Synodic Day and Sidereal Day If given the sidereal day, is it possible to derive the synodic day of Earth? If a derivation is possible, could anyone illustrate it or point me in the right direction?
Wikipedia's Synodic day begins:

A synodic day is the period it takes for a planet to rotate once in relation to the star it is orbiting (its primary body).

Question: How can one calculate that?

Surprisingly your linked Wikipedia article does not link to an explanation of how to calculate it, and neither does this.
The equation for a synodic period is based on subtracting two frequencies, the same way a radio down-converts to an intermediate frequency by mixing two frequencies nonlinearly and using the difference in frequencies as the new signal:
$$f_{\text{diff}} = f_> - f_<$$
where the > and < indicate the larger and smaller frequencies.
In orbital mechanics we usually don't talk about orbital frequency, but instead orbital period where $T = 1/f$, so:
$$\frac{1}{T_{syn}} \ = \ \frac{1}{T_<} - \frac{1}{T_>} \ = \ \frac{T_> - T_<}{T_> \ T_<}.$$
Use that if the two rotations are in the same direction which is the case for Earths rotation around its axis and around the Sun.

could anyone illustrate it or point me in the right direction?

Put the sidereal day for $T<$ and a year for $T>$ and you'll have your answer.
But what if they are rotating in opposite directions?
In that case you can define one as having negative frequency or a negative period, and replace the minus sign with a plus sign.


Detector Pixel Geometry? When analyzing an image I'm used to thinking of the pixels as a grid of boxes the photons have fallen into. I assume that there's zero space between the boxes. That seems obviously wrong, since there needs to be some separation on the detector to prevent electron bleed between pixels. So, what's the real geometry of pixels in a detector like (especially for: CCDs and infrared arrays)? Is it closer to a grid of points than a grid of boxes?

Usually the separation exists, but is negligible compared to the size of the pixels.  Hence, the loss due to photons "falling through the cracks" is usually very small, and can be well accounted for using either Monte Carlo-based methods or experimentally using a focused source of known brightness. For example, attached is the user manual for the WIYN One-Degree Imager (ODI), a CCD detector that I have done some work on. There is a section on detector geometry that will probably help.

http://www.wiyn.org/ODI/ODIUserManual.pdf


What happens to the momentum of a star as its mass is converted to energy? Stars move relative to many frames of reference.  We can use the CBR for this question, I suppose, but it may not matter.  Their momentum p=mass x velocity.

Stars lose mass as it is converted to energy. Percentage-wise it isn't much, but it is tons per second, so it isn't negligible.
What happens to the momentum of the star as it loses mass? Presumably it is conserved, so does the velocity of the star increase?  Or does the overall system carry off some momentum with the energy, leaving the stellar velocity constant?

Momentum is conserved. But it is the momentum of the system as a whole. The energy released by nuclear fusion escapes as a result of neutrinos and radiation. If these emissions are isotropic (the same in all directions), or at least with no hemispheric asymmetry, then NO net momentum is removed from the star, it is not accelerated and therefore its velocity with respect to any reference frame remains unchanged. It's momentum in that rest frame is then reduced by the percentage of mass that it has lost. The lost momentum exists in the escaping radiation and neutrinos which will appear anisotropic in any frame of reference other than the rest frame of the star.

In practice most radiation loss is fairly symmetric, if not spherically, then at least not in the sense that more radiation does not emerge from one hemisphere. I am not sure about neutrinos, but I see no reason for asymmetric emission and these are a fairly minor component in any case.


How would a Fast Fourier Transform Telescope work without a mirror, dish or lenses? I recently came across a short article mentioning the idea for a telescope built without mirrors or lenses that instead uses an array of antennas and massive computing power to resolve an image. Such a thing will apparently become possible if "Moore's law continues to hold", but doesn't explain how it will work.

Here's a link to the original paper: Tegmark and Zaldarriaga 2009

  We will show that it acts much like a single dish telescope with a
  dramatically larger field of view, yet is potentially much cheaper
  than a standard interferometer with comparable area. If a modern
  all-digital interferometer such as the MWA 1 is scaled up to a very
  large number of antennas N , its price becomes completely dominated by
  the computing hardware cost for performing of order N2 correlations
  between all its antenna pairs. The key idea behind the FFT Telescope
  is that, if the antennas are arranged on a rectangular grid, this cost
  can be cut to scale merely as N log2 N using Fast Fourier Transforms.
  As we will see, this design also eliminates the need for individual
  antennas that are pointable (mechanically or electronically),

Unlike a traditional interferometer, which consists of individual telescopes that focus light (or radio waves) to a point before ultimately combining their images, the antennas don't have a dish that focuses the radio waves from a certain point in the sky. Instead it's just an array of antennas that images the entire sky at once:

  The results above also show that the FFTT can be thought of as a cheap
  maximally compact interferometer with a full-sky primary beam.  To
  convert a state-of- the-art interferometers such as MWA 1, LOFAR
  [2], 12 PAPER[4], 21CMA [3] into an FFTT, one would need to do three
  things:

  Move all antenna tiles together so that they nearly touch.
  Get rid of any beamformer that “points” tiles to- wards a specific sky direction by adding relative phases to its component antennas, and
  treat each antenna as independent instead, thus allowing the array to
  image all sky directions simultaneously.
  Move the antennas onto a rectangular grid to cut the correlator cost from N2 to N log2 N .

Can someone explain how a computer can resolve an image without a lens, dish or mirror to make parallel rays of light converge on a sensor? I thought that a sensor without a lens, dish or mirror will just produce a blurry mess.

Edit: I think I understand it now. Watching this video was massively helpful for me in understanding how Fourier transformations allow you to isolate individual waveforms from a signal containing many different waveforms.

From one perspective, the purpose of a telescope lens, or mirror is to delay the light arriving in the centre of the instrument a bit relative to the light arriving at the edges. A lens is thicker in the middle, and light travels more slowly in glass, the path to the centre of the mirror and back to the detector is longer, because of the curve of the mirror than the one to the edge, and so on. The result is that all the waves which hit any part of the lens or mirror from a point source end up arriving at the detector in phase so they add up, while waves from anywhere else cancel out.

A phased array antenna (common for short wavelength radio operations) achieves the same delay by capturing the signal at lots of points on the (flat) antenna and then artificially delaying it before combining it with the signals from the other points.

An interferometer (such as the Event Horizon telescope) does the same thing digitally. The full waveform of the signal (more or less) is recorded in multiple places and then they are combined in a large computer with various slightly different delays to separate waves coming from slightly different directions (I am over-simplifying greatly).

What this article proposes is mainly a way to do that for more antennae at less computing cost by placing the antennae in a specific layout and using a bit of clever maths (the FFT) to do all the combining in one go.


Should the sun appear very low in the horizon if its path is always between the two tropics? If the sun never rises above the tropic of cancer and never goes bellow (southern) of the tropic of capricorn then it seems that it should appear to the observer lower and lower in the horizon for as we move north. But from what I have seen this is not what I have observed traveling in northern and southern countries.
Am I not remembering right or is my expectation wrong?

Your explanation is correct, so you are not remembering correctly. The farther north you go, the lower the sun is to the horizon at noon.

Of course it depends on how far north you go. If you only travel 5° farther north in latitude, then the sun is only 5° lower in the sky. 5° is a "long" ways to travel on the earth, but in the sky 5° lower is not that noticeable.


How to derive the redshift of GW150914? We all know LIGO has detected gravitational wave(GW) directly. prl

The LIGO team argue they detected the merger of two solar mass black holes($36M\odot+29M\odot-3M\odot$) at z=0.09.

My questions are

It seems GW has Doppler shift too. There is a GW spectrum with emission or absorption lines like optical spectrum?

Figure.2 of the PRL paper seems to be very simple. Generally, people take their masses derived seriously? There may be other models which can give different masses. It is different from Taylor binary which was observed many years.

What does a 100HZ GW mean?  That is the frequency of the two arms shrink and extend?

What does the frequency stand for in the binary system?

The discovery can not be confirmed. DO you think the team win a Nobel prize only after other facilities like LISA find GW again?

That is a lot of questions, but I can take them in order:

  It seems GW has Doppler shift too. There is a GW spectrum with emission or absorption lines like optical spectrum?

Yes, GW has Doppler shift too, as they are travelling at a finite speed, but no, there are no emission or absorption lines we can detect it from. In EM radiation those are caused by the fact that light is quantized, such that certain wavelengths can be absorbed or emitted by atoms. As far as we know, gravity is not quantized, or the effect is too small to be measured. Any irregularity in the signal is then going to be continuous, and not wavelength specific. Detecting a spectra in gravitational waves is unlikely, but would strengthen the theories of quantum gravity and the graviton.

  people take their masses derived seriously?

The method for the mass estimate is described well by @Rob Jeffries, but an even simpler point can be made: There are no other ways to measure their mass more accurately.

  What does a 100HZ GW mean?

Hertz is the frequency, 1/s. That is the number of waves per second. 100Hz is comparable to normal sound waves.

  What does the frequency stand for in the binary system?

As two objects revolve around each other, they would make two waves per orbit,resulting in a double frequency of the system's rotation.

  DO you think the team win a Nobel prize only after other facilities like LISA find GW again?

This is almost certainty ending in a Nobel prize, and they do not necessarily have to wait till after the results are confirmed by other experiments. That said, verification by peer researchers is one of the core principles of science.


Classification of a Comet Google's definition of a comet is:

  a celestial object consisting of a nucleus of ice and dust and, when
  near the sun, a ‘tail’ of gas and dust particles pointing away from
  the sun.

This doesn't cut it for me. Neither does wikipedia's similar entry.

It pertains to the sun, even though comets are present in other parts of the galaxy other than the solar system and probably other galaxies.

Whats more dwarf planet can be classified as follows:

 * orbits a host star
 * rounded by its own gravity
 * not massive enough to induce thermonuclear fusion
 * not massive enough to clear its region of debris

Which pretty much fits the bill of a comet.

Do you think the following is a fair / recognised classification of a comet:

Comet:
* not rounded by its own gravity
* not massive enough to induce thermonuclear fusion
* not massive enough to clear its region of debris
* composition of ice and dust

But then that's pretty similar to an asteroid:

Asteroid:
* not rounded by its own gravity
* not massive enough to induce thermonuclear fusion
* not massive enough to clear its region of debris

However, this means if an asteroid is comprised of ice and dust, it becomes a comet?

Or are asteroid never comprised of ice and dust? Does my asteroid classification need beefing up, regarding an asteroid's composition?

I'm feeling the comet and the asteroid both need some classification pertaining to their orbit as well? Though thats tricky because only some comets and asteroid, so they're similar in that regard as well! Maybe more details on each objsts composition is needed, like I say...

You've answered your own question there, the difference between the two is that asteroids are rock whereas comets are made up of ice and dust.  An asteroid doesn't heat up when it gets close to the Sun and doesn't produce a tail.  Comets can orbit the Sun or any star in a long elliptical (Comet West) or short elliptical (Comet Encke 3.3 Years) period whereas asteroids are found in the Asteroid Belt between Mars and Jupiter and in the Kuiper Belt.  Sedna is a long period asteroid further out any other object in the solar system but it is still an asteroid according to what we know. https://www.universeguide.com/fact/comets  https://www.universeguide.com/fact/asteroids


Are there other planetary systems where gas giants are on the inside to rocky planet orbits? I understand that formation theories for gas giants suggest they should be born further away where there is more gas for them to monopolize vs the sun, and then to form Hot Jupiters they need to migrate inwards, which would disrupt the orbits of rocky planets.

Are there any quibbles with this? Out of all the extrasolar systems is there even one where you have hot or warm gas giants and then cold Earth or Super-Earth sized terrestrials? Bodies like we have in the Kuiper belt lie beyond our gas giants but they are mostly made of ices and represent a small leftover fraction of the system mass, so they don't really count here.

The Kepler-20 system has planets with masses in the following order, going outwards from the star:

Kepler 20b:  $\approx 10 M_\oplus$
Kepler 20e:  $\approx M_\oplus$
Kepler 20c:  $\approx 16 M_\oplus$
Kepler 20f:  $\approx 1.5 M_\oplus$
Kepler 20g:  $\approx 20 M_\oplus$
Kepler 20d:  $< 20 M_\oplus$

If Wikipedia is to be believed, Kepler 20b may be a rocky world;  but at the very least, Kepler 20c is a hot Neptune and has a rocky planet (Kepler 20f) outside its orbit.


Astronomical databases for machine learning? I was reading this question about open problems in astronomy. The bounty-receiving answer says that

  I have just left the field where some colleagues are trying to develop some software to use image subtraction techniques to isolate individual stars in the centre of clusters.

I find this problem interesting. Where could I find databases of astronomical images for testing different kind of ML techniques to solve such problems?

The European Southern Observatory has catalogues with image data available from http://www.eso.org/qi/, you will have to register before you are able to access them.

I'd suggest you look at other observatory's websites for their data. You will have to look past the pages targeted at the general public and find links for data or science, or user portal or something like that. They are sometimes difficult to find.

Ideally you would have a standard annotated data set of images for classification so that you will be able to compare your results with others. Unfortunately I'm not aware of any standard sets.

For literature on using pattern recognition on astronomical images, http://astrometry.net/biblio.html is a good resource. They've created an ML system that is able not only to distinguish stars from other object classes but also able to identify which stars are visible in an image! Very interesting research.

NB For classifying stars in clusters, the problem is likely not classification, but more segmentation.


What is the condition for the number density of a gas to be ultra-relativistic or non-relativistic and degenerate or ideal What is the condition for the number density ($n$) of a gas to be ultra-relativistic or non-relativistic and degenerate or ideal?

I found problems in this subject when I was reading about White-Dwarfs.

A book said $n \gg n_{QUR}$ or $n \ll n_{QUR}$ is the condition for being degenerate or ideal (classical), respectively. And $n_{QUR} = 8 \pi (kT/hc)^3$. Is this condition right? I know where $n_{QUR}$ comes from but I don't know what it really shows.

In that book it is said that $n \ll (mc/h)^3$ is the condition for being ultra-relativistic for a degenerate gas and $kT\gg mc^2$ is the condition for being ultra-relativistic for an ideal gas. Where do these equations come from? What is $mc/h$ ?

In another way, my question is mainly about this diagram and about how we can get the lines in this diagram.

Credit: The Physics of Stars, A.C.Phillips

$k$ Boltzmann constant
$h$ Planck constant
$c$ light speed
$n$ is number density, not the number of particles
$n_{QUR}$ is the quantum concentration number for ultra-relativistic particles

This is fairly standard bookwork that I am not going to reproduce and you should understand that the lines on your diagram are "fuzzy" in the sense that they mark the loci where you can neither use one approximation or another.

So first, the division between a classical gas and a quantum gas (note that all the gases in your plot are ideal - the term ideal refers to the particles being point-like and non-interacting, which is as true for the electrons in a white dwarf as those in the atmosphere of the Sun). The condition here is that for a classical gas, the density/temperature combination should be such that the phase-space (3d momentum $\times$ volume) occupied by each particle is much greater than $\hbar^3$. This ensures that there are more available quantum states than indistinguishable electrons to fill them.

The second criterion is to compare the momentum of particles with $mc$. Clearly if these are comparable then the particles are relativistic. In a classical gas, the momentum depends on the square root of temperature, as you quote. In a Fermi-Dirac quantum gas, the maximum momentum only depends on the density of particles (as $n^{1/3}$). Hence the differing criteria.


What happens to photons that don't interact with something? Say a photon left one of the first stars ever created in our universe.  This photon has been traveling for 13+billion years through the vastness of space without ever hitting anything.  Sure, it's probably been tugged at by gravitational interactions, but it's still happily traveling at the speed of light without direct interaction with matter.

As we know, most of the universe is composed of mostly empty space.  The odds of any single photon hitting something (like the Earth as starlight, etc) would seem to be very small.

What will eventually happen to this photon?  Will it eventually reach the other end of the universe, and if so, what would happen to it there?

Your statement "most of the universe is composed of mostly empty space" is very vague. In reality, the Universe is embedded in diffuse background photon fields, from low energy (like the so-called cosmic microwave background which is relic from the big bang) to very high energy (from the extreme compact objects, like AGNs). Of course, there is also unknown dark matter. It seems that dark matter probably only takes part in weak interaction. Thus, it probably won't interact with photons (the same argument with cosmic background neutrinos). The chance of photon-photon interaction and photon-electromagnetic-fields interactions depends on the energy of the photons. For high energies, it is not low as you expected. Even if a photon gets lucky and doesn't interact with the other particles, still it won't reach the edge of the Universe. The reason is that the Universe is expanding.


Finding the number of stars n light-years from the sun The nearby stars seem homogenously spaced out enough to give a general "stellar density."
Does there exist some kind of mathematical expression that can determine the number of stars n light-years from the sun?
Or—
Is there a rough value for this supposed stellar density idea? (Ex: 0.3 stars per cubic light-year.)

Wikipedia, quoting Erik Gregersen. The Milky Way and beyond. gives a local value of 0.004 stars per cubic light year, or 0.14 stars per cubic parsec.

So within n light years you would expect $0.004\times 4\pi n^3/3\approx 0.016n^3$ stars.

This predicts the nearest star to be about 4 light-years away, consistent with observation.


What would the effects on or around Earth if Betelgeuse went supernova? There's been a lot of news lately about Betelgeuse possibly exploding sooner than perhaps was expected. Some examples:

Guinan, Edward F.; Wasatonic, Richard J.; Calderwood, Thomas J. (8 December 2019). "ATel #13341 - The Fainting of the Nearby Red Supergiant Betelgeuse". The Astronomer's Telegram. Retrieved 27 December 2019. (Link)
Guinan, Edward F.; Wasatonic, Richard J.; Calderwood, Thomas J. (23 December 2019). "ATel #13365 - Updates on the "Fainting" of Betelgeuse". The Astronomer's Telegram. Retrieved 27 December 2019. (Link)

What would the effects on Earth be if Betelgeuse were to explode within our lifetimes?

Of course, this would be highly unlikely to happen during our lifetimes, but I'm curious what sorts of effects, if any, it might have on our planet (including our upper atmosphere). I imagine the effects would be measurable but negligible.

I'm looking for imaginative but scientifically-based answers.

tl;dr - The main measurable effect may be minor climate cooling, but in day-to-day life, the only difference would be that we see a cool, bright explosion in the sky, and eventually, Orion becomes "incomplete".

The effects would likely be quite minimal. What Will Happen When Betelgeuse Goes Supernova? by  Corey S. Powell, former editor in chief of Discover and reproduced from their Quora answer addresses this exact question quite well. The supernova itself would be quite bright in the sky for a while, and it would certainly be something we'd all want to get nice photos of before it goes away, but the actual "shock wave" would take much longer to reach us, and given the distance between us and Betelgeuse, it will have largely dissipated.

The article describes a period of climate cooling a couple million years ago which corresponds with evidence of a nearby supernova, but we have yet to establish a firm, causal link. As of now, we don't expect to really measure any effects, also partly since the shock wave from the supernova would take much longer to reach us than the light. So even if it we saw it go supernova in the next 20 seconds, we would not be able to observe the effects in our lifetimes.

As for our upper atmosphere, as you mentioned, the radiation and material from the supernova would not nearly be strong enough to do anything like destroy ozone (or as the article mentions, ionize the atmosphere), so I imagine the upper atmosphere would stay relatively the same (this assumes the lack of causal link between supernovae and climate cooling - if a link is established definitively, then perhaps that cooling could have some effect on the upper atmosphere - that goes more into meteorology and weather than astrophysics)


Roche Limit for Earth-like body around Earth-like body In a discussion elsewhere about binary planets, it came the question.

How near can a planet like Earth (maybe Earth itself) be to another planet like Earth before both of them tearing apart due to tidal forces?

In other words: Which is the Roche limit for Earth, if it were orbiting (another) Earth?

According to wikipedia, the roche limit for an object with the same density as the Moon would be around 10,000 kilometres. Since the density of the Earth is about 166% of that of the Moon, if you don't take the atmosphere, or oceans into account, then the binary planets could be much closer to each other. However, when you take into account the oceans, and atmosphere, it would probably be a lot further away.


Was the solar system flat before the planets formed? Or does the Kozai-Lidov mechanism flatten orbits after the gas giants form? An astronomer living on Jupiter measures the orbits of the inner planets against Jupiter's ecliptic. Their inclinations are; Mercury 5.7 deg. Venus 2.1 deg. Earth 1.3 deg. Mars 0.6 deg.   It looks like Jupiter attracts smaller planets towards its own ecliptic.   If the original nebula was Oort cloud shaped, then the Sun and planets were assembled from components that had random orbital inclinations.  The "missing" momentum in the angular momentum problem was used up in the assembly of the solar system

The solar system was "flat" before any planets formed.

When a roughly spherical cloud collapses, it conserves angular momentum but loses energy. This means it tends towards a minimum energy configuration for a fixed angular momentum, which is a flat disk. The planets formed in this disk and that is thought to be why their orbits are, by and large, in the same plane and nearly circular. The timescale for forming a disk from a collapsing gas cloud is much less than a million years and is a prerequisite for forming planetesimals, that then go on to form planets on timescales of 10-100 million years.

The Kozai-Lidov mechanism enables cyclic interchange between eccentricity and inclination. It cannot be responsible for making a system flat and stable.


If a pulsar's rotational and magnetic axes were aligned and both aimed at us, would we see a steady radio source? In other words, a non-pulsing pulsar?

Two things would be required. First, your line of sight would have to be close to looking along the magnetic axis of the neutron star. Second, that magnetic axis would have to be closely aligned with the rotation axis of the pulsar.
If both of these are pseudo-randomly distributed and the pulsar beam is narrow, then this is inherently a very unlikely scenario. The cumulative probability of a randomly oriented axis in space having an inclination $<i$, where $i=0$, means aligned, is $1 - \cos i$. And if both line of sight and magnetic-rotation axes require close alignment, you are dealing with the product of two of these factors.
One factor that may work in favour of seeing such objects is that it seems likely that pulsars drift towards alignment of the rotation and field axes as they get older (e.g. Young et al. 2010), but on the other hand, the pulsar beam is expected to get narrower and the pulsar activity weakens and eventually switches off on a similar timescale. It is also possible that young pulsars are more likely to have misaligned, almost orthogonal rotation and field axes, in which case the scenario is impossible for young pulsars.
Still, this is an interesting question. If pulsars are born with random alignments, you might expect a small number to be born with aligned rotation and field axes, and for a small number of those to have beams aligned with our line of sight. They would be characterised by very little modulation of their emission.


Exoplanet rotation period Is there any way through which we could determine the ROTATION period of an exoplanet based on the data we observed?(from transit)

Do you mean the time it takes for the exoplanet to rotate around its axis? The transit method is used to detect the presence of an exoplanet as it passes in front of the star (as seen from Earth). After several transits you can get the orbital period, i. e. the time it takes for the planet to orbit around the star.

There are methods to measure the rotation period of astronomical objects (exoplanets included). You can use the Doppler effect to measure the change of wavelengths of the planet's light as it rotates. This technique has been used to measure the rotation period of exoplanet Beta Pictoris b.

Suppose the planet (as seen from our point of view) rotates from right to left. If so the right-hand edge of the planet's disc will be moving towards us, while the left-hand edge will be moving away from us. Therefore, the wavelength of the light coming from the right will be slightly shorter (i. e. it will be blue-shifted) than that coming from the left (which will be red-shifted). The greater the difference in wavelengths, the faster the planet is rotating. The method cited in the article above seems to be a bit more complicated, but that's the basics of it.

Alternatively, if you can image the planet with enough resolution, you might be able to spot different terrains or patterns on its surface, and by measuring the time it takes for those patterns to appear again in the same apparent position you might be able to calculate the rotation period. But this is probably unfeasible for most exoplanets, at least with the instruments we have right now.


What visible star is closest to the ecliptic? If we were to project the ecliptic out as an infinite plane, what visible star (besides the sun) would lie closest to it?

Regulus ($\alpha$ Leo) is the one to beat, with ecliptic latitude $\beta$ = +0.46$^\circ$ and apparent V magnitude +1.4.
Some dimmer stars are closer to the ecliptic, e.g.
$\alpha$ Lib ($\beta$=+0.33$^\circ$, V=+2.7),
$\delta$ Gem ($\beta$=-0.18$^\circ$, V=+3.5), and
$\delta$ Cnc ($\beta$=+0.08$^\circ$, V=+3.9).
For more suggestions, poke around in Stellarium with the Ecliptic turned on.

If you want linear distance from the ecliptic plane, multiply the distance from us by $\sin \beta$.


if the big bang only expanded the universe when and how did it originate? If the big bang expanded the universe from a singularity point, that means that there was matter before, so where and how did matter form for the first time if it's even possible to know such a thing?

"before" is not the right word. It suggests that there was an arrow of time which allowed to trace events. But time was created only with the big bang, there was no before.

We don't know where the matter came from, or what caused the Big Bang. One hypothesis suggests that it was caused by the collision of branes in other universes, but there's no way to verify this, so in a strict sense it isn't science. Many researchers think we'll never have an answer this.


Is it possible to do planet observation during the day? Keep in mind I'm just a newbie. So.... I have a new Newtonian telescope. It has 150mm opening and 1400mm Focal Distance on top of an equatorial mount. I have 25 and 10 mm eyepieces... and a 2x Barlow. I sat down to do observation yesterday and was able to see Saturn and Jupiter (warming up for the alignment on the 21th). Now, I did notice that as they sank closer to the horizon, the effect of the atmosphere was more apparent (and they were less sharp) so I'd like to see if it is possible to see them earlier when they are higher in the sky to try to minimize the effect of the atmosphere... but that might mean to watch them when the sun hasn't set yet. I know that we can't see them with the naked eye when it is daytime but given that the telescope is letting a lot more light in, then it should be possible to see them if I point in the right direction? I intend to use the sun to get RA/DEC alignment (using the telescope shadow, don't intent to look at the sun with the telescope.... yet! :-D) so that then I can point the telescope in the direction of the planets (and I am using Stellarium on my Debian box to get the numbers and stuff).

Jupiter can be seen during the day.

This image is by Philip Crude. Philip is an experienced astro-photographer. On his webpage http://www.billionplanetsquest.com/p/planets.html he gives details of the equipment and settings used:

This image was captured through my [i.e., Philip Crude's] Celestron CGEM-800 using a ZWO ASI120MC camera at prime focus. Resolution 640 x 480, 35fps best of 900, Brightness @ 1, Gamma @ 48, Gain @ 11. The image was acquistioned with SharpCap and processed with RegiStax and CS6.

Philip also tweeted the image here.
The main difficulty is pointing the telescope in the right direction, as you won't be able to align it against the background of stars, so you need to predict the position and point the telescope accordingly.


What would be the difference in distance between earth and moon on supermoon night (14 November 2016) As we all know the usual distance between earth and moon is 384,400 km.
But i was thinking how much closer it will be on 14 November 2016, as it will be a supermoon, the brightest and biggest moon in 60 years.

Based on some information from this article

http://www.space.com/34515-supermoon-guide.html

The Nov 14 2016 supermooon's expected peak of full phase
is on the morning ov Nov 14 at 8:52 AM EST

According to some quick calculations that I performed
using PyEphem

and assuming the moon would be viewed by an observer in NYC at 08:52AM EST

0.00236372323707 AU from observer
An AU is 1.5 * 10^km

Edit: I needed a better precision for km to AU

http://www.iau.org/static/resolutions/IAU2012_English.pdf

According to this, the AU can be more precisely defined as
149 597 870 700  meters +/- 3 meters

0.00236372323707 * 149 597 870 700 =

353 607 963.19 meters or

~ 353,608 km from earth

here is a quick run through of the inputs I used for the program

>>> moon = ephem.Moon()
>>> nyc = ephem.Observer()
>>> nyc.long, nyc.lat = '-74.0059', '40.7127'
>>> nyc.date = '2016/11/14 08:52:00'
>>> moon.compute(nyc)
>>> print moon.earth_distance

    0.00236372323707

Learn more about PyEphem package here
http://rhodesmill.org/pyephem/index.html

Note I'm just an amateur and these calculations might not take in certain critical factors. I'm uncertain if PyEphems earth_distance property calculates distance to the moon or the moon's center.

If it does calculate distance to the center of the moon this number could be about ~1700 km smaller.

Given your figure of 384,400 km average, this would put the moon just under

~30 792 km closer to the earth or about 91.98 % of it's normal distance


Do pulsar beams interact with the interstellar medium? Pulsars are rotating objects that emit a beam in a direction that is not aligned with the axis of rotation. Because of this, we see them as a periodic pulse. If the beam isn't towards us, we don't see it. Since we need to be on the path of the beam to see it and the beam is only a few degrees wide, it is believed that there are many objects that are pulsars, but that we see as ordinary neutron stars (or can't see at all).

Does the beam from a neutron star interact with the surrounding medium in such a way that we might see the result of this interaction? The same way you can't see a laser beam, but in a very dusty room, you can see the light from the laser bouncing off of the dust particles, making it possible to indirectly see the beam.

Yes. the emissions from a pulsar do interact with interstellar medium - this effect is called dispersion. When the emission from the pulsar hits free electrons (and dust) in the interstellar medium, different frequencies in the pulse become "delayed" by different amounts corresponding to their frequencies - specifically, lower frequencies are delayed more than higher frequencies. So, when we receive the signal from a pulsar in a radio telescope, the pulse is "dispersed", or curved. Software is used to de-disperse these pulses by testing out many different "dispersion measure" values to line up the emissions correctly at different frequencies. You can find that here.

This is the main way we can be sure the pulses interact with the interstellar medium. Based on how dispersed those pulses are, and how much "stuff" we think the interstellar medium contains, we can get a very basic idea of how far away the pulsar is when we accurately calculate its dispersion measure.

I don't, however, know of any effects like what you referred to with the laser, resulting in an ability to "indirectly" observe pulsars... the pulses themselves are already quite weak and "seeing" a pulsar directly takes a decent amount of time on a telescope along with large amounts of data processing. From my personal experience in the field along with taking many observations on radio telescopes designed to observe pulsars, I highly doubt that the laser-pointer effect you described is observable.

For more details on pulsar dispersion, take a look here.


A question about the distribution of mass among the planets in the Solar System Is it coïncidence that the mass of a planet has about its maximum in the middle planets (Jupiter), while the mass of the closest and furthest planets has a minimum (Mercury and Pluto)?

First, Pluto isn't a planet, and hasn't been for about a decade. It's a dwarf planet, and is better grouped with other small, rocky objects beyond Neptune's orbit, such as Haumea. These objects, and other minor planets, have more in common with objects in the Kuiper Belt and Oort Cloud than the eight planets of the Solar System.1

Therefore, we see a different pattern: Four relatively low-mass planets (Mercury, Venus, Earth and Mars) in the inner Solar System (the four terrestrial planets) and four relatively high-mass planets (Jupiter, Saturn, Uranus, and Neptune) farther out (the four giant planets). The question now is why the two groups are arranged as they are, why the terrestrial planets are close to the Sun while the giant planets are further away.

I wrote about this in my answer to Why do the gas giants in the Solar System have comparatively large orbits compared to the inner planets?. It boils down to something called the frost line, a rough boundary before which ices of water, ammonia, methane, and other compounds cannot form. Beyond it, they can form ice crystals (not just water ice, of course), which made it easier to accrete large amounts of material from the protoplanetary disk. This leads to the more massive gas giants.

In general, this explains the mass distribution of the planets, although there are exceptions (e.g. Hot Jupiters), which often result form planet-planet interactions, rather than forming at distances inconsistent with the frost line.

1 I exclude Planet Nine from this list, as its existence has yet to be verified.


Can there be an infinity of stars in the Universe? I have a mind puzzle.

Can there be an infinity of stars in the Universe ?

I have 2 opposed reasonings, yet I don’t find any flaw in them. Can you help me here ?

Answer A : No, the number of stars cannot be infinite.

In the whole Universe, at an instant t, there are W molecules of water. This number may be very big, but it has a definite value. The same goes for stars. Each of all the stars at the instant t has her own size, and we can give her a name. We can list all of them, and we can sort them from nearest to farthest — if we are omniscient, of course.

Answer B : Yes, the number of stars can be infinite.

Space has no finite volume. Space has no ending wall. If I advance into space, I can always go further. Wherever I go, space is by default filled with… vacuum. But it could be otherwise. [I am not sure of this point.] Space could be by default filled with… air. So there would be +∞ molecules of air in the Universe. We can apply this reasoning to stars. Let’s say there is, in average, 1 star / 1060 m3. So, in the total infinite volume of space, there would be +∞ stars. When I advance into space, I would keep meeting new stars, without end. Like in these video games in which mountains keep appearing as far as you walk.

Theoretically yes, there can be infinitely many stars. Since this space you're talking about is better described by general relativity and Einstein's field equations. One of its solutions describes the universe as being spatially infinite.


False-color wavelength assignments in this "drop-dead gorgeous" image of NGC 2903? C|Net's Hubble spots drop-dead gorgeous spiral galaxy tucked into Leo links to NASA's Hubble Spots Stunning Spiral Galaxy which shows the image below.

The caption on the NASA page doesn't mention the color coding, nor does it mention a reference to a technical page for the image's construction or history.

Question: Is this close to a "straight RGB" image of NGC 2903, or were filters used at certain wavelengths, possibly including IR or UV, in order to highlight certain regions and emissions, then assigned false colors. If the latter, is it a standard color coding?

The NASA page says:

  NGC 2903 is located about 30 million light-years away in the constellation of Leo (the Lion), and was studied as part of a Hubble survey of the central regions of roughly 145 nearby disk galaxies. This study aimed to help astronomers better understand the relationship between the black holes that lurk at the cores of galaxies like these, and the rugby-ball-shaped bulge of stars, gas and dust at the galaxy’s center — such as that seen in this image. Text credit: ESA (European Space Agency)

Click to view full size!

Spacetelescope.org
indicates under "Colours & filters" that the blue channel of the image is from a 658 nm filter (red)
and the red channel is from an 814 nm filter (near infrared).
The cyan and orange rows of the table also list these filters; presumably the green channel is a blend of the two.

The ACS instrument handbook says filter F658N is narrow, passing only NII or slightly redshifted Hα emission (Table 5.1, Figure 5.4).
Filter F814W is wide, passing wavelengths between 710 nm and 960 nm (Table 5.2, Figure 5.1).

NGC 2903 is one of the brighter galaxies not in the Messier catalog.
Here is a wider angle, more natural RGB image by Bob Franke, rotated for easier comparison.
The reddish emission nebulae here appear blue in the HST image.


Can fish and other sea creatures survive in watery planets like Neptune or any other planets that has water? I'm a bit curious since Neptune does not have a land surface and it is covered with water, I mean water has oxygen right? What will happen if we put some sea creatures in these watery planets? Can they survive or live? I'm not referring to Neptune alone, I mean all the planets that has some large amount of water in it. Thanks.

According to Wiktorowicz & Ingersoll (2006), Neptune is too hot and too dry to form liquid oceans at the present time. This may seem counterintuitive for a planet with a far higher bulk fraction of water than the Earth which is located so far from the Sun, but the water is going to be mixed up with hydrogen, helium and various other materials in the planet. The water is going to be in a supercritical state, transitioning to even weirder forms in the extreme temperature and pressures of the deep interior.

If Neptune survives the post-main sequence evolution of the Sun, it may be possible that it forms oceans of some kind in the distant future. In any case there is very little free oxygen so life that requires it (e.g. fish) would not be able to survive. The known extrasolar Neptune-like planets are much hotter than Neptune as they are much closer to their stars, so the "too hot" problem would also apply there.

A better bet might be an ice moon such as Europa. In particular, the radiation environment of Europa might have led to oxygen being produced and dissolving in the ocean. The pressure would be extreme, though life has been found at the bottom of the Mariana Trench. The precise conditions in the Europan ocean, and the oceans on other ice moons is not well known, so there may well be additional problems. It is not clear that there would be the right availability of nutrients, for example.


Pluto's orbit overlaps Neptune's, does this mean Pluto will hit Neptune sometime? We know that the orbits of Pluto and Neptune overlap.
This means that pluto sometimes crosses the orbit of Neptune; will Pluto hit Neptune in any circumstance?

No, Pluto is a so called resonant trans-neptunian object; the orbital period of Pluto is almost exactly 3:2 (1.5) times that of Neptune. This means that every time Pluto nears perihelion and is therefore closest to the Sun and also closest to the orbit of Neptune, Neptune is always at a specific angle (50° according to Wikipedia) in front or behind Pluto.
(See for instance Figure 3 and surrounding text in Jewitt, D., Morbidelli, A., & Rauer, H. (2007). Trans-Neptunian Objects and Comets. Springer.)


Luminosity in polar regions I tried to find the habitable planets, thus I suppose to calculate the effective temperatures that incident on the planets from their host stars, I used

$$
T_\mathrm{eff}=\frac{L}{4\pi\sigma_\mathrm{SB}R^2}
$$

but I don't know how to calculate the effective temperatures in polar regions?

The formula you have written resembles the Stefan-Boltzmann law, and its correct form is $$T^4 = \frac{L}{4\pi R^2 \sigma_{SB}}$$.  It relates the luminosity of a blackbody of a given radius to its effective temperature. Earth as a whole can be assumed to be a black body at an effective temperature. So you can calculate the effective temperature of the "earth". This is very well explained in the example section of this wiki page: https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law

For calculating the effective temperature of the polar region, first assumption is that polar region is a black body. Then the question to ask is: what is the effective teperature of the polar region given that the it reaches equilibrium when the Sun light falls on it. To calculate this further, you would need to calculate the angle at which the sun light fall on the polar region you are interested in and also the distance travelled by the sun light in the atmosphere. These parameters will vary with respect to the time of the year, time of the day and the region you are interested in (North pole, South pole). Some calculated temperatures are given on the same wiki page. I hope that helps.


How do pressure, temperature, and gravity affect a smaller body than Earth with the same bulk composition? Earth has a density of about ${\rm 5.513 ~g/cm^3}$ with one Earth radius and one $g$.  Venus, a very similar planet to Earth and likely with about the same bulk composition, has a density of about ${\rm 5.24 ~g/cm^3}$ with 0.95 Earth radii and $0.9 \cdot g$.  How does this scale?  If a planet has 0.5 Earth radii, how would the lower core pressure, lower temperature, and lower gravity impact the density based on the same bulk composition?

If you know for sure and can prove it: write a paper, submit it to Nature or Science. They will accept it and will give you the cover image.
We can only look at the planetary interiors superficially. We know the bulk mass, we can do a bit tomography with seismics to determine boundaries in the speed-of-sound, with magnetotelurics in the magnetic and electric permeability and maybe a bit with gravity and rotational information on the distribution of masses inside and their surface compositions.
We can make reasonable assumptions on the chemistry and our knowledge how these materials behave. We can even make experiments where we try to re-create the conditions within the interior - but that's where the difficulty starts: temperatures of a few 1000°C and especially pressures over long duration are something which concurrently are hard to re-create in a laboratory (e.g. Dorogokupets et al) , thus many information are from shock-experiments where the pressures only persist very shortly.
In essence the equations-of-state for the materials inside the planets cores are only approximately known, deduced as well as possible, interpreting available information as consistently as possible. This level of detail, in getting all the internal structure right is of course tremendously easier on Earth with all our seismic, magentotelluric and gravity stations and plethora of satellites measuring all things also from the exterior. Data on other planets is MUCH more sparse, so information on the internal structure must be taken with many more grains of salt than the ones of Earth. Many is assumed by analogy, taking into account the few information available while matching the physical models we can make of the chemistry and their assumed equations of state.


Are there equations to simulate creation of a solar system? I've been wanting to create a simulator to run tests based on properties of a star to find patterns in how planets form. Are there any already made equations for any of these countless relationships? Here are a few examples to further explain what sort of information I am looking for.

For example, how does the mass of a star relate to the distance of the frost line, and is there an already existing equation to estimate this?

More general example; If I wanted to find out how to estimate, say, a planets surface temperature based on a stars temperature and distance, what is the best way to do this?

Another example; To find the most likely average mass of a potential planet, what information would I need? Distance to star? Stars mass? Distance to nearest planets?

Another; How would I go about finding likely measurements of compositions of any hypothetical planets (surface/atmosphere/etc)

Absolutely any information is very welcome. I do not require any form of proven or precise measurements. If there is no equations, then all I need is what factors are involved in specific attribute development.

I'd like to emphasize my understanding that this is an extremely complex subject to go into detail. All I am asking for is any information or recommendations.

Here is a related question to my dilemma, in addition to this information, I am looking for more. Absolutely anything that may help, probably will. I am attempting to make as highly detailed of a model as possible.

I may have to answer this in parts, a bit at a time, so I apologize if I'm not able to answer everything at the moment.

  If I wanted to find out how to estimate, say, a planets surface temperature based on a stars temperature and distance, what is the best way to do this?

Ah, I just studied this about a week ago! What you're looking for is the effective temperature of the planet, which can be calculated as
$$T=\left(\frac{L(1-A)}{16 \pi \sigma D^2} \right)^{\frac{1}{4}}$$
where $L$ is the star's luminosity, $A$ is the planet's albedo, $\sigma$ is the Stefan-Boltzmann constant, and $D$ is the distance from the star.

  how does the mass of a star relate to the distance of the frost line, and is there an already existing equation to estimate this?

I wasn't able to find an explicit equation for this, but it's calculable. The frost line is a certain distance from a star in a stellar nebula such that the temperature is about 150 Kelvin.

If you know the mass of the star, you can use the mass-luminosity relation:
$$\left(\frac{L}{L_{\odot}} \right)=\left(\frac{M}{M_{\odot}} \right)^a$$
Re-arrange that to solve for the luminosity. The inverse-square law says that the intensity of power drops off as the distance from the source increases:
$$I=\frac{P}{4 \pi r^2}$$
You can solve for $r$ if you can figure out the required intensity:
$$r=\frac{1}{2}\sqrt{\frac{P}{\pi I}}$$
Here, $P \approx L$. Figure out the necessary intensity to heat the nebula to 150 K and you've found the frost line.

  To find the most likely average mass of a potential planet, what information would I need?

I can't help you much there, but I can give you a starter. The density $\rho$ of a circumstellar disk at a distance $r$ (as given by Michael Woolfson in On the Origin of Planets) is
$$\rho (r)=C \exp \left[-\frac{(r-r_{\text{peak}})^2}{2 \sigma ^2} \right]$$
where $\sigma$ is one standard deviation and $r_{\text{peak}}$ is the peak density, and $C$ is a constant.

As promised, here are some more, set forth by Woolfson.

The disk decays over time; its density at time $t$ is approximated by
$$\rho_t=\rho e^{- \gamma t}$$
where $\gamma$ is a decay constant. If you've taken certain math course, you've probably seen this equation many times before, just in with $\rho=y$ and $\gamma=k$.

Another equation - really, a couple equations - deal with accretion by a small dust particle in a circumstellar disk. The variables involved are: $m$ (mass of the particle), $t$ (time), $s$ (the radius of the particle), $\rho_s$ (the density of the particle), $\rho$ (the density of the disk), $f$ (the fraction of the medium that is dust), $T$ (the temperature), and $k$ (Boltzmann's constant, not be confused with the Stefan-Boltzmann constant).

The relevant pair of equations are:
$$\frac{dm}{dt}=4 \pi s^2 \rho_s \frac{ds}{dt}$$
$$\frac{ds}{dt}=\frac{3f \rho}{4} \left( \frac{kT}{4 \pi \rho_s^3} \right)^{1/2}s^{-3/2}$$
Using these, you can find the size and mass of a given dust particle at any given time.


How poor was our tally of objects that could produce potential extinction-level events back in 1998? A negative comment below my SciFi SE question Why did the Armageddon (1998) plot require a “rogue comet that jarred loose a Texas-sized asteroid”? says:

I don't understand the question. They needed an asteroid to hit the earth and they needed a reason for NASA to have plausibly not seen it, hence they knocked it out of its normal orbit with something.

to which I replied:

@Valorum This is 1998! That NASA might not have "seen it" is completely plausible. One of the side effects of this film was to help get NASA and ESA to step up their game and start searching for objects that may one day intercept Earth.

While a lot of asteroids were known in 1998, certainly there has been some additional effort to look at near Earth asteroids in the last few decades using existing telescopes, not to mention efforts to pu new ones in space for this purpose.
Question: How poor was our tally of objects that could produce potential extinction-level events back in 1998? Could there have been one that NASA didn't "see" at that time?

Related questions about efforts to look for dangerous objects, most of this is post-1998:

Why has the Earth-Sun libration point L1 been chosen over L2 for NEOCam to detect new NEOs?
What is the “interesting story” on the limitations of NASA's current near Earth object survey network"?
Has the National Near Earth Object Preparedness Strategy and Action Plan seen any action?
How will the “fleet of small asteroid hunters” proposed by the B612 Foundation & York Space Systems work?
Do all dangerous asteroids first pass through keyholes?
Understanding gravitational keyhole analysis for Near-Earth Objects
What is the name of the new NEO searching telescope “heavily based on” NEOCam?
Why did the Herschel Space Telescope need helium coolant?
How will the “fleet of small asteroid hunters” proposed by the B612 Foundation & York Space Systems work?

It was fairly poor.  Wikipedia has a list of large, potentially hazardous asteroids, and 9 on that list were known before 1998:

2201 Olijato
1620 Geographos
4183 Cuno
1981 Midas
3122 Florence
3200 Phaethon
4486 Mithra
4197 Toutatis
(4953) 1990 MU

The largest of these, 3200 Phaethon, is 5.8 km, so considerably smaller than the Yucatan impactor, but big enough to do plenty of damage if it does eventually get nudged into our orbit
At that time we hadn't observed (53319) 1999 JM8 a 7 km asteroid, so it is not inconceivable that we might not have observed a 10 km asteroid (similar in size to the low end of estimates of the Yucatan impactor).
Now Armageddon has a "Texas sized asteroid", and Texas is huge, 1300 km from Brownsville to the Rita Blanca grasslands.  It is considerably bigger than Ceres. There is essentially no chance that an asteroid the size of Texas would have gone unnoticed by NASA, indeed it would probably have been a naked eye object and known by the ancient civilisations.


Does the Reflection Method for Looking at the Sun Work For Refractors? I own a newtonian reflector and occasionally use it (using the reflection method with paper) for looking at the sun. I happen to also own a refractor, but I'm not exactly sure if the reflection method will damage the equipment (notably the lens on the refractor) if I do.

Actually, your Newtonian may run the risk of suffering from the projection method. The secondary, planar mirror in the Newtonian will be subject to heating. Although it is not in the prime focal point, the light rays are already pretty much bundled up, and this may lead to heating and ultimately to failure of the secondary mirror!

With refractors, you don't have anything in the optical path up until the eyepiece, so in a sense it is more safe.

However: I strongly suggest against using simple projection methods. One alternative is to use a Herschel-wedge. However, these are again not that useful on Newtonians, because they filter the light at the eyepiece end of the telescope. It is probably better to buy a solar filter which is rather inexpensive and can be fitted to any kind of telescope. Without wanting to promote any specific brand, the company Baader has a good explanation on the topic on their website. The important thing is that the filter is securely fastened to the optical tube assembly, and that there are no scratches or cuts in the filter foil.

Another much more expensive method is to use a specialized solar telescope. These usually use very narrowband filters and coatings on the telescope lenses. Often they are specialized to one of the hydrogen or calcium spectral lines, showing great amount of structure in the suns atmosphere.


Are there lightning bolts on Mars? Although the Martian atmosphere is thin, there are many atmospheric phenomena that occur on Mars: storms, dust devils, (carbon dioxide) snow,... Some storms even stir up the atmosphere of the entire planet!
During Martian storms, is there any lightning and thunder, like what we see during Earth storms? What about in the past, when atmospheric conditions were different, has there ever been any lightning on Mars?

Lightning may have been detected on Mars, although it currently appears to be rare, and attempts to replicate the 2006 results have failed so far.
New research found that the low pressure on Mars might be the reason why lightning is uncommon.  The atmospheric pressure on Mars appears to be at a minimum for making lightning, as opposed to just coronal discharge. With less atmospheric pressure than it has now, more static charge would build up on each grain, making a sudden lightning bolt more likely.  More pressure, and there would be more grains flying around, building up charge more quickly and increasing the chance of lightning.  Thus the chance for a lightning bolt to happen, rather than just coronal discharge, is much lower than it is on Earth, and is in fact at a minimum.  This could explain why there has been only one detection so far.  Source


Is a kilonova bigger than a supernova? My question is straight forward.  Is a kilonova bigger than a supernova?

Update

How does a kilonova differ from a nova in terms of...

luminosity
duration
energy
volume of space occupied by the ejecta
speed of the ejecta
other metrics?

Although it's a little tricky to say what "bigger" means in this context, the answer is, in most senses, no.

A supernova puts out about ten to a hundred times as much energy in the form of light, and hundred or more times as much matter is ejected. (A core-collapse supernova undoubtedly puts out much more energy in the form of neutrinos as well.) What matter is ejected by a kilonova does go out faster (30-60,000 km/s, versus about 10,000 km/s for supernova ejecta).

On the other hand, a kilonova puts out much more energy in the form of gravitational waves, so they're bigger in that sense.


What is the magnitude of variation of Earth's orbital inclination? Please excuse if this has been asked and I just didn't find it.

I have found several references that indicate that the inclination of Earth's orbit varies over a period of ~70,000 years, but I can't seem to find anything that tells what the range of variation is. I know that it's currently at 1.57° to the invariable plane, but I'm looking for the value of the maximum tilt.

Thanks in advance!

In a solar system model by Varadi Runnegar and Ghil Earth's inclination varies between 0 and about 0.05 radians (or about 3 degrees)

The variation is rather chaotic, but you may note that there seems to have been a qualitative change in the pattern about 60-70 million years ago


Can the paper narrowing Solar System's barycentre to within 100m help find Planet Nine? I've just read this recent news article, Astronomers Have Located The Centre of The Solar System to Within 100 Metres, reporting on a paper[1] that claims narrowing the Solar System barycentre to within 100 meters, just outside the sun, and I wonder, could this result help prove or disprove the hypothesis of a massive Planet Nine as proposed by Batygin and Brown[2] in 2016?
I imagine that any unknown mass concentration would change the barycentre and generate a error, and if we could measure this error, it should be possible to pinpoint the position of said mass concentration. Can we just use probe tracking data from the Voyagers or Pioneers to see if their hyperbolic trajectories match the barycentre reported in the paper, for example?
[1]. Vallisneri, M., et al. “Modeling the Uncertainties of Solar System Ephemerides for Robust Gravitational-wave Searches with Pulsar-timing Arrays”. The Astrophysical Journal, vol. 893, no 2, abril de 2020, p. 112. DOI.org (Crossref), doi:10.3847/1538-4357/ab7b67.
[2]. Batygin, Konstantin, e Michael E. Brown. “Evidence for a Distant Giant Planet in the Solar System”. The Astronomical Journal, vol. 151, no 2, janeiro de 2016, p. 22. arXiv.org, doi:10.3847/0004-6256/151/2/22.

Short answer
Short answer is no, it can't prove or disprove the existence of Planet 9. The reason is because, even if there is a significant difference between the barycenter of the solar system with and without Planet 9 we wouldn't be able to tell without hundreds if not thousands of years of precise data. If we don't have Planet 9 as a reference and are limited to the Sun and the eight known planets for reference then without another reference in the same frame we have no way to observe the motion of our reference.
It is perhaps useful to consider the solar system (i.e. the Sun plus the eight known planets) as orbiting together around the barycenter of the solar system plus Planet 9 system. Without seeing Planet 9 as a reference how would we have any notion that this other barycenter even exists?
Of course Planet 9 would still be exerting gravitational influence and over time this would cause a deviation between where the barycenter should be and where it actually is -- that is the only way we could use the motion of the barycenter to infer anything about individual bodies. Essentially, the barycenter is a convolution of the gravitational influence of all the mass,5 its deconvolution requires analysis of temporal variation.

 Unnecessarily long answer
Note   -   When I discuss variations in the barycenter I will be referring to the barycenter of the solar system in heliocentric, not barycentric, coordinates (in barycentric coordinates of course there is no variation in the barycenter, by definition). This simplifies some of the qualitative analysis.

Rigorously answering this requires answering several separate questions (some which aren't necessarily relevant in answering the broader question but do provide useful background):

Does the location of the barycenter over time contain enough information to recover information about the contributing masses?
How precisely can we predict the location of the barycenter?
How much deviation would a $5$-$10~M$⊕ planet at $400$-$800$ AU cause to the center of mass of the solar system?
If no such deviation exists, does that rule out the existence of Planet 9? Conversely, if such a deviation does exist, does it rule in the existence of Planet 9?

 1 - Location of the barycenter over time
The short answer to question 1 is yes (sort of). There is information on the orbits of the contributing masses which can be backed out from simple analysis of the location of the barycenter over time.
The emphasis here is because at any given point in time the location of the barycenter contains no information about the location of individual bodies, only the entire mass distribution. However, since most of the mass in the solar system is discretized into large bodies (as opposed to a gas or dust cloud, protoplanetary disk, etc), the influence of each massive body on the location of the barycenter will vary with a frequency proportional to its orbital period -- and, critically, this is fairly measurable (but only if we can also know the relative positions of the bodies). Consequently, the influence of the individual masses can be retrieved (this is not necessarily the case for gas or dust clouds and protoplanetary disks).
In a simple system, retrieving information about the orbits is fairly straightforward (although I will neglect the rigorous demonstration for brevity). The issues arise when the system becomes more complicated.
One source of confound is resonant orbits. Certain resonances would defy deconvolution because the frequencies of their respective influences on the barycenter are synchronous. Furthermore, multiple resonances can't necessarily be disambiguated: the frequency pattern in the motion of the barycenter caused by 2 bodies in a resonance could be replicated by 3 bodies in a resonance. Pluto and Neptune are in a 2:3 resonance, so I don't think we can immediately dismiss the possibility of a theoretical Planet 9 being in a resonance of some kind and that this could hinder deconvolving its effect on the barycenter.
Another significant challenge is the qualification that the barycenter must be observed over time, and more specifically how much time is needed to deconvolve the various influences. As I mentioned, each body's influence on the barycenter is cyclic based on their orbital period and eccentricity. Using the Bulirsch-Stoer integrator from Vulcan and tracking an energy analog of the barycenter's motion1 as a function of time we can analyze the effect of the different orbital frequencies.
Let's consider a three body system of the Sun, Saturn, and Jupiter. In this system the influence of Saturn and Jupiter should clearly be visible in the resulting sinusoid. After integrating this system for $2$ Saturn years ($60$ Earth years) the energy-analogy over time looks like this:

It should be pretty clear that the above behavior can be created by two sinusoids with different frequencies, to any who find this unclear I recommend having a look at this tool. If wanted to we could use Fourier analysis to deconvolve those two frequencies and recover the orbital periods of Jupiter and Saturn (I will also neglect this for brevity).
Let's consider a slightly worse case: a three body system comprised of the Sun, Jupiter, and a hypothetical hot jupiter with a semi-major axis of $1.55^{}$x$10^{10}$ m on a relatively circular orbit. Our energy-analog after $3$ hot-jupiter years ($12$ Earth days) looks like this:

Where did the other sinusoid go?
Its effect is still there, but we haven't sampled sufficient time to see it. We would need to look at $1,095$ hot-jupiter years ($12$ Earth years) in order to see one orbit of our toy Jupiter2 Fortunately we don't have to integrate quite that long to see the effect of the toy Jupiter on the energy-analog. After $320$ hot-jupiter years ($3.5$ Earth years):

This becomes even more complicated with non-zero eccentricities and unstable orbits. For a similar three body, hot-jupiter system but with the toy Jupiter having an eccentricity of $0.65$ over a period of $7$ Earth years the energy-analog is:3

There are other ways to draw inferences about the orbital parameters from the relative motion of the barycenter (e.g. the shape of individual cycles), but that is for a different question.
In sum, the location of the barycenter over time could certainly be used to find a mass imbalance, but but not necessarily argue for the existence of Planet 9 -- that would depend on the specifics of the observations.

 2 - Precision of barycenter predictions
There are number of significant barriers to accurately calculating the barycenter of the solar system but the most troublesome are the uncertainties regarding the interiors of Saturn and Jupiter. Specifically, the behavior of liquid metallic hydrogen at such massive pressures and (more so for Saturn) understanding their gravitational moments (Fortney 2004).
The issue this raises is that without knowing the centers of mass for Jupiter and Saturn (which contain 92% of planetary mass in the solar system) with sufficient precision we don't know well enough what the solar system barycenter should be to determine whether or not the true barycenter differs enough to indicate the existence of Planet 9.
With information from the Cassini spacecraft combined with data from the VLBA radio telescope the Saturn ephemerides was constrained to $4$ km in 2015. This represented an improvement of about an order of magnitude. The Juno spacecraft provided a similar improvement to the Jupiter ephemerides, constraining it to within $10$ km in 2019.
What is less well constrained is how constant these parameters are, so it bears noting that an inaccuracy of $350$ km in the location of the center of mass of Saturn's core within the planet corresponds to an uncertainty of $100$ m in the position of the solar system barycenter. Jupiter lacks a dense core like Saturn's, so it is more difficult to quantify a direct relationship between inaccuracies, but suffice it to say that an inaccuracy of $100$ km in Jupiter's center of mass corresponds to an uncertainty of $100$ m in the barycenter.4
So assuming Saturn's center of mass has not drifted much since September of 2018, I would estimate the precision of the calculated solar system barycenter to be roughly $\pm15$ m.

 3 - Deviation from predicted barycenter
Calculating the effect Planet 9 would have on the barycenter is actually quite trivial thanks to superposition (see this answer for a more rigorous treatment of the limitations of this concept when it comes to gravitational interactions). The equation for calculating the distance from the primary body to the barycenter in a two-body system,
$$
r_1 = \frac{a}{1 + \frac{m_1}{m_2}}
$$
can also be applied to calculating the barycenter of two barycenters where $a$ is the distance between the two barycenters, $m_1$ is the mass which contributes to the primary barycenter, $m_2$ is the mass which contributes to the secondary barycenter, and $r_1$ is the distance from the primary barycenter to the shared barycenter. And since $99.86$% of the mass of the solar system is contained in the sun it is a rigorous approximation to use
$$
r_1 = \frac{200 \cdot 1.496 \text{x} 10^{11}}
           {1 + \frac{1.989e30}{5 \cdot 5.972 \text{x} 10^{24}}}
    = 4 \text{x} 10^8 \text{ m}
$$
as a lower bound, and
$$
r_1 = \frac{1200 \cdot 1.496 \text{x} 10^{11}}
           {1 + \frac{1.989e30}{10 \cdot 5.972 \text{x} 10^{24}}}
    = 5 \text{x} 10^9 \text{ m}
$$
as an upper bound for the distance from the barycenter of the solar system without Planet 9 to the barycenter with it.

 4 - Interpretation
Even the lower bound of $4\text{x}10^{8}$ m is a huge deviation, so did we just demonstrate that Planet 9 can't exist? Unfortunately no. This goes back to number 1 - the problem is time. There are a number of ways to think about this, but the salient point is that the influence on the barycenter of the solar system would have been persistent since well before the earliest observational records. Planet 9 would have to have an orbital period on the order of $10,000$ years. Over periods of thousands of years it would have obvious and readily measurable effects on the motion of the other planets (and that motion would be reflected in the motion of the true barycenter), but otherwise the effects are simply too small to disambiguate from noise.

1 I create this analog by calculating a hypothetical energy of the system if the center of the sun was the barycenter of the system and comparing it to the initial value. It is not tremendously physically meaningful but it does reduce the parameter-space of the temporal position of the barycenter to two dimensions, thereby making it easier to show the effect I discuss graphically.
2 This is one of the reasons simulating hot jupiters is so difficult (and one of the things the Vulcan project is intended to remedy): the limiting factor for orbital integration is always the fastest-orbiting body (although multi-step integrators mitigate this somewhat) so systems with large differences between orbital periods take proportionally longer to simulate. Using the Mercury code, a gigayear solar system simulation with a hot-jupiter on an 8 day orbit would require roughly 10 months of simulation time on a high-end desktop computer.
3 Note that the energy analog is degenerate here because the hot-jupiter will be ejected after approximately 10 Earth years.
4 These uncertainty relationships are from my own calculations and may be flawed as they have not been rigorously tested or reviewed.
5 And I do literally mean all the mass. As in all the mass in the universe, gravity has an infinite range after all....


Calculate latitude and longitude based on date and sun I can't seem to find this algorithm, if it exists. Can you calculate the longitude and latitude of where you are standing based on the date and the angle of the sun? Also, would the algorithm work for any date past or future?

Thanks for any help, AD

What you are looking for is the navigation method used by ships and aircraft before the advent of GPS.  It requires not only an instrument for measuring the angle between the sun or a star (such as Polaris) and the horizon, but also an accurate time measurement, and charts that can be used to interpret the numbers.  And of course an accurate chronometer -- you'll need to know the time at Greenwich Obervatory, in London.

The traditional instrument used for determining this angle is the sextant.  Here is an article describing the use of a sextant for deterining one's position:

http://www.ehow.com/how_7562747_use-nautical-sextant.html

I suppose that an algorithm for doing what you are asking is available where good sextants are sold - as to the charts, well, perhaps there are downloadable charts somewhere.

Also, here is a good article on determining latitude and longitude by the stars.  Latitude is what a sextant will tell you, longitude is determined differently.  For that you need two clocks: one set to GMT; one set to your local UTC time (NOT your timezone time).  The method is described here: How to Calculate Longitude.  Which brings us to the question of how to calculate your local solar time, which is described in this article How to Calculate Solar Time.  Clearly you can't use this for navigation - it's convenient for human economics, but not finding your location.

Just to make it clear why you can't use local civil time for determining longitude is that the solar time is nearly the same on both Hawaii and Kodiak islands, but Hawaii's civil time is one hour after Kodiak's.  And China stretches some 5,026 kilometers across the East Asian landmass, which is about 4 hours of solar time, but it all has the same civil time.

This question might better be asked in the Sailing SE.  Oh, wait, there isn't one, yet.


How many minutes does a sundial change every day? I'm planning to create a sundial in Ghana (10° 28.852', 0° 26.032') in July 2014.
I will stay there for about 3 weeks.
Do I need to adjust the numbers anywhere in these 3 weeks?
I know the shadow's path changes through the year, but how much does it change?
Is the change significant for a sundial?

(btw, is this the correct site to ask or should I use the astronomy site?)

You need to look up "the equation of time" - this is a correction to the time reported by the sun on any given day - the apparent solar time - (and measured by your sundial) and the standard civil time - the mean solar time - it varies throughout the year and the maximum your sundial will be out is about 15 minutes - but the difference and the rate of change varies. http://en.wikipedia.org/wiki/Equation_of_time

If your sundial is merely for educative/demonstration purposes, this might not matter so much of course - as the sundial will surely demonstrate that the sun's shadow can be used to tell the time.


What is meant by matter distribution? Frustratingly, I find this could be a very simplistic question and may leave me feeling a little stupid for asking it. But I have to ask this.

My lecturer keeps refering to "matter distributions" for various rotation cases, e.g., Keplerian rotation, solid body rotation, flat rotation etc etc.

I was wondering if anyone knew exactly what this term means, or the specifics for each case?

I assume you are referring to the rotation curve of galaxies, i.e. the speed of matter (stars, gas clouds, etc.) as a function of its distance from the center. This speed will depend on the matter distribution, that is, how the galaxies' constituents are distributed throughout the galaxies.

Keplerian rotation

If (almost) all of the mass $M$ were located in the center, then a star at a distance $R$ would have a speed $V = \sqrt{GM/R}$. This is called Keplerian rotation, because this is how planets in the Solar system move, which were described by Johannes Kepler. In this case, the mass inside a radius $R$ is (almost) constant, i.e. $M'(<R) = M$.

Flat rotation

Most disk galaxies have a large concentration of visible mass in the center (the "bulge"), so one might think that Keplerian rotation would be a good description of the stars' speeds. But this is not what is observed. In fact, the speed is seen to rise steeply the first few kpc, and then stay constant out to distances where no light is detected. This will happen if the mass $M'(<R)$ inside $R$ increases with $R$, i.e. if the matter is distributed throughout the galaxy. This is one of several pieces of evidence that 1) there exists some matter that we cannot see, in addition to what we can see, 2) that this matter dominates the kinematics of the galaxies, and 3) that it resides in a halo that extends beyond the visible matter. We call it dark matter.

Solid body rotation

If in addition to a given matter distribution the distances between the individual parts are fixed, the object we are discussing would describe a solid body. Of course this can't be the case for a galaxy, but it will be the case for a rocky planet. In this case the velocity of a given point will depend linearly on its perpendicular distance to the axis of rotation, i.e. $V \propto R$.

Density profile of the matter distribution

Excactly how the matter is distributed in the galaxy is not easily deducted, and varies from galaxy type to galaxy type. An early description of the matter density as a function of distance from the center was the so-called NFW profile, but other models may be better, e.g. a Sérsic profile or an Einasto profile.

So, to summarize, the term "matter distribution" refers to density as a function of location, i.e. $\rho(\mathbf{r})$, for some object. This can be for any object — solid, liquid, gaseous, or consisting of several components that are held together by gravity — but in any case it is true that its form determines its rotation, and the three different rotation forms you mention are often discussed in the context of galaxies.

The three different cases are illustratred below. Keplerian, solid, and flat rotation are drawn in green, red, and blue, respectively. For comparison, a typical spiral galaxy velocity curve is drawn in magenta.


Inverse of the sunrise equation - finding locations with a given sunrise time on a given day I'm working on a project for fun where I represent some sleep data geographically. For a given day, I have a date, a time for falling asleep that night, and a time for waking up the next day. The idea is something like this xkcd comic. My attempts at taking the inverse of the sunrise/sunset equations I've found online have not been fruitful.

Basically, I would like to find an equation or algorithm to give a rough location on Earth with a specific sunset time on some specific day and a specific sunrise time the next day. I am not too concerned with accuracy since this is just for fun, so I'm hoping elevation information isn't necessary.

I'm a programmer, so the goal is really to be able to type something like

>>>> locate(date = '2014-08-27', sunset = '10:00 PM', sunrise = '7:30 AM')

and get back

Latitude: 49.887, Longitude: 96.141

I could program it myself if I just understood enough astronomy. Any help is greatly appreciated.

To avoid the headaches associated with finding roughly two dozen similar solutions in the zig-zagging time zone boundaries , you might want to stick with UTC. If you want to deal with local time - it's going to get a little crazy if you also try to include daylight savings toggling on and off.
The points on the earth where the center of the sun coincides with the horizon (ignoring topography, oblateness, atmospheric refraction, the finite speed of light and other small effects) is just the circle on the earth where a cone drawn from the Sun intersects a spherical Earth.

above: A sphere inside a cone, from http://mathcentral.uregina.ca.
But the next step is hard - getting from that circle to the lat/lon coordinates on the surface, because the axis is tilted, and because the motion around the sun speeds up and slows down as the Earth revolves closer and farther from the Sun. These can be approximated by periodic functions with some coeficients, and that's what you will find in the math behind the first answer. For the second answer, I'll list two Python solutions - PyEphem and Skyfield. Both are easy to use, but you are separated from the actual math (in one case it's an ephemeris/table. The third answer is really a collection of NASA/JPL routines that are highly regarded but may involve more time for you to get up to speed compared to the Python packages.
Answer 1: Astronomical Algorithms
This is something you'll have to dig into a bit, but if you like to program, it may be exactly what you're looking for. The Gaisma website is one of my favorites on the internet - easy to use and presents a bunch of information in easy-to-understand graphics. Click around!
I believe that this site uses algorithms from the collection found at this NOAA site. Click around there as well. They provide Excel spreadsheets which contain the algorithms and other resources. The "main" resource is a collection of algorithms published in the book Astronomical Algorithms - Jean Meeus. From that Amazon page you can see that there are many similarly titled books. I'd recommend going to a library if possible, because it's (in my opinion) always good to go to libraries. However parts of these can be found on-line. For example, a few pages shown from the book Astronomical Formulae for Calculators (1988) include an interesting table of contents.
Answer 2: Python packages PyEphem and Skyfield
I'll copy a bit of the text from this answer:

The Python package PyEphem has been around and well supported, and is the pythonic reincarnation of XEphem. I haven't used it, but I believe it keeps enough information about orbital parameters at certain epochs to generate an ephemeris, including some gravitational perturbations. In other words, it's much more than planets moving on fixed elliptical orbits around a fixed sun. So I believe it runs without internet connection.

I never used it because I was recommended to look at Skyfield and it's exactly what I needed. It downloads a standard JPL ephemeris that you choose, and then just uses it from your hard drive after that. However, in order to deal with leap seconds and other time related effects, it occasionally needs to check the internet for leap second information updates, since these are arbitrary.

I don't know if Skyfield has a mode to avoid that. Actually that's a good question. If you work with a timescale that doesn't have leap seconds, I am not sure if it will run in its current version.
Both Skyfield and PyEphem Python packages have been written and are maintained by @BrandonRhodes.

I've included a simple Python script just to illustrate how Skyfield can be used. If you are comfortable with coding without curly braces, I would strongly recommend you give this a try. It's incredibly powerful and Pythonic.
This is just a starter - you need to add some better housekeeping to detect sunrises vs sunsets, and maybe a more global-type search in some cases. Actually there is some slightly tedious housekeeping necessary to make this work robustly.
note: you can turn on atmospheric refraction using the arguments in the apparent() method. See the Skyfield API documentation for more info, and for a discussion about iterating using Skyfield methods - especially solving for times, see this helpful answer.

def alt_lonlat(lon, lat, t):

    topo = earth.topos(lat, lon)

    alt, az, dist = topo.at(trise).observe(sun).apparent().altaz() ## apparent() args for atmospheric refraction

    return alt.degrees

from skyfield.api import load
import numpy as np
import matplotlib.pyplot as plt
import scipy.optimize as spo

data  = load('de421.bsp')
ts    = load.timescale()

# your example:  '2014-08-27', sunset = '10:00 PM', sunrise = '7:30 AM'

trise  = ts.utc(2014, 8, 27,  7, 30, 0)
tset   = ts.utc(2014, 8, 27, 22,  0, 0)

earth = data['earth']
sun   = data['sun']

zerozero = earth.topos(0.0, 0.0)   # gotta start looking somewhere!

alt, az, dist = zerozero.at(trise).observe(sun).apparent().altaz() ## apparent() args for atmospheric refraction

print "at trise, JD = ", trise.tt
print "at (0N, 0E) Sun's altitude: ", alt.degrees, "azimuth: ", az.degrees
print "at (0N, 0E) Sun's distance (km): ", dist.km

# Find points on equator where sun is on horizon (rise or set) at t=trise

limits   = ((0, 180.), (180, 360.))
lonzeros = []

for a, b in limits:

    answer, info = spo.brentq(alt_lonlat, a, b,
                              args=(0.0, trise),
                              full_output = True )

    if info.converged:
        lonzeros.append(answer)
        print "limits ", a, b, " converged! Found longitude (deg): ", answer
    else:
        print "limits ", a, b, "whaaaa?"
        lonzeros.append(None)

# make some curves

lats = np.linspace(-60, 60, 13)

longis = []
for lon0 in lonzeros:
    lons = []
    for lat in lats:

        answer, info = spo.brentq(alt_lonlat, lon0-90, lon0+90,
                                  args=(lat, trise),
                                  full_output = True )
        if info.converged:
            lons.append(answer)
        else:
            lons.append(None)

        lons = [(lon+180)%360.-180 for lon in lons]  # wraparound at +/- 180

    longis.append(lons)

plt.figure()

for lons in longis:
    plt.plot(lons, lats)

for lons in longis:
    plt.plot(lons, lats, 'ok')

plt.xlim(-180, 180)
plt.ylim(-90, 90)

plt.title("at trise, JD = " + str(trise.tt))

plt.show()

Answer 3: SPICE Kernels
As pointed out by @barrycarter in this comment below this answer, the JPL SPICE Kernels are available. I'm not familliar with them but it's what NASA uses so it must be pretty good :)

Appendix:
Here are some screenshots for London, UK (Gaisma from here):


Is a year really 365.24 days, or is it 365.2564 days like I remember? The NPR News item and podcast Spring Starts Today All Over America, Which Is Weird includes the following:

But why isn't the time of the equinox the same each year?
The short answer is that the time and the date are imperfect human constructs that we use to keep track of our planet's movements.
The longer answer involves leap years.
All of this is caused simply by the fact that the spin of the Earth doesn't divide evenly into one year," says Michelle Thaller, an astrophysicist turned space communications expert at NASA.
One spin of the Earth around its axis is one day. "The problem is we're happily spinning on our axis, and the Earth is going around the Sun, but one year — one complete path around the Sun — isn't an even, exact number of days. In fact, it's 365.24 [days]."

Wikipedia gives Earth's orbital period as 365.256363004 days and I have always remembered it to be 365.2564 which is the same value as Wikipedia, just to fewer digits.
So is a year really about 365.24 days or closer to 365.2564 days?

Both are correct.

The Sidereal Year is the length of time it takes Earth to complete an orbit around the sun, relative to the fixed stars. It is 365.2564 days.

The Tropical Year is the length of time it takes for the Sun to complete a cycle around the Ecliptic and return to the position in the cycle of seasons; e.g. from Vernal Equinox to Vernal Equinox.  It's about 365.24217 days, about 20 minutes shorter than the Sidereal year because of the precession of the equinoxes. This is the year that the Gregorian Calendar is attempting to emulate.


Why no Polaris distance in GAIA DR2? Querying GAIA DR2 for Polaris (for example using https://gea.esac.esa.int/archive/) yields empty parallax columns. Anyone knows why it was not measured/published?

My query for stars north of +75° and brighter than magnitude 4 returns only 6 stars, brightest of which is γ Cephei.
Polaris may be too bright for Gaia.
Even if it doesn't saturate the detector, a bright star will probably have larger astrometric errors than a star between magnitudes 6 and 12.
DR2 coverage of magnitude 5 and brighter stars is better than DR1 but still rather sparse.
Section 3.1 of this paper plots parallax coverage vs. magnitude and explains why some stars have positions but not parallaxes.


Would a 1000 km telescope be able to resolve microscopic detail? I have calculated that a 100m telescope can resolve 16m on the moon. Does that mean a 1000km telescope is 10000x better and can resolve 1.6 mm objects on the moon and even smaller closer? Or is there some limit to using a telescope as a microscope?

Ignoring the effects of the atmosphere and the engineering issues you would need to overcome to build such a monstrosity, yes you could resolve details that well.  There is no physical reason you couldn't.

There are many optical interferometers, but their baselines are currently only tens to hundreds of meters wide. See this List of astronomical interferometers at visible and infrared wavelengths.

In fact, radio astronomers regularly achieve such spatial resolution by combining multiple far apart radio telescopes together to act as one giant telescope. This is a technique called very long baseline interferometry, and it was used recently to create the first image of a black hole by the Event Horizon Telescope using sites distributed across the Earth.


Could dark matter be "inside-out" reality? Where we are, and in the conventional galaxies etc. that we see, there is a whole lot of relatively uniform space but in isolated places within that space, there are pockets of concentrated "mass" which is highly organised into all the objects we observe and study, and out of which we are made.

Now suppose that there were regions of space where that is all turned on its head, and there were huge vast swathes of relatively uniform matter, and within that matter there are pockets of "space" which is highly organised into some other form of object, out of which other objects and beings are made.

Obviously on the first, and simplest test - the question of "how much" dark matter there is, this theory passes the empirical test of reality with flying colours, i.e. astronomers seem to observe much more mass in the regions of hypothesised dark matter in accordance with what this conjecture implies.

What other testable predictions does this conjecture make, and how well do they fit our observations?

As a disclaimer: This answer does not include references, and is mostly based on my first intuition when reading the question. I can see that it can potentially be an interesting question to discuss.

First, we live in a solar system, in a galaxy, in a galaxy cluster. That in itself means that there is a lot of matter around. We are also (at least very likely) living in an overdensity in the universe, meaning that there is extremely much matter where we are compared to the average in the universe.

If we assume that we do live in an underdensity of the universe, that would mean that there is more matter present in the universe on average compared to where we are. However, if pockets of space in a distribution of regular matter was dark matter, then it sounds like there would be a larger amount of regular matter than dark matter in the universe, something which does not agree with our estimates.

If we assume for a minute that it is the other way around (that we live in an area with a lot of matter, not a lot of space), then this theory will struggle with explaining some of the things that point towards the existence of dark matter. One such thing is the rotation curves of galaxies, which indicate that there is more mass presence than there is seen. I expect that there would not be enough space within a galaxy to create such "pockets" of space. Say, however, that we assume this was possible, then the voids we observe between galaxies and galaxy clusters would be even more gigantic pockets of space, which would mean that there would be an enormous amount of dark matter, which I would think would be inconsistent with what we see from the CMB.


Is it possible to see stars that are gravitationally lensed by the sun from the ground? Roughly 100 years ago, Eddington confirmed Einstein’s general relativity theory by photographing stars behind the obscured Sun during a total eclipse.

A coronograph is a device that mimics what happens during an eclipse and blocks the light that comes from a star so that nearby objects (which would otherwise be hidden by the star's glare) can be seen. This device was invented in 1931 by Lyot, 12 years after the aforementioned eclipse.

Would it be possible to use a coronograph with a ground-based telescope to reproduce the observation done by Eddington in 1919? Or would the light from the Sun scattered in the atmosphere be enough to hide the stars for which the gravitational lensing is visible?

The answer is no. But there's an interestign alternative.

The Eddington's experiment was not about seen a gravitationally lensed star but to see a small displacement of its position (just below 2 arcseconds) due to the solar gravitational field. Lensing involves considerably magnifying the source and that can only be done if the star is directly behind the Sun (or very close to that direction).

Only close to the Observer-Sun line the star would be considerably magnified and the flux increased. The result would be an Einstein ring, an hyperdistorted version of the image of the star built with the light-collecting area as large as the Sun (at least).

For this to happen you need to locate yourself at the solar gravitational focus. Theoretically this is not a single point, in fact it is an entire line extendign from the Sun outwards (we call it the gravitational focal line). The thing is that if you are too close to the Sun the Einstein ring is "located inside the solar disk". So you need to get very far away to be able to see the focused light. The point at which the Einstein ring grows larger than the apparent diameter of the Sun is located $550$ astronomical units away from the Sun. This is the minimal distance you would be able to exploit the gravitational lensing effect of the Sun.

A distance of $550\; AU$ is insanely huge. The farthest probe we have sent, the Voyager 2, is currently located $147\; AU$ from Earth, so you would need almost 4 times that to reach the point where this could become feasible. Also you would probably prefer to position yourself a bit farther since the Einstein ring would still be located inside the Solar Corona and a lot of information could be washed out by the intense light (and in the case of a radio telescope you would see an opaque wall). There are a lot of important things that could be done at the focal line of the Sun, you would have a crazy huge magnification (just as if the object was as close to you as the Sun from that same distance! which equates to tenths of millions magnifications in many cases), and the flux would be equivalent to the one obtained by a $1.000.000 \; km$ lens or even more.

It has been said that exploring the solar focal line will be the most important milestone between the exploration of the outer Solar System and the firsts attempts to explore the closest stellar systems. The main problem is that you need to position yourself with incredible alignment accuracy to get the Einstein ring of an object behind the Sun. A probe like that would be usefull to observe just a single object during the entirety of the mission (maybe an exoplanet wich might be crucial for our understanding of habitability), since the delta-v needed to change position and get to another part of the Solar System $550 \; AU$ from the Sun would be insane and it would be more economically feasible to lunch a another probe to the next direction just to observe the other target. A good place to place a telescope like this would be an possible moon around the hypothetical Planet 9 (which is considered to reach $800\; AU$ at perihelion), the problem been again that you would only observe breifly objects that happen to fall at the opposite position relative to the Sun.

A mission to the focal line has been under study for years now. You can read about the details in a presentation that was given by one of the main proponets, astrophysicist Claudio Maccone. There's also a nice talk about the topic on Keck's YouTube channel.