Seconds of Arc and the Unaided Eye I've been presented with this problem:

  Say that Jupiter, with its diameter of 142,000 km, was located where Mars now orbits.  What would be the angular size of (the newly-relocated) Jupiter during a close approach, when its distance would be 79,300,000 km?  Would we be able to see Jupiter as a round object with our unaided eye, or only as a point of light?

The angular size is easy enough to calculate given $ \frac{Angular \space Size}{206,000} = \frac{Linear \space Size}{Distance} $.
So $\frac{Angular \space Size}{206,000} = \frac{142000}{79300000}$ meaning Angular Size = 369 seconds of arc.

However, would the unaided eye be able to see this? Knowing that the eye can detect 1 minute of arc across, surely  it can detect 6.15? Any help appreciated, thanks!

Yes, you could definitely see Jupiter as a round object if it's
angular diameter was 369 arcseconds. The limit for seeing something as
round appears to be a diameter of around 1 arcminute. Notes:

The most direct comparison is M16, the Eagle Nebula. According to
http://www.nightskyinfo.com/archive/eagle_nebula/

The M16 complex is visible to the naked eye on clear nights as a hazy patch

(and thus not a single point), and http://messier.seds.org/m/m016.html
notes that M16 has an angular diameter of about 7 arcminutes.

http://www.astropix.com/doubles/ notes that:

In the real world, folks with excellent eyesight can resolve down to
about 2 arcminutes. Folks with average eyesight can resolve about 3
arcminutes.

Of course, resolving a double star as two separate points of light
isn't the same as seeing something as a circle, but provides a good
upper limit.

And
https://en.wikipedia.org/wiki/Phases_of_Venus#Naked_eye_observations
notes that:

The extreme crescent phase of Venus can be seen without a telescope by
those with exceptionally acute eyesight, at the limit of human
perception. The angular resolution of the naked eye is about 1 minute
of arc. The apparent disk of Venus' extreme crescent measures between
60.2 and 66 seconds of arc,[4] depending on the distance from
Earth. Nevertheless it is possible for observers with extremely acute
eyesight to see a crescent Venus under ideal atmospheric
circumstances.

Note also that Venus appears circle-like (not point-like) when it transits the Sun, but the same is true of Mercury, so this may be a special situation that doesn't count.


How would you relate the respective solar gravitational pulls of two planets? I was asked this question:

  How much weaker is the gravitational pull of the Sun on Neptune at 30 AU, than it is on Earth at 1 AU?

How would I even begin to determine this?

For the most part, you can assume Newton's law of universal gravitation, (see link for more information). Basically, this law can be stated as the following.

For the gravitational force between two celestial objects (or really, any two objects with mass), the force is given by the equation,

$$ F_G = G\frac{m_1 m_2}{r^2} $$

Given the mass of the two objects, $m_1$, $m_2$, and the distance between them, $r$. The constant of proportionally given by $G$, the gravitational constant (not to be confused with $g$, the acceleration of an object with respect to the conditions on the surface on the Earth).

For the specific problem(s) that you are given, you can say that, using the information from Wikipedia, for the mass of the Sun, Earth, and Neptune,

$$ m_\text{Sun} = 1.988 \times 10^{30} \text{ kg} $$
$$ m_\text{Earth} = 5.972 \times 10^{24} \text{ kg} $$
$$ m_\text{Neptune} = 1.024 \times 10^{26} \text{ kg} $$

Your question gives us a preliminary distance. The $r$ for the distance between the Sun and the Earth is given as $1 \text{ AU}$, the $r$ for the distance between the Sun and Neptune given as $30 \text{ AU}$. Note that $1 \text{ AU} = 1.496 \times 10^{11} \text{ m}$.

Therefore, for the gravitational pull (really another way of saying the gravitational force) between the Sun and Neptune,

$$F_{N \leftrightarrow S} = \left( 6.674×10^{-11} \text{ N m$^2$ kg$^{-2}$} \right) \frac{\left( 1.024 \times 10^{26} \text{ kg} \right) \left( 1.988 \times 10^{30} \text{ kg} \right)}{\left( 5.983 \times 10^{12} \text{ m} \right)^2} $$

Therefore, for the gravitational pull between the Sun and the Earth,

$$F_{E \leftrightarrow S} = \left( 6.674×10^{-11} \text{ N m$^2$ kg$^{-2}$} \right) \frac{\left( 5.972 \times 10^{24} \text{ kg} \right) \left( 1.988 \times 10^{30} \text{ kg} \right)}{\left(1.496 \times 10^{11} \text{ m}\right)^2} $$

Solving for these quantities (note that the answer that one might get may be numerically different depending on the sources of the planatary constants),

$$F_{N \leftrightarrow S} = 3.796 \times 10^{20} \text{ N}$$

$$F_{E \leftrightarrow S} = 3.541 \times 10^{22} \text{ N}$$

A bit of extra information, the gravitational force felt by the Earth or Neptune from the Sun is exactly the same as the force that the Earth or Neptune acts upon the Sun, as per Newton's third law of motion. But, by now, I suspect that the problem is answered to where taking the ratio of the two is a trivial manner.

Suggested by uhoh:
The force exerted by the Sun on Neptune (and vice versa) is about 93.3 times weaker than that of the force exerted by the Sun on the Earth(and vice versa).


Could we verify the structure of a black hole by observing an orbiting object? Can we determine the physical structure of a black hole by observing its gravitational effect on objects in orbit?  There are three possibilities that I see, and would like to test:

Singularity at the center - The traditional view that there is a singularity at the center (e.g. all mass is concentrated in the center).
A structure where the black hole is filled throughout with matter inside the event horizon - My proposal, due to gravitational time dilation slowing / stopping time (e.g. a singularity will not form until an infinite amount of time has passed).  If the answer to my question on this subject here is correct, we should observe either possibility 2 or 3:  Does matter accumulate just outside the event horizon of a black hole?
A hollow shell, with matter only on / near the outside edge of the event horizon - My proposal, combined with the "vapor bubble" behavior, as hinted at in some of the answers to the linked question.

Would these three possibilities differ gravitationally in ways that could be observed (e.g. by observing orbiting stars), or would they be identical from the outside?  If they would produce different results, I would love to see what result it is that we actually observe!  We have observed stars orbiting the super massive black hole at the center of the galaxy.  http://scitechdaily.com/astronomers-discover-star-orbiting-the-black-hole-at-center-of-the-milky-way/

The comments from @userLTK, and @Lacklub are correct.

Lets assume there is an object of radius $R$ and mass $M$, from a Newtonian point of view, if you are at another radius $r$, such that $r > R$, then there is no difference in the gravitational field  experience by an object at $r$ if the mass spread across a shell of radius $R$ or if its concentrated anywhere between $r=0$ and $r=R$. GR doesn't do much to change this, and in fact if $r >>R$, then the result is of course exactly the same.

Now to black holes, all taken from from Wiki https://en.wikipedia.org/wiki/Black_hole

Physical Properties
"The simplest static black holes have mass but neither electric charge nor angular momentum...This means that there is no observable difference between the gravitational field of such a black hole and that of any other spherical object of the same mass."

So basically if you're outside the event horizon (not that'd you know where it was) you're experience with the black hole is the same as with a planet or star of that mass.

Again from same wiki article:

Singularity
"At the center of a black hole, as described by general relativity, lies a gravitational singularity, a region where the spacetime curvature becomes infinite... It can also be shown that the singular region contains all the mass of the black hole solution".

The "it can be shown" references page 204 of Carroll, Sean M. (2004). Spacetime and Geometry. I don't have my copy with me right now so I can't look it up, but I would say I remember reading at in Carroll back in the day.

Finally, let me just add that, once something passes into the event horizon of a black hole, there's no getting back. So we really have to ask ourselves how any information about the "stuff" inside the event horizon would come to us? I like your idea of getting some indirect evidence and perhaps gravitational waves will shed some light on this topic but I don't suppose there is any known way to getting direct access to anything beyond the event horizon.


How can I make the fringing pattern visible in my flat (twilight) images for filter i? I have a bunch of twilight flats that have been preprocessed (dark, bias, overscan etc.) and need to remove the additive fringing from the (multiplicative) twilights. I normalized each one by dividing by the mean and then combined them using ccdproc in astropy. I then divided the normalized flats by this "new" flat and viewed them in DS9 but I still cannot see the fringing. I know it's there because I've seen it clearly before. I could use some helpful tips. TIA

PS I also tried in old skool Iraf and don't see anything obvious.

I was finally able to make the fringing visible by block averaging after taking the steps above. If anyone else has any helpful tips I'd love to hear them!


Is there any update on the Availability of Event Horizon Telescope Images? There has been no news on the obvious public channels from the EHT project since last (Northern) summer. Does anyone have more information on when any results from the April 2017 run might be released?

A secondary question is has the data from the April 2018 run been successfully gathered? I know they have to wait for Antarctic spring to ship the hard drives from the South pole, but that should have arrived a couple of months ago.

Today, April 10th 2019, there was a press conference where finally an image of M87 was released:

Scientists have obtained the first image of a black hole, using Event Horizon Telescope observations of the center of the galaxy M87. The image shows a bright ring formed as light bends in the intense gravity around a black hole that is 6.5 billion times more massive than the Sun. This long-sought image provides the strongest evidence to date for the existence of supermassive black holes and opens a new window onto the study of black holes, their event horizons, and gravity. Credit: Event Horizon Telescope Collaboration


Is it a coincidence that both the Sun and the Moon look of same size from the Earth? The Sun is huge, when compared to the Moon. Despite the huge difference in their size and distance from Earth, is it purely coincidental that they both look almost the same from Earth?

The coincidence isn't so much that they appear very similar sizes from Earth, but that we are alive to see them at the point in time in which they appear very similar sizes.  The moon is slowly moving away from the Earth, and at some point in the future the moon will be unable to totally eclipse the sun and conversely, if you could step far into prehistory, you would be able to see the moon with a much greater angular diameter than you see it now.

Most research I've found on the topic seem to be unavailable through my institute, however I did find one paper, "Outcomes of tidal evolution", which references results from Goldreich's research on the subject.

  This qualitative description of the eventual disruption of the Earth-Moon system is confirmed by the results of Goldreich's numerical integration, which showed that the moon will recede to 75 Earth radii, when spin-orbit synchronism will be reached; then the Moon's orbit will decay steadily inward because of the influence of the Sun.

For reference, the Moon is currently at a distance of approximately 60.3 Earth radii.  As such, the moon will steadily move away until synchronism would be reached, and from that point begin to recede towards the Earth due to the tidal affects of the Sun on the Earth disturbing the synchronization. It would seem that at some eventual point in the far distant future, it will return to this coincidental position once again.

  Counselman III, Charles C. "Outcomes of tidal evolution." The Astrophysical Journal 180 (1973): 307-316.


Finding hour angle, altitude I know the latitude and longitude of a city ($41^\circ 87'$ N, $87^\circ 62'$ W), the declination and right ascension of a star($16.51^\circ$, $68.98^\circ$) and the local time, date (27.12.2017, 1:20 AM). Could you please show me the exact calculations for hour angle, altitude?

I will use the formula $LST=100.46+0.98d+15UT$, where d is the number of days since 1.1.2000 at 00:00 and UT is the universal time.

At 1:20 AM, $d=6570.055$.

UT is $UT=7:20$ or in hours $UT=7.3$.

$LST=6648.0639$ or $LST=168.0639$

$HA=LST-RA=168.0639-16.51=151.5539$

How is my calculation?

This should really be a comment, but... you can usually check your calculations using a program like Stellarium (screenshot above).

From the latitude/longitude and right ascension/declination you give, it appears you are viewing Aldebaran from Chicago, IL at 7:20am UTC on 2017-12-27 (since Chicago is on Central Time and not observing Daylight Saving Time, which is 6h behind UTC).

As you can see from the above, the hour angle is about 3h16m19s (or 49.16 degrees), the azimuth (which is what I think you actually wanted instead of hour angle) is about 254 degrees, and the altitude is about 41 degrees.

The process for converting right ascension and declination to azimuth and elevation isn't difficult, and I considered posting a general answer, but I wasn't sure what format to use. Most programming languages have an astronomy library (which will be more accurate), and there are plenty of sites that already walk through how we get the formula, not to mention my https://astronomy.stackexchange.com/a/14508/21 which gives sourced formulas.


Why the Circumstellar Habitable Zone is defined as it is, if life could be possible outside of it? According to Circumstellar habitable zone

  In astronomy and astrobiology, the circumstellar habitable zone (CHZ), or simply the habitable zone, is the range of orbits around a star within which a planetary surface can support liquid water given sufficient atmospheric pressure

But then you have moons in the solar system such as Titan, Europe, Enceladus where some scientists believe life could be possible.

The most likely cradles for life inside our solar system

Life on Titan

Why the CHZ is defined as it is, if life could be possible outside of it? Is it pourly defined or scientists believing there could be life outside the CHZ in the solar system are having too much wishful thinking?

From your first link, the definition is:

  "The circumstellar habitable zone (CHZ), or simply the habitable zone, is the range of orbits around a star within which a planetary surface can support liquid water given sufficient atmospheric pressure.[1][2][3][4][5] The bounds of the CHZ are based on Earth's position in the Solar System and the amount of radiant energy it receives from the Sun."

but as you will see further down in that same article:

  "In subsequent decades, the CHZ concept began to be challenged as a primary criterion for life, so the concept is still evolving. Since the discovery of evidence for extraterrestrial liquid water, substantial quantities of it are now thought to occur outside the circumstellar habitable zone. The concept of deep biospheres, like Earth's, that exist independently of stellar energy, are now generally accepted in astrobiology given the large amount of liquid water known to exist within in lithospheres and asthenospheres of the Solar System. Sustained by other energy sources, such as tidal heating or radioactive decay or pressurized by non-atmospheric means, liquid water may be found even on rogue planets, or their moons."

So really, it's just historical - it is accepted now that it is a limited view, and there is just as likely to be liquid water elsewhere. That doesn't necessarily mean life is as likely, as with lower solar radiation, there may be insufficient energy to support life even with liquid water.


Mathematical Analysis of Blackhole Mathematically, for black holes old enough that the stellar material has collapsed all the way into
the singularity, the region between the horizon and the singularity is occupied by a spacetime
where the time and space coordinates are reversed from those of the outside world. What this
means in terms of what you experience is unknown. Other more complex conditions can occur of
the black hole is rotating. In that case the singularity becomes a ring around the center of the
black hole. You can pass through the center, but the tidal gravitational field would be lethal in all
likelihood. In nearly all cases there would be gravitational radiation rattling about, and this would
cause distortions in spacetime that would probably lead to spectacular optical distortions.
My question is Black hole absorbs everything. Is there any mathematical proof regarding this?

The Schwarzschild metric can be written as
$$ c^2 d\tau^2 = \left( 1 - \frac{r_s}{r}\right)\ dt^2 - \left(1 - \frac{r_s}{r}\right)^{-1}\ dr^2 - ...,$$
where $r$ is the radial coordinate, $t$ is the coordinate time, $\tau$ is the proper time (that measured on an observer's own clock) and $r_s = 2GM/c^2$ is the Schwarzschild radius. I have left out the angular terms on the right hand side which contribute a further negative term independent of whether $r$ is greater or less than $r_s$.
For an observer with mass, $d\tau>0$; for a massless particle $d\tau=0$ (e.g. a photon).
When $r<r_s$ the first term on the RHS is negative, while the second term becomes positive. In order for the LHS to be $\geq 0$, then
$$ \left(\frac{r_s}{r}-1 \right)^{-1}\ dr^2 \geq \left(\frac{r_s}{r}-1\right)\ dt^2 + ...$$
$$ \left| \frac{dr}{dt}\right| \geq \left(\frac{r_s}{r} -1\right)$$
What this means is that $dr/dt$ can never be zero, which means the direction of radial travel can never reverse. ie. Anything that enters a black hole (i.e. for which $r<r_s$) and has $dr/dt<0$, can never have $dr/dt >0$.
There is a slightly more satisfactory "proof" using Eddington-Finkelstein coordinates, that shows that all future light cones point inwards and that $dr<0$ when $r<r_s$.


Escaping a black hole I often hear that nothing can escape a black hole because its "escape velocity" is greater than c. If that is accurate, what about the following? I know that the following has a lot of most likely impossible assumptions, but I'm trying to understand whether the escape velocity explanation makes sense.
Suppose you've got some sort of space ship that's inside the event horizon, and it hasn't been destroyed by the gravity or tidal forces. Also suppose that you've got a whole lot of reaction mass and a fantastic power supply.
Could this space ship, through applying enough constant force escape the event horizon?

One of the clearest way to see what's going on is to look at a Penrose diagram of a Schwarzschild black hole. A Penrose diagram is like a map of the spacetime drawn in such a way as to preserve angles and put the light rays diagonally at $45^\circ$ angles, forming the light cones.

Since we're mapping all of an infinite space(time) into a finite drawing, distances are necessarily distorted, but that's a small price to pay.

Time goes up on the diagram, and a typical infalling trajectory is in blue.

Because every massive object must be locally slower than light, it must stay within those light cones. Hence, no matter how you accelerate, at every point of your trajectory you must go in a direction that stays within those $45^\circ$ diagonals at that point. But once you're inside the horizon, every direction that stays within the light cones leads to the singularity.

Accelerating this way or that just means that you get to choose where you hit the singularity--a little on the left on the diagram or a little to the right. Trying to escape using high acceleration brings you closer to the light's $45^\circ$ lines, which because of time dilation will actually shorten your lifespan. You'll actually get to the singularity sooner if you struggle in that manner.

This particular image came from Prof. Andrew Hamilton. Note that it pictures an eternal Schwarzschild black hole, i.e. one that has always and will exist. An actual black hole is formed by stellar collapse and will eventually evaporate, so its diagram will be slightly different (in particular, there will be no "antihorizon"). However, in all respects relevant here, it's the same situation.


LLR & Orbital Motion As I understand it, light that is emitted from a source is not imparted with the motion of the source and so always follows a "straight line". If this is correct, I am having a difficult time conceiving how the Lunar Laser Ranger experiments can detect photons.

In this experiment, a pulse of light is aimed towards a retro-reflector and it is reflected back. It takes about 2.5 seconds round trip and due to diffraction, the returning pulse covers a circle of approximately 20kms in diameter.

The Earth's orbital velocity is approximately 30km/s and in the time the pulse takes to make the round trip, the detector would be 75kms further along the Earth's orbital path. If my first paragraph is accurate, and the light pulse is not imparted with the Earth's orbital velocity, then how is detection of the returning light pulse achieved?

I hope this makes sense and thank you.

Well it kind of makes sense, but what is important here is the Earth's velocity relative to the Moon, not the Sun!

Roughly: the Moon-Earth separation is 400,000 km and a complete orbit takes 27 days. Thus the relative orbital speed is more like 1 km/s and so a 20km diameter beam easily covers the expected relative motion of about 2.5 km during the 2.5s round trip.


Stars in star clusters in SMC and LMC Is there any catalog or any paper published in any journal that lists the stars discovered under whichever star cluster of the Small Magellanic Cloud (SMC) they belong to? There is one for Large Magellanic Cloud (LMC) in the paper Efremov 2003 "Cepheids in LMC clusters and the period-age relation". But I can't find any for SMC.

It would be better still if the stars happen to be Cepheids.

Finally, after searching quite a deal, I found this paper which would aptly answer the question.

The paper from 1999 published by Pietrzynski and Udalski in Acta Astronomica lists the Cepheids in the star clusters of Magellanic Clouds.


Mechanisms of binary/multiple star formation What are mechanisms of binary/multiple star formation in different mass ranges (low, intermediate and high stellar masses)?

I don't have enough reputation to comment...

I think this might help you understand the formation or binary and more stars systems. This of course is not the only possible method but it might explain the systems with big mass differences.

As the initial rotation speed increases (marked in the videos as beta) you will see how the protoplanetary disk breaks up and form more than one protostar.

http://www.astro.ex.ac.uk/people/mbate/Animations/prestellar_discs.html


Where have all the Vulcanoids gone? This answer to Does anything orbit the Sun faster than Mercury? explains that while Vulcanoid asteroids may have been plentiful in the past, large ones have currently been ruled out though smaller ones less than about 6 km might still be there. It's difficult to observe them because from Earth it requires pointing close to the Sun, and from spacecraft that are much closer (i.e. Mercury and lower) it's really hot and difficult to do.
I think that the current limits are set by analysis of historical STEREO images, space telescopes in orbit at 1 AU designed to look at and around the Sun. A Search for Vulcanoids with the STEREO Heliospheric Imager
My understanding is that during solar system formation there were asteroids all over the place as things formed and then collided, but as some bodies became large large swaths were swept clean and some bands remained.
Have any theories been put forth explaining why the vulcanoid belt seems to be depleted of large asteroids? Is this band, while quite stable, simply not stable enough? Do some think that it simply was never populated by large asteroids to begin with for some reason?
Question: Where have all the Vulcanoids gone? (sung to the tune of Where have all the flowers gone?)

The effects of solar radiation are the main suspect for clearing out the vulcanoid region of whatever was there to begin with. Radiation pressure will tend to blow small dust out of the region. Larger objects tend to get cleared out due to the Yarkovsky effect and the YORP effect.
The Yarkovsky effect occurs as a result of temperature variations across a rotating object and the time lag for regions of the object's surface to heat up and cool down as they move into and out of the sunlight. This affects the distribution of the radiated photons (which carry momentum), and over time will act to shift the asteroid's orbit. Vokrouhlický et al. (2000) estimate that this effect would clear kilometre-sized objects out of the vulcanoid region on a timescale of a few billion years.
A further effect that is relevant is the YORP effect (short for Yarkovsky, O'Keefe, Radzievskii and Paddack, see also the question What is the difference between the Yarkovsky effect and YORP effect?). This affects the rotation of irregular objects like asteroids and can result in them spinning up to the point where they break apart. This is thought to be a significant contributor to the population of binary asteroids. In the vulcanoid region, this would be a mechanism for breaking up larger asteroids into fragments small enough for the Yarkovsky effect to rapidly clear them out of the vulcanoid region: Collins (2020) puts this right in the title: "The YORP Effect Can Efficiently Destroy 100 Kilometer Planetesimals At The Inner Edge Of The Solar System".

... The YORP effect destroys Vulcanoids by spinning them up so fast that the gravitational accelerations holding components of the body together are matched by centrifugal accelerations, this causes the body to rotationally fission. i.e break apart. We calculated the timescale of this fission process for a parent Vulcanoid and for each of their subsequent generational fragments. We show that objects with radii up to 100 kilometers in size are efficiently destroyed by the YORP effect doing so in a timescale that is much younger than the age of the Solar System...


How far can we detect lightning in radioastronomy? The wikipedia article on whistlers has this information:

Voyager 1 and 2 spacecraft detected whistler-like activity in the
vicinity of Jupiter known as "Jovian Whistlers", implying the presence
of lightning there.

This surprised me, because this implicates a short range for detection of lightning, and I thougth atmospheric discharges produced a lot of radio noise, and so radiotelescopes on Earth should have picked signs of jovian lightning before the voyagers. As it seems not to be the case, what makes lightining hard to detect at distance?

Frequencies of terrestrial whistlers are 1 kHz to 30 kHz, while radio telescopes work at 30 megahertz to 300 gigahertz.
Radio telescopes would need to be 1000x larger in order to resolve the direction of extraterrestrial whistlers.


In the 1950's how were radio-astrometric positions with portable dishes so precise they could be assigned to their dim optical counterparts (Quasars)? In my question Why are quasars so far away that they couldn't be optically resolved in the 1950's? I included the following short paragraph, but then added strikethrough to the second sentence and added a parenthetical based on @DavidTonhofer's comment.

So today instead of "quasi-stellar" we might simply say "unresolved". The light is star-like because the light from galaxies has a strong stellar component. (this second sentence is argued against in comments)

I then read further and found this short but informative historical account in University of Manchester's Jodrell Bank Centre for Astrophysics page The MKI and the discovery of Quasars:

Initially the radio sources were observed with the telescopes relativly close together. In this case the fringe pattern is very broad and greater than the typical angular size of radio sources so virtually all are observed to give fringes. However as the telescopes are moved apart, the fringe spaceing becomes finer and the fringes from many of the radio sources first reduced in amplitude and then disappeared. The separation of the telescopes when the fringes from each source reduced in amplitude (and the wavelength of the radio waves observed) allowed the angular size of the radio sources to be found.
To the surprise of the astronomers quite a number were found to have exceedingly small angular sizes - so small in fact that their images on a photographic plate would look like the images of stars. Much work then went into finding the precise location of these objects and finally it was possible to obtain their spectra. They were nothing like the spectra of stars or galaxies and remained a puzzle for some time. Finally it was realised that the reason that the spectra were so different was that the objects were so far away that their light was greatly red shifted. They were at very great distances away from us and so it was not surprising that their images were so small!
Because their images looked liked those of stars they were called Quasi-Stellar-Objects or QUASARS for short.

Question: In the late 1950's how were astrometric positions of these radio sources made so precisely that they could be assigned unambiguously to dim, star-like spots on photographic emulsions?

The 25ft Transportable Radio Telescope

In many cases in astronomy, there’s a difference in finding the position of something precisely vs. seeing detailed structure in that object.  In the former case, what you need to do is to find the centroid of the image.  So even with relatively coarse spatial resolution, you can often get very precise positions because you can centroid the images very well.  (EDIT: However, this may not apply to early radio observations because single-dish radio telescopes don’t form images.)
Specifically for 3C273, the first quasar identified, Wikipedia says that its position was refined with lunar occultation. This technique involves observing a source as the Moon approaches it and determining the precise moment when it disappears behind the edge of the Moon. Here you need to measure time precisely and to know the Moon's path very well.
Also see this answer to Has radio astronomy ever been done on objects that appear very close to the Moon? Is this avoided?
The first radio source to have a star-like optical counterpart associated with it was 3C84, and that was done by measuring the radio position precisely using interferometry, i.e. combining signals from multiple telescopes.
So I think that this was not in fact possible with single-dish radio telescopes in the 1950s, and required more advanced techniques that came a bit later.


Could asteroid mining affect gravitational balance? I'm just a guy who is interested in all topics around space, so forgive me if it is a dumb question.

I have read some things about asteroid mining in the past and I think it's a really interesting topic, as it opens up a lot of possibilities.

But one thing came to my mind. Humans tend to pushing things a little too far sometimes when we want to reach a goal. So, imagine, we are getting really good at off-earth mining at a larger scale and in a larger timespan, we continue to mine thousands and thousands of asteroids. Wouldn't that affect the gravitational balance of our star system at some point, or is the gravitational influence of an asteroid belt negligible? I am talking about extreme examples, like removing nearly the whole asteroid belt beyond Mars. Also, what would happen to all the stuff we don't need from an asteroid?

I'm sure I'm not the first one who thinks of something like this, but I found nothing about this particular topic.

Something to remember is that the total mass of asteroids in the asteroid belt is tiny compared to the masses of the planets. Ceres, the largest body in the asteroid belt by a wide margin, is only 0.00015 times the mass of Earth, Ceres itself accounts for about a third the belt's mass:

Asteroid belt facts from ScienceDaily

I haven't been able to find any papers or articles that suggest the residents of the asteroid belt have any measurable perturbation on the orbits of Jupiter and Mars (the other way around is a different story!). If the asteroids that spend most of their time near Mars and Jupiter were to disappear, I doubt there would be any observable effect on the orbits of the planets.

Consider that comets and asteroids commonly pass close to Earth and the Moon and their influence on Earth's orbit is negligible.

As for what we would do with what we don't need from an asteroid, there are a few options I could think of based on the size of the target.
With smaller asteroids, if there is a risk that any changes to its orbit (such as an impact or accidental explosion) that would send it close to Earth, a complete controlled demolition of it might be warranted, or simply attach large chemical rockets to it and push it away. In the near-future, it is likely that even if we do start mining asteroids, provided we haven't dangerously altered its orbit, they will simply be abandoned afterwards.


Could mirrors be replaced with CCDs? Why do telescopes use mirrors that simply reflect photons, when they instead could be covered with large sensors to register them? Reflection is all good and well, all thanks to silver and beryllium for that. But wouldn't it be better to electronically register the photons directly instead of after having them bounce around between stupid mirrors? Would any data be lost in a pure CCD-telescope without any mirrors or lenses?

Couldn't a large wired CCD light sensor send the detected signals further in a smarter way than a stupid physically reflecting surface can do? It's the same photons and the telescope itself doesn't generate any new information about the distant galaxies it reflects. Why physically bend mirrors for adaptive optics, instead of bending the raw binary data with an algorithm for the same effect?

The CCD has no way of recording the direction, the point in the sky, from which a photon is coming.

Say you point your mirror-less telescope at the Moon.  Every point on the moon's surface would be reflecting photons onto every part of the CCD at the same time.

You've just created an expensive, sensitive, ambient light meter. There would be no image information whatsoever.


Are there any planets or moons denser than Earth? Earth has the highest density out of all planets, planetoids and moons of our planetary system, and also has a higher density than the Sun. Do we know any exoplanets or moons denser than Earth?

I feel it's a cheap answer but heavy Jupiters can get much denser than Earth because planets with Jupiter's mass stop adding size as they add more mass.   A planet with Jupiter's size and 10-12 times Jupiter's mass would be over twice Earth's density.

As far as Earth-like planets, there's a cool property of terrestrial planets, more mass means more tightly packed in their cores.  Basically a similar effect to the heavy Jupiters but not as pronounced.    If you double the mass and keep the element ratio the same, the density should increase.    For example, a planet like Mercury with a very high iron content, but much greater mass should easily surpass Earth's density.

Kepler 10b, appears to be a super-earth and its estimated density is greater than Earth's at 5.8 ± 0.8 g/cm3.  It has a mass of about 3.7 Earths.   The ± 0.8 g/cm3 offers some room for uncertainty, but some of the more massive terrestrial planets should be more dense than Earth.

I'd take this list with a grain of salt, but you can sort by density.

Kepler 131c which you mention appears to be a super Earth but its mass has a high margin for error.  I would add that a mass 8 times that of Earth and a radius smaller than Earth is probably impossible, so I'm highly skeptical of some of those numbers.


Is cosmic shear generally agreed to have been observed? A review by Martin Kilbinger, 2015 says:

  Cosmic shear is the distortion of images of distant galaxies due to weak gravitational lensing by the large-scale structure in the Universe. Such images are coherently deformed by the tidal field of matter inhomogeneities along the line of sight.

As a rough analogy: imagine looking at an image full of galaxies through a rippled piece of glass. While a circular lens might make a galaxy look smaller or larger, a ridge would have a preferential direction of magnification, and the galaxies under that ridge would have a statistical tendency to have slightly different sizes on average parallel vs perpendicular to the ridge.

Is such a phenomenon generally agreed to have been observed yet? Are there any plots (of any kind) that show such experimental data?

If I take this rough analogy one step further, while the peaks and troughs of the ripple magnify and demagnify, the midpoints between would be more prismatic - they would tend to deflect the centers, or locations of objects systematically, rather than affect the shapes. Applying this to cosmic shear, wouldn't the existence of shear essentially suggest that - to put it simply - stuff is, in fact, "not quite" where it appears to be?

above: from Free Ripple and Frosted Glass Overlays, Bojan Živković.

Yes it has been observed. There are numerous papers covering this concept. Before I get into weak-lensing (because that is what you reference in your question) I'll note that cosmic shear is simply a specific type of distortion of an object by a lensing event. It is not limited to the weak lensing regime and cases of strong lensing exhibit cosmic shear as well. So images like this one show cosmic shear.

As far as cosmic shear by weak lensing goes, the first published and recorded case of a weak lensing measurment was in 1996 and published by Brainerd, Blandford, & Smail (1996). The cosmic shear you speak of is referred to as polarization in that paper, but they're essentially trying to measure cosmic shear. Note also, while this is weak lensing, most of that lensing occurs by one galaxy lensing another. As such it is also termed galaxy-galaxy lensing.

Since the first detection, there's been steady, but limited work on galaxy-galaxy lensing and shear measurments by papers such as Mandelbaum et al. (2006), van Uitert et al. (2011), and Velander et al. (2013). I think the Velander paper is the latest and greatest measurements you can get and they have a lot of good material in there. Because you requested an image, I'll pull out one from this paper.

  Caption: Galaxy-galaxy lensing signal around red lenses which have been split into luminosity bins according to Table 1, and modeled using the halo model described in Section 3.2. The black dots are the measured differential surface density, $\Delta \Sigma$ , and the black line is the best-fit halo model with the separate components displayed using the same convention as  in Figure 3. Grey triangles represent negative points that are included  unaltered in the model fitting procedure, but that have here been moved up to positive value s as a reference. The dotted error bars are the unaltered error bars belonging to the negative points. The grey squares represent distance bins containing no objects.

Note that in this paper, they're ultimately trying to measure the tangential component of the cosmic shear (you can't measure the radial component), denoted by $\gamma_t$ and by equation (1) in their paper, they relate this to the surface mass density $\Delta \Sigma$. So in their plot above, when they say they're measuring $\Delta \Sigma$, they're really measuring $\gamma_t$ and converting through a multiplicative constant.

The limiting factor in progressing further in this field is the lack of detailed observations required to do a true cosmic shear measurement. Since this measurement can only be observed statistically and not directly, you need observations of the sky which are both deep and wide. It is hoped that the WFIRST satellite will provide these measurements and thereby expand ten-fold the weak lensing work astronomers can do which in turn may some day allow us to make important statements about the content, distribution, and properties of Dark Matter.


It moved in a straight line with sparkling track.What could it be? Today I saw something that passed like a jet. But am sure it's not so because jet will leave a long track of its passage. It moved in a straight line with sparkling track and what I saw was observed even from  200kms away from my destination. What could it be?

It is most likely a jet contrail. Not all contrails are persistent, and a high altitude jet can be seen over a great distance.

A fireball may leave a trail, but if it does, then it is likely to persist for as long or longer than a jet contrail. A fireball appears to move much faster than a jet, crossing the sky in a few seconds. It also appears to be very bright.

The image and description both point to this being a jet.


Quadrantids Meteor Shower 2020 viewing times It's my first time going to see the Quadrantids Meteor Shower, or any meteor shower in fact. I just want to know if I'm getting the right info for viewing this event. I got my info from this site Timeanddate.com. My location is Penang, Malaysia. Peak time: 5th Jan 2020

To my knowledge,

I won't be able to witness the event earlier than 4am because the position of the radiant is below the horizon.
I will need to be at a dark area (using this as a guide)
I will need to be facing 32-41 degrees northeast of True North. For this I will be using Google Maps instead of a compass because compass is magnetic North?
I will need to look for it at 6 to 33 degrees above horizon (depending on the time)
A local article such as this can't be trusted because it says you can view it from 11pm (there are many sites other local sites claiming you can see the event starting from 11pm)

Good to go?

Quick answer because the shower is tonight.

For any given meteor shower in a given year (e.g. Quadrantids, now) where can I see a rough prediction of how many days it will last?

The PDF linked on the page linked in the answer there says the shower is unusually short (FWHM  of 4 hours) and the maximum is at 08:00 UTC which is 4 PM where I am and must be similar there, so its already getting weaker by nightfall, so you should start watching BEFORE the radiant rises because while their trails all appear to "radiate" away from the radiant, they go in all directions. I'd say start checking around midnight looking in the direction where the radiant will rise later, which I suppose must be at least roughly EAST.

Try to find a viewing location with an unobstructed and dark eastern sky.

Keep watching all night of course, as the radiant rises more could be seen  geometrically even though the number may be decreasing for other reasons, and after all that's only a prediction and not guaranteed.

At least it's not as short as this one! What makes some meteor showers continue for days, while the "Unicorn shower" can be shorter than one hour?


Why does the eclipse in this video look annular? In this article about the AS870 flight that flew through the umbra of the March 8-9 2016 solar eclipse, there is a link to a YouTube video showing the eclipse recored by a handheld camera in the cabin.

Of course this is personal video and not recorded for analysis, but the images - especially during the zoomed part - remind me of an annular eclipse, rather than a total eclipse. Below are some screenshots - the weak Fresnel reflections in the cabin window act a bit like attenuators (ND filters) and give less saturated, though distorted views. Basically they all look like donuts.

My first thinking was that if it was a total eclipse on the ground, it couldn't be annular at 35,000 feet closer to the moon. But thinking and looking again, the sun is near the horizon, and the plane at about 150 West longitude is not necessarily closer to the sun than observers on the ground were in Indonesia circa 120 East.

My question is - is that just corona, or is the plane farther away enough from the moon at this time and location to make the eclipse annular?

It may be helpful to read about hybrid eclipses.

note: I'm looking for a quantitative answer, not an opinion.

There is another version of the video on YouTube with more views, but the narrative doesn't hold a candle to double rainbow guy original and musical version.

Here is a screenshot of a typical AS870 flight from flightradar24.com, pretty much between 152 and 158 degrees west.

This is from the second link in the top line - the Alaska Airlines blog entry for the flight.

How an eclipse could possibly appear annular from a plane over one place on the planet while appearing total on the ground at another place on the planet 5,000 kilometers away (note, these occur at different times - the shadow is moving relative to earth and the earth is rotating):

It is quite clear from the film (and the awful soundtrack) that a total eclipse was observed. The Sun's corona and prominences cannot be seen during an annular eclipse. Nor is there a sudden fading or drastic brightening at the beginning and end of an annular eclipse.

The accompanying article talks about "totality" and removing safety glasses to look at the Sun during totality, and that the period of totality was 1 min 53 seconds; the (frankly, annoying) narrator shouts the word "totality" about 23 times.

The eclipse was observed towards the end period of totality (what I mean by that is the shadow is tracking west to east across the Earth's surface. That means the Sun was above the western horizon as seen by the plane, and at 5:35pm, would be 10 degrees above the horizon. That means that in order to observe this event the plane must have been closer to the moon than a point on the surface of the Earth where totality would have been observed at the same time. That means the period of totality would have been slightly longer and the moon would have had a slightly larger angular size as seen from the plane.

If I had the exact latitude and longitude I could tell you just how much longer the eclipse was. Messing about with Stellarium, I can get a reasonable match of the eclipse time at W157 58 35, N31 22 00. At 10,000m the eclipse is total and lasts about 70s. At sealevel at the same latitude it lasts maybe a few seconds longer (it's difficult to be sure). But the sea level region of totality observed at the same time would have been somewhat further east (but still along the path where totality was observed), where the period of totality would have been shorter.


Between Mercury and Venus, which planet produces a longer transit? Assuming both planets describe same path on the solar disk Is it even possible to give a definite answer to this question? There are a lot of factors involved in this like Earth's position and motion relative to the planets and of course their own motion and apparent sizes in earth's sky.

Or is it that obvious that Venus transit will be longer because its orbital speed is less than Mercury?

This is a cool question; I'll give it a go.

  Or is it that obvious that Venus transit will be longer because its orbital speed is less than Mercury?

No, nothing here is obvious!

tl;dr: Times for the transit of Mercury can range from 5.37 to 8.30 hours, and for Venus they range from 7.94 to 8.12 hours under assumptions of coplanarity but allowing for eccentricity.

Let's assume all orbits are in one plane. We can get the orbital speed of body $i$ with semimajor axis $a$ and current distanc $r$ from the vis-viva equation;

$$v_i = \sqrt{GM\left( \frac{2}{r_i} - \frac{1}{_i} \right)}$$

where $GM$ is the standard gravitational parameter of the Sun which is about 1.327124E+20 m3/s2.

               a          ecc        pei           apo       v_peri   v_apo
Earth    1.49598E+11   0.016709  1.470984E+11  1.520976E+11  30286.6  29291.1
Venus    1.08208E+11   0.006772  1.074752E+11  1.089408E+11  35258.8  34784.4
Mercury  5.79091E+10   0.205630  4.600125E+10  6.981695E+10  58976.4  38858.6

Imagine Venus and Mercury crossing a line fixed at one end to the Sun and the other end to the center of the Earth. The linear speed of a planet $i$ relative to that line will be $v_i$ minus the speed of the line at the distance of planet $i$ or $v_E r_i / r_E$, and the angular velocity seen from Earth will be that divided by ($r_E - r_i)$ or

$$\dot{\theta}_i = \frac{v_i - (r_i/r_E) v_E}{r_E - r_i} = \frac{r_e v_i - r_i v_E}{r_E^2 - r_i r_E}$$

(seconds)     Venus          Mercury
Earth       peri   apo      peri   apo
peri       28570  29242    19335  29884
apo        28060  28626    19384  29646

(hours)       Venus          Mercury
Earth       peri   apo      peri   apo
peri       7.936  8.123    5.371  8.301
apo        7.794  7.952    5.384  8.235


Help with Converting Right Ascension and Declination into 3D Vectors Using the Right Ascension and Declination values from the HYG Database from The Astronomy Nexus:

Right ascension is measured in hours and declination is measured in degrees.

1 - Convert right ascension and declination into degrees

float getAngle(float value, char units)
{
    if (units == 'h')
    {
        // 15 degrees = 1 hour
        // 360/24 = 15

        return value * 15;
    }
    else
    {
        return value;
    }
}

When right ascension is inputted, the input will get multiplied by 15, because RA is measured in hours.

When declination is entered, the input value is returned, this is because Dec is already measured in degrees.

2 - Work out the vectors

Two isosceles triangles can be derived, one for the horizontal plane and another for the vertical plane. The angles are alpha and delta, right ascension and declination respectively. Using these angles and knowing that some of the distances are equal to the distance between the centre of the celestial sphere to the star, the following code should work.

Vector3 getVectors(float RA, float Dec)
{
    float x = distanceStars;
    float z = (2 * distanceStars * (Mathf.Sin((getAngle(RA, 'h'))/ 2)));
    float y = (2 * distanceStars * (Mathf.Sin((getAngle(Dec, 'd'))/ 2)));

    return new Vector3(x, y, z);
}

Unfortunately, my code doesn't seem to work as this is the output that I am getting:

As you can see, the stars (white dots) are not spreading out evenly around the Earth (which is the black sphere in the middle).

It would be great if someone could help me with this.

The stars should fit $r = \sqrt{x^2 + y^2 + z^2}$ but seem to be plotted in the $x = r$ plane instead.

Conventionally the x axis is at (α=0h, δ=0°),
the y axis is at (α=6h, δ=0°),
and the z axis is at δ=+90°.

Using right triangles instead of isosceles triangles,

Then $h = r \cos \delta$, and
$$\begin{align}
x &= r \cos \delta \cos \alpha \\
y &= r \cos \delta \sin \alpha \\
z &= r \sin \delta
\end{align}$$

Most math libraries use radians for angles.
Passing angles in other units to trig functions may result in a celestial sphere covered somewhat evenly but incorrectly.


Is Zeta Reticuli within the Orion Arm? My research has told me astronomers have labeled the galaxy arm that our system is located the Orion Arm but I have yet to find out if that also applies to Zeta Reticuli or not (not asking about aliens or anything, just trying to get my locations straight).

Let's take a look!

While I can't quite track down the source of this image, I've found it throughout the web, so hopefully it's roughly accurate. I'm pulling it from the Wikipedia article on the Orion Arm.

In this image you can also see the Orion Nebula sitting (relatively) cozily to the Sun.  The distance between the two is believed to be around 1350 light years.  Zeta Reticuli, on the other hand, is believed to be a mere 39 light years from Earth. Given the orientation of these two locations and the relative distances, it would be hard to place Zeta Reticuli anywhere but firmly within the Orion Arm.

Not exactly rock solid evidence, but sometimes some back of the envelope deduction will do.


Official degrees of earth’s rotation per day What is the official degree to one decimal point please, of the earth’s rotation in one single day. Can it be confirmed that it is exactly 360.0 degrees using official data? Thank you in advance.

First, we need to decide which definition of "day" to employ. There are several types of days:

Apparent solar day: the time between two successive culminations of the Sun (apparent Noon) from an fixed Earth-based observer;
Mean solar day: a more uniform, averaged solar day without seasonal variations;
Stellar/Sidereal day: the time needed for the Earth to rotate once relative to the stars;
SI day: a unit of time containing exactly 86,400 SI seconds defined by caesium atoms.

Since we generally refer to the traditional day/night cycle when we say "day", this means a form of solar time. To get a more averaged value, let's use the mean solar day.

The current formula linking the Earth Rotation Angle (ERA) to the modern approximation of mean solar time, UT1 (basically the Earth's clock following the mean day/night cycle), is by definition :

$$ERA = 2π(0.7790572732640 + 1.00273781191135448 T_u) \text{ radians}$$

Where $Tu$ is the Julian UT1 Date - 2451545.0

So according to this formula, a (UT1) day is 1.00273781191135448 Earth rotations, which multiplied by 360° is about 360.98561°. However, the Earth's rotation and revolution are not constant, and are always changing at somewhat unpredictable rates, so the angle is not perfect, but the changes are very slow. So this is more a modern approximation rather than an exact value. Rounded to one decimal place, this gives you 361.0°, a figure that will likely remain true for at least several millenia.

If you want to know the amount of Earth rotation for every SI day, you're in luck: it is possible to consult reports of the Earth's orientation (rotation and polar motion) thanks to the IERS. Values are tabulated for each 0h UTC every day in the IERS publications, allowing to derive the angle that Earth has rotated every 86,400 SI seconds, allowing to scientifically monitor variations in Earth's rotation compared to a very constant unit of time realized by atomic clocks. Nowadays, the Earth's rotation is measured and reported thanks to radio telescopes and VLBI observing distant objects in the universe. Official reports of its orientation, the Earth Orientation Parameters are published on the IERS Website.

As an example, here is a graph showing the amount of Earth rotation every 86,400 seconds constructed with IERS data of the last year:

As we can see, there are several seasonal, periodic and unpredictable variations in Earth's rotation.


Can CME destroy planetary rings? Is Corona Mass Ejection able to reach the outer planets such as Jupiter and Saturn? If so can it blows away the icy rings or the distance is simply beyond the reach of maybe like Earth it is protected by a magnetic field?

Yes it can reach both Saturn and Jupiter. But it's not powerful enough to remove their rings. Even Earth's radiation belts don't get blown away.

Edit: Jupiter's magnetic field is $\approx 770 \mu T$ (at the surface of its equator) which is way more powerful than Earth's ($\approx 40 \mu T$ at the equator). Saturn's magnetic field on the other hand is a tad less powerful than that of the Earth ($\approx 20 \mu T$ at its equator). So Earth's magnetosphere shields it's particles from CMEs. Jupiter is further away and its magnetosphere is way stronger than Earth's. So it shields everything effectively. Saturn is even further away and although its magnetic field is comparable to that of the Earth, the distance from the Sun assures us that a CME wouldn't have a significant impact of any sort there.


Does Ceres have an exosphere similar to the Moon? Thinking about the question and answers for Does the Moon have any oxygen in its atmosphere? begs the question, despite being considerably smaller than the Moon and further away from the sun, does Ceres have a similar atmosphere as the Moon?

In general, asteroids experience the same fundamental processes as the moon: solar wind, micrometeorites  and photons from the sun are striking them and sputtering material off to form an exosphere.  So there must be some sort of exosphere, but its density and composition will vary with distance from the sun, as well as the gravity and composition of the asteroid.

A 1992 ultraviolet study of Ceres found $OH$ emission, consistent with escaping water, and suggesting a ice cap.

But a more recent study with the more sensitive VLT  did not detect $OH$, limiting the maximum amount of water produced  per second to $7 \times 10^{25}$ molecules per second at the time of the observation.


Any information on the salt content of Mars' atmospheric dust? Answers to and comments on How to drink water on Mars? suggest that one would not want to drink melted water ice obtained from exposed surface deposits near the poles because it may contain various salts and other dangerous chemicals (perchlorate salts are one particularly nasty suspected component of mars soil and it's water content).

Mars' atmosphere contains dust even when there are no dust storms, and in this answer it is explained to be the cause of the reddish color of the Martian sky as has been photographed by several Mars landers and rovers.

Are there any measurements or even models or predictions on how much of the particulate matter in Mars' atmosphere could be tiny particles of salt, or salt mixed with other minerals? Put more simply, how salty might Mars' atmosphere be?

Martian sunset (credit: NASA/JPL-Caltech/Texas A&M/Cornell) borrowed from this answer.

[WARNING: I may be wrong. Astronomy is my hobby and I will answer to the extent that I know of. I may be a bit off.]

Mars' atmosphere contains I guess what you can say 2 major parts. Gases, and debris picked up by winds on the lower levels of the atmosphere (like Earth).

Gases

Mars' atmospheric gas composition mainly consists of CO$_2$.

More specifically, in an article page by NASA, Mars' atmosphere contains

  Carbon dioxide: 95.32 percent

  Nitrogen: 2.7 percent

  Argon: 1.6 percent

  Oxygen: 0.13 percent

  Carbon monoxide: 0.08 percent

  Also, minor amounts of: water vapor, nitrogen oxide, neon, hydrogen-deuterium-oxygen, krypton and xenon.

    http://www.answers.com/Q/What_are_the_contents_of_Mars%27_atmosphere

Dust/other stuff

Mars' atmosphere, although 99~100 times thinner that that of Earth's, can support winds, weather/storms, and clouds. The winds are a major criterion to Mars' permanent atmospheric dust composition. Dust particles, airborne or not absorbing sunlight to some extent, which in turn warms up the atmosphere in and near their vicinity. Warm pockets of air flow toward colder regions, generating winds; the same effect here on Earth. Strong winds lift more dust off the ground, which in turn heats the atmosphere, raising more wind and kicking up more dust. This is amplified by Mars' axis tilt and momentum, which can contribute to Mars' massive dust storms encircling the planet.

In addition to the winds, the Martian atmosphere is dense enough to support vortexes. Dust devils routinely kick up the rusty dust that covers Mars' surface.

So we can see now the causes for such permanent content of dust in the atmosphere of Mars, with...

  ...higher amounts of dust in the northern fall and winter, and lower
  amounts in the northern spring and summer.

    https://www.space.com/16903-mars-atmosphere-climate-weather.html

Let's now take a look at Mars' surface dust composition.

Its dust mainly is composed of Perchlorate salts; Chlorine; Iron(III) Sulfide; NiFe; Sulfur; Iron Sulfide; Nickel; Silicon; Oxygen; Iron; Magnesium; Aluminum; Calcium; Potassium; Pyroxene; Plagioclase; Glass; Olivine; and Feldspar. I would imagine that the dust in the air would contain many of these same components as the surface dust, although some heavier salts and chemicals would only last a brief time in the air.

According to Wikipedia,

  The NASA Phoenix lander first detected chlorine-based compounds such as calcium perchlorate. The levels detected contain 0.6% of the salt perchlorate.

and

  The soil has a basic pH of 7.7

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

At such high levels, there should be definitely perchlorate salt in the air, but not necessarily permanent though, as the atmosphere may not be able to support the density of the salt.

However, a recent study of terrestrial dust devils has shown that around two thirds of the fine particles lifted by these vortices can remain suspended in the atmosphere and be transported around the globe. I would theorize that lighter salts would be able to remain suspended in the air, while heavier ones would slowly settle to the surface, given enough time. But hey, this is MARS! With those huge dust storms and continuous vortexes, I doubt there is enough time.

All of this suggests that during the times when there are rather windy conditions, such as in a close proximity to a dust devil, or during one of the dust storms, the atmosphere should contain almost the same concentrations of salts as in the soil itself.

During calm periods, the air dust content should still remain the same.


Uranus' axis of rotation-when discovered? Who discovered it, and how was that accomplished.  I had assumed it was known before Voyager 2 arrived at the planet since it isn't mentioned in  JPL voyager 2 Uranus Approach

The high inclination of the Uranus system to its orbit around the Sun has been known since the satellites were discovered.
Lassell 1851 shows Ariel and Umbriel moving at least as far north and south of the planet as they do east and west of it.
William Herschel probably noticed the same thing about Titania and Oberon in 1787.

When Uranus is at an equinox, its equator appears edge-on to the Sun, making good opportunities to observe the planet's oblateness from Earth.
Mädler 1842
and Young 1883 measured significant polar flattening.
Barnard 1896 thought the satellite orbits might be inclined 20°-30° to the equator.
It took longer to determine that the equator was within 1° of the four largest satellites' orbital planes.


Convert Stellarium video capture to 360 for youtube I am wanting to capture video of Stellarium (I can already do this with CamStodio) and somehow convert this into a video usable for the YouTube 360 (or VR!) capability.

Even if it can be converted into 180 and I need to add an image of the ground or spaceship to fill the rest of the image, that would still be super helpful!

I have found this, which gives a good overview of the maths of converting various projections (I haven't gone through it in detail yet), but I was wondering if someone has already written something to do this, before I go and write my own code for it?

I already use Davinci Resolve for my video editing, and I'm on a Windows 10 Pro (with access to various Linux machines).

Cheers!

360° panoramic media conventionally use equirectangular projection
and a 2:1 aspect ratio.
I made a still panorama this way:

Configuration > Tools > Screenshots: Custom size 2048 x 1024
View > Sky > Projection: Cylinder (i.e. equirectangular)
View > Sky > Stars: Relative scale 0.70 (limits brightest stars to a reasonable size)
FOV 180° (Ctrl+Alt+1)
Other view options as desired (e.g. turn off ground, atmosphere, labels)

Then I saved a screenshot (Ctrl+S) and injected the appropriate metadata.
The result is viewable as a 360° panorama.

Constellations near the center of view look OK in any view direction,
but stars near the zenith look rather dim and rice-shaped.
I guess this could be remedied by something like the "Spheric mirror distortion" option (Configuration > Tools > Planetarium options) but within the "Cylinder" projection.
I don't know of a Stellarium plugin which does this yet.


Absorption of light by matter What determines whether or not photons can be absorbed by matter? Intuitively, the answer is that a charged particle can absorb photons (whether it's positively or negatively charged), and neutral particles can't. But light is often absorbed by atoms in bulk matter, increasing their thermal motion. The problem with this is that atoms are neutral, and so you wouldn't think that they could absorb light.

It is the other way around.
An isolated, point-like charged particle like an electron cannot absorb a photon. It is forbidden, because energy and momentum cannot be simultaneously conserved. Only scattering is permitted in such circumstances.
However, a charged particle that is within the electric field of another (e.g. an electron in the electric field of a nucleus) can absorb (or emit) a photon.
The way to think about this classically is in terms of electric dipole moments. An atom has no net charge but is can have an electric dipole moment, because the positive and negative charge are not in the same place. This electric dipole moment can interact with incoming electromagnetic waves and can be made to oscillate by extracting energy from the incoming wave.
In the quantum mechanical extension to this picture, the electric dipole will only oscillate in certain modes with discrete energies. Only photons within a certain narrow range of frequencies/energies will be able to excite (or de-excite) these oscillations.
The extension to bulk matter (and by that I assume you mean solids) is that the atoms themselves are arranged into lattices and other structures. These have electric dipole (or higher order multipole) moments that can be excited into oscillation by electromagnetic waves/photons in a similar way. i.e.  They offer further vibrational modes that act as a pathway to the absorption and emission of photons.
PS.
Nuclei themselves can have electric dipole or higher order multipole moments and are able to directly absorb (or emit) gamma rays.


Understanding gravity Forgive me if my question has an obvious answer, but I need to know the answer. I always thought that more massive/energetic objects had a stronger force of gravity than less massive ones; that is, the Sun would create a stronger gravitational pull than the Earth does.

Upon looking into Newton's law of universal gravitation, I began to wonder whether I misunderstood this. Does the gravity between the Sun and Earth have an equal force on both? As in, does the Earth pull the Sun with as much force as the Sun does on Earth? If so, would the Sun just resist accelerating to Earth because it is extremely massive? Mass is just an object's resistance toward accelerating from a force, right?

If this is the case, would it apply to GR? Sorry for hypothesizing, I'm honestly trying to get a grasp on it.

The magnitude of the force of gravity between two bodies is proportional to the product of their masses:
$$F=G\frac{m_1m_2}{r^2}$$
This doesn't change depending on which body you're applying the force to, i.e. if you interchange the masses. The magnitude is the same.

What does change is the direction of the force. Force is a vector quantity, denoted as $\vec{F}$ or $\mathbf{F}$. If we write the equation for gravity using proper vector notation, we have
$$\mathbf{F}=G\frac{m_1m_2}{|\mathbf{r}_1-\mathbf{r_2}|^2}\frac{\mathbf{r_1}-\mathbf{r_2}}{|\mathbf{r_1}-\mathbf{r_2}|}$$
Here, the positions of the objects are represented by vectors, $\mathbf{r}_1$ and $\mathbf{r_2}$. Additionally, $|\mathbf{x}|$ denotes the norm of a vector $\mathbf{x}$ - its magnitude.

Now, if you interchange the masses, the direction of the force changes, although $|\mathbf{r_1}-\mathbf{r_2}|=|\mathbf{r_2}-\mathbf{r_1}|$, because this refers to the magnitude of the vectors. So the force applied on one object is the opposite of the force applied on the other object. This is Newton's third law.

The acceleration is more interesting. The force on object $1$ due to gravity is
$$F_1=m_1g_1$$
Here,
$$g_1=\frac{Gm_2}{r^2}$$
where $m_2$ is the other mass. This should tell you that $g_1\neq g_2$, except when $m_1=m_2$.

I'm not an expert in general relativity, but I do know that it describes how spacetime curves due to the presence of one body. The solution to the Einstein Field Equations, the metric, is different for different bodies, because one piece of it, the stress-energy tensor, is different for objects of different mass/energy/etc.


How do I obtain the brightness of a star in the night sky? I am planning to conduct a research which involves observing the brightness of a star. So far, I haven't been able to obtain much information as to how I may go forward with the observation process. I'm willing to buy a telescope and other equipment. What telescopes are ideal for this and what other equipment must I require to obtain the apparent brightness value of the star?

I don't know about the telescope, but I know that you will need the telescope that has CCD technology to be able to convert from counts/second (as brightness) that CCD detects to quantitative values easily.

For the observational procedure, here is the standard regardless of what kind telescope you use:

Take a picture of your object (let's name it ImObj).
Take a picture of a standard star (let's name it ImStd) or stars (i.e., stars that have real brightness in a database, let's call it MagStdReal).
Take pictures for calibration (let's name it ImCal). If CCD telescope, these images include dark and flat images.
Use ImCal to calibrate both ImObj and ImStd (i.e., removing noise and etc.).
Convert image brightness of calibrated ImStd and ImObj to quantitative values (let's call it MagStdFake and MagObjFake). Note that you can use any arbitrary zero reference point for MagStdFake and MagObjFake, but they use the same reference.
Find conversion equation to MagStdReal = f(MagStdFake). The simplest form of f() is just linear shift: MagStdReal = a + MagStdFake.
Apply MagObjReal = f(MagObjFake). And now you get to do some science.

Also note that, a telescope filters certain wavelengths with certain response function. So, you have to observe everything in one setting of telescope. You cannot mix images or products of different setting of a telescope. Only MagObjReal or MagStdReal those are telescope insensitive (ideally).

I provide very short and conceptual procedure to give you the idea. There are a bunch of details that are important. If you use keywords in the list procedure here and search in Google, you should get more relevant hits to each topic.


Does lack of atmosphere on the moon make earth-shine more, or less, bright? The earth acts like a light-globe in the sky for half of the moon, the side that is facing the earth. In what way does lack of moon atmosphere influence earth-shine. Does it make earth-albedo brighter, or dimmer? Either way, do we know with how much? Put differently; would an atmosphere make it easier, or more difficult, to read a newspaper on the earth facing lunar surface, after sunset?

The lack of a lunar atmosphere doesn't affect Earth's albedo at all, since albedo is only a measure of the diffuse reflection of a body.
However, the apparent brightness of the Earth as viewed from the surface of the Moon would be affected if the Moon had an atmosphere.  This, of course, would depend on the atmospheric composition of the Moon.
If the atmospheric composition of the Moon was such that it absorbed visible light, then the Earth would appear dimmer, or perhaps not even be directly visible (similar to how we can't directly see the Moon when the weather is cloudy here on Earth).
If the atmospheric composition of the Moon absorbed almost no visible light, and was highly refractive, the Earth-Shine might be overall brighter than if the Moon had no atmosphere at all. The refraction would bend some of the Earth-Shine into the surface of the Moon which otherwise would have passed by. This picture shows the analogous situation for Earth and the Sun:


Help me identify a cluster as I was browsing Nasa's Eyes few days ago, I saved a screenshot,  but it seems text is blurrry, supher-485, first letter is S ends with er-485 from as I can read
it kinda looked like Cosmic Web, Indra's net, so i saved cool screenshot
Help me identify this cluster?

Since this appears to be a screenshot of Eyes on Exoplanets, I searched for "exoplanet 485" and found Kepler-485 b.
With the colors exaggerated, the label area could be "Kepler-485" in yellow overwritten with "Sun" in orange.
If this is correct, it's not a cluster, just a star with a planet, plus some artifacts from the instrument which took the image.


Help w/ Hubble's Law + Doppler Effect. I had a question about the following chart. What mathematical observation or equation could be made about the data given in respect to the doppler effect? I'm confused if this means that planets are moving at an exponential rate away from us? Would these galaxies appear blue shifted? Thank you in advance for any sort of help.

Look up the laboratory wavelengths of the Calcium H and K lines
Calculate the redshift implied by the observed wavelengths of the H and K lines.
$$ v/c = (\lambda - \lambda_0)/\lambda_0,$$
where $v$ is the implied recession velocity, $c$ is the speed of light, $\lambda$ is the observed wavelength and $\lambda_0$ is the laboratory wavelength.
Plot recession velocity versus distance and see what you get.
Read about the Hubble redshift-distance relation.


Technique for folding sparse event data in order to detect an underlying periodicity? @JamesK's answer to When will the next transit of Earth be visible from Mars? Was the last one really on May 11, 1984? lists dates of 18 transits which I've converted to days since 1500-Jan-01 as
[35011, 44319, 73176, 109885, 138742, 148050, 176908, 213616, 242474, 251782, 280639, 317347, 326656, 346205, 355514, 384371, 421078, 430388]

Answers to Why do transits of Earth across the Sun seen from Mars follow a pattern of occurring after 26, 79 then 100 years? explain why this happens and why if they happen at all, the happen in May and November.
As an exercise I'd like to use folding to see if I can detect those or similar periodicities in addition to the 6 month periodicity.
I could

for each period of N days recalculate those modulo N
histogram the results

but since there are so few event, they rarely occur when their time comes, and so many days between events, there won't be obvious peaks in the resulting histogram.
Question: Is there a technique for folding sparse event data in order to detect an underlying periodicity?

Events convoluted with σ=1000 day gaussians purely for illustration purposes. I thought about applying some Fourier transform to this but I don't think that that makes sense, and would be cheating rather than folding.

While this is not exactly how folding works, one-dimensional data encourages trying some simple statistical tallying first.

What very easily reveals those periodicities is just tallying up the deltas, with an appropriate fuzziness for what periodicities are relevant:

Deltas

[0, 9308, 28857, 36709, 28857, 9308, 28858, 36708, 28858, 9308, 28857, 36708, 9309, 19549, 9309, 28857, 36707, 9310]
Already here, some patterns should be apparent. But we have some extra information that could be used to filter out the short period noise.

Rounding at an appropriate sampling frequency

Earth transits aren't going to happen every day. If there was an Earth transit yesterday, it can't happen again today, or next day, or even the next month. In fact, we know they can only happen once for every synodic period of Mars-Earth (780 days).
To not miss out in any "legitimate" part of the signal, we should sample at at least twice of that rate. "one year" happens to be pretty close to half of 780 days.
The periodicity is then readily visible:
[26, 79, 101, 79, 26, 79, 101, 79, 26, 79, 101, 26, 54, 26, 79, 101, 26]
Which also reveals an even higher extra periodicity of about 206 years.
With a strong 206 year periodicity found in a set of events only covering a little over 400 years, there's not enough data to with any certainty detect any higher order periodicities that that.


Decreasing rate of Earth's rotation: where does the power go? I understand from Wikipedia that, "The length of the day, which has increased over the long term of Earth's history due to tidal effects, ...".  If the angular velocity of the earth is decreasing then the stored rotational energy is decreasing.  Where does the power/energy go?

You're correct in assuming the net angular momentum of the system in question here will remain constant. The Moon's orbit around Earth is responsible for the slowing of Earth's rotation. This effect is extremely small.

The decrease in Earth's angular momentum is transferred to the moon, which resultantly sees it's orbit accelerate. This acceleration also causes the moon to move further and further from the Earth. This trend will continue until they reach a common speed.


Parallax due to the motion of the sun Generally, the parallax is contributed to the motion of the earth around the sun. However, at the same time, the sun is moving too. It carries the earth with it. Presumably, the shift of the sun is bigger than the earth-sun distance. So, that will also cause apparent shift of a star in the background.

I am not so familiar with astronomy. Is my argument right?

You are referring to what is known as proper motion, which is the apparent motion on the sky due to the relative motion of the solar system and the stars.

To measure an accurate parallax you have to observe for long enough that you can disentangle the parallax motion, which is of course modulated on a 1 year cycle, from the proper motion of a star, which produces an approximately linear, monotonic drift of the star's position with time.

This is perfectly illustrated in this graphic produced by HST/NASA, which shows the track on the sky taken by Proxima Centauri (a nearby star with a large parallax and a large proper motion). The small, periodic wiggles are caused by the parallax shift, whilst the linear trend in the path is due to proper motion. The two are easily separated if you observe for at least 18 months with several precise position measurements during that time.


In what form was the space at singularity? It is somewhat imaginable for all the matter or energy to be condensed in sub-atomic size. But what about space? The space is neither 'matter' nor 'Energy'.

If there were no space at singularity where did it come from?

A singularity in the strict sense is not assumed in the beginning of the big bang, but instead a Planck epoch governed by a yet to confirm quantum gravity. General relativity and quantum theory, are not consistent in this phase. This means, that space and time, if they existed at all in some sense, didn't follow the laws of physics we inferred from experiments.

This epoch is thought to have been followed by a grand unification epoch, where gravity is already separated from a unified strong, weak and electromagnetic interaction.

Then inflation is thought to have followed, "driven by a negative-pressure vacuum energy density". That's when spacetime expansion is thought to have accelerated rapidly, weak and electromagnetic interaction still unified.

All this is thought to have happened within the first $10^{-32}$ seconds.

Most of the physics in those early phases is not (yet) accessible to experiments. A few cosmic ray experiments may try to look into the energy ranges, thought to have been present in the early epochs of the universe.

More on the chronology of the universe.


Why does the exoplanetary system TOI-178 challenges current theories of planet formation? I am excited reading the news about ESA's latest exoplanet discovery using CHEOPS. The system TOI-178 consists of (at least?) five planets, following a 18:9:6:4:3 pattern. In the press release, the authors write:

But while the planets in the TOI-178 system orbit their star in a very orderly manner, their densities do not follow any particular pattern. One of the exoplanets, a dense, terrestrial planet like Earth is right next to a similar-sized but very fluffy planet – like a mini-Jupiter, and next to that is one very similar to Neptune.

In the few systems we know where the planets orbit in this resonant rhythm, the densities of the planets gradually decrease as we move away from the star, and it is also what we expect from theory.”

My question: What exactly does "the theory" state exactly? And why exoplanets in orbital resonance necessarily imply decreasing density of the planets if the system did not collide with any other system after its formation?

The original research paper is published is found on A&A and available via arxiv, too.
One of the mysteries to understand in planet formation is the distinction between terrestrial planets and gasous planets - there is no continuous distribution curve. That means that different processes must be involved. Understanding how the involved processes interact, what conditions are required to trigger the individual processes, and how they shape a final system is important for the overall understanding. And this is easiest, the more complete information you have about a system and the better you can deduce its past evolution from the current state. The laplace chaining of the MMR of the planets helps in this endevour.
Thus using this somewhat surprising stability of the system, we can tackle further issues with the formation of the individual planets itself and possibly also the planetesimal-disk interaction. In their summary they state:

In this respect, the different planets of the TOI-178 system lie on
both sides of the radius valley (Fulton et al. 2017). Therefore,
reconstructing the past orbital and atmospheric history of this planet
may provide clues regarding the origin of the valley.

The conjected 'decreasing density' of planets from center outward directly comes from the density of available solid material in the protoplanetary disk: further inward at higher temperatures only the heavier elements can condense. Barring any migration of planets and swapping radial order, this defines the radial density distribution. You see the very same density variation inside our own solar system. The equation of state of matter at high pressures in the planets centres somewhat slightly distorts this, but the general trend applies.


Period of unique horoscopes? What is the period of unique horoscopes? In other words how long does it take before any configuration of the classical planets repeats?

The seven classical "wandering stars" are: Sun, Moon, Saturn, Jupiter, Mars, Venus, and Mercury.

Assume a horoscope is created by dividing the ecliptic into 12 equal regions and then for a particular day noting which region each of the 7 "wandering stars" is occupying relative to the earth. For example, a horoscope could be: Sun in Aries, Moon in Capricorn, Saturn in Taurus, Jupiter in Libra, Mars in Pisces, Venus in Aries, Mercury in Aries.

Since Saturn revolves around the sun once every 29 earth years, Jupiter every 12 years, and Mars 1.88 or so, multiplying together we have 654. So, relative to each, they would seem to repeat their configuration every 654 years.

This is potentially a complicated question which might require a simulation to determine the answer. For example, today (5 December 2016), the following is true for the tropical zodiac:

Sun in Sagittarius
Moon in Aquarius
Mercury in Capricorn
Venus in Capricorn
Mars in Aquarius
Jupiter in Libra
Saturn in Sagittarius

So, one instance of this problem would be to determine the next date at which this same configuration will next recur.

I wrote https://github.com/barrycarter/bcapps/tree/a944a8d8315fd6169907f73e47698f030befe3bb/ASTRO/bc-zodiac.c to solve this. I may've tweaked bc-zodiac.c since then, so you'll need the exact link above (not the most recent version of bc-zodiac.c) to confirm/replicate my results. In the directory https://github.com/barrycarter/bcapps/blob/master/ASTRO/

The file "houses.txt.bz2" contains a list of all house changes for now +- ~15000 years in a mostly-English-readable format.
The file "house-states.txt.bz2" is a list of all "horoscopes"(house states for all planets) in machine readable format for the same period. Please see https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-zodiac.txt (which also contains the text of this answer) for details on how to read this file.

Important caveats below, but first, some notes:

You'll notice that lines like:

BCE 13200-JUL-12 09:48 SUN ENTERS ARIES PROGRADE -479626193452 S0+

suggest the vernal equinox was on July 12th about 15000 years ago (of course, no one actually used our calendar back then), which seems a bit off. Further, you'll notice:

CE 17191-FEB-23 00:27 SUN ENTERS ARIES PROGRADE 479386211293 S0+

suggesting the vernal equinox will occur on February 23rd (about a month early) 15000 years from now (assuming they continue using our calendar).

See https://astronomy.stackexchange.com/a/13009/21 for why this happens (my answer there has slightly different dates/times, because I used a less accurate model).

In the 11,100,017 days (about 30,391 years) covered, there are 6,359,874 horoscope changes, for an average horoscope change roughly every 1.75 days. Broken down by planet:

The moon changes house 4,875,276 times, about once every 2.3 days.
Mercury changes house 450,523 times (including retrograde house changes), about once every 25 days.
Venus changes house 384,971 times (including retrograde), about once every 29 days.
The Sun comes in third with 364,691 changes, about once every 30 days. A more exact number is once every 30.4367697585 days. When you multiply this by 12, you get 365.2412371020 days, which, as expected, is a good estimate of the length of the tropical year (the exact value given by http://hpiers.obspm.fr/eop-pc/models/constants.html is 365.242190402, which is 82 seconds different).
Mars changes house (including retrograde) 209,501 times, about once every 53 days.
Jupiter (w/ retrograde) 49,215 times, once every ~225.5 days.
Saturn (w/ retrograde) 25,697 times, ~432 days.

As expected, most of these house changes (6,283,141) are in the forward or "prograde" direction and very few (76,733) are in the retrograde direction. Of course, the Sun and Moon always move prograde, another reason the prograde number is so large.
Since there are 12 houses and 7 planets, it would seem there are 12^7 = 35,831,808 possible horoscopes. However, there are actually only 3,732,480:

Since Mercury's greatest elongation is 28 degrees, it must remain within one house of the Sun. Given the Sun's house, there are only 3 possibilities for Mercury's house (either the Sun's house or one of the two adjacent houses).
Since Venus' greatest elongation is 47 degrees, it must remain within two houses of the Sun, so there are only 5 possibilities given the Sun's house.
Combining these, we see there are 12*12*3*5*12*12*12 or 3,732,480 possible horoscopes.

Of these possibilities there are 2,565,570 (about 69% of the total possible) actual horoscopes. Of these,

969,067 horoscopes occur only once
The horoscope "4045424" (translated below) occurs 18 times (most recently in 1799 and next in 5391), but two others occur 17 times, 10 appear 16 times, and many appear 15 times, so it's only the most frequent by coincidence (probably just because of where DE431 happens to start and end).
"4045424" translated is Sun in Leo, Moon in Aries, Mercury in Leo, Venus in Virgo, Mars in Leo, Jupiter in Gemini, Saturn in Leo.

Note that this horoscope is probably only the most frequent because of how I chose the start and end dates (based on the DE431 ephemeris).

Important caveats:

My standard disclaimer: please do not rely on anything I do/post anywhere. I can and do make mistakes. Feel free to check my work and alert me to mistakes.
If you actually want to check my results, you will need to download kernels from NASA. See http://naif.jpl.nasa.gov/naif/tutorials.html for more details on how this all works.
Note that NASA can't predict planet positions (or even the Earth's own orientation) into the far future: they solve differential equations to predict planetary positions and then publish an approximation of those solutions. The further away we get from today, the less accurate these predictions are.
We have a pretty good approximation of how the Earth precesses, but we can't even predict THAT super accurately more than a few months in advance. Note that https://naif.jpl.nasa.gov/pub/naif/generic_kernels/pck/ shows the high precision precession kernel is only good through 6 Mar 2017, and even THAT is updated twice a week (see https://naif.jpl.nasa.gov/pub/naif/generic_kernels/pck/aareadme.txt for details). Since precession affects the location of the first point of Aries, this adds another uncertainty.
Precession doesn't change the ecliptic itself, but nutation does. We don't have a super-accurate model of nutation, and I am using the undocumented SPICE functions zzmobliq_ and zzwahr_ as per a private reply to my https://naif.jpl.nasa.gov/pipermail/spice_discussion/2016-January/000423.html
The calendar on https://cafeastrology.com/astrologyof2017horoscopes.html (and the 2016 equivalent) should agree with my results within +- 1 minutes. Remember to convert between ET on their site and UTC in my file.
http://aa.usno.navy.mil/data/docs/EarthSeasons.php can also be used to check some of my calculations (within +- 1 minute), recalling that, in the northern hemisphere:

"SUN ENTERS ARIES" (as in "first point of Aries") marks the vernal equinox.
"SUN ENTERS CANCER" (as in "Tropic of Cancer") marks the summer solstice.
"SUN ENTERS LIBRA" marks the autumnal equinox. As nearly as I can tell, there is no well-known associated phrase with this point in the zodiac.
"SUN ENTERS CAPRICORN" (as in "Tropic of Capricorn") marks the winter solstice.

If you use HORIZONS (http://ssd.jpl.nasa.gov/?horizons) to check my work, you'll notice my results are accurate to the minute for about +- 1000 years from now, but drift in accuracy after that. I'm not exactly sure why this is.
I used a "stepsize" of 86400 seconds (1 day), so if a planet changed house and then changed back in under a day, my program may miss it. Someone w/ more computing power may wish to re-run my program with a smaller stepsize.
The moon's position varies with your location on Earth. I used a geocentric model, giving the moon's position for a theoretical observer located at the center of the Earth.
I actually computed planetary barycenters (the center of mass of the planet and its satellites) because NASA doesn't provide planet position information for the entire 30,000 year period I computed (see http://ssd.jpl.nasa.gov/eph_spans.cgi?id=A for details).

For Mercury and Venus, this no effect (no moons), and the Mars' system barycenter is within a few inches of Mars' own center (Phobos and Deimos have very little mass). Jupiter is fairly massive, so its system's barycenter is fairly close to its own center, and even the effect Titan has on Saturn isn't that great.

The human-readable time in houses.txt.bz2 is the output of the C function http://naif.jpl.nasa.gov/pub/naif/toolkit_docs/C/cspice/timout_c.html (format ""ERAYYYY##-MON-DD HR:MN ::MCAL ::RND" modified slightly by Perl). They are reasonable for current dates, but not necessarily for past dates because the Gregorian Reformation of the calendar occurred at different times for different countries, and the BCE/CE time system didn't exist prior to 1 CE (and probably not even then). For precision work, see bc-zodiac.txt for details.
As above, I use "ARIES" to mean the "1st house", ie, the constellation that starts at the vernal equinox ("the first point of Aries", where the ascending ecliptic meets the celestial equator) and ends 30 ecliptic degrees of ecliptic longitude later.

This was reasonably accurate in 56 BCE, but is certainly not true today as the vernal equinox is in Pisces and drifts towards Aquarius, heralding the "Age of Aquarius".

In the 30,000+ year period covered by my program, the first point of Aries makes a complete circuit (and a bit more) of the Zodiac, even ending up in Ophiuchus for a while.

Of course, the constellation boundaries I mention above are those defined by the IAU. I don't know how (or if) the ancients drew their constellation boundaries.

I am leaving my old answer below, because NASA is a trusted site for astronomers, and WebGeoCalc and HORIZONS are important resources. The step-by-step method I show below can help people here make their own calculations or help others. I am in no way affiliated with NASA or the government of the United States, except as a citizen.

EDIT: As I should have guessed, people have already done this: https://cafeastrology.com/astrologyof2017horoscopes.html (and I'm guessing there are many many other sites too, just google "Mercury enters Aries" "Venus enters Taurus" type stuff).

I hope to provide a better answer later, but for now:

Visit http://wgc.jpl.nasa.gov:8080/webgeocalc/#CoordinateFinder and do this:

and click "Calculate". This will tell you every time Mercury has an ecliptic longitude of 30 degrees, the cusp between astrological Aries and Taurus.

If you do this for all 7 planets and every multiple of 30 degrees (12*7 = 84 times), you can use the results to see when the combination of constellations change. However, this is has several problems:

It's extremely inefficient
If you hover over "Solar System Kernels kernel set" above, it says the kernels are valid from 1950 to 2050. However, with the options I chose, you can only get results from 1966 to 2034, possibly because the ECLIPDATE frame is only defined in that interval (when I choose years outside this range, I get an unhelpful and inaccurate error-- you may get different results for other planets).
The results are in calendar times; however, this may be my fault. You may be able to get the results in a more numerical and useful format.

The "correct" way to do this is to use CSPICE (http://naif.jpl.nasa.gov/naif/tutorials.html) which lets you make calculations +- 10000 years from now.

Leaving that aside, let's check our results. If you scroll down a bit in the results window, you'll see:

Let's check the result for 2017-04-20 by visiting http://ssd.jpl.nasa.gov/horizons.cgi and entering these parameters:

to get these results:

As you can see, Mercury's ecliptic longitude crosses 30 degrees between 2017-04-20 and 2017-04-21, so our results are reasonable. Interestingly, Mercury is in retrograde at this time, so it actually crosses from Taurus into Aries, not vice versa.


How to make a telescope for viewing planets, moon and DSOs using a convex lens of aperture 100 mm and focal length 200 mm and other lenses at home? I wanted to make a telescope with DIY things lying around in home. I read up that the aperture was a very important aspect and thus bought a convex lens with 100mm aperture and 200mm focal length, to use it as an objective lens. Apart from that, I have the following lenses lying around:-

1 convex lens    -   focal length 20 cm             aperture - 10 cm
2 convex lenses  -   focal length 17 cm             aperture - 5 cm
1 convex lens    -   focal length 2.2 cm            aperture - 2.5 cm
1 concave lens   -   focal length 8.33 cm           aperture - 5.5 cm
1 convex lens    -   focal length 200 cm            aperture - 5.5 cm

I can also buy lenses of range of Power(In Dioptres) from (-14D) - (+14D) of aperture range of 5.5 cm to 7 cm, if suggested.

I made a telescope using lens 1(as objective) and lens 3(as eyepiece) which on calculation, gave a magnification of 9x, but according to stellafane.org and http://skyandtelescope.com the minimum magnification is 14x and that I was 'wasting' the light gathered by the objective lens. This telescope showed me a bright Mars, Jupiter and Saturn, but as dots. Upon seeing the Moon, I could just see it a little bigger than what would be seen with the naked eye.

I would like to know the way to make the telescope, which I would like to use to observe planets and DSOs if possible. I am ready to make separate telescopes also, if suggested.

I would like to know which lenses should I use, or buy and the way to use it. I live in a suburb and it is not very lighted up.

Is there any way I could use the lens 1 in a telescope which I could use to view the planets?

[Edit]

I have accepted @JamesScreech 's answer and am on my way to making it. Apart from that, I would also like to use the big aperture of lens 1, for viewing DSOs, But, it has a very short focal length(just 200mm), and therefore, I am finding it hard to get any eyepiece lens which can "completely accommodate the exit pupil of my scope"(as said on skyandtelescope )

Any way to get such a lens, or make an eyepiece barrell sort of, using a combination of lenses mentioned above or buying. And how to use it?

Thanks.

You would be better off using lens 5 and 3. Though don't expect much, a simple single lens for the objective and for the eyepiece will give a lot of chromatic aberration (false colour). Refracting telescopes usually use at least two lens of different glass types (crown & flint often in a doublet) and eyepieces usually have 4+ lens.

You should be able to see craters on the Moon, the 4 Galilean Satellites of Jupiter, a couple of belts on Jupiter and Saturn's rings. You wont be able to reduce chromatic aberration as simply as that, the type of glass (it's refractive index) and curvature of the lens all need to be matched. You would be much better off buying a ready built objective and eyepieces if you want to make a refractor yourself, though it would probably be cheaper to buy a ready made telescope.