiving about their true position.

Do astronomers take lightspeed into account when they map stars ? If so, how do they figure out the true position of a star.

With all that in mind do galaxies really look like what we see in our photographs? If a galaxy has a diameter of bilions of lightyears, what we see could be a really skewed and warped image of the real galaxy. Can we computationaly reverse the warping ?

EDIT
Image 1 and 2 explain what I mean by "Warping". Stars closer to the core move faster and in combination with their distance from the viewer, information by light should give the impresion of a more "spiraly" galaxy as it whould take less time to reach the observer. Image 2 whould be the "true" shape of the galaxy if the effect could be reversed (I applied a counter clock-wise warp distortion)

An other factor of false position are gravitational lences. Although gravity from black holes do give the maximum distorsion, gravity from other stars could divert light reaching us. The diversion is negligent but can get big for astronomical distances (Image 3). Am I right ? If I am, do Astronomers even bother to make simulations to map the known sky as it whould be in real time ?

The effect is small, but not negligible. It is not accounted for in astronomical catalogues.

Let's work it out. We can start with the visible stars. Most of these are closer than 1000 light years; let's say 100 light years. The typical velocity of these stars with respect to the Sun is tens of km per second; let's take an extreme example of 100 km/s in a direction at right angles to the star-Sun line.

100 years at 100 km/s is 300 billion km. This sounds huge, but it is a tiny fraction of the size of the Galaxy and small compared with the distances between stars; but by how much does it change the apparent position in the sky? The change in angle (in radians) is given by the tangential distance moved in 100 years, divided by the distance to the star. This is the ratio of the speed of the star to the speed of light and is actually independent of distance.

For the example given, the change in position is $3.3\times 10^{-4}$ radians or 69 arcseconds. This is an enormous change in astronomical terms (not an angle you could easily resolve with the naked eye), but it is not accounted for in astronomical catalogues, since we are interested in where to look at stars now and neither do we know the accurate tangential velocity of everything in the sky.

What is accounted for in telescope pointing software is relativistic aberration, which is a change in the apparent position of a star due to changes in the relative motion as the Earth orbits the Sun. These corrections can be as large as 22 arcseconds.

As for galaxies - you have your dimension estimates wrong. A large galaxy is 100 thousand light years across, with stellar speeds limited to a few hundred km/s. I don't understand what you mean by "warping". Individual stars will have moved whilst the light travels towards us, others will have taken their place, stars will have died and others born, whole galaxies will have disintegrated or merged. All this is known. We can only observe what we can see now.

A dynamical timescale is typically tens of millions of years in a Galaxy and far longer than it takes light to travel across the galaxy. To all intents and purposes, when we look at a distant galaxy we are seeing everything at roughly the same epoch in that galaxy's history.


Why is the Moon receding from the Earth due to tides? Is this typical for other moons? After reading the Q&A Is the moon moving further away from Earth and closer to the Sun? Why? about the tides transferring energy to the Moon and pushing it from Earth, I have a question:

How is that energy actually being transferred to the Moon? The creation of tides requires energy, so I would expect that this should take energy from the Moon, slow it and cause it to eventually fall back to the Earth. Why isn't that happening?

Finally, if this is the general mechanism, would other moons that orbit around planets with a liquid surface and causing tides, be receding from their parent planets?

It's pretty simple, actually.

The Moon creates tides. Due to tides, the water bulges out towards the Moon (and also on the opposite side).

But the Earth also rotates pretty fast (once a day), faster than the Moon orbits the Earth (once a month). There's friction between the rotating Earth, and the watery bulge created by tides. The rotation of the Earth "wants" to rotate the bulge faster.

In effect, the rotation of the Earth drags the tidal bulge forward - the bulge is always a bit ahead of the Moon. When the Moon is at meridian, the tide is already decreasing.

So there's a bit of extra watery mass on Earth, a little bit ahead of the Moon. This watery bulge interacts gravitationally with the Moon.

This has two effects:

it slows down the rotation of the Earth, gradually sucking energy from it (the Moon pulls the bulge, and therefore the Earth, "back")
that energy is dumped into the Moon's orbital motion, effectively "pulling" it forward

When you dump energy of motion into an orbiting body, it settles into a higher orbit - higher orbit means more energy. Therefore, the transfer of energy from Earth' spin to Moon's orbit gradually makes Moon's orbit larger and larger.

This only happens because the Earth is spinning faster than the Moon orbits it. If the Earth was tidally locked to the Moon (spinning exactly as fast as the Moon orbits it), then no transfer would happen. If the Earth was spinning slower than the Moon's orbit, then the transfer would be opposite (from Moon's orbital motion to Earth's spin).

Note: Counterintuitively, a satellite with more energy actually moves slower, but in a higher orbit. The extra energy goes into raising the orbit, not into making its speed faster. Why this happens exactly is a whole 'nother discussion.


Farthest Orbit around the Earth? What is the farthest orbit that can be achieved around the Earth? How fast is that?

Not the question below:
Feel free to answer in a comment

I ask because of this question: https://worldbuilding.stackexchange.com/questions/104593/can-you-get-a-orbital-mass-driver-to-work-this-way

Is there an orbit exist where 2 objects going in opposite directions can intersect repeatedly with minor course corrections?

The farthest Orbital Point from Earth is regulated by the fact that at what point the gravitational pull of the Sun overcomes the gravitational pull of the Earth.

I found a post for this on Quora

This was the most up voted answer by user Paul Olaru on Quora:-

  Approximately 1.5 million kilometers. It is limited by the point at which the influence of the sun's gravity becomes sufficiently stronger than Earth's that any object at that distance would become unbound from Earth & end up in independent orbit around then sun.

  That limit applies in general to any system of two bodies, and is called the Hill sphere. It's not actually perfectly spherical, but close enough for most practical purposes, and lies between the L1 & L2 Lagrange points. For circular orbits, the radius of the Hill sphere for any planet or moon can be calculated from

  r = a*(m/(3M))^(1/3)

  Where r is the radius of the Hill sphere, m is the mass of the planet or moon, M is the mass of the sun a planet orbits, or the planet that a moon orbits, and a is the distance from the sun (or planet) to its planet (or moon).


Why can we sometimes see the moon during the day? At some points in the year, the moon at my location can be clearly visible during the day. Why is this?

If it is any help, I live in Alberta, Canada.

The Moon makes a full orbit around the Earth once per month. From full moon to full moon, it is 29.53058886 days (synodic month).

Full moon happens when the Moon is at one side of the Earth (right side in the image), and the Sun is at the other side. This way, Moon is fully iluminated by Sun (except in the cases where it is exactly in the opposite side of the same line, where you have a Moon eclipse). In this phase, Moon is at its highest altitude at midnight.

Then, as the Moon orbits, it comes closer to the Sun in the sky. This means both that it shows us less iluminated percentage and that its rise and set get delayed. At Last Quarter, Moon is at its highest altitude at sunrise. This means that you will be able to see the half moon figure on the sky during the morning (and last half of the night).

Then it comes New Moon, where the Moon is nearest to the Sun in the sky, and almost invisible, since it shows us the dark side during the day. Its highest altitude is at noon, but you will not be able to see it. You can, if you are lucky, see a Sun Eclipse.

Finally it comes First Quarter. To put it short, highest altitude is at sunset and you are able to see Moon during all the afternoon (and first half of the night).


Can solar luminosity & activity be predicted? Can a star's luminosity and its activity, solar flares and sunspots, be predicted?

If so, with what accuracy & precision?

All of these effects are related to the 11 year solar cycle. And while we know there are times when the sun is "active" versus "inactive" we aren't necessarily predicting exactly where sunspots or solar flares will occur, but how many we see in total.


Given that Kepler's Third Law as stated on most websites doesn't include mass, how does it work for planets in other star systems? Kepler's Third Law as stated on websites I found while searching for "how to convert between a planet's orbital period and semi major axis" is $P^2 = A^3$ where P is the orbital period in years and A the semi major axis in AU.

However, why is mass of the central body not included?  If Earth was orbiting at the same distance around a star with more or less mass than our Sun, surely its orbital period would be different?

Though the question may be closed as a duplicate, here's an answer that shows how to calculate periods using Kepler's $\color{blue}{\text{3rd}}$ law but with all the constants and units explained.

Here's the equation from Wikipedia's Orbital_period; Small body orbiting a central body. The idea here is that the the size difference is so big that we can assume the central body doesn't move. In reality there is always at least a tiny motion. The Sun-Earth system rotates around a point about 450 kilometers from the Center of the Sun, and outer planets (especially Jupiter and Neptune) move the Sun so much that the center of mass is sometimes completely outside the Sun!

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

Which is Kepler's $\color{blue}{\text{3rd}}$ law:

$$\frac{T^2}{a^3} = \frac{4 \pi^2 }{GM}$$

  What are the units?

While it is convenient to use AU and years, as you've found out that only works for orbits around the Sun. The reason is in that case you're really simplifying the equation to a ratio; period divided by another period around the same body, and semimajor axis divided by the semimajor axis of that same other body.

For general calculations I recommend you use meters, seconds and kilograms. You can use any set of units, but you have to be careful that everything is in the same units.

  What is $GM$?

It's the gravitational constant $G$ times the mass $M$ of the central body. You can look them up separately, but it turns out that you can find $G$ times $M$ listed as the standard gravitational parameter. These are more accurate than multiplying $G$ times $M$ because each one alone is known to much less accuracy. That's because we can't (easily) measure the mass of a planet by comparing to a standard kilogram. Instead we look at periods and distances of orbits and determine the product $GM$ together.

Here are a few of the numbers from the Wikipedia article, I've rounded them to three decimal places. You can see more decimal places and discussion in the question Where to find the best values for standard gravitational parameters of solar system bodies?.

body      GM (m^3/s^2)
-------   ------------
Sun       1.327E+20
Earth     3.986E+14
Moon      4.905E+12
Jupiter   1.267E+17

So to get the period of the Earth around the Sun:

$$T = 2 \pi \sqrt{\frac{(\text{1.496E+11})^3}{\text{1.327E+20}}} = \text{3.156E+07 sec} = \text{1 years.}$$

To get the period of Europa around Jupiter:

$$T = 2 \pi \sqrt{\frac{(\text{6.709E+08})^3}{\text{1.267E+17}}} = \text{3.067E+05 sec} = \text{3.550 days.}$$

Wikipedia gives 3.551 days, which is the same considering we're only using three decimal places here. If you take advantage of using $GM$ instead of multiplying $G$ times $M$ and use many more decimal places, you should get agreement to the known periods to many more decimal places as well.

  But why don't I get the right period for the Moon?

The Moon's mass is more than 1% that of Earth, so the approximation treating the central body as not moving doesn't work.


Why is ice on Mars not covered in dust? There's exposed water ice on Mars, like in craters like this

At the same time, Mars has dust storm, covering the solar panels of rovers, and occasionally covering the entire planet

So why is this ice not covered in dust?

I understand that dry carbon dioxide ice can maintain a clean surface, as it's seasonally sublimated into the atmosphere and re-deposited, but how come water ice (at least some of it) is also left clean?

A lot of the water ice on mars is covered with dust.  NASA's Phoenix mars lander only had to scrape off some of it to find the layer of ice underneath  as seen here: https://www.nasa.gov/mission_pages/phoenix/images/press/sol_020_024_change_dodo_v3.html


Humans surviving in space To my (admittedly limited) understanding a human being in outer space wouldn't last very long, mainly due to three factors:

the extreme cold
the intense radiation
and the vacuum of space

So if my understanding is correct, humans would die within seconds (if anyone could put an approximate number on this that would be great!).

What I want to know is that if a human being was exposed to space for half that time, would they suffer lasting and life threatening after effects?

In other words, is there any amount of exposure to space that wouldn't result in your imminent or eventual death?

Would exposure of just a hand kill you, or would you just lose that hand?

Would being exposed to space from inside a room with a single open door help much (I am assuming here the room would shield you from much of the radiation in space), or would the other two factors still kill you off?

Please feel free to correct my assumptions wherever I am wrong.

Many thanks for any help offered to my slightly morbid musings on space.

One would not, under normal conditions, die within seconds. This is a common trope in science fiction, but there has been quite a bit of real research into the subject.

Let's speak on each point.

The extreme cold

This isn't really a huge problem. Space isn't necessarily cold. The extraordinarily low density of matter in space means that the movement of energy through induction is extremely limited. If one were to step into 'space' one wouldn't feel cold, or even cool.  For engineers, often the greater problem is not how to deal with cold, but how to deal with heat, as even though one can't easily release heat through induction, one can certainly gain it through radiation or internal chemical processes.

A person, however, is likely to experience cooling due to the process of liquids in the body evaporating and escaping, taking some of their heat with them. It's unlikely that this would prove fatal though, or at least it's unlikely to be the first thing to prove fatal.

The Intense Radiation

Radiation is a problem in space, but it's not going to kill you so quickly.

  During a 360-day round trip, an astronaut would receive a dose of about 662 millisieverts (mSv), according to RAD measurements. National space agencies limit exposure to about 1000 mSv or less during an astronaut's entire career; NASA's limit corresponds to a 3% risk of exposure-induced death from cancer. (Kerr, 1031)

Currently, orbital missions get some protection from radiation as they are within the Earth's magnetic field, but for those travelling beyond this there isn't a lot of protection. Part of the reason is that things that provide good protection are also very heavy, which can be quite difficult in the economies of reaching escape velocity. As such, one wouldn't necessarily be exposed to that much more radiation outside of a shuttle or their suit in space than they would otherwise.

The Vacuum

This is where things can get messy, but there has been some research on the subject. Obviously one can't breath in a vacuum, in fact the lungs actually begin to dump oxygen, essentially working in reverse due to the lack of air pressure. Our victim is likely to lose consciousness within 10-15 seconds. If they happen to be holding their breath when they become exposed, the pressure difference could cause some real damage to their lungs.

People often theorize that blood would boil in space. Geoffrey Landis disagrees with this on his blog, in which he speaks on this subject. I highly recommend reading through it as I would consider him both thorough and authoritative on the subject.

I couldn't find the official report, but there are frequent references to a 1965 (Earth based) pressurization incident, in which a technician's suit depressurized within a chamber. He remarked afterwords that the last thing he recalled before blacking out was feeling his saliva begin to boil on his tongue.

While your blood arguably won't boil, you would experience the boiling of other liquids in your body, water particularly.  This can result in rapid swelling of the body. While not pretty, this isn't likely to kill you. However, nitrogen can also become gaseous due to this, potentially resulting the bends, a condition more closely associated with pressure differences in diving. This condition could prove fatal, but AFAIK there hasn't been a lot of research into this in reference to vacuum exposure (This condition tends to become dangerous during the return to normal pressures).

All of this can be survived, as long as pressure is returned relatively quickly. Within a minute is probably enough to save a person's life with little lasting damage. After 2-3 minutes, it's likely too late. The difficulty is that the unfortunate victim may be able to survive 1-2 minutes of exposure, they really only get 10-15 seconds to do something about the problem. Of those who have survived vacuum pressurization incidents, their survival is mostly credited to the quick thinking of those around them.

To respond to your more specific questions:

Exposure for half of what would generally be fatal

This could vary a lot, like depending on whether they were holding their breath. However, generally it's believed that a person could recover from this with little to no permanent damage. Probably the biggest risks are lung damage due to pressure differences, the bends and other related complications, and brain damage due to oxygen deprivation.

Exposure to space that wouldn't result in imminent or eventual death

A little exposure without other complications will generally be survivable. We don't have a ton of data on the subject, so it's possible that unpredictable complications could arise, and frankly it's not exactly advisable, especially considering the person will lose consciousness quite quickly.

Partial exposure

I can't really rely on any specific source for this, but presumably exposure of an extremity would be perfectly survivable. I'm not quite sure how this would work as it would be very difficult to partially expose someone to a vacuum. The human body isn't exactly leak proof. In theory, you would begin to exhibit symptoms in the exposed body part, but you would remain conscious (assuming your respiratory system isn't exposed) and generally be in control of what's going on. Eventually other complications would likely set in and need to be managed carefully.

Exposed to space from inside a room with a single open door

As noted the radiation and temperature is not terribly dangerous, but with no way to repressurize the room this can quickly turn fatal. The Russian Soyuz II mission ended catastrophically when the capsule carrying the cosmonauts depressurized on way for reentry, killing all three crew members on board.  A valve malfunctioned during the reentry, causing depressurization. The craft was recovered after reentry and had not taken any significant damage. It was believed that Patsayev had made attempts to fix the valve in order to save the crew, but unfortunately there was simply too little time to accomplish this. Initial reports suggested that the crew had died of asphyxiation, however autopsies showed that the cause was hemorrhaging in the brain - reportedly caused by oxygen and nitrogen in the blood boiling. Not sure how this works out with Landis's claim that blood does not boil.

Radiation Will Make Astronauts' Trip to Mars Even Riskier
Richard A. Kerr
Science 31 May 2013: 340 (6136), 1031. [DOI:10.1126/science.340.6136.1031]


Are space telescopes completely out of the earth's atmosphere? Is the Hubble Space Telescope, and every other space telescope for that matter, completely outside the Earths atmosphere?

A handful of space telescopes are located in Langrange point L2, 1.5 million km from Earth. This is much farther away than the Moon, and far outside Earth's atmosphere.

WMAP and Planck, which measure the cosmic microwave background (CMB), are located here because Earth is a hundred times brighter than the CMB in this wavelength region. Herschel observes in the infrared, which is heat radiation, so it was also put in L2 to avoid Earth's heat. Others can be found in the list provided in Keith's answer.

Another advantage of putting a telescope in L2 is that the Earth and Sun are always in the same direction, making it easier to observe most of the sky. The cost of putting it here is probably more than an order of magnitude higher, though.

Whether or not low Earth orbit telescopes like Hubble at 550 km are outside Earth's atmosphere is a bit subjective. The atmosphere doesn't stop at some particular point, instead gradually thinning. These space telescope are definitely far enough away that they are not heated up. Neither does the atmosphere affect their observations (by seeing). They are, however, slowly decelerated and thus need a boost once in a while not to fall down.


the most accurate system that can show the positions of planets at my birthday My birthday : 4 December 1982

If you have such astronomical calculator on your computer please send me the result !

Alternatively if you know an online astronomical service for my purpose please link it here ?

I am so curious to discover geometrical proportions between distance and angle of planets at that date ...

The most accurate system is the ephemeris service from Jet Propulsion Laboratory (JPL): http://ssd.jpl.nasa.gov/horizons.cgi


How to deal with data that's billions of years old? We look at galaxies that are billions of light years distant and draw conclusions based on the red shift of the light. How is the age of the light accounted for. We see the reality of a galaxy perhaps a billion years ago. Why not conclude that redshift is related to time instead of distance?

The redshift you're referring to is a cosmological redshift, which is fundamentally different from a Doppler redshift. At the heart of it, it has nothing to do with neither time, nor distance. It is a consequence of light traveling through an expanding space, no more and no less.

Because space expands, the emitting source is carried away to a certain distance from us, so we can relate the redshift to a distance. And because it takes time for the light to reach us, we can relate the redshift to a time. But this requires a model for how the Universe has expanded through time, and how light travels.

The model that best describes our reality is the so-called standard cosmological model, or ΛCDM ("Lambda-CDM") model, which is a mathematically based, physically motivated, and observationally constrained model. It uses the so-called Friedmann–Lemaître–Robertson–Walker metric, which is a description of space itself, the Friedmann equations, which describes how space expands, and an equation of state, which describes the relation between pressure and density. The various parameters that enter the equations are then observed and constrained in a number of different ways.

With this model, we can relate the distance to an object, and its age, to the redshift we observe.


What was the composition of the primordial asteroid belt? According to one theory, the region between Mars and Jupiter was once populated by several primordial planets with a combined mass of roughly the earth.  These planets provided the material that made up the original asteroid belt.  Since then, 99.9% of this material was lost.  The current composition of the asteroid belt is about 10% metallic and 90% silicate/carbonaceous.  Could we assume that the original asteroid belt had the same composition?  Or, at least a very similar composition?

We can be pretty sure that the asteroid belt once held one or more sizable planetesimals,as there is evidence that it or they were large enough for gravity & internal heating to drag the heavy metals (mainly iron & nickel) to the centre,forming a metallic core. Bodes Law decrees that there should be a planet where the asteroid belt is now,but it seems that as this planet was forming it was shattered by an impact & the remnants scattered. Some reached Earth,which is how we know that their materials had differentiated into metallic core & rocky mantle. It is likely that some of the huge asteroids which have struck the Earth in the distant past were fragments thrown out of the asteroid belt & that some hit Mars & Venus as well. There may be more material in the asteroid belt than we think,because some fragments are so small & widely dispersed that they don't register in our telescopes. I think we can assume that the primordial planetesimals had a similar though not identical composition to the asteroid belt of today.


What error bar / confidence level is generally reported in galaxy distances? In reporting distances of galaxies, what are the typical error bars and associated confidence levels?

The error bars are typically very large.
There are several ways of determining the distance of galaxy:
Cepheid Variables
Cepheid variable stars have a known relationship between luminosity and period.  If we can observe these variables as individual stars in a galaxy, we can determine distance.
But individual stars can only be observed in the nearest galaxies. This method can't determine the distance of galaxies in which we can't resolve individual stars.
There is about +-7% uncertainty, plus possible systematic errors (wikipedia gives a distance modulus uncertainty of 0.16 for an individual galaxy, and states that the uncertainty in distance can be calculated as 0.461 x distance modulus uncertainty = 0.07)
Type 1a supernovae
These have a known and fixed luminosity (of about magnitude -19.3) If we can observe a supernova of the right type in a galaxy, we can tell how far it is.
But to use this method we have to wait for a supernova, and even in the larger galaxies they are rare.  Also the "known luminosity" has to be calibrated against Cepheid variables, so there might be systematic errors.
A well-observed supernova can fix the distance with uncertainty of about 5%, however there may be systematic errors.
Association with other galaxies
If two galaxies are interacting then we know that they must be about the same distance,  if we can measure one, (by supernova) we get an estimate of the distace of the other.
Redshift
Distant galaxies are receeding from us and E. Hubble observed that there is a correlation between the redshift caused by the galaxy moveing away from us, and the distance.  So for more distant objects we can use the red-shift as a proxy for distance.
However the peculiar motion of the galaxy also causes red (or blue) shift and this is not know. The calcualation of distance depends on the observed constant of proportionality, and while this is about 70 the exact value is not known (and probably not constant)  This method can't be used for local galaxies.
Other methods
There are other methods but they are less reliable, See Extragalactic_distance_scale These methods may have uncertainties five times greater than measurements of supernovae.
Systematic errors are due to uncertainties in the models being used.  For example the nearby cepheid variables tend to be from stars with lots of carbon, nitrogen and other elements (besides H and He). These have a different luminosity from stars formed with little except H and He.  Cepheid variables were used to measure the size of the milky way, assuming that the stars being measured were like the nearby stars.  They were not.  Correcting this systematic error resulted in distances that were double what they had been previously.  The calculated size of the milky way doubled!
So if someone says This galaxy is 50 million light years away, you can be reasonably confident that it is somewhere between 30 and 80 million light years away.  And perhaps more confident if there has been a recent supernova there.


Is there anything like a timelapse showing the Alpha Centauri stars orbiting around each other? It is my understanding that the angular separation between Alpha Centauri A & B is more than enough for even amateur telescopes to resolve as 2 separate stars. Going by the plot below, over the past 20 years there should be noticeable movement showing them orbiting. Do we have any images/videos that capture this movement?

Source click for full size

Unfortunately, I couldn't find any infos on this specific case.
But as I remember, there was a similar study from NASA a while ago, observed with the Hubble telescope:
https://hubblesite.org/contents/news-releases/2013/news-2013-22.html#section-id-2
There is a nice animation that show the proper motion of Proxima Centauri relative to background stars (not specified) between 2000 and 2010. The attempt was to detect the presence of planets around Proxima Centauri.


Universe as 2d sphere surface - a round trip? If Universe's analogy is 2D surface of the sphere, could we come around in circle? Say on Earth you decided to travel straight in one direction, after completing one full circle you'd find yourself at exact spot where you started.

Now, if we up that up by 1 dimension and apply it to our Universe, would the same thing happen if we pointed the rocket in a straight line? I.e. would the rocket come back to earth?

This is another case of taking an analogy too far (the other one I can think of is assuming that the rubber sheet analogy indicates that there is some extra spatial dimension$^1$). In this case, I'm assuming you're talking about the "balloon" analogy - the universe is expanding, like a balloon, but since there's no center, all bits of matter are on the surface of the balloon. This naturally leads to the idea that if you travel in one direction, you'll eventually come back to where you started. This is, unfortunately, not true.

Scientists often talk about the shape of the universe - how it is curved on a local or global scale. There are three possible categories, depending on one parameter, $\Omega$:

Positive curvature $\to$ $\Omega>1$
Zero curvature $\to$ $\Omega=1$
Negative curvature $\to$ $\Omega<1$

This diagram gives a good visual of each curvature:

I should add that these are merely examples of universes (more accurately, manifolds) with each $\Omega$. There are other cases with these curvatures, and so other universe where you can start at one point and end at the same point, as in a 3-torus.

All of these shapes could exist. However, in our universe, the global curvature is one where $\Omega \approx 1$ - a flat universe. The measurements that essentially clinch this are from the WMAP probe, which has analyzed the CMB and determined the age of the universe and its curvature. Measurements show, to a decent degree of accuracy, that our universe is flat - and so if we started in one direction, we would never return to where we started. Check out the NASA page for a full rundown of WMAP's results.

$^1$ Good questions dealing with this can be found here, here and here; John Rennie's answer on Physics is especially good.


Are all gas planets guaranteed to produce their own magnetic field? All gas giants in the solar systems have two things in common: abundant hydrogen gas and a magnetic field.

I'm wondering: could there be a gas planet that somehow can't produce any magnetic field on its own?

I learned there many kinds of bizarre and unfamiliar worlds from reading up terrestrial exoplanets and moons, so I thought maybe there is a unique gas planet that doesn't produce any magnetic field - or maybe the magnetic field disappeared after a while.

BTW I'm aware of hydrogen becoming metallic under a specific condition.

It's not technically impossible, but it does seem incredibly unlikely that a gas planet would lack a magnetic field. This is because hydrogen (by far the most abundant element) attains metallic properties when put under extreme pressure. And a spinning metal/conductor generates magnetic fields via the dynamo effect. This is why earth has a magnetic field, except instead of metallic hydrogen spinning in our core, its our liquid iron outer core that does the trick.

Any collection of gas large enough to form a planet would certainly create the needed pressure to transform hydrogen in its metallic form. For a gas planet to have a magnetic field, it really only needs to meet 2 conditions:

The planet needs abundant free hydrogen (which is extremely
likely given that almost all matter is hydrogen).
The planet needs to be spinning (also extremely likely because a
planet retains the net angular momentum of all the stuff that fell
onto it while it was forming... the only way a planet avoids
spinning is by tidal locking, such as the moon always facing one
side towards the earth... and even then it's still spinning
slowly... it's just spinning once every time it completes an orbit.)

Ultimately, given the vast and possibly infinite size of the universe, and the great abundance of planets, I'd bet there is a planet somewhere that happens to be made of gasses other than hydrogen, or happens to not be rotating, and thus there would be a gas giant without a magnetosphere. But I'd be very surprised if we find one in my lifetime, or even in the the next 1000 years!


HR diagram x-axis (color) and redshift? I've read several articles and watched some videos on the HR diagram, but no-one seems to mention the effect of redshift on the color of a star. Is there some calibration step that needs to be applied when looking up the Luminosity for a star that you've observed the color of?
For instance, if I observe a star a being a certain color then using the x-axis on the HR diagram I can find that observation's color and find a corresponding location in the HR diagram. However if the star I'm observing is redshifted the color I've observed is not the actual color of that star.
I'm super-new at reading the HR diagram but it really fascinates me and I cannot find why redshift isn't mentioned in relation to it.
Edit: it's not answered by the suggested
Relationship between photometric colour and redshift
as my question isn't so much about about spectroscopy or the types of redshift but simply if the HR diagram applies to far away stars or not.

The short answer is that the redshift of stars that are close enough to put into H-R diagrams (or "color magnitude diagrams", if we're being precise, given that you're talking about using colors) is so small that the effects are minimal.
The correction to color you're thinking of (called "K correction") depends on the redshift: the smaller the redshift, the smaller the effect.
But in order to distinguish and measure individual stars (so as to construct an H-R diagram), the galaxy has to be close enough -- the upper limit (for observations with the Hubble Space Telescope) is about 10 Mpc, or a redshift of $V \sim 1000$ km/s. For that kind of redshift, the K corrections will be $\sim 0.01$ magnitudes or smaller (you can test out different values using this online calculator -- note that it uses $z = V/c$ for redshift). This means the effect will be smaller than typical observational errors.
Put simply, galaxies that are far enough away to have large enough redshifts to worry about the effects on colors are too far away for you to observed individual stars and make H-R diagrams.
For stars within our own galaxy, the redshift (or blueshift!) effects will be due to relative velocities $< 200$ km/s or so, which means the effects will be really small.


Amateur observing targets for binary star systems? Are there any easily resolvable, binary star, observing targets visible from ~N40°?

I'd like to be able to show, in one observing session, Mizar and Alcor (naked eye resolvable), another with binoculars, and finally one in a small (say, 4 to 6" reflector) amateur scope.

Okay! Here are some of the popular targets, search for the season of visibility yourself please.

The above list describes the angular seperation between the stars and also the magnification required to resolve it, both explicitly. So, this list is not only for amateurs, but there are some intermediate ones which you can try if you have a good enough telescope.


Is there a term for stars without any planets? Quite simply, is there any used term for stars that don't have any planets? While contemplating this and doing numerous searches to no avail, I think I found my own answer, but I'm asking here for more input.

The reason I think there is no real term used as such is because there is yet to be confirmed of any stars that do not consist of any planets. I'm aware that we know of stars at this point in time where we haven't identified any planets YET, but that is not the same as actually confirming if planets do or do not exist around the star.

Am I correct in assuming this is the reason why no term has came to fruition yet? Or is there a term used in relation to talking about the possibility of none existing?

I've searched a fair bit for this but haven't been able to find anything. I know that current models show that it's highly unlikely for this to occur, but I'm curious.

A star is described as such whether or not it has planets.

There doesn't appear to have ever been a need, and to be honest, I can't think of any need for one. When it was assumed (in historical times) that we were on the only planet, stars were not thought to have planets. The title worked then. Now, as we know most stars are likely to have planets as part of normal formation, we can still call them stars.

Typically words are invented as there is a need for them.


Time period in which a planet rotates I notice that Wikipedia give two contradictory terms for the time period in which a planet rotates in relation to an infinitely-distant object. In one place this is called a "sidereal day" and in another "stellar day". Which of these are correct? For that point, is this an IAU definition? I cannot find the definitions on the IAU website.

From the Sidereal Day page:

  A sidereal day is the length of time it takes a planet to rotate from the perspective of a distant star.

From the Day page:

  stellar day - an entire rotation of a planet with respect to the distant stars
  sidereal day - a single rotation of a planet with respect to the vernal equinox

Since etymologically sidereal references the stars (Latin sidus), one might expect a sidereal day to correspond to the rotation of the Earth with respect to distant "fixed" stars. But this does not seem to be the case, and because the the difference is not significant, the two definitions are sometimes used interchangeably.

I don't know of an explicit IAU definition of sidereal day. However, since sidereal time is defined as the hour angle vernal equinox (which is a local definition, although Greenwich is conventional), defining the (mean) sidereal day in terms in reference to the equinox is practically the only sensible choice.

Effective 1985, UT1 is now computed using very long baseline interferometry of distant quasars, and so can be taken to be authoritative regarding the "fixed stars". The coordinated universal time (UTC) time approximates UT1 with atomic clocks. Anyway, UT1 derived the rotational period of the Earth as $p = 86164.09890369732\,\mathrm{s}$ of UT1 time and the mean sidereal day (in 2000) would be $86164.090530833\,\mathrm{s}$ of UT1 time, following the explanation:

  The length of one sidereal day is defined by two successive transits of the mean equinox; while the Earth is rotating eastward, the mean equinox is moving westward due to precession. Therefore, one sidereal day is shorter than the Earth's rotational period by about $0.008\,\mathrm{s}$, the amount of precession in the right ascension in one day.

See: Aoki, S., et al., The new definition of universal time, Astron. Astrophys. 105, 359-361 (1982).


Temporary family member 2020-CD3 So, recently I came to know that this asteroid is orbiting Earth from past 3 years and still it is orbiting. It is kind of a mini Moon to the Earth know to have come from asteroid-belt.

  Is there any threat of collision by 2020-CD3 on Earth and how long it will take to escape the Earth's gravity?

There is no chance of a collision in the short term.  2020 CD3 is in a rather chaotic orbit of the Earth that extends well beyond the moon and doesn't come very close to Earth Each orbit is different, but no orbit brings it closer than the moon. The orbit is irregular because it is perturbed both by the moon and by the gravity of the sun. At its furthest point from the Earth. The gravity field that it moves in is not a simple "inverse square law" because of the combination of Earth, Moon and Sun.

It is likely to escape Earth in the next few months, and it will go back to be a sun orbiting asteroid, we know this because we understand how gravity works, so we can predict its future path. Its future path involves it leaving the neigbourhood of the Earth.  However as its orbit does bring it close to the Earth's orbit there is a chance of it colliding with the Earth at some point in the future. It would make a bright fireball, but would not be dangerous, as it is only a couple of metres across. Objects like this collide with the Earth frequently.

Its orbit is not yet well enough known to determine exactly how it will return to a solar orbit, and so we can't yet predict whether it will be temporarily captured or even collide with Earth on its next close approach. But the probability of a collision is low.


Moonrise after sunrise but moonset before sunset? Today (1/24/2020) for Boston MA timeanddate.com lists the following four times:

Sunrise  7:05 AM
Moonrise 7:06 AM
Moonset  4:33 PM
Sunset   4:47 PM

Now, if moonrise is after sunrise, I'd have thought the "new moon" conjunction already happened. But since moonset is before sunset, I'd think it hasn't happened yet.

What gives? Is this a rounding error? Usually the moon "moves" across the sky "slower" than the sun does (as in a solar eclipse where the moon moves from west to east back across the sun).

Right now, the moon is south of the ecliptic (for those of us in the northern hemisphere).  This means it encounters the horizon before the sun.


Would a red dwarf star resemble our own Sun at sunset to an observer on a nearby planet? Suppose you're observing a red dwarf star at Noon from the surface of an Earth-like planet orbiting in the red dwarf's habitable zone, so that the red dwarf's luminosity is equal to our own Sun.
Aside from the red dwarf's greater perceived size in the sky, would its color and brightness closely resemble our own Sun at sunset in Earth's sky on a clear, dust and smoke-free evening?
Is it fair to say that since the Sun's color and brightness changes as it dips closer to the horizon due to scattering of blue light, there could conceivably be a close color/brightness match to any G, K, or M class star at some point along the Sun's trajectory in the sky?

Your question may ulitmately be about the physiology of the eye, which is off-topic here.
The spectrum of the Sun seen low on the horizon is quite different to the spectrum of an M-type red dwarf. The reason that a red dwarf is red, is not just that it is cool, but that there are great chunks of the spectrum that are absorbed by molecules in the photosphere of the star. These molecules start to form when the temperature of the photosphere falls below 4000 K.
In contrast, the Sun has an almost continuous spectrum all the way into the ultraviolet, with relatively narrow absorption lines caused by atomic species in the solar photosphere. This is then modified, mainly, by Rayleigh scattering, which smoothly scatters blue light out of the line of sight towards the Sun.
Whilst any sort of spectrograph or prism could distinguish these differences, whether your eyes could is another matter. So whilst it is conceivable that the colours would appear similar for some particular angle above the horizon, the detailed spectrum would be quite different.


How could eclipse happen on the same day in a different calendar? (45 AD) According to the eclipse calculator I use (Alcyone), a partial eclipse 0.314 magnitude occurred August 1, 45 A.D. in Rome starting at 8:35am local time.

This would seem to correspond to an eclipse which was predicted to fall on the birthday of the emperor Claudius in the consulship of Marcus Vinicius and Statilius Corvinus (45 AD), as described in Dio Cassius Book 6, Section 26. The birthday of Claudius was Kalends Augustus (the first of August).

If this is the case, it would seem that there is an exact correspondence between the Gregorian and Julian calendar going back to 45 AD. However, I would expect that small differences in leap years and other irregularities would make this exact match on the same day unlikely.

Another example is the eclipse of 1140 at London which the Saxon Chronicle says occurred on 13 Kalends April (March 20th) and Alcyone says occurred March 20, again a perfect match. Yet another example is the eclipse of 809 which passed over the Faroe islands and the Saxon Chronicle says occurred 17 Kalends Augustus (16 July) on the second day of the week. Now, of course, 16 July 809 is a Thursday in the Gregorian calendar, not the second day of the week, so something would seem to be fishy somewhere.

Is Alcyone/NASA making some kind of adjustment so that these pre-Gregorian dates match up?

From the documentation for your program, http://www.alcyone.de/plsv/documentation/overview.html,

  Dates are in the Julian Calendar through 4 October 1582, and then in the Gregorian Calendar, which begins on 15 October 1582. (Compute for 1582, and you will see that 1. oct to 1. nov. is short.)


Which quantum reality will be seen in a far away galaxy I recently watched Brian Greene documentaries on parallel universes and am trying to wrap my brain around multiple quantum universes that form at every event and Einstein's relativity that in space-time, past present and future coexist so we can observe the future of a far away galaxy from Earth by moving.
https://www.youtube.com/watch?v=MO_Q_f1WgQI - Around 7th minute - explains how an alien moving towards earth has a now slice cutting in Earth's future

My question is which quantum universe will be seen because future events in that galaxy would create multiple such universes.

My question is regarding many worlds interpretation of quantum mechanics according to which multiple alternate futures are possible https://en.m.wikipedia.org/wiki/Many-worlds_interpretation

The question has been answered by James K below.

So, the presenter explains that two observers who are not moving relative to one another will agree on the meaning of "now".  However if one observer is moving, then they will no longer agree, and what one considers to be now will be in the past or future of the other.

He observes that a sufficiently distant observer, in another galaxy 100 million light years away that is moving at "bicycle" speed relative to us would have a 200 year discrepancy in their understanding of "now".

But this alien would have absolutely no way of observing our universe now. They could not see us in our "now", nor in their "now" of 200 years ago, because they are 100 million light years away. They can only observe how our galaxy was 100 million years ago.  They can certainly not observe our galaxy in the future, even if they travel towards us at speeds approaching the speed of light.

So the explanation of different "now" is about agreement on clocks, not about actual observations.

I think then the program errs by treating Special Relativity as a full description of nature, rather than as a model. Relativity is a classical theory, in that in Relativity, if you know the initial state, you can predict the future state. There is no uncertainty principle in general relativity. It is correct to say that in Relativity the future is as real as the past. It is unclear if the same is true in Quantum Mechanics, it depends on how you interpret the maths. Whether the future exists or not cannot be answered by theory.

Since different interpretations of QM are based on the same maths, there is no way of testing which interpretation of quantum mechanics is "correct" or even whether calling an interpretation "correct" has any scientific meaning. In the "many worlds" interpretation, the notion of a single space-time which you can take slices of is lost completely.

I might note that I'm not a great fan of the explanations in that series, which seem to me to be confusing on a number of levels, and are more aimed at creating "wow" than being clear. For example, the mixed metaphor of "slices of bread and aliens riding bikes" seems to have confused you into thinking that there are aliens who are already seeing how our planet will be 200 years from now.


What calculations show that Comet Shoemaker-Levy 9 orbited Jupiter for several decades before its spectacular impact? (Chodas, Sekanina & Yeomans) This answer to A moon in eccentric orbit dipping below Roche limit includes the following about Comet Shoemaker–Levy 9

Here is a nice figure of its last passes:

I looked up "fragment A" in JPL's Horizons and it provides a modeled trajectory only between 1992-07-15 and 1994-07-17 but the link cited above says

Computations by Paul Chodas, Zdenek Sekanina, and Don Yeomans, suggest that the comet has been orbiting Jupiter for 20 years or more, but these backward extrapolations of motion are highly uncertain. See "elements." and "ephemeris." at SEDS.LPL.Arizona.EDU in /pub/astro/SL9/info for more information.

If I paste some of that into a search engine I get http://spider.seds.org/sl9-list.html with lots and lots of links, but now I'm in way over my head.
Question: What calculations show that Comet Shoemaker–Levy 9 was orbiting Jupiter for several decades before its spectacular impact? (Chodas, Sekanina & Yeomans) Where can I read about these calculations? Have there been further refinements since, based on any subsequently discovered historical precovery data?

Chodas and Yeomans published a paper called The orbital motion and impact circumstances of Comet Shoemaker-Levy 9.  In it they describe the algorithms and techniques they used to reconstruct the orbit backwards from the discovery of the comet.  SL9 had an incredibly chaotic orbit in the mathematical sense in that small perturbations in the position would result in large effects in reverse-time orbital simulations.  In the words of the paper authors:

Unfortunately, SL9'sorbit about Jupiter was among the most chaotic of
any known solar system body, with an effective Lyapunov time on the
order of 10 years (Benner and McKinnon 1995). As a result, a single
backward numerical integration does not provide definitive answers on
the orbital history of this object. A better approach is to account
for the uncertainties in the initial conditions of the backward
integrations, and to investigate the motion in a statistical manner
using a Monte Carlo analysis (Chodas and Yeomans 1995).

For comparison, the Lyapunov time for the Solar System is on the order of millions of years.  An orbit trajectory plot for SL9 is the most complicated I've ever seen (excluding human-built craft)!

I don't know if precovery data was ever used to refine the orbit, though papers are still being published on SL9's orbit.  Unfortunately, Invariant Manifolds and the Capture of Comet Shoemaker-Levy 9, published in 2019, is behind a paywall.


Strong force and metric expansion If the expansion of the universe (the metric) continues (and perhaps is accelerating), in a very large but finite time the expansion of the metric will clash with the effects of the strong nuclear force. What happens then? Would the energy required halt the expansion and if so what is the slowing effect of the "stretch" on the strong force today - too small to be measured?

Update: Thanks for the answers so far - I hadn't even considered the big rip scenario. I was merely thinking that as the metric expanded the strong force inside baryons would require unlimited resources of energy to be overcome - but I now understand that is wrong however it still looks to me like a potential perpetual motion machine - continually creating matter/anti-matter pairs. Perhaps that should be a separate question though...

Excepting a Big Rip scenario, there is no eventual 'clash'.

Consider a Friedmann–Lemaître–Robertson–Walker universe:
$$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\left[\frac{\mathrm{d}r^2}{1-kr^2} + r^2\left(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2\right)\right]\text{,}$$
where $a(t)$ is the scale factor and $k\in\{-1,0,+1\}$ corresponds to a spatially open, flat, or closed cases, respectively. In a local orthonormal frame, the nonzero Riemann curvature components are, up to symmetries:
$$\begin{eqnarray*}
R_{\hat{t}\hat{r}\hat{t}\hat{r}} = R_{\hat{t}\hat{\theta}\hat{t}\hat{\theta}} = R_{\hat{t}\hat{\phi}\hat{t}\hat{\phi}} =& -\frac{\ddot{a}}{a} &= \frac{4\pi G}{3}\left(\rho + 3p\right)\text{,} \\
R_{\hat{r}\hat{\theta}\hat{r}\hat{\theta}} = R_{\hat{r}\hat{\phi}\hat{r}\hat{\phi}} = R_{\hat{\theta}\hat{\phi}\hat{\theta}\hat{\phi}} =& \frac{k+\dot{a}^2}{a^2} &= \frac{8\pi G}{3}\rho\text{,}
\end{eqnarray*}$$
where overdot denotes differentiation with respect to coordinate time $t$ and the Friedmann equations were used to rewrite them in terms of density $\rho$ and pressure $p$. From the link, note in particular that $\dot{\rho} = -3\frac{\dot{a}}{a}\left(\rho+p\right)$.

If dark energy is described by a cosmological constant $\Lambda$, as it is in the standard ΛCDM model, then it contributes a constant density and pressure $\rho = -p = \Lambda/8\pi G$, and so no amount of cosmic expansion would change things. Locally, things look just the same as they ever did: for a universe dominated by dark energy, the curvature stays the same, so the gravitational tidal forces do too. Through those tiny tidal forces, the dark energy provides some immeasurably tiny perturbation on the behavior of objects, including atomic nuclei, forcing a slightly different equilibrium size than would be otherwise. But it is so small that that it has no relevance to anything on that scale, nor do those equilibrium sizes change in time. The cosmological constant holds true to its name.

On the other hand, if dark energy has an equation of state $p = w\rho$, then a flat expanding universe dominated by dark energy has
$$\dot{\rho} = -3\frac{\dot{a}}{a}\left(\rho+p\right) = -3\sqrt{\frac{8\pi G}{3}}\left(1+w\right)\rho^{3/2}\text{,}$$
and immediately one can see that there is something special about $w<-1$, leading to an accumulation of dark energy, while the cosmological constant's $w = -1$ leads to no change. This leads to a Big Rip more generally, as zibadawa timmy's answer explains.

  ? If the metric expands surely objects get further away from one another and that would include the stars inside galaxaies as well as the galaxies themselves?

Not at all. It wouldn't even make any sense: if you have an object like an atom or a star in gravitational freefall, by the equivalence principle only tidal forces across it are relevant. The tidal forces stretch the object until the internal forces balance them. But for a Λ-driven accelerated expansion, dark energy contribution to tidal forces is constant. Hence, an object already in equilibrium has no reason to further change its size, no matter how long the cosmic acceleration occurs. This also applies to galaxies, only that the internal forces are also gravitational and balance the dark energy contribution.

Looking at this in more detail, imagine a test particle with four-velocity $u$, and a nearby one with the same four-velocity, separated by some vector $n$ connecting their worldlines. If they're both in gravitational freefall, then their relative acceleration is given by the geodesic deviation equation, $\frac{D^2n^a}{d\tau^2} = -R^\alpha{}_{\mu\beta\nu}u^\mu u^\nu n^\beta$. The gravitoelectric part of the Riemann tensor, $\mathbb{E}^\alpha{}_\beta = R^\alpha{}_{\mu\beta\nu}u^\mu u^\nu$, represents the static tidal forces in a local inertial frame comoving with $u$, which will drive those particles apart (or together, depending). Hence, keeping those particles at the same distance would require a force between them, but it's not necessary for this force to change unless the tidal forces also change.

Galaxies don't change size through cosmological expansion. Stars don't either, nor do atoms. Not for Λ-driven expansion, at least. It would take an increase in tidal forces, such as those provided by a Big Rip, for them to do so.

A related way of looking at the issue is this: according to the Einstein field equation, the initial acceleration of a small ball of initially comoving test particles with some four-velocity $u$ (in the comoving inertial frame), which is given by the Ricci tensor, turns out to be:
$$\lim_{V\to 0}\left.\frac{\ddot{V}}{V}\right|_{t=0}\!\!\!\!= -\underbrace{R_{\mu\nu}u^\mu u^\nu}_{\mathbb{E}^\alpha{}_\alpha} = -4\pi G(\rho + 3p)\text{,}$$
where $\rho$ is density and $p$ is the average of the principal stresses in the local inertial frame. For a positive cosmological constant, $\rho = -p>0$, and correspondingly a ball of test particles will expand. That's cosmic expansion on a local scale; because of uniformity of the FLRW universe, the same proportionality works on the large scale as well, as we can think of distant galaxies as themselves test particles if their interactions are too minute to make a difference.

Again, we are led to the same conclusion: if the ball has internal forces that prevent expansion, then those forces don't need to change in time unless the dark energy density also changes, which doesn't happen for cosmological constant. If they're not 'clashing' now, then they won't need to in the future either.


How long do hot X-ray sources last? Some X-ray sources in nebulae are caused by hot materials in the millions of degrees that formed in supernovae. What's the cooling rate of this stuff? How long does it stay this hot?

If we would take our solar system for example, the time from the formation of the nebula (by a supernova?) until it was cool enough to form solid planet was only a few million years. That's very quick, for instance relative to the age of the Earth. How ephemeral are these X-ray sources? I would expect them to not last as long as IR sources.

Is there an order of magnitude for the lifetime of the emission? Does it take millions of years? Tens of millions? Tens of thousands of years?

A comment suggests it doesn't really cool down, but diffuse to the interstellar medium. How long does it take for it to diffuse until there isn't a clear X-ray source?

This is not an answer, but it is too long to put in the comment.

There are formulas for cooling timescales of different radiative processes. Cannot recall right now about sources for reading.

One thing about nebulae. Some of them might have compact objects such as pulsars or black holes feeding energy from the centers. In those cases, you have to consider the energy input from the central sources and radiative cooling.

Another thing about X-ray emission processes. There are thermal X-ray emission (i.e., bremmstrahlung) and nonthermal ones (i.e., synchrotron and inverse Compton; IC). Each process dominates in different environment. Regarding to SN nebulae, nonthermal processes are the dominant ones, the electrons involving in the processes are relativistic and thermal process requires dense electron medium (for collision) to maintain the thermal equilibrium. IC normally dominates at early ages of nebulae due to the compactness of the system, while synchrotron self-absorption is also strong at these ages. At older ages, synchrotron dominates, while the IC can interact with the background photons (i.e., cosmic background radiation) and results in X/gamma-ray photons as well.

Last about IR emission. Since IR emission is mostly from thermal process while X-rays are from nonthermal processes, X-rays tend to last longer.


Could anything consume a small black hole? Whenever I read about black holes, it is normally involving how they devour anything that comes too close... regardless of how big anything is.

But what if it were a really small black hole vs something really large? How about something like a large star or neutron star... could a black hole be small enough to get devoured / consumed by something like that or would it just not be possible?

Black holes don't "devour".  They can't "eat" things.  But things can "fall into" a black hole.  It really is just gravity, but really really intense gravity.

A really small black hole would have a mass of about 3 times the mass of the sun. All neutron stars are smaller than that (or they would turn into black holes).  A really large star could be more massive and would be much much larger in volume.  A black hole could orbit with such a star, but if it got close enough it would pull gas off the star which would then fall into the black hole, causing it to get larger.

There is no way that a large star can in any way "pull apart" or otherwise "consume" a black hole.

There is a theoretical notion of the quasi-star. These are similar to a Thorne–Żytkow objects.  In normal stars, the outer layers are prevented from collaping by energy from nuclear fusion in the core.  In a quasi-star, the core collapses into a black hole and the release of energy from matter falling into the black hole prevents the collapse of the star. A steady state can be achieved, as if more matter starts to fall in, more energy is released causing the star to grow, and reducesing the amount of matter falling in. Nevertheless, ultimately all the matter from the star would fall in.

Smaller black holes can't form from stars. We don't know if they exist at all, but if they do, very very small black holes could be too small to interact much with regular matter. A very small black hole could be smaller than a proton, and even if one fell into the Earth, it could pass through the gaps between and inside atoms. Such a black hole would be very hot, due to Hawking radiation.


Why is the opening in the Anglo-Australian Telescope's dome so small? Many older or "classic" telescope domes have a horizon-to-zenith opening in the dome, and this helps speed up the thermal equilibration between the inside and outside air, decreasing turbulence and its effects upon resolution.

Source

Source

But the Anglo-Australian Telescope has second mechanism that tightly constrains the vertical extent of the opening as well, leaving just a tiny hole barely big enough for the telescope to see the sky.

What are the benefits of each of these, and how do the minimal aperture domes deal with temperature differences?

below: Screen shot from the video A 2dF night at the Anglo-Australian Telescope found here.

This is a two part windscreen designed to minimize the effects of windshake on the telescope and to avoid the deterioration in image quality that wind would cause. The AAT is in a tall 6 story dome on a pretty exposed part of Siding Spring Mountain and so is likely more affected by wind gusts. Initially there were issues with the mount being too flexible and having an lower resonant frequency than expected, which was in the range that can be excited by wind buffeting (a mention in the bio of Harry Minnett, a CSIRO engineer who worked on the AAT). This may have also lead to a more extensive windscreens but windscreens, particularly upper ones, are fairly common and exist at several other observatories. As an example, the pointing limits of the Isaac Newton Telescope page discusses the sky area that is available with and without lowering the upper windshield.


Calculating Local Sidereal Time When calculating local sidereal time, 100.461 "is the value needed to yield the correct value of GMST" - David Hammen (left as comment on my previous question). How is this value calculated?

If (hypothetically) I wanted to use January 1st 12:00 (noon) 2010 as my epoch, how would I calculate this number?

The USNO page refers to USNO Circular 163 which in turn refers to a paper by Aoki et al. and one by Moyer. These papers explain how the expressions created by Simon Newcomb at the end of the 19th century were updated to work with the new system of constants introduced in 1984. If you want to change the epoch, you could redo these calculations with the epoch of your choice.

Further changes to celestial mechanics calculations have occurred since then, but apparently the USNO has decided that for the accuracy stated in the USNO web page, 0.1 second, it is not necessary to overhaul the calculation to account for all the changes since 1984.


At the Big Bang, when everything was close together, why did it not "collide", violating Planck length or Pauli Exclusion Principle? How could so much matter, or "all" in fact, have been concentrated in a smaller universe without being in the actual same place? Why did this not result in undercutting the Planck Length or violating the Pauli exclusion principle?

The Pauli Exclusion Principle forbids two indistinguishable fermions occupying the same quantum state. It does not prevent them getting arbitrarily close together so long as they have very different momentum states.
The big bang model relies on classical General Relativity. When we go back to scales where quantisation of space might become important (i.e. Planck length scales, corresponding to the first $\sim 10^{-43}$ s of the universe) then GR needs to be replaced by some quantum theory of gravity (which we don't have).


How can Mercury's sodium tail be imaged? I recently learned that Mercury has a sodium tail. Can this sodium tail be imaged with a ground based telescope? Is so, what is the minimum setup required to be able to take a picture of it? Do you need any specific filters?

Yes, it is possible; see https://twitter.com/sevospace/status/1328082050889490433
However, M. Voltmer used a “custom-made sodium filter,” without giving any details…


Is the bright red body a star? I was helping classify galaxies at Galaxy Zoo when I came across this picture of a galaxy. The dull red body looks huge compared to the galaxy. Is it because it is closer to the earth? It would be interesting to know what is going on here.

It is a bright star. You can see the "diffraction spikes" that are the signature of a point-like object.

The star will be in our Galaxy, so is certainly closer than the galaxy in the image. However, this is not why it appears to be big.

The image appears big because the starlight will be spread by the telescope objects over a finite area - the point spread function. In addition, when a star is very bright, the central CCD pixels saturate and charge can spill into adjacent pixels. Even if that were not the case, there is enough light in the "wings" of the point spread function to produce a large image.


Did this paper just argue that there is surely life on Titan? I just read the popular press version of a new paper, and now I have a question about Titan.

Here are some quotes from the pop article:

  A rocky, ocean-bearing planet with more than 0.1 percent methane in its atmosphere should be considered a potentially inhabited planet, the researchers say. And if the atmospheric methane reaches levels of 1 percent or more? In that case, “potentially” doesn’t cut it—such a world would “likely” be home to alien life.

...

  Thus, methane and carbon dioxide together, unaccompanied by carbon monoxide, on a rocky, ocean-bearing world would best be interpreted as an airtight sign of anoxic life.

Now, some information about Titan from the Wikipedia page about Titan:

Titan's atmosphere
    Stratosphere:
        98.4% nitrogen
        1.4% methane
        0.2% hydrogen
    Lower troposphere:
        95.0% nitrogen
        4.9% methane
        -OR, depending on who you ask-
        97% nitrogen
        2.7% methane

  The density of Titan is consistent with a body that is about 60% rock and 40% water.

  There are trace amounts of other hydrocarbons, such as [...] carbon monoxide [...].

If I understand correctly, 'trace amounts' of CO is in line with the requirements stated in the Catling paper.

So, is Titan covered in microbes, or what?  Since the article didn't mention Titan, I presume I missed something.  What was it?

The underlying assumption in the paper is that the planet has water in liquid form.

The paper suggests that finding both CO2 and CH4 in the atmosphere of a rocky planet in the "habitable zone" (in which water can exist in liquid form) would be hard to explain, without life. Thus one doesn't need to detect atmospheric oxygen. If you can detect both CO2 and CH4, this should be taken as evidence of anoxic life (probably in the form of microbes)

Titan has CH4, but it doens't have much CO2 and there is no liquid water.
Titan is not in the habitable zone. On Titan, H2O is a forms part of the "rocks", and the oceans are made of hydrocarbons. Titan is not a good location to look for life as it doesn't have free liquid water. We have a photo from Titan's surface and it isn't covered in Microbes


Have all Wolf-Rayet stars evolved off the main-sequence? From what I've read from various sources, Wolf-Rayet stars are massive, evolved stars that typically are the hottest, most luminous stars known. They experience high mass loss, taking their mass from $\sim 100 M_\odot$ to only $\sim 20 M_\odot$. However, Wolf-Rayet stars like R136a1, 2, and 3 are still supermassive, at around $150-220 M_\odot$, and are about only 1 million years old.
Does this mean that these stars already have burned their hydrogen and have evolved off the main sequence, or are these stars still on the main sequence but are classified as WN5h as a result of its high surface temperature and luminosity?

Wolf-Rayet stars have long been a subject of controversy in astronomy. Observationally, they are the class of stars that astronomers identify as being very luminous, having very little hydrogen, and are mostly composed of helium, carbon, nitrogen, and oxygen. This class of stars remained an enigma until the 1980s, when Wolf-Rayet stars started to be considered as the end result of high-mass stellar evolution (due to better modeling and observations, specifically improvements in spectroscopy and better telescopes). Nowadays, WR stars are suspected as the likely stellar progenitor to black holes and high mass neutron stars (depending on the mass and metallicity, i.e. see Heger et al. 2003).

Does this mean that these stars already have burned their hydrogen and have evolved off the main sequence, or are these stars still on the main sequence but are classified as WN5h as a result of its high surface temperature and luminosity?

Regardless of whether you interpret the Wolf-Rayet phenomena as the end result of high-mass stellar evolution, a Wolf-Rayet star is off the main sequence because it is a type of core-helium burning star: it exhausted the hydrogen in its core and thus is not on the main sequence.
Wind mass-loss of Wolf-Rayet stars is strong for their relatively small size. That is, the mass loss rate is actually comparable to that of main sequence O-type stars (the likely stellar progenitor of Wolf-Rayet stars), however the Wolf-Rayet star has smaller radius and smaller radius of gyration than main sequence stars so the comparison becomes complicated. Such mass-loss rates depend on the star's luminosity (and thus mass), as well as the metallicity. The star WR R136a is located in the Tarantula nebula, which is not a high-metallicity environment, meaning the mass-loss rate for this star is not particularly high, which is consistent with it still having such a large mass. Recent analysis of this nebula has actually revealed that the existence of such very high-mass stars might not be as rare a once thought, implying there may be many stars similar to R136a in the cosmos.