Why is the Cosmic Microwave Background at the same distance no matter the direction we look? I've read from different sources that:

The CMB is visible at a distance of 13.8 billion light years in all
directions from Earth, leading scientists to determine that this is
the true age of the Universe.

And that got me thinking, if this is true, there are 2 possible scenarios:

These are not really the true limits of the universe, but it's the farthest we can watch with our actual technology.
Or we happen to be the center of the universe, because we see the "beginning" of it at the same distance no matter the direction we see at (which I think this a dumb supposition)

So, why do we actually see the CMB at the same distance no matter what direction we look at?

The CMB is visible at a distance of 13.8 billion light years in all directions from Earth, leading scientists to determine that this is the true age of the Universe.

This is wrong in a few ways. First, we do have good reason to think that the CMB was produced around 13.8 billion years ago, but that doesn't mean it's 13.8 billion light years away. The light is right here (that's how we detect it), and the matter that emitted the light is roughly 47 billion light years away in the present era, if our cosmological model is correct.
Second, we don't work out the age of the universe from the distance to the CMB (or the matter that emitted it). We can't see how far away it is. We work out the age of the universe by fitting a cosmological model to a variety of evidence (including the detailed spectrum of the CMB), and then we say that the CMB was emitted 13.8 billion years ago because that's what the model implies.
Third, the CMB doesn't date back to the beginning of the universe – although the time from the big bang/end of inflation to the emission of the CMB (around 380,000 years) is so much smaller than the time from the big bang to now (around 13.8 billion years) that this makes little difference.
We aren't at the center of the universe. The universe is just homogeneous (the same everywhere), and it's large enough that light hasn't had enough time to cross it since the CMB was emitted. The CMB we see is from the same distance in every direction because it's all been traveling for the same amount of time. So your first scenario is the correct one.


Wouldn't the rings of Saturn experience tidal effect? The "Lord of the Rings" has more than 60 moons, some of which are larger than Mercury. My question is, wouldn't the ring, which is mostly comprised of rocks, experience tidal effect whenever a moon is near?

Most of the 60 moons in the Saturn system are far away from the rings and very small, so their effect on the rings is negligible. But larger ones that are closer in (Enceladus) do have a rather significant effect on the rings, but as the gravitational pull of these moons is radially outward, it is hardly visible. On the other hand, small moons inside the rings that are a few kilometers in diameter cause significant changes in the ring system.

(source: magicgate.eu)
The orbits of those giant rocks are slightly inclined relative to the rings, which causes the rings to create a bulge in the direction of the small moon as seen in the picture. And these effects are fairly significant as the rings are mostly just a few hundred meters thick or less without a moon present, but with one, they are possibly hundreds of kilometers spread out.


Optical emission line data archives I am trying to do exhaustive optical emission line analysis of some active galaxies using archival data.
What are some of the publically available data archives from where one could obtain optical emission line data? I am new to this, and so far I've found the HST MAST Archives and SDSS DR16.

A few sources come to mind to add to MAST and Sloan.
Data releases from the SAMI Galaxy survey, which contain (or will eventually contain) emission line spectra for about 3500 galaxies. The papers for the first two releases can be found [here] and [here].
Another (smaller and older) dataset can be found at the Penn State Center for Astrophysics. It contains spectroscopy of KISS emission line galaxy candidates. The download links to the dataset are on the CASt page [here].
I'll continue to update if I find more of a substantial size, but I think this should expand what you have already quite a bit.


Question about total absolute magnitudes of galaxies - negative or not? I have seen this in a couple of places, but when the authors in the paper - http://chandra.as.utexas.edu/~kormendy/kfcb-accepted.pdf -
wrote:

  "All 10 of our ellipticals with total absolute magnitudes M_VT ≤ −21.66 have cuspy cores – “missing light” – at small radii. Cores are well known and naturally scoured by binary black holes formed in dissipationless (“dry”) mergers. All 17 ellipticals with −21.54 ≤ M_VT ≤ −15.53 do not have cores."

Since absolute magnitude is the magnitude of that object at 10 pc away from Earth, then this is why the galaxies have such negative (meaning, very bright) values for total absolute magnitude, correct?

However, in this paper - http://arxiv.org/pdf/astro-ph/9602044v1.pdf - the author writes:

  "The morphological properties of galaxies between 21 mag < I < 25 mag in the Hubble Deep Field are investigated using a quantitative classification system based on mea- surements of the central concentration and asymmetry of galaxian light."

The galaxies here are between 21 to 25 absolute magnitudes (?) in the infrared (I?) band, but why are these magnitudes not negative?

The morphological properties of galaxies between 21 mag < I < 25 mag
  in the Hubble Deep Field are investigated using a quantitative
  classification system based on mea- surements of the central
  concentration and asymmetry of galaxian light.

Those are apparent magnitudes, not absolute.

The convention is to use $M_X$ for absolute magnitude and just $X$ for apparent mag (or m$_X$), where X is the bandpass symbol (U, B, V, R, I, J, L, H, K, etc).

And, the $V_T$ and $M_{V_T}$ (note the correct subscripting, which the authors failed to do) means the V band radial profile is extrapolated to infinite radius to give a total for the whole galaxy.


We are sure that there are no brown dwarfs closer than ______ I don't quite know how to word this question.

How close could a brown dwarf be to us and have gone undetected until now?

Or maybe, with what probability are we sure that there is no brown dwarf within X light years of the sun? (For different values of X)

(Don't worry about addressing Nemesis or a hypothetical binary partner for the sun. I'm just asking about undetected brown dwarfs..., even outside of the Oort cloud but within Alpha C. I don't know how sensitive the WISE survey was.)

I think the answer is roughly 3-5 light years.

The recent WISE survey in the near-infrared should have been capable of detecting even a very cool nearby brown dwarf (and indeed it has detected some very cool brown dwarfs - e.g. a 250K brown dwarf only 6 light years away Luhman et al. 2014). Since WISE was an all-sky survey and there is great interest in finding these ultracool brown dwarfs, then it is reasonable to assume that any similar, object within a few light years would already have been found. The caveat to this is that the point spread function of the WISE survey s not that great an so near the galactic plane it could be that source confusion might have hidden the brown dwarf. On the other hand, nearby objects usually have high proper motion, so it would probably have been uncovered in combination with the earlier 2MASS survey, since as I show below, such an object may have been detected in both.

Some details: As an example let's take a very low mass brown dwarf - say 20 Jupiter masses and 5 billion years old - and place it 4 light years from the Sun. According to the evolutionary models of Saumon & Marley (2008), such an object has an intrinsic luminosity of $10^{-7}$ times (1 ten millionth) that of the Sun and a temperature of 400 Kelvin and would appear to have a spectral type of late T or early Y. This is similar to the coolest brown dwarfs yet found.

From the calibration of absolute magnitudes versus spectral type for cool brown dwarfs in Marsh et al. (2013) we know that at 4 light years, such a brown dwarf would have magnitudes of $H=16$ and $W2=10$. The former is just about bright enough to have been seen in the 2MASS all-sky survey and the latter is easily detected by WISE. The combination of data would also easily have revealed the large parallax of such an object. We can conclude that an object would have to have much lower mass than this to remain undetected and thus would arguaby not be called a brown dwarf, but would really be a "free-floating planet".

Now you might ask, what about a cooler brown dwarf? Well the theoretical predictions of magnitudes and colours become much more uncertain, but I would simply refer to the Luhman discovery. This brown dwarf is about as cool as a brown dwarf could be given the age of the Galaxy, may only be 10 Jupiter masses, yet was still easily detected in the WISE survey. Therefore I think this sets a fairly firm empirical limit.


How often does Earth cross the galaxy regions with higher supernovae activity? According to Summa Technologiae, a book by a Polish author Stanisław Lem that based his science fiction novels on scientific research of the era, the Earth crossed in the moment of forming of life the arms of the Galaxy, with high supernovae activity, which triggered life creation. Afterwards, it moved to regions with low supernovae activity, which enabled the preservation of existing life.

I'd like to test the validity of this claim against current research. How often does the Earth cross regions with higher concentrations of stars (and therefore, much greater chance of gamma burst wipe)? Additionally, how many times the probability of nearby supernova explosion is higher in that regions?

The paper "Frequency of nearby supernovae and climatic and biological catastrophes" by Clark, McCrea, and Stephenson published in Nature estimates (at 50% probability) that the Solar System passes within 10 parsecs of a supernova every 100 million years. This supernova would be part of a 20-parsec strip in which an estimated 50 supernovae occur.

They do speculate that a connection between the Solar System passing through these regions and climatic and biological events (such as ice ages) is possible.

You can read the paper yourself for more details. It's only two pages and you can purchase 48 hour access for $5 from Readcube.

Source:

"Frequency of nearby supernovae and climatic and biological catastrophes" - Clark, McCrea, and Stephenson


Solar flares/CME dangers to life? I know a repeat of the 1859 Carrington Event (coronal mass ejection) would pose a severe risk to power grids and telecommunication networks but wouldn't pose an immediate threat to life.
If a large solar flare can release 10e32 ergs of energy, is there an energy level where a solar flare or CME would pose an immediate threat to any life on the Earth that doesn't live below sea level.

Abrevaya et al. (2020) in "The UV surface habitability of Proxima b: first experiments revealing probable life survival to stellar flares" (abstract, PDF) say:
"Our results show the impact that a typical flare and a superflare would have on life: when microorganisms receive very high fluences of UVC, such as those expected to reach the surface of Proxima b after a typical flare or a superflare, a fraction of the population is able to survive. Our study suggests that life could cope with highly UV irradiated environments in exoplanets under conditions that cannot be found on Earth. "
That suggests that total-extinction conditions aren't found on Earth, either.  The paper has a lot more detail of their assumptions and modeling, but that's my quick takeaway from the abstract.


Would a human body float in the dense atmosphere of Venus? To survive high in the atmosphere of Venus, all you would nead to wear is suit that protects you from the sulfuric acid vapors in the air and a supply of breathable air. Assuming (for simplicity) that this gear has the same density as the human body (the air supply would probably much heavier), could you float on top of denser gas layers below you?

I believe the density of the gas would only have to be about as high as the density of water on the surface of the sea. And would that mean that when you pour out a bottle of water, the water would float in the air (given the temperature/pressure ratio is below the boiling point)?

No. According to the Nasa Venus fact sheet the density of the atmosphere at the surface is ~65 kg/m^3.

For comparison, water is 1000 kg/m^3. We only just float in water.

So if you were there, in a suit that could somehow withstand the heat (464 C), the "air" would feel thick.

Maybe a person could strap on some wings and fly ...


How long is the Sun closer to Earth then Mars is? A recent question at Space.se has generated a few comments that are interesting.

But the answer to one particular comment to might address a lot of misconceptions.

  I imagine the actual halfway point is close to the sun Source

As the two planets orbit the sun at different speeds, I assume it must be a significant amount.

How long is the Sun closer to Earth then Mars is?

Let us first consider the Earth a fixed point and Mars moving around the Sun on a circular orbit with angular velocity equal to the relative angular velocity ($\omega_\bigoplus - \omega_♂)$
The distance between Mars and the Earth can be described as the square root of $R_E^2+R_M^2-2R_ER_Mcos(\theta)$, and when this is equal to $R_E^2$:  $cos(\theta) = \frac{R_E^2-R_E^2+R_M^2}{2R_ER_M} = 0.75$, meaning this happens when the elongation of Mars is around 0.72 and -0.72 radians (40 and -40 degrees). For smaller angles than 40, the cosine is bigger making the distance smaller (same goes for angles between $-40$ and $0$). Therefore, Mars is closer for $80/360 = 22$ percent of the synodic period of Mars, and for the remaining 78% or around 600 days.


Can two planets in an empty universe meet/be pulled together? For this question assume that the entire universe is completely empty. The universe is not expanding or contracting, it is completely motionless since time immemorial and has never moved before.

Only two identical earths without moon are left in this entire universe and they are 5 billion light years apart. From their starting position the earths do not spin or move at all they are completely motionless.

Would these planets ever meet or at least be pulled together?  Or are the their respective gravities too small to have any meaningful effect?

Yes, they would experience gravitational attraction. It would take a long time for them to collide... the formula (derived here and shown here) is:

$$t = \frac{\pi}{2} \sqrt{\frac{d^3}{2G(m_1+m_2)}}$$

where $d$ is the initial distance between the two planets of mass $m_1$ and $m_2$ and $G$ is the Gravitational constant. This gives a time of about $10^{23}$ years, much much longer than the universe has existed. This assumes Newtonian mechanics. Relativity would not change the conclusion much.

There is no known upper limit to gravity, and plenty of indirect evidence that it has no upper limit.


76mm reflector vs 70mm refractor? Both are at sale at $200 I'm starting my astronomy hobby and looking for a beginner telescopes to start with. I've searched couple websites and found these 2 within my budget.

I know for a fact that refractors gives you sharper and clearer images but its original price is actually cheaper than the reflector telescope. I will list the specifications below. It would be really appreciated if you guys can tell me the relative benefits and drawbacks of each.

Refractor: Saxon 70mm Telescope

OPTICAL DESIGN: Refractor

APERTURE: 70mm

LOWEST PRACTICAL POWER: No

HIGHEST PRACTICAL POWER: 165x

FOCAL LENGTH: 400mm

FOCAL RATIO: F/10.0

EYEPIECES: 20mm & 10mm

FINDERSCOPE: 5x24

DIAGONAL: 1.25" 45° Erect Image Diagonal

MOUNT TYPE: Alt-Azimuth

TRIPOD: Aluminium

This telescope includes a pre-assembled tripod, two 1.25” eyepieces, a cleaning cloth and a bonus padded backpack.

Reflector: Saxon 767AZ Reflector Telescope

OPTICAL DESIGN:Newtonian Reflector

APERTURE: 76mm

LOWEST PRACTICAL POWER: No

HIGHEST PRACTICAL POWER: 152x

FOCAL LENGTH: 700mm

FOCAL RATIO: F/9.2

EYEPIECES: H25, H12.5 and SR4

FINDER-SCOPE: 5x24

BARLOW LENS: 3x

DIAGONAL: No

MOUNT TYPE: Alt-Azimuth

TRIPOD: Aluminium

OPTICAL TUBE DIMENSIONS: 12.5 cm x 68.5cm

OPTICAL TUBE WEIGHT: 1.76kg

SHIPPING DIMENSIONS: 74 x 25 x 23cm3

SHIPPING WEIGHT: 5.4kg

The accessory pack includes three 1.25" eyepieces, a finder-scope, an erecting eyepiece for terrestrial viewing, an adjustable aluminium tripod with accessory tray and a moon filter, which reduces the brightness of the moon and makes surface details sharper and clearer.

Personally, I would go for neither of those. $200 is a bit steep for such a small refractor or reflector telescope (I am assuming this is USD?)

For a budget of $200, I would be looking at some second hand telescopes, as you will likely be able to pick up a 90mm refractor or 130mm reflector second hand with that budget.

It also depends what you want to do with it. For simple observations of planets and the moon, it will be fine, but don't expect to see much detail with such a small aperture. The optics on these aren't fantastic either so DSO's will not look great, although some of the brighter ones will be viewable.

So my answer would be don't get either of them, shop around second hand places and most importantly, read reviews on telescopes before buying them. If you can, join an astronomy group in your area and have a look through their scopes and see how you get on.


How are pulsars detected from Earth? This paper on Arxiv indicates that the signal period of known pulsars is 1 ms-15 s (0.07-1,000 Hz since F=1/T). I have an EE background and it's a long story how I ended up here, but suffice it to say I'm weak on astronomy.

This chart on Wikipedia implies that the entire range of pulsar frequencies should be absorbed by the atmosphere, but this guide and even this budget RF telescope writeup clearly state that pulsars can be detected from Earth's surface.

My questions is: how are pulsars detected from Earth--what am I missing here? Are RF telescopes simply that receptive? Any insights appreciated.

the signal period of known pulsars is 1 ms-15 s (0.07-1,000 Hz since F=1/T).

That's the frequency of how often the signals occur, not the frequency of the waves they emit. Just compare it to a lighthouse, which has a signal period of a few seconds but a wave frequency of $10^{15}$ Hz.

The first known pulsars, for example "Little Green Man" or PSR B1919+21, were radio pulsars; your graph shows those waves are detectable from Earth's surface. I stumbled upon this article which lists the frequency of some of the pulsars:

  After the discovery of pulsars at 81.5 MHz (Hewish et al. 1968), the first pulsar detected at 40 MHz was PSR B1133+16 (Arecibo Observatory, 1968, unpublished). In 1969–1970 the pulsars PSR B1133+16 and B0809+74 were detected at 25 MHz (Bruck 1970).

Below is the full detection range for the telescopes:

  Therefore joint observations (simultaneous or not) will be highly beneficial, providing a continuous frequency coverage from 8–240 MHz.

which is roughly equivalent to wavelengths of 40m to 1m.

X-ray pulsars like Centaurus X-3 were discovered later, and not by telescopes on the surface:

  Centaurus X-3 was first observed during experiments of cosmic X-ray sources made on May 18, 1967. These initial X-ray spectrum and location measurements were performed using a sounding rocket.


Does the summit of a small mountain offer better seeing? I have a small amateur telescope, and would like to defeat the atmospheric seeing conditions as much as is possible.

Fortuitously, I happen to live fairly close to this lovely patch of bare elevated rock:

It's a granite peak about 3,155ft (960m) above sea level.

Would this peak be expected to have improved seeing, as compared to the seeing from an adjacent valley on the same night?

In terms of steadiness of the air and hence stability of the image (ie ability to see fine detail on planets or split close double stars), it could go either way, depending on the atmospheric conditions, including wind speed and direction, and the local topography.

In contrast, valleys can sometimes provide better seeing because they provide some protection from atmospheric conditions.

However, and all other things being equal, you will probably have a higher limiting magnitude (ie be able to see fainter objects) from the top of the mountain.


How to reduce star trailing when taking long exposure photographs? My telescope's sidereal tracking is lousy: it just doesn't work. To overcome this, when taking long-exposure photographs through it, what's the best way to reduce star trailing?

Sounds like you should get a new mount. :-)

Being more serious, even expensive mounts have periodic tracking errors (due to manufacturing tolerance of gears and moving equipment) that need to be corrected. These are the basic ways to do the corrections:

Add a guide scope to the side of your scope, use an eyepiece with cross-hairs, and watch a star while you take a photo. When the scope drifts, you push buttons to move the scope to keep the star centered.
Do the same thing with a guide scope but use a dedicated guide camera that provides the corrections to the tracking.
There are also guide cameras that you can insert just in front of the camera. They use a small "pick-off" mirror to look at a star in the field (off axis guider). The guide camera provides the corrections to the tracking.

Of course, all of these methods require a mount that can be moved slowly and smoothly in both right-ascension and declination to keep the guide star centered. Also, the mount needs to be an equatorial mount and accurately aligned. (More expensive mounts can be alt-azimuth mounts and use a gadget to rotate the camera during the exposure.)


Can we use Breakthrough Starshot spacecraft to detect any unknown massive object in our solar system? For example, the hypothetical Planet Nine.

Since we just want to prove or rule out the existence of Planet Nine in our solar system, we can avoid most of the technical challenges for StarChip (camera not required, lower propulsion power, etc). So StarChip would become much cheaper, and we are able to launch thousands of the probes, covering a large area of the sky. If a massive object exists within the area, it will affect nearby passing probes and lead to trajectory anomaly, which maybe detectable and the trajectory data can be used for calculating the possible position of the object, and then we can do some follow-up observations and studies using telescopes.

Actually we can use other probes, but StarChip seems cheaper and has faster speed so that we don't have to wait too long if the targets are thousands of AU away from us.

So could this technique be used for detecting Planet Nine or any other unknown massive object in the solar system?
What's the appropriate speed for StarChip so it can accumulate enough gravitational effect to be detectable?

This is an interesting question but the answer is that this is not feasible nor efficient at all.

You can regard each photon of light coming from outside the Solar System acting as a probe. We can measure the deflection angle of them by analysing the gravitational microlensing effect a planet would produce on them. For that you need two things; a planet massive enought to be spotted by this effect (predicted mass for Planet 9 is in place) and enough photons coming from different directions to have a chance of some of them passing close to Planet 9. And here's the thing: the net flux of incoming photons from all the stars and galaxies from all directions into the Solar System is still not enough to have even the slightest chance of some particular ray to be deflected or blocked on its way to an observer on Earth that would be just looking in that precise moment in that precise direction (as Planet 9 misaligns with that particular star from which the photons are coming after a few minutes). This could be changed by continuous observation of the entire sky, so maybe when LSST opens in 2022 we might have a chance to apply this method of detection.

So, if all the photons coming to Earth from outside the Solar System could be used as natural probes that might show Planet 9 by microlensing or occultation, immagine how puny would humanity's effort be in comparison if we tried to launch even just a fraction of a millionth  of the amount of photons currently traveling towards us from all directions.

There's also the inverse method (which is more similar to the 'launching probes to the outside' idea): photons coming from the Sun (which are a lot more than those coming from the stars and are been "launched" in all directions with higher coverage of each square degree of sky) could hit Planet 9. We need the information of those "hitting probes" to get to Earth and this is difficult, many of our solar photon probes are going to be reflected in directions unaccessible to us and many more are going to be absorbed by the atmosphere and surface materials of Planet 9. But since the Sun is emitting so many photons a few percent might actually come back to the inner solar system in just a few days. Those reflected photons are our best chance and indeed they are the ones we are searching for when we try to find direct evidence of Planet 9.

No human attempt to launch trillions of probes to the region of the sky we expect to find Planet 9 (even if we manage to reduce it by many orders of magnitude) will be as productive as the Sun launching so many photons continuosly and Planet 9 reflecting them back to Earth. To compete with that we would probably need more mass than the mass of the entire Solar System to make the tiny probes and no gain at all compared to the reflected photons that are in place right now.

I guess there is a minor detail we could talk about: reflected photons have to actually hit the target and our tiny probes could just pass close to it and signals would reveal, by some Doppler shift, a change in speed and thus the gravitational influence of the body. So actually we have to talk about the cross-section of the planet in a different way in both situations. For the photons coming from the Sun to be reflected back the considered cross-section would be the cross-area of the planet which for a planet the size of Neptune is $1.9 \cdot 10^9 \; km^2$. In the case of the nano-probes the cross-section can be expanded to the region where the gravitational influence of the planet ceases to be sufficiently large to perform a detectable change in speed for the tiny probe. I'm not going to calculate this but let me make a liberal guesstimate by using the expected Hill sphere for Planet 9, which is $15 \; AU$ in radius and thus would make for a cross-sectional area of $1.6 \cdot 10^{19} \; km^2$. That might be great for having some chances but the probes would have to be moving extremely slow for the gravitational influence to be able to operate for enough time as to make the deflection be apparent, and this is a constraint that makes the project again unfeasible since our probes would be moving at some percents of the speed of light (if not then we would have to wait centuries or even millennia for the experiment to be completed).


How is it known that Pillars of Creation are destroyed? We can observe Pillars of Creation with 7000 years delay, but it is destroyed only 6000 years ago. How do we know that? How the information about explosion reached the Earth before light do?

Much of the evidence for the destruction of the Pillars of creation lies in observations of dust clouds in the region (Flagey et al. (2009)). The morphology and structure of a cloud in the dust are suggestive of a shock wave, which seems to be indicative of a recent supernova. The shock waves from supernovae do often interact with both the interstellar medium and local gas clouds. The original observations supporting this hypothesis were made by the Spitzer Space Telescope.

It should be noted that the SST uses infrared light, and as the dust wasn't seen in the visible portion of the spectrum, we wouldn't necessarily have realized from the Hubble photos that there were any effects from a hypothetical supernova.

Some dissenters believed that something completely different is responsible for the behavior of the dust - possibly the winds of massive stars. Now, more recent x-ray observations (see e.g. Guarcello et al. (2012)) have failed to see either the x-ray spectra expected of O-type stars or the x-ray emission that would be indicative of a supernova, leading some to conclude that the Pillars of Creation have not yet been destroyed.


Do photons slow down as they approach the event horizon of a black hole? So, if a black hole is so dense, that light cannot escape, what happens to the photons as the approach and go past the event horizon (or some other horizon, about which I am ignorant)?  Do photons slow down at some point, or is it somehow binary, where at one moment they are moving at the speed of light and the next they aren't moving at all?  Or, do they continue to move at the speed of light, just never moving beyond the black hole?

Photons travel on null geodesics through space time. In the curved space-time near a non-spinning black hole, a distant observer could infer that the speed of light (travelling on a radial path) is given by
$$ v_{\rm light} = c \left(1 - \frac{r_s}{r}\right),$$
where $r_s$ is the Schwarzschild radius.

As $r \rightarrow r_s$ then we see that $v_{\rm light} \rightarrow 0$. So yes, according to an observer far from a black hole, the speed of light approaching a black hole appears to slow down and asymptotically approaches zero at the event horizon.

However, the locally measured speed of light is always $c$.


Can a laser pointer ruin a long exposure photo? I am a hobby astronomer and usually out with people that aren't familiar with the night sky.
In order to help them find constellations or show them stuff in the sky, I was thinking of buying a laser pointer.  Now because I do not want to ruin any photos of people that might be around my observation place, do some of the more experienced folks of you know about the effects of these laserpointers to photos?

Can laser pointers be used?
Yes and no.  If you are attending a Star Party hosted for astronomers and astrophotographers, then laser pointers are usually among the items listed that you may not use.  Some star parties hosted for astronomers (not the public), may schedule an event when a tour of the night sky (via laser pointer) is performed and all astrophotographers know this and aware that they may not want to be imaging during that particular session.  Laser pointers are not used or permitted at any other time.
If the star party is being held for the public (not for astrophotographers) then laser pointers are often used during sessions on each night and astrophotographers know that this.  They might be used at any time and without warning.
Will they ruin images?
This is a "it depends" answer.  Laser pointers used by amateur astronomers are green (typically 523nm) lasers.  This particular wavelength is used to because human eyes are very sensitive to it (it is smack in the middle of our visual spectrum) and it reflects off the particles in the air showing an obvious beam.  Other colors can be used... but unless the laser is especially powerful, the beam isn't visible.
Professional astronomers will use lasers to create an artificial star for purposes of adaptive optics.  These lasers excite gasses in the sodium layer (about 80-100km above sea level) and since the gas is sodium, they use a sodium laser to excite these gasses to create the artificial star.  Sodium lasers are yellow.  These are especially powerful lasers.
Meanwhile, the lasers amateur astronomers use are typically not very strong.  Though laws & regulations may vary by country or local government rules, these tend to be fairly low power lasers.  Here in the United States they are typically 5 milliwatts or less.  It turns out your ability to see these lasers really depends on how close you are to the source.
A 5mw laser is easily visible to people just standing a few feet away, but people a bit farther away will struggle to see it.  People on the far side of a field wont see it at all.
When I do astronomy outreach, I typically use a 5mw laser and those standing near me can see the beam.  If I were to do a tour of the night sky using a laser for a moderately larger crowd (say... 25-50 people) then I'd need a stronger laser ... maybe 50mw ... otherwise those farther away will struggle to see the beam.
If you're out with a few friends and worried you might be ruining images for someone taking images 10 blocks away... no worries.  Their cameras will not detect the laser beam.
Safety
As always, use proper care when operating a laser and avoid pointing them at people or aircraft.  This keeps the local constabulary from having to deal with the situation.


When was the last eruption on the Moon? The lunar mare are basins that filled up with magma over the course of the Moon's history. So there were eruptions on the surface of the Moon at some point in time. When did the last eruption happen on the Moon?

A long time ago.
The major mare forming eruptions occurred more than 3.5 billion years ago, but a paper  Ages and stratigraphy of mare basalts finds that some volcanic basalts are as young as 1.2 billion years. However, it seems that impacts have been the only active geological process for a billion years, giving rise to the observation that just as complex life on Earth got started, the moon (geologically) died.


If a massive object like Jupiter flew past the Earth how close would it need to come to pull people off of the surface? I understand this is a silly hypothetical but I'm asking for a 7 year old so please bear with me.

Imagine an interstellar stray gas giant comes flying through our solar system.

If we were not concerned that it would also steal our atmosphere and create tidal forces that destroyed everything... How close would it need to come to us to exert enough gravity to lift people off the ground and pull them into its own orbit?

TL:DR Jupiter isn't dense enough for its gravity gradient over Earth's radius to produce a 1g tidal acceleration, even right at Jupiter's surface.

thanks to PeterCordes

Jupiter's gravity will pull on the Earth itself, as well as everything on it.

It's not like a vacuum cleaner that selectively lifts small and light objects, the gravitational force will scale with the mass of each object; if the Earth is a zillion times more massive than we are, then Jupiter's gravitational force will also be about a zillion times larger.

What that means is that Earth will accelerate towards Jupiter, and we will accelerate along with it, and so we won't "feel the tug" anywhere near as strongly as one might suspect.

Instead, let's think about the size of the Earth, and the fact that people on the near side will be closer to Jupiter than the center of mass of the Earth, and people on the far side will be farther away.

Since people nearer to Jupiter will feel a slightly stronger acceleration than the center of mass of the Earth, they will feel a quite gentle tug. We'll calculate that in a minute.

But believe it or not, people on the far side of the Earth, feeling less of a tug than the Earth's center of mass, will believe they are being pulled in the opposite direction! They won't really be pulled away from Jupiter, but they will not accelerate towards Jupiter as fast as the Earth, and so it will feel like they are being repelled.

This kind of force is called a tidal force and this is the picture that's often used with the concept:

 Source Replace "Satellite" with "Jupiter"

The acceleration we feel due to gravity is expressed as

$$a_G = \frac{GM}{r^2}$$

where $G$ is the gravitational constant and equal to about $6.674 \times 10^{-11}$ m^3/kg s^2 and M is each mass that's pulling on you.

If you put in 6378137 meters and the mass of the Earth ($5.972 \times 10^{+24}$ kg) you get the familiar 9.8 m/s^2.

If Jupiter were 114,000,000 meters or 114,000 kilometers away, the Earth would accelerate at 1 g towards it, but people on the close and far side would accelerate very differently. On the close side, being 6,378 kilometers closer, would feel an acceleration 1.2 m/s^2 greater, so they would feel that they weighed 12% less. And people on the far side would also feel about the same amount lighter because they felt less acceleration than the Earth.

If Jupiter were so close that it were practically touching the Earth, it still wouldn't pull is off of Earth, assuming that Earth remained intact. But that wouldn't last very long!!! Earth would be accelerating towards Jupiter at about 20.9 m/s^2, and people on the near side would feel acceleration of 24.8 towards Jupiter, but relative to Earth that's only 3.9 m/s^2, so not enough to overcome Earth's gravity of -9.8 m/s^2.

On the far side of Earth it's similar; the acceleration towards Jupiter would be 17.8 m/s^2 but minus Earth's acceleration of - 20.9 it's -3.0 m/s^2 away, but that's also not enough to overcome the attraction to Earth of in this case +9.8 m/s^2.

When Earth touches Jupiter, we will feel about 40% lighter on the near side and 31% lighter on the far side of Earth, but we would not leave the surface.

However, in just minutes we'd be pulled so deep into Jupiter that we would be crushed by Jupiter's internal atmospheric pressure.

It would certainly be fun, but it wouldn't last long!


If a comet strikes a glacier, does it make a print on the land below? The Younger Dryas Impact Theory describes an asteroid impact in Northern America, possibly the eastern part of Hudson Bay, which contributed to the end of the last ice age.

If a comet strikes a glacier, would it make a print on the land below? If specifically the comet that made this impact crater in Hudson Bay, and a glacier that was maybe comparable to Antarctica so 2 km thick.

While it's a clever suggestion, the answer is almost certainly that they are not related because the Nastapoka Arc "crater" is too big.   It's 450 km in diameter.  By comparison, the Chicxulub crater is 180 km.   Even if an impact through mile of ice could, somehow, spread the impact out wider, the proposed Younger Dryas Impact object was almost certainly too small to leave a 340 km crater.

There's also the problem that glacial ice carves the landscape and a neat circular crater is unlikely, even if the size had been a closer fit, an impact crater that was run over by a glacier should look disturbed by the glacier, not a neat circle visible on the surface.

What geological evidence might remain for a large impact on a glacier is an interesting question given that the ice would significantly buffer any impact and continue to reshape the landscape below it as the glacier moves and/or melts and possibly floods the region.

Mile high ice glaciers are relatively rare events, both localized and periodic, and large meteor impacts are also rare events.   A combination of the two, large enough to leave clear evidence through the ice, would at best, be a hugely rare event and there's a good possibility that the right combination of impact and ice hasn't happened at all over the last 2.5 million years (northern Hemisphere) or on Antarctica (3% of the surface area of the earth) - last 20-30 million years.

I think the safer bet is that there's no impact crater from the proposed Younger Dryas impact specifically because of the thickness of the glacier.   It remains a debated hypothesis, so there's some uncertainty, but Ice should significanly buffer all but the largest impacts.

For Younger Dryas, there should be other physical evidence such as debris from the impacting meteor or comet, but debris cast over a glacier that first re-freezes over and later melts can be spread very far and be very hard to identify.   Much of it could have washed into the ocean.   A somewhat related example, it took 105 years before debris from the Tunguska event was identified.

(and this question should probably be moved to Earth Science).


Determine north just by one shadow I am not sure if this is the right place to post at, but I tried to find the best place to do so.

To my point, I wonder if it is posssible, to determine where north is, just by one shadow, that means no Sun, no time, just a shadow in a picture or video...

Do you get my point? Like this picture, can I tell where north is?

Like this image:

With only one shadow in a picture, no, you can not.

In order to get North's position from a shadow you need to know where you are and which solar time and date is it there. These are two extra data and you can not disentangle them, even if the picture is perfect.

e.g. suppose a picture in which you see exact measurements of shadow and people, and they are both exactly of the same size. This means Sun is 45º above Horizon. But that may be local noon at Bourdeaux on March 20th (thus shadow pointing North), or local noon at Christchurch on same date (thus shadow pointing South), or local 3PM solar on Quito on March 20th again (thus shadow pointing East).

I chose March 20th because it is easier to visualize where the Sun would be (exactly over the Equator), and these cities due to their latitudes:

Bourdeaux: 44º 50' N (almost 45ºN)
Christchurch: 43º 31' S (almost 45ºS)
Quito: 0º 13' N (almost on Equator)


Why is the Oort cloud presumed to be spherical? Most descriptions of the Oort cloud depict it as a mostly spherical distribution of planetesimals, with occasional allowance for an inner component that is more donut-shaped. This is slightly at odds with the fact that most protoplanetary clouds and their derivative objects - planets, asteroids, comets and dust - will collapse to a fairly well-defined plane relatively early in a stellar system's evolution.

What evidence is used to postulate this? Does it come from numerical simulations of the solar system? Or does it help account for observed orbital inclinations of real comets?

The latter.  Long-period comets appear to come from random directions.


How actively is Deep Learning being used in exoplanet identification? Papers, like this one from Google, show that Deep Learning and Machine Learning are effective in detecting exoplanets from luminosity data provided by telescopes like TESS or Kepler.

These models seem extremely quick and lightweight. Has it become somewhat of a standard practice to use these on freshly obtained TESS data? If not, why, and what is done with the data instead? Are there drawbacks to using these seemingly accurate models?

Also called Astronet-K2, the deep learning A.I is efficient at spotting data that could find exo-planets, but that's were the issue is. That's all it is capable of doing, so it misses or ignores other potentially interesting data that human astronomers would classify as curious or odd, and could lead to other potentially new discoveries. It's missing the human factor, more or less. In both the source linked in the question and mine, it has only found 2 new exo-planets, so it doesn't look like is has free reign In the field. Even if the A.I is currently used automatically, it's work would be monitored and checked by humans.  My source material was from 2019, so if I missed a more recent accounting of the system that has different or upgraded methods, please let me know.
https://www.google.com/amp/s/www.technologyreview.com/2019/04/01/136239/deep-learning-has-found-two-exoplanets-that-human-astronomers-missed/amp/


From which country or area is the new moon visible first? Being able to see the Moon is dependant on various factors like sunset timing etc. Considering all these factors, which country or area of the world sees the new moon first? I am asking about the new moon.

The duration of one lunation (the period between one new moon to the next one) isn't neither constant as the Moon rotates around the Earth and it around the Sun (changes between 29.272 and 29.833 days due to the perturbing effects of the Sun's gravity on the Moon's eccentric orbit), nor integer divisible by 24 hours or one Earth's rotation around its axis. So this position of the Moon on the skies where the next new moon as the first of its lunar phases will be first observable constantly changes.

     Phases of the Moon, as seen looking southward from the Northern Hemisphere. The Southern Hemisphere will see each phase     rotated through 180°. (Source: Wikipedia on Lunar phase)

Saying it differently, by the time the Moon completes one lunation (or it's synodic period) and starts the next one, it won't be positioned exactly above the same Earth's longitude (East to West) as the one it started at.


Milky Way position on the sky I'm looking for some sort of boundary data to be able to render the milky way on a star map, as visible from Earth. Something that looks like this:

For that, I need something like a collection of RA hours and DEC degrees of the "boundary points" of what's visible from Earth (technically the Galactic Center), possibly with proper motion too. I'm not looking for precise luminosity data or anything like that, just the points of the blob on the sky that most resembles the Milky Way's shape and position from the Earth. It's important that I want to render the sky map for any given surface point on Earth, for any given time (within the last 100 years at least).

Do you know of a database like that? I've been looking on VizieR but I couldn't find what I was looking for.

There is a very nice project called d3-celestial by Olaf Frohn on github. In contains a data file describing the Milky Way as polygons, see here. A demo showing this Milky Way can be found here. And even better, the source for this data is cited, pointing to the Milky Way Outline Catalog by Jose R. Vieira.

Depending on your project, the json format from d3-celestial might be easier to read than the one from Jose R. Vieira.
Note that you don't have to worry about these "contours" moving on a time scale of hundred years, but this is another question.


Naming of the planets of the solar system Planets of the solar system have been named with Roman mythology gods names. I have a few questions on my mind concerning that subject for a while:

Who decided to name them like this?
When did the planets got named?

Finally, the ultimate question is: if the planets got named by the Romans, implying they had not technology to observe the solar system whatsoever, why some of the names match the color of the planets?

Mars is red - Mars is the god of war
Neptune is blue - Neptune is the god of the oceans

Or maybe I am wrong (most likely) and the planets have been named more recently.

Despite of the explanation of your question, it is a valid question to ask why planets all have Roman names. First of all, the Romans could, like the Greeks and Sumerians, could only see Mercury, Venus, Mars, Jupiter and Saturn. These planets can be seen with the naked eye. However, the fact that the Romans could see these planets, didn't give all the planets latin names.

If the Names of the planets were given by the ones that discovered them first, the planets would have Sumerian names. The answer to the question why the planets have latin names is because they don't. The first western scientist took over the Roman names, because latin was they scientific language of the middle ages and the renaissance. After the discovery of Uranus and Neptune these planets were not given these names by their discoverers. Only after longer controversy western astronomers have standardized the names of planets and moons according to Greek and Roman mytholohy, but this was only around the second half of the 19th century.

Since the beginning of the 20th century the IAU is setting the standards for naming celestial bodies including planets. However, the IAU doesn't set the names, it only sets the standards. Newly discovered objects are now named according to these rules by their discoverers. Older planets are still living with their old names and for western astronomers those are still the names from Greek and Roman mythological figures. However, the IAU allows you to consistently name the planets by their Arabic or Chinese names. So, the question is valid, the text around your question is unfortunately not true, since the planets don't necessarily have Roman names. If you are An Asian person and correctly think the planets should have their Chinese names, because they were earlier in discovering these objects than the Romans (although not earlier than the Sumerians), you are allowed by the IAU (sic) to call them by their Chinese names.


Is the sun any brighter during a solar eclipse I keep hearing that you can go blind during a solar eclipse.

Is the sun actually any brighter during a solar eclipse? This is how the media seems to frame it, and they suggest you need special glasses to view the eclipse.

But after some reading, it seems that the warning is about just not staring at the sun in general, and there is nothing special about the eclipse.

Is this the case?

The Sun is the same as ever. It's just that people tend to stare at it for long times during the eclipse. A quick glance doesn't do much damage, but prolonged staring could be bad.

Those who see the phase of totality have an additional risk. After the Sun goes completely dark, and it then finally shows up again, some people feel compelled to keep watching that tiny sliver of light without protection for a long time. Don't. When the actual surface of the Sun is visible again after totality is over, it's time to put the googles back on.

Do not use regular sunglasses. They typically don't block the kinds of radiation they should block to allow you to watch the event safely. There are several options here:

Very dark welding glasses. Ask a welder or go to a welding shop. They have these special glasses to protect themselves from the radiation from the welding arc, which is similar to the Sun's radiation. These glasses are numbered to show how dark they are. For viewing the Sun, use the darkest ones, labeled #14. A welding mask with a #14 glass is perfectly safe no matter how long you're looking at the Sun. #13 glass might be safe too for short durations, but it's not recommended. Anything lower than that is not safe.

Goggles made specifically for watching the Sun. There are many kinds of them. As long as they pass the ISO 12312-2 international safety standard they're fine.

I recommend the goggles made of Baader solar film. They show the Sun in its original color (not red, green, orange, etc). There are several kinds here, too; the Baader AstroSolar Silver/Gold Film passes the ISO standard, these are what I use. You can buy them online in various places and they're very cheap.

More info:

https://eclipse2017.nasa.gov/safety

https://eclipse.aas.org/resources/solar-filters


Why are stars so far apart? So, this is mere musing, but it seems that stars are quite extremely far apart. I tried to determine the mean distance between nearest neighbors for stars (in just our galaxy) but I'm not sure what it is, though a first approximations I would guess that the mean is around 5 ly with some 0.5 standard deviation or so!

However, assuming I didn't screw up, the distance between two sun-massed objects required to generate 1,000N of force between them is an incredible 54.3 million ly. [Edit] If you change that to 1 billion Newtons, the distance is one galactic radius -- 54000~ ly!

I'm probably mis-estimating how much even 1kN of force would effect solar trajectories over time, but I still can't help but feel that our solar neighbors are very far away given the spheres of interaction that I would expect.

Any insights?

Most stars are of a solar-mass or below. The average number of companions that each stars has (in the sense of being part of binary or higher multiple systems) systems ranges from 0.75 for stars of a solar mass to approximately 0.35 (not a well-established number) for the more numerous M-dwarfs. Let's take a compromise value, say 0.5.
The separation distribution of these multiples peaks at around 50 AU for solar-type stars, reducing to about 5 AU for low-mass M dwarfs. Again, lets take a compromise value of 20 AU. See Duchene & Kraus (2013) for all the details.

So if we take 1000 stars, then 333 of them (roughly speaking) are companions to another 333 stars, while 333 are isolated single stars. (NB This does not mean the frequency of multiple systems is 50%, because some of the companions will be in higher order multiple systems)

Thus, taking your calculation of the separation between stellar systems of 5 light years ($= 3.2\times10^{5}$ AU), then the mean separation is:
$$\bar{D} = 0.667\times 20 + 0.333 \times 3.2\times10^{5} \simeq 10^{5} AU, $$
but the median separation is 20 AU!

This is of course sophistry, because I'm sure your question is really, why are stellar systems so far apart?

Stars (and stellar systems) are born in much denser environments. The number density of stars in the Orion Nebula Cluster (ONC - the nearest very large stellar nursery) is about 1 per cubic light year. The equivalent number for the solar neighbourhood is 0.004 stars per cubic light year. Thus the average interstellar separation in the ONC is 1 light year, but in the solar neighbourhood it is about 6.3 light years.

The reason for this separation at birth is the Jeans length - the critical radius at which a clouds self-gravity will overcome its thermal energy and cause it to collapse. It can be expressed as
$$\lambda_J = c_s \left( \frac{\pi}{G\rho } \right)^{1/2},$$
where $c_s$ is the sound speed in a molecular cloud and $\rho$ its density. For star forming giant molecular clouds $c_s = 0.2$ km/s and $\rho=10^{-23}$ kg/m$^3$. So clouds of scale hundreds of light years could collapse. As they do, the density increases and the Jeans length becomes smaller and allows the cloud to fragment. Exactly how far the fragmentation goes and the distribution of stellar masses it produces is an area of intense research, but we know observationally that it can produce things like the ONC or sometimes even more massive and dense clusters.

From there we know that a new born cluster of stars tends not to survive very long. For various reasons - outflows, winds and ionising radiation from newborn stars are able to heat and expel the remaining gas; star formation appears to have an average efficiency of a few to perhaps 20-30%. Expelling the gas, plus the tidal field of the galaxy breaks up the cluster and disperses it into the field, which gives us the lower field star (or system) density that we see around us.

Once the stars are part of the field they essentially don't interact with each other; they are too far apart to feel the influence of individual objects and move subject to the overall gravitational potential of the Galaxy.
 so your consideration of the force between stars is not really relevant.


How do scientists know there are about 300 billion stars in a galaxy and there are about 100 billion galaxies? Just to think about this is mind boggling. But how do scientist get these numbers? What technology/system/theory do they they use?

The way it works is as follows. We do detailed studies of stars in the solar neighbourhood. This establishes the local density of stars and the mix of masses they possess (called the stellar mass function). We compare that with the mass function of clusters of stars and note that to first order it appears invariant.

We can then triangulate the problem in various ways: we can make a model for the stellar density of the Galaxy, assume it all has the same mass function and hence get a number of stars. The model may be based on crude light-to-mass conversions, but more often would be based on deep surveys of the sky - either narrow pencil beam surveys from HST, or broader surveys like SDSS, The key is to be able to count stars but also estimate how far away they are. This is highly uncertain and relies on some assumptions about symmetry to cover regions of our Galaxy we cannot probe.

Another method is to count up bright objects that might act as tracers of the underlying stellar population (eg red giants), compare that with the number of giants in our well-studied locale, and from this extrapolate to a total number of stars, again relying on symmetry arguments for those bits of the Galaxy that are distant or obscured by dust.

A third way is to ask, how many stars have lived and died in order to enrich the interstellar medium with heavy elements (a.k.a. metals). For example, it turns out there must have been about a billion core-collapse supernovae to create all the oxygen we see. If we assume the mass function is invariant with time and that supernovae arise from stars above 8 solar masses, then we also know how many long-lived low-mass stars were born with their high-mass siblings and hence estimate how many stars exist today.

The number, whether it be 100 billion or 300 billion is no more accurate than a factor of a few, but probably more accurate than an order of magnitude. The main issue is that the most common stars in the Galaxy are faint M dwarfs,that contribute very little light or mass to the Galaxy, so we really are relying on an extrapolation of our local knowledge of these objects.

The number of galaxies problem is easier, though the number is less well defined. We assume that on large scales the universe is homogeneous and isotropic. We count up how many galaxies we can see in a particular area, multiply it up to cover the whole sky. The number must then be corrected for distant faint galaxies that cannot be seen. The difficult here is that we are looking into the past and the number of galaxies may not be conserved, either through evolution or mergers. So we have to try and come up with a statement like "there are n galaxies in the observable universe today that are more luminous than L". I think this number is certainly only an order of magnitude estimate.


In Chromoscope, what is the X-ray source between Ophiuchus, Libra and Scorpius? Chromoscope is a nice toy. Pressing 'L' shows the constellations as guide.

There's a bright X-ray source between Ophiuchus, Libra and Scorpius. In Hydrogen-Alpha there are two circular formations to the right and left of it. Are the latter formed by polar jets from the former, and what is that object?

The source is Sco X-1. It is a point source of X-rays.

I'll have get back to you on why it looks that way in chromoscope, but it looks incorrect to me.


New NEOs are found daily. Is there any evidence gravity is pulling them to a centralised location? With all of the new NEOs (near-earth objects) are they being pulled in or are they interacting with each other?

No, they are being discovered where they are in their (usually fairly stable) orbits. They are not discovered because they are arriving or going to some particular location.

The one exception is long-periodic comets that have fallen in from the Oort cloud and now pass through the inner parts of the solar system, maybe to be captured by the gravity of Jupiter in a closer orbit.

(Strictly speaking all these objects are all pulled towards a central location, the Sun, but because they already have a sideways velocity that just means an orbiting motion where they are constantly falling around it.)

NEOs are too light and small to interact much with each other - the gravitational influence between them is microscopic compared to the influence from the planets which is usually microscopic compared to the sun's influence.

The reason we are seeing a lot more NEOs now is that new instruments are better and more searches are being conducted. LINEAR for example found  231,082 small Solar System bodies (of which 2,423 are NEOs) in the early 2000s and the Catalina Sky Survey has found 6,515 NEOs since 2009.


applying nutation to ecliptical coordinates (apparent position of star) I'm trying to calculate RA and DEC with nutation correction. The direct calculation in MEEUS Page 151 can't be used for stars close to celestial poles and I thought I'd turn RA/DEC into ecliptical longitude and latitude and then add delta psi to the longitude and transform back to RA/DEC. Unfortunately my calculations are wrong somewhere and I can't figure out where I made a mistake.
You can find my spreadsheet here: https://1drv.ms/f/s!AukrJ2RhCvoHh0w8MygI0qajeoXt

The standard methods of applying nutation retain equatorial components of position, and for near-polar objects they avoid the drawbacks of the abbreviated approximation by converting the equatorial components of the position into rectangular components and applying the nutation matrix for the date.  In principle this method need not involve approximation.

The topic is explained in principle in chapter 3 of 'Explanatory
Supplement to the Astronomical Almanac' (ed. P K Seidelmann, 2nd ed, 1992/2006), where the full nutation matrix is specified on p.115 in terms of classical components of nutation in longitude and obliquity, dpsi and deps.

The calculations to evaluate these components has undergone much development over the years, the latest version has a very long list of >1000 trigonometrical components, but a selection of specifications and an approximation is given in the suite of algorithms and codes available at the website for the IAU SOFA software ('Standards of Fundamental Astronomy'), at  [http://www.iausofa.org] , and of course any user can apply further shortening approximations to the series.

The IAU suite has three versions of the calculation of the nutation components in longitude and obliquity, routines iau_NUT80 (older standard),  iau_NUT00A (full and very long post-2000 standard calculation), and iau_NUT00B (abridged version of the modern calculation).  There is also a routine iau_NUMAT for calculating the nutation matrix.

(An intermediate approximation for the components of nutation, useful for applications where abridgment 2000B is too drastically approximated but 2000A is too long, is given in the US Naval Observatory's 'NOVAS' software suite (Naval Observatory Vector Astrometry software) [http://aa.usno.navy.mil/software/novas/novas_info.php]. The NOVAS suite also offers a more integrated approach to calculating astrometric problems , where IAU-SOFA provides individual short modules to be selected and assembled by the user.)

Lastly, the question doesn't mention applying a precession correction as well, but in case that is wanted (which it very often is, in conjunction with nutation), the references already given above provide the needed resources.


Solar maximum and minimum What are solar maximum and solar minimum?
Is there a relation between sun spots and solar maximum and minimum?
Thank you!

Solar maximum
Solar maximum or solar max is a normal period of greatest solar activity in the 11 year solar cycle of the Sun. During solar maximum, large numbers of sunspots appear and the sun's irradiance output grows by about 0.07%.1 The increased energy output of solar maxima can impact Earth's global climate and recent studies have shown some correlation with regional weather patterns. source

Solar minimum
Solar minimum is the period of least solar activity in the 11 year solar cycle of the sun. During this time, sunspot and solar flare activity diminishes, and often does not occur for days at a time. The date of the minimum is described by a smoothed average over 12 months of sunspot activity, so identifying the date of the solar minimum usually can only happen 6 months after the minimum takes place. Solar minima are not generally correlated with changes in climate but recent studies have shown a correlation with regional weather patterns. source


What should be the declination of a star to be marginally circumpolar given the latitude & height from ground? If I'm given the latitude, I can easily find the declination by using
declination = 90 - latitude.
I can also find the angle of dip from the height but should I add or subtract it from the declination I found?
Also I have to consider atmospheric refraction but I don't know how.

Just to be clear, the "height from sea level"  (mentioned in the title) does not change what star is circumpolar. Whether the ground is at 0 units above sea level or 1500 units above sea level does not make a difference. I think what you intended to write is the height above the ground makes a difference ("angle of dip from the height" mentioned in the question). At 0 units above the ground, a circumpolar star has a declination of D1. At 1500 units above the ground, a circumpolar star has a declination that is farther from the pole, or D1-angle. (I assume that you are in the northern hemisphere.)

Refraction raises the apparent altitude of an object, so you can see a circumpolar star that is farther from than the pole than is possible without refraction. Thus, a circumpolar star has declination = D1-angle-Refraction. The adopted value for refraction of an object on the horizon is 34 arcminutes (Ref 1: The Astronomical Almanac, 2001, page A12). The actual amount of refraction depends on the atmospheric pressure and temperature. An approximate formula for refraction (applicable for observed altitudes below 15 degrees) is R=P(0.1594 + 0.0196a + 0.00002a^2)/[(273+T)(1 + 0.505a + 0.0845a^2)] where T is the temperature (degrees C), P is the barometric pressure (millibars), and a is the altitude (degrees) (Ref 1: page B62)


Why is there a black stripe in the Hubble images of Pluto? While reading reports about the New Horizons misson, I noticed an odd vertical, black stripe in the images of Pluto.

Here is an example:

Source: Hubble Discovers a Fifth Moon Orbiting Pluto (07.11.12). Credit: NASA; ESA; M. Showalter, SETI Institute

Why is that black stripe there?

That photograph is a composite of two images taken with different exposure times.
To be correct we'd have to say that the exposure of the two photographs is different, i.e. the outer photo was created by absorbing more light. In this case we can assume that the focal ratio (derived from Hubble's lens aperture) and the luminance of the scene (how much light is travelling in the lens' direction) are identical for both photographs, which leaves only the exposure time as a free variable when it comes to determining exposure.
This is necessary because we're photographing objects with very different brightness. For Pluto to show up a relatively short exposure time is required, but its moons reflect much less light and would need a longer exposure time to visible. As long as the sensor is exposed, Pluto would continue to increase in brightness to the point that it becomes washed out. Objects that are significantly brighter become over-exposed resulting in a loss of detail and fidelity, known as blown-out highlights in photography. In our case Pluto would turn into a solid white dot compared to the more detailed picture that is now possible. You can draw a parallel with false color images rendered from infrared: this composite is not what the human eye would see if it was capable of picking up this level of light and detail.
In another Hubble image, NASA explained the reason why composite imaging is used:

This is a composite image because a single exposure of the stellar background, comet Siding Spring, and Mars would be problematic. Mars actually is 10,000 times brighter than the comet, so it could not be properly exposed to show detail in the Red Planet. The comet and Mars also were moving with respect to each other and could not be imaged simultaneously in one exposure without one of the objects being motion blurred. Hubble had to be programmed to track on the comet and Mars separately in two different observations.
Source: Hubble Sees Comet Next to Mars

Very long exposure times are often necessary since relatively little light is coming our way from distant planets and stars. As the Hubble website explains for its Deep Fields images:

Hubble has made a series of very deep observations taken in very dark parts of the sky. Like using a long exposure on a digital camera, these long exposure shots (up to several weeks) reveal very faint details that are not normally visible in shorter exposures.
Source: "What are the Hubble Deep Fields?", Spacetelescope.org FAQ.

Wikipedia summarises a paper by Robert E. Williams and the HDF team, "The Hubble Deep Field: Observations, Data Reduction, and Galaxy Photometry" as follows:

Between December 18 and December 28, 1995—during which time Hubble orbited the Earth about 150 times—342 images of the target area in the chosen filters were taken. The total exposure times at each wavelength were 42.7 hours (300 nm), 33.5 hours (450 nm), 30.3 hours (606 nm) and 34.3 hours (814 nm), divided into 342 individual exposures to prevent significant damage to individual images by cosmic rays, which cause bright streaks to appear when they strike CCD detectors. A further 10 Hubble orbits were used to make short exposures of flanking fields to aid follow-up observations by other instruments.
Source: Hubble Deep Field, Wikipedia, retrieved 2014-12-09


Energy deposition in planet Earth due to gravitational waves The gravitational wave detected by the LIGO observatory acted on planet Earth streching and shrinking it a little bit. I assume that not all of this action was fully elastic and that some energy of the gravitational wave was lost by the passage through planet Earth.

Are there some scientific estimates of how much energy was deposited in planet Earth by this gravitational wave?

Are there estimates how much energy is in total deposited by gravitational waves in planet Earth per year?

Don't know about the energy transfer but from a rough assumption we can calculate the energy amount that actually passed through Earth depending on the mass converted and the distance of the event. The energy absorbed by Earth would be far less than that obviously.

The measured event was two black holes merging with a total mass loss of $3 M_{\odot}$. While the distance for that event was calculated to $1.3*10^9 ly$.

With comparing the total energy released (assuming full conversion of mass to gravitational wave energy) over the released area:

$$
E_{Earth} = \pi r_{Earth}^2*\frac{m_{conv}c^2}{4\pi r_{dist}^2}
$$
$$
E_{Earth} \approx (6378km)^2*\frac{3M_{\odot}*c^2}{4*(1.3*10^9years*c)^2}
$$
$$
E_{Earth} \approx 4.0*10^{13}m^2*\frac{5.7*10^{30}kg}{6.72*10^{33}s^2}
$$
$$
E_{Earth} \approx 3.39 * 10^{10}J
$$

So the maximum amount of energy passing through earth would have been roughly $34 GJ$. That is about the kinetic energy of three Boeing 747 in flight - or neglectable on a planetary scale.

So even IF the total energy would have been absorbed by Earth, the amount would still have been - not much.


Is a black hole a 5 dimensional vortex? We know that a black hole behaves like a whirlpool or a tornado or any of the other rotating phenomena we experience on Earth.But the thing is, all these phenomena, except the black hole, are 2-Dimensional (plus 1 dimension of time) rotating vortices, they move objects from 2D planes to other 2D planes instantly, but through a 3D medium. My question is this, since a black hole is a rotating sphere(3D for that matter, plus 1D of time), does it mean that its endpoint in the vortex is 5D?

The short answer is no, with some caveats to the effect of sort of, depending on how loose an analogy you want to make.

Sound propagation in a fluid is limited by the speed of sound, which can be used to define a "sound cone" structure analogous to the causal light cone structure in spacetime. This is a described by an acoustric metric, which could have an acoustic horizon when the speed of the fluid exceeds the speed of sound, and even an analogue of Hawking radiation.

A general acoustic metric for a perfect fluid has the form
$$g_{\mu\nu}=\alpha^2\begin{bmatrix}-(c^2-v^2)&-\vec{v}\\-\vec{v}&\mathbf{1}\end{bmatrix}\text{,}$$
where $\alpha$ is a conformal factor. The Schwarzschild black hole can be put in this form, as in the Gullstrand–Painlevé chart is not just spatially conformally flat, but exactly Euclidean at every instant of time. One can imagine that a Schwarzschild black hole is like a drain sucking space down to the singularity at the local escape velocity.

The rotating Kerr black hole cannot be put in this form, although its equatorial slice can be. See Visser and Winfurtner (2005) for details, as well for a discussion as to why interpreting the metric as corresponding to a physical fluid is problematic even in the Schwarzschild case due to the the conformal factor.

However, Hamilton and Lisle (2008) found that the Doran chart of Kerr spacetime can be interpreted as a six-dimensional "Lorentz river" characterized not only by a velocity but also a twist. This really quite different from the acoustic fluid analogy, as the 'river' does not spiral inwards, but rather has an intrinsic twist that rotates infalling objects. Still, it is interesting in its own right.

Image by Andrew Hamilton. (Twist not shown.)

References:

Visser, M., Weinfurtner, S. "Vortex analogue for the equatorial geometry of the Kerr black hole", Class. Quant. Grav. 22:2493-2510 (2005) [arXiv:gr-qc/0409014]
Hamilton, A. J. S., Lisle, J. P., "The river model of black holes", Am. J. Phys. 76:519-532 (2008) [arXiv:gr-qc/0411060]


What are the white reflections on Ceres? What are the white reflections on Ceres?

https://solarsystem.nasa.gov/planets/dwarf-planets/ceres/overview/

According to the Wikipedia page linked by @BowlOfRed, they're ice and/or salt.

  a material with a high level of reflection, and [the scientists] suggested ice and salt as possibilities.

  ...

  On 9 December 2015, scientists reported that the bright spots on Ceres may be related to a type of salt, particularly a form of brine containing hydrated magnesium sulfate (MgSO4·6H2O); the spots were also found to be associated with ammonia-rich clays. Arizona State University scientists have proposed that the bright spots are best explained as resulting from briny water erupted from Ceres's interior that subsequently sublimated, leaving behind only the salt deposits. Near-infrared spectra of these bright areas are consistent with a large amount of sodium carbonate, (Na2CO3) and smaller amounts of ammonium chloride (NH4Cl) or ammonium bicarbonate (NH4HCO3). These materials have been suggested to originate from the recent crystallization of brines that reached the surface from below.


Are red dwarfs (M dwarfs) known to be strong infrared sources? If so, why? Are they known to be surrounded by large amounts of dust? Do they typically flare?

Yes, red dwarfs are strong in the infra-red. This is because they are much cooler than our sun, and just like a red hot poker which has cooled from being yellow hot, their emissions are redder. It takes less energy to emit red light than to emit longer wavelengths, and still less energy to emit in the infra-red. Being much smaller than the sun and with less fusion taking place, red dwarfs have less energy.They are known to emit flares, sometimes quite large ones.


How does time "add up" for observers outside of and travelers inside of a Krasnikov tube? A ship travels 99% of c from Earth (leaving at Time A, say 2016) to Destination Star (arriving at Time B, say 2020, accounting for relativistic effects) to stretch the tube, leaves a buoy at Destination Star, then uses the tube's topology to return to Earth, effectively faster-than-light, moments after it left, according to how the Krasnikov tube is supposed to work (Time A plus a few minutes). Can the ship now return to Time B (2020) from "Time A (2016) plus a few minutes" using the tube just as quickly? If so, would its return to Time B mean "Destination Star in the ship's year 2020, the moment when it arrived / first stretched the tube to this destination?" And if so, would it mean returning to the system "before" the ship dropped the buoy?
My understanding of a Krasnikov tube is that it's a theoretical means for FTL travel that doesn't violate causality, because a ship traveling along the return path of a tube "rolls back" time to just moments before the ship originally left.

Now, I understand that there is a way to construct a return tube that would violate causality (http://arxiv.org/pdf/gr-qc/9702049.pdf), but let's assume for a moment there's some "chronology protection" in place that doesn't allow that to happen (the tube implodes when you try to do that), and that the question of the exotic matter needed to create such a tube isn't an issue for the ship's makers.
The questions I have then are:

A) Am I even thinking about this correctly? Where have I messed up?

B) I don't understand how a system of Krasnikov tubes, as some sources suggest, could be used to create an interstellar network for FTL travel. It seems to me that the traveler in the ship can never actually have an impact on his exit or return points, since he will always be returning to Time A (2016) moments after he left, or Time B (2020), the moment he arrives?

I think one of the most important things you missed in this article is the bit where they say the energy requirements are "unphysical".  This is no different from saying it's impossible from our point of view.
A couple of quotes from the article :

Here as well, it was found that the wall of the “warp bubble” surrounding a spaceship must be unphysically thin compared to the bubble radius
Estimates of the total negative energy required to construct Krasnikov tubes of even modest dimensions were shown to be unphysically large.  Similar difficulties have been recently shown to plague warp bubbles and wormholes.  The Krasnikov tube suffers from some of the same drawbacks as these other pro
posed methods of faster-than-light travel, and is hence also a very unlikely
possibility.

The model you're discussing is really best thought of as a mathematical exploration, rather than a practical proposal of any kind.  I would suggest the authors are disproving the viability of Krasnikov tubes rather than anything else.  What was done was to try and fix problems with the previous model, but in doing so the authors ( to their credit ) show that they cannot bypass the fundamental problems making such a system unrealisable in practice.


How long do solar eclipses last on Metis? How long does Jupiter obscure the sun, from the perspective of Metis, the innermost moon of Jupiter?

In other words, how long does Metis remain in the shadow of Jupiter?

If it varies by time of year, what are the minimum/maximum values?

I am not sure of any direct measurement, but it can be figured out rather easily and accurately from theory with a little bit of maths. We have the following parameters for Metis and Jupiter (taken off Wikipedia):

Diameter of Jupiter: about 140 Mm (megameters, 1 Mm = 1000 km), so radius ~70 Mm. Compare Earth at ~6.5 Mm radius, 13 Mm diameter.
Orbital distance for Metis from Jupiter's center: 128 Mm
Eccentricity: effectively zero, so we can treat it very accurately as a circular orbit.
Orbital period: 25.4690 ks (kiloseconds, 1 ks = 1000 s). Compare against Earth's day, 86.4 ks, and orbital period of its Moon, about 2500 ks (more accurate: 2360 ks).
Also, one more relevant point is the distance to the Sun and its diameter, to find the apparent size: average distance to the Sun is 779 Gm ($7.79 \times 10^5\ \mathrm{Mm}$), and Sun's diameter roughly 1 Gm ($10^3\ \mathrm{Mm}$).

The way we will reason now is as follows. Jupiter "moves" in one circuit across the sky dome of an observer on Metis in one orbit. Hence, the time of eclipse is the time required for Jupiter to "move", from arrival of its leading edge to departure of its trailing edge, across the image of the Sun on the same sky dome. Equivalently, then, it is the time required for it to move through an angular distance equal to its own angular diameter plus two angular diameters of the Sun (one on first contact, the other before last contact):

$$\mbox{Eclipse time} := \frac{2 \cdot (\mbox{Angular diameter of Sun}) + \mbox{Angular diameter of Jupiter}}{\mbox{Angular speed of Jupiter on sky dome}}$$

Now the denominator is easy: a full revolution is $2\pi\ \mathrm{rad}$ or about $6283\ \mathrm{mrad}$, so we just divide that by 25.4690 ks, 4o get

$$\mbox{Angular speed of Jupiter on sky dome} \approx 246.7\ \frac{\mbox{mrad}}{\mbox{ks}}$$

Hence, now all we need is the angular diameters of the Sun and Jupiter in milliradians and we'll be done.

This is easily doable with a little bit of trigonometry: while I do not have a diagram on hand, you can easily find one. The angular size of an object of girth $D$ emplaced at a distance $d$ from an observer is

$$\mbox{Angular size} := 2 \cdot \tan^{-1}\left(\frac{D}{2d}\right)$$

Note that you'll often see the approximation

$$\mbox{Angular size} \approx \frac{D}{d}$$

but this is only good when the object is very far, i.e. $d \gg D$. In this case, though, that is not the case at least for Jupiter, so we have to use the full formula with the inverse tangent. Using the figures just given, we have

$$\mbox{Angular size of Jupiter} = 2 \cdot \tan^{-1}\left(\frac{140\ \mathrm{Mm}}{2 \cdot 128\ \mathrm{Mm}}\right) \approx 1.000\ \mathrm{rad}$$

or accurately(!) $1000\ \mathrm{mrad}$. The Sun is far enough that we can just use the approximation and get 1/779 rad or about $1.28\ \mathrm{mrad}$. Hence the total angular distance to cover is 1.28 + 500 + 1.28 ~ 503 mrad, and the eclipse time equals

$$\mbox{Eclipse time} = \frac{1000\ \mathrm{mrad}}{246.7\ \mathrm{mrad/ks}} \approx 4.08\ \mathrm{ks}$$

or for so many still not fully accustomed to the SI, a bit longer than an hour (3.6 ks), and in local measures, a bit less than 1/5 of a Metian day-month.

NOTE: I should point out a caveat of this is is that that this doesn't take into account the inclination of orbits - both Jupiter's and Metis's which may make the Sun in effect "encounter" Jupiter at a different point than simply cutting along its equator. But it should still give you an idea - 4 ks or so, plus-minus some amount that'd be too much work to get.


What is the detector size of LANDSAT 8's detector array? According to the suggested tags, this must be the right forum:
I found  that for each band of Landsat 8 over 7000 detectors are on the array. Does anyone know how big each of the sensors is?

The answer can be deduced from the main landsat 8 page, where you find the following picture of  QWIP detector arrays, and QWIP is short for Quantum well infrared photodetector, I added a gray arrow pointing to one of the 3 arrays.

I cannot give you the exact dimensions in mm, but it is clear from the SMD chip elements (SMD = surface mounted device) - one of them I highlighted with a gray ellipse, that each sensor is probably coin-sized, if not smaller. Or in numbers: I guess a side length $1.5\pm1 {\rm cm}$ seems reasonable, it depends a bit whether the sensor has the same size as the silver square or whether there is a border around.
References

Some introduction to SMD resistors


Are Trans-Neptunian Objects (TNOs) asteroids? Are Trans-Neptunian Objects (TNOs) asteroids?

Please, base your answer. If possible, give sources. Thanks.

The IAU does not give a definition of "asteroid" (in contrast to its defintion of "planet") The definition of "asteroid" is therefore determined by actual use, rather than a specific authoritative source.

The IAU in its guidelines on "naming" doesn't use the term "asteroid" (except qualified as "near earth asteroid". The term is somewhat ambiguous.  Instead it considers "planets", "dwarf planets" and "minor planets".

Minor planets can be dwarf planets, asteroids, trojans, centaurs, Kuiper belt objects, and other trans-Neptunian objects.

Initially solar system bodies were classed as "comet" (appears fuzzy due to a coma of dust and gas) "asteroid (appears star like, a single point of light with no fuzziness). The planets appear disc-like.

TNOs have no coma, but their surfaces are dominated by ices (and would start to evaporate if ever brought to the inner solar system).  Thus they are rather different from the traditional rocky asteroids.

So you should avoid using an unqualified "asteroid" as it may be ambiguous. You can say "minor planet" if you want to include both TNOs KBOs and classical asteroids.  Or "Main belt asteroid" for those between Mars and Jupiter.

If you read "asteroid" it would normally exclude the outer solar system bodies. But as it may be ambiguous, so read the context.


Is it possible to measure galactic red shift using consumer telescope equipment? I was wondering if it would be possible to measure red shift from distant galaxies using consumer grade telescope equipment and spectroscopy filters. (like this one)

I imagine it would require a tracking mount, and enough aperture to gather data on distant faint objects, but I wonder if it would require more sensitive equipment then a consumer grade tools available for under 5-6 thousand.

A long term goal would be to replicate Hubble's work, and directly measure the expansion rate of the universe from backyard equipment.

Looks like you can measure the cosmological redshift of quasars using that equipment and an 14" reflector:

http://www.rspec-astro.com/sample-projects/
(halfway down the page)

So the answer appears to be yes, you can do it, assuming that page isn't a total fabrication. Seems plausible to me: the redshift of that quasar is sizable at 0.15 (though that's small for a quasar), which means that the spectral lines will be shifted by something like 15%. So you  don't need to have a very fine diffraction grating to see the shift. And that quasar is pretty bright for a quasar, at magnitude 13, so you can easily spot it with a good scope. Sounds like a fun project!


Why is this telescope so short? How hard is it to make such a fast primary? The recently discovered object C/2019 Q4 (Borisov) is in the news because it might have an eccentricity greater than one, which means it might be from outside the solar system, though not necessarily.
MPEC 2019-R106 estimates e=3.08.

I looked at Sky and Telescope's Is Another Interstellar Visitor Headed Our Way? which shows the image and caption below.

Although it is impossible to be sure from only this photograph, the 0.65 m telescope looks like the distance from primary to the small secondary is only about 1 meter, maybe 1.2 at most.

Questions:

Why is this telescope so short?
How hard is it to make such a fast (and therefore deep) primary?

  Gennady Borisov with the 0.65-meter telescope he built and used to discover the new comet. G. Borisov

I don't have enough room in the comments for this, so I'm writing here, although it's probably not a true answer since I know nothing about that particular telescope.

Anyway, if you look at many SCT systems, and their derivatives such as Ritchey-Chretien, Dall-Kirkham, etc, the distance between primary and secondary is often not too big.

If the secondary has negative curvature, that means one of its focal points is behind it. That's where the focal point of the primary also is. So the focal length of the primary is bigger than it seems just by looking at the picture. There's a focal point geometrically located in front of the telescope; the distance between it and the secondary or the primary depends on the design parameters of the system. It can't be too close to the secondary, or else the secondary would have to have extraordinary amounts of curvature.

Also keep in mind that there must be a gap between the edge of the primary and the inner surface of the OTA, so the primary is smaller than the visual estimate of the hole.

We also don't see the bottom of the instrument, so we don't know how far the primary mirror cell is protruding out the back of the instrument - though that's limited by the mount's big fork.

Also, look at a small SCT, such as the Celestron EdgeHD8 OTA. Visually, the tube seems a bit longer, compared to its diameter. However, the EdgeHD is a much smaller instrument. The relative amount "wasted" at the bottom by the primary cell is different. Also, the secondary is "buried" into the OTA, whereas the 0.65m scope has its secondary sticking out of the OTA.

Anyway, this is a bunch of handwaving based on visual estimates. It is possible that this is a system with a primary somewhat more strongly curved than usual.


What is our universe expanding into? From inflation theory and other theories explaining the birth of the universe, the universe is expanding and recently it's been discovered that the expansion has been accelerating. My question is simple, what is happening on the boundary of our inflating universe and what is that 'space' that is experiencing that boundary?

It's a rather difficult concept to grab but, in short, the image of expansion you have in your mind is wrong. It's not your fault, it's everywhere out there, the inflatable balloon etc. And it's terrible :)

Another way to look at it is that the universe properties are changing. Whether it is infinite or finite, its size (as measured within itself) changes. Things are pulled appart. You could imagine that free vacuum keeps being generated between all objects but that would be a misrepresentation of the theory.

The most simple way I see it is actually what the equations say. It's lame but sometimes the math are the simplest way to understand a phenomenon. The equation say the metric is expanding. It means two point in the fabric of the universe are measured at increasing distances over time. As a result, unless objects are pulled to each other with sufficient force, they tend to drift appart. It's a property of the universe that has not been explained yet (at least not completely), and we still are on the fence as to what influences the expansion rate.

Ultimately, this concept is very easy to accept when you cease to represent the universe from outside of itself. The universe is everything that is so if any image in your mind represents the universe from outside, it won't give you a good picture. Balloons are especially bad :). If you need an image, draw a grid or take a notebook, draw some points here and there on it and imagine the grid keeps getting bigger and bigger. But don't look at the whole page, forget the page limits, focus on the grid. And keep in mind that the points do not expand, only the grid does...


How bright are the outer planets, when at close range? Seven probes have passed by or orbited outer planets. I wonder how their photos differ from the naked eye view one would have if one were actually there.

Saturn is 10 AU from the Sun, which means that it has just 1% of the sunlight we have on Earth. Would an astronaut orbiting Saturn, looking out of the window, even notice that Saturn is there? A clouded winter night on Earth can be really very dark, so you can't see where you put your foot while walking. But the outer planets at least have star light and a wage glimpse of the sun.

Can the light conditions out there be compared to a clear night on Earth with a full moon, half moon or new moon?

There is a rule about average surface brightness: it is conserved as you change distance or magnification.  For the gory details see.  Telescopes make unresolved objects brighter, but for resolved objects, a telescope shows you the actual surface brightness (assuming that the exit pupil diameter is well matched to your eye's pupil (~7mm) and is not throwing away light, dimming the view).   The surface brightness of any resolved object through a telescope is the same as it would be if you were there.
Also, the 0.01 of earth sunlight received at Saturn is way more than the lighting provided by a full moon which is 400,000 times fainter than the Sun.
What you also need to take into account, is that when you look at Saturn through a telescope, it is generally (not always) done at night when your eyes are dark adapted.


Could Venus or Mercury have a moon that we haven't detected? Let's consider Mars' two moons for a moment. They're small, not even spherical, and most likely are captured asteroids. They weren't discovered until the late 1800's. Telescopes are much more powerful today, so you'd think we'd have found any similar moons at Venus or Mercury by now, if they exist.

The trouble is, there are some very dark asteroids out there. And of course, there are some much smaller ones too.

How do we know Venus or Mercury doesn't have a small dark asteroid moon? Did any probes actually try radar to find them?

It's unlikely that either Mercury or Venus could have moons to begin with. Both of these planets are pretty close to the Sun — and in general, this prevents moons from finding stable orbits.

If a moon were too close to the planets, it would fall within the Roche limit and be torn apart by tidal forces. If a moon were too far from the planets, it would fall outside the Hill sphere and be pulled into the Sun.

The zones where moons around these planets could be stable over billions of years is probably so narrow that no body was ever captured into orbit when the planets were first being accreted.

Now, it gets even more complicated in Mercury's case. Its Hill sphere extends about 3000 miles (~4828 km) — that is, any satellite more than 3000 miles from the planet will be pulled away by the Sun's gravity.

However, its Roche limit extends to 3600 miles (~5794 km), so any satellite within this would be ripped apart. Venus lacks this excuse, but the region between where the Sun would pull away a satellite and the Roche limit is probably too small for a satellite to happen to have formed there.

Next, you have to realize that inner planets generally don't have moons. Solar wind pressure usually sends dust and debris flying away from the planets, so they don't have any spare material for moons.

Remember that Earth's Moon is thought to have formed from a collision with the Mars-sized planet Theia. This makes Earth an oddity in itself. Meanwhile, Mars is thought to have captured Phobos and Deimos from the asteroid belt — so Mars and its moons didn't condense from the same dust mass.

With all this, it'd be extremely surprising if we found moons on either Mercury or Venus.


Triple stars with retrograde orbits Where can I find the observed frequency of triple star systems where the outer star is in a retrograde orbit?

The usual recipe used in the population synthesis literature is that triple stars comprise ~10% of all stellar systems and the mutual inclination is uniform in the cosine.  Most of these systems have a large enough semi-major axis ratio that the inner binary is dynamically decoupled from the outer binary, so retrograde orbits will be just as likely as prograde orbits.  So, to an order of magnitude, ~5% of stellar systems should be triples with retrograde orbits.

Things get more complicated if you are interested in more details.  More massive stars have higher multiplicities, so the fraction is probably somewhat higher for them.  For more compact systems, dynamical effects like the Kozai-Lidov mechanism will become important and can lead to prograde orbits becoming retrograde and vice versa.  Moreover, in compact systems, retrograde orbits are slightly more stable than prograde orbits, so they are likely to be a little more overrepresented, though I'm not aware of any observational evidence that there is any great difference between the two.

If you want more details on the observations of multiple stellar systems you should read Raghavan et al. (2010).  However, I don't think that they measured mutual inclinations.  In general those measurements are very difficult to do.  Andrei Tokovinin has done a lot of great observational work on triple stars as well, so you might want to look at Tokovinin (2014) and Tokovinin (2008) among other papers.  There's also a good review by Duchene & Kraus (2013).


How come the northern and southern lights are the same? Not an expert, hence the question. To my knowledge the magnetic field of the earth travels south to north. Wouldn't there be a bias on whether the charged particles causing the auroras are negatively or positively charged? If each pole attracts the opposite, how come they're the same?

As most people are aware when they play with bar magnets, magnetic poles repel magnets of the same polarity and attract opposites. Magnets have two polarities: north and south.

Electric charges are classified as positive and negative, and also attract opposites and repel charges of the same sign.

However, magnetic poles and electric charges do not match up in the way your question suggests - it's not like North is positive and South is negative and they attract electric charges. Electric and magnetic fields are connected and interact (in fact, both are produced by moving electric charges) but the signs/polarity do not interact in the way you are imagining..

Aurorae (northern and southern lights) are produced because charged particles are streaming toward the Earth all the time in the solar wind. Charged particles (positive or negative) are blocked by magnetic fields - the easiest way to imagine this is to draw the Earth's magnetic field like a bar magnet and the charged particles cannot cross the field lines. At the poles the field is not parallel to the Earth's surface (blocking charged particles) but perpendicular to it, so the charged particles can move into the atmosphere without crossing magnetic field lines. In the atmosphere they interact with particles in the air and create the beautiful lights that we see.

So, in short, the lights are the same at the north and south poles because it is the same kind of particles interacting with out atmosphere.


Is it possible make artificial gravitational waves? Is it possible to generate gravitational waves (creating low or high disturbance in space-time) from very heavy bodies spinning at very high speed artificially?

Typical accelerating masses will generate gravitational waves.  but because gravity is so weak of a force; and current/plausible near future detection apparatus will be unable to detect signals from anything less dense than gravitationally stable masses of degenerate matter (ie neutron stars or white dwarfs) moving at near relativistic speeds.