How to compute the sun's (upper) transit As detailed in https://physics.stackexchange.com/questions/41450/is-there-a-simple-accurate-formula-for-calculating-transit-times-from-rise-and, the sunrise, and set times can accurately be used to compute the sun's transit. Yet, in some cases, these do not exist, e.g. in Finland in the summer.

Is there an easier and/or more general way to compute the transit time (and only the transit time) given the UTC and observer position (latitude, longitude, height).

The library at http://www.jstott.me.uk/jsuntimes gives sunrise and sunset, but not the transit time. find_transit at riset_cir.c in https://github.com/brandon-rhodes/pyephem/tree/master/libastro-3.7.7 might do this, but seems too general and complicated.

Here is the page code for my BASIC-to-php implementation of the algorithms to calculate the  (geocentric) RA & Dec of the Sun for a given date from Jean Meeus' book Astronomical Algorithms as well as calculating the angular distance between two celestial objects.  I can't remember the accuracy of the calculations and I didn't make the corrections for observer lat & long.

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">

<META HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=ISO-8859-1" />
 <base href="sionnagh.com" />
<title>Mick Todd Distance Calculator</title>

<?php include("mtheader.php"); ?>

Mick Todd Distance Calculator

<?php
$earthincl=23.44; //Earth's axial tilt
$sollat=0;

$fdate = $_POST["fdate"];
if ($fdate=="") {
 $fdate=getdate(date("U")-date("Z"));
 $fday=$fdate[mday];
 if (strlen($fday)==1) {$fday="0".$fday;}
 $fmonth=$fdate[mon];
 if (strlen($fmonth)==1) {$fmonth="0".$fmonth;}
 $fyear=$fdate[year];
 //$fdate=$fdate[mday]."/".$fdate[mon]."/".$fdate[year];
 $fdate=$fday."/".$fmonth."/".$fyear;
}
else {
 $fday=substr($fdate,0,2);
 $fmonth=substr($fdate,3,2);
 $fyear=substr($fdate,6,4);
}
$fra1 = $_POST["fra1"];
$fra1 = strtolower($fra1);
if ($fra1<>"") {
  if (strpos($fra1,"h")==0) {
  $frad1=$fra1;
  $fra1=converttohms($fra1);
  }
  else {
    $frad1=converthtod($fra1);
  }
}
$fdec1 = $_POST["fdec1"];
$fdec1 = strtolower($fdec1);
if ($fdec1<>"") {
  if (strpos($fdec1,"d")==0) {
    $fdecd1=$fdec1;
    $fdec1=converttodms($fdec1);
  }
  else {
    $fdecd1=convertdtod($fdec1);
  }
}
$fra2 = $_POST["fra2"];
$fra2 = strtolower($fra2);
if ($fra2<>"") {
  if (strpos($fra2,"h")==0) {
    $frad2=$fra2;
    $fra2=converttohms($fra2);
  }
  else {
    $frad2=converthtod($fra2);
  }
}
$fdec2 = $_POST["fdec2"];
$fdec2 = strtolower($fdec2);
if ($fdec2<>"") {
  if (strpos($fdec2,"d")==0) {
    $fdecd2=$fdec2;
    $fdec2=converttodms($fdec2);
  }
  else {
    $fdecd2=convertdtod($fdec2);
  }
}

function JulianDay ($fday, $fmonth, $fyear, $fUT) {
    $JDpartsign=(100*$fyear+$fmonth-190002.5)/abs(100*$fyear+$fmonth-190002.5);
 $JD = 367*$fyear - floor((7*($fyear+floor(($fmonth+9)/12)))/4) + floor((275*$fmonth)/9) + $fday + 1721013.5 + $fUT/24 - 0.5*$JDpartsign + 0.5;
 return $JD;
}
function converttodms($inval) {
 if ($inval==0) {
   $fdsign=1;
 }
 else {
   $fdsign=$inval/abs($inval);
 }
 $inval=abs($inval);
 $fddeg=floor($inval);
 $remain=($inval-$fddeg)*60;
 $fdmin=floor($remain);
 if (strlen($fdmin)==1) { $fdmin="0".$fdmin; }
 $remain=($remain-$fdmin)*60;
 $fdsec=round($remain*100)/100;
 if (strpos($fdsec,".")==1) { $fdsec="0".$fdsec; }
 $fddeg=$fdsign*$fddeg;
 $fdms=$fddeg."d".$fdmin."m".$fdsec."s";

    return $fdms;
}
function converttohms($inval) {
    $inval=$inval/15;
    $fddeg=floor($inval);
    $remain=($inval-$fddeg)*60;
 $fdmin=floor($remain);
 if (strlen($fdmin)==1) { $fdmin="0".$fdmin; }
    $remain=($remain-$fdmin)*60;
 $fdsec=round($remain*100)/100;
 if (strpos($fdsec,".")==1) { $fdsec="0".$fdsec; }
    $fdms=$fddeg."h".$fdmin."m".$fdsec."s";

    return $fdms;
}
function converthtod($inval) {
  $posh=strpos($inval,"h");
  $posm=strpos($inval,"m");
  $poss=strpos($inval,"s");
  $temph=substr($inval,0,$posh);
  if ($posm) {
    $tempm=substr($inval,$posh+1,$posm-($posh+1));
  }
  if ($poss) {
    $temps=substr($inval,$posm+1,$poss-($posm+1));
  }

  $fdms=($temph+$tempm/60+$temps/3600)*15;

  return $fdms;
}
function convertdtod($inval) {
  $posd=strpos($inval,"d");
  $posm=strpos($inval,"m");
  $poss=strpos($inval,"s");
  $tempd=substr($inval,0,$posd);
  if ($posm) {
    $tempm=substr($inval,$posd+1,$posm-($posd+1));
  }
  if ($poss) {
    $temps=substr($inval,$posm+1,$poss-($posm+1));
  }
  $fdms=$tempd+$tempm/60+$temps/3600;

  return $fdms;
}

function SolarCoords($fday, $fmonth, $fyear, $fUT) {
//Julian Day of 1991/ 5/19 at 13 UT         JD = 2448396.04167
//Julian day of 2000/01/01 at 12 UT         JD = 2451545.0
//number of Julian days since 2000/01/01 at 12 UT               -3148.95833
//number of Julian centuries since 2000/01/01 at 12 UT  T = - 3148.95833/36525
//used by the algorithm for L                                                           = 0.086213780

    $k = 2*pi()/360;
  $T = (JulianDay($fday, $fmonth, $fyear, $fUT)-2451545.0)/36525;
    //echo "\$T: ".$T."
";
 //mean anomaly, degree
 $M = 357.52910 + 35999.05030*$T - 0.0001559*$T*$T - 0.00000048*$T*$T*$T;
    //echo "\$M: ".$M."
";

    // mean longitude, degree
    $L0 = 280.46645 + 36000.76983*$T + 0.0003032*$T*$T;
    //echo "\$L0: ".$L0."
";

    $DL = (1.914600 - 0.004817*$T - 0.000014*$T*$T)*sin($k*$M) + (0.019993 - 0.000101*$T)*sin($k*2*$M) + 0.000290*sin($k*3*$M);
 //echo "\$DL: ".$DL."
";

    // true longitude, degree
    $L = $L0 + $DL;
 if (abs($L)>360) {
        $divsign=($L/360)/abs($L/360);
  $div=floor(abs($L/360));
  $L=$L-$divsign*$div*360;
  if ($L<0) { $L=$L+360; }
    }
    $L=round($L*10000)/10000;

    return $L;

}

?>

<form METHOD="POST" ACTION="mtdistcalc.php" name="mtdistcalc">

Date <input type="text" name="fdate" id="fdate" size=10 value="<?php echo $fdate; ?>"> <input type="button" value="Now" onclick="fdate.value=settoday();">

Object coordinates (#1) RA <input type="text" name="fra1" id="fra1" size=15 value="<?php echo $fra1; ?>">  Decl <input type="text" name="fdec1" id="fdec1" size=15 value="<?php echo $fdec1; ?>">

Object coordinates (#2) RA <input type="text" name="fra2" id="fra2" size=15 value="<?php echo $fra2; ?>">  Decl <input type="text" name="fdec2" id="fdec2" size=15 value="<?php echo $fdec2; ?>">

<input TYPE=SUBMIT VALUE="Calculate distances">

</form>
<?php
echo "

Date:	".$fday."/".$fmonth."/".$fyear." "."0000UT
";
echo "
Julian Day: 	".JulianDay($fday,$fmonth,$fyear,"0")."
";
echo "
Ecliptic latitude of the Sun: 	".$sollat."&deg; (assumed)
";
$sollong = SolarCoords($fday,$fmonth,$fyear,"0");
echo "
Ecliptic longitude of the Sun: 	".$sollong."&deg; (".converttodms($sollong).")
";
echo "
Earth's axial inclination: 	".$earthincl."&deg;

";

//echo "\$feclong ".$feclong."

";

$deltaSol = round(asin(sin($sollat*M_PI/180)*cos($earthincl*M_PI/180) + cos($sollat*M_PI/180)*sin($earthincl*M_PI/180)*sin($sollong*M_PI/180))*180/M_PI*10000)/10000;
$alphaSol = round(acos(cos($sollong*M_PI/180)*cos($sollat*M_PI/180) / cos($deltaSol*M_PI/180))*180/M_PI*10000)/10000;
if ($sollong>180) {
 $alphaSol = 360-$alphaSol;
}
echo "

Sol

RA: 	".$alphaSol."&deg; (".converttohms($alphaSol).")

Dec:	".$deltaSol."&deg; (".converttodms($deltaSol).")

";

//spherical angular distance
//cos(theta) = sin(delta_a) sin(delta_b) + cos(delta_a) cos(delta_b) cos(alpha_a-alpha_b)
if ($fra1<>"" AND $fdec1<>"") {
  $theta1 = round(acos(sin($deltaSol*M_PI/180)*sin($fdecd1*M_PI/180) + cos($deltaSol*M_PI/180)*cos($fdecd1*M_PI/180)*cos(($alphaSol-$frad1)*M_PI/180))*180/M_PI*1000)/1000;
}
if ($fra2<>"" AND $fdec2<>"") {
  $theta2 = round(acos(sin($deltaSol*M_PI/180)*sin($fdecd2*M_PI/180) + cos($deltaSol*M_PI/180)*cos($fdecd2*M_PI/180)*cos(($alphaSol-$frad2)*M_PI/180))*180/M_PI*1000)/1000;
}

if ($fra1<>"" AND $fdec1<>"" AND $fra2<>"" AND $fdec2<>"") {
  $theta3 = round(acos(sin($fdecd1*M_PI/180)*sin($fdecd2*M_PI/180) + cos($fdecd1*M_PI/180)*cos($fdecd2*M_PI/180)*cos(($frad1-$frad2)*M_PI/180))*180/M_PI*1000)/1000;
  echo "
Angular distance between Object #1 and Object #2 is: ".$theta3."&deg;
";
}

if (($fra1<>"" and $fdec1<>"") or ($fra2<>"" and $fdec2<>"")) {
  echo "

&nbsp;	RA	Dec	Solar elong.
";
}
if ($fra1<>"" AND $fdec1<>"") {
  echo "
Object #1:	".$fra1."	".$fdec1."	&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;".$theta1."&deg;
";
}
if ($fra2<>"" AND $fdec2<>"") {
  echo "
Object #2:	".$fra2."	".$fdec2."	&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;".$theta2."&deg;
";
}

?>


Does matter accelerate to the speed of light as it approaches the singularity? $$v_\mathrm e=\sqrt{\frac{2GM}r}$$

If we understand the escape velocity to be the speed needed to escape from the 'surface' of a gravitational object described by the above equation.

It is equally true that the escape velocity is the speed an object pulled from rest will reach when it hits a certain point in the gravitational field.

Since the event horizon is defined as when the escape velocity reaches the speed of light for a black hole, would matter accelerate near the speed of light as it approaches the singularity? Accretion disks are known to move at relativistic speeds near the event horizon, but perhaps oddities occur beyond this point?

The event horizon can be derived from setting the escape velocity of an object to the speed of light, it is a point of no return. I understand there are better definitions, but this is not really germane to the question at hand. How about from the perspective of another in-falling observer?

There is indeed a sense that infalling objects accelerate to the speed of light as it reaches the event horizon, but you need to be careful what you mean by the speed of the object because the speed is observer dependent.

This is explained in detail on the Physics site in my answer to Will an object always fall at an infinite speed in a black hole? An observer watching from far from the black hole would see the falling object initially accelerate towards the black hole but then decelerate to a halt at the event horizon. However an observer hovering a distance $d$ above the event horizon would see the falling object pass them at a speed:

$$ v = c \sqrt{\frac{r_s}{r_s + d}} \tag{1} $$

where $r_s$ is the radius of the event horizon. As the distance above the horizon $d$ goes to zero the speed calculated from equation (1) goes to the speed of light $c$.


Difference of sunset and sunrise of two cities are different For 26 Nov 2016:

Consider Tehran with 06:51 sunrise and 16:52 sunset.
Another city is Rafsanjan in 850KM southeast of Tehran, sunrise is 6:21 and 16:45 for sunset.

Time zone of two cities is +3.5

Why difference of sunrise is 30 minutes while sunset is 7 minutes?

Tehran (35.7$^\circ$ N) has 23 fewer minutes of daylight than Rafsanjan (30.4$^\circ$ N) on Nov 26 because the Sun is 21$^\circ$ south of the celestial equator and Tehran is 5.3$^\circ$ farther north than Rafsanjan.
For the same reason, the north pole has 24-hour night and the south pole has 24-hour daylight at this time of year.
On May 26, when the Sun's declination is 21$^\circ$ north, Tehran has 25 more minutes of daylight than Rafsanjan does, and the north pole has 24 more hours of daylight than the south pole does.


Why are there no gamma-ray bursts detected in our galaxy? I found from Wikipedia and other sites that there are no GRBs detected in the Milky Way. Can someone give a feasible reason for that? Why are there no GRBs detected in the Milky Way galaxy?

All models of gamma-ray bursts involve extremely energetic phenomena: particular types of supernovae, the coalescence of binary compact objects, strong magnetar flares, or tidal disruption events. It turns out that these events are quite rare - so rare, in fact, that GRBs would be expected to occur in a low-redshift Milky Way-like galaxy at a rate of only one every few million or tens of millions of years (Zhang & Meszaros 2003). If the Milky Way was undergoing a period of intense star formation, yielding more massive stars and therefore more supernovae, then this rate would go up, but still not significantly. Bear in mind that the Galactic supernova rate, for example, is believed to be only several supernovae per century, and the vast majority of supernovae don't lead to gamma-ray bursts.
We know of many extragalactic GRBs for several reasons. It helps that we can observe large numbers of galaxies (thanks to how bright the bursts are), and if we could look at millions of Milky Way-like galaxies, it wouldn't be surprising if we could detect $\sim$1 event per year. (There's also the advantage that star formation peaked at a redshift $z\sim2$, and so high-redshift objects would be more likely to produce more GRBs!)


How do isolated binary compact systems survive their formational supernovae? In a broad sense I'm trying to understand how a binary black hole or neutron star systems can form from isolated binary massive star systems (i.e., other than from migration within rich BH/NS clusters), and especially, how these can form within the constraints that allow for observed merger rates. I was hoping that this question would answer me, but it doesn't, because I realised I've never seen an explanation of how a binary system could survive a SN of one of its stars in the first place.

As far as I understand, we think one route to observed BH/NS mergers is a binary system of massive stars. This gives a scenario of 2 massive stars in an isolated binary system, one of which undergoes core collapse and type 2 SN. The two stars must be quite close (or later perturbed) and not greatly "kicked" on formation of either compact object, otherwise they couldn't later merge in the observed time. They probably also have similar ages (formed in the same cloud).

But in this scenario

How does the second-to-collapse star, survive being ripped apart by a nearby SN at such short distance, and remain in a small enough orbit, which we know happens since we later observe the merger of two compact objects? How does it survive structurally and regain stability in a form capable of future SN, given the intense flood of neutrinos and other energy/particles, some of which can penetrate to its core? Or does the first SN directly trigger the second SN?
Even assuming the second star survives with enough mass and undisrupted structure for SN (or is triggered directly into SN), wouldn't the transferred momentum from the first SN ejecta "on its doorstep" be sufficient to simultaneously unbind the binary system, or the momentum transferred from the second SN be enough to eject the first NS/BH, so that the two resulting compact objects no longer form a binary system?

OK, here's what I've been able to piece together with a little research (not really my field, so I may get a few things wrong).

The short answer is a combination of: A) It's pretty hard to actually disrupt a massive star via a neighboring supernova, and the two stars aren't intially that close to each other; and B) Lots of complicated effects due to the binary interaction.

I'll focus on what seems to be the best current model for forming a compact double neutron-star (DNS) system -- one with a small enough separation in the resulting neutron stars so that gravitational-wave emission can cause them to merge within a few billion years -- which is discussed in this 2017 article by Tauris et al.. It looks a bit like this:

Massive binary star system forms, with moderate separation (not as close as the final DNS binary will be).
Primary star (initially more massive) evolves faster, to the point where it expands and fill its Roche lobe, leading to mass transfer to the secondary.
This means the primary loses mass while the secondary gains mass. So when the primary reaches the supernova stage, it has less mass in its envelope to eject -- and the companion star may now be the more massive of the two.
Primary's core collapses, producing first SN. This probably won't destroy the companion; it might disrupt the system -- not so much because the companion is kicked a lot but because the NS remnant is lower mass than the pre-SN primary star, and it probably gets a kick from explosion asymmetries. But recall that the companion star may now be more massive, so the gravitational disruption is less.
Companion star evolves, expanding to overfill its Roche lobe, leading to a "common envelope" phase where the NS is actually inside the companion star's outer envelope.
Common envelope phase leads to

A. Significant shrinking of the orbit as the NS experiences dynamical friction;

B. Ejection of much of the companion star's outer envelope.
System is now much more compact, with a NS and a stripped helium-star companion.
Possible further mass transfer from companion onto neutron star, leading to even less of an envelope around the companion and an even lower mass.
Companion's core collapses, causing second SN. This has little effect on the first NS (i.e., the former primary star), since the NS is a very small target with extreme density and surface gravity: very hard to disrupt. In addition, the companion has very little envelope left to eject at this point.

Some key points:

The binary starts off with a significantly wider separation than the final DNS will have, so the effects of the first SN on the companion will be less extreme than you might imagine;
Mass transfer (and ejection) during the Roche overflow and common envelope stages removes mass from the stars, which means there's less envelope to eject when a SN kicks off (so it does less damage to the other star/NS) and also that the SN ejection of a star's envelope doesn't have a strong effect on the gravitational binding energy, since there's less of an envelope to eject.

This paper by Liu et al. (2015) presents some modern simulations of the effect of a SN blast wave on a nearby companion star. It turns out to be pretty hard to disrupt the companion: even for relatively small separations, the companion only loses about 10% of its mass, and this goes down to a few percent or less as the separation increases. (Remember that the companion is itself a massive star most likely still on the main sequence, and so relatively compact, with high surface gravity.)

Can the explosion push the companion star away? Probably not very much: again, the companion star is itself quite massive -- maybe more massive than the original primary because of the mass transfer in stage 2 (above) -- so it's hard to push it around.


How gravity works in a black hole? In astronomy class, we have to state how gravity work in the black hole? I know for one fact is that black hole is strong enough to stretch you thin enough like a noodle. Any ideas?

Gravity works exactly the same around a black hole as it does everywhere else in the universe:

The mass of the black hole distorts spacetime, and then objects follow the shortest route in the curved spacetime. The only difference between the gravity of a black-hole and the gravity of Earth is that the Black hole has a much larger mass in a much smaller volume, so the gravitational field is much stronger.

The strength of gravity has several consequences I'll mention three:

First close to the black hole spacetime is so curved that light can't escape. Now all gravity bends light a bit, but you usually don't notice it. A black hole bends light a lot.  The edge of the black hole is called an event horizon and it is the closest distance that light can go to a black hole and not fall in. Nothing can ever pass from within the event horizon.

Next there are tidal forces. All gravity can cause tides: the force of the earth gravity is a bit more on your feet than on your head. Usually you don't notice. However a solar mass black hole has such extreme tides that you would be pulled apart by the difference in force (this is the spaghettification you mention)

Finally, inside the black hole, space works like time. You can't change your direction in time (unless you have a Delorean) Inside a black hole you can't change direction in space.

But remember the only difference between the gravity around a black hole and the Earth is a matter of degree.


A Noob's question on declination The Sun's position at noon is perpendicular to the local x-y plane on the equator during equinox. So the declination of Sun on that day is 0. Today the declination of Sun is close to 23, can this be taken as a virtual equator and hence the tropic of Cancer is in 46.5N (by adding 23.5 to it) and tropic of Capricorn is in 0.5S? Or my assumption is wrong?

I don't think this particularly useful way to think about it. The equator is the equator, Cancer is 23.4 North, and Capricorn is 23.4 South. The equator is a great circle on the sphere of the sky, but the Tropic of Cancer is not a great circle.  These are fixed.

Now (in Northern Hemisphere summer) the sun is directly above a point on (or near) the tropic of cancer.  This doesn't make the tropic of cancer a "virtual equator". It just makes it the tropic of cancer in summer.

It makes a difference because the path of the sun seen from the equator on midsummer is not the same as the path of the sun seen from the tropic of Capricorn on the equinox.  On the equinox, the sun traces a great circle in the sky (from wherever you view it) On midsummer's day the sun doesn't trace a great circle (anywhere).


how to assess a periodical signal? After folding a light curve, we may  get some kind of shape.
How to tell whether there is a real periodical signal quantifiably?

The fluctuation may be small, and the folded shape may not be perfect.

You have it the wrong way around. You have to establish a periodicity before doing the folding.

There are various ways to do this, but the most frequently (pun intended) used is the Lomb-Scargle periodogram.

Start with a datset consisting of a set of points $(t_i, m_i\pm \sigma_i )$, where $t_i$ are times of observation and $m_i$ and $\sigma_i$ are the fluxes and their uncertainties at those times.

You then calculate an array of chi-squared statistics, by fitting a mean level $\bar{m}$ and sinusoidal functions over a range of discrete frequencies $\omega$. The aim is to maximise the statistic
$$p(\omega) = \frac{\chi^2_{0} - \chi^2(\omega)}{\chi^2_{0}},$$
where
$$\chi^{2}_{0} = \sum (m_i - \bar{m})^2/\sigma_i^{2}$$ and
$$\chi^{2}(\omega) = \sum  (y_i - \bar{m})^2/\sigma_i^{2},$$ with $y_i = A\sin(\omega t_i + \phi) + A_0$, where $A$, $A_0$ and $\phi$ are free parameters that are optimised for each frequency (i.e. you find the minimum possible $\chi^2(\omega)$ at each frequency).

Once a period is obtained from a peak in the power spectrum, $p_{best}$, then it must be established whether this peak is "real" or significant.

Theoretically, the probability that $p>p_{best}$ is $(1-p_{best})^{(N-3)/2}$ for $N$ independent measurements (see Cumming, Marcy & Butler 1999). This is the "false alarm probability" - the chance that you would get $p>p_{best}$ from random, aperiodic data with those uncertainties.

However, real data has correlated noise and the points are not independent. So what you do instead (see Collier Cameron et al. 2009) is shuffle all the times of the data points (i.e. randomly reallocate the times of each observation) and go through the process again. Do this a squillion times with different random reallocations and see how many times you end up with $p>p_{best}$. This gives you some idea about how confident you can be that your detected periodic signal is real. This is called a "bootstrap Monte-Carlo method".

There are other ways to do this; my answer is not complete; but it's the one I'm familiar with.


Why is the Moon smooth near the equator and bumpy near the poles? It looks as if it was polished by applying emery to its spinning equator.

My initial guess was that it was simply an illusion caused by light direction, but the terrain map shows the same phenomena.

Google Moon photo and terrain maps (source)

North pole mosaic assembled from photos by NASA's Lunar Reconnaissance Orbiter (source):

South pole shot by NASA's Clementine (source):

There are a few factors that create this effect:

First in the visual light images, the view around the equator is seen back lit, with almost no shadow, the few around the pole is seen with low lighting, and low lighting brings out features of the terrain. The Polar mosaics have an unreal quality, since they are made of images all taken when the sun is locally at its maximum above the horizon, so the light appears to be coming from all directions at once.

The first mosaic and the topographic map have a projection that is "conformal" this means that circular craters appear circular on the map, but it does introduce very large distortions of scale this, combined with shading which has been added to the map, makes regular-sized craters near the poles appear huge. The shading scales with the map, so craters near the pole have much more dramatic shading than those near the equator, that is an artistic effect, not real.

Finally on one face of the moon, that which faces us, and is in the middle of the pictures, there are large "maria". These are relatively flat lava plains. There are no maria at the poles. Most are on the near side of the moon: see Why are most lunar maria on the visible side?

So this is a combination of low angle of light at the poles, the way that google has projected and shaded their map and the distribution of lava plains on the moon. The poles are not more bumpy than the equatorial region in general. They are more bumpy that the lunar maria, but the apparent effect you describe is an illusion.


Why is the magnetic reconnection(of sun) effect many orders of magnitude faster than predicted by standard models? Why is the sun's magnetic reconnection effect many orders of magnitude faster than predicted by standard models?

We don't know for sure.

If we knew the answer, then we'd have a new standard model and then the reconnection speed would be nicely comparable to what's predicted. (zibadawa timmy)

However, the typical models of magnetic reconnection do not account for the extra resistivity that results from small-scale electromagnetic waves, which can be upwards of ~12-15 orders of magnitude larger than that caused by Coulomb collisions. (honeste vivere)


Question about orbital direction on binary systems I wonder if is possible to find a binary system in which one star is orbiting the center of mass in one direction and the latter orbiting in the opposite direction.
Would this transgress angular momentum conservation?

Stars don't "orbit the centre of mass", they orbit each other.
What is true is that in a two-body system, the orbit of each star is an ellipse with the centre of mass at the focus.  What is not true is that the acceleration is towards the centre of mass.  There may be nothing at the centre of mass to attract!  The acceleration of each star is directed towards the other star. This is the "law of gravity". So it is quite accurate to say "Each star orbits the other star".
An orbit in which one star moves clockwise and the other moves anti-clockwise would break the law of gravity, because the acceleration would not be directed towards the other star.


Which lens is right to see which planets with an Orion StarSeeker 150 mm? I'm an amateur astronomy lover and I have an Orion StarSeeker IV 150 mm, together with a bunch of lenses, which are the following:

1x 23mm (32.6x) and 1x 10mm (75x) wide-field 60-degree eyepieces 1x Plössl 32mm 1x Plössl 12,5mm 1x Plössl 6mm 1x Plössl 4mm 1x Barlow lens 2x 1x green filter No.56 1x red filter No. 23A 1x yellow filter No.12 1x orange filter No.21 1x blue filter No.82A 1x polarizer filter 1x lunar filter Crystal View.

I've been able to see the moon using the Barlow Lens and a 10mm lens, but I can't see other planets, star clusters and other stellar objects. I've searched on the internet, but haven't been able to find information that is understandable for an amateur, so here I am.

Which lens is best to see which planets and star clusters?

Thanks to whoever will help me :)

You SHOULD be able to see Mars, Jupiter, and Saturn easily with your scope with any eyepiece.  In fact, You should be able to see them easily naked-eye.  But, I assume you want to see more detail, not just a bright dot in the sky.

Your telescope is a 150 mm Newtonian reflector with a focal length of 750 mm (giving you a focal ratio of f/5).  It looks like you have six eyepieces, each with different focal lengths, and a 2x Barlow.  Based on that, you should get the following magnifications:

Eyepiece        Mag         With Barlow
----------      -------     ---------------
32 mm           23x         47x
23 mm           33x         65x
12.5 mm         60x         120x
10 mm           75x         150x
6 mm            125x        250x
4 mm            188x        375x

Realistically speaking, anything over 200 or 250x is going to be a waste in this telescope.  The maximum useful magnification of any telescope depends on a fe factors, such as aperture, eyepiece and telescope optics quality, atmospheric steadiness (seeing), humidity, air pollution, etc...  While you may often hear people say the maximum useful magnification is about 50 or 60x per inch of aperture, in reality, on an average night, it's more like 25 to 30x.  Based on this, you shouldn't expect anything over 200x to be useable very often, maybe 150x, unless you live in a place with unusually clear, steady air (like living at high altitude in the mountains).  Actually, I just looked, and Orion says the maximum useful is about 177x, which would be based on 30x per inch of aperture.

Still, with even 177x, you can get some pretty good views of planets.  The 4mm will give you 188x, which should be a nice view, though probably a little fuzzy.  dropping to 6mm will give you 125x, and that shouldn't be too bad.  For Jupiter, you should be able to make out cloud bands and possibly the Great Red Spot, if it's in view.  And, of course, you can see the four Galilean moons.  With Saturn, you'll be able to see the rings, but I wouldn't expect to see the Cassini division - but you might get a hint that it's there.  With Mars you can make out hints of the polar caps, and just maybe a large feature like Syrtis Major.  The filters may help with planetary viewing, helping you bring out the details by increasing contrast.

Here's a couple links to useful discussions of filters for planetary viewing:

Sangamon Astronomical Society Filter Advice
Agena Astro Planetary Filter Info

You will find it easiest to start out with a wider view eyepiece, like your 32mm.  Get the planet centered, then switch out eyepieces. Trying to get right-on a target with a high-power eyepiece is much more difficult, even with GoTo.

For other objects, things are a bit different.  Deep Sky Objects (DSOs: star clusters, galaxies, and nebulae, basically anything outside of our solar system), are many, varied, and can be quite difficult.  Even through larger telescopes, they are often nothing more than fuzzy patches.  They will NOT look like you see in pictures.

Your scope has an alt-az GoTo mount.  This is not always a good thing.  It must be properly aligned, and even then, a lot of alt-az GoTo mounts just aren't all that accurate. When you do your startup routine, it should ask you for at least two alignment stars.  You need to get these aligned as exactly as possible.  If it gives you the option to add more, do so. The more you add, and the more accurately you align them (meaning centered in the field of view, not just somewhere in the eyepiece), the better the GoTo and tracking performance will be.  I've had observing sessions with GoTo scopes where they were off by several degrees.  If you don't know your way around the sky to begin with, you have no idea where you're looking.

For DSO's, aperture is important, but so is getting the right exit pupil.  A guy I know on Reddit wrote up a great discussion of exit pupil you can read here.  Simply put, you should tailor your exit pupil to the need.  This partly will involve measuring your own pupil (this Sky and telescope article includes how to measure your own pupil size).

Finding them, then is the issue.  Even with GoTo, it's not always easy. And just because your GoTo scope points at the object (if it's properly aligned) doesn't mean you can see it.  The description on Orion's Website of your telescope's GoTo capability says: "Huge GoTo database of over 42,000 objects, and handy built-in Tour mode lets you enjoy great views on any clear night."  Unfortunately, a lot of these objects simply aren't going to be viewable in your scope, or from where you're observing.

And observing site is an important consideration.  If you're in an area with a lot of light pollution, the sky background will be so bright that you can't get enough contrast between it and the object you're trying to view to make the object visible.  Still, that said, there are a lot of things you should be able to see from even moderate light pollution.  I would start with the Messier objects, as they are some of the brightest DSOs in the night sky and fairly readily found (a couple are even naked-eye objects).  There should be a menu option in your scope's hand control for these.  If you're properly set up and aligned, it should be able to find them for you.

I mentioned above, if you don't know your way around the sky, it makes it harder.  You should definitely take the time to learn the constellations, at least the more common ones, as well as the names of some of the bright stars.  Most GoTo alignment systems ask you to locate 2 or 3 bright stars by name, and they mostly use the same stars.  There's between 150 and 200 commonly used bright stars for this purpose.  I wouldn't say you have to memorize them all, but if you can find a good chunk of them, say maybe 50-75 of the most common, it makes it easier to find your way around the sky.  If someone says "it's 5 degrees south of Betelgeuse" or "Find Arcturus and move East a few degrees," you'll have a better understanding of what you're looking for.  Also, learning to estimate angular measurements (there's some great tricks out there for estimating degree measurements using fingers on an outstretched hand) can make a difference.

One of the best resources for the beginner is the book Turn Left At Orion by Guy Consolmagno and Dan M. Davis. It's written to help people new to the hobby learn their way around.  You can find it on Amazon, Barnes & Noble (bn.com), or order it from any good bookstore (if they don't stock it already).

Probably the BEST resource, however, is to join an astronomy club or society.  In the US, membership is usually $50 (USD) or less per year.  I suspect they are similarly priced in other countries.  They are a great place to meet people with the same interest.  The best part is, most such groups have people who enjoy teaching newcomers the tricks to finding things and how to really observe them.  They also usually have people who can teach you how to properly collimate, maintain, and even repair your telescope.  And a lot of clubs have their own observing sites, or at least have scouted out good observing sites in the area.  It's really a worth it to join a club.

My club, the Houston Astronomical Society, only costs $36 per year, and we have some great benefits for members (including loaner telescopes, a dark site for observing, and even small observatory).

Good luck and clear skies!


How can a star's rotation speed be measured? How can a star's rotation speed be measured?

If stars could be resolved, then you could use the Doppler shift to see that half of the star is bluer and the other half is redder,  but stars (other than our Sun) are seen as point objects. With only one point of measurement, how can the rotation be inferred?

This article mentions that stellar rotation speeds have been measured, but doesn't explain how.

The projected equatorial velocity of a star (commonly written $v \sin i$, where   $i$ is an inclination angle for the rotation axis to the line of sight) is measured by observing the Doppler broadening effect on spectral lines.

There is no need to resolve the star. Light from one limb is approaching at $+v\sin i$, whilst light from the other limb is receding at $-v\sin i$. The net effect is that the spectral lines are convolved with a broadening function that has a full width of $2 v\sin i$.

Providing that this rotational broadening makes a significant contribution to the overall line width, then it can be measured (by direct fitting, Fourier transform or cross-correlation methods).

True rotation speed requires a measurement of rotation period and knowledge of the stellar radius. Whilst the former is known, the latter usually isn't.

Note in your article, I suspect the rotation speeds are actually the inverse of rotation periods. Angular velocities are often referred to as rotation rates and can be derived without spectroscopy by measuring periodic light variations from surface inhomogeneities on a star.


Orbital resonances and position of planets I am interested in quick generation of planetary systems for programs similar to Space Engine or Pioneer Space Simulator. As far as I understand, formation of planets is a complex process and the planet positions are to a great extend given by the orbital resonances, particularly with the gas giant planets. It is very clearly visible in this talk of Sean Raymond.

My question is: How exactly do the orbital resonances affect the stability of the planet trajectories? Which resonances are stable, and which are unstable in the long term evolution of the planetary system?

Watched the video, and I thought it was really cool seeing the modeling of different young solar-systems.  The orbital resonances they were talking about were for smaller objects that form debris fields in resonance with a giant planet.   Since they only looked at models for the inner planets, the closest example to the bit in the video with our current solar-system would be Jupiter's Hildas.  For that specific part of the video, they only showed one inner most giant planet.  For more than one, Resonances like that are unlikely to form between giant planets, like in our solar system where most of the resonance objects fall inside Jupiter or outside Neptune.

The mathematics gets too complicated for me to know for sure, but as I understand it, planets in resonance is neither stable nor unstable, unless the aphelion of one planet is in resonance with the perihelion of another, in which case the planets could drawn together I would think.  That's my amateur answer to your question.  I wouldn't think orbital resonance affects stability.

When one planet is dominant and the other quite small, then resonances appear largely stable (Neptune/Pluto for example), provided they never get close to each other, which Neptune-Pluto don't.

Jupiter and Saturn were thought to have entered orbital resonance at one point in the early solar-system and that, from what I've read, affected other objects in the solar system but didn't affect their relation to each other much.  I think the only way to get any kind of answer is to run long term simulations on various model.   In short, planets can be stable both in resonance with each other, and not in resonance and planets can be unstable, again, either in resonance or not in resonance, though if they have unstable orbits, they won't stay in resonance for long.


Do parsecs scale as an inverse proportion with parallax angle? high school student here.
In astronomy my teacher says that if the parsec angle ($\alpha$) of a star is 1/50 arcsecond, then that star is 50 parsecs away.
I'm not really convinced - calculation of distances from parsecs requires using the formula $$\text{distance in AU}=\frac{1 AU}{\tan(\alpha)}.$$
This means that distance is inversely proportional to $tan{\alpha}$. Distance is not inversely proportional to $\alpha$.

I know you can use a small angle approximation, but that feels to me as very dodgy - you can get fairly large discrepancies in the calculated distance depending on whether you use $tan(\alpha)$ or $\alpha$ to find the distance.

So, what is the best practice when calculating distances in parsecs? Is my teacher's approximation acceptable in "real life" astronomy?

For angles of magnitude 1/50 arcsecond (of the order $10^{-7}$, the difference between $\tan\alpha$ and $\alpha$ is truly negligible (if alpha is in radians) and only a matter of a choice of units if you don't. It is quite correct to say: distance (in AU) = 1/angle (in radians)  to get the distance from the angle in radians.

Using parsecs a means that we don't need to use a different unit for the angle, and the point of parsecs is that we can use "arcseconds".
So it is correct to use distance (in parsec) = 1/parallax angle (in seconds)


Why could I so easily see and photograph such a bright totally eclipsed Moon from a bright city road? (May 26, 2021 total lunar eclipse) Due to scheduling and geometry I could only snap last night's lunar eclipse with an older model cell phone on a pedestrian overpass of a brightly lit city street, but surprisingly the Moon was quite bright even during the totality phase (U2 to U3 (14.5 minutes) also see below) when the Moon was within the geometrical umbra of Earth's shadow.
   2021-05-26
----------------
U1 = 09:44:57 UT     first contact with umbra
U2 = 11:11:25 UT     begin totatity
U3 = 11:25:55 UT     end totality
U4 = 12:52:22 UT     last contact with umbra

This eclipse just barely grazed the inside of the umbra.
The last total lunar eclipse I remember was quite a long time ago. It was more central and the Moon barely visible with night adapted eyes viewed from the countryside, so it's not a fair comparison.
But I'd like to ask anyway:
Question: Was the radial gradient of Earth's umbra unusually bright for the May 26, 2021 total lunar eclipse? Might there have been more light scattered from Earth's atmosphere into the umbra than average, due to weather or perhaps the recent eruption Mount Nyiragongo in the Democratic Republic of Congo? Or was the brightness of the northern edge of the Moon pretty much normal for this geometry?

above: Snapshot at 2021-05-26 11:23:58 UTC North is left (altered only to obscure a phone number on a sign)
below: from eclipse.gsfc.nasa.gov's Total Lunar Eclipse of 2021 May 26 found here. North is up

Based on your description of your experiences, my guess is that it’s not that this eclipse was especially bright, but rather that the previous one you saw was unusually dark.  My experience (late 1990s to present) has always been that a totally eclipsed moon is easy to see even under bright urban conditions, but this page says that there were some eclipses in the 1980s and early 1990s that were hard to see due to recent volcanic eruptions.  Do you know the date of the previous eclipse you saw?


Lightkurve nan values after bin light curve I am using lightkurve 2.0.2 library with Python 3.8.5 and astropy 4.2 for processing exoplanet transits. However when I want to bin light curve to fixed number of points, all values in light_curve.flux except first two are nan. What I am doing wrong?
The purpose of code bellow is to obtain transit of Kepler-10 a from light curve of Kepler-10.
import lightkurve as lk  # Import lighkurve package

tp = lk.search_targetpixelfile("Kepler-10", mission="Kepler", exptime="long", quarter=1).download()  # Download target pixel for Kepler-10
lc = tp.to_lightcurve().flatten().remove_outliers()  # Convert target pixel to light curve
fold = lc.fold(0.837)  # Fold light curve to period of Kepler-10 a (0.837 days)
bin = fold.bin(n_bins=101)  # Bin light curve, so it should contains only 101 points.

print(bin.flux)  # [0.99999749 0.99999977 nan nan nan nan nan nan nan ... nan nan nan nan]

lc and fold are valid light curves without NaN values. E. g. fold looks like this:

So if there is any problem, it should be on line with fold.bin. Also, this is not problem only for Kepler-10.

Thanks to all for your help. Because of the planetmarker's comment I found out that it is also necessary to set the time_bin_size parameter. After upgrade to lightkurve 2.0.6, there is also parameter bins, which sets time_bin_size automatically.
bin = fold.bin(bins=101)  # bins instead of n_bins


Was the "green star" event in NGC 3314 ever figured out or named? from NGC 3314 Variable Object by By Bill Keel (University of Alabama, Tuscaloosa) and Lisa Frattare (Hubble Heritage Project, STScI). The object in question is circled.

Comments below this answer led to Dust in Galaxies which shows the image of the two visually overlapping galaxies together called NGC 3314 which says

In a March 2000 observation of the galaxies, a prominent green star-like object was seen in one of the arms. Astronomers theorized that it could have been a supernova, but the unique filtering properties of the foreground galaxy made it difficult to decide definitively.1
1Keel, Bill; Frattare, Lisa. "NGC 3314 Variable Object". heritage.stsci.edu. Retrieved 2015-11-05.

which explains the story, including:

Tracking down a stellar explosion
We first noticed a new object in NGC 3314 while discussing the details of color rendition for the Hubble Heritage release, incorporating data from both the 1999 and 2000 observations in all four filters. There was a prominent green starlike object in one of the spiral arms. "Prominent" is relatve here; the visual magnitude was 21.6. Green stars are not found in nature, because the spectral band we see as green is narrow compared to the amount of light stars put out in various wavelengths, so this was a signal of something quite unusual. This was quickly traced to an object appearing only in the March 2000 observations, so that combining data from some filters in which it was there and some in which it wasn't produced the giveaway color. The fact that we had blue-light images from both sets of data let us show that it had indeed newly appeared, and was not some object with a bizarre spectrum that vanished in some filters. In asking what kind of new object this was, we could quickly eliminate several possibilities:

and ends with

To get this observation on the record, and seek confirming or prediscovery observations, we reported it to the Central Bureau for Astronomical Telegrams, who included it in International Astronomical Union Circular 7388. To date we've gotten no reports of other observations, or of helpful non-detections in 1999 or 2000. So if you happened to take a deep image of NGC 3314 during this period...

Question(s): Rather than ask a half-dozen questions that likely won't be answerable individually, I'll lump them here. If there are answers to some but not others I'll accept them and edit the question to narrow it to what's answerable.

Were any more images found?
Was some part of a light curve established?
Was the object ever confirmed to be a supernova?
Does this "little green object" have a name or identifying designation of some kind beyond "NGC 3314 Variable Object"?
I can not understand the explanation for why it was thought to be green, is there a way to express it in simpler terms?

IAU circular 7388 says the following but doesn't give a name

POSSIBLE NOVA IN NGC 3314
W. C. Keel, University of Alabama; and L. M. Frattare, Space
Telescope Science Institute, on behalf of the Hubble Heritage Team,
report the detection of a possible nova in the overlapping galaxy
pair NGC 3314A and 3314B.  HST WFPC2 observations with the F450W,
F555W, and F675W filters on Mar. 10.47-10.57 UT show a new stellar
object that was not present on WFPC2 F450W (to B about 25.5) and
F814W images from 1999 Apr. 4.  Standard (approximate)
transformations to the UBV system give magnitudes B = 22.42, V =
21.64, R = 20.87.  The new object is located at R.A. =
10h37m12s.82, Decl. = -27o40'51".5 (equinox 2000.0), which is 1".2
west and 10".3 north of the optically brighter foreground nucleus
of NGC 3314A (itself at position end figures 12s.91, 41'01".8).

tl;dr No additional data of interest, but I explain how I searched, and I can explain the green color.
It seems that there have not been any additional observations of that object, which perhaps isn’t too surprising given how faint it is - 21st magnitude is possible only with big telescopes.
Simbad lists that object under the name of “EQ J103712-274051” but doesn’t list any other names, nor any references besides the IAU Circular already mentioned.  That’s a purely coordinate-based designation, with the “EQ” meaning it is based on equatorial coordinates.  Simbad says that that designation is internal to Simbad for objects of unknown nature and is to be avoided in the literature.
Looking at ADS, the only citation for the IAU Circular is a 2019 paper by Bonanos et al., which is a catalog of HST variable sources.  That seems promising on its face, but digging into the paper it turns out that the sources they flag are different ones, and the citation is in a footnote that says, “This object is unrelated to the transient in NGC 3314 reported in the IAU Circular 7388 by Keel & Frattare (2000).”
I also looked for further observations in Vizier- that can be a good way to find observations of a given source, even if it wasn’t called out specifically in a publication. While that search yields lots of objects within 5”, especially in HST observations, none of them seems to quite match up in coordinates and magnitudes with what Keel & Frattare report in the IAU Circular.
Regarding the green color that drew their attention to the object in the first place, here’s my interpretation of what they are saying there.  First, note that HST filters are (usually) named with “F” for filter (as opposed to “G” for grism), a number that gives the filter effective wavelength in nm, and then “N”, “M”, or “W” for “narrow”, “medium”, or “wide”, indicating the width of the filter bandpass.  So the F450W filter is similar to Johnson B, for example, and the F555W and F675W filters are green and red, respectively.
From 1999, when the object wasn’t bright, they had blue and red images.  From 2000, after it brightened, they had blue, green, and red images.  (The 1999 “red” data, F814W, are actually near-infrared, but I’m guessing they used those in the red channel for the Heritage image.)  So in combining to make an RGB (color) image, they had two each of red and blue images (one before and one after brightening), but only one green image (after brightening), which they would have needed to weight doubly in the composite to get a reasonable color balance. That’s why that object ended up looking green - they weighted the “bright” green data twice as much as the “bright” red and blue data.
EDIT:
I just realized that it would be possible to test the idea above
by downloading the individual filter images from the HST data and
looking at them directly.  Fortunately, the Hubble Legacy
Archive makes this pretty easy to
do, and even provides pre-combined color images for datasets where
multiple filters were taken.  So we can directly download images made
from data only before and only after the outburst of that object and
see what they look like.
This
search
gives all WFPC/2 images of NGC 3314, and we can see that there are
some from 1999 and some from 2000.  Those are the pre- and
post-outburst data, respectively, that were combined in the original
press release images shown above.   If you look in the column that says "Level", the Level 4 data products are color composite images.  The two options there are composites of the 1999 data and of the 2000 data.  Let's look.
Here are the 1999 (pre-outburst) images, which are only in red and
blue filters.  Presumably the color composite fills the green channel
with some combination of the red and blue:

And here are the 2000 (post-outburst) data, combining images in red,
green, and blue filters:

Below I've zoomed in and taken a screenshot, trying to get the images
centered as close to the same as possible.  I've positioned them so
the object in question is just to the left of the red crosshairs.  Top is 1999, bottom is 2000.
1999:

2000:

You can see that the newly-brightened object is basically white, not
green.  The individual, isolated red, green, or blue pixels are noise.  In any image, you will have some pixels that are extra-bright, either from noise in the detector, or from a cosmic ray hit that was not removed entirely.  Since they are random noise, they aren't in the same place in every image.  So if you make a color composite image, then those randomly-bright pixels will show up as the color of whatever filter they appeared in, since they are bright only in that image and dark in the images taken in the other filters.


Use of type-I a supernovae as standard candle Why only type -Ia supernovae are used as standard candle to calculate age of universe and why not type II, type Ib, type 1c ?

A type 1a supernova forms when a white dwarf grows through accretion to a certain size, at which it becomes unstable. This means that the precursor object is always a white dwarf of mass 1.39 solar masses. As the precursor object is always of the same type and the same size, the supernova is thought to be the same brightness.

On the other hand, type II supernovae form from the collapse of giant stars. The precursor object can have a range of masses from about 10 Solar masses, to a maximum of over 100. The brightness of the supernova depends on the mass of the precursor (and other factors such as the metallicity) and so type II supernovae are not all equally bright.

The other types also have precursors which vary in size or other factors, only type 1a supernovae are thought to always have a white dwarf of mass 1.39 as a standard precursor, so only type 1a are suitable as a standard candle.


How would an exoplanet be found from earth if our view of its star system is "from the top"? When we find an exoplanet by observing the dimming of its parent star, this means that the planet passes in front of the star from our view.

What if our view of a star system is from its "top" or "bottom"? Then from our view, none of its planets would pass in front of the star so we could observe a dimming. How would we know, then, whether such a star has or doesn't have planets?

First of all, it is not that you can always find the planet (if it is there).

Anyway, if this is the case, when we observe the system not totally face-on, we resort to radial velocity measurements. The principle is simple, the actual much less.
The orbiting of the planet around the star causes the star to "oscillate" around the center of mass.
Here you can see a good example:

However, when the system is exactly face-on, the Doppler shift is useless, but still the gravitational effect of the planet onto the star is useful and used.
Here is what happens in the latter case:

It is clear that the star undergoes a change in position.
Then usually, the radial measurements go along with astrometrical measurements, to identify the position of the star along the planet's orbit.

Of course, you need some conditions are verified to allow this measurement.
First of all, you planet must be massive enough to bring a sensible change in the position of the star. Then your instrument must be sensible enough to resolve the tiny effect originated by this change.
Possibly, also other conditions play a role, but I am not an expert (you can take a look at here).

For a general comprehension, also a wiki for Methods of detecting exoplanets.


How can astronomers determine the difference between "hydrostatic equilibrium" and "just happens to be spherical"? This is relevant for the definition of a dwarf planet.

I presume the answer will be, well, if we can tell the mass of the body and guess the material.   I don't find this very satisfactory because (1) may be impossible and (2) will have large error.

I think you're asking, "If we know an object's shape, can we determine if it is in in hydrostatic equilibrium?" If so, one might wonder if astronomers classify basketballs or ball bearings as being in hydrostatic equilibrium, since they're so spherical.

Below about 100 km in radius, the answer is in general no. Given some population of randomly lumpy objects (like asteroids), some of them will be close to the shape of a sphere purely on accident. The composition matters too -- an object of this size made of hydrogen gas would assume a spherical shape from hydrostatic equilibrium, but an object made of rock might not (like Mathilde below). We could make better predictions given detailed knowledge of the materials and environment of the object, but this is not always possible, as you mentioned. For small objects, intermolecular and atomic forces dominate gravity.

Once you get to a certain size of object, it becomes much easier to make a prediction about hydrostatic equilibrium. This is still very much context-dependent, and you still get complications from material composition, temperature, etc. However, atomic binding forces have some fixed strength, but gravity scales like mass. Given ordinary astrophysical materials, we can be very sure that a body like Jupiter is under hydrostatic equilibrium.

You can make some order of magnitude estimates by assuming that the atomic interaction energy must be at least as large as the thermal energy (Hughes and Cole 1995). If you check out equation 5 of that paper, you'll see an explicit expression for a radius that divides spherical objects and nonspherical ones. At some mass, the binding atomic energy becomes dwarfed by the gravitational potential, and you always get a spherical object.

tl;dr -- small objects no, big objects yes, medium objects may require detailed modeling.


What is the minimum mass required so that objects become spherical due to its own gravity? Is this minimum mass known? or maybe, is it given in terms of density? If so, how much density is the minimum to have an spherical object due to its own gravity?

Icy objects, such as most in the Kuiper belt can reach an equilibrium if they are about 400km across, whereas the rocky asteroid Pallas, at 572km clearly has an irregular, non spherical shape. All rocky objects larger than Pallas (and there aren't many) are spherical.

Rock tends to be stronger than ice. Rocky objects are able to withstand their own gravity for longer than icy ones. Pallas is a reasonable cut of point. The next smaller asteroids (Vesta, Hygiea etc) are round-ish, but not in hydrostatic equilibrium. On the other hand Small, icy moons such as Miranda and Mimas are in, or close to equilibrium. Mimas has a diameter of just under 400km.


How much greater is the star density in our galactic bar? How much denser is it in the galactic bar than the "normal" density at the same radius?

Is it just a few percent? or is it, say, "three times" as dense?

Or are other factors at play: star brightness, gasses?
Or do we really not know?

In other barred galaxies which are vaguely similar to the Milky Way, the contrast in (projected) stellar surface density between the bar and the inter-bar region at the same radius (e.g., along the bar's minor axis, perpendicular to the bar) is typically a factor of at least two; in particularly strong bars it can be as high as six (see, e.g., Figure 5 in Ohta et al. 1990). Similar contrasts are seen in N-body models of disk galaxies that form bars.

It's much harder to figure this out for the Milky Way, because we're not looking down it from above. The best attempt to derive a model of the bar's 3D stellar density from star counts and distance estimates that I know of is Wegg et al. (2015). From the face-on projected view of their model (their Figure 14), I would guesstimate the maximum contrast as a factor of 4 or so.

Figure 14 of Wegg et al.: projected face-on view of stellar density for the Milky Way (full model in right-hand panel).

The 3D density (which is maybe what you're really asking about) in the inner part of the bar is not quite as great as this suggests, because the inner part of the bar is vertically thick, forming a "boxy/peanut-shaped" bulge (this would correspond to the red region in the figure above). So the contrast would be a little less compared to the (less thickened) inter-bar region. But the outer part of the bar is roughly as thin as the rest of the disk, so the projected surface density contrast would mean a similar contrast in 3D stellar density.


Planetary gas giants I've read that a gas giant, like Jupiter, doesn't have a rocky surface. But planets must start as a rocky conglomerate of flotsam.

Is the interior so hot that all the material is molten and there is not a crust?
Does the atmosphere creates enough pressure to create a supercritical fluid?

What is the situation on Jupiter?

Short answer: 1) Yes and no; 2) Yes, there is a supercritical fluid, of hydrogen.

Long answer:

It's fairly hot deep down in Jupiter; estimates range from 10,000 K to 24,000 K. You would think that anything in the core would be liquefied, and you could be right. Many models predict that Jupiter's core is rocky, but others predict that it is liquid. Still others say that it is "liquefying" - i.e. parts are liquid and parts are solid. We don't know for sure which are right.

In the early solar system, Jupiter would indeed have to have had a large core of rock (and potentially ice). It would have accreted gases, and eventually the planet we know today would have formed, with thick cloud layers complete with hydrogen, ammonia, and lots of other elements. But after Jupiter accumulated its mantle, there would have been a lot of pressure on the original core. It seems like much of the matter inside it would have been liquefied and carried away ("redistributed") into other parts of the planet. Still, some might have remained.

Surrounding the core, there is a layer of liquid hydrogen. Yep, liquid hydrogen, just as you said in your question. Temperatures and pressures are so high that hydrogen has passed its critical temperature (33 K) and become a supercritical fluid.

So, in summary, we don't know for sure the state of Jupiter's core is, but parts or all of it may be liquefied, and we also think that there is indeed a layer of hydrogen as a supercritical fluid surrounding the core. The same may also be said for other gas giants, although, still, we don't know for sure.

I hope this helps.

Sources:

http://www.dailygalaxy.com/my_weblog/2013/02/jupiters-dissolving-core-sheds-light-on-alien-planets.html
http://www.universetoday.com/47966/jupiters-core/
http://www.nasa.gov/audience/forstudents/5-8/features/what-is-jupiter-58_prt.htm
http://phys.org/news/2011-12-jupiter-core-liquefying.html
https://en.wikipedia.org/wiki/Jupiter#Internal_structure
http://arxiv.org/pdf/1111.6309v1.pdf (You might find this particularly interesting, as it considers gas-giant exoplanets)


Could not slow neutrinos orbit galaxies and clusters, thus comprising a large component of even COLD dark matter? Cold dark matter is the type of dark matter that is most eminently neutrino-free. But neutrinos themselves suffer a large survivorship detection bias (“all the neutrinos you can detect necessarily have to have relativistic speeds”) https://physics.stackexchange.com/questions/267035/where-are-all-the-slow-neutrinos
By the same principle that there are more pebbles than boulders, and more red dwarfs than blue giants, slow neutrinos should be more abundant than the fast, detectable ones.
Non-relativistic (slow) neutrinos - in particular those going at below the galactic escape velocity - could be a major competent of cold dark matter: slow neutrinos could be slow enough to orbit galaxies and galactic clusters and thus form a significant amount of cdm, which should be shaped as a galactic halo.
If neutrinos are produced at all speeds, those stellar (modern) neutrinos below galactic escape velocity will accumulate indefinitely. This should generate a halo-shaped cloud, and this could comprise a large part (of course 15% are MACHOs etc https://ned.ipac.caltech.edu/level5/Sept17/Freese/Freese4.html - I am assuming here a multi-component DM) of cold dark matter.

These neutrinos would have to be really cold. The cosmic neutrino background is at 1.9K, and they are considered hot dark matter, because they would have been highly relativistic at the epoch of structure formation. To be considered cold dark matter, and also to be captured in orbits in galaxies, the neutrinos would have to be much colder - totally non-relativistic now.
Let's assume an average energy then of about 0.1 eV for each neutrino (similar to their likely rest mass energy). In order to account for $\Omega_{\rm CDM}\sim 0.3$ there would need to be $5\times 10^{10}$ per cubic metre, or about $10^{10}$ per cubic metre per flavour, on average over the universe.
To account for dark matter in galaxies (e.g. the Milky Way) we need $\sim 10^{12} M_\odot$ within about 100 kpc, meaning a number density of neutrinos of $10^{14}$ per cubic metre.
These neutrinos are spin 1/2 fermions and thus would have a Fermi energy of about $5\times 10^{-3}$ eV. That would mean if they were any colder than 6 K, they would be degenerate and exert a degeneracy pressure. This would be sufficient to prevent the halos forming - Tremaine & Gunn (1979) showed that cold dark matter halos cannot be made from low mass leptons like standard neutrinos.
Edits:
A back of the envelope Tremaine-Gunn limit (see also Boyarsky et al. 2009) is to assume that the escape speed of the galaxy is $v$, it's dark matter halos has radius $r$, the total mass is $M$ and the neutrino mass is $m$.
The number of available quantum states to spin half fermions in this volume, up to a momentum $mv$ is
$$ N = \left(\frac{4\pi r^3}{3}\right) \left(\frac{8\pi}{3}\right) \left(\frac{mv}{h}\right)^3$$
We can write
$$ v = \left(\frac{2GM}{r}\right)^{1/2}.$$
The mass contained in these particles cannot be greater than if every quantum state is filled by one fermion of mass $m$ and if this is to explain dark matter, then this mass must be $\sim M$. Thus
$$M < m\left(\frac{4\pi r^3}{3}\right) \left(\frac{8\pi}{3}\right) \left(\frac{m\sqrt{2GM/r}}{h}\right)^3$$
and
$$mc^2 > 8.9\left(\frac{r}{\rm 100 kpc}\right)^{-3/8} \left(\frac{M}{10^{12}M_\odot}\right)^{-1/8}\ {\rm eV}\ .$$
Thus there are not enough quantum states to accommodate a halo of fermions unless their rest mass energies exceed about 10 eV. For neutrinos, there are 3 flavours and anti particles, which reduces this number by $6^{1/4}$, but conversely it must be increased because particles in the halo cannot be uniformly distributed in velocity between 0 and $v$.
10 ev is about two orders of magnitude larger than the likely rest masses of the known neutrinos.
The idea that neutrinos from stars can make any contribution to dark matter halos is untenable. The vast majority of solar neutrinos have energies above 0.1 MeV, and so for an assumed neutrino rest mass energy of $\sim 0.1$ eV, they have Lorentz factors that exceed $10^6$ - i.e. they travel exceedingly close to the speed of light and are not confined to galaxies. The neutrinos emitted during supernova explosions are even more energetic. Secondly, even if you invent a magical process that can produce neutrinos with kinetic energies less than 0.1 eV, you still have the Tremaine-Gunn limit to overcome and even if all the rest mass of all the stars in a galaxy was converted into neutrinos, that mass would still fall short by more than an order of magnitude of that required to explain dark matter in galaxies. You cannot imagine that the neutrinos would "pile up" over time because the vast majority of the mass that has ever been turned into stars is still in the form of stars today and the stars that have already lived and died are a tiny percentage of the required dark matter mass.


Why can't we observe a solar eclipse and the preceding/following lunar eclipse from the same place? Lunar and solar eclipses go in tandem so the two often happens two weeks apart.

However I noticed if there is a solar eclipse in my area then the preceding/following lunar eclipse is not visible from my area because the Moon is set when it happens.

Similarly when lunar eclipse is visible in my area, the corresponding solar eclipse is not visible at my area because the Sun is set.

Recently the same thing happens in July 2 when there were a total solar eclipse on the other side of Earth, and soon we will see a partial lunar eclipse from my area.

The only exception was when the eclipse happens near the sunrise or sunset.
For example on May 31 in 2003 a partial eclipse could be seen from Hungary at the morning when the Sun rose already partially eclipsed just few minutes before the maximum occultation.

Two weeks before that there was a total lunar eclipse where we could only see the beginning of the eclipse as the Moon was already set when totality occurred.

These eclipse pairs seem to occur almost exactly at the same time of day.
Why?

For a solar eclipse, the Moon has to be between the Earth and Sun. For a lunar eclipse, the Earth has to be between the Moon and Sun.

Since it takes the Moon about 4 weeks to orbit the Earth, it will take it about 2 weeks to go from a lineup of Moon-Earth-Sun to a lineup of Earth-Moon-Sun.

To make this more complicated, the Earth is rotating so that it takes about 24 hours to go from midday to midday.

So, say we have a solar eclipse at midday - there's a line between the centre of the Earth, through you, the Moon and the Sun. Two weeks later, when the Moon has moved around to the other side of the Earth this means that when you go out at midday, the line is now Moon-Earth-you-Sun. So you can't see the lunar eclipse, because the Earth is now in the way.

You can see both eclipses two weeks apart when they are around sunrise/sunset because you're no longer on the line through the centre of the Earth between the Sun and Moon.

Update:

I was going to do the math - but I think that this map of eclipses in 2019 demonstrates it better. If you compare the solar eclipses with their paired lunar eclipses, we see that there is a bit of overlap (since eclipses are not instantaneous) - but since these are at the edges of the eclipse, it's practically impossible to see a full solar eclipse two weeks after seeing a full lunar eclipse from the exact same location.


Why are larger hex mirrors round and the smaller aren't? Per this diagram why are the larger (30 meter and European) mirrors filled out to nearly a full circle, while the smaller mirrors composed of hexagons retain a hexagonal border limit instead of filling out and approximating a full circle?

Start by  drawing the largest inscribable circle (circle completely contained within the mirror assembly).  Then  draw, e.g., the smallest superscribing circle (completely contains all the mirrors), and calculate how many more hexagons are needed to fill the additional area covered.

Finally, draw the circle showing the design entrance aperture of the telescope to see how much of the aperture is lost vs. how much of the outer hexagons' area is unused in each case.

Then do a cost vs.  quality tradeoff.  Take into consideration (for the smaller telescopes) the delta cost of using more, smaller hexagons.


How to interprete function value of get_sun in astropy As I know, spring equinox of 2016 is on 2016/3/20 4:30 GMT.

I execute following statements in python:

```

from astropy.time import Time

from astropy.coordinates import get_sun

t=Time('2016-03-20 4:30:00')

s=get_sun(t)

```

I expect to get s.ra.degree and s.dec.degree very close to 0.

But I got s.ra.degree=0.7052336 and s.dec.degree=0.30536325.

Why?

Hope some one can help me. Thanks.

Looking at the astropy documentation, the Sun's position is calculated in GCRS co-ordinates these are intern expressed in J2000.0 and not JNow.

The difference is therefore due in part to precession between J2000.0 and JNow.

You will also find that the time of the equinox is not exactly 4:30 but has been rounded. The algorithm used to get to 4:30 itself may also have precision limitations due to all sorts of reasons. The same will be true of the calculated position of the Sun at a given time.

These errors will all add up to give a slightly incorrect figure.

Cartes du Ciel (a well known planatarium application) calculates the Sun's location at 4:30 as being:
JNow
RA: 00h00m01.71s
DE:+00°00'01.4"
J2000
RA: 23h59m11.85s
DE:-00°05'23.6"

So it looks like most off the error is due to limits of the Astropy library.


Is there a term for "path of the Sun at a given point of the Earth" I want to buy a house and I'm asking myself what quantity of daylight the various sides of the house receive.
I know the house is roughly oriented in the east-west direction.
But I'd like to know the exact apparent path of the Sun around the house depending on seasons, with (ideally) the elevation angle of the Sun.
Is there a technical term for "the apparent path of the sun at a given point of the Earth, at a given time of the year"?
Also, is there an online tool that allows one to visualize this path on a map?

Merriam-Webster defines diurnal arc:

the portion of the diurnal circle of a celestial body that is above the observer's horizon

and diurnal circle:

the apparent circle or parallel of declination described by a celestial body in consequence of the earth's rotation

The NOAA Solar Calculator does not implement the exact feature in question but can show the Sun's azimuth for any given location, date, and time.


Bright star or flare near orion? This morning at 6:12 am looking south from Georgetown, Texas, there was a brief very bright stationary object what appeared to be a star that once I saw it within the next few seconds quickly vanished.  Did not appear to be other than a stationary object.  What was it?

Iridium satellites are known for predictable flare events but are gradually being de-orbited and replaced.
In particular, Iridium 66 would have flared at 6:04 several degrees east of Orion,
but its orbit has almost certainly changed since its de-orbiting process started on August 2.
Other Iridium satellites have taken 3 to 5 weeks to finish this process.

Other satellites can flare too if a shiny surface catches the Sun just right.
Heavens-Above shows two satellites passing south of Orion at 6:12 on the morning of August 19:
USA 215, an American spy satellite;
and Cosmos 2084, a failed Russian early warning satellite.

Another possibility is a head-on sporadic meteor.

Update: Iridium 66 finished de-orbiting on August 23.


Why are the magnetic poles of a pulsar so far off the rotational axis, yet stable? My understanding is that a pulsar is a neutron star whose magnetic poles don't coincide with the rotational poles and whose magnetic poles sweep through Earth (allowing us to hear radio noise).

If we keep hearing pulsars, the polar offset must be very stable.

Why is such an alignment offset so stable?

The magnetic field of a star is not entirely a result of the global spin of the star. The global spin is part of it, but there are other mechanisms as well. Within the star, there are convection zones, meridional flow, etc.

http://solarscience.msfc.nasa.gov/dynamo.shtml

All these flows generate their own field components. The overall field is simply the sum of all little fields. Its general orientation might be close to the global spin, if the strongest components are aligned to it, but there are many smaller components with different orientations. Therefore, the total field of the star can be somewhat slanted.

And then the star collapses into a neutron star, and its field is compressed. The collapse itself may be slightly asymmetrical, and may further deviate the magnetic field axis.

As a result of all of the above, it's by no means unusual that the magnetic field of the neutron star is not aligned with the spin.


Where do gas giants end up? I want to create a semi realistic star system generator but I am not sure where to "put in" gas giants. My guess is they can form pretty much anywhere within the sphere of influence of a star at the right circumstances.

I am still debating how to exactly go about generating the system but it won't be super complicated. I probably have plop in a star with a random mass and based on modifiers like distance, reflection, greenhouse and random chance plop in certain types of planets. It does not need to be realistic but it needs to feel realistic for none astronomy experts.

What you're asking about is not known at the moment, but I can briefly comment on how flawed our knowledge is.

1.) Theoretical predictions:

As one comment mentioned, the question "where do planets end up?" is dependent on the answer to "where do planets form?" and the answer to "how fast do planets migrate?" which is not well understood at the moment.
There are of course predictions for both those processes, but those essentially don't agree on anything and especially not with the observations.

A recent overview and a very short video by the authors of that paper summarizing it.

2.) Observations:

We have only very few multi-planet systems, and not a single one where we can claim we know all planets down to a certain size or mass limit.
I suggest you take a look at the data on http://exoplanets.org/table and use the filter "NCOMP >= 5" (also sort by name), this way you get current candidate systems with equal to or more than 5 planets.
You will see there are not many systems that we can base any systematic understanding on.

Plots like the first one in the answer to this question that seem to indicate that we know quite many systems, hide the fact that those systems are all single or double planet-systems.

This prohibits basing your generated systems based on data or planet formation theory. But I suggest that you can use a different method:

3.) Long-term stability:

We don't know all systems that are possible, but we certainly know that all systems under certain conditions become unstable. There is an overview article or also a long video outlining that there should be a simple stability criterion for planetary systems:
If any two planets come closer to each other than ~10 mutual Hill-radii, their orbits will become unstable and possible send the whole system into chaos.
The mutual Hill radius of two planets with masses $m_1$ and $m_2$, and central star $m_s$ and semi-major axis distances $a_1$ and $a_2$ from the star is

$$R_{hill,mutual} = \left( \frac{m_1+m_2}{3 \cdot m_s} \right)^{1/3} \cdot \frac{\left( a_1 + a_2 \right)}{2} $$

So my idea of a simple solar-system generator would be to choose some first planetary mass, place it somewhere.
Then build up the star system from the inside out planet by planet, by chosing the next planetary mass from one of the above mentioned 1-or-2-planet distributions (this one could be a realistic one, like in Mayor et al. 2011, if you know how to re-shape random distributions).
You calculate $a_2$ for them such that their mutual distance is $> 10 \cdot R_{Hill,mutual}$.
In this way you'll probably get pretty packed systems, but you can of course take their mutual distances to be larger than $10 \cdot R_{mutual, Hill}$.

So in this way the properties of your generated star systems would be:

Mutual distances: Stable physical foundation
Planetary mass distribution: Would be on shaky feet, as we don't know enough about multi-planet systems, and our data on low-multi systems is heavily biased towards the methods used to detect them.
System architecture / order of the planets: So far we've seen many unexpected things and there is no data or reason to assume there should be any strong constraints. I think here you can let your fantasy or the random masses generator roam free.

You could even go one step further and include one of the (half-way) realistic mass distributions for stars, to generate your $m_s$ values.

Your code (e.e. if you code in C) will only be able to give you uniformly distributed random numbers. If you want them to follow a specific distribution, you need to transform them. Google "transform probability density function" or ask someone with math education that you know, if you don't know how that works.


Could pure iron from a star make it to Earth? Since iron is a stars waste could a star that goes Super nova eject pure iron to travel across the galaxy?

https://space.stackexchange.com/questions/35101/99-99-pure-iron-meteorite

Sure. Iron is not made in significant quantities in the Big Bang, so 100% of the iron on Earth today was synthesized in a star. There's no other source.

It escapes in many ways:

Supernovae
Mergers between compact stars like white dwarfs and neutron stars
Stellar winds from more ordinary stars

Supernovae are the main source of modern-day iron, with r-process nucleosynthesis during the explosion producing a lot of iron and elements near iron.


What was James Bradley's calculation to calculate the speed of light? Looking into the history of how the speed of light came to be determined, to what we know it to be today, James Bradley is often mentioned. He was credited with the discovery of the aberration of light and having used the aberration of light to calculate the speed of light. There are many online sources that confirm this, though the year of discovery and calculated speed of light differs here and there.

I've read through Bradley's "A Letter from the Reverend Mr. James Bradley Savilian Professor of Astronomy at Oxford, and F.R.S. to Dr.Edmond Halley Astronom. Reg. &c. Giving an Account of a New Discovered Motion of the Fix'd Stars" published in "Philisophical Transaction vol 35" which explains his findings of the aberration of light but makes no mention of calculating the speed of light.

How did James Bradley calculate the speed of light and what publication of his shows these calculation?

Despite many online sources crediting James Bradley with having calculated the speed of light to a 295.000 km/sec (britannica) or 301.000 km/sec (wikipedia), a small nuance ought to be added. Like Ole Romer before him in 1676, James Bradley calculated the speed of light relative to something else but both never provided a value in Earth-based units.

He wrote,

  This therefore being 20’’,2, AC will be to AB that is the Velocity of Light to the Velocity of the Eye (which in
  this Case may be supposed the same as the Velocity of the Earth's
  annual Motion in its Orbit) as 10210 to One from whence it would follow
  that Light moves or is propagated as far as from the Sun to the Earth
  in 8’ 12’’.

  (source)

The calculation to get to the ratio of 10210 is,

size of a radian (206265’’) / angle of aberration (20’’,2)

This ratio also holds for the speed of light to the speed of the earth. Not sure though exactly how britannica and wikipedia arrive at their respective values.


Has lunar opposition surge ever been observed from Earth? From Earth orbit? This thorough answer to How long does lunar opposition surge last? Are there measurements of the full Moon getting suddenly brighter? details observations of lunar opposition surge by the Clementine and Lunar Reconnaissance orbiter.
But we don't have to go to lunar orbit to see the sunlit Moon, we can observe the lunar disk right here on Earth.
Question: Has lunar opposition surge ever been observed from Earth? From Earth orbit?

While I couldn't find any quantitative information about views from Earth or orbit down to zero phase angle (especially since views direct from Earth are limited in phase angle due to Eclipse), this website gives an unreferenced data point suggesting Apollo astronauts observed that a zero-phase full moon is approximately 30% (0.2 magnitudes) brighter than we can ever observe from Earth.
This site cites a value stating that the moon's brightness at zero phase angle may be as much as 14x brighter than an ideal diffuse reflecting disk, with 40% of the increase within the last 5˚ of phase angle (which cannot be observed from Earth).
Finally, there is one data set I have found which seems to show the opposition surge effect, but as noted, it is not what I would consider complete, and only has a few data points outside of Earth's viewing capability. That said, the source does not provide a value for maximum reflectance based on their model.

Robotic Lunar Observatory data, Reflectance vs Phase Angle


Constant interval between pulses from all pulsars It was found from the observations of the radiopulsars that the intervals between their pulses are changing
periodically, besides, for all pulsars this period is the same and equal to one stellar (sidereal) year. Explain
this phenomenon

The variation comes from the Earth's movement around the Sun. When the Earth is moving away from the pulsar in question, each subsequent pulse has slightly further to travel (up to 30km for a 1 second pulsar on the ecliptic down to zero for a pulsar located due North or South) and so is slightly delayed compared to the one before. Six months later, each subsequent pulse has slightly less far to travel, and so arrives a little earlier compared to the one before. The variation ranges from zero at the North or South poles to 1 part in 10000 on the ecliptic (this just comes from the ration between the Earth's orbital velocity and the speed of light).


What is the extent of the "worldwide campaign" to observe near-earth object 2012 TC4? This nifty answer describes and provides links to the data of the recovery of near earth object 2012 TC4, and the answer also shows the first actual image of the body for this pass. At apparent magnitude +27 (at about 52 million km) it may be the faintest asteroid to have been imaged to date. It's worth visiting that answer first before reading on here.

This comment points out that 2012 TC4 appears on the Goldstone Asteroid Schedule, a list of upcoming planned observations of asteroids by antennas at the Deep Space Network. Here are the entries for 2012 TC4 in the Upcoming Observations and Past Observations tables:

Upcoming Goldstone Observations
DATES           TARGET    H     Notes:
2017 Oct  9-14  2012 TC4  26.7  Close approach = 7.9 Earth radii on Oct. 12. NHATS

Past Goldstone Observations
DATES           TARGET          Notes:
2012 Oct        2012 TC4        Echoes not detected. Likely equipment issues.

I noticed at the top of the page there is a link to the The Arecibo asteroid radar schedule, and in the fifth line of the list shown in the screen shot below, I see that for 2012 TC4 there will be what is labeled there as a "WORLDWIDE CAMPAIGN!" of observations. The all-caps and exclamation point are a direct quote from Arecibo's table.

Question: What is the extent of the worldwide campaign to observe 2012 TC4? How many facilities? Is it coordinated? Perhaps simultaneous reception of reflected signals by multiple sites? Will it be coordinated with other observations such as infrared or optical wavelengths?

The plot below the Arecibo table is made from JPL Horizons data in the Earth-Moon barycenter frame. It shows +/- 24 hours from closest approach.
The hyperbola (red) is the calculated trajectory of 2012 TC4 with dots every 2 hours and the short, (roughly) circular arc is the Moon, as descried in this question. The closest approach to Earth's center is projected to be about 51,000 km. These are preliminary calculations based on the first observation and may be refined a bit as the time of closest approach nears.

The JPL press release mentions that more than 12 observatories, labs and universities will be involved but doesn't mention which.
https://www.jpl.nasa.gov/news/news.php?feature=6906

The University of Maryland has a website on the 2017 apparition of 2012 TC4
http://2012tc4.astro.umd.edu/index.shtml
Here's a snapshot of the list of planned observations shown there


How big will Apophis appear? How big will Apophis appear in the sky at its closest point of approach in 2029? Will it compare  to the moon? How long will it linger as it passes?

Current estimates put Apophis's diameter around 325 m and its 2029-04-13 approach about 38000 km from the center of the Earth. I figure an angular size <= 2 arcsec, almost starlike even if you manage to track it in a telescope.
Ephemerides show it at apparent magnitude 3.5 or so just before closest approach - visible but not outstanding to the unaided eye at night - and spending about 36 hours within 1 lunar distance of the Earth.


Bounded Observable Universe and General Relativity The book General Relativity for Mathematicians by Sachs and Wu defines a "spacetime" as a connected 4-dimensional, oriented and time oriented Lorentzian manifold.  I emphasize the word "connected."  There are those that say the observable universe is bounded, because the universe is expanding faster than the speed of light.  A bounded observable universe that is isolated from the rest of the universe, strikes me as a disconnected manifold, and would not seem to comport with general relativity's notion of spacetime.  Is this correct?

Following up on the above point, wouldn't there have to be some sort of discontinuity, at the moment when the universe splits off into isolated pieces?  How many discontinuities are being hypothesized?  Does general relativity contemplate discontinuities in spacetime and gravity?

The universe may be bounded (or it might not be) and parts of it are receding faster than the speed of light (in some sense) but these two notions are independent.  It would be possible to have a universe that unbounded and expanding, or bounded and stationary etc.

Similarly it seems that "connected" means connected in the topological sense: the only simultaneously open and closed non-empty subspace of the universe is the whole universe. This is independent of any physical laws such as "information can't travel faster than the speed of light" which may make parts of the manifold inaccessible to us.

The universe could be said to be the disjoint union of the observable and non-observable universe, but these sets are not both open in the topological sense.


What is the solar system orientation to its direction of travel? I saw the picture and seem incorrectly drawn. Its shows the the edge of the solar system leading facing the Heliosheath. It would not be like NASA to post something as incorrect as this. The solar system should at 60 degrees to the direction of travel?
I'm not sure maybe this picture is correct?
https://solarsystem.nasa.gov/news/784/nasas-voyager-2-probe-enters-interstellar-space/

This is Venus, high in the morning sky, amid the faint pillar of light called the Zodiacal Light. The glow is sunlight reflected off cometary dust in the inner solar system.
  Above is the centre of the Galaxy area of Sagittarius.  Alan Dyer

Since the  Zodiacal Light lies in the ecliptic plane, the orientation of the Solar system in the Galaxy is something like this:

The Sun moves through the local interstellar gas cloud approximately into the direction of Scorpio with a speed of about 25 km/sec.

Since Scorpio is fairly close to the galactic center (Sagittarius), the Solar System is moving approximately edge-on through the gas cloud as in the NASA diagram.

(Image: NASA, ESA, Z. Levay (STScI) and A. Fujii)


How difficult is adaptive optics computationally? What kind of computer do you need to do adaptive optics of big telescopes? Are we talking supercomputer clusters or can my desktop do it, or my smartphone? I suppose it depends on the size of the telescope and the wavelength that is targeted. I'm interested in both the kind of computers that are used and the actual number of FLOPs per second that is required.

If you're only trying to correct a small field, using a single guide star (or a single laser guide star), then the requirements aren't very arduous. The MACAO adaptive-optics module used on the Very Large Telescope at the European Southern Observatory dates from the early 2000s and is based around two 400 MHz PowerPC 604 chips (one for the "supervisory computer", the other for the "real time computer"). You can read more about it here (PDF file).

For next-generation multi-conjugate adaptive optics, which tries to correct a larger field using multiple guide stars, the requirements are bit more stringent: my impression is that you need several GFLOPs in the real time computer (some of this may be handled by using specialized DSP chips instead of general CPUs), and additional processing power for external monitoring and analysis.

So, crudely speaking, your smartphone could probably handle single-guide-star AO (except it would need to be re-engineered a bit and running a real-time operating system, and you'd need an extra computer for external monitoring); for multi-guide-star AO, you'd end up needing the equivalent of a powerful (multi-CPU) server or (small) cluster.


Why is there a zig-zag in elemental abundances? In this answer to Is lithium considered a metal in astronomy? this image from Wikimedia of the abundances of elements in the universe was shared:

Credit: Wikimedia Commons user 28bytes, under C.C.-by-S.A.-3.0.
I thought it was fascinating that there is a near-constant zig-zag of abundancies.
Are these values accurate, and if so, what would cause this zig-zag?

Nuclei are more stable if they have an even number of protons Z, and are also more stable if they have an even number of neutrons N. This is because the particles form pairs. (Almost all nuclei with both N and Z odd are unstable with respect to beta decay.)
If a nucleus is going to have significant cosmic abundance, it must be either absolutely stable or at least have an extremely long half-life. This is more likely to happen when Z is even (and also when N is even).
Stability also effects probabilities of different types of decay. E.g., if a nucleus can undergo both beta+ and beta- decay, then it is usually more likely to do the decay that results in the more stable daughter.


Were there any images of a transitary event of Jupiter or Saturn as imaged by a deep probe mission I'm quite certain that there isn't any actual images of a transitory event of Jupiter or Saturn across the solar disk from say e.g.,s the Voyager probe, Pioneer 10/11 or even recent New Horizons mission?

Another related question I have, what would the approximate transit of Jupiter look like as seen from Saturn. The distance between Saturn and Jupiter is approximately 4-5AU, and the diameter of Jupiter is one tenth that of the Sun's. Are we looking at perhaps a transit which covers 10-15% of the solar disk?

N.B. Please note I understand that to observe a transit of Jupiter we have to be superior to it in the Solar System, hence I am asking for images taken by deep Solar System probes only.

Cassini is supposed to have used its VIMS spectrometer to observe a Venus transit from Saturn in December 2012. NASA said it was a first. They were more interested in an absorption spectrum than an image.

Celestia can model such events.
Here is a video of a simulated grazing Jupiter transit from Saturn.
There are similar videos of Saturn transits from Uranus and from Neptune.

Transits are rare events because two planets must be near the intersection of their orbital planes at the same time.
A numerical integration study by John Walker found only 1413 transits of Jupiter and 201 of Saturn from outer planets in 250000 years.
A similar, smaller study could determine whether a given spacecraft could have caught a transit on its way through the interplanetary void, but I would not expect a positive result.


Would LIGO Detect Head-On Collision? Assume two black holes collide head-on.  In other words, they were not orbiting one another before the collision.  I know this is unlikely.  Further assume that their sizes and distance from Earth are similar to past collisions detected by LIGO.

My understanding is that detectable gravitational waves are caused by massive objects undergoing massive accelerations and that detections so far have been of colliding objects orbiting one another many times within a fraction of a second.

I assume that a head-on collision would create a single gravitational "pulse" instead of a wave.  I don't know if "pulse" is the right word for this.  Would this pulse be detectable by LIGO?

There are two issues: Would there be gravitational waves to be detected and would LIGO detect them.

On the first issue, gravitational waves are quadrupolar, and a cylindrically symmetric system will not produce any. (Specifically, the second time derivative of the quadrupole moment of an isolated system's stress–energy tensor must be non-zero in order for it to emit gravitational radiation.) So a head-on collision of equally massive black holes would not produce significant gravitational waves.  If the BHs were of different masses, or the collision not quite head-on, gravitational waves would be produced.

On the second issue there are at least three ways LIGO might might miss them.  First, they might simply be too faint.  A merger through rotation is one of the best ways to produce gravitational waves, and a head-on collision is one of the worst (during a BH merger, of course).

A second way it might miss is if the merger is in one of its blind spots. Each LIGO detector has blind spots and there's no rule that says a BH merger can't be in one of them. (In fact, the neutron star merger was located by use of the blind spots. The third gravitational wave (GW) detector, Virgo, in Italy, didn't pick up the neutron star merger.  It was bright/loud enough that it should have, which meant that the merger was in one of Virgo's blind spots, which helped to narrow down its location.  A GW detected in only one detector would be listed as a possible GW, but not as a detection, because it might simply be terrestrial interference.

But another way it might miss is if LIGO had not been looking for the waveform of that kind of merger. General Relativity allows us to compute in very good detail the GW signal from any type of collision that we think likely. Because the GW signals are so faint, the first step of detection is matching computed GW waveforms to the data and looking for matches. Once a match is found, the signal is analyzed in more detail, but they have to know where to look. If no one expected the waveform and it was not strong enough to pop out of the data and yell "Here I am, dummy!" it might be missed. Since head-on collisions are expected to be very rare, it's possible that they are not in the library of waveforms that LIGO checks.


Is antimatter also attracted by gravitational field? As we all know matter is attracted by the gravitational field of black hole. So, my question is: Is there any possibility of a black-hole attracting antimatter?

Can you explain how/why please?

Yes, common believes are because gravitation has effect on particles with mass. Antimatter are equivalent to there counterpart, sharing the same mass but opposite charges. Therefore, antimatter is attracted by gravitation.
But, this is still not experimentally proven, therefore a more theoretical explanation of the gravitational interaction.
Some researches indeed suggest anti-gravitational interaction between matter and antimatter.

source