Strange square-like formation on Mars Recently I've explored an area on Mars' surface which one looks like huge hatch cap & hole, about 10 kilometers, between tons of circular craters.

The formation is very unique and strange to me. Somehow it has moved from original location and broke into two pieces then half part disappeared.

What do you think about it?

Google Maps link(25.3717187,166.6337854)

I looked at the image and annotated it based on what it most likely is:

In black is the formation itself.
In blue is a depression underneath it.
In grey is part of the boundary of what you believed to be the disappeared half.
In green is a ridge to the north.
In red are several craters.

Here's a more likely hypothesis for what this actually is.
The depression
The blue oval contains a slight depression. It could be a number of things, including an old crater, a basin, or the result of the collapse of part of the surface. I'm actually a little bothered by the idea of it as an impact crater. We can see from the shadows on the western walls that it is only a little shallow, especially compared to other similarly-sized impact craters in the area. It's possible that it's a crater that was filled in with sediment.
If it's a natural basin, it could once have been a Martian lake - although this is quite speculative. It could also be the remains of a pit crater (there are many pit craters in the area), formed possibly from the cave-in of a series of lava tubes. Again, this is also speculative, and the depression is quite large, about 10 miles long on its major axis.
If one of these options is correct, the depression is not very mysterious. You've identified the grey part of it as the location of the other half of the formation, partly because it appears to be roughly the same size. I find this unlikely, for a couple reasons:

The ridges look similar in height to the edge of the depression on the other side of the formation, indicating that it is actually a part of that feature.
Half of an outcropping does not simply vanish by itself. I know of no mechanism for this to happen. Erosion could take its toll over time, and perhaps that caused weathering of the formation, but it couldn't move it.

I think it would be clearer that this is actually part of the depression if the ridge enclosed in green were not there.
The mesa
So if we assume that the outcropping is all that there even was of the object, we simply have to explain why it's so square. This really isn't that mysterious. There are plenty of other mesas in this cartographical area of Mars, designated the Elysium quadrangle by the USGS for mapping purposes. Here are some examples, which are similarly shaped:

The craters
Finally, we have the three large craters. They aren't too mysterious; craters are normal on Mars, and there are plenty, of varying sizes, in the Elysium quadrangle. The one reason the smallest stands out is because it is near the center of the northeast edge of the mesa. I'm reasonably certain this is just a coincidence.

Here's my hypothetical and likely not entirely correct timeline for this formation:

The large basin forms, from either a crater or some other process.
The mesa and ridge form, possibly as a result of the erosion of some larger feature.
Craters form through impacts. One happens to be on the mesa itself.

I think your identification is simply a case of pareidolia, which is the human tendency to find shapes and patterns where none really exist. A well-known Martian example is the Face of Mars.


How many stars would it take to draw a line across the middle of the sky that appeared solid? I was talking with a friend about how slowly the star field changes (based on the speed that we are moving through the galaxy) and I started to wonder about a star's visible size.  They are basically the pixels that make up our sky.  This made me wonder, how many stars (let's use the north star as a reference.) would it take (lined up side by side) to draw line that appeared solid across the middle of the sky, perpendicular to the horizon which appeared solid?

This is more of a question of human perception, than astronomy. I was going to answer your question with this: "One, it just needs to be close to Earth." But, I decided it wasn't THAT funny. Anyway, stars are essentially point sources as far as our eyes are concerned. Typical visual resolution is about 0.02° or 0.0003 radians. Assuming from horizon to horizon is 180° (or π radians) that calculates out to roughly 10,000 stars. I'd probably increase that by 50% or 100% to be sure. You do understand that you can't line stars up side by side, I hope. There's no need, its about the angular distance, not absolute distance between them, that matters. They can be light years apart as long as they appear to be within about 1 arcminute of one another, our eyes will see them as a single object - subject to the psychological aspects of keeping color and apparent magnitude roughly uniform as well.


Asteroid locations at the time of the crucifixion of Jesus I am interested in learning the locations and velocities of the near earth asteroids at the time of the crucifixion of Jesus Christ.
 I would like to back-calculate the paths of all known near earth asteroids to find out if any was between the earth and the sun at the time of Jesus' crucifixion on Good Friday.  The Horizons system of NASA JPL only permits back calculations to about 1400AD.

I want to see if there was a clustering of asteroids on Good Friday between the earth and sun.  If so they could be the remnants of a gigantic collision that would have been capable of making a dust cloud in interplanetary space that could have produced a shadow on the entire earth which would account for the observations reported by the gospels and the Chinese astronomers and St. Dionysius and others.

In other words if any asteroids are calculated to have been between the earth and sun on Good Friday they may be left-over debris from the possible collision that produced the dust cloud.

My question is this.  Do you know of any available computer program like the JPL  Horizons program that will work to give asteroid positions back in the time period between 29AD and 35AD?  (I give a range of years because biblical scholars give a range of possible years for the Crucifixion of Jesus.  The determination of a debris asteroid would help historians know the timing of the Crucifixion better, also.)
Thank you for your attention and assistance.
Best Regards,
Dr. Francis Kelly
President, The Catholic Association of Scientists and Engineers

The computer program you are asking for is literally impossible:  Even ignoring the affect of the greater universe and transient visitors such as comets on hyperbolic orbits, the Solar system is a chaotic complex system.  We have a pretty good handle on the motion of the bodies we are, and have been aware of, during the time modern astronomy has been studying the skies, i.e., over the past few hundred years.  (Remember that in the earlier years of telescopic astronomy, we were unaware of most of the currently known members of the Solar system, including all of the asteroids, and every massive object beyond Saturn.)  As you deviate further from the present (either past or future), the chaotic nature of the system comes more into play, especially on smaller bodies such as asteroids.  As a result, stating orbital positions either far in the past or far in the future requires a larger margin of error as the time differential grows.  Eventually you get to a point where the error is so great that we cannot say reliably where an object was or will be.  Perhaps if someone had come up with the laws of planetary motion sooner than Kepler, we would have more reliable data further into the past, and therefore more accurate orbital elements over a longer range than we do, but that didn't happen.

The asteroid belt (white) and Jupiter's trojan asteroids (green)

From Wikipedia, linked from the Asteroid page

In addition to the chaotic nature of the Solar system, there's also the fact that the vast majority of the asteroids revolve around the Sun outside of the Earth's orbit, so the chances of any asteroids we know of having "a gigantic collision ... that could have produced a shadow on the entire Earth" (i.e., between the Earth and the Sun) are between slim and none.  (Note also that the amount of material present in said "dust cloud" would have to be immense in order to have a noticeable affect on the amount of sunlight falling on the Earth.)  The only way such a collision could have occurred would be if two objects traveling through the Solar system happened to collide in such a manner that the "dust cloud" the collision produced happened to fall between the Earth and Sun - and also left the Solar system without a trace afterwards.  This would have required the objects being in similar trajectories crossing the plane of the ecliptic at a high angle and speed - a chaotic event if there ever was one!

So, basically you're looking for a way to prove that a chaotic event happened sufficiently far in the past that chaotic errors make our knowledge of most of the Solar system (asteroids and such) and outside influnences questionable - no, that's not going to happen.  Math doesn't work that way.


How to find star location knowing latitude and LST Newbie learning Astronomy and was curious how to locate a star giving a Latitude and LST.  For example if my LST is 4:00 and my latitude is 20*N, how would I find where the Horsehead Nebula appears in the sky?  The only formula I can think of is the Hour Angle formula..

Welcome !. I think you still need the co-ordinates of the star that you are interested in, its Right Ascension (the sky equivalent of longitude on the Earth) and the Declination (equivalent to latitude). Then as you say, you can use the hour angle (HA) formula:

HA = LST - RA

The Hour Angle tells you how many hours west or east of the meridian your star is (and can be further converted into an azimuth).

For the altitude of the star, you need a bit of trig:

altitude = arcsin(sin(latitude) * sin(declination) + cos(latitude) * cos(latitude) * cos(HA))


Where was this photo of Nancy Grace Roman taken, what is the display shown? Where was this photo of Nancy Grace Roman taken, what is the display shown, is there any technical information on this display and where is it now? If this is a NASA photo then there should be an original ID or catalog number and hopefully explanation, but it may not be a NASA photo.
I spotted this in Nancy Roman Telescope’s Primary 2.4-Meter Mirror is Ready when looking for background information related to the question What “improvements in technology” allowed the primary mirror of RST (WFIRST) to be less than 1/4 the weight of Hubble's?

Source which links to this higher resolution version

See also

Hubblecast 113a
YouTube: Hubblecast 113: Nancy Roman — The mother of Hubble

The display is shown when the narration says:

Nancy followed her dream and became a professional astronomer, working in the new discipline of radio astronomy.  And in 1961 she became the first chief of astronomy in the office of Space Science at NASA.

I'm guessing that this display is for the position and other information for a radio telescope, but if so, which one?

Her Wikipedia page also includes this picture (public domain), where she wears the same dress, and where a similar display can be seen in the background:

Roman sits at the control console for the Orbiting Astronomical Observtory satellite, launched in 1972 and nicknamed Copernicus. This is a publicity picture; she never actually worked in the Goddard control room.

This page from the University of Chicago also includes a picture of OAO-3 (Copernicus) control room, where a similar display can be seen (I won't paste the image here because of copyright). However, after careful examination, they look similar, but the layout is actually not the same.
This page from the University of Wisconsin has a (poor) picture of OAO-2 (Stargazer) control room (again, not pasted here), where a similar display can be seen. Despite the poor quality, this one seems to fit the one where Nancy Grace Roman is featured.
From these evidence, I'd say:

It is definitely a display from an Orbiting Astronomical Observatory, either OAO-2 (Stargazer) or OAO-3 (Copernicus).
At least one of the three sites is wrong in its picture caption—I'd guess Wikipedia rather than the university sites, which would make OAO-2 the "winner".


What would a kHz gravitational wave detector look like? (mountains on millisecond magnetars) How would it differ from LIGO/Virgo? The abstract of Gravitational waves from mountains in newly born millisecond magnetars says:

In this paper we study the spin-evolution and gravitational-wave luminosity of a newly born millisecond magnetar, formed either after the collapse of a massive star or after the merger of two neutron stars. In both cases we consider the effect of fallback accretion, and consider the evolution of the system due to the different torques acting on the star, namely the spin up torque due to accretion and spin-down torques due to magnetic dipole radiation, neutrino emission, and gravitational wave emission linked to the formation of a `mountain' on the accretion poles. Initially the spin period is mostly affected by the dipole radiation, but at later times accretion spin the star up rapidly. We find that a magnetar formed after the collapse of a massive star can accrete up to 1 $M_{\odot}$ , and survive on the order of 50 s before collapsing to a black hole. The gravitational wave strain, for an object located at 1 Mpc, is $h_c \sim 10^{-23}$ at kHz frequencies, making this a potential target for next generation ground based detectors. A magnetar formed after a binary neutron star merger, on the other hand, accretes at the most 0.2 $M_{\odot}$, and emits gravitational waves with a lower maximum strain of the order of $h_c \sim 10^{-24}$ , but also survives for much longer times, and may possibly be associated with the X-ray plateau observed in the light curve of a number of short gamma-ray burst.

Question: What would a kHz gravitational wave detector look like? (mountains on millisecond magnetars) How would it differ from LIGO/Virgo?

Figure 1. Pictorial description of the millisecond magnetar. There are two coordinate systems, one having the rotation axis ($\Omega$) and another having the magnetic moment ($\overrightarrow{\mu}$) axis. These axes are inclined at an angle $\alpha(t)$. The blue dots show matter falling on the two polar caps and forming two accreting columns. As the star rotates, it radiates energy in dipolar radiation and grav- itational waves. The red dotted lines represent the escaping neutrinos which carry away heat and angular momentum in the form of a wind.

In principle, not much different to LIGO, since LIGO is sensitive to kHz gravitational waves. But you need to find ways to increase the sensitivity at kHz frequencies, without compromising response at lower frequencies, or perhaps having a tuneable response. These improvements are likely to come with increased laser power and developments in the optics and mirror coatings.
Detecting gravitational waves (GWs) is limited at high frequencies by the length (or effective length, with some cunning deployment of mirrors) and at low frequencies by various sources of noise, including "radiation pressure noise" which is an inescapable consequence of using powerful lasers to do the detection.
Roughly speaking, to get maximum sensitivity at 1kHz (LIGO is most sensitive at 100Hz), you make your interferometer arm length equal to a quarter of the GW wavelength (or the total path length of the light to be $\lambda_{\rm GW}/2$). This ensures that perturbations to the arm length all "sum up" whilst the laser light is in transit. Thus for a GW frequency of 1KHz that means $L \simeq 75$ km (or a total path of 150 km).
That's fine - something of LIGO's size can be made effectively that long by bouncing the light up and down the arms about 10 times, which is much less than the $\sim 270$ times that LIGO uses in order to get sensitivity at lower frequencies.
Sources of noise at 1kHz are mostly limited to simple "shot noise" due to the finite number of photons in the laser beam. So you would need to boost the power at the beam splitter by using a more powerful laser to begin with or by employing clever recycling techniques to amplify that power within the interferometer arms.
In LIGO, they use a laser with a power of about 40W, split between two arms. The 4km arms form Fabry Perot resonators that boost this by a factor of 270, but that isn't the power at the beam splitter. From the beam splitter's point of view, the FP arms are about 0.97 times reflective. To amplify the power at the beam splitter you put a recycling mirror prior to the beam splitter, with a reflectivity of 0.97. This allows all power to transmit into the FP arm cavity, but also amplifies the power between the beam splitter and the FP input mirror by a factor of about 35.
At present LIGO is tuned to have a high amplification factor in the FP arms and a modest power recycling gain, which is limited by the effective reflectivity of the FP cavity due to losses at the reflective surfaces.
To get more power at the beam splitter, whilst maintaining sensitivity at 100 Hz, would need big improvements in the mirror and mirror coatings technology; improvements in laser stability at high power and the clever used of "squeezed vacuum" techniques which can to some extent shape the noise frequency profile, allowing modest improvements in the shot noise at the expense of radiation pressure noise at lower frequencies. For instance, by minimising losses in the mirrors, one could make the FP arms more reflective, which allows a bigger recycling gain (which goes as $(1-R)^{-1}$).
Even without these improvements, you could tune LIGO to work better at 1kHz, at the expense of its response at lower frequencies. The gain of 270 is not realised at high frequencies. You could drop this to around 20, which would increase the reflectivity of the FP arms  to about 0.997 by reducing the losses (which increase with how many times the light bounces around) and could then use a more reflective power recycling mirror and get a power recycling boost of 330. This would gain you a factor of $\sqrt{10}$ in SNR at 1kHz at the expense of losing a factor of 10 sensitivity at 100Hz.
There is lots more detail in the technical documentation for the Einstein Telescope. This proposed third generation GW detector operates on the same principle as LIGO, but with improvements. It will have several "V" shaped interferometers to improve its directional response and remove dead spots; it will use more powerful lasers; it will operate cryogenically to improve sensitivity at low frequencies; it will have 2 interferometers within each set of arms, tuned for low and high frequencies respectively, as I've indicated above; it will have 10 km arms and use more massive mirrors to reduce the effects of radiation pressure and other displacement noise at low frequencies; and it is proposing big developments in optics technology (e.g. mirror losses at 1 ppm) to boost power and get high frequency response.


Launch windows for Venus transit; how to convert heliocentric longitudes to date? Being a bit familiar with celestial mechanics, I know that the Hohmann transfer orbit is the quickest way to transfer between two circular orbits of different radii around a central body in the same plane. JPL's Let's Go to Mars! Calculating Launch Windows explains the geometry and offers a very intuitive guide how to calculate the travel time to Mars, and this can easily be repeated for Venus.
Just for fun, I would like to mark all launch windows for a travel to Venus on a calendar, similar to the list compiled by Don P. Mitchell
1991, Jun 5     Unused Type I window
1993, Jan 5     Unused Type I window
1994, Aug 9     Unused Type I window
1996, Mar 20    Unused Type I window
1997, Oct 15    Cassini flyby

Question: I have a probably minor (astronomical) issue: I am struggling how to convert heliocentric longitudes into real dates? I guess I am missing the English terms here. I would appreciate a small reminder how to proceed.

Heliocentric longitude is a direction, not a date.
A Heliocentric longitude of 0 is the direction from the sun towards the vernal equinox (currently in Pices)
You can ask for the date on which the Earth is at a particular Heliocentric longitude.  A careful calculation would require accounting for the eccentricity of the Earth's orbit, and also adjusting for the oddities of the calendar, but since when the Sun in on the vernal equinox (About March 21), the Earth will be at 180 degrees of heliocentric longitude, and longitude will increase by 360/365.24 degrees per day.


Looking for a routine to correct for Local Standard of Rest (LSR) I've been scanning the Milky Way with a Small Single Dish Radio Telescope, obtaining Spectral data with a Span of ~ 2 MHz, centered at the 21 cm line. With this information I can derive a relative velocity (due to Doppler shifted emission) between the observer (telescope, at lat ~ 33° South) and gas clouds in the line of sight. To study the position of this clouds relative to the Galactic Center, I know that the way to correct for this is subtracting our motion relative to this 'fixed' point: correcting by Earth's rotation, Earth motion on the Solar System plane and the System's motion itself (everything projected onto the line of sight), so my data is comparable to any other observation.
I've been reading on related questions such as

The Solar Motion and the peculiar velocities of stars

What are galactic speeds measured against?

How spectrographs that measure radial velocities manage to translate variations in the stars' spectrum lines into the “speed” of the star?

I also looked on "Tools of Radio Astronomy", T.L Wilson, K. Rohlfs, S. Hüttemeister, 5ed., pg. 189. that briefly describes this procedure.

The thing is I can't find any routine explained in detail, about this:

So I wonder how this is really done in practice and if there's any, e.g., Python Package that does it (maybe in Astropy?).
EDIT: Thanks to @ELNJ 's response and help I managed to get agreeable results. This is data from a Standard Region, particularly S9.
def Freq_to_v(freqval, freq0, mjd_time = 59244.8, obs_ra = 268.1, obs_dec = -34.43):

    #https://docs.astropy.org/en/stable/api/astropy.coordinates.SkyCoord.html#astropy.coordinates.SkyCoord.radial_velocity_correction

    #This is the observation for S9, 30-01-2021, 16:34 hrs CL.
    #S9 Std. Reg. Obs. time = 59244.8

    t = Time(mjd_time, format='mjd', scale = 'utc')

    #Obs. Location and height in GEODETIC coordinates (and not geocentric a X Y Z quantities tuple!!).

    loc = EarthLocation(lon= 289.47, lat =-33.27, height = 1450)

    sc = SkyCoord(1*u.deg, 2*u.deg)

    vcorr = sc.radial_velocity_correction(kind='barycentric', obstime=t, location=loc).to('km/s')
    #####################################################

    #Regular Doppler Shift Calculation. Here are two conventions: optical and radio doppler shift.

    shift_opt, shift_rad = (freqval - freq0)/freqval, (freq0-freqval)/freq0 #shifting factor, no dimensions.

    v_opt, v_rad = const.c.to('km/s')* shift_opt , const.c.to('km/s') * shift_rad #velocities.

    #################################################### To barycentric:

    rv = v_rad - vcorr #corrected to barycentric w/o Special Relativity terms.

    #This is the observation for S9 Std, Region., 30-01-2021, 16:34 hrs CL.

    ################################################### To LSR:

    my_observation = ICRS(ra=obs_ra *u.deg, dec= obs_dec *u.deg, \
     pm_ra_cosdec=0*u.mas/u.yr, pm_dec=0*u.mas/u.yr, \
     radial_velocity= rv, distance = 1*u.pc)

    new_rv = my_observation.transform_to(LSR()).radial_velocity

    return new_rv

You can compare Williams' 1972 result

with mine:

and check that it's in fact consistent.

There are several corrections here, and as you suspect, all of them can be done in astropy.  You'll need to do something like this:

Transform your observed velocity to the solar system barycenter (thus correcting for Earth's rotation on its axis, and its orbit around the Sun.  There is an example here in the astropy docs that shows how to do that.  If you follow that example, you'll first need to define your radial velocity as an astropy "quantity" object with units, e.g. something like this:

import astropy.units as u
# Define RV of 30 km/s
rv = 30 * u.km / u.s

You can ignore the discussion in that example of the relativistic correction, and just add the barycentric correction to your velocity, e.g. corrected_rv = rv + vcorr

Then, you can correct for the fact that the Sun isn't moving quite in step with the average of the nearby stars, and convert from the barycentric radial velocity to the local standard of rest, or LSR.  Astropy has some examples here but they don't do anything quite like this, so here's some example code.

Set up an observation that has the coordinates you observed, and the corrected radial velocity from step 1.  (I'm continuing with 30 km/s here, but you would substitute whatever you got in step 1.)
from astropy.coordinates import ICRS, LSR
my_observation = ICRS(ra=17.75*15*u.deg, dec=-29*u.deg, \
     pm_ra_cosdec=0*u.mas/u.yr, pm_dec=0*u.mas/u.yr, \
     radial_velocity=30*u.km/u.s, distance = 1*u.pc)

For specificity, I've chosen coordinates that are toward the Galactic Center. Here we're specifying zero for the proper motion terms since astropy won't do the velocity transformation if it doesn't have all three velocity components; that's also why we need the distance.  (You can experiment and convince yourself that the values don't matter in terms of what you get for the radial velocity.)  Now, you can transform that to the LSR frame and print out the new radial velocity:
new_rv = my_observation.transform_to(LSR()).radial_velocity
print(new_rv)

which gives 41 km/s, since the motion of the Sun relative to the LSR has a component of 11 km/s (the "V" component in the UVW system), for example in this derivation of the solar motion from Gaia data.


Seeing comet NEOWISE? Today the NY Times published an article that explained how to see NEOWISE at night.
The article explains it like this:

To catch NEOWISE yourself, look up at the northwest skies about an hour and a half after sunset. Experts suggest going to the darkest area you can for best viewing. Find the Big Dipper and follow its ladle as it arcs in the direction of the horizon.

NEOWISE will appear under the Big Dipper about 10 degrees above the horizon and be about as bright as that constellation’s stars. If you hold out your arm, 10 degrees is roughly the part of the sky covered by your fist. Over the next few days, NEOWISE will move higher in the sky and be easier to spot, reaching its apex on July 23, when it makes its closest approach to Earth.

It says that I'll be able to see the comet "under the Big Dipper about 10 degrees above the horizon".
What "horizon" is the article referring to? It says above the horizon and under the Big Dipper. Is the Big Dipper just above where the sun disappears and the comet will be seen in that area?

Horizon
The horizon is Earth's horizon ... absent any obstructions such as buildings, trees, etc.  Or another way to think of it is the horizon you would observe looking out across the ocean while standing on a beach.
The notion that it is "10° above the horizon" means that from the horizon line, it would be 10° up.  You closed fist (as if you are about to punch someone) measure about 10° (from your thumb to your pinky finger) if held at arm's-length.  Close your fist and extend your arm ... placing the bottom (the edge along the side with your pinky finger) on the horizon.  10° up is roughly where the knuckle of your thumb is located.
Caveats:  Time and Date
But a couple of cautions should be mentioned regarding the article...
Objects in our Solar System move from night to night and this motion can sometimes be noticed even from hour to hour.   Tonight (July 21, 2020) it is still below the "bowl" section of the Big Dipper asterism.  But in a few days that will no longer be true.  One week from today it will no longer appear "below" any part of the Big Dipper asterism.  It is heading through the Coma Berenices constellation.
Also, the 10° measurement... this assumes both a time in the evening AND a latitude on Earth.
I'm located at roughly 42° North latitude.  At 10:30pm for me and tonight (because time and date both matter) the comet is 22° above the horizon.  But if I wait an hour and check 11:30pm ... comet is only 14° above the horizon.  This is because the comet is setting ... sort of.  Technically if you are far enough North the comet doesn't "set" below the horizon.  Since I am located at roughly 42° North, anything with 42° of the celestial pole (near the North Star) doesn't actually "set" below the horizon.  It will be up all day and all night.  But as dawn approaches and the twilight brightens, the sky glow will wash out the comet until it is no longer possible to view from any "daytime" part of the planet.
Cheating
Rather than trying to imagine where the comet is, there's a much easier way to just nail it.
Technology has come a LONG way over the years.  These days, you can get an app for your smartphone that will tie into the GPS, compass, and level/tilt/gyro sensors on your phone.  It's amazing how well this stuff works.  You hold the phone up to the sky, and it knows everything in the sky that should be visible from your current location and in that direction.  It will literally draw the sky and show you the precise location of the comet.


Need help with a simple Blackhole question Suppose the amount of mass in a black hole doubles. Does the event horizon change? If so, how much does it change?

Let's have a look at the Physical_properties subsection of the Wikipedia article on Black holes:

The Schwarzschild radius which is defined only for a non-rotating black hole is given there as

$$r_S=\frac{2GM}{c^2}\approx 2.95 \frac{M}{M_{Sun}} \text{km}.$$

  The Schwarzschild radius was named after the German astronomer Karl Schwarzschild, who calculated this exact solution for the theory of general relativity in 1916.

So the Schwarzschild radius increases linearly with mass, by 2.95 kilometers for every solar mass you add, provided you add it carefully and don't spill any.

Trying to express the "size" of the event horizon for an arbitrary black hole is going to be more complicated, so I'm not going to adventure into that territory. However, @Chappo's comment suggests the dependence on mass will also be approximately linear as well.


What percentage of the hydrogen today has never been in a star It stands to reason that some of the hydrogen and helium that formed directly as a product of the big bang might never have fallen into a star to re-ejected when that star explodes.  My question is, given the best theory, what percentage of that matter has managed to drift without being sucked into a star.  Do we have any idea?

About 70% of the baryonic matter in the universe is hydrogen, with a mean density of about $4\times 10^{-29}$ kg/m$^3$.

Most of the stars that have ever been born are still alive, since an average star is only about $0.25 M_{\odot}$ and has a lifetime much longer than the age of the universe (so very little material has actually been recycled).

If we assume there are $10^{22}$ stars of $0.25 M_{\odot}$ in an observable universe of radius 47 billion light years, that are 70% H by mass, the "stellar" hydrogen is only one part in 73.

So, there is only one hydrogen nucleus (a proton) in a star for every 73 in the universe. This ratio would have been smaller in the past (E.g. when theSun was born). But as I mentioned, most of this hydrogen (about 90%) is in stars that are longer lived than the universe. Therefore my very rough estimate is that around 1 hydrogen atom in every 1000 on Earth has been inside a star. This contrasts markedly with say that 100% of carbon and oxygen atoms have been inside a star.

EDIT: To be fair, this calculation hinges a lot on how many stars there are in the observable universe. This number is very uncertain and could be higher - perhaps $10^{23}$ (see here), in which case my numbers are somewhat pessimistic and it might be more like 1 H atom in 7 is inside a star and 1 H atom in about 100 on Earth were inside a star. However, I don't think there is any argument that the majority of hydrogen in the universe is not, and has never been, in a star, but whether that is 90% or 99% is still moot.


Is there a type of star that emits relatively monochromatic visible light? I was thinking about this question on Worldbuilding and the answers involving monochromatic ambient light, which got me wondering if there were any stars that emitted relative monochromatic light (a "very" narrow band, although I don't have a definition, but something narrow enough to produce the effect in that photo, atmospheric effects notwithstanding).

So my question is: Is there any type of star that emits an extremely narrow band of visible light (not necessarily exclusively visible, non-visible wavelengths can be included) that we have either observed or theorized to exist? "Narrow" as in, on par with a low pressure sodium-vapor lamp.

My Google searches are hindered by a lot of noise, mostly things like monochromatic star patterns, and a few hits regarding non-visible radiation.

In some sense, yes.
UV continuum $\rightarrow$ UV line
This is probably not the answer you're looking for, but massive stars (O and B stars) emit a very hard UV spectrum of light (a UV "continuum", i.e. a broad range of wavelengths). Since these stars don't live for long (because they burn their fuel fast), they tend to be located in the gas clouds from which they were born. The UV light ionizes the enshrouding hydrogen, but the protons and electrons quickly recombine. If the electron goes directly to the ground state, another UV photon is emitted, capable of ionizing another hydrogen atom. In most cases, however, the electron "cascades" down multiple levels, emitting photons of different energies, which may excite other atoms.
It turns out that for every 3 ionizing photons emitted by the star, 2 will eventually — after several interactions — result in the photon corresponding to the energy difference between the first excited state and the ground state; the so-called "Lyman $\alpha$" photon, with a wavelength of 1216 Ångström (121.6 nm). Although there is some broadening due to thermal motion of the atoms and, in particular, due to the resonant scattering of Lyman $\alpha$ on neutral hydrogen, the result is that most of the light from these stars is converted into a single, very narrow (of the order of a few Ångström) emission line, i.e. very close to monochromatic light.
These stars are probably rarely surrounded by planets (because the radiation pressure will tend to blow away the particles used to build planets), and even if they were, these processes happen farther away from the stars than the planets would be. But if you observe a young galaxy, whose spectrum is dominated by young stars, the Lyman $\alpha$ emission line is often the only light visible.
UV line $\rightarrow$ visible line
The Lyman $\alpha$ line is still in the UV, and thus invisible to humans. However, since light is redshifted as it travels through the expanding Universe, Lyman $\alpha$-emitting galaxies sufficiently far away will have their emission line carried into the visible range. Since the shortest wavelength we can see is 400 nm, it must be redshifted by a factor $400/121.6 = 3.3$ (i.e. $z=2.3$), corresponding to a distance of 18.6 billion lightyears (but if it's farther away than 25.5 billion lightyears, it will shift into the infrared and be invisible again). Note, though, that this is only in principle; such galaxies are far too dim to be visible to the naked eye. Too see them, you must use a telescope and a camera.


Choosing a diffraction grating I am interested in doing some solar spectroscopy on my own for fun. How do I go about choosing the appropriate diffraction grating?

There isn't a lot of engineering here. Just use a grating dense enough to give you a comfortable angle. Less dense grating (fewer lines per mm) means a more shallow angle and narrow spectrum, and that can be harder to use. Super-dense grating might be expensive and fragile. Stay somewhere in the middle - you'll have to do a few trial runs on your own, see what works for you.

I made this spectrograph for my kids, works well enough:

http://sci-toys.com/scitoys/scitoys/light/spectrograph/spectrograph.html

I used the cheap grating that they provide on their own store, 1000 lines / mm, not bad:

http://www.scitoyscatalog.com/product/DIFFRACTION.html


How to fix a Direction in the context of universe? My question is regarding the direction/orientation of earth. Is it a constant with reference to universal parameters?

I had a nice long answer all mapped out, then I came across a similar (identical?) question...
refer answer here:
How does the earth's plane of orbit (ecliptic) vary over time?

Basically it is changing about 1.4 degrees per 10000 years about now, but that will change chaotically over longer periods.


Is our supercluster part of a galaxy filament? It was discovered recently that the Milky Way galaxy lies inside a supercluster named Laniakea Supercluster. On the other hand, wikipedia says that galaxy filaments are "the largest known structures in the universe"
https://en.wikipedia.org/wiki/Galaxy_filament

My question is: is the Laniakea Supercluster part of a galaxy filament, and if so, what's the name of the galaxy filament?

The Laniakea Supercluster contains several galaxy filaments, even just reading the wikipedia articles you provided you can see the difference in size of the two:

Galaxy Filament

  They are massive, thread-like formations, with a typical length of 50 to 80 megaparsecs, (163 to 261 million light years) that form the boundaries between large voids in the universe.

Laniakea Supercluster

  The Laniakea Supercluster encompasses 100,000 galaxies stretched out over 160 megaparsecs (520 million light-years).

The reason why Laniakea isn't the largest "structure" in the universe is because it inst a single object, rather an arbitrarily assigned group of galaxies. From this image below you can see how the supercluster is given an imaginary edge based on the peculiar velocities of the galaxies within it, there is no observable edge to the supercluster, unlike galaxy filaments that have visible edges.


Why have brown dwarf classes been dubbed L, T and Y? The classes used to categorize stars (O, B, A, F, G, K, M) are in a bizarre order for historical reasons. Stars were labeled based on the spectral lines that were visible, then the categories were put in a different order to account for their temperature. When discovering the temperature of stars, the categories were not renamed, so that the old catalogs could still be used, and everyone just learned "Oh be a fine guy/gal kiss me".

The first seven classes were updated as knowledge improved, but brown dwarfs were discovered more recently, so why have the letters L, T and Y been chosen, rather than the easier to remember P, Q and R (or some other set of  letters), for instance?

The choice of L and T is explained in Kirkpatrick et al. (1999) 'Dwarfs Cooler than "M": The Definition of Spectral Type "L" Using Discoveries from the 2 Micron All-Sky Survey (2MASS)'. The principles behind the choice are given at the start of section 5.1:

In choosing a letter designation for the new spectral class,
three important points must be considered : (1) The letter
must be unambiguous, having not been used for any currently recognized spectral type. For example, though "N"
follows "M" in the alphabet, it would be a poor choice of
letter since it is used for a class of carbon stars. (2) The letter
must create a taxonomy that is clearly distinguished from
other types of astronomical objects. In this case, the letter
must be accepted by the entire community, both by
researchers involved in low-mass star and brown dwarf
science and by astronomers in general. Although stellar
spectroscopists might find "E0," "E1," "E2," etc., perfectly
acceptable as new spectral subclasses, extragalactic morphologists already recognize these designations as elliptical
galaxy types. (3) The letter must stand the test of time. For
example, choosing "D" to mean that these objects are
"degenerate" brown dwarfs would be flawed reasoning.
Some of these dwarfs certainly are substellar (see § 7), but
such a designation cannot be tied uniquely to any particular
spectroscopic trait. The designation should apply to spectral features alone and be free of physical interpretation.
Our understanding of the underlying physics may change
with time ; our choice of letter should be impervious to such
changes.

Applying these, they find the letters H, L, T and Y to be ok: the rationale for excluding the other letters is given in their Table 5:

Letter
Status
Notes

A
In use
Standard spectral class

B
In use
Standard spectral class

C
In use
Standard carbon-star class

D
Ambiguous
Confusion with white dwarf classes DA, DB, DC, etc.

E
Ambiguous
Confusion with elliptical galaxy morphological types E0–E7

F
In use
Standard spectral class

G
In use
Standard spectral class

H
OK

I
Problematic
Transcription problems with I0 (10, Io) and I1 (11, II, Il)

J
In use
Standard carbon-star class

K
In use
Standard spectral class

L
OK

M
In use
Standard spectral class

N
In use
Standard carbon-star class

O
In use
Standard spectral class

P
Problematic?
Incorrect association with planets?

Q
Problematic?
Incorrect association with QSOs?

R
In use
Standard carbon-star class

S
In use
Standard spectral class for ZrO-rich stars

T
OK

U
Problematic?
Incorrect association with ultraviolet sources?

V
Problematic
Confusion with vanadium oxide (V0 vs VO)

W
Ambiguous
Confusion with Wolf-Rayet WN and WR classes

X
Problematic
Incorrect association with X-ray sources

Y
OK

Z
Problematic?
Incorrect implication that we have reached "the end"?

They go with "L" as it is the closest letter to "M" that is still available. They prefer it to "H" because of the occasional use of the term "hydride dwarf" to refer to M subdwarfs which are dominated by CaH and MgH bands.
After that, the sequence is going alphabetically through the remaining available letters, so they put Gl 229B, which has a notably different spectrum to the other (L-class) brown dwarfs known at the time, into spectral class T. Spectral type Y took a few more years to show up in observations, and was chosen by the same principle.
If observations get to the point where it is worth erecting a new spectral class beyond Y then things will get interesting.


I've seen some stars tonight. What could I possibly have been looking at with these pictures? The weather tonight here is great and I decided to shoot some hand-held (too bad I don't have any tripod) night mode pictures shot with P10. I have no experience in stars observation therefore I can't tell what was I looking at. But I'd definitely love to know what are those. Forgive me if the quality of pictures are bad; I've tried my best. At some point I found some of the stars look clear on my phone but fainter over my laptop monitor.

Picture 1 - Taken under hand-held night mode: ISO/400 S/0.5s F/2.2

Picture 2 - Taken under hand-held night mode: ISO/800 S/0.5s F/2.2

Latitude, Longitude & Compass details: Right is for P1; left belongs to P2. Keep in mind that it may not 100% accurate.

Time: 22 Dec 2018 / 22:52 P.M. UTC+8
Location: Ampang, Malaysia.

Star labelled 4 is the only brightest among the others. It shines and flickers so bright that visible very well with naked eyes. The rest are barely to be seen if not focus well.

in-the-sky.org planetarium mode is your friend. It will pick up your location and time zone from your ip address, but you can manually set it to anywhere else.

I've added screenshots at both 20:52 where the Moon is about 26 degrees above the horizon, and at the stated time of22:52 where the Moon is more like 53 degrees.

20:52:

22:52:

Here's a plot of altitude and azimuth of the Moon on that night. Python script as well.

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Topos, load, Loader

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads = 180/pi, pi/180

load = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files

data    = load('de421.bsp')
earth   = data['earth']
moon    = data['moon']
ts      = load.timescale()

print earth.at(ts.now()).position.km   # for no reason at all

local_hours = np.linspace(18, 24, 13)
UTC_hours   = [h-8 for h in local_hours]
times       = ts.utc(2018, 12, 22, UTC_hours, 0, 0)

KL     = earth + Topos(3.1333, 101.6833, elevation_m=66)
Ampang = earth + Topos(3.1255, 101.7588, elevation_m=66)

alt, az, dist = KL.at(times).observe(moon).apparent().altaz()
KLalt, KLaz   = alt.degrees, az.degrees

alt, az, dist        = Ampang.at(times).observe(moon).apparent().altaz()
Ampangalt, Ampangaz  = alt.degrees, az.degrees

if True:
    plt.figure()

    pts = [6, 8, 10]

    plt.suptitle('Moon  22-Dec-2018 KL & Ampang', fontsize=16)

    plt.subplot(4, 1, 1)
    plt.plot(local_hours, KLalt)
    plt.plot(local_hours, Ampangalt)
    plt.plot(local_hours[pts], Ampangalt[pts], 'ok')
    plt.ylabel('Altitude (degs)', fontsize=16)
    plt.xlabel('hours (local)', fontsize=16)
    plt.ylim(0, 90)

    plt.subplot(4, 1, 2)
    plt.plot(local_hours, KLaz)
    plt.plot(local_hours[pts], Ampangaz[pts], 'ok')
    plt.ylabel('Azimuth (degs)', fontsize=16)
    plt.xlabel('hours (local)', fontsize=16)
    plt.ylim(0, 90)

    plt.subplot(2, 1, 2)
    r = (1. - Ampangalt/90.)
    r[r>1] = np.nan
    theta = rads*Ampangaz
    x, y = r*np.sin(theta), r*np.cos(theta)
    plt.plot(x, y)
    plt.plot(x[pts], y[pts], 'ok')
    th = np.linspace(0, twopi, 361)
    xc, yc = np.sin(th), np.cos(th)
    plt.plot(xc, yc, '-k', linewidth=2)
    d = 0.05
    plt.plot([-d, d], [ 0, 0], '-k')
    plt.plot([ 0, 0], [-d, d], '-k')
    plt.xlim(-1.1, 1.1)
    plt.ylim(-1.1, 1.1)

    plt.show()


Why do years become shorter with time? I've read (in Gates to the Hebrew calendar p.128   and other sources) that the Earth orbits the Sun in 365.256 days, but each 1000 years the year becomes shorter by 5.5 seconds.

  "The accurate length of the year is 365 days, 5 hours, 48 minutes, and 45.216 seconds. However this time is going and becoming shorter by 5.5 seconds each 1000 years.

Does this mean that in 10000 years it'll become shorter by 55 seconds, and in 100 years it'll become shorter by 9.16 minutes? What's the explanation for the year shortening with time?

[N.b. I've not read the first book mentioned above and the cite is a translation was presented by a lecturer from which I took notes. Therefore, it could be that the original book contains the explanation. I have no way to check it and that's why I'm seeking here the explanation].

First we need to decide what you mean by "year"

The first time period you mention "365.256" is the Orbital period of the Earth. It is equal 365 days, 6 hours, 8 minutes, 38 seconds.

The second period is the tropical year, the time between two equinoxes. They are not equal because this period depends on the tilt of the Earth. As the tilt changes (called precession) the tropical year is about 20 minutes shorter than the orbital period of the Earh. Since this is the time between seasons, this is the length of the year that is more important practically, and is the basis of most calendars.

These periods do change over time. This is due to solar system dynamics. The Tropical year can change because the rate of precession is not constant.

The Orbital period also changes due to perturbations from other planets (mostly Venus and Jupiter) The Earth can gain some orbital Energy from other planets, causing it to move slightly further from the sun and increase the length of the year. It could also lose some energy, causing it to approach the sun and shorten the length of the year.

This rate of change is not constant. And over the long term the Earth sometimes gains energy and sometimes loses. Currently, the Earth is gaining energy (mostly from Jupiter) and the distance to the sun is increasing by 0.0005% each year

However if you look at the average changes over a period of 6000 years, there is almost no change in the length of the orbital period. In fact, on average over a period from 3000BCE to 3000CE the Earth has actually lost energy and the distance to the sun has dropped by 0.000003% per year. (Figures from https://ssd.jpl.nasa.gov/txt/aprx_pos_planets.pdf)

Your value of 5.5 seconds relates to the tropical year and so includes second-order precessional effects.

While these can be usefully modeled by a linear formula over 1000 years, this is just an approximation and over the longer term the actual orbit of the Earth is pretty stable.

Over the very long term (billions of years) other factors, such as loss of mass by the sun need to be considered, but that is not the cause of the short term variation in year length.


Can one approximate the semi-major axis of an orbit as the average orbital distance for eccentric orbits? Kepler's 3rd Law (regarding the relation between orbital period and the semi-major axis of an orbit) applies to all elliptical orbits. But as I understand it, the only reason it is safe to use the approximation that the semi-major axis of an orbit is approximately equal to the average orbital distance is because the planets of our solar system are generally not very eccentric.

As an example, the Earth is about 1 AU away from the Sun and has a very low orbital eccentricity, so one can say that the semi-major axis of the Earth's orbit is also approximately 1 AU in length. Since P^2 (yr) = a^3 (AU), one can deduce that the orbit of Earth should take about a year. But, what if the Earth had a higher eccentricity - If the orbital eccentricity of the Earth were higher, would it still be wise to approximate the orbital distance as the semi-major axis of an orbit?

The maths says that the semi-major axis is not a good measure of average distance for high eccentricity (elliptical) orbits.

There are basically two ways to measure this : (1) an average over the entire orbit on a purely geometric basis, and (2) the average over time.  These give quite different results - qualitatively different.

Average distance on geometric basis.

In a way this is the most obvious one.  The mathematics is :

$$R_g = \frac a {2\pi} \int_0^{2\pi}r\,d\theta$$

And we use for an ellipse :

$$r = \frac \rho {1+\epsilon\,cos\theta}$$

The result is (eventually :-) ) :

$$R_g = a\,\sqrt{1-\epsilon^2} \tag{1}$$

Average over time.

This is a tricker calculation, but in fact it's the most "human" average.  We want :

$$R_t = \frac 1 T \int_0^T r\,dt$$

The maths on this is tedious and not at all useful, but the result ( with a little help from Wolframalpha.com ) is :

$$R_t = a\,\frac 2 {\pi} \frac{2+\epsilon^2}{(1-\epsilon^2)^{\frac 5 2}}tan^{-1}\left( \sqrt{\frac {1-\epsilon}{1+\epsilon}} \right) \tag{2}$$

Now by any definition that's a messy equation, but the main point is to note the key difference between the time averaged expression (2) and the geometric one (1).

The geometric expression gets smaller as eccentricity gets larger.

The time averaged one gets larger as eccentricity gets larger (a slight simplification).

In fact for both $\frac R a$ starts off at $1.000$ but when $\epsilon\to 1$ then  $R_g\to 0$ and $R_t\to \infty$ !

Why ?

The geometric expression doesn't allow for the orbital speed of the object.  As you increase eccentricity an orbits ends up with having a faster close approach, so less time close up to it's primary.  As it moves away from the primary, the orbital speed also decreases, so it spends much more time at large distances than at short distances.


Is it possible to extend the life of a star? Is it true that when a star 'runs out' of its fuel, it hasn't really used up all its hydrogen, it still has hydrogen left, but the hydrogen is not at its center where the fusion happens. If something crashed into it rearranging its content a bit, could its life be extended?

If something crashed into it rearranging its content a bit, could its life be extended?

Yes! This is precisely what we think causes the creation of blue stragglers.

When we look at a cluster of stars, we expect them to all be of roughly the same age.  This is usually borne out by observations.  Because larger stars evolve faster, the age of a cluster can then be roughly determined by working out the mass of the most massive star still on the main sequence (i.e. still burning hydrogen in the core).  We call this the main-sequence turnoff.

But in some clusters, we see and handful of stars that are more massive, but still on the main sequence, even though all the evidence says they should have already evolved into giants and other more-evolved objects.  The consensus view is that these stars are at the very least interacting with binary companions, and many are thought to be merger products. i.e. something left over from two stars merging into one.  Either two companions in a binary have orbited so close that they've become one star, or two stars have simply collided (which is quite possible in the dense cores of clusters).  Depending on what types of star collide, you can basically get a new star where everything is mixed up again, allowing the star to be "born again".


What is the definition of "deep space"? I need to know the definition of deep space.  I am an amateur astronomer, and am experiencing some difficulty understanding this.

Deep space could be simply defined, as a part of the universe, which is lying beyond the Earth-Moon system. However, when speaking about deep space objects we generally mean stars, star clusters and nebulae, which are outside the Solar System.


SETI search for extraterrestrial life In all these years of listening, why hasn't SETI detected anything ? Is there no intelligent life out there or is it too far away to detect ?

There are many reasons why nothing has been detected. We could be too far away, no one is listening, listened at wrong time, and many more. But this does not mean there is no life on other planets. We just haven't discovered them yet.


Our Sun already a White dwarf? I watch the discovery science channel a lot & enjoy the material about space, stars, planets, etc.  In one episode the narrator states our Sun will one day become a white dwarf star.  The size of our Sun will double and ultimately the Earth will be inside this enlarged Sun. This the end of all life on Earth.  Then I read a recent Snapple bottle cap that stated our Sun is really white.  I read it again and chuckled: is it already a White dwarf star at this point?  I take it that the original color of the star happens to be white.  How will we know when the Sun will become a White dwarf and increase in size the engulf the Earth? Basically what is the time frame in this context.

First. In about 6 billion years the Sun will become much larger than it is now, maybe a factor of 20. It will be a red giant. Then a little while later it will get really big, as large as the earth's orbit and perhaps the Earth will be engulfed.

Second. Only after this does the Sun lose about half of its mass and the remainder is exposed as a small, hot, white dwarf star, about the same size as the Earth but half the mass of the Sun. It will have a density about 1 million times that of water and it is held up by a pressure that only arises because of the quantum mechanical nature of the electrons within it.

The Sun is not currently a white dwarf. It is a G-type dwarf star, supported by "normal" gas pressure and will remain reasonably stable, though will become about 50% more luminous over the next 5 billion years or so.


Does Earth's magnetic field become stronger when a solar flare hits? This curiosity arises as follows: when ejecta from a strong solar flare hits Earth, will the Earth somehow counter its impact with a stronger magnetic field.

Or, earth's magnetic field as limited power strength to resist?

No, but the magnetic dipole of Earth has a field strength that increases with closer distance. On the direct line between Earth and Sun the magnetic field strength is always $B(r) = M/r^3$ where $M$ is the magnetic moment of the terrestrial dipole and $r$ is the distance to the dipole, aka Earth.
When the solar wind conditions change, the boundary between the wind and the terrestrial magnetosphere changes in location, until it finds a new pressure equilibrium.

The approximate location of the boundary shock can be found by equating the ram pressure of the solar win $p_{\rm ram}= \rho u^2$ with the magnetic dipole pressure $p_{\rm dip}=M^2/(2\mu_0 r^6)$, where $\mu_0$ is the vacuum magnetic permittivity. If you solve this for $r$, you will find a closer location $r$ for the shock for stronger conditions, and then checking at which magnetic field strength this occurs via $B(r) = M/r^3$, you'll find that the magnetic field at the shock has of course increased, but only because the shock has moved closer, not because  $M$ has changed.


Stellar Mapping and Coordinates Why are the Galactic Coordinates not aligned so that the Axes go through the Long Bar Y-axis aligned to it's longer side, positive with Sol, and the X-axis align to the center of the short length with Sol being in the negative? Z should obviously be on the galactic plane where it generally is I think...

I'm curious about whether there is a reason we shouldn't align it like this? It seems to make the most sense to me where as the current system, while I understand it, to me is just to Earth-centric, especially when we have something that stands out so much on the galactic scale you could use to figure out where you are with where as Where Sol is being 0,0,0 isn't really helpful to anyone unless you're already on Earth ^.^

The other part of my question is less "real" and more of a "I would like to know if there are any problems with doing this that I am unaware of" thing...

There are 3 question areas...

1
I use a unit I call a Lida, It is 24 terameters long and roughly equivalent to a Light Day.
On this map distances will be measured in Lida, more specifically, KiloLida.
1 Sector = $20 kL^3$
1 Block = $200 kL^3$
1 Grid = $2,000 kL^3$
1 Octant = $20,000 kL^3$

Each of the bigger areas contains 1000 of the next smaller unit down and can easily be referenced by 3 digits. So if I said Grid 346 that would reference the Grid in that Octant where each of those digits are xyz starting from the Galactic Center. Lower numbers are always closer, higher are always further from the Galactic Center.

Why not use parsecs and light years? They're really not practical to use. Likewise, AU are weird, and other units you either end up with very long numbers or poor conversions. Also with this unit, I find it very easy to work out distance and time in my head. The answer is 1.07% off, but that's not really an issue when you're not trying to be 100% precise.

Why use 20kL rather than 10kL or 1kL not many stars in that volume from what I looked at before so it makes sense to use that as division and it also divides up nice to the higher divisions.

2
I use another unit.

1 brut = $1*10^{19} kg$

The largest reason I'm using this is that scientific notation doesn't work in emergencies or common parlance so it makes sense to me to have a larger base unit when dealing with these larger scales. $1*10^{19} kg$, if I remember right (cuz I came up with this a few years ago and haven't used it since), comes from when the reference point numbers on the wiki for scale started getting fishy looking to me. Not a hard science thing, but seemed like a good place to have something like this.

3
This is the way I label Stars, planets, moons, etc... because it makes sense. I keep saying that, but you know, it's true ^.^

SO I - Star (or black hole)
SO II - Spheroid
          Brown Dwarf (Star?)
          Gaseous Spheroid - 12,000+ km radius and above (Most of its size is from gases)
          Terrestrial Spheroid - 1,500 to 12,000 km radius (Has a rocky mantle)
          Icey Spheroid - 200m to 1,500 km radius (Has an ice-water mantle)
SO III - Debris
          Asteroid - Comprised mainly of rock
          Comet - comprised mainly of ice-water

Child Star (?) - SO I that orbits a SO I
Planet - SO II that orbits a SO I
Moon - SO II that orbits a SO II
?? (currently also called a moon it seems) - SO III that orbits a SO II

SO means Stellar Object.
both SO and Spheroid are kinda bad names considering nothing but stars are Stellar and stars are Spheroids, but whatever, I think it gets the point across. Though I wish I could think up something better.

Anyways, do any of these 3 things create problems that I don't know about or that could cause a problem that I possibly can't see due to being a pleb in these matters?

The reason why we centre our galactic coordinates on the sun is that is where we are.

What actually is the purpose of a coordinate system‽ I'd like to be able to point to a body and say that it is at coordinates X, Y, Z (with uncertainties in each direction)

Now we know the direction of the centre of the galaxy, but we don't know it's distance with accuracy. If we were to use the location of the centre of the galaxy for the galactic coordinates we could not give accurate coordinates for any of the local stellar objects.

If we had a warp drive and could travel halfway across the galaxy before lunch then it might make sense to use a different set of coordinates. However, as things stand there is nothing wrong with being Earth centric.

I have mentioned the remainder of the question in comments.


Could a gas giant orbit a star with a 30 year period at a distance of just 1 AU? Assuming a star with similar properties to our sun, could a gas giant orbit it with an orbital period similar to that of Saturn, but, at a much closer orbital distance, more similar to 1 AU?

Gas giants can be found very close to their suns. The closest ones are then called Hot Jupiters.
But not with a orbital period of 30 years. Everything that orbits a star with one solar mass at 1 AU will also have an orbital period of 1 year, according to Keplers 3rd law:
$$
T^{2}=\frac{4 \pi^{2}}{G M} a^{3}.
$$


Is Mercury's core liquid? A very basic question, but one to which I keep finding different answers: does Mercury have a liquid core, or is it all solid? Whatever the reason, what are the causes of it being so?

It's liquid.
As detailed here,

To figure out whether Mercury's core was liquid or solid, a team of scientists led by Jean-Luc Margot at Cornell University measured small twists in the planet's rotation. They used a new technique that involved bouncing a radio signal sent from a ground telescope in California off the planet and then catching it again in West Virginia.
After 5 years and 21 such observations, the team realized their values were twice as large as what would be expected if Mercury's core was solid.
"The variations in Mercury's spin rate that we measured are best explained by a core that is at least partially molten," Margot said. "We have a 95 percent confidence level in this conclusion."

The NRAO has another article on it, which goes slightly more in-depth into the subject.
The official site of the Messenger mission is slightly more cautious:

However, these constraints are limited because of the low precision of current information on Mercury's gravity field from the Mariner 10 and MESSENGER flybys. Fundamental questions about Mercury's core remain to be explored, such as its composition. A core of pure iron would be completely solid today, due to the high melting point of iron. However, if other elements, such as sulfur, are also present in Mercury's core, even at a level of only a few percent, the melting point is lowered considerably, allowing Mercury's core to remain at least partially molten as the planet cooled. Constraining the composition of the core is intimately tied to understanding what fraction of the core is liquid and what fraction has solidified. Is there just a very thin layer of liquid over a mostly solid core or is the core completely molten? Addressing questions such as these can also provide insight into the current thermal state of Mercury's interior, which is very valuable information for determining the evolution of the planet.

At this point in time, though, all evidence indicates that Mercury has a molten core.
As userLTK pointed out, lower pressure inside Mercury makes it easier for the core to stay liquid at lower temperatures.


Would we be able to feel the acceleration of a planet with a highly eccentric orbit? Even though our Earth moves on average 67,000 miles per hour around the Sun, we cannot feel its movement since its speed is fairly constant. More precisely, its change in speed is very low and our bodies cannot detect that low of acceleration.

What if instead a hypothetical human observer were placed on a planet with a very eccentric orbit, say 0.99? Would that human be able to feel the planet's acceleration as it approached its host star, and as it receded from its host star? And if a human observer would not be able to feel any acceleration even at e=0.99, what is the minimum eccentricity, if any, that a planet must have in order for a hypothetical human observer to feel the effects of its acceleration and deceleration?

No.

While you're on that planet, you are also in the same orbit around the star as the planet is. As long as you're in the same orbit as your vehicle (planet, spacecraft), you don't experience any acceleration relative to the vehicle. Whether you're on the planet, or you're just tagging along on the same orbit, is the same thing.

Similar question: in a satellite in a very eccentric orbit around the Earth, is any break in microgravity experienced in various parts of the orbit? Same answer - no. The satellite and everything in it follow the same orbit, so microgravity conditions are maintained.

Now things would be different if you had a very eccentric orbit around a source of much more intense gravity, like a neutron star or a black hole. At the lowest point in the orbit, you could experience tidal forces: the things closer to center experience stronger gravity compared to things further away from center.

But for this to be experienced within a small spacecraft, or within the limits of one human body, the gradient of the field would have to be very strong indeed; the orbit would have to reach very low, very close to the neutron star or black hole.

However, what you would experience would not be orbital "acceleration", but just tidal force. Even so, in extreme cases the tidal force could be strong enough to destroy the ship or kill the passengers.

See the short story "Neutron Star" by Larry Niven for a literary depiction (fairly accurate physically) of this phenomenon. Unfortunately, the story is spoiled by the explanation above, but hopefully learning about gravity compensates for spoiling the story. :)


Are there any common characteristics of habitable planets? Recently, over a thousand exoplanets were confirmed in one huge sweep which is a huge achievement and in the mix a dozen or so possibly habitable planets were confirmed as well. This made me wonder, Are there any common characteristics of (possibly) habitable planets? If so, what are they? Could scientists use these characteristics to locate them? Or are they using these characteristics already?

NOTE: Answers can be theoretical as well!

I think I will answer your question from two vantage points. Firstly, what do we have to measure about a new exoplanet to consider it potentially habitable? Keep in mind that in detecting these 1000 or so new exoplanets, we've only gone so far as to show that they (statistically) exist and to measure some very very basic properties of them. It is on these very basic properties that we make our judgement of habitability and further research and observation is necessary to confirm these suppositions. When a new exoplanet is announced, we generally know the following information, which we use to determine habitability:

Distance from the star. This is a crucial piece of the puzzle because it tells us generally how hot or cold the planet is. We presume that life cannot exist on planets which are boiling hot or freezing cold. This information can be ascertained from either the radial velocity method or the transit method.
Size of the planet. Note here, size generally refers to diameter or radius. This is important because we want to generally know what type of planet we've detected. Gas giants are, well ... giant and inhospitable to life as we know it. We want to make sure the planet we've found is small enough that it is likely terrestrial. Although size is not a guarantee that it is terrestrial, its a piece of the puzzle. We can measure size by the transit method, however there is some degree of uncertainty to this process because we need to know the size of the host star to a good degree of accuracy, which is not always possible.
Mass of the planet. This is a crucial component in determining habitability. Size alone may tell us that a planet is what we call a "Super Earth" or "Sub Neptune" in that it is slightly larger than our own planet. The problem with that is, if the planet sits on the size border between Earth and Neptune, how are we to know whether it is rocky or gaseous? This can be resolved by looking at the mass, which is ascertained by the radial velocity method. The mass combined with size, gives us density and density is a surefire way to know if the planet is made of dense material or light gases.

That's really about it for basic information gleaned from simple planet detection methods. If you have distance from star, size, and mass (all of which you can get from the RV and transit methods) then you can presume that the planet is Earth-like and at a good temperature. That's enough to claim potential habitability, but not enough to actually show it can support human life. For that, you need to investigate the planet further and show it meets a variety of other conditions. As of know, some of those conditions are measurable with enough work, some aren't. These conditions may include, but are not limited to:

An atmosphere with water in it. We have the technology to measure this fact, but I'll say that it is very very difficult to do so and I don't know if anyone has really successfully shown water exists in an exoplanetary atmosphere (some have convincing evidence though). Generally, the process to do this is to measure the radius of the planet as a function of wavelength. Different wavelengths of light will refract through the atmosphere differently (just like in a prism) and so you measure a different radius of the planet (radius being the physical size of the planet plus the extent of the atmosphere) at different wavelengths. From this information, you can model what the atmosphere must be made of to cause the radius to change depending on wavelength. The issue with this process is that is requires removing all possible false detections and having a great signal to noise which is very hard.
A magnetosphere. We believe this is a necessity for life. Every second of every day our planet is being bombarded with high energy particles from our Sun. Most of these particles are harmlessly deflected by our magnetosphere and so we're protected. Planets without this protection would have a hard time hosting life as that life would be destroyed by radiation (or at least life like us). There is currently no way to detect whether a planet has a magnetosphere, although I believe there are potential ideas on the chalkboard.
The presence of a moon. This one is a bit iffy. Some people will argue our Moon has been instrumental to our development and for life in generally. The argument generally goes that the Moon has helped stabilize our tilt and thus kept our seasons and global weather well regulated. While we might be able to exist on a planet without such conditions, some believe life cannot begin on such a harsh planet without that regulating moon. There is also the benefit that the Moon has shielded us from a large number of impacts by taking the blow itself. Again, we could live on the planet, but the lack of a moon might not foster growth of life naturally. Moons, at this time, are out of our reach to detect.

There are numerous other factors for what helps define whether a planet is habitable, be it by us or in hosting new life. The list could go on and on, but these are the major points I think.


If the axis of the Earth was through Greenland, what would the north celestial pole be? If the axis of the Earth was through Greenland, what would the north celestial pole be?

If you assume that the shift was due to natural process, it would still point more-or-less towards Polaris.  (If the shift is due to magic or advanced technology, all bets are off.)

I think your question may be based on a false premise, that the celestial sphere is fixed relative to the Earth.  It isn't at all.  Over the timescales we're dealing with it's basically just fixed and unchanging.

The rotating Earth acts like a gyroscope and its axis points in some direction.  To a good approximation, that direction is fixed in space due to conservation of angular momentum.  So in the absence of external torques, the Earth just sits there and spins like a top.

(To a better approximation, there are some external torques and the gravitational effects of the Sun and Moon causes the Earth's axis to slowly precess. It wobbles (a bit like a top slowing down) once every 26,000 years, actually chaning where it points to in space.  I think this is probably an irrelevant detail though, for your question.)

So, if the Earth's axis passed through Greenland, what would happen?  (Other than the climate changes due to the polar and equatorial regions moving around, of course.)

We have in fact seen something like this in the Earth's past, and it's called Polar Wandering. We observe that the Earth's north pole has moved relative to the Earth's surface and actually has moved quite a lot. But what happened was that the Earth's axis stayed fixed relative to the stars and the Earth's bulk shifted relative to its axis of rotation.

The True Polar Wander page referred to above has an excellent diagram:

So, the answer is that if the shift occurred because of changes to the Earth's moment of inertia, the axis of rotation would still point towards Polaris.  If it happened due to outside influences (a giant impact, aliens, magic) it might wind up pointing most anywhere.


Can I see "my own hand" in the void space? If I were floating in a void (far away from any galaxy or cluster), would the light from these objects at such distances be enough to see for example, my hand?

Yes, and no. You will be able to see your hand's silhouette as it blocks incoming light from other sources, but you won't be able to see any features on it. For example, illuminance of starlight overcast moonless night sky is quoted at $0.0001\ lux$ which means that in absence of a brighter light source, human eye simply isn't that sensitive to such low illuminance, even if your hand reflected all of the incoming light (it doesn't). What is the minimum lux rating that a human eye can resolve is not as straightforward to answer though, as it varies greatly by specific wavelengths. This is explained in more detail for example in this Human eye sensitivity and photometric quantities (PDF) document, or perhaps Wikipedia on Scotopic vision. But in a nutshell, no, you wouldn't be able to see any features on your hand in outer space, unless closer to a brighter visible wavelength light source. You'll still be able to infer that it's still there and all fingers on it accounted for, though.


Comprehensible full manual for bolometric light curve reconstruction Could anyone point me to some manual or tutorial, possibly step-by-step one, on reconstruction of (quasi-) bolometric light curve using UBVRI-photometry data?

Both links and names are fine for me.

If you want a bolometric light curve from photometric points, you don't need to reconstruct its SED inside the range you already have data points. Use effective central wavelength of each band with its magnitude. Do trapezoid. This is an approximation, but it is better than reconstructing the SED that makes you rely more on assumptions for most circumstances, especially that you have UBVRI.

If, under some circumstances, you really need some more data points in between, or you really want a SED, blackbody is the first-order approximation, and is commonly used in SNe (at young ages). Less known issue in optical VBRI. There might be some issues with U like UV excess/suppressed. In this case, just fit the data to a blackbody.


How to calculate luminosity in g-band from absolute AB magnitude and luminosity distance? How can I calculate the (non-bolometric) luminosity $L$ of a galaxy (or a star for that matter) over a given band from its AB apparent magnitude $m_{AB}$ over that band and its luminosity distance $d_L$?

For instance, consider the g-band which typically has a $\lambda_{eff} = 467 \text{ nm}$ and a $\Delta \lambda = 100 \text{ nm}$. Given this galaxy has an apparent AB magnitude of $m_g = 22.5$ and luminosity distance of $1991 \text{ Mpc}$ (i.e. $z = 0.355$ if you are curious), what is its luminosity? I know I shold use the following equation, but I don't know what value to pick for $M_\odot$.

$$L/L_\odot = 10^{0.4(M_\odot - M)}$$

I tried 5.12 from a website which is the value of $M_\odot$ in g-band but it gives me a different answer compared to when I calculate luminosity using flux:

$$L = 4 \pi d_{L}^2 f_\nu \Delta \nu$$

where $f_\nu$ is flux density and $\Delta \nu$ is the frequency width of the band (this is an approximation to the integration over frequency, assuming bands are step functions). So, how can I find luminosity using absolute magnitude?

Please provide your references.

Here is my attempt to reconcile your calculations.

If the AB g-band apparent magnitude is 22.5, then the flux density in the g-band is given by
$$f_{\nu} = 10^{(-48.6-22.5)/2.5} =  3.63 \times 10^{-29}\ {\rm erg\ cm}^{-2} {\rm s}^{-1} {\rm Hz}^{-1}$$

If the distance is 1991 Mpc, then the absolute g magnitude is
$$ M_g = m_g - 5\log d + 5 = -19.0$$

The absolute AB magnitude of the Sun in the g-band (from your source) is 5.12.

The latter tells us that the ratio of the "g-band luminosites" of the galaxy and the Sun are given by
$$\frac{L_g}{L_{\odot,g}} = 10^{0.4(M_{{\odot},g} - M_g)} = 4.45\times 10^{9}$$

One cannot say more than this, in particular one cannot calculate the luminosity of the galaxy, without knowing more about its spectrum.

Also note that the equation above cannot be used to find the ratio of flux in one band to bolometric flux, as I think you are trying to do. To see this, consider that the absolute V-band magnitude and bolometric magnitude of the Sun are almost the same. This does not mean that all the flux from the Sun emerges in the V band!

Your second method requires a figure for $\Delta \nu$, but you haven't said what you have used. The g-filter has a width of around $\Delta \lambda =100$ nm. Using $\lambda \nu = c$ we can say
$$ \Delta \nu  = |c\Delta \lambda/\lambda^2| = 1.28\times 10^{14}\ {\rm Hz}$$
$$ L_g = 4\pi d^2 f_{\nu} \Delta \nu = 2.2 \times 10^{42} {\rm erg/s}$$

The two figures would agree precisely if the "g-band luminosity" of the Sun were $4.9 \times 10^{32}$ erg/s, or in other words if 13% of the solar luminosity emerged in the g-band as defined above. This does not sound unreasonable, but requires a detailed integration of the solar spectrum over the actual g-band filter profile.

Additional: Just to check the numbers above, we can estimate the flux density of the Sun (in wavelength units, above the atmosphere) to be about 1.7 W m$^{-2}$ nm$^{-1}$ at 467 nm (just search on 'solar spectrum' for many examples) at a distance of $1.5 \times 10^{11}$ m. Putting these numbers together, I calculate a "g-band luminosity" for the Sun of
$$ L_{\odot,g}=4\pi d^2 f_{\lambda}\Delta \lambda = 4.8\times 10^{25} \rm W$$
which is indeed 13% of the solar bolometric luminosity.


Where are the Moon rocks from Apollo? There were so many Apollo Missions, and they landed on the moon (so they say). When astronauts came back to Earth from the moon, they brought Moon rock samples. Where are these moon rocks/samples now, and where can we see them?

It's well-explained on NASA'a web site ... http://curator.jsc.nasa.gov/Lunar/index.cfm

About 380 kg of rocks and soil were brought back.

Note that ...

  "The lunar sample building at Johnson Space Center is the chief repository for the Apollo samples."

http://curator.jsc.nasa.gov/Lunar/laboratory_tour.cfm

About three hundred moon rocks were given as gifts by the US to cities and countries around the world, and you can easily find that list, and hence go see them all. (More fun than PokemonGo!)

List showing where to see moon rocks

You can usually very easily see one at the local "national museum" in your country.  I've seen maybe five all over the world at the places listed there.

It seems that at least 100 have gone missing!

There are a few moon rocks in the wild and there were cases of theft - they would be worth millions of bucks ... TV show about that.


Does CIBER Experiment from Caltech suggest that there can be lots of stars which are not in any galaxy? My question is about the implications of the observations recently made by the Cosmic Infrared Background Experiment, or CIBER, from Caltech. I've read at Caltech web site:

  "The total light produced by these stray stars is about equal to the background light we get from counting up individual galaxies."

Does that suggest that about a half of the stars in the observable universe could not belong to any galaxy?

Does that suggest that about a half of the stars in the observable universe could not belong to any galaxy?

Not really.  A key sentence in the article is "The best interpretation is that we are seeing light from stars outside of galaxies but in the same dark matter halo".  So the stars are still within the dark matter halo of a galaxy, but are outside the boundary of the galaxy if the dark matter halo is not considered.

Furthermore the "intrahalo light" explanation is just one of two possible explanations according to Updated analysis of near-infrared background fluctuations which explains:

  Two scenarios have been proposed to interpret the clustering excess. The first advocates the contribution from intrahalo light (IHL), i.e. relatively old stars stripped from
  their parent galaxies following merging events. These stars
  therefore reside in between dark matter halos and constitute a low-surface brightness haze around galaxies. The IHL
  is expected to come mostly from low redshifts (1 + z <
  ∼ 1.5)
  systems (Cooray et al. 2012b; Zemcov et al. 2014).

  The second scenario is instead based on the presence
  of a class of early, highly obscured accreting black holes
  of intermediate mass (∼ 10^4−6M⊙) at z >
  ∼ 13 (Yue et al.
  2013b, 2014). As a suitable mechanism to produce such
  objects does exist – the so called Direct Collapse Black
  Holes (DCBH, for a concise overview of the problem see
  Ferrara et al. 2014), and the interpretation of the super-
  massive black holes observed at z = 6 seemingly requires
  massive seeds (Volonteri & Bellovary 2011), such hypothesis seems particularly worth exploring.

  Both scenarios successfully explain the observed clustering excess, albeit with apparently demanding requirements. In fact, if the excess is to be explained by intra-
  halo light, then a large fraction of the stars at low-z must
  reside outside systems that we would normally classify as
  “galaxies” (Zemcov et al. 2014). On the other hand, in the
  DCBH scenario the abundance of seed black holes produced
  until z ∼ 13 must represent a sizeable fraction of the estimated present-day black hole abundance, as deduced from
  local scaling relations (Kormendy & Ho 2013) and recently
  revised by Comastri et al. (2015). However, it is important
  to outline that both scenarios are not in conflict with any
  known observational evidence


Is the age of the universe relative to an observer's location in that universe? According to Wiki the age of the universe is 13 billion years old, and I was taught that background radiation made the universe uniform in all directions.

Doesn't this define a sphere of space in the universe with Earth at the center? This sphere would have a radius of 13 billion light years.

We (on Earth) could not see anything outside this sphere. As light did not exist more than 13 billion years ago. So for us this defines some kind of edge for the universe.

But what happens when an alien in another galaxy 5 billion light years away from Earth also looks towards the stars. Don't they see a uniform background radiation and measure the same age of the universe?

That alien would also have a sphere of 13 billion years.

So we have two spheres, one for Earth and one for our alien friend.

These two spheres would overlap each other by 5 billion years. Which means, if we look in the opposite direction from that alien's galaxy. We can see an additional 5 billion years further than they can.

If we measure the total width of the two overlapping spheres we get a distance of 18 billion light years.

I now grasp my hair in confusion. Is this possible that two visible spheres could overlap to create a distance longer than the oldest ray of light? And from there I fall into more questions. Can two people measure the same universe age from different spots in the universe.

That's why I think my idea is is wrong, but how could it be wrong and why?

You are labouring under the misapprehension that how far we can see directly gives the age of the universe. Whilst it is true that the oldest light we can see was emitted some 13.7 billion years ago, the stuff that emitted that light is now roughly 46 billion light years away, thanks to expansion of the universe.

The universe itself probably extends significantly beyond this and may be infinite. That does not conflict with the big bang model or relativity. If the universe is infinite then it has always been infinite. Places that are separated by more than the oldest light ray as you put it, are not in causal contact and, depending on the cosmological parameters, may never be.

In a homogeneous, isotropic universe (and don't forget we can only make assumptions about regions beyond our observable universe), all observers would agree on the age of the universe and that the universe was once very much smaller and all would see (on average) the same sort of universe in all directions.

To ensure that places that are outside of causal contact now are homogeneous requires that they were in causal contact in the past. This is the nature of the "horizon problem"".

I took the picture below from the "slideplayer" website, I am unsure who the originator is, but it serves its purpose here and I think encapsulates your question. It appears in this diagram that A and B can never have "communicated" and can never have originated from the same place. The solution provided by inflation is to have a massive exponential expansion of space in the first fraction of a second. In short, the distances between points in space (or galaxies if you like, although there were no galaxies at the epoch of inflation) are suddenly increase by many, many orders of magnitude. This give the appearance of faster-than-light motion, although the cosmic speed limit only applies to local measurements and not to the expansion of space itself. The end result is a universe that appears homogeneous well beyond the limits of a radius of the age of the universe in light years.


Python: Find brightest pixel in fits-file and ignore brighter artefacts I have 4000 fits-files images from SPITZER, each containing the same star over a time period. Each image is a photmetry cube of 32 pixel * 32 pixel * 64 consequent time frames. I need to look through all the single frames and order the pixel by brightness. Not hard to do with Python, but sometimes I have in the frames short-lived events/artefacts which are brighter than the brightest pixel of the star (see below). These artefacts last only over one or two consequent time frames

How can I easily correct for the artefacts? Is there an astropy/pyfits package that does this? Or some kind of bad pixel flagging in SPITZER (I know that Kepler-images have flagging of bad pixel)?

Optional: It would be convenient to have the code in vectorized form, avoiding loops, so that I save computtional time

Fig: In the left frame everything is normal, in the right frame, in the top right corner you see one of the artefacts, having a value below a saturation treshhold

Following Mick's comment, I found an astropy-function - DAOStarFinder - that does what I was looking for. It scans an image for light-sources and one can set threshhold for a detection, e.g. the "roundness", "sharpness" or brightness of an object have to be between [-0.05,0.05], [0.6,0.8] or >5*standard_deviation, respectively. I must say, that I have the impression the roundness and sharpness measure don't work consistently and sometimes values for the same object change enormously from one frame to the other. But DAOStarFinder gives you also a rough estimate of the flux density of the object, which I could use for sorting the sources and filtering out the relevant one. Here is a link to example script and to the documentation of DAOStarFinder. And here the example script explicitly:

>>> from astropy.stats import sigma_clipped_stats
>>> from photutils import datasets
>>> from photutils import DAOStarFinder

>>> hdu = datasets.load_star_image()
>>> data = hdu.data[0:400, 0:400]
>>> mean, median, std = sigma_clipped_stats(data, sigma=3.0, iters=5)
>>> daofind = DAOStarFinder(fwhm=3.0, threshold=5.*std)
>>> sources = daofind(data - median)

>>> print(sources)


How can an infinite universe expand? I understand the expansion of the universe as actually an increase in the ratio of space to matter. Is this a correct understanding?  Otherwise, I don't understand how an infinite structure can expand.

Expansion means that distances are increasing as a function of time. Say if the distance between two galactic clusters is $D$, then in an expanding Universe the distance is governed by some strictly increasing function of time $a(t)$ called the scale factor where

$$D=a(t)D_0$$

where $D_0$ is the distance at the present time and by definition $a(t_{0})=1$.

Cosmology assumes that the Universe is on large scales the same everywhere (homogeneous) and the same in all directions (isotropic) so the above applies to all distances above a certain scale. The scale factor $a(t)$ can be found from the Friedmann equations and initial conditions.

Expansion is possible in Universes of both finite and infinite spatial extent.

As the volume of a (large enough) region of space increases in proportion to $[a(t)]^3$, but the amount of matter remains constant, the matter density changes in proportion to $[a(t)]]^{-3}$. Expansion however also decreases the kinetic energy of of its contents, so the energy density decreases by a greater factor if the contents has kinetic energy.


How spectrographs that measure radial velocities manage to translate variations in the stars' spectrum lines into the "speed" of the star? Like ESPRESSO or CARMENES, for example. I just don't understand the process or the way these spectrographs manage to obtain the velocity of the "wobble" of the star (due to the presence of another companion such as an exoplanet) just by analyzing variations on spectral lines.

There are many ways to measure a star's motion, radial velocity (wobble) can be measured using doppler spectroscopy. The first exoplanet discovered by this method was 51 Pegasi b by Michel Mayor and Didier Queloz, who discovered the planet on December 1995.

A drawback of this method is that it can only detect the movement of a star towards or away from the Earth. If the orbital plane of the planet is "face on" when observed from the Earth the wobble of the star will be perpendicular to an observer's line of vision, and no spectrum shift will be detected.

In most cases a distant planet's orbital plane is neither "edge-on" nor "face-on" when observed from the Earth. Most likely it is tilted at some angle to the line of sight, which is usually unknown. This means that a spectrograph would not detect the full movement of the star, but only that component of its wobble that moves it towards the Earth or away from it.

The ESPRESSO instrument uses two 90x90 mm CCD detectors, one red and one blue sensitive. The detectors view the light after it is reflected off of an echelle grating, which is optimized for use at high incidence angles and therefore in high diffraction orders. Higher diffraction orders allow for increased dispersion (spacing) of spectral features at the detector, enabling increased differentiation of these features.

The equations are relatively simple. The observed Doppler velocity is $K=V_{\mathrm {star} }\sin(i)$, where $i$ is the inclination of the planet's orbit to the line perpendicular to the line-of-sight.

Reference:

"A Jupiter-mass companion to a solar-type star", Nature volume 378, pages 355–359 (1995), by Michel Mayor and Didier Queloz

An improvement to Mayor and Queloz's equations is offered in:

"The Rossiter-McLaughlin effect and analytic radial velocity curves for transiting extrasolar planetary systems" (Mar 25 2005), by Yasuhiro Ohta, Atsushi Taruya, and Yasushi Suto

  "7. Conclusions and discussion
  We have discussed a methodology to estimate the stellar spin angular velocity and its direction angle with respect to the planetary orbit for transiting extrasolar planetary systems
  using the RM effect previously known in eclipsing binary stars (Rossiter 1924; McLaughlin 1924; Kopal 1990). In particular we have derived analytic expressions of the radial velocity
  anomaly, $\Delta_{vs}$, which are sufficiently accurate for extrasolar planetary systems. If the stellar limb darkening is neglected, the expression is exact. We have extended the result to the case with limb darkening and obtained approximate but accurate analytic formulae. For a typical value of $\gamma = R_p / R_s \sim 0.1$, the formulae reduce to a simple form (eqs. [40], [43], [44], [45], [48], and [49]):

  $$\Delta v_s = \Omega s X_p \sin I_s \frac{\gamma^2 \{1-\varepsilon(1-W_2) \} }{ 1-\gamma^2-\varepsilon \{ \frac{1}{3}-\gamma^2 \} } \tag{56}$$

  during the complete transit phase and (the following is scrollable):

  $$\small { \Delta v_s = \Omega_s X_p \sin I_s \frac { (1- \varepsilon) \{ - z_0 \zeta + \gamma^2 \cos^{-1} (\zeta / \gamma) \} + \frac { \varepsilon }{ 1 + \eta_p } W_4 }{ \pi (1- \frac {1}{3} \varepsilon ) - (1 - \varepsilon) \{ \sin^{-1} z_0 - (1 - \eta_p) z_0 + \gamma^2 \cos^{-1} (\zeta  / \gamma) \} } \tag {57} }$$

  during the egress/ingress phases, where

  $$\begin{align}
W_2 & = \frac { (R^2_s - X^2_p - Z^2_p)^{1/2} }{ R_s }, \tag {58} \\
W_4 & = \frac {\pi}{2} \, \gamma^{3/2} (2 - \gamma)^{1/2} \, (\gamma - \zeta) \; x_c \frac {g(x_c ; \eta_p , \gamma) }{ g(1- \gamma ; - \gamma , \gamma) }, \tag {59}
\end{align}
$$

  where $g(x; a, b)$ is defined in equation (A17). The definition and the meaning of the variables in the above expressions are summarized in Table 1.

  The numerical accuracy of the above formulae was checked using a specific example of the transiting extrasolar planetary system, HD 209458, and we found that they are accurate
  within a few percent. Our analytic formulae for the radial velocity anomaly are useful in several ways. One can estimate the planetary parameters much more efficiently and easily,
  since one does not have to resort to computationally demanding numerical modeling. Furthermore, the resulting uncertainties of the fitted parameters and their correlations are easily evaluated.

  ...

  Table 1. List of notation
  Variables    Definition     Meaning

  Orbital Parameters
  $m_p$        Sec.2        Planet mass
  $m_s$        Sec.2        Stellar mass
  $a$          Fig.1       Semimajor axis
  $e$          Fig.1        Eccentricity of planetary orbit
  $\varpi$         Fig.1       Negative longitude of the line of sight
  $i$            Fig.2        Inclination between normal direction of orbital plane and y-axis
  $r_p$         Eq.[1]       Distance between star and planet (see Fig.1)
  $f$          Eq.[2]        True anomaly (see Fig.1)
  $E$         Eq.[2]       Eccentric anomaly
  $n$          Eq.[3]       Mean motion
  $M$        Eq.[4]       Mean anomaly

  Internal Parameters of Star and Planet
  $Is$        Fig.2         Inclination between stellar spin axis and y-axis
  $λ$          Fig.3         Angle between $z$-axis and normal vector $\hat{n}_p$ on $(x, z)$-plane
  $Ω_s$         Eq.[12]       Annular velocity of star (see Fig.2)
  $R_s$         Sec.4      Stellar radius
  $R_p$         Sec.4      Planet radius
  $\varepsilon$         Eq.[38]       Limb darkening parameter
  $V$           Sec.6      Stellar surface velocity, $R_sΩ_s$

  Mathematical Notation
  $X_p$         Sec.4      Position of the planet
  $γ$         Eq.[25]       Ratio of planet radius to stellar radius, $Rp/Rs$
  $η_p$          Eq.[28]       See Fig.6
  $x_0$          Eq.[33]       See Fig.6