Thinking of buying a new telescope. What aspects should be considered? I'm new to astronomy and I have a Celestron 50Z but I'm thinking of buying a better telescope.
I see that there are many options out there but I'm not really sure as to which aspects I should consider that make a good telescope.

Can anyone point me to the right direction? If you ask how much I'm willing to pay, I'd say up to US $300.

Any suggestions not just on telescopes but as I said, things I should be considering which make a good telescope worth buying.

There are many aspects that should be considered, but a few very important ones are:

New or secondhand? With your budget, buying a new telescope will only get you so far. But if you try to look into secondhand products, you might be able to find something more advanced—well-kept telescopes last years. Of course, you must be careful with whom you are buying it from, and make sure to try it before buying as well as asking about the last time the mirror was cleaned, what eyepieces are included, etc.
Dobson or not? There are many upsides to having a Dobson-type telescope: you get a much larger mirror for the same price, it's easy to set up and maintain... However, it does take more room than other telescopes, and you will have to do more maneuvers to aim and track things with it.
On the same vein, does size matter? If you are looking for the most compact and lightweight telescope possible? If yes, that will have an impact on the kind of telescope you should buy: maybe you will want a Maksutov instead of a Newton, maybe you will want a tripod mount because it can fold and takes the least room.
What do you want to see? As StephenG said earlier, it is important to know your expectations as for what you want to observe with your telescope. If you live in an urban environment and/or are unable to go to spots free of light pollution, that severely limits your possibilities to (mainly) planets and the Moon. To that effect, you might be more interested in a telescope with a bit more focal length. If you are indeed able to go to spots free of light pollution, then you probably want to maximize the diameter of your mirror, because if you get a mirror that is 20% larger, you will get 44% more light.

Some other questions that may influence your choice are: is this a long-term investment or are you looking into making an upgrade in the next few years? Do you want to try astrophotography? How much time are you willing to take to set up your telescope before observations? Do you enjoy using a sky atlas to find the exact object you are looking for? Etc.

To give you an example, I was a similar situation a few years ago. I had a budget of $400, and I wanted a telescope for visual observation only. Because I had access to good observation spots and size was not an issue, buying a Dobson was an easy choice. I joined an astronomy club and a member was going to upgrade his own Dobson (https://www.astroshop.eu/gso-dobson-telescope-n-250-1250-dob/p,6680) to a bigger one, so I was able to test it and buy it secondhand knowing exactly what I was getting.

You should also have in mind the fact that a telescope will often require a few accessories (eyepieces, filters,...) that you might need from day one, so do look at what is included with the telescope you buy, as good eyepieces are things you can keep with your next telescopes, and bad accessories can affect the quality of your observations.

Astronomicably,


Why does Earth have a dip in the CO2 absortion spectrum from 14 to 16 micron? This paper shows that there is a dip in the CO2 absorption spectrum of Earth. In essence the trough of the absorption of CO2 for Earth is cut into two separate troughs instead of 1 large trough. Why is that?

original without arrow
As there have been some comments on whether this is planet related or molecule related, here is an additional image which shows the absence of the double trough in the CO2 absorption pattern for Venus whilst showing it for Earth:

Eric Jensen has already provided a nice link to a description of the basic structure of the ${\rm CO}_{2}$ spectrum, so I'll focus on the question of why there's a "spike" at 15.0 microns in the Earth spectrum, but not in the Venus or Mars spectra.
If you look at the link in Eric's answer, the very first image shows a high-resolution version of the ${\rm CO}_{2}$ absorption spectrum, including the main features to the right and left. There is also a very strong absorption peak in the middle (the "Q-branch") at about 15.0 microns (667 ${\rm cm}^{-1}$ in wavenumbers). The question then becomes: why does this feature show up as an extra dip in the Venus spectrum (and, barely visible, in the Mars spectrum) -- as one might expect -- but apparently as a peak in the Earth spectra?
The answer hinges on the fact that the spectra shown in the question are not, strictly speaking, absorption spectra. They are actually radiance/intensity measurements of infrared light emitted by different layers in the atmosphere of the planet in question, as seen from orbiting spacecraft. (In the first figure, it's narrowed to the atmosphere above certain locations on the Earth.) They thus show a combination of the molecular absorption spectra and the temperature structure of the atmosphere.
In simple terms, the spectra show thermal (blackbody) emission from layers in the atmosphere where upward-going photons can escape to space without being absorbed by higher layers in the atmosphere. The altitude of these layers is determined by the amount of the absorbing gas (e.g., ${\rm CO}_{2}$) and the strength of the absorption at a given wavelength. The stronger the absorption, the higher up is the layer that can emit freely to space.
(Note that for wavelengths with effectively zero absorption from any molecule, the "emitting layer" is actually the surface of the planet.)
How much infrared light is emitted at a given wavelength then depends on the temperature of the emitting layer, and it is here that the vertical temperature structure of the atmosphere becomes important. For Venus and Mars -- and for most of the emitting altitudes for Earth -- the temperature decreases with altitude, so that higher emitting layers have lower temperatures and thus emit less radiation. The net effect is to reproduce the absorption features, even though the whole thing is not, technically, an absorption spectrum.
But in the case of the Earth, the strongest absorption -- e.g., in the Q-branch part of the ${\rm CO}_{2}$ absorption, at 15.0 microns -- forces an emitting altitude that is so high it is actually in the stratosphere. And in the stratosphere, the temperature increases with altitude. So the Q-branch absorption peak forces the emitting layer to be at a higher temperature than is the case for the rest of the ${\rm CO}_{2}$ absorption regime.
This University of Chicago MODTRAN web page, which plots predicted emission from the Earth's atmosphere as seen from an altitude of 70 km (default setting), shows this rather nicely. The default parameters include a modern ${\rm CO}_{2}$ concentration of 400 ppm, which gives the feature seen in the Earth observations (i.e., the local emission peak in the middle of the 15 micron trough, which I've labeled "Q-branch reversal"):

But if you dial the ${\rm CO}_{2}$ concentration way down (e.g., to 10 ppm), the emitting altitudes are lower for all ${\rm CO}_{2}$-absorbing wavelenths, so much so that the emitting altitude for the Q-branch is down below the stratosphere, and so is at a lower temperature than the rest, and thus emits less flux, giving you the expected profile (i.e., with the weakest emission at the Q-branch wavelength, instead of a reversal):

(Finally, I believe the peculiar spectrum for the "Antarctic" case in the first figure is due to the unusual temperature profile of atmosphere in arctic regions, which includes the temperature increasing with altitude in the first few km above the ground; the fact that the interior of Antarctica is high altitude might also be relevant.)


Can a system of three stars exist? Just like binary stars can a system of three stars mutually equidistant from each other?
What I mean is three stars at the vertices of a equilateral
If yes than what will be the orbit of a planet out there
If no what will be the problem for it's nonexistence?

It is possible in a Trojan configuration:

In the place of the "Planet" on the image, also a small star could exist. The third star would be at $L_4$ or at $L_5$. This configuration could be made stable.

However, as this link shows,

  In unnormalized units, this criterion becomes

  $$\frac{m_2}{m_1+ m_2} < 0.0385$$

  We thus conclude that the $L_4$ and $L_5$ Lagrange points are stable
  equilibrium points, in the co-rotating frame, provided that mass $m_2$
  is less than about $4\%$ of mass $m_1$.

Thus, the mass of the second star should be at most 3.85% of the central star.

As far I know, no such known star system exists, but if it would, it would be stable.

Stable planetary orbit is possible to

either very close to one of the stars (compared to the size of the triangle)
in the other Lagrange point
or very far from all of them.

If the triangle is big, then a planet even in the habitable zone is possible.


Why is the core of a gas giant supported by electron degeneracy pressure instead of nuclear fusion? After a Sun-sized protostar forms, its core will become denser over time due to radiation. The core eventually gets dense and hot enough for hydrogen fusion to take place. In the late phases of the star's life, the core will continue fusing hydrogen until it runs out and outer layers begin fusion hydrogen, while dumping helium into the core. The core will get massive enough that it will begin to contract, and it will eventually become electron degenerate due to the high densities.

However, as the core became denser and denser, it reached temperatures at which fusion could take place, but it did not become dense enough for the matter to become electron degenerate, right? Only later in its life, when the star's hydrogen ran out, would it become dense enough for that.

So then why would gas giant cores be electron degenerate, but not become hot enough for nuclear fusion? Shouldn't they start fusing before they can become electron degenerate, as stars do? Am I misunderstanding this concept entirely, or is there more to it than what I described?

The test to see whether degeneracy pressure is going to be significant is to compare $kT$ with the Fermi energy $E_F$

The Fermi energy is the energy level up to which all energy states would be occupied in a completely degenerate fermion gas. It is given by (for non-relativistic conditions)
$$ E_F = \frac{h^2}{2m}\left(\frac{3}{8\pi}\right)^{2/3} n^{2/3},$$
where $m$ and $n$ are the mass and number density of the fermions (in this case, electrons).

If $E_F \gg kT$, then the gas can be considered completely degenerate and the electrons exert a non-relativistic degeneracy pressure. If $E_F \sim 10 kT$ the electrons are partially degenerate.
In both these circumstances, the pressure exerted by the electrons is much higher than they would have in a perfect gas, because a large fraction of electrons occupy high energy (and momentum) states.

Degeneracy can be achieved either by having a low temperature or a high number density of fermions. In a white dwarf, electrons are completely degenerate because the number density of electrons is extremely high. In a gas giant, the electron density is nowhere near as high, but then the temperature is a lot lower and so they reach a state of partial degeneracy.

The pressure of a partially degenerate gas is higher than that of a perfect gas at the same density and temperature. More importantly, the pressure is only very weakly dependent on temperature. As a gas giant radiates away its potential energy, it is able to do so whilst only contracting very slightly and the core does not become hot enough to initiate nuclear fusion.

Note that degeneracy pressure and thermal pressure are not two different things. Both arise merely as a consequence of particles having a distribution of momenta and obeying (in this case) Fermi-Dirac statistics to determine how they occupy the possible energy states. The expression $P = nkT$ is simply a convenient approximation for fermions that holds only when $E_F \ll kT$ and the quantum nature of the particles is not apparent.


Do astronomers and astrophysicists more often use diameters or radii when discussing about planets, dwarf planets, exoplanets and stars? Mathematicians much more often use radii over diameters when discussing about circles and spheres, because in mathematics the radius is more fundamental than the diameter (the sphere is defined using its radius).

But what about in astronomy? Do astronomers and astrophysicist more often use diameters or radii when characterizing planets, dwarf planets, exoplanets and stars?

Definitely radii, with one notable exception...

When observers talk about how large an object is on the sky, they usually discuss angular size, which is related to the diameter of an object, not the radius.  So when discussing the angular size of, say, Alpha Centauri A (a few milliarcseconds), this is related to the diameter of the star, not the radius.


Can this picture of the moon directly above the sun from the horizon be real? I recently saw this image, but somebody told me that it is fake, possibly photoshopped. The moon and the sun appear to be on the same perpendicular from the horizon. Is there any way to tell if this shot was actually captured?

No, it's not real.  Given that the moon appears to be nearly full, it would be near to the horizon opposite the sun, which is clearly not the case.  Also the framing of the sun and moon between the trees means that the sun and moon are on the same side of the trees, which is not possible when the moon is full.


Can a donut-shape planet or star be formed? How stable is a donut-shaped star or planet configuration and under what conditions such object may form?
Is there any evidence suggesting that such objects might exist?

In principle, yes. In practice, no.
The question has been studied for a long time. The classic treatment is Dyson's papers on "anchor rings" in 1893 (paper I, paper II) but it goes back to the study of "figures of equilibrium" starting with Newton's considerations of the oblateness of the Earth and then continuing with Maclaurin and Jacobi's analysis of rotating ellipsoids of rotating homogeneous self-gravitating fluids (see Chandrasekhar's book Ellipsoidal Figures of Equilibrium 1969 for everything on the topic; a shorter overview).
The basic story for ellipsoids is that as you add angular momentum they become more oblate (the Maclaurin ellipsoids), undergoing an instability beyond a critical point to become general three-axis ellipsoids (the Jacobi ellipsoids), that may also contain fluid motions (the Dedekind ellipsoids). Eventually there is too much angular momentum and it breaks up, but Cartan showed that the mode that grows fastest is the low-frequency harmonic making it just split into two. So this way you will not get a torus-shaped planet or star.
Dyson considered the instabilities of a ring already existing. He found that for nearly circular rings:

The annular form of equilibrium of rotating gravitating fluid is stable for disturbances symmetrical about the axis, and for disturbances  which alter the shape o,f  the central curve,  but is unstable for long beaded disturbances.This result  was, perhaps,  to  be  expected, as  by  means  of  beaded  waves, the mass would naturally be broken up into spheroidal masses.

Similar results seem to hold for more ellipsoidal rings.
The condition for instability is $R>3r$, that is, thin hoops are unstable while really thick doughnuts may be stable.
On the other hand, an accreting disk may somehow shed angular momentum. Could that produce a fluid or solid torus (albeit with some body in the middle)? This is the problem of why Saturn retains its rings. Tisserand showed that $N$ satellites could orbit in the same orbit if $m/M < 2.3/N^3$ -- the more satellites, the more they would tend to clump unless they were very tiny. (Sheeres & Vinh 1993) tighten the bound a bit. So unless there are other forces than gravity keeping things in place a dust or gas ring will tend to clump.
Since then others have studied the question, such as (Wong 1974). They found that in principle it looks like there is a pathway of unstable intermediate shapes from the Maclaurin ellipsoids to the torii (see Ansorg, Kleinwächter & Meinel 2003), and if there is enough dissipation of the right kind it might counteract the beaded instability splitting it. But it all looks unlikely to happen unless everything is set up perfectly.
Wong's paper speculates in various astrophysical cases where a torus may be possible. The closest to reality is likely ring galaxies, where the stellar ring remains for a relatively long time. There is also a fair bit of interest in toroidal clouds around black holes (with or without accretion disks), since they make sense as models for active galactic nuclei. But here there are gas pressure and luminosity from disks complicating things: it is far from a ring-shaped star or planet.
(Which is a shame, since I had so much fun modelling them!)


How to get error out of astropy constants How do I get the error value out of an astropy.constants quantity?

In [87]: from astropy import constants as c

In [88]: c.M_sun
Out[88]: <Constant name='Solar mass' value=1.9891e+30 error=5e+25 units='kg' reference="Allen's Astrophysical Quantities 4th Ed.">

In [89]: c.M_sun.value
Out[89]: 1.9891e+30

In [90]: c.M_sun.error
AttributeError: 'Constant' object has no 'error' member

While I'm not familiar with the package, a very quick look at the documentation suggests that you want

In [90]: c.M_sun.uncertainty

instead.  I've just checked and this appears to be correct.

> python -c "from astropy import constants as c ; print c.M_sun.uncertainty"

5e+25


Is the list of the most abundant elements in the universe a fact? I've heard  Neil deGrasse Tyson giving a short list of the most common elements in the universe on a video, which went on something like:

hydrogen
helium
oxygen
carbon
nitrogen
etc

As I understand astronomers use spectroscopy to determine an object's chemical composition.
My question is how can they determine the most common elements for certain, when most of what we see is a just a fraction of the stars, nebulas and an even smaller fraction of planets out there?

Neil de Grasse Tyson is extrapolating.
Chemical abundances can and have been measured in a huge numbers of stars in our own Galaxy and considerably fewer in local galaxies. The chemical abundances of glowing gas clouds can also be measured and this technique has a much greater reach. In addition, there are ways in which the summed light from all the stars in a galaxy can be used to give crude information about their average chemistry, and again, this can be used out to large distances.
From these measurements we have a pretty good idea of the chemistry of our local part of the universe. One can then construct an inventory to make a table like the one in your question. This is dominated by stars and gas - planets cannot be measured but are a negligible fraction of mass.
Now there is a complication. The chemistry of the universe changes with time, because hydrogen and helium are gradually being turned into heavier elements inside stars, and then much of the products are distributed into the interstellar and intergalactic medium when stars die. Thus to get an "up to date" inventory one should exclude the older stars and perhaps focus more on the gas, which gives an idea of current chemistry in the interstellar medium.
Having done all this, you get the table in your question - which applies to the local universe.
It is then a fundamental assumption in cosmology that the universe is homogeneous on large scales. There is thus no reason or evidence to suppose that things are different elsewhere. Indeed, given that we now pretty much understand why the chemical league table looks the way it is - a simple consequence of the physics of star formation, nuclear fusion and mass loss in stars - then it is difficult to imagine any scenario in which it could be very different anywhere else.


Two body orbit of equal masses Given two bodies of equal mass in an elliptical orbit:

I know they will be orbiting about a common center of mass, i.e. the barycenter. But, do the velocities have to be equal in magnitude and opposite in direction (normal to R when R is their distance from each other) for the orbit to be stable? I think, if the velocity of one mass were to vary with respect to the other it would create a moving barycenter in which the two masses would collide or throw on another out of the orbit, but I can't find any mathematical verification of this.

I know they will be orbiting about a common center of mass, i.e. the
  barycenter. But, do the velocities have to be equal in magnitude and
  opposite in direction (normal to R when R is their distance from each
  other) for the orbit to be stable?

The orbital velocities do not have to be and in general are not equal in magnitude. What is equal is the angular velocity, that is the angular rate (e.g., in $rad/sec$) that two bodies will be orbiting their common barycenter. The orbital radius, $r$, orbital velocity, $v$, and angular velocity, $\omega$, are related by the equation

$$\omega = v/r$$

Note that these are all scalar quantities and $v$ can be thought of as the component of the velocity vector which is perpendicular to $r$. Since conservation of angular momentum implies that $\omega$ remain constant for each body, individually, then we know that $v/r$ must also be constant which implies that bodies which orbit farther away from the barycenter must necessarily be orbiting faster and vice versa.

Now of course that simplifying explanation works for two bodies. Once you throw in more than two bodies, things become more complicated and the barycenter can move around, causing more complex motions.

  I think, if the velocity of one mass were to vary with respect to the
  other it would create a moving barycenter in which the two masses
  would collide or throw on another out of the orbit

This is a subtly different question. If one of the two masses had a varying orbital velocity, that would imply it was gaining or losing energy by some mechanism. This can occur through things like tidal interactions as it does for our Earth and Moon. As stated above, the angular velocity must remain constant which implies that the body whose velocity is changing must also migrate towards or away from the barycenter, potentially resulting in a collision or escape. Since the Moon's orbital velocity is being sped up through tidal interactions, it is moving farther away as a result. Another example might be two black holes emitting gravitational waves as they orbit which propagates away energy and thus they orbital velocities, causing them to get closer together until they collide.


Good ways to safely view an Eclipse? August 21 is going to see a total eclipse across most of the US, and I'm planning a trip to what I hope is an ideal viewing location.

What are good ways to safely view the eclipse?

I don't want to make a big trip out of it, only to be looking into a poor quality pin-hole shoebox, or $2 mirrored cardboard glasses.  How can get I get a good, detailed clear view of it while staying safe?  What about ways to photograph it effectively?

Projection is a safe way to view the partial phase of the eclipse.

A pinhole produces a small image of the sun, but doesn't require any special equipment. A larger image can be obtained by projecting through a telescope:

Point the scope at the sun, but never look down a scope that is pointing anywhere near the sun, neither look along the tube. Use its shadow to line it up with the sun.  Hold a white card half a metre or so behind the eyepiece, and focus the telescope to make a sharp image of the sun. You will want to provide some shade for the image.

Image from Bakersfield College

This works well with binoculars, you need some way to mount, or prop up the binoculars, and you will need to cover up one of the lenses (or you will get two overlapping images.

If you have a larger telescope, you will want to cut down the aperture, to prevent overheating.

You will be able to see sunspots on the disc of the sun and the growing bite of the sun hidden behind the moon (which will be otherwise unseen). You can practice on any sunny day, You should be able to see some sunspots, and perhaps faculae on any day.

Other alternatives are specially designed solar viewing glasses. They must be unscratched and meet ISO 12312-2:2015 standard. Welders glass is also possible if rated "14" (the darkest possible).

It is not possible to use any form of glasses with a telescope. It is possible to get an aperture filter, which may be useful if you have the right telescope, but the projection method is better if you want more than one person to view at the same time.

Any kind of improvised solar filter is unsafe. This includes: stacks of sunglasses, pop-tarts wrappers, exposed film and viewing reflections in water.

During totality, the sun is hidden by the moon, and so it is not dangerous to look at the solar corona during totality with the naked eye. If you had been wearing solar viewing glasses, be ready to replace them. You MUST put them back on before totality ends.

Photography is hard. The images of the solar eclipse are made with professional telephoto lenses and good deal of skill. Instead you can photograph the scene. A timelapse of the eclipse can be a very effective way to record the event, especially if you have some interesting foreground and background. The eclipse itself will be a tiny dot you can still document how the scenery and shadows change during the event.


Gravitational wave detection time difference between LIGO Livingston and LIGO Hanford Quote from LIGO's news release:

By looking at the time of arrival of the signals—the detector in
Livingston recorded the event 7 milliseconds before the detector in
Hanford—scientists can say that the source was located in the Southern
Hemisphere.

GPS coordinates of LIGO Livingston: 30.5630018,-90.7763949
GPS coordinates of LIGO Hanford: 46.4554032,-119.4092701
Distance: 3030.13 km
Now if the gravitational waves propagate outwards at the speed of light a difference of 7ms implies the following maximum possible detector distance:
299792458 * 0.007 / 1000 = 2099 km

Where am I wrong ?
Update:
As pointed out by james-k in the comments below, the calculated 2099 km is the minimum distance and not the maximum ... these waves seem to have caused some mental derangement ;)

Updated to use reported timing confidence intervals instead of trying
  to infer those from reported sky location resolution uncertainties.
  The latter approach was misleading because the sky location has been further
  resolved using more information than just the arrival time difference.

Beware that I am neither an astronomer nor involved with aLIGO.
I just use classic euclidean geometry here,
combined with occasional transformations between cartesian
and spherical coordinates.

We have two observatories, $L$ (Livingston) and $H$ (Hanford)
and a far-away point $E$ emitting a (gravitational wave) signal.
Let us assume that the signal propagates at constant speed $c$
through low-density matter, including planet Earth.
The straight-line distance between $L$ and $H$ (passing through Earth)
is $|HL| \approx c\cdot 10.0\cdot10^{-3}\,\mathrm{s}$.

$L$ receives the signal from $E$ about
$t_{ELH}\approx 7\cdot10^{-3}\,\mathrm{s}$ before $H$.
This implies that $E$ is a distance
$d = c\,t_{ELH}$ closer to $L$ than to $H$.

Now assuming large-scale euclidean geometry and looking at the
plane containing $E,L,H$, the set of all candidate points $E$
where $|EH|-|EL| = d$ is one branch of a hyperbola, namely
the branch closer to $L$:

In fact, setting
$$\begin{align}
    r &= \frac{d}{2}    & r_1 &= \frac{|HL|}{2} & R &= |LE|
\\  L &= (r_1,0)        & H &= (-r_1,0)         & E &= (x,y)
\end{align}$$
we get the cartesian hyperbola equation
$$\frac{x^2}{r^2} - \frac{y^2}{r_1^2-r^2} = 1$$
and in polar coordinates, viewed from $L$:
$$R = \frac{r_1^2 - r^2}{r - r_1\cos\alpha}$$
For very large $R/r_1$, $|\alpha|$ must be very close to its asympotic limit:
$$\begin{align}
    \cos\alpha &\approx \frac{r}{r_1} = \frac{d}{|HL|} = \frac{t_{ELH}}{t_{HL}}
\\  \text{where}\quad
    t_{HL} &= \frac{|HL|}{c}
    \approx 10.0\cdot 10^{-3}\,\mathrm{s}
    \qquad\text{(time directly from $H$ to $L$)}
\end{align}$$
In 3D this means that $E$ lies on one of the two sheets of a hyperboloid
of revolution whose axis of revolution is the line $HL$.
In particular, in $L$'s view of the sky, partially hidden under the horizon,
there should be a fairly thin circular band that covers all projected
candidate points for $E$.

The solid angle of view enclosed by the circular band is
$$  A_{\text{ste}} = 2\pi(1 - \cos\alpha)
\approx 2\pi\left(1 - \frac{t_{ELH}}{t_{HL}}\right)$$
Uncertainties in $t_{ELH}$ yield uncertainties in $\cos\alpha$
which widen the circular band.
Thus the circle becomes an annulus.
The solid angle of view covered by that annulus measures
$$\begin{align}
    A_{\text{ste.max}} - A_{\text{ste.min}}
    = 2\pi(\cos\alpha_{\text{min}} &{}- \cos\alpha_{\text{max}})
    \approx 2\pi\frac{t_{ELH\text{.max}} - t_{ELH\text{.min}}}{t_{HL}}
\\  \text{briefly:}\quad
    \Delta A_{\text{ste}} &\approx 2\pi\frac{\Delta t_{ELH}}{t_{HL}}
\end{align}$$
Substituting reported data for $t_{ELH}$, with $\text{min}$ and $\text{max}$
referring to 90% probability intervals, we have:
$$\begin{array}{rcrrr}
    \text{entity} & \text{unit} & \text{nom} & \text{min} & \text{max} & \Delta
\\\hline
    t_{ELH} & 10^{-3}\,\mathrm{s}   & 6.9   & 6.5   & 7.4   & 0.9
\\  \cos\alpha & 1                  & 0.69  & 0.65  & 0.74  & 0.09
\\  90^\circ - \alpha & {}^\circ    & 44    & 41    & 48    & 7
\end{array}$$
And thus the annulus, when projected to a unit (celestial) sphere, covers
$$\Delta A_{\text{ste}} \approx 0.18\pi \approx (1.9\cdot10^3)^{\circ\circ}$$
(The ${}^{\circ\circ}$ means
square degrees.)
In contrast, the reported value of the credible region varies between
$140^{\circ\circ}$ (50% probability) and $590^{\circ\circ}$ (90% probablity).
This is because large parts of the annulus could be ruled out based on
additional information such as signal strength ratios.
But those refinements are not the topic of this thread.

For easier visualization, let us move to a point $B$ on Earth's surface
where the center of the annulus is in the zenith,
and the annulus itself is at constant altitude $90^\circ - \alpha$,
as given in the above table.
To that end, let $O$ be the location of Earth's center and require
that the 3D ray $OB$ points in the same direction as the ray $HL$.
This determines the latitude and longitude of $B$.
Note that the parallel shift from $HL$ to $OB$ incurs a parallax,
but that is negligible because the distance to $E$ is so large.

Idealizing our planet to a sphere and doing the calculations
(transforming the locations of $H$ and $L$ from spherical coordinates to
cartesian coordinates, computing the difference vector, normalizing it,
and transforming back to spherical coordinates), I find $B$ to be at
$$B\colon\ 27^\circ 21'54.22''\mathrm{S}\ 38^\circ 36'33.45''\mathrm{W}
= 27.365061^\circ\mathrm{S}\ 38.609293^\circ\mathrm{W}$$
which is some 700km southeast of Rio de Janeiro in the Atlantic.
If you had been there at 2015-09-14 09:50:45 UTC
and had spun around yourself, drawing a band at constant altitude
$90^\circ - \alpha \approx 44^\circ$ and width $7^\circ$ in the sky,
that would have marked all the directions
where the signal source $E$ could have been.

To get the sky view simulated, I have written a tiny script gw150914.sts for
Nightshade:

# Display the sky at the time of the aLIGO GW150914 observation
# from an Earth-based position chosen such that the set of possible sky
# positions of the source form a circular annulus with azimuth-independent
# altitude = asin(delta_t/10ms), about 44 deg.
clear
timerate rate 0
date utc 2015-09-14T09:50:45
flag show_tui_datetime on
set home_planet "Earth"
moveto lat -27.365061 lon -38.609293 alt 0 heading 0 duration 2
wait duration 2
flag azimuthal_grid on
zoom fov 120 duration 2

This gives the view below. I have added dashed green circles to delimit
the annulus.

But do not expect to find likely candidate spots within that annulus.
The annulus hides the radial depth of the projected space volume.
A range for that radial depth is given by the estimated bounds
for the distance $R=|LE|$. The corresponding flat space volume is
$$\Delta V = \frac{1}{3}\Delta A_{\text{ste}}
\left(R_{\text{max}}^3 - R_{\text{min}}^3\right)$$
Plugging in the reported values, again from 90% probability intervals,
$$\begin{align}
\Delta A_{\text{ste}} &\approx 590^{\circ\circ} \approx 0.0572\pi
&   R_{\text{min}} &\approx 0.23\cdot10^{9}\,\mathrm{pc}
&   R_{\text{max}} &\approx 0.57\cdot10^{9}\,\mathrm{pc}
\end{align}$$
yields
$$\Delta V \approx 10^{25}\,\mathrm{pc}^3$$
To get a very rough estimate of the number of galaxies in that volume,
let us multiply $\Delta V$ with a rough galaxy density estimate of $10^{-20}\,\mathrm{pc}^{-3}$,
which results in about $10^5$ galaxies in that volume.
My conclusion from this back-of-the-envelope calculation is
that we cannot hope to localize $E$ further, unless given more information.

Things would have been a bit different if we had had a third aLIGO observatory
distant from the line $HL$ operating and capable of detecting the signal
from $E$.
Then we could draw three annuli, take intersections and thus
determine the direction of the signal within one reasonably small spot
on the celestial sphere.

Further reading

As this LIGO post tells
us, parts of the annulus could be ruled out by considering additional features
of the data such as varying signal strength.
A more detailed description of the probability distributions for the results
is given in this LIGO paper.
More information about the methods used to further narrow the credible
sky regions of the source location is given
in this paper.

Addendum

An equivalent script gw150914.ssc
for Stellarium:

// Display the sky at the time of the aLIGO GW150914 observation
// from an Earth-based position chosen such that the set of possible sky
// positions of the source form a circular annulus with azimuth-independent
// altitude = asin(delta_t/10ms), about 44 deg.

var obs_lat = -27.365061;
var obs_long = -38.609293;
var obs_alt = 0;
var obs_locname = "Atlantic, 700km SE from Rio de Janeiro";
var obs_planet = "Earth";
var obs_date = "2015-09-14T09:50:45";
var duration = 2;

core.clear("deepspace");
core.setTimeRate(0);
core.setDate(obs_date, "utc");
core.setObserverLocation(obs_long, obs_lat, obs_alt, duration,
                         obs_locname, obs_planet);
core.wait(duration);
core.moveToAltAzi("90", "0", duration);
core.wait(duration);
GridLinesMgr.setFlagAzimuthalGrid(true);
StelMovementMgr.zoomTo(120, duration);

And yes, core.setObserverLocation expects longitude before latitude.


What complementary observation campaigns have looked at Kepler's field of view? It seems to be common practice to double up observations. For example, shortly before New Horizons flew by Pluto, several of its moons were discovered by telescopes as a concerted effort was put in because of the occasion.

Have other telescopes made systematic and dedicated observations of the field of view of the Kepler telescope? I imagine speculatively in this paragraph that such observations could've been part of a survey process to determine where to point Kepler out of a set of candidate FOVs to begin with. And once decided upon, that field more closely examined. Then during the Kepler mission persistently observed in order to help rule out noise from passing asteroids and binary and variable stars and to gather complementary data. To characterize transiting planets, it is important to characterize the stars they are orbiting, something which telescopes with other kinds of instruments can be helpful with. Those are just my speculations about why I would think that Kepler's primary mission was "doubled up" by other kinds of telescopes.

What are some of the preparatory, complementary or follow up observation campaigns that have been made of the Kepler field of view? Is Kepler's FOV now one of the most and best observed parts of the sky?

The Kepler Field had been covered in the near IR by the 2MASS catalog since that covered the entire northern sky.  There was also a Smithsonian Institute program called the Kepler Input Catalog program that observed the Kepler field with the 48-inch telescope at the Whipple Observatory on Mount Hopkins, Arizona in the SDSS g, r, i, and z bands plus a custom filter centered on the Mg b lines.  Also many individual objects studied by Kepler were observed in follow up studies.  So, yes this is a well studied region of the sky now.  Now, however, the Pan-Starrs survey has imaged most of the whole sky with several optical filters.

After the reaction wheel failure on Kepler, it was repurpose as the K2 project.  In this project, it observes a different part of the ecliptic plane every 3 months.  Catalogs of stars and galaxies in each of these locations are created ahead of time with the EPIC survey.  In addition, there is a SN/AGN search ongoing with K2 that includes simultaneous groundbased observations using various wide field imaging telescopes around the world such as the SkyMapper in Australia.  A workshop on this will be held in February.


I'd like to photograph NEOWISE from Los Angeles, what is the optimal time? I live in Los Angeles and have a camera adapter for my telescope -- nothing fancy just an Astromaster 114. What is the optimal time, in terms of dark sky and bright comet, to take a photo of Comet NEOWISE? I.e., when will it be brightest relative to the surround sky? When will it be largest?
And, lastly, what is the best time tonight to view it?
EDIT: I've spent the last two nights at sunset trying to locate it, but it doesn't really get dark until about an hour after sunset around 8pm. I also spent over an hour on the roof this morning from about 4am til 5:15am. I couldn't see it despite clear skies (or what passes for clear in LA).

I don't know if this is helpful but using the Python package Skyfield one can calculate the positions of things including comet C/2020 F3 (NEOWISE) and the times that they rise and set for the month of July:

Here are the rise and set times of the Sun and the comet for reference for the next week
sunset:  2020-07-12T03:06:17Z
sunrise:  2020-07-12T12:51:20Z
sunset:  2020-07-13T03:05:55Z
sunrise:  2020-07-13T12:51:56Z
sunset:  2020-07-14T03:05:32Z
sunrise:  2020-07-14T12:52:33Z
sunset:  2020-07-15T03:05:07Z
sunrise:  2020-07-15T12:53:10Z
sunset:  2020-07-16T03:04:41Z
sunrise:  2020-07-16T12:53:48Z
sunset:  2020-07-17T03:04:13Z
sunrise:  2020-07-17T12:54:27Z
sunset:  2020-07-18T03:03:43Z
sunrise:  2020-07-18T12:55:06Z

cometset:  2020-07-12T04:15:12Z
cometrise:  2020-07-12T10:46:43Z
cometset:  2020-07-13T04:34:10Z
cometrise:  2020-07-13T10:48:01Z
cometset:  2020-07-14T04:54:27Z
cometrise:  2020-07-14T10:51:29Z
cometset:  2020-07-15T05:15:33Z
cometrise:  2020-07-15T10:57:33Z
cometset:  2020-07-16T05:36:48Z
cometrise:  2020-07-16T11:06:45Z
cometset:  2020-07-17T05:57:17Z
cometrise:  2020-07-17T11:19:29Z
cometset:  2020-07-18T06:16:07Z
cometrise:  2020-07-18T11:36:02Z

Here's the python script:
from skyfield.api import Topos
from skyfield.api import Loader
from skyfield.data import mpc
from skyfield import almanac
import numpy as np
import matplotlib.pyplot as plt
from skyfield.constants import GM_SUN_Pitjeva_2005_km3_s2 as GM_SUN

from skyfield.api import load
loaddata = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files

ts = loaddata.timescale() # include builtin=True if you want to use older files (you may miss some leap-seconds)

with load.open(mpc.COMET_URL) as f:
    comets = mpc.load_comets_dataframe(f)

comets = comets.set_index('designation', drop=False)

print(len(comets), 'comets loaded')

row = comets.loc['C/2020 F3 (NEOWISE)']
print(row)

eph = loaddata('de421.bsp')

earth, sun, moon = [eph[x] for x in ('earth', 'sun', 'moon')]

comet = sun + mpc.comet_orbit(row, ts, GM_SUN)

times = ts.utc(2020, 7, range(1, 32))
ra, dec, distance = earth.at(times).observe(comet).radec()

RA, Dec = [thing._degrees for thing in (ra, dec)]

plt.plot(RA, Dec, 'ok')
plt.gca().set_aspect('equal')
plt.show()

t0 = ts.utc(2020, 7, 12)
t1 = ts.utc(2020, 7, 19)
LA = Topos('34.05 N', '118.25 W') # 34.05, -118.25

times, is_sunrises = almanac.find_discrete(t0, t1, almanac.sunrise_sunset(eph, LA))
for time, is_sunrise in zip(times, is_sunrises):
    if is_sunrise:
        print('sunrise: ', time.utc_iso())
    else:
        print('sunset: ', time.utc_iso())

times, is_rises = almanac.find_discrete(t0, t1, almanac.risings_and_settings(eph, comet, LA))
for time, is_sunrise in zip(times, is_rises):
    if is_sunrise:
        print('cometrise: ', time.utc_iso())
    else:
        print('cometset: ', time.utc_iso())


Which eyepiece should I use to view planets with a Slokey Skyways 50080 telescope? I am new to Astronomy. My husband bought a Skyways 50080 for me for my birthday. It's an f=500mm refractor made by Slokey. The 80mm aperture makes it a fast f/6.25.
It came with three eyepieces; 25, 10 and 6 mm, and a 3x Barlow lens.
I can see the moon very clearly, however I can't see Mars very well. I use the 10 mm eyepiece when viewing the planets, I am not sure if this is too high a magnification or not.
Which eyepieces should I be using to see the planets with this type of telescope? Any suggestions will be greatly appreciated.

General guidance seems to be to use a magnification of  30 to 50 times the diameter in inches of your telescope's aperture. An aperture of 80mm is just over 3", so we are looking at magnifications of 90x to 150x.
Your telescope has a focal length of 500mm without the Barlow lens, and 1500mm with the Barlow lens. Since magnification is simply the focal length of the telescope divided by the focal length of the eyepiece, we can see that using the 6mm eyepiece without the Barlow lens will give you a magnification of 83x, and using the 10mm lens with the Barlow will give you a magnification of 150x.
Using a Barlow lens will increase the magnification, but it will also reduce the image quality and also the effective aperture, so the image will not be as bright and contrasty. I would suggest forgetting about the Barlow lens (at least initially), and just work your way up from the 25mm eyepiece to the 6mm. There is no point in using the 25mm eyepiece with the Barlow, since that will only give you 60x magnification, and trying to use the 6mm eyepiece with the Barlow is likely to prove frustrating, if not unworkable.
If you enjoy using your telescope, maybe treat yourself to a 3mm eyepiece, so you can get 167x magnification without the Barlow, but read up on telescope lenses (especially "eye relief"), first. Don't just buy the first lens that you find on eBay.
How to See Mars in 2020
Which lens and magnification is best to see planets?


What is a typical polar angle of a pulsar beam? I've found many articles on the geometry of pulsar beams, but I have not been able to find what would be a typical angle for the beam cone.  Illustrations I've found seem to look like it might be 1 or 2 degrees.
What range of angles might you expect to find for the polar angle of a pulsar's beam cone?

Typical angles can span the range from several degrees to close to $90^{\circ}$ in some extreme cases. For pulsars with narrow emission cones, the opening angle of the emission cone is approximately
$$\rho\approx1.24^{\circ}\left(\frac{r_{\text{em}}}{10\;\text{km}}\right)^{1/2}\left(\frac{P}{\text{s}}\right)^{-1/2}\propto P^{-1/2}$$
with $r_{\text{em}}$ the emission height and $P$ the period (Lorimer & Kramer, Handbook of Pulsar Astronomy). The emission height is the distance from the center of the pulsar at which the radio emission occurs; it is typically much larger than the radius of the pulsar. The expression can be derived by making some assumptions about the location of the last open magnetic field lines. The relation works quite well for long-period pulsars with $P\gg100\;\text{ms}$, which, accordingly, fall in the range $1^{\circ}\lesssim\rho\lesssim10^{\circ}$, but fails for millisecond pulsars, some of which have opening angles quite a few times larger.
To give a better visual perspective: The angle of the emission cone, $\rho$ depends on several things. In the classical Julian-Goldreich model, the pulsar is surrounded by plasma, which is forced to corotate with the neutron star by its magnetic field. If we take this to its logical extreme, there must be a radius at which rotating plasma would have to travel faster than the speed of light; this radius defines a surface we refer to as the light cylinder.
Any magnetic field lines which cross the light cylinder's surface do not close. Assuming that the emission cone must fall within the larger cone of open field lines, we can determine the largest allowed angular radius of this cone-within-a-cone if we know the emission height - the point above the pulsar at which radio emission occurs. Not surprisingly, the size of the light cylinder is determined by the rotational period; from the argument above, we can estimate $\rho$ from that and the emission height.


Do we expect life to develop only in certain regions of the galaxy? If I understand correctly, the Earth lies in an area of the galaxy - the "local bubble" - that is relatively free of dust and gases and where no new stars are being born.

What is the current theory on the relation between life on Earth and this location? Or in other words: Would we expect life to exist anywhere in the galaxy, e.g. also inside some nebula, or only in specific areas, and how are they characterised?

There's a short answer here and a long answer.

Short Answer: It isn't special.

Long Answer: Yes and No.

The Local Bubble really isn't so special with regards to where it's placed in the galaxy. Location-wise, it's perched on the inner edge of the Orion Arm, one of the Milky Way's spiral arms. The Orion Arm, too, isn't incredibly special. It's relatively small - it's often referred to as the "Orion Spur" - and near 1/4 of the way out from the galactic center. Not too far away, but not too near. The donut-shaped area 4,000-10,000 parsecs out from the galactic center is, however, sometimes called the Galactic Habitable Zone. In theory. life as we know it could exist quite nicely any where in this area (I should note that the scientific community is divided on whether the Zone is important, with some saying that it is arbitrary).

But let's dive in deeper. Actually, we're in a decent spot in the "Arm." We're on the edge of it - not too near large groups of stars that could effect our system via gravity, or near any supernova candidates. And the "Arm" itself isn't really a spiral arm. It is, as I said earlier, more of a spur, or bulge. There aren't as many stars to watch out for. Also, we're near the corotation circle - i.e., we orbit the galaxy at such a distance that we move at a similar speed to the arms, reducing further the likelihood of crossing them. Is this necessary for life? Nope. But does it help? Oh, yes.

There are, of course, other factors that make this area a nice spot: No black holes nearby, gamma-ray bursts, etc. But could life survive elsewhere? Yep. These same characteristics can be found in many other places in the galaxy - and life can probably survive in even more dangerous spots.

By the way, I'd avoid using the term "nebula". Technically, we're not in one, or anywhere near one. The term can refer to a multitude of objects - planet-forming areas, "clouds" surrounding white dwarfs, and other rather photogenic places - none of which we inhabit.

I hope this helps.


How do I plot orbit planets and satellites? I need to find the way to plot orbit planets and satellites in a coherent way. Let's say I want to generate a chart of the Moon's orbit around the Earth, or the jovian satellites around Jupiter.
The hard way would be the use of a CAD software, manually. However, I'm not sure if I will be able to represent the orbits excentricity. And it would be a slow process, one by one...
My idea is to do it programmatically and recursively, so I can generate hundreds of charts for my articles. Can AstroPy do it? I'm using Debian 9 and I checked poliastro, but I had problems with the requirements... I also checked Xephem, and I could not compile it (it's too old, I think). Could Celestia or Kstars do that?
Any Linux/Unix tool would be fine! Especially if it works from the command line.
EDIT: I tried to compile ORSA without luck. Also failed.
I'm discovering the combo of python3-ephem with gnuplot. I can generate the data with Ephem and plot it with Gnuplot. What do you think about this combo?
EDIT 2:
Not only python3-ephem... Skyfield and REBOUND are good options!

One keyword to search for is "spice kernel".
SPICE stands for Spacecraft Planet Instrument C-matrix Event. That is a type of data format in which orbital data are commonly provided by ESA, NASA and other institutions dealing with detailed data; it is the recommended format by IAU to use. Often the so-called kernels for objects and missions are also readily available for download and digestion of tools (e.g. here or here).
Quite a few poplular solutions are python modules like HelioPy, SpicePy or PoliAstro and possibly this tutorial might do what you want. They usually have good documentation with extensive examples for simply copy&paste on how to use them. That said, python is not the only option, there are solutions in other languages as well.
It's also worth to note that some missions come with their own dedicated tools like Cassini's JCSN or Cassini orbit plotting tool.
For natural bodies AstroPy, possibly along with JPLEphem might suffice and prove quite versatile.


Was the Geocentric Model correct at all? It's easy to find resources stating that the heliocentric model is right and geocentric is wrong.
But how wrong was it? Was it correct in any way?
It was built on incorrect assumptions, but despite that - was it of any use to describe the apparent motion of celestial bodies? Was it more accurate for some things, but less accurate for others? Or was it altogether a flop and astronomers couldn't get anything out of it either way?

I can only find multiple articles proving the heliocentric model, explaining the geocentric one or claiming that it was simply wrong - but I can't find anything about its accuracy and usefulness.
(1) (2) (3)

Edit 1:
I incorrectly used geocentric model when it seems I wanted to say Ptolemaic model - the one with deferents and epicycles, with Earth as its origin.

Ptolemaic model (click for full size)

Thank you for clarifying, and sorry for the confusion. The answers provided regarding any other geocentric models are still valid and useful, so this is just a minor errata.

Ptolemy's epicyclic, geocentric model, in use until the Renaissance, was very accurate in terms of predicting the positions of planets and the times of eclipses. What it couldn't account for were things like the correlations between apparent size and phase of Venus, or to properly account for the variation in brightness of the planets.
Thus the reason for discarding the geocentric model was not really because it lacked precision, but that it failed to explain various other observational facts, especially after the development of telescopes.
No doubt you could tune the Ptolemaic system even further (more epicycles?) to iron out some of the small errors that were revealed by Tycho's positional measurements at the turn of the 16th century, which had a precision unavailable to Ptolemy. However, the advent of Kepler's laws and subsequent explanation by Newton, rendered the geocentric model obsolete.
As you can judge from (well written) articles like this one, geocentrism is actually quite hard to kill-off observationally, if you are prepared to accept that the universe is arranged "just so".


How are East and West defined on other bodies of our solar system? I noticed that East and West appear swapped on maps of the Sun, even though the Sun rotates around its polar axis in the same sense as Earth.

See for example below Sun disk with a Venus transit diagram or also here.

So, at least for the Sun, East and West are defined from an Earth observer viewpoint. This makes me wonder: How are East and West defined on other bodies of our solar system?

Actually, it is a bit contentious. The IAU defined the North Pole of a planet to be that axis of rotation which lies north of the solar system's invariable plane. Which means that it is possible for bodies to rotate retrograde, and Venus and Uranus are considered to rotate retrograde with this definition. This definition is not universally accepted.

From early periods when solar maps were drawn, they have been labelled such that features appear on the "Eastern" limb of the sun and move to the "Western" limb, though this means that they are moving from Earth's East to Earth's West. This convention hasn't changed, though things have changed for the moon, just to confuse things.


How often are there lunar eclipses on Jupiter For the Earth-Moon system, the orbit of the Moon is at a slight incline compared to the plane of the ecliptic. This incline is enough for there to be eclipses roughly twice a year rather than every lunar month.

But Jupiter is much bigger, so its shadow is bigger also. So it seems more likely for a satellite to go into its shadow. How often does this happen?

Every Galilean moon and inner moon go into lunar eclipse once per orbit.

How often is there a lunar eclipse of the Jovian moons?

I set up a little animation of the Galilean moons (I didn't include the inner moons), and made a few assumptions (negligible moon size, sun at infinity, perfectly circular orbits) and found that there is a lunar eclipse in the Jovian system just under 10% of the time.

Here's a bigger version of the second graph.


Invisible in visible light but visible in non-visible light? This may be a stupid question, but I've been wondering this for a while and haven't yet found a solid answer. I'm worried the answer will be something along the lines of "of course not, you dummy" but it's bothering me and I haven't found the answer so I figured I'd take the risk.
Is it possible for matter to ONLY be visible in an area of the spectrum other than visible light? For instance, can any matter be invisible to us but visible in the UV or IR range?

While individual atoms or molecules can emit energy at discrete or narrow ranges of wavelengths, once you have larger chunks of matter (and they are above absolute zero) they will emit, at least, a continuous black body spectrum. While black body radiation can be observed at all wavelengths, far away from the peak wavelength it may be below our ability to detect its emission.


What can we expect from the first accurate image of a black hole? From recent news from the Max Planck Institute for Radio Astronomy:

  The European Research Council (ERC) has awarded 14 Million Euros to a
  team of European astrophysicists to construct the first accurate image
  of a black hole. The team will test the predictions of current
  theories of gravity, including Einstein's theory of General
  Relativity.

I've seen computer depictions of black holes in the past, are they similiar to what we can expect from this new endeavour?

In the next image we can see a computer model of a black hole of ten solar masses as seen from a distance of 600km with the Milky Way in the background with a horizontal camera opening angle of 90 degrees:

Image Source

Since the astronomers are using radio telescopes and not optical telescopes, I'd like to point out why they are doing so - The centre of the Milky Way is a very dusty place. Wavelengths from the millimeter to optical get easily absorbed by all this dust, so it's very difficult to see the centre of the galaxy in the optical spectrum. But radio waves do not get absorbed, and the centre of our galaxy is a very strong source of radio waves. Therefore looking in radio will give us the clearest picture of what's happening over there.

They mention in the article that they use a technique called VLBI (Very Long Baseline Interferometry) to image the blackhole.

The images you get from VLBI aren't images in the traditional sense, like you would get from an optical telescope. VLBI measures the phase difference of waves arriving at two different antennas (on two different continents, perhaps) and use this phase difference to infer the size of the source in the sky. Contradictorily, even though they can resolve sources of less than an arcsecond in size, it's very difficult to localize the absolute position in the sky to a very high degree.

So hopefully we will be able to resolve the event horizon of the black hole, but we will (almost definitely) not be able to see a picture like the one you posted.


How unusual is the solar system? In Isaac Asimov's Foundation series, the main characters are looking for the long-lost Earth. There are two major features of the solar system that seem to make it unusual among the millions of inhabited systems: the rings of Saturn and the size of the Moon.
The conversations below are mostly between Golan Trevize, ex-military, well-versed in astronomy, and Janov Pelorat, a naïve and recluse history teacher.
The rings of Saturn

"That's what the poem was speaking of. Three wide rings, concentric, wider than the planet itself."
Trevize said, "I never heard of such a thing. I don't think rings can be that wide. Compared to the planet they circle, they are always very narrow."

"Is that sort of thing common?" asked Bliss, awed.
"No," said Trevize. "Almost every gas giant has rings of debris, but they tend to be faint and narrow. I once saw one in which the rings were narrow, but quite bright. But I never saw anything like this; or heard of it, either."
Pelorat said, "That's clearly the ringed giant the legends speak of. If this is really unique-"
"Really unique, as far as I know, or as far as the computer knows," said Trevize.

The Moon

"A giant satellite is more difficult to accept. No other inhabited world in the Galaxy has such a satellite. Large satellites are invariably associated with the uninhabited and uninhabitable gas-giants. As a Skeptic, then, I prefer not to accept the existence of the moon."

"Yes. It's rather farther from the planet than one might expect but it's definitely revolving about it. It's only the size of a small planet; in fact, it's smaller than any of the four inner planets circling the sun. Still, it's large for a satellite. It's at least two thousand kilometers in diameter, which makes it in the size range of the large satellites that revolve about gas giants."
"No larger?" Pelorat seemed disappointed. "Then it's not a giant satellite?"
"Yes, it is. A satellite with a diameter of two to three thousand kilometers that is circling an enormous gas giant is one thing. That same satellite circling a small, rocky habitable planet is quite another. That satellite has a diameter over a quarter that of Earth. Where have you heard of such near-parity involving a habitable planet?"
Pelorat said timidly, "I know very little of such things."
Trevize said, "Then take my word for it, Janov. It's unique. We're looking at something that is practically a double planet, and there are few habitable planets that have anything more than pebbles orbiting them. Janov, if you consider that gas giant with its enormous ring system in sixth place, and this planet with its enormous satellite in third—both of which your legends told you about, against all credibility, before you ever saw them-then that world you're looking at must be Earth. It cannot conceivably be anything else. We've found it, Janov; we've found it."

Quotes are from Foundation's Edge (1982) and Foundation and Earth (1986).
My question is in two parts:

Are the various descriptions and facts given above generally correct?
How unusual are Saturn's rings and the Moon compared to other systems we've observed?

First, Azimov was writing (Edge 1982, and Earth 1986) before the first extra solar planets were tentatively detected (1989), or definitely detected 1991 or 1992. So what he wrote could not be based on any real data on extra solar planets, and must have been an educated guess.
However both books were written after the discovery of Charon, so he may have been ill advised to claim that large moons were unlikely around non-gas giants.
Personally I think both the large moon and broad planetary ring guesses are unlikely to pan out. Uncommon maybe but not uncommon enough for Azimov's purpose.
If you believe it, we have already discovered an exoplanet with a more impressive ring system than Saturn's


What astronomical observations would give conclusive proof of alien life? There are several ways we are coming closer to answering the question "is there life elsewhere in the universe?". One is by first understanding very well the origin life on our own planet. Another is by trying to look for radio signals from extraterrestrial civilizations. Another is to try to search for habitable exoplanets, and eventually examine the habitable ones in detail via direct imaging. Another is too look for evidence of life or life that existed in the past right in our own solar system (for exmaple on Mars, Titan, Europa, etc).

1) Have I missed any other astronomy-related approaches to answering this question?

2) Short of radio contact from an ET civilization, or a visit by aliens, or the discovery of life/past-life in our solar system, what would give conclusive evidence of alien life in the galaxy?

For example:

In 20 or so years when we develop the technology to directly image exoplanets, we will be able to determine their chemical composition and characterize their structure. If astronmers who study exoplanets found what they are looking for -evidence of oxygen and other biomarker gases - this might be highly suggestive of life, but not conclusive evidence. What kinds of observations of an exoplanets could enable us to say with ceartainty "there is life on this planet"? Short of telescopic observations of truly unbelievable detail, I cannot think of a way that we can look at a planet and say "yes there is life on it, or no there is not" other than perhaps if seeing lights on the dark side of the planet.

Edit: Actually, I forgot the obvious scenario where there are trees/grass on the exoplanet.  But I don't think the plant life is required to be a different color from the rest of the terrain, and a planet with aliens does not have to have plant life. So we might not be able to detect life visually.

Lots of frontier science lectures to watch on nearby topics here:
https://astrobiology.nasa.gov/seminars/

Your very question seems to be the hottest topic in astronomy today.

My impression is that if oxygen gas and methane were on the same exoplanet, then it certainly must be living biological processing creating that there. Nothing else could explain the coexistence of the two. And even if one of those two gasses could occur abiologically, and a range of exoplanets with one of them would be found, statistically one could deduce a frequency of planets which must have life, because the abiotic processes could only explain a certain maximum part of them.

It is a pitty that most exoplanets are in size between Venus and Neptune, which are at the same time the least explored planets in our Solar system. When a sub-Neptune or a super-Venus has its atmosphere examined for the first time, we won't have as good references as we could've had.


How does one use SPICE to calculate sunrise and sunset? I've seen it mentioned here that SPICE can be used to calculate sunrise and sunset. How?
According to the official documentation, SPICE is used to calculate:

Time system conversions
Positions of spacecraft and natural bodies
Orientation of spacecraft and instruments
Reference frame transformations
Illumination angles

I know sunrise and sunset can be calculated from the Sun's coordinates and the sunrise equation.
I was wondering if SPICE had a specific algorithm for this purpose?

I'm no SPICE expert but here are some potential solutions (unless of course you want to try Skyfield's Almanac methods!)
Firsrt possibility, but all these may not work
this answer links to this comment links to SPICE Tutorials, Updated December 11, 2019 links to the 69 slide presentation SPICE Geometry Finder (GF) Subsystem; Searching for times when specified
geometric conditions occur dated January 2020.

Much SPICE software computes a geometry parameter at a given time, t, i.e. x = f(t). Example: on 2011 MAR 30 14:57:08, what is the spacecraft’s altitude
above Mars?
The Geometry Finder subsystem does the inverse: it finds times when specified geometric events occur

GF provides two primary types of event-finding APIs

Boolean: a geometric condition (an event) is true or false
– Numeric: a geometric quantity has a given value, is within a given range or has achieved a local or global maximum or minimum

In your case you would specify a position on a body, then look for times when elevation was 0 or 0+/- small.
Another possibility, this seems to address exactly what you are looking for

https://naif.jpl.nasa.gov/pub/naif/toolkit_docs/C/cspice/spkcpo_c.html
Examples
The numerical results shown for these examples may differ across
platforms. The results depend on the SPICE kernels used as
input, the compiler and supporting libraries, and the machine
specific arithmetic implementation.

Compute apparent solar azimuth and elevation as seen from a
specified surface point on the earth.
Task Description
In this example we'll use the location of the DSN station
DSS-14 as our surface point.
We'll perform the solar azimuth and elevation computation two
ways:

Using a station frame kernel to provide the
specification of a topocentric reference frame
centered at DSS-14.

Computing inline the transformation from the earth-fixed,
earth-centered frame ITRF93 to a topocentric frame
centered at DSS-14.

Note that results of the two computations will differ
slightly. There are three sources of the differences:

The station position is time-dependent due to tectonic
plate motion, and epochs of the station positions used
to specify the axes of the topocentric frame are
different in the two cases. This gives rise to different
orientations of the frame's axes relative to the frame
ITRF93.

The two computations use different earth radii; this
results in computation of different geodetic latitudes
of the station. This difference also affects the
topocentric frame orientation relative to ITRF93.

The station movement between ET and the epoch at which
the DSS-14_TOPO frame is specified contributes a very
small offset---on the order of 10 cm---to the station-sun
position vector, expressed in the ITRF93 frame.

and there's a lot more there.
It seems you can do the same thing in Python using SpicyPy

Find the azimuth and elevation of the apparent position of the Moon as seen from the DSN station DSS-13 by the following steps:

Find the apparent position vector of the Moon relative to the DSN station DSS-13 in the topocentric reference frame DSS-13_TOPO at epoch ET. Use light time and stellar aberration corrections.

For this step, you'll need to have loaded a station SPK file providing geocentric station position vectors, as well as a frame kernel specifying topocentric reference frames centered at the respective DSN stations. (Other kernels will be needed as well; you must choose these.)

Convert the position vector to latitudinal coordinates. Use the routine spiceypy.reclat for this computation.

Compute the Moon's azimuth and elevation as follows: azimuth is the negative of topocentric longitude and lies within the range 0-360 degrees; elevation is equal to the topocentric latitude. Display the results in degrees.


Is it possible to disprove geocentrism without telescopes? The development of telescopes enabled the discovery of moons orbiting around Jupiter, and the existence of a full set of phases of Venus, as opposed to a limited set which would be predicted by geocentrism.

But is it possible to disprove, or at least make implausible, geocentrism without a telescope? Preferably without calculus or more advanced maths.

I'm curious for two reasons. One is knowing whether it was lack of technology, or humanity's attitude towards the universe, that was responsible for geocentrism. The other is that I've heard that some astronomical facts, such as the size of the earth, were calculated before the advent of telescopes, and I'd find it interesting if more facts could have been deduced even with a limited set of technology.

I would argue that the Apparent retrograde motion of the planets over time, is evidence against geocentrism, and probably why Aristarchus of Samos proposed a Heliocentric model of the Solar System.


Do orbital resonances always form naturally? For example, if I throw two planets to orbit a star at random direction, would they form an orbital resonance?

If the question is "if I throw two planets to orbit a star at random direction, would they form an orbital resonance?" -- then in general, no. A resonance is an integral ratio (1/1, 2/1, 3/5, etc.) between the periods of motion of objects -- i.e., the ratio of their periods forms a rational number. Formally speaking the odds of getting a integral ratio (let alone a strong, low-order ratio, since those are the dynamically interesting ones) if you set the system up "randomly" should be infinitesimal, because irrational numbers are (infinitely) more abundant that rationals.

However, if the orbits of one or both of the planets can change over time, then the ratio between their periods changes, and they can end up in a resonance. (Which is maybe answering the title question.) How often this happens depends on whether the planets happen to start near a strong resonance, and on how rapidly the orbits change. (If the orbit of a planet changes slowly, then it won't encounter new resonances very often; on the other hand, rapid orbital change can overwhelm the effect of weak resonances, so that the planet passes through the resonance without being caught.)

For example, it's thought that Neptune and Pluto were originally not in resonance; but the gradual outward migrations of Neptune (due to various gravitational encounters between planetesimals and the giant planets) changed its orbital period and meant that eventually it reached 2/3 resonance with Pluto, and Pluto was "captured" by the resonance, after which it stayed in resonance with Neptune.

The vast majority of objects in the Solar System are not in resonance with anything else, which is perhaps another way of answering your question. (I.e., in practice it doesn't happen very often.)


periodic inclination change circumbinary planet I am studying circumbinary planets and I made a python model to calculate certain properties of the orbits, and when I calculate the inclination I get a sine wave.

I do not see a way to predict the shape of the curve in advance, despite it being a rather simple curve. Furthermore, when I plot the eccentricity of the orbit the frequencies of the curves do not align.

Would anyone know why it oscillates at this rate and amplitude?

And a side question, would anyone know some literature on the subject of circumbinary planets?

The period of the stars' orbits is likely so much shorter than the planet's orbit that from the planet's perspective the pair looks like two donuts, or toroidal-shaped masses, possibly overlapping depending on the masses and sizes.

In orbital mechanics you can think of this roughly as a single star with a really huge J2 term (gravitational quadrupole moment).

In addition to apsidal and nodal precession (which you would have seen) J2 is well known to induce oscillations in inclination. For example see J2 long-period perturbations in the inclination.

Also from Two-Body, J2 Perturbation & J4 Perturbation Propagators:

  The solutions produced by the J2 Perturbation and J4 Perturbation propagators are approximate, based upon Keplerian mean elements. In general, forces on a satellite cause the Keplerian mean elements to drift over time (secular changes) and oscillate (usually with small amplitude). In particular, the J2 and J4 terms cause only periodic oscillations to semi-major axis, eccentricity and inclination, while producing drift in argument of perigee, right ascension and mean anomaly.


How do I remove fungus from a telescope mirror? I use a Celestron NexStar 130SLT Azimuthal Newtonian Reflector. I noticed that recently, due to the wet season, some tiny spots of what appears to be fungus have appeared.

Having taking appropriate precautions to prevent the same in future, how can I presently get rid of these fungi colonies on my primary reflector?

Several points need to be made.

1. Performance

The performance of the whole optical stack is incredibly resilient w.r.t. small spots on the primary mirror. The mirror might look visually very, very dirty, but the performance of the whole instrument will remain essentially the same. Even if you chip the mirror at the edge, it shouldn't matter. Take a marker and scribble a dozen lines on the mirror - it doesn't make a large difference for performance, not greater anyway than what the 4 vanes of the secondary mirror already do (we can start a separate discussion about diffraction artifacts from spots and lines on the mirror - but the bottom line is, don't worry about it).

With that in mind, you should be conservative when deciding to clean the mirror. Let it accumulate some amount of dirt, the scope will be okay.

2. Permanent damage

That being said, mold or other living things may chemically corrode the mirror. Again, a few spots of corrosion do not impact performance, but you don't want to allow this process to continue.

Dust will generally not corrode the mirror. Salt (from ocean breeze) will usually corrode it in time. With mold it's hard to say, but it's best if it's removed.

3. Prevention

Mold grows because the mirror is wet. The mirror fogs up usually when you bring the scope back into a warm room from outside; the mirror is cold, and water in the warm air condenses on it.

Procure a cheap, small fan, and point it at the rear end of the scope while it's warming up after you bring it back in. The air current will reduce water condensation on the mirror. Turn the fan off after 30 ... 60 minutes.

Also, keep the OTA (the big tube of the telescope) capped off while not in use. This will reduce dust deposits, and therefore mold will find less food to grow. At least cap off the top end and store the scope in a vertical position; that will reduce dust deposits greatly. The bottom end is less important.

Avoid storing the scope in a place with large day/night temperature variations. When the mirror is colder than air, condensation may form. This can happen spontaneously with scopes stored in a shed outside. The mirror fogs up in the morning, when air is getting warmer but the scope is still cold. This is much less of an issue with small scopes, but it's something to keep in mind.

You could cap off both ends while in storage, and place a bag of silica gel inside the OTA (don't drop it on the mirror). That will suck up all humidity inside the OTA, but you have to bake the silica gel in the oven every so often to keep it working effectively. You could buy silica gel online, it's very cheap. Avoid temperature variations and the gel is not necessary.

4. Clean up procedure

This is the procedure recommended by Optic Wave Labs:

https://www.youtube.com/watch?v=9Y8xFnXFVGQ

I make mirrors but I don't coat them myself (yet) - however, I've found that procedure to work very well on both coated and non-coated glass.

A few minutes of initial soaking in water makes everything easier later on.

After soaking it, add a few drops of detergent and rub very gently with your clean fingertips or with a soft, clean, lint-free cloth. Don't put pressure on the mirror, just try to detach the dirt from mirror in the horizontal direction and let it wash off. Don't drag dust particles across the glass, just detach them and let them go; this is why fingers are better than cloth - because you can feel large particles coming off.

When done, wash off the detergent with lots and lots and lots of water.

Using tap water for most of the cleaning procedure is absolutely fine. If you can drink the water, it's fine for the mirror too. Don't use hot water if the mirror is cold, there's a very small chance of cracking it, especially with cheap, mass produced glass.

When you're done cleaning, rinse the mirror with distilled water. Again, a few water spots on the mirror don't matter to performance, but the mirror looks better if it's spot-free. Tap water tends to leave water spots if you don't follow up with distilled water rinsing, and most people find the spots not pleasing from an esthetic perspective (but performance is same). So just splash a bottle of distilled water on it at the end and you'll be fine. No touching after rinsing.

Some dirt or mold may not be removed by the usual water-based procedure. In that case, use isopropyl alcohol, the highest concentration you can find (over 90% is fine). Rub gently with your clean, non-greasy fingertips or a lint-free cloth. If it doesn't come off quickly, leave it alone. Alcohol has probably killed the mold anyway.

Acetone might be used for things that resist even alcohol. But again, you have to use your judgement: a few little spots won't affect performance, so is it worth the trouble?

In any case, rinse with distilled water at the end, then let it dry off in an inclined position (carefully to not tip the mirror over on its face). Put a towel on the table in front of the mirror, so if it does topple over, at least it falls on something soft. Do not let it dry in a horizontal position, you'll get water spots and dust.

If there's some permanent damage in a few little spots, don't worry about it. The mirror will work essentially the same.

It is recommended that you do the water-based procedure every 1 or 2 years, depending on how dirty the mirror is. If dust contamination is moderate, leave it alone.

5. Post-cleanup

After everything is done, re-collimate the telescope carefully. But this should go without saying, since collimation is part of the regular maintenance of newtonian telescopes.


Adaptive Optics Shack-Hartmann Sensor and Deformable Mirror I have a technical question regarding the Shack-Hartmann sensor in deformable mirrors.  What algorithm is used to take the Shack-Hartmann measurements of local tilt and turn them into a full phase estimate that can be passed to the  deformable mirror?

(Source: Wikipedia1)

I see that the the sensor itself can generate an estimate of the derivative of the wavefront $w(x,y)$ by observing the offset of the centroid of the guidestar. If the delta in from center of the guidestar is $\delta_x$ and $\delta_y$ then the derivative of the wavefront in that region is proportional $\delta$:

$\frac{dw}{dx} \propto \delta_x $ and $\frac{dw}{dy} \propto \delta_y $

Geary's Introduction to wavefront sensors2 has the following diagram:

with the comment: "To reconstruct the wavefront, the local tilts must be
stitched together".

Is it enough to connect them end-to-end, perhaps make them non-negative, and pass them on? The values transmitted are usually Zernike Polynomial modes, so do we we desire to make them positive or zero mean? What algorithm is actually used to take the Shack-Hartmann measurements of local tilt and turn them into a full phase estimate that can be passed to the deformable mirror?

1 By 2pem - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=15279624

2 Geary, Joseph M.. (1995). Introduction to Wavefront Sensors.

Speaking as one who has dealt with those equations (and those problems) since 1984.

Yes, the simplest approach is to use matrix methods to force the phases at the edges of every subaperture to be the same.  The assumption is that there is no phase tearing , and if there were, adaptive optics pretty much would be unable to handle such disruptions.   Once the full "map" is constructed, the mean value is removed (aka the "piston" term, zeroth Zernike) to reduce the overall stroke requirement on the deformable mirror (DM).  In most systems, the mean tip & tilt values (first Zernike terms) are removed as well and passed off to a full aperture tip/tilt correction mirror.  Again, that reduces overall stroke requirements on the DM.

There's lots more in practice, because the subaperture actuators on the DM (other that segmented ones, which have a host of other problems) have what's known as the "influence function".  That is, moving an actuator essentially creates a "hill" or "valley" over a decent region of the mirror.  These influence functions may or may not overlap neighboring actuators, depending on the mirror design, but in any case the final correction matrix has to take into account the difference between an idealized subaperture tip/tilt correction and what the mirror actually can do.

Follow that up with spatial and temporal (history) noise reduction filters, and off you go.

Now, back to the start: the constant of proportionality between Hartmann spot motion and phase tilt (or tip) is a simple function of the lens focal length. Draw the triangle consisting of the optic axis distance from lenslet to image plane (i.e. lens focal length), the distance from the origin to the centroid of the spot image, and the hypotenuse from the center of the lens to the centroid of the spot.  You can thus easily see that the tilt is the difference between the length of the hypotenuse and the focal length.  Convert that from meters (or microns) to wavelengths of the sensing light, and there you are.


Dark matter hypothesis and visible light Dark matter is hypothesized as the mass based on gravitational observations is much higher than that observed via measuring how bright a galaxy is.

Why could it not be that multiple stars in the distant galaxy are in a straight line, blocking one another from view and so the brightness from the hidden stars can't be measured? Similar to how during a solar eclipse, the sun is not visible behind the moon? Granted the moon is much closer than the sun and blocks out most of the sun, but cumulatively the brightness from multiple stars in the line of sight would not add up.

Nice thought, but as the saying goes, space is really big. Stars have a tiny angular size when viewed from even their nearest neighbours (except perhaps in clusters & galaxy cores), so not much starlight actually gets intercepted by other stars, relative to the amount of light that gets absorbed by interstellar (or intergalactic) gas and dust.

And even that absorbed light isn't actually lost: it increases the kinetic energy and the temperature of the gas & dust. So some of the energy is re-radiated at a lower frequency, typically as infrared or microwave radiation. By looking at the spectrum of that radiation we can determine its temperature, and use Doppler techniques to estimate the kinetic energy of the gas & dust.

When a star intercepts light from another star, the light energy isn't destroyed. Some of the light is actually reflected, the rest of the energy is assimilated by the receiving star's atmosphere, where it will contribute to the receiving star's own light emissions.

Astrophysicists take these things into account when evaluating the brightness of galaxies. Plenty of people have tried all sorts of alternative hypotheses to dark matter, but so far dark matter is the best explanation that we have.


What have we learned from the observation of most distant galaxies We have seen galaxies around 13 billion years old. Since these galaxies formed so early in the history of the known universe, are there observables about these galaxies that are apparently different from much younger galaxies. I suppose all current physics laws had condensed out by then but was just curious about the formation and 'aging' of such old galaxies.

Nearby galaxies are seen in their old age. Distant galaxies are, on average, the same age as the local ones. But due to the finite speed of light, the farther you look, the younger you observe them. That means that the most distant galaxies are observed in their infancy, and they look much different from the old ones.
Galaxy evolution
This is one reason to observe galaxies at all distances: We get a statistical picture of the galaxies through different epochs, and hence probe their evolution. From such observations, we see e.g. how they grow with time, how their star formation increase in the early Universe until the Universe was roughly four billion years (Gyr) old and the decreased (Madau et al. 1996), and how the metals and dust built up over time.
Galaxy formation
Observing the most distant ones is not just a matter of breaking a previous record in distance, or time. When we look at the most distant (spectroscopically confirmed) galaxy to date, GN-z11, we see 13.4 Gyr back in time. What's the big deal if we find another one that is 13.5 Gyr back in time? Well, since the Universe at that time was only a few hundred million years (Myr) old, that would give us quite good constraints on the minimum timescale of galaxy formation. Can a galaxy form in only 300 Myr or 200 Myr, or does it need 400 Myr?
Physical properties
As I said above, the most distant galaxies look vastly different from the local ones. For instance, disks hadn't formed yet, both because it takes some time to settle into a disk, and because galactic mergers were more common at that time, which disturbed the galactic dynamics. The physical properties of the interstellar medium were also different, e.g. because it was more metal-poor, since metals are formed by stars over time. Dust, which is made from metals, was also sparse. But how long time does it take for stars to form metals, and dust? Previously it was thought that most dust was produced in AGB stars which die peacefully and hence have good conditions for making dust (high density and cool gas). But these stars typically live for billions of years, and observations of the most distant galaxies showed us that they contained dust as well. So people started wondering if supernovae, which are the violent deaths of massive and hence short-lived stars, could do it, even though it was thought that the gas was much too hot too form dust right after the explosion, and much too dilute after it had cooled down. But indeed it seems they can (e.g. Kozasa et al. 2009).


Relative black hole? When some object is moving, it appears smaller and more massive. What if it's moving so fast it should become a black hole? Some calculations:

Let observed radius $R_0\gamma$ ($R_0$ - radius at rest) be equal Schwarzschild radius for object with observed mass $M_0\gamma$ ($M_0$ - mass at rest):

$$R_0 / \gamma = \frac{2GM_0\gamma}{c^2}$$

$$\gamma^2=\frac{R_0c^2}{2GM_0}$$

$$\frac{v}{c}=\sqrt{1-\frac{2GM_0}{R_0c^2}}$$

For Earth it gives $v=0.999999999304c$ and for Sun $v=0.999997877903c$. Is my reasoning correct? How would we see such objects?

To answer quickly, the answer is that no, an object cannot be made to collapse as such.

There are multiple reasons as to why. Pela brings up a great argument here, reminding us that if a black hole could be formed this way, then I could be made to collapse into a black hole just by something moving by me at extreme velocities. Clearly, this cannot occur.

To understand why this is, we must first completely throw away the idea you have of relativistic mass. The entire concept of relativistic mass was created to allow for Newtonian mechanics to still hold in relativistic situations. By modifying the mass of an object at extreme velocities, equations such as $F=ma$ can be made to still hold. To paraphrase Philip Gibbs, gravity couples not only to mass but also to momentum and itself. Therefore, a description of spacetime in the vicinity of an object will be incomplete without momentum considerations. From this relativistic momentum, we can compute the relativistic energy.

While these quantities are useful for giving detailed descriptions of the body, the answer is actually somewhat simpler. We must remember that an object must behave the same in all inertial reference frames. If the object does not become a black hole in its own frame (as it clearly will not), then it cannot appear to be one in any other.

If you are wondering how to reconcile this with a relativistic mass that approaches infinity, the answer is simply that relativistic mass is not real mass. The metrics which describe spacetime in the vicinity of these objects refer to the invariant mass, which is the rest mass of the body.


Computation of mean sidereal time at Greenwich using solar position algorithm Using Equation 28 in Section 3.8.1 of Solar Position Algorithm (Reda, I. and A. Andreas, Jan. 2008), I computed the mean sidereal time at Greenwich for 2003 October 17 for four different times (UT), using a Casio fx-991MS. The computed values were cross-schecked online using the URL Computation of GAST, GMST, and ERA

Equation 28 used to compute GMST is:

v0=280.46061837 + 360.98564736629*(JD - 2451545) + 0.000387933*JC^2 - JC^3/38710000

JC is computed as:

JC = (JD - 2451545)/36525

The computed values of JD and JC are as follows:

1. 00:00:00 UT: 2452929.5   (JD) & 0.037905544 (JC)
2. 05:30:00 UT: 2452929.729 (JD) & 0.037911818 (JC)
3. 11:00:00 UT: 2452929.958 (JD) & 0.037918083 (JC)
4. 19:21:00 UT: 2452930.306 (JD) & 0.037927611 (JC)

However, there is a clear offset between the computed values and the values provided by the aforementioned URL. For example:

1. 00:00:00 UT: 22.32738 hour    (Computed) & 1.6726265 hour (URL)
2. 05:30:00 UT: 16.7231 hour     (Computed) & 7.1960077 hour (URL)
3. 11:00:00 UT: 11.30528667 hour (Computed) & 12.702743 hour (URL)
4. 19:21:00 UT: 2.93042 hour     (Computed) & 21.075605 hour (URL)

Moreover, if I subtract 22.32738 hour from 24.0, I will get 1.67262, which is the value provided by the URL.

At the same time, using the same set of equation , if I compute GMST for 1987 April 10 at 00:00:00, the computed and URL values are the same.

April 10 at 00:00:00:

JD=2446895.5
JC=-0.127296372
GMST = 13.18 hour (Computed) & 13.17954 hour (URL)

Kindly help me to understand where exactly lies the problem. If the problem lies with the concept regarding the interpretation, please help for a better understanding.

Note: Knew gmst is now deprecated , but since i am a beginner thought of understanding gmst foremost and then will dwell into current scenario.

Thanks.

The formula 28 from SPA that you linked is given with units and in a form that do not match official versions. It takes a little work to 'translate' it: also, it is needlessly susceptible to roundoff error (because all significant figures for both the date and time-of-day are strung together in one very long number).  The authors of SPA do not appear to have been striving for full official accuracy, probably their application did not require that.

I can't quite work out from the data in the question what might have happened to the calculations, online or otherwise, but maybe the following reference data could be of help.

GMST for 0h UT is still tabulated in the Astronomical Almanac, and there are also two rather easy formulae derived from the official definitions that you can use (see below), they make somewhat better preservation of end-figures than a calculation in the SPA form.

In the meantime, just for reference, the Astronomical Almanac for 2003 gave the GMST for 0h UT on 2003 Oct 17 as 1h 40m 21.4554s (1.6726265 h).  That is for 0h UT, and during the following 24h, the GMST for x hr UT is is easily found, it is very close to  $ [ (GMST for 0h)  +  x * 1.00273791 ] $ .

2003 was a changeover year for the official (IAU) calculations that relate GMST and UT1. Before 2003, the formula of Aoki et al (1982), "The new definition of universal time" (http://adsabs.harvard.edu/abs/1982A&A...105..359A) was the one adopted. From 2003 Jan 1 the formula of N Capitaine et al. (2003), "Expressions to implement the IAU 2000 definition of UT1", Astronomy & Astrophysics 406, 1135-1149 (http://adsabs.harvard.edu/abs/2003A%26A...406.1135C)  was (and still is) adopted. The difference between the results during 2003 was small (+0.0003 time-secs to get the result of the newer calculation), but it looks as if AA for 2003 was still using the older 1982 formula.

The Aoki et al (1982) formula gave :

GMST of 0h UT (in time-secs) =

$ 24110.54841 + 8640184.812866 *Tcy + 0.093104 *Tcy^2 -0.0000062*Tcy^3 $

plus a very small correction for DeltaT: this can be had from the version of the Aoki formula given in the Capitaine reference at B.3 on page 1149, (which seems to be the one that was used in official calculations) but it can practically be left out at the precision of the data in the question. Tcy means the time interval from J2000 (Julian Date 2451545.0) reduced to centuries of 36525 days. (The JD for 0h on 2003 Oct 17 0h UT was JD(UT) 2452929.5 days and DeltaT was close to +64.55 secs.)

The current formula from the Capitaine (2003) reference (most conveniently from formula B.2 on page 1149) gives

$ 24110.5493771 + 8640184.7945360 *Tcy + 0.0931118*Tcy^2 -0.0000062*Tcy^3 $

plus a very small fourth-order term (given in the reference but omitted here) and again a very small correction for the slight effect of DeltaT which can be left out or included if desired as

$ +307.4771600 * DeltaT/(86400.0*36525.0) $ (time-secs), where DeltaT is still in seconds.


How many planets have we discovered that can support human life? I have heard a lot of buzz about distant planets that could potentially be second homes for human existence, but what is that approximate number?

There is currently only one planet known to be capable of supporting human life, and you're on it.

Several planets have been found in the region in which we expect water to be liquid on much of the planet. Of these, only one fits the criteria of being Earth-sized and well placed in the habitable zone: Kepler 186-f

However we know nothing about it's atmosphere (or lack of one). The star is a red dwarf, so it could be subject to dramatic solar flares. The planet is rather colder than earth, so could be in a perpetual "snowball world" state, depending on the composition of the atmosphere and the strength of the greenhouse effect. The atmosphere would be very unlikely to be even close to breathable, and it is nearly 500 light-years from Earth, so could not be reached in a reasonable amount of time, even with much more advanced propulsion.

At the moment we can't usually detect most Earth-like planets in the habitable zone of brighter stars like the sun, though the probably do exist and may be common.


Why is the ring of light around the M87 black hole bigger than the photon sphere? Today we know that a black hole represents a region of space where the gravitational forces are very intense, this consequence of a concentration of matter in a very small region of space. On April 10 of this year, scientists from the EHT project (event Horizont Telescope) presented to the world the image that I believe everyone has seen. The first image of a black hole in the galaxy M87. We know that what we are really seeing is the "shadow" of the black hole.

But the "photon sphere", where light can orbit a black hole, is actually at $1.5R_s$, (Schwarzschild radius), whereas the observed ring is a factor of 2 wider than this. Why we not see a ring of light at the radius of the photon sphere?

The apparent radius of something residing in a Schwarzschild metric, when viewed from infinity is given by
$$ R_{\rm obs} = R \left(1 - \frac{R_s}{R}\right)^{-1/2}\ ,$$
where $R_s$ is the Schwarzschild radius $2GM/c^2$.

This enlargement is due to gravitational lensing and the formula is correct down to the "photon sphere" at $R =1.5 R_s$.

Most of the light in the EHT image comes from the photon sphere. It is therefore observed to come from a radius
$$ R_{\rm obs} =\frac{3R_s}{2}\left(1 - \frac{2}{3}\right)^{-1/2} = \frac{\sqrt{27}}{2}R_s\ .$$

There are small (<10%) corrections to this for a spinning black hole governed by the Kerr metric.

A sketch which illustrates why gravitational lensing leads to an enlarged image can be found in Pela's answer to a related question.