How does an object at L1 stay at L1? I've recently been reading about the DSCOVR satellite, and I'm confused about it's orbit.
According to NOAA

A million miles away, DSCOVR orbits a unique location called Lagrange point 1, or L1. This point is a gravity neutral point in space, allowing DSCOVR to essentially hover between the sun and Earth at all times.

With it being at L1, it's orbit around the sun is slightly shorter than Earth's. So my uninformed gut feeling tells me it should orbit the sun in a few days less than Earth does, but it clearly doesn't.
How is this object at L1 kept at L1?
Could it be that the gravity of the Earth acting on it in a really convenient way? Or does it use fuel? What fuel? Or is it a property of the L1 point that I have misunderstood that an object will naturally stay there once in position?

I'll give a more grasp the concept answer if you don't mind.   Consider the discovery of the planet Neptune.
Neptune was discovered because it was observed that Uranus wasn't moving as fast as orbital mechanics suggested it should be moving.  By calculating how much Uranus had slowed down, an estimate was made for where an undiscovered planet should be and Neptune was identified very soon after.
Uranus is well inside Neptune's orbit, nearly twice as close to the sun and very far inside of Neptune's L1, but Neptune was still able affect Uranus' orbit enough to get noticed.  Props to Newtonian mechanics.  It works exceptionally well for a lot of measurements.
Loosely speaking, this type of orbital variation, L1 semi-stability or orbital perturbations like the slowing of Uranus' movement is a subset of the 3-body problem and that's actually more real.   Observed orbits don't precisely follow Newton's or Kepler's formulas because of other massive bodies and perturbations.  3-body problem math gets crazy complicated, but the specific example of L1 can be explained using basic arguments without much math.
Earth's L1 point is about 1% closer to the sun or about .99 Astronomical Units or "AU".   1% closer gives an orbital velocity of a little over 0.5% faster and an orbital period of a bit over 1.5% faster or an orbital period of about 359.8 days, about 5.4 days shorter.  That's what .99 AU circular Kepler orbit should do.
Consider an object at .99 AU, and a period of 359.8 days orbiting the Sun just inside the Earth's orbit - assuming both with circular orbits.  Now imagine that object passes the Earth.  It's relatively small so it doesn't change the Earth's orbit much, but passing that close to the Earth would give it a considerable gravitational tug, basically tossing it out of it's orbit.   .99 AU is too close for a 2nd Kepler orbit and as a result, it's quickly tossed into a new orbit.
Two bodies can't orbit a massive central body with just 1% variation in their distance unless both bodies are very low mass (for example, Janus and Epimetheus are a rare exception to this).
Earth doesn't have low mass, so an object with a Kepler orbit at .99 AU would get tossed about after just one pass.  In other words, if the object orbited as fast as your "gut" tells you it should, then it wouldn't be stable at all.
Now consider an object at .99 AU but orbiting the sun too slowly to maintain it's orbit.  Slower velocity means the object falls in towards the sun, speeding up as it falls, entering a new, shorter period and more elliptical orbit.  That's where the balance comes in.  The slower velocity means the object wants to fall towards the Sun but being between the Earth and the Sun, Earth's gravity pulls it towards the Earth.   The point where those two forces are equal is the L1 point, where it's no longer a Kepler orbit, but a 3 body orbit, slower than a Kepler orbit, but pulled in opposite cancelling ways by the Earth and the Sun.   The "too slow" orbit is still unstable, but it's considerably more stable than a Kepler orbit in the same location would be.
With Earth's perturbed elliptical orbit, L1 isn't really a point so much as a smallish region, but on a computer, if we give the Earth a perfect circular orbit around the Sun with no other planets, L1 becomes a point of perfect balance.  For actual orbits, L1 is always unstable.
Despite it's instability, Lagrange points are very useful low maintenance areas to station telescopes because much less energy is needed to keep the telescopes there.  This only works if the central body (the Sun) is at least 24.96 times the mass of the orbiting, less massive body (the Earth)


Can stars be seen in the background when looking directly at our solar system? Imagine viewing the solar system from the outside:

Considering the sun is quite bright, would you still be able to see stars in the background, or does the sunlight disable us to do so?

Consider this photo:

Does the light from the sun disable us from seeing stars in the background or have they simply been removed in an editor after the fact of being taken (the photo)?

Obviously there needs be some way of seeing stars from space, since looking at photos from space stations (or even from Earth) those are full of stars.

But, viewing the solar system from above (from the outside), can you see stars in the background or does the sunlight simply block them out?

Stars are much less bright than the earth in full sunlight as per your picture. So with the exposure required to give that picture of earth, no, you won't see stars.

But move the camera a bit to the side, so earth is not in shot and you can see stars.

The same goes for the sun - make sure sunlight is not entering the lens, and you can easily see stars.

Sunlight doesn't block any light out (unless you want to see a star that is actually behind the sun) it's just so bright that you need to make sure it isn't in shot.


Finding the SN explosion time by t-square fitting In a couple of papers I've read that it's usual practice to fit early rise of supernova luminosity curve by $t^2$. Let me describe how I see it:

I take the early fast rising part of luminosity curve and then fit them to a curve:

$L = a\cdot\left( t - t_0 \right) ^ 2 + L_0,$

Where $t_0$ is the moment of explosion.

Is this approach right?

I think you misunderstood.

$(y-k) = a (x-h)^2$ is a parabola with its apex at $(h,k)$. Therefore, the formula you mentioned above has $t_0$ as the time at peak, not at the explosion.

For the case of SN, we normally do not know when the explosion happened. Mostly, we have an image before SN detection that can set a constraint for the explosion happening in between the last non-detection to the first detection.

However, most analytical models require the explosion time as the reference, so the parabola fit is applied as an approximation.


Beginner telescope for my daughter My husband and I are trying to figure out what type of telescope to get our young daughter. We would like to be able to see planets but nothing too extreme. Just something to get her started. She has shown a big interest in astronomy and we would like see if it sticks then we will get her a much better one. Any suggestions?

I have been looking to get into astronomy myself and I have been impressed with the reviews on the Celestron Firstscope, which is around $50. It has a decent optic size and is small enough to pack up and get out to a clear area if you live in a high light pollution area. The only drawback I find with the scope (without actually using it) is it doesnt have a viewfinder but you can find one of those easily.


Coordinate transformations between reference frames in spherical astronomy Suppose there are two observational frames of reference with origins $O$ and $O'$, respectively, separated by some constant distance. A body located at point $P$ has Cartesian coordinates $\left(x,y,z\right)$ and $\left(x',y',z',\right)$ in $O$ and $O'$, respectively; and similarly, the spherical coordinates of $P$ in the unprimed and primed frames are $\left(r,\theta,\phi\right)$ and $\left(r',\theta',\phi'\right)$.

Say $O$ determines the Cartesian coordinates of $O'$ to be $\left(a,b,c\right)$. Then the Cartesian coordinates of $P$ in $O$ are simply related to the primed Cartesian coordinates: $$\left(x,y,z\right)=\left(x'+a,y'+b,z'+c\right)$$

My question, then, is how does one compute the affect of translation on spherical coordinates. That is, given $\left(a,b,c\right)$, how can one write $\left(r,\theta,\phi\right)$ as a function of $\left(r',\theta',\phi'\right)?$

There is kind of an answer over at Math. All you can do in spherical coordinates is to change the position of your "pole", i.e. you have $(1,0,0)$ in your first coordinate system, which is mapped to some $(r', \vartheta', \varphi')$ in the second coordinate system. The two angles represent a rotation, and the $r$ represents a scaling.

I think since we still deal with a vector space (for the angular part), you can simply add the origin of coordinate system two to vectors of the first coordinate system. The radius is a scaling, and needs to be multiplied. Hence for a point $p=(r, \vartheta, \varphi)$ in coordinate system 1 you get:

$$
p'=(r \cdot r', \vartheta+\vartheta', \varphi+\varphi')
$$


Are there ways to estimate size of the "whole universe"? Words escape me, but by "whole universe" (I think) I mean everything that's spatially connected to the observable universe in a conventional sense. If there is a better term for it, please let me know!

I was surprised to find out that theories and measurements about the big bang seem to mostly relate to the size of the observable universe via expansion, and if I understand correctly don't say much about the size of the whole universe.

Have I got this right? Are there any ways at all to try to estimate size of the whole universe? Does the term even mean anything?

See @RobJeffries' excellent answer to When will the number of stars be a maximum? and @Acccumulation's comment for background on this question.

tl; dr The universe is probably infinite, but if that's the case it's impossible to verify. If the universe is finite, and small enough, and the global curvature is equal to the curvature of our observable universe, then we will be able to estimate its size.

If the global curvature of the universe isn't positive, then the size of the universe is infinite, (and it's always been infinite since the dawn of time, at the Big Bang), assuming the topology of the universe is trivial.
Measurements of the observable universe indicate that the curvature may be zero (giving a flat universe), or even negative; if the curvature is positive, then its value is very small. We assume that the observable universe is typical of the whole universe, but of course that's impossible to verify.
Wikipedia says that

experimental data from various independent sources (WMAP, BOOMERanG,
and Planck for example) confirm that the observable universe is flat
with only a 0.4% margin of error.
[...]
The latest research shows that even the most powerful future
experiments (like SKA, Planck..) will not be able to distinguish
between flat, open and closed universe if the true value of
cosmological curvature parameter is smaller than $10^{-4}$. If the true
value of the cosmological curvature parameter is larger than $10^{-3}$ we
will be able to distinguish between these three models even now.

So even if the total universe does have positive curvature, its size is much larger than the observable universe, assuming that our observable universe isn't a patch that's abnormally flat.
According to How Big is the Entire Universe? by Ethan Siegel, if the curvature is positive, then the diameter of the total universe is at least 14 trillion lightyears, but that article is from 2012, more recent calculations may give a larger value.


Is there anything special about our position in the galaxy thats necessary for life? It's a fact that our planet's location from our parent star (among other factors) is of crucial importance to the existence of life. But can our position in the galaxy be considered a factor when considering potential planets to carry life?

In any given position in the galaxy for instance there are distinct radiation levels, distinct gravitational effects, space-time, etc. These factors could contribute to the existence of life on earth or even its continuity, wouldn't they?

Certain regions of the galaxy are more favourable for life, others less. Life takes some time to develop, and it needs chemical elements heavier than helium, e.g. carbon or silicon.

To get sufficient time to develop, the solar system must not be disturbed too much by neigbouring stars. Therefore inner densely populated parts of the galaxy are less suitable than the periphery we are located. (The central star of the planetary system must not be too young, as well, to allow evolution.)

Heavier chemical chemical elements are sufficiently abundant mainly in parts of the galaxy with relatively young stars (population I). This kind of stars is found mainly in the thin disk close to the galactic plane.

Very old stars tend to expand, and may eventually destroy potential life on their planets, if they are surrounded by rocky planets at all.

More discussion on a potential galactic habitable zone on Wikipedia.


What is the underwater temperature of Europa? If Europa has an oxygen atmosphere with a water mantle, is it possible that there could be life under the crust of Europa ice on the ocean where it is warmer? What could be the temperature under the ice?

Image of all the water of Earth beside it in a sphere.

No, nothing on Europa could possibly be photosynthesizing as we know it. Jupiter doesn't emit light, and what it reflects from the sun is not enough, plus there's no significant amount of carbon dioxide there.

However, the first life on Earth was not plants, it was chemotrophs. These bacteria gained their energy from the heat of hydrothermal vents, and if there's life on Europa, it would almost certainly take this form.

About the temperature: Europa's heat comes from friction due to tidal forces, so it comes from within. The water temperature increases as you go down, but the pressure is crushing down there and it's unknown whether life could form under those conditions.

The tenuous oxygen atmosphere comes from charged particles hitting it's cold surface and creating vapor. It does not indicate biological processes.


What if the black hole in the center of the galaxy grew faster? Imagine that hypothetically the black hole in the center of the milky way gradually increased in mass by for example 50% every year. That is exponential increase in mass.

Which visual effects would we see on for example Alpha Centauri, stars more distant and on M31? Red or blueshift? Lensing?

How about linear increase in mass?

Are there software I can use to visualize it as seen from earth?

this is not really an answer to your question, but to the assumption you made.

The BH in the galactic center cannot grow that much. There are two problems. First, there is not enough food close by. Any potential food (gas clouds, stars, and dark matter) will have some non-zero angular momentum preventing it from coming close enough. Loosing this angular momentum is difficult (it cannot be radiated away like energy), the only way is to exchange it with other objects either via impact (of gas clouds) or gravitational interactions with other objects (not the BH).

The second problem is that a feeding BH is surrounded by an accretion disc of hot gas. In that disc, angular momentum is slowly transported outwards and mass inwards via viscosity (that viscosity most likely originates from turbulent magnetic fields that become unstable -- the magneto-rotational instability). This process inevitably heats the accretion disc to very high temperatures ($10^{6-9}$K) such that it emits a wind of raditation and particles (similar to the Solar wind, but much much stronger). If the BH feeds too much, this wind becomes so strong that it pushes away any further infalling material (potential food).

This second process limits the growth of any BH to double in mass in no less than about $10^6$ years, I think (I'm not too sure--if you want a precise value, consult the literature), even if the first problem was no issue.


Solar Wind and Asteroid orbital behavior Can the solar wind affect the orbital elements of a nearby asteroid?

I would like to know if solar activity has an effect on the asteroid's orbit or gravitational stability and or can cause a gravitational disturbance to a meteoroid near the earth.

Sort of, via radiation pressure and heating.
The Sun emits photons, which carry with them energy and momentum. Any flux of photons applies pressure to an object it hits; this is the basic principle behind solar sails. In general, the larger the object, the more force it feels, since pressure is force per unit area. There are two main ways light from the Sun can significantly affect minor planets, such as asteroids:

The Yarkovsky effect: For a rotating body, there is a delay between when a portion of it receives heat and when the heat is re-radiated. The spinning object then has an offset between the angles of absorption and emission, which means that the radiation applies a net force. This can cause slow but steady changes; over 12 years, the asteroid 6489 Golevka deviated from its predicted orbital path for 15 kilometers.
The Yarkovsky–O'Keefe–Radzievskii–Paddack effect: The YORP effect occurs when irregularities in the surface of an object causes incoming light to be scattered in different directions. This changes the rotation direction and rate of the body. This has been observed in asteroids 54509 YORP (a rotational change of 250 degrees over four years) and 1862 Apollo.


Is a white dwarf hotter than a Red Giant? From what I've read, white stars are hotter than red ones. But a white dwarf would have just heavy elements to fuse, so shouldn't it be less bright?

"White" stars are typically much brighter than Red stars, as both the "color & brightness" of a star are directly proportional to the temperature. The only reason there are "bright" red stars is that their radius is incredibly large. Note that the "color" of a star is directly linked to the temperature.

The equation that best demonstrates this is the luminosity equation of a black body. Stars aren't perfect black bodies, but they are close enough that they are treated as such.

  L = 4πR²σT⁴

this equation tells us that the Luminosity (L) is proportionate to the Radius Squared (R²) and the Temperature to the Fourth power (T⁴). The bigger the brighter, or, the hotter the brighter. Meaning that for a given radius the hotter the star, the more luminous, and the same goes for stars of the same temperature, the larger the radius the more luminous.

White dwarfs on the other hand are not stars in the sense that they do not fuse anything, they simply glow due to the lingering heat that was generated during their time as stars.

As shown in the HR-Diagram, White Dwarfs are some of the hottest objects in the universe, and as stated by agtoever there has not been enough time for even the oldest white dwarf to have cooled passed something like 4800K.


The relation of the lit side of the moon and the sun During different days when both the sun and the moon are seen I notice that the axis of the lit, illuminated, side of the moon does not pass through the sun there is some degree of diversion. Why?

I'll assume you mean the lit side of the Moon (this wasn't clear from the original question, but I've edited it).

Suppose there's a crescent Moon visible in the sky during the day. Plot a straight line between the tips of the crescent, then plot a perpendicular straight line across its center across the sky. Intuitively, that line should pass through the Sun. You're saying it doesn't, and you're asking why.

In fact, that line does pass through the Sun, and what you're seeing is an optical illusion.

If you imagine the sky as a perfect hemisphere, with you at the center, the geometry will match up. But we we tend to imagine it as a flattened dome, probably because the horizon is obviously far away, but we have no reference points for the zenith so we imagine it being closer. This is (part of) the source of the "Moon illusion", where the Moon appears bigger when it's close to the horizon. In fact, it's very nearly the same angular size, but since we imagine it being farther away it must be "bigger".

The same mention distortion of the sky's geometry makes it look like the axis of the Moon's lit side doesn't point directly to the Sun.

Next time you see this, try holding up a straight rod intersecting both the Sun and the Moon. I think you'll find that the rod passes directly across the axis of the Moon's lit side.


How to find the distance between two galaxies from ra and dec and redshift I have ra and dec and corresponding redshift of galaxies. I would like to calculate the distance between two galaxies.
I did like this to find the nearest neighbour galaxies.
Suppose I am trying to find the nearest neighbour of galaxy A(which has RA,DEC)  and the target galaxy has ra1,dec1,
so to find the distance between two galaxy, I did like this
dist_sqd= sqrt( (RA-ra1)**2 + (DEC-dec1)**2 )
#RA, DEC are in degree

Is this the correct way to find the distance.
As there is projection in 2D, Since, I already have redshift, Is there any best way to find the distance between two galaxies properly?

Since you are presumably using Hubble's law to get the distance to each of the galaxies, the approximate estimate would the "the greater distance ± the smaller distance".
But the geometric solution is useful too. Your strategy appears to be something like:

Calculate angular separation $\Theta$ from RA and DEC.
By trigonometry, the distance is then $d_{a,b} = \sqrt{d_a^2\sin(\Theta)^2 + (d_b - d_a\cos(\Theta))^2}$

The problem is that you can not calculate the angular separation on a sphere by the Pythagorean formula, instead, you need the haversine formula. To keep this post self-contained, that is:
$$\Theta = 2\sin^{-1}\left(\sqrt{\sin^2\left(\frac{\phi_2 - \phi_1}{2}\right) + \cos(\phi_1)\cos(\phi_2)\sin^2\left(\frac{\lambda_2 - \lambda_1}{2}\right)}\right)$$
Where $\phi$ is declination and $\lambda$ is right ascension.
People are also linking this answer from the comments, which deals with the problem by converting everything to Cartesian coordinates first, which may or may not be easier for your use case.


Natural units of star clusters angular momentum and energy I'm calculating the total angular momentum of a bunch of theoretical stars of a few solar masses, separated by a few light years, moving at velocities of about a few dozens of km / sec.

Mechanical angular momentum relative to an arbitrary origin (the cluster's center of mass is a good choice) is this :
\begin{equation}
\vec{L} = \sum_i \vec{r}_i \times m_i \, \vec{v}_i,
\end{equation}
so the SI units are $\mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{sec}$.  But in real astronomy, what should be the most simple/natural units for star clusters ?  $M_{\odot} \cdot (\mathrm{ly})^2 / \mathrm{year} \equiv M_{\odot} \, c^2 \cdot \mathrm{year}$ ?

What are the angular momentum units used by astronomers ?

And what about energy units for a cluster of stars ?  I'm interested in kinematical energy and gravitationnal potential energy of the whole cluster.  Certainly not joules !?  And $M_{\odot} \, c^2$ would certainly be too much, I suppose !

Using unit analysis, I've found that a natural unit of angular momentum should be
\begin{equation}\tag{1}
L_0 = \frac{G M_{\odot}}{c} \approx 8.816 \times 10^{41} \, \text{kg} \, \text{m}^2/\text{s}.
\end{equation}

In the case of a bounded system of two equal masses, circularly moving around their center of mass, the total angular momentum (relative to the center of mass) is
\begin{equation}\tag{2}
L_{\text{bound}} = 2 M \, v \, r = \frac{2M \, v^2 \, r}{v}.
\end{equation}
Now, using Newton's equation and the gravitationnal force, we have
\begin{equation}
M \, v^2 = \frac{G M^2}{4 \, r}.
\end{equation}
Thus, equation (2) becomes :
\begin{equation}\tag{3}
L_{\text{bound}} = \frac{G M^2}{2 \, v}.
\end{equation}
The classical theory cannot be trusted when $v \sim c$, and since $v < c$ (forgetting the 2 factor) :
\begin{equation}\tag{4}
L_{\text{bound}} \approx \frac{G M^2}{v} > \frac{G M^2}{c} \equiv L_0.
\end{equation}
So $L_0$ appears to be a good unit.  When $L \sim L_0$, we then need relativity.  We're in a classical regime when $L \gg L_0$.

The system isn't bounded when $L \gg L_{\text{bound}}$, which implies $L \ggg L_0$.

I don't know if this is a good way to introduce the $L_0$ unit defined by (1).  Any opinion/suggestions/comments on this ?

Do astronomers use $G M^2 / c$ as natural unit of angular momentum, when they study star clusters and globulars ?


Stellarium script star of bethlehem Does anyone out there have a script to replicate the star of Bethlehem.  In this case meaning a conjunction of Venus and Jupiter near Regulus.

I like sharing the following quote from Wikipedia on the Star of Bethlehem

Astronomers have made several attempts to link the star to unusual celestial events, such as a conjunction of Jupiter and Saturn or Jupiter and Venus, a comet, or a supernova.
Some modern scholars do not consider the story to be describing a historical event but a pious fiction created by the author of the Gospel of Matthew.

This said, luckily you specify that you are after a conjunction of Jupiter and Venus. WolframAlpha can determine the conjunction of Venus and Jupiter closest to the year 0:

8:07:26 pm LMT  |  Thursday, October 14, 2 BC (extrapolated Gregorian calendar)

I have never used Stellarium, but other software, where entering that date and searching for Jupiter did the job.


Identify this 2 axis motorized tripod I am NOT an astronomy person, so this may be a dumb question, however I have recently found a tripod that appears to be for star tracking as far as I can guess. I would like to know the type of tripod and if possible the brand as it is not labeled and I do not know how to operate it.

It appears to be an older version of the Bushnell Voyager Sky Tour 800x70mm Refractor Telescope

Full manual here


Have we measured or found evidence of a detectable origin point of the big bang? Based on reading the wiki page of the big bang, a simple question with a probably not a simple answer came to mind.

The statement on the wikipedia page was.

  Since Georges Lemaître first noted in 1927 that an expanding universe could be traced back in time to an originating single point, scientists have built on his idea of cosmic expansion.

My question is:  Have we measured or found evidence of a detectable origin point of the big bang?

There is no single origin point.

Before the big bang, the entire universe was contained in an extremely dense singularity - there was no "space" for the universe to expand into, the very fabric of spacetime was contained entirely in the singularity. There is no single "origin point" in the universe where the big bang took place where we'd see the entire universe receding from us. Rather, from any point in the universe we can see everything else receding. The big bang happened everywhere, since the whole universe, including space itself, came from that singularity.


US observations relevant in UK? When I listen to the Planetary Radio podcast, there is a section at the end where they suggest objects to look out for in the night sky. If I am in the UK, but they are talking about the USA, will I still be able to find the objects they are referring to?

If they mention times of night then you have to figure out to what extent the time zone (BST/GMT vs the US time zones) makes any difference, but it shouldn't make more than an hour difference either way for an object that is moving sidereally with the rest of the night sky.

The main difference would be that the USA is at lower latitudes than the UK. Therefore there could be instances where an object with a small or southerly declination is better seen from the USA than the UK. Conversely, more of the sky is circumpolar from the UK.

Finally, if you are trying to see a specific event at a given time (UT), then of course it may well make a difference. The Sun could be up in the UK and down in the USA, or vice-versa.


How to compare a reflector telescope to a refractor one? I already have a small 60mm Tasco refractor (with 300mm focal lenght) and a homemade 40mm refractor (with 1300mm focal lenght) and I want to buy a new telescope to have a better experience (I don't have an EQ mount yet)

I face the current choice: for the same price, I can have a 70mm refractor (900mm fl) with a great EQ mount, or a 114mm reflector (900mm fl) with a less great EQ mount...

How can I compare those 2 two? They are not the same type, and I feel like comparing oranges with apples.

I know the technical difference between them, and I know the details about collimation. I just don't know how to rate them and chose one.

In short: Is it better a big reflector or a small refractor?
(If I want to extend the information to other choices I might encounter)

Small and effective refractors are easier to make than small and effective reflectors, but reflectors are cheaper in general. Unless you have some special reason to want to avoid a larger aperture (eg because you are engaged in solar observation) or a special reason why you want the optical qualities of a refractor, you almost certainly want a bigger aperture and so for a fixed price you are likely to be better off with the reflector.

Not having a EQ mount may be a bit of an issue though.


How much oxygen did the Warwick/Cambridge study find in the rocky debris around white dwarf GD 61? There are several articles out saying that a Warwick/Cambridge study of ultraviolet spectroscopy data from the Cosmic Origins Spectrograph on the Hubble Space Telescope found a large amount of oxygen in rocky debris around the white dwarf GD 61. They say that this amount of oxygen is indicative of a larger mass (asteroid), that was once composed of the rocky debris, which was 26% water.

(source: sciencedaily.com)
Image credit: copyright Mark A. Garlick, space-art.co.uk, University of Warwick and University of Cambridge
What I cannot seem to find is how much oxygen the spectroscopic examination found. Also, what methods did they use to determine how much water there would have been based on the level of oxygen?

In Farihi et al. (2013) (it's a Science paper, unfortunately I'm not sure its content is freely accessible), they actually measured metal excess in the white dwarf GD 61 (for an astronomer, everything that is nor hydrogen neither helium is a metal). Due to high surface gravity in white dwarfs, any heavy element should sink rapidely in its atmosphere; therefore, they infer that, if you find traces of metals, it means that it comes from accreted material, "polluting" its atmosphere.

So if you find oxygen in a white dwarf spectrum, you know that it has to come from somewhere else, and if the white dwarf has a circumstellar disk, it has to come from planetesimals that are accreted from the disk.

Now, what they did was to measure abundances (or upper limit for their abundance) for a bunch of chemical elements (O, Mg, Al, Si, Ca, Fe) and carbon. Then, they could infer different things to determine the total oxygen budget:

There is a carbon deficiency, that means that carbon abundance has no impact on the total oxygen budget;
Elements Mg, Al, Si, Ca, are assumed to be carried as MgO, Al2O3, SiO2, and CaO at their maximum abundance, so once they had these abundances, they could substract it to the total oxygen budget;
The remaining oxygen excess, found in FeO and debris, is interpreted as water-rich (if you want to form FeO, you need water somehow).

That's how they were able to determine this water abundance in debris around the white dwarf GD 61.

I'm not sure the oxygen abundance would be very meaningful here, but since you asked for it, it is of -5.95+/-0.13 dex.


View of other planets from the Earth There appears to be varying positions/spots of view of other planets from Earth, if I'm not wrong. Now, given that all planets are on the same plane, why do their positions change on the night sky for similar seasons, when they come back to earth's view? How is this possible?

Shouldn't all the outer planets be observed within the horizon's range (and not in the high skies), now that they lie on the earth's tropical plane? Give or take a short distance from where the horizon 'touches' the ground?

The Earth's equator is not in the same plane as the planets. The planets, including the Earth orbit the sun in a plane called the ecliptic. The equator of the Earth is tilted by 23.5 degrees to the ecliptic.

This means that the planets will be seen on the same great circle of the sky that the sun is on.  The planets move relative to the sun and the stars, and so when the ecliptic is high in the sky, the planets will appear high above the horizon. It will tend to be during winter that the planets will appear highest.


What are iron stars? I have a question about iron stars. I was fascinated about the theory about the inability of protons to decay, and it led me to a point where I read something about some special kind of star called an "iron star". Can anybody enlighten me a bit, if possible, on what exactly it is and why it may be significant? And how do iron stars form?

"Iron stars" are a hypothetical kind of stellar remnant - similar to a neutron star or white dwarf - that are hypothesized to form in the far future of a Universe undergoing heat death, under the assumption that protons do not decay, which is something that we do not know if is or is not the case. Some of the more promising theoretical models for proton decay (a fair number of so-called "Grand Unified Theories" [GUTs] which one should note, despite confusion in the popular press, are not the same thing as unified quantum gravity theories like string theory) have already been experimentally rejected, but that doesn't mean it doesn't occur at a rate slower than we can currently detect (or may ever be able to).

In particular, the idea rests on the fact that, given infinite time, there is no such thing as a stable element except, perhaps, for iron. What we call "stable" elements are really just radioactive elements with a half-life too long to observe, or too long to matter, whichever you prefer. For example, up into maybe about 16 years (300 megaseconds) or so ago, the element bismuth (element 83) was widely believed to be entirely stable, but then its most stable isotope was discovered to be unstable to alpha decay with a half-life of around $10^{19}\ \mathrm{yr}$ ($10^{20}\ \mathrm{Ms}$).

Feynman (I believe) once said that "everything not forbidden is compulsory". What this means is that, according to the laws of quantum mechanics, there is a nonzero probability that an agent can register any process not forbidden by, say, a conservation law, as having occurred, and will register such as having occurred given infinite time. In the case of atomic nuclei, if there is a nucleus with less nuclear energy than a given one, there is a probability to register the latter as now being the former, i.e. as having undergone a decay to it. (It cannot go the other way, because that would imply a violation of the law of conservation of energy.)

All elements heavier than iron have more nuclear energy than iron. Thus, by this principle, ALL isotopes of them are radioactive: we just don't know the half-lives. E.g. even common copper is radioactive (unstable) and will likely undergo alpha decay to iron, two atomic numbers below it. It's just that the timescale for this decay is likely so inordinately long that we cannot ever have any hope of observing it directly.

Thus, given unlimited time, anything made entirely of such elements heavier than iron will decay until it becomes purely composed of iron.

However, now one might be wondering something: clearly, stars are not made of heavier elements, and white dwarfs are typically also made of light elements like carbon and oxygen, and iron cores collapse into neutron stars if not black holes, neither of which can reasonably be called "iron", so how can we have "iron stars"? Well, it turns out that something opposite to radioactive decay will also happen for at least aggregate concentrations of light elements too: they will undergo cold fusion. Helium has less nuclear energy than 4 (protonic) hydrogens - that's why the Sun and main-sequence stars are able to perform hot fusion to a net energy gain to begin with. Contrary to what you may have heard, cold fusion is not impossible: it's just really damned slow, as in, taking far, far longer than the elapsed age of the Universe, just like it is for the decays just mentioned. But in any case, in "eternity", cold fusion will likewise raise the atomic number of the nuclei in a white dwarf, to again, converge on iron. This is an "iron star".

But, as said, this all depends on two things:

that the ultimate fate of the Universe is heat death,
that there is no proton decay.

Neither of these are things we can "know" to good confidence. In particular, there are still major open questions in fundamental physics that leave the door wide open to these being things that could go differently. In the case of the ultimate fate of the Universe, it is intimately bound up with the nature and governing laws controlling the still as-yet poorly understood "dark energy". Heat death will occur if the dark energy remains constant, i.e. acts as a true cosmological constant as per Einstein. However, if the dark energy either rises or falls with time, we may end up with two other "catastrophic" scenarios respectively called the Big Rip and Big Crunch. We have no evidence to rule these out because we don't have any regarding the physics governing future evolution of dark energy. The formation of iron stars by cold fusion and radioactive decay of so-called "stable" heavy elements is estimated to take a half-life of $10^{1500}$ (years, megaseconds, pretty much doesn't matter at this scale as that exponent itself is likely approximate).

Likewise, there are potential mechanisms - such as quantum tunneling via "virtual black holes" - that could still cause the decay of protons with half-lives on the order of $10^{200}$. A detector of the type we use now to look for proton decay (basically a big tank of water) could not detect these directly, as it would require many orders of magnitude more mass than there is in the current observable Universe and, moreover, would collapse into a black hole long before one got that far.


Which astronomer set his beard on fire because he left the cap off his telescopes finder? There is a famous story of an experienced astronomer setting his beard on fire because he left the cap off his telescopes finder. I know it's said to be true of Galileo, but I thing it's relatively recent - i.e. 20th century.
Does anyone know his name? Or can someone cite a source?

From the book Turn Left at Orion by Guy Consolmagno and Dan M. Davis, Galileo might have set his beard on fire with a telescope pointed at the Sun.

  And you must keep your eyes -- and your clothing -- out of the path of the projected light. It is said that Galileo set fire to his beard doing this!

In his defense, he was the first person to look at the Sun with a telescope, so you can't blame him for not knowing about the associated risks!

But this story might also be apocryphal. Many people set various things on fire with a telescope (some even doing it on purpose).


Fate of the Spiral Arms of the Milky Way and Andromeda In a few billion years, the Milky Way and Andromeda will collide. Chances are, there won't be many spectacular stellar collisions, and there's a pretty good chance that the Solar system (and most other stellar systems) won't be flung out into deep space, or be torn apart by incoming stars. But what about on a larger scale?

The product of the collision - dubbed "Milkomeda" - will be an elliptical galaxy, unlike the Milky Way and Andromeda, which are both spiral galaxies. My question is this:

How will the structure of the spiral arms be effectively destroyed by the collision? What mechanisms will cause them to coalesce, and what structures will take their place?

It indeed appears the Andromeda galaxy (M31) and The Milky Way (MW) are en route to a collision. This will lead to a merger of the two galaxies forming an elliptical galaxy. The flattened disc structure of M31 and MW comes about because their gas dissipated energy but conserved momentum: a disc is the minimum energy state at given angular momentum. The gas in the disc then forms stars, which are more or less on the same orbit as the gas, i.e. circling the respective galaxies. A disc galaxy is dynamically cold, i.e. the speed of the mean motion of the star around the galaxy is much larger than their random motions.

The merger destroys all these structures and replaces them with a much smoother and dynamically hot structure, when the density is roughly constant on (concentric) triaxial ellipsoids. There are several types of stellar orbits in such galaxies, but most important are the so-called box orbits, when the stars oscillate in a large box-like volume and have no preferred mean direction of motion. In the early phases, the elliptical galaxy will have some temporary structures (so-called shells), which are remnants of the merger process itself.

The gas in M31 and MW will most likely be swept into the inner parts of the elliptical where it may form many new stars and contribute to the feeding of the AGN which emerges from the coalescence of the two supermassive black holes of M31 and MW. But energy fed back from the new formed stars via supernovae and stellar winds as well as from the AGN will drive the remaining gas out of the elliptical galaxy, leaving it "red and dead", i.e. without star formation and young blue stars.

The merger of M31 and MW will hardly increase the already neglible rate of stellar collisions. Given the vastness of galaxies compared to the size of stars, such collisions are extremely unlikely. The only places where such collisions occur are the very dense cores of globular clusters.


Ceres is spherical, Vesta isn't – is it the largest non-spherical solid body in the solar system? I was showing a recent video about Ceres to a friend and we looked up asteroids. Ceres has attained a fairly spherical shape, but Vesta still has a way to go. From Wikipedia:

  Vesta's shape is close to a gravitationally relaxed oblate spheroid, but the large concavity and protrusion at the southern pole (see 'Surface features' below) combined with a mass less than $5×10^{20}$ kg precluded Vesta from automatically being considered a dwarf planet under International Astronomical Union (IAU) Resolution XXVI 5.

I don't know the proper term to use here. Perhaps it is "hydrostatically relaxed" or something similar – I'm looking for the term that would suggest a body has reached a fairly round shape without major excursions, under the influence of its own gravity.

I think Ceres meets this criterion, and Vesta doesn't. Is Vesta the largest such body (by dimension or volume, not sure which would be the most reasonable measure)?

Among the asteroids 4 Vesta is the largest non-spherical object in the Solar system. But far out there in the Kuiper Belt is 136108 Haumea, and maybe yet undiscovered objects, that are very elongated. Haumea is 1,960 × 1,518 × 996 km. That's larger than the 960 km diameter Ceres and hardly spherical with 2:1 proportions. Its fast rotation and composition of malleable volatiles makes this possible. Maybe Haumea is "hydrostatically relaxed" without being spherical?

Haumea images below from Wikipedia.


Is it possible neutron stars are actual elements? As you go up the periodic table (more protons), the ratio of neutrons to protons steadily increases as well. Are we sure there are absolutely no protons and electrons in a neutron star, or could there be so much more neutrons that we cannot measure any protons and electrons? Perhaps then a neutron star is a nucleus of some huge element with a neutron:proton ratio higher than we can distinguish.

To be considered an element, they would have to have a positively charged nucleus. They don't. The neutron star is largely neutral.

They would have to have a cloud of surrounding electrons. These electrons would have to share "orbitals" with other electrons around other, nearby neutron stars. That doesn't happen.

Finally, at these scales, gravitational interaction predominates. Even if neutron stars did have some positive charge, and did have electrons orbiting them, interaction with other neutron stars would still depend a lot on gravity.

Finally, even if a neutron star was charged, it's not clear whether electrons around them would actually fall into some kind of orbitals governed by quantum mechanics. QM tends to not happen at such a scale.

In conclusion, no, they are completely unlike atoms and their nuclei.


What equipment and techniques were used to study Betelgeuse's diameter in 1920? Comments below What is the maximum distance measurable with parallax? discuss challenges associated with parallax measurements of Betelgeuse and link to Wikipedia’s Betelgeuse; Distance measurements which contains the intriguing sentence

  In 1920, when the first interferometric studies were performed on the star's diameter, the assumed parallax was 0.0180″.

How exactly were the first interferometric studies performed on Betelgeuse's diameter? What actual equipment and technique was used to do this in 1920?

Was this two optical telescopes and a long baseline?
Are they still around today?
Was this the first attempt to measure a star's diameter directly?
Was this attempt at least partially successful?

Addressing item #4 further, the paragraph goes on to say

  This equated to a distance of 56 pc or roughly 180 ly, producing not only an inaccurate radius for the star but every other stellar characteristic. Since then, there has been ongoing work to measure the distance of Betelgeuse, with proposed distances as high as 400 pc or about 1,300 ly.

Had they used the correct parallax and distance, would they have arrived at a diameter that would have been roughly correct?

These measurements were made by Michelson & Pease (1921) using the 100-inch Cassegrain telescope at the Mount Wilson Observatory. Modern-day interferometry typically uses multiple telescopes (e.g. IOTA and the Keck Observatory interferometer), as this is typically needed to achieve long baselines (with long baselines admittedly being less important for short-wavelength observing than for, say, radio waves). However, the interferometer the two astronomers used was slightly cruder: a set of four mirrors mounted above the 100-inch mirror. These mirrors were movable, reaching separations of up to 20 feet. Light would reach the outer two mirrors (spaced 20 feet apart) and then be reflected to the inner two mirrors, which would then send the two beams to the main mirror, turning the telescope into an interferometer.

This was indeed the first time the diameter of a star (excluding the Sun) was measured; Michelson had built the instrument three decades earlier, but tested it by observing the Galilean moons. Betelgeuse is an obvious choice, as its large envelope and proximity to the Solar System mean it has one of the largest stellar angular diameters at any wavelength as seen from Earth.

I believe the interferometer is still in existence, in the observatory's museum. Here, however, is the diagram presented by Michelson and Pease in their 1921 paper:

With regards to the star's physical diameter: The pair measured an angular diameter of 47 milliarcseconds, which is certainly in line with the accepted values today. There is variation in Betelgeuse's angular diameter, of course, due in part to the star's variability, but 47 milliarcseconds falls right in the middle of the range expected at optical wavelengths. Had they had the correct parallax, they would likely have arrived at a very reasonable radius.


If thousands of exoplanets have been discovered, why haven't we discovered planet 9 yet? According to NASA, we've discovered 3,453 exoplanets so far [1]. With some of the farthest being at 13,000 light years away [2]. This would suggest today scientists have a really advanced capability for this kind of discoveries. However, the existence of a ninth planet in the solar system (which has not only been proposed but its characteristics are already predicted with great precision[3]) has not yet been discovered. If a ninth planet existed in our solar system, shouldn't it have already been discovered? How can exoplanets thousands of light years away be discovered weekly, while a giant planet right in our backyard still lays undiscovered?

There are two primary reasons Planet Nine (if it exists) is extremely hard to detect: it's very dim and its position is not well constrained.

Brightness - We are able to see planets in our solar system because they reflect light from the sun.  The amount of light they reflect depends on how large the object is, how reflective it is, and (most importantly) the amount of sunlight that hits it.  The hypothesized distance at which Planet Nine is orbiting is so far away that it will receive only a tiny amount of sunlight. At this distance, its apparent magnitude would be greater than 22.  In comparison, Pluto's apparent magnitude is about 15.  Magnitude is a logarithmic scale, so that means that Planet Nine would be $631$ times dimmer than Pluto, making it incredibly hard to see.
Unknown Position - The estimates of Planet Nine's characteristics are based on how it gravitationally disrupts the orbits of some trans-Neptunian objects.  Based on these observations, scientists can calculate where it is likely to be, but these calculations depend on its size.  Estimates of its mass range from 2 to 4 times the mass of the Earth.  And all of these are just estimates, giving us a relatively large patch of sky to search to find Planet Nine.

Combining both of these reasons, it is very difficult to search for Planet Nine.  We're looking for an incredibly dim object in a large area of the sky.

Now, when we look for exoplanets, things are actually a lot easier.  To detect exoplanets we hardly ever look for the planets directly.  We've only detected about 20 planets via direct imaging.  Instead, we look at the light from the star.  The stars are many orders of magnitude brighter than the planets, and are very easy to see.  All we have to do is see how the light from the star changes over time.  An early method detect shifts in the radial velocity of the star (its motion towards or away from us) using Doppler spectroscopy.  The more prevalent method today, the one that the Kepler mission used, is transit photometry.  We measure the light output of a star, and look for dips in its intensity, caused by the planet orbiting in front of the star and blocking some of the light.

Because we're looking at a star, we don't have the problems we have when looking for Planet Nine.  Stars are very bright, their positions are known very accurately, and we don't have to wait too long to make a detection.  Plus, surveys like Kepler are able to scan many, many stars at once.


Turn celestial coordinates into light years I have a question about celestial coordinates. I have been working on a calculator that takes in different stars and tells you how far apart they are, and their coordinates (x, y, z). But I didn't realize there was no such system. Is there any way to turn celestial coordinates into a coordinate plane, i.e. (x [ly away from earth], y [ly away from earth], z [ly away from earth])?

Rectangular celestial coordinates are used regularly. The most common examples are the JPL ephemerides for solar system objects. Wikipedia's page on geocentric equatorial coordinates includes the equations required to convert right ascension and declination to xyz.

The most commonly used reference system for astrometry is the ICRS coordinates that are closely aligned with the J2000 equatorial system. Implementaions of ICRS were used for the Hipparcos and Gaia astrometry data.


Brown Dwarf transcending past the sun with the naked eye I seen with my naked eye in mid 2014 from Ca. right after sunrise a black spot in front of the Sun.I observed it for about a hour with my naked eye. Here is a picture that closely resembles what I observed, except the object was a third the size of the black dot in the posted picture and not in the middle of the sun but off to a angle.Since Venus did not transcend with the Sun in 2014 my speculation would have to say it might have been a Brown Dwarf.The link provided below of civilizations from the past documenting Venus transcending with the Sun throughout history could be observed with the naked eye,So a Brown Dwarf would be easier to see with the naked eye.The picture is from google and is the closest image I can give to what I seen.I have no knowledge of the picture and can not determine if it is real or fake.

Proof of venus or a sunspot seen with naked eye transending past the sun
http://sunearthday.nasa.gov/2012/articles/ttt_73.php

Possibly an aircraft or flock of birds, or maybe an insect really close to your eye, but more likely an illusion or misperception. It's not very "bright" to look directly at the Sun with your naked eye!


Why does Gaia use only calcium NIR lines for stellar radial velocity measurements? I was reading this overview article about the Gaia spacecraft and I saw the following statement:

These spectra provide radial velocity information that are used to study the kinematic and dynamic evolution of the Milky Way. Radial velocities are derived from three isolated Calcium lines at 849.8, 854.2 and 855.2nm. Other lines in the 847 to 874nm range can provide data on star composition, surface gravity, and metal abundance.

note: as pointed out in the comments below, the third line is at 866.2nm, not 855.2nm - this is suggested to be a typo - the '855' number also shows up on this ESA page.
The high resolution spectroscopy only seems to operate between 847 and 874nm, and "three isolated Calcium lines" are used to measure radial velocity.
Do all stars have enough calcium in their atmosphere to produce strong enough features to measure the radial velocity so accurately? I had thought that there are some stellar populations have very little besides hydrogen and helium in their atmosphere.
Are these always emission lines or absorption lines, or will there be some stars with one and some of the other? What fraction of stars simply won't have significant amounts of calcium?

above: Gaia's Radial Velocity Spectrometer from here, credit: ESA.

above: Gaia's Imaging system, including mirrors 4, 5 and 6, prisms, diffraction gratings, and CCD array, from here, credit: EADS Astrium.

above: Gaia's Optical Module, including Radial Velocity spectrometer (gratings) and a focal field corrector, from here, credit: SAS Astrium.

The Ca triplet in the near infrared are extremely strong resonance absorption lines. They are by far the strongest features in the near infrared spectra of cool G,K,M type dwarfs and giants, which will be the majority of the stars observed by the Gaia RVS.
The Ca triplet lines are so strong that even in low metallicity halo stars, that have little Ca in their photospheres, these lines are still strong enough to measure radial velocities.

The lines are much weaker and much broader for hotter O, B and A stars, and measuring radial velocities for these will be difficult and much less precise.

You can have a look at an atlas of the Gaia Ca triplet region for stars of  different spectral types in Figure 2 of Munari et al. (2001). http://cds.cern.ch/record/531022/files/0109057.pdf

I should also add that these three lines are not the only features used to determine the velocities, they are just the strongest features in the spectra of most stars.


Are exoplanets at dwarf stars less likely to have super-rotating atmospheres or asynchronous tidal locking? Dwarf stars have terrestrial sized planets orbiting in habitable zones very close to them. These exoplanets are often said to be tidally locked to their star, like the Moon is to Earth, and that they thus have one hot hemisphere and one cold.

But in the Solar system there's only one tidally locked planet, and that is Mercury which is asynchronously locked in a 2:3 relation. It does rotate relative to the sun anyway. And only one known planet which practically doesn't rotate at all, Venus. But Venus has a super-rotating atmosphere that distributes heat from insolation all around anyway.

Are there reasons to believe that asynchronous tidal locking and super-rotating atmospheres are less common for planets with a dwarf stars than in a planetary systems with a Solar like star?

I'm not exactly sure what you're asking, cause you touch on a few related points.

The reason planets around red-dwarf stars are thought to be tidally locked is because the tidal force is comparatively much greater for the habitable zone.   Take, for example, a star with half the mass of our sun.   It would have, roughly speaking, 1/16th the solar output, so a planet, to get the same amount of heat, would need to be 4 times closer or 1/4 AU.

The tidal force increases by the cube of the distance, or, 4^3, or 64, and the mass, 1/2, means that for just 1/2 the solar mass, the tidal force on a planet getting equal heat is 32 times greater.   That's a ballpark ratio to the 5th power, which is highly significant.    When you get into solar masses 25% the mass of the sun or less, the tidal force for a habitable zone planet can be thousands of times greater than the tidal force for the habitable zone in our solar system.   Because of the high tidal forces, planets in the habitable zone of small stars are very likely tidally locked, at least for any near-circular orbits.  That's pretty darn close to certain when the star is below a certain size.

There are some theories that thermal heating of a planet's atmosphere, which takes time, so it doesn't happen precisely at mid-day, but a couple hours after that, there's expected to be a tidal torque in the upper atmosphere and an equilibrium where the constant rotation of the atmosphere drags on the surface and keeps the planet rotating, so the planet never becomes tidally locked.    Venus might be an example of that.    But the articles below suggest that this is predicted to only happen in the larger red-dwarfs, 50% the mass of the sun or larger.  Article here and here for more details.

In summary, for smaller red dwarf stars, tidal locking is probably very common within the habitable zone.   For larger red dwarf stars, you might get a mix, growing less common for further out planets.

Rotations maintained by wind torque are probably, at least, to my thinking, not very rapidly rotating, similar to Venus, a slow rotation makes super-rotating trade winds easy.  (I say trade winds because I don't think weather-wind or local wind is considered super-rotating), so I think we should just consider relatively permanent wind speed and direction).

And for any slowly rotating planet, Super-rotating wind should be quite common.

For the very smallest stars, like Trappist 1, which at .08 solar masses isn't far off from the minimum size for a red dwarf, it's inner most planet has a period of just 1.5 days, it's corresponding rotation (Sidereal, not solar) has a 1.5 day period too.   That's probably a fast enough rotation to generate a significant Coriolis effect and some interesting weather (provided the planet hasn't lost it's atmosphere - which is also possible with close orbits around small stars).   That's an equatorial surface velocity of over 600 km/h, so for very close orbit planet around the smallest red dwarfs, super-rotating winds may not happen.

Looking at the planets orbital periods around Trappist 1.  The inner 2-3 planets might rotate too fast to have super-rotating winds.   The 5th one, for example, with an orbital period of 9 days, corresponding to a sidereal rotating speed at the equator of not much more than 100 km/h, it's probably likely to have super-rotating winds.  The longer the orbital period, the more likely super-rotating winds are to happen.  Any tidally locked planet with an orbital period over 10 days, should be statistically likely to have super-rotating winds (compared to sidereal rotation).   Solar rotation, well, by that measurement, all tidally locked planets with atmosphere have super-rotating winds.


Which is the largest planet ever observed? I don't know much about planets outside our solar system. Do we even know the limit of the size of how large a planet or satellite can be? The largest planet I know of is Jupiter, but could there be planets as large as our Sun or even larger? Is there a theoretical limit to the size of a planet?

Yes, there is a limit. Anything with a mass larger than about 13 times that of Jupiter would be called a brown dwarf (a failed star), though whether such an object would consist entirely of gas, or had a rocky/icy core as is probable for most giant planets, is not presently observable. Any larger than about 75 Jupiter masses and we would just call it a star. The exact definition of what a planet is (especially, the 13 Jupiter mass boundary) is still disputed.

Of the bonafide planets that have been detected and confirmed, the catalogue at exoplanets.org lists Kepler-435b as the one with the largest measured radius (although its radius error bar overlaps with that of other planets). The quoted radius is $1.99 \pm 0.18$ times that of Jupiter.

Most giant planets have very similar radii for masses between about 0.5 and 10 times that of Jupiter. The reason for this is that they are largely supported by electron degeneracy pressure. The diversity in the radii (between about 0.7 and 2 times that of Jupiter) of such planets is not yet fully understood.

The plot below shows mass vs radius for "planets". The smaller (probably rocky/icy) planets do show a trend of increasing radius with mass (the solid line is where a theoretical relationship for rocky/icy planets has been used to estimate the mass from the radius). The gas giants above about 0.5 Jupiter masses show no trend and a small scatter.


What is an Ideal apparent field of view for viewing planets through a zoom eyepiece? I was reading the specs of a few zoom eyepieces, and in every eyepiece, there is a term called 'apparent field of view'. I looked up and searched about it and figured out that if you divide this by the magnification caused by the eyepiece, you get the real field of view which is basically the diameter of the circle visible in the telescope. But one thing that I was unable to find that was what would be a good field of view for viewing planets through a zoom eyepiece, so this is my question.

Planets are very small objects, so they'll fit into whatever AFoV the eyepiece provides.
One highly-rated planetary eyepiece, the Vixen HR, has an AFoV of only 42°.
The main problem with low fields of view and high magnifications, however, is keeping the planet in the centre of the field of view.
The hardest type of mount to do this with is a manual altitude/azimuth mount, because you need to move in two axes at the same time.
In your case, with a manual German equatorial mount, you only need to worry about tracking with the right ascension (RA) axis, and your mount may well have a slow-motion control to help you out with that.

So, when looking to buy a zoom eyepiece for looking at planets, don't worry about the FoV it has. Almost all zoom eyepieces have quite small fields of view when compared with wide-field fixed-length eyepieces. I believe there's only one zoom - a Leica - which has a comparable FoV.

(And if tracking planets at high magnification manually still isn't for you, then consider a wide-angle eyepiece (82° or greater) to give yourself more "drift time" before you have to nudge the telescope)


Is it correct to say that only binary and solitary star systems are stable? Recently it has been shown that a large number of binary stellar systems exist.  It seems to me that the orbits of stars in systems containing three or more stars will inevitably be chaotic, resulting in either mergers or ejections.  Binary stellar systems, though, are effectively stable (neglecting the very slow decay of orbits due to gravitational radiation and perhaps tidal interactions).  So, it seems that in the long run we should expect most stellar systems to be either binary or solitary; the first due to ejection (eventually) of all but two of the stars from a multi-stellar system, and the second due to the ejection of single stars from multi-stellar systems.  Is this correct?

No.
It isn't proven that any real orbital systems are stable on sufficiently long time scales.  Even our (single star) Solar System is not stable.  In their paper: On the Dynamical Stability of the Solar System, Batygin and Laughlin, use the numerical techniques pioneered by Laskar to show scenarios in which various planets fall into the Sun or are ejected from our Solar System on a timescale of 20Gyr.
The area of math that dominates long term orbital stability questions is Chaos Theory, which studies the types of systems of differential equations for which very tiny changes in the initial conditions cause very different resultant solutions.  Over long time scales, tidal forces, stellar radiation, and other energy sources cause orbits to evolve.  When resonances are reached, orbits can begin to de-stabilize.  Examples of this can be found in the Kirkwood Gaps of the asteroid belt.
Given the overwhelmingly high populations of binary star systems in the Milky Way (table below from wikipedia), we could probably call binary and single star systems stable at least at 10-100Gyr time scales.

However,Correa-Otto and Gil-Hutton point out that merely being in a galaxy causes binary orbits to be perturbed:

The main influence of the Galaxy on the WBS [Wide Binary Star systems] are
twofold: first, the disruption of the binary system, which produces
gaps in the orbital elements distributions of the three sets; and
second, the changes in the orbital configuration of the surviving WBS,
which modify the orbital element distributions of the population in
each set.

Given the paucity of discovered n-star systems where n is greater than 2, I think we can safely say these systems are mostly unstable at timelines on the order of 1-10Gyrs. In fact, many of them are unstable on much shorter timelines.


If Earth didn't rotate, would we feel heavier? Suppose the Earth's rotation slowed for some reason. Would the lack of centrifugal force cause us to feel heavier than normal?

Likewise, if Earth's rotation increased, would we feel lighter as centrifugal force lifts us from the ground?

The centrifugal force at the equator is only 1/289 of normal gravity, or 0.35%. That's too small to be perceived.

http://books.google.com/?id=mE4GAQAAIAAJ&pg=PA358&lpg=PA358

It gradually diminishes as you move away from the equator, and becomes zero at the poles.

In any case, if Earth stopped rotating, the added weight at the equator would be the least of your worries. There would be major perturbations in terms of weather, etc, that would amount to global catastrophe.

  Likewise, if Earth's rotation increased, would we feel lighter as
  centrifugal force lifts us from the ground?

In theory, yes. But the rotation increase would have to be huge. Again, in that case, there would be other, much bigger effects, that would be much more important than your loss of weight. E.g. the whole planet would bulge at the equator and become more flat at the poles, with incalculable effects on plate tectonics, weather, etc.


Creating a rectangular star map I'm trying to find/create a star map between the declinations of -45 to +60, and need it to be a rectangle. Is there any software/website that would let me create this? Is there a name for this kind of map/projection?

(I've attached a diagram to try and make myself clearer)

Thanks for any help :)

Equirectangular
or equidistant cylindrical projection
is the simplest way to meet these requirements.
Plate Carrée is the case where distortion is minimal at the equator.
Dominic Ford's Map of the Constellations uses it,
as do the Deep Star Maps
by Ernie Wright of the NASA Scientific Visualization Studio.
In free software,
Stellarium calls it "Cylinder," and
Cartes du Ciel calls it "Cartesian."
If you plot your own, you could scale the horizontal dimension by
$\cos 30^\circ \approx 0.866$ to minimize distortion at
$\pm30^\circ$ declination.

Since you are not plotting the polar regions,
other cylindrical projections are possible.
The Mercator
projection is available in both Stellarium and Cartes du Ciel.
Stellarium also offers the
Miller cylindrical projection.


What would be the average distance between each 1-degree latitude on Jupiter? Do planets like Jupiter have a system of latitudes and longitudes (I believe yes)? So if yes, what purpose do they serve? Is there anything different in their purpose from Earth lat-longs? and What would be the average distance between each 1-degree latitude on Jupiter?

Yes, they have coordinate systems as well. For example, the Wikipedia page on the Great Red Spot mentions it lies 22 degrees south of the planet's equator. Jupiter has a polar radius of 66,854 km; a latitude degree is $2 \pi / 360$ times that, so almost 1,167 km.

Note that because Jupiter lacks a solid surface, longitudes are less well-defined. In fact, there are three different systems for this:

  Because Jupiter's visible features do not rotate uniformly at all latitudes, astronomers have defined three different systems for defining the longitude.

Rocky planets like Mars do have a well-defined coordinate system, as shown in this map and on this page.

These coordinates are used to mark and find interesting places on either the surface (like on Earth) or in the atmosphere of a planet. There are also tables available showing which latitude/longitude is directed towards us at a certain moment, so that you can compute whether a particular region of the planet is visible from Earth or not.


What is visible light colour output of different stars? I know the colours of the stars relate to their heat, however if I were standing on a different planet identical to Earth, but orbiting a blue Bright Giant or a Red Giant, or perhaps even a Red Dwarf, what colours in visible light would I be able to see? Is it the same as with our star, i.e. would any be missing such as if I'm in a room with a black light?

Compare the effective temperature of the stars with Planck's law, and with a light bulb (2000 to 3300 K): The color of a light bulb corresponds to spectral class L (red brown star) or class M (red star, including dwarfs and giants); violet, and some blue are missing.

Light of Brown Dwarfs (spectral class T) would look red from their black body radiation, but due to their chemical composition the actual color may vary a bit over different tones of red. They would mainly feel hot, and look dull red at the same time, similar to glowing iron. Colors can hardly be distinguished under these lighting conditions.

Some of the very hottest stars (spectral class O) may ressemble "black" light, looking bluish, and causing fluorescence due to to the shift of the spectrum into the ultraviolet, including x-rays.
Without protective ozone/atmosphere layer, this wouldn't be healthy.

This may apply to some degree also to spectral class B (blue white stars); plasma cores of lightnings are of a similar temperature (up to 30,000 K or more) as the effectve temperature of class B or class O stars.

Any spectral class in-between (A, F, G, K) would look almost white, a little bluish for hot stars, yellowish for cooler stars, after some adaption of the eyes.


How does the number in stellar classification indicate properties within the range of that class? I am doing some research for a video game and I wanted to store stellar data on a star, such as its Stellar Classification, and then from that derive a physical appearance in the game.  These are fictional stars.

My original thought had been to store the star's classification, For example G2V and then at runtime I could use that data to determine things like surface temperature, brightness, radius, and mass.  From temperature I can also derive color.

However when I sat down to start coding a first pass on this I noticed something that confused me.

According to Wikipedia, our Sun, Sol, is a G2V star.  The number is supposed to indicate a "sub" classification.  A G2 star would be 20% between a G star and a K star.  A table there lists a G type star as having a surface temperature between 5,200 K and 6,000 K.  Therefore I figured a G2 star would have a temperature of about (I know nothing here is EXACT) 5,360 K (20% of the way between 5,200 and 6,000) but apparently it actually has a temperature of 5,800 K which is about 75% between the given range.  So why is Sol not a G7 of G8 star?

And then once I hit that irregularity, I had to assume that things like Radius and Mass are probably also not linearly interpolated along the range.

Is there a feasible way to derive properties such as I am looking at based on the Stellar Classification?


Speed of blast from supernova How fast does the blast front of a supernova expand at? Is it close to the speed of light or is it less than a quarter of the speed of light?

The speed of the blast front depends on the initial energy release and the density of the medium into which it is expanding, see here.

Theory suggests and measurements confirm expansion rates of the order of thousands of km/s or a few $\times 10^6\ \mbox{m/s}$ or  $\sim 1\% \mbox{c}$.


How can Halley's Comet be predictable? How can we know that Halley's Comet is going to come again every 76 years? I understand that the comet has an orbit around the sun but how can scientists know that it won't be thrown off course for example by passing Jupiter and Jupiters gravity giving it a tug in a different direction? Its not a very big object in space so it seems to me that its orbit could be changed very easily.

Each time Halley's comet passes us, we can make a pretty good estimate of its current orbit, and determine how close it will get to the massive bodies of the solar system like Jupiter, Saturn, Uranus, or Neptune on its next orbit. We can make good estimates of gravitational perturbation effects, and thereby know where to look for it.

In fact, astronomers have been doing exactly that for centuries with remarkable accuracy:

  Halley thus concluded that all three comets were, in fact, the same object returning every 76 years, a period that has since been amended to every 75–76 years. After a rough estimate of the perturbations the comet would sustain from the gravitational attraction of the planets, he predicted its return for 1758.

  Halley's prediction of the comet's return proved to be correct, although it was not seen until 25 December 1758, by Johann Georg Palitzsch, a German farmer and amateur astronomer. It did not pass through its perihelion until 13 March 1759, the attraction of Jupiter and Saturn having caused a retardation of 618 days. This effect was computed prior to its return (with a one-month error to 13 April) by a team of three French mathematicians, Alexis Clairaut, Joseph Lalande, and Nicole-Reine Lepaute.

However, Halley's position turns out to be hard to predict for time scales much beyond that of a single orbit; over the long term, the orbit is chaotically unstable and it's possible that it will collide with a planet, or be ejected from the solar system, within some hundreds of thousands of years.

You mention:

  Its not a very big object in space so it seems to me that its orbit could be changed very easily.

As it turns out, the mass of an object in space doesn't affect how the gravity of other bodies changes its trajectory. A feather, a baseball, a comet, and a small moon making a close flyby of Jupiter along the same trajectory, one after another, will all exit Jupiter's neighborhood in the same direction and at the same speed. The small moon will have more effect on Jupiter than the feather will, but aside from that there's no difference.


Capability of observing transits with terrestrial telescopes of various sizes? I have access to my university's telescope, Dearborn Observatory, an 18.5 inch refractor on the shore of Lake Michigan, just north of Chicago (yes, it's an atrocious location, but the telescope still works fantastically given this), and I was wondering if it would be possible to detect exoplanet transits.  I have access to a CCD.  Obviously, normally this would take quite a while, but I was looking at Wikipedia's list of transiting exoplanets and several of them have periods of less than 10 hours or so.

Are there any good resources about the math for this?  Or is the approximation that size of the change in light observed by the transit is just using the ratio of the area that the planet occludes to the area of the host star a reasonably good approximation?

Or is observing the transits of any known exoplanets completely unfeasible with this size of telescope?  From previous nights of photographing the sky, it seems like the telescope's limit is roughly magnitude 13-15, beyond which the signal-to-noise ratio is just too low to get meaningful data.

A bright, nearby star with a large exoplanet would be best. For example, the first star to have a transiting planet observed was HD 209458, in Pegasus. It has a magnitude of 7.65.  When the large "hot jupiter" transits the star it dims by a relative flux of 0.984. That corresponds to a change in magnitude of about 0.016. I.e. it changes from a magnitude 7.65 star to a 7.67 star.

If you know the radii you can calculate the relative flux during a transit by assuming the planet blocks all the light.

The radius of the planet HD209458b is about 100000km and the star 800000km, so the planet has an area 1/64 of the area of the star's disc, and as expected 1-1/64 = 0.984: which is in good agreement with the light curve.

That won't be noticeable with the naked eye, and is challenging for an amateur. A couple of links:

http://www.britastro.org/vss/ccd_photometry.htm general information on photometry. (summary: you need some kit)
http://brucegary.net/book_EOA/EOA.pdf (summary: it can be done, with a 14 inch telescope and Arizonan dark skies.)


Where is the flaw in this hypothetical concept of escaping event horizon using another's blackhole flyby? imagine two black holes approaching each other in a way that they will perform just very close flyby. You have a spaceship just below the event horizon of one black hole in line between centers of both black holes all the time.
When the black holes are closest to each other you should find yourself outside of the event horizon for a moment as the influence of one black hole should cancel out some part of the influence of the another black hole ( in similar fashion as the influences cancel out in the exact middle of a planet ); therefore, enabling you to escape. However, current physics says that nothing can leave the event horizon. Where is the flaw in this scenario?

In short, it doesn't work.

the event horizon is defined as the region of space which you can't escape from, If you are between two black holes, and you can escape then by definition you are not inside the event horizon. In the case when there are two black holes, the black hole is not a Schwazchild black hole (which is a solution of Einstein's equations for a single point, uncharged non-spinning mass) and the event horizon is not spherical. To determine the shape of the event horizon for two masses you need to use numerical solutions of the equations, you can do this yourself with the Einstein Toolkit, but the computing requirements are very high. A supercomputer is required to get very far.

In the exact situation you draw, it is likely that the release of graviational radiation would lead to a merger rather than a pass but you would need to run the simulation to confirm this.

Now you can't just remain stationary inside the event horizon. If you started inside it you can't stay in the same place and wait for a the pass of the other black hole to pull you out. Inside the event horizon spacetime is flowing towards the singularity and even if the shape of the event horizon is changed by the passing of another black hole, nothing passes from inside to out.


Save current state in ds9? After fooling around with a FITS file(s) (e.g. setting scale, changing color) is it possible to save the session such that there is a file associated with all those tweaks the next time I open it up?

Save and Save Image just seem to keep whatever I set it to, but when I load those new saved files back into ds9 the settings are all back to whatever default load settings I have enabled.

I would also like to be able to do this if I have FITS files loaded into multiple frames so I don't have to go through the process of loading each individual image.

Perhaps you are looking for the backup function

  File -> Backup...

and to restore the settings

  File->Restore...


Why does the Sun deviate from a typical blackbody spectrum in the S band? This is sort of a follow-up to this question, and my answer to it.
The graph I see here details the radio (and other frequencies) emission of the sun. What's most notable and interesting to me is the deviation from the typical blackbody spectrum of a body at around 6000 K. The sun seems much brighter in the S band (and at some larger wavelengths) than a typical blackbody.
My question is: why does this happen? Is there an underlying physical/astrophysical phenomena that we know of that causes this? Or is this something that's undergoing research right now?

There are other ways of getting emissions than just direct thermal radiation. Most of it happens through plasma interactions in the solar corona and atmosphere than in the chromosphere. This review paper names bremsstrahlung, gyroresonance, cyclotronmaser, and plasma radiation as sources, each with their own brightness temperature way above 6000 K. (See also these lecture slides for a bit more in depth math of the sources).
My own summary would be that there is plenty of energy in the magnetic field and plasma, and this can be released through various wave modes producing radio signals. Further, fast charged particles are spinning around field lines or colliding, producing electromagnetic emissions of various types. None of these are blackbody sources, but they add to the blackbody spectrum to produce the observed curve.


Retrograde Motion: Stars in the background I was reading about the retrograde motion of Mars. All explanations give reference to the stars in the background. But I do not understand why this is necessary. I say this since stars themselves also appear to move and can see and observe this without any references. This brings up the second point: if the stars also appear to move, why do explanations have these background stars frozen?
I am new to all this so this might be a naive question :)

Go outside.
Pick a star, any star and watch it for a while.  It moves.  As the Earth spins all the stars move across the sky, taking 23hr 56 min to move all around the sky to the same place.  This is due to the rotation of the Earth (less than 24 hours due to the orbit of the Earth). We sometimes say the "celestial sphere" appears to rotate around the Earth, due to the Earth's rotation.
The planets also move across the sky every day, taking just less than 24 hours to all around the sky.
But the planets also move relative to the celestial sphere, and since th stars don't move relative to the celestial sphere we can say that the planets move relative to the stars.  It is this motion that is of interest, since it is not due to the rotation of the Earth, but to the actual motion of the planet relative to the Earth. This motion is much slower than the apparent motion of the celestial sphere; Jupiter takes 12 years to make a full journey around the sky. But this motion is special, complex and interesting.  It is this motion that earns planets the name "wandering stars" in ancient Greek.
The stars are at a very great distance, so to all intents and purposes they are "fixed". Relative to each other, the stars hardly move at all. You would need to wait centuries before you noticed any difference. So the background of fixed stars makes a very convenient way to refer to the motion of the planets


The fastest Probe we could build now I found this at
http://www.nature.com/news/the-exoplanet-next-door-1.11605

  Even when launched by one of the most powerful rockets on Earth, boosted by a gravitational slingshot around Jupiter, and further accelerated by a radioisotope thruster, that probe would take about 28,000 years to reach α Centauri. Quoting British author Douglas Adams, McNutt quips: “Space is big. Really big.”

Is this still the fastest we can get right now?

For small values of practical an interstellar Orion craft could achieve ~3.3% c (~133 years to Alpha Centauri, one way with no deceleration at the end of the trip)). Which is in principle within reach of present technology if we chose to pursue it. But cost and aversion to nuclear technology on such an extravagant scale make it unlikely that we will see Orion space craft of any ilk built in the foreseeable future.


Does the Sun twinkle? Does our solar system have enough bodies in orbit to appear to twinkle to another distant solar system viewed from the side?

https://en.wikipedia.org/wiki/List_of_asteroid-discovering_observatories

https://phys.org/news/2018-11-nasa-lucy-sky-asteroids.html

While objects blocking the sun may have an effect on the twinkle if viewed on another planet, a twinkle is mainly caused by the effect the atmosphere of a planet has on the light, or something interfering with the light that is passing to your eyes.
From Wikipedia's article on Twinkling:

Twinkling, or scintillation, is a generic term for variations in apparent brightness or position of a distant luminous object viewed through a medium. If the object lies outside the Earth's atmosphere, as in the case of stars and planets, the phenomenon is termed astronomical scintillation; within the atmosphere, the phenomenon is termed terrestrial scintillation.[2] As one of the three principal factors governing astronomical seeing (the others being light pollution and cloud cover), atmospheric twinkling is defined as variations in illuminance only.
Stars twinkle because they are so far from Earth that they appear as point sources of light easily disturbed by Earth's atmospheric turbulence, which acts like lenses and prisms diverting the light's path. Large astronomical objects closer to Earth, like the Moon and other planets, encompass many points in space and can be resolved as objects with observable diameters. With multiple observed points of light traversing the atmosphere, their light's deviations average out and the viewer perceives less variation in light coming from them.


What would the night sky look like if Earth orbited an intergalactic star? If the Earth and its host star were located exactly in-between the Milky Way and Andromeda would the night sky be completely void of light?

Would stars be visible to the unaided eye?

So there could be a civilization that exists in which the night is completely void of light? …that would suck

Since Andromeda is already visible to the naked eye, to a civilization located at half the distance from the Milky Way, Andromeda would be still be visible. Its total brightness would be four times higher, but since its area would grow by the same factor, its surface brightness would stay constant. The Milky is less bright by a factor of ~2.5, but also smaller, so the surface brightness and hence visibility would be similar.

Individual stars would not be visible except if 1) one exploded as a supernova or 2) the civilization has evolved into beings with extremely large pupils, so that more light is being recorded by their retinas. In fact, we can calculate how large:

According to this list, the brightest star known in the Milky Way is called WR 25, and has an absolute magnitude of $M = -12.25$. Half of the distance from the MW to Andromeda is $d \sim 390,000$ pc, so the distance modulus is
$$\mu = 5\log d - 5 = 22.95,$$
and the apparent magnitude of WR 25 would be
$$m = M + \mu = 10.7.$$

Under the darkest conditions, humans are able to see objects down to an apparent magnitude of ~6, i.e. a factor of
$$f = 2.5^{10.7 - 6} = 75$$
smaller. Thus, to be able to see WR 25, the diameter of the aliens' pupils would have to be a factor of $\sqrt{75} \sim 9$ larger. Since human pupils are roughly 6 mm wide, their pupils would be 5 cm wide, and the eyeballs somewhat larger, incidentally roughly the size of a teacup.

Thus, it seems that my compatriot H. C. Andersen predicted the existence of these beings in his fairytale The Tinderbox.


Calculating Angular Distance Given the horizontal coordinates of two objects in the sky, how does one go about calculating the angular separation between them?

You use Spherical Trigonometry

Given $A_1$ and $A_2$ are the respective azimuthal coordinates of the two objects, and $a_1$, $a_2$ their respective altitudes,

the angular seperation $\theta$ is given by

$$\cos \theta = \sin a_1 \sin a_2 + \cos a_1 \cos a_2 \cos (A_1-A_2)$$


What's special about Saturn that it has distinctive rings? This may overlap with other questions, but I'm asking what is unique about Saturn in location, properties, etc that is has far more pronounced rings?   It's between Jupiter and Uranus and Neptune in size and position.  So what's special?  Is there some Goldilocks zone of rings formation?   Or is it just an accident of history?

Saturn's rings are unique in that a lot of the rings compositions contain new and recycled material. Much of the uniqueness is complementary from Saturn's moons, which play a critical role in shaping and recycling the rings materials.

A great example is the moon Enceladus, which contributes largely to the E Ring. Icy material is ejected from Enceladus, a process called cryovolcanism (spewing plumes of icy materials instead of silicate rock); due to the lack of atmospheric pressure and low surface gravity, Enceladus doesn't hold the material that it ejects, but rather feeds material into the E Ring. Discovered by Cassini, the material is comprised of simple hydrocarbons such as methane, propane, acetylene and formaldehyde.

A contributing factor for the unique shaping of the rings and much of their orbital stability stems from the moons also. Another well known example is the Cassini Division and a process called orbital resonance. Essentially the moon Mimas orbits once for every two orbits for material inside the Cassini Division. As a result of that orbital resonance the material receives accumulation of radial momentum, the effect of such accumulation forces a tug of war with the particles. Mimas attracts the material pulling it into the A Ring as the material increases in velocity, resulting in the division.

Other moons have also been documented to have this effect, such as the moons Pan and Daphnis in the A Ring, however it is thought that the effect from these smaller moons only create a division around 10km in width. Stability can be exaggerated from the F Ring, with the moons Pandora and Prometheus. These moons twist the material into a spiraling vortex, like a horizontal tornado encircling the planet. Processes of stability arise from moon ringlets, such as the Titan Ringlet. Titan has orbital resonance with material in the C Ring and provides radial momentum with the Colombo Gap, the gap is slightly elliptical rather than circular (just as Titan is) and the material is attracted towards Titan in the same pattern, in a process called apsidal precession. This creates a compression and expansion of the rings giving them stability while accumulating velocity due to orbital resonance.

Distortions from the moons on an independent inclination to the rings as they come close to the rings, also pull material from the rings and spread it around in chaos, additionally recycling material.

Each ringlet has an individual story as to the forces which make that ringlet unique, comprising of the orbital parameters of the moons and the unique positioning of the forces around them. Along with material that is recycled in a unique fashion not evident with other planetary rings system.

The uniqueness stems from these factors.


Movement of the Sun in space Does the Sun undergo any cycles or revolutions approximately every 24,000 years or 8.6 million years? In Hinduism it is believed that the Sun revolves in an elliptical orbit around a large mass at these rates.

The precession of the Earth's spin axis orientation takes place on a 25,700 year cycle. But this isn't connected with the Sun's movement through space.

The Sun orbits the Galaxy about once every 220-240 million years. It also oscillates about the Galactic plane with a period of about 70 million years.


What is the physics of the "spinning dust" contribution to Cosmic Microwave Background measurements? The lengthy paper Nine-Year Wilkinson Microwave Anisotropy Probe (WMAP) Observations: Final Maps and Results mentions "spinning dust" about 51 times. It is a potential contributor to the measured spectrum and the details of its emission spectrum has more than one proposed model.

question: Is there a simple way to understand the physics behind the emission spectrum of spinning dust and how it would differ from non-spinning dust?

"Spinning dust" is a mechanism proposed to explain a particular feature in the foreground emission of CMB; a bump around $\nu\sim20\,\mathrm{GHz}$.

Dust grains acquire charge through photoelectric emission and collisions with electrons and ions (Draine & Lazarian 1998). As zephyr comments, if the dust is a poor conductor, its charges will, in general, be unevenly distributed, causing small dust grains to exhibit an electric dipole moment $\mathbf{\mu}$. But the molecules making up the grains may themselves have a dipole moment, and even a perfect conductor
will, in general, have its charge centroid displaced from the mass centroid (Purcell 1975).

Collisions and radiation may cause the grains to start spinning, and in the presence of magnetic fields (which are very common in the interstellar medium), this spin in turn causes the particles to emit radiation with a power $P$, according to Larmor's formula which for a rotating dipole can be written as:
$$
P = \frac{2}{3} \frac{\mu_\perp^2 \omega^4}{c^3}.
$$
Here
$\mu_\perp$ is the component of $\mu$ perpendicular to the rotation axis,
$\omega$ is the angular frequency of the rotation, and
$c$ is the speed of light.

The emitted radiation matches the rotation frequency which lies in the (tens of) GHz-region, corresponding to wavelengths in the microwave region.

In contrast, the radiation from non-spinning dust will be thermal, thus lying in the infrared.

The paper you link to (Bennett et al. 2013) shows the difference in Fig. 22 (although thermal radiation peaks outside the observational range of WMAP): The spinning dust peaks around $\nu\sim20\,\mathrm{GHz}$, while the thermal radiation peaks around $\nu\sim2\,000\,\mathrm{GHz}$.

Note that spinning dust also emits thermal radiation, and in fact thermal fluctuations within the grains cause the charges to shift around quite rapidly, changing $\mu_\perp$ and $\omega$ (Hoang et al. 2015), thus effective smearing out the spectral lines.

A symmetrical dust grain will arguably be hit by particles/photons, on average, as much from one side as from another. A mechanism to accelerate an asymmetrical grain's spin is proposed by Purcell (1975). Below is Fig. 1 from his paper. A particle that strikes in a concavity is more likely to interact with the grains twice. If the gas it colder than the grain, it will be heated and leave the grain with a larger velocity than it entered with, causing the grain in the figure to start spinning counter-clockwise; it the gas is hotter (which is more often the case), it will cause the grain to spin clockwise.


Location of the milky way in August in Death Valley I am going to be in Death Valley around August 8th. I was hoping to take a milky way shot with my camera when there. I would be shooting around 10pm.

I am using google maps to scout out popular viewpoints in Death Valley for potential locations for the shot, as I will be getting terrestrial objects in the frame.

From what I have read , I believe the milky way will be visible most of the night, and best viewable in the earlier hours of the night. In addition it will be in the south sky.

So will it be precisely due south? Or will it be slightly south west or south east? If I could pinpoint the location, it would be easier to scout out a spot.

Thanks!

The Milky Way will be in a straight line that goes from SSW, (a bearing of about 200 degrees) almost directly overhead and to the NE. The brightest part of the Milky way will be low in the sky in the SSW.

In future, you can check the visibility and direction of any astronomical object with planetarium software, such as the (free) Stellarium.