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Calculate planet's surface temperature by distance from star
If the planet and star in question are very similar to the Earth and the Sun, how can I calculate the planet's surface temperature by knowing the distance from the star?
This question was originally asked on the Worldbuilding stack exchange, and someone suggested I ask it here.

How closely firmly can we assume a planet is Earth-like? This is especially concerning considering being 'Earth-like' is also going to be a function of the planet's distance from their star. Generally though, this is not a terribly solid method though - in our own solar system Mercury is roughly half the distance from the Sun as Venus, but Venus has a hotter surface temperature.

Like, exactly the same in atmosphere, size and core temperature.

At work so I can't really go through the math, but my suggestion would be to use the inverse-square law and the angular diameter of the planet to calculate how much total energy is reaching the planet from the star.

Depends on whether or not your planet has an atmosphere (I'm guessing it does). If it doesn't the calculation is easily performed by the question linked by FJC above. If it does have an atmosphere, then it is vastly more complex and heavily depends on the type of atmosphere.

@zephyr In addition to type of atmosphere, whether it has liquid oceans and what they are made of, how much of the surface is ocean to land, how well the oceans circulate, how easily they evaporate, whether it can form glaciers, surface color or albedo, clouds and/or dust in the upper atmosphere, location of land masses.  It goes beyond just atmosphere, though atmosphere is a big factor.

Neat and tidy calculations for this aren't possible.
The math behind effective temperature of a planet can be found here and that's a straight forward calculation where you can input distance, solar energy, planet's albedo and get an average temperature.   Earth's effective temperature calculation comes to -21 °C, which obviously isn't accurate, but that's as far as the calculation can take you.
Feedback mechanisms:
With essentially no change to distance from the sun, and only relatively small changes in tilt and eccentricity, Earth's temperature can fluctuate from a current 15 °C global average to about 5 °C as it goes in and out of ice ages.
Feedback mechanisms like albedo (ice cover is the key driver, but also deserts, surface area of the oceans), CO2 capture, which increases with colder oceans, can amplify small changes, leading to significant changes in surface temperature.   An orbital change, which should account for maybe 1 degree C, or an increase in CO2, which, by itself, captures less than 1 degree C, through feedback mechanisms, can lead to much larger temperature changes.
More on feedback mechamisms here.
Land placement and Ocean circulation
5 or 10 million years ago, Earth's average temperature was about 18-20 °C and that change was driven, primarily, by the closing of the Isthmus of Panama which affected ocean circulation.  Just the closing of the Isthmus made ice ages possible.  That's a pretty small change in the grand scheme of things, whether your planet has a near equatorial passage where oceans can flow or whether it doesn't, but that one thing can mean more than 10 degrees.
As Antarctica drifted over the south pole (the last 30-35 million years or so) glaciers formed and the earth cooled, because as ice forms, more sunlight is reflected into space and oceans levels drop.  Oceans having the lowest albedo and ice the highest.  Also, as Antarctica driffs away from the south pole over the next 20-30 million years, Earth is expected to warm, unless another change occurs, like the continued absorption into land and the oceans of CO2.
Earth can only enter an ice age when there's land near the poles.   The glacial period of 440-460 million years ago most of the land was over the south pole.  The sun was about 4% less luminous and the earth had much more CO2 (though CO2 levels did drop during that time, they didn't drop close to where they are now) and the orbit of the earth, 440 million years ago has some unknowns, but a large land mass over the south pole is thought to have played a key role in making that glaciation possible.  The tipping point of ice formation at the poles is particularly sensitive and a significant variable.
So, even with very similar to earth like planets.  Just where the land is and how well the oceans circulate and whether the planet can form glaciers and sea ice, or not, can fluctuate perhaps 20 degrees C in global average temperature.  How much CO2, how much active volcanism, how much reflective material in the upper atmosphere - all variables.
Even with identical earths, the feedback mechanisms make it difficult to predict what Earth's temperature would be if you pushed Earth 1 million or 1/2 million miles further from the sun.  The calculations for surface temperature wouldn't follow a simple mathematical formula.   Now, you'd still get colder as you moved the planet away, but one push might give you 1-2 degree and the next push might give you 5-10 and the one after that 0-1.  There would be no way to do a clean formula.

I think it is important to mention that your link for the surface temperature is really the temperature at the top of the atmosphere (despite what the section title says). It does not take the atmosphere into account.

I thought it was surface temperature if the earth had no atmosphere, which, granted, would change quite a few things, but it's just a calculation.

Well yes, that's an equivalent scenario. Either it represents the temperature at the top of our atmosphere or else the temperature on our surface if the atmosphere didn't exist. Either way, it doesn't take the atmosphere into account.

@zephyr I probobly should have covered the atmosphere a bit more.   I'll look at editing to add that, but the question was on planets "very similar to Earth", which, I think, implies at atmosphere like ours, water oceans, etc.


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What is at the edge of the galaxy?
The solar system is thought to have a heliopause which can be considered its edge for many purposes:
The heliopause is the theoretical boundary where the Sun's solar wind is stopped by the interstellar medium; where the solar wind's strength is no longer great enough to push back the stellar winds of the surrounding stars. This is the boundary where the interstellar medium and solar wind pressures balance.

Heliopause: the boundary between solar wind and interstellar wind where they are in equilibrium.

Is there a similar phenomenon at the interface between space dominated by the Milky Way and intergalactic space?
What is at the edge of the Milky Way galaxy?

@SirCumference I think that is pretty much the question asked.

@Hohmannfan I'd say he's asking about what lies at the edge of the Milky Way. Or maybe his title isn't related to the actual question.

Also, the Solar System is usually defined by the objects orbiting the Sun — that is, everything within the Sun's Hill sphere.

@SirCumference As stated in the question body, I'm (tentatively) thinking of "the edge of the galaxy" as the interface between space dominated by forces in the galaxy and space beyond that / intergalactic space. It may be a fuzzy boundary; there may not a precise line. Or perhaps it is precise at a point in time but it fluctuates over time due to dynamic interactions between the 2 regions.

@SirCumference This is by analogy to the "heliopause" which is considered for many purposes the edge of the solar system. For example NASA announced Voyager 1 had reached interstellar space when it had crossed the heliopause: Voyager 1 Reaches Interstellar Space

@SirCumference I don't know whether there is an analogous phenomenon in the outer galaxy not. I'm open to other definitions of "the edge of the galaxy" if there are better ways of thinking about it.

@lycan the heliopause is based on a field which has a precise size that we can measure linearly from the source. The galaxy as far as i know has no such thing. Also it depends on what you call "part of the x" A lot of people wouldn't consider the Oort Cloud part of the solar system for example whether you extends that far or not and you might have the same thing with the galaxy, but worse because you might not consider a star part of the galaxy even though it's inside what you consider the galaxy and galaxies are pretty much defined by stars...

The short answer is: probably nothing much, because galaxies are very fuzzy objects without "edges".
If you look at the stellar disk, it just fades out in density, with no evidence for a sharp cutoff (this is true of disk galaxies in general). There may or may not be a "break" at some radius, but this just marks the beginning of a steeper falloff in density, not an "edge" in any real sense. The gas disk extends even further out (that is, further out than we can easily trace the stars in the disk), and doesn't show much evidence for a sharp cutoff either.
There's also a round halo of stars, which is low density but extends to at least 400,000 light years; two stars were recently discovered  which are plausibly part of the halo and located at distances of about 800,000 light years. (Note that there are other galaxies -- such as the Large and Small Magellanic Clouds -- which are closer to us and to the center of the Milky Way, so you have to accept that individual galaxies can be found "inside" other galaxies.)
And then there's the halo of dark matter...
I can think of two ways in which you might talk about the outer boundary of an isolated galaxy (which the Milky Way sort of is) from a theoretical perspective. The first is the idea of a halo "accretion shock", in which relatively cold gas falling onto the galaxy from intergalactic space runs into the galaxy's halo of hot gas ("hot" = millions of degrees), creating a shock where the infalling gas slows down and the local density of gas goes up. (This is a little bit analogous to the heliopause, though the direction of gas flow is the other way around: the heliopause is where the outflowing solar wind shocks into the more or less stationary interstellar gas.) The problem is that extensive, hot halos of gas are thought to form only around massive galaxies, and it's not clear the Milky Way is massive enough to qualify. It also won't work for galaxies in massive groups and clusters, where the individual galaxies aren't accreting gas any more (the cluster is what has the hot halo and the accretion shock).
The other possibility is a region where the infalling dark matter particles merge into the dark-matter halo (whose particles are orbiting within their common gravitational potential, and so are not as a whole falling towards the center any more). Simulations suggest that as one moves inward the local density of dark matter starts to increase rather rapidly at this point (and the average velocity of the dark matter particles changes), but it's still a fairly smooth process (unlike gas, the dark matter particles aren't "shocking" into each other). There's no way of determining where this is for the Milky Way, other than something really vague like "maybe 1 million light years"?

Great answer, thank you. Exactly what I was looking for. Have searched for this multiple times in the past but never been able to find a discussion of it.

It is argued that the edge of the galaxy has the halo of dark matter at around 100,000 light years, but there are arguments that extend the galaxy from anywhere between it being 120,000 to 200,000. The major factor though for putting the edge at around 100,000 light years is that there is a major star density fall off for no known reason.
Similarly the depth of the galaxy is said to be 10,000 ly at the bulge and 1,000 ly for the rest of the galactic plane. There is similarly no reason for the fall off in star density in that direction known. And there are massive xray and gamma ray structure coming out of the galactic center that is 50,000 ly on either side which you could argue is the real edge of the galaxy along that axis.
So what lies at the edge of the galaxy depends on how you define the edge of the galaxy which could be:

100k x 100k x 10k
100k x 100k x 100k
200k x 200k x 10k
200k x 200k x 100k
120k x 120k x 10k
120k x 120k x 100k

Or you could use some other arbitrary number, but you'd need to make that argument.
I however think that a star density definition should be how we should define things "officially" because it is more or less how most people think of the galactic limits in the first place. Of course if you define it like that then you're dealing with issues like the New Outer Arm where the density falls off and then increases, the actual density isn't a perfect shape, and we might have to say that things we call 2 or 3 galaxies are actually 1 galaxy, such as the 2 dwarf galaxies orbiting the Milky Way or the where there exist 2 galaxies with a stream of stars between the two as they merge over millions/billions of years.

Thanks for an interesting overview. What about the interstellar medium within a galaxy? Is there a point where it is overpowered by the intergalactic medium that could be considered an edge of the galaxy? That seems to be how astronomers think of the edge of the solar system: the point where the interplanetary medium is overpowered by the interstellar medium.

@Lycan I dunno. If that's how they think of it then they define it so I guess that's right, but you're largely talking about plasma and radiation that would be more fluid, flows, ripples, twirls, etc and so you still have the smae problem, it's not a perfect shape and you'll get wonkey shapes that result in multi-galaxies becoming 1, or structures that "part of the galaxy" that stretch far out, making the galaxy longer, but noone thinks of them as defining to the size of the galaxy in shape.

While the Solar wind is pretty much homogenous in all directions, the galaxy puffs out stellar winds from discrete and temporary star forming regions. Protostars and giant stars which go supernova young, punch "chimneys" through the interstellar medium in the galactic disk here and there. The Local Fluff is btw our region inside the Local Chimney, a relatively low density region formed by pressure from the young and exploding stars in the Scorpius-Centaurus Association 400 light years from here. And the supermassive black hole turns its jet on and off depending on when it accretes material. I don't think there's a pan-galactic Milky Way wind that looks anything like the Solar wind. But around an intensely starbursting quasar galaxy maybe?

Very interesting! Had no idea about the chimneys.

@Lycan   I'm not an expert, but I don't think you're question was really answered. Another way of asking this would be, "Does the Milky Way move through the local intergalactic medium or do they move together?" It's been published that the Milky Way moves toward the "Great Attractor" at about 600 km/s (the speed of the solar wind is about 400 km/s). The density of the intergalactic medium has been estimated at about 1H per cubic meter (1H per cubic cm for the ISM). If you say that the ISM meets the IGM at about 300 km/s then the equal pressure at the "galacticopause" would be about 10E-22 N/m2

@Lycan cont. By comparison, the solar wind pressure where there is an equilibrium with the ISM is on the order of 10E-13 N/m2. I arrived at this value since the solar wind pressure at Earth varies from about 1-6 x 10E-9 N/m2 and the heliopause is estimated to be between 80 and 120 AU.

@JackR.Woods Definitely sounds like what I was visualizing. I don't really have the understanding to evaluate your numbers here. Just to throw something else in, I since found on the Interstellar medium page on Wikipedia this note: "[The ISM] fills interstellar space and blends smoothly into the surrounding intergalactic space." Possibly meaning it has no abrupt edge but blends out in a smooth gradient. Unfortunately it's unsourced.

Maybe the local IGM is moving toward the "Great Attractor" at about the same speed as the galaxy.

@JackR.Woods Yes, I would think that everything in the entire Local Group and much more is moving towards it at similar speed. It's hundreds of millions of light years away. The IGM between us and Andromeda is within 2½ millions. And the star bursts winds that leave the MW disk mostly, if not entirely, rains back by gravity, like lava spewed out from volcanoes. That would be the wind our galaxy is traveling through. On average the same from all directions, but regionally temporarily spotty.

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How is a day measured on a gas giant?
After watching this JPL video of a Cassini time lapse of Saturn, I wondered since being mostly gases, how do we know how long a day—one rotation period—is?
Is there a usable relation between how fast the upper most layer of gases is and the planet's solid core?

I believe a tag rotation-period applies here, but I can't create one.

That the rotation period of the bulk mass of a planet is estimated through something with the magnetic field is true.
But let me maybe elaborate a bit in-depth on that.
No planetary magnetic field is a dipole. Earth has strong magnetic anomalies over the Atlantic, there may be ionic disturbances by moons like Io, or your magnetic field is just really crazy, like Uranus and Neptunes.

(A simple, low-order model of Earth's magnetic field, from GFZ Potsdam)
But no matter how it looks, it will usually be constant on lines of latitude and longitude. This in particular signifies that the field and its distortions are created somehow by the bulk mass of the planet, else the field would move relative to a surface map, in the case of Earth.
So all this talk just to establish the following: We can take the shapes of the magnetic field as if they were surface shapes, and by following them determine a rate of rotation.
Now the story doesn't end there. While this sounds nice in principle, to do this in practice, we would need to send a probe and do in-situ measurments of this changing magnetic landscape.
But what can be done from afar, and from Earth is observe ionospheric patterns. Because the ionosphere of a planet is well, ionized gas, at super-low pressures (roughly on the order of microbar to nanobar) it is frozen-in with the magnetic field that makes the particles of the plasma gyrate. This means, the ionospheric plasma follows the magnetic landscape, now translating into a plasma landscape. This plasma landscape is then usually observable from Earth by the radio signals they emit. It is also possible to see this in the ultraviolet, if there is strong plasma activity through auroral bombardment of a planet's upper atmosphere.
For your second question, that's a simple no.
First of all there is no 'uppermost' layer of atmosphere, as the atmosphere is continously decreasing in density with increasing distance from the planetary body in question.
What you're possibly thinking about are the upper cloud layers, as they visibly end at some height.
But wind speeds are much more determined by pressure gradients created by the central star heating the planet, and the atmospheres ability to cool.
There is one class of planet, however where a relation between the wind speeds at roughly 0.1 bar pressure and magnetic dynamo might exist. Those are Hot Jupiters which host atmospheres that might be sufficiently ionized for its winds to be braked down by the planetary magnetic field.
Then there would exist a limiting speed, given a certain planetary field. However this is still a field of active research, but it would give you 'a' relation between the two.

@Cody: Thank you. But apparently OP preferred the simplified variant, which ofc is his right.

I thought for a long time that day's on gas giants were measured by waiting for a certain feature come around, like Jupiter's great red spot.  However, looking it up just now I learned that that it is done by measuring the magnetic field.  This article is specifically about Jupiter, but I imagine measuring Saturn is much the same,

Scientist were finally able to use radio emissions from Jupiter’s
  magnetic field to calculate the planet’s rotational period and speed.
  While other parts of the planet rotate at different speeds, the speed
  as measured by the magnetosphere is used as the official rotational
  speed and period.

I'm afraid I don't have an answer to your second question.

I'm not an expert in this so I don't know for sure, but my guess would be yes.  Since Earth has solid, (relatively, at least over the course of a day) unchanging features, I bet a measurement of land features and a measurement of the magnetic field would give approximately the same answer.  I hope someone with expertise can come along and weigh in on that.

Earth's inner core rotates easterly (a teeny tiny bit faster than the surface) and it's outer core rotates westerly (again, a teeny tiny fraction).   http://www.livescience.com/39780-magnetic-field-pushes-earth-core.html    I found a 5 year old article that said 1 degree every million years, but I think that should be confirmed before posting given the opposite rotation rate article came out more recently.

There's already good answer provided, but it is worth noting that the rotation rate of a planet is ill defined, as many planets are unlikely to be uniformly rotating. For example, Hot Jupiters present zonal flows, so should the rotation rate of the planet be measured taking into account the flows or not?
We don't even know if there's a solid core in gas giants, so the question of the rotation rate should be 'is there a way to measure the rotation rate profile of a planet' and the answer is yes, but it's really hard. This is for example one of the long-term objectives of the Juno Mission. By measuring the pressure, density and composition of the atmosphere at different radius, you can try to stick a general circulation model that would give you the local speed hence the rotation rate. Of course, the picture is made harder when you consider magnetic fields.

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Mathematically calculate if a Planet is in Retrograde
I am trying to find out if a planet is in an apparent retrograde motion with respect to the earth at any given point in time. Given the time in Julian days.
I already have the geocentric and heliocentric coordinates and velocities of the planet using the JPL ephemeris, but not sure of what mathematical formula to use to identify the retrograde motion.
I read this article "https://physics.stackexchange.com/questions/249493/mathematically-calculate-if-a-planet-is-in-retrograde" which had a similar question but I am not sure of:

Is the answer correct as it has not been marked as correct?
How to represent earth in the XY plane only when it has XYZ planes and why do we even need to do that? Can't we just use the heliocentric equatorial plane which I think is the same as the one used by NASA JPL ephemeris.

This is a really interesting question, and it needs a clear definition of "retrograde motion". I think motion where Right Ascension (RA) decreases is the standard. So you need the position of the planet with respect to the earth, projected on to the celestial sphere. It's a little complicated if neither orbit is exactly in the plane of the ecliptic. The answer in physics stack exchange doesn't give enough to actually get what you need.

If you have the planets dx, dy, and dz with respect to Earth, this would just be when dy is negative. However, depending on what you have and the level of precision you require, the problem might be more difficult. I assume you know you can find this information online and are just interested in calculating it yourself for fun. Feel free to google talk me at carter.barry@gmail.com for help- I do stuff like this for fun too.

@barrycarter I tried to do using both the approaches the one mentioned by you to look for dy and also tried the one suggested by Mark in the post for which I added the url were he suggested d (tanarc(dy,dx))/dt . Even though I see trend reversals happening in both the approaches but the dy change is not that accurate and happens few days earlier than what is suggested by the almanac, the other approach gives a more close answer

@barrycarter what coordinate system is that?? I think RA is the proper place to define retrograde and prograde. dy can increase or decrease for retrograde depending on which side of the sun you are talking about. But I agree that migration is a good idea, the apparent retrograde motion of a planet really is more "Astronomy" than "Space Exploration".

@uhoh I have no idea what I was thinking when I wrote that. The correct way is to of course look at the change n the angle, as ADUser notes.

Rough answer: a planet is retrograde equally before and after its opposition. For the times between oppositions and the length of retrograde, see https://en.wikipedia.org/wiki/Apparent_retrograde_motion#From_Earth Using this, you could come up with simple (though not super-accurate) formulas for when a given planet was in retrograde.

Here is a quick answer. Apparent retrograde motion of a planet as seen from Earth means that its ecliptic longitude is decreasing. (Thanks to @barrycarter for correcting my previous statement.) You have to calculate the apparent angular position of the planet from the Earth in ecliptic coordinates (as explained here and here) and find places where the ecliptic longitude is decreasing.
You can subtract the positions of Earth from the position of the planet in either coordinates. Then you express that vector in spherical coordinates. It doesn't matter if it's heliocentric or geocentric. What matters is the direction that the axes of the coordinate system is pointing, not where the center is.
If you want RA and dec, use ICRF coordinates, and these can be transformed to ecliptic coordinates as well. An easy way to do this is to use the python package Skyfield.
Here's a sample script and a plot. This is Mars from 2010.0 to 2020.0 and the thick red line represents retrograde motion.

import numpy as np
import matplotlib.pyplot as plt

from skyfield.api import load

planets = load('de421.bsp')

earth   = planets['earth']
mars    = planets['mars']

year_zero = 2010
days = np.linspace(1, 3650, 10000)
years = year_zero + days / 365.2564
ts = load.timescale()
t = ts.utc(year_zero, 1, days)

# thanks to @barrycarter's comment, do it the right way!
eclat, eclon, ecd = earth.at(t).observe(mars).ecliptic_latlon()

eclondgs      = (180./np.pi) * eclon.radians
eclondel      = eclondgs[1:] - eclondgs[:-1]

eclondel[eclondel < -300] += 360. # this is a fudge for now
eclondel[eclondel > +300] -= 360. # this is a fudge for now

prograde   = eclondel > 0.

eclon_prograde   = eclondgs.copy()[:-1]
eclon_retrograde = eclondgs.copy()[:-1]

eclon_prograde[~prograde]  = np.nan
eclon_retrograde[prograde] = np.nan

plt.figure()
plt.plot(years[:-1], eclon_prograde,   '-g', linewidth=1)
plt.plot(years[:-1], eclon_retrograde, '-r', linewidth=3)
plt.ylim(0, 360)
plt.title('Mars Ecliptic longitude (degrees) AD 2010.0 to 2020.0',
          fontsize=16)
plt.show()

I actually think retrograde means the ecliptic longitude is decreasing, but that's just me being nitpicky :)

@barrycarter You are right and it's not nitpicking at all. Good catch - thanks! After face-palming I fixed the text and python. For the math I just changed .radec()to `.ecliptic_latlon(), renaming the variables and fixing the text took longer.

@CarlWitthoft thanks! I guess I've "settled" for something that's easy enough even I can use it. And there's all those darn python packages...

@LouisMaddox thanks for the edit! In 2016 I started with Python 2, so there's going to be other issues like this with older scripts here; please feel free to fix anything you find. Welcome to Astronomy SE!

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text:
Darkness in Outer Space
I am trying to imagine how dark outer space is. Looking at images of cosmonauts in space proved not very useful, as the exposure makes space look completely dark.
Thinking about it, I would imagine an object would be quite well illuminated, being bombarded by provided it's not in the shadow of a planet or some other celestial body. Is that correct? How about light scattering? Is cosmic dust abundant enough for the purpose?
EDIT: @zephyr made me realize that the quantity of light is a function of position, so let's say the question is with regards to the Milky Way. Still, I would appreciate some comments on what's the situation outside the Milky Way too.

I assume by space you mean in orbit around Earth or perhaps on the Moon, not in some random void of the universe, far from anything else.

I am not sure if I mean that. :) It was my understanding that light is ever-reaching and that generally such voids would be a rare occurrence? If that's not the case, does ``anywhere inside a galaxy'' (like the Milky Way) sound more reasonable?

Light is certainly pervasive everywhere, but the amount of light you might receive at any one location is highly dependent on how close you are to a light source. Thus the answer would be different if you were near Mercury versus Pluto, or inside our galaxy vs outside it. Just wanted to determine which situation you were referring to in your question. Feel free to edit if you want to provide something more specific.

Measurement of what is known as the extragalactic background light is an entire area of study. Here's a recent review paper on the topic by Asantha Cooray.

Just to put in a simple answer:
if you mean (A) for astronauts just in orbit (at the space station and so on) or on the surface of the moon.  Yes, it is incredibly bright.  Just like on Earth in the Sahara at noon!
A point of confusion is that THE >> SKY << IS BLACK, but it is incredibly bright.  On Earth in daytime, the sky happens to be blue, and it is incredibly bright.  On Moon in daytime, the sky happens to be >> black <<, and it is incredibly bright.  (Same deal in orbit as on the Moon.)
(If you are, say, explaining to a young person "is there daytime/nighttime on the moon" that's the salient point to explain: on the moon, confusingly, the sky is black during both daytime and nighttime.  But everything's just as bright on the moon in daytime as on Earth in daytime, there's the same amount of light, i.e. you can trivially read a book or see the ground, just the same as on Earth.)

If your question is about (B) in the solar system - say "on Uranus" or the like.  There is remarkably less light there than on Earth, the sun is much smaller.  Many young people don't realize this; everything's basically dark all the time from about Saturn outwards!
Outstanding page explaining this:
http://www.astronoo.com/en/children/sun-apparent-size.html

Finally if your question is about either (C) inside our galaxy but not near a star, or, (D) midway between the galaxies:
That's too difficult.

Actually even though the Sun appears smaller the further out you travel in the Solar System, there is still a significant amount of sunlight. NASA have even created an app were you can calculate "Pluto Time" which is the time just after sunset when the sky is as bright as it is on Pluto.

Hey @Dean great app suggestion!  Yes for any young folks reading there's a picture right here: http://www.astronoo.com/en/children/sun-apparent-size.html showing how dark it is on Pluto, cheers

Also in relation to your point (D) I did a rough estimate for the brightness of Andromeda half way from the Milky Way and I got an apparent magnitude of 1.4, which is about as bright as the star Regulus. So if you assume the milky way is just as bright then you would effectively have two bright diffuse objects around 1-2 magnitude in the sky and not much else visible to the human eye.

Some very good point raised here, especially that black can be bright too.

Let's be clear here - the sky is black, and the illumination level is bright. But the black sky itself is not bright. It's like being in a room with a very bright overhead light but black walls and a black ceiling. It's plenty bright in the room, but the walls and ceiling are still dark.

For sure Mark.  So, where it says "IT IS BRIGHT" that means, well, it is "bright" - so, if you hold up a book you can easily read it, the pages are blazing white.  For sure.   Yes, this sort of thing has to be clearly explained for our cohort of younger readers here!  (Man, if only this existed when I was a kid.)

I think you can actually experience the answer to your question by going out into the deep wilderness late on a night where there is no Moon, make sure no artificial lights are anywhere nearby, and simply observe.  That's how dark it is in a reasonably random location in the Milky Way!  It will be very dark indeed, but you might still be able to see something (other than the stars, of course).  It's a good question-- can you see your hand in front of your face by starlight alone?  I don't think you can, but I haven't done the experiment!

It is an outstanding point that, indeed, that is a random point in the Milky Way !!  You simply hide our Sun (by putting the Earth there) and then it's pretty much identical to being anywhere in the Milky Way, not close to a star.  Great post.

Good analogy. Actually, when writing the question, I was thinking precisely of the woods at night. It's so dark you most likely won't be able to see anything! But then again, I would imagine that the situation is a bit different in outer space, as there are no trees or Earth to obstruct light.

There could be a factor of  2 or so difference, it's true, but that kind of variation you'll also get just from choosing random places in the Milky Way.

The reason the night sky is dark is our distance from the nearest stars (over 4 light years) which would be as bright as our sun (93 million miles) if they were not over 2500 times further away and only the brightness of a star, not the sun.

Whilset superficially appealing that isn't a valid reason for why the sky is dark - please see https://en.wikipedia.org/wiki/Olbers%27_paradox for reasoning

@adrianmcmenamin  I read it before I posted and after. Some stars we cannot see. If we are in a city many more we cannot see. The nearest and brightest are pinpricks and all more than 24 TRILLION miles away.

It doesn't really matter how far away they are - if the universe was of infinite age and infinite extent and matter was uniformly distributed we'd still have a bright night sky: individual photons do not grow dim over distance, so an infinite number of sources, no matter how far away, would burn your retinas out!


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Massive degraded impact crater on Mercury - chemical evidence?
The planet Mercury has a myriad of impact features - this has been known for some time. However, observe in the following images, a relatively large 'High-Mg region' detected on the planet (below):

Image source: Messenger website
From the group's press conference website, they state that:

The unusual composition of the high-Mg region and the relatively thin crust in this region suggest that it may be the remains of an ancient and highly degraded impact basin.

How is the 'unusual composition' of this region suggestive of being the remains of an ancient, highly degraded, impact basin?

The composition is "unusual" in that it is not an expected composition for Mercury and the source of the "unusually" high amount of Mg is likely from outside Mercury, namely asteroids/comets. Its the same principle with water on Earth. We don't expect Earth to naturally have as much water as it does, so we assume much, if not most, was brought in by comets impacting the Earth.

@zephyr okay, thanks for the explanation - I literally can't tell, but are the quotation marks supposed to be sarcastic?

No, I was merely quoting the term from the press conference. I can try to expand my comment into a full answer if you'd like but I'm not sure how much more there is to say.

There is still some research to do to definitively answer this question. Having said that though, according to research published in the paper "Evidence for geochemical terranes on Mercury: Global mapping
of major elements with MESSENGER’s X-Ray Spectrometer (Weider et al. 2015), the high Mg region (HMR) could be due to the consequences of an impact event - the authors from the article suggest that:

therefore, that the HMR may also be an impact related
  feature, and that its high Mg/Si ratio is evidence of
  mantle excavation during a large impact event early in Mercury’s
  history (>4.1 Ga).

and later, partly resurfaced by subsequent volcanism and smaller impact events.
The authors elaborate further that the mechanism behind it could be

If an ancient large impact at the HMR site penetrated into a
  vertically differentiated mantle (with additional lateral variations
  possible) and reached a deep, otherwise unsampled layer (i.e., the
  lherzolitic source proposed by Charlier et al. 2013*), the HMR composition
  would be the signature of high-degree partial melting of
  that mantle source.

Additional resource cited:
Charlier et al. 2013, Phase equilibria of ultramafic compositions on Mercury and the origin of the compositional dichotomy, Earth and Planetary Science Letters


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Are sunspots vertically displaced from the surrounding photospheric plasma?
While reading a paper on the helioseismology of sunspots (Cameron, et al.), very brief reference is made to the "vertical displacement of a blob of plasma." This caused me to wonder if the emissive surface of the sunspot itself were perhaps depressed from the remainder of the photosphere. If this does occur, is there any literature on mechanisms? I would be very interested to learn more. There was one paper I came across around formation of sunspots from a toroidal field (Parker); but, it seemed to focus more on the vertical displacement of magnetic field lines rather than the plasma itself.
Thanks.

You sure that's referring to the sunspot itself, rather than a "side-effect" ?  IIRC, temperature, magnetic, etc. steep gradients lead to a 'burp' of plasma being ejected at high speed.

Yes.
The idea that sunspots are depressed slightly came as a possible explanation for the Wilson effect. The Wilson effect was discovered as the shape of sunspots as viewed from Earth changed as the Sun rotates, in a way consistent with the change in perspective looking onto a slightly depressed region. While this isn't the only explanation for the effect, it's certainly the most prevalent.
More specifically, as Solanki (2003) writes, the depressions indicate a lowering of the layer where the optical depth $\tau=1$ (keep in mind that the bottom of the photosphere is the layer where $\tau=2/3$). There are two causes mentioned: lower temperature and magnetic effects.
Sunspots are cooler than the surrounding areas, as is well known, and thus appear darker. We see the same thing apparent at the boundaries of solar granules: Cooler, darker gas sinks and lets the hotter gas (which is less dense) move upward. Additionally, the opacity $\kappa$ is temperature-dependent, which may impact how far one can see into the star.
Not only is temperature a factor, but so is the magnetic field. Sunspots are, at heart, a magnetic phenomenon, and thus the radial force equation is substantially different. Normally, in a star, the equation of hydrostatic equilibrium is
$$\frac{\mathrm{d}P}{\mathrm{d}r}=-\rho g$$
for pressure $P$, density $\rho$ and gravitational acceleration $g$. However, when the magnetic field becomes important in a sunspot on the solar surface, the force balance is
$$\frac{\mathrm{d}P}{\mathrm{d}r}=\frac{B_z}{4\pi}\left(\frac{\mathrm{d}B_r}{\mathrm{d}z}-\frac{\mathrm{d}B_z}{\mathrm{d}r}\right)$$
where $r$ and $z$ are the radial and vertical coordinates (note the change of coordinate system - $r$ is along the surface, and $z$ is perpendicular to it!). The force from the magnetic field implies a lower gas pressure and a greater depression.

Astounding knowledge!

Thanks for this! The paper from Solanki is exactly what I was looking for.

Is Kramer's law relevant to the photosphere?

It was a genuine question. I actually don't know how bulk opacity varies with temperature at a few thousand K, but molecules and atmosphere are more opaque than ions, so I'm sure opacity sharply increases with decreasing temperature.

@RobJeffries (I deleted my previous comment; $H_{-}$ ionization might not be as dominant in relatively cool sunspots as I thought.) I couldn't tell you. My knowledge of stellar atmospheric models is minimal, especially at such low temperatures. I've always just naively assumed Kramer's law is valid roughly up to the photosphere or beyond, which I now don't think it is.

@RobJeffries I haven't been able to find anything conclusive; if you want to ask that as a new question, perhaps someone has some information.


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Exactly how elliptical is Mercury's orbit, visually, without exaggeration?
Reposting question from https://www.reddit.com/r/askscience/comments/5e1zhy/what_does_mercurys_orbit_really_look_like/ which reads:

First the full question: What does the orbit of Mercury really look
  like visually, without artistic liberties taken to amplify the
  eccentricity of the orbit, or simplifying it to appear as a perfect
  circle? What eccentricities are needed for orbits that readily appear
  elliptical?

I originally did this as a self-answer, but my answer was of such poor quality I've now deleted it. I hope to do a better answer backed by some calculations, but, as always, anyone else is free to answer.

Strange wording:  the orbit just is , and "visually" really has nothing to do with it -- other than the angle of observation.

Some things we know about Mercury's orbit:

Semi-major axis: 0.387 AU, about 57.9 million km
Eccentricity: 0.205

We can calculate the semi-minor axis, $b$ from the equation
$$e=\sqrt{1-\frac{b^2}{a^2}}\to b=0.379\text{ AU}$$
We can also calculate the distance to the focus from the center of the ellipse, $f$, as
$$f=ae=0.078\text{ AU}$$
I used Mathematica's ParametricPlot to plot Mercury's orbit using these parameters, along with Earth's. Earth's semi-major axis is 1 AU, and with an eccentricity of 0.016, its semi-minor axis is about 1 to three significant figures; I've treated it as a circle. Mathematica gives

ParametricPlot[{{0.387*Sin[x] - 0.0793, 0.379*Cos[x]}, {Sin[x],
  Cos[x]}}, {x, 0, 2*Pi}, AxesLabel -> {AU, AU}, PlotLegends ->
  Placed[LineLegend[{Red, Blue}, {"Mercury", "Earth"}], {1, 0.25}],
  PlotStyle -> {Red, Blue}]

I'd consider the deviation from a perfect circle to be noticeable insofar as the focus is shifted from the point $(0, 0)$, but it's not incredible.

I would argue that it's not noticeable and that the primary effect of the eccentricity is that the Sun is noticeably not the center of the orbit.

@barrycarter That's basically what I meant, but I can make that clearer.

Looks OK compared with the NASA version.

Here you go. Image is from this NASA website and shows Earth, Mercury and Mars, with circles for comparison.

@rob-jeffries and @hde-226868 have both given excellent answers: after remembering that I was the OP, I upvoted both and approved one.This answer is mostly my attempt at redemption:

The image above (Sun depicted at 3 times actual diameter) shows the
orbit of Mercury with the Sun as the center. This agrees with two
other images that are also meant to show Mercury's orbit to scale:

https://www.reddit.com/r/askscience/comments/5e1zhy/what_does_mercurys_orbit_really_look_like/dab8g7q/ which links to:

http://m.teachastronomy.com/astropedia/article/Mercurys-Orbit which embeds:

Both of these look similar to my answer, so I'm reasonably confident
that I'm correct this
time. https://github.com/barrycarter/bcapps/blob/master/STACK/bc-mercury.m
if anyone wants to check my work
Things to note:

The primary effect of the eccentricity is that the center of
Mercury's orbit is quite far from the Sun.
The orbit's center is about 0.0796 AU or 11.9 million km from the
Sun, and the other focus is 23.8 million km from the Sun.
Mercury's semiminor axis is 97.86% the length of its semimajor
axis, so is looks very much like a circle. To see this, we plot
again, this time using the midpoint of the two foci as the origin:

The x diameter is 587 pixels and the y diameter is 575 pixels. If you
draw a blue circle with a radius averaging Mercury's semiminor and
semimajor axes, you get:

I'll leave it to the reader to decide if Mercury's orbit does or does
not look like a circle.


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Celestron Astromaster 114eq Star pointer finder scope help required
Just got a Celestron Astromaster 114eq as a christmas present and I have tried setting up the Star Pointer in the daylight. I did not use the red dot method as in the instruction manual, instead, I tried to do this by viewing a object through the main scope (placing the object in the center of the main scope) and then I made some slight adjustments to the Star Pointer finder scope aligning both circle and dots on the finder scope using the two adjustment screws.
This seems to work fine and finds the targets quite well,but when it turns dark and I turn on the the red dot in the finder scope and aim at a star I have yet to manage to focus in on it. Should the method I used to set up the scope finder or do I need to set up the scope finder in the dark?
Any help would be appreciated.

Per the guidelines, this question is absolutely on-topic here.

when you say " I have yet to manage to focus in on it", do you mean that you can't see any star, or you can't get the star in focus?  Make sure to use the  20mm eyepiece intially.

For an astronomical object, a 1000 mm focal length objective forms an image 1000 mm away. However, the image of a terrestrial object focuses farther out: 1010 mm for a target 100 m away, 1020 mm for a target 50 m away. When switching from a terrestrial target to an astronomical one, the eyepiece must move closer in.
Aligning the finder on a terrestrial target is bound to produce some parallax error. After focusing the telescope on a bright planet or the Moon, aim the telescope at Polaris and fine-tune the finder.
Finally, even with good finder alignment, sometimes you will have to sweep to locate your target. It gets easier with practice.


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How does the length of the minor semi-axis of the moon's disc's inner terminator vary through the cycle?
The curve that separates the bright and dark parts of the moon's disc as viewed from the earth is called the inner lunar terminator, which is semi-elliptical in shape. We can model the disc as a unit circle centred on the origin, in which the inner terminator runs from (0,1) to (0,-1) and its major semi-axis always has length 1.
How does its minor semi-axis vary during the lunar cycle? Please assume that orbits are circular.
In other words, if the point halfway along the inner terminator is X, what is the equation for the motion of X back and forth along the line-segment from (-1,0) to (1,0)?
Note that the shape of the visible disc is not a lune, except in the two trivial cases.

If the lunar period is p, and X starts at the origin at t=0 and moves right, perhaps the motion is harmonic and the answer is that X = $\sin(\frac{2\pi t}{p})$, which implies that the shape of the moon's disc changes fastest at half moon? But are things that simple?

This reads like a homework problem, but one of the moon phase videos at https://svs.gsfc.nasa.gov/4537 may help you figure it out.

It's not a homework problem. But it's true that a good guess at the equation could probably be made by spending a few hours stepping through one of the NASA videos and repeatedly measuring the semi-axis on the screen.

A sine function would oscillate both ways. The lunar terminator does not do that.

@MikeG - Good point! As you say in your answer, we need to flip from sin to -sin at full and new moon.

The moon is rotating wrt to the sun. As the moon rotates, the longitude of the terminator increases linearly. Treating the moon as being at "great distance, so the moon appears as an orthographic projection, the position of the middle of the terminator varies sinusoidally, it moves fastest when the moon is at first or third quarter.
In the orthographic projection, the longitude and latitude $(\lambda,\varphi)$ are mapped to (x,y) by
$$
\begin{align}
x &= R\,\cos\varphi \sin\left(\lambda - \lambda_0\right) \\
y &= R\big(\cos\varphi_0 \sin\varphi - \sin\varphi_0 \cos\varphi \cos\left(\lambda - \lambda_0\right)\big)
\end{align}$$
For points on the equator, and treating the central longitude $\lambda_0$ as the lunar prime meridian, and a scale of R=1, that simplifies to
$x = \sin(\lambda)$. As the terminator moves, its longitude increases linearly, so x varies sinusoidally.
Correcting for elliptical orbits, librations of the moon and a general perspective view would only marginally alter this.

Many thanks for this, but please can you expand it. I don't understand how you arrive at the conclusion that it varies sinusoidally, or why, having assumed the moon is at great distance, you then say you can introduce a general perspective view, which would put the observer only a small distance away, without changing the sinusoidal nature of the motion.

Extended. I didn't say that the general perspective view wouldn't change the motion, but it doesn't change it substantially. Our distance from the moon is much greater than it's radius.

Thanks. I understand the answer now and have accepted it. It's as if a plate is rotating at constant speed on a large table and we are looking at it from the table's edge. Which is close to a definition of sinusoidal.

I think there are discontinuities.

As this NASA visualization video shows, the terminator moves almost sinusoidally, always from lunar east to west, never from west to east. We see the sunrise half of the terminator in the waxing phases and the sunset half in the waning phases. At the new and full phases, our view of the terminator switches from one half to the other, making the position vs. time function a discontinuous, piecewise-sinusoidal sawtooth wave.


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How does the dense atmosphere of Venus affect the planet's rotation?
Reading through questions about my favourite planet, Venus, I came across this answer to the question What is the current accepted theory as to why Venus has a slow retrograde rotation?, specifically where the answer suggests one of the possible reasons for the slow retrograde motion as being

The spin slowed to a standstill and then reversed, caused by the sun's gravity, the dense atmosphere and friction between core and mantle

Based on a fairly recent study published in the Nature article "The four final rotation states of Venus (Alexandre C. M. Correia & Jacques Laskar, 2001).
Given the relatively high density of the Venusian atmosphere, this leads to the question:
How does the dense atmosphere of Venus affect the planet's rotation?

How does the dense atmosphere of Venus affect the planet's rotation?

I don't know all the details, but I can give a partial answer, with links to further reading if interested.
Tidal locking, and orbits slowing down due to a tidal bulge and tidal forces and/or a denser or greater gravitational gradient on one side of the tidally locked object are all pretty straight forward, and I don't think that part needs to be explained too much.  The tidal bulge has mass and that mass creates drag on the rotation (or speeds it up if the bulge is behind the object causing the tides).
And it's worth adding that rocky worlds are gravitationally lumpy, where the heavier side (gravitational gradient), tends to, over time and in tidal locking, face the object it orbits.   The near side of the Moon, for example, is slightly denser on average than the far side.  See here for a better explanation and here for gravitational map of the moon (the Earth side is on the left in the picture below).

Venus' atmosphere, particularly it's upper atmosphere, behaves very differently.   There's still an atmospheric tidal bulge, but because of the heat from the sun, the hot gas on the sun side of Venus upper atmosphere has lower density, so the tidal bulge may have less mass, not more mass, even with greater size.   Atmosphere doesn't behave like a typical tidal bulge orbiting close to a star because the hot atmosphere is less dense and the night-side of Venus' upper atmosphere is more dense.  The (by mass) true tidal bulge is on the far side of Venus, away from the sun, not the near side.
Another effect is that the heat from the sun causes something like a jet-stream on Venus, a very powerful jet stream.   The wind speed on Venus' upper atmosphere is faster than any hurricane on Earth and the upper atmosphere tends to rotate in the same direction around Venus.   This solar driven rotation (may be) a factor in Venus' surface having a slow retrograde rotation where, without an atmosphere it might be tidally locked.
It's difficult to imagine an upper atmosphere wind could "spin" a planet, and it would only do so very very slowly, (Venus' atmosphere is about 1/10,000th it's total mass, not too far off from the ratio of Earth's oceans to Earth's total mass) and the tidal slowing of a planet is very gradual, at Venus distance taking perhaps 100 million years to slow a rocky body of that size to tidal locking.     So the rotation may be a balance between Venus' solar driven jet stream and it's gradual tidal locking.
At least, that's what I came up with doing some reading on this.    From Wikpedia:

The rotation period of Venus may represent an equilibrium state
  between tidal locking to the Sun's gravitation, which tends to slow
  rotation, and an atmospheric tide created by solar heating of the
  thick Venusian atmosphere

and link to the source, which is in French but English translation is available.   (see footnote 100 and 101)
On Venus "jet-stream" like upper atmosphere, note that while the jet-stream on Earth is driven by surface temperature variation perhaps more than solar and by Earth's Coriolis effect, and it varies by seasons.   Venus doesn't have surface temperature variation to speak of and essentially no Coriolis effect due to it's slow rotation, so it's jet-stream tends to be much more along latitude lines and much less curved than Earth's.    That's not to say that there's no variation.   Venus' wind seems to operate on a 5 day cycle and it appears to have grown stronger in the years since we've been observing Venus.  Neither of those is well understood.
Article here

At the very top of the cloud layers on Venus, wind speeds reach 355
  km/hour (or 100 meters/second). This is the same the jet stream here
  on Earth. As you descend through the cloud layers, though, the wind
  speeds pick up. In the middle layer, the winds can reach speeds of
  more than 700 km/hour. That’s faster than the fastest tornado speed
  ever recorded on Earth.
But then as you descend further down through the clouds, the
  thickening atmosphere slows the winds down, so that they act more like
  currents in the ocean than winds in the atmosphere. Down at the
  surface, the winds only move at a few km/hour. That’s not much, but
  the thick atmosphere can still kick up dust and push around small
  rocks.

and

The winds on Venus travel in a westerly direction, the same backwards
  direction that Venus rotates. Seen from above, Venus rotates in a
  clockwise direction. This is backwards from the other 7 planets, which
  rotate counter-clockwise.

another article here.
and, related, quote from Wikipedia and source article.

Venus's rotation has slowed down by 6.5 min per Venusian sidereal day
  in the 16 years between the Magellan spacecraft and Venus Express
  visits

So, short summary, Venus atmosphere may play a key role in it's reverse rotation, or, maybe it's rotation was reversed by an impact a long time ago and it's still slowing down from that impact.   The impact theory might make the most sense, but impacts large enough to spin a planet are quite rare so, it shouldn't be assumed.   I don't believe it's known 100% whether Venus' atmosphere caused it's backwards rotation, but it seems reasonable and possible.   My personal favorite idea, though it may not be likely, is the gyroscope flipping hypothesis.  (Posting video cause it's really cool, but it has nothing to do with your question).
https://www.youtube.com/watch?v=1n-HMSCDYtM

The idea must have to do with tidal locking of Venus' rotation with its orbit.  That also happened to Mercury, except that Mercury has maintained its elliptical orbit so it actually switches the face toward the Sun each time it passes perihelion.  Venus' orbit is circular (probably because of the stronger tidal deformations of its malleable atmosphere), so tidal bulges on its atmosphere would tend to slow the rotation to match the orbital period and keep the same side facing the Sun.  That goes a long way toward stopping the rotation, though what reverses it a little must be something else, a final step that has to do with the mantle rotating differently from the atmosphere.  That last bit must be complicated.


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What could be the possible origin of boron found in the Martian soil if not from ancient groundwater?
In a recent observation reported in the ScienceAlert article 'Boron has been detected on Mars for the first time" reporting that

The Curiosity rover has found boron on the surface of Mars, indicating that, at some point, the Red Planet had long-term habitable groundwater.

Making the analogy that

Boron here on Earth is associated with sites where there was once lots of water, but it's since evaporated away - like California's Death Valley.

However, they state that this may not necessarily be the case for Mars, so
What could be the possible origin of boron found in the Martian soil if not from ancient groundwater?

Meteoric source?

That, or Martian soil just tends to be richer in boron compared to Earth soil anyway.

Either are possibilities, also possibly volcanic - there is bound to be research somewhere about this.

Research published in the PLoS One paper "Boron Enrichment in Martian Clay" (Stephenson et al. 2013) suggest a volcanic origin for the boron, in the chemical analysis of a Martian meteorite (found in Antarctica), they explain:

A straightforward geochemical interpretation of our results is that boron, a relatively volatile and soluble element, was first concentrated in the fluid dregs of lava (4–7 ppm boron has been detected in the late stage mesostasis of other nakhlites) and then became further concentrated by any groundwater or hydrothermal fluids that came into contact with the rock.

Further modelling reported in the paper "Partitioning of light lithophile elements during basalt eruptions on Earth and application to Martian shergottites" (Edmonds, 2014), also point towards a volcanic origin. Their modelling is based on analyses of shergottite pyroxenes (Martian meteorites), stating that observed elevated boron alongside other light lithophile elements or LLE are to do with the earlier mantle chemical processes, extrapolated from those observed on Earth:

The range in LLE and trace elements observed in shergottite pyroxenes are instead consistent with concurrent mixing and fractionation of heterogeneous melts from the mantle.

Also, a volcanic origin is suggested in the Goldschmidt Conference Abstract "Boron isotopes in the nakhlites: Implications for crustal fluids on Mars" (Spivak-Berndorf et al. 2008), also based on analyses of nakhlites:

The lack of isotopic fractionation between the igneous and
  secondary alteration phases in the nakhlites indicates that (1)
  the B in the fluid that deposited the iddingsite was derived
  from the primary igneous minerals and (2) there was minimal
  B isotope fractionation between the fluid and the iddingsite


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How could a neutron star cause repeated 'fast radio bursts' detected coming from outside of the galaxy?
According to the ScienceAlert article Mysterious repeating radio signals have been detected coming from outside our galaxy, based on the Letter to Nature "A repeating fast radio burst", there has been a series of signals detected that the article reports that scientists have not seen before.  From the article:

Astronomers have detected repeating blasts of radio signals coming from deep space. These short-lived signals are known as 'fast radio bursts' (FRBs), and although we've heard them before, they were always thought to be one-off events coming from random locations. But for the first time ever, researchers have now heard repeating signals, all emanating from a single unknown source outside our galaxy.

Emphasis mine
In the Letter to Nature, it is hypothesised that the source could be that

these repeat bursts with high dispersion measure and variable spectra specifically seen from the direction of FRB 121102 support an origin in a young, highly magnetized, extragalactic neutron star

This leads me to the question:
How could a neutron star cause repeated 'fast radio bursts' detected coming from outside of the galaxy?

Neutron Star, I choose you! use Rapid Spin aim towards Earth... [accelerating charged particles emits photon(see synchrotron radiation) and although the beam is pointed towards Earth it's suffers from axial tilt syndrome BTW it isn't 100% neutron.]

The progenitors of FRBs have been tied to soft gamma-ray repeaters (SGRs), and are likely young magnetars displaying irregular atmospheric and magnetic activity. The earliest connection between the two - as far as I know - was made by Lyutikov (2002), who compared the outbursts to solar flares (the magnetar bursts are sometimes referred to as "giant pulses"), based on models by Thompson et al. (2002).
The theory is that currents dissipate into the neutron star's magnetosphere, at rates coupled to the star's crust. Slow deformation can allow dissipation to proceed at a constant rate, while "fractures" in the crust can lead to quick outbursts. The release of magnetic energy through reconnection may then be responsible for the radio outbursts. The evolution of the field is similar to the behavior of the Sun's magnetic field in the corona, leading to the analogy to solar flares.
Popov & Postner (2007) state that Lyutikov's model's predictions agree with several observed bursts. While the arXiv version of the Letter concerning FRB 121102  does not specifically delve into the model, it seems to support it, although there are doubts as to the feasibility of repeated giant flares.


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What causes the abundance of dust to 'float' in the thin lunar atmosphere?
Several missions to the Moon have revealed a lunar horizon glow, such as the ones seen below by the spacecraft Surveyor 7 in 1968:

Image source: NASA
According to the NASA article "Model Helps Search for Moon Dust Fountains" written in 2010 suggests the cause of the phenomenon is due to sunlight shining through a dust layer suspended in the very thin atmosphere (exosphere) of the moon.  From the NASA webpage:

"It has been suggested that electrostatic forces play a role in the ejection of dust from the lunar surface, and its dynamics in the atmosphere, but we really don’t understand how it gets there in such high abundances."

Given that there is no wind or precipitation in the lunar sky, my question is:
What causes the abundance of dust to 'float' in the thin lunar atmosphere?

A recent study in Geophysical Research Letters — "Dust charging and transport on airless planetary bodies" (Wang et al. 2016) and reported in the ScienceAlert article "NASA just explained why Moon dust is 'levitating' above the lunar surface" — seems to explain the phenomenon through laboratory simulation experiments which observe what occurs when micron-sized dust is exposed to ultraviolet (UV) radiation or electrically-charged gases known as plasmas.
From these experiments, the researchers observed that:

In both cases, the dust particles would leap several centimetres above the surface.

The experiment observations can be seen below:

Image source: Science Alert article linked above
Thus, one can conclude that on the Moon light would scatter through these levitating particles to create the horizon glow.
The mechanism behind the 'levitation' is stated as being:

This is due to the reaction [with the plasma and UV light] causing an emission and re-absorption of electrons inside 'micro-cavities' formed between neighbouring particles, which can generate unexpectedly large electrical charges and intense repulsive forces.
These forces cause the dust particles or clusters to lift off the
  surface, or 'levitate'.

Although the effects are observed with the lunar horizon glow, the researchers expect that effect would cause a suspension of dust across the entire lunar surface to some degree.
A significant implication of this research is stated in the NASA article "Dust “Floats” Above Lunar Surface: Electrostatic Dust Transport Reshapes Surfaces of Airless Planetary Bodies" as:

The laboratory observations also showed dusty surfaces becoming smooth as a consequence of dust mobilization. These electrostatic dust processes could help to explain the formation of the “dust ponds” on asteroid Eros and comet 67P, and the unexpectedly smooth surface on Saturn’s icy satellite Atlas.


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Exo Planets and Planets difference?
After hearing that it has been confirmed that we found seven similar exoplanets like earth. I thought exoplanets were dead stars. After doing a little bit of research I found out that exoplanets are planets that are orbiting a star, NOT the sun. But isn't a star like the Sun? Can someone please explain this to me?

ANOTHER star than the Sun. It's as simple as that. "Sun" is just the name of one star.

"Planets" are in our solar system. "Exoplanets" are in other star systems (or possibly associated with no star system at all).

The 2018 meeting of the IAU may see two competing redefinitions of what constitutes a planet. One will remove the concept of roundishness as being completely superfluous and will change the qualification of orbiting the Sun to orbiting a star. This will make all of the exoplanets discovered to date "planets". The other will focus solely on the concept of roundishness. This alternative will preclude calling those exoplanets "planets" (we can't observe how round they are, at least not yet) but will add hundreds of objects to the list of planets in the solar system.

Not all planets are exoplanets, but all exoplanets are planets.
When we differentiate the two, we use "planets" to refer to the eight planets revolving around our Sun. The Greek prefix "exo" (outside) should indicate that exoplanets orbit other stars.
In reality, you can call exoplanets "planets", and you'll often hear "Astronomers have found x planets around star y". The word "exoplanet" is used interchangeably with planets from other planetary systems.

A better way to put this: Exoplanets are planet-like objects that orbit some star other rather than the Sun.The current definition of "planet" means that there are eight planets in the entire universe. The qualification that planets orbit the Sun is one of the two parts of the definition of "planet" that needs to be modified or removed. This needs to be modified. The part that needs to go, much to the consternation of Alan Stern, is the concept of roundishness. An object that orbits a star, that hasn't undergone fusion, and that has "cleared its neighborhood" should be deemed to be a planet.

@DavidHammen I guess, if we want to be pedantic. In practice people rarely acknowledge that difference and use exoplanets and planets interchangeably for one's outside the Solar System.

Exoplanets are planets that are orbiting any star that is not our SUN. Planets are those which orbit the Sun namely: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune. However, the closest exoplanet to Earth, that is, Proxima Centauri B orbits its parent star named Proxima Centauri.

How does it add to the already existing answer?


text:
Are all planets/galaxies moving away from *us*?
From what I can see in a physics textbook page (on redshift), it seems to imply that all planets/galaxies emit red, rather than blue waves (for red-shift).
Wouldn't this mean that everything is moving away from us and that we're the centre of the Big Bang? (See quote from the book below)
On a similar note, does everything move at the same speed away from the centre of the Universe? - If not, then wouldn't that mean that we'd see a few planets/galaxies that emit blue, as they are traveling with a greater velocity from the centre than we are (with it being closer to the centre) and that they are 'catching up' with us?
It's not a big deal and I don't need to know it, but I would like to know, if possible.
~StratusGold
Quoting the textbook: "When we look at light from distant galaxies, we find that the wavelength has increased. The wavelengths are longer than they should be - they're shifted towards the red end of the spectrum. This is called red-shift. (...omitted...) Measurements of the red-shift indicate that these galaxies are moving away from us (receding) very quickly - and it's the same result whichever direction you look in" - CGP GCSE AQA 9-1 Physics Revision Guide, Topic 8 'Space Physics', Pg 102

I don't have time for a full answer, but the quick version is this.  Not all galaxies are moving away, there are a couple of small ones orbiting the Milky Way, and Andromeda is on a collision course straight for us.  Red shift means that the frequencies are stretched out, blue shift means they are compressed.  It does not mean exclusively red or blue light.  Also, as weird as it sounds, everywhere was the "center" of the big bang.  Its complicated...  In general everything isn't just moving away from us, everything is moving away from everything else (with some exceptions, see Andromeda).

Ok, thanks @Cody I think I understand it a bit more now :)

This answer explains why every galaxy can be said to be in the center. This answer explains why Big Bang wasn't a point-like explosion in space.

No, not all of them. Only the ones very far away.
When you walk your dog, the dog is tied to you via the leash. It does not recede into the distance.
Planets in the Solar System are all tied up to the Sun via gravity. They don't recede into the distance.
All stars in our galaxy are bound together in a big clump (the galaxy) and so they stay together.
Finally, all galaxies in our neighborhood are also bound by gravity, so they stay together.
Beyond that range, things are not bound by gravity, so they keep running away from each other.
Gravity is like a leash. If things are close enough, they stay together.

Keep in mind that the universe does not have a center. It's infinite (as far as we can tell), and infinite things don't have centers. Rather, if you take any two distant galaxies, they all recede from each other. But the expansion does not go back in time to a common center.
Here's a comparison. Imagine you're a bi-dimensional creature; instead of having 3 dimensions, you only have 2. Like a stick figure drawn on a paper. You only exist in the 2 dimensions of the paper, and can't even imagine the 3rd dimension.
Now imagine a balloon. You draw a bunch of stick figures on it. Now blow air into the balloon. All stick figures recede from each other, but for them there's no center for that expansion - because all they know is the surface of the balloon. They don't know anything that's not on the surface. So they all see each other running away, but there doesn't seem to be a center for that expansion.
It's just an analogy, because the universe is not a balloon, but it will get you started.

Finally, red-shifted galaxies don't emit only red light. They emit the usual radiation like everything else. But because us and them are running away from each other, radiation from one side reaches the other side a little bit stretched out. For the visible part of the spectrum, it's like taking the spectrum and shifting it towards the red part.
That's all there is to it. It's still the usual radiation, but all wavelengths are a little bit blown up.


text:
Sidereal time and right ascension
I'm new to astronomy and have to make a star map that displays the stars above a current location. From the information I've pieced together I understand that declination is equal to the current latitude of the observer, however I am slightly confused about right ascension. My two quesions are;

If I work out GMST using the equations here; http://www2.arnes.si/~gljsentvid10/sidereal.htm will the answer be the current right ascension above Greenwich or are there more steps I've missed?
If I want to find the sidereal time of, for example, Sydney, Australia, do I need to account for daylight savings at either location or do I simply add 11 hours to GMST?

Thank you for your help.

I can't vouch for the formula, but you can try it out and check with some internet source, eg USNO, to see if it works.  If the formula is correct, that's it.
As for the sidereal time in Sydney, you need the difference in longitude between Greenwich and Sydney, and use the conversion rate of 15 degrees longitude is equivalent to 1 hour.


text:
Appearance of sun at geologic past
After origin of Earth ( 4.54 billion years ago ) to today; what are the changes  through-which the sun undergone with time? , ( if seen from Earth. )
I'm doing some palaeobiology-landscape rendering; so I'm thinking if I want to add the sun in such pictures; should I draw it just like present day? Or the look/ appearance (colour, size, flames, position, halo etc) varied in observable amount through the geologic time-scale(precambrian-cambrian-ordovician-silurian-devonian-carboniferous-permian-triassic-jurassic-cretaceous-palaeogene-neogene etc). (the timescale terms provided just to hint the answerer the level of details may NOT to go )
Update: Here I attach a  small version of geologic time table to indicate time ranges.

from here .
To see the entire chart; see international commission of stratigraphy charts

Can you add some more astronomically-friendly information, like millions of years?  recambrian-cambrian-ordovician-silurian-devonian-carboniferous-permian-triassic-jurassic-cretaceous-palaeogene-neogene is not so helpful. Also.... flames???  :)

@uhoh okay I'm editing the question. By "flame" I tried to mean was there solar storms , volcanoes etc that may be required? (don't know is there any term for that. In many cosmology-related popular-science books I've seen telescopic images of specialized stars that have radiating flame-like things)... but no idea if there any term for it. Will provide an image when I'll found out the book.

OK thanks for the edit, it's a big improvement. Are you talking about illustrations of the appearance as it might be from the Earth's surface with something similar to human vision, or from space? The reason I ask is threefold. 1) Changes in the atmosphere might have caused tiny changes in "color" of the sun or the sky, 2) Solar flares, sunspots, halos (what are those?) would probably be almost impossible to see during the day with human vision without various optical tricks or instruments, and 3) illustrations of views from space have a lot more flexibility because they aren't always accurate

Can you just say explicitly the number of millions of years, don't make readers open up your chart in a new tab and try to read all those words and numbers please? Over the last 100 million years? Or the last 4 billion years? Most readers will not want to read about the Changhsingian/Wuchiapingian transition for example.

@uhoh 1. Yeah I meant as would look from earth surface. 2. I added a more simplified chart. I meant the more or less the "periods". That is why I used -cambrian-ordovician-silurian-... . It doesn't seems to me in a new epoch or stage a so big stellar change would happen (just guessing)...

I'm going to gently suggest that you not post the vision-chart anywhere in your question here in Astronomy Stackexchange - it is just distracting, and some times that can trigger people to vote to close. You really just want to ask for some general trend over the last 4 billion years I think. Just ask that.

Okay so should I delete both charts and include numeric values?

Let us continue this discussion in chat.

Maybe just keep the smaller one if you want, at the bottom. But in the main question just state in the first one or two sentences "4 billion years" or whatever number. First make it clear and simple just what specifically you are asking - very clear, and very early in the question.

I'm doing some palaeobiology-landscape rendering; so I'm thinking if I want to add the sun in such pictures; should I draw it just like present day ?

Although the Sun was less luminous 4.5 billion years ago (about 70% of it's current luminosity, it is thought), and while this would affect it's color, making it perhaps a little more orange than yellow, the effect is small.
The Sun is still going to be a blazing near-white object in the sky of roughly the same size.
Although there was probably a color shift to a slightly more orange tint, the composition of the Earth's atmosphere was radically different (see for example this page on Paleoclimatology) and this would have had a much more significant effect on the appearance of the Sun (and the lighting of the landscape) than any minor change in the Sun's appearance.
There are some issues relating to the development of life and this change in the Sun's luminosity (like the Faint Young Sun Problem) but these won't make much difference to your rendering, I think, compared to the atmospheric and climate issues.
However the climate changes are not fully understood and it's rather speculative to try and make specific claims about what the overall effect would have been on the illumination levels and color of light reaching the surface in these periods.

Thanks for this very-well-written, simple and nice answer. :)

You can divide this into two astronomical epochs, which are most easily understood in terms of the Hertzsprung-Russell diagram.
The first is the pre-main-sequence (PMS) phase, where the protosun contracts towards its main sequence, hydrogen burning, configuration. During this phase the (proto-)Sun declines in luminosity at roughly constant temperature for several million years, then heats up, with a slightly rising luminosity, before settling onto the main sequence with a slightly lower luminosity (see diagram below) after about 20 million years. The diagram gives you a rough idea of the "colour" that the protosun would have during this phase. The Sun is of course changing in brightness accordingly, but the radius of the Sun also decreases throughout this phase, so would cover a progressively smaller angle on the sky as seen from the Earth.
$$ R = \left(\frac{L}{4\pi \sigma T^4}\right)^{1/2}$$

A plot showing the paths (solid lines) of stars of different masses from birth (the Sun is still $\simeq 1M_{\odot}$ at birth) in luminosity and temperature, until they begin the H-burning main sequence. The dashed lines are "isochrones" at the indicated ages. The Sun "begins" its life (or at least emerges from its natal gas and dust less than a million years after birth) at the top of the $1M_{\odot}$ track.
The second phase is of course much longer, but the changes are far less dramatic. For the next 4.45 billion years the Sun just gets a little brighter and a little larger, but stays more-or-less at the same temperature. Over that time the Sun has got about 30% more luminous and 15% larger at a roughly constant rate.
If you need more detail than that, then you need to access a proper stellar evolutionary model. There are lots of those available. e.g. the Geneva models (the Schaller et al. 1992 set are appropriate for the Sun).

What's the point in discussing the proto-sun? The OP says they're doing "palaeobiology-landscape rendering" of periods during Earth's history which would have been long after the proto-sun phase. As far as I read the question, only the last two paragraphs are relevant.

The chart in the question begins at -4.6 billion years. @zephyr I just went from that. The question also includes the "pre-Cambrian" period, a subset of which occurs whilst the Sun was a PMS star. I don't see the problem in providing a full answer.

Part of answer is amazingly clear; but some portion I'm in doubt... does the 10^4 yr, 10^5 yr etc means years ago (past) from today? or starting from sun's birth towards future? (i mean what is the year-0 in your graph?) Also could you help a bit more by mentioning where in this above graph the Earth was borne(4.54bya)? I'm sorry I do not have astronomy/physics/maths background after high-school.

It is also surprising that how the life line of a stellar object can go to-and-fro along time. Maybe it is a drawing error? or maybe I've understood something wrong?

The ages are times since "birth" (a rapid event that takes less than a million years and leaves the protosun at the top of the 1 solar mass, $1M_{\odot}$ track). The plot only shows the first 20-30 million years of the Sun's life. The to-and-fro is no mistake and based on well understood physics.

@AlwaysConfused -- The $10^4$ year, $10^5$ year, etc. curves represent the time from when a star is "born". Before this, a protostar collapses due to self gravitation. This collapse slows down markedly when a protostar becomes opaque, changing a protostar into a pre-main sequence star.

Oh i see... the isochrones (10^4 yrs etc) not indicating a universal time... it is indicating a particular age of all stars and not simultaneous. The time-direction (past->future)is shown with the arrow signs on black tracks; and each black track indicates a separate, individual star. Am I now correct on understanding the graph?

The times are the times since the birth of a star. Tracks are shown for 7 stars with different masses. The appropriate one for the Sun is labelled as $1M_{\odot}$. The direction of the track is indeed in the direction of the arrows and towards increasing age. Ignore the thing labelled "birthline", it is an unnecessary complication here and is basically when the star becomes possible to observe.

Thank you for your response; no... birthline is not creating any unnecessary complication; I was initially reading the graph in a completely wrong way. Now it is clear.


text:
Need help on Springel model for galaxy simulation
I am working on a code for galactic dynamics simulation. Initial conditions (positions and velocities) are based on a model with a massive dark halo and a thin exponential disk.
Now, from the following paper "Tidal tails in CDM cosmologies", I am studying the response of the dark matter profile.
Here's a capture of the article part that confuses me :

Firstly, I don't understand how one gets equation (16) from the fact that angular momentum of individual dark matter orbits is
conserved, i.e :
\begin{equation}
r_{i}\,M(r_{i})=r_{f}\,M_{f}(r_{f})\quad\quad\text{(eq 16)}
\end{equation}
" where $r_{i}$ and $r_{f}$ are the initial and final radii of some dark matter mass shell, $M(r)$ gives the initial NFW mass
profile, and $M_{f}(r)$ is the final cumulative mass profile after the disk is formed."
Question 1) Is (equation 16) got by the momentum theorem ? ($\dfrac{\text{d}\vec{L}}{\text{d}t} = \vec{OM} \times \vec{F}$)
And how to prove it, i.e what's the demonstration to get (equation 16) ?
I don't understand what's the difference between $M(r)$ and $M_{f}(r)$.
After, it is said that $M_{f}(r)$ is the sum of the cumulative mass of the disk and the dark mass inside the initial radius
and author writes (equation 17 on paper) :
\begin{equation}
M_{f}\,(r_{f})=M_{d}\,(r_{f})+(1 - m_{d})\,M(r_{i})\quad\quad\text{(eq 17)}
\end{equation}
I would like to connect this equation with the total mass of galaxy which could be written (according to me) as :
\begin{equation}
M_{total}\,(r) = M_{d}\,(r) + M_{DM}\,(r) = m_{d}\, M_{DM}\,(r) + M_{DM}\,(r) = (1+m_d)\, M_{DM}\,(r)
\end{equation}
with $m_{d}$ the ratio between disk mass and halo mass.
Question 2) Could you help me to connect the total mass with this equation (17) ?
After with equation (18), author of paper writes that the final profile $M_{h}(r)$ of the dark matter halo is then given by :
\begin{equation}
M_{h}\,(r) = M_{f}\,(r) - M_{d}\,(r)
\end{equation}
Question 3) So should I conclude that $M_{f}\,(r)$ is the total mass of galaxy (disk + halo) as a function of $r$ ?
Any help would be great,
Regards
UPDATE 1 :
I have contacted the author of paper "Tidal tails in CDM cosmologies", i.e Volker Springel and here are its comments :

Question 1*) Answer: For simplicity, you can consider that all particles are on circular
  orbits. The initial circular velocity of a particle is $$v_i = \sqrt{(G M(r_i)/r_i)}$$, the final one is $$v_f = \sqrt{(G M(r_f)/r_f)}$$. The
  corresponding angular momenta are $$L_i = m r_i v_i$$ and $$L_f = m r_f v_f$$.
  Now, the angular momentum is an adiabatic invariant when the halo is
  slowly compressed, hence $$L_i = L_f$$. The latter equation then gives eqn
  (16) when one plugs in the velocities.
Question 2*) "I don't understand what's the difference between $M(r)$ and
  $M_f(r)$ ?"
Answer:
$M_i(r)$ is the initial and $M_f(r)$ the final cumulative mass profile.
Question 3*)
I would like to connect this equation with the total mass of galaxy which could be written as :
$$ M_{total}(r) = M_d(r) + M_DM(r) = m_d M_DM(r) + M_DM(r) = (1+m_d) M_DM(r)$$
Answer : This can only be written for the initial distribution in
  this form, where the assumption is that the mass that will later make
  up the disk is distributed proportionally to the dark matter.
Question 4*)
  Could you help me to connect the total mass with this equation (17) ?
After with equation (18), you write that the final profile M_{h}(r)$ of the dark matter halo is then given by :
$$M_h (r) = M_f(r) - M_d(r)$$.
So should I conclude that M_f(r) is the total mass of galaxy (disk + halo) as a function of r ?
  < Answer : "Yes."

In conclusion, I can solve my question 1*) but not question 2*) nor question 3*)
Right now, my goal is to demonstrate how to get  :
$$\begin{equation}
M_{f}\,(r_{f})=M_{d}\,(r_{f})+(1 - m_{d})\,M(r_{i})\quad\quad\text{(eq 17)}
\end{equation}$$
Any help to acheive this is welcome, thanks

Yes, that is simply the angular momentum of the shell written down, and using the fact that it's conserved.

To prove it, should I write : $$\vec{L} = \vec{constant} = $$ $$\vec{OM} \times \vec{p} = M(r_{i})r_{i}^{2}\dot{\theta}=M(r_{f})r_{f}^{2}\dot{\theta}$$ ?? But how to introduce the two different masses $M_{f}(r)$ or $M(r)$ ? If you could detail the demo. thanks a lot

I'm unsure what you're confused about. You already wrote the identity down with two different masses. $M(r_i)$ will in general be different from $M(r_f)$. And if you then call $M(r_f) = M_f$ is just a matter of notation.

$M(r)$ is called "the initial NFW mass profile" and $M_{f}(r)$ is called  is "the final cumulative mass profile after the disk is formed". Do you think that solution could be found by putting notation $M(r_f) = M_f$ ??

Given the relation above of $\vec{L}$, I can't find equation 16. But if I put circular velocities like $v_i = \sqrt{(G M(r_i)/r_i)}$ and $v_f = \sqrt{(G M(r_f)/r_f)}$, then I think I have to set : $\vec{L_i}=\vec{L_f}$, don't I ?

It's hard to understand the wider picture with just a snippet of your source, but it looks to me like you're trying too hard here. Angular momentum is $L=Mvr$ and angular momentum conservation is just $M_1v_1r_1 = M_2v_2r_2$. If you can say that $v_1=v_2$ (it's unclear to me if that is the case from your snippet) then you're left with $M_1r_1 = M_2r_2$.

I think it is combining centripetal acceleration (because equations and abstract "compare circular velocities to escape speed") for a low eccentricity orbit $m v^2 /r$ and center of mass for equations 16 and 17. Just a guess though.


text:
What's the influence of a tilted orbital plane, when observing an exoplanet transit?
I made an illustration to explain what i mean : If we assume we have two similar planet/star systems (similars in size/mass/period..) but tilted differently relative to us.
How can we predict that differences in dimming measurements (time & intensity) are caused by the tilting of orbital plane? (and not by the diameter of the exoplanet)

If I understand the question, the (tiny) apparent angular size of the star from Earth is not relevant, It is the size of the star relative to the size of the planet and its orbit. I think there is a good question here.

The depth of the transit tells you the size of the planet (relative to the star) in both cases.
The difference in total eclipse width tells you the impact parameter (how far from the centre of the disk, the transit crosses). This in turn depends on the orbital inclination and the ratio of the stellar radius to the radius of the orbit.
A second constraint on inclination and the ratio of stellar radius to orbital radius comes from the ratio of total transit duration to the time when the planet is fully eclipsing the star. A more inclined transit has a shorter period of total eclipse but longer ingress and egress.
To get a full solution then requires some estimate of the stellar mass, which leads via Kepler's third law to the orbital radius.
There are also some important additional complications due to limb darkening and eccentricity that mean, in practice, this problem is not solved analytically, but by fitting a detailed model to the light curve. For instance, the transit depth becomes smaller for transits that don't cross the centre of the stellar disk, because of limb darkening.
The best picture I can find is copyrighted material, but you can find it here from Carole Haswell's book on Transiting Exoplanets. The inclinations range from 90 degrees to 80 degrees.


text:
Is there a connection between Jupiter's orbital period and the Sun's solar cycles?
The cycle of solar maximums and solar minimums are each about 11 years, which is close to the orbital period of Jupiter. I thought it was a numerical coincidence, until google turned up some results of a possible connection. A lot of it is beyond my head though.
I was under the impression that there are gaps in our knowledge regarding Jupiter's core, which would likely be the driver of its magnetic field. And since solar cycles are related to sunspot activity, which in turn is related to the magnetic field of the Sun, I am struggling to find even a basic understanding of the cause-effect relationship (if there is one).
Is there a connection between Jupiter's orbital period and the Sun's solar cycles? Or is it blind speculation?

Related questions (with answers) Are we sure the solar cycle is not related to the orbit of Jupiter ? and What are some theories to explain the 11 year solar cycle ?

Jupiter's own magnetic field is highly unlikely to affect this. However, apart from "numerological" studies, there are other connections examined: https://arxiv.org/abs/1702.00641


text:
Stars in constellation patterns by catalogue number?
I need a resource that can tell me the stars that make up a constellation (pattern), preferably by Henry Draper or some catalogue number? All i can find is lists of every star in a constellation (area) or incomplete lists of stars by name.
If this doesn't exist can you suggest a way to work it out myself?

There is no standardised way of drawing the constellation figures.

Maybe, I can use the most brilliant stars.

It's not always the brightest stars, though.  I would recommend just going through Wikipedia, looking at the images they have, and matching the stars at the vertices in those images.  It'll take a bit of time, but probably less time than it would take you to do it any other way.

You must manually do this, Vans - - there's no "logical" way (such as brightest) to get them.  (Note that even if it was "brightest", that wouldn't tell you how to connect them.)

In other prograns such as Google's, they have an overlay from some famous drawing of the constellations.

Have you looked at Stellarium's sky cultures? They provide several different sets of lines, including those from H.A. Rey, who famously tried to make constellations look more like what they're supposed to represent.

What you ask for does not exist. There is no standard for what the patterns are. In modern astronomy, the constellations are areas of the sky. You can find lists of the stars in each area, but it seems that you want only the stars in the "pattern".
When you see a star map, lines are sometimes drawn between the brighter stars. These help people to compare the stars to the map, our brains like to look for connections, and so linking the stars together helps us to make sense of the sky. However different stellar-cartographers will draw the lines in different ways. Consider Gemini:

The same constellation is represented in different ways. Neither presentation is official, it is an artistic choice which stars to include and how to link them. It is therefore impossible to say which stars "make up a pattern" and which do not.
However, if you look at the catalogues, and see which stars have received a Bayer designation you will have a list of the visible stars in each constellation. This won't, of course, tell you how the stars may be linked.

If you visit https://github.com/Stellarium/stellarium/tree/master/skycultures you see several sky cultures that include various constellation shapes (you also get this directory when you download stellarium).
As an example, lets look at https://github.com/Stellarium/stellarium/tree/master/skycultures/western_rey, in particular the file https://github.com/Stellarium/stellarium/blob/master/skycultures/western_rey/constellationship.fab and line number 127 which reads:
Aql 10 98036 97649 97649 97278 97278 96229 96229 95501 95501 97804 99473 97804 95501 93747 93747 93244 95501 93805 93805 93429
The "Aql" stands for the for the constellation Aquila, the Eagle (https://github.com/Stellarium/stellarium/blob/master/skycultures/western_rey/constellation_names.eng.fab verifies this)
The "10" means there are 10 lines that make up the markings for Aquila. Note that these lines are specific to the H.A. Rey skyculture. Different cultures will have different lines.
The next 20 number represent the lines themselves. For example "98036 97649" is a line from star 98036 to star 97649, and "97649 97278" is a line from star 97649 to star 97278. Since these two lines share star 97649, they are connected, but that's not always the case with adjacent lines. For example "95501 97804 99473 97804" represents a line from star 95501 to star 97804 and another unconnected line from star 99473 to star 97804.
The next question: what exactly IS star 98036?
If you look at https://github.com/Stellarium/stellarium/blob/master/stars/default/name.fab you'll see it's Beta Aquilae, but not all stars have names in this file, and the name isn't particularly helpful in finding the star itself.
Stellarium itself stores this data in the "catalog files" at https://github.com/Stellarium/stellarium/tree/master/stars/default with the format described at http://stellarium.sourceforge.net/wiki/index.php/Star_Catalogue_Format
However, this can be ugly. Since Stellarium uses the standard Hipparcos catalog, you can visit ftp://cdsarc.u-strasbg.fr/pub/cats/I/239/ and download "hip_main.dat.gz". The format is described at http://cdsarc.u-strasbg.fr/viz-bin/Cat?I/239
The line relating to star 98036 starts:
H|       98036| |19 55 18.77|+06 24 28.6| 3.71|...
which tells us star 98036 is at J2000 right ascension 19h55m18.77s, declination +06 degrees, 24'28.6" declination, and has a magnitude of 3.71. This agrees almost perfectly with https://en.wikipedia.org/wiki/Beta_Aquilae (the declination is slightly off, but this may be because Beta Aquilae is a double star, and Hipparcos references the brighter star and Wikipedia references the midpoint of the double star-- or something).
There are other files in the stellarium github (eg, culture-specific star names) that also use the Hipparcos numbering designation.

I did this manually some years ago by listing the BSC numbers of stars and drawing lines between them.  The resulting data file can be found at the bottom of this page: BSC constellation lines.  As @james-screech mentioned, there is no standard way to draw the constellations, so this is just one (my) interpretation.  To give you some idea, maps (in Dutch) using these data can be found here: NP, Eq 0h, Eq 12h, SP.

You can find the constellations here:
http://www.iau.org/public/themes/constellations/
Take a look at the gif for each constellation and you will find out which stars are connected (there is also a text file for the boundaries of each constellation in RA and Dec, though this is not what you want).

I have compiled exactly the kind of list that you're looking for.
The stars are identified by their Yale Bright Star Catalogue HR designation. This is the best catalogue to use to define the IAU constellations because it is conveniently-sized (only 9,110 stars, down to a magnitude of about 7.5) and all the IAU constellation patterns can be defined in terms of these stars.
It is based on canonical IAU constellations. I also throw in J2000 positions, magnitudes and names for each of the stars in each pair of the constellation.
For example, Cassiopeia appears as 4 lines (below shown wrapped)
Cas 21  2.294583    59.149722   2.27    Caph|bet Cas|11 Cas
    168 10.127083   56.537222   2.23    Schedar|alf Cas|18 Cas
Cas 168 10.127083   56.537222   2.23    Schedar|alf Cas|18 Cas
    264 14.177083   60.716667   2.47    BU 499A|BU 1028|gam Cas|27 Cas
Cas 264 14.177083   60.716667   2.47    BU 499A|BU 1028|gam Cas|27 Cas
    403 21.454167   60.235278   2.68    Ruchbah|BUP 19A|del Cas|37 Cas
Cas 403 21.454167   60.235278   2.68    Ruchbah|BUP 19A|del Cas|37 Cas
    542 28.598750   63.670000   3.38    Segin|eps Cas|45 Cas

Download my list here.
Note that the asterisms defined by the IAU may be slightly different from what you see in planetarium programs or star charts. However, the IAU should be seen as an authoritative source and any variation on the patterns is up to the "poetic license" of the user.
Most of these stars (possibly all, I haven't actually checked) will have a Hipparcos (HIP) and/or Tycho (TYC) index entry as well and also