Are satellites orbiting around earth visible to the naked eye? I was just lying under the sky trying to possibly see some meteorites, unfortunately never seeing any I might add, but I saw three objects all moving at about the same speed( all at different times). They were far too faint and small to be aircraft or anything inside of the atmosphere I would think and there were no blinking lights. They were also moving fairly quickly it seemed. Could these have been satellites in low earth orbit? It's the only other thing I would think that they could be.

A lot of satellites are visible under the right conditions. Usually up to 2 hours after sunset and 2 hours before sunrise. This allows the sun to strike the satellite when you are on the dark side. Depending on the orbit, it will take between 1 and 5 minutes to traverse most of the sky. Usually, they will enter the shadow and you lose sight of them.


Can anyone one show how speeds greater than c cause a paradox on a space time diagram? So according to Relativity, time moves at different rates for different observers, and distant observers will actually disagree on the order of events.  Now from what I gather this gets sorted out for objects with relative speeds less than c, but would ultimately cause a paradox for anything moving faster because it would time travel.  However, I don't really see how.  On a spacetime diagram light always moves at a 45° angle, or at least some angle between 0° and 90°, and the only way I see anything effecting the past is if something moves at an angle greater than 90°.

So can someone please give a visual representation that explains the issue?

Suppose there was a magic gun that fired a bullet at ten times the speed of light relative to the firer.

If I have the only such gun, and I don't move then there is no paradox.

But now suppose these guns are generally available.

At time zero, an observer flies past me at velocity 0.5 c and I give them one of these guns. 7 years later, I fire the gun after them, so in my frame of reference the bullet will catch them at about time  7.36 years when they are about 3.68 ly away.

To get these numbers I just need to solve $10*(t-7) = 0.5*t$ to calculate when the bullet overtakes the spaceship.

In their frame of reference though, that occurs about
time 6.38 years after they pass me, at which point, I am about 3.19 ly away from them (we both agree on our relative velocity).

This is calculated using the Lorentz transformations. $$t' = \gamma(t - vx)$$ taking $c = 1$. Here $t = 7.38$, $\gamma = \frac{1}{\sqrt{1-v^2}} \sim 1.15$ $v = 0.5$ and $x = 3.68$.

They now fire back and we can do a similar calculation, but in their frame of reference. So in their frame the bullet will pass me at time $t_1$ when
$(t_1 - 6.38)*10 = 0.5 t_1$ which tells us that $t_1 = 6.71$.

Now we use the Lorentz transformation again to determine what time that is in our frame. We find that it is 5.81 years. A bit more than a year before we fired our bullet.

The key thing is that $n$ times the speed of light in one frame will be backwards in time in another frame.

The diagrams below illustrates this scenario.

Both diagrams show exactly the same events. The one on the left uses the time and space coordinates of the original observer ("you") the one on the right uses the coordinates of the other observer. In bith cases time is on the y axis, space on the x axis. Units are years and light years.

The black diagonal lines are some possible world lines of light rays.
The red lines are lines of constant x (solid) or t (dashed) as determined in "your" frame of reference. The thick red arrow (at x = 0) is your world-line.

The blue lines are lines of constant x (solid) or t (dashed) as determined by the ship. The thick blue arrow is the ship's world-line. As can be seen you meet at the event that you both agree is x = 0, t= 0.

The green line is the bullet that "you" fire. It crosses three solid red lines and goes a bit less than half way from one dashed red line to the next (travels 3+ ly in one 0.36 y). The cyan line is the return bullet. It crosses 3 solid blue lines and goes rather less than half way from one dashed blue line to the next (travels 3+ ly in about 0.3y).

Assuming I've got this right, all the intersections and crossing patterns of red, blue and black lines, and the four world-lines should agree between the two diagrams, since they really do depict the same events.  In particular both diagrams show a causal loop, in which the cyan bullet arrives back at the red observer before the green bullet is fired. They disagree about which bullet travels back in time, but that's relativity for you.

What this tells us practically is that any physics which allows FTL travel but not time travel must not be Lorentz invariant, so your gun will work differently in different frames of reference. We know of no such physics at present, and see no clues which suggest it might exist.


Why isn't the asteroid belt affected by Jupiter's gravitational field? Jupiter's mass is just about a 1000th of the sun's and the asteroid belt is slightly closer to Jupiter than it is to the sun.

If the heavier the object, the more curvy space is around it, why isn't the asteroid belt's movement more turbulent owing to the effect of both Jupiter and the sun?

The asteroid belt is affected by Jupiter's gravity.

There are stable orbits inside of Jupiter's orbit. Jupiter's Hill Sphere has a radius of 53 million km. If you are more than 53 million km from Jupiter, then the Sun's gravity dominates and you can orbit the sun. But Jupiter orbits 780 million km from the Sun, so there is plenty of space between Mars and Jupiter for asteroids to have stable orbits.

As they pass Jupiter on their orbit they will get first a pull forward, and then a pull back but as these pulls happen at different locations on each orbit they tend to cancel out and result in effects like precession of the asteroid's perehelion or periodic variations in eccentricity. These effect act on the Earth and other planets too.

However if the time the asteroid takes to orbit the sun happens to be an exact fraction of the time that Jupiter takes, then it might receive the same pull at the same location. Resonant orbits like this tend to be unstable. The asteroid will be nudged out of the resonant orbit and into one that doesn't match up with Jupiter.  This creates the Kirkwood Gaps in the asteroid belt.

Source


Why can't we point the centre of the universe from inflation graph we see? This question is not about whether it makes sense to have a centre of the universe or not instead, it's about the inflation graph we used to see while describing the Big Bang. From the inflation graph (drawn above), it's clear that the universe is supposed to have a centre at somewhere around "The yellow point". Still why do we say that the universe has no centre or universe has it's centre at each point of it.
What's the problem with the figure or my understanding?

That diagram does not depict the entire universe. At most, it depicts the history of what is now our observable universe (specifically, a 2D slice through it), with us at the center only because we're observing it. Someone at the furthest reaches of that portion of the universe would see us at the furthest reaches of their observable universe, and themselves at the center.
And like us, they would observe the universe to be expanding away from them, and doing so equally in all directions. This, as far as we can determine, is true for anywhere within the observable universe. The simplest assumption is that it continues to do so outside it as well, and there is no evidence that it does anything else.


Why is Io's volcanism chemically so different from Earth's? Volcanic gases on Earth are mainly water, carbon dioxide and sulfur dioxide, but on Io there is a notable absence of water and carbon dioxide in volcanic gases, since sulfur dioxide is the dominant gas expelled by volcanoes in this satelite, why does this happen? It seems strange to me that water is not an important volcanic gas on Io when it's quite common in the composition of the other Jovian satellites.

In addition, the Io's surface is very rich in sulfur, which comes from volcanoes, but other bodies of the solar system that also experienced intense volcanic activity at some time have a surface dominated by basalt, such as Mercury or Mars. So, if Io has had intense volcanic activity for a long time, why doesn't it have a dark gray basalt surface like Mercury but a yellow one because of sulfur?

Two main reasons:

Location, location, location

The protoplanetary disk from which planets and satellites formed was not homogeneous.  Lighter elements and more readily dissociated molecules will be blasted away and/or apart if they are sufficiently close to the sun.  The sun emits a great deal of energy, and UV radiation in particular, which is great at breaking up molecules and providing an energetic kick to lighter atoms and molecules.

So objects like Mercury, which formed close to the Sun, will be composed of different material than will objects that form further away.  It's possible for objects to move from close to far, or far to close, though.  This is basically why the Earth has any meaningful amounts of water: the late heavy bombardment consisted of a relatively large number and volume of icy objects that formed further out having moved further in (or developing a large eccentricity that brought them inward) and ultimately colliding with objects such as the Earth.

To the best of my understanding, Io is believed to have formed in the general vicinity of Jupiter, so would have had access to a lot of materials that Mercury simply did not.  Sulfur and sulfur compounds in particular.

Why it has volcanism

Earth has volcanism, and a molten outer core, because of latent heat from its formation, radiocative materials in the mantle, and possibly other sources we don't completely understand.  That's a simplification, of course, as I'm not even touching upon plate tectonics and all that, but we just want to contrast it with an object like Io.

Io, on the other hand, has volcanism due to internal friction generated by tidal heating from its elliptical orbit around Jupiter, which is maintained as elliptical (instead of dissipating energy and circularizing) via an orbital resonance with other Galilean satellites.  The changes in the strength of Jupiter's gravitational pull on the planet variably stretches the planet, which stresses it and creates heat.

About that location...

So you might be thinking that if Io formed near Jupiter, which is beyond the (water) frost line, then why doesn't it contain large amounts of ice, or at least water?  Well, the location is important for this, too.

Io is the innermost of the Galilean moons.  Europa, the next closest Galilean moon (and one of the ones Io is in a resonance with), is more than 1.6 times as far from Jupiter as is Io.  This means it receives significantly more incident radiation from Jupiter than the other moons do (Europa would receive only approximately 37% of the radiative energy that Io does).  This radiation was sufficient in the early period of their lives to do what the sun's radiation does: blast water apart into hydrogen and oxygen (and then starts blasting the hydrogen away).  As such, while Io would have had access to a number of ices, including water, for its original formation, most of it would have been lost to this radiation.  Moons that are further out, or were captured by Jupiter much later in Jupiter's life, can retain their (water) ices, but Io was not so fortunate.

Thus you end up with a rocky, silicate-heavy world that's devoid of ices but has significantly more sulfur than an inner planet like Mercury.  The sulfur becomes pronounced in the volcanism for chemical reasons: it's volatile and pretty easily becomes gaseous.  And it doesn't need to have volcanism that behaves like volcanism on Earth because the source of its heat is fundamentally different.


How many galaxies have been discovered? I know Andromeda is our nearest galaxy. But how many known galaxies have been discovered and what are they?

First, the nearest galaxy to ours is not Andromeda, according to the NASA based page "The Nearest Galaxies", it was until recently considered to be the Canis Major Dwarf Galaxy at approximately 42,000 light years away.  However, recently, there is suggestion that the closer Omega Centauri Globular Cluster may be a disrupted dwarf galaxy (Another source).

Now, in answer to your other query of how many galaxies there are - there is a global project called Galaxy Zoo which is attempting to catalogue and classify all observed galaxies, so far, according to the Galaxy Zoo for astronomers page, thus far they have

  the entire Sloan Digital Sky Survey spectroscopic sample and all existing Hubble Space Telescope surveys (around 1.5 million galaxies in total).

Pictured: The Hubble Ultra-Deep Field which contains over 10,000 objects, the majority of which are galaxies. The image is only a bit over 3 arcminutes across--a tiny sliver of the sky.


If the Moon were impacted by a suitably sized meteor, how long would it take to impact the Earth? An answer to the question of How well would the Moon protect the Earth from a Meteor? mentions as a possibility that the Moon could get knocked into the Earth.

What is the smallest change to the orbit of the Moon from being impacted by a large meteor that would cause it to eventually impact the Earth (i.e. "circling the drain")? What timeline would that look like (minutes, hours, days, years, etc)?

As several people have said, this is incredibly unlikely. Part of the reason why is that the "circling the drain" effect you describe doesn't really happen for solid objects much less dense than black holes. Orbits are not "precarious" in that way.

So, suppose something large enough and fast enough to change its velocity noticeably, but not large enough or fast enough to shatter it, did hit the Moon. The effect would be to shift the Moon from its present almost circular orbit around the Earth, into an elliptical one. Depending on the direction of the impact, it would either get a bit nearer to the Earth than it is now, once per orbit, or a bit further away (it also might swing North and South a bit). What is important though, is that this elliptical track is stable at least for a while. Suppose it gets knocked into an orbit that is 220000 miles from the Earth at its closest and 240000 miles at its furthest, that is where it will stay. It will not "spiral in".

Over a long enough period the gravity of the Sun also comes into play and things may shift a bit, but that is a relatively small effect.

Now, suppose that the impact was really big, or perhaps there were a long series of impacts (starting to look like enemy action..) so that the innermost point of the ellipse was eventually driven down to within a few thousand miles of the Earth, somehow miraculously not smashing the Moon to fragments in the process. At this distance it starts to matter that the near side of the Moon is closer to Earth than the far side, so that Earth's gravity pulls on it more strongly. If it orbited closer than about 3000km to the surface of the Earth for long (the Roche limit) these forces would eventually pull it to pieces, and Earth would probably have a pretty set of rings for a short time before internal collisions between the bits caused them to rain down on Earth and kill everyone.

Finally suppose the impact(s) was(were) so big that they actually put the Moon into an elliptical orbit whose innermost point was so close to Earth that the Earth and Moon touched. This is manifestly impossible without shattering the Moon, but in that case, the Moon would indeed hit the Earth. The time for the impact would be about 1/4 of the Moons current orbital period, which is to say about a week.


How much does cloud cover affect viewing the total eclipse As I look for a spot for viewing the total eclipse, how much will cloud cover affect the viewing experience? Is there a threshold list like:

Under x % won't have a noticeable difference
Over x % you won't be able to see the stars
Over x % you won't be able to see the sun's atmosphere
Over x % you will only notice the darkening

I don't know if there is a list but I'me sure your local weather surface will give you some idea. Cloud cover could have a big effect of course. But it will be the same as it is for any day or night. Total overcast will allow you to notice the darkness only. Just use your common sense on what cloud cover you deem too risky for travel. There is an internet site: https://clearoutside.com/forecast which is supposedly developed by astronomers to help determine viewing conditions. Just input a latitude and longitude and it gives you info related to viewing conditions. Different days are on left after local forecast. Click the triangle in the lower right of the date and it gives you a detailed forecast for the location you entered. This includes predicted cloud cover by the hour.


Can stellar black holes cancel each other? Hypothetical question: when a pair of identical stellar black holes are very close to one another wouldn't the side facing each other experience lower gravitational force? How is this logic un/sound?

Your logic isn't unsound. By symmetry there would be a gravitational neutral point between them.

Of course, the black holes could not then be described by the Schwarzschild metric, since that only applies to a spherically symmetric situation.

As you brought the black holes closer together, the isopotentials would become distinctly aspherical, the event horizons would be complicated shapes and ultimately would merge - although as the black holes would presumably be in orbit and rotating, the spacetime geometry would be even more complex than that.


What would be the scientific impact of locating the Thirty Meter Telescope at its alternate site? The NPR.org article Hawaii Protesters Block Access Road To Stop Construction Of Massive Telescope briefly summarizes the situation, one paragraph says:

  The TMT, as it is called, has been the object of intense opposition for nearly a decade and the building project has faced numerous delays, protests and lawsuits. But a decision by the state's Supreme Court in October cleared the way for construction to begin after reinstating a building permit that was revoked years earlier. And last week, Gov. David Ige announced several roads would be closed beginning Monday to allow for the movement of large equipment onto the mountain.

and another says

  "Hawaii is a unique place," Meisenzahl said, noting that protesters gathered on the mountain were engaging in a peaceful, nonviolent form of protest called Kapu Aloha. "It's a small island and everybody knows each other so law enforcement and opponents are working together. Despite what is happening today it's a positive environment," he added.

Question: What would be the scientific impact of locating the Thirty Meter Telescope at its alternate site? Is it a loss of resolution, or infrared spectrum, or both, or something else?

I think the Canary Island site would be the alternate to Mauna Kea, since the E-ELT would be located in Chile and putting both (extremely) large telescopes in the same hemisphere is undesirable.

The Canary Islands have slightly poorer seeing on average, but the main difference would be the higher column density of water vapour. Mauna Kea is a  higher and "drier" site, which means you can observe into the mid-IR.

Looking at the first light instrument package I would guess that WFOS (wide field imaging at 0.3-1 micron) would not be badly affected. But there would be some impact on IRIS (near-IR spectroscopy) towards the long-wavelength end of its range.

The near-IR adaptive optics would also be somewhat affected by the slightly poorer natural seeing, but I'm not sure I can quantify this.


Will the night sky ever be more stars than darkness? Given enough time, will enough light from distant stars reach Earth so that our night sky will be more than 50% stars?

Nope! The universe's expansion is accelerating, which means there's a "horizon" beyond which light can never reach us, no matter how long it travels. Although the light always travels at the speed of light, the "amount of space" between photon and us can increase faster than the speed of light. This isn't a real "speed" - it's not any one object moving faster than the speed of light, it's all of space between us and the photon locally expanding a little bit - but it means the photon will never reach us. The acceleration means that this horizon will eventually shrink, as the expansion speeds up.
This means there's a finite amount of stuff that we can ever reach, and that can ever reach us. So you won't build up more and more distant stars & galaxies in the night sky forever.


(Sound) Echo from X takes 5-7 days, Hey I want to describe an event A happening roughtly 5-7 after event B. I'd like to describe it as the 'A happened after the sound of B echo'd off X', where X is recognisable space thingy...

(I'm talking mildly poetically, so I'm aware there is no sound in space, but I'm interested in what X might be, if such a thing was possible - if it makes you feel better you can use light and I'll describe 'A happened before X saw B') :)

If there would be air at room temperature between Earth and Moon, sound would take about 28 days from Earth to Moon and echoed back, with helium instead of air 9 days, and with hydrogen about 7 days.

Calculated from http://www.phy.mtu.edu/~suits/SpeedofSoundOther.html and http://en.wikipedia.org/wiki/Moon

You could also reflect Earth light from Pluto back to Earth, 11 times this double Earth - Pluto distance to get about 5 days of travel of light, much depending of the orbital position of Pluto.


Brightest star visible from Italy and Argentina A friend of mine tends to ask crazy question about everything. Today he ask me what is the brightest star that is visible on both the hemisphere and in particular from Italy and Argentina. I tried to ask wolframalpha but nothing came up. Thanks in advance.

Sirius is the brightest star in the sky (excluding the Sun, of course), with a visual apparent magnitude of −1.46, and is visible to everyone in both Argentina and Italy, depending on the season.

In fact, Sirius is visible from anywhere in the Southern Hemisphere, including all of Argentina (which, I might add, is an enormous country stretching from just above the Tropic of Capricorn, all the way south to the Drake Passage separating Tierra del Fuego from Antarctica).

Sirius doesn't rise above the horizon for locations in the Northern Hemisphere north of 73°, but this is well beyond the northernmost point of mainland Europe at around 71°N, so Sirius is visible in all of continental Europe. As the northernmost location in Italy is around 47°N, Sirius at its zenith appears well above the horizon to anyone in Italy.


Why moon sighted only in Kerala but not in rest of India? I know for most of you the question may look silly, but Kerala is a state in India where the moon (Ramadan moon) will be sighted but not in rest of India.

I was reading the news article where I got a little hint but was not clear for me since I am newbie, tried to google but couldn't find much info :) can someone explain me.

  Eid is likely to be celebrated in Kerala tomorrow. Due to the
  geographical location of the coastal state, the lunar moon is sighted
  in the state a day before the rest of India.

Here is the article link Moon sighted in Kerala

Think of it like this:

Each hour after New Moon, the Moon is moving away from the Sun. Depending on the position of the Sun, Moon, and observer, it is unlikely to be visible until at least 15 hours after New Moon, and in some cases not until 24 hours after New Moon.
Someone will be the first to see the thin crescent after the New Moon. This occurs when the Sun is far enough below the horizon to make the sky "dark" and the Moon is far enough above the horizon (and away from the Sun) to be "clear" of the sky's glow. ("dark" and "clear" are relative. The sky will still be quite bright but not like the middle of the day.)
Observers that are farther east than the first person would not be able to see the Moon because the time would have been earlier. That is, the Moon would have been closer to the Sun and not visible.
Observers that are farther west should be able to see the Moon more easily because it is later, and therefore the Moon is farther from the Sun. In other words, the longitude affects how easy it is to see the new Moon.
Whether it is easy or difficult to see the new Moon also depends on the latitude because this can affect where the Moon is relative to the Sun and horizon. This variation changes throughout the year.

It looks like Kerala is on the west coast, so they may have been in the position to see the Moon in that 15 to 24 hour time range whereas it would have been too early for people further to the east to see it.

Kerala is also farther south than the rest of India. I have not done the calculations for this New Moon, but it is possible that being south and west increased the possibility of seeing the Moon before people further north or east would see it. If this is the case, then Kerala would have been in the best place to be the first to see the new Moon -- assuming the sky was clear :-).

The image shows that, according to the calculations by https://www.moonsighting.com/visibility.html, the new moon might be visible from only the far South-Western coast of India.


Can a star orbit around multiple planets or a planet with massive moons? Note: I'm not talking about a star orbiting around a single or lonely planet :)

I know a star orbiting a planet is almost impossible because if a planet is more massive than a star, that "planet" would probably be a star. But how about if the planet is not lonely like:

A multiple planetary system which consists of free floating planets only but the planets are so massive so that their total mass is larger than a star
A planet itself is not more massive than a star but it has so many massive moons

Is it possible for a very small star to orbit around planets in such situations?

There's theoretical ways to do it but it's so unlikely as to probably not exist.   Assuming you want a system where the planets are in stable orbits around each other.   The basic difficulty is the 3 body problem or n body problem.  More on it here and here.

For example, a massive planet could (in theory) have a single super-massive moon of similar mass to the planet, though that would probably be considered a 2 planet system.  It's not possible for one planet to have two super-massive moons because that wouldn't be stable.   In general, planets are many times the mass of all their moons combined, similarly stars are many times the mass of all their planets combined and when that stops being true the system is no longer stable.   It's very difficult to generate sufficient mass by adding moons to a system, or by adding enough planets to get more massive than the sun, unless the planets crash into each other but past a certain mass, they'd stop being planets and become a kind of star when enough planets combined.

If we look at gravitational systems, like our solar System, something like 99.7% of the mass is in the sun, so the sun dominates and everything orbits the sun.   A few of the larger objects have moons, and curiously, but only cause they're in relatively calm orbits far away from other planets, a few asteroids orbit each other, but the system is very structured around the sun with much smaller stable orbital zones around the planets.

Pluto also has a rather curious orbital system of it's own, likely caused by an impact, where Pluto and Charon are comparatively large and 4 tiny moons orbit around them.

Source:

But these kind of structured systems are only possible if you have a significant mass differential.   When you have 3 or more bodies of similar mass and similar distance you get a high degree of mathematical chaos and instability.   There are creative mathematical tricks to make it work, but none of them are stable or likely.

This is what 3 similar mass bodies look like, and in such a system, with constant changes, the most likely scenario is that one of the bodies eventually gets ejected.  (source, N-body problem above)

There are star systems with several stars, but they are either unstable or contain significant differences in mass.   The gravitational structure applies for large planets the same as it does for stars, and there's an article about that here.

You can create stability by having 2 objects orbiting each other and a 3rd massive object quite distant.  (Alpha Centauri is that kind of set-up though Proxima Centauri is quite a bit smaller, but it's the same system).

You can even create the heirarchy where you have 2 objects orbiting each other and then 2 more, orbiting each other, but distant so the 2 co-orbitals orbit around each other, and if you do this enough times, you can kind of create enough planetary mass where a more massive star could distantly orbit the entire thing, but it gets very structured and very distant.   It's not what I'd consider a normal orbit.

You could also cheat and have several planets in a wildly unstable general proximity orbit and have a star some distance away orbiting the chaotic mess in the middle, but it wouldn't be stable for long.


Interaction of stellar winds in binary star systems Binary star systems are fairly common but I am unable to find any information on how the stars interact with one another on aspects other than gravitational affects.  Using Alpha Centauri as the candidate system, what possible effects could happen where their stellar winds reach one another?
Would a cloud of magnetized gas form? Would there be a highly radioactive area between them? Could it form a sort of micro-nebula between the stars?

Systems like this are known as colliding-wind binaries (CWBs), and they do produce some interesting effects. When winds collide, they create shocks, which in turn heats up gas. The most notable result is x-ray emission, which tends to happen whenever you have non-negligible shock heating in stellar winds; there is certainly additional thermal and non-thermal (e.g. synchrotron) radiation at other parts of the spectrum. I don't know if this would lead to significant structure formation - bear in mind that hot stars with strong winds may also generate more prominent outflows and nebulae through unrelated mechanisms; Eta Carinae is a great example.
This actually poses a problem for people trying to study stellar winds in x-rays. We can derive a lot of information from spectroscopic observations of winds - composition, temperature distribution, mass-loss rates - for simple cases when x-ray emission arises from the line deshadowing instability (Owocki, Castor & Rybecki 1988), leading to embedded wind shocks (EWS).
Magnetic fields complicate matters by forming shocks as plasma is forced to travel along field lines, and colliding-wind binaries make it very difficult to study the individual winds because much of the emission is now coming from the CWB shock front and not purely the EWS model. I don't really know how folks studying CWBs disentangle the effects - when I was working on stellar winds, we mostly just ignored all magnetic and O+O binary stars in the sample!


Size of Sun as seen on Pluto What would be the size of Sun as seen from surface of Pluto? It obviously depends on the distance and one would expect it to be very small. However, due to different optical phenomenon, the size of Sun seen on surface of Pluto could be larger. Can it can be estimated from the annular eclipse Sun-Pluto eclipse picture or other pictures taken by New Horizons probe ? Thanks for your insight.

The picture is also posted on this forum question.

Edit: From the answer given by @Conrad Turner below, Sun should be just pin point at that distance. However, when seen by New Horizons probe from behind Pluto, its size is seen to be almost as large as Pluto. The atmosphere of Pluto must be bending the light so much to make this happen.

Update: This picture was taken from a distance of 2 million km from Pluto. At this distance Sun will have an angular size of 0'49" (distance 5804550000 km, diameter 1392000 km), while Pluto has the angular size of 4'5" (distance 2000000 km, diameter 2372 km). However, the picture is showing Sun to have an angular size slightly larger than Pluto (or nearly the same size). Hence, high refraction or scattering or some other phenomenon are at work here.

Considering sizes and distances, if atmosphere of Pluto does nothing to light, the ring of Sun should have been visible around Pluto once probe was 10 million km beyond Pluto. However, the ring of Sun is seen in this picture when probe is  only 2 million km beyond Pluto. The scientists must have known the optical characteristics of Pluto's atmosphere and hence they got the probe early in proper position to take this picture.

New Horizons encountered Pluto when it was about 33.5 au from the Sun.

The solar diameter is $1.39\times 10^{9}$ m.

An astronomical unit is $1.496\times 10^{11}$ m.

The angle subtended by the Sun at Pluto (in radians) is the ratio of the diameter of the Sun to its distance from Pluto. Converting to degrees, minutes, seconds, gives 0.01589 degrees, or 0.95 arcminutes.
The atmosphere of Pluto is way too thin (much thinner than the Earth's) to make any significant difference to this at all.

The diameter of Pluto is $2.37\times 10^6$ m. In order for Pluto and the Sun to subtend a similar angle in an image, you just need to move far enough away from Pluto to make the same angle.

Thus
$$ D_P/R_P  \simeq (D_P + D_{PS})/R_S,$$
where $D_P$ would be the distance from New Horizons to Pluto, $D_{PS}$ the distance from Pluto to the Sun, $R_P$ the radius of Pluto and $R_S$ the radius of the Sun. Because I will show that $D_P \ll D_{PS}$ we can say that
$$ D_P \simeq \frac{R_P}{R_S} D_{PS} \simeq 0.057\ au$$.

New Horizons is travelling at about 14.5 km/s. So a simple bit of Maths tells us it would takes about 6.5 days to get into this position.

The picture I think you are talking about was taken on July 15th 2015 only 7 hours after the flyby. The Sun and Pluto do not share the same angular size in this photo. You are not seeing the Sun at all in this picture. It is eclipsed by Pluto completely. All that is being seen is sunlight refracted (or more likely scattered) through Pluto's atmosphere. The Sun acts more-or-less like a point source of light at more-or-less infinite distance.

The angle that the light needs to deviate in order to be seen through the atmosphere is about $0.19$ degrees (the half-angle subtended by Pluto at the Sun [negligible] plus the angle subtended by Pluto at New Horizons after 7 hours). If we were to model the atmosphere as a glass prism. The deviation angle is $\delta \simeq (n-1) \alpha$, where $n$ is the refractive index and $\alpha$ is the opening angle of the prism. If you let $\alpha$ be quite large (let's say 1 radian) to simulate rays coming through the atmosphere, then $n = 1 + \delta$, where $\delta$ is in radians. So $n\simeq 1.003$. NB: This assumes there is no scattering in Pluto's atmosphere, but I suspect that is not true and that my estimate of the refractive index is a large overestimate. In fact I'm sure it is an overestimate because this is larger than the refractive index of air on Earth. i.e. I am sure that what we are seeing is scattered light, not refracted light.


Centerless expansion or a matter of perspective I get this question a lot:

  Where is the center of the Universe?

I see this question on the internet all of the time:

  Where is the center of the Universe?

Now I have a question regarding all of the answers that I always either give or read:
The question linked is the last question and answer that I saw before posing this question here, and is in no way the same question, as this question asks how the concept of a centerless universe can work, whereas the question below asks about how a universe can be perceived to be centerless even when it might not be.

Is the Universe centerless? Or does it simply appear to be because of our perspective?

$\\\\$
This is about to be a bad example, but please bear with me.

If you were a bird floating in the ocean, among a myriad of other birds, and you let time pass such that every bird floated away from the other birds in some random fashion. Would you, at some later time when everybirdy was roughly separated from one another, say that based upon your observations of the present conditions, and some evidence that at one point all of the birds were together, that the ocean is centerless? Or could you assume that because of your position in the flock, that from your perspective you can't tell where the center is, nor do you think that from a different birds position would you be able to tell? Creating the illusion of a centerless ocean.

Obviously this requires you not be at or close to the very edge of the flock.

Edit: Also see examples in KenG's posed answer and comments.

The first principle of cosmology is called the "cosmological principle", which asserts that the universe is more or less the same everywhere spatially, but changes with time.  This is a fundamentally new way to think of the history of the universe, and has revolutionized our ability to understand long-range observations.  We do not, however, know that this principle is true, we only know it is both useful and consistent with all observations.  So we adopt it as a working hypothesis, and all of cosmology follows.  Some day we may discover it is inadequate, but until then, we will continue to use it.  The same could be said for other principles, like conservation of energy-- we don't know that's true either, but it's useful and consistent with observations.  The cosmological principle rules out that the universe has a unique center.

Your question is whether the apparent lack of a unique center stems from our location not being special enough to be able to tell that it does have such a center (say, by encountering a boundary).  Ironically, the cosmological principle turns that question on its ear, and points to the possibility that one's perspective always puts themselves at the apparent center.  So we would say the illusion of having a center is all about perspective-- not the illusion of not having a center!


White dwarf's impact on orbiting bodies Recently the Kepler telescope in its study of white dwarfs detected the first planetary object transiting a white dwarf in the data from the K2 mission. It was consistent with earlier theories' prediction that a planetary object orbiting a white dwarf would slowly disintegrate. Why would a planetary object that is orbiting a white dwarf disintegrate? I read it here.

The original paper published in Nature (preprint): A Disintegrating Minor Planet Transiting a White Dwarf

I think Aabaakawad's link gives a complete answer, but to give an astronomy for dummies answer, there's nothing about a white dwarf that causes a planet's orbit to decay at least, not directly.  Your article (I've pulled quote below the caption):

  Slowly the object will disintegrate, leaving a dusting of metals on
  the surface of the star.

That's only talking about this particular situation and there's a difference between disintegrate and decay.    This planetoid is enormously close to the white dwarf.  So close, that what we think of as normal white dwarf/planet dynamics (very cold) is no longer true.  This planetoid is slowly being vaporized.

Looking at the orbital period of 4.5 hours (about 1948 orbital periods in 365.25 days).   The orbital distance to orbital ratio relation is exponential to the power of 2/3 (this varies a bit due to eccentricity, but it's generally correct), so an orbital period 1948 times faster means about 156 times closer, and giving the white dwarf equal mass to our sun, that puts the planetoid at a bit under 1 million KM.  If this white dwarf is lighter than the sun, the planetoid would need to be even closer.  That's close to the Roche limit and would be inside it if the planetoid wasn't dense and rocky/metallic.

If we estimate the white dwarf to be about the size of the Earth, which is a common size given for white dwarfs, An Earth sized object from 1 million KM would be larger in the sky than the sun appears from Earth, and presumably quite a bit hotter than the surface of our sun too, so this isn't a tiny white dwarf in the sky from the perspective of the planetoid.   It's a blazing furnace of a sun, so hot, it vaporizes metallic gas and dust off the surface of the planet.

The article mentions this (end of page 3), that Poynting-Robertson drag see here and here, and that may be a factor in any orbital decay in this scenario.  The article is clear that there's a good deal of uncertainty on with that effect, and that only affects tiny particles, but enough tiny particles could create a drag over time. . . . (maybe).   The general scenario with this orbit is a planet scorched and as a result, is losing material.  It's likely the very high heat that's driving any orbital decay, not gravity.

Gravitational decay / orbital decay does happen, usually much more slowly.   That's probably not what's happening here.

There are some interesting orbital effects that can happen when a main sequence star goes red dwarf and later when it creates a planetary nebula, significant increases in tidal forces due to the star's greater size in the first case and increased drag in the 2nd, but at the white dwarf stage, there's no significant orbital decay effects.

Update:

  Why not Poynting Robertson drag and Orbital decay effect the planetoid
  when white dwarf was a star or even red giant? Is there any
  "interesting orbital effects" when a star undergoes red giant?.Can you
  update your answer to summarize the forces and their effect on the
  planetoid in each phase of the star.
  and also what do you mean by orbital decay? Does it have something to deal with the Roche limit.

OK, I think, having read more about it, Poynting-Robertson effect only matters when the orbiting objects are very small.   I've linked it twice above, but the simple explanation is that objects in orbit move and so any light or debris from the sun hits the moving object at an angle, not direct on.    If the object is small enough, this over time drives the dust and maybe grain of sand sized particles into the sun.   This doesn't affect larger objects, so it's not really relevant to any planets or planetoids.

As far as "interesting red dwarf" effects.   That really has to do with tides.   Using the Moon/Earth example, the Moon creates tides on the Earth, a tidal bulge towards the Moon, but because the Earth Rotates faster than the Moon orbits, this tidal bulge is always ahead of the Moon and this creates a gravitational tug on the moon that pulls it away from the Earth - very slowly.

The same thing happens with planets around stars, but even more slowly,   lets pretend it's just the Earth and the Sun - a 2 body system (in reality, with several planets it's much more complicated), but just Earth and sun, teh Earth creates a tidal bulge on the sun, the sun rotates ahead of the earth, this causes the earth to very slowly spiral away from the sun - so slowly that it might take a trillion years for the Earth to spiral away.

Now when the sun goes Red Giant, the sun is essentially the same mass but much more spread out and parts of it, much closer to the Earth and less gravitationally bound to the sun.   This creates a far larger tidal tug.   Also, as the sun expands it's orbital velocity drops, because orbital momentum is concerved, so when the Sun is Red Giant, the tidal bulge will be behind the Earth which drags it in towards the Sun.    Due to the size and proximity of the Red Giant star, this draws the remaining near-by planets towards the sun fairly quickly, at least compared to main sequence stages which, provided the sun rotates faster than the planets orbit, has a much smaller outward tidal pressure on the planets.

And when the sun goes planetary nebula, any debris in the planet's path can also cause the planets to slow down slightly - the precise process there I'm less clear on, but in general, any orbital debris creates drag and can slow down a planet's orbit.   This may be a key factor in the formation of hot jupiters, cause they can't form close to their suns but enough orbital debris can drive them in closer to their suns.  (or planet to planet gravitational interactions can too).

That's the gist of the Sun-Planet orbital relation.  When the sun is young, planets are mostly driven outwards, and young suns can have far greater solar flares and stronger solar wind.   How much that effects the planets, I'm not sure.

During the Main sequence stage, stars tend to push planets outwards (unless they rotate very slowly, in which case the tidal effect is reversed), but this tidal effect is very small and very gradual.

During the Red Giant stage, stars tend to drag planets in wards, and I assume, during the planetary nebula stage as well.   This effect is larger for closer planets.

You also asked about Orbital Decay - if you click on the link, there are examples of that.  That probably gives a better explanation than I could.   In general, Orbital decay happens very slowly unless you're talking Neutron Star or Black hole in which case the relativistic effects can cause orbital decay to happen quite fast.  There's nothing about a white dwarf star that would cause faster than normal orbital decay but a white dwarf star would lose any tidal bulge tugging that a main sequence star has, so there would be essentially no tidal outwards pressure either which could in theory speed up decay cause you've lost a small outwards pressure but you would still have any debris or space dust clouds causing a small inwards pressure.  (if that makes sense?)

That's my layman's explanation anyway.


what kind of telescope eyepieces is best for viewing planets up close ok so I am a total amateur when it comes to telescopes and lens, and this is probably a pretty stupid question, but I am trying to find a lens/eyepiece that is good for viewing Jupiter and Saturn up close, and I have no idea what I am looking for, I have an Orion Sky Quest XT8 plus reflector telescope, (not sure if that matters) but if someone knows what I am looking for pls lmk, thx Noelle:)

To get good views of Jupiter and Saturn you need quite high magnification - probably 100 times or more. The XT8 has a focal length of 1200 mm and so to get that magnification will need an eyepiece of 12 mm focal length. I believe the scope comes with a 10mm eyepiece which will give 120 times magnification, which should give decent views.

You could go for higher magnification with shorter focal length eyepieces, but if you do you will find that the object you are observing scoots across the field of view rather quickly and you will spend more time moving the telescope than actually observing.  As you get more experienced you may find you get more adept at moving the scope, and so could use more magnification.

It’s important that the mirrors of the scope are properly aligned (collimated) so as to get the sharpest views. The manual should tell you how to do that.


What causes "North-South ambiguity" when doppler radar imaging a planet surface equator on? I'm having a hard time understanding why there would be this ambiguity. Also, why would this be not the case if the planet was viewed pole on?

Edit: "NOrth SOuth ambiguity" is a term used in a paper that seemed most concise but I guess not. When using radar to doppler image a planetary surface which happens to be viewed equator on, we have no way of distinguishing whether the surface feature read in return signal belongs to the northern or southern hemisphere. WHat causes this?

It's a great question!

Delay-Doppler imaging is a technique to produce a kind of image that represents the topography of a rotating body, usually when it is so far away that there's no way to resolve the object with traditional imaging, radar or optical.

A powerful radar signal is sent from Earth to an asteroid or a planet, and the reflected signal returning to Earth is recorded and analyzed. The "special sauce" that allows the technique to work is the pseudo-random sequence encoded into the modulation of the outgoing signal. Using correlation techniques, one can extract reflected components from a range of different delays due to different distances to each part of the planet, and different Doppler shifts due to different projected (radial) velocities of each part of the planet.

The raw data is pretty messy, but after you record the body rotating a substantial fraction of a full turn, using modeling, you can recover a reconstructed image of the planet's topography.

The hard part is that while Doppler shift gives you east-west discrimination since one side moves towards you while the other moves away from you (assuming the rotation axis is "up-down"), there's no easy way to tell the difference between the two hemispheres in the reflected signal.

Thus the north-south ambiguity means that there is no temporal difference between a signal reflected from upper hemisphere and the lower. Say at 50 degrees North and 50 degrees South. The signals would have BOTH the same delay (since they are the same distance away) and the same doppler shift (since they are both rotating at the same speed.

Remember that for a spherical-like object of radius R who's axis is perpendicular to us, the projected line-of-sight (radial) velocity will be:

$$v_r = \frac{2\pi R}{T} \cos(\phi) \sin(\lambda)$$

where $T$ is the period, $\phi$ is the latitude and $\lambda$ is the longitude. It's symmetrical about the equator (cosine behavior) but it varies monotonically east-west (sine).

There are solutions to this if the object is large enough though. If you have a huge dish and the object has an angular size on the same scale as the beam width of your dish, you can record data with the beam illuminating first one hemisphere more strongly, then the other. That provides some additional information, and with subtraction and sufficient modeling one can overcome the North-South ambiguity.

Of course, if the body's rotation axis is oriented obliquely towards or away from us, then the symmetry is broken and the ambiguity can also be resolved in some cases.

Related and possibly helpful:

Why do radar maps of the surface of Venus have missing slices?
Emily Lakdawalla's How radio telescopes get "images" of asteroids

Left: A very slow GIF Source.

Right: A radar image shows asteroid 3122 Florence and tiny echoes from its two moons. Here is an animation that shows them more clearly. The direction of the radar illumination (and thus the direction toward Earth) is at the top." From here. NASA / Jet Propulsion Laboratory. This is a small subset of the frames contained in the original 36 MB GIF, and the size has been decreased by a factor of 2 in order to fit in SE's 2 MB limit. From the question What is the physical geometry of this apparent "eclipse" of a tiny moon of Asteroid Florence?

Note: These two images do not necessarily have the same orientation.


Invisible Mass Affecting the Expansion of the Universe Are there enough neutrinos & other invisible particles in a cubic light year of space to significantly affect the expansion of the universe,& at a very rough estimate without taking into account WIMPS & other exotic particles for which there is no firm evidence,what would be the total mass of this quantity of matter? Only a tiny proportion of all the neutrinos emitted since the Big Bang have been absorbed,the rest are still flying around in search of a home.

Yes, there is enough of these things, especially dark matter which is accelerating the expansion of the universe.


Researching mechanics without “fixed stars” In the history of humanity, easily observable extra-(Solar System) objects greatly helped understanding certain phenomena inside the Solar System. Importantly, the “precession of the equinox”, and also true periods of heliocentric orbits lying near the ecliptic.
Imagine a system like our Solar System with indigenous astronomers who are unaware of far-away objects (or cannot determine their bearings). How can astronomers possessing some Early Modern technology learn of absolute (inertial) spacial directions then? Ī̲ see several possibilities: looking at highly inclined subsystems (think Uranus with its rings and satellites), noticeably eccentric orbits (such as of Mercury or, occasionally, of comets), or just inclined orbits (think Pallas). But do other reliable references exist?

Say I have a single observatory, and the only things I measure are the times and azimuth angles for the high magnitude solar system bodies (the Sun, Moon, Mercury, Venus, Mars, Jupiter, and Saturn) as they ascend into the sky and descend out of it (for a fixed elevation angle).
If I gather these for a few years, I will have enough data to have an overdetermined system to calculate the latitude of my observatory, the Earth's axial tilt, the Earth's rotation rate, and the orbital elements of all the bodies.
The trick would be to realize that all these bodies (including the Earth) can be approximated by rotating spheres that are revolving around the Sun (with the Moon also revolving around the Earth).  Solar and Lunar eclipses certainly would have helped with this notion.
Though our own ancient astronomers used the relatively fixed locations of the stars as a reference to better track the motions of the planets, they certainly aren't necessary. In fact, perhaps we would have understood the motion in our Solar System sooner in our development if we hadn't been distracted by the rest of the Universe.


How much does the orbit of Mars deviate when it is in linear conjunction with Jupiter and Saturn? Is the gravity of the planets Jupiter and Saturn strong enough to alter the orbit of Mars, or do they just give a small wiggle after which Mars is following his former orbit?

To properly answer this question, you should apply the 3 body problem, which, gets very long and mathy.  It's technically not solved but it can be approximated with very good accuracy.

Applying basic Newtonian formulas can give you an estimate.

The sun has about 1,048 times the mass of Jupiter, and if we line up Mars at it's Aphelion (1.67 AU) to Jupiter at it's Perihelion, (4.95 AU).  Mars at it's furthest and Jupiter at it's closest, Mars is still about twice as close to the Sun as it is to Jupiter.

For simplicity, I'm going to call Mars at Aphelion an "MAU" - Mars Aphelion Unit, so Mars, furthest from the sun, 1 "MAU", Jupiter closest to Mars, about 2 "MAU" and Jupiter closest to the Sun, 3 "MAU"

So, in perfectly lined up conditions, Jupiter never has more than 1/4,000th the gravitational pull on Mars than the Sun does and we also have to factor in the tidal force.   It gets tricky here, cause all 3 objects exert a force on the other (or stretch space time which tells the objects where to go if you like the relativity version), but I'm going to stick with forces.

Jupiter tugs on Mars and the Sun at the same time so how far it pulls Mars from the sun is more of an an applied tidal force.  At closest point, Jupiter's tug on Mars from 2 "MAU" and it's tug on the Sun from 3 "MAU" means it tugs on Mars 2.25 times stronger.  Take the inverse, the effective gravitational pull - at closest, is the inverse of that, or 44%, so you can only apply 56% of Jupiter's gravitational pull towards Mars relative to the sun.  And that's at peak.  As Mars moves closer to the sun and/or further from Jupiter that number drops.

The effect of Jupiter on Mars orbit relative to the sun is never more than about 1 part in 7200 of that of the sun.

As Mars moved towards Jupiter in it's orbit it speeds up and effectively moves away from the Sun, and as it moves away from Jupiter, this effect is largely cancelled out over time.   In effect it's a slight stretching when moving towards Jupiter and squashing when moving away, of Mars' elliptical orbit every time around the sun.  These are sometimes called orbital perturbations.

1 part in 7,200 works out to about 0.14% and that's the peak Jupiter effect.    Rounding down, a .01% variation on Mars' orbit seems a plausible estimate and that might even be a little high.  Applying that to the diameter of Mars' orbit, that works out to a stretching/squashing of about 20,000 to 25,000 km.  Compared to Mars natural variation based on it's eccentricity of over 26 million miles, that's pretty insignificant.   This stretching and squashing is also largely balanced out as Jupiter circles around the sun.

Now if we apply the acceleration formula, I get a somewhat different number.   g = G*M/r^2.

Gravitational Constant:   6.67E-11
Mass of Jupiter 1.898E27 kg
r = distance between the two.   Closest pass is about 3.28 AU, or 4.91E11 meters

The peak gravitational acceleration Mars feels from Jupiter is about 5.26E-7 m/s^2.   That sounds negligible, but Mars takes 187 days to orbit the sun and it spends about half that time closer to Jupiter than the Sun is to Jupiter.

As a (bad) estimate, because that's the maximum gravitational force at closest pass, if we estimate that over a Mars year, that works out to about 90 days of "falling" towards Jupiter and 90 days falling away.  Obviously the period is longer than 90 days but the force is less when Mars is further away so that's a guestimate.   Apply (d=1/2at^2) to 90 days or about 7.8 million seconds.  The distance using that formula comes to about 16,000 km.

So, both estimates are pretty close, one of 16,000 the other of 20-25,000 km.   This also works well with the 4 or 5 significant figures usually given for Mars' obit.   4 figures on distance works out to nearest 100,000 km, which wouldn't be affected by Jupiter's stretching and squashing of Mars' orbit.

Now as for Saturn, at 30% the mass of Jupiter and roughly twice the distance from Mars and the Sun, you're looking at less than 8% of the gravitational force and with tidal effects figured in, maybe just 4%-6% the orbital perturbations.   Jupiter is the one that really matters.

Orbital stretching and squashing is what lead to the discovery of the planet Neptune, because astronomers noticed that Uranus orbit was effected in a similar way to how Jupiter affects Mars.  Observing those perturbations told them about where to look for Neptune and it was found not long after.

I want to point out that over tens of thousands of orbits, these perturbations can lead to changes in Mars' orbit, primarily it's eccentricity.   It's semi-major axis and period of the orbit remain mostly unchanged, but it's eccentricity operates on two cycles.  Mars' eccentricity varies on a 96,000 year and a two million year cycle.  Like Mars, Earth's eccentricity also has cycles driven by the other planets, 95,000, 125,000 and 413,000 years.   Mars' eccentricity is prone to greater variation, presumably (I think) because it gets closer to Jupiter.

As a rule, these eccentricity oscillations move back and forth unless two planets end up in resonance with each other.   When that happens the variations can continue to grow leading to significant movement of one or both of the planets over millions of years.  This is more likely to happen shortly after the solar-system forms.  Our solar system is thought to be long-term stable with no orbital resonances between planets now, and none expected for at least a couple billion years.   There's a small chance that Jupiter and Mercury will fall into resonance in a few billion years.   Here's a nice detailed article on orbital eccentricity variation and stability of the solar system.

That's a ballpark answer anyway.   Corrections to my math are welcome and if anyone can do the 3 body problem, feel free.  That would be much more accurate, but I think 16,000 - 25,000 km per orbit is in the ballpark.


If space was n>3 dimensional, would clouds of particles still turn into 2D disks due to gravity? I don't fully understand why clouds of particles which turn into galaxies and solar systems usually result in an (approximately) 2D shape, but whatever. Would they still result in an almost 2D shape if our space was n dimensional with $ n > 3 $ and the laws of physics were the same (adapted for that $ n $)? Or would it be an (approximately) $ n-1 $ dimensional shape?

It would be of interest to have an answer to this question under the assumption that the attractive force was still proportional to $1/r^2$ even when the dimensionality is greater than 3.

It just so happens that this problem has been rigorously analyzed.  Max Tegmark wrote an interesting paper on it which can be gotten from the arXiv.  It is not super-technical, so I recommend you look at it.  Basically, with more than three space dimensions there are no stable orbits. Particles either disperse to infinity or collapse to a point (presumably a black hole).  (Tegmark also analyzes the effect of extra time dimensions -- weird.)

Added: A comment asks if a cloud of particles behaves differently and might still be stable. The answer is "no".  Here's why.

There are two cases here "gas" and "dust". The difference is that gasses have internal pressure which does affect the cloud's dynamics, while the particles in dust are big enough that collisions between them become infrequent and can be ignored for their effect on the dynamics. (Note: the only important difference between "gas" and "dust" is if the particles interact often enough to produce enough pressure to significantly affect dynamics.  It's not their composition. The terms "gas" and "dust" are simply terms of art used to describe the two cases.)

The dust case is trivial: The dust particles spend nearly all of the time in their individual orbits without colliding and if the orbits are not stable, the dust cloud isn't either.

Gas is more complicated, but it's probably easiest to look at what happens in three dimensions.  A non-rotating gas cloud collapses -- that's how galaxies and stars form!  A gas cloud that's rotating enough to support itself against collapse is relying on the centrifugal force to support it against gravity -- which doesn't work in four or more dimensions.


Earth's magnetic field vs. refrigerator magnet According to this article:

  The Earth's field ranges between approximately 25,000 and 65,000 nT
  (0.25–0.65 G). By comparison, a strong refrigerator magnet has a field
  of about 10,000,000 nanoteslas (100 G)

The Earth's magnetic field is between 150 and 400 times weaker than a refrigerator magnet.

Question: Can any magnet (or even one much stronger) of the Earth divert the cosmic rays as Earth's? Do these magnets somehow "interfere" with Earth's? If not, Why? Maybe because the size (not the power)?

Can any magnet (or even one much stronger) of the Earth divert the cosmic rays as Earth's?

No, because even if the strength of the field may be higher, the size of magnetic field is too small.  Although the force that your local fridge magnet exerts may be larger than the one of the Earth, it does so only for a much smaller area.  Even if you brought your magnet outside of the magnetosphere, charged particles would pass straight by it and only have a tiny deflection, before the particles are outside of the fridge magnet's reach of influence again.  The deflection would be the integral of force over path length of the particle inside the field.

  Do these magnets somehow "interfere" with Earth's?

In the immediate surrounding of the fridge magnet, the fridge magnet magnetic field is stronger than the one of the Earth, but at any larger distance, the Earth magnetic field is stronger.  The larger Earth magnetic field is not significantly affected.

  Maybe because the size (not the power)?

Yes, size does matter.

In the magnetosphere, the only significant magnetic field is that of the Earth.  The field of your fridge magnet is non-existent there, as its range is very small.  Charged particles are diverted by the magnetic field in the magnetosphere.  When they do enter the Earth atmosphere, many of the remaining one are absorbed by the atmosphere (forming the northern lights).


How to track orbit of an exoplanet? I was wondering if there is an established method to keep track of the orbit of an exoplanet assuming we know a - the semi-major axis of the orbit, e - the eccentricity of the orbit, and i - the inclination of the orbit. Can we track its position once a month?

To describe the position of an orbiting body you need 6 numbers.  There are different ways to do this:
You can give the position $(x,y,z)$ and velocity $(\dot x, \dot y, \dot z)$.  At a given time $t_0$, and then use Newtons laws to work out the position of the planet at any time in the future.
You can give the orbital parameters:

Eccentricity (the shape of the ellipse)
semi-major axis (the *size of the ellipse)
(inclination, Longitude of ascending node, argument of periapsis) The 3d position of the ellipse.  These depend on the particular choice of reference frame.  For exoplanets it is common to take the reference plane to be the the plane through the star and perpendicular to the line of sight from Earth. This means that many exoplanets have an inclination of about 90 degrees.
Mean anomaly at epoch (The position of the planet on the ellipse at time $t_0$)

And then use Keplers laws to work out the position of the planet at any time in the future.
So we can track them theoretically.
However determining these six values from observations is not easy, and in many cases we don't know all six at all well.  If we can detect a transit we know that the inclination is close to 90, and we might be able to combind transit observations with radial motion observations to get some limits on the other values.  Nevertheless, these six values can't be simply measured from the data.
For most exoplanets they can't be directly imaged (they are too close to the glare of the star, and to dim to be seen) There are a few exoplanets that can be tracked They are usually very large, warm planets that are far from their star.
cf. Transformation of Orbit Elements, State and Coordinates of Satellites in Two-Body Motion


What exactly are "H30α and He30α images"? (ALMA) I ran across this paper's title η Carinae: high angular resolution continuum, H30α and He30α ALMA images (arXiv) and see that the body of the paper also mentions "H40α, H30α and H29α".
What notation is this, in which contexts is it used, and what exactly to the numbers and greek letters denote? (I'm assuming that H and He are elements).

These are the names of recombination lines. In terms of the Bohr model of the atom:

The electron in a Bohr atom can fall from the level ($n+\Delta n$) to $n$, where $\Delta n$ and $n$ are any natural numbers (1, 2, 3, …) by emitting a photon whose energy equals the energy difference $\Delta E$ between the initial and final levels.  Such spectral lines are called recombination lines because formerly free electrons recombining with ions quickly cascade to the ground state by emitting photons. Astronomers label each recombination line by the name of the element, the final level number $n$, and successive letters in the Greek alphabet to denote the level change $\Delta n$: α for $\Delta n=1$, β for $\Delta n=2$, γ for $\Delta n=3$, etc.  For example, the recombination line produced by the transition between the $n=92$ and $n=91$ levels of a hydrogen atom is called the H91α line.

So H30α indicates the transition from levels 31 to 30 of the hydrogen atom.
As noted by Peter Erwin in the comments, the series with low $n$ have traditional names. In particular, the Balmer series (transitions to $n=2$) are traditionally known as Hα, Hβ, etc. The first four of these are in the visible part of the spectrum.


How were the first stars formed? In class, we read a Scientific American article, "Cloudy with a Chance of Stars" that explained how the cores and the dust around them formed stars(he called them eggs for a chicken/egg analogy, where cores are eggs and I suppose novae are the chickens because the novae cause the core to start to collapse), but we don't know how the very first stars were actually formed.

What are the leading hypotheses in this area, or am I misinterpreting the article? If you have a Scientific American account, here's the article I am referring to.

It is a question of where, but also to better understand, you must know when it happened. Because earls structure was different from today's.
From wiki on Universe's chronology,
By NASA/WMAP Science Team - Original version: NASA; modified by Ryan Kaldari, Public Domain, Link
you will learn that first stars are thought to appear when the Universe is about 3% of its current age, i.e. 400 million years old. This is the time it took for first atoms to cool down sufficiently and gather together in the clouds in question.
To see where first star formation happened, we have the possibility to look back in time.

we can see the star being born, and for this we need to use radio frequencies, because infrared was the only thing coming trough the surrounding gas clouds, from that time is so red shifted...
if we are very lucky, we can see these stars exploding - since they are so big, they will quickly come to an end and explode - but because of surrounding gas clouds it is regarded as a rare event.
there are simulation tentative like this one where we use currently know physic laws to calculate the evolution of matter since it first emitted light (CMB - the bluish-greenish thing on above image); the very small differences in this primordial light we are able to see allow the simulation to explain densities differences and the apparition of filaments, planes and clusters (nodes at the intersection of planes and filaments). These things can be artistic, as you can see here filaments and a Protogalaxy simulated:

from Sky & Telescope
Clusters where the denser places of the universe, and due to this highest matter density, is where first stars where created.

Here is an images where it is thought a filament is actually seen, lightened by a catastrophic event it hosts:

The Lyman-alpha blob (blue fuzz) surrounding the quasar UM 287 (white dot in center) extends 1.5 million light-years long, far too big to be contained within the quasar's host galaxy or the host galaxy's halo. The blue fuzz is likely to be part of a cosmic web filament.
S. Cantalupo / UCSC

I invite you to get a better idea of what we can see of the distribution of stars from our planet here, where you can zoom in and out and get an understanding of scales.


Could a cross-ringed planet exist? My title doesn't do my question justice in explaining, but the kind of satellite system I'm thinking of is two rings on two separate axis, much like (if you've seen it) the film 'Treasure Planet' shown below.

My guess is no, as the ring of a planet matches it's axis, but I was wondering whether or not some other phenomenon or influence could cause this formation.

While such a scenario would be unlikely, all you'd need is an existing ring and an icy moon to enter inside the Roche limit at a measurably different inclination. I wouldn't want to guess how long a set-up like that would be stable, but it's possible to have one.


Can I use the Tokyo Skytree as clock? After my observation today at the Skytree in Tokyo, I am wondering if this building could be used as a giant sundial?

As an aside, it would be interesting to consider how the Skytree might be used to display the time.

Anything which casts a shadow can become a sundial, as JH's comment points out.

The other thing you need to know is the latitude of the building, as that affects the change in shadow direction per hour of elapsed time.  Worse, due to the eccentricity of Earth's orbit (and rotation axis), there's an offset between clock time and sundial time which varies over the year.  You can get most of the formulas at the Wikipedia page .

There's a handy calculator at blocklayer.com


Why Milky Way and Andromeda are being drawn together if there was 'Big Bang'? Scientists used to use following reasoning:

most galaxies are red-shifted $\implies$ there was a 'Big Bang'

Why this is being considered valid since not every galaxy has such property?

The evidence for expansion is that the redshift is proportional to distance.

The redshift of a galaxy can be divided into two components: that due to the cosmological expansion, which stretches the wavelength of light whilst it travels towards us; and a peculiar motion with respect to the cosmological expansion, which causes a straightforward doppler shift.

The former term increases with distance - this is known as Hubble's law. The redshift here is always positive (ie always a stretching of wavelength). It is this that tells us the universe is expanding. The latter term is caused by the gravitational effects of other nearby galaxies and clusters of galaxies on the galaxy in question. It is typically of order a few hundred km/s and can be positive or negative (ie it can cause a redshift or blueshift).

Redshift due to cosmological expansion completely dominates at distances greater than a few hundred million light years. Up till then (and Andromeda is only 2 million light years distant), peculiar motions can result in blueshifts for some galaxies.


How to know the orbital and physical characteristics of the solar planets with only a non-professional telescope? I have always been fascinated to see that with so little knowledge, scientists have succeeded in discovering the planets of the solar system, and have succeeded in defining them by their masses as well as by their orbits, temperatures, and so on.

I would like to try to trace this process myself but I do not know how to do.

My starting postulate is simple, I have at my disposal a simple telescope and DSLR camera.

My goal is to define the solar planets, ie to know their masses, their sizes, their orbits, their revolutions.

How to achieve it? Do you have any leads?

EDIT :
Thank you for your awesome answers and sorry for my late reply.

I think that I have to separate this problem into 3 different points :

earth's variable : I can find it by reproducing the Greek calculus
Sun's variable : with the two-body problem
planets' variable : with the Gauss' method & observation

I keep you update about my advancements ;)

This is a fascinating goal, and very difficult. It would make a book on recreating the history of science.

It took many people spending lifetimes collecting data to get anywhere. However, you don't need to rediscover that the planets go around the Sun I would think.

There are some key milestones that need to be reached. For example, you can distances to planet orbits in terms of ratios of a reference distance, like the Earth Sun distance, WITHOUT knowing the actual Earth-Sun distance. You only know that Mars' orbit is x times bigger than Earth-Sun orbit.

The first calculation of Earth-Moon distance was done by the Greeks. Aristarchus or Eratosthenes perhaps. One of them found the diameter of the Earth using shadows  of monuments and in wells at noon, and the other estimated the arc of the Earth by Its shadow on the Moon, and thus the size of the Moon.

I think then it goes something like Newton measure the mass of the Earth by measuring g for a dropped object. Then using estimates of the Moon's distance and its orbital period he could determine the constant G in his law of universal gravitation. Then you can calculate the Earth-Sun distance and the rest of the planets fall into place based on their periods.

Now the other challenge is collecting useful data for determining orbits of planets. There are a couple of ways to "determine an orbit from observations". And this is a particularly useful project. Getting a mental picture of what to do can be difficult. But basically, you record the time and the position of a planet in some Earth centered coordinate system, like Right Ascension and Declination. Or relative to the know position of nearby stars.

Now, the orbits are ellipses with the Sun at one focus, so the Earth based system is not much good. You convert all your observations to Sun centered coordinates in the form of lines or vectors that point to the planet. But they point to the Planet from where the Earth was when you took the measurements. These are a set of lines in space and they don't go through the Sun or anything like that. They are just from some point in space (where the Earth was) pointing to the Planet.

C.F. Gauss (I think Laplace improved on this - not sure) showed that you only need three observations to find an orbit because only one ellipse will have points on the three or more lines and have a focus at the Sun. Two points are enough if you have other information like is it getting closer or further and which way is it going (it might be in retrograde so you can't tell).

More measurements taken further apart will increase accuracy and refine the orbit.

Also, the distance to Jupiter or Saturn could be found by watching their moon's orbits and noticing how they stray from prediction with distance due to the speed of light at opposition versus being further away.

So, you can see that the steps are not too daunting if you don't need to wait for a lunar eclipse and take a few other things as known.

I think you could do the temperatures with a small reflecting telescope. It is by detecting the infra-red radiation from the planet. A reflector is used so as not to absorb the IR. The instruments used were a bolometer that responds to heat and a galvanometer. The galvanometer is an incredible precise and sensitive instrument with no electronics.

Search on "calculate orbits from observations" and there are bunch of PDFs from good sources.


Is there a formula to calculate the angular speed at which the Sun apparently moves around the Earth? Is there a formula for how to calculate the angular rate at which the Sun apparently moves around the Earth?
I know it takes 0.25 degrees/min, but as I understand it depends on other parameters as well (such as how close a certain place to the equator, months of the year etc.), but I'm looking for a formula that makes it easy to calculate based on my location proportional to the equator and different time of the year (unless if I'm mistaken and it's constant)

If you don't have to have that precise value, then the angular speed of the sun is simply this:
$$\frac{360°}{24\text{ h}}$$
We can simplify this to SI units (we won't be using radians here, for simplicity):
$$\frac{360°}{24\text{ h}}=\frac{360°}{24\text{ h}}\times \frac{1\text{ h}}{3600\text{ s}}=\frac{1}{240}\frac{°}{s}$$

Let's suppose, that the angular speed of the sun is not constant on Earth. What would happen? The time between two same noons would be different. That would mean, that Earth would rotate with different angular rate on different points. We know that this isn't true, so the supposition is false. (But the Sun does rotate on different latitudes with different angular rates, so the angular speed of some star, looking from the Sun, depends on your location. Of course, this is theoretical and nobody has tried that yet.)
The angular speed doesn't depend on the latitude, but it depends on the current distance between the Sun and the Earth. For that you need true anomaly $\nu$, which is given by Kepler equation, which is transcedent (it can't be solved with finite steps = you have to use computer). Why does it depend on the current distance? If the Earth would not circle around the Sun and it would be stationary, the day would last for around 23 h 56 min. But it does circle around the Sun, thus the some point on the Earth has to wait a little more, to reach the culmination of the Sun, so it lasts for 24 h. The angular speed around the Sun is through the year changing, so the true day deviates from the mean day by a really tiny amount (i think that around 30 s).


GW190521's Luminosity Distance is greater than the age of the universe - how? According to this article about the gravitational wave observation, GW190521, the estimated luminosity distance is appx 5+/-2 Gigaparsecs, which is appx 16 billion light years +/- 6:
https://dcc.ligo.org/public/0165/P2000020/012/LS17910.pdf
While I understand that this is a crude measurement, is it not true that the distance cannot be greater than the appx 13.7 billion light years of the universe?  If so, should not there be another constraint on the distance? Or is it just more accurate to report the calculated luminosity distance and leave it at that in the report?

Luminosity distance is simply the distance that the object would have to be at in a static special-relativistic (or Newtonian) universe to have that luminosity. It's not the distance in the expanding general-relativistic universe that we actually inhabit.
Out to distances of roughly 1 billion light years, spacetime is roughly flat, the Hubble speed is not too large, and the luminosity distance is not too inaccurate. At substantially larger distances, it's way off.


How Can I View KIC 9832227? I've heard that KIC 9832227 is a contact binary star system that is expected to merge and explode as a "red nova" sometime in the next ~5 years or so.

http://www.sciencemag.org/news/2017/01/colliding-stars-will-light-night-sky-2022

Before that happens, I'd like to see it.  Where would I need to be / what sort of equipment would I need in order to be able to view this before it goes nova?

Right now it doesn't look like much. It has a magnitude of about 12. And appears as a single point of light. So it requires the upper end of amateur equipment to observe. If you want to measure the variation, which is only 0.2 magnitudes, you would need to take images. So you would need a proper photo-imaging system.

In other words it is a tricky amateur target. You could try using an online telescope. The star is within the range of the Bradford robotic telescope.  (Except this is offline and doesn't have an expected date to be back in service :(

In the Digital sky survey (using professional equipment), the star appears in the centre of this image. This has very high magnification, this image is about 4 arcminutes across, about 1/8 the apparent size of the moon.

It is hard to give an exact spec. There is a formula: $m=\log_{10}(D)\times 5 + 7$ which calculates the limiting magnitude $m$, from the diameter of the mirror in cm $D$. This suggests you need at least 13cm mirror. However it assumes clear and stable skies, so in practice you would need something clearly in excess of that.


What does the Sun look like from Eris at its aphelion? At its aphelion (most distant location from the Sun) the outermost-known planet Eris is about 100 times as far from the Sun as Earth. What does the Sun look like from Eris' surface? Is there still a proper day or does the Sun look more like the brightest star in a night sky? Would it hurt to look into the Sun from Eris? And would it still be dangerous to look into the Sun with average binoculars?

Light flux decreases as one over distance squared. So if Eris is 100 times further away from the Sun than the Earth is, then the amount of light that reaches Eris is 10 000 less than that on Earth.

Since a difference of five magnitudes is a factor of 100, The Sun would have an apparent magnitude -16.7 seen from Eris' aphelion, as opposed to -26.7 magnitude at Earth's 1 AU. For reference, this is still brighter than a full Moon, which is listed by Wikipedia's Apparent_magnitude; table of notable celestial objects at a magnitude of -12.9.

So you would still tell the difference between night and day, but the day would be rather dim and the Sun would appear point-like.

Solar eclipse glasses reduce the Sun's brightness by a factor of 500 000, so it might still be dangerous to look at the Sun with average binoculars (or even with the naked eye).


Would space-faring vehicles in interstellar space be in pitch darkness, show reflections from the nearest star or be bathed in light completely? This is not a science fiction question, but I was considering the visual accuracy of ships in space depcicted in Battlestar Galactica and the like. They are always strongly lit.

If they are in between stars, would they be visible at all or bathed in light equally on all sides?

Pursuing it further, at what distance from a star does an object no longer reflect it's light as the dominant light source and after that is it dark?

The brightness of the spaceship follows (almost) the inverse square law, meaning twice the distance from the star, the brightness will be a quater.
In the middle of nowhere, but within a galaxy, it would look like in a moonless, and cloudless night, far away from any artificial light source.
It wouldn't be pitchblack, but much too dark to read a newspaper. You would see the stars in the background, and the fuzzy band of the Milky Way. The starship would occult the stars; that way you would see the black silhouette of the starship.

At 630.1-fold the distance of the Earth from the Sun (about 3.65 light-days), it would look like at full moon (derived from apparent magnitudes of full moon and Sun)

But maybe the starship is illuminated artificially.


How do you convert an observed magnitude to a flux? For a series of supernovae I have the observed magnitude and I want to convert these values to flux.

I understand that I need to apply:

$m - M_r = -2.5\log_{10} \frac{F}{F_r}$

where $M_r$ is a 'reference magnitude' with corresponding known 'reference flux'.

What exactly should I use as my reference values?

The reference magnitude and flux really depends on what filter (i.e. range of wavelengths) the observation was made at (e.g., see the UBV system, or a more complete set of photometric passbands here). If you know the filter, and it is in one of the common ultraviolet, optical, or near infrared filters, then this website and this website provide useful reference fluxes. Using the common AB magnitude system then a source with a measured magnitude of zero in any band by definition has a reference flux of 3631 Jy (where 1 Jansky = 10-26 W Hz-1 m-2).


What is the significance of the velocity resolution in spectral line observations using radio interferometry? I've always struggled to conceptualize the term "velocity resolution" in spectral observations. In principle, it is simple to calculate. For example, one can take the spectral width and divide it by the number of channels. ie. A 16 MHz wide window with 4096 channels yields a 3.906 kHz / channel velocity resolution. For an observation centered on the H1 spectral line, we get 0.82 km / s / channel utilizing the speed of light and the photon energy at the 21-cm H1 line ~ 1.42 Hz.
So, why is there a "velocity" defined in the resolution term? I understand dimensionally it comes from the speed of light, but what does the final velocity signify? Does it have anything to do with how the source moves (ie. object of interest is moving) with respect to the observer? Is it better to have a larger or smaller velocity resolution or does it depend? In general, why is the value important? Thanks in advance!

Short version: velocity resolution is the smallest velocity difference you can measure between two moving objects, using a given spectrum.
More details:
As you probably know (based on your implicit use of the formula in your question), we can measure velocities by using the Doppler shift.  To do that, we need to measure a feature (an absorption or emission line) in an object's spectrum, and then determine how much it is shifted from the known "rest" wavelength or frequency.
Now imagine that there are two gas clouds along the line of sight when you take a spectrum, both emitting light in some spectral line.  (This is pretty common when observing neutral hydrogen in the plane of the Milky Way, for example.)  If they have the same velocity, then their emission spectra will lie exactly on top of each other, and you won't be able to see that there are two clouds there.  If they have slightly different velocities, then the different Doppler shifts of the two lines will give them slightly different wavelengths.  You can measure the two different velocities if the shift between them is larger than the resolution of your spectrograph, i.e. if the finite width of spectral features imposed by your optics and detector doesn't smear them together too much.
You can probably see from this that a smaller velocity resolution is better.  A resolution of 1 km/s means that you can distinguish features as close as that, while 10 km/s means that a 1 km/s difference in two features would be undetectable.  It is similar to spatial resolution in this sense.  A spatial resolution of 1 arcsec is better than a resolution of 10 arcsec, since it means you can distinguish finer details.
Note that a given velocity resolution does not mean that that is the smallest Doppler shift velocity you can measure for a single object.  That is because you can generally find the central wavelength of a given spectral feature (which is what you need for a Doppler shift measurement) very precisely, even if the feature is broad.  For example, the spectrographs used to measure radial velocities in searches for extrasolar planets (e.g. HARPS) typically have spectral resolution on the order of 1 part in 100,000, i.e. the resolution is 1/100,000 of the observing wavelength.  If you put that in the Doppler shift formula, that translates to a velocity resolution of 1/100,000 of the speed of light, or about 3 km/s.  Yet those instruments routinely measure velocities of just a few meters per second, which is 1,000 times smaller than their nominal velocity resolution.
As a final technical note, in your question you calculated the velocity width of one channel in the spectrum, but the velocity resolution would be twice as large as this, because of the need to Nyquist sample the spectrum.  One way to think about that would be to imagine a series of spectral features that were separated by exactly the velocity width of one channel.  If that were the case, then every channel would have a feature, and your spectrum would just appear flat. (Even if you just had two features in consecutive channels, they would blend to appear as one.)  But if the features were separated by twice that velocity, then only every other channel would have a feature and you could see that there were distinct features there.  So the spectral resolution is always at least twice your channel width, but could be limited by other parts of your instrumentation to be larger than that.


A day in earth, a thousand years somewhere else By followig the laws of physics, is it possible, that in a place far far away that one day (at that place), is counted as 1000 years in the earth.

I am familiar with black holes having impact on time. But would it be possible without black holes and such,

I think i've heard that the gravity has impact with the time too?

Assuming you mean a day in terms of an Earth day in both cases, the phenomenon I believe you're referring to is called time dilation, which affects the speed at which time is experienced by one body relative to another. This rate is affected by velocity and gravity, each of which can cause the bending of spacetime that results in the discrepancy between objects. Both gravity and velocity slow one's time as they increase.

The situation you describe is probably not naturally occurring, however is not outside the realm of reality.  As referenced by Wiki from Calder's book Magic Universe, Calder claims that an acceleration of a constant 1G (what we feel on Earth all the time) would result in the affects of time dilation allowing you to traverse the entirety of the known universe in a single human lifetime (for the traveler). Conversely, during your trip, the rest of the universe would be aging 'normally'. There are plenty of technical problems with the feasibility of achieving this, but is conceptually possible.

I'm not particularly qualified to explain the equations behind it, so I'll leave that to our more capable members. Coincidentally, time dilation is the excuse I give when asked why I walk so quickly.


Is an elliptical orbit around a black hole, or an orbit consisting of three bodies (including the black hole) possible? If this is possible, would it then lead to an object undergoing varying time dilation (predictable variation with elliptical and random variation with three body)?

Elliptical Orbit

Yes, an elliptical orbit around a black hole is possible.  Provided the orbit doesn't cross too near the event horizon, the orbital mechanics are the same as orbiting any massive object.  Gravitational time dilation is a function of distance only, so the time dilation experienced at any point in the orbit should be easily calculated and presumably predicted ahead of time or calculated for past times.

Three Body Orbit

When you mention an orbit consisting of three bodies, do you mean the black hole, another massive object, and a much smaller object?  Or do you mean three massive bodies?

If the orbit consists of the black hole, another massive object, and a smaller body, the system should be predictable as a restricted three body system.  This is made even easier if the smaller body finds its way to a Lagrange point, or resides in a stable repeating orbit.  There are limits to the predictability of the restricted three body cases, which differ based on requisite constraints (e.g. circular vs. elliptical orbit for the smaller body).  If the constraints are not met, or near their limits (not hard limits), the predictability of the orbit will suffer and thus the predictability of the time dilation will suffer as well.  Constraints for the elliptical restricted three body problem include the mass of the small body being "negligible" relative to the two larger bodies, and the ability to approximate the co-orbit of the larger bodies as Keplerian between point masses.  Thus if the mass of the smaller body is some appreciable fraction of either larger body, or if the larger bodies orbit very closely to one another, this may not hold.  The circular restricted three body problem adds the constraint that the co-orbit between the larger bodies must be circular, so if it is near-circular that condition may or may not be said to be met.  Whether constraints are met would be most easily judged by setting a time over which predictability of the system may be maintained.  So if the goal is to be able to predict the system for 1000 years, and the smaller mass is small enough that the system is predictable under the restricted three body problem, the system can be said to meet the constraints for those purposes.
If the orbit consists of the black hole and two other massive bodies, the system will not be perfectly predictable (this may not be totally true, apparently solutions exist in the form of convergent power series though they may not be practical due to their size).  If this is the case, time dilation will not be predictable, but will be calculable for the present and past (provided there is position data from the past).


What factors influence a star's temperature and density? Assuming a relatively even proportion of mass and radius, a 0.25 solar mass and radius star would have a density of 22.5003 g/cm³, or about 16 times our Sun's density.
However, there is obviously a range of densities that this sized red dwarf could have.  One of the primary factors is its temperature, with cooler meaning a higher density.
My question is three-pronged: what would increase or decrease this sized star's temperature, how would this impact its density, and what factors other than temperature would increase or decrease this star's density?

For main sequence stars there is not much choice. If you pick mass, then all the rest is determined; any variation you see is due to different chemistry (thus fractions of different elements).
For the main sequence one generally assumes that $L \propto M^{3.5}$ and $R \propto M^{2/3}$ (and similarily the density is given for a given mass). The Hertzsprung-Russel diagramme is expression of this.
Any stars not on the main sequence, be that the white dwarfs or any giants, are there due to their peculiar chemistry. White dwarfs are degenerate stellar remanents, slowly cooling. The giants have evolved away from main sequence. They have some other fusion process(es) than hydrogen burning in their core so that their luminosity is much bigger than hydrogen-burning stars; at least they have a hydrogen-depeleted and helium-enriched core and started hydrogen-shell burning.


Why is declination positive in the northern hemisphere? Simple question -- why is declination positive in the northern hemisphere?

Seems like it should've been called "inclination" since the word originated north of the equator.

After quite an extensive search of dictionaries and books on astronomical nomenclature I came across this article that has a comprehensive review of celestial coordinate systems. And upon reading page 84 (8/14) I came across the answer to your question.

It would appear that the reason declination is used, as opposed to inclination, is that declination was first interpreted as a "deviation from standard" not an inclination of angular distance from a given point as we know it today.

This "deviation from standard" was first used to describe the motion of the sun  along the ecliptic that "deviated" from the then widely accepted notion of all heavenly bodies following the line of the equator. It goes into more detail in the article but it would appear that when translated from Greek to Latin and then Latin to Arabic, the precise meaning of the word changed and evolved so that now the term declination refers to the angular distance of any celestial body.

Source:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1942PASP...54...77W&db_key=AST


How is the difference of bolometric magnitudes not dependent on the stars' radii? The difference of 2 bolometric magnitudes is given by:
$$M_{bol, ★} - M_{bol, ☉} = -2.5 \cdot \log \left( \frac{L_★}{L_☉} \right)$$
But Pogson's equation is:
$$M_{bol, ★} - M_{bol, ☉} = -2.5 \cdot \log \left( \frac{F_★}{F_☉} \right)$$
where $F_★=\frac{L_★}{4\pi R^2}$, so how come the first equation isn't dependent on the radius?

The R in that equation is the distance from the star to observer, not the star radius. The light emitted from the star is distributed uniformly on a sphere of radius R, and when the light arrives to the Earth, that sphere will have a radius equal to the distance Earth-star.
Therefore, the second relation for the two fluxes is about the apparent magnitudes (which describe the brightness of an astronomical object observed from Earth),
$$m-m_\odot = -2.5 \log F/F_\odot$$
The first relation is alright. The absolute magnitudes are related with the luminosity of the star (the overall energy flux emitted by the star) and they are not dependent on the distance to the observer.