Different radio process band ranges If we want to observe a galaxy for which we have optical data, what parameters we need to know to calculate its total radio flux at 3 cm?
Specially I am interested to know how to differentiate among the different process in radio, like synchroton, free-free and inverse compton? Specially when I know its Hydrogen recombination line luminosity is in the power of 42, and electron temperature 10000 K with distance 20 Mpc

From literature, I assume 3 cm is 10GHz, which should fall under either synchroton or free-free emission. But how do I know exactly?

Without a lot of expensive equipment the best you can do is make a rough estimate. By incorrectly assuming that the light that you see describes the black body temperature of the object, by assuming that the system is in thermal equilibrium with itself and that the temperature within the system is spatially and temporally uniform, you can use Planck's law to calculate the spectrum.

Planck's law describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature $T$.

The spectral radiance of a body, $B_ν$, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. Planck showed that the spectral radiance of a body for frequency $ν$ at absolute temperature $T$ is given by:

$$ B_{\nu }(\nu ,T)={\frac {2h\nu ^{3}}{c^{2}}}{\frac {1}{e^{\frac {h\nu }{k_{\mathrm {B} }T}}-1}}$$

where $k_B$ is the Boltzmann constant, $h$ is the Planck constant, and $c$ is the speed of light in the medium, whether material or vacuum. The spectral radiance can also be measured per unit wavelength $\lambda$ instead of per unit frequency. In this case, it is given by:

$$B_{\lambda }(\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}{\frac {1}{e^{\frac {hc}{\lambda k_{\mathrm {B} }T}}-1}}.$$

The law may also be expressed in other terms, such as the number of photons emitted at a certain wavelength, or the energy density in a volume of radiation. The SI units of $B_ν$ are $W·sr^{−1}·m^{−2}·Hz^{−1}$, while those of $B_\lambda$ are $W·sr^{−1}·m^{−3}$.

The "Planck Calculator for Infrared Remote Sensing" can be used to make some calculations:

Temperature T(K)                 10,000
Center Wavelength μm             11.0
Spectral Radiance (W/m^2-sr-μm)  5292.388800699331

With a temperature of 10K the center frequency is 11 $\mu m$.

That's the oversimplified (inaccurate) layperson's calculation. More accurate calculations are made by allowing for everything from: relative velocity, the composition of the object's atmosphere (and our own), the effect of distance, even other elements, upon the spectral distribution (short list).

Consequently, after correcting for light travel time effects the response of the Balmer and optical helium lines should generally be strongest during low continuum luminosity states. Responsivity that depends upon photon flux and continuum state may explain a number of outstanding problems currently under investigation in broad line variability studies of these and other emission lines. [Abridged]

Once calculations are made for the blackbody curve in accordance with Planck's law it is necessary to transpose that to Rayleigh–Jeans law which is applicable at large wavelengths (low frequencies) but strongly disagrees at short wavelengths (high frequencies) where Planck's law is more useful.

See Wikipedia's "Comparison [of Rayleigh–Jeans law] to Planck's law".

Further reading:

"Introduction to Millimeter/Sub-Millimeter Astronomy" 3 Mar 2009, by Wilson
"Free-Free Spectral Energy Distributions of Hierarchically Clumped HII Regions" 30 Mar 2004, by Ignace and Churchwell
"What the Optical Recombination Lines Can Tell Us About the Broad Line Regions of Active Galactic Nuclei" 17 Jun 2005, by Korista and Goad.

If you want to include the effect of quantum gravity in your calculations a recent paper is: "Black body radiation in a model universe with large
extra dimensions and quantum gravity effects" 8 Jan 2018, by Nozari, Anvari, and Sefiedgar

  Abstract: "We analyze the problem of black body radiation in the presence of quantum gravity effects encoded in modified dispersion relations and in a model universe with large extra dimensions. In this context, modified form of Planck distribution, Jeans number, equipartition theorem, spectral energy density, Stefan-Boltzmann law and Wien's law are found and the corresponding results are interpreted. Then, entropy and specific heat of black body radiation are obtained in this setup. Finally, the modified form of Debye law and Dulong-Petit law are investigated in this framework.".

This article from the PrimaLuce Lab's website: "Radio astronomy at school with SPIDER radio telescopes" explains backyard radio astronomy.

Their 3M diameter SPIDER 500A's H142-One receiver has a 1.42 GHz superheterodyne type radiometer/ spectrometer, double conversion (type UP/DOWN) with 50 MHz bandwidth (RF=1.395MHz-1.445MHz) and 14-bit analog to digital converter. It has a spectrometer with 1024 channels (each 61 KHz) and is able to record many radio sources in the Universe, with a theoretical flow of at least 5 Jy. Price available upon request.

Their SPIDER 230C Compact Radio Telescope measures 2.3 meters and is priced at U$16K, from what I can determine from reading the literature it features the same receiver and software; dish diameter and mount capacity being the only difference, along with beam width and sensitivity.

The mesh on their dishes is good up to 5 GHz but the feedhorn normally supplied is for 1.42 GHz, I imagine that they are willing to do some amount of customization.

An explanation of the atmospheric window can be found here: "The Effects of Earth's Upper Atmosphere on Radio Signals", this image is from NASA's website:

  Question: From literature, I assume 3 cm is 10GHz, which should fall under either synchroton or free-free emission. But how do I know exactly?

3cm is 10 GHz and can be received with a 10 GHz receiver. The difference between synchrotron and free-free is explained on NRAO's webpages  Free–Free Radiation and Synchrotron Radiation.

Acceleration by a magnetic field produces magnetobremsstrahlung, the German word for “magnetic braking radiation.” The character of magnetobremsstrahlung depends on the speeds of the electrons, so these somewhat different types of radiation are given specific names. Gyro radiation comes from electrons whose velocities are much smaller than the speed of light: $v≪c$. Mildly relativistic electrons whose kinetic energies are comparable with their rest mass $m_ec^2$ emit cyclotron radiation, and ultrarelativistic electrons (kinetic energies $≫m_ec^2$) produce synchrotron radiation.

Acceleration by an electric field accounts for free–free radiation, the resulting emission is called free–free radiation because the electron is free both before and after the interaction; it is not captured by the ion. If the ionized interstellar cloud is reasonably dense, the electrons and ions interact often enough that they quickly come into local thermodynamic equilibrium (LTE) at some common temperature, so free–free radiation is usually thermal emission.


How can we detect water on Mars-like exoplanets? According to data from Curiosity, Mars' dust holds about 2% water by weight

This wasn't previously detected, so the impression we have had of Mars being incredibly dry may need to be altered. Okay it is still very dry, but there is potentially extractable water in amounts usable by colonists.

But this sort of thing would be good to discover before going to another planet, so how can we detect this sort of thing without actually going to the surface and heating up the local dirt?

It would be difficult to detect the water-by-weight in the dust of another planet to the precision that has been done for the Martian soil.

However, having said that, there are a couple of ways to detect the presence of water, and that would be, according to "60 Billion Alien Planets Could Support Life, Study Suggests" (Gannon, 2013), to detect the presence of clouds.

To confirm that the clouds are water clouds, not something the iron-vapour etc, according to web article "Hunting for Water on ExoPlanets" (bolding mine),

  Using ESO’s Very Large Telescope (VLT), a team of astronomers have been able to detect the telltale spectral fingerprint of water molecules in the atmosphere of a planet in orbit around another star. The discovery endorses a new technique that will let astronomers efficiently search for water on hundreds of worlds without the need for space-based telescopes.


How is the redshift of starlight precisely measured? When galaxies move away from us (caused by the expanding spacetime) their light seems to show a redshift. But what is really needed for this?

What time lap is for example needed to see a difference in frequency, and how does this precisely work?

Is this also possible for gravitational waves, or are amplitudes the only solution to derive the redshift from?

I don't really think I fully understand the first part of your question.

Redshift is measured by comparing the wavelengths of a redshifted pattern of absorption or emission lines with the wavelengths they would have in an object at rest. All lines are shifted by the same factor of $1+z$, where $z$ is the redshift.

Gravitational waves cannot be used to obtain a redshift because the intrinsic wavelength of the waves is not known. An inspiralling binary emits waves with a frequency equal to twice the orbital frequency. This in turn depends on the mass, but the observed frequency is also redshifted. There is thus a "degeneracy" between the mass and redshift.

What you can do with gravitational waves is estimate the physical distance to the source using the wave amplitude and how quickly the amplitude changes with time. If the host galaxy of the source can then be identified, then you can use the redshift of that galaxy to get an estimate of the Hubble parameter (e.g. see Abbott et al. 2017).


I saw something unusual in the sky So I'm a beginner at astronomy and I was trying to look at the moon but while trying to aim at it I saw a ring-like object in the sky. Does anyone know what this would be? all I could see was a ring with nothing in the middle of it. Was this a special type of star?

If you were using a telescope with a central obstruction (like a newtonian reflector (including dobsonian mounted ones) or an SCT or MCT, then what you saw is most likely an out of focus bright object (such as a star).

When things are in focus, a star shows as a point (actually, at very high magnifications, a tiny disc with concentric light and dark rings around it - this is called an Airey disc - caused by diffraction effects - but usually the magnification is low enough that it looks like a point).

When things are out of focus, you seen the shape of the telescopes aperture that gets bigger the more out of focus you are. For a refractor, with no central obstruction, this is a circle. For things with a central obstruction, you see a ring (donut) shape. This is what causes donut shaped out of focus highlights on pictures taken with mirror lenses. (and why normal out of focus highlights with normal lenses are round or close to it (the camera iris blades often produce a polygon shape when the lens is stopped down)).


Is the total energy of the CMBR diminishing with time? Is the total energy of the CMBR diminishing with time? Since it's existence the volume of the Universe (at least the visible part) has grown by a factor three, which is to say, the number of photons per unit volume is diminished by a factor 27. But the wavelength of these photons has grown by a much bigger factor. So you should think the total energy of all the CMBR photons has become less over time.

Yes. The energy density in the radiation field scales like $a^{-4}$, and the volume scales like $a^{3}$. Since the total energy is the density times the volume, the total energy scales like $a^{-1}$. Note that $a$ is called the "scale factor" and $a = \frac{1}{1+z}$, where $z$ is the redshift, assuming $z=0$ now.


Can the energy transport by radiation occur in the convection region of a star? I am new to stellar astrophysics and trying to understand the energy transports in the interior of stars.
Can the energy transport by radiation occur in the convection region of a star?
Here are my thoughts:
Since the gas pressure in the convection region dominates, $\nabla_{\rm gas}$ $<$ $\nabla_{\rm radiative}$, so energy transport by radiation will not occur in the region.
Please correct me if I am wrong.

Radiative energy transport continues. The point is that the radiative flux, which is proportional to $dT/dr$ can be overtaken when the temperature gradient achieves the adiabatic value and convection starts. Once convection is started, it is very efficient and the majority of energy flux will be transported by convection.
Details
Broadly speaking, radiative flux is ever-present and proportional to $T^3(dT/dr)/\kappa$, where $\kappa$ is the opacity.
Convective flux begins when the temperature gradient exceeds the adiabatic value and is proportional to the difference between these two to the power of $3/2$ and to the temperature, local gravity and other thermodynamic properties of the gas.
In most cases what happens is that convection is very efficient when it turns on. This means that most of the energy flux is transported by convection and the temperature gradient is almost equal to the adiabatic value. Thus you can estimate a fraction of the flux carried by radiation by assuming the adiabatic temperature gradient$^1$.
I think this fraction is small throughout the bulk of most convection zones.
There are however some circumstances where you need to be a bit more careful. These are where the density/temperature/gravity/opacities conspire to stop convection. So at the edges of convection zones there will be more complicated regions where the fraction of radiative flux grows and the temperature gradient can be superadiabatic. An obvious example is the outer parts of cool stars where the convective envelope changes into a photosphere dominated by radiative heat transport.
$^1$ An example:
In the convection zone
In the convection zone in the Sun at $r\sim 0.9R_{\odot}$, with an enclosed mass of $m\sim 1M_{\odot}$, the temperature $T \sim 10^6$ K, density $\rho \sim 20$ kg/m$^3$ and pressure $P \sim 10^{11}$ Pa.
The adiabatic temperature gradient
$$ \left(\frac{dT}{dr}\right)_{\rm ad} = \frac{T}{P}\left(\frac{\gamma-1}{\gamma}\right)\frac{Gm\rho}{r^2} \simeq  3\times 10^{-2}\ {\rm Km}^{-1}\ ,$$
where the ratio of specific heats $\gamma=5/3$.
The radiative heat flux is given by
$$F_{\rm rad} = \frac{16\sigma_{sb} T^3}{3\kappa \rho}\left(\frac{dT}{dr}\right)\ , $$
where $\kappa$ is the opacity. From a plot in Turck-Chieze & Couvidat (2011), I estimate
$\kappa \simeq 30$ m$^2$/kg at this radius.
Plugging the numbers in: $F_{\rm rad} \sim 1.5\times 10^7$ W m$^{-2}$.
This is to be compared with the actual energy flux of $L_\odot/4\pi r^2 = 8\times 10^7$ W m$^{-2}$.
Thus, the convective flux must be about a factor of 5 times larger than the radiative flux.
In the radiative zone
Then to check the methodology:
Consider somewhere well inside the radiative zone at $r=0.2R_\odot$, where $\rho \sim 4\times 10^4$ kg/m$^3$, $T\sim 9\times 10^6$ K, the enclosed mass $m \sim 0.35M_\odot$ and $\kappa \sim 0.2$ m$^2$/kg.
The adiabatic temperature gradient is $\sim 0.1$ K m$^{-1}$ and using this in the radiative flux formula gives $F_{\rm rad} \sim 3\times 10^9$ W m$^{-2}$ compared with the actual flux of $1.6\times 10^9$ W m$^{-2}$.  i.e. In this case the temperature gradient must be about half of the adiabatic gradient, such that all of the flux is carried by radiation.


Astrophysical "unitised" version of the Gravitational constant This question relates formalising the gravitational constant, $G$, (which in S.I. units is usually quoted as $\sim6.67\times10^{-11}\,\mathrm{m}^{3}\,\mathrm{kg}^{-1}\,\mathrm{s}^{-2}$) in units which are more appropriate to astrophysics, such as in the field of galaxy dynamics.

What would $G$ be in more appropriate units to astrophysics?

Useful formulations which have taken some time to dig out:

$G\sim{4.3\times{10^{-6}}}\,\mathrm{kpc}\,\mathrm{M}_{\odot}^{-1}\,\mathrm{km}^{2}\,\mathrm{s}^{-2}$

or

$G\sim{4.3\times{10^{-3}}}\,\mathrm{pc}\,\mathrm{M}_{\odot}^{-1}\,\mathrm{km}^{2}\,\mathrm{s}^{-2}$


GW from merging of neutron stars and black holes If it had been detected gravitational waves from merging of black holes (BH), and neutron stars (NS), why there are no detection of black holes merging with neutron stars?

Why a BH-NS merging should be a rarer event than BH-BH and NS-NS?

I don't think we have a clear answer yet.  LIGO hoped to detect black hole (BH) mergers and expected to detect neutron star (NS) mergers, yet the BH mergers are observed more frequently, partly because they are louder and thus LIGO can see them from further away and partly because the 20-40 solar mass BHs LIGO observed are more common than we expected.

BHs this size can be formed by the supernova implosion of a massive star, but BHs formed this way should be quite rare  They can also be formed by the merger of two smaller BHs -- which is exactly what LIGO was observing!  It seems likely that these larger (but still stellar mass) BHs were produced by previous mergers.

So in the end, the frequency of LIGO detections depends on two things: The real distribution of objects in binaries: NS-NS, NS-BH, and BH-BH and the loudness/brightness of the gravitational waves emitted by the merger.

In general, black holes form from more massive stars than neutron stars, and that means rarer stars. (The moreso since the more massive a star is, the quicker it sheds mass through stellar winds.)

So NS-NS systems come from lesser-massive stellar pairs.  NS-BH systems form the same way, but with one of the original stars being more massive than would lead to a NS.  On the other hand, the BH-BH systems we're detecting probably come from BHs that themselves were the result of mergers in very dense stellar environments.

Consequently, we wouldn't expect that NS-BH mergers would be especially correlated with BH-BH mergers, since they precursor systems are formed through different mechanisms.  But we would expect NS-BH mergers to be rarer than NS-NS mergers since the precursor systems are rarer.

There's a bit of handwaving in that!  I'll try to find some papers on the subject -- there must be a bunch.


Do stars twinkle when seen from Mars' surface? Mars' average atmospheric pressure is 0.006 atm (0.088 psi). Is that enough to make fixed stars on Mars' night sky twinkle? Do we know an air pressure or density limit for that?

Following howstuffworks and my own metereological intuition, the key effect for stars to twinkle seems less the pressure or size of the atmosphere but rather if different layers (usually of different temperature) exist, and if the gradiant between the layers is steep enough.
Light passing through zones of differently dense air on its way to the observer lead to small (time-dependent) distortions occuring for instance over warm asphalt of a road. Here, the light travels horizontally through the atmosphere, rather than vertically like in the case of twinkling stars.
I like to imagine the parcles of hot, less dense air rising from the ground as the bubbles in a lava lamp.
In the case of a hot road, a few dozends of meters (on Earth's surface) is enough to cause the flickering of e.g. cars at a distance.
The visible image then becomes somewhat blurred, and since the air is constantly (and irregularly) moving upward over the hot street, it looks as if the air is flickering.
Applied to your question: I know that dust devils are likely occuring on Mars, which need atmospheric instabilities in order to exist, so I would assume that there are indeed atmospheric conditions where one could observe twinkling stars on Mars.


Is the moon "perfectly" tidally locked and, if not, how long would it take us to observe it's rotation? I have a general understanding of how and why a body in space can be tidally locked to it's planet or sun and I'm aware that our moon is in such a state.

My question is, if our moon once had a rotation, has it slowed down to a point where across the course of our normal lifetimes we can't observe it's spin but across the span of hundreds, if not thousands, of years we could observe its current rotation?  In other words, if it has a rotation and if I looked at the moon today and jumped into the future, how far would I have to go to see a noticeable difference?

If the question is to relative then I'll ask it like this-- at the moons current rate of rotational speed and deceleration, how long would it take for the moon to rotate, say, 15 degrees on it's current axis?  I assume this would be enough to make the moon look "different" to the naked eye so I'll go with that concrete figure.

...or is the moon in either a state of near-equilibrium or is it "wobbling" due to other forces outside the gravitational pull of the earth (e.g. pull of the sun, asteroid bombardment, comets passing by, etc.) having minor effects on its rotational state and thus making it's rotation inconsistent and unobservable?

The question is interesting, but I suspect the answer is that the Moon will never show its "far side" to the Earth, because there are differences between the side that faces us and the far side that suggest there is something quite permanent about its orientation.  So while the rotation was locking, it either settled into a state of minimum potential that it has maintained ever since, or it built up the differences between the side we see and the side we don't after it tidally locked, but either way, this means it has showed us the same side for billions of years, so will likely continue to do that.  I don't think anything would likely knock it out, but we can't rule out chaotic orbital effects so I don't really know.  The Moon's orbit varies a lot with time, so maybe it's possible that the orbit itself could change, showing us the other side of the Moon.  Or, an asteroid hit or some such thing might occur.  A billion years is a long time, but I'll bet humanity will always see the same side from Earth.


View of moon from South Pole? What would the moon look like to someone at the South Pole?  To a first approximation, it's bisected by the horizon. But how far above and below does it get?

The moon orbits close to the plane of the ecliptic, which is tilted at 23.5 degrees to the equator, the moon is slightly off the ecliptic, and can be about 5 degrees above the ecliptic.

So, on the winter solstice (June 21) if full moon happens to be furthest from crossing the ecliptic, the moon would be 28.5 degrees above the horizon: roughly a third of the way between the horizon and the zenith. It can get the same distance below the horizon, and during summer the full moon would not be seen at all.

Our moon is unique in being close to the plane of the ecliptic, and not in the plane of the equator, which suggests its formation was not like that of other moons in the solar system.


How to get the zeropoint magnitude? I'm working with some data and I have some values named fluxmag0 and fluxmag0err that should correspond to the zero point values and error in flux units.

The values are around $10^{13}$ for fluxmag0 and $10^9$ for fluxmag0err

I need to have the zero point error in magnitude units. If I take $2.5\log_{10}(\mathrm{fluxmag0err})$ the values are huge so I suppose it's not the way to compute this.

How can I convert these values in terms of magnitudes?

Thanks a lot in advance for your help

Magnitude is defined relative to a standard flux of a magnitude 0 object $F_0$.

$$M = -2.5\log(F/F_0)$$

if the error is $F_E$, then $F =F_0\pm F_E$, so the magnitude M is

$$ -2.5\log((F_0+F_E)/F_0) < M <-2.5\log((F_0-F_E)/F_0)$$

The errors here are tiny in comparison to the measured flux, only one part in 10000, so the range of magnitudes is similarly tiny +-0.0001


Why does Saturn have a more prominent equatorial bulge and higher flattening ratio than Jupiter despite rotating slower? It is my understanding that a planet's equatorial bulge is caused by the centrifugal force produced by its rotation, and the faster a planet rotates the bigger the bulge and flattening ratio of a planet is.

Jupiter's rotational period is 9.925 hours while Saturn's is 10.56 hours. However, Saturn not only has a higher flattening ratio than Jupiter, but also has a bigger equatorial bulge (11,801 km vs. 9,276 km) despite rotating slower and being smaller. Am I missing something?

Consider the dimensionless parameter $k$ for some planet, given by
$$k \equiv \frac{\Omega^2 a^3}{G M}$$
where

$\Omega$ is the planet's equatorial sidereal rotation rate,
$a$ is the planet's equatorial radius,
$G$ is the universal gravitational constant, and
$M$ is the planet's mass.

The values for this dimensionless parameter are

0.003461 for the Earth,
0.08919 for Jupiter, and
0.1576 for Saturn.

The factor $k$ is essentially the ratio of centrifugal effects to gravitational effects at the "surface" at the equator of the planet, surface in scare quotes because giant planets don't quite have a surface. Ignoring higher order terms, the flattening of a uniform density planet in hydrostatic equilibrium can be shown to be $\frac54k$. The mathematics that go into this derivation are a bit hairy, even for a planet of uniform density, so for now I'll just state this as a fact. (But see Rotational Flattening, for example. Note well: The derivation at this site involves a bit of hand-waving. Overcoming the hand-waving involves even more mathematics.)

The assumption of uniform density is not that good. Even if one does allow for density variations as a function of radial distance from the center of the planet, the parameter $k$ defined above plays a significant role in the computation of what the flattening should be. The observed flattening of Saturn versus that of Jupiter is consistent with the fact that the value of $k$ for Saturn is 1.76 times that of the value for Jupiter.


Convolve a SED with a filter. Is convoluting the mathematical operation? I know that in order to get the Flux of a star (or something else) in a particular filter from its SED (luminosity per unit wavelength), I need to convolve the spectrum (SED) with the filter response. Most of the formulas I see to do this are
$$F_{b}=\dfrac{\int f(\lambda) T_{b}(\lambda)d\lambda}{\int T_{b}(\lambda)d\lambda}\,,$$
where $f(\lambda)$ is the SED, and $T_b$ is the filter response in band $b$.

This formula seems very different from a mathematical convolution that I would write as
$$f*g(\lambda)=\int_{-\infty}^{\infty} f(x) g(\lambda-x)dx\,.$$

Are these two convolutions the same thing? Or is the "astronomy convolution" a different thing (e.g. SED Fitter python package)?

To get the flux of an SED through a particular filter, you actually multiply the the SED by the filter's response. Talking about convolution in this context is a bit of a misnomer.

Basically, for each wavelength, you look at what fraction of the light will go through the filter, and you sum up the values you get at those wavelengths.
The division by $\int T_{b}(\lambda)d\lambda$ is just a normalization term.


If a rocky world is large enough (not a gas or ice giant) can it have helium from the primordial disk in its atmosphere? When planets are formed do only gas and ice giants accrete hydrogen and helium? Does this accretion have to be beyond the "snow line"?

In other words, would all hydrogen or helium in the atmosphere of a super-earth come from outgassing (radioactive decay, etc)?

The snow line is the inner radius from the star where the water condenses onto dust and larger particles as ice (ie snowballs).  The hydrogen and helium never freezes out.  Planets start forming by accreting these particles (with or without ice but definitely not with H or He), but when a growing planet reaches a sufficiently high mass, it starts to pull in H and He (and more importantly, can hold onto them as they have low atomic mass and thus tend to escape quickly unless there is enough gravity). Exactly what mass that occurs at depends on a lot of factors (temperature, pressure, etc), but it happens below Neptune's mass and in part of the super-earth mass class.  The upper range in mass of the super-earth are also called mini-Neptunes or gas dwarf planets and from their typically low densities it is clear that they are mostly composed of H and He.

However, beyond the snow line it is more likely for a planet to grow to a large mass since more of the surroundings has condensed out of the gas into small particles, therefore accretion is more efficient.  So, it may be that mini-Neptunes are usually forming beyond the snow line and then some of them migrate closer to its star.  But, I don't see any reason to doubt that some of the many mini-Neptunes with H and He found by Kepler/K2 inside the snow-line were formed in situ.  I am not an expert in this field, and perhaps the experts would disagree.


How long until the Earth and Moon become a binary planet? With the rate at which the moon is further receding into a higher orbit, how long until the barycentre between us and the moon leaves the earth, and going by the IAUs 2006 definition update, we become a double planet?

I'm not sure I agree with the double planet POV, but the calculation is pretty simple.   The earth weighs 81 moons, so for the Barycenter to be outside the earth, the distance (center of Moon to surface of earth), = 81 earth radii.

or about 515,000 KM.   It's current farthest distance is 405,000 KM, average distance 384,000 KM and closest 363,000 KM  Source, so, it depends on whether you mean, temporarily outside, outside more than 50% of the time or always outside - each would provide different answers.

But moving away 4 CM per year (same source) or 1 KM every 25,000 years, it would need 2.75 billion years to move away the necessary 110,000 KM necessary to have the barycenter move outside of earth at the moon's furthest point.  (less if the orbit gains eccentricity which is possible).   But probably more than that, cause as the moon moves away from the earth, the tidal forces that continue to push it away grow weaker and the earth slows down a little.    If the earth ever gets tidally locked with the moon, the reverse will happen and the sun's tidal forces will draw the Earth and Moon slowly towards each other, so a precise answer is too mathematically difficult for me, but longer than 2.75 billion years seems a pretty good guestimate.


What effect does a natural satellite have on a planet's rotation and revolution? how is the rotation and revolution of a planet that does not have a
 natural satellite different from  a planet that has say 1 natural satellite
(ex. earth and moon system).

A body and its satellite in truth form a system and both rotate around their common centre of gravity. Hence both the Earth and Moon rotate around this centre - however because the Earth is significantly more massive than the Moon this can be approximated as the Moon orbiting the Earth.

The Moon does have a measurable impact on the Earth - tides are the obvious example, and tidal friction is also slowing down the Earth's rotation by a very small amount (around 2.3ms per century).

In the Moon's case, though, the Earth's gravitational pull has already effectively locked its rotation period with its orbital period around the Earth - so we essentially only ever see the same face of the Moon on Earth.

The Earth and Moon are much closer in mass than most other planet/satellite systems in the Solar System - think of the huge mass of Jupiter - and so it is difficult to show how these planets' satellites have a similar impact on the planet - but the same effects are there, albeit in much smaller scale.


Can Earth host another natural satellite than the Moon? Are there orbits with long-term stability (astronomical time scales) around Earth? Low altitude orbits will decay due to friction with Earth's atmosphere. Inside Earth's Roche limit, a Moon could not form anyway due tidal forces. At altitudes closer to that of the Moon, the Moon would presumably destabilize an orbit such that the object would either collide with Earth or the Moon or be ejected.

Many of the other planets in our solar system have multiple natural satellites; where this occurs, they are always vastly smaller than the primary - think of Jupiter, Saturn, and even Mars. On the other hand Earth, and apparently Pluto, each has only one natural satellite, relatively large in comparison to the primary. Is this because a large satellite will tend to clear the entire "sphere of influence" around its primary of any other material?

So my question is: does the existence of our Moon effectively prevent the formation or capture of any other natural satellite around Earth? (Considering only true satellites, not "companion" objects or quasi-satellites)

Short answer:   It might be impossible "long term".  It's problematic because of the size of the Moon and the proximity to the Sun.

Long answer:

The Earth-Moon system in empty space could hold several more moons (like the Pluto-Charon system).    The problem is proximity to the sun in addition to the size of the Moon.    Planets closer to their star have a harder time holding onto moons in part cause the Hill sphere is smaller, in part cause the tidal forces are that much bigger (and those 2 factors may go hand in hand, not in addition to).   But it's not just coincidence that Mercury and Venus don't have moons, they're also less likely to hold onto Moons.

Mars has a smaller Hill Sphere than Earth due to it's lower mass but it's 2 moons are very tiny.  Also, one of Mars 2 moons might not be there that much longer.  It may crash into Mars in as little as 100 million years.

Earth's proximity to the Sun gives the Earth a Hill Sphere of about 1.5 million km, and the true region of stability (long term) is about 1/3 to 1/2 of that, so about 500-750,000 km which isn't all that much further out than the Moon is currently.

Our Moon's average distance is 384,000 km, but our moon has a slightly eccentric orbit so it's distance varies between about 363,100 and 405,700 km.  Source.

The 2nd factor is the relative mass between the Moon and the planet.   Mars has 2 moons but it's moons are very small and they have very little gravitational effect on each other.    Our Moon is 1/81st the mass of the Earth which is the largest Moon to planet ratio in our solar-system.

Using the simplified hill sphere formula, the cube root of (the ratio of the masses/3).   Basically the cube root of (1/243) gives Earth's moon's a hill sphere of about 16% it's distance to the Earth.   Any Earth orbiting object that passes within the Moon's hill sphere is obviously not going to be in a stable orbit.   In fact the region of instability probably extends well past that 16%.   How far past, I'm not sure, but I can safely say that anything even remotely close to the Moon's orbit couldn't orbit the Earth for any period of time.

Using 16%, the clearly unstable region is between 300,000 km and 470,000 KM.   It would be impossible for an object to orbit the Earth at a greater distance than the Moon.   Solar and Lunar perturbations would be too big.

If it's possible for Earth to have a 2nd moon at all, it would need to orbit quite close to the Earth.   Probably well inside 300,000 KM.  The closer to the Earth the more the Earth's gravitation would be dominant, so if I was to guess, I think the Moon would need to be quite close, maybe in the 50,000 km range - but that's just a guess.

For comparison, Jupiter's 3 moons in orbital resonance; Ganymede, the largest of the 3 has a hill sphere is about 3% it's distance from Jupiter.   The 1:2:4 orbital resonance works out to the 1.5th root of 2 or about a 1.59 to 1 orbital distance ratio.  Those 3 moons also have very nearly circular orbits.    If the Moon's orbit was nearly circular, there might be a stable resonance 1.59 times closer to the Earth, but because of the Moon's orbit is measurably eccentric, I don't believe the 2:1 orbital period ratio would be long term stable because there's too many wobbles in it.   I think your best bet for a stable orbit for a 2nd moon would be very close to the Earth, probably in the 50-100,000 KM range (but that's a rough guess).

It's worth noting that "long term stable" isn't a precise term cause there's no clear cut-off.  A thousand orbits?  A million?   A billion?   Lifespan of the Sun?

I realize that "maybe, maybe not" isn't an answer, but I wanted to touch on some of why it's difficult.  A captured satellite couldn't get captured into a relatively circular near earth orbit because the incoming velocity is too great.  In theory, if there was enough debris, a satellite could form close to the Earth and perhaps be stable for relatively long term, but the Moon would to some extent prevent such a formation in a similar way that Jupiter prevents the asteroid belt from coalescing into a tiny planet.   (the low mass of the asteroid belt plays a role in that too, perhaps a bigger role), but Jupiter is a factor. Article here


How to make motion of the Sun more apparent at seconds scale? Each time I look at a shadow cast by a sunlit object it seems as if the shadow doesn't move at all. But if I look at it after say 5 minutes, I can notice that the shadow has actually moved a bit. This allows to say "yes, the Sun has moved in the sky". But this doesn't make it all that obvious that "the Sun is continuously moving all the time".

So I'd like to make some optical configuration which would allow me to watch a spot of sunlight or a shadow moving much faster — say, 1 mm/s — still being from the Sun, not from an artificial light source. I've tried doing this with a cylindrical silvery surface like that of a kitchen rail system, sticking black tape on it and leaving only a lateral strip of reflecting surface, then then letting Sun rays to hit it almost tangentially, and to observe the reflection. But either this doesn't give good enough magnification to make Sun's motion apparent, or the reflected spot just becomes too faint for the naked eye.

So, what can I do to effectively magnify angular motion of the Sun to observe it (almost) directly at the scale of seconds? I's OK if it could be visible for e.g. a minute and required me to adjust my instrument to view the next piece of motion. But I'd like to do it as simply as possible, without too expensive hardware.

The obvious answer is to use a telephoto lens -- or any other magnification scheme.  As an extreme example, if you had a post 1 km tall, the shadow of its top (think sundial) would move pretty clearly.  If you re-image a short post, or just re-image the sun itself, which enough magnification, you'll see the motion.

This is kind of the same as trying to observe a celestical object with a high-power telescope. If you don't have an equatorial mount, everything in your field of view will move quite noticeablly.


Is the sun flipped on the other side of the world? As the orientation of the moon is different in the northern and the southern hemisphere, is the orientation of the sun different in both hemisphere?

Does a sunspot appearing in the Sun northern hemisphere from the Earth northern hemisphere will look as located in the Sun southern hemisphere from the Earth southern hemisphere ?

If I understand well, an equatorial mount can compensate for the Earth tilt, but how is possible to compensate from one hemisphere to the other one? Is the image flipped?

thank you!

Yes, the orientation of the Sun will be different from the Earth's northern and southern hemisphere, just like your example of the Moon.

I would not say that a sunspot in the "northern" hemisphere would appear to be in the "southern" hemisphere just because of the change in orientation. North is fixed on the Sun and Moon, just like they are on the Earth. (If you turn a globe of the Earth "upside down", the northern and southern hemispheres do not switch, just the orientation."

The view through a telescope may change the orientation, but the directions North, South, East, and West on the Sun and Moon remain the same. The user of the telescope needs to determine which directions those are (up, down, left, right). It changes with the design of the telescope (reflector versus refractor) and the number of optical elements.


What makes the Moon and Venus shine? Is there some substance on Venus or some kind of material, because neither bodies have any source of light.

The sun is an excellent source of light for all the planets and moons, although it gets a little dodgy way out near Pluto. The moon reflects about 10% of the sunlight that hits it, that's why we see it. Venus' albedo is about 0.75, what with its clouds and all. That means 75% of the sunlight that hits it is reflected back into space for us to see.


How useful are filters for spotting nebulae? Recently I've been trying to spot a couple of nebulae, IC59 and IC1318. The skies at the site are modestly dark (4.5 on the Bortle scale), I'm using an 8" scope, and I've slowly scanned around the areas endlessly with a low magnification/wide angle eyepiece but I haven't been able to find either of them.

Is a filter likely to significantly help in this situation? Would a nebulae filter be more helpful than a light pollution filter?

With an 8" scope, a filter will very likely give you better results than observing without a filter. Although a filter does block light, the crucial aspect is that a filter increases contrast (by blocking light pollution and extraneous wavelengths of light more than the nebula), thereby allowing you to spot low contrast diffuse nebulae (like IC59 and IC1318) much more easily. This is in fact more critical for visual observation than for photography because it is possible to increase contract in post-production with photography. You will find that visual astronomers go to great lengths to increase contrast--baffling, flocking, premium mirrors, etc.

Light pollution filters are broadband, meaning they allow all light except light emitted by streetlights. From darker skies (like bortle 4.5), this will not give dramatic results. I would recommend using an OIII filter to start with. The OIII filter is a narrowband filter, meaning it cuts all light except at a very narrow wavelength range from ionized oxygen. Very often OIII filters make the difference between being able to see an object and not being able to see it, even from very dark skies. As an excellent test of the OIII filter, check out the Veil Nebula too; from your skies, this object would be rather difficult without an OIII but rather stunning with an OIII filter.


How to use a telescope to find a specific celestial body? I am a beginner interested in astronomy. I bought a Celestron AstroMaster 130EQ telescope. It's a Newtonian Reflector and it's not computerised. I did manage to see the Moon in some magnifications but my real question is this:

Suppose I want to see Proxima Centauri and I know it's coordinates, like Right ascension is 14h 29m 43.0s and Declination is -62° 40' 46". Having three analogue dials on my telescope (picture from the front) (zoomed in), how do I adjust my telescope with the help of the dials with the coordinates so I can view them from my telescope?

If I am wrong that these are the dials I have to use to adjust my telescope to the coordinates then please explain how.

I hope this question will also be useful to other amateur astronomers. Thanks.

You have to set your equatorial axis to tilt to match your latitude eg if you were at 5 degrees North then the axis needs to tilt to 5 degrees. Then you need to find an object of known RA and declination (at my latitude I always used Polaris as it didn't move) and then set the setting circles (the dials) to those, then direct the telescope to the RA and declination you seek.


What prevents consequent eclipses? I know there are a number of parameters that determine the occurrence of a solar or lunar eclipse. The number should be 18; 6 degrees of freedom for 3 rigid bodies. But can't figure out why eclipses are not repeated.

When we have a night with a full moon, the next night we again have a close-to-full moon, so it can be considered a distribution over several days. What prevents the same happening for eclipses? Is it possible to have two semi-eclipses in two consecutive days? If not, why does such a big difference occur over just a day?

Thanks,

You have to remember that the alignment for an eclipse is a very rare situation. It's not as simple as the occurrence of a new moon. The planet and the moon have to align just so in relation to the sun for an eclipse to occur. Sorry, no numbers here but just trying to emphasize that several factors have to occur between 2 fast moving bodies.


Determine distance from the galactic center in DS9 I have a list of coordinates of black hole/neutron star systems in decimal degrees (ie. {(266.416,-29.009), (266.418,-29.009),...} from DS9 and would like to determine the distance from SgrA* in parsecs. To my knowledge, 25.8 parsecs ~ 1 degree, but the distance I calculate do not agree with values previously determined by a colleague.

For example, I have a list of nine very faint X-ray transients (VFXTs) and their coordinates, one of which is (266.4191667, -29.0101111). Using the distance formula, I estimated the degree distance from SgrA* - from Wikipedia, in decimal degrees, (266.4168371,-29.0078106) using the Euclidean distance formula and multiplied my value to get

  (25.8 pc/deg)(0.00327404 deg) = 0.0844702 pc

from the center; but this value does not agree with any previously determined values of the VFXTs offsets.

The list I received from my colleague for VFXT offsets is

"vfxt_d = [.7209, .8605, 1.0233, 3.1395, 3.1628, 10.4651, 14.8372, 25.6744, 29.2093] "

I would appreciate some pointers to complete this project sometime today. Thanks in advance.

You cannot estimate the distance to anything or from anything in 3-dimensional space using a 2-dimensional coordinate alone.

You may calculate an angular distance (as per Mike G's answer), then assume they are all at a common distance and translate that angular distance into a physical distance, but since they aren't all at the same distance, that won't work.

To do what you want requires an estimate of the distance to these sources.


Why doesn't the solar azimuth vary uniformly For the past few dates, I've been observing the rotation of the earth around the sun and noticed that the azimuth of the sun doesn't vary uniformly throughout the day. The azimuth varies faster during some parts of the day, depending on the position
link.

Why does this happen? I've been thinking about the cause, but can't come up with a proper explanation.

The Sun's hour angle, measured around the celestial equator, changes at a nearly constant rate.
The gnomon of a sundial is aligned with the polar axis to take advantage of this.
The celestial equator is oblique to the horizon except at the Earth's equator and poles.

The lines of azimuth converge at the observer's zenith and nadir, so the Sun's azimuth changes more rapidly near one of those points (e.g. summer midday, winter midnight) and more slowly near the horizon.
The amplitude of this fluctuation depends on the Sun's declination and the observer's geographic latitude.

For an observer in the tropics, on a day when the Sun passes directly overhead, its azimuth swings abruptly from 90° to 270° at noon.
For an observer near the north or south pole, the Sun remains near the same altitude all day, and its azimuth rate stays near the average 15°/hour.


What could have made a satellite appear to move side to side while near the horizon? Yesterday while coming out of a restaurant my wife and I noticed what appeared be a satellite passing quite high in the sky. Based on the location, time, and brightness this appears to have been a passing of the GEOS 3 Rocket. What made this unusual is that when it got low (perhaps 25-30 degrees off the horizon) it suddenly appeared to dart side to side (over 15 degree sweeps) multiple times over perhaps a second, while continuing its general western trend, before vanishing.

Since this all happened rather quickly I had to estimate the distances after the fact using my hands and what few reference points were available. It is certainly possible that it was lower in the sky than my estimate as that is always a tricky part to judge. I'm quite certain of the sweep distance but an answer that could explain a smaller sweep could still be very useful.

What could cause a satellite to appear to move side to side in this fashion?

Your link shows estimated magnitude, and it gets quite dim when it gets close to the horizon.

above: from heavens-above, (from url: lat=52.2681&lng=-113.8112)

As a spacecraft in low earth orbit appears to approach the horizon, the line-of-sight distance from it to you can change from a few hundred to a few thousands kilometers, which is why the brightness in your table changes so much.

The path in the sky shown in the link is going to be almost vertical (reaches 70 degrees). Wikipedia says GEOS-3 is a satellite put into orbit in 1975. These days there is a big effort to de-orbit the "leftovers" so they burn up in the atmosphere (space junk mitigation) but back then it was common for each launch to leave other stuff up there for decades or longer.

Wikipedia says the COSPAR-ID for the actual GOES-3 satellite is 1975-027A. That means it was the 27th internationally recognized launch in 1975. "A" is assigned to the GEOS-3 satellite itself, but if you look up 1975-027 in the satellite catalog (SatCat) at Celestrak.org you can see there are were four other objects associated with the launch being tracked. One of them has burned up, but the others are still there.

You can double check if 1975-027B is the one that your link is talking about. Anyway, since they are so old, none of them is capable of a spectacularly sudden engine burn and change in orbit. So I'd have to say that whatever you did see, it isn't associated with the GEOS-3 launch.

Another point to consider; if you just coming out of a restaurant, the chances are low that you could see a 5th magnitude object. Near the horizon the apparent speed would be much slower than that of a satellite overhead.


Calendar year versus Tropical year I'm wondering about the convention for keeping the calendar year in sync with the tropical (exact) year.  I did a preliminary calculation, but, it seems to be out by more than the average calendar year's additionally added leap second, every year or two.  Where have I gone wrong?  A bit of my reference material follows.
(365.2425 days/year - 365.24219 days/year) * (24 hours/day * 60 minutes/hour * 60 seconds/minute) = 26.7840000011347 seconds/year.  So, shouldn't this many seconds be subtracted, from the calendar year, each year?
Calendar year
The Gregorian calendar, as used for civil and scientific purposes, is an international standard. It is a solar calendar that is designed to maintain synchrony with the mean tropical year (Dobrzycki 1983, p. 123). It has a cycle of 400 years (146,097 days). Each cycle repeats the months, dates, and weekdays. The average year length is 146,097/400 = 365 97⁄400 = 365.2425 days per year, a close approximation to the mean tropical year of 365.2422 days (Seidelmann 1992, pp. 576–81).
Gregorian calendar
The Gregorian calendar attempts to cause the northward equinox to fall on or shortly before March 21 and hence it follows the northward equinox year, or tropical year. Because 97 out of 400 years are leap years, the mean length of the Gregorian calendar year is 365.2425 days; with a relative error below one ppm (8·10−7) relative to the current length of the mean tropical year (365.24219 days) and even closer to the current March equinox year of 365.242374 days that it aims to match. It is estimated that by the year 4000 CE, the northward equinox will fall back by one day in the Gregorian calendar, not because of this difference, but due to the slowing of the Earth's rotation and the associated lengthening of the day.
Leap seconds
Between 1972 and 2020, a leap second has been inserted about every 21 months, on average. However, the spacing is quite irregular and apparently increasing: there were no leap seconds in the six-year interval between January 1, 1999 and December 31, 2004, but there were nine leap seconds in the eight years 1972–1979.

Leap seconds don't correct for the problem of the calendar years cumulatively drifting out of sync with the seasons, which is what the first source means with 365.2425 being inequal to 365.242374. Instead, the purpose of leap seconds is to correct the problem that 24 * 60 * 60 = 86400 seconds is not the length of the day, and the reason we don't get leap seconds nicely on schedule is that the slowing of the Earth's rotation is nonuniform.
What the second source means when talking about the slowing of the Earth's rotation, is that when the day gets longer, the length of the tropical year in days gets shorter, which means that the relative error between the mean calendar year and the tropical year gets even worse.
The cumulative error in the calendar is not corrected by any means currently, and it will take a long time for the error to be meaningful in the sense of impacting everyday life, so it is not viewed to be a real problem now. This might change in a couple of thousand years, or then people even then might value the algorithmic stability of an unchanging calendar more than the spring equinox's occurence near March 20th.


Stellar data for data mining Where could I find detailed data from stars? Like spectrum, x ray and so on ?

Because I would like to do some data analysis on it.

The Sloan Digital Sky Survey website has a good page with lots of links on it to retrieve spectra in different ways, you can access it here. The most useful of these is probably this one. This is just one of many places you can search but its a good starting point if your just curious about analysing spectra and not too picky about the source.


Is there any way to seperate astronomical and instrumental objects? Here is a image of hcg79 taken from APO. I'm just curious if there is any way to seperate astronomical and instrumental features in this picture? Thank you.

A priori it is very difficult to distinguish the origin of any particular feature from just one image.

For that reason it is established workflow, especially in astronomical context, to create 4 kind of images:

the light frame R. That's the actual image of what you are interested
in
the dark frame D. That's an exposure of identical length and at
identical conditions, but with closed apperture. That takes care to
assess the sensor reading in the absense of anything visible
the bias frame B. That's a short-duration image which allows to assess the
read-out noise
the flat field F. That's an image of a perfectly white /
grey area which allows to assess the different sensitivity of the
individual pixels.

Then you can obtain the corrected image C via $ C = (R-D-B) \cdot \frac{<F - D - B>}{F - D - B}$

https://en.wikipedia.org/wiki/Flat-field_correction


How are Galaxy Super Clusters Generated I have seen pictures of clusters of galaxies, usually used in regards to theories of dark matter and galaxy formations. One of the most famous ones has the perceived shape of a stick-figure. If I am not mistaken some of these clusters seem to be bigger than our observable Universe horizon of $14$ billion light-years, which is as far as we can see because of the age of the Universe.

My assumption is that these are just coordinates plotted into a computer simulation. However, are these clusters the size of our observable Universe horizon or are these more theoretical abstractions?

EDIT: Interesting new find on the subject. Laniakea Supercluster.

No, the structures are not of the size of the observable Universe. In fact modern cosmolgical models rely on the Universe being homogeneous over scales above $100 \textrm{Mpc}$. This is an observational fact, which serves as a base for the so-called cosmological principle. See also a wiki article on large-scale structure, where the scale is mentioned in the section "End of Greatness".

The scale corresponds to a few hundred million light years. The largest structures, therefore, are clearly way smaller than the size of the observable Universe, which is of order of ten billion light years.

Note, that because it takes time for the light to reach us, the current size of the Universe we observe is larger than its apparent size due to cosmological expansion.


How can we see Venus at night? According to the Globe Earth Theory, the Sun is in the middle of the Solar System and the Earth is the 3rd planet from the Sun, making Mercury and Venus between Earth and the Sun. If this is true, Mercury and Venus should only be seen from the lit side of the planet, since it is always between the Sun and Earth. How can you see it at night?

Before we begin: how far a planet is seen from the Sun is called elongation, and it's measured in degrees. 0° elongation means it's right on top of the Sun (or behind); 180° elongation means it's opposite to the Sun (it's highest in the sky at midnight, when the Sun is on the other side of the Earth).

https://en.wikipedia.org/wiki/Elongation_(astronomy)

See the image below, where the Sun and the Earth are shown, along with an inferior planet (such as Mercury) and a superior planet (such as Mars). Elongation is the angle marked ε.

  How can you see it at night?

Both Venus and Mercury are always seen close to the Sun. You can never see them at midnight, true. But you can see them during twilight, or, in Venus' case, shortly after nightfall (and then it sets quickly), or shortly before dawn in the morning (and then it's masked by the light of day). This is not a paradox or an impossibility. Here's why:

Mercury in particular can only be seen during twilight, very close to sunset or sunrise. Its orbit is so close to the Sun in space (only 39% the size of Earth's orbit), it does appear very close to the Sun even in the sky as seen from Earth. You could not see it during the day, of course, because the sky is so bright. But during twilight you can sometimes see it peeking just to the side of the Sun, very briefly - then it either sets in the West, or is swallowed by the light of day.

The greatest elongation that Mercury can achieve is 28°. Hold your arm straight out in front of you and make a fist; twice the size of your fist is nearly 28°. But even that's quite exceptional; most of the time Mercury stays closer to the Sun, closer than half of that angle (less than one fist).

So I should make it very clear: it's very rare that you can actually see Mercury with your naked eyes. It pretty much has to be right at maximum elongation. I've observed its phases (just like the Moon, or Venus) in a small dobsonian telescope a couple times; at max elongation it's close to the first quarter phase, or last quarter, and it looks like a very tiny replica of the Moon seen at that same phase. Kind of like this (except in my small telescope no craters or mountains were visible, because the image was just too tiny):

https://en.wikipedia.org/wiki/Planetary_phase

Venus' orbit is quite a bit wider, about 72% the size of Earth's own orbit. So it can swing ahead or behind the Sun quite widely. But that's all it can do - swing ahead or behind. It can never go "all the way around" the Earth. You can never see it at midnight, when the Sun is on the other side of Earth.

Venus' maximum elongation is 47°. Stretch your arm out and make a fist; now stick your thumb and your pinky sideways as far as they go (like the "hang loose" gesture, but stick them out all the way). The span from tip to tip is about 25°; twice that span is the maximum elongation of Venus.

Now imagine Venus at maximum elongation - two "hang loose" hand spans. The Sun has already set in the West about 1 hour and 40 minutes ago, so this is during the night, it's dark already, and now the Sun is one "hang loose" hand span below horizon. But Venus is at maximum elongation, which means it's one handspan above horizon. That's quite hight in the sky actually, you can see it pretty well.

So, there you have it. You can actually see Venus during the night. Not for long, because it still can't wander off too far from the Sun. It must remain within two "hang loose" hand spans from the Sun, or less than that distance, at all times. So Venus itself will set in the West soon after you see it. It cannot hang around until midnight.

Venus also shows phases, just like the Moon or Mercury. I've watched those too in my telescopes.

You can get more precise angle measurements if you can use an old-school marine sextant, like captains on ships back in the day. The fist and "hang loose" gesture techniques are just approximations. Or just get a protractor and tape two straws to it at the desired angle, and sight along the straws to estimate the angle in the sky.

The superior planets (Mars, Jupiter, etc) are not like that. Their orbits are larger than Earth's own orbit. So they can actually be seen, once in a while, high in the sky at midnight, when the Sun is on the opposite side of Earth.

Please refer to the diagram at the top of this reply, it's quite self-explanatory.

So, as you can see, the standard solar system model in astronomy is in excellent agreement with what you can see with your own eyes. There is no disagreement between observation and the model - if there was, the model would be fixed. That's how science works.


Are there ways other than the collapse of a star which have been observed to form black holes? Every time I hear about a black hole, it's always in conjunction with the collapse of a star. Have any other processes been observed to create a black hole?

There are currently three classifications of black hole and these are stellar, super massive and miniature.

Stellar

Stellar black holes are formed by a massive star collapsing and its mass being great enough to form a black hole of moderate size

Super-massive

These black holes can have the mass equivalent of a billion suns, these most likely exist at the centre of galaxies and are probably as old as time itself.

Miniature

Miniature black holes are theoretical and are thought to have been created at the early stages of the universe. these require masses less than that of our sun and are thus very unlikely.

Other

"But Rhys" I hear you cry, "these are all from stars you fool!"

And yes, yes they are, this is because black holes are formed from mass, and lots of mass (except miniatures), coincidentally large amounts of mass tends to gravitate towards each other and form stars long before it forms black holes.

Therefore theoretically, a black hole does not even need stars, it just needs lots and lots of mass, but having that much mass without forming a star is no easy task.

Short answer, you can make a black hole from anything as long as it has enough mass, but good luck stopping this from forming a star beforehand!


What would be the apparent magnitude of Betelgeuse if it were as close to Earth as Sirius? How large and bright would Betelgeuse appear if it were as close to Earth as Sirius, before and after it goes supernova?

Given a distance of $d_1 = 640~\mathrm{ly}$ for Betelgeuze and $d_2 = 8.6~\mathrm{ly}$ for Sirius the difference in the apparent magnitude for Betelgeuze at distances $d_1$ and $d_2$ that is given by

$$m_1 - m_2 = 5~\mathrm{mag} \cdot log_{10}(\frac{d_1}{d_2})$$

is $9.4~\mathrm{mag}$. With magnitude $m_1 = 0.45~\mathrm{mag}$ we get $m_2 = -8.9~\mathrm{mag}$ for Betelgeuze at the distance of Sirius.

This is about $860$ times brighter than Sirius ($-1.46~\mathrm{mag}$). It would be visible during daytime


Are there more stars in the universe than grains of sand in the Earth? My ten year old daughter asked me this question at breakfast and I initially said yes, but on second thought I was also puzzled.

A quick google gave me these (approximate) figures:

7.5 x 1018 grains of sand in all the beaches and deserts of the world
7 x 1022 stars in the observable universe

If these are reasonable estimates, then there are approximately nine thousand stars in the observable universe for each grain of sand on Earth. (By observable universe, I mean in all the galaxies that we can detect with our most powerful telescopes.)

Robert Krulwich: Which Is Greater, The Number Of Sand Grains On Earth Or Stars In The Sky?


how to plot sky map like www.nakedeyeplanet.com with dates tags with planetary motion path I am curious how to plot planetary motion star chart with date tags like on www.nakedeyeplanets.com by Martin J. Powell.

Is there any way to generate images similar like that. What software should be used MATLAB or what.

Also should have following features:
 - constellation boundaries
 - non-overlapping naming of star,constellation
 - different color for planet path for each year

Example, click for full size:

The Python program Skyfield can generate the data for most of this. It can for example generate curves for planet trajectories, provide star positions and magnitudes from a catalog, and even generate boundaries for the constellations.

https://rhodesmill.org/skyfield/api.html

Then Matplotlib or Plotly can be used in the same python script to make plots as nice-looking as you like.

There may be other canned programs that can do this as well, just for example Stellarium and Celestia.

For further resources, see answers to Where can I find/visualize planets/stars/moons/etc positions?

and the animation in this answer


Is there a pattern between the mass of a body and the mass of orbiting objects around it? I was looking at Wikipedia's Solar system page, and it says that Sun represents 99.86% of the whole solar system mass. I found that pretty huge.

So i calculated ratio of masses : Earth / (Earth + Moon) and it's about 98.78%.

I did the same with Jupiter : Jupiter / (Jupiter + Io + Europa + Ganymede + Callisto) and it's about 99.97% (I ignored small satellites).

Why is it such a high ratio in this 3 examples?
Is 1% a regular number for satellites' mass?
It is simple gravitation maths ?

I understand that one body has to be much more massive in order to be a planet-satellite system, otherwise it's a double planet system. But I would have thought 90% would be enough.

No, there is no pattern on 1%. There are cases of double stars and (allegedly) of double planets.

In our Solar System, Charon has a mass of 11.6% of that of Pluto.


Data for red-shifting I am looking for basic data regarding red-shifting that  comes with reliable measure of distance of the emitting star.

Red shift is usually measured for galaxies rather than individual stars. Unless a star has just gone supernova, it's usually not bright enough to be seen even w the world's most powerful telescopes at the distances where cosmological redshift comes into play. Hubble's law operates over large distances; the expansion constant being 67.8 km/sec per megaparsec (3.3 million light years)
Andomeda galaxy (M31) is 2.54 million light years away, and although some individual stars are visible at that distance, the galaxy has a net blueshift of 0.001001 (303 km/sec) due to its peculiar (i.e. non-Hubble) velocity.
Further out is where you start to see redshifts in excess of peculiar motion; and there you need to start using standard candles of some sort.
Cepheid variables when visible, Supernova, globular cluster brightness functions, are all used, as well as some spectral methods: Cosmic distance ladder

Supernova measurements go the furthest out, highest redshift, but are still a bit of a can of worms as far as interpretation goes.


Breakthrough Starshot; could we detect similar probes flying by Earth? If extraterrestrial civilizations were sending probes by Earth similar to our plans for Breakthrough Starshot (where the probes are only a few meters in diameter and will be moving at up to 0.2c when they fly by the planet), would we be able to detect them, technologically speaking? And if so, what are the odds we would catch a glimpse of one?

EDIT/ADDITION: An additional question I thought of is whether astronauts would potentially be able to perceive such objects with the naked eye, or would the probes be moving far too fast to register?

I'm not in a position to toss real numbers around at the moment, but here's a thought experiment:

Iridium satellites reflect sunlight toward Earth's surface, producing bright flares, up to magnitude -8. That's about 1 million times brighter than the faintest light source a fully dark-adapted eye can perceive -- a five-magnitude difference in brightness is a factor of 100, and under really dark skies humans with excellent vision can see down to magnitude 7.

I think these reflections are coming off a panel of about 3 x 1.5 meters, so it's comparable in size to your probe.

These satellites orbit about 800km above Earth's surface. How far away could one be, and still produce a visible reflection? Brightness falls off as the inverse square of distance, so the greatest distance at which you could barely see a reflection would be 1,000 times greater, or 800,000km.

Light travels at 300,000km/s. A probe traveling at 0.2c would cover 60,000km/s. Traveling 800,000km would take about 13 seconds.

That means that something the size of an Iridium satellite, reflecting sunlight directly toward an observer during its entire passage, could be visible for up to almost half a minute -- blue-shifted for the first part of that, and red-shifted for the last part. (Unless it's quite far away at closest approach, which would make it dimmer, the transition from "basically heading straight for you" to "basically heading straight away from you" would take place in the literal blink of an eye, a matter of milliseconds.)

Now, this is probably an upper bound -- probes won't have the same shape and reflectivity as an Iridium satellite, and won't be oriented to reflect directly toward an observer during their entire transit. But, somewhat to my surprise, it looks like you might be able to see one approaching and receding.

On the other hand, it's also possible that an object making 0.2c might have a visible interaction with the solar wind. (That requires more numbers than I'm able to run at the moment, but having thought on it a little bit, it seems unlikely.)


What are these objects on the night sky generally between 8:00 and 11:00 PM? Many time I saw a moving star in the night generally been 08:00 to 11:00 PM. It is not any satellite, because it move in all directions, although it moves in a straight line. Sometimes from north to south, and sometimes from east to west and vice versa.

It is also not a comet, because its brightness is continuous, and it moves continuously in a direction. It is also not a plane ✈, because in the night a plane is travelling with three-four coloured light. And they generally do not move on the way of planes.

What are those moving stars?

Your objects sound like satellites to me. Why do you think they are not satellites?

Where I live at 40 degrees north latitude, these are the paths that we see the satellites moving (just a rough estimate: not a scientific count)
• 90% are in a polar orbit, so they generally move south to north or north to south (different parts of the same orbit). These generally follow lines of right ascension.
• 9% are inclined 50 to 70 degrees to the equator, so they cross the sky from NW to SE or SW to NE (different parts of the same orbit)
• 1% go across the southern half of the sky from west to east .
•0% go from east to west because there are very few satellites in such orbit.

As Jan mentioned, what is your latitude? That can affect which objects you see and where you see them.

Also, how long are they visible? They ones you see with your unaided eyes will cross the sky in 4 to 7 minutes although the time they are visible can be less. You will see more in the eastern half of the sky after sunset than in the western half because of the phase angle.


What does the Sun's orbit within the Solar System look like? Everything in the Solar System revolves around the "barycenter": the overall center of mass. This barycenter is not in the center of the Sun. Some articles and essays I've read go so far as to suggest that the position of the barycenter does not have a set of fixed coordinates within the System: it fluctuates.

Well. Since everything, including the Sun, revolves around this barycenter, the Sun must have its own orbit around it. What does it look like? How large is it? How elliptic?

(In my research, I have tried and failed to establish whether the barycenter is within the Sun or outside it. Either way, an orbit is an orbit).

In case anyone can't follow the links in my comment (above), here are the two pictures I mentioned. From: here and here.

The first claims to show the track of the solar system barycentre in the heliocentric reference frame. The outer yellow circle marks the photosphere of the Sun. The second plot claims to show the track of the centre of the Sun in the barycentric reference frame. The yellow circle shows the photosphere of the Sun to scale. As you can see, the plots are actually (almost) the same! Given that to go from one frame to the other is just a translation, I suppose they can both be right providing the x and y axes are defined appropriately.

To answer the questions posed: "What does it look like" - it looks like these two pictures. "How large is it?" As you can see, the maximum separation between the barycentre and the solar centre appears to be about 2 solar radii over the timescale covered by these plots, but is as small as a tenth of a solar radius (e.g. in 1950). "How elliptic?" Not at all really, it is a complicated superposition caused by the orbits mainly of Jupiter and Saturn, but all the planets contribute to a greater or lesser extent.

The barycenter is calculated from the instantaneous positions of all the discrete masses in the solar system. I do not know for sure, but I assume that it includes all of the planets, and that everything else is negligible at the scale of the thickness of the line.


Where would an object fly to if it were to leave the gravitational field of Earth? And would it stop somewhere? What would happen to it? My 3,5 year old son is asking :-) He can take answers for a 15 yo I think. Many thanks for your help!

The gravitational field of the Earth has not a border, what you can leave and there is weightlessness. Instead, going more and more far away from the Earth, so will its gravity weaker and weaker.

At some point, the gravity of other bodies (Moon, Sun, other planets) will be stronger than the Earths.

What will happen to a body which leaves the space where not the gravity of the Earth is the strongest, depends on the gravitational force of the other bodies of the solar system. These are also moving (mostly, orbiting the Sun and the moons are orbiting their planets), thus there is no general answer to your question.

For example, in the case of the Starman car (Elon Musk's Tesla shot into the space) will have this orbit in some centuries. Where will it be in millions of years, there are only probabilistic calculations. If the object did not left the Earth too quickly (it went only a little faster than the escape speed), then it it will likely start a roughly "random walk" between the orbits of the Venus and the Mars. This is what also most known asteroids are doing.


What apparent sizes of the sun an Earth like planet can have? What is the smallest and largest sun an Earth like planet could have in its sky? What kinds of stars are those?

I saw this video and thought, yeah, some of those stars are really big in absolute terms, but Earth would have to be orbiting them proportionally further to keep the climate, making them smaller in the sky, so the video is not accurate.

I'll make a very simple dimensional calculation here.

Total power from of a star with radius $R$ and temperature $T$:

$$ P_{tot} \ = \ R^2 \ T^4$$

Flux $I$ (power per unit area) at an orbital distance $r$:

$$I \ = \ \frac{R^2 \ T^4}{r^2} \ = \ T^4 \ \left( \frac{R}{r} \right)^2$$

Apparent size (radians, small angle approx):

$$\theta \ \approx \frac{R}{r}$$

So:

$$\theta \ \approx \ \frac{\sqrt{I}}{T^2}$$

Surprisingly, for a fixed incident flux of light (ignoring different planetary warming effects of different wavelengths) the apparent size of a sun viewed from a star varies roughly as the inverse square of its temperature, and does not depend on the actual size of the star.

Your milage may vary.

Using Wein's dispacement law which relates the wavelength of maximum spectral radiance of a blackbody $\lambda_{max}$ with its temperature $T$ using Wien's displacement constant $b$:

$$\lambda_{max} \ = \ \frac{b}{T}$$

The apparent size scales with the square of the peak wavelength of a star's blackbody spectrum.

$$\theta \ \approx \ \lambda_{max}^2 \ \sqrt{I}$$

Yay! Reddish stars will appear much larger than bluish stars - the result we were all expecting.

And if you'd like do some landscaping, the 2008 Scientific American article The Color of Plants on Other Worlds is a good read in stellar spectra, atmospheres, and photosynthesis.


What is causing this strange shimmering pattern of light during a total solar eclipse? Below this question was left a comment linking to the Smarter Every Day video Space Station Transiting 2017 ECLIPSE, My Brain Stopped Working - Smarter Every Day 175.

At about 04:30 they show a short clip of a video recording of what looks like a white bed sheet hanging vertically and a shimmering pattern of light on it, I assume related to the total solar eclipse.

What's going on? What causes this? What's it called?

These are called shadow bands.

As the sunlight is reduced to a very narrow strip, in the last few moments before totality, turbulence and refraction in the atmosphere will cause shimmering bands of light and shadow.

What you are seeing here is essentially the same as the twinkle that you see from stars.


Getting Solar Noon from Sunset and Sunrise I've seen the solar noon formula described as snoon = 720 – 4*longitude – eqtime. I'm curious if it is equally reliable within a relatively small margin of error (+-60 seconds) to simply get the midpoint between sunrise and sunset, given that data is already calculated and available.

Is there any occasion when solar noon wouldn't be equal time from sunrise and sunset, give or take the aforementioned margin of error?

The approach of using the midpoint of the sunrise and sunset times is reasonably accurate. I do not know what the accuracy is, but probably within 1 or 2 minutes. It would not be as accurate in polar regions where the sun is close to being visible all day.

The main reason it is not precise is that the sun is moving north (December through June) or south. This changes the time of sunset with respect to sunrise if the Sun were not moving north or south. Solar noon would be earlier or later, respectively, from the midpoint.


How to calculate a planet's current position within the solar system I'm making an application that needs to calculate a planet's current position within the solar system - I don't want to calculate its position in the sky.

For example, what I want to do is exactly what this website does:

http://www.theplanetstoday.com/

I want to create an overhead view of all the planet's positions in the solar system.

Where do I start? Is there a latitude/longitude coordinate system for the solar system? Everything I've found seems to calculate a planet's position in the sky from the observer's point on earth.

Buy Astronomical Algorithms 2nd ed. by Jean Meeus. It is published by Willmann-Bell Inc. If you buy it direct from the publisher, you will get a corrected printing with all errata up to August 10, 2009. There are a number of systems for describing the location of a planet. This book provides a method to find the heliocentric longitude, latitude, and distance from the Sun. It also explains the different coordinate systems and how to transform from one to another.

The method you use depends on how accurate you need to be. If you state your needed accuracy some of us might be able to suggest other alternatives.


Is there such a thing as a black dwarf Does such a heavenly body exist called a black dwarf? It would be a Brown dwarf that has cooled to the point it no longer gives off heat or light.

A black dwarf is a white dwarf that has cooled to the point that it is no longer emitting light. However it takes a long time for a white dwarf to cool down. The exact timescale is rather uncertain, but at least hundreds of billions of years, perhaps hundreds of trillions of years.

The upshot is that that no black dwarfs exist in the universe today. The Universe is only about 14 billion years old, so there hasn't been time for any black dwarfs to form.

The paper A Dying Universe: The Long Term Fate and Evolution of Astrophysical Objects discusses brown dwarfs, black dwarfs, and other objects that may exist in an aging universe.


TIME TRAVEL-Can it be really done? Is it posssible for us to time travel?
Is light essential factor or the dark matter and dark energy?

In theory it is possible to do time travel. There are several possibilities from theoretical point of view. Two aspects are there, travelling to the future and travel back to past. Although many scientists including Stephen Hawking said that travelling to the past is not possible (as there are several paradoxes including "the grand father paradox") that results in some weird phenomena (like in grandfather paradox, if you are allowed to travel back in time and you went to the time of your grandfather's youth and kill him then you are not going to born in the first place in future, thus a contradiction). But travelling to the future is seems possible. There are several ways to do that, thanks to The Theory of Relativity. As Einstein said, time flows at different speeds in different part of the universe, faster in Earth as compared to the near vicinity of a black hole, due to the stronger gravitational pull of the black hole then the Earth. So if you travel near to a black hole and keeping a safe distance and starts revolving around it for few hours and return back to Earth, then you will find the many decades has been passed here, due to the different speed rate of time in the two places, so essentially you are in future (but alas you cannot go back!). Another way is to travel at a very high speed (almost near the speed of light) in a spaceship for few years also take you to the future back on Earth, this is because time slows down in case you speed up. And many more concepts of time travel exists involving Wormhole and others.

Grandfather paradox
Time travel
See this


Correct relation between metallicity (z) and iron content ([Fe/H]) The Wikipedia entry on Metallicity states that:

$\log_{10}\left(\frac{Z/X}{Z_\mathrm{sun}/X_\mathrm{sun}}\right) = [\mathrm{M}/\mathrm{H}]$
where $[M/H]$ is the star's total metal abundance (i.e. all elements heavier than helium) defined as a more general expression than the one for [Fe/H]:
$[\mathrm{M}/\mathrm{H}] = \log_{10}{\left(\frac{N_{\mathrm{M}}}{N_{\mathrm{H}}}\right)_\mathrm{star}} - \log_{10}{\left(\frac{N_{\mathrm{M}}}{N_{\mathrm{H}}}\right)_\mathrm{sun}}$
The iron abundance and the total metal abundance are often assumed to be related through a constant A as [citation needed]:
$[\mathrm{M}/\mathrm{H}] = A\times[\mathrm{Fe}/\mathrm{H}]$
where $A$ assumes values between 0.9 and 1. Using the formulas presented above, the relation between $Z$ and [Fe/H] can finally be written as:
$\log_{10}\left(\frac{Z/X}{Z_\mathrm{sun}/X_\mathrm{sun}}\right) = A\times[\mathrm{Fe}/\mathrm{H}]$

This all sounds reasonable, but I have two questions:

Where does the $[\mathrm{M}/\mathrm{H}] = A\times[\mathrm{Fe}/\mathrm{H}]$ relation come from? I could not find a proper source.
This article actually says (see Eq 9) that the general relation is $\log_{10}\left(\frac{Z/X}{Z_\mathrm{sun}/X_\mathrm{sun}}\right) = [\mathrm{Fe}/\mathrm{H}]$ i.e., they seem to equate [M/H] with [Fe/H]. I assume that it is because the aforementioned A parameter is between 0.9 and 1, but again that leaves me with the need for a proper source to state that.

$$[\mathrm{Fe}/\mathrm{H}] = \log_{10} \left[\frac{f_\mathrm{Fe}(\alpha)}{f_\mathrm{Fe}(0)} \frac{m_z(0)}{m_z(\alpha)} \left(\frac{Z}{X}\right)_\ast \left(\frac{X}{Z}\right)_\odot \right]$$

You can use this relation to estimate [Fe/H] from [M/H] of a star. From [M/H] you can get the (Z/X) ratio.


Rendering asteroid orientation as seen from Earth I am writing a program to render an asteroid as seen from various parts of the solar system.  My goal is to roughly reproduce the image of Cacus according to information on its DAMIT detail page.

Here is my approach:

Create a solar system at JD0=2443568.0.  The axes of the visualization are arranged according to the ecliptic with positive X axis in the direction of the vernal equinox and other axes according to the right-hand rule.  Earth is positioned at t=JD0.  (image)
Place the asteroid at (0, 0, 0) and begin to rotate the object according to the matrix multiplication described by the Description page and Kaasalainen & Torppa 2001:

φ0=0 and t=JD0, so there is no initial Z axis rotation
Latitude β is -63, so rotate 90 - (-63) = 153 degrees on the Y axis (image)
Longitude λ is 251, so rotate the object 251 degrees anticlockwise around the Z axis of the ecliptic (image)

Using ephemeris from the NASA/JPL Small Body Database, move Cacus to its position in orbit at JD0. (image)
Move the camera to Earth's position at JD0 and point it at the asteroid. (image)

My final rendering does not match the projection from DAMIT.  I understand that there are factors like light-time correction that I'm not including, but it looks completely different.

What is wrong with my approach to rotating the asteroid on its spin axis?

(The work-in-progress visualization is online here)

From the paper by M. Kaasalainen, J. Torppa, and K. Muinonen, equation (1) is:
$$ \vec r_{ast} = R_z(\omega(t-t_0)+\phi_0) R_y(\hat\beta) R_z(\lambda) \vec r_{ecl} $$
To reverse this equation you must reverse the order of the rotations and the rotation angles:
$$ \vec r_{ecl} = R_z(-\lambda)R_y(-\hat\beta)R_z(-\omega(t-t_0)-\phi_0) \vec r_{ast} $$
From your description, you reversed the order of the rotations but not the rotation angles.


Rate of star collpse It is believed that our sun will become a red giant with a diameter of about 1 AU. When the fusion slows down, gravity will collapse the sun. Since the energy release from the fusion diminishes slowly (maybe hundreds of years for a significant change of 1%).

Let's assume that only a small fraction of the sun (10%) is needed to collapse (to 1/1000 the volume) to produce a nova. Would it still be possible for a star to collapse in a few hour (instead of decades)?

Well, I'm not sure exactly what you mean by "collapse" (if it takes $10^9$ years to die, the last few hours satisfy what your asking about). However, maybe some explanation would help:

You're asking about the interaction between two timescales - the Dynamical Timescale, which is the time it takes for a star to free fall into a nova, and the Nuclear time scale (I think), which is the rate at which the sun exhausts it's fuel. The dynamical timescale for the Sun is like 1000 s, and the nuclear timescale is like a Gyr (usually what we think of when we think "lifetime of the Sun").

Any nuclear effects would totally dominate over dynamical effects (since they take much, much longer), and without some specific collusion of behavior (angular momentum, metalicity, what have you), collapse would occur with the burning of fuel and not with the dynamical timescale until the Sun exhausted all it's fuel, and would collapse in a matter of minutes.


So, what exactly is an 'ultra-cool' dwarf star? The TRAPPIST-1 system is around an ultra-cool dwarf star. I went looking for more information on that kind of star, and found very little. The Wikipedia article on it lengthened from a minimal stub to one paragraph in the 15 minutes from when I began the search, after the press conference on the discovery of 7 Earth-sized planets in orbit around TRAPPIST-1.

What kind of star is this?

Breaking the phrase down:

Dwarf star - a term I will never understand - is used to describe relatively small, dim stars. Unfortunately, this encompasses most main-sequence stars, which are indeed dwarfs compared to large giants and supergiants.
Ultra-cool, as called2voyage already discussed, means that the star has an

  effective temperatures of less than 2,700 kelvin.

This is about half of the Sun's surface temperature. Spectroscopically, it means that these stars are of class M7-L8, i.e. really, really cool stars that just barely reach the threshold for nuclear fusion. Actually, some L dwarfs will never fuse hydrogen as normal stars do, and will become brown dwarfs. Therefore, "ultra-cool dwarf" doesn't necessarily refer to just main sequence stars.

Other characteristics in some cases (from Cruz et al.):

Lithium in some L-type objects, indicative of low temperatures, low masses, and young ages in those particular objects.
Very low surface gravities in late M-type dwarfs (though likely not as much in an M7 dwarf like TRAPPIST-1).
Very low metallicity in certain other L-type objects.

TRAPPIST-1 actually appears to be a rather massive ultra-cool dwarf, then, in comparison to some of these other objects.

Here's an annotated Hertzsprung-Russell diagram, with a box for ultra-cool dwarfs and a circle for where TRAPPIST-1 is, approximately:

Image from Wikipedia courtesy of user Saibo, under the Creative Commons Attribution-Share Alike 2.5 Generic license.


Dating very old objects/events After reading this space timeline, I have wondered how they managed to make these assumptions about various events like when the first black holes formed. This is obviously far beyond the good ol' carbon-14 dating procedure.

One property of distant objects, that follows from the limited speed of light, is that we are observing them in the past; the light that they emit takes time to reach us. For very distant objects, we are effectively observing the universe at a very early time.

Objects that are very far away, such as distant galaxies, are receding away from us, which causes the light that they emit to be doppler shifted. The recession velocity is proportional to the distance away, and in turn so is the magnitude of the redshift. This redshift allows us to put a number to the distance away; combining this with the speed of light limit allows us to determine how old the light we are observing from them is.

These two effects, the limited speed of light and redshift, effectively allow us to determine an evolutionary timeline for the universe, and place constraints on when certain objects appear.


Why could a gas planet not get much larger than Jupiter? What is the maximum size for a gas giant?

Usually they can get is Almost twice.  Here's the evidence:
 It is difficult for them to become even bigger than that as once they reach 15 Jupiter masses or so there is enough pressure in the core to ignite deuterium fusion, so those are considered “brown dwarfs” rather than planets.
An example is TrES-4, which is located 1,400 light years away in the constellation Hercules. The planet has been measured to be 1.4 times the size of Jupiter, but it only has 0.84 times Jupiter’s mass. With such a low density, it is dubbed the puffy planet.
It was discovered in 2006 orbiting a brightish star 1500 light years from Earth. Its diameter of about 1.8 times Jupiter’s makes it the largest accurately measured planet. Strangely, though, TrES-4 is very light for its size. It has only 88 per cent of Jupiter’s mass, giving it a density of roughly 0.2 grams per cubic centimetre, less than that of cork. Just how a planet can be as fluffy as TrES-4 remains a mystery.


Why does the convective core in an intermediate to high mass star shrink? The image below shows the evolution of the hydrogen mass fraction profile for a 5 solar mass star in the main sequence. I would expect that the size of the convective core stays roughly constant as the hydrogen is been fused, leading to a final step-like profile when the core hydrogen is exhausted. However, all the literature and simulations show a clear slope in the final step-like profile which results from the shrinking of the convective core.

Does anyone have any insight about the reason behind the shrinking?

(source of the image: http://astro.if.ufrgs.br/evol/evolve/hansen/StellarEvolnDemo/m5z02evoln.html)

Edit 1:

Below is the evolution of the temperature profile for the same star. Feel free to comment on it.

Personally, I am surprised how little the temperatures change. Since the main nuclear reaction is the CNO cycle which scale as $~T^{16}$, I was expecting a much more violent change. However, the central temperature only increases in about 30% during the whole main sequence. Interesting.

(Source: http://astro.if.ufrgs.br/evol/evolve/hansen/StellarEvolnDemo/m5z02evoln.html)

Edit 2:

I thought a nice explanation could be in the Ledoux criterion for convection. This criterion states that chemical gradients have a stabilizing effect against convection (i.e., it impedes convection) which would lead to the conclusion that, at the interface between the radiative and convective zone, the radiative transport would take over. However, I simulate the evolution with and without the Ledoux criterion and in both cases the convective core shrinks.

Whether convection exists depends on whether the interior radiative temperature gradient reaches the adiabatic temperature gradient.

The interior radiative temperature gradient is proportional to the opacity and the outward energy flux, and inversely proportional to $T^4$. As the star evolves on the main sequence, the central temperature goes up and the opacity (e.g. Kramer's opacity goes as $T^{-7/2}$ goes down. You are also removing free electrons (combining them with protons to form He) which reduces the Thomson/Compton scattering opacity. This means the radiative temperature gradient goes down and can fall below the adiabatic gradient, meaning that energy transport reverts to being radiative.

It stays convective right at the centre for longest because that is where the radiative temperature gradient is still largest (driven by the extreme temperature dependence and high outward energy flux of the CNO cycle nuclear reactions).


Get date and time by position of the sun and the observer position I am working on an astrology project (don't panic, my question is about astronomy or math), where I need to figure out some values by the longnitude of the planets.

I have a program for this, the Swiss ephemeris.

With this, I can get the position of the planets by the time and the geolocation of the observer.

What I need to do is to do it backward, so I need to get the exact date and time by the given position of the sun and the position of the observer.

For example, today (2015.12.16 00:00:00 UT) the longitude of the sun is 263.6893755 from postion (lng,lat,alt) 19.2,47.29,0 on Earth.

I need a formula, what tells me, what was the UT time from this longitude of sun -88 degree.

Can anybody give me a clue, how can I calculate this?

I made some research of course on the net, and I found the related wiki article, but if I am honest, there are too many unknown terms and phrase here, what I really do not understand, and I do not want to dig in too deep.

I think you can very easily use the transformations available in the wiki link to obtain the Julian day, then exploit the position of observer to get local sidereal time.
Lets start by computing the Julian day
you have a given longitude $L$ which gives you $n$ number of days since Greenwich noon as
$$n = \dfrac{L-280.4860}{0.9856474}$$
giving the corresponding Julian day as
$$JD = n+2451545$$
once you have Julian day you can use the following link to obtain UT
http://ssd.jpl.nasa.gov/tc.cgi


Relatable comparison of VY Canis Majoris to the Sun? I was trying to describe how vast the largest known star is to someone and felt I wasn't quite able to relate the scale difference. I know it's roughly 1500 times larger than the sun. Anyone know of a good analogy?

After some playing around with wolfram alpha and google my best comparison has been

The sun compared to VY Canis Majoris is like a donut compared to the London Eye.

The London Eye is about 120m in diameter, this divided by 1500 is about 8cm which is roughly the diameter of a ring donut.