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Stellarium 0.10.4: planet orbits change over time?
If I draw the orbit of Mars in stellarium 0.10.4 for 23 Sep 2015, it passes south of Chi Leonis:

If I advance the date to 17 Oct 2015, however, the orbit of Mars correctly goes right through Chi Leonis (Mars will occult Chi Leonis on 18 Oct 2015):

Is this a known bug in 0.10.4 that's been fixed? (I unfortunately can't upgrade)

That's not the orbit, it's the projection of Mars' location on our celestial sphere. Since our orbit and Mars' are not in the same plane, some wobbling like that is expected.

Probably this is due to the fact that orbits of the Earth and Mars are inclined to the ecliptic differently so our viewing angle of the orbit of Mars changes slightly over the year.  The program is probably showing the orbit of Mars projected onto the Celestial Plane as seen by us on the one specific date. Note, this is different from the ephemeris orbital data which gives the apparent position as seen by us at each date/time.   Just a guess though.

Edit (I looked, unfortunately most of the sources are astrology, not astronomy), but Mars last went in Retrograde March 2014 and it's not due to go again till May 2016 (every 2 years, 2 months).    My guess below was incorrect (though it's kind of cool, I'll leave the picture up).
Mars does have a tilted orbit compared to earth and that could be the reason.  That and the angle from the earth changes too.
Source

Mars, from our point of view, slows down, temporarily reverses direction then returns to it's original direction and continues on.   The entire "S" motion takes maybe 3-4 months and your 2 photos are 24 days apart.  (but this won't happen till May 2016)

Source of photo and explanation here:


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If I were to point into the sky, how many galaxies would be in this line?
If I point in the sky and draw a line from the tip of my finger to the end of the universe, how many galaxies would this line cross? I had a discussion with my dad and he said due to the universe being so empty the line wouldn't cross any solid matter. I said that the universe is so enormous in size that there would be many galaxies and stars crossing the line.
Is this calculable? If yes which one is correct?

Take a look at the "Deep field" image from Hubble: http://www.spacetelescope.org/science/deep_fields/
You can see in this tiny part of the sky, hundreds of galaxies, but between there is lots of dark sky. Most lines of sight don't end on a galaxy. I'd say dad was right. I would estimate, from the deep field, that you have about a 1/10 chance of pointing at a galaxy.
Even if your finger does point at a galaxy, the distances between stars are vast. Think of two footballs separated by the distance between Europe and China as a model for the distances between stars. A finger pointing at a galaxy would be unlikely to point exactly at one of the stars.
This is probably a Good Thing. Olber's paradox supposed that if each line of sight ended on a star, the night sky would be a uniform blaze of light.

How thick is the line? An extended finger is about 30 arcminutes across, the size of the moon, but a line is a line, 0 arcminutes.
The Hubble Deepfield, as mentioned by Kilfiger, is only 2.5 arcminutes across, 1/12th of a finger-width, yet contains hundreds of galaxies.
If you want to be a stickler for true lines, the image does contain dark space too. At a guess, there's not easily accessible software for counting such things, it looks like 70-80% blackness (ignoring obvious stars), so 1 chance in 5 your line is pointing at a galaxy.

On average, you will point at one galaxy.
The argument goes as follows:
Looking at the spectrum of quasars, which are effectively point sources, and which lie at distances of the order of the size of the observable Universe, it turns out that typically, roughly one damped Lyman $\alpha$ system (DLA) is detected. DLAs are huge reservoirs of neutral gas, which are early stages of galaxy formation.
At least in a handful of cases, the galaxy counterpart of the DLA has been confirmed. These observation are very challenging, since they are a faint sources at very small projected distances from the very bright quasar. Here's the setup:
So, accepting a DLA as a galaxy, I'd say that you will typically point at one galaxy.

Actually we can't have an specific answer to that. Because we don't have enough knowledge about the size of the universe and the density of galaxies in other parts of the universe.
You both might be true.
What happens if we are in a low density of matters but far from us we have a high density of galaxies and stars but we can't see them because of being too much far or being hidden from us behind something like Dark Matter or anything else.
Human are just guessing and are not too sure about anything.

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Online wide-field star cameras?
Where can I find a large list of online star cameras with wide fields of view?
http://www.itelescope.net/all-sky-cameras is one example, but the images are fairly low resolution and there aren't many of them.
Is there a better site or sites to see wide-field star images?
I realize there are several online telescopes, but they all have a small limited field of view (which makes sense), so you can only see small portions of the sky at once, not the entire sky as above.

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two black holes orbiting each other within their event horizon
Is there any argument against this situation: two black holes, one within the event horizon of the other, and the system is stable.
It is interesting to me because that if this works, we could have the system's event horizon not in the shape of a perfect sphere. (The event horizon would probably be moving with the orbit)

No clue, but boy that would be interesting to think about.  Wouldn't any assymetry in the size of the two black holes cause the other to consume the smaller one?

A very interesting question, but shouldn't the size of black holes be taken into account?

Inside the event horizon is a weird place to be. Inside a black-hole space-time is flowing towards the singularity faster than the speed of light (as measured from outside the black-hole. The effect is that any thing inside a black hole will, in a finite (and usually short) amount of time end up at the singularity.
Models of merging black holes exist such as video from nasa. It requires super-computers to solve Einstein's equations numerically.
As I noted in a comment, no stable orbit can exist within 1.5 times the radius of the event horizon. The orbital velocity is c at that distance (1.5 Schwarzschild); it's also called the "photon sphere". Within that radius all orbits are unstable. And the radiation of gravitation waves means that as two black holes orbit each other, they will lose energy, and so their orbits will decay. As the black holes approach, the event horizons are distorted, and merge into an droplet shape

two black holes, one within the event horizon of the other

The premise is wrong. There can be no such thing. What you and me, external observers, understand to be a "black hole" is actually the whole volume inside the event horizon. It is a chunk of spacetime causally disconnected from where we are now.
When two black holes get close enough together, the event horizons bulge towards each other. When they touch and become a single event horizon, the merger process has begun in force (sometimes this is called "collision"). The only possible outcome at that point is that the two BHs will merge and become a single, bigger black hole.
See this video for an example:
https://www.youtube.com/watch?v=p647WrQd684
(note: the two red spheres at the end of the simulation have no physical reality, please ignore them)

It's a technical point, but your discussion of "event horizons" here isn't precise.  The event horizon is a global property of the spacetime.  You can choose a sequence of spacelike surfaces (as was done in the movie) where the intersection of the event horizon on those slices will appear to merge when taken in sequence, but that's not the same.

The event horizon is a global property of the spacetime.  There is only one event horizon in this case.  For there to be "two" black holes, we usually mean that their apparent horizons are separate at some time, and the ultimately merge.
This technicality aside, there's a ton of research in this are because black hole mergers are a likely source of gravitational radiation, and there are multiple experiments around the world attempting to directly detect such radiation, including LIGO in the US.  There are also several space-based experiments proposed but not currently funded.
A lot of the theoretical work in this are uses large numerical simulations on supercomputers to compute important quantities, including event and apparent horizons.  Due to the emission of gravitational radiation, the black holes that you describe will slowly in-spiral and will eventual merge completely. (The gravitational radiation carries away energy, so their orbital radius will shrink.) Eventually they will settle (asymptotically) into either a Kerr or a Schwarzschild solution.

It is an unlikely scenario.
The event horizon is the point where orbital velocity exceeds C (the speed of light).
Matter is incapable of equalling or exceeding C within its frame. Therefore, a stable orbit is implausible, since the second black hole would need to be moving faster than C. Thus, the orbit is inherently unstable, and decaying.
We should be able to detect a velocity shift in the orbits of objects outside the event horizon in the accretion disk, and the event horizon should also have a bulge, but depending upon the current spot in orbit, that may become undetectably small as the smaller black hole spirals in closer and closer.
Note that I'm not up on all the details of frame dragging and other black hole related weirdnesses, so this is only a first order examination of the principles, hence implausible rather than impossible.

The orbital velocity is c at 1.5 Schwarzschild, its called the "photon sphere". Within that radius all orbits are unstable.

There's a lot wrong with this answer, and there's a lot of speculation. The science of BH merger is well understood - no need to speculate.

@FlorinAndrei And the progenitor of much of it now thinks it mostly wrong... Hawking has changed his views.


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How is the Lithium Depletion Boundary used to determine the age of a stellar cluster?
According to my understanding of Soderblom et al. (2014), lithium ages of stars are determined as follows:

Determine lithium abundance from equivalent width measurement of Li$_{\mathrm{I}}$ transitions at 6708 angstroms for all stars in a coeval group.
Plot all stars in group on color-magnitude diagram, with each star labeled as Li-rich or Li-poor.
On plot, look for the boundary where the stellar population transitions from depleted to undepleted. That is the Lithium Depletion Boundary (LDB).
Use the luminosity of the LDB and the age-LDB luminosity relationship to determine the age of the stars in the group.

What is the age-LDB luminosity relationship, and what are its physical origins?

This was very helpful in explaining how the physics of the stellar interior explains the age-LDB luminosity relationship. I see now that the mass of the star is the essential factor, which makes intuitive sense. Thank you!

When low mass stars are very young, they are termed pre main sequence (PMS) stars. These PMS stars have larger radii than main sequence stars of the same mass, and energy transport in their interiors occurs primarily through convection. The convection ensures that the star is thoroughly mixed and chemically uniform.
As the PMS star radiates away its gravitational potential energy, it contracts. The virial theorem tells us that as it does so, its interior becomes hotter. Roughly speaking, the core temperature is proportional to $M/R$, where $M$ is the mass and $R$ the radius.
Nuclear fusion of hydrogen will not commence until the core temperature reaches more than 10 million K, however there are other fusion reactions that become possible at lower temperatures - namely deuterium burning at around 1 million K and then lithium burning at around 3 million K. The latter reaction is not energetically important in the star's life because there is not much lithium in the star to begin with (about 1 part in a billion), however this lithium can be observed in the photosphere of the PMS star (via the 670.8 nm Li I resonant absorption line) and the convective mixing means whatever we see at the photosphere also represents the abundance at the core.
The Li burning reaction is extremely temperature dependent (like $T^{20}$ or thereabouts), so it turns on like a switch once the core reaches the appropriate temperature (e.g. see Bildsten et al. 1997). The time it takes for a PMS star to reach this core temperature basically depends on its mass. More massive, and hence more luminous stars contract faster and reach the Li burning temperature quicker. Once they do so then the Li in the star is rapidly and thoroughly consumed by fusion. The relationship between the age at Li destruction and the luminosity of the PMS star at that time is the age-LDB luminosity relationship that you refer to.
The result is that if you look at a bunch of stars in a cluster (assuming they all have the same age), then the more massive, more luminous PMS stars will have destroyed their Li, whilst the lower mass, lower luminosity stars will still contain their original Li content. The luminosity at the reasonably sharp transition between these two regimes is known as the LDB.
LDB ages are arguably the most accurate way to find the ages of stars in clusters. All age determinations are to some extent dependent on what physical ingredients are in stellar evolution models, but the sensitivity of the age-LDB luminosity relationship to various uncertainties is quite weak (e.g. Burke et al. 2004) - we basically understand the physics of a contracting, fully convective ball of gas quite well. The LDB ages can also be precise, because the very sharp turn-on of Li burning and its rapidity should lead to a sharp transition between stars with low luminosities that have Li and stars with only slightly higher luminosities that don't.
Interestingly, in the last couple of years it has become apparent to us (and others) that there are some model ingredients that are not completely understood, namely the effects of dynamo generated magnetic fields and dark starspots. Both of these may lead to the suppression of heat transport, either throughout the star or just at the surface, slowing the PMS contraction so that at a given age, the PMS star has a cooler core temperature. This might delay the onset of Li burning and mean that the currently determined LDB ages are underestimates by 10-20% or so (e.g. Jackson & Jeffries 2014; Somers & Pinsonneault 2015 ).


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Planet-timer "device" from 1970s or 1980s possibly by Edmund Scientific
Many years ago (early 80s or late 70s), I bought a cardboard "device" which I think was called "Planet Timer" and I think was sold by Edmund Scientific. Does anyone have any information on this device? Google was unhelpful.
The device had concentric revolving circles around a fixed Sun at the center. By revolving the circles to the current time (via tables provided with the device), you could presumably see what constellation each planet was in. You could also use it to approximate planetary conjunctions.
The circles were equal distance from each other, so I'm pretty sure the device couldn't actually have worked, but I'm curious to see if I can find out more about it.
Note: this is not a "planisphere" device, which did work, and for which I can still find plenty of information.

I ended up finding this on page 14 of the Edmund Scientific Holiday Sale catalog, with sale ending on December 31, 1983:

except that it's "Planetimer" (and not "Planet-timer"), catalog number H9454.
A 1200dpi image somewhat shows how it's supposed to work:

My original question was going to be: could this device have worked?
I would also appreciate any other information on this product (including a better scan, ideally of the product itself, if anyone still has one).
For reference, here is the page on which the ad appears and the cover of the catalog in which I found it:


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How to determine atomic number density of an element in a star based on equivalent width measurements
Given an equivalent width measurement $W$ of a spectral line of element $X$ and the effective temperature $T_{eff}$ of a star, how can you determine the atomic number density of $X$ in that star?
According to Warner 1965, for the 6707.8 Lithium line we can apply the approximation
$$\log{\frac{N^*(X)}{N^{\odot}(X)}} = \log{\frac{W^*(X)}{W^{\odot}(X)}} - 4.93 (\gamma -1)$$
where $\gamma = \frac{T^{\odot}}{T^*}$ and $\log{W^{\odot}(X)}$ must be measured. The same paper says that $\log{W^{\odot}}(Li)=-6.15$ (my goal is to solve for the abundance of Lithium given a measured W of the 6707.8 $\overset{\circ}A$ line.
Under what regime is this approximation valid? Also, I am unable to solve this equation because I don't know what $N^{\odot}(Li)$ is. How can I determine a reputable value for $N^{\odot}(Li)$?

The measurement of a chemical abundance is not a question of using a simple equation.
The simplest it gets is using a "curve of growth", which relates equivalent width to abundance and assumes you already know the temperature of the star and its surface gravity.
For the Li 6708A line, the relevant tables, that can be interpolated, are found in Soderblom et al. (1993). http://adsabs.harvard.edu/abs/1993AJ....106.1059SS . These are LTE curves of growth based on Atlas atmospheric models.
As an aside, the equivalent width of this line in the Sun is 3 mA. The formula you quote looks like some sort of linear approximation, but this resonance line rapidly enters the saturated (non-linear) portion of the curve of growth for equivalent widths of 50 mA or more. As the equation is couched in terms of ratios, then the units you use are entirely up to you.

Thank you for this. I am wondering under what regime the linear approximation I wrote above applies. While I am familiar with the curve of growth abundance analysis, I am currently measuring the equivalent width in the spectra of sunlight-reflecting asteroids, so was thinking such analysis would not be necessary. Is the approximation valid for this?

@user44816 Have a look at the curve of growth I referenced. If you're looking at sunlight reflected from asteroids why do you think the equivalent width of the Li line will be different from the solar spectrum?? By definition, such a spectrum is used as a proxy for the Sun and calibration standard in most chemical abundance analyses.

Thanks, will do. I did not expect the equivalent width of the Li line to be different from the solar spectrum, however my code is returning an value that is not 3 mA, which is why I was curious. Thanks again, I will definitely check out this paper.

@user44816 Your code? You mean your spectrum? 3 mA is very hard to measure unless you have extremely high S/N and resolution.


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How Would a Neutron Star Actually Appear?
Having seen many pictures produced by artists of neutron stars and planets that orbit some of them, I was wondering how a pulsar would appear to a human being, in visible light (assuming the intense radiation etc. doesn't kill us in the process).
As I understand, the pulsar's beam is projected from the star's magnetic poles rather than rotational poles, which are not necessarily in line with each other. Given that pulsars rotate extremely quickly and the beam could be visible across vast distances - such as if it were shining through the pulsar's nebula - would it appear as a straight line, curved line or perhaps a cone? This is assuming the beam can be seen in visible light.
Given the incredible density of neutron stars and their small physical sizes, would the night sky be visibly distorted to the point where (for example) just after sunset on a hypothetical planet, one could possibly observe other planets near or behind the star that would otherwise be blocked by it?
Given their small surface areas, would a neutron star still appear as luminous as say, the Sun, at a similar distance? How close would you have to get to a neutron star for its apparent magnitude to match the Sun's from Earth?

Not related to your question, but what things would look like on the surface of a Neutron Star is much more interesting.  Because of the way light bends, the sky when standing on the surface of a Neutron Star would be squeezed into a tiny circle and the planet would visibly appear to rise up around you, taking up most of what you can see.    http://apod.nasa.gov/htmltest/gifcity/nslens_ul.html

@userLTK It's a fascinating link, and a negatively curved horizon would be amazing to see to say the least!

Does anyone know if such "ultracompact" neutron stars actually form?

Your question is too general, you need to get to specific examples.
First, very few neutron stars are pulsars. Pulsars are either a brief phase during a pulsar's spin-down at the start of a neutron star's life, or they are the product of the spin-up of a neutron star in a binary system. Most neutron stars fall in neither of these categories.
A standard neutron star will look like any other star at a similar temperature. Most of them will be very hot indeed - 100,000 K or more, though the cooling histories of neutron stars are still uncertain and depend on some exotic physics. Such an object is "white hot" - it emits black body radiation at all frequencies visible to the eye (as well as lots more at UV wavelengths).
How close would you have to get for it's apparent luminosity/magnitude to match the Sun? Well that depends on the size and temperature of the neutron star. Most are thought to have a diameter of 20 km. The way you would do the calculation is equate the blackbody radiative flux per unit area at a given distance to the solar radiation constant of about 1300 W per square metre. However, there are two wrinkles for a neutron star: First, the radiation is gravitationally redshifted, so the temperature we measure is lower than the temperature at the surface. Second, General Relativity tells us that we can see more than just a hemisphere of the neutron star - i.e. we can see around the back - and this increases the flux we observe. These are roughly factor of two effects, so just to get an order of magnitude estimate, ignore GR and assume a 10 km radius NS with $T=10^{5}$ K.
Using Stefan's law for a blackbody, then at a distance $d$, we have that
$$\frac{4 \pi r^2}{4\pi d^2} \sigma T^4 = 1300\ W\ m^{-2},$$
where $\sigma$ is the Stefan-Boltzmann constant.
For $r=10$ km, then $d=7 \times 10^{8}$ m, which is coincidentally about a solar radius. Of course this distance depends on the square of the temperature, so a younger NS with $T=10^6$ K, then $d \sim 1$ au.
These are the distances where the total flux at all wavelengths would be similar to that from the Sun. To do the calculation just for the visible range we need to account for the bolometric correction, which converts a visual magnitude to a bolometric magnitude. The bolometric correction for the Sun is $\sim 0$, whereas the bolometric correction for a very hot star could be -5 mag. This means that only 1% as much flux from the hot neutron star emerges in the visible band compared with sunlight. This means that the distances calculated above, if we require the visual brightness of the neutron star be similar to the Sun, must be reduced by a factor of 10.
To turn to pulsars. Note that the pulsed radiation does have an optical component and pulsed optical radiation has been seen from a number of pulsars. Optical synchrotron emission would just appear to be a periodic, intense brightening of the pulsar, as the beam sweeps across the line of sight. If you were not in the line of sight, then you would not see the pulsed optical emission. If you could observe the beam passing through nebulosity or some other medium around the pulsar then yes there may well be some effects you could see in terms of ionisation or scattered light coming from along the beam path.
Lastly, the gravitational lensing effect. Yes, this should be strong close to a neutron star. The deflection angle (in radians) is given by
$$ \alpha  = \frac{4GM}{c^2 b},$$
where $b$ is how close the light passes to the neutron star and $M$ is the neutron star mass. Expressing $b$ in terms of the 10km radius of the neutron star:
$$ \alpha \simeq 0.83 \left(\frac{M}{1.4M_{\odot}}\right) \left(\frac{b}{10 km}\right)^{-1},$$
where strictly speaking this formula is only valid for $\alpha \ll 1$.
So consider a planet directly behind the neutron star at a distance of 1 au. The light from this would only need to be bent through an angle of $\sim 2 \times 10\ km/1\ au \sim 10^{-7}$ radians in order to be seen from a planet diametrically opposite at a distance of 1 au. So this is easily possible. However, the image would likely be highly distorted, especially if the neutron star was spinning. It would not look dissimilar to this simulated black hole image, but with a bright neutron star in the middle rather than a black disc.

A very interesting answer. I had imagined a neutron star's luminosity would be higher than what would be calculated due to light emitted from its 'far side' being bent towards an observer, but I did not realise it would also be redshifted in such a way as to make the star appear cooler.

Does the lensing increase the observed flux in this case? Thinking in terms of light rays emitted from the surface, some emitted nonradially from the back hemisphere will be seen, but this also means that some emitted from the front hemisphere that "would have been" observed won't be, because they will bend to miss the observer. ... For a hypothetical nonrotating neutron star, spherical symmetry implies only the redshift matters due to energy conservation. For a more realistic one, it would depend on relative orientation.

@StanLiou that does sound correct. It can't be brighter in all directions.

The statement that a Pulsar will look like a black body with a high temperature does is not supported by the evidence. Optical measurements of the Crab Pulsar show a flat spectrum see this. This is a result of the optical emission being from synchrotron radiation rather than the hot surface.
The recent Gaia DR2 results include the Crab Pulsar as DR23403818172572314624 this has a BP-RP colour of 1.0494 which equates to a temperature of around 5,100 K from the DR2 HR diagram. This is very similar to the temperature shown in the DR2 data. This needs to be used with caution as the calibration is for a star with a 'Black Body' atmosphere rather than an 'atmosphere' radiating due to synchrotron Radiation. See this for the full DR2 data.
We don't know how large the radiating 'atmosphere' is but a rough idea could be calculated from the DR2 data in the link above. However the parallax (distance) uncertainty is quite large so would need a better distance measurement.

I can kind of give an answer, but I welcome correction.

I was wondering how a pulsar would appear to a human being, in visible
  light

It wouldn't look like much in the visible light spectrum unless there was a significant nebula, then we might see the effect of the pulsar on the nebula, but not the pulsar itself.  X-rays and radio waves aren't visible, and if the pulsar wasn't directed at us, we wouldn't see it pass through empty space.
Neutron Stars are generally too hot for us to see.  If one was to cool down significantly, to maybe 10 or 20 thousand degrees on the surface, then it might glow visibly blue and look like the brightest star in the sky, still just a point in the sky, but the brighest point in the sky at 1 AU.
But mostly they're too hot to glow in visible light.
What you might see from 1 AU from a Neutron Star could be the accretion disk.   Matter that falls into a Neutron Star gets very hot and the energy if impact is far greater than the energy of fission, so as matter gets close to the Neutron star and spirals in, you're probably talking x-rays and gamma rays, but you might see a visibly glowing accretion disk at some distance out, perhaps in a gradually decaying orbit.   In effect, what you could see would depend on what's around the Neutron star than it would depend on the star itself.

As I understand, the pulsar's beam is projected from the star's
  magnetic poles rather than rotational poles, which are not necessarily
  in line with each other. Given that pulsars rotate extremely quickly
  and the beam could be visible across vast distances - such as if it
  were shining through the pulsar's nebula - would it appear as a
  straight line, curved line or perhaps a cone

The problem here is, you can't see the beam.   You see light as it's pointed towards you, you can't see a light beam in space (even if it's visible light).
You can see a beam not pointed at you in the atmosphere because of reflection off dust and water molecules in the air.
(see little picture)

In space, matter is far more spread out.   It's true that a pulsar can light up part of a nebula, though the nebula may also glow on it's own anyway (I'm not 100% sure on that), but a Nebula is very large and very spread out.   To see it from the naked eye, I don't think you'd see much other than perhaps a large glow.
If you could see a pulsar beam, it takes light 8 minutes to for light to travel 1 AU, and a pulsar can rotate hundreds of times, perhaps thousands of times in 8 minutes, so if you could actually see the beam, it would be enormously curved, like a spiral.  The light itself would travel in a straight line but since the source of the light was rapidly rotating it would appear like this (picture below), if there was sufficient material for the light to reflect off of (which there probably wouldn't be, not within 1 AU).

In reality, it would look nothing like that, but if you could see the the beam, that's what it would look like.   What that spiral looks like from a single point is a pulsar, off, on, off, on, off, on, etc.
Also, the light never travels in a spiral, it travels in a direct line away from the Pulsar, but like the water spiral here, which falls down in a straight line, but it looks like it falls in a spiral (if that makes sense).

Given the incredible density of neutron stars and their small physical
  sizes, would the night sky be visibly distorted to the point where
  (for example) just after sunset on a hypothetical planet, one could
  possibly observe other planets near or behind the star that would
  otherwise be blocked by it?

Well, for starters, without a sun there, planets would probably not be visible.   If the Neutron Star glowed brightly due to a hot accretion disk you couldn't see anything behind it cause the brightness of it would make seeing light bent around it pale by comparison.
Now if the Neutron star was dark, to our eyes, then we could see gravity lensing around it, but stars, not planets cause planets would be dark.  (The moon would be very dark too, visible more by what it blocks than what it shines).   The lensing would be quite small however.   Visible lensing would only be a few times the diameter of the Neutron star, maybe 100 miles across, which, 93 million miles away is really tiny.  You might see some odd warping of a star here or there when properly lined up, but to see any interesting visible lensing you'd need a pretty powerful telescope.

Given their small surface areas, would a neutron star still appear as
  luminous as say, the Sun, at a similar distance? How close would you
  have to get to a neutron star for its apparent magnitude to match the
  Sun's from Earth?

Kind of touched on this above.   The Neutron Star can give off a lot of energy in it's pulsar beam, but it's mostly x-rays, not visible light.  How bright it is would depend on how much material is falling into it at the time, so there's no right answer to how close the Earth would need to be to have equal brightness.   It's a different kind of brightness too, mostly not visible light.   But there's no way to answer that question cause it depends on too many things.
When a Neutron star is just formed (which usually happens after a supernova so there's enormous energy released), but when the star just forms, it's maybe 12-15 miles in diameter but it's surface temperature can be (guessing) perhaps a billion degrees, though it cools very quickly.    A very young Neutron Star might emit more energy to our sun, though much of it would be in Neutrinos that would largely pass through the Earth.   But that level of energy output wouldn't last long.   It would cool down to about a million degrees within a few years.    Source.

Mostly wrong. Just picking up on a major point. A blackbody at a hot temperature radiates more energy at all wavelengths than a cooler object with the same emitting area. As they cool down, neutron stars become less visible.

Visible to x-ray telescopes or visible to the Human eye?   The question was about visible to the Human eye.

At all wavelengths.

If we assume that the pulsar's surface is like that of other neutron stars, unless the beam is pointed at you, it will look like other neutron stars. RX J1856.5-3754 (https://en.wikipedia.org/wiki/RX_J1856.5-3754) is one of very few neutron stars we can see at optical wavelengths.  It has a visual magnitude of 25.6 at ≈61 parsecs (the Sun's apparent visual magnitude at that distance would be about 8.75).
   Turning the cranks I get an absolute visual magnitude MV of 21.67 and a  visual luminosity of ≈.00000018.  Taking the square root, I'd need to be about .00043 AU away, or about a tenth the diameter of the Sun for it to be as bright as the Sun is from Earth, visually. At only 14 km or so in diameter, it would very small, about 4.7% the apparent diameter of the Sun--not much more than a point.
   But as noted above, the actual, bolometric, luminosity of the neutron star would be much, much higher.  A person looking at it (unprotected) from that distance it would be blinded and fried in short order.  One might also be far enough down the gravity well at that distance that the relativistic effects that dim the star would be less and the star would appear even brighter.  And one might note some tidal effects as well.  This situation calls for the "General Products Hull" Larry Niven used for his story, "Neutron Star!"

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why is it always planets orbiting stars?
In our solar system, there are 8 planets orbiting a star, the Sun.
And I understand that there are about 500 confirmed solar systems out there.
But why is it always planets orbiting stars? Why can't it be several stars orbiting a planet, or a star orbiting a star? Why is a star by definition stationary as opposed to planets which are moving.
Of course, in that case, it wouldn't make much sense to call it a solar system, but still.

We have Galaxies, stellar clusters, 50% of all stars are double stellar systems. There are many asteroids and comets orbiting our star. Stars are not stationary, but move around the galaxy and wobble when their planets tug on them. That is partially how we detect them. So no, your premise is wrong. It is absolutely not only planets that orbit stars.

"A solar system" isn't the correct term. You mean "a planetary system". The Solar System is the name of our planetary system.

"Why is a star by definition stationary as opposed to planets which are moving." This just isn't true. Both star and planets orbit a point in the system known as the barycenter (a.k.a. the center of mass of the system).
Because stars are much more massive, this barycenter is much closer to the center of the star than it is to the planets. Hence it appears to the casual observer that the less massive object orbits around the more massive object. In fact both objects orbit the barycenter. Like so:

The following diagram (seen on the wikipedia page above) shows the motion of the solar system barycenter with respect to the center of the Sun. Note that the barycenter is close to the center of the Sun, but often spends time outside the visible surface of the Sun. It follows this complicated path, mainly because of Jupiter but all the other planets make smaller contributions too.

The gravitational pull from these stars. There is so much matter compacted into these stars that the gravitational pull is strong enough to hold all these planets together.

Is it instrinsically impossible for a very big star to have a very small star circle it?

It is possible that one star could devour another, and the two mesh..

Planets often circle stars because they have a very strong gravitational pull. Infact, our moon orbits the earth and we can look at a solar system in the same way. The earth has enough matter inside it to keep the moon in orbit, and the sun has enough matter inside it to create a gravitational pull strong enough to keep a entire solar system in its orbit.
You can't have several stars circling a planet, the pull of a single planet wouldn't be strong enough (the planet's center would be so dense nuclear reactions would start occurring and it would be a star anyway) . You can have binary star systems though, which are where 2 stars orbit around their common barycenter.
So basically, things orbit other things with a stronger gravitational pull.


text:
Has someone ever captured footage of the collision of a meteoroid and a meteoroid?
For a video game I'm looking for footage of a collision of meteoroids or maybe asteroids, not planets. Even though I have searched for a while I couldn't find anything. So has anyone ever captured footage?

Deep Impact, the NASA comet probe, not the movie, may be as close as you can get to what you want now.
Here's a video of the action when 370 kg of copper hits a comet at 10.2 km/s.
There may be better video out there, but the movie trailers really mess up the search.

I'm fairly certain it has already happened a lot. But the closest footage I found from what you are looking for is the aftermath of a collision, right here. However, I don't think we'll ever get to record an actual collision. Space is big, you know. Chances of a collision are quite rare. But you can find a lot of simulations on Youtube.

It's a Needle in a Haysta... In The Pacific Ocean
Sometimes galaxies' gravities catch each other and they slowly merge. Even with billions of stars and planets merging with billions of other stars and planets, its theorized that almost no collisions take place. Point being that although it does happen:

Space is mostly "empty" and statistical odds of collision are close to nil
Unlike planets, astroids often have complex trajectories that aren't shared with other solid bodies (two asteroids very rarely even have a similar route, like planets and moons do)
Humanity's ability to monitor the night sky is limited. We see very little of what happens out there.

Sorry to say but I think you're unlikely to find what you're looking for!
Next Best Bet:
Try Googling asteroid belt collisions? Or collisions in planet's rings? Those areas are rocks and ice, so a collision might be similar to an asteroid/asteroid collision.
No two asteroid collisions would be alike anyhow. They would vary widely. If one is huge and one relatively small, that collision would be very different from a collision where they were the same size. Or head on versus rear-ending, etc. So, you've got some artistic freedom for your video game.
Lastly, you could Google collisions of asteroids that hurdled into terrestrial  planets or moons without atmospheres. Watching a collision with Earth or Titan does you no good because the atmosphere changes the dynamic of the impact.
Just remember to think of the physics. If the two sides colliding are well lit, there shouldn't be ice. Well-lit means the sun likely melted the ice. The ice is on the dark side and in the crevasses.


text:
Delta-v from Mercury surface to Venus surface
What it says on the tin: using the most efficient launch opportunity afforded by orbital mechanics, what delta-v would you have to impart to a payload to send it from the surface of Mercury to crash on the surface of Venus?

Assuming circular, coplanar orbits, Trans Venus Insertion (TVI) is 8 km/s from Mercury's surface. That's with horizontal take off and no gravity loss. That's for the minimum energy Hohmann transfer.
Exiting the Hohmann transfer and soft landing on Venus would take 11.9 km/s if Venus had no atmosphere. But much, maybe all of that 11.9 km/s could be done by aerobraking.
I used my Hohmann spreadsheet to get these numbers.

What do you mean when you say "no gravity loss"? My calculations suggest 8km/s wouldn't be fast enough, but I might be misunderstanding.

@barrycarter during vertical ascent, gravity subtracts from from the rocket's thrust. That's gravity loss. The trajectory you show in your answer is not Hohmann so it would take more delta V than the minimum energy ellipse. I have not examined your math. I use the Vis Viva equation, a widely used method in orbital mechanics

If you launch from the side of Mercury furthest from the Sun,
parallel to the horizon in the direction of Mercury's orbit at
13 km/s when the Mercury-Sun-Venus angle is 53 degrees, your
payload will reach Venus in 35 days, and its path would look
something like this (each black dot = 1 day). IMPORTANT CAVEATS BELOW IMAGE.

This doesn't really answer your question, since I'm not sure
13 km/s is the minimal required speed. I also made several
simplifying assumptions (perhaps too many), so the answer
above isn't exact.
Here's the heavily commented Mathematica script I wrote, which
also explains the simplifications I made.

(* cleaned up and well-commented version for SE *)

(*

I assume that Mercury's orbit is planar, and that Venus' orbit lies in
the same plane, or at least intersects it at the appropriate time.

This allows me to use a two-dimensional equation for acceleration due
to gravity, instead of a 3 dimensional one.

The equation (below) is the acceleration imparted to an object of mass
m1 at {x1,y1} by an object of mass m2 at {x2,y2}, given that the
gravitational constant is g.

Note that the mass of the object being accelerated (m1) is actually
irrelevant; however, I include it as a parameter for symmetry

*)

accel[{x1_,y1_},{x2_,y2_},m1_,m2_,g_]=g*m2/Norm[{x2-x1,y2-y1}]^3*{x2-x1,y2-y1}

(*

The mass, semimajor axis, period, and radius of Mercury, in kg, m (not
km), and s

*)

mercsma = 57909050000;
mercper = 87.9691*86400
mercrad = 2439700
mercmass = 3.3011*10^23

(* solar mass, in kg *)

sunmass = 1.98855*10^30

(* gravitational constant of universe, in kg-m-s system *)

g = 6.6740*10^-11

(*

Heliocentric, so Sun is always at origin. In theory, the positions of
the other planets (eg, Jupiter) could help boost your payload, so you
might be able to launch with a lower speed than I find below

*)

sun[t_] = {0,0}

(*

I assume Mercury's orbit is circular. Since the actual orbit is
elliptical, you could get a boost for your payload by launching it
when Mercury's distance from the Sun is increasing the fastest (in
other words, solar radial velocity is greatest)

I've chosen the x axis to be the line connecting the Sun to Mercury at time 0.

*)

merc[t_] = {Cos[t*2*Pi/mercper],Sin[t*2*Pi/mercper]}*mercsma

(*

I also ignore Venus' own gravity: you can do slightly better by noting
that Venus will pull the payload towards itself once the payload gets
close enough.

I do want to plot Venus' orbit, so I use the semi-major axis and
period values below.

Venus' starting angle (vsa below) was found by trial and error to make
sure Venus was at the right place when the payload crossed its orbit.

*)

vensma = 108208000000
venper = 224.701*86400;
vsa = 53*Degree;
ven[t_] = {Cos[t*2*Pi/venper+vsa],Sin[t*2*Pi/venper+vsa]}*vensma

(*

If we launch from side of Mercury furthest from the Sun, the payload's
starting position will be Mercury's position plus Mercury's radius in
the x direction

NOTE: This start position is completely arbitrary; you may get better
results by starting at different positions on Mercury's surface.

*)

s0 = {mercsma+mercrad,0}

(*

The initial velocity of the payload (with respect to the Sun) will be
Mercury's velocity + whatever velocity (delta v) we impart to the
payload.

Note that both the direction I choose for initial velocity (in the
same direction as Mercury's orbit) and the magnitude are
arbitrary. You may get better results by aiming the payload at a 45
degree angle or straight up or something.

NOTE: If I change 13000 to 12500 below, Mathematica will refuse to
solve the differential equation. This doesn't necessarily mean 13000
is a minimal velocity, but there is apparently some sort of important
change between 12500 m/s and 13000 m/s

*)

v0 = merc'[0] + {0,13000}

(*

Mathematica won't close-form integrate this problem, so we integrate
numerically, which requires a start time (0) and an end time (below).

I chose 35 days after confirming that's how long it takes the payload
to reach Venus.

*)

timelimit = 86400*35;

nds = NDSolve[{s[0]==s0, s'[0] == v0,
 s''[t] == accel[s[t],sun[t],1,sunmass,g] + accel[s[t],merc[t],1,mercmass,g]
},s,{t,0,timelimit}]

(* The use of [[1,1,2]] below is just Mathematica nesting weirdness *)

g= ParametricPlot[{nds[[1,1,2]][t],merc[t],ven[t]},{t,0,timelimit},
 Mesh -> timelimit/86400, AxesOrigin->{0,0}, PlotStyle -> {Blue,Red,Green},
 MeshStyle -> {Black}
]

For minimum energy, I think you want it to reach Mars as it reached the opposite side of the sun from where it was launched.   In other words, you want an elliptical orbit with perihelion at Mercury at launch and aphelion at Mars.

Did you mean Venus? And I'm still not sure that's correct? Wouldn't it be aphelion for Mercury and reaching Venus at perihelion for minimal orbital distance traveled?

I meant aphelion at Venus.   Mercury is closer to Sun so it has to be the perihelion.  This is minimum Energy orbit that connects the two points.

Oh, you mean the payload should assume an elliptical orbit where Mercury is its perihelion and Venus is its aphelion? I'm still not sure that's true, since it ignore Mercury's own gravity.

@barrycarter Outside of the planetary spheres of influence, transfer orbit is modeled as an ellipse with a focus at the sun center. Within planetary sphere of influence orbit is modeled as a hyperbola with a focus at planet center. Speed of a hyperbola is $sqrt(Vesc^2+Vinf^2)$. This is the patched conics approximation.

@barrycarter Minimum energy ellipse (a.k.a. Hohmann) would indeed have perihelion at Mercury and aphelion at Venus.

Am voting this down. A Mercury to Venus Hohmann would take about 76 days. From Mercury's surface, insertion to a Hohmann orbit would take less than 13 km/s.

@HopDavid So you're saying the Hohmann orbit would take longer, but would require a slower initial speed?

@barrycarter Yes.


text:
What constellations touch the 9-degree wide Zodiac?
Say Barry, could you abuse the answer-your-own-question feature of
this site, and answer the following question:
As noted in How many constellations in the Zodiac? the
ecliptic itself touches 13 constellations, but the Encylopedia
Brittanica defines the Zodiac as "a belt around the heavens extending
9 [degrees] on either side of the ecliptic":
http://www.britannica.com/topic/zodiac
The Old Farmer's Almanac for 2014 (and previous years) notes (page
114) that Moon occasionally crosses into 5 "non-Zodiac" constellations
at times:

So, exactly which constellations are "in the Zodiac" if the Zodiac
extends 9 degrees on either side of the ecliptic?
And why 9 degrees? I know Venus can be as much as 8.25 degrees from
the ecliptic,

While I can't answer the 9 degrees part,
http://www.space.com/5417-ecliptic-zodiac-work.html notes, the band of
the zodiac actually passes through 22 constellations:

In fact, as pointed out by the well-known astronomical calculator,
Jean Meeus, along with Ophiuchus, there are nine other constellations
that occasionally can be visited by the Moon and planets: Auriga, the
Charioteer; Cetus, the Whale; Corvus, the Crow; Crater, the Cup;
Hydra, the Water Snake; Orion, the Hunter; Pegasus, the Flying Horse;
Scutum, the Shield; and Sextans, the Sextant.

So in truth, there really aren't twelve zodiacal constellations, but
twenty-two!


text:
Absolute/apparent magnitude and distance for HIP31978 inconsistent?
EDIT: I am still following up, but here's the latest response from Wolfram:

Paclet servers are updated and optimized for each version of
  Mathematica, so I suspect you may be working on an older version which
  we no longer update. Many times we get reports that data from the
  paclet data for version 9 was incorrect, only to find that it has
  already been fixed for version 10.

According to Mathematica and several other sources, the
known information about HIP31978 (also known as "S
Monocerotis" and "15 Monocerotis") includes:

The star system is about 101.06 light years from our own.
The star's absolute magnitude is -2.79 (some sources say even lower).
The star's visual magnitude from our solar system is 4.66.

The star is slightly variable, but not enough to explain
the following discrepancy:
If the star's magnitude at 32.6 light years is -2.79
(definition of absolute magnitude), it would be about 9.6
times fainter at 101.06 light years (it's actual distance
from us). However, that's only about 2.5 magnitudes
fainter, nowhere near enough to bring it down to magnitude
4.66.
What's happening here, and is this a one-off special case,
or does this happen a lot?
For reference, my calculation of the apparent magnitude:
-2.79 - Log[100, (32.6/101.06)^2]*5 == -0.333192

Further, a star that is 3.1 times further away is NOT 3.1 times fainter, it is 9.61 times fainter.

Are you sure that's true for magnitudes, not just intensity? It seemed counter-intuitive to me too, but I think magnitude is a measure of light intensity at a given point, not the total light intensity emitted. Example: Sirius has apparent magnitude -1.44, absolute magnitude +1.45 (so apparent is 10^(2.9/5) = 3.8 times brighter). Its distance is 8.60 light years, which is about 1/3.8 of the 10 parsecs used for absolute magnitude.

Not sure what your point is. 3.1 times further away is 9.61 times fainter, corresponding to an increase of 2.46 magnitudes. No debate.

My mistake. I incorrectly computed 5 magnitudes = 10 times brighter, not 5 magnitudes = 100 times brighter. Will fix my post shortly

The absolute magnitude you quote appears to correspond to the Hipparcos parallax distance estimate, which is $\sim 1000 \mbox{ ly}$, not the $\sim 100 \mbox{ ly}$ distance you have used.

This makes the numbers work, so I'll accept the answer, but it still bugs me. The whole point of having an absolute magnitude is to compare relative brightness. Using two different definitions seems inconsistent with this goal.

@barrycarter There is only one definition of absolute magnitude - the magnitude the object would have at 10 pc.

Oh, OK, I get it. @ConradTurner is saying that the distance estimate I have is way off from the distance estimate used to compute the absolute magnitude... by a factor of 10. So, Mathematica is using one source for distance estimate and another for absolute magnitude, and the two sources don't agree closely on distance.

@barrycarter More likely is that Mathematica has a typo in the distance, or some other error, rather than another source for the estimate.


text:
Why do proto stars on the Hayashi track get dimmer as they contract?
Why do proto stars on the Hayashi track get dimmer as they contract?
My expectation is that they would get hotter and brighter the more they contact. Why is this wrong?

see https://astronomy.stackexchange.com/questions/13260/why-do-pms-stars-on-the-hayashi-track-remain-at-a-constant-temperature-while-the?rq=1

The Hayashi track has an almost constant effective temperature. The pre main sequence star (not protostar) is powered by the release of gravitational potential energy caused by its contraction.
If the star is getting smaller at constant effective temperature, then of course it's luminosity ($\propto R^2 T^4$) decreases.
So perhaps your question is why do PMS stars contract at constant temperature? In other words, why is the Hayashi track almost vertical in the HR diagram? The answer to this is derived in some detail on the relevant Wikipedia page and arises from the fact that low mass PMS stars are fully convective and have atmospheric opacity dominated by H- ions.

Thanks! But yeah, why don't they heat up?

@Dawn Read the Wikipedia page and then ask a question about which part is unclear. The derivation is quite involved and I see no purpose in cutting and pasting it here.

Nevermind then :(


text:
Why do PMS stars on the Hayashi track remain at a constant temperature while they contract?
Why do pre-main-sequence stars on the Hayashi track remain at a constant temperature while they contract?
I've read the Wikipedia article, so no need to repeat the derivation.
What I took away from that is that the extra heat generated by contraction is given off almost exactly because the whole star is convective and the atmosphere is very capable of radiating that heat away if the opacity is mainly caused by hydrogen ions, which is the case in the cooler of the pre-main-sequence stars.
Is that correct?

At least one statement here is incorrect. Only half the heat generated by contraction is radiated away. The rest makes the star hotter.

The surface temperature of fully convective stars is set by their opacity mechanism, often emission when the H minus ion is created.  The rate the star emits light is not set by the contraction rate, it is the other way around-- the star loses heat at a rate set by its structure, and that's what sets the contraction rate.  (As said, the rate of gravitational energy release is always double the rate the star loses heat to starlight.)  The reason the surface temperature always has to be something like 3000 K is pretty complicated, but it seems to be due to the fact that this temperature is a kind of peak in the surface's ability to emit light.  Convection sets the temperature structure, but there are a range of possible solutions with different internal energy, so the history of heat loss picks out whichever solution has the right internal energy.  But just why the surface is always about the same T is still the trickiest part to understand, I am hoping to get a better handle on that eventually.  Ignore anything you see in introductory textbooks, they never get it right.


text:
We are sure that there are no brown dwarfs closer than ______
I don't quite know how to word this question.
How close could a brown dwarf be to us and have gone undetected until now?
Or maybe, with what probability are we sure that there is no brown dwarf within X light years of the sun? (For different values of X)
(Don't worry about addressing Nemesis or a hypothetical binary partner for the sun. I'm just asking about undetected brown dwarfs..., even outside of the Oort cloud but within Alpha C. I don't know how sensitive the WISE survey was.)

Possible duplicate of Is it possible that the Sun has a binary partner (the Nemesis Theory) that has eluded detection?

The summary answer is a few light years, maybe 5, since ultracool brown dwarfs only a few light years away have been found in all-sky surveys.

I think the answer is roughly 3-5 light years.
The recent WISE survey in the near-infrared should have been capable of detecting even a very cool nearby brown dwarf (and indeed it has detected some very cool brown dwarfs - e.g. a 250K brown dwarf only 6 light years away Luhman et al. 2014). Since WISE was an all-sky survey and there is great interest in finding these ultracool brown dwarfs, then it is reasonable to assume that any similar, object within a few light years would already have been found. The caveat to this is that the point spread function of the WISE survey s not that great an so near the galactic plane it could be that source confusion might have hidden the brown dwarf. On the other hand, nearby objects usually have high proper motion, so it would probably have been uncovered in combination with the earlier 2MASS survey, since as I show below, such an object may have been detected in both.
Some details: As an example let's take a very low mass brown dwarf - say 20 Jupiter masses and 5 billion years old - and place it 4 light years from the Sun. According to the evolutionary models of Saumon & Marley (2008), such an object has an intrinsic luminosity of $10^{-7}$ times (1 ten millionth) that of the Sun and a temperature of 400 Kelvin and would appear to have a spectral type of late T or early Y. This is similar to the coolest brown dwarfs yet found.
From the calibration of absolute magnitudes versus spectral type for cool brown dwarfs in Marsh et al. (2013) we know that at 4 light years, such a brown dwarf would have magnitudes of $H=16$ and $W2=10$. The former is just about bright enough to have been seen in the 2MASS all-sky survey and the latter is easily detected by WISE. The combination of data would also easily have revealed the large parallax of such an object. We can conclude that an object would have to have much lower mass than this to remain undetected and thus would arguaby not be called a brown dwarf, but would really be a "free-floating planet".
Now you might ask, what about a cooler brown dwarf? Well the theoretical predictions of magnitudes and colours become much more uncertain, but I would simply refer to the Luhman discovery. This brown dwarf is about as cool as a brown dwarf could be given the age of the Galaxy, may only be 10 Jupiter masses, yet was still easily detected in the WISE survey. Therefore I think this sets a fairly firm empirical limit.


text:
Ninth planet - what else could it be?
So recently it was announced that there might be a ninth planet out there.
Some people (from NASA) say the odds are 50% that it is a real planet, since it hasn't been discovered by telescopes yet.
Let's say there is no ninth planet. Is there another explanation for those orbits that require a big mass (8 times earth mass) out there?
Could it even be a very small black hole (a few centimeters in radius) or maybe a clump of dark matter?

I'm not sure enough to state this as an answer, but the perturbations needed to affect orbits might be so small that it could be caused by things like out-gassing, non-uniformities in the heliosphere environment, or hitherto unstudied effects from proximity to interstellar space (i.e., shifts in stellar wind, etc.). The simple fact is that there needs to be proof that there is a planet out there -- Neptune was predicted similarly but not accepted until confirmed by observation!

What else could it be?  Very simply, it could all be a coincidence...
What they have done is noted that objects in the outer solar system have their orbits grouped in interesting ways. Someone suggested this might indicate there was a heavy planet (or core of a giant protoplanet) that influenced their orbits in this way; the recent announcement is the result of examining this idea in detail.  It is likely to be caused by gravitational influence of such a body.
Of course the TV and newspapers have picked this up and said a planet HAS been discovered. But there might be nothing at all; the loose clustering could all be a coincidence - though that seems, statistically, unlikely at the moment.
Here is the paper for anyone who wants to check.  (Beware, lots of mathematical terminology.)

The introduction of the paper mentions some alternative interpretations put forward in earlier papers. It seems as if this "problem" has been noted earlier. This new solution to the problem beats earlier solutions. But maybe it is an evolving discovery process which again will invent new better explanations?
In addition to those mentioned in the paper, I suggest the following candidate alternatives:
-) Too few observations in order to hold up in a soon dramatically increased discovery rate. 4 or 5 or maybe even 8 TNOs which, as they claim, give a 0.007% significance level is certainly resting on some assumptions which will be challenged if thousands of similar objects are found in the next decade. So, as Andy has answered: possibly a coincidence due to small sample. (The paper's 0.007% probability is, well, we'll see)
-) Weird unthought-of science bias which, in math more than in lens, tends to select inadvertently among the (candidate) observations made and their characteristics. Since the authors are at the top in their fields, and Michael Brown has discovered loads of distant objects, including Sedna, because of that maybe one could suspect that some one thought has created a bias somehow.


text:
Which planets have abundant amounts of photoluminiscent matter?
Which planets have abundant amounts of photoluminiscent matter? Photoluminiscent or chemoluminiscent whatever the glow-in-the-dark stuff it may be. Planets or bodies which have good enough the photoluminiscence(light absorb and then release) property.

Sorry, but none of the planets have substantial amounts of phosphorescent chemicals.
Perhaps the closest to what you are thinking is the aurora, which can be seen on all the planets with atmospheres (to a greater or lesser degree) notably on the Gas Giants and on Earth. However, while the mechanism of aurora is similar to florescence, it is not phosphorescence.
No planets outside of the solar system have been directly imaged beyond a dim dot, so there is no information about exoplanets.

"so there is no information about exoplanets." Spectra?

exoplanets are detected by their transits, or other indirect methods. To get a spectrum one would need to "see" the planet directly. There are few direct observations of exoplanets, and none of terrestrial expolanets. Not enough to get a spectrum. Rocky planets are probably composed of silicates, olivines, feldspars etc much like the planets of the solar system.

You can do transits in different bands and compare depths and widths. This is called transit spectroscopy. A reference is https://www.eso.org/sci/meetings/2014/exoelt2014/presentations/LopezMorales.pdf, this team publishes nice papers about that. Emission spectro exists as well, but I agree with you that probably we wouldn't see anything from terrestrial bodies yet. Maybe even not with JWST.

For the first time, astronomers detected the fluorescent signature of HD 189733b which is an exoplanet

Earth
That is the only place in the universe we know there is stuff that glows in the dark.

More serious though, the only possible place I can think of is Io. There may be some reactions taking place in the sulphur material ejected from the volcanoes that produces light.

You might find phosphorescent metal sulfides http://www.google.com/patents/US2544507 on Io. I've tried making my own phosphorescent Calcium sulfide, and it isn't easy. The stuff is likely to be rare on most planets and moons.

Other moons which glows are Europa and Titan.


text:
What Would it Look Like Inside Jupiter's Great Red Spot?
From what I've read, Jupiter's 'Great Red Spot' is an unbelievably powerful storm. Do astronomers and scientists have any idea of what that might look like if it were possible to be inside/below it? Would it look like a storm on Earth (thunder, lightning, winds, etc.), or would it look completely different?

You would not actually be able to see that much, since sunlight there is less than 4% of its strength at Earth's distance to the Sun. That is still enough to see though (much brighter than moonlight at night in fact). The problem is the transparency of the atmosphere: we can not look in, so you could not look out. You would only see a weak brown or red glow below the cloud cover.
However, there is plenty of lightning! Strong winds and thunder exist too.

Lightning on Jupiter

4% sunlight is plenty of light to see with, but the transparency (or lack of transparency) of the gas would be the key factor.   It looks red to us, but I don't know how transparent / how much visibility there would be from within.

There would definitely be high winds, and I assume that means very fast moving clouds.
 The clouds are a reddish orange, as we can tell from images of the Great Red Spot.  From what I understand, it would look something like this, with lightning happening if you were in the cloud tops of the Great Red Spot, looking at the sun (granted, I could be off):
YouTube video of Jupiter Red Spot Development
Screenshot:

I am making this as a level for Planetary Settlers, and trying to get as accurate as I can based on what we know.  Please correct me if any of the details are off.  I made the sun smaller according to the size difference we would see due to the further distance Jupiter is from the sun.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Where can I find the positions of the planets, stars, moons, artificial satellites, etc. and visualize them?
What resources are available to find the positions of planets, stars,
moons, artificial satellites, asteroids, and other heavenly bodies?

I was sleeping under the stars with a program called redeye... taught me star trivia like Vega and all the constellations. It tilts and zooms and has labels.

There are many resources online, so this is a community wiki
answer. Please feel free to add to it!

If you want to visualize the stars/planets/etc (as viewed from Earth
or another location), you are looking for planetarium software:
https://en.wikipedia.org/wiki/Planetarium_software

If you want accurate positions for stars/planets/etc, you are looking
for horizons:

https://en.wikipedia.org/wiki/JPL_Horizons_On-Line_Ephemeris_System
http://ssd.jpl.nasa.gov/?horizons

If you want to compute star/planet positions yourself, you have
several options:

SPICE (http://naif.jpl.nasa.gov/naif/tutorials.html) will give you
results that match HORIZONS very closely. You can also use some of
SPICE's functionality online at
http://wgc.jpl.nasa.gov:8080/webgeocalc/#NewCalculation

SPK (Spice kernel) files are available at https://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/ -- while these are primarily intended for use with CSPICE, the format is documented and you can use it directly. It's described at https://github.com/skyfielders/python-skyfield/issues/19 and implemented in Python at https://github.com/brandon-rhodes/python-jplephem/tree/master/jplephem

If you want to do the SPICE computations yourself (numerically solving the differential equations), see https://astronomy.stackexchange.com/a/13491/21

You might also want to use an n-body simulator to do the computations yourself: https://physics.stackexchange.com/questions/25241/what-open-source-n-body-codes-are-available-and-what-are-their-features

IAU SOFA (http://www.iausofa.org/) will give you the International
Astronomical Union's "official" libraries to compute positions.

VSOP theory (https://en.wikipedia.org/wiki/VSOP_%28planets%29) is
another option.

PyEphem (https://en.wikipedia.org/wiki/PyEphem), which derives
from MIT's libastro library
(https://stuff.mit.edu/afs/athena/project/xephem/src/xephem-3.5.2/libastro/)

Skyfield: https://github.com/skyfielders/python-skyfield

libnova (http://libnova.sourceforge.net/)

If you want to treat planetary orbits as simple ellipses and ignore perturbations, you can find orbital elements at https://ssd.jpl.nasa.gov/txt/p_elem_t1.txt but you'll probably want to visit https://ssd.jpl.nasa.gov/?bodies#elem first

VSOP87-multilang Implements the VSOP87 theory in over a dozen different programming languages.

Note that these computational libraries don't always agree with each
other or with planetarium software:
https://stackoverflow.com/questions/16293146

If you want a catalog of stars/etc, NOMAD
(http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/nomad) and GAIA (http://gea.esac.esa.int/archive/) are
the largest such catalogs (about 1 billion entries each, lots of overlap). The NOMAD link also provides references to smaller catalogs.

For star names, https://www.iau.org/public/themes/naming_stars/ has an official list and Stellarium has several lists in the subdirectories of https://github.com/Stellarium/stellarium/tree/master/skycultures in files named star_names.fab (including the official IAU list)

For high-resolution astrophotography try ALADIN (http://aladin.u-strasbg.fr/#AladinLite)

If you want to find artificial satellites and spacecraft, start at
https://en.wikipedia.org/wiki/Two-line_element_set

Transits of Mercury and Venus across the Sun? http://eclipse.gsfc.nasa.gov/transit/transit.html

Planetary conjunctions?
http://laps.noaa.gov/albers/ast/conjun/conjun.html

Occultations of stars by asteroids? The International Occultation
Timing Association (IOTA) at http://occultations.org/

Create your own Sun-Earth-Moon model by using a ~17 inches (~44 cm) beach ball or box as the Sun, 4-5 mm object like a pencil eraser as the Earth and a 1 mm sized drill bit or other object to represent the Moon. Then, find a 150-foot string or tape measure. The sun and earth are ~154 feet (~46m) apart. The earth and moon are ~4.7 inches (12 cm) apart.

I also use SIMBAD (http://simbad.u-strasbg.fr/simbad/sim-fid) if I want to do a quick search for a star or galaxy based on its identifier, and you can also do the reverse and searching for coordinates too.

I like this one for seeing what's overhead http://www.ucolick.org/~bolte/AY4_00/weblab/project/chart.html .

That appears to be a framed verison of http://www.fourmilab.ch/yoursky/ -- feel free to add the latter link above.

@barrycarter I wonder if astroquery mentioned here and here could be added as well?

@uhoh This is a community wiki answer, so please feel free to add them yourself. But, if you run into trouble, let me know and I'll add them.

@barrycarter okay I'll give it a try tomorrow, thanks!

For Python, astropy might be useful here as well: https://docs.astropy.org/en/stable/coordinates/solarsystem.html

To add to the excellent answer by barrycarter, there are 2 planetarium-like codes, that I know of, that run on a mac and would make excellent tools for viewing certain astronomical objects. The codes are Stellarium and Celestia. Both turn your computer into your own planetarium where you can search and view objects in space.

It's actually a public wiki answer (not all my content), so you can add to it directly if you'd like. I had considered listing individual planetarium software (Stellarium is my favorite), but thought linking to the wikipedia list page may be more useful and neutral.

The second program is called "Celestia" not "Celestial".

Celestia even lets you move around different from Stellarium which is more like a planetarium-theatre


text:
What Are Some Good Southern Hemisphere 7x50 Binocular Targets?
I've been interested in astronomy for a while but only recently purchased my first set of binoculars (7x50). I've been learning various constellations as well as spotting some other objects with the help of my planisphere. I live in the suburbs of a big Australian city, but far enough out that the light pollution isn't too bad.
I've already been looking at objects like the Orion Nebula and Pleiades, but hoped some experts might be able to give me a list of objects that would be worth hunting down with my binoculars. Ideally some that would be visible now (it's Summer here), but some for other times of the year would also be fine.

Found a pile for southern binoculars here: https://books.google.com/books?id=foTRJ8yQcwEC&pg=PA154&lpg=PA154&dq=tarantula+nebula+binoculars&source=bl&ots=ozuWf6ap2o&sig=23A3IfrdJy98rmURKQg-ss1mM6Y&hl=en&sa=X&ved=0ahUKEwjlqujKpPDKAhXG4iYKHTVcAmYQ6AEIMzAG#v=onepage&q=tarantula%20nebula%20binoculars&f=false

related: Binocular highlights from Tasmania next few weeks?

For anyone else who might be asking a similiar question, I found a great resource for this. The PDFs found at the URL below contain a good list of binocular, naked-eye and telescope targets for each month of the year in both hemispheres.
http://www.skymaps.com/downloads.html

Pleiades -  a jewel box (open star cluster)

Coal Sack - in Crucis, there's a globular cluster I recall there somewhere. Hours of fascination.

Messier objects

Jupiter moons

Saturn's rings and moons,

Lyra (constellation)

Large and Small Magellanic Clouds - extra-galactic remnants on the southern pole of the Milky Way

Just for starters. By the time you work through all these, you'll start to gain an appreciation of your own.
Happy Hunting!!!


text:
Is it true that the sun goes down to the west?
I know the sun doesn't go down but we are moving and is it true that the sun goes down to the west? I've read on many webpages that this is not entirely true.

"I've read on many webpages that this is not entirely true" - can you give a link to illustrate what you mean? Are you asking if the Earth rotates on its axis, or something else?

@Andy on this page for example: http://solar-center.stanford.edu/AO/sunrise.html

That is talking about the exact east and west points on the horizon. If you imagine the horizon as a circle from 0 to 360 degrees, with 0 due north, 90 east, and 270 west, then the sun rises between 67 and 113 degrees at different points of the year, if you are on the equator.

"Actually, the Sun only rises due east and sets due west on 2 days of the year -- the spring and fall equinoxes! On other days, the Sun rises either north or south of "due east" and sets north or south of "due west.""
Although this is true, when we talk about the sun setting in the West, we mean West of the meridian (i.e. not East) not due West at the cardinal point.

Round here the sun rises in the NE in the summer, and sets in the NW. In winter it makes its way all the was from ese to wnw before giving up.

At the poles the sun just circles round the horizon during summer, and setting is a long process that takes a chunk of Autumn and has no specific direction :-)

Earth rotates. That gives us the feeling of Sunrise and Sunset. Actually there are no such things. See Below animated image. Earth completes one rotation in a day (24hrs). (Actually 1 rotation of earth is called a "Day").
This image also depicts the cause of Solstice and Equinox, earth's tilted axis.

Minor nit: the Earth completes on rotation or sidereal day (360°) in a bit under 24 hrs. Because the Earth is also moving it its orbit about the sun and has moved about 1 degree of arc during the course of a day, Earth has to rotate about 361° to complete a solar day, which is the day we measure on our calendars.

I live in an 'L' shaped building on the 14th floor, so I have a pretty clear view of the horizon. Up until March 20th of this year, the sun was setting behind the other wing of the building. My living room window faces roughly NW, and on March 21st, I was able to see the sun at the horizon for the first time this year. Each day, the sun will set a little further northwest until it's setting almost directly opposite my window. Then it will begin to retreat until one day in September, when it will begin to set behind the other wing of the building again.This is the first year I took note of the day that sunset was first visible, and I'm planning to note which day is the last day I can see the sun set.

Today's the 5th of June. On the 21st or 22nd, the sun will set the furthest to the north it will get. After that, it will begin the 3 month series of sunsets that will take it behind the other wing of the building. Right now, it's setting around 8:30 (EDST). I try to watch the sunset each day. Somehow it's dramatic, almost as if the next sunrise is in question. Some sunsets are plain, but sometimes the last rays of the sun light up the clouds in awesome glory.
https://drive.google.com/open?id=0By3-fBUSwRxMYjhfdWhLel9oLWc


text:
Which of Jupiter's moons did I most likely see?
I was gazing through my binoculars at Jupiter last night and noticed 3 (maybe 4) tiny dots in line with it. A first I thought they might have been background stars but quickly realised their colour was similar to that of Jupiter and their brightness was larger than expected considering their size. As a result, I'm pretty certain I was looking at Jupiter's moons.
Presumably, I was seeing the largest of its moons (Io, Europa, Ganymede, Callisto). They were far too tiny to see any details, obviously, but would it have been only those moons I could see? I assume the rest are too small. And is seeing those 4 moons as tiny bright dots through my 7x50s feasible?

Please do not write last night - [edit] your question and enter exact time and location.

The galilean moons are visible in binoculars, indeed. Yesterday evening (assuming you're in the western hemisphere), 3 of them were visible as separate dots. See the chart here:
http://www.wolframalpha.com/input/?i=jupiter+galilean+moons+on+march+6+2016+at+9pm+pst
You can adjust the query to reflect your time zone, and the actual date/time of the observation.
There are also smartphone apps that allow you to determine the position of the moons at any given time. On Android I use the Moons of Jupiter app.

Yes, you have definitely seen the Galilean moons, as they are named after their discoverer — Io, Europa, Ganymede, and Callisto. Those four moons are so bright that you would in fact be able to see them with the naked eye if they hadn't been just next to the much brighter Jupiter.
In contrast, the fifth brightest Moon, Amalthea, requires a telescope of at least 25 cm diameter.


text:
How does gravity really work
I am only 12 years old and I'm constantly wondering and trying understand how gravity really works. On YouTube everyone always talks about objects warping space time around themselves and uses the analogy of a trampoline. I still don't understand gravity because if space were like a trampoline, then earth would be spiraling in towards the sun along with all the other planets, right? So could someone explain to me how gravity really works without the trampoline analogy?

Duplicated on physics.SE: http://physics.stackexchange.com/questions/243317/how-does-gravity-truly-work

I'm just can add what I assume and it may be wrong. But as far as I know your assumption earth would be spiraling into sun isn't that wrong. AFAIK there are 2 ways. the orbiting bodys may escape or crash into the body they orbit. But this will take a long long time. Considering all the objects which tended to escape or crash into the sun did this allready a long long time ago. So earth and other planets are those which shift on such a low magnitude that they did it for billions of years and will keep doing for additional billions. but for infinite given time all objects will some"day".....

either crash into sun or escape its orbit. But I have to clarify again. This is what I know so far. and it might be not correct at all. (In what case I'd like to get corrected.)

The "real world" trampoline has friction acting on things on its surface, so they gradually lose energy and spiral inwards. In space there is no friction, so the planets stay orbiting for near-forever.

@pjc50: But isn't there still taht little bit of friction, which is the reason of why its just "near-forever" and not "forever"?

@Zaibis Quite the opposite, actually. As the orbiting body loses orbital speed (due to tidal forces etc.), the radius of its orbit gets bigger, not smaller. This is still a massive simplification, of course - there's plenty of other things happening.

Mandatory XKCD reference.

@Luaan: Woh... gotta digest that. this actually doesn't fit into my imagination of gravity. But thanks for enlighten me!

Mandatory Feynman reference. (Talks about magnets, but the lesson about how you think about things applies in any science.)

https://www.youtube.com/watch?v=MTY1Kje0yLg
This video is one of my favorite visualizations.

@Zaibis Don't feel too enlightened -- I think Luaan is simply wrong. If you brake an orbiting object it will indeed lose energy and altitude. In the new lowest position (the perigee) it will have a higher velocity, but at all times the sum of new velocity and "potential energy" from the gravity field is smaller than it was in the original orbit, demonstrated by the fact that it will be too slow at apogee to sustain the original higher orbit (we slowed it down there!). Cf. my reply to a comment Luaan made under my post below. He confuses the effect of two unrelated tidal forces on the moon.

@Zaibis As a side note, the trampoline metaphor of a gravity well works nicely for this confusion. Try to get the ball higher in the well by slowing it down...

@PeterA.Schneider: " Try to get the ball higher in the well by slowing it down..." What won't work?! If so, I guess, I got it.

@Zaibis To avoid confusion: You cannot get the ball out of the "gravity well" (the trough in the trampoline surface) by slowing it down. You need to accelerate it. That it will lose kinetic energy climbing out of the trough notwithstanding -- the border case is that it is at standstill at an infinite distance (big trampoline, eh). All the kinetic energy is now stored in the gravitational field (much like separating opposite electrical charges stores energy in the electric field). In both cases the energy can be recovered by falling back.

@Luaan If you want to experiment with orbits I can recommend the free iOS game Space Rider. It's not always intuitive, but always logical ;-). (What does that say about us?)

@PeterA.Schneider Well, of course you need to add the potential energy. I was only talking about the orbital speed, perhaps that's what makes it sound so weird. A circular orbit's speed is inversely proportional to the radius. I thought it's obvious just slowing the object down isn't enough - if that were the case, it would be kind of hard for our spaceships to return back to Earth's surface :D When you slow down, you're making the opposite side of your orbit faster (ignoring the atmosphere) and vice versa. That's the "massive simplification" part - I was only talking about circular orbits.

I found the PBS Space Time videos pretty good quality (though I haven't watched them all by far). They still have that annoying american documentary habit of trying to talk as fast as possible, but they're surprisingly accurate. https://www.youtube.com/watch?v=YycAzdtUIko&list=PLsPUh22kYmNAmjsHke4pd8S9z6m_hVRur deals with relativity in particular. It goes far beyond the usual popular "explanations", and should ultimately give you the "Aha" moment, because it doesn't distort the theory to the point of being just confusing and useless.

@Zaibis Sorry for the confusion - I was talking about circular orbits (that's the "massive simplification" part). When you're on a circular orbit and you slow down, the other side of your orbit will go deeper into the well ("down"), and it will be faster - in other words, you're no longer on a circular orbit. Conversely, to get to a higher orbit, you need to add speed, but once you get to the highest point, you're traveling at a lower speed (and need to add more speed to circularize, though still less than on a smaller orbit).

The 'Trampoline' requires gravity to explain gravity...and falls apart when a two body problem is modeled... therefore it is not a valid explanation. the graviton theory explains gravity but falls apart with the problem of inertia. the best explanation is neutrino string induction .. fulfills the gravity issue and inertia. fits with relativity both the special and general theory.  try looking at  www.themechanismofreality.com for a full discussion.  easy read ... hofmannfan seems to have read it in 15-20 min.

First of all: "How gravity really works" is a deep question, and any serious scientist would quickly concede that all we have is an incomplete working model. You certainly have heard about  General Relativity; the first image on the page is your trampoline.
Our working model, General Relativity, is working because it explains a lot of observations very nicely. (Careful, here is another deep question lingering: "Explains" means that we can predict some observations from other observations with the model of gravity we have in our  mind. It does not necessarily mean that we understand the "real nature" of the underlying issues.) But we are very confident that the model is working over a wide range of observations. One of the last "first-time" observations which followed the predictions and thus gave us more confidence in the model was the two black holes colliding lately. Lately? Well, billions of years ago. We just learned about it lately.  Here is a link to a New York Times article with an impressive video. (I think one can still read a limited number of Times articles for free, so try it out.)
Our model of gravity is incomplete because it doesn't connect well to the model of nature we have for other things (elementary particles, quantum physics). For a while (like 70 years or so) it didn't connect at all; Einstein himself completely failed to connect the dots, which was probably not encouraging since he had received the Nobel Price for laying one of the foundations of quantum physics and was the obvious authority about gravity. If he couldn't do it, who could?
If I'm not mistaken, the physicists today are making progress, slowly. This connection between quantum physics and gravity is one of the main unresolved problems in modern physics.
Last, let me address your concern about the planets spiraling into the sun. This idea probably comes from actual balls on an actual trampoline spiraling in, I suppose. You probably know that the balls lose speed due to friction, much the same way you slow down on your bike when you stop pedaling. Some of the kinetic energy is transformed into heat.
And you know what? You are right. Given enough time, the planets would eventually fall into the sun. Low-flying satellites fall back to earth after a few years, because there are still traces of atmosphere slowing them down out there. The reason is that there is "friction" in the wider sense involved in all large-scale processes in the universe. That is actually one of the fundamental physical principles making up the world we know. It's just that the near-vacuum between the planets doesn't provide that much friction, and the planets are fairly massive bodies with an enormous mass and kinetic energy. It will take a long long time for them to lose enough energy that they'll be so close as to touch the sun. (Perhaps too long to happen at all.) In fact, over human life times the planets, moons and stuff are almost perfect examples for movement without friction. But in the astronomical time scale -- billions of years --, there certainly is friction. For example, the moon is showing us always the same side because friction slowed its rotation so that the rotation is now "locked" with its orbit.
Bottom line: The idea that gravity bends space and time "explains" all large-scale observations so far; the "trampoline" is a good model for a 2-dimensional "space", i.e. a surface, if you ignore friction.

The Moon is also much further than it was in the past. Tidal friction decreased its orbital speed, which increases the orbital radius. The radius is increasing by about four centimeters a year, nowadays.

+1 for "We're not entirely sure, but here's some of our best guesses based on observations."

Please correct the statement about moon. The moon is not synchronized ("locked") by accident but due to tidal forces of gravity - the closer side of moon suffers higher gravity than the other one. This force can actuall force object to increase its rotation, if it would rotate slower than the orbiting speed.

@libik I cannot see anything which would need correction (in particular, I did not say or imply "by accident" -- on the contrary, I mentioned friction as a cause) . One could mention tidal forces but I thought friction is good enough without detouring too much. You make an interesting point with a possible accelerating rotation due to tidal forces; but it's still a slowing down (to close to 0) relative to it's orbital reference frame.

@Luaan There are two tidal forces in play. (1) The moon receives energy from earth's rotation by the tidal forces the rotating earth exerts on it, accelerating it in the direction of earth's rotation. This lifts it (slowly) higher in earth's gravity well, as you say correctly. (2) The moon's cyclical deformation ("kneading")  caused by the moon's rotation in earth's inhomogeneous gravity field convert(ed) some of the rotational energy into heat, eventually synchronizing orbit and rotation, at which point there are almost no moon tides any more (short of those due to libration, I believe).

@PeterA.Schneider Sure, no argument over that. Perhaps I should just stop simplifying too much. Reading what I've written again... I can see why you'd think it's misleading. I ignored your (2) because, as you noted, today it only serves as a way to maintain the equillibrium - locking. A more accurate way of saying what I was trying to say would be "The orbital speed increased while maintaining a roughly circular orbit, which leads to a bigger orbital radius, but a smaller orbital speed."

Why are objects not escaping?
Consider first an object with velocity, and no gravity in action:

Then, that blue object will become more and more distant, if it continues in the same direction.
But it does not continue in the same direction, after a while, the gravity of the big black object has changed its course:

That happens again, and again and again:

Your question is: Why does not the object spiral in? You are perhaps thinking that as it comes closer, the gravity becomes stronger, and therefore the object is forced to come even closer.
But when it falls closer, its velocity increases. As we have seen, the velocity of the objects tries to make it escape. So when it is closer, it has more velocity to counteract the increased gravity.
Edit: Just in case of a more literal interpretation of your question, the trampoline in the original analogy causes friction, and therefore spiralling, but space is a vacuum.

I think the key as to why it doesn't fall in is that in space we don't have friction - on a trampoline energy is constantly being removed from the ball via friction, whereas in space there's nothing to slow our planet down, so it just keeps going

@Jeff Edited in

My physics teacher in high school said "the earth falls towards the sun all the time, but continues missing it because of its speed."

The trampoline analogy is useful if you understand gravity within a General Relativity framework. The conceptual problem there is that actually the space-time is wrapped in 4, not in 3 dimensions, i.e. including time.
In fact, when the Earth rotates around the Sun, it loses a very tiny amount of energy in form of gravitational waves. So, the Earth is actually spiraling towards the Sun. The thing is that this gravitational wave emission is so small, that by the time we observe any considerable spiraling, the Earth and the Sun would have already ceased to exist. Much before that, the Solar System becomes unstable due to chaotic effects already contained in classical Newtonian mechanics.

Great question!
Have you heard of Newton's First Law? It says that an object in motion continues moving at the same speed and in the same direction unless acted upon by a force.
When we roll a ball along the ground, it will eventually stop. Before Newton, many people believed that everything slows down by itself. Newton's insight was that this is not true, and actually the only reason a rolling ball will slow down is because the ground and the air rub or push against the ball to slow it down.
On a trampoline, a ball will rub against the trampoline material and against the air, which slows it down. This is the only reason that the ball ends up spiralling towards the centre.
When there is nothing to slow the object down, it won't spiral towards the middle, it will just keep going round and round forever. In space there is (almost) nothing to slow an object down.
If you find this hard to believe, you can write a computer program to do all the calculations and see what happens! I have made an example simulation for you. You will see that without friction, the planet will end up where it started each time it goes around the sun. If you change the planet's initial yspeed from 20 to 40 and then click "Run" up the top you will see a more circular orbit. You can change other things and see what happens. I hope you find this useful!

Nice simulation. (Although the planet escaped the sun after it got close. :-) )

It's an easter egg ;) Actually it's a good discussion point---it reminds us that the simulation is ONLY simulating gravity, not collisions, and also that when the planet gets really close to the sun, the time-step of the simulation causes big inaccuracies. This can be reduced by more sophisticated numerical methods like Runge-Kutta, but now I'm well beyond the scope of the question!

I don't know if it's the same simulation anymore when you do this, but if you change the for-loop condition to i < 1 rather than i < 5 and you change the timeout parameter to setInterval to 10 rather than 100, the simulation gets a whole lot more pleasant to watch. It runs slightly faster, but the framerate is much higher, so the movement of the outer body isn't so jagged.

Thanks Alex! Actually the timeout parameter should be 20 and then (assuming your CPU is fast enough) it is the same simulation. On my computer this slows the simulation down by 25%, presumably because my CPU isn't fast enough. Still, it does look smoother; here is a new simplified version: https://jsfiddle.net/0erknpk8/38/

Neutrino String Induction-Refraction is the cause of gravity. Some will say that neutrinos are insignificant, but Dirac, Hawking, and Tyson think otherwise and most discount the effect of a charged particle traveling at the speed of light. Keep in mind that no one can, or has proved that mass is a property of matter, more of an effect.
Go to www.themechanismofreality.com , this site explains exactly how gravity works. Every Physicist who examines this agrees that this is correct. From CERN to the University of Beijing's Physic Department agree that this is a 'Fantastic connection between graviton physics and string theory"! This was also confirmed , indirectly, by LIGO and the gravity wave announcement. Enjoy!

This is a link only answer. (not encouraged), also, that paper seems a little weird towards the end.

I have the nagging feeling that the lack of math, references and collaboration indicates it's not revolutionary science but at most a popular science article. It's tough; one should, of course, not gratuitously add math just to appear serious. But this kind of lone paradigm-changing breakthrough (which, I think, is claimed here, since I never heard of it before) is exceedingly rare.

In order to make the theory in the article more palatable you could try to put it in context. Like, start with what the conventional theory (and its famous proponents) thinks a Neutrino is and how it interacts, and why a different assumption could explain gravity.