How could we detect planets orbiting black holes? Detecting planets orbiting around stars can be made with several methods, transit photometry seems to be by far the most popular method nowadays.

I wonder how we could detect planets orbiting black holes. Which methods are we left with? Has it been done before?

Gravitational microlensing is a way of finding planets that does not care how luminous or otherwise the hosts of the exoplanets are.

The way it would work is that you stare at a dense background field of stars; then when a foreground black hole passes in front of a background star, the light from the star is magnified by gravitational lensing. Typically, the lensing event takes a few weeks for the magnification to develop and subside.

If the black hole has a planet, it may be possible to "see" its graviational potential, which will manifest itself as an asymmetry in the lens light curve or possibly even a little extra spike in the light curve lasting a few hours.

This technique is well established and is already used to detect planets around unseen objects. It is more sensitive than the transit and Doppler methods to planets orbiting a fair distance from their parent star.

The difficulty here is not finding planets around a black hole, it is proving that the planets you found were actually orbiting a stellar sized black hole. The microlensing event would only be seen once and the black hole system would likely be invisible. There is a possibility I suppose that it may undergo some sort of accretion activity after having been found by lensing, and this accretion might then be picked up by telescopes trying to identify the lensing source. I guess a good place to start would be events where a stellar-mass lens is inferred, but no star can be seen after sufficient time has elapsed that the lens and source star ought to be resolved.


Is there a relation between mass, radius, and luminosity in deuterium-burning brown dwarfs? I'm a worldbuilder and my setting has a lot of brown dwarfs.

While I'm trying to keep my setting as science-compliant as possible, I can't seem to find any resources for brown dwarfs like there are for main sequence stars, just basic encyclopedia pages describing what brown dwarfs are.

I do know that brown dwarfs undergo deuterium fusion for a short period of time after they form, cf. Wikipedia's Brown dwarf; deuterium fusion.

Does that process lead to a mass-radius-luminosity relation like hydrogen fusion does in main sequence stars?

Yes, there are monotonic relationships between mass and luminosity and radius on the "deuterium burning main sequence".

Deuterium "burning" begins when the core temperature exceeds just over $10^6$ K. This happens very early in the life of a contracting protostar and, because it is fully convective and thoroughly mixed, all the D is "burned" in less than a couple of million years. The process takes much longer (5-50 million years) in lower mass brown dwarfs. Thus only systems younger than 50 Myr could have a D-burning brown dwarf. Below about 13 Jupiter masses, the brown dwarf core never gets hot enough to ignite D.

D burning mimics hydrogen burning in that it is possible to stabilise the contraction of a protostar if the D burning rate can match that which would be released by gravitational contraction. The burning process then acts as a thermostat, keeping the core at roughly constant temperature, and the protostar/brown dwarf at constant luminosity, until all the D has been depleted. The big difference between D and H burning though, is that the initial D/H ratio is something like $2\times 10^{-4}$, so D burning does not last long.

The rate of gravitational contraction of a protostar/brown dwarf increases with mass. Hence the rate of D burning, and hence the luminosity, must increase with mass. The virial theorem used with an ideal gas approximation then tells us that since the central temperature is roughly proportional to mass/radius, and since the D burning reaction rate is very temperature sensitive ($\propto T^{12}$), the D burns at almost the same temperature, regardless of mass, and so the radius during D burning will be roughly proportional to the mass. This is only rough because the cores of such objects are close to degenerate and ideal gas approximations stop working.

The plot below (from [Tout et al. 1999]) 1) shows the H-burning (lower) and D-burning (upper) "main sequences" for stars as dotted lines in the Hertzsprung- Russell diagram. You can use this plot to estimate the luminosity and radius for a star of a given mass on the D burning sequence. Note the isochrones (line of constant age, running across the diagram, labelled in Myr). There is no convenient formulae that I am aware of.

For brown dwarfs I can't find anything similar. Below are cooling curves from Burrows et al. (1997), showing how luminosity changes with time. D-burning manifests itself as a plateau at a luminosity that increases with increasing mass, and with a duration that increases with decreasing mass. The brown dwarfs are the green curves.


Has the sun become way brighter the last years? Up to last year, I never got blinded by the sun in everyday situations (e.g. running towards it), at no time of the year I was. But since last christmas, it occured to me more often that I cannot watch in the direction where the sun is because it is so bright that it really hurts me.
My mother and my roommate have observed the same phenomenon independently of me.

This, of course, is only anectodal evidence, which is practically worthless. We all get older and we may have the same kind of "photophobia", or we just forgot about being blinded the last few years and just now consciously felt an remembered it.

Therefore I want to ask this question: is there any objectively observable increase of brightness over the last few months and if so, why?

The Sun's luminosity is stable to 0.1% - it varies slightly in response to, or associated with, levels of solar magnetic activity (Solanki & Unruh 2012).

This variation would not be perceptible and is completely outweighed by the $\pm 3.4$% variation caused by the non-circular orbit of the Earth (the Sun is closest in January).

Obviously, if you live in the northern hemisphere, then the Sun is generally higher in the sky now (in May) than it was in December. Its direct light is less absorbed/scattered by the atmosphere. This is a much more perceptible effect. It could change the visual brightness by tens of per cent, and by even more in the short wavelength blue part of the spectrum. That is the only physics-based explanation I can offer.


How would parallax affect an object at 200-1000 AU (for example, the 9th planet)? We know from this question that an object is hypothesized to be at between 200-1000 AU that is large enough to detect, known now as the 9th planet. It should move about 40 arcseconds per year. It seems to me that the parallax distance would be more than this distance per year. Would it be easier to detect the 9th planet using parallax than its movement around the Sun?

The parallax is found from a triangle with the Earth's orbit at its base. The annual parallax here would then be between 1/200 and 1/1000 radians ($\sim$ 1000 to 200 arcseconds) which is indeed enormous and much bigger than the likely proper motion during a year.


Why don't we see purple stars I know that we don't see green stars because in blackbody radiation star doesn't produce one spectrum.

The stars that have peaks in the green spectrum produce other spectrum in nearly same amout. This make them appear white.

If stars have peaks in the infrarad spectrum that has wavelength little longer than red.

We see those stars red because they produce red spectrum more than other visible spectrum.

But if stars have peaks in the ultraviolet spectrum that has wavelength little shorter than violet. Why don't we see those stars purple ?

Purple and green star exists. For example, a blue supergiant star can have its surface temperature to 50000 K, which peaks at purple in the blackbody spectrum. However, our eyes are more sensitive to blue than purple, and the contrast between blue and purple in the spectrum is insignificant, therefore we see blue.

Also, check this article: https://parade.com/295384/marilynvossavant/can-stars-be-green-or-purple/


What happens when two black holes collide? There are many hypothesis saying that there will a collision between the milky way and Andromeda galaxy...so what happens when two black holes will collide?? And how will it effect the other objects revolving around them??

There are many hypothesis saying that there will a collision between
  the milky way and Andromeda galaxy...so what happens when two black
  holes will collide??

There's a few answers in the links in my comment above copied, here and here, but it's such a fun question I thought I'd answer it anyway.

With Andromeda and the Milky way's super-massive black holes in their respective centers, those objects are so massive, they are likely to be largely unaffected by any stars in their path and they will just fly towards each other at whatever rate and direction gravity dictates and by the time they get fairly close to each other, their respective gravity should have them moving quite fast towards one another.   I don't know if anyone knows if they will orbit around each other for an extended period of time or spiral in fairly quickly.   That will depend on the angle of approach.  It's possible they could pass by each other, miss and each could get sent far out away from the other, part of an enormous orbit around each other, taking perhaps billions of years to lead to an eventual collision.

There's a few simulated videos out there on what happens when 2 super massive black holes spiral into each other.   Here's one.

As Super-Massive black holes approach, each will pass through roughly half of the other galaxy, disrupting the orbits of any stars they pass hear-by, though they'd likely need to pass within less than a light year to have significant effects on the star's direction, which wouldn't happen that often, but it would happen.

Sagittarius A is huge for a black hole, but quite small compared to the space between stars.  It's estimated to be about 44 million KM in diameter, which means it would fit (Just barely) inside the orbit of Mercury and our Sun.   Mostly it will fly past stars.   Andromeda's super massive black hole may be several times larger, but still fairly small compared to the distance between stars it's likely to pass on the way towards their mutual collision.

It's possible that one or both stars will pass through each other's galaxy relatively collision free but it's possible that one or both of them will get close enough to a star to generate a large accretion disk.   While black holes don't have charge, their accretion disks do, and if that happens, that could be an interesting and not very well understood interaction between the 2 super-massive objects and their accretion disks.   Also, as they spiral in towards each other, several of their near-by stars will be cast any-which way, some of them, inevitably inside towards one of the 2 black holes.   It could be a very impressive show.

Finally, as they merge, which, might take quite a bit of time if they end up orbiting each other, perhaps millions, even billions of years.   If/when they do merge, there could be some curious and not very well understood gravitational wave effects. bending and stretching of space like ripples.  (Gravity already bends space, but not in measurable ripples.   Our observation's of gravity warped space is a smooth curve.

I'll re-post this article here from comments that says it's possible for 2 black holes to repel each other if they bend space in opposite spin-directions and approach each other on a level plane.   Less close and they could still easily disrupt outer planet orbits, casting some planets every which way.

  And how will it effect the other objects revolving around them??

Imagine jellybeans in a salad spinner going as fast as you can spin it, and you remove the top.  That's basically what will happen to any near-by stars.   Both galaxy centers are quite crowded with stars (and perhaps several black holes) orbiting their centers.   The gravity assists from 2 super-massive objects moving towards each other will be significant and, basically stars will be flying all over the place.   Stars have tiny masses compared to those objects and they could be sent in any direction.


Do massive bolide entries coincide with meteor showers? Are exceptionally large, bright, meteors (such as Tunguska or Chelyabinsk) more common during common meteor showers like the Perseids, or the Leonids?

If not, why is that? Do the meteor showers simply not carry any more massive pieces?

In historical times we do not have enough data to do serious statistics about meteors of the size of the Chelyabinsk meteor.

In Wikipedia's list of meteor air bursts there are only six events within two orders of magnitude of the Chelyabinsk meteor.

$$\begin{array}{lll}
\hline
\text{Date} & \text{Location} & \text{Energy [TJ]}\\
\hline
\text{1490-04-04} & \text{Qingyang district, China} & unknown\\
\text{1908-06-30} & \text{Tunguska event} & 42000–63000\\
\text{1930-08-13} & \text{Curuca River Area, Brazil} & 38–20920\\
\text{1963-08-03} & \text{Prince Edward Islands} & 740–1490\\
\text{2009-10-08} & \text{South Sulawesi, Indonesia} & 130–210\\
\text{2013-02-15} & \text{Chelyabinsk meteor} & 2100\\
\hline
\end{array}$$

Note that for the Tunguska event it is not proven to date that it had been caused by a meteor.


Is the North star going to stay in the North sky? Considering how the our solar system, and how every other star in the galaxy is traveling through the cosmos, it seems like it's a miracle that the stars and constellations in the sky stay as constant as they are.

How is it that the North Star seems to be fixed in space? Is it following our star through space? are we following it? Is it actually moving? Are we getting further away from it? Closer to it? What is our relationship to Polaris in the Galaxy?

The relative direction to an object depends mainly on two things:

The location of the observer and the object
The coordinate system you use to determine direction.

So the first part of that concerns where we are compared with Polaris (and by extension of your question, where the other stars are as well).  Yes the stars and other components of the galaxy all have certain motions.  But this motion is not large compared to the distances between stars.

Most of the stars we can see in the sky are quite nearby, and are orbiting the galaxy at similar speeds.  The speed of the solar system around the galaxy is around $230 km/s$.  The relative speed between our sun and nearby stars is much less than that.  If we assume that most stars are farther than $5ly$ from us, then we can calculate how long it might take to make a $1\%$ change in distance.
$$t = \frac{d}{v}$$
$$t = \frac{4.7x10^{23}m}{2.3x10^5m/s}$$
$$t = 2.06x10^{11}s$$

That works out to over six thousand years.  There's a lot of caveats with that number, but it gives you a good idea that for the most part it takes too long for stars to move relative to each other for you to notice during your lifetime without very careful observation.

The second point is that things can move if we change what we mean by "north" or any other direction.  North, both on earth and in astronomy, is defined by the rotational axis of the earth.  But the axis changes its direction with respect to other stars over time.  Besides some smaller scale wobbles, the largest change to the direction of the axis is due to precession and has a period of about 26,000 years.  This change will take "north" from a direction that passes relatively near Polaris in a large circle around the sky.  With such a long period, again this isn't something to be seen by individuals without detailed measurements.

So the answer is that people alive today will be able to always think of Polaris as the pole star, but this will be less true for future generations as changes are easily visible on the scale of thousands of years.


Seeing a line rather than a dot when looking through refracting telescope I bought an amateur(and rather cheap) refracting telescope and while I can see moon quite clear and crispy I am having difficulties in observing some larger stars.

All stars appear as lines rather than dots. It is always one longer line. I read something about those lines but this stands only for reflecting telescopes that have some kind of structure inside that holds a mirror and this structure interfers thus producing these lines. But I don't have a reflecting telescope, mine is refracting.

I cannot seem to find anything on google regarding this. Maybe it is because I lack the vocabulary.

EDIT #1:

This is the telescope I have.

Edit #2:

This is my depiction of what I see. Forgive my MS pain skills. As I focus out(or in, not quite familliar with the naming) it goes more clearer and clearer but I never get to see a simple shiny dot. The closes to dot is the first image but this is so blurry that I barely see it. The red line is the one I perceive as the clearest. The orange lines is what I perceive as "outline" and the yellow inside is that shiny effect one usually gets when looking at objects that emmit light.

First off I want to say I am a complete amateur when it comes to stars and even bigger when it comes to optics.

I was looking at sky from indoors and my windows have double glass. It turns out that this was the problem or maybe the glass on the window is not some good quality.

In any case, all works well now and I am thrilled at what I'm seeing in the night sky.


When was it worked out/discovered that our sun can't go supernova? As the title says, when did we realise with reasonable confidence that our star is not going to be going out in a supernova blaze of glory?
I ask because a while ago I read The Songs of Distant Earth by Arthur C. Clarke which was from 1986 (but based on an earlier story) and a plot point is that the sun went supernova. I was suprised that this was still consdiered a possibility in the 80s but accepted it.
However today I began reading The Gods Themselves by Isaac Asimov, written in 1972, and in it a character states that the sun is too small to go supernova.
This confused me as I thought both Clarke and Asimov where both scientifically knowledgable and informed.
I can find plenty of information online about how our sun will likely end and why it won't go supernova but not any imformation on how long we have known (and therefore whether Asimov was ahead of his time or Clarke was uninformed in a way that would suprise me)

I think the definitive work is that of Hoyle & Fowler (1960).
They argued that supernovae were produced by two possible mechanisms - what they called an implosion/explosion or an explosion within degenerate matter. Both of these mechanisms required very high internal temperatures ($>2\times 10^{9}$ K) and they argued that this could only be achieved for both mechanisms if the star is more massive than the Chandrasekhar limit of $5.8M_{\odot}/\mu_e^2$, where $\mu_e$ is the number of mass units per electron (1 for hydrogen, 2 for helium, carbon, oxygen, 2.25 for 56Fe etc.); otherwise it could achieve a degenerate state, supported by electron degeneracy pressure, and cool indefinitely without contracting any further.
In addition, they successfully identified these two types of supernovae with Type I and Type II supernovae and argued that very high mass stars accounted for Type II supernovae via the implosion/explosion mechanism, whereas the Type I supernovae arose from the ignition of degenerate material in less massive stars, but still larger than the Chandrasekhar mass, which is $>1.2M_{\odot}$ for all plausible compositions. They even then conclude that stars up to $10M_{\odot}$ may lose enough mass to avoid a Type I supernova and become stable white dwarfs.
Though it seems not to be explicitly stated in the paper, it is implicit that the Sun could not explode as either type of supernova.
As an aside - I reminded myself of the plot of "Songs of Distant Earth" (I read it loooong ago). Clarke's story posits that the "neutrino problem" (the lack of detected solar electron-neutrinos), which was a thing back in the 60s, was actually because something odd was going on in the Sun's core. So I don't think that Clarke was suggesting that under normal stellar evolutonary circumstances the Sun would imminently explode. Indeed it was well known at that time that the Sun should continue on its main sequence life for billions of years.


Secondary maximum in SNIa lightcurves for near infrared Type Ia supernovae (SNIa) are thermonuclear supernovae that generates a lot of Nickel 56 which emits photons while disintegrating.

How is the secondary maximum in lightcurves for reddish bands (see I, J, H, K bands in the figure below) explained ?

The main power driving a supernova (of type Ia) light curve is the explosive detonation of a carbon/oxygen white dwarf. This produces a large amount of iron-peak elements (particularly Ni) via nuclear fusion of carbon, oxygen and alpha particles via magnesium, neon, sulphur and silicon.

The decay of the light curve is then largely governed by the radioactive decay of $^{56}$Ni via $^{56}$Co to $^{56}$Fe, which releases gamma rays that "reheat" the ejecta. The half-lives of these decays are about 6 and 77 days respectively, and this is the basic reason for the decay of the light curve happening on timescales of tens of days.

However, as you have noticed, in the (rest-frame) infrared part of the spectrum, there is a secondary peak.

Now, whilst the emission in the visible part of the spectrum is largely of a thermal continuum nature (the ejecta are typically at temperatures of $\sim$10,000K during the early parts of the decay), the infrared portion can be dominated (and certainly is after $>20$ days) by line emission (that is then smeared out by turbulent doppler broadening).

The species that emit this radiation are ionised iron-peak elements (mainly cobalt and iron). As the ejecta cools, the species that are present recapture electrons, becoming less ionised. At around 7000K there is a fairly abrupt transition between the gas consisting of mainly doubly ionised species to mainly singly ionised species. This results in an order of magnitude increase in emissivity of the gas due to radiative recombination of the ions and electrons. It turns out that this radiation emerges mainly at near infrared wavelengths.

As the gas cools further and becomes wholly singly ionised, or even neutral, then the emissivity falls drastically again. There is therefore a "sweet spot" where the infrared emissivity of the ejecta is maximised, that occurs about 20-30 days after the explosion, that manifests itself as a secondary bump in the infrared light curves.

Much more detail can be found in Kasen (2006).


where to obtain ephemeris data for geodesy satellites? Is anyone aware of where one can obtain ephemeris data in Geocentric Celestial Reference Systems (GCRS) coordinates for a geodesy satellite such as LAGEOS (It doesn't really matter which mission)? Or, alternatively, simulated data in GCRS coordinates. Specifically, Cartesian position and velocity coordinates. I can't seem to find them anywhere.

Update: My confusion has increased. I will try and rephrase my question.

Is it possible to obtain an ephemeris file and perform a transformation such that the ephemeris data can be expressed in GCRS coordinates? And if so, what is the process?

Purpose of question:
What I am interested in is ephemeris data for a satellite (doesn't really matter which mission) that is subject to classical Newtonian acceleration plus relativistic corrections as recommended by the IERS such as Schwarzschild, Lense-Thirring and Geodetic accelerations.

The JPL Horizons website gives ephemerides for a large number of Earth satellites as well as deep space spacecraft, planets, moons, asteroids, etc.

If you configure it as shown below, selecting the units you'd like, the full state vectors (x, y, z, vx, vy, vz), use geocentric coordinates (@500) you will get the following output. You can choose various time step sizes (down to 1 minute) and time frames. You can choose direct download as csv text rather than HTML if you like - recommended for long files.

It looks to me that because of the special status and use of the LAGEOS satellites, the ephemeris only predicts a relatively short period into the future, unlike other spacecraft which have predictions for years sometimes.

This should get you started! If you have more questions about this post as comments and I or someone else will try to help.

Here are the Table settings. I believe that the ICRF/J2000.0 with Ecliptic and Mean Equinox settings will give you something close to GCRS but right now I am not sure how close. I'll keep reading and ask for help in comments.

Here is a sample output from the above configuration. It scrolls left-right and up-down.

Revised: Sep 11, 2017                 LAGEOS-2                         -122195

LAGEOS-2 (1992-070B, "Laser Geodynamics Satellite", NASA/ASI)
60-cm diameter sphere, 426 corner cube reflectors (VIS), 4 germanium (IR)
Deployed: 1993-Oct-22 from Space Shuttle Columbia (STS 52)

*******************************************************************************
Ephemeris / WWW_USER Mon Sep 11 21:51:05 2017 Pasadena, USA      / Horizons
*******************************************************************************
Target body name: LAGEOS-2 (spacecraft) (-122195) {source: LAGEOS-2}
Center body name: Earth (399)                     {source: DE431mx}
Center-site name: BODY CENTER
*******************************************************************************
Start time      : A.D. 2017-Jan-01 00:00:00.0000 TDB
Stop  time      : A.D. 2017-Jan-05 00:00:00.0000 TDB
Step-size       : 1440 minutes
*******************************************************************************
Center geodetic : 0.00000000,0.00000000,0.0000000 {E-lon(deg),Lat(deg),Alt(km)}
Center cylindric: 0.00000000,0.00000000,0.0000000 {E-lon(deg),Dxy(km),Dz(km)}
Center radii    : 6378.1 x 6378.1 x 6356.8 km     {Equator, meridian, pole}
Output units    : KM-S
Output type     : GEOMETRIC cartesian states
Output format   : 2 (position and velocity)
Reference frame : ICRF/J2000.0
Coordinate systm: Ecliptic and Mean Equinox of Reference Epoch
*******************************************************************************
            JDTDB,            Calendar Date (TDB),                      X,                      Y,                      Z,                     VX,                     VY,                     VZ,
**************************************************************************************************************************************************************************************************
$$SOE
2457754.500000000, A.D. 2017-Jan-01 00:00:00.0000,  5.049112760756619E+03,  9.246781656953866E+03,  5.722633641721932E+03, -2.978199703429669E+00,  3.689975896939779E+00, -3.350502047340569E+00,
2457755.500000000, A.D. 2017-Jan-02 00:00:00.0000, -6.140907600397200E+03, -8.148707798015072E+03, -6.918983669661846E+03,  2.434461456961566E+00, -4.250750074318191E+00,  2.808377534161328E+00,
2457756.500000000, A.D. 2017-Jan-03 00:00:00.0000,  6.849580412560958E+03,  6.106324592083777E+03,  7.740446107133596E+03, -1.906840987637830E+00,  4.973834339923835E+00, -2.290764139470047E+00,
2457757.500000000, A.D. 2017-Jan-04 00:00:00.0000, -7.504132672898423E+03, -4.707277560275819E+03, -8.539479988453819E+03,  1.304309397636298E+00, -5.242849004192330E+00,  1.679103108590806E+00,
2457758.500000000, A.D. 2017-Jan-05 00:00:00.0000,  7.740851830184564E+03,  2.245705288312970E+03,  8.942780571589567E+03, -6.155844902136471E-01,  5.667380276723395E+00, -9.687665040214430E-01,
$$EOE
**************************************************************************************************************************************************************************************************
Coordinate system description:

  Ecliptic and Mean Equinox of Reference Epoch

    Reference epoch: J2000.0
    XY-plane: plane of the Earth's orbit at the reference epoch
              Note: obliquity of 84381.448 arcseconds wrt ICRF equator (IAU76)
    X-axis  : out along ascending node of instantaneous plane of the Earth's
              orbit and the Earth's mean equator at the reference epoch
    Z-axis  : perpendicular to the xy-plane in the directional (+ or -) sense
              of Earth's north pole at the reference epoch.

  Symbol meaning:

    JDTDB    Julian Day Number, Barycentric Dynamical Time
      X      X-component of position vector (km)
      Y      Y-component of position vector (km)
      Z      Z-component of position vector (km)
      VX     X-component of velocity vector (km/sec)
      VY     Y-component of velocity vector (km/sec)
      VZ     Z-component of velocity vector (km/sec)

Geometric states/elements have no aberrations applied.


Why argon instead of another noble gas? I noticed that the atmospheres of Earth and Mars have a little bit of argon in them (1% to 2%). I checked Venus, too, which has 0.007% argon, but that's still more than any other noble gas in the Venusian atmosphere.

I checked the 4 outer gas planets, plus the moon Titan, and for the most part could not find Argon listed in their atmospheric composition. Jupiter is listed as having a very tiny trace amount of Argon that is hard to interpret because it's listing it as relative to helium and then relative to Jupiter/Sun, whatever that means. The moon Titan is listed as having trace argon without specifying any numbers. The others did not mention Argon.

So why do the rocky planets tend to have argon, but the gaseous planets don't?

Why argon specifically? Presumably helium is too light and will float to the top and get knocked away by the solar wind. However, there is one noble gas between helium and argon, namely, neon. So why don't we have neon in our atmosphere instead of argon?

Edit: Maybe the outer planets do have Argon, but it all sank to the bottom and thus we don't detect it? I'm also curious about where Argon lies in the "food chain" of nuclear fusion for an average star like our Sun. I have a hard time seeing how Argon-40 would be more populous than Neon-20 though.

Doing a bit of reading up on this, I might have an answer, though credit where credit is due, the answer isn't really mine:

https://www.reddit.com/r/askscience/comments/3wsy99/why_is_neon_so_rare_on_earth/

When the planets coalesced, it's likely that there was very little ices/gas around the inner planets when they formed and the Earth's atmosphere and water (CH4, NH3, CO2 and H20 being the 4 most common outside the frost line ices).  These likely came from asteroids and meteors that formed outside the frost line and later crashed onto earth.

Neon is the 5th most common element in the milky-way but because all noble gases have very low freezing points, it's likely not be very common even on comets or meteors for the same reason that water or CO2 aren't common inside the frost line, Neon, and other noble gases likely stay free and don't collect on comets or meteors in high amounts.   (I looked, but couldn't find an article to verify that).

But if comets have low noble gas content, then we have to look for an alternate source.  With that in mind, and going back to the first link, Argon is produced by radioactive decay of Potassium 40 and that would explain it's relative abundance compared to the more common noble gas, Neon.   Helium (Alpha particles) is also produced inside the earth and Radon is is too in small amounts but Radon also decays - that's not related to your question though.

If Argon on planets comes primarily from Potassium 40, you should expect the amount of Argon to have a roughly similar ratio to the amount of potassium on a planet and not be relative to the percentage of atmosphere.   A 2nd factor, how much gets blown off the planet over long periods of time is a factor too.   Venus in general should be able to retain much of it's Argon based on atomic weight (40) similar to CO2 (44), but if it loses even a tiny percentage of it's Argon over time, that would be a factor too.

Now, to see if this is possible, I should run some numbers, but I warn you, my math can be a little rusty.

Potassium is the 7th most common element in the Earth's lithosphere at about 0.26% and about 0.0117% of that Potassium is Potassium 40.   Using a very rough estimate of $2.3 \times 10^{19}$ tonnes for the Earth's crust, $(2.3 \times 10^{19}) \times (2.5 \times 10^{-3}) \times (1.17 \times 10^{-4}) =$ about $6.7 \times 10^{12}$ or 6.7 trillion tons of Potassium 40 currently in the Earth's crust. (There's probably a fair bit more in the mantle, so these numbers are rough)

With a half life of about 1.248 billion years, that's sufficient time for over 3 half lives if we start after the late heavy bombardment, which suggests a bit over 7/8ths of the original Potassium 40 in the Earth's crust has decayed into Argon 40, so there should be, given the age of the Earth and abundance of Potassium 40, a bit over 7 times 6.7 trillion tons or, lets ballpark and say a bit over 50 trillion tons of Argon that formed on earth by Potassium decay.  (I'm ignoring any that might have been produced prior to the late heavy bombardment, cause I assume that could have blown some of the atmosphere off the earth or heated the atmosphere enough for the sun to blow some of it off).    Also, doing a bit of research, only 11% of the Potassium 40 decays in to Argon 40, 89% undergoes beta decay into Calcium 40, so for this to work, there would need to be a fair bit more Potassium in the earth than I estimated, but that's still likely the case.

The mass of the atmosphere is about 5,140 trillion tons, and 1.288% of that (By mass, not volume) = about 66 trillion tons, so the Argon we should expect from Potassium 40 decay and the amount of Argon in the atmosphere are pretty close.   Some Argon gas might have escaped and some should still be trapped inside the earth but the numbers are close enough to work and I think that's very likely the answer.  It also suggests that the Earth has lost relatively little Argon to space, which also fits with the Atmospheric Escape article.

A 2nd way to look at this is that Argon 40 makes up 99.6% of the Argon in the atmosphere and Stellar Nucleosis likely wouldn't account for a ratio anywhere close to that (not a typical stellar link but Wikipedia says Argon 36 is the most common isotope).   The decay of Potassium 40 does explain the 99.6% Argon40 ratio.

If we apply a similar estimate to Venus, with Venus atmosphere about 94 times the mass of Earth's, and we assume a similar amount of Argon-40 being produced in Venus' crust we could roughly expect 1.28%/60 or about 0.02% Argon by mass in Venus's atmosphere or perhaps, if the Earth lost a pretty high share of it's lighter crust elements after the giant impact, we might expect a bit more than that on Venus, perhaps 0.03% or 0.04% as a rough estimate.   Using your number of 0.007%, that's lower than I calculate it should be, but Venus could have lost a higher share of it's Argon than Earth and it also might be slower to release trapped gas inside it's crust than Earth because it doesn't have plate tectonics, so the number for Venus looks "about right" too.    It's the Potassium 40 in the crust.   I'm convinced.

Interesting question.  I learned something researching it.


Could dark matter assist with star formation? This is related to the question: "Can dark matter decrease the Jeans length?" If dark matter assists in any way with star formation, shouldn't it be detectable by precisely measuring velocities in young open clusters? In particular, could the motions of stars in the Pleiades all be explained without the presence of dark matter?

Strange you should ask - I am currently working on a paper on NGC 2516 - "the southern Pleiades" and looking at the dynamical status of the stars compared with the distribution of visible mass.

Our conclusion is that the radial velocities of the stars are in virial equilibrium, with velocity dispersions that are entirely consistent with the mass that is present in the stars that are visible. So, no need for dark matter there - you could probably rule out any contribution of more than about 50% of the visible stars. (NB the point of the work is not to look for dark matter, is is not expected - see below).

This is one of the first very detailed measurement for an open cluster (like the Pleiades), because the velocity precisions required have to be considerably better than the velocity dispersion. And for a sparse cluster like the Pleiades, this is $<1$ km/s, which is challenging.

There is one old paper by Jones (1970) that uses proper motions in the Pleiades to estimate a dynamical mass based on virial equilibrium of $690\ M_{\odot}$, which compares with $470\ M_{\odot}$ from directly counting stars. But Jones points out that the latter number is a lower limit (they certainly couldn't see very low mass stars and brown dwarfs in their study) and the former number has large uncertainties. So there is no real evidence for any "dark matter".

The dynamics of globular clusters are easier to measure - they are more massive and have much higher velocity dispersions. The general picture is reasonable agreement between velocity dispersions and inferred mass distributions from counting the visible stars. There are of course uncertainties, but there can be no large dark matter component. Limits as small as $<6$% dark matter have been set in some clusters (e.g. Ibata et al. 2012).

The first Gaia satellite results, due in about 18 months time, will blow this field apart. We will have exquisitely precise tangential velocities for $\sim 1000$ stars in the Pleiades.

Dark matter is not expected to play a role in small scale (star cluster) formation. When clusters form, the dissipation of energy in gas interactions is what allows them to end up as bound systems. Dark matter is dissipationless. As the escape velocity of these clusters is 1-10 km/s, compared to velocities of hundreds of km/s for datk matter, there is not even expected to be any gravitational concentration of dark matter. This is quite different from galaxy formation, which is associated with pre-existing dark matter structures.


Why does the Moon appear gray when passing between the Sun and the Earth? Shouldn't the Moon appear as bright as a full Moon seen at midnight from Earth?

The photo was taken by DSCOVR at Lagrange point 1.

In the picture, The Moon appears dark gray. Of course the Earth appears bright, reflecting sunlight from clouds and water. The Moon's surface is gray and should reflect less light than the Earth.

It should be irrelevant that we see the far side, since the reflectivity of the Moon's surface should be the same on the far side as the side that faces the Earth.

The midnight full Moon appears much, much brighter as seen from Earth than it does in this picture, despite the fact that the amount of sunlight reflecting from the surface of the Moon is the same in both instances.

I understand the photo was taken with 3 separate exposures of red, blue and green, but this should not affect the brightness.

So why does it appear so dull?

That's what it really would look like if you were there with DSCOVR. The albedo of the Moon is only about 0.136, about half of the Earth's average albedo. Of course the part with clouds is higher.

I was shocked too, but it was explained in written copy that accompanied  the release of the original image.

  Shouldn't the Moon appear as bright as a full Moon seen at midnight from Earth?

It does. If the moon were a diffuse, white ball, a full moon would be about seven times brighter!

If you watch the image or GIF, the Moon is roughly the same brightness as central Australia or the Sahara region.

Phil Plait explains well in Bad Astronomy.

There's a lot to read here.

EDIT: I just ran across these images of astronauts on the surface while reading this answer. Their suits are not 100% white to begin with, but the Lunar soil - at least in these locations - is significantly darker. It is close to the same color as the (presumably) nearly-black radiator fins for the heat sink of the RTG unit (2nd photo) at the astronaut's foot.

above: "Buzz Aldrin carries the EASEP." from here

above: "Astronaut Alan L. Bean from Apollo 12, put the Plutonium 238Pu Fuel from the Lunar Module into the SNAP 27 RTG" from here.

above: Image from NASA/NOAA from Bad Astronomy


How is Mercury tidally locked if the ratio is not 1:1? The ratio for a planet to be tidally locked has to be 1:1, but the ratio for Mercury is 3:2. How is Mercury tidally locked if the ratio is not 1:1?

The simple answer to your question is that Mercury is not tidally locked. You may have seen old books (before 1965) that said it was tidally locked, because it was once assumed to be so. Alternatively, as zephyr said, your source may have been referring to the 3:2 resonance, but that is also not really the same thing.


Is there a strict chronological order that determines the shape of galaxies? Assuming no collision happens, do all elliptical galaxies eventually become spiral? Also, when a galaxy becomes spiral or barred spiral, can it switch between these shapes or is it set forever?

Galaxies evolve from spiral to elliptical.  The spirals are formed by patterns of new star formation in the disk surrounding the bulge, which contains mostly old stars.

As galaxies run out of hydrogen gas clouds, which is the raw material from which stars are formed, then no new stars form, and with no new star formation, the complex structures of the arms are lost. Elliptical galaxies are made of old stars and have little new star formations. They were formed when spiral galaxies merged, causing a burst of star formation, and leaving a bulge of old stars and little gas to make new ones. The ultimate fate of the milky way and the andromeda galaxy is to merge and form a giant elliptical galaxy.


Can a telescope detect a blackbody? I'd like to determine the distance at which a telescope with a certain set of parameters (sensitivity, collector area) can detect a blackbody with certain other parameters (temperature, emitting area).

For example, let us suppose we have an infrared telescope with a sensitivity of '1Jy/sr' and a collector area of 10m^2 at a distance D from a 300m^2 perfectly black body at a temperature of 30 Kelvin.

How do I determine D? What additional information do I need to perform the calculation?

Let's assume a uniform laminar emitter, oriented so that it is at right angles to the line to the detector.

The specific intensity emitted by a blackbody, at the surface of the blackbody is
$$B_{\nu} = \frac{2h\nu^3}{c^2} \frac{1}{\exp[h\nu/kT] -1}\ {\rm Wm}^{-2}{\rm Hz}^{-1},$$
where $B_{\nu}$ is the "Planck function".

Using the definition of specific intensity we can calculate the flux (energy per unit time, per unit frequency) received at the detector as
$$ f_{\nu} \simeq B_{\nu}A \Delta \Omega \cos \theta\,$$
where $\Delta \Omega$ is the solid angle of the detector subtended at the laminar surface, $A$ is the area of the laminar surface, and $\theta$ is the angle with respect to a normal from the surface and here, $\cos \theta \simeq 1$.

The solid angle subtended by the detector at the surface is $\Delta \Omega = a/D^2$, where $a$ is the area of the detector, but we can also define a solid angle $\Delta \omega = A/D^2$, which is the solid angle of the surface at the detector. Thus
$$\frac{f_{\nu}}{\Delta \omega} =  B_{\nu}a  $$

If the detector has a fixed sensitivity per unit solid angle then unless the object is unresolved, then it does not matter how far away it is; the value of $D$ does not come into it. It is a well-known fact in astronomy that resolved, extended objects have an observed surface brightness (flux density per unit solid angle) that is independent of distance.


Can the interferometer called "Gravity" measure "a few centimeters on the Moon"? Phys.org's Very Large Telescope sees star dance around supermassive black hole, proves Einstein right links to several ESO videos, including Interview with Reinhard Genzel (in English). After 08:51 in the video professor Genzel says:

So we take four 8 meter class telescopes of the European Southern Observatory, the Very Large Telescope in Chile, and combined it optically to an equivalent 130 meter telescope, and the bigger the telescope, the better the resolution.
So with this interferometer, called Gravity, we can measure basically a few centimeters on the Moon if you like, and that gives us the precision to really you know nail the motion of a star around a black hole.

To the question Is seeing the apollo moon Landers via earth telescope that hard? I recently wrote Yes, it still is. and then asked Will the Magdalena Ridge Optical Interferometer be able to image extended objects like the surface of the Moon?, but if Gravity can resolve a few centimeters on the Moon then it should be capable of photographing all of the Moon landing sites.
Question: Did professor Genzel mis-speak, perhaps meaning that the resolution is a few centimeters at the distance of the International Space Station, or does Gravity at least theoretically have a resolution of a few centimeters at the distance of the Moon?
I know that three orders of magnitude is fairly small for astrophysicists, but since the reference is The Moon and there's stuff on it that we'd like to see, in this case enquiring minds want to know!
FYI there is also Interview with Reinhard Genzel (in German); it might be interesting to compare.

A 130m baseline operating at 2 microns gives a theoretical resolution of $2\times 10^{-6}/130$ radians. At a distance of 400,000 km this translates to 6m.

My guess is that Genzel is referring to the accuracy with which the position of a point source of light can be measured. This not the same thing as the smallest thing you can image. The centroid of a Gaussian can be measured much more precisely than its FWHM.

The moon landing sites have been photographed by the lunar reconnaissance orbiter. If the 'foil hat brigade don't believe that, they aren't going to believe a reconstructed interferometric image in any case.


Difference of Gas Giant Planet and Ice Giant? What is the Difference of or the Structure of a GAS PLANET AND AN ICE GIANT Why they called ICE GIANTS like Neptune and Uranus. Do they have a similarity ?

Gas Giants consist primarily of the gases hydrogen and helium, with a rocky/iron core.

The interiors of Ice Giants consist primarily of what astronomers call 'Ices'. These are compounds like water, methane and ammonia, and are so-named presumably because they are solid at cold (-200 C) temperatures, unlike hydrogen and helium, which remain gaseous up until very close to Absolute Zero.

Hydrogen and helium are very light, and it only takes a small amount of heat energy to make them fast enough to escape the gravitational field of most planets. Jupiter and Saturn are the only planets heavy enough to retain a significant fraction of hydrogen and helium (though the atmospheres of ice giants Uranus and Neptune consist mainly of hydrogen and helium, it's mainly ices further down).


Converting equatorial coordinates to Cartesian coordinates for extragalactic distances I'm trying to search for galaxy clusters with the friends-of-friends (FoF) algorithm. I have the equatorial coordinates ($\alpha, \delta$) and redshifts ($0.5<z<2.5$) of galaxies that I wish to convert to Cartesian coordinates, with the following:
$$
X = R\cdot \cos(\delta)\cos(\alpha) \\
Y = R\cdot \cos(\delta)\sin(\alpha) \\
Z = R\cdot \sin(\delta)
$$
The problem I'm having is deciding the correct $R$ to use: should it be comoving distance, angular diameter distance or luminosity distance, given that I only have redshifts to work with? How do we determine the best LOS distance measure to use for this case? Is there a better way to do this?
For example, in astropy, their default Distance quantity computes the distance from redshift (and a given cosmology) using luminosity distance.
References

Some notes on the Friends-of-Friends Algorithm

I find this discussion of cosmological distances by David Hogg to be very useful in answering questions like this.  In Section 4 he says:

The line-of-sight comoving distance between two nearby events (ie, close in redshift or distance) is the distance which we would measure locally between the events today if those two points were locked into the Hubble flow. It is the correct distance measure for measuring aspects of large-scale structure imprinted on the Hubble flow, eg, distances between “walls.”

That would suggest to me that it’s the distance measure you want for your problem, but I encourage you to read further in that article to be sure.


Will a Ball placed close to Sun fall into it? Like a ball falls on the earth due to gravity, if suppose we are able to place a ball close enough to the sun. Will the ball fall into the sun due to its gravity or will it be pushed away due to its energy?

Short answer is yes. The only way radiation pressure would be a factor was if the ball was in orbit around the Sun, this would mean that the slight pressure it feels would cause it to drift away slowly (assuming its made of the most heat resistant material on Earth!). But in your question you said "placed" near the sun not in orbit around it, so the strength of the sun's gravity would overcome all other forces and pull it in.


Will a planet's core always be very hot? Including formed & forming planets, will the core of a planet always be very hot, or are there any planets with cold cores?

We would expect all large bodies, such as planets, to have a hot interior.

The interior of a planet is heated through two main mechanisms: Heat released from gravitational energy as the planet is forms, and heat released from the decay of radioactive elements, particularly uranium.

The outer layers of the planet are a very good insulator, so the any heat is trapped in the interior for a long time.

In gas giants, gravitational energy is the main source of heat, Saturn has an internal temperature of about 21000°C, Jupiter is higher, 35000°C in some models. Terrestrial planet's cores are heated by radioactive decay. Earth has a core temperature of about 6000°C. Smaller bodies have cooler temperatures. Dwarf planet Pluto may have a core temperature less than 1000°C, and asteroids would have reached thermal equilibrium, and have roughly the same temperature throughout.

No planet in the universe can be much more than about 10 billion years old, and this is not old enough for a planet like the Earth to have formed a completely cold core.


How massive can a star be at birth? We have discovered some incredibly massive stars. R136a1, the most massive known star is estimated to have 265 times mass as our Sun. Yet it has been burning for at least a million years, and must have shed a considerable mass as stellar wind. In its infancy, R136a1 must have been much more massive.

Which got me thinking about the question: What is the theoretical limit of mass that a newborn star can have?

I found a text that suggests a Population III star can have a mass between 200 and 10^5$M_{Sun}$. There you can read:

  "...This would apply for VMOs larger than $M_c$ $\approx$ 200 $M_{Sun}$. Stars larger than 150$M_{Sun}$ are termed supermassive objects (SMOs)...".

It seems supermassive stars like R136a1 (with masses in excess 150$M_{Sun}$, the accepted upper limit for a non-pop III star) can be formed by the collision of massive stars.

So, It is not difficult to imagine, say, two 150$M_{Sun}$ monsters colliding and forming a 300$M_{Sun}$ object.


attractors of Mercury's spin-orbit relation Mercury rotates at such a rate that the Sun appears to stand nearly still at perihelion, when the tide is strongest.

Is there a mechanism that tends to adjust the orbital eccentricity to improve the match – minimize the tidal torque (integrated over time), or some such?

If so, might there be other stable combinations of eccentricity and rotation coupling, other than the obvious 1:1?

EDIT: I'll rephrase, since most of you seem to think my question was something elementary like “how does Mercury rotate?” or “what the heck is tidal locking anyway?”.  I suppose it's not a bad heuristic to assume that the newcomer is completely ignorant, but I wish that assumption didn't override a literal reading of the question.

Each planet's orbital eccentricity varies over time.  Mercury's is now suspiciously near to what it would need to be for the sun to stand exactly still at perihelion.

So it's conceivable that in the dawn of time Mercury orbited at some arbitrary eccentricity and rotated at roughly 3/2, then resonance brought the rotation rate to exactly 3/2, and then something adjusted the eccentricity so that Mercury's long axis – the line of the lumps on which the tidal torque acts – tracks the sun more closely around perihelion.  (I wouldn't expect a perfect match at perihelion, because though tide is strongest then it doesn't vanish at other times!)

If that adjustment effect does happen, how does it work?

And if that happened for 3/2, could it also happen for other resonances, say 2/1 or 5/4?  For an odd denominator (other than 1) does the effect, whatever it is, behave differently (unless Mercury has a weird shape)?

Over enough time there is a tendency of all orbiting objects to fall into resonance with the object they're orbiting.   1:1 being is most common.

1:1 usually called Tidal Locking, is quite common. The Moon to the Earth.   Pluto and Charon to each other.  All four of the Galilean moons are tidally locked to Jupiter, likely others.

Mercury's 3:2 ratio is more unusual but not necessarily uncommon, especially for a planet with an eccentric orbit.

See Here

An eccentric orbit likely helps create a spin-orbit resonance other than 1:1 and resonance other than 3:2 should certainly be possible (the Wiki article implies that) though there are no known examples that I know of.

For a really crazy orbital spin - check this out:  http://www.businessinsider.com/pluto-moons-weird-behavior-2015-6


How (the heck) was this photo of Venus at inferior conjunction (between us and the Sun) taken? This great answer by @gerrit discussing planetary phases seen in visible light contains the image I've included below. According to the Wikimedia Commons link these are ESO images from the Venus transit 2004 VT-2004 observing campaign.

The ESO, or European Southern Observatory:

  Organisation Européenne pour des Recherches Astronomiques dans l'Hémisphere Austral) is a 16-nation intergovernmental research organization for ground-based astronomy. Created in 1962, ESO has provided astronomers with state-of-the-art research facilities and access to the southern sky. (emphasis added)

These images are very high quality, and the last one, dated 8/6/04 (08-June-2004) appears to be taken at/near inferior conjunction, when Venus is almost directly between the observer and the Sun.

Question: I'd like to understand with what instrument this image was taken, and how the technical challenges imaging Venus with such a small angular separation between it and the Sun were overcome! I did a quick check and during the UTC day of June 8th, 2004, Venus moved from about 0.6 to 1.1 degrees from the Sun as viewed by Earth. How was Venus imaged so nicely? Was this really imaged from the ground through Earth's daytime atmosphere, or from space? In either case, how was a high quality telescope pointed so close to the Sun and shielded from stray light and protected from damage?

below: Phases of Venus, from here.

I think you'll have to live with an incomplete answer. It appears this was taken by an amateur astronomer using a simple 4.5 inch telescope. You can find, on the ESO website, a collection of many images taken during the 2004 transit. On this page, is the image you are asking about (along with some other images by the same person). The credit for that image is given as:

  The Phases of Venus

  Statis Kalyvas
  4.5-inch Mizar telescope
  Nikon Coolpix 750
  February 27 - June 8, 2004
  Thessaloniki, Greece

I'm assuming (because I can't find any reference to state otherwise) that the name "Mizar telescope" is just the personal name Statis Kalyvas gave to his telescope. Although you can see that the equipment was nothing an amateur astronomer couldn't buy. A pretty typical telescope with a standard camera. Nothing "professional" or extremely high tech.

We can probably guess at some other pieces of the puzzle though. From the image below, you can see that the 2004 transit began (from Thessaloniki's local time) at 7:13 AM and ended at 1:25 PM.

Given that the observation period of the transit itself happened during the day, Statis definitely would have needed a solar filter for his telescope, which would have blocked upwards of 99.999% of the light from the Sun, allowing him to view the transit safely.

For the other images leading up to the transit, my guess is that he made his observations during in the evening when the Sun was near the horizon and not very bright (due to much of the light being scattered). Venus should've been higher in the sky than the Sun and still visible at that time. An eyepiece which gives a small field of view would've been useful here.

You can possibly guess that the image was taken near the horizon because a 4.5 inch telescope should have a resolution of $1.21\ \mathrm{arcsec}$ at best whereas, near conjunction, Venus is almost an arcminute in size. Venus should be well resolved, but it looks a bit fuzzy. This is likely attributed to the high airmass near the horizon. The drastic difference between the clarity of the image on 3/4 and 13/4 (just ten days later) leads me to believe the telescope was capable of better resolution and was simply impeded by the atmosphere near the horizon.


Could this estimate of the size and mass of the Chicxulub Impactor be accurate? My understanding has always been that the Chicxulub Impactor was in all likelihood an asteroid with a diameter of about 8 - 12 km, but then, the other day, I came across this paper. In it, they say that the diameter of the impactor could have been as large as 81(!) km, and that it was probably a comet and not an asteroid, a claim they base on the amount of iridium found in the layer associated with the impact. I'm not an astronomer so I'm not really in a position to evaluate the veracity of their claims and how they arrived at them, which is why I'm wondering if what they say could plausibly have been the case?

Well, that article was never accepted for publication in any peer-review journal apparently.

That said, estimates may vary widely depending on assumptions about the composition and velocity of the asteorid. One could estimate the mass of the object assuming compositional ratios similar to a certain class of objects and integrating the whole Iridium deposits but that's still just a very indirect method with high uncertainties and a lot of assumptions to be made.

One may also estimate the impact energy by using the size of the crater. Since that is related to kinetic energy it is a way to estimate the mass and velocity of the impactor, but this are coupled. Not only that, the kinetic energy depends on $m$ and on $v^2$ thus is very sensitive to the velocity of the object. The impactor might have had $m = 1$ and $v = 1$, or $m = 0.25$ and $v = 2$ (in arbitrary units) and the kinetic energy would have been the same. As you can see this dependence is a source of high uncertainty in the mass. The assumption of the impact velocity is needed and the precission of this assumption is very important if you don't want a wide variety of masses for the impactor. Once you got the mass of the impactor you need to know the density of it to estimate its size (another assumption). If it was a comet then the density is as low that for that mass the object would have been gigantic (but then one have to explain the low impact speeds since a comet would have been traveling faster in general). A purely rocky body is denser and thus for the same mass you need a smaller diameter. We in fact have done geochemical measurements on the impact site and there is strong evidence that the object was an asteroid and in particular a carbonaceous chondrite, which means that we have a good estimate of the density of the object. Detailed analysis of the structure of the impact crater can also decouple the mass-velocity constraint whithin the $E_{kinetic} = \frac{1}{2}mv^2$ relationship. Overall there best estimate is still around 12 km in diameter (+/- 3 km).


Advice on a thread size and fitting for a Celestron binocular tripod mount I have an older pair (8 years old) of Celestron 10X50 binoculars with what looks like a 0.25" screw hole. The closest product on the Celestron website now is this.

I have taken to looking at the moon and planets lately, with the binoculars handheld. I have a photo tripod. I got an Opticron binocular mount (this one) which fits my quick release plate perfectly. It also claims a 0.25" screw.

The problem is the the screw goes in to the hole (behind the little cap with 10X50 on it) and sticks. It goes in quite easily for 2 threads and stops. I've cleaned both, and used a tiny amount of lubricant on both, but with no joy.. Once in the mount does just about hold the binoculars, although they slew readily around the axis of the screw.

Any suggestions?

The standard thread for these is 1/4 - 20 UNC.  The 20 means 20 threads per inch - you could try measuring that. The same size fastener should be what's on the Opticron, so it should "just work".  It's possible that there is something in the hole in the binocular that's blocking the thread.

However (and if you are not the first owner),  someone might have tried to use the wrong fastener and damaged the thread.  Threads can be recut with the right tool.


Do white dwarfs produce stellar wind? If not, why is that? I've been told white dwarfs don't produce stellar wind like the sun produces solar winds. I don't quite understand why that is and can't really find any evidence that confirms or denies that.

My understanding is that solar wind is the product of the sun's variable magnetic field and consists of charged particles. Does nuclear fusion have to be occurring at the star's core for these charged particles to exist?

The strong gravity of the white dwarf star makes it difficult for stellar wind to escape. Unglaub (2008) shows that for surface gravity $\log g > 7.0$ hydrogen-helium winds are not possible for solar or sub-solar metallicity white dwarfs and only a weak metallic wind would be possible (note that in astronomy, "metal" tends to refer to any element heavier than helium). There is a group of hot white dwarfs with temperatures above 60000 K called "hot wind white dwarfs" showing ultra-high excitation absorption lines. According to Reindl et al. (2019), these features arise from wind material trapped in the magnetosphere of the white dwarfs.


How many stars are visible to naked eyes around Equator area? In the best sky conditions, we can see about 44,000 stars in the sky. But what if I am located around Equator area? How many stars I can see?

From what I know, the number of stars I can see should be far less ( I don't know why, though). Is it true? If yes, why?

The numbers you're quoting are extreme, more like a theoretical limit, and meaningless in practice.

The 44k number would be true if you assume a limiting magnitude of 8. There aren't many people alive today who have seen at least one object at mag +8. This is not so much due to vision problems (although that may be important for some people), as to the fact that a mag +8 sky is an incredibly rare occurrence today due to light pollution.

If you live on a continent with a low population density (such as North America, Australia, or the remote parts of Asia), then yes, maybe you have a chance of catching a mag +8 sky - if you drive far out into the desert, many hours away from any city or town. But if you live in Europe or the like, forget it.

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

http://www.jshine.net/astronomy/dark_sky/

Furthermore, the 44k number is obtained by assuming the mag 8 conditions, and then counting all known objects in the sky whose brightness exceeds that limit. It is very unlikely that any one person would see them all in one lifetime. Seeing even a single star at mag 8 can be a difficult undertaking for many people, even in ideal conditions.

It is very hard to quote a number, since it is so much dependent on light pollution, transparency, visual acuity of the observer (nearsighted people can't focus well enough to see very faint stars), the observer's age (acuity gets worse with age), the observer's experience (visual astronomers with years of experience will be able to see much fainter stars because they know how to look), etc.

If you're on a farm, away from cities, in a place with reasonably low light pollution, and your eyes are good, and you've been sitting in perfect pitch black darkness for at least 30 minutes prior - when you look up you can reasonably expect to see a few thousand objects, mostly stars. Keep looking, and after a while you will distinguish one or two thousand more stars, very faint, that you could not see at first sight. Practice this steadily for a few years, and you'll add maybe another thousand; but you won't be able to see those all at once - only one at a time.

Now travel to the Cerro Tololo site in Chile, up in the mountaineous desert, zero light pollution, excellent transparency, and you'll multiply all those numbers by a factor of 2x ... 5x.

As you can see, the numbers are very flexible because there are so many factors involved. You can't just slap a 44k label on it and call it a day; that doesn't make any sense in reality. Astronomers know that the pure magnitude number doesn't mean much by itself, because it is just one factor among many.

In a place with very high light pollution (like where I live, in the middle of a large, dense, sprawling urban area in California), you'd be lucky if you can see a hundred stars at night.

Or, in a place with zero light pollution, shine a flashlight into someone's face, and you've temporarily blinded them. You've reduced the number of stars they could see by an order of magnitude for the next half hour (night vision gradually recovers, and it takes 30 minutes to fully recover, according to US military manuals and visual astronomers practice).

Now, Equator versus poles.

The spin axis of the Earth has a fixed direction in space. The north - south axis always points at the same places in the sky, near Polaris in the north, and near the Southern Cross in the south.

For this reason, at the North Pole, you only ever see half the sky - the northern half, centered on Polaris. At the South Pole, you only see the other half. You'll never see Polaris from Australia.

At the Equator, you pretty much see the whole sky - not all at once, of course, but as the Earth rotates, it gradually exposes all parts of the sky at different times. However, the parts near Polaris and near the Southern Cross are always low in the sky and a bit hard to see.

So, bottom line is this:

In any place on Earth, in any given instant, you see exactly half of the sky. At the poles, that half is all you ever see. At the Equator, if you're patient, you'll eventually see the whole sky in time, as the Earth rotates and gradually exposes different parts of the sky at different times.


How do CMB photons 'gain energy when they pass through normal regions of space with matter' and 'lose energy when they pass through voids'? The Space.com article Huge Hole Found in the Universe says:

The gargantuan hole was found by examining observations made using the Very Large Array (VLA) radio telescope, funded by the National Science Foundation.
There is a "remarkable drop in the number of galaxies" in a region of sky in the constellation Eridanus, Rudnick said.
The region had been previously been dubbed the "WMAP Cold Spot," because it stood out in a map of the Cosmic Microwave Background (CMB) radiation made by NASA's Wilkinson Microwave Anisotopy Probe (WMAP) satellite. The CMB is an imprint of radiation left from the Big Bang, the theoretical beginning of the universe.
"Although our surprising results need independent confirmation, the slightly colder temperature of the CMB in this region appears to be caused by a huge hole devoid of nearly all matter roughly 6 to 10 billion light-years from Earth," Rudnick said.
Photons of the CMB gain a small amount of energy when they pass through normal regions of space with matter, the researchers explained. But when the CMB passes through a void, the photons lose energy, making the CMB from that part of the sky appear cooler. (emphasis added)

Question: How do CMB photons gain energy when they pass through "normal regions of space with matter" and lose energy when they pass through voids?

It is the late time integrated Sachs Wolfe effect.

As they travel towards us, apart from the general expansion, photons from the CMB gain energy when they fall into potential wells (where matter is). Of course, they lose it again as they emerge on the other side of the well, but the cosmic expansion means that the well isn't quite as deep by the time that happens. As a result, photons gain a little energy compared to the average CMB sightline.

The opposite happens for voids. Energy is required to enter a void (the potential is higher than average), but all this energy is not retrieved because the potential flattens due to expansion as the photon traverses it. Hence the directions of voids would be a bit cooler.


Spicules vs Prominences/filaments What is the difference between spicules and prominences/filaments. I know that they are both in the chromosphere, and I know the difference between provinces and filaments (depends on the view), but how are spicules different?

Spicules are much smaller, more dynamic, and possibly hotter. They are also much more abundant. While the term spicules is strictly defined for objects off the solar limb, their disk counterparts (let's call them fibrils for simplicity) cover the whole Sun, all the time. Spicules have lifetimes of a few minutes and extend up to ~15 Mm, with the most energetic of them being ejected with speeds of around 100 km/s.

While prominences/filaments have a relatively stable temperature around 10,000 K, spicules/filaments are also seen in filters sensitive to temperatures around 100,000 K, meaning that some of them will undergo a thermal evolution to at least the transition region.

Spicules/fibrils are a candidate for energy and mass deposition in the corona.


Can a star eject a liquid or a gas that has not been ionized? Can a star eject a liquid or a gas that has not been ionized? I am implying that these gases could be a source of reusable energy.

There are a variety of scenarios in which a star can emit gases, although that's probably not the best way to think about it. A better way to visualize it is in terms of mass loss rather than gas emission. Inevitably, most or all of a stars mass must eventually go "back to the galaxy in which they are in."

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

In a star the size of our Sun or thereabouts, towards the end of its life it will form a red giant. After this stage, its outer layers will "puff out" creating a planetary nebula. This is essentially the star emitting most of its material out to its surroundings, contributing its mass to the formation of new stars later on.

The other route, the supernova, happens when a star much more massive than the sun starts fusing iron and suddenly collapses, rebounding and creating a large shockwave which expels most of its material out.

Lastly, certain kinds of stars like Wolf-Rayet stars or (possibly) R Coronae Borealis variables can expel matter due to the fact that they are unstable and can suddenly eject part of their outer layers.


Determine the moons of Jupiter through a telescope I was doing my own "space exploration" last night with a telescope. Being a space noob I can't visually determine stars or planets (I know the moon, though), but I focused on one particularly bright one when I saw 4 smaller "stars" on each side. After a quick Google I learned I had discovered the moons of Jupiter!

I now know the moons I could see were Io, Europa, Ganymede, and Callisto. But which is which?

Is there an easy way to determine which Galilean moon is which?

Use an orrery that will let you specify a specific epoch and vantage point. There are some quite fancy ones online, for example this Solar System Scope:

What you do in this particular tool is click on the calendar bar below and enter date and time of your observation (if you forgot that, there's a good chance your photograph has a time stamp, either of the file or in its EXIF data). You can adjust epoch later on, too. Then double-click on Jupiter, then its Orbit button, then click on link As seen from Earth below, then click Go back button and use mouse wheel to zoom out so all four Galilean moons display. Then compare with the photograph that you took. Note that it might take a bit of rotation for you image to align perfectly. I use the old neck twisting and turning technique, but photo editing tools or even rotatable screens will do just as well, if not better.

Note that this mentioned tool does offer settings to switch between orrery and realistic model and large and realistic sizes of displayed celestial bodies. But there are many such tools online or downloadable for both computers and smartphones. I think even this one is available as an app for smartphones for free, but last time I tried to install it, it didn't work for me. Maybe they fixed that. Do check because it's a lot easier if you can check orbital alignment of Galilean moons as seen from Earth directly on the field, as you're doing your observations.

Have clear skies and I wish you happy hunting for planets and moons of our Solar system and beyond!


Can we see asteroid 1998 OR2 with unaided eye? Will I be able to see the asteroid 1998 OR2 with the naked eye on April 29?

If so, where can I find more information on when and where I'll be able to see it?

According to the page on The Sky Live the asteroid is currently at a visual magnitude of +11, which is also what Sky Safari says. It is not due to get much brighter so no, it will not be visible with the naked eye. According to the predicted light curve, there is a chance it will get to naked eye brightness (from a dark sky site) around April 2079.... So there is quite a wait! And these predictions are based on current models which can change with new information.

You can use theskylive to check on things like comets and asteroids, or if you prefer a mobile app, I always recommend Sky Safari


Telescope on drones to escape light pollution? Would it be possible to mount a telescope on a high quality drone and fly above the light pollution to get better star shots (if the wind allows it of course)?

Yes, but not in the exact way you think of.
To avoid light pollution it is better to go sideways.  If you are in a light-polluted city, there would still be light pollution at an altitude of 400ft (the maximum operating altitude of drones) And, moreover, amateur drones are not powerful enough to carry a telescope of any useful size. So you will find that driving a few miles out of town will be much more effective, and much cheaper.
On the other hand, water vapour in the atmosphere absorbs infra-red light. If you can take a "drone" up to 40000 ft, you would have a good location for infra-red astronomy.  You would just need a big "drone". . . A 747 would be big enough. SOFIA is just that: a telescope mounted on a jumbo jet to get above the water vapour in the lower atmosphere.
So, it's not practical to use an amateur drone to get above light pollution. It is practical to put a telescope on a plane but only for quite specialist observations.


what is the difference between luminosity distance and distance? This is a very basic question but I am little confuse the difference between luminosity distance and distance. Can anyone tell me that what is the difference between luminosity distance and distance?

The luminosity distance is the apparent distance of an object based on how bright is it and how bright it appears to be.  If you have two stars of equal luminosity, the more distant one will appear dimmer.
For local objects, the intensity of the light will obey an inverse-square law. So, for local objects (stars) the luminosity distance is equal to the actual distance.
But for very distant objects, like quasars or distant supernovae, the curvature and expansion of space-time will also affect luminosity. The luminosity distance is the distance obtained by ignoring these factors and just assuming that the intensity of the light from a distant object obeys the inverse square law.
As an example: a type 1a supernove is observed to have a magnitude 18.7, and is assumed to have an absolute magnitude of -19.3
It's luminosity distance in parsecs is $$10^{(18.7+19.3)/5+1} = 398 \text{Mpc}$$
If it has a redshift of z = 0.086 then its co-moving distance (what you would measure with a ruler) is 398/1.086 = 366 Mpc.


How frequently does a comet drastically change its orbit in the Solar System (like 67P did in 1959)? Comet 67P/Churyumov–Gerasimenko flew by Jupiter at only 0.05 AU distance in 1959. This encounter lowered its perihelion from 2.7 to 1.3 AU (almost becoming an Earth crossing comet). Since it happens to be the target of the Rosetta mission (and was picked only because of a launch delay), I wonder if it is as common as it seems that a 4+ billion year old comet shifted its orbit drastically less than 60 years ago?

How often does Jupiter do something like this?

How many known comets are known to have had their orbits substantially changed by a single flyby?

How many known comets are known to have had their orbits substantially
  changed by a single flyby?

Probably nearly all if not all of the (traditional) short period comets were deflected in closer to the sun by one of the gas giant planets.  The reason for that is simple.  Comets can't form in orbits with a near-sun fly-by so nearly all of them had to be deflected inwards by a gas giant planet.   From the Wikipedia article quoted above.

  Based on their orbital characteristics, short-period comets are
  thought to originate from the centaurs and the Kuiper belt/scattered
  disc

Short Period Comets also have a limited lifespan.  Halley's for example, with a perihelion of 0.6 AU, which is closer to the sun than Venus, it loses a significant amount of ice every time it flies past the sun and Halley's may lose all of it's ice within a hundred or so more fly-bys.  Source

Most Short Period comets are also planet crossers because they have highly elliptical orbits and planet crossing orbits aren't long term stable either.   So of the 586 known short period comets, it's safe to say they have an lifespan of (probably) less than a million years when averaged out.  Also, if you consider that Jupiter can work both ways, it can pull comets out of the inner solar system as well as toss them in, then you can work out that they would need to be replaced at least fairly often, perhaps on the scale of every 100 years or so, perhaps on the scale of decades.  (there's probably more accurate estimates out there, my estimate is on the rough side)

There's a great deal of variation with individual comets life spans though.   For example, a Halley-type orbit (see picture) doesn't cross Jupiter's orbit due to the 18 degree inclination, but Halley type comets are more rare (75 known per Wikipedia article above).   Jupiter family short period comets are more common (511 known), and those are also more likely to get jostled by Jupiter since they cross Jupiter's orbit.

Source of picture here.

Wikipedia also mentions Main-belt comets, but those aren't what I think of as traditional comets as they orbit within the asteroid belt and don't have highly elliptical orbits.   Because the Asteroid belt is mostly stable in a kind of gravitational zone between Jupiter and Mars, those "comets" may have been there for billions of years, and are only beginning to show comet like features as the sun has increased in luminosity.

The most common type of comet, long period comets, per the article, some 4,600 of them and about 10 new ones discovered every year and many many more exist but have not been discovered.  The Oort cloud has perhaps more than a trillion "potential" comets, or cosmic icy objects 1 KM or greater in diameter.   Source.

I don't know enough to say how Oort cloud objects are directed toward the inner solar system on average.   With so many objects, it's possible that some Oort cloud objects naturally formed with high enough eccentricity to pass through the inner solar-system once an orbit.  Some might be deflected by objects like Planet Nine (assuming it exists) or by repeated passes of dwarf planets or by the occasional pass of a rogue planet or the very rare close pass of another star, but it's way above my paygrade to try to estimate by which method Oort cloud comets are usually directed into the inner solar-system.  A percentage of those probably pass close enough to Jupiter or one of the other gas giants for a measurable deflection but I couldn't even guess what percentage.   Since these are the most common comets, it's very hard to give an estimate to your question in relation to these.

Loosely related, but Shoemaker-Levy was captured into Jupiter's orbit around 1960 or 1970.  It's hard to know what it's orbit was before then because the calculations are difficult, but it had to be a Jupiter crosser of some kind.  It's apogee (.33 AU) extended outside of Jupiter's Sphere of influence, which works out to about .22 AU, but it was a stable enough orbit to make a few cycles before crashing into Jupiter in 1994.

Jupiter's sphere of influence (.44 AU in diameter) is barely over 1% of it's orbital journey around the sun, but over hundreds or thousands of orbits, 1% is enough to clean out or deflect most of the objects that cross it's path.

Comet Ison, a long period comet was also (virtually) destroyed when it passed around the sun close enough to break apart in November 2013.   Comets seem to die fairly frequently.   Fortunately (or perhaps unfortunately), there's trillions more where they came from.   :-)

Hope that wasn't too vague an answer, but I think the short answer, based on deduction is that events like the one you mention with Jupiter deflecting Comet 67P are reasonably common, maybe every few decades, maybe once a century.   They're not rare events.   Earth will measurably deflect a small asteroid in 2029.   Asteroids are much more common than comets, but Earth is also much smaller than Jupiter.  Icy bodes are likely to pass close enough to Jupiter to have a significant deflection every so often and there's about a 50/50 chance on whether that deflection will push them closer to the sun or further away with a small chance that they will be caught in Jupiter's orbit.

Hope that wasn't too general.   Since this had gone unanswered, I thought I'd give it a shot.


Are there any bright non-stellar objects which make up asterisms? Hopefully the title is very self-explanatory. I was wondering if there were any non-stellar objects, such as bright galaxies, which when looking with the naked eye were mistaken for stars in the time when asterisms were documented and catalogued...

For example, within the Big Dipper I believe it is Mizar which is in fact a binary star system. (Correct me if I am wrong!)

I was hoping perhaps for distant yet bright Messier objects being mistaken for stars...

Most of Messier's objects are too dim to be seen by the naked eye. The few that are visible were mostly listed as stars in older lists.

Edmund Halley lists 6 "luminous patches" which "discover themselves only by the telescope, and appear to the naked eye like small fixt stars" (Halley).

These include M42, in Orion, part of the sword (or what ever that is between the hunter's legs), and Omega Centuari, listed as a minor star in the centaur (roughly on the horse's back) in Bayer's list.

(The other objects on Halley's list were either known to be "cloud-like" [M31] or too faint for naked eye astronomy.]


What percentage of habitable-zone planets are detectable by transit? I realize that the probability of detecting a planet by transit depends on the size of the star, the size of the planet's orbit, and the size of the planet; and ranges from ~10% to a small fraction of one percent.
However, the habitable zone of a star is a function of the star's brightness, and thus at least statistically connects the size of the star to the size of a planet's orbit if the planet is assumed to be in the habitable zone.  There is also a known distribution of the relative number of stars of different sizes.
With all that put together, can we form a reasonable estimate of the percentage of habitable-zone planets that can be detected by transit methods? The overall percentage has to be between 10% and about 0.1%, which suggests that if we are able to detect, say, 5 habitable-zone planets within 100 light years, there are probably actually at least 50.

Back of the envelope time.
First, we have to assume perfect data, so the only factor at play here is whether there is a geometric eclipse or not. Of course if you have poorer data then you will miss some planets because they are too small. i.e. We are looking for the fraction that can be detected in principle.
Let's assume the planets are small enough that their size does not really influence the transit probability, which is given by $\sim R_*/a$, where $a$ is the semi-major axis.
Let's assume circular orbits.
Let's assume a bare planet with no albedo so that the equilibrium temperature is given by
$$ T_{\rm eq} = T_* \sqrt{\frac{R_*}{2a}}\ .$$
Let's assume that a typical star in the Galaxy is an M-dwarf with a temperature $T_* \simeq 3500$ K and a radius of $R_* = 0.5 R_{\odot}$, and let's assume that the likelihood of planet occurrence is independent of stellar mass, so that the properties of an M-dwarf can be assumed for statistical purposes (in practice the answer will depend on what kind of star you are considering).
Let's assume that a habitable planet needs the equilibrium temperatures to be between 273K and 350K (arbitrary I know, and ignores the issue of atmospheres). The range of $a$ for this temperature range, around our fiducial M-dwarf, is between $50R_{\odot}$ and $82R_{\odot}$, with a probability of detecting a transit of between 0.6-1.0%.
So that is my answer 0.6-1%
The probability for higher mass stars is smaller and there are fewer of them. That is because although they are larger, the planets have to be much further away to be in the habitable zone (e.g. the transit probability for Earth is 0.4%). The main uncertainty is the probability of close-in planet occurrence for very low-mass stars where the transit probability can be much higher even though there are fewer host objects.


Hypothetically, would we be able to see the moon from Saturn's North Pole? Hypothetically speaking, if Saturn had a solid surface that we could stand on, would we be able to see our Moon from the North Pole of Saturn? Would the rings of Saturn obstruct the view?

First of all, at that distance seeing the Moon and seeing the Earth amounts to the same thing. At its closest, Saturn is around 3000 times as far from Earth as the Moon is, so viewed from Saturn, the Moon is never more than about a minute of arc away from the Earth.

If you can see Earth from the North Pole of Saturn, you can see the Moon, also.  (Though it would be a lot dimmer.)

But if you can See the Earth from Saturn's pole, then you can also see Saturn's pole from Earth.  So your question can equally well be phrased as "Can Saturn's north pole be seen from Earth and if so, when?"

If you look at pictures of Saturn from Earth, the rings are sometimes tilted so far that they cover one polar region, though it's also true at those times that polar region tilted away from Earth, anyway!  Most of the time the rings are either not tilted enough to cover the poles or tilted to cover the south pole.

So the answer is that during most of Saturn's orbit around the Sun, the rings would not obscure the sightline from Saturn's north pole to the Moon.


How can the oblateness of Earth impart torque to change the longitude of ascending node of a satellite? I've read that you can change the longitude of ascending node of a satellite (i.e. changing the orbital plane without changing the inclination) to change its orbital plane by using the torque imparted by the oblateness of Earth. How does this happen exactly? I mean how does the oblateness of Earth cause a torque induced precession of a satellite?

PS. I think I don't understand how torque induced precession works and that's why I am not able to understand this. So if any of you kind gentlefolk are willing to explain, I thank you in advance.

EDIT: I just came across Nodal Precession. And from what I understood that due to Earth's equatorial bulge, the gravitational force vector is offset from its center, thus producing torque. I can visualize how the longitude of ascending node will change, but can't visualise how will the change in the torque caused by slightly offset gravitational force vector will cause it.

I'd like to start by quoting from the Wiki page on Nodal Precession you mentioned:

  A non-rotating body of planetary scale or larger would be pulled by
  gravity into a sphere. Virtually all bodies rotate, however. The
  centrifugal force deforms the body so that it has an equatorial bulge.
  Because of the bulge the gravitational force on the satellite is not
  directly toward the center of the central body, but is offset toward
  the equator. Whichever hemisphere the satellite is in it is
  preferentially pulled slightly toward the equator. This creates a
  torque on the orbit. This torque does not reduce the inclination;
  rather, it causes a torque-induced gyroscopic precession, which causes
  the orbital nodes to drift with time.

(emphasis mine).  Note that this is a small force and the precession is slow compared with orbital speeds.  Precession towards (or coasting away from) the equator thus is a change in latitude compared with what would be expected, and this "twist" changes the expected longitude at a given point in time because the satellite's vector velocity changes (I believe in both direction and magnitude, but I'm not certain about the magnitude).


How small a star can provide Sun-level illumination to its planets? ...or how small can a solar system be, to sustain an Earth-like planet?

Sun is not really small, and 1AU is pretty far when you look at orbital radii of exoplanets and size of their stars.

How small a star can be, to still provide Earth's equivalent of 1.36 kW/m2 above its own Roche Limit (of Earth-sized planet)? What would the star's diameter and the orbit providing such irradiance be in such a case?

I guess a Brown Dwarf wouldn't be enough to give that much within its Roche Limit, meaning any planet receiving enough light would be torn apart by tidal forces, and likely stepping up to smallest Main Sequence stars would step up the energy output enough that a planet at the Roche Limit would be scorched like Mercury, so its orbit would need to be considerably larger... but I don't quite know how to perform the needed calculations.

I consider small stars to mean main sequence stars (or brown dwarfs) with mass less than the Sun, and then also consider compact stellar remnants - white dwarfs and neutron stars.

Main sequence stars and brown dwarfs

What you need is a mass-luminosity relation combined with an expression for the tidal radius in terms of the stellar mass. The latter also depends on the mass of the planet, so when you say earth-sized, I'll assume that means mass and radius.

So going through the calculation.

Flux at the planet is $L/4\pi r^2$, where $L$ is the luminosity and $r$ the orbital radius (assumed circular).

Let's next use an approximate relation that
$$\frac{L}{L_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^3$$
A more accurate numerical relationship could be obtained from evolutionary model calculations. There is a complication that the luminosity of very low mass stars and brown dwarfs does depend on age.

For a rigid body the tidal radius (Roche limit) is
$$ r_t\simeq 1.4 R_E\left(\frac{M_{\odot}}{M_E}\right)^{1/3},$$
where $R_E$ and $M_E$ are the radius and mass of the Earth.

So setting $r=r_t$, the flux to $1400$ W m$^{-2}$ and replacing $L$ in terms of mass, we find
$$\frac{M}{M_{\odot}} = \left[\frac{4\pi \times 1400}{L_{\odot}} (1.4 R_E)^2 \left(\frac{M_{\odot}}{M_E}\right)^{2/3}\right]^{1/3}.$$

All that remains is to put the numbers in and I get $M = 0.026 M_{\odot}$.

So, a small brown dwarf - but there are caveats. First, and most importantly, as I said, the luminosity-mass relation is a bit rough and ready, and is certainly age dependent for something as low mass as this. Second, the expression for the Roche limit depends a bit on structural properties of the planet, but I think this introduces relatively little uncertainty.

If you use that mass and the tidal radius formula, you find that the orbital radius is only $5R_E$. As the minimum size of a brown dwarf is about a Jupiter radius, it appears that tidal breakup would not be the limiting factor for an Earth-like planet. Also note that the spectrum of radiation from a brown dwarf might contain the same power per unit area as sunlight, but it has a very different spectrum; mostly infrared with almost nothing in the visible band.

Compact stellar remnants

Smaller "stars" are possible in the form of the collapsed remnants known as white dwarfs or neutron stars. Planets have been detected around neutron stars.

White dwarf stars are "born" very hot, cool rapidly and then more slowly. To answer your question requires an addendum, which is a minimum time you want this illumination for. A newborn WD has about a solar luminosity and has a similar diameter to the Earth. So you planet could be 1 au away from this and still receive a sun-like power per unit area. BUT, it would be short-lived and the spectrum of the light would be nothing like the Sun. The WD would have a temperature in excess of 50,000K and most of the light would be in the UV.

Cooling curves (and observations) suggest they cool to $10^{-4}$ solar luminosities in less than a billion years and have more sun-like temperatures, but to receive sun-like illumination at this point, your planet would need to be only 0.01 au from the white dwarf. White dwarfs have a mass a little less than the Sun. Using the same tidal radius formula as above we get $r_t \sim 0.004$au. So I conclude that yes, it might be possible to have a long-lived planet with Earth-like illumination around a white dwarf. But note (a) conditions change rapidly (compared with the Sun) and (b) you have to arrange to get the planet that close, since an orbit of that size would have been inside the red giant progenitor of the white dwarf.

Neutron stars are born in the cores of supernovae explosions and have a radius of about 10 km and mass about 1.5 times that of the Sun. If your planet can survive the supernova (or be constructed afterwards) it faces a similar set of problems to a planet around a white dwarf.

Neutron stars are born with surface temperatures of billions of degrees, emitting copious gamma ray radiation. They cool very rapidly - within a million years the surface temperature is down to an X-ray emitting million degrees and perhaps to UV-emitting temperatures of tens of thousands within 100 million years. It may be that they never cool much below this, since they are heated by their rotational spin-down, the decay of their strong magnetic fields and accretion of gas from the interstellar material. Even at 50,000K, because of its small size, the neutron star has a luminosity of only a millionth that of the Sun. That means your planet would have to be moved within 0.001 au of the neutron star to receive the same power and this would be well inside the tidal breakup radius of $\sim 0.005$ au.

I conclude that there really is no long-term possibility for sun-like illumination from a neutron star. Note also, that I have ignored magnetospheric and pulsar-like emission from the neutron star. These are also short-lived phenomena only seen in the first million years or so of a neutron stars existence.


How was all the matter curled up inside a singularity during big bang ? If the singularity had infinite mass, temperature and density, then it should curve space and time but there was no space-time before big bang. So where did the singularity come to existence without space-time ?

The very short answer is "no one knows". The available evidence very strongly suggests that the Universe 13.8 billion years ago was hot, dense and expanding and the further back we look, the more indirect the clues we get are, but the hotter and denser it seems to have been.

There are multiple theories for what happens as we follow this process back and back, but theories is all they are. Scientists work to make testable predictions that would follow from some of them but not others, and allow us to whittle down the pool, but it's hard.

All that said, at the broadest level there seem to be two common current views as to how to address the question "what happened before the big bang".

One is to argue that it is simply a bad question: time, like space, is "curved" by the presence of a lot of mass or energy and the question may be analagous to asking "what is North of the North pole". There simnply is no "before" direction from "the Big Bang", or to put it another way, it is the nature of this universe to have a time dimension which only goes a finite distance in the "past" direction. This, very loosely, is the "no boundary" approach of Hawking and others.

The other view, which has names like "eternal inflation" or "cosmic inflation" is that the original state of the universe was a smooth and fairly featureless, but extremely rapidly expanding (exponentially rapidly) one, which was only meta-stable. Part or all of this universe, spontaneously "changed phase" to produce normal space and time, and as a result of the energy released by the phase change, was filled with very hot, very dense, matter and energy, still expanding, but no longer exponentially. Indeed, gravity then started to slow the expansion.  In some theories just a small region would undergo this phase change, in others it would be the whole universe. Either way, all of the universe we see comes from a small region of the inflating universe.