What caused gravitational waves detected by LIGO? I understand that in recent discovery gravitational waves have been detected. But knowing space time is not static it probably changes constantly in many places in the universe due to changes in distribution of massive bodies. For example, could the detected wave just be a slight oscillation in Gravity due to Jupiter fly by, or moon, or something similar?

It couldn't be Jupiter or the moon because

The frequency of the detected waves is too fast. The detector found something that rotated several hundred times a second, and then stopped.
The amplitude is too big. Even though Jupiter is close, the gravitational waves it produces in its orbit are extraordinarily weak. Far too weak for us to detect them (good thing too, because if it was producing powerful gravitational radiation, it would start to spiral into the sun)

We know one thing that can produce a chirp of this frequency and amplitude, and that is a black hole merger. In the last few moments before merging, these two black holes converted about three solar masses into gravitational radiation: That is a colossal amount of energy. About $$500,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000$$Joules.  That is why for one moment it was the "brightest"  point in the gravitational sky

(Note. standard form is for wimps)


Gas Giant temperatures Sudarsky's gas giant classification predicts the visual appearance of gas giant planets based on their temperatures.

But what determines their temperatures in the first place?  Is it just the irradiation from their star, or is there more to it than that?  If there's more to it, what else is involved?

The other thing(s) involved are:

Contraction and the release of gravitational potential energy. Gas giants are born bigger than they are now. They gradually contract towards a completely degenerate configuration, and as they do so half the released potential energy is radiated away, whilst the other half heats the planet.
Gravitational diffusion and separation also releases gravitational potential energy. Heavier elements tend to settle towards the centre, falling through the majority hydrogen.
Radioactive decay of heavy elements that may form part of the giant planet's core.
If the "planet" is more than about 13 times the mass of Jupiter, then it will likely attain a central temperature sufficient to fuse its deuterium. Some would refer to such objects as brown dwarfs at that point.


Is the "mean orbit" centered on ellipse center or ellipse focus? Mean longitude and mean anomaly of a body are calculated as the angle covered by the body along a fictious circular orbit, and this is ok.
But where is this orbit centered? Five different sources say five different things:
Wikipedia: circle center = ellipse focus

University of Texas: who knows?!? No ellipse at all:

University of California: who knows? Mean anomaly mentioned but not drawn; only eclipse shown, no circle:

Britannica Enciclopedia: who knows? Both circle and ellipse shown, but mean anomaly not shown:

Max-Planck-Institut für Sonnensystemforschung: circle centered on ellipse center

Theory & formulas: longitudes and anomalies summed up together, no pictures at all.
I am very confused.

But where is this orbit centered?

The true anomaly is the angle as measured from the central body between periapsis passage and the object's current location. The orbit is an ellipse with one of the two foci at the central body. This concept is central to Kepler's laws and Newtonian mechanics.
The eccentric anomaly is the angle as measured from the center of the ellipse between periapsis passage and the projection along the minor axis of the point in question onto the minimal bounding circle that encompasses the elliptical orbit. That minimal bounding circle is necessarily centered on the center of the orbit rather than the central body.
Mean anomaly is a fictitious angle. Unlike true anomaly and eccentric anomaly, mean anomaly does not indicate where the object is. Mean anomaly instead is the angular displacement that a fictitious object in a circular orbit about the central body with the same semi-major axis length as the object in question would have passed through in the time since periapsis passage.

Aside: The image from wikipedia is flat-out wrong. The center is correct, but the size is incorrect. The circular orbit should intersect the ellipse, twice. It should be outside of the ellipse at periapsis and inside the ellipse at apoapsis.


How bright is it on Pluto in the middle of the day? I just saw a documentary about New Horizons and Pluto with plenty of photos New Horizon had taken of Pluto. All these photos were very bright, like Earth would be during the day. Considering the distance between Pluto and the sun I suspect the brightness of the photos has been manipulated.
So, if you would be standing on Pluto's equator when the sun is in zenit, how bright would it be?

Back when New Horizons was preparing for its flyby back int 2015, NASA's website set up a tool to allow you to experience the brightness of light at High Noon on the subsolar point on Pluto.
From Space.com: NASA's 'Pluto Time' Shows You How Bright It Is on Dwarf Planet

To an observer on Pluto's surface, the sun would be about 1,000 times
dimmer than it is here on Earth, NASA officials said. At noon, the
sunlight would be strong enough for you to read a book, they added.
(Though it might be hard to keep your eyes from straying upward to
take in the exotic Pluto sky, with the big moon Charon looming just
overhead.)

The NASA Pluto Time website can be found here:
https://solarsystem.nasa.gov/planets/dwarf-planets/pluto/plutotime/
In general, the time it will return for you will be a little bit after sunset (or before sunrise), but still bright enough for pretty good vision. For my current location, at about 41° N latitude , on October 31, 2020, it's giving me a local time of five minutes after sunset.


Which moon has a scar as if it was torn in half and put back together? When I was a kid I had several beautiful encyclopedias. In one of them there was a description of moon orbiting on of the gas giants that had huge ripples all along its surface.

It said in the text that it's plausible to think that the moon was torn apart and then got stuck back together.

Could you remind me which one is it? I would like to read a little more about it.

I believe the moon in question is Miranda, as James K suggested. It is hypothesized that it suffered a collision with another large body sometimes in the past. There are several large features that suggest this:

Three large grooves, each > 200 km long, cover parts of the surface (the "scar" you talked about). These are the coronae.
Observations indicate that some parts of the surface appear less than 100 million years old.
Verona Rupes, a 10 km-high cliff, is the largest in the Solar System. Other, smaller, rupes exist elsewhere on Miranda and across the Solar System, but not quite as high as this one.

Alternate explanations include heavy crustal activity in the past. Tidal effects from Uranus may also have played a role (either at different times than the putative impact, or instead of it entirely). Gas giants strongly interact with their moons through tidal forces; cases include Io's volcanism, driven by Jupiter, and the covered oceans of Europa and Enceladus.

Here is a partial map of Miranda, displaying the coronae and several other features:


Are gas giants actually rare? Kepler data suggests that the vast majority of planets are smaller than 4 Earth radii, with larger planets quickly becoming vanishingly scarce. Yet somehow, our own solar system harbours not just one, but two giant planets. This seems suspicious.

Are giant planets expected to be as rare as the Kepler data suggests? Or is there some bias inherent in the transit method that makes detecting giant planets more difficult? Is it perhaps the long orbital period of a planet that formed outside the frost line that has caused this apparent dearth of giant planets?

There is an observational bias and it is taken into account when you see inferences about planet frequency.

The methods to find planets are inherently biased towards finding large, close-in planets. Both the transit and doppler-based methods suffer from this bias.

The paper you reference takes into account this bias to arrive at the statistics you highlight in your question. Note however that the figure for Earth-sized planets is that 16.5% of stars have at least one such planet in an orbit with periods less than 85 days. There are no planets of any kind at all in our solar system that have periods less than 85 days. The figure quoted for Neptune-sized planets is 31% with orbital periods less than 245 days and there are no such objects in the solar system. Finally, the figure quoted for giant planets is that 5.2% of stars have a giant planet in an orbit with a period less than 418 days. Again, no such planet exists in the solar system.

You are correct that it should be unlikely for giant planets to form close to their parent star and this is one of the reasons that they are rarely found there. It is widely believed that the close-in gas giant planets that are seen, have migrated inwards from a formation site further out.

That said, it is perhaps true that our solar system is a little unusual.
Whilst we do not yet know what fraction of stars have gas giants in Saturn-like orbits (because observations have not been going on long enough), some work (e.g. Rowan et al. 2015) suggests that only $\sim 3$% of Sun-like stars have an analogue of Jupiter (periods of 5-15 years and masses of 0.3-3 times that of Jupiter).


Is the diameter of the observable universe a relative quantity? The diameter of my observable universe is 90 billion ly measured in proper distance. But isn't lenght a relative quantity in the theory of Relativity? Could an observer moving at a different velocity than me measure a different diameter for my observable Universe?

If the observer has always been moving at a high speed, then yes, s/he will measure a different value than you. But if s/he accelerated to a high speed just now, you would obtain the same result.

The reason is that the radius of the observable Universe is defined as the distance light has had the time to travel since the Big Bang, and that does not depend on your instantaneous velocity.

However, if an observer started moving at, say, 0.866 times the speed of light when the Universe was born, then in the reference frame of that observer, the Universe would only be $\sqrt{1 - 0.866^2} = $ 1/2 the age measured by you, i.e. 6.9 Gyr. At this age, the radius of the observable Universe was only ~39 Gly, i.e. ~0.8 times the present value.

Note that in order for this result to be exact, the observer would have to slowly accelerate. The reason is that in an expanding Universe, a given "peculiar" velocity wrt. the surrounding matter will slowly decrease, because the comoving coordinate system expands with the Universe. For instance, an observer starting out with $v = 0.866c$ at the time the cosmic microwave background was released 380,000 yr after the Big Bang, would only be moving at 80 km/s today if s/he didn't keep accelerating.

If the observer were clever, s/he would probably realize that s/he did't live in an asymmetric universe where all matter seemed to be whirling by at close to the speed of light, but rather that the Universe were quite isotropic, but s/he had somehow acquired a high speed with repect to everything else. It would then be possible to deduce the "true" age of the Universe.


Dataset for globular star cluster Where can I find the
dataset(s) for a globular star cluster?

The datasets should have the coordinates, color index (for different wavelengths) , magnitude etc. of every star in the cluster.

I want to do data analysis on it just for fun.

I would start by doing a search on VizieR, select "Globular_Clusters" from the right-hand list box (this tells me there are around 900 catalogs in this category) and click "Find Catalogs".

Some of the returned results will be catalogues of stars within a given cluster, the density map on the right-hand side of the list will show these as a single point. Other catalogues will be lists of multiple globular clusters, the density map will show there is data from all over the sky.

I hope that's useful for you!


Length contraction of a star Something moving toward or away from us will be contracted in length according to special relativity.

As the M31 galaxy is moving toward us at great speed it's "depth" should appear slightly flattened for us. A sphere moving toward us at the speed of light will appear "pancaked" in lack of a better word.

This was predicted by Lorentz and has been proven in experiments.

Does this also mean that electromagnetic radiation (light) from a flattened star is focused in the direction of motion relative to the observer?

Note that for cosmic expansion, special relativity only applies when the spacetime region in question is approximately flat, which in our case happens when it is small.

  As the M31 galaxy is moving toward us at great speed it's "depth" should appear slightly flattened for us. A sphere moving toward us at the speed of light will appear "pancaked" in lack of a better word.

No, that's not correct. You have a worldline in spacetime, and its direction at an event is your four-velocity. The orthogonal complement of your four-velocity, i.e., all directions perpendicular to it, is your space. What Lorentz-FitzGerald length contraction says is that if you project a moving sphere onto your space, the result is contracted along the direction of motion.

The difference between this and what you say is that there is no mention of how the sphere will appear. The sphere is contracted in the inertial frame comoving with you at that instant. It doesn't appear contracted--in fact, if you took a picture of it with a hypothetical super-fast camera, the relativistically moving sphere will appear... still completely spherical. But not quite the same as a non-moving sphere if it has features on the surface, as Penrose-Terrell rotation makes some of its "back parts" visible.

  Does this also mean that electromagnetic radiation (light) from a flattened star is focused in the direction of motion relative to the observer?

Yes. Just as a vertically falling rain comes at an angle when you're riding in a car, the angles at which any signal comes are different in a moving frame compared to a stationary frame. This is described by relativistic aberration. Additionally, Doppler shift changes the intensities of a radiating source along different directions, making the overall effect relativistic beaming.


What is wrong with my calculations of Jupiter's apparent magnitude? I was searching for a formula to calculate the apparent magnitude of a planet knowing its physical characteristics and stumbled upon this answer.
The final formula given is:
$$m_{planet} = V_{planet} + 5 \log_{10}\left( d_{e-p} \right) - 5$$
In which,
$$ V_{planet}=-2.5 \log_{10}\left(a_p \frac{ r_p^2 }{ 4 d_s^2 } \right) - V_{sun}$$
where:

$m_{planet}$ is the planet in question's apparent magnitude,

$V_{planet}$ is the planet's absolute magnitude,

$a_{p}$ is the bond albedo of the planet,

$r_{p}$ is the radius of the planet,

$d_{s}$ is the distance of the planet from its star,

$d_{e-p}$ is the distance between the planet and the observer (here on Earth) in parsecs, and,

$V_{sun}$ is the star's absolute magnitude.

To test it out I tried to calculate the apparent magnitude of Jupiter as seen from Earth at an average opposition. The values then are:

$a_{p} = 0.343$,

$r_{p} = 69,911,000$ m,

$d_{s} = 778.57 \times 10^9$ m (5.204 AU),

$d_{e-p} = 628.97 \times 10^9$ m $= 0.0000203835294968$ parsecs (4.204 AU)

$V_{sun} = 4.83$

All the above values were taken from NASA's Jupiter fact sheet.
Plugging in the numbers and I get $m_{planet} = -10.38$, which is much brighter than the actual apparent magnitude of Jupiter at average opposition of $-2.x$. Where did I go wrong?

I think that should be $+ V_{Sun}$ in your second equation, not $- V_{Sun}$.  Otherwise, a fainter Sun (larger magnitude) would give you a brighter Jupiter (smaller magnitude).   Those need to go in the same direction.
Applying that change (assuming the rest of the calculation is right) would increase your answer by $2 \times 4.83 = 9.66$, which gets you much closer to the right magnitude.
Your estimate will now be a bit on the faint side - not sure why - but at least a lot closer than before.


Naked-eye limiting magnitude: mismatch between Wikipedia, online converter and Shaefer's formula I'm trying to make sense of the difference between the reference Sky Quality Meter (SQM) readings given in the Wikipedia article on Bortle scale, online converters like [1] that are based on Shaefer's paper [2], and an equation from [2] itself.
Wikipedia article gives naked-eye limiting magnitude (NELM), as well as corresponding SQM readings for each Bortle class. The original paper [3] by Bortle only specifies NELM, so the SQM column in Wikipedia must have been derived from the NELM values.
But in the very first value, SQM reading for Bortle class 1, the Wikipedia article fails to match what the online converter yields (other such converters use the same formula). Namely, the "excellent dark-sky site" NELM is 7.6–8.0, for which Wikipedia gives lower bound of SQM reading $21.99\,\mathrm{mag/arcsec}^2$, while the converter [1] yields $25.50\,\mathrm{mag/arcsec}^2$. The upper bound, NELM=8.0, is converted to $22\,\mathrm{mag/arcsec}^2$ in Wikipedia, while the online converter simply fails, yielding NaN. This failure is because the formula used by the converter has 7.93 as the maximum possible NELM.
If we use equation $(18)$ from [2] instead (with $k_v=0.3$ and $F_s=1$, as suggested in the text under the equation), we'll get the range of SQM readings $24.35\,\mathrm{mag/arcsec}^2$ to $26.32\,\mathrm{mag/arcsec}^2$, which doesn't match Wikipedia either.
Also, these values $\ge22\,\mathrm{mag/arcsec}^2$ seem to also conflict with [4]:

The natural level is around $21.6\,\mathrm{mag/sec}^2$ at a solar activity minimum.

Besides, an answer on Astronomy.SE implies that values of NELM larger than 6 don't make any sense at all.
So, how to make sense of the conflicting results? Which of these are right, which are wrong? Or do they correctly describe different things?
References

K. Fisher. "Conversion Calculator - NELM (V) to MPSAS (B) systems Rev. 8/2006"

Schaefer, B. E.. "Telescopic limiting magnitudes". Astronomical Society of the Pacific, Publications (ISSN 0004-6280), vol. 102, Feb. 1990, p. 212-229.

Bortle, John E. (February 2001). "Gauging Light Pollution: The Bortle Dark-Sky Scale". Sky & Telescope. Sky Publishing Corporation.

Dark Skies Awareness: "Sky Brightness Nomogram"

Being by no means an expert of light pollution, I still find my intermediate results worth sharing.
On the one hand, there is the Bortle Scale, and I find the flow chart from darksky.org the most accesible way (and probably it is also a reliable source):

This flow chart refers to the original publication by Bortle  from 2001 which was mentioned in the question. For all objects mentioned in a flow chart, one could determine their magnitude and obtain a magnitude-Bortle-scale conversion table (I might do that in a later edit).
On the other hand, there is a conversion site directly by a manufacturer of SQMs which converts ${\rm visual\,mags}/{\rm arcsecond}^2$ to ${\rm cd/m^2}$ according to the following formula:
$$[{\rm value\,in\,cd/m^2}]=    10.8\cdot 10^4 \cdot 10^{-0.4 \cdot [{\rm value\,in\, mag/arcsec^2}])}$$
The impression I got from my research is that there is no single, golden source publication linking NELM with physical properties such as ${\rm cd/m^2}$.
Last but not least, I stumbled up Roy Bishop's visibility factor (see German wikipedia) as more advanced (aka quantitiative) approach.


How are image credits expressed in Astronomy presentations? Sorry if this is a stupid question. I'm preparing an astronomy presentation, and I want to make sure I give credits where credits are due. But I see credits appear in different formats, for example:

ESA/Hubble & NASA, V. Antoniou; Acknowledgment: Judy Schmidt - a slash and an ampersand
ESA/Hubble & NASA, B. Nisini - a slash and an ampersand
NASA, ESA, and R. Humphreys (University of Minnesota), and J. Olmsted (STScI) - commas
ESA/Hubble & NASA, L. Stanghellini - commas and ampersand

If possible, I want all of the credits in my presentation to have the same format. So essentially, my question is: do a comma and a slash mean different things in image credits? Or can I write everything using the same format, e.g. for the four photos above:

ESA / Hubble / NASA / V. Antoniou / Judy Schmidt
ESA / Hubble / NASA / B. Nisini
NASA / ESA / R. Humphreys (University of Minnesota) / J. Olmsted (STScI)
ESA / Hubble / NASA / L. Stanghellini

There is no conventional format, really. For publications purposes if one uses data from a telescope or image or code created by someone else or similar, the source usually requests being cited or acknowledged in a form specific to their own taste. Some examples include what you quoted or “This paper makes use of the following ALMA data: ADS/JAO.ALMA#******. ALMA is a partnership of ... “.
Answering your question.

For ESA and NASA there is probably a difference between “/“ and “&”.
For a publication in a journal you must follow the exact wording.
If you just give a presentation no one minds if you use your favorite formatting while acknowledging all parties involved.


Do far away planets reflect light in the same way our moon/earth reflects light? Here’s a scenario:

Say you’re up in space and you’re far enough way that you can see the planet earth. Whilst looking down, you can see  blue sky, blue water and green land and white clouds and darker shades of different colors. This is because our sun is reflecting light off our planet that we can see these colors.

Now you’re up in space and you’re far enough way that you can see the planet Jupiter. If you looked, would you be able to see colors in the same way we’ve seen from Juno’s pictures. I see red, white, orange… The sun is so much further away, how can the colors we see be just as bright/visible to us? The same question holds true for planets outside our solar system. They’re so far way from there light source, yet I can see colors on them. How come the distance of the light source doesn’t affect the brightness of the colors I see on these planets?

Like if I was to turn on a light bulb in a huge dark room and shine that light in the middle, the surrounding areas would be much brighter than the furthest point of that room/and at some point there would be no light reaching an even further point.

Bear in mind I’m a basic idiot when it comes to this stuff.

The short answer is yes, the outer planets of the Solar System reflect light in much the same way that the Earth does, but exoplanets are too far away for us to see clearly.

If you looked at Jupiter, you would be able to see the same colors that show up in Juno's photographs.  Even though the sun is less bright out there, it is still bright enough to illuminate the surface.  At Earth, the sun's irradiance (its intensity) is ~1361 W/m$^2$.  At Jupiter, the solar irradiance is ~50.26 W/m$^2$.  This is about the intensity of a common lightbulb.  Going further out, the solar irradiance at Pluto is ~0.873 W/m$^2$.

Now, this is not a very high intensity.  When you move away from the sun, things get darker and darker.  However, we have special instruments to counteract that.  We use telescopes to gather up large amounts of light and focus them, so that we are able to get a clear picture.  That is how telescopes work.  We have pictures of Jupiter from the Earth, most notably from the Hubble Space Telescope, and you can see that they are, indeed in color. (The black dot in the middle of the Great Red Spot is the shadow of Ganymede, one of Jupiter's moons).

There are plenty of resources online to learn how telescopes work.

As for planets outside our Solar System, we can't actually get pictures of them.  Any images you've seen of exoplanets are actually drawings made by artists.  They're so far away that we just don't have the resolving power to see them clearly.


Why wasn't there any oxygen in the beginning of the Earth? In the total beginning of the Earth there was CO2, N2 and H2O. So there was O2 but always in bond with Carbon and hydrogen, but what is the reason for that? Is O2 always reacting with free carbon atoms?
The temperature of the Earth was is those beginning days around 1200degree Celsius. Is that also a factor of those bindings?

The 1200 C temperature is a big clue, given the auto-ignition temperatures of most "stuff". https://en.wikipedia.org/wiki/Autoignition_temperature

Any "free" oxygen would immediately "burn" with carbon and just about anything else, yes.


Perihelion dates and parameters I have to find the perihelion and aphelion data for the 20th century for Earth. I found this table http://www.astropixels.com/ephemeris/perap2001.html, but no such table for the past century.

I think it can be found using JPL Horizons web interface, but can not use the code (the boundaries of the task I was given) and do not want to check 36000+ entries by eye to find maximums and minimums: this is not the general way to solve the problem.

I believe that there is some relationship between  Keplerian elements and perihelion (argument of periapsis or mean anomally) but after checking the file from JPL Horizons I could not find any such relationship. Therefore, I have two questions:

1) Is there an archive with the dates of peryhelion and aphelion in the past?

2) What is the relationship between Keplerian elements and the perihelion (of course, if such exists)? Unfortunately, I was not able to find it from the picture on https://en.wikipedia.org/wiki/Orbital_elements. More precisely, is it true that argument of periapsis or mean anomally is equal to 0 at the perihelion?

In googling I stumbled across the following list of Perihelion dates and times back through the beginning of the twentieth century, and forward to then end of the twenty-first.

http://cococubed.asu.edu/data/perihelion_earth_1900-2100.dat

This seems to agree with the values generated by the US Naval Observatory's Perihelion and Aphelion page, which accepts individual years as values. (Site is kind of fickle on Mondays for some reason; perhaps scheduled downtime.)

https://aa.usno.navy.mil/data/docs/EarthSeasons.php

Mean Anomaly, for an elliptical orbit, can be thought of as the quasi-angular parameter representing the angle swept since the time of periapsis of a hypothetical object in a circular orbit with the same radius as the semimajor axis of the elliptical orbit. As such, yes, it will always be 0 when your actual object is at periapsis.

Argument of Periapsis, on the other hand, is the angle measured in the plane of the orbit, from the Ascending Node to the Periapsis of the orbit, in the direction of travel around the orbit.  It remains the same regardless of the position of the object in the orbit. As such, it will only be 0 at periapsis when the Ascending Node and the Periapsis are the same point.


How are sulfuric acid clouds able to completely cover Venus? It is a well-known fact that an opaque layer of highly reflective clouds of sulfuric acid completely shrouds Venus preventing its surface from being seen from space in visible light. This is mentioned in the official NASA site and also in some papers:

The atmosphere of Venus is made up mainly of carbon dioxide, and
thick clouds of sulfuric acid completely cover the planet. (nasa.gov)

The clouds that completely shroud the planet are located in the altitude range of approximately 50–70 km with upper and lower haze extending up to about 100 km and 30 km correspondingly, thus occupying the upper troposphere and the entire mesosphere.
Titov, D.V., Ignatiev, N.I., McGouldrick, K. et al. Clouds and Hazes of Venus. Space Sci Rev 214, 126 (2018). https://doi.org/10.1007/s11214-018-0552-z

The sky of Venus is fully covered by thick clouds of sulfuric acid that are located at a height of 45-70 km, making it hard to observe the planet's surface from Earth-based telescopes and orbiters circling Venus. (eurekalert.org)

But in this question, user @Cornelisinspace came to the conclusion that "there is much less sulfuric acid than water !".  So, if there is less sulfuric acid than water (which is already present in trace amount):

how is it able to "completely spread over" and "fully cover the planet"?

why is the cloud cover so thick (so thick that it reflects most of the solar radiation (except in some dark patches where it gets absorbed) making the planet opaque)?

The Venusian cloud deck is highly acidic. How much water there is in Venus's atmosphere above and below the cloud deck is irrelevant. What matters is how much sulfuric acid there is within that cloud deck.
And it is a lot, with the cloud deck droplets being between 75% to nearly 100%  sulfuric acid by weight. Since the atomic weight of sulfuric acid is about 5.44 times that of water, this weight range corresponds to between 35.5% to nearly 100% sulfuric acid by volume, or by molecule. It's only in the polar regions where the sulfuric acid content of the cloud droplets is at the low value of 75% by weight.
The top of Venus's thick cloud deck is marked by where incoming solar radiation can dissociate the sulfuric acid in the cloud deck. The bottom of Venus's thick cloud deck is where the high temperature (temperature increases with increasing depth in the cloud deck) dissociates sulfuric acid. Sulfuric acid is stable within these bounds.
The dissociation at the top and bottom of the cloud deck means there has to be a source of sulfuric acid to counterbalance the dissociation. The source is the dissociation of other molecules at the top of the cloud deck, plus the large amounts of sulfur dioxide ("large" compared to other trace components) in Venus's atmosphere. The dissociation above the cloud deck creates atomic oxygen. This atomic oxygen is very volatile and readily combines with sulfur dioxide to form sulfur trioxide. This in turn readily combines with water to form sulfuric acid.
In short, conditions just above the cloud deck replenish the sulfuric acid in the cloud deck.
References:
Barstow, J. K., et al. "Models of the global cloud structure on Venus derived from Venus Express observations." Icarus 217.2 (2012): 542-560.
Lee, Yeon Joo, et al. "Vertical structure of the Venus cloud top from the VeRa and VIRTIS observations onboard Venus Express." Icarus 217.2 (2012): 599-609.
Young, A. T. "Are the clouds of Venus sulfuric acid?." Icarus 18.4 (1973): 564-582.


How are people converting intensities in Janskys to Kelvin? I'm reading some ALMA proposals and I am often seeing a conversion from Jy to Kelvin when, for example, people quote noise levels or source flux.

For example on their sensitivity calculator
(https://almascience.nrao.edu/proposing/sensitivity-calculator) 1 mJy is equivalent to around 1 K when the angular resolution is 1". I can tell it somehow depends on the resolution because it won't give me a conversion unless I enter a resolution.

What's happening under the hood here? And why does their documentation treat sensitivities entered in Kelvin differently from when they're entered in Jy? Is this an interferometry thing?

Thanks

This is, indeed, a result of how we measure things in radio astronomy.  It's not just interferometry, but radio astronomy in general.  The thing they're referring to is a concept called "brightness temperature".  In the low frequency limit (valid for radio astronomy), we can use the Rayleigh-Jeans approximation, which gives us the expression
$$
T_B = \frac{S_\nu \lambda^2}{2k \theta_s^2}
$$
where $S_\nu$ is the flux density, $\lambda$ is the wavelength, $k$ is Boltzmann's constant, and $\theta_s$ is the size of the source (or the resolution in this case).  As long as your source and wavelength are constant, brightness temperature and flux density can be interchanged.

The reason astronomers talk about temperatures instead of fluxes sometimes is twofold.  First, a large number of objects examined by radio astronomers are not blackbodies.  Therefore we cannot calculate their temperatures like we normally would.  Instead, we calculate what their temperature would be if they were blackbodies.  This at least gives us a value to compare them to other things.

Secondly, when discussing radio telescopes and antennae, we usually say that various parts of the antenna contribute different 'temperatures'.  We can quantify system temperatures to help us understand noise.


Are there any Stars we know don't have planets? We have been looking for planets around other stars for a while (see related https://space.stackexchange.com/questions/1484 ) and we are finding lots of planets, some of them are Earth like According to an the answer at https://space.stackexchange.com/questions/4062 there are over 14,000 stars within 100 light years of Earth, the answer goes on to say there are "1,500 potentially habitable planets within 100 light-years" of Earth

Based on current trends it looks like there are lots of stars with lots of planets.  I am beginning to assume that our solar system is not unique and that  every star has several planets.

Are there any Stars that we know are without any planets?

I am beginning to assume that our solar system is not unique and that every star has several planets.

Not quite, but indeed a study published in Nature in 2012 found that, based on our observations so far, roughly 17% of stars host Jupiter-mass planets, 52% host "Cool Neptunes" and 62% host Super-Earths. (Note that these percentages do not add up to 100%, because they are not mutually exclusive possibilities). This was particularly surprising, because half of all visible stars are believe to be in binary systems, which would make planetary systems very unstable, but some binary systems have been found to have planets too.

So indeed it seems the majority of stars have planets, but it's very unlikely that all of them do.

However, the exact answer to your question "do we know of any stars with no planets" has got to be "no", because there remains a possibility that they have planets that we simply haven't been able to detect, because of limitations in our techniques to detect them.


Gravitational lensing I was reading about gravitational lensing. I was just curious to know if there are any formulae or methods to find the angle at which the light deflects due to strong gravitational lensing?

And how far should that massive object be present so that we can observe multiple images of the observing object?

Absolutely:
$$\theta = \frac{4GM}{c^2b},$$
where $\theta$ is the angle of bending, $G$ is the gravitational constant, $M$ is the mass of the star/blackhole/whatever, $c$ is the speed of light, and $b$ is the "impact parameter", the distance from the light ray to the center of the mass on its closest approach if it was NOT deflected (i.e., $b=0$ for an impact).


How was Cruithne's orbital time calculated? How was it determined that Cruithne takes 770 years to complete its cycle with Earth? Janus and Epimetheus were disovered long enough ago, and have short enough cycle times, that we've had time to directly observe their 8 year cycle more than 6 times over. But Cruithne was only discovered 30-ish years ago, so certainly it's cycle time wasn't determined by simple observation.

So what method was used to arrive at this 770 year number?

It's a good question!

I went to the JPL Horizons web page and looked up 3753 Cruithne and saw that the current orbital solution is based on observations from 1973 to 2018, about 45 years. That's a fraction of an orbit, but very careful measurements of it's position on the celestial sphere (RA, Dec) over time will allow astronomers to calculate its position and speed by fitting those observations to many possible trajectories and finding the one that's the best fit.

JPL/HORIZONS               3753 Cruithne (1986 TO)         2019-Feb-14 23:48:43
Rec #:    3753 (+COV) Soln.date: 2018-Sep-27_06:26:23    # obs: 677 (1973-2018)

The orbit is elliptical but has a nearly-identical semimajor axis as Earth, so it's period around the Sun is one year just like Earth. That means that it in 45 years it has been possible to observe about 45 orbits around the Sun, allowing astronomers to nail down its orbit with great precision.

I plotted the distance from Earth to Cruithne from 1973 to 2018 below from the Horizons data. At a half of an AU and 0.1 arcsec astrometric position precision (averaging over many observation) that gives the position to tens of kilometers over a span of decades.

Then using accurate orbital modeling using the gravity of all the solar system's bodies as described here it is possible to project the position with great accuracy, as long as non-gravitational forces like unpredictable outgassing aren't too strong, and predictable non-gravitational forces like solar radiation pressure can be modeled as described here.

Over time the orientation of the elliptical orbit slowly precesses around the Sun due to the gravitational effects of Earth. This is the ~770 year period you've mentioned.

In the video Near-Earth Asteroid 3753 Cruithne they hold the position of the Earth fixed and you can see how Cruithne's orbit sloshes back and forth.

This GIF is too large to embed here, but it demonstrates the sloshing nicely: https://en.wikipedia.org/wiki/File:Animation_of_3753_Cruithne_orbit.gif


Could space time curve from the vantage point of earth and out into the universe, and would d = 1 / P in that case give false estimates? The seventeenth century saw a revolution in astronomy. The invention of the telescope and the acknowledgement of the heliocentric system triggered a race amongst astronomers to measure the parallax of stars - the annual displacement of stellar positions due to Earth's motion around the Sun. In the late 1830s these measurements enabled astronomers to determine the distances to a handful of stars for the first time. From the 1850s onwards, the application of photography to astronomical observations transformed the practice of charting the sky, allowing the compilation of larger and larger catalogues of stellar positions and distances.

That robust marvel of trigonometry applied to observations of other stars, lets me look up into the sky for 6 months, measure the apparent motion, then divide 1 by that, and voila, the distance to the star.

For example, the distance to Alpha Centauri, https://www.wolframalpha.com/input/?i=1+%2F+.77

These distances are used in all textbooks, all databases, all conversations, and does not correct for any space-time curvature whatsoever.

To draw a line on the sky, and use trigonometry, seems almost a bit too simple. Anyone have any thoughts on if space-time might curve in weird ways sort of like how a fish eye lens on Instagram distorts an image?

Yes, but the amount of curvature is too small in scale proportionally to affect such things, especially when one thinks of how away from us the objects are that we observe.  Even the moon is a quarter of a million miles away, and the Earth's gravitational well is quite small.


Time-density and time-temperature functions of the Universe since the Big Bang I am looking for these... functions. I want to be able to say, "at a 1873. second there were around 54230000000 K temperature and 0.00435 kg/m^3 density".

On the net I could find only tables or diagrams, as this: .

(Here is a link to the page it accompanies: http://www.astro.ucla.edu/~wright/BBhistory)

I am practically interested in the interval between the first seconds and the first some ten millions of years.

Is it possible? Some simple formula were also enough (although I think, there are different formulas for the different eras after the Big Bang).

Both the temperature and the density of a toy universe can, in a way, be linked to the age of the universe ($t$). Certain properties of the universe are dependent upon a scale factor $a(t)$ that is a function of time. The expansion of such a universe can be written in the Friedmann-Lemaitre-Robertson-Walker (FLRW) metric, generally written as
$$ds^2 = dt^2 - a(t)^2(d\Sigma^2)$$
where, in spherical coordinates, $d\Sigma$ is given by
$$d\Sigma^2 = \frac{dr^2}{1-kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2$$
where $k$ is a constant. This shows that, in a toy universe, $r$ and $dr$ is proportional to $a(t)$, and so in a perfectly spherical toy universe, you could relate the volume to $t$. We know that, for a sphere,
$$V=\frac{4}{3}\pi r^3$$
and because $d\Sigma^2$ (and by extension $r^2$) is related to $a(t)^2$, we can relate volume to time. If you knew the mass-energy of the universe, you could then relate the change in density to $t$. Unfortunately, our universe does not appear to be perfectly spherical, and so this may not apply.

You can relate something else to the age of the universe: the temperature of the cosmic microwave background. It has cooled over time, and is presently at about 2.7 Kelvin. While the temperature of the CMB may not be governed by the scale factor $a(t)$, it is certainly governed by time, and thus one you could in fact write the temperature of the universe as function of time. Unfortunately, I haven't been able to find data on the evolution of the CMB, but I can try, and perhaps this will lead you to an answer.

To summarize: In a perfectly spherical universe, assuming you know the matter-energy content and the scale factor $a(t)$, you could write an equation for the density of the universe. In any universe, you can also write an equation for the temperature of the CMB - the "temperature of the universe".

I hope this helps.

Source for FLRW metric: http://www.universe-galaxies-stars.com/Robertson-Walker_coordinates.html


Generate an uniform distribution on the sky How to generate an uniform distribution of stars on the sky ? What I want is make a simulation of random points following uniform distribution on a part of the sky (making the assuption that we have the same amount of stars in any direction). The problem is how I do that because of the cos(dec) on the RA ?

What I did in python is this

import numpy as np
RA = np.random.uniform(-180,180)
Dec = np.random.uniform(-90,90)

But it is totally false, since we will have more points on the poles (if I do many points, this will clearly not be uniform)...

Thanks a lot in advance!

Random points on the surface of a sphere can be generated by allowing the azimuthal angle $\phi$ to take a uniformly distributed random value between 0 and $2\pi$. To convert this to RA in degrees you multiply by $180/\pi$. To convert to hours, minutes and seconds you divide the $\phi$ in degrees by 15, which gives the hours, divide the remainder by 60 which gives the minutes and then divide the remainder of that by 60 to give the seconds.

The declination angle $\delta$ is not uniformly distributed, because the area covered at an angle $\delta$ goes as $\cos \delta$.

Note the potential confusion here between declination and the polar angle $\theta$, which is conventionally measured down from the pole. The area of a strip of width $\Delta \theta$ at a given $\theta$ on a unit sphere is $$\Delta A = 2\pi  \sin \theta\ \Delta \theta = 2\pi  \cos (\pi/2 - \delta)\ \Delta \delta = 2\pi  \cos \delta\ \Delta \delta$$

To translate this into something we can use, we say that the probability of finding a star in a small range of declinations
$$ dP \propto \cos \delta\ d\delta,$$
so the cumulative probability of finding a value of $\delta$ between $-90^{\circ}$ and $\delta$ is
$$ P(\delta) = \frac{\int^{\delta}_{-90^{\circ}}\cos \delta^{\prime}\ d\delta^{\prime}}{\int^{+90^{\circ}}_{-90^{\circ}}\cos \delta^{\prime}\ d\delta^{\prime}},$$
where the denominator makes sure the proportionality is correctly normalised. From this, we obtain
$$P(\delta) = \frac{1}{2}(1 + \sin \delta) \tag*{(1)}$$

The procedure for then obtaining a random declination is to assign a uniform random number to the cumulative probability $P$ between 0 and 1. Then from equation (1) we have
$$ \sin \delta = 2P -1$$
$$\delta = \sin^{-1}(2P-1),$$
where $P$ is a random number between 0 and 1.


How (the heck) does Astrometry.net work? What magic is this?

This answer to my question Astronomy detective question: what part of the sky are these photos of? What are a few of the stars? nails it, and the output (linked there) contains the following images along with a whole lot more information.

The images are analyzed by astrometry.net

How does it do what it does? If I understand correctly you can upload random star images with a wide variety of fields of view (from telescopes to cell phones) and random orientations and no further information, and it figures it out.

Is there a way to explain how the algorithm works with more detail than "a computer matches the dots to known stars"?

Related question about these cell phone images: Milky Way with a cell phone, how is this possible?

I've been trying to figure out the technical details of astrometry.net for quite some time. As others already pointed out, the main input to the whole process is a list of stars. I will not go into details on how astrometry.net does it, just note that you can either use its internal simplexy algorithm or use SExtractor. In the end you need a list of coordinates for stars (plus optional flux/intensity/brightness and background).
The .xyls files are used to store these input values (small example):
 x-coord(px)   y-coord(px)        flux  background
 1008.911987,   557.925659,  10.556271,  32.320175,
 1449.509277,   643.280212,   6.580036,  27.963276,
  185.978119,  1253.869751,   5.525373,  27.713015,

Background is approximated before/during star extraction (SExtractor has a function for that).
Now the algorithm goes into creating quads. The cited paper always talks about 4 stars that form a quad. But by my own experiments they seem to be just triangles. Maybe they should have better named them "Asterism" (the code has a lot of very bad naming, as we soon will discover, took me hours to figure out).
So we have a triangle and want to know if a similar triangle exists anywhere in the selected astrometry.net index. So this is where the real magic happens IMO. The papers talks about geometrically hashed lookup. IMO the word "hash" is a bit misleading here as it has not much in common with real hashes. Real hashes should normally give very different results for small changes in the input (e.g. checksum or hash-table bucket distributions). I would rather say that they perform a geometric transformation where the result must be a one-dimensional value (e.g. a double value).
Abstractly speaking we want to map numerous properties to a single value. The value should vary very little if the overall properties also vary little, so from looking at the delta of both translation results we can deduct how similar the two objects were. One such property could e.g. be the angle between two lines.
From here it should be "obvious" how one can use this approach to drive a search. In this simple one-dimensional case we could e.g. use a binary search. For astrometry.net this problem gets technically more complicated, as we need to search for two "hashes". This is done with a KD-Tree. The one used by astrometry.net is optimized for the pre-built index files, so they can be accessed very fast and without much (memory) overhead. Basically a KD-Tree can optimize the question "give me the closest point(s) to X/Y".
In the picture below I tried to visualize how we can search for similar triangles by reducing the question to two numbers. We basically search for the normalized blue vectors (or ones that are pretty close). It should be obvious that this eliminates any rotation and scaling in the question asked, so any similarly shaped triangle will match, regardless of orientation or size. IMO it simply boils down to the fact that the shape of a triangle is defined by two parameters, e.g. not sure why they didn't use two angles.

In reality we have to make the lookup multiple times to also search for flipped and/or inverted variations. After a similar triangle is found (this will happen a lot), the process goes into the verification step. The resulting triangle will give the program the hint on how all other stars must be translated to match the given triangle. With that it can try to match all other stars with its know star catalog.

Query images may contain some extra stars that are not in your index catalogue, and some catalogue stars may be missing from the image.

These can be seen as distractors and conflicts? in the debug output of the program:
verify: logodds -1.38629, 0 matches, 0 conflicts, 1 distractors after 0 field objects.
verify: logodds 333.123, 99 matches, 0 conflicts, 115 distractors after 213 field objects.
108 matches, 424 distractors, 2 conflicts (all sources)

I believe these numbers are not "really" accurate as the checker will probably bail out once it thinks it is impossibly a match (e.g. after certain distractors are found without any matches). Anyway, this is more or less to whole process in detail. I left out some of the magic, e.g. there are a lot of probability checks for speed. Also the way potential quads are chosen from all stars is quite elaborate in reality (and should probably match closely to how this was done when the index files were created).
Side note on indexes: As they contain x/y positions for known stars and stars wander a little over time (a few faster, most very slow), the index can get outdated and start to not match (no idea if this would be in a decade or a millenium). Regular star catalogs give x/y position at a given epoch time plus x/y-speeds to calculate the actual position in any given point of time. IMO with modern CPUs this can be done in seconds for millions of stars (so time independent indexes could be doable, although probably not for the initial triangle matching, but surely for the verify phase). Also with the new ESO Gaia data releases there shouldn't be any gaps anymore (as noted by astrometry.net as small holes in the USNO-B1.0 catalog). But it seems the official available indexes haven't been updated yet to use the new gaia catalog.
Disclaimer: This knowledge was acquired mostly by reading and testing with the astrometry.net source code. So any conclusions I made could be wrong. But I would say it all makes sense if put together. Below I'll give a few more details into the actual implementation inside astrometry.net.
Edit: After reading https://iopscience.iop.org/article/10.1088/0004-6256/139/5/1782 I came to the conclusion that real quads with four stars are probably used in indexes with smaller view angles (zoomed in). I used a picture taken with a regular 55mm DSLR lens for my tests. This would make sense and it basically is exactly what an n-dimensional tree is made for (the question for the closest neighbors now involves 4 parameters).
Each index_t contains two KD-Trees, namely codekd and starkd. The tree codekd contains the information of all "quads", while starkd contains regular star coordinates for later verification.
The "hashing" bits can interestingly be found in "solver.c" in the function "check_inbox".
static void check_inbox(pquad* pq, int start, solver_t* solver) {
    int i;
    double Ax, Ay;
    field_getxy(solver, pq->fieldA, &Ax, &Ay);
    // check which C, D points are inside the circle.
    for (i = start; i < pq->ninbox; i++) {
        double r;
        double Cx, Cy, xxtmp;
        double tol = solver->codetol;
        if (!pq->inbox[i])
            continue;
        field_getxy(solver, i, &Cx, &Cy);
        Cx -= Ax;
        Cy -= Ay;
        xxtmp = Cx;
        Cx = Cx * pq->costheta + Cy * pq->sintheta;
        Cy = -xxtmp * pq->sintheta + Cy * pq->costheta;

        // make sure it's in the circle centered at (0.5, 0.5)
        // with radius 1/sqrt(2) (plus codetol for fudge):
        // (x-1/2)^2 + (y-1/2)^2   <=   (r + codetol)^2
        // x^2-x+1/4 + y^2-y+1/4   <=   (1/sqrt(2) + codetol)^2
        // x^2-x + y^2-y + 1/2     <=   1/2 + sqrt(2)*codetol + codetol^2
        // x^2-x + y^2-y           <=   sqrt(2)*codetol + codetol^2
        r = (Cx * Cx - Cx) + (Cy * Cy - Cy);
        if (r > (tol * (M_SQRT2 + tol))) {
            pq->inbox[i] = FALSE;
            continue;
        }
        setx(pq->xy, i, Cx);
        sety(pq->xy, i, Cy);
    }
}

The pquads are the potential "asterisms" we are currently creating, and this function's job is to set the "query" parameters via "set[xy]" on the bottom. Those are the actual values later looked up in the codekd tree.
A pquad is basically a line between two points (also called backbone-stars) called fieldA and fieldB (which are actually indexes to get an x/y position). Additionally it must have at least one additioanl xy point (exactly one in our triangular case). The pquad also contains sintheta and costheta (set in check_scale).
    double dx, dy;
    dx = field_getx(s, pq->fieldB) - field_getx(s, pq->fieldA);
    dy = field_gety(s, pq->fieldB) - field_gety(s, pq->fieldA);
    pq->scale = dx*dx + dy*dy;
    pq->costheta = (dy + dx) / pq->scale;
    pq->sintheta = (dy - dx) / pq->scale;

As we see both code parts translate and scale the vector AC in relation to the line AB.
I hope this info is useful to somebody, even if it got a bit long!


Calculate Atmospheric Extinction Using Source Altitude Angle I am trying to code up the following article http://dl.acm.org/citation.cfm?id=2349512 (alternative free access version) to calculate the light reflected from the moon to a given point on earth at a given time. However I occasionally get a gross error when calculating the distance of the light path through the atmosphere (Figure 2. of the article.)

The approximation given is to take the inverse sine of right ascension in radians, however whenever RA approaches pi, the number blows out to 8e15. Can you suggest a better method?

You seem to be referring to the computation of $m$ in equation (15) for the extinction as the light from the source passes through the atmosphere. The author intends to say that $m\approx \frac{1}{\sin(\alpha_s)}$, where $\alpha_s$ is the altitude of the source (in radian), so restricted to the range $0$ to $ \pi/2$. (Note this is not the inverse of the sine function as the author says but the reciprocal of the sine function)  The altitude is the angle measured from the horizon to the source, which is $=\pi/2-\theta$ where $\theta$ is as in figure 2. Note this blows up when $\alpha_s=0$ ($\theta=\pi/2$) because in the simple approximation being used the path length through the atmosphere in infinite.

The approximation being used here is that of a flat Earth, and $m$ is the number of equivalent atmosphere heights that the light path travellels through


Is a black hole a perfect sphere? I know all about how black holes form and why their gravity is so strong. However, is the gravity equally powerful in all directions? Will the event horizon appear as a perfect sphere?

You cannot see the event horizon. That being said:

A non-rotating black hole, free of external influences, has perfect spherical symmetry. All its properties are exactly the same in any direction, period. This is the Schwarzschild metric.

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

Even if it is electrically charged, if it's non-rotating, and free of external influences, it is still perfectly spherically symmetrical - this is the Reissner–Nordström metric.

https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric

As soon as it begins rotating, the black hole is no longer in perfect spherical symmetry. It acquires an ergosphere, which is shaped like a flattened ball. This is the Kerr metric.

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

Significant external factors can distort the shape of the event horizon. Two black holes merging will go through a process where spherical symmetry is lost temporarily:

https://www.youtube.com/watch?v=JOQMvyLYmd4

Technically, anything near a BH would slightly distort the metric of spacetime, but in practice this would not be easily measurable unless the external factor is very large (another very massive object). So for most practical purposes, the metrics described above would apply.


Can we find rocks from Venus or Mercury on Earth? As we know, we can find rocks from Mars on Earth, how about Venus and Mercury? Is that not found yet? Or it is impossible to find them because they are closer to the Sun, and debris won't go far away from the Sun?

You can think of it in terms of Hohmann transfer orbits, which define the minimum $\Delta v$ that needs to be applied to bring something from one orbital radius to another orbital radius when orbiting a massive body. This calculation takes into account that the two objects have Keplerian orbits where the objects begins with at least the orbital speed of the initial orbital radius.

The Hohmann $\Delta v$ is given by
$$\Delta v = \sqrt{\frac{GM_{\odot}}{r_1}} \left(\sqrt{\frac{2r_2}{r_1 +r_2}} -1 \right),$$
where $r_1$ and $r_2$ are the initial and final radius from the Sun.

Conceptually what you have to do is give a rock enough energy to escape from the planet and then give it an additional $(m/2)(\Delta v)^2$ of kinetic energy to get it to transfer into the other orbit. If the ejection speed is $v_{ej}$ then
$ v_{eg} > \sqrt{(\Delta v)^2 + v_{esc}^2},$
where $v_{esc}$ is the escape velocity.

The numbers for Mercury $\rightarrow$ Earth are $\Delta v = 9.2$ km/s and for Mars $\rightarrow$ Earth $\Delta v = -2.6$ km/s (you have to slow it down to allow it to fall inwards).

The escape velocities for these planets are 4.35 km/s and 5 km/s respectively (so almost the same).

This means you need to give a rock more kinetic energy to get it to Earth from Mercury as from Mars. In the case of Mars, the transfer kinetic energy is almost negligible once the rock can escape Mars' gravity. In the case of Mercury, the rock needs to be given an initial ejection velocity of $> \sqrt{9.2^2 + 4.3^2}= 10.1 $ km/s. This compares with $> \sqrt{2.4^2 + 5^2}= 6.5$ km/s for Mars. At lower ejection speeds most of the ejected objects will be reaccreted by the planet.

Against this, the leading theory to explain migration of rocks between planets is high velocity impacts. Objects falling from much further out will hit Mercury with greater speeds than Mars and impart greater energies to the ejecta.

A paper addressing the possibility of Mecurean meteorites was presented by Gladman & Coffey (2008). They concluded that once ejection speeds are large enough ($\sim 10$ km/s) to produce Earth-crossing ejecta, that significant accretion of meteorites should take place. Several per cent of high speed ejecta should impact the Earth (or its atmosphere at least) within 30 million years. This compares with an efficiency a factor of 2-3 higher for Mars.

There are various reports and speculations that at least one meteorite in existing collections (NWA7325, pictured) may have come from Mercury. See here for example. It appears that the main problem is getting agreement on what the chemical signatures of such meteorites are.

Accretion of material from Venus is a different matter. The required ejection velocities are higher because the escape velocity for Venus 10.4 km/s. But more importantly, drag in the dense Venusian atmosphere would prevent anything emerging from the planet with anything like these speeds.


Has life been discovered outside of Earth? I've just watched a movie that made me curious as to whether life has really been discovered on another world such as Europa??

No life has been discovered out of the Earth. It is still discussed that bacteria once existed at Mars.


How are non-visible electromagnetic waves detected that originate from space? How are non-visible electromagnetic waves from space detected and interpreted by the receiver? Is it in the same manner, as say radio waves produced on Earth, where the EM wave interacts with the telescope antennae and causes electrons to oscillate and as such is recorded as an electrical signal which can then be interpreted as an image?  I have tried to research exactly how the receiver converts EM waves into a "signal" we read but generally, information just states that the em wave is detected, amplified, and stored by a computer.

All electromagnetic radiation is fundamentally indistinguishable - i.e., all rays of light (electromagnetic radiation) are composed of photons - and they only vary in their properties, such as the energy and phase of the light. Thus, detecting light works fundamentally the same way for any electromagnetic radiation, e.g., the light is typically concentrated by some kind of system (for example, in telescopes and in our eyes this system is composed of lenses), though this is not completely necessary always, and then the light is absorbed by a detecting device; however, in practice light of certain energy/frequency requires different detecting devices. In your eye/brain system, the detection device is an array of the photoreceptor cells (rods and cones) in the retina. For observing light from objects out in space, we use telescopes, where there is a range of such detection devices that have been invented over the years.

(https://www.nasa.gov/directorates/heo/scan/communications/outreach/funfacts/txt_radiowaves.html)
Modern telescopes utilize photodetectors (e.g., a CCD chip) that convert light into electrical current which can be read into a computing device and interpreted, and many more complicated optical systems (e.g., adaptive optics) are used to study the light received from distant objects in space. These detectors/receptors are based on solid-state physics (another example of how quantum theory helped revolutionize astronomy research), since they usually are composed of a P-N junction (which is just one application of P-N junctions for harvesting light and converting it to energy). As explained above, all electromagnetic telescopes fundamentally use photodetectors, but how the light is harvested at the photodetector depends on the energy of the light.
The wiki article for solid-state photodetectors explains the various types and the mechanisms of their operation, if you have more questions about this please start a new question. Note, that biological photoreceptors operate on different mechanisms than solid-state ones.
For visible light, obviously, the telescope focuses the light onto the viewer's eye! But you asked about non-visible light specifically. Visible light is only a very small sliver of the electromagnetic spectrum. On the low-frequency end of the spectrum are radio waves, which require very large antennas to collect enough radio energy to study them and extremely sensitive receiving equipment.
As the energy (frequency) of the light increases we next have infrared which is often detected with the same telescopes used to detect visible light, just with a different photoreceptor. The light that has higher energy than visible light - UV, X-ray, an gamma-ray - where most of the light at these wavelengths is absorbed by the Earth's atmosphere, so observations at these wavelengths must be performed from the upper atmosphere or from space.


Methodology for size of Universe I understand that the Universe could be either finite or infinite and that it is, at the least, "very large". :)

My question is, why doesn't the math involved in making the CMB and other predictions, in effect, tell us the size of the Universe? We know how hot it was then and how cool it is now. Does scale not affect these calculations?

In other words if a given volume is a million degrees on "day" one and it's now 2.5 degrees don't those temperatures change if you make the volume 200 times larger?

The calculations make the Observable Universe, the CMB and the temperatures then and now all come out right and validate the Big Bang theory. How can they do that with both a smaller Universe (Observable Universe) and the much larger actual Universe?

Is it just because whatever the size of the Universe is, it was all 1 million degrees (example) at one point? Otherwise, it seems the predictive math involved would produce different numbers for different volume.

There's a subtle misunderstanding there, which is rarely made clear in discussions of the Big Bang and the current model of the universe.

Big Bang did not explode to fill the universe. Matter and energy has always filled the entire universe pretty uniformly, but 13+ GYears ago, the universe was very small and its energy density was very high. But it was still completely filled. Our theories combined with our observations don't tell us whether it was infinite or finite, just that whatever it was, it was was very dense and very energetic.

From that point -- what came before, if anything is utterly beyond the current scope of physics and is pure speculation -- space itself started to expand very, very rapidly and since the mass-energy was conserved, the density of mass-energy rapidly decreased.  This hasn't stopped, though it has slowed.  Our best measurements to date suggest that it will never stop and probably is speeding up some.

When we look at the CMB, we're looking back in time to about 380,000 years after the Big Bang, to the point where the universe had cooled so that it was no longer a plasma and became transparent to light. Prior to that point the plasma was essentially opaque and there is no way to use electromagnetic radiation to see earlier.

We see the CMB 13 GYears away in every direction, but we're not seeing to the "edge" of the universe (if there is one). We're seeing to the most distant point from which light has has time to get to us travelling at lightspeed. So 13 GYears back is 13 GLightyears out.  There's stuff (probably stars and planets and galaxies and all the stuff we know and love) further away, but we can't see it because light hasn't had time to get here from there.

Basically, the universe that we can in principle observe is a 26 GLightyear bubble in a much larger universe. (BTW, all these measurements are light-travel distances. When you take GR into account it gets more complicated, but those complications are not critical to this discussion.  See Observable Universe, Distance Measures and Comoving and Proper Distances for typically Wikipedian incomplete-and-confusing but still useful articles.)

The one thing based on observation that we can say is that the universe is probably at least ten times the diameter of what we can observe. Assuming General Relativity and assuming that the larger universe is about as homogeneous as what we can see, the fact that we observe our bit of it to be very flat means that the edge or the radius of curvature or whatever makes the whole thing finite has to be at least ten times further away -- otherwise we'd see curvature in our own observable patch of space.

So the problem is that the part of the universe that we can see is observed to be too small a part of the whole shebang to make any kind of estimate of how big the entire universe is.  (It's a whole lot like trying to survey an area of a few square miles accurately enough to estimate the size of the Earth. There's so much noise in the measurements due to local topography, that all you can do is say "It's a lot bigger than a few miles.")

Caveat: If I understood you correctly, your question is about the size of the whole universe, not just the visible universe.  If I'm wrong, let me know and I'll frantically re-write this!

Note that even the parts which rely on observation necessarily rely on those observations filtered through the lenses of theory. I've tried to use only the parts of cosmology which seem to be solid and to avoid the wild speculation that produces popular science books and TV specials.  But it's still quite possible that it will all be overturned as we come to learn more. (I'll be surprised, but that's happened more than once!)


Calculating 2004 BL86 position using software Is it possible to view asteroid 2004 BL86 using Stellarium or the iOS Star Walk app?

I'm going to describe the steps that I followed to show 2004 BL86 in Stellarium in my notebook:

Open the Configuration Window by pressing F2
Select the Plugins tab and from the left list select Solar System Editor
Click in the configure button at the bottom
Go to Solar System tab in the opened window
Click the button Import orbital elements in MPC format...
Click in the drop-down menu called Select bookmark... and select MPCORB: near-Earth asteroids (NEAs)
Click the button Get orbital elements
In the search field, type 2004 BL86 and select it once it appears below
Click the last button called Add object
Close the windows and then press F3 to open the Search window
Start typing 2004 BL86 and once it is showed below, click in the magnifying glass to the right, that will get you to the position of 2004 BL86

Please let me know if that answers your question. I'm using Stellarium version 0.13.1... and sorry if my English is not very precise :)


CoRoT FITS files read error I downloaded fits file using wget script from https://exoplanetarchive.ipac.caltech.edu/docs/datasethelp/ETSS_CoRoT.html
I am not able to view the file. I tried atsromagic and DS9 .  After File->Open I get an error message "..error while reading.." .It happens with both softwares.
Is anyone able to view these files? If yes, which software did you use?
Another possibility is that using using wget script on windows it got corrupted. Is there a command line tool to check integrity/validity of fits file?

These are FITS binary tables not FITS images so the normal FITS image tools very likely  won't read them. You have a couple of options to examine and manipulate them:

Use the 'fv' tool which works on FITS images or tables (and integrates with ds9)
Use TOPCAT, a very powerful tool for examining, plotting and many other operations on all formats of tables (not just FITS ones)
Use Python and AstroPy Tables e.g.

from astropy.table import Table

table = Table.read('EN2_STAR_CHR_0102990364_20070206T133547_20070401T235934.fits')
table
<Table length=71042>
          DATE                DATEJD            DATEHEL       STATUS ...     BG    CORREC_RED CORREC_GREEN CORREC_BLUE
  yyyy-mm-jjThh:mi:ss    COROT JULIAN DAY   COROT JULIAN DAY         ... ELECTRONS ELECTRONS
        bytes23              float64            float64       int32  ...  float32   float32     float32      float32
----------------------- ------------------ ------------------ ------ ... --------- ---------- ------------ -----------
2007-02-06T13:35:47.000   2593.06626508258 2593.0705519515977     16 ... 373.35654        0.0          0.0         0.0
2007-02-06T13:44:19.000  2593.072191038021 2593.0764776164892      0 ... 378.81073        0.0          0.0         0.0


Which of the planets would be detected if they were exoplanets? An (almost-)omnipotent being selects the solar system, presses Ctrl+C, and then Ctrl+V several times, creating copies at distances of 5, 500, 50.000, and 5.000.000 light-years away. All in a direction where the instruments we have are currently directed that way.

So we see these four star systems at various distances.

With our current technology, of 2020, which of the planets will we discover in each of these distances?

The distance moduli at 5, 500, 50,000 and 5,000,000 are -4.1, 5.9, 15.9 and 25.9 respectively. We add that to the absolute visual magnitude of the Sun, 4.8, to get apparent magnitudes of 0.7, 10.7, 20.7 and 30.7 respectively.

Most planets are detected by the Doppler wobble in the parent star or if the planet transits in front of the parent star. Some are detected by direct imaging, but these are planets much larger than Jupiter, orbiting at much greater distances than Jupiter (although see below).

Both the main detection techniques require relatively bright stars. The Doppler wobble technique relies on measuring small velocity shifts. These would be of the order of 10 m/s caused by Jupiter, but only 7 cm/s caused by the Earth. To claim a detection requires observing the star over two orbits (at least). Current technology would easily detect Jupiter, probably Saturn, but possibly not the Earth, if you observed for long enough. But to collect enough photons requires a star brighter than about 13-15 mag. So only your closest distances are possible.

The transiting technique has found (large) planets around stars a little fainter than this, but these tend to be in short period orbits. No Jupiter-like (in terms of size and orbit) have been founds by transits and no Earth-like ones either. You could just detect a Venus-like planet for your closest two distances. There is also the factor of orientation to consider. The probability of alignment such that a transit occurs is something like stellar radius/planet orbit radius. For Venus this about $6\times 10^{-3}$, and even more unlikely for the more distant planets.

So my conclusion is yes, you would detect planets at the first two distances using the Doppler wobble technique (providing you observed for 20 years or so). But you wouldn't see anything by looking for transits, unless you were very lucky and spotted Venus, again at your closest two distances.

More information and detail
.
If Alpha Centauri A's solar system exactly mirrored our own, what would we be able to detect?

Note added.

There are currently instruments probing Alpha Centauri for planets, by direct imaging. This corresponds to your first distance. I need to check their sensitivities...

There is also a possibility in the near future (with SKA) of detecting radio emissions from an Earth-like planet/civilisation at the first of your distances. See https://astronomy.stackexchange.com/a/10638/2531


Can someone help me to identify this section of the night sky? I have this image from a treasure hunt, I'm fairily certain it's a real piece of the sky, but I've been unable to identify it, can anyone help?

I've just submitted a cropped section to Astrometry.net (that's astrometry, not astronomy). Click the "web" to upload an image and start the search, then take some time to read How (the heck) does Astrometry.net work? while waiting for it to compute...

No quick solution. That doesn't mean that this isn't real, it's just that the first attempt didn't succeed.
better luck next year!


Is oxygen really the most abundant element on the surface of the Moon? I found this infographic that seems to say that oxygen is the most abundant element on the surface of the Moon. Is this really the case? If so, under what form is this oxygen?

Yes, that's correct; it's also true for the Earth's crust. The reason is that "rocks" are typically made up of components containing combinations of silicon or one or more metals (e.g., magnesium, aluminum, iron) and oxygen, such as silica ($\mathrm{SiO}_{2}$); alumina ($\mathrm{Al}_{2}\mathrm{O}_{3}$); lime ($\mathrm{CaO}$); iron oxide ($\mathrm{FeO}$); and magnesium oxide ($\mathrm{MgO}$).
Examples of common lunar minerals formed from these components includes plagioclase feldspars (mixtures of NaAlSi$_{3}$O$_{8}$ and CaAl$_{2}$Si$_{2}$O$_{8}$), pyroxene (typically XYSi$_{2}$O$_{6}$, where X and Y are metals such as calcium, sodium, iron, magnesium, and aluminum), and olivine (made up of Mg$_{2}$SiO$_{4}$ and Fe$_{2}$SiO$_{4}$), along with oxide minerals like ilmenite (FeTiO$_{3}$). (Source)
Since in all these cases you have between one and two oxygen atoms for every non-oxygen atom, you end up with oxygen as the most abundant single element.


Is there a name for the star/supernova hypothesized to have been a precursor to the Solar System? I found an article suggesting a supernova injected debris which helped nudge the formation of the Solar System 4.5 billion years ago.

I couldn't find a name for this hypothetical star/supernova after some research. Is there one? What is it?

I couldn't find a name for this hypothetical star/nova after some research.

First off, the existence of this former object is hypothetical. Others working in this field have different hypotheses / conjectures; the science is far from settled. Secondly, if true, this star formed a supernova at the time the solar system was born and hence it no longer exists as a star. Thirdly, what's the point? Naming this hypothetical object would add zero credence to the hypothesis, and arguably would detract as being a bit premature.


Recent research regarding if the local group of galaxies will remain bound to the Virgo Cluster? The question of if the local group of galaxies will remain bound to the Virgo Cluster has been asked before (Is the local group bound to the Virgo cluster?). I am wondering if there is any more recent research regarding this? I haven't found anything more recent than the 2002 paper mentioned in the post I linked above: https://arxiv.org/abs/astro-ph/0204249
This is a subject that I know very little about, so I'm looking for advice on how to get more informed about it. There is a comment on the aformentioned post that says, "The confusing part is that there seems to be a gravitationally bound Virgo cluster, of which we are not part, and a gravitationally unbound Virgo supercluster, of which we are part." Does current research suggest that we are permanently bound, or only temporarily?
This more recent post (Time-frame regarding cosmic expansion and the bound state of the Virgo Cluster), for example, claims that we may be bound gravitationally to the Virgo Cluster enough to resist cosmic expansion and asks for a timeframe for which we are inexorably bound, but the answer seems insufficient to justify this assumption.

The most recent paper I have found is Hoffman, Y., Lahav, O., Yepes, G., & Dover, Y. (2007). The future of the local large scale structure: the roles of dark matter and dark energy. Journal of Cosmology and Astroparticle Physics, 2007(10), 016. They also found that the Local Group gets separated.


How does the concept of a universe with no center work? I understand basically that the universe is homogenous (looks the same from every point) and I was told in my astro class that it's supposed to have no center, but how that works boggles my mind. I can't get over the feeling that from some point in the universe a civilization could look in a certain direction and see a sign indicative of the direction towards the edge or the center, like younger galaxies or a lack of galaxies moving a certain direction.

EDIT: Thanks to Sir Cumference for highlighting some things that are unclear about my question. Sure, to a certain extent you might be able to think of the universe as being without a center, but if you think of a ball growing larger from nothing, and you pick a random point in the growing sphere, you can always say that the point is further from some points on the sphere than others, or in other words, that there is a direction towards a "center".

Any other details you can give that are related at all would be extremely welcome.

When we talk about the universe, we are really talking about one of two things:

The observable universe, which is everything we can possibly see.
The Universe, which is everything that has ever existed, currently exists, and will exist.

The observable universe has its own center, usually the Earth. It is a spherical region of everything that we can see, essentially anything whose light has reached us. We usually refer to this when we say things like "there are $10^{86}$ atoms in the universe."

In reality, everyone has their own observable universe, and it can change depending on where you are. An exoplanet far away has its own observable universe, and can receive light from different places. Essentially, you are the center of your own observable universe.

I assume you're talking about the latter, though. The Universe (notice the capital "U") is all of space and time and its contents. Anything that has existed, will exist, and currently exists is part of it.

The Universe is thought to be infinitely large, so it can't have a center. The center of something is the point equidistance from the edges, but if something spans infinitely long, it would just keep going. It wouldn't have an edge, and thus it wouldn't have a center. You couldn't find the point equidistant from the edges if it just spans infinitely.

You might ask, "then where did the Big Bang start? Surely it must've been the center of the Universe, right?" Well, you can say the Big Bang happened everywhere. Before the Big Bang, matter filled most of the Universe's emptiness. It was essentially dense, and so got extremely hot to the point where no hadrons could form.

It is thought that this temperature caused space itself to expand. Essentially, more space was created in between all the matter, until everything was able to cool down. This is what we call the Big Bang. It didn't happen at a certain point, but rather it happened everywhere.

EDIT: I understand your confusion. Let me just clear some common misconceptions:

The Universe is not like a ball. Rather, you can think of it like a flat grid, and its "expansion" just means that the distances between objects on the grid are getting larger. In essence, more space is being created between the objects. That's what we mean by expansion — that objects are moving away from each other, since more space is being created between them.

Here's an easy analogy: imagine you are walking your dog. Suddenly, the ground begins expanding between you. You and your dog will separated and continue receding away from each other.

That's essentially happening everywhere: space is expanding between everything, so we are drifting away from other galaxies. The Universe is infinite, and we can constantly drift apart from other objects because space is being created in between us. Here's a GIF I made that might help you get it:

You can see how the galaxies drift apart as the space between them increases. And this happens everywhere in the Universe. So let's keep that in mind and clear up the Big Bang.

The Big Bang did not happen at a single point, nor did the Universe begin at a single point. The Universe is, and has always been, infinite. The Big Bang was just when the Universe's expansion really began — that is, when objects started drifting away from each other. The Universe was still infinite, but there was less space between the matter.

This density caused the Universe to get extremely hot and expand. So here, space itself was distorted and began to expand. More and more space was created between the matter, and still is now (although now it is mainly due to dark energy instead of heat).


Creating an HR diagram from tabulated stellar data I teach statistics and study Astronomy.  I'd like to create an activity for my statistics students using star data.  They could enter the data into Excel and create an HR diagram.  We are currently studying Correlation & Regression.  I wanted to show them that a scatter-plot can be VERY interesting.

Where can I get stellar data that is already in a format that my students can use excel to create a scatter-plot?  Unbeknownst to them, they will create an HR diagram.

With the Gaia Data Release 2, you can plot an HR diagram with a few million stars. From the Gaia website :

The data can be accessed in the search tab, where you can specify the columns you want to display (phot_bp_mean_mag and phot_rp_mean_mag). There is a button at the bottom to export the results to a file on your computer. You can then plot one of the columns as a function of the difference between the other two.


How to find a Cepheid's pulsation period using its graph? Basically, I've been seeing these graphs that say: Cepheid Period-Luminosity Relation and the Period axis is supposed to be "logarithmic";

How does this graph work? Since the luminosity of a star doesn't just increase, as we know, but it increases and decreases periodically, why does the magnitude keep raising?
How do I find the pulsating period of such a star through a graph like this?

This graph is not the light curve of a cepheid. It is used to find the average brightness of the star, and from that, estimate its distance.

The rate at which cepheids pulse is related to their average luminosity (and so their average absolute magnitude). This graph relates the period to their average absolute magnitude.

To use the graph, observe a cepheid for long enough to establish a period (say 30 days, for example). Then read the graph (an absolute magnitude of just under -5.5 in the example)

You now know how intrinsically bright the star is on average. By looking at its average apparent brightness, you can calculate an estimate of its distance. See How does the cepheid variables help in finding distance


Are there predictions of the distribution of matter at the end of time Are there any theoretical predictions or "educated speculation" by cosmologists on how matter will be distributed at very, very distant times in the future?

For example, will matter be:

agglutinated into a big cold ball?
diluted across the universe in tiny bits?
large clusters of matter here and there?
something else?

The basic answer to your question is "yes to all of the above, in different eras".

If we use the concordance $\Lambda$CDM model, with the cosmological constant remaining constant, then the universe will keep on expanding at an accelerating rate forever. The first big effect of this is that gravitationally bound clusters of matter such as galaxy clusters will become separated by ever larger distances (but they themselves are not changed noticeably by this). The current cosmic web of galaxies and voids disintegrates into little separated droplets over the next few hundred billion years.

Inside these "island universes" local gravitational interactions tend to make galaxies merge with each other, eventually ending up as a single big elliptic galaxy.

Over long time (around $10^{20}$ years) close-encounters between stars lead to some stars being flung out of the galaxy into the outer darkness, and the others bunching up closer to the central black hole. Eventually this reduces the galaxies to big black holes and scattered stars. The scattered stars cool down to low temperatures.

If proton decay happens, then somewhere beyond $10^{33}$ years or so the scattered stars quietly and slowly evaporate into elementary particles. The black holes are also expected to evaporate, and after $\sim 10^{99}$ years or so the only thing left will be separate stable elementary particles - electrons, positrons, neutrinos, photons, maybe some stable dark matter. Due to the expansion each particle will be alone in its visible universe.

A good overview is Adams and Laughlin's "The Five Ages of the Universe" (or their paper, for more detail).


How can I predict space directions? Did the Neptune and Pluto trajectories ever cross or was it a mere reclassification? Do studies say that I can predict an intersection between 2 trajectories  or movements based on data now? Do I have to consider at least 3 solid or larger space bodies moving in a way like the 3-body problem?

If you project the orbits onto a plane, for example the plane of the ecliptic, the projections will cross. But that's only because you're looking at a 3D problem in 2D. If you look at the orbits in 3D, you'll see that Pluto's orbit is highly inclined (17º) from the ecliptic, so it never actually passes through Neptune's orbit. Each time it seems to cross (in the 2D view), it's actually well above or well below Neptune's orbit. Pluto and Neptune never come closer than about 17AU.

At least for now. Minor perturbations over tens of millions of years may change that. Or might not. Pluto and Neptune currently have a 3:2 resonance, and such resonances tend to be stable.

You might want to look at http://en.wikipedia.org/wiki/Pluto, especially the section "Relationship with Neptune".


Average rate of supernova x number of stars If we would take the average number of supernovae per year and multiply that by the number of stars, could we figure out how many years it will be before all stars are gone?

Not that easy way. Only stars heavy enough will undergo a supernova explosion. The majority of stars is too light. The lifetime of a star is mostly determined by its mass. In some cases (supernova type Ia) a companion star provides mass to white dwarf, which originally has been too light to explode as a supernova.

Hence your technique can only work, if it is applied to stars of a certain mass and composition separately, and then summed or integrated up over all these restricted classes of stars. All together you'll get results for stars which eventually will end up as supernova, i.e. the heavy ones with more than about 8 solar masses.

... And in the meanwhile new stars will form.


Moving lights in the sky looks like they were moving coordinated I was out with a couple friends a night and we looked up to the sky just watching the stars. I noticed what looked like a star moving then told my friends and they saw it too.

After we saw the one moving star we noticed 2 more, it seriously looked like 3 stars just decided to have a meeting. It was the weirdest thing I've ever seen

These are almost certainly satellites. Keep in mind that satellites look like slowly moving stars; well, they move more slowly than meteors do at least! If they were moving somewhat slowly and constantly, and if they had no additional flashing lights and no bright colors, they were definitely satellites. NOSS satellites used to be known for flying in groups of three, but it looks like none of the generations of those triplets are still flying in formation. It looks like the Chinese Yaogan satellites are the only ones flying in triplet formation these days; you probably saw one of them.

See here.


What would happen if a person was out in space? would they get picked up by the pull of gravity from inbetween our planets, or sun? I'm curious, if someone were to be floating out in space. Would they eventually be captured by the gravitational pull from another planet or sun. To be more specific the person isn't anywhere near said planet or sun they are just passing between them. My question, is there some sort of current in between planets or sun that can have the slightest effect on an object passing through?

There are no currents.

It all depends on the initial speed of the object, and the distance to the sun and planets.

If it's moving pretty slowly initially, it will eventually get captured in a loop orbit around some planet or maybe around the sun.

If it's moving pretty fast initially, it will described a trajectory through the system and eventually leave, never to return.

This is no different from what happens to asteroids, comets, the planets themselves, etc. This is all the play of gravity. This is called orbital mechanics. It's part of what is popularly known as "rocket science".

There is an app called Universe Sandbox. Install it on your computer and play with some solar system simulations. Setup a simulation with a central star, and just one body near it. Launch the body with various speeds. See what happens. The various scenarios unfolding from there are the answer to your question.


Can we measure/detect GR light bending during solar eclipse using ordinary equipment? Suppose an amateur astronomer finds himself watching Solar Eclipse. He can see the stars around totality. Is Sun's GR effect strong enough for the amateur astronomer to be able to tell that a particular star is appearing a few arc-degrees (or arcminutes, or arcseconds) away from where it should?

If not, what is the minimum requirement to be able to witness GR for yourself during a Solar Eclipse?

The deflection you are looking for is $\sim 1.7$ arc seconds, which should be within the seeing limits at a good site. It should also be within the resolution limits of a telescope with $\ge200$mm diameter primary.

Having said that you will not be able to see it with the naked eye, you will need to be able to photography the star field during the eclipse and again when it is visible at night and then do some arduous plate measuring and data reduction to detect the shifts of stars in the field.

So amateur equipment is up to detecting the bending of light around the Sun in a total eclipse, but if by witness you mean see the stars move with the naked eye the answer is almost certainly no.


What is the orbital velocity of Pluto around the Pluto-Charon barycenter? I know the formula for orbital velocity but that doesn't seem of much help when I'm trying to calculate how fast the primary wobbles, or since the barycenter is outside Pluto in the Pluto-Charon system, how fast it orbits.

From the NASA Pluto Fact Sheet:

Mass of Pluto $m_1 = 1.303\times 10^{22}\ \mathrm{kg}$
Mass of Charon $m_2 = 1.586\times 10^{21}\ \mathrm{kg}$
Mean distance from Pluto $r = 19596\ \mathrm{km}$
Orbital period $P = 6.3872\ \mathrm{d}$

The orbit is nearly circular (the fact sheet gives an eccentricity of 0.0, other sources give very small values). So treat as circular motion.
The radius of the orbit of Pluto around the barycentre is given by:
$$r_1 = r \left(\frac{m_2}{m_1+m_2}\right) = 2126\ \mathrm{km}$$
Then use the usual formula for circular motion:
$$v = \omega r$$
where $\omega = \frac{2\pi}{P}$, giving a value of around $24.2\ \mathrm{m\,s}^{-1}$.


relation between azimuth angle of the sun and angle of the shodow What is the relation between azimuth angle of the sun the bearing of an vertical object's shadow?

For example if the bearing of the shadow is 75° can we conclude that azimuth angle of the sun is 195°? Is the diagram below correct?

Your diagram is correct, your maths isn't.

The bearing of the object's shadow is, as your diagram shows, away from the sun. If the bearing of the shadow is 75°, then the sun must have an azimuth of of 75°+180°, or 255°.

Similarly if the bearing of the shadow was 300° (pointing towards the northwest) the sun would be at an azimuth of 300 - 180°, or 120° (in the southeast)


What is the air pressure in the heliosphere (Sun's atmosphere)? Specifically, I am wondering what the pressure is at a distance of around 1 AU. Does it it decrease quadratically with distance?

I am going to assume that what you mean here is "what is the pressure" in the solar wind? There is no "air"!

The solar wind is pretty complex, consisting of a "fast wind" observed predominantly at high solar latitudes and a slower more variable wind a low latitudes. Both components essentially consist of an expanding stream of protons, electrons plus a small fraction of helium nuclei.

I think all the information you want is contained in this paper by Ebert et al. (2009), but I will attempt a broad-brush summary.

The fast wind has a typical proton number density of $n_p=3$ cm$^{-3}$, a temperature of $T=2\times10^{5}$ K and a speed of $V=700$ km/s. Note that "temperature" is a slippery concept here since the velocity distribution of the particles is non-Maxwellian. However, ignoring this and assuming the protons are matched by numbers of electrons, then the "thermal pressure" $P_{th} = (n_p + n_e)kT \simeq 10^{-11}$ Pa. However, this is negligible compared with the dynamic pressure $P_d = \rho V^2$, where $\rho$ is the mass density given roughly by $\rho = n_p m_p$. Thus $P_d \simeq 2\times 10^{-9}$ Pa (i.e. about $2 \times 10^{-14}$ bar).

The "slow wind" is quite variable, but mean values could be $n_p = 6$ cm$^{-3}$, $T=8\times 10^{4}$ K and $V=400$ km/s. Here again, the thermal pressure $P_{th} \simeq 10^{-11}$ Pa is dwarfed by the dynamic pressure $P_d = 1.6 \times 10^{-9}$ Pa.

The thermal pressure component falls as distance from the Sun cubed, ie. as $P_{th} \propto R^{-3}$. This is because for mass conservation, then a sherically expanding wind must have a density that falls as $R^{-2}$, but it is experimentally found that the proton temperature also falls as $\simeq R^{-1}$.

The dynamic pressure does fall almost as $R^{-2}$. This is because there is very little radius dependence for the proton velocity (i.e. the protons do not slow down, right out to the heliopause - as measured for instance by Voyager 2).


Could an astronomical process create elements in the islands of stability? I was reading about heavy elements recently, and I took note of the theorized islands of stability, and how some the elements involved are considerably heavier than anything we've been able to synthesize so far. Is there any known astronomical process that could create elements well above the 120-proton mark?

To this point, this is purely theoretical.  We have no solid evidence of islands of stability.  They may or may not exist to begin with.  It's hard to speculate on the methods of creating something we don't even know exists.

That said, currently, the heaviest elements found in nature appear to be synthesized in the mergers of neutron stars.  It might be possible that a similar process could create even heavier elements.  It's possible that some of the discovered transuranic elements we've discovered might be formed in such mergers, however their half-lives are so short that they break down before they're detectable.  If there are stable transuranic elements, then they might be formed in this manner, but we've never seen any evidence of them.  Me might not know what to look for, but so far we've detected no suggestions of such elements.

Another option is the accretion disks of black holes.  It is believed that nucleosynthesis occurs in accretion disks, and one could speculate that there may be conditions in accretion disks that could make it possible.

But without any evidence of the islands of instability, this is entirely conjecture.