Do stars start with rocky core? Is it true that stars start with a rocky core, then collect hydrogen by the gravity of that rocky core, just like planets collecting materials?

No, stars do not have a rocky core. They start as clouds of gas, become gravitationally unstable and collapse.

Some planets are thought to start this way too - though probably not the ones in our solar system, because the outer three three giant planets have solid cores and Jupiter is likely to also. The two modes of giant planet formation are reviewed in detail by D'Angelo et al. (2010).


Assumption of our universe being the surface of a 4-D sphere in order to describe relative uniformity in the cosmic microwave background? It seems as if this is necessary to assume in order for the CMB to make an sense. Otherwise, an instantaneous infinite Euclidean space model would work. The issue is, with any visualization of the Big Bang, let's take a Big Bang visualization in 3 - dimensions, the radiation emitted from the creation of hydrogen atoms would not necessarily be all around us, as it is. Our position in space would dictate what we would observe in terms of background radiation.

To a high degree of accuracy, the cosmic background radiation we observe is homogeneous and isotropic... except for the CMBR dipole anisotropy, which is there because we are moving at $369\pm 0.9\;\mathrm{km/s}$ relative to the CMB rest frame, in which this anisotropy would vanish.

Relevantly, the Friedmann-Robertson-Walker family of solutions of general relativity used to model the Big Bang cosmology assume that the universe is, on the large scale, homogeneous and isotropic.

  It seems as if this is necessary to assume in order for the CMB to make an sense. ... Our position in space would dictate what we would observe in terms of background radiation.

That's not the case. There are four distinct homogeneous and isotropic spatial geometries: the Euclidean space $\mathbf{E}^3$, the hyperbolic space $\mathbf{H}^3$, the real projective space $\mathbf{RP}^3$, and the sphere $S^3$. In any of those cases, the universe would look the same all around us and be independent of our position. Thus, the sphere is possible but not necessary. The first two of those cases correspond to the "flat" and "open" Big Bang cosmologies, respectively, and the last two are variations of the "closed" cosmology.

And if we don't require the universe to be completely homogeneous and isotropic, but merely look like it in the part of it we observe, then much more exotic geometries are also possible. With cosmological inflation, virtually any large-scale geometry can be "blown up" to look flat, homogenous, and isotropic within our observable portion--even though it might not be s beyond the our cosmological horizon.

  Otherwise, an instantaneous infinite Euclidean space model would work.

It does work, in the sense of being consistent with observations. In fact, in the standard ΛCDM cosmological model, space is infinite and Euclidean.


If you lived on the far side of the Moon, how could you infer the existence of Earth? Suppose that you deposit an astronomer, armed with our current knowledge of orbital mechanics, on a dome on the far side of the Moon, so that the Earth is perpetually hidden from them.

(And, of course, assume that this person has no specific knowledge about the system they're in beyond what they can glean from observations. If you will, imagine that they learned all our modern orbital mechanics and related physics in alpha centauri, and then got teleported to our Moon.)

Now, it is reasonable to expect that this person should be able to deduce from observations of the sky that the body they are on is one half of a binary system, and they should be able to measure the orbital characteristics (semi-major axis, ellipticity, inclination) as well as the  position of the barycentre (much closer to the other body, corresponding to a much more massive partner). What observations are needed to deduce this? What level of observational accuracy is needed for those observations, and to what historical epoch does it correspond to? (I.e. would Tycho Brahe's kit have been sufficient? Would Galileo's? Would the ancient Greeks'? Or would this require a late-19th-century (or even later) observatory?)

(As pointed out in MartinV's answer, our astronomer might find it hard to distinguish between situations with an orbiting pair vs one single huge body. Thus, if convenient, you can assume that, via short ~100km forays from the dome, our astronomer is able to measure the lunar radius by measuring solar inclinations at different points with known distances between them, à la Erathostenes.)

A body tide seismometer on the lunar far side would pick up both the solar tide, and the 20 inch body distortion produced by the earth.
While "tidally locked", the moon is not in a perfectly circular orbit, and also wobbles a bit; libration. Your seismometer should pick up both effects.

Watching the parallax of mars cycle every 28 days, as suggested in comments above, might be a simpler way to go.


Is there a way to estimate the age of M dwarf stars? Wolf 1061 (only 13.8 ly away) was recently found to have three rocky planets one of which is in the habitable zone. It was stated on the english wikipedia site that it has a very stable light curve and was not found to have "any significant activity such as sunspots or flares."
Would this imply that this star is very old? The expected lifetime of this M dwarf with a mass estimated to be 0.25 solar mass could potentially reach a trillion years.

Really difficult.

Very young ($<50$ Myr) M dwarfs may still exhibit lithium in their photospheres. Any older and it will have been depleted.

For older dwarfs you then move on to looking at rotation and activity. Both decline with age, but in an M dwarf as cool as this, the decay timescale is many Gyr, so it is only weakly constraining and not very well calibrated for such low mass stars. A very slow rotation rate or very low level of magnetic activity might indicate it was older than 5 Gyr.

The lack of flares and the lack of light curve modulation by starspots may indicate that it has spun down, is magnetically inactive and is therefore older than 5 Gyr ( I cannot immediately locate the source of this Wikipedia claim). The ratio of X-ray to bolometric luminosity would be more definitive and could probably be derived from the ROSAT all-sky survey (or a constraining upper limit would be found).

A useful paper is by Stelzer et al. (2013). http://arxiv.org/abs/1302.1061
She finds various activity diagnostics for nearby stars, including this one. Its projected rotation velocity is 1.5 km/s, which suggests a rotation period of $\sim 10$ days for a star of this size. This could not be less constraining on the age! The X-ray to bolometric flux ratio is $10^{-4.45}$ which suggests it is neither very young or very old - say between 1 and 5 Gyr for a star of this mass.

Another possibility is to look at the Galactic space motion and position. Older stars tend to have higher velocities, especially perpendicular to the Galactic plane, and also tend to be found further from the Galactic plane. The velocities for this object do not look large, suggesting it is not very old, ie younger than perhaps 10 Gyr and likely a young disk object with age of $<5$ Gyr.

You can also look at the metallicity. This is difficult to measure in M-dwarfs. Maldonado et al. (2015) give [Fe/H]$=-0.05 \pm 0.03$. This is very close to the solar value and does not constrain the age beyond suggesting the star is younger than 10Gyr.

Beyond this, it is basically impossible. Even if you had a very precise distance and could place the star on a HR diagram, low mass M dwarfs evolve so little in luminosity and temperature over many Gyr that this offers no constraint.

So my conclusion - the age is between 1 and 5 Gyr, with the probability peaking somewhere in the middle.


Redshift for gravitational waves? According to general relativity, would gravitational waves experience the same sort of redshift that electromagnetic waves experience due to the expansion of the universe?

And are there astrophysical processes that would be expected to have a characteristic 'frequency signature' analogous to the lyman or balmer series of hydrogen which would enable a measurement gravitational wave redshift.

I'm thinking of processes like baryon accoustic oscillations, certain types of supernovae, neutron star mergers, etc.

Yes, gravitational waves can be redshifted. The answer of this question explains why: https://physics.stackexchange.com/questions/137292/can-gravitational-waves-be-red-shifted
The answer to the question above basicly states that doppler effect applies to any wave that propagates at light speed.

Actualy measuring the redshift is much more difficult, because neither redshift nor rest-frame mass can be extracted from the signal. Only the combination of the two can be obtained - M(1+z). This is called the redshifted mass. Deriving actual redshift from redshifted mass is quite complicated as this problem is relatively new and no efficient solution exists yet. This article here describes the derivation of the redshift in detail:
http://arxiv.org/pdf/1312.1862v2.pdf
From it you can see that it is indeed possible to derive redshift from redshifted mass, but the method is really imprecise as it has a massive 10 - 20% uncertanity.


How to chose a comfortable 2 inch eyepiece for a Skywatcher Dobsonian 150 mm f/8 reflector? Totally newbie here. I got the Skywatcher Dobsonian 6 inch (150 mm) f/8 Newtonian reflector. I had a brief chance to test it yesterday when the skies here in Melbourne cleared a little bit. I was able to see the Moon that was not that high in the sky.

With the telescope, it came with two 1.25" eyepieces; 10 mm and 25 mm, and a 2" adapter.

I found out the best view with the 25 mm although it wasn't that clear. I still have to check the mirrors alignment and collimate.

However, I found a bit uncomfortable to use it though. It was hard to keep the entire view clear. If I moved slightly I couldn't see the entire view.

After reading some articles, it seem that a 2" eyepiece offer a more comfortable view experience.

I'd like to ask; how can I choose a 2" eyepiece that would be more comfortable for observing the Moon and planets? What should I look for in terms of types or specifications?

Short version:
The reason that 2" eyepieces will offer more comfortable views is because they are lower powers and therefore have larger exit pupils.

Long version:
In a telescope, the objective is the lens or mirror which collects light and focuses it at a distance equal to the focal length, creating an image at that point.
(You can test this out most easily on the moon: leave the dust cap in your focuser and use your finder to point the telescope at the moon. You should be able to see a clear (after focusing!) image of the moon projected on the dust cap)
When you insert an eyepiece, you're using it to inspect the image created by the objective.
This is why eyepieces with smaller numbers actually magnify the image more: with a 5mm eyepiece you're looking at a smaller central portion (imagine 5mm, though that's not correct, as it also depends on the FoV of the eyepiece) of the image which the objective has collected; and with a 30mm eyepiece you're inspecting a larger part of the image.
And this results in the exit pupil - a function of telescope and eyepiece - which is the size of the image you're looking at. It's the beam of light which enters the pupil of your eye. At lower magnifications (higher eyepiece focal lengths) the exit pupil is larger, and therefore easier to view comfortably.
You can test this out: cut a series of small holes (from 0.5mm up to 7mm) in a piece of paper and try looking through them. You'll find that the larger holes are much easier to see things through.
To calculate the exit pupil, divide the focal length of the eyepiece by the F-ratio of your telescope.
(And, related, to calculate the magnification you'll get from an eyepiece, divide the focal length of your telescope (in mm) by the focal length of the eyepiece (in mm))
For example, your telescope is f/8, so a 30mm eyepiece would give an exit pupil of 3.75mm - nice and sufficiently wide for you to be able to find the view and not lose it.
But if you want to look at a planet then you'll want more magnification, and you'll have to use a shorter focal length eyepiece. Take a 5mm for instance. That would give an exit pupil of only 0.625mm.
Another effect of decreasing exit pupil is decreasing image brightness. You'll find that, apart from the brightest objects, you probably won't want to go below a 1mm exit pupil very often.
Here's an image from the Wikipedia article on the subject which nicely illustrates the concepts:

(Image credit: By Evan Mason - Own work, CC BY-SA 3.0)
You'll see another term in the image: eye relief.
Eye relief is the distance from the surface of the top lens of the eyepiece to the point where the image is formed (the point where you position your eye). This distance increases as the diameter of the lens increases.
If you wear glasses while observing then it's important to buy eyepieces with an eye relief of at least 15mm (actual figure varies, maybe ±5mm, depending on the shape of your face and glasses you wear) so that you can see the whole view.
(Also, if you wear glasses while observing, are you sure that you need to? Simple long- or short-sightedness can be corrected by focusing, and mild astigmatism isn't visible at higher powers.)

You ask

how can I choose a 2" eyepiece that would be more comfortable for
observing the Moon and planets? What should I look for in terms of
types or specifications?

The Moon and planets are objects which you'll generally want to use a high magnification to view. High magnifications are served by 1.25" eyepieces (there's no need to make them as large as 2" - they're looking at a smaller part of the objective image - and making them in 2" barrels costs more to make) and give a smaller exit pupil.

In your telescope, a good starter set of eyepieces, in my opinion, would be:

A low power wide field eyepiece, with a focal length of 30-40mm, and a FoV of 68°-82°.
A medium power wide field eyepiece, with a focal length of 10-20mm, and a FoV of 76°-100°
A high power wide field eyepiece, with a focal length of 5-10mm, and a FoV of 76°-110° (This one's a bit more tricky, as the maximum magnification which will be usable will depend on your local conditions - the "seeing". If seeing is generally good you might be able to go higher than 5mm, but if it's bad then you might not be able to go much higher than 8 or 9mm)

(Notable brand names which come to mind are APM, Baader Morpheus, Explore Scientific, Pentax, Tele Vue)
You don't really need more than three good eyepieces.
You could also get away with only one high-power, an 8-24mm zoom, and a Barlow lens. But I think the wide field eyepieces are the better option in your telescope, because with a wider FoV you don't have to nudge the telescope so often to keep the object in view.
(The "good" is quite important: almost all eyepieces look good in the dead centre. Then the view starts to degrade towards the edges. Generally, the more money you pay, the less the view degrades (or starts to degrade further away from the centre). Also, and fortunately for you, the degradation of the view is greater as the F-number gets lower. At f/8, you won't need to spend loads of money to get great views, but you'll have to do your research and find out what other people with the same telescope like to use)

Finally, a couple of tips for comfortable viewing:

Make sure you're sitting down while observing. It makes a huge difference to your ability to keep your head still. If you must stand, then try and have something to hold on to or lean against.
Practice makes perfect: the more you use your telescope and eyepieces, the better you'll get at using them, and you'll find it easier to maintain a steady view.

These related posts Eyepiece diameter for small focal length planetary viewing on a large dobsonian? and What are the advantages and disadvantages of a 2-inch eyepiece versus a 1.25-inch eyepiece? have good answers to similar questions.


What allows WFIRST to have similar resolution to Hubble over a 100x larger solid angle? In the video linked in the Space.com article What Would It Mean for Astronomers If the WFIRST Space Telescope Is Killed?, (available in YouTube as WFIRST: The Best of Both Worlds) after 02:00 NASA astrophysicist Neil Gehrels says of WFIRST:

  It has the same image precision and power as the Hubble Space Telescope, but with one-hundred times the area of sky that is used.

Looking at the crude drawing on this NASA page shows only a Cassegrain-like geometry, but there must be additional goodies to obtain a similar resolution with a 10x wider field. Does it have three, four, or five mirrors like the E-ELT or the LSST (cf. What is a quaternary mirror and why does the E-ELT need one? and answers therein) or corrector lenses? What is the magic sauce that makes this possible, and is there at least a schematic diagram of the optics somewhere?

above: From NASA.

The ultimate resolution of a telescope depends upon its diameter. The larger the telescope aperture, the finer the resolution.
Hubble and WFIRST are both the same diameter 2.4m so they both have the same theoretical resolution.
However the image quality on any telescope degrades the further you are away from the optical axis. In WFIRST there is a 3rd mirror that corrects for these inherent distortions for a much wider field of view.
The last factor is the improvement in detector technology. The individual pixels are smaller, smaller gaps, and there chips are much larger. Also WFIRST has mosaic of these  CCD chips to cover the larger corrected field of view.
The paper Optical Design of the WFIRST-AFTA Wide-Field Instrument (Pasquale et al. 2014, Proc. SPIE-OSA Vol. 9293, 929305) discusses the design of the optical path (cycle #4). In addition to a small figure adjustment to the primary, the figures of the secondary and a tertiary mirror are optimized to provide high resolution over a much wider field than the HST. This design also provides a pupil at which filters can be located as well as an intermediate focus (1:1 with the final focus) as shown below.

The basic optical design is decidedly simple for fabrication, integration and test purposes. This approach begins with the base optical design; all three powered mirrors are optically co-axial and simple conics. We considered more complicated designs, including the use of tilted and de-centered mirrors and Zernike and/or anamorphic surface figures.[2,3] TMA designs are widely preferred for high A*Omega telescopes such as JWST and ATLAST.
The primary mirror, referred to here as Telescope mirror 1 (T1), is a fast f/1.2 primary with a linear central obscuration of 31%. It is a light-weighted mirror using a hollow honeycomb core. T1 and Telescope mirror 2 (T2) together form an ~f/8, intermediate focus. This focus is uncorrected, with a large caustic image and strongly curved image surface. You will note the off-axis field bias (seen in Figure 1); this is typical in all TMA systems.
Within the instrument, Mirror 3 (M3) is almost a 1:1 magnification relay of the intermediate image to the focal plane. Working in concert with T1 & T2, these 3 powered mirrors form the large corrected field of the TMA. A pupil is formed between M3 and the focal plane. With a diameter of approximately 100mm, the pupil allows for the insertion of bandpass filters and spectral dispersion elements via an element wheel. Finally, in order to package the system into the volume constraints, two fold mirrors were used.


What is this web on the surface of the Sun? I was going through my Social Media Feed and found the attached post too frequent. The caption reads this is the best image of our Sun. Just as an example, the Universe Today's This is the Highest Resolution Image Ever Taken of the Surface of the Sun

Why? What exactly are the black lines that appear to be like kind of a web, and will such patterns be observed if in case the star was not the Sun but some other star? Are they thought to be common?

The dark lines are colder areas at the edge of the convection cells, where the cooled down plasma sinks towards the inside of the Sun. Now "colder" for the surface of the Sun, is still pretty hot, as explained here.

The yellow parts are where the plasma rises to the surface.
Each yellow spot (which is actually the size of a country) is called a granule, and this web-like appearance is called granulation.

In the outer part of the Sun (the convection zone in the image below), there is convection, that is hotter plasma floats towards the top, cools down at the surface, and sinks back down, like in a lavalamp.

The existence of a convective zone in the outer part of the star is determined by the mass of the star, and all stars with a convective zone in their upper layer are thought to have such granulation patterns. So stars like our sun, or smaller have these patterns.

For larger stars, though, the convective zone is in the inner part of the star, and the outer part of the star is the radiative zone, so there might not be the same patterns on the surface.


Why does the Fourier transform of this CMB image have a hole in it? The BBC's Desert telescope takes aim at ageing our Universe contains the image below of the Cosmic Microwave Background from the Atacama Cosmology Telescope or ACT. It looks like this is plotted with a diverging colormap which makes sense since they've likely subtracted off the average temperature and probably at least its dipole component if not more (CMBR dipole anisotropy (ℓ = 1)
But looking at it closely, I thought I noticed a strange regularity to the spatial scale, so I downloaded it and imported into Python and took the Fourier transform to see what I'd find.
Despite this being an ugly and very unscientific analysis, I'm still seeing both a sharp cutoff below a spatial frequency of 1.0 inverse degrees, and "ringing" or $J_1$ Bessel-like ripples at higher frequencies.
Are either or both of these profound and important to the analysis, or an instrumental artifact, or something else?

Image in the BBC news item is a JPEG and you can see the subfield discontinuities (part of the way the JPEG wavelets are implemented) once the the thee RGB color channels are separated, but this is a square grid and not related to the clearly circular patterns in the Fourier transform.

Python script: https://pastebin.com/t52VbqeD
Exported image converted to PNG: https://i.stack.imgur.com/BC87Y.png

Having now looked at the paper by Aiola et al. (2020), it emerges that for that map, they filtered the data to exclude low frequency multipoles with $|l|<150$, corresponding to about 1 degree. This filtering was done to all the maps in the paper and will be responsible for the dramatic "hole" in your Fourier transform.
As for the high frequency rings, well they might correspond to high $l$ peaks in the angular power spectrum of the CMB, but other things to consider:
I don't know if the image has already gone through some deconvolution process. If not, then your CMB signature will also be multiplied by the FT of the angular response function of the telescopes (probably a Bessel-like function with some high frequency features).
The finite size of the map will introduce apodisation effects, whereby your Fourier transform is the convolution of the CMB transform with the FT of the apodisation function (a narrow 2D ${\rm sinc}$ function.
You also have the pixel size to worry about. Pixelation will be convolution of the true signal with a square top hat kernel. This will also introduce a multiplicative function in to the net Fourier transform of the FT of the pixel kernel (i.e. multiplying by a broad 2D ${\rm sinc}$ function). I think I see this as a cross-shaped pattern on the image.


Higgs Bosons in the centre of Black holes or extremely large stars With the discovery of the Higgs Boson as the mass carrier, what is the possibility at the centre of a black hole being so dense and gravity so infinite that the structure of a proton or neutron are ripped apart into their fundamental elements by these extreme energies that Higgs Bosons are massed together at the centre, while the remaining massless subatomic particles are just outside this layer, similar to an onion  (Similar to a star as more elements fuse together they become more dense, and gravity forces them to the centre again like an onion)

No. That's not what a black hole is. A black hole is a vacuum solution to general relativity. In other words a black hole is just mass, without a "thing" left to be massive. The mass is collapsed to a singularity, and is surrounded by an event horizon.

If you pass over the event horizon, every path you can take in will lead you to the singularity. In this sense the singularity is more like a point in time than a point in space. You can't "see" the singularity, since it is always in your future (you can only see things that are in the past)

There is no "thing" at the singularity, There are no protons neutrons or higgs bosons, and any that are formed will find themselves at the singularity, and ceasing to exist in a short time.

So black holes are very very weird. They can't be easily understood my analogy. They are understood in terms of Einstein's mathematical theory of general relativity.


How much delta-v does a typical geostationary satellite carry? I'm wondering - how much delta-v do typical geostationary satellites carry with them once they get into the orbit for things like orbital adjustment or movement to the graveyard orbit?

There's a mountain of readily available information, of which the following is one example.  From   GOES page
edit
There is more information there about the thrust produced, so I suspect you can calculate the total delta-kinetic energy available.

A2100 has been used for over 40 missions and is available in different versions catered to specific mission needs in terms of
dimensions and power availability with satellite mass varying between
2,000 and 6,740 Kilograms.
Each GOES-R-class satellite is 6.1 by 5.6 by 3.9 meters in size and has a dry mass of 2,857 Kilograms. At liftoff, the satellite
weighs 5,192 Kilograms, loaded with propellants to support a mission
of at least 14 years.
GOES-R, S, T, U launches with a propellant and pressurant load of
2,335 Kilograms with Mixed Oxides of Nitrogen and Hydrazine acting as
main propellant components and Helium gas used for the pressurization
of tanks.


Options for first time astrophotography w/ 80/900 refractor? Recently I bought a telescope and I started to take pictures of my observations. The most basic solution, I'm using a smartphone adapter and my Xiaomi Mi 6 camera to do that but the images (except for the Moon sometimes) is too bad and Jupiter and its moons not even show in my screen. So I thought in use a more serious device to take the pictures.

I have in my personal finance control a monthly quota to "astronomy", but to start I think in spend about 200€ (~$230,00).

I work with programming for 10 years (C#), so I can use something more "crude" if needed (like Raspberry Pi or Arduino). You guys can suggest something to start take decent shots of  Jupiter and its moons?

There is no "best" camera as it really depends on the subject you intend to image and the optical instrument you intend to use with it.  This could turn into a very long answer, so I'll try to offer a short response first and then offer a bit more to consider.

For planetary imaging, use a camera capable of doing video frames and this does not need to be a camera with a large chip or high resolution.  Jupiter, for example, is just fractionally under 44 arc-seconds diameter when viewed at opposition. (compare to the moon at roughly 30 arc-minutes or more than 40x the angular width).

Typically you'd capture perhap 30 seconds to a couple of minutes worth of frames and then feed those frames into stacking software.  Ideally the video frames use a "lossless" format.  Several image processing applications can stack the frames (e.g. AutoStakkert & Registax are popular and free).

A camera such as the ZWO ASI120MC-S is very popular and in your budget.  There's the ASI290MC for those with deeper pockets (and versions for those with much deeper pockets).    From QHYCCD there's a QHY5R-II ... more to the entry-range with lower resolution and a smaller chip.

Ideally your scope would be on a motorized equatorial mount so that it keeps the object in your field of view (otherwise the object will quickly drift out of the camera frame.  For this scope & field of view combination, that would happen in about 45 seconds for a planet in the "center" of the frame to drift out of frame completely).  If using a 2x barlow, it would happen in about half the time.

A barlow is usually used for planetary imaging (the simulated field of view did not use a barlow).  To understand why, you'd want to read about Shannon-Nyquist sampling theory and how it applies to astrophotography.


How to calculate solar coordinates. (Right Ascension, Declination, and ultimately the Altitude and Azimuth of the Sun) As the title says I am trying to calculate solar coordinates of the sun for a given location and time. I have spent many 10's of hours on this without success so I will sincerely appreciate help on this.  My RA value seems to be right on but the Dec, Alt and Az are off.

Alt value is within 6 degrees but I cannot figure out why it is that close because the latitude is in radians instead of degrees like the other values. If I change latitude to degrees the values don't even make sense.

I have included all the values of the intermediate variables to help see at a glance what might be off. I hard coded the JD and the Local Hour Angle just to take those out of the equation for debugging.  Links are shown as to where I got those values.

The C# code is below although the programming language doesn't matter. The formulas are the same in any language.

  //for formulas
            double D = 2458599.125000 - 2451545.0; //the  first  term is the jd of interest; 2nd term is jd of 1-1-2000 at 12:00:00.0
            double Latitude = 41.369341;
            double Longitude = 81.885493;

            double gTmp = 357.529 + 0.98560028 * D;  //mean anomaly of the sun  = 7310.0765751550007
            double gTmp1 = (gTmp % 360);     //subtract multiples of 360 (returns remainder in degrees)  =  110.07657515500068
            double g = gTmp1 * Math.PI / 180;   //Convert to radians = 1.9211986657737494
            double qTmp = 280.459 + 0.98564736 * D; //mean longitude of the sun = 7233.33868336
            double qTmp1 = (qTmp % 360); //subtract multiples of 360 (returns remainder in degrees) =  33.338683359999777
            double q = qTmp1;   //don't convert to radians here; convert after terms in degrees are added in the computation of L; =33.338683359999777
            double L_tmp = q + 1.915 * Math.Sin(g) + 0.020 * Math.Sin(2 * g);//geocentric apparent ecliptic longitude of the sun(adjusted for aberration) = 35.124421105295383
            var L = (L_tmp % 360) * Math.PI / 180;   //Convert to radians; //subtract multiples of 360 = 0.61303679614439033
            double eTmp = 23.4393 - 0.00000036 * D;  // = 23.436760515
            double e = eTmp * Math.PI / 180;   //Convert to radians;   = 0.40904863698815186

            //http://www.neoprogrammics.com/de405_usno_ae98/DE405_Sun.php
            // formulas per https://aa.usno.navy.mil/faq/docs/SunApprox.php
            double RA = (Math.Atan2(Math.Cos(e) * Math.Sin(L), Math.Cos(L))) * 180 / Math.PI; //right ascension converted from radians to degrees = 32.838762087228496
            double dec = (Math.Asin(Math.Sin(e) * Math.Sin(L))) * 180 / Math.PI;  //declination converted from radians to degrees =  3.228747909189217
            string[] sunCoordinates = new string[2];
            sunCoordinates[0] = ToHMSStr(RA);    // = "2h 11m 21.3s"
            sunCoordinates[1] = ToDMSstr(dec);  //  = "13d 13m 43.49s"

            //compute alt and az per https://aa.usno.navy.mil/faq/docs/Alt_Az.php and https://aa.usno.navy.mil/faq/docs/GAST.php
            //LHA for testing is from https://ssd.jpl.nasa.gov/horizons.cgi#results and https://www.vercalendario.info/en/how/convert-ra-degrees-hours.html
            //from http://www.neoprogrammics.com/de405_usno_ae98/DE405_Sun.php
            double LHA = 307.39;   //Local Hour Angle
            double lat = Latitude * Math.PI / 180; //Convert to radians; //Latitude is defined above; We get sign (below) after converted to Rad????  = 0.72203120983028346
            double alt = (Math.Cos(LHA) * Math.Cos(dec) * Math.Cos(lat) + Math.Sin(dec) * Math.Sin(lat)) * 180 / Math.PI; //convert from radians  = 53.266598643452909
            double az =Math.Atan2(-Math.Sin(LHA) * Math.PI / 180, (Math.Tan(dec)*Math.Cos(lat) * Math.PI / 180 - Math.Sin(lat) * Math.Cos(LHA)) * Math.PI / 180); //convert each term to radians   = 2.4585704411338072
            az = az * 180 / Math.PI;   //convert from radians   = 140.86570991258415

```

For the Sun at JD 2458599.125, RA = 32.8° and dec = 13.2° are correct regardless of geographic location.
In the alt/az calculation,

LHA should be 323.6° and converted to radians, and
the expression for alt needs a call to Math.Asin.

Trigonometric expressions are easier to verify if they are free of inline unit conversions.
Store angles in radians where possible; otherwise make it obvious with a name suffix such as _deg.

When computing the local hour angle, you can approximate GAST with GMST.
The difference eqeq is smaller than the approximation error of the Sun position algorithm you're using, and the alternative GMST formula is valid for D within a few centuries of J2000.0.


If blackholes are spherical, what is the prespective of the blackhole image? If black holes are spherical, so shouldn't it be absorbing light from all the directions? How would one explain the light ring. which seems more like a 2D ring than a sphere?

One can never actually see the event horizon, one sees the gas that is about to enter the event horizon and gets hot along the way.  But this gas has a history, it doesn't just fall straight in.  The history imparts some angular momentum to it, like the water in a draining bathtub.  So it spirals in, not falling straight in, and this creates a disk shape called an accretion disk.  The disk does not have to be as thin as a pancake, but it's far from spherical, and that gets further distorted by the strong gravity (creating the brightness variations and the central dark sphere).

The disk does have a hole in the center because at some point the orbits go unstable and fall quickly below the event horizon.  It could be argued we are not actually seeing the event horizon where it goes black, we are seeing the drop in gas density where the orbits are no longer stable and the  slow spiral turns into a very rapid fall, but the way the strong gravity bends the light rays is always very important as well.

The distortion by the strong gravity is what provides the test of general relativity that everyone is so excited about.  The strong gravity produces a "shadow" effect, so you don't have to see the accretion disk from above to get a spherical blackness at the center.


Why does it make sense to say the universe has no centre? If we suppose we and everything in the universe is perfectly flat.  That is, we have width and depth, but no height.  And we are situated on the surface of a balloon blowing up with air and expanding "outwards".  Us flat people can move around on the surface of our balloon but not through it, in much the same way we navigate around the earth, in a plane say. No matter where something is situated on the surface of the balloon, one point is no more central than any other point.  In other words, the surface of a balloon (the "universe") has no centre.  Is this why we say it makes no sense to say the universe has a centre, or edge for that matter, or is this analogy stupid?

The answer to this question lies in cosmology.

It's because of relativity in positions.  Every galaxy appears to be moving away from us and the universe we see is only the observable universe.  Further points are more in the distant past, when you observe the deep field you observe the past.  This may seem irrelevant to your question, but the idea of a "center" to a universe where any position in the universe, be it from our galaxy or from a galaxy in the Sloan Great Wall is the true center is not consistent with observation.  That's because galaxies/clusters of galaxies are moving away from each other, relative to each other.

Every galaxy is slowly being pulled apart from other galaxies, save for those in gravitationally bounded clusters.  (The whole architecture of the universe, with massive voids of absolute nothing and bunched up clusters of galaxies reflects this quite a bit)  If the universe had a central point, this would be easily observed because there would be variation in the color shift of galaxies that could be analyzed to deduce a central point where the galaxies are moving away from.

That is not what is observed.  Instead, pretty much most galaxies (save for exceptions like Andromeda or the Triangulum, who prove the rule since those galaxies are gravitationally connected to our galaxy via the local group) are moving away from us.  The Great attractor being not a path to the center of the universe, but the center of the recently discovered Laniakea super cluster.  So if there's attraction between galaxies, it's because of gravitational interactions.  The universe is also without any real edge unless you count the edge of the observable universe.  The observable universe being more a historical record than what the contemporary universe around us looks like and is a byproduct of light speed being so limited.  The speed of light in a vacuum is also the speed of information.  The whole universe is far more massive by comparison.

So to speak about a center of the universe would require the following:

A region of space that galaxy is moving away from or towards. (Which does not exist, instead relative trajectories and gravitational clustering is all regional at most.)
Edges that are defined enough to pinpoint a center (If such edges exist, they are impossible to observe due to the limited speed information travels in the universe)
Absolute positioning to pinpoint where this center would be. (All positions are relative, the galaxy, the sun and our planet all are moving at extremely fast velocities we don't notice thanks to relativity.)


Is it possible that a star system ejects an object out of the galaxy? I know it is possible that a star system ejects a planet out of the star system and the planets becomes rogue planets and orbit around the galaxy. Is it also possible that a planet gains enough speed because of ejecting and escapes out of the galaxy?

Intergalactic planets have never been observed.  They're too small and it's too great a distance, but intergalactic stars have, and if stars can be ejected from a galaxy, planets can to.  The orbital mechanics is the same for both.

The escape velocity of the galaxy is impressive.  For solar-systems, where most of the mass is in the star in the center, the escape velocity is just about the orbital velocity times the square root of two.   Source.

For the Milky way excape velocity calculation is larger and more complicated because most of the mass is in the dark matter halo and very much spread out, not localized in the center.   The center is still more dense, the mass is far more distributed so the square root of two rule doesn't apply.  Using the Sun as an example, the orbital velocity of our sun is about 220 km/s and it's escape velocity is 537 km/s.  Source.   As objects moves further towards the edge of the Milky way, the escape velocity decreases and the orbital velocity generally increases, so at close to the edge of the Milky way, you'd get something close to the square root of two ratio again.  That means an object close to the edge of the Milky way only needs to add 41.4% of it's velocity to escape.

Gravity assists depend on the relative velocity of the larger object to the "stationary Milkyway", so, if our solar-system was to fly close enough to very massive star and get a big gravity assist, assuming the massive star has a similar orbital velocity around the Milky way, 220 km/s, that could add as much as 440 km/s to our solar-system's velocity which would be enough to escape the Milky way.   Only large stars have enough gravity for that kind of kick, but there's enough large stars in our galaxy that they shouldn't have a hard time ejecting objects that fly reasonably close to them.   You probably need at least a half dozen or dozen or so solar masses and a close fly-by, but, statistically, it should happen from time to time.

The Wikipedia article points out that galactic collisions could be responsible for these stellar ejections, and galactic collisions probably do eject a great deal more material compared to a galaxy in a peaceful spiral arm rotation, but it's still possible for gravity assists to cause objects to be ejected even from a galaxy not currently colliding with anohter, because twice the orbital velocity of the large star plus the orbital velocity of the approaching star or planet, when combined is greater than escape velocity, at least for the outer parts of the spiral galaxy.   Needless to say, most gravity assists don't line up perfectly, and half of them push an object further in, not further out, but statistically, and with multiple assists, it's very likely that stars and rogue planets get ejected from the Milky-way, it's probably happening as I write this.

By some estimates, there might be a HUGE number of objects that doesn't belong to any galaxy.


Better than Earth habitability Earth undoubtly has very good conditions for supporting life. Although it is expected that many other planets on the outer space have conditions at least as good as Earth, the vast majority doesn't, making them unhospitable to life or probably being able to support only very simple lifeforms. Earth itself for some billions of years until the Ediacaran or Cambrian could only support very simple lifeforms.

There are many parameters that may influence the habitability of a planet and its ability to support complex life: Star type; star temperature; star luminosity; stellar activity; stellar stability; star age; planet age; planet composition; planet size; orbital excentricity; orbital length; rotation axis inclination; planet tectonics; planet magnetosphere; presence and influence of satellites; abundance of water; planet atmosphere; interactions with other planets; presence or absence of asteroids, comets and minor planets planets belts and their position, distribution and composition; galactic orbit; galactic neighborhood; mass-extinction events rate, probability and intensity; and hundred of other possible variables including some based on pure luck and random chance.

Many of the parameters are modeled after Earth itself, since Earth is the only place so far that we know that life exists, and even if we found some alien life somewhere, it will probably be limited only to very simple forms of life.

But, what combinations of those parameters could lead to a planet with better life support than Earth itself?

Ok, you may argue that the question is too broad, so by "good life support" we could say something that allows the planet to evolve plenty biodiverse multicellular life ranging from simple microscopic creatures to complex dozens-meters long creatures with many body-differentiated parts and organs in a short timespan. So, a planet that has an environment which allows the development of richly-diverse and complex plant-like and animal-like creatures in a billion years after formation and stay like this for another 10 billion years is expected to be more life-friendly than Earth.

Further, lets restrict the biochemistry to what we know: water-based and carbon-based life, but not necessarily oxygen-breathing.

By the way, I am not asking anything about intelligent life or humans, just complex multicellular and biodiverse life.

By the way: two important links about the subject: http://www.nature.com/scientificamerican/journal/v312/n1/box/scientificamerican0115-32_BX1.html and http://www.nature.com/scientificamerican/journal/v312/n1/full/scientificamerican0115-32.html

Note: I am self-answering my own question in hope that someone post another answer that beats this one.

Earth is near the inner edge of the Sun's habitable zone. And since the Sun is expected to grow and increase it luminance, Earth might be unhihabitable for any life somewhere between 1 or 3 billion years in the future. So, a planet that have longer time to develop before its parent star moves it beyond the inner edge of the habitable zone is more favorable than Earth.

Since Earth itself always was inside the habitable zone since it formed 4.6 billion years ago and will still be for lets say more 1.4 billion years, with a large uncertanity factor, this gives roughly 6 billion years of time for complex life to develop. Given that it is unlikely to form early due to the time needed for evolution to take place and due to an elevated level of large bollides collisions, we could discount the first 2 billion years from any life bearing planet, including Earth, as unlikely to develop complex life. Further, it is unlikely that complex life would finally evolve out from simpler forms when the planet is already overheated and already crossing the inner edge of the habitability zone, so lets take out the finishing 10% of that period for any planet (probably something more than 10%, but lets keep this as a conservative estimative). So, for Earth, this gives a window of a size of 3.4 billions years to complex life evolve. Similar planets with larger windows have better probabilities.

Stars larger and more luminous than the Sun tends to be more unstable and live shorter. As a result, it is expected that planets around stars larger than the Sun has less time to develop complex life, and thus a shorter time-window. On the other hand, this means that stars smaller and less luminous than the Sun gives a larger time-window to the planets to develop life.

For stars smaller than the Sun (a G-type yellow star), we could consider the K-types (aka, orange dwarf) and the M-types (aka, red dwarf) as specially favorable.
An orange dwarf star may live for 10 to 30 billions years in the main sequence. A red dwarf star may live in the main sequence for trillions of years.

However, planets in the habitable zone of red dwarfs are likely to become tidally lock, and we don't know if this is really that bad or not for life biodiversity. Lets assume that this is really bad, so a planet orbiting an orange dwarf in the habitable zone is likely to have a better habitability than Earth.

Accordingly to this, a planet with two times the mass of the Earth, will have stronger gravity, and thus it is likely to be flatter. Further, it is likely to have a ticker atmosphere that would protect the surface from UV radiation better than Earth. It would be geologically active for a longer time, resulting in more carbon cycling. With the right quantity of water (not a desert nor a global very deep ocean), it might be an archipelago world, since its flatness would not allow the ocean to be very deep nor the continents to be very large. As a result, life would flourish in a number of rich biologically favourable environments significantly larger than Earth. Further it magnetic field is likely to be stronger than Earth's one, protecting the surface from cosmic rays.

As a result, a planet with two Earth masses orbiting an orange dwarf star in the habitable zone has a good chance to be more habitable to life than Earth itself.

Needless to say, near-circular orbits are more favourable than excentric ones, since excentric orbits may make the planet enter in periods of freezing or boiling. However, a reasonably excentricity that periodically changes the environment in a significant manner, but not as too much that it would extinguish non-extremophile life, might give to the planet life a selective pressure needed for developing rapid evolution to face the always changing climate.


Meteor shower predicted to be best viewed BEFORE midnight (for reasons other than the moon)? Reading that the Leonids meteor shower is almost upon us, I clicked some links (e.g. EarthSky.org and NatGeo) to read more of the prediction; especially the part about when to best view them.

I had a Déjà vu when I saw "early morning hours" because it seemed to me that "that's what they always say".

I'm guessing that it's not a coincidence that the constellation for which the shower is named always seems to rise around or after midnight at the time of year when the shower occurs.

So I'd like to ask if this is true, that they pretty much always are best viewed after midnight (except when issues of weather or the Moon interfere), and if there has ever been a shower where this wasn't the case — one that was best viewed before midnight.

Not that I know of, and it would somewhat surprise me.

The number of meteors from a shower you will see depends on the time of night through a number of reasons (not considering the weather and the moon):

The distribution of meteoroids on Earth's orbit is not homogeneous. The highest potential occurs when Earth moves through the peak of the distribution. This is associated with a certain time on Earth.
Meteors from a shower appear tot come from the same spot in the sky. This spot is called the radiant and moves through the sky along with the stars. The higher the altitude (height) of the radiant, the more meteors you will be able see.
Meteors enter Earth's atmosphere because Earth moves into the meteoroid cloud. This means that you will see more meteors when the spot on Earth where you are standing is facing the meteoroid cloud. This is always the direction in which the Earth is moving. If you are on the other side, the meteoroid cloud is 'behind' you and possible meteors will have to move faster than the Earth in order to enter the atmosphere. As pointed out by user Florin Andrei in the comments, this effect is also called the "bugs on the windshield" effect, because

  Earth is like a car moving fast, and meteors are like insects that get splattered. Of course you get more bugs being splattered on the piece of glass that's facing forward.

Depending on your location, 1. and 2. could combine to maximise their effect some time before midnight. However, the effect of 3. is quite big. Because of the way the Earth rotates, this will always happen early in the morning, before sunrise.


How does the Moon move in the "night" sky as seen from the poles? A photo in the NYTimes article Ice Surveys and Neckties at Dinner: Here’s Life at an Arctic Outpost has got me thinking.

During the winter months at either the north or south pole, when it is constantly dark, how does the Moon behave?

Would a nearly-full moon sometimes circle for days without setting, or would it still rise and set, or would it be all together missing, or some combination of all of the above?

The moon is quite interesting as you get near the poles. It still moves round due to the Earth's rotation once a day, but it also orbits with a period of a month so what you get is two weeks of moon above the horizon followed by two weeks where you cannot see it.

What changes through the year is what phase you see when the moon is visible, and in the middle of winter you have two weeks around full moon, whereas in summer you have two weeks around the new moon - not so exciting. Spring and autumn obviously have two weeks around first and last quarter, respectively.

This simulation using Stellarium shows the effect quite well.

From astro.cornell.edu - this wording may help:

  The second movement caused by the Moon's orbit around the Earth is analogous to the movement of the Sun over the course of a year only it repeats over the course of a lunar month. Near the new Moon phase, the Moon is near the Sun and therefore never rises during the winter. As the Moon approaches full, it will start to pop up above the horizon. Eventually near the full Moon phase it will be high enough in the sky to stay up all day and circle like the Sun in the video above. The elevation of the circle will rise as the Moon becomes completely full and then start to decrease until it begins to dip below the horizon. Eventually the Moon will stop rising at all as it gets close enough to the new phase. The cycle then repeats.

And another useful reference at planetarium.madison.k12.wi.us:

  At the beginning of Winter, when it's nighttime all of the time, the moon would be in the sky for the 2 weeks closest to Full Moon, and then below the horizon for the next 2 weeks. And at the beginning of Spring (click on the graphic), when the sun is at sunrise all of the time, the moon would be up in the sky for the 2 weeks closest to First Quarter (waxing), and then below the horizon for the next 2 weeks.

  The animated graphic above shows what we would see from the North Pole if we went out every day at noon, for 14 days in a row, from March 1st to March 14, 2006. We start with a thin crescent moon near the horizon, and end with a full moon near the horizon. Halfway through, the First Quarter moon would be when the moon is highest above the horizon.

  Keep in mind, that if you were observing the moon constantly, throughout a 24 hour period, the moon would seem to move to the right in the sky along with the sun, stars, and planets due to the Earth's rotation. Nothing would seem to rise and set: they would just seem to circle around you.


Conundrum involving distance to object, universal expansion, age of universe, etc I was recently reading about a galaxy or quasar (not sure which, so lets just say quasar) where the estimated distance to this thing was some very significant fraction of the age of the universe.  What I mean is this:  I can't recall the exact particulars, but assuming for argument's sake that the age of the universe is 13.8 billion years old, this thing was like 13 billion light years away.  (Again, some detail here might be slightly off; the thing might have been estimated at 12 billion light years away - so focus on the general principle I am layout out here, not any particulars).

What I don't understand is this.  If the thing is 13 billion light years away, then the light reaching my eyes as I look at it now took 13 billion years to get to me.  Yet the age of the universe is 13.8 billion years old (again if I'm slightly wrong in some of these numbers, just accept them for now for argument's sake).  It means the light must have left this object when the universe was much "smaller" or more compact, and things were much closer together.  Which means the light should have passed my position a long time ago if I have been either "moving" or "expanding" (or some combination) away from the object at anything except some rather large percentage of the speed of light.

To try to boil this down, or restate it in simpler terms:  1) either it seems I am moving/expanding/red-shifting away from this object at some significant fraction of the speed of light (haven't done the numbers, but might be 80% or 90% or so) and have been for a while, or 2) something is "off" (the age of the universe, or the distance of this object).

or 3) there is something I'm not understanding.  Which is why I came here.

I have never seen it stated that we are "expanding" or "redshifting" or [insert whatever term you like] at some significant fraction of the speed of light; I have only ever seen it stated that for some brief period of time after the big bang the universe could have (or did) expand faster than light, but again that was for like less than a second if memory serves.

At any rate, anyone care to explain?

Thanks.

The 13 billion lightyears distance of the quasar mean 13 billion lightyears light-travel distance. In other words, the light took 13 billion years to reach us, independent of the distance the quasar is now away from us.

The current proper distance (a chain of rulers would measure) is called comoving distance; it is much larger than the light-travel distance for large redshifts.

The quasar is moving away from us in an accelerated fashion, and is now moving away faster than light. This means, light which is emitted now (in the sense of cosmic time) by the galaxy, the quasar may have evolved into in the meanwhile, will never reach us.
The light, the quasar emitted 13 billion years ago, left the quasar just in time to leave the regions of the universe which later have been receding faster than light from us, to eventually reach us.

As a thought experiment, imagine walking with 5 km/h (our playground speed of light) on a rubber band which is expanding with a constant rate of 1 km/h per meter actual length of the band. (This leads to an exponential acceleration of the distance between the two marks on the band). If you start closer than 5 meters away from your goal, e.g. from a start mark (our quasar) 4.50 m away, you'll finally reach the goal mark. The start mark will soon be further away from the goal mark than 5 meters, therefore receding with more than 5 km/h from the goal shortly after you left the start mark.
At the moment you arrive at the goal mark, the start mark will be much further away (comoving distance) than the distance you needed to walk (light travel distance). And you've been walking a longer distance than the (proper) distance between the marks was at the time you started walking.

Btw.: Acceleration is only felt as a force, when the velocity to the local rubber band (mataphoric space-time) is changed.

Example calculations with a protogalaxy of redshift $z=11.9$:
Based on the Cosmology Calculator on this website, the cosmological parameters $H_0 = 67.11$ km/s/Mpc,
$\Omega_{\Lambda} = 0.6825$ provided by the Planck project, and the scale factor $d(t) = d_0 / (1+z)$, setting $\Omega_M = 1- \Omega_{\Lambda} = 0.3175$, the
age of the universe is $13.820$ Gyr, and the comoving distance of the protogalaxy is $d_0 = 32.644$ Gly.

The age of the universe, we see the protogalaxy (at redshift 11.9), was 0.370 Gyr, light-travel distance has been 13.450 Gly, proper distance was 2.531 Gly.

After the protogalaxy has been emitting light 0.370 Gyr after the big bang, the light travelled towards us through space of redshift beginning with 11.9 shrinking to 0; the light arrived at us 13.820 Gyr after the big bang. The comoving distance (to us) of the space traversed by the light started with 32.644 Gly shrinking to 0. The remaining distance, the light needed to travel, started with 13.450 Gly shrinking to 0. The proper distance between the protogalaxy and us started with 2.531 Gly increasing to 32.644 Gly due to the expansion of spacetime.

Here some intermediate states described by
a couple of tuples, consisting of

redshift $z$,
according age $t$ of the universe (Gyr),
comoving radial distance (at age $t$) of the emitted light, we can now detect from the protogalaxy (Gly),
remaining light travel distance of that emitted light (Gly),
proper distance of the protogalaxy at age $t$, according to $d(t) = d_0 / (1+z)$:

$$(11.9,  0.370, 32.644, 13.450,  2.531),$$
$$(11.0,  0.413, 32.115, 13.407,  2.720),$$
$$(10.0,  0.470, 31.453, 13.349,  2.968),$$
$$( 9.0,  0.543, 30.693, 13.277,  3.264),$$
$$( 8.0,  0.636, 29.811, 13.184,  3.627),$$
$$( 7.0,  0.759, 28.769, 13.061,  4.081),$$
$$( 6.0,  0.927, 27.511, 12.892,  4.663),$$
$$( 5.0,  1.168, 25.952, 12.651,  5.441),$$
$$( 4.0,  1.534, 23.952, 12.285,  6.529),$$
$$( 3.0,  2.139, 21.257, 11.680,  8.161),$$
$$( 2.0,  3.271, 17.362, 10.549, 10.881),$$
$$( 1.0,  5.845, 11.124,  7.974, 16.322),$$
$$( 0.0, 13.820,  0.0  ,  0.0  , 32.644).$$

The Hubble parameter, meaning the expansion rate of space per fixed proper distance, is decreasing with time. This allowed the protogalaxy to recede almost with the speed of light, although it was just about 2.5 Gly away from us (proper distance) in the time, when it emitted the light we detect now.
Nevertheless distant objects in this space accelerate away from us, since their increasing distance is multiplied with the expansion rate of space.


How does gravity affect pressure under water? How different would being under water in Europa differ from being under water on Earth? Would it be possible to dive deeper on Europa with a submarine than on Earth and by how much?

For a liquid, hydrostatic pressure is $\rho g h$ where $\rho$ is density (this is always the same for all water) g is gravitational acceleration and h is depth.

The gravitational acceleration on Europa is 1.3 $\text{m/s}{}^2$ (compared with 9.8 $\text{m/s}{}^2$ on Earth). But on Europa there is 20km of ice floating on the water.

As a rough estimate, the gravity in Europa is 1/10 that on Earth, so the pressure at the bottom of the ice/top of the water would be comparable to the pressure at 2km in the Earth's Oceans.

Since we can build submarines that can operate at 10km, we could reasonably suppose that we could operate submarines down to about 100km on Europa, which takes us to the rocky surface. However we have no way to pass through 20km of ice crust to reach the water layer. Building the submarine may be the easy part!


Can a Neutron Star become charged? If you keep adding electrons or protons to a Neutron star, what would happen? Are there decay processes that convert them to neutrons in the absence of the oppositely charged particle?

Charge is a conserved quantity. If you add a net charge to an object, then it becomes charged.


Could celestial coordinates be encoded into three-word combinations? I came across this cool tool which encodes geographic coordinates into 3 dictionary words - table.lamp.chair - for example. You can see their map here https://map.what3words.com/. It divides the world into a grid of 57 trillion 3mx3m squares and each one is assigned with one of these three-word addresses.
It made me curious as to whether this sort of thing could work for communicating the locations of stars.

It would be pretty simple to assign unique three-word addresses to stars in stellar databases. What would be even more powerful is if we could encode celestial coordinates into three-word addresses, and create an interactive "what3words" space map. But would this be at all useful?

The finite surface area of the earth allows us to create the grid of 57 trillion squares. Can we work with the infinite nature of the celestial sphere to create something similar?

Possible, yes. In fact since you can just use their algorithm. Sky locations expressed a RA and Dec map directly to coordinates expressed as Longitude and Latitude. The celestial sphere is no more infinite than the terrestrial one.

Each square seems to take about 0.1 arcseconds, which is very accurate, but it does raise a problem: The stars move. Barnards star has proper motion of over 10 arcseconds per year. This is an extreme example, but all the nearby stars move, so their addresses would change. The algorithm is designed not to give nearby locations similar names. So each year Barnards star would have 10 completely different addresses. Other stars would change address due to parallax. If you don't account for precession, all stars would be being renamed all the time.

Now the purpose of the three word addresses is to give addresses to the address-less, in a way that is memorable. Is there a need for addresses of sky locations? We have a cute way of naming larger regions of the sky (the 88 constellations) and stars have names (or catalog numbers) that don't change due to their real or apparent motion. If you need to refer to a location you can use the numeric coordinates.

So: this would not be useful, either for science (since the stars would keep moving address) or for outreach (the Constellations do very well for the whole sky)


How many satellites orbit the Earth? At what rate do they orbit the earth? I was looking at a full moon the other night with my telescope and noticed a black dot move across the moon quite rapidly. This happened a few times and others who were with me were able to confirm.

There is no way the ISS would make its orbit around the Earth that quickly for us to be able to see it a few times. Therefore there must have been others. How many are there, and how long does the average satellite take to make a full orbit around the earth?

The dots we saw looked much like this:

There are over 1000 operational satellites, and probably as many again defunct or non-operational satellites in orbit, plus a great deal more small pieces of space junk.

About half of these are in a low earth orbit (LEO). Satellites in LEO have a period of between 90 minutes and 130 minutes. But you would not expect a satellite to follow the same path in the sky in two orbits, as the earth will have moved between orbits of the satellite. So it is very unlikely that you will see the same satellite pass in front of the moon twice.

Satellites in LEO are moving fast! They will cross the moon's disc in less than a second, and transits are rare, as each is a path only 7km wide. at any one time, a transit is visible from less than 100th of 1% of the Earth's surface. To obtain the awesome photo that you have linked, the astrophotographer would have had to use some orbital prediction software to calculate the exact time of a solar transit by the ISS and travel to a carefully calculated position (and prayed for clear skies). You could be observing the moon all year and never see a transit.

Most of the remaining satellites are in Geostationary orbits, They orbit at a great distance over the equator. As their orbital period is exactly one day, they appear to hang in space above a point on the Earth. Such satellites are too distant to be seen as "dots".


Is it ever possible that the graviation from the mass of kinetic rotational energy will overcome centrifugal force? This is, I think, somewhat intuitively obvious, but this comment got me thinking about this:

what is gravitational force?

"Rotation speed can create centrifugal force opposing gravity and making things lighter," Rotation also stores energy, and energy is mass, per $E=mc^2$. A body spinning sufficiently fast will exert higher gravity, e.g. a slowly-spinning neutron star will have a weaker gravitational pull than equivalent neutron star that spins very fast."

Intuitively, my answer is "no way", added mass by rotational velocity I would think, could never be extensive enough to counteract the "flicking off" or centrifugal force of very fast rotation.  Even with a Neutron Star I would think it's impossible, but I'm not 100% sure, so I thought I'd ask.

If we ignore relativity from motion but not $E=mc^2$, as that's the crux of the question

Centrifugal force = $m v^2/r$

Kinetic energy of rotation = $1/2 I w^2 = 1/5 m v^2$

Mass equivalent of kinetic energy = $1/5 m v^2/c^2$

gravitational force $g=\frac{Gm}{r^2}$

so if we apply the gravitational force of kinetic energy

$g = G(1/5 m * v^2/c^2) / r^2$, or, simplified, $G*m*v^2 / r^2 c^2$

and we compare the two equations
Centrifugal  $m v^2/r$
additional gravitational   $G*m*v^2 / r^2 c^2$

we can remove $m*v^2$ from both on the top and 1 $r$ from the bottom

additional gravitation ratio to centrifugal force = $G/r * c^2$

$G$ and $c$ are numerical.    $G$ is very small, $c$ is very big and the ratio grows smaller as the radius grows larger.

gravitational constant:   $6.67408 × 10^{-11} m^3 kg^{-1} s^{-2}
c = 2.998 x 10^8 m s-2

the ratio, unless my math is broken, centrifugal force to additional gravity from added mass by kinetic energy of motion = $1/r * 1.35 * 10^{27}$, so you'd need a hugely small r, almost a plank length or a singularity where the added gravity from kinetic energy of rotation would overcome the "flicking off the surface" or centrifugal force.

When I work out the units, I get meters per kilogram, which I don't think is right.   The units should cancel out with a ratio of two forces in opposite directions, so I think I made an error, but I don't see where I made it.

My question is two fold.    1) is my math broken? and if so, where?   and 2) is the added mass from kinetic rotational energy ever relevant, say in a very rapidly rotating Neutron star?   Could it ever assist the Neutron star in collapse or add gravity?, I can see how it could add to flattening, as rotation flattens objects naturally and perhaps, add a speck of gravity on the polls where centrifugal force is zero, but logically, I think, rotational energy would end up spinning any non black hole object apart, long before the added energy of rotation had enough mass to make a measurable difference on gravity.    Is my sense right or is there a situation where added mass from energy of rotation could overcome the centrifugal force?

In some sense, the answer is trivially 'yes':
Due to spacetime curvature, there exist no circular orbits below $\frac{3}{2}r_s$, with $r_s$ the Schwarzschild radius.
So whenever you get an object sufficiently dense, that its radius is below its $\frac{3}{2}r_s$, the mass equivalent to any nonzero rotational energy overcomes the centrifugal force of an orbit at $\frac{3}{2}r_s$.
Below $\frac{3}{2}r_s$ even zero rotational energy overcomes the centrifugal force, since the latter is then pointing inward.

This doesn't require a Black Hole.


How can I measure the mass of eclipsing binary stars? Is there any method to measure the mass of eclipsing binaries without using Doppler shift effect?

It depends.

For many categories of stars, it is possible to determine the approximate mass of the star from its non-redshifted spectrum. Stars on the main sequence: this table of spectral classifications lists the approximate mass for the various classifications. A more precise mass estimation should be possible with a more precise classification. Sol, for example, is a G2, near the high end of G, putting it near the top end of the range (0.80 - 1.04 stellar masses).


What causes the aurora of other planets to have different colors? All I could come up with is it depends on the atmospheric composition of the planet as they will interact with the charged solar particles which might be true but is there is there a definite reason behind the different colors of aurorae on other worlds (Saturn's blue aurora for example)?

Ultimately, you are correct. However, there may also be some clarification in order: what blue aurora of Saturn are you referring to? The most well-known auroral images of Saturn like this one, are not true color images. That is, what we see in the photo isn't what you'd see with the naked eye; instead, the photo has been enhanced to emphasize something interesting, in this case the aurora. In reality, that beautiful blue you see in the picture isn't even of light that's visible with our eyes. It's UV light which would be completely invisible to us. People working with data taken by a specialized camera have to give the UV light some visible color or we wouldn't see it at all. The Saturn aurora images I am aware of all have been modified in this manner.

However, aurora themselves are the result of charged particles funneling down a planet's magnetic field and crashing into the planet's atmosphere. On Earth, the aurora occurs because of the atoms in the atmosphere getting energized by these collisions, and then emitting light when they de-energize. Here, it's the oxygen and nitrogen in the atmosphere that are responsible for aurora light (primarily greens from oxygen, but also reds at higher altitudes from oxygen and occasionally blues from nitrogen). In other planets' atmospheres dominated by different atoms and molecules, the reactions will be different, and the wavelengths of light emitted will be different as well. For example, Saturn's aurora are due to energized H2 molecules. On Jupiter, it can be the infalling particles themselves (sulfur and oxygen) that are responsible for auroral emission.


Orbital eccentricity variation of the other planets? On Earth, it's fairly well published, mostly in climate change related articles, that the Earth's orbital eccentricity operates on a 413,000 year cycle with roughly 90,000–125,000 year variation within that cycle.

Source

The cause of this variation is primarily the other planets, with Jupiter and Saturn usually mentioned as the primary causes (same link).

Question is two fold.   How stable is the Earth's orbital eccentricity cycle?  413,000 years sounds enormously precise, but logically, I would think small changes in planetary orbits would create some variability.   Is the 413,000 years well established and repeating or is it more uncertain?

And, do we have a good estimate of other planets orbital eccentricity variation?  I looked, but couldn't anything at all on other planets eccentricity cycles.   The closest I came was this gravity simulation chart below.

90,000 year eccentricity chart of the 4 inner planets here,
and method used.  Source.

You may take a look at the lates parametrization file by JPL-NAIF for the precession, nutation and pole orientation of the largest known bodies.

Although, for the large time scales you are asking, I expect you will need to propagate the data and make your own wild guess, or dig into appropiate literature about solar system physics.


How does the field decay of a magnetar power the emission of high-energy electromagnetic radiation? In Space.com's Dr. Pulsar and Mr. Magnetar? 2 Star Types May Turn into Each Other (and in the linked video) "Tom Prince, a physics professor at the California Institute of Technology and a senior research scientist at NASA's Jet Propulsion Laboratory" is quoted as saying:

  First, magnetars don't last long — just a year to a few years, before colossal waves of X-rays dissipate the magnetic energy...

and Wikipedia's Magnetar says:

  A magnetar is a type of neutron star believed to have an extremely powerful magnetic field (∼1013 to 1015 G, ∼109 to 1011 T)1. The magnetic field decay powers the emission of high-energy electromagnetic radiation, particularly X-rays and gamma rays2.

  1https://arxiv.org/abs/1703.00068

  2Rare Earth: Why Complex Life Is Uncommon in the Universe. Springer. ISBN 0-387-98701-0.

Question: How does the field decay of a magnetar power the emission of high-energy electromagnetic radiation?

More specifically:

Is it as simple as $\partial \mathbf{B} / \partial t$ produces an electric field which accelerates charged particles that happen to be there, and then those accelerating particles radiate photons?
Does the energy radiated by these photons cary any significant fraction of the total energy in the field?
Is the word done producing these photons the reason that the magnetic field decays?

First, I feel like the article is confusing, so it would be better if you read peer-reviewed articles instead.

(If I recall correctly, Kaspi & Beloborodov 2017 per your reference discusses more towards observations. It might be easier to understand if you check theoretical papers instead).

  How does the field decay of a magnetar power the emission of high-energy electromagnetic radiation?

A young magnetar (i.e., strongly magnetized neutron star) has two main energy components: rotational (aka. spin) and magnetic. Rotational energy has shorter timescale as shown in Eq. 2, i.e. ~ one year (https://ui.adsabs.harvard.edu/abs/2010ApJ...717..245K/abstract). Magnetic energy would dominate later (see https://ui.adsabs.harvard.edu/abs/2012Ap%26SS.342...55G/abstract; note, I have never read this paper but I notice a good number at its abstract). Therefore, this gives you some sense that timescale matters. You might want to also check how a magnetar/pulsar links to the pulsar wind nebula, accelerated particles, and radiation (e.g., https://ui.adsabs.harvard.edu/abs/2014MNRAS.437..703M/abstract).

  Is it as simple as ∂/∂ produces an electric field which accelerates charged particles that happen to be there, and then those accelerating particles radiate photons?

Check Eq. 14 of the second paper. Simply yes. Note: not all particles radiate.

  Does the energy radiated by these photons cary any significant fraction of the total energy in the field?

I don't understand this question. If you mean the total radiation field, significant fraction of energy in the radiation depends on factors such as timescale.

  Is the word done producing these photons the reason that the magnetic field decays?

Not sure what you are referring to, but it sounds like a "not quite so" to me.

noted:

Pulsars are objects when observed we see pulse of light. Some pulsars are white dwarfs, while most of them are neutron stars.
The misalignment of rotation and magnetic-dipole axes + beam direction relative to observers are important.
While a young magnetar might produce strong jets, which can create the pulse, we might not observe the pulse because its environment does not support. You might want to check "choked jets" (https://arxiv.org/abs/1906.07399).


Why are explosions always used to represent the Big Bang? In documentaries the Big Bang is often represented as blackness followed by a big explosion with loud noises. This always feels cheap and falling short of explaining the origin of the Universe. Is this really the best way to get green-thumbs thinking about the origin of the Universe? Are there other, better, representations or analogies of the Big Bang? If so, how do they better describe the current cosmological model than the "explosion" analogy?

No, it's not the best reproduction of explaining the big bang. Generally when you think logically, it was an explosion of matter, however it was not a fireworks show.

To best explain it simply, it is conceived as a singularity that rapidly expanded. An important thing to understand is that within the beginning of the expansion, the matter was quark–gluon plasma and opaque, meaning there was no 'BANG'. Intensity of the heat was too great to allow quarks to form hadrons. Hadrons comprise of baryons such as protons and neutrons, at the earliest stage of inflation protons had not formed, being a representation of the 'big bang' they aren't implicitly going to introduce these factors. A general representation if the bang factor itself, if the bang is presented as an inflating balloon filled with gas, it's not going to attract. To be a true analogue, the balloon would simply appear into existence (between 10^-6 sec and 1 sec), already expanded and have pockets of matter now cool enough to begin producing light, within the next 10 seconds electrons begin to form attracting photons and neutrons, the first atomic bonds (hydrogen) would continue for the next 380,000 years. Much of the earlier matter has been annihilated by antimatter and we begin to see matter in the form it is today.

In a nutshell, it's hard to present this theory when something just 'becomes', so I believe they use the explosion as a way of demonstrating the nature of forces rather than an analogue of mere inflation.


When I see Mars, I see only red light I have a 700 mm focal and 76 mm aperture telescope. I bought it a few days ago. I tried to spot Mars for 3 days but I only see red light. I use an 20 mm eyepiece for finding it and 4 mm eyepiece for seeing but only red light is visible. Please help me.

Mmm... it's a pretty small scope using a fairly high power. The disc of Mars is not going to be very large (just a few seconds of arc across) so I don't think you should expect to see something that looks like a HST image! Mars is pretty red as a rule, so 'red light' is what you should be seeing because the disc is too small to show much detail.


What is the most dense object in the universe? Inspired by this answer to the question, Why is the Sun's density less than the inner planets?, what is the most dense object in the universe?

Let us define this as the largest observable density of a stable object, in order to exclude black holes which may have  a very large (infinite) density at their centers or objects collapsing towards a black hole status.

If we restrict the definition in this way, then the answer should be the core of the most massive neutron star that we know about.

At present there are a couple of neutron stars with mass of about $2M_{\odot}$ (Demorest et al. 2010; Antoniadis et al. 2013. Depending on the exact composition and equation of state at their centres these should have densities of around $2 \times 10^{18}$ kg/m$^{3}$ at their centers and average densities of $\sim 10^{18}$ kg/m$^3$.

Note that these densities are around 3 times the density of a proton or neutron or 5-10 times the density of nuclei at zero pressure.

In principle, the density of a single electron is much higher.


Fraction of initial mass lost (radiated) by neutron star mergers compared to black hole mergers? GW190521 black hole merger total mass calculation and missing mass, how does this happen? notes that there are about 9 solar masses missing from the final black hole.
GW170817 is the first observed merger of two neutron stars, detected in several ways including a weak gravitational wave.
Do neutron star mergers also radiate several percent of their mass as gravitational waves, or is the fraction much smaller. They comprise ordinary matter rather than being singularities in spacetime, so my guess is that the fraction is much smaller, but I have no idea.
My question is motivated by this answer.

Related:

"Who saw" the binary neutron star merger first? What was the sequence of events? (GRB/GW170817)
What does "GPU-accelerated butterfly matched filtering over dense bank of time-symmetric chirp-like templates" mean? (GW170817)

Dimensionally speaking, the luminosity of a gravitationally radiating binary system, consisting of two objects of mass $M$, separated by $R$, goes as $(M/R)^5$.  The timescale of the chirp for such a system goes as $M^{-3} R^4$. (Schutz 1999).
Thus the total energy released goes as $M^2/R$, i.e. it is proportional to the gravitational potential energy of the system.
Because of the $R^{-1}$ dependence, it is basically the mass and radius of the "final" state that determines the energy lost. For the black hole case, the final mass is just less than $2M$ and the final configuration event horizon is $4M$ (with $G=c=1$). So $M^2/R = M$. Thus I would expect the mass-energy released in gravitational waves to be a fixed fraction of the combined mass of the black holes (note that unequal mass black holes will lead to complications).
Looking at the data for the list of mergers this model seems reasonable, with the fixed fraction being about 5%.
Extending this to neutron stars, well the "final" radius is going to depend on the physics of the neutron star material and so will be model dependent. However, that radius will be $>4M$ (i.e. probably several times the Schwarzschild radius). Another way of saying this, is that neutron stars can't get as close together before the merger takes place.
So from that point of view I would expect $<$5% of the combined mass-energy is radiated as gravitational waves.
Observationally, there is no accurate estimate of the final remnant mass for GW170817.


Orbits using Newtons laws I am making a small orbit simulator. I figured out Kepler's laws and know how to work with them to "update" an orbit based on time. So now I want to know how I can describe orbits with newton laws. Given a position vector and a velocity vector of a satellite, What I currently do is this:

Calculate gravitational acceleration: (g*mass)/distance^2
Add this number to the velocity vector.
Multiply the new velocity vector with the DeltaTime
Add it to the position vector
Go back to 1.

However when converting this to Keplerian Orbit Elements doesn't really make them constant.(Calculations from here: Cartesian To Kepler)

Also the orbit isn't really stable, and also wobbles around. So can anyone help me fix this? Or am I doing something wrong?

I like to classify solutions of the problem of the time evolution of the
complete initial state of a set of objects at some epoch time, where the
objects are subject to Newtonian gravitation into two main groups.
One approach is to use orbital elements of some sort.
The other is to use a numerical initial value problem solver,
aka a numerical integrator. The latter is the primary subject of this answer.

Note well: This classification isn't quite perfect as hybrid approaches are
also possible, wherein one uses a numerical integrator to integrate
time-varying orbital elements.

Keplerian orbital elements work quite nicely in the case of two point
masses, or more generally, two objects with a spherical mass distribution.
The anomaly is the only Keplerian orbital element that changes over time.
Keplerian elements can be used in situations where the underlying
assumptions are approximately correct by developing a model of how
those supposedly unchanging elements vary with time. One way of doing
this is to use Lagrange's planetary equations. (There are other related
approaches such as Gauss' planetary equations, Delaunay's planetary
equations, etc.)
Lagrange's planetary equations yield expressions for how Keplerian orbital
elements vary over time given a set of perturbing forces.

Another approach is to use something akin to those Keplerian elements
such as Delaunay elements), coupled with planetary equations for those
alternative elements. Yet another approach is to use orbital elements
(e.g., Brouwer-Lyddane elements, SGP4 elements) in which the planetary
equations are embedded in the orbital element to Cartesian state
transformation algorithm. This final approach is used to this day to
describe vehicles in Earth orbit.

The other approach is to use numerical integration.

I'll start with a discussion of how to solve for the value of a scalar
function $x(t)$ at some time $t_1$ given an initial value $x(t_0) = x_0$
and some well-behaved (continuous and bounded) derivative function
$f(x(t),t) = dx(t)/dt$ that describes the time evolution of $x$.
This falls in the very broad category of initial value problems.

Suppose the ordinary differential equation cannot be solved analytically
and cannot be expressed terms of a useful power series. This doesn't mean
nothing can be done. There are a number of techniques for solving this
problem numerically.

Note the dependence of the derivative function on the dependent variable $x$.
This becomes the much simpler problem of numerical quadrature if the
derivative function can be expressed independent of $x$.
The discussion that follows assumes that the derivative function $f$ does
indeed depend on $x$.
Note well: Newtonian gravitation falls in this category.

It also assumes the derivative function is well-behaved.
Numerically integrating across a discontinuity is a bad idea.

The foundation of the integration-based techniques for a scalar function
is the mean value theorem, which says that at some time $t_c$ between
$t_0$ and $t_1$, the value at $t_1$ is exactly
$x(t_1) = x(t_0) + (t_1-t_0)\,f(x(t_c),t_c)$. If only we could find that
magical $t_c$ and the derivative at that point. There's a chicken and
egg problem here: that magical point in time is not known. Even if it was,
the derivative function depends on state, and that too isn't known.

A very simple approach around this problem is to assume that this magical
point is the initial point:
$$x(t_1) = x(t_0) + (t_1-t_0)f(x(t_0),t0)$$
This works quite nicely for values of $t_1$ that are very close to $t_0$.
It doesn't work very well at all where $|t_1 - t_0|$ is not small.
This suggests splitting the interval $(t_0, t_1)$ into a number of smaller
intervals. This results in Euler's method: Apply the above to advance
state to time $t_0+\Delta t$, then to $t_0+2\Delta t$, and so on, eventually
reaching the desired time. Euler's method is rather lousy, even for a simple
first order scalar ODE. We can do much better than this. The key reason for
discussing Euler's method is that it is the basis for many other integration
techniques. Learn how it works, then toss it.

One approach to approving on Euler's method is to somehow correct the
result from Euler's method. For example, take an Euler step and compute
the derivative at the end point. Then use the average of those two
derivative values (the original value used to make the Euler step, and
the other from the end of the Euler step) to recompute the step from
$t$ to $t+\Delta t$. THis is Heun's method.

Another approach is to guess that the magical point $t_c$ lies
somewhere between $t$ and $t+\Delta t$. Perhaps the middle? We can use
Euler's method to advance state to the midpoint, and then use the
derivative at that point to advance state from $t$ to $t+\Delta t$.
This is the midpoint method.

Both Heun's method and the midpoint method appear to be steps backwards,
computationally. While Euler's method requires but one evaluation of the
derivative function per time step, these improved methods require two.
However, the error growth is in general so much smaller with either
Heun's method or the midpoint method compared to that from Euler's method.
This means that those "improvements" most definitely are improvements.
The expense of calling the derivative function twice per step is more than
offset by the fact that imprpovements enables take steps that are orders
of magnitude larger than one can make with Euler's method.

Both Heun's method and the midpoint method are simple improvements.
This problem has been studied in many guises. There are many more advanced
techniques. One is the class of Runge-Kutta integration techniques.
Both Heun's method and the midpoint method fall into this class.
The most popular of these, classical Runge-Kutta 4, is a significant
improvement on those two methods. There are even higher order Runge-Kutta
integrators than RK4.
Heun's method also falls into the broad class of predictor-correctors,
wherein one method (the predictor) advances state to the end of the interval
and another method (the corrector) uses the derivative at this approximate
endpoint to correct the guess made by the predictor.

The above focused solely on first order ODEs involving a scalar function.
What if the problem is multidimensional or involves higher order derivatives?
The mathematics of the techniques described above can easily handle
multidimensional data: Simply use the vector-valued time derivative.
Since a higher order ODE can be converted to a first order ODE via an
augmented vector-valued state, the same approaches used to address
multidimensional data can also be employed to address higher order ODEs.

There's a problem with doing this: It throws out geometry. For example,
consider the rather simple first order ODE $\dot x = -y, \dot y = x$.
The solution to this multivariate ODE is uniform circular motion.
Applying Euler's method to this results in
$$\begin{aligned}
x(t+\Delta t) &= x(t) - \Delta t\,y(t) \\
y(t+\Delta t) &= y(t) + \Delta t\,x(t)
\end{aligned}$$
The square magnitude of this new vector is
$(x(t)^2+y(t)^2)(1+\Delta t^2)$, which is always greater than magnitude
of the vector at the start of the step. This is not uniform circular motion.
The solution obtained via Euler's method instead spirals out. Other techniques
spiral inward. A geometric integrator on the other hand will somehow
maintain the constraint that $x^2+y^2$ is a constant of motion.

The above example showed why we don't want to toss geometry in a very
simple problem. The geometry of Newtonian gravitation, along with much
classical mechanics in general, is symplectic geometry.
This is why symplectic integrators are of great concern. A simple example
again, with Euler's method: Suppose the second derivative of position
is given by some function $\ddot {\vec x}(t) = \vec f(x(t),t)$.
Applying the basic Euler method against 3+3 dimensional phase space
dictates that
$$\begin{aligned}
\vec x(t+\Delta t) &= \vec x(t) + \Delta t \vec v(t) \\
\vec v(t+\Delta t) &= \vec v(t) + \Delta t \vec f(\vec x(t),t)
\end{aligned}$$
A simple change makes this symplectic:
$$\begin{aligned}
\vec v(t+\Delta t) &= \vec v(t) + \Delta t \vec f(\vec x(t),t) \\
\vec x(t+\Delta t) &= \vec x(t) + \Delta t \vec v(t+\Delta t)
\end{aligned}$$
As is the case with the scalar techniques discussed at the start,
Euler's method is a starting point rather than the end with
regard to symplectic integration techniques. Symplectic Euler's method
is rather lousy. But at least orbits don't spiral outward.

In making an N-body gravitation simulation, the size of N (the number of
bodies) is a key concern. Simulating a galaxy is a very different concern
from simulating a star system. The techniques used in simulating the
formation of a galaxy are very different from those used to develop a
solar system ephemeris. Galactic scale simulations cannot afford to
calculate all of the N2 gravitational interactions amongst all
the particles, and because N is so large, it cannot afford anything
more complex than very simple integrators. A star system model that does
not calculate all N2 of the gravitational interactions
or that uses very a simple integrator will by viewed in disdain.


Information on Planetary nebula formed from the death of our star the 'Sun'! I want to know the expansion rate, composition, mass and density of hydrogen present of a Planetary nebula similar to that which will be formed after the death of our star the sun.

Knowing these data is important for me to do some calculations.

A Study to Show : Jupiter : Our Next Star
My 11th grade research paper 2018
Answers under -
Planetary nebula, pg 8-9
Total mass of the planetary nebula formed
= 0.49 M⊙ (99% hydrogen)
Revisiting mass loss, pg 12
Density of CSE1 on reaching Jupiter’s Orbit after Thermal Pulse 1, pg 17


Tidal influences of planets on the sun How big are the tidal influences of planets (say, Jupiter) on the sun? Do they lead to any measurable effects, for example, do they influence the sunspots, or are they observable in any other way?

The basic tidal acceleration felt by a test mass near the surface of the Sun, due to a body of mass $m$ at a distance $r$ is given by
$$a_{\rm tidal} = 2\frac{Gm R}{r^3},$$
where $R$ is the radius of the Sun.

You can work out that Jupiter has the largest tidal effect on the Sun, very closely followed by Venus, and both produce tidal forces that are only a factor of $\sim 2.2$ times greater than that produced by the Earth on the Sun.

The distortion of the Sun could be modelled as following the consequent equipotentials (well inside the Roche lobe) in a binary system, but this is not analytically solvable and I could not locate any straightforward approximations. One way to judge the size of the effect would be to compare the tidal acceleration with the centrifugal acceleration at the Sun's surface $R\omega_{\rm Sun}^2$.

So $a_{\rm tidal}$ for Jupiter is $3.7\times10^{-10}$ m/s$^2$, whereas the centrifugal acceleration (at the solar equator) is 7 orders of magnitude larger at 0.006 m/s$^2$. Given that the Sun is almost spherical to 1 part in 100,000, because the centrifugal acceleration is only 20 millionths of the Sun's surface gravity, we can safely assume that tides due to planets that are 7 orders of magnitude weaker do not distort the shape of the Sun to any degree worth worrying about.

I was surprised to find that the potential influence of planetary tides on solar activity is still an area of recent research. Suggestions that planetary tides might be responsible for sunspots or the sunspot cycle on the Sun go back a long way. Grandpierre (1996) suggested that planetary tides influence flows in the Sun and could be responsible for the 11-year solar cycle. A NASA report by Hung (2007), although not peer-reviewed, does suggest a link between solar activity and planetary positions. The claim is made that major solar flares happen "underneath" the locations of the major tide-inducing planets (Jupiter, Venus, Earth, Mercury [in that order]).

I very quickly found a peer-reviewed paper in "Solar Physics" by Seker (2013), which discusses all this and reviews the evidence on sunspots. They conclude that the appearance of new sunspots is not correlated at all with the positions of the planets. However, they do raise the caveat that because the magnetic fields producing the spots are produced at the tachocline between the radiative core and convective envelope of the Sun and take several days to buoyantly rise. Thus the longitude at which the spots appear and where they are generated, with respect to the positions of the planets, can change significantly.

I guess the jury is still out, but this is quite "fringe" material. The solar cycle is certainly thought to be a product of the dynamo mechanism that produces the magnetic field. However, the location of sunspots definitely cannot be predicted with such models at the moment. Interest in this topic has been revived recently by claims of correlations between magnetic activity and the orbital positions of hot Jupiter-type exoplanets and even the suggestion that tidal effects or interactions between stellar and planetary magnetic fields might be responsible for increased magnetic activity (see Lanza 2014 for a recent review). For these close-in exoplanets of course the tides are many orders of magnitude larger than in the solar system (see equation above).


Can objects enter hydrostatic equillibrium through processes other than the influence of gravity? Wouldn't a droplet of liquid (say water) floating around in space become spherical and "in hydrostatic equilibrium" due to surface tension but not gravity?

Strictly speaking (as far as I know), hydrostatic equilibrium applies whenever a fluid balances external body forces with the pressure gradient.  From Wikipedia:

  In continuum mechanics, a fluid is said to be in hydrostatic equilibrium or hydrostatic balance when it is at rest, or when the flow velocity at each point is constant over time. This occurs when external forces such as gravity are balanced by a pressure gradient force.

I think the concept happens to be most frequently used in areas where gravity is the external force, but it could in principle be anything else. So, though I stand to be corrected, I think a droplet isolated in space long enough could be said to be in hydrostatic equilibrium, even though the most relevant force is the surface tension, rather than gravity.