What constraints are there on solid material in the ice giants? Recently there has been experimental verification of superionic water ice (suggested to be called ice XVIII, not sure if this nomenclature is official yet), which turns out to be a solid (or at least, solid-ish). This has been proposed to comprise a substantial fraction of the volume of ice giant planets like Uranus and Neptune, which is something of a departure from the idea that the bulk of these planets is fluid, though potentially the presence of substances other than water may modify things.

My question is whether there are any observational constraints on the presence of solid material in Uranus and Neptune, e.g. from the gravitational moments, tidal parameters, or global oscillations ("seismology") - I'm aware that the K2 mission failed to detect any global oscillations on Neptune so I'd guess this latter method may not help much.

(There is a previous question about solids in gas giants, but that one seems to be more focused on rock rather than ice and has ended up addressing Jupiter rather than the ice giant planets.)

NASA conducted a study on the Ice Giants in 2016 after recognition as a high-priority flagship mission target in the 2013-2022 Planetary Decadal Survey (overview of the Decadal Survey from The Planetary Society), a community document which helps set priorities and goals. The full report and summary are available from LPI's Ice Giants Mission Planning page and is intended to review the science and mission options leading into planning for the next decadal survey report. In the Science section starting on page 29, the introduction has a table of the major components of the different planet types (terrestrial, gas giant and ice giant) where ice giants are listed as ~10% gas, ~65% ice, ~25% rock with a total mass of ~15 times that of the Earth. This is in comparison to terrestrial planets which are almost 100% rock and gas giants which are ~85% gas. This is illustrated in the figure below taken from the report (their Figure 3-1):

In the lead-in to the subsections on the various science questions and subsections, the report goes on to say:

  We  therefore  consider  the  objectives  of determining  the  interior  structure  of an ice  giant  (Section  3.1.1)  and  the  bulk  composition (including noble gases and key isotopic ratios, Section 3.1.2) as the highest-priority science to be done.

Section 3.1.1 on interior structure states that there is a presumption that these "ice giants" are dominated by water but there is very little direct evidence. It then explains that ice giants are normally modelled with three layers: an outer Hydrogen/Helium envelope, below that a thick fluid ionic water layer (or water mixed with other ices like ammonia and methane), and at the center, a small rocky core. However the limited amount of gravity field data from the Voyager 2 flybys showed that both planets are not as centrally condensed (less mass in the center) as expected in the simple 3 layer structure and also:

  "there is no strong motivation, beyond computational convenience, for a 3-layer structure (Helled et al. 2010)." Link to Helled et al. 2010 journal article (free access)

Both Uranus and Neptune have bulk densities a little under compressed water, implying that there has to be Hydrogen/Helium (H/He) envelope and which is indeed seen in spectra of the visible surface. Below this envelope though little is known apparently. The low density and the limited gravity data would imply a mostly water interior with other gasses such as ammonia or more H/He to keep the density lower than water and a very small amount of rock in the core Nettelmann et al. 2013 (arXiv paper link). This model however predicts an ice:rock ratio of more than 10:1 which is not seen in any of the other icy moons and the model does not explain well the heat flux coming from the interior. This presentation by Nettelmann at a 2014 ice giants meetings talks more about these models, the constraints from the gravity data from Voyager 2 and the heat flux problems and shows possible internal structure diagrams.

The interior structure section of the Ice Giants report ends by calling for measurements of the higher order gravitational moments J2, J4 & J6 to higher precision, particularly for Neptune. However this is limited to the bulk structure and doesn't constrain the layering which would need a Doppler imager on a spacecraft to measure planetary oscillations analogous to solar helioseismology and/or an entry probe. The Doppler imager is discussed more in Sections 3.3.1 and 3.3.2 of the report which note that these new instruments have the potential to provide great information on the interior structure, but that Doppler imagers are technologically immature and that even with the largest telescopes on Earth or near-Earth spacecraft it has not been possible to see if the oscillations exist and the oscillation amplitude may be too small to detect. The report also notes such an instrument is also large, power hungry and produces a large data volume, all risks on a constrained mission to the outer planets but the benefits outweigh the risks. (Using up most of your mass and power budget on an instrument that may fail, or not produce a measurable result, makes people very nervous...)

In summary, there are some limited constraints on the bulk components of the ice giants but this is still a poorly constrained and understood problem which is a subject of ongoing research but is need of additional constraints and data from a dedicated mission. Such a mission would need to be flagship class which makes it expensive and would need broad support from the majority of the planetary community and would likely need to be the top-ranked mission from the next decadal survey covering the 2023-2032 timeframe to stand a chance of being funded and flying.


At what distance could a supernova damage the Earth's ozone layer? As from my latter question it seems Betelgeuse might be much closer than the usually presumed 640 light years. It might be as close as ~440 ly. Suppose it is, would this have any dangerous effects on the Earth's ozone layer (damaging it dangerously) and on orbital spaceflight? How close would a star like Betelgeuse have to be to have dangerous effects on the ozone layer and above (and thus on humans and other life on Earth)?

I came across your question and figured I would reply to it because doing a project on the subject. Explanations here are draw from the website astronimate.com

Would a supernova of Betelgeuse only 440 light years from Earth have any dangerous effects on the Earth's ozone layer (damaging it dangerously) and on orbital spaceflight?

No. A star would need to explode within 30 to 50 light-years to jeopardize Earth. Fortunately, no stars even remotely close to such distances are prime to explode any time soon. In fact, the most likely bomb candidate, IK Pegasi, is a safe 150 light-years away. And, at such distances, we would see the spectacular celestial object, but avoid virtually all danger!
However, were a star 30 light-years away to explode, Earth would be in major danger. Some, or all, of our ozone layer, would disintegrate, leaving us vulnerable to lethal radiation from our Sun.
Not to mention, phytoplankton and other aquatic food chain staples would completely parish. Ultimately, our ocean life and primary food chain would die off, creating massive evolutionary problems.
Plus, essential gases in our atmosphere, like nitrogen and oxygen would likely be ionized by bombarding radiation. Eventually, this would cause obvious dangers to most living organisms. Alas, Earth seems to be quite safe from supernovae for several million, or billions of years. Yet, hopefully, we will experience a supernova from a safe distance. After all, this would be one of the most notable sights in recent human history!


Black holes and material interaction I was wondering about what happens when a black hole forms.

I mean, a black hole forms when a star collapses, so when an enormous mass of protons, electrons, neutrons etc collapses together.
But when it does, all those particles should come "in contact", so what happens when electrons and protons come in contact?
Shouldn't they annihilate each other releasing energy?
In the end, shouldn't the singularity be composed only by the mass of particles without an electrical charge?

I'm not a professional, I just find astronomy fascinating and I like to read about it, so forgive me if I said something obviously wrong :)

Electrons and protons do not annihilate each other, since protons have a much greater mass than electrons. Only matter and antimatter does, since a particle and its antimatter counterpart have the exact same mass.

However, under the right conditions, an electron and proton can combine. This forms a neutron, since the charges will cancel each other out. This process is called electron capture, and it's the reason why neutron stars form.

I'm not an expert on black holes (I haven't studied GR), but I know that a lot of black holes form from neutron star collapses.


How much solar UV radiation would someone get who were on the surface of Titan? The Saturn system is about 10 times as far from the Sun as the Earth. This question is concerning Titan for two reasons: Titan has an Earthlike atmosphere so you don't have to account for a higher radiation due to vacuum. Titan's surface pressure is 1.45 atm (21.5 psi) but imagine you're standing on the summit of a mountain where the pressure were 1 atm (14.7 psi). The 2nd reason is that due to Titan's Earthlike atmosphere you don't need a spacesuit but an oxygen mask and warm clothes only (that should cover your entire body if you don't wanna get frostbites).
Compared to the average UV dose on the surface of the Earth, how much radiation would you get on a mountain peak on Titan? Since Titan is 10 times farther from the Sun, does this mean you get 10 times less the radiation on Earth, or does it decrease with the square so you get a 100 times less radiation of what you get on Earth? Or something else?
Also, do you know a calculator that can determine UV radiation and account for all factors?

You are correct that the incident radiation density is 100X less - it's a matter of computing the area of a spherical surface, which goes as the square of the radius. So even in the absence of atmosphere, and assuming you wear some kind of spacesuit that's transparent in the UV, you won't get much of a tan.


Is lunar elevation at a given location for a given day unimodal? Is lunar elevation at a given location for a given day unimodal:

Unimodal function

In other words, once the moon's elevation reaches a minimum (which may
be above or below the horizon depending on if the moon is
circumpolar), does it increase continuously to its maximum, or can it
decrease briefly and then increase again (resulting in a local minimum
that's not a global minimum) before reaching the maximum.

And also similar to the above as it goes from the maximum elevation
(which may also be above or below the horizon depending on whether the
moon rises at all on a given day).

Why this is important: I think I found a bug in libnova that assumes
the moon is circumpolar if it is above the horizon when it's due
north. This is untrue

My fix
does not assume this, but does assume that lunar elevation is
unimodal.

If that's untrue, I'll need to further fix my fix.

EDIT: Here's how far I've gotten:

The moon's declination and hour angle (at a given location) are
both functions of time. Thus, we can implicitly consider the moon's
declination to be a function of its hour angle. This only works
because the moon never experiences retrograde motion. Bodies that
experience retrograde motion could have two declinations for a given
hour angle, so the declination would no longer be a function of the
hour angle.
Using the standard formula, we see the moon's elevation is:

$$
   \sin ^{-1}(\sin (\text{lat}) \sin (\text{dec}(\text{ha}))+\cos (\text{ha})
    \cos (\text{lat}) \cos (\text{dec}(\text{ha})))
$$

where dec(ha) is the declination at hour angle, ha is the hour angle,
and lat is the latitude.

Since the moon's elevation is a differentiable function (except
potentially at the poles and equator), we find the derivative with
respect to the hour angle:

$$
   \frac{\text{dec}'(\text{ha}) (\sin (\text{lat}) \cos
    (\text{dec}(\text{ha}))-\cos (\text{ha}) \cos (\text{lat}) \sin
    (\text{dec}(\text{ha})))-\sin (\text{ha}) \cos (\text{lat}) \cos
    (\text{dec}(\text{ha}))}{\sqrt{1-(\sin (\text{lat}) \sin
    (\text{dec}(\text{ha}))+\cos (\text{ha}) \cos (\text{lat}) \cos
    (\text{dec}(\text{ha})))^2}}
$$

The moon's elevation will reach a min/max (in the sense of
unimodality) when this derivative is 0. That occurs when:

$
   \sin (\text{ha})\text{ = } \text{dec}'(\text{ha}) (\tan (\text{lat})-\cos
    (\text{ha}) \tan (\text{dec}(\text{ha})))
$

So the question becomes: can this equation have more than two
solutions for 0 < ha < 2*Pi

Which I hope someone answers at: https://math.stackexchange.com/questions/587136/

Yes, it is unimodal.

Earth's movement is a simple rotation, thus if the Moon were fixed on sky it will describe a circumference like the fixed stars.

From that basic movement we need to introduce Moon's own movement on the fixed stars sphere, which makes Moon not to complete the circumference in 1 day. Instead, Moon performs an arc similar to that of the fixed stars but shorter.

This circumference-like movement does not wobble, so it has just one maximun and one minimum in elevation.


Do we finally have a decent set of parameters of 162173 Ryugu? "Give a man a watch and he will know the time; give him two and he will never be sure" - this was my impression when reading the Wikipedia article on 162173 Ryugu.

Mean diameter:

0.865±0.015 km[3]
0.87 km[4]
0.90±0.14 km[5]
0.92±0.12 km[6]
0.980±0.029 km[7]
1.13±0.03 km[8]

And no mass estimates at all.
Now that Hayabusa2 has arrived, do we have solid measurements of Ryugu's size and mass yet?

A science paper from 2019 by K. Kitazato et al. shows a shape model of the surface in Fig. 2, so  we now indeed know the exact size and  the exact mass, even the surface temperature distribution:

The quoted paper refers to S. Watanabe et al., Science, 2019 which has all the details

Ryugu has an oblate body, with an equatorial radius of 502 ± 2 m and polar-to-equatorial axis ratio of 0.872 ± 0.007. The total volume obtained from the SPC-based shape model is $0.377 {\rm km}^3$, with an uncertainty of 1.3%. We conducted a gravity measurement during a spacecraft ballistic descent down to 0.85 km from the asteroid surface and a subsequent ballistic ascent up to 5.4 km. The estimated mass is $4.50\times 10^{11} {\rm kg}$, with an uncertainty of 1.3%, mainly because of uncertainties in the solar radiation pressure on the spacecraft.

The highlighting in the quote is from me.


Do objects look larger the further away they are, beyond z=1? I stumbled upon this text which illustrates and says:

  We find that as the object is moved to higher redshifts its angular
  size first decreases (as naively expected) but soon begins to increase
  after passing through a minimum value.

Is this correct for the mainstream standard model? What would an English language explanation be?

Yes this is correct for the standard model. I believe the English language explanation would be that the angular size of a galaxy does not depend on its current distance to us, but its distance when the light we currently see from it was emitted (here using "distance" synonymous with "proper distance"). Though I also believe there are other complicating factors such as the spatial geometry of the Universe.

Look at the first diagram in Figure 1 in this paper, notice how the size of the past light cone in terms of proper distance firstly increases as we go back in time, before reaching a maximum and decreasing to zero at t = 0.


How many of the whole zodiac constellations can be seen in one single night? As far as I understand, since the the 12 zodiac constellations (astrological signs) 'surround' the earth, hence it's possible to see by naked eyes a group of 6 astrological signs (only) simultaneously - from one hemisphere.

Assuming it's correct, my question is how many of the 12  zodiac constellations are possible to see in one single night?

Since the earth moving constantly, it seems that the answer depends on the time and season, but assuming that the day and the night are equal in time (March 21th), what will be the answer? To my logic it should be 12 but I am not sure about it.

Running a simulation in Stellarium, on March 21st, you can just about see Pisces at about 7pm, but it is setting, and at 5am Aquarius is just rising, before being hidden by the sun. The sun is "in Pisces" on the 21st (the astrological houses have not been updated) but it is a pretty big constellation!

This means that you can see, or partly see all the zodiac constellations.  Although Pisces and Aquarius are not fully visible, but you can see the other 10 at some point during the night.


Why do Seyfert Galaxies emit less energy than Quasars and Blazars, and how is the emitted energy measured? When I was reading information about QSOs on a website, it said that Seyfert Galaxies emit less energy than Quasars and QSOs.

Why do Seyfert Galaxies emit less energy than quasars, and how is this energy measured?

The linked website seems to summarise generally accepted theory.

Seyfert galaxies appear less energetic than quasars or blazars. The energy is measured by observations of their brightness in the sky (at different wavelengths of light) combined with their distance (inferred from their red-shift). There are also Radio-galaxies

We think that active galactic nuclei don't release the same amount of energy in all directions, instead they have a "beam". They also have an accretion disc, the inner parts of which are extremely hot If we happen to be looking directly down the beam, they appear exceptionally powerful, and we call them "blazars" If we see them off centre we can see into the extremely hot central part of the accretion disk and they will be a quasar or (if less powerful) a Seyfert galaxy, and if we see them side-on they appear less powerful, but the two lobes of high-energy gas and particles emitted along the beam will be visible, and we will observe a two-lobed radio galaxy.

See these related answers: Quasars and SMBH and Difference between quasar and Active Galactic Nuclei?

http://www.astronomy.ohio-state.edu/~ryden/ast162_9/notes37.html


How to use spectral flux density data while using Planck's law giving spectral radiance? I'm currently working on a fun project in my free time where I'm trying to calculate the temperature, among other things, of stars based on their spectra. Since I have essentially no prior experience in astronomy, except for quite a strong interest, this may sound like quite a dumb question, but I would appreciate an answer. The data provided from SDSS (Sloan Digital Sky Survey) that I'm using plots the spectral flux density, Fλ, over wavelength, λ. Now my problem is, I'm trying to do a curve fit in Python using Planck's law, (I'm aware that it is also possible to use the proportions between red and blue light, but I first want to see how close I can get using this method) but Planck's law (as below) gives spectral radiance. This means that while the data I'm using is in different units than the numbers I get from Planck's law. Therefore, (as you can see below) they are on greatly different scales. So my question is, how do I fix this? How can I use Planck's law to be able to fit the data? Is there any relationship between units that I have not been able to find? I want to clarify that I have tried my best to find the answer on my own, using Wikipedia and other sources, but my attempts have, obviously, not been rather successful.

The data that I'm using (The y-axis is 10^-17 erg/s/cm^2/Å):

The graphs I have got using Planck's law:

As I understand it, spectral flux is defined as
$ F_{\nu} = \int_\Omega I_{\nu}(\theta, \phi)\cos{\theta} d\Omega $
where $d\Omega$ is the solid angle element over which the integral is performed. Here $I_{\nu}$ is the specific intensity. The $\nu$ subscript denotes frequency dependence.
According to A. Choudhuri "Astrophysics for Physicists" p. 25, blackbody sources are isotropic emitters therefore we can drop the angular dependence in the spectral intensity and $I_{\nu}(\theta, \phi)$ becomes your Planck formula $B_{\nu}(T)$.
Hence
$ F_{\nu} = \pi B_{\nu}(T)$


How would one figure out the rate of axial precession of a planet? I want to start with the fact that I am by no means an astronomer, or even a hobbyist in the field. I am attempting to build a fantasy world which still reflects accepted physical laws.

My problem comes from the fact that I have no idea how to go about figuring out the period of axial precession for a fantasy earth-like planet. For the Earth, I believe the period is roughly ~26,000 years. I've spent a few days trying to find resources, but in  all honesty I can't wrap my head around half of the equations I've seen presented.

How would I go about doing this calculation for a given planet, and which of its properties would I have to specify in order to do this calculation?

Edit Important? Details:
My Earthlike planet has a mass 1.2597x the mass of our earth, and two fairly large moons (actually, one of them may be large enough to classify it as a binary planet but I'm not sure).

Wikipedia's page on Axial Precession has a good deal of mathematics on it and, unless you have absolutely got to have precise numbers for some obscure story reason (which is probably what I'd call the tail wagging the dog - change the story to avoid that problem) then making up the numbers or just forgetting all about Axial Precession is the way to go.    The periods involved will be very long by the standards of any story you're likely to write, so why burden yourself with something you probably don;t need.

The maths that follows only gives a rough approximation anyway, and you don't want to even contemplate the kind of things you need to do to get a better one : it's not worth it.

That said let's have a look at the very basic theory result that Wikipedia gives :

There are two component to axial precession that matter (for Earth) : the one due to the Moon and the one due to the Sun.  The Moon's effect is actually larger, but these numbers are very sensitive to the values you use.

The Solar Contribution

$$\frac{d\psi}{dt}=\left[ \frac {GM_s}{a_s^3\left(1-e_s^2\right)^\frac 3 2} \right] \left[ \frac {C-A}C \frac {cos\epsilon}\omega \right]$$

Lots of symbols so what do they mean ?

$G$ - the Universal Gravitational Constant also famous from $F = \frac {GM_1M_2} {r^2}$ Newtons law for gravitation.
$M_s$ - The Sun's mass - in your case you need the mass of your planet's star, of course.
$a_s$ - The semi-major axis of the orbit of the planet around it's star
$e_s$ - The eccentricity of the planet's orbit around it's star.

Now that second term in square bracket, which is also in the expression for the Lunar contribution.  This one is trickier.

$C$ - moment of inertia (of Earth) around the axis of rotation
$A$ - moment of inertia around the equator
$\epsilon$ - the angle between the equatorial plane and the ecliptic plane (see below)
$\omega$ - Earth's angular velocity (due to it's rotation, not it's orbit)

Now this expression is really poorly dealt with in Wikipedia because of two problems.

$epsilon$ in Wikipedia is assumed to be the same for both the Solar and Lunar contributions.  This is not (AFAIK) correct.  The angle should be the angle between axis of rotation of the body and the plane of the orbit of the other body (which means it's different for the Sun and Moon).

The $C-A$ and $A$ terms are really hard to deal with for mere mortals (and frankly just messy for anyone else).  For your purposes I would propose the following compromise term instead.  It's based on modeling the planetary bulge as an ellipsoid of constant density compared with the $A$ value for an ideal sphere - both objects have the same mass and density, which I'm taking as constant .  I'll spare you the derivation :

$$\frac {C-A} C \approx 1 - \frac{R^2} {a^2}$$

where in this case $R$ is the average radius of the planet and $a$ is the equatorial radius of the planet.

The Lunar Contribution

$$\frac{d\psi}{dt}=\left[ \frac {GM_l}{a_l^3\left(1-e_l^2\right)^\frac 3 2} \left(1-\frac 3 2 sin^2i\right) \right] \left[ \frac {C-A}C \frac {cos\epsilon}\omega \right]$$

Not much change here expect that the masses and so on refer to the Moon and not the Sun (hence the different subscripts).  There is one additional term which is the factor :

$$\left(1-\frac 3 2 sin^2i\right)$$

This corrects for the effect that the angle of inclination of Moon's orbit to the ecliptic is not zero.  The ecliptic being the plane with the Sun and the Earth's orbit in it.

You have to decide these numbers for yourself.

The total effect :

The total effect is simply the sum of the two other effects so :

$$\frac {d\psi}{dt} = \frac {d\psi_l}{dt} + \frac {d\psi_s}{dt}$$

If you had multiple moons you would need multiple lunar correction terms.

Just for clarification that $\frac {d\psi}{dt}$ means the rate of change of the angle $\psi$ with respect to time $t$.  To get how big an angle you'd move through in a century you do this :

$$\Delta \psi \approx \frac {d\psi}{dt} \Delta t$$


How do comets' nuclei stick together? Comets are commonly called icy dust balls or dirty snowballs and as such rather light:

Known comets have been estimated to have an average density of $0.6 {\rm g}/{\rm cm}^3$. Because of their low mass, comet nuclei do not become spherical under their own gravity and therefore have irregular shapes.

I read about their live times, I learned that they can rotate, how they can be classified by their type. NASA summarizes the key facts as follows

Comets are frozen leftovers from the formation of the solar system composed of dust, rock and ices. They range from a few miles to tens of miles wide, but as they orbit closer to the sun, they heat up and spew gases and dust into a glowing head that can be larger than a planet. This material forms a tail that stretches millions of miles.

What interests me has to do with their formation. I know that comets are not spherical and rather small in size. I am now wondering how comets' nuclei stick together, more specifically if there is some empiral knowledge or even some theory I should know about. Obviously, gravitation is not strong enough, so it must be merely the structural stability of the material, but how can that be estimated?
References

Press Release of the Max Planck Society: Are comets more like "Icy Dust Balls" or "Dirty Snowballs"? (2005)
The creation of comets
Can two comets travel together as one?
How big do comets get?
Why did the Comet Shoemaker-Levy-9 fragments cause such large explosions on Jupiter?
The article from the Scientific American If comets melt, why do they seem to last for long periods of time?

Van-der-Waals forces and gravitation are the main answer.
The main property keeping together bodies smaller than a few $100 {\rm km}$ are their constituents surface forces between the grains (van-der-Waals) and chemical/ molecular bonding within the grains themselves. It's the same mechanic which makes the dust and grease in your flat stick to surfaces and the fluffy dirt clumps under the bed stick together, thus cohesion/ adhesion.
The typical internal strength of such dirty iceball as comets are, or rubble piles as some minor asteroids like Ruygu seem to be (e.g. Wada et al, 2018), does not need to be large - there is nothing in deep space which pulls it apart. And even the few µg of self-gravitation help to keep smaller fragments and ejecta to return to the parent body when hit by the inevitable micro-meteorites after their intial formation (Blum et al, 2017) (so yes, there is still self-gravitation involved which helps form them; they are just too small to achieve hydrostatic equilibrium and differentiation).
67P Churyumov-Gerasimenke provided a nice cometary physics lab which allowed in-situ investigation. The strength of the material could be analysed directy (Heinisch et al, 2018), (Grousin et al, 2018) with the lander Philae, even though it landed somewhat precariously (see also this overview by Blum et al (2018) focussing on planet formation and lessons learned from 67P in particular).
In summary: most measurement, both in-situ and lab experiments, indicate that the internal strength of comets is within the range of $1 {\rm Pa} \ldots  ... 10 {\rm kPa}$ and most likely at the lower end of this range. Their strength is thus determined by their small gravity and the inter-particle forces, thus van-der-Waals which in part is possibly enhanced a bit by sintering together the different constituents grains.


Why is it not always light? Hopefully this is not a silly question, but I was wondering why earth is not always in the light (and any other planet for that matter). I mean, we have billions of stars just in our galaxy that all give off light from all directions. That light has obviously reached is as we can see them shimmering in the night sky.

With that in mind, what has happened to the light from the stars? Has it weakened somehow? (didn't think that was possible) Surely if we are being hit with light from all directions, it makes sense that we should always be illuminated. So what's going on here?

Olber's Paradox (Heinrich Wilhelm Olbers, 1758–1840): If the Universe were infinite, or there were infinitely many stars, then no matter which direction you look, you'd see a star. Thus, the night sky would be as bright as the Sun. Therefore, the Universe is not infinite, and does not contain an infinite amount of stars.

Also, the light from distant stars does 'get weaker' because of the inverse square law; this is the same reason lights look dimmer when they're further away.

I wonder if gravitational redshift has anything to do with it, too. Anyone?


What is a quaternary mirror and why does the E-ELT need one? In this recent BBC article I read the one-sentence paragraph:

  "Development on the mirrors - in particular, a very complex quaternary mirror - continues apace."

and became interested in the following questions:

What is a quaternary mirror
Why does the E-ELT need one
What is it about this mirror that is "very complex"?

note: the E-ELT is the European Extremely Large Telescope nearing final design, and to be installed in Chile's Atacama Desert (host to many other telescopes).

A drawing of the E-ELT from from Wikipedia (note scale of humans):

An image from the BBC article:

You may know that a standard Newtonian telescope has two mirrors, they are called the primary and secondary mirror.
The E-ELT has five mirrors: The quaternary mirror is simply "mirror number four", counting in the direction the light enters the scope.
It's complex because that's where the adaptive optics sits:

The quaternary mirror has an approximate diameter of 2.4-m (2380x2340mm). It is a flat adaptive mirror, with up to 8000 actuators, thereby allowing the surface to be readjusted at very high time frequencies. [...]
This mirror will correct in real time for high order wavefront errors (e.g. atmosphere, wind shake, low spatial frequency telescope errors) and small amplitude residual tip-tilt corrections.

Source: The European Southern Observatory's page on the E-ELT optics


Stellar mass database I need a stellar mass database that also has the distance of the star/system from the sun. Can anybody tell me the name of such of database and where I can find it. I need to do a project on this. Thank you for the answers.

Check out the on-line SIMBAD astronomical data base. You can even make separate lists based on criteria (like: list all of the stars between 10 and 20 parsecs (have to convert parallax to light years/ parsecs = 1/parallax (in arcsec/ 1 parsec = 3.262 LY) with class between G0 and G9 from 0 to +30 deg declination). It's pretty cool and shows numerous listings of distance values, brightness, metallicity, etc. complete with date of study and source. I just looked again and masses aren't listed (sorry). Masses aren't directly measured anyway (except in the case of binaries where we know the distance). Masses are essentially determined from the Hertzprung -Russell (mass/ luminosity) diagram and can be refined if we can determine age by looking at spectra. Back to Sol-Station or Wikipedia or I should say their references.


Do radio telescopes see other stars better at night? As you know, we don’t see stars with the naked eye during the day because of the visible glare of our Sun.
As I understand our Sun emits radio frequency waves, too. Is there a radio ‘glare’ from the Sun such that radio telescopes can better focus on other stars during the night?

The Sun doesn't substantially impact radio observations during the day, because radio telescopes operate at long wavelengths. In general, light at longer wavelengths scatters less than light at shorter wavelengths, and so visible light from the Sun scatters much more than radio waves from the Sun.$^{\dagger}$ The former effectively fills the daytime sky, while the latter does not, leaving the radio sky dark even during the day. Couple this with the fact that the Sun - though bright in radio waves - is not extremely bright compared to other radio sources,$^{\ddagger}$ and you find that daytime radio observing isn't significantly different than nighttime observing.
Strictly speaking, you could argue that there may be indirect effects from the Sun. Water vapor is an important source of opacity, causing several extremely opaque bands, and atmospheric emission can add noise to the system temperature of your instrument, causing complications to observations. Humid summer days, for example, can be problematic for observing astronomical water vapor emission. As the Sun does impact weather and the climate, there are minute differences between day and night, but I don't believe those are significant when compared to how much the Sun affects optical observers.
As a final note: radio telescopes don't commonly observe stars, which aren't usually strong radio sources. There are occasionally highly-publicized stellar bursts or flares - BLC1 is an example, assuming it's natural - and star-exoplanet interactions can lead to radio emission, as in the case GJ 1151. But radio astronomers rarely look for radio emissions from stars, in comparison to other objects like radio galaxies, pulsars, and clouds of neutral hydrogen.

$^{\dagger}$The same is true for light traveling through interstellar space, which is why infrared and radio emission can travel through dust while visible and ultraviolet light can't; you need wavelengths much larger than the sizes of dust grains for minimal extinction.
$^{\ddagger}$A commonly quoted value is that around $\sim1\text{ GHz}$, the Sun has a flux density of $S_{\nu}\sim10^6\text{ Jy}$ when it's quiet, with an increase of 1-2 orders of magnitude during active periods. That's only a couple of orders of magnitude brighter than the next brightest sources in the sky, and the difference shrinks substantially in the $\sim100\text{ MHz}$ regime.


Identifying common galaxy spectral lines How does one identify the common galaxy absorption lines (e.g. Na, Mg, K, etc.) and emission lines (H-alpha, O III, S II, etc.) by just looking at a galaxy spectrum (like the one below)? I need to know how to do this because I want to calculate the redshifts of galaxies using the emitted wavelengths from the SDSS table.

You need to compare it with the spectrum of a similar galaxy at a known redshift, that would probably enable you to identify features with known rest wavelengths.

If you can find such a template, then the best way of estimating a redshift for a galaxy spectrum like this, consisting of mostly weak and blended absorption features, is to cross-correlate your template galaxy with this one. The cross-correlation "lag" gives you the difference in redshift.


Why is Starlink polluting the night sky a big concern if we have space telescopes? There's a lot of concern in the Astronomy community over the deployment of Starlink satellites. For a good discussion, see the related question How will Starlink affect observational astronomy?
But why is there so much concern over this problem, given that there are numerous space telescopes? Presumably a lot more will be launched in the upcoming years thanks to satellite launches becoming cheaper. Aren't space-based observations superior in the first place thanks to the lack of an atmosphere?
I understand this sucks for amateur astronomers but is it also a big problem for professional researchers?

It's a problem because there are still lots  and lots and lots of ground-based telescopes.
Ground-based telescopes are still (by far) the biggest optical telescopes, and the cost of space telescopes is prohibitive for many research projects. It will be a long time before a telescope anywhere close in size to the VLT can be launched.
Most space telescopes are specialist devices, observing in a particular part of the spectrum that is blocked by the atmosphere (so there are infra-red, ultraviolet and X-ray telescopes in space) Or doing a specific task (looking for exoplanets, or mapping the positions of stars)
Space is getting cheaper, but it will be a long time before it is as cost-effective as ground telescopes in the optical range. And so it will be a long time before all professional telescopes are in space.
So it is a problem for professional astronomers.


Gravitational field for oblate spheroid? Consider a planet, described as an oblate spheroid. Assume that the spheroid is uniformly dense but not a point source. Outside of the object, do all vectors in the gravity field point through a single point inside the planet?

If you have a self gravitating sphere than all force vectors at the surface are normal to that surface (and other equipotentials interior to the surface). These force vectors only point towards the centre in the case of a sphere (or at the poles and equator of an arbitrary spheroid). However, even if you remove any centrifugal effects and just have a rigid, non-rotating spheroid, the gravity vector still does not point to the centre of the spheroid except at the poles or the equator and that is true either inside or outside the surface.

The gravitational potential outside of a uniform spheroid of mass $M$ can be expressed to high accuracy with
$$\Phi = -\frac{GM}{r} + \frac{kG}{2r^3}(3\cos^2\theta -1),$$
where $r$ and $\theta$ are the usual spherical coordinates and $k$ is a constant equal to the difference in the moments of inertia about axes parallel and perpendicular to the rotation (symmetry) axis (note that the object would have to be rotating in order to be a self gravitating spheroid, but that the potential above is just the gravitational potential and does not include any centrifugal, non-gravitational component and is equally appropriate for a rigid oblate spheroid).

Taking the gradient of this potential (since gravitational field  $\vec{g} = -\nabla \Phi$), you see there is a radial term in $\hat{r}$, but also a $\hat{\theta}$ component
$$g_{\theta} = \frac{3kG}{r^4}\sin \theta\cos\theta$$
that is not directed towards the centre of the spheroid.

In the limit of an almost disk-like spheroid, then at large distances from the centre and reasonably close to the disk plane, this term will dominate and the gravity will tend to act towards the disk midplane rather than towards the centre of the disk.

The above analysis only includes the monopolar and quadrupole gravity fields. There are (smaller) higher order terms that can be included in the potential, but the result is the same -- that the gravitational field outside an oblate spheroid is not a central force (Hofmeister et al. 2018).


Need help understanding the recent observation of extra-galactic planets! The new paper Probing Planets in Extragalactic Galaxies Using Quasar Microlensing (also in ArXiv) is too mind-boggling for me to understand exactly what is happening.

I have some very basic understanding of gravitational microlensing from other Q&A here (and Wikipedia) but I can't figure out what is happening in this paper.

There are thousands of planets seen in another galaxy, which do not orbit stars? There is discussion of "emission close to the event horizon of the supermassive black hole of the background quasar"... there are too many unusual things going on at once for me to fathom.

Is it possible for someone to sort out the basics of what's happening? The Quasar is in front, and the galaxy containing these planets is behind? What emission?

Instead of reading half-comprehensible newspaper articles, I found it more helpful to go to the original source.

There, the authors explain their method very clearly:

There is a background galaxy, the Quasar galaxy. This object, being a Quasar, has an actively accreting SMBH. The SMBH, being an actively accreting one, will emit strongly in X-Rays. Those X-Rays, being emitted from the rotating accretion disc, deep inside the SMBH's gravitational well, will suffer from strong red- and blue-shifts.
The paper works with only one line here, the Iron $\rm K\alpha$ line, observed by Chandra in 2009, 2012 and 2017.
The red- and blue-shifted light of the SMBH, the 'image' will then travel through space and time.
Eventually this light will be lensed by a foreground galaxy, and in lucky cases the foreground galaxy can create Einstein rings, or at least multiple paths so that the same quasar image crosses the lensed galaxy at multiple positions through the foreground galaxy.
The object of interest now, (RXJ1131-1231) consists of a quasar at $z=0.658$ and a foreground galaxy at $z=0.295$ that lenses the background quasar four times, giving four images in X-rays that can be spectrally resolved, to measure the distributions of red- and blueshifted $\rm K\alpha$-lines in each pixel.

And where the 'micro' in microlensing comes in is

  Light from these Quasar images crosses different locations of the foreground galaxy and is further lensed by nearby stars in the region in the lens galaxy.

So what is important here is:

No individual stars or planets are actually resolved
The amount of microlenses must therefore be inferred by spectroscopy (that's why all the talk about red and blueshift).

Now comes physics into play. The authors argue that they find anomalously high red- and blue-shifted $\rm K\alpha$-line distributions, that must be explained by planetary-mass objects.

To understand how, let's consider fig. 1 from the paper:

This shows the redshift distribution (The factor $G = \frac{E_{observed} }{ E_{rest-frame} }$) around the background Quasar accretion disc. Also shown is an example trace of a foreground object that could lens the emitted $K \alpha$ light from the disc.

** The rest of the problem can now be understood in a very visual way:

If the passing foreground object can actually lens the light in a way that would be seen in the spectrum is now determined by the ratio of the SMBH Gravitational radius and the foreground's object lens radius. This works as follows:

The gravitational radius is just the Schwarzschild radius $r_g = GM/c^2$ and sets the scale of the above image. The lens radius $r_l$ is a more involved function of cosmological distance and lens mass. One can imagine it however, being a circle traveling on the green trajectory in fig 1. that will lens all the background light that it hits into one observational data point.
The latter point must be emphasized, as the situation in fig 1. will be vastly under-resolved in the actual observation.

So if $r_l \gg r_g$ the lens will lens red- as well as blue-shifted $\rm K\alpha$-light into the image pixel and no net energyshift will be detected. If $r_l \le r_g$, then the lensed light will probably be very red or blue-shifted, according to the G-values it hits on it's trajectory.

** We're nearly done:

Now the lensing configuration of RXJ1131-1231 and their cosmological distances conspire to make the case $r_g = r_l$ correspond to Jupiter mass lens masses. This means objects of Jupiter masses and below will be able to contribute to a measurable Fe $\rm K\alpha$ redshift, but only if they are not buried in the larger lensing radius of a star (which would have $r_l \le r_g$)!

But, as we already said, the authors only can spectroscopic information about the lensed images. So they try to reproduce the redshift-distribution of the spectrae of their images with a star-only distribution, which fails.

Subsequent tests they make, indicate that they have to inject a large number of objects with $r_g=r_l$ into their test-lensing distribution to reproduce their data.

Summarizing:
This is a purely statistical argument for microlensing by a certain number of planetary-mass bodies in the X-Ray spectrae of 3 of their 4 macrolensed images. It's not a discovery and as another Q&A pointed out, there are no planets to name or point to.
It is even not the only explanation for the X-Ray data seen, as there are other competing ones.
But the presented planetary explanation is the only one that can reach sufficient redshifts seen in the data.

(P.S. I hope I didn't make it more complicated than the original paper, I've actually tried to summarize it...)


Why aren't stars like Eternally Collapsing Objects? We know that stellar radiation pressure balances the gravitational compressive forced of a star. Are there other factors which resist such a collapse? Also, if radiation pressure balances gravitational compression, shouldn't the process of shrinking be a quasistatic process making stars Eternally Collapsing Objects (ECOs)?

We know that stellar radiation pressure balances the gravitational compressive forced of a star.

We do not know that. Degeneracy pressure, thermal pressure, and radiation pressure are what collectively balance gravitation in a star. Degeneracy pressure dominates over the other two in white dwarfs and neutron stars. Thermal pressure dominates in low mass stars such as the Sun. It's only in very massive stars where radiation pressure dominates.

  Shouldn't the process of shrinking be a quasistatic process making stars (Eternally Collapsing Objects)?

You are forgetting about fusion. A star is more or less in equilibrium thanks to a number of negative feedbacks. (Negative feedback in a control system is good. Positive feedback is very bad. The "screech!" sound you occasionally hear from microphones is positive feedback.)

Suppose the fusion rate increases slightly in a region near the center of a star. This increases the temperature and thereby the pressure, which in turn causes the core to expand, which in turn causes the density and temperature in the core to fall, which in turn decreases the fusion rate. The reverse applies for a slight decrease in the fusion rate. Similar negative feedbacks apply for changes in temperature, pressure, and density. A main sequence star is self regulating.

Things get a bit trickier when a star leaves the main sequence, but that is a different question.


Are red moons possible without eclipses or smoke from fires? This morning I was on the bus and as I looked out of the window I saw that the moon was larger than usual and red. I live in eastern Norway and it was about -20ºF / 30ºC outside. There's no forest fires, and I live in a small village. There's no eclipse scheduled either. The moon went behind the mountain line before I got to school ( bus ride happened  7:03 - 7:20 ), and nobody there had seen it. Can you explain this?

What causes the moon to be red is our own atmosphere.  It's the same effect that causes a sunset to be red.  As the sun nears the horizon, the amount of air that we're looking through increases - and with it everything in the air such as dust, pollution, humidity, etc....  If you look at the moon near the horizon, you'll notice it is significantly yellower/oranger than when it's high in the sky, where it is normally whiter and bluer.  Again, this is all caused by the air and the stuff in it.

When the moon turns read during an eclipse, it is because it is passing through the shadow of the atmosphere, which filters the light passing through and on to the moon, leaving more red light to make its way to the moon and, then, to reflect off of it.

A similar action happens when smoke from fires gets into the air - it filters out the shorter wavelength, bluer light and lets the red pass.  Other pollutants besides smoke particles can have the same effect.

So, the only way for this to happen is for something to pass between the sun and the moon, or the moon and the earth - between the source of light and where it is received.

As for the moon appearing larger: this is an optical illusion.  No one is entirely sure why it happens, but the moon does APPEAR larger on the horizon than when it is high in the sky, but this is merely an illusion.  To prove it, next full moon, go outside with a ruler and hold it up at arm's length and try to measure the apparent size of the moon (this will vary from person to person due to the length of your arm).  Then, wait until it's high in the sky and do the same thing.  It will measure the same size (give or take a small amount due to the change in position affecting the distance from your eye to the ruler).

If you have the right kind of telescope and mount, you could connect a camera, take a picture on the horizon, wait for it to pass close to its highest point and take another picture, then compare.  Assuming no changes were made to the camera or telescope, the images should show the same apparent size.


How to determine the orbital period of an asteroid using Kepler laws given its orbit's radius? Given that the mean distance of an asteroid from the Sun is $450 \times 10^6$ km is it possible that the asteroid completes one rotation around the Sun in two years?

According to the third law of Kepler the farther a planet if from the Sun the longer is its orbital period. Because the distance from the Sun to earth is $149 \times 10^6$ km and it takes a year for earth to complete the rotation then for the asteroid it would take 3 years.

I tried to actually calculate how much time would take for the asteroid to complete one rotation combining Newton's law of gravitation and Kepler's third law and given $G=6.673 \times 10^{-11} Nm^2/kg^2$ and converting all distances to meters:

$$
T^2=\frac{4\pi^2r^3}{G\cdot M_{Sun}}=\frac{4\pi^2\cdot 450^3\cdot 10^{21}}{6.673\cdot 10^{-11}\cdot 1.989\cdot 10^{30}}=\frac{4\pi^2\cdot 450^3\cdot 10^2}{6.673\cdot 1.989}=2.99\cdot 450^3\cdot 10^2=27246375\cdot 10^3.
$$

Then $T=\sqrt{27246375\cdot 10^3}=165064$ seconds $=45$ hours. Is there something wrong in my calculations or I'm not using the correct formula?

Here's how I would do it. I'd convert everything to a single, standard set of units as recommended in the comments, and also stick to one digit before the decimal in scientific notation:

$$ T^2=\frac{4\pi^2r^3}{G\cdot M_{Sun}}$$

Using all numbers in the same units:

$r \ 4.5 \times 10^{11} \ (m) $

$G = 6.674 \times 10^{-11} \ (m^3 \ kg^{-1}  s^{-2}) $

$M = 1.989 \times 10^{30} \ (kg)$

$$ T^2=\frac{4 \cdot \pi^2 \cdot 9.11 \times 10^{34}}{6.674 \times 10^{-11} \cdot 1.989 \times 10^{30}} = \frac{3.60 \times 10^{36}}{1.33 \times 10^{20}} = 2.71 \times 10^{16} (s)$$

$T = 1.65 \times 10^8 \ (s) $

$\frac{1.65 \times 10^8}{365.25 \times 24 \times 3600} = 5.21 \ (years)$

Just fyi you can also call the product $GM$ the standard gravitational parameter and those numbers are more accurate than $G$ and $M$ are separately, as discussed in this answer.

The comments under your question also mention that it is sometimes easier to use AU and years as units and use Earth's orbit for scaling to the 3/2 power, but only for things orbiting our sun.  With $r = 3$ AU:

$T^2 = 3^3$

$T = 3^{3/2} = 5.2 \ (years)$


Does the milky way have relativistic mass against galaxies which are moving away from it at high speeds? According to the article cited behind, and to a post here in Astronomy SE , there are galaxies moving away from Milky Way faster than light, even at speeds of 2.3c . According to this article
Can two galaxies move away from each other faster than light

  The two galaxies we've been discussing are not travelling through
  space, it is the space between them that is expanding. Or put in
  another way, they are stationary and all the space around them is
  being stretched out. This is why it doesn't violate the theory of
  relativity, because it is not motion in the traditional sense.

So does this motion between galaxies "in not a traditional sense" count at all to calculate relativistic mass?

What the mass increase from special relativity tells you is that if you see a physical system (say, a rocket with a physics laboratory on board for convenience) flying past your location at a certain relative velocity and you measure the mass of an object in that laboratory, for instance by measuring its gravitational effect on another mass in the same laboratory (using lasers to measure distances and velocities and so on), you will get an answer which is higher than the one an observer in that laboratory would get.

In and of itself this doesn't even tell you much about what you'd measure about a far distant laboratory, especially one far enough away that the expansion of the universe significantly effects your measurements.

Furthermore, since these two galaxies on opposite sides of the observable universe cannot actually  exchange messages and observe one another, the mass increase is not really a consideration -- you can't measure the mass of an object in the other laboratory, because you can't get a signal there and back.


How far away is the nearest compact star remnant likely to be? Neutron stars and black holes are hard to detect when they are solitary, and there seems to be big uncertainties about how common they are. White dwarfs are much easier to detect and the nearest one is Sirius B only 2.6 parsec from here. Should we expect to have exotic company even closer by? How likely is it that we have a yet undetected compact star remnant nearby, as in as close as our nearest active star? What are the odds of spotting one of them nearby?

How could a yet undetected one be discovered? Could one of the upcoming sky surveying telescopes catch it or would one have to rely on a rare microlensing event? How would it then be observed? Would such an exotic object, say, just one parsec away, give important insights in physics given its relativistic effects and strange composition?

There can be no closer white dwarf. The coolest, oldest white dwarfs (3000K), would be rare, but are still luminous enough $6\times10^{-6} L_{\odot}$ to have been easily detected at distances closer than Sirius. At the distance of Sirius, such an object would have a visual magnitude of around 12-13 and would be brighter at near infrared wavelengths where all sky surveys such as 2MASS would definitely have spotted it from its parallax.

Neutron stars and black holes could be almost undetectable but are expected to be roughly 10 and 100 times rarer respectively. Calculated as follows:

Let us assume that $N$ stars have ever been born in the Milky Way galaxy, and given them masses between 0.1 and 100$M_{\odot}$. Next, assume that stars have been born with a mass distribution that approximates to the Salpeter mass function - $n(m) \propto m^{-2.3}$. Then assume that all stars with mass $m>25M_{\odot}$ end their lives as black holes, all stars with $8<m/M_{\odot}<25$ end their lives as neutron stars and about half the stars with $0.9<m/M_{\odot}<8$ end their lives as white dwarfs (the other half are still alive as main sequence stars, as are all stars born with lower masses).

So, if $n(m) = Am^{-2.3}$, then
$$N = \int^{100}_{0.1} A m^{-2.3}\ dm$$
and thus $A=0.065N$.

The number of black holes created will be
$$N_{BH} = \int^{100}_{25} Am^{-2.3}\ dm = 6.4\times10^{-4} N$$
i.e 0.064% of stars in the Galaxy become black holes. NB: The finite lifetime of the galaxy is irrelevant here because it is much longer than the lifetime of black hole progenitors.

In a similar way, the number of neutron stars
$$N_{NS} = \int^{25}_{8} Am^{-2.3}\ dm = 2.6\times10^{-3}N$$
and the number of white dwarfs
$$N_{WD} = 0.5\times \int^{8}_{0.9} Am^{-2.3}\ dm = 0.027 N$$

Now we use these results as scaling factors to apply to the local stellar population. There are about 1000 "normal" stars in a sphere of 15 pc radius, thus a density of 0.07 pc$^{-3}$. Thus one uses the results above to calculate the density of compact remnants and then take $(3/4\pi n)^{1/3}$ as an estimate of the average distance to one of them. This gives an expectation value of 18 pc to the nearest black hole, 11 pc to the nearest neutron star and 5 pc to the nearest white dwarf.

Thus the distance to the nearest white dwarf is roughly as expected. For reasons discussed in my answer to this related question the distance calcuated to the nearest black hole and neutron star remnants is likely to be an underestimate because many escape from the Galaxy or have very high velocity dispersions and much larger Galactic scale heights than normal stars. So whilst it is possible that an unseen one exists closer than Sirius, it is highly unlikely.

How could such an object be detected? An old, cold neutron star or black hole could be completely undetectable at all wavelengths of electromagnetic radiation - though it could be fruitful to examine carefully any candidate detections [see below] for signs of X-ray emission due to accretion from the interstellar medium). But your question has I think the correct suggestion. The objects would likely have a substantial proper motion and so there is a decent chance that you would see a "moving" gravitational lensing signature. This would still be very small unless the object just happened to pass directly in front of a background star - but such a microlensing event would be transient and may not be observed. More likely is that Gaia would pick up the subtle shifts in the positions of background stars changing over the 5 years of its mission. As per your other question: Will Gaia detect inactive neutron stars?


Specific word for the point of the terrestrial surface "under" a star or other object in the sky The title says it : I am looking for the word to use for a point, located on Earth surface (maybe an hypothetical perfectly spherical Earth), that has the same coordinates as a specific star, and therefore can be considered to be "under" the star in layman's terms.

Positions are usually given relative to the observer, not relative to the object. So instead of saying "I am under the star X", you would say "star X is overhead". The word for an object directly overhead is zenith: "the star X is at the zenith".


What color distribution does a protoplanetary disk have? A kids question indeed, but I do not have a conclusive answer. An usual artist's impression uses red and yellow as dominating colors, and a rather homogenous, dense disk, but how reasonable is that?
If you follow e.g. Sciencedirect's teaser page on Protoplanetary disks

Dust grains are a relatively minor constituent of protoplanetary disks, but they represent the starting point for the formation of rocky planets like Earth, and possibly also gas-rich planets like Jupiter. These grains are small, typically 1 m in diameter or less.

Later on the same page:

However, the fact that roughly half of young stars have debris disks of dust thought to come from asteroids and comets implies that growth of large solid bodies occurs in many protoplanetary disks, even if the mechanism remains obscure.

This two statements for me seeded some doubts that homogenous red is really the correct color to associate with the inside of a protoplanetary disk.
I also found a paper by Alan P. Boss stating

Theoretical models of disks undergoing the accretion of mass from an infalling cloud envelope predict disk temperatures in good agreement with these constraints: a moderately warm (500–1500 K) inner disk, surrounded by a cool (50–150 K) outer disk.

This would at least support the color scheme if you assume black body radiation.
Related

What makes protoplanetary disks start rotating? (Initial energy needed to rotate)

The answer seems to be a clear "it depends":

Homogenity of the disks depends on the age of the protoplanetary disk.
Temperature is a function of the radius and the color-scale used in (old) artists' impressions is indeed correct.

On Wikipedia on the topic there is a gallery of different observations, also in the visual spectrum. An impressive picture of the density fluctuations within the protoplanetary disk is provideded by the Disk Substructures at High Angular Resolution Project (DSHARP) as explained e.g. by the ESO.
Edit Remarks following a suggestion by Rob Jeffries:

Please note that the DSHARP images are not from the visual part of the spectrum.


Is there still a red shift when moving perpendicular to the direction of incidence? This answer suggests that when there is zero radial velocity... there will be no Doppler shift but it's not likely to be intended as precise and absolute.

I remember reading about something like this in a post here but I can't find it. While to first order there is no frequency shift when moving perpendicular (we wouldn't expect there to be be due to symmetry i.e. which way? Up or down?) I think there is a second order effect which is a Doppler shift associated with the small change in angle due to astronomical aberration.

Question: Is this a thing? If so, what would be an expression for the frequency shift and what is it called? If the magnitude of each effect is $|v/c|$ is it always a red shift and simply

$$\Delta f/f = -\frac{v^2}{c^2}?$$

Yes, there is a transverse relativistic Doppler shift. You can think of it as being caused by time dilation. https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect

There can be a redshift or a blueshift depending on when, where and who does the measurement.

e.g. a receiver with a source going around it in a circular orbit. The receiver sees a lower frequency (redshift), by a factor $\gamma = (1 -v^2/c^2)^{-0.5}$. On the other hand, a receiver orbiting the source would receive a blueshifted signal by the same factor. When $v \ll c$ then
$$\gamma \simeq 1 + \frac{v^2}{2c^2}\ ,$$
so whilst the standard (longitudinal) Doppler shift is of order $v/c$, the transverse Dopper shift is of order $v^2/c^2$.

NB: This scenario is chosen so that the relative motion of the source and receiver are perpendicular to the line between them. All other scenarios are complicated by the usual Doppler shift you would see because there is a component of the velocity along the line joining the source and receiver.


Does this video catch an Eta Aquarid - Earth skimming meteor? I did a time-lapse animation looking towards the sunrise from Sydney, Australia on 25th of April.  You can see it here around 1m 25s in (that link should jump to just before it appears very near where the Sun is about to rise):

http://www.youtube.com/watch?v=lqr_KuLd2-c&t=1m25s

It then proceeds towards the right of screen over around the next 21 seconds (around 10½ minutes) before disappearing off the right. You might need to send the video to full screen to see the small white 'bullet of haze' that is moving against a very wide background.

One of my first thoughts was that it might be a meteor skimming the upper atmosphere, but I was less confident of that given it seemed to be coming from the general direction of the Sun, and I'd thought that was a direction unlikely to produce meteors.

That was until I read Eta Aquarid Meteor Shower in 2016 which states:

  The best time to view the Eta Aquarids is in the early mornings, right before dawn.

Well, at least that pins down the same time of day, if not the direction from which it came.

What is the likelyhood that this feature is a piece of comet debris that skimmed Earth's atmosphere?

Things entering the atmosphere (whether a meteor or an atmosphere-skimming object) travel too fast to be visible for 10 minutes.  Normally a pass of a satellite (at above atmosphere height) takes 2-3 minutes and most meteors are over in seconds.

Even the famous daylight fireball from the 1970s was over in a very short time.

I would guess what you have there is a plane at high altitude, though it's moving a bit slower than I would expect. (I'm open to suggestions.)


When a rogue brown dwarf star hits a white dwarf star, will the collision at most cause a nova? When a rogue brown dwarf star hits a white dwarf star, will the collision at most cause a nova?
Or can the explosion be heavier?

Collisions between a rogue brown dwarf and any other star would be very rare because the space between them is so vast.   I don't want to say it'd never happen, but it would be a rare event.   It's much more common for two stars that are already in the same system to collide by spiraling into each other, usually by tidal decay.

A collision with a brown dwarf and a white dwarf has been recognized.   The nova of 1670 didn't behave like an exploding star and recent research identified it as spiraling in collision between a brown dwarf and a white dwarf.  Original article and abstract here.

As for what kind of explosion is likely.  I would imagine something like a type-A supernova would be possible if there was sufficient mass to exceed the Chandrasekhar limit of about 1.4 solar masses.   Because brown dwarfs are much less mass than that, about .013 to about .08 solar masses, the white dwarf would need to be at least 1.32 solar masses prior to the collision.  Most white dwarfs are less than that.

CK Vulpeculae, the 1670 nova wasn't a type 1a supernova.  If it had been, it would have been over 100 times brighter than Venus (ballpark estimate) based on it's distance 2280 light years (plus/minus 490).   It would have been many times brighter than it's visible but not very bright peak of 2.6.   SN 1604, for comparison was about 20,000 light years distant and it was brighter than any star in the sky and visible during the day at it's peak.

The mass of the two colliding objects and energy of collision would have considerable effect on the brightness, but, unless the Chandrasekar limit is reached, and keeping in mind that direct collisions would be rare, and it wouldn't hit all at once, the result should be much less energetic than a supernova, but if it's close enough, it could still be a visible nova in the sky, like CK Vulpeculae in 1670.


Is looking through a telescope to do astronomy a thing of the past? Do astronomers actually look through a telescope with the naked eye or do they mostly look the the data generated by telescopes on computers?

Are they mostly working behind the computer and pencil and paper or is it still stargazing?

Do they actually point a red laser to observe constellations or do they do it all on the computer, and lasers that can point us at constellations are more of a gimmick for amateurs?

I’m sure astronomers would need to look under the lens to confirm the computers aren’t processing data that’s wrong. Do astronomers actually trust the algorithms that automatically do data entry from telescope observed data to their notebooks enough that they usually don’t bother to under the telescope until they’ve confirmed something?

Scientifically there's little to gain when you look through a telescope with your own eyes. Attach a camera to the telescope and you immediately document what you observe and take out the subjective factor from it at the same time and allow careful analysis and more detailed. Additionally you gain improved sensitivity enhanced observation duration and broaden (or narrow) the range of possible wavelengths which allows to gain additional insights.

That said, many professional astronomers still enjoy looking at the sky with their own eyes, be that with or without telescope.


Is moisture harmful to a Newtonian Reflector? It is often cold outside, especially in the evening. So when I take my Newtonian Reflector back in my house, it starts condense.

Can the water destroy my telescope? or the mirrors? Do I have to clean it?

My telescope is fully painted, but I'm afraid of the screws. Screws often rust.

The tube itself is not that much of a problem, just wipe it with a soft cloth. The screws should be stainless steel, surface rust can be wiped off as well.

But moisture, which contains humic acids from plants and pollen as well as mold damage the reflective surfaces of mirrors. If it is an expensive one, leave it outside or in a shelter until the next day and store it with the main mirror pointing down.  Don't expose it to a temperature change of -10 to +20°C. Do not clean optical surfaces mechanically if your unsure about the coating and the cleansing stuff. Wiping the aluminium surface will scratch it if it has no protective coating.

If it is a particular expensive bucket and you're in an area that experiences much moisture (where i live, water sometimes runs off the tube, really), consider heaters. Or try another day :-)

Amateur astronomer forums like e.g. cloudynights.com have much more info on this.

Hope that helps.

Edit: also: How do I remove fungus from a telescope mirror?


Shall we say now that Pluto is "larger than Mercury" ? http://pluto.jhuapl.edu/News-Center/News-Article.php?page=20150717-3

"Scientists working with NASA’s New Horizons spacecraft have observed Pluto’s atmosphere as far as 1,000 miles (1,600 kilometers) above the surface of the planet, demonstrating that Pluto’s nitrogen-rich atmosphere is quite extended."

I note that conventional descriptions of the gas giants' sizes include mostly atmosphere.

The implication of the question is that this extra 1000 miles should be added to Pluto's radius. The answer is no. For all of the solid planets, it's that solid surface (or solid+liquid surface in the case of the Earth) that counts, not the outer reaches of the atmosphere. The surface is a clear-cut, non-arbitrary boundary. The atmosphere? They can extend a long way out. A non-arbitrary boundary is always going to be preferred over an arbitrary one.

That's not possible in the case of the Sun and the giant planets. A somewhat arbitrary boundary is needed for those bodies. That somewhat arbitrary boundary explicitly excludes the upper reaches of the atmosphere, which extends out for many thousands of kilometers. For the giant planets, some use a tenth of the Earth's atmosphere as defining the arbitrary "surface", others, one atmosphere, yet others, ten atmospheres. That factor of 100 variance in pressure amounts to about a hundred kilometers. That's not that much considering how big the gas giants are.


Venus presents same face toward Earth at closest approach. What do you make of that? Is reference 6 in this Wikipedia article "dubious" and this passage in need of excising or clarifying?

  Venus's 583.92-day interval between successive close approaches to Earth is equal to 5.001444 Venusian [sic] solar days, making approximately the same face visible from Earth at each close approach. Whether this relationship arose by chance or is the result of some kind of tidal locking with Earth is unknown.

Gold T., Soter S. (1969), Atmospheric tides and the resonant rotation of Venus, Icarus, v. 11, p 356-366

Is it still an "unknown" or "crazy idea that didn't develop with further investigation"?

The idea that Venus is tidally locked with Earth sounds far-fetched. We understand the slowing of Earth's rotation, and I don't recall ever reading of Venus's resistance to the change!

I wouldn't call the reference dubious so much as very much out-of-date. The paper in question was written in 1969, which was still pretty early in our understanding of the dynamics of the Solar System, including Venus. There would have been a lot of uncertainty in many of the measurements, such as rotation speed, which would have been determined by radar. Consequently, the possibility of resonance was there, at least until better measurements were available. However, the wikipedia article on Venus, and the attendant 1979 paper citation (#90) indicate the idea of a resonant Venus was rejected decades ago once sufficiently accurate radar observations of the planet had been made.

So, clarifying of the wikipedia article you mention is probably in order.


Difference between J2000, FK5 and ICRS coordinate systems? Which one does the Yale Bright Star Catalog use? I have no background in astronomy. I have been wanting to write code to make star charts for a given location and time, which led to this question.

I figured out that I would need to convert the coordinates of stars from the equatorial coordinate system to a horizontal coordinate system defined by a time and place of interest.

I decided to use the Yale Bright Star Catalog for this purpose.

From what I could gather, the Yale Bright Star Catalog does not explicitly mention the coordinate system it uses, it mentions the epoch to be J2000 frequently though and reports RA, DEC. So, it is my understanding that the coordinate system should be an equatorial one with J2000 as the equinox.

As a novice, I could understand the horizontal, equatorial and ecliptic coordinate systems. However, when I started using an astronomy library to write the code, I realised that in practice, there are equatorial coordinate systems such as FK5 and ICRS with minute differences; and that ICRS is the coordinate system adopted by the IAU.

So, which coordinate system is the YBSC catalog in, exactly?

I have noticed that mentioning the epoch, but not the exact coordinate system is the somewhat confusing, but standard way to report coordinates. Example: coordinates in M33's Wikipedia Page, or CDS name resolver.

So, in general, which equatorial coordinate system should be considered 'default' when just the epoch is mentioned as in the examples above? Should it be FK5, ICRS, or something else?

FK5 is based on optical observations. ICRS is based on radio data. Radio observations (ironically) are much higher accuracy than optical through the application of long baseline interferometry.
Other than extremely accurate astrometry (fractions of an arcsec), they are essentially identical.


Lagrange points and the radius of the Hill sphere (Disclaimer: I know the "Hill sphere" is just an approximation of something which isn't genuinely spherical.)

In a two-body system, the approximate formula for the Hill sphere radius of the smaller body is

$$r_{\mathrm{H}} \approx a (1-e) \sqrt[3]{\frac{m}{3 M}}$$

where $a$ is the semi-major axis of the small body's orbit around the larger, $e$ denotes that orbit's eccentricity, $r_H$ is of course the radius of the Hill sphere, $m$ is the mass of the smaller body and $M$ is the mass of the larger one.

If and when the eccentricity is negligible, this is approximated further to

$$r_{\mathrm{H}} \approx a \sqrt[3]{\frac{m}{3M}}$$

This formula is often used to approximate the distance between the smaller body and the L1 and L2 Lagrange points, although these distances are not in fact identical (L2 being further distant than L1.)

The distance $r_1$ between the smaller body and L1 is obtained by solving this equation (which appears also to assume negligible eccentricity):

$$\frac{M}{\left(a-r_{1}\right)^2}=\frac{m}{r_{1}^{2}}+\frac{M}{a^2}-\frac{r_{1}\left(M+m\right)}{a^3}$$

Similarly, the distance $r_2$ between the smaller body and L2 is obtained by solving:

$$\frac{M}{\left(a+r_{2}\right)^2}+\frac{m}{r_{2}^{2}}=\frac{M}{a^2}+\frac{r_{2}\left(M+m\right)}{a^3}$$

As noted above, $r_H$ as obtained using the approximate formula approximates both $r_1$ and $r_2$.

Various online sources use language such as "The Hill sphere lies between the L1 and L2 Lagrangian points" or "the Hill sphere extends somewhere between the L1 and L2 Lagrangian points". I find this language unclear, and can't tell which of the following these sources are saying:

$r_H < min(r_1, r_2)$

$r_H \leq min(r_1, r_2)$

$r_H < max(r_1, r_2)$

$r_H \leq max(r_1, r_2)$

$r_1 < r_H < r_2$

$r_1 < r_H \leq r_2$

$r_1 \leq r_H \leq r_2$

$r_1 \leq r_H < r_2$

or something else.

As I write this, I'm not sure how trivial a task it would be to work through the approximations for the L1 and L2 points to determine if the Hill sphere radius formula overestimates or underestimates each, so I apologise if this is something other users of Astronomy SE would find trivial.

Since the question is about inequalities and ratios, I took the script from this answer and made it better by normalizing it, now $a=1, M1=1$ and $m \equiv M2/M1$

For physical situations in the spirit of Hill spheres and Lagrange points, I think the answer is always going to be:

$$r_1 < r_H < r_2$$

but I can't prove it, that would need math and it's too late tonight for me for math.

Without normalizing $a=1$, the distances and Hill sphere radii for the Sun-Earth and Sun-Mars systems are as follows:

a_Earth:     149598023  km
Sun-Earth L1:  1491524  km
Sun-Earth L2:  1501504  km
Earth r_Hill:  1496531  km

a_Mars:      227939200  km
Sun-Mars L1:   1082311  km
Sun-Mars L2:   1085748  km
Mars r_Hill:   1084032  km

We now apply the normalizations mentioned above:

Python 3 or 2

import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import brentq

#unitless   a = 1, M1 = 1, m = M2/M1
# so for the Hill sphere and L1/L2 points of the Earth orbiting the Sun
# m = 5.9724E+24/1.9886E+30  = 4.88987E-07 and
# L1 = 0.0054525 AU, r_Hill = 0.0054625 AU, L2 = 0.0054724 AU

def solve_L1 (r, m):
    return m/r**2 + 1. - r*(1. + m) - (1.-r)**-2

def solve_L2 (r, m):
    return 1 + r*(1 + m) - (1.+r)**-2 - m/r**2

def r_Hill(m):
    return (m / 3.)**(1./3.)

ms = np.logspace(-8, -1, 61)

answers = []
for m in ms:
    r = r_Hill(m)
    L1 = brentq(solve_L1, 0.5*r, 1.5*r, args=(m,))
    L2 = brentq(solve_L2, 0.5*r, 1.5*r, args=(m,))
    answers.append([L2, r, L1])

L2s, rhills, L1s = np.array(list(zip(*answers)))

m_Jup = 1.8982E+27/1.9886E+30
rhill_Jup = r_Hill(m_Jup)
L1_Jup = brentq(solve_L1, 0.5*rhill_Jup, 1.5*rhill_Jup, args=(m_Jup,))
L2_Jup = brentq(solve_L2, 0.5*rhill_Jup, 1.5*rhill_Jup, args=(m_Jup,))

if True:
    plt.figure()
    fs = 14
    plt.subplot(2, 1, 1)
    plt.plot(ms, rhills)
    plt.plot([m_Jup], [rhill_Jup], 'ok')
    plt.text(0.001, 0.03, 'Jupiter', fontsize=fs)
    plt.ylabel('r_Hill / a', fontsize=fs)
    plt.xscale('log')
    plt.yscale('log')
    plt.subplot(2, 1, 2)
    plt.plot(ms, L1s/rhills, '-')
    plt.plot(ms, L2s/rhills, '--')
    plt.plot([m_Jup], [L1_Jup/rhill_Jup], 'ok')
    plt.plot([m_Jup], [L2_Jup/rhill_Jup], 'ok')
    plt.xlabel('m = M2/M1', fontsize=fs)
    plt.ylabel('L1 or L2 / r_Hill', fontsize=fs)
    plt.xscale('log')
    plt.show()


What are the effects of using non-spherical lenses in refracting telescopes? Non-spherical (or non-circular) mirrors for reflecting telescopes are common and discussed in many places...
But what about elliptical, parabolic or hyperbolic lenses?

Short answer: they exist, but they aren't cheap (unless made of molded plastic).
Long answer:
Telescopes use parabolic, hyperbolic, elliptical, and less frequently, non-conic aspheres to achieve optimal performance for their design intent.  A reflective telescope using 1 or 2 mirrors simply cannot achieve acceptable performance with spherical mirrors.
Also worth noting, the traditional hand-grinding method of shaping a telescope mirror tends to form a parabolic shape anyway, so it just works out.
Transmissive optics (i.e. lenses) however, have the ability to put many elements/surfaces in a single axis, and thus can achieve better performance quality.  Having said that, designs often limit the number of lenses for various reasons: size, weight, total transmission (since each lens absorbs some light), simplicity of assembly, and sometimes cost.
Thus, engineers will use aspheric surfaces to achieve these performance goals. In mass production, plastics and some glasses can be molded into aspheres with relatively low cost. Some can be formed by polishing a flat or spherical shape while the glass is under stress, and then when the stress is removed, it relaxes into an asphere. A lot of military/aerospace equipment use aspheres even if they are not low-cost, because it makes the equipment meet the size- and weight-to-performance goals.
There is even a common telescope design that implements an aspheric lens.  The common Schmidt-Cassegrain telescope (SCT) has a front glass element which looks flat, but is actually a very slight 4th-order asphere ("w" shaped).  This is manufactured commercially using the stressing method mentioned before.


How did single dish (or single receiver) radio telescopes originally generate images? That actually sums up my question nicely: How did single dish (or single receiver) radio telescopes originally generate images? - or at least 2D intensity maps or contour plots.

Early radio telescopes were actually not much more than a Directional Antenna pointing up, connected to a sensitive receiver with various filters, maybe an IF stage or two, but no demodulation. You measured the "noise" signal strength as a function of time on a chart recorder - pen and ink.

I think there were radio images generated and published well before there were high granularity interferometers and computational correlators. The early ones were contour plots, on 2D plotters - again pen and ink.

How was this done? How did, say, a single dish antenna with a single feed generate radio maps?

edit: It doesn't necessarily have to be a dish antenna - it's the single receiver part I'm interested in.

They scan the object, if you point the dish a a point in the sky as the Earth rotates the dish scans across astronomical objects, then move the dish to point at a slightly different position and let it scan across the object again, and again. After a while you can re-construct an image from the scan lines in a similar way to analogue TV.


Two moons and one Earth? Can any one explain to me What will happen to Earth if its has two moons?

Same Distance & two sides of earth (North & south)?

Will it be good for humans or bad?

An orbit as you describe is not stable. Even if the two moons are intially exactly the same distance from the Earth, small perturbations from the sun and other planets will cause one of them to speed up and the other to slow down. Eventually either one of them will hit the other, or one of them will hit the Earth. Whichever happens, it's not good.


How fast do the poles of Jupiter and Saturn precess? As Jupiter and Saturn have very fast rotation periods, about 10 and 11 hours, respectively, I am assuming that their poles will precess much faster then Earth. Since Earth's poles precess every 26000 years, I think Jupiter and Saturn would have a precession time of about ~10000 years or so. Is this estimate correct or am I way off?

You’ll find the answers you want in the Report of the IAU Working Group on Cartographic Coordinates and Rotational Elements 2015, published by Archinal et al in 2018. I found a copy at https://astropedia.astrogeology.usgs.gov/download/Docs/WGCCRE/WGCCRE2015reprint.pdf. A new version might be available this year, as the committee meets every three years… and take about three years to publish its report!
What you’re looking for are on pages 22 and 23 and are the mathematical formulas for the direction of the North poles of Jupiter and Saturn. These start by $ \alpha_0 $ and $ \delta_0 $ and are rather long for Jupiter, quite short for Saturn.
As for how long it would take their pole to fully precess by 360°, it’s exactly 1 million years for Saturn… as for Jupiter, because the formula is more complex, it’s more difficult to say, but I’d guess around 5.5 million years from the first term should be a good ballpark.
Clear skies!


How does the Helioseismic and Magnetic Imager (HMI) compute the Magnetic Activity Index (MAI)? I am looking for a description about how this number called Magnetic Activity Index (MAI) is calculated. All I know is that the processing pipeline of the instrument HMI calculates it. But I would like to know how this quantity is defined and computed! I have not yet found anything on google about it, not even an article.

I found a quote from K. Jain et al.  which defines a Magnetic Activity Index might be pointing you in the correct direction - for me it still sounds rather abstract:

To obtain the MAI, we convert magnetogram data to absolute values, average over the length of a ring day and apodise them into circular areas to match the size of the Dopplergram patches used in the ring-diagram analysis.

However, I also found http://hmi.stanford.edu/teams/rings/modules/maicalc.html which contains a link a source code repository. The man-page of maicalc states:

maicalc takes as its input a dataset of records representing solar magnetograms during a selected interval, and produces as its output the values of the Magnetic Activity Index (MAI) appropriate for a selected set of target locations as they would be mapped and tracked by mtrack for the duration of the input data interval and apodized by pspec3.


Can life survive on the equator of cooled and fast rotating white dwarf or neutron star? As I know, some stars produces carbon, nitrogen and oxygen when they are old and become white dwarfs or neutron stars. And even their surface gravity is strong, it may rotate very fast so that the surface gravity is diminished, hence materials such as rocks and water can present at the surface normally.

My question is, is it possible for life (eg:bacteria) to survive on the equator of white dwarf or neutron star if it is cooled and rotate very fast? Or is it even possible for the equator of white dwarfs or neutron stars become the origin of life?

I am going to attempt a weak answer, mods feel free to delete it, but I'm fairly certain I'm right.

Shortly - no. It's not possible. Even if you balance gravity and centrifugal force perfectly at ground level at the equator, they will very, very quickly become imbalanced as soon as you move north, south, or up from there. So quickly in fact that the gradients may be too big even for a human being not moving at all. Maybe if you're laying down, with your body oriented along the equator, but even then I think the gradients would be too big.

Maybe bacteria would survive, briefly.

I'm sure the math could be done quite easily to estimate the gradients. This is based entirely on intuition.


How to calculate galaxy bolometric luminosity? I am a bit confused by bolometric corrections. If I have an x-ray luminosity in the 2-10 keV band, how does one convert that to $L_{bol}$?
From Netzer's book The Physics and Evolution of Active Galactic Nuclei I got these bolometric correction factors:
Optical: $BC_{5100} = 53 - log(L_{5100})$
and x-ray: $log(L_{5100}) = 1.4\times log(L_X) - 16.8$
where the bolometric correction for the x-ray luminosity ($L_X$) is obtained in two steps, using the equation for the optical BC again. The index $5100$ stands for the optical continuum measured at $5100$ angstrom.
I can't figure out what I have to do with this $BC_{5100}$ once I've got it. Multiply by $L_X$? The book says "(...) BCs, that can be used to convert a single-band measurement of $L$ into an approximate $L_{bol}$."
I'm happy to use correction factors defined elsewhere instead of the ones I quoted. I just want to calculate an estimate for $L_{bol}$ for my galaxies.

The bolometric correction is the difference between a bolometric magnitude and the magnitude in some band.
$$BC = M_{\rm bol} - M_{5100} = -2.5\log\left(\frac{L_{\rm bol}}{L_{5100}}\right)
$$
$$\log L_{\rm bol} = \log L_{5100} - 0.4BC \ ,$$
$$ \log L_{\rm bol} = \log L_{5100} +0.4\log L_{5100} - 21.2  \ ,$$
$$ \log L_{\rm bol} = 1.4(1.4 \log L_x -16.8) - 21.2\ , $$
$$ \log L_{\rm bol} = 1.96\log L_x  -44.72\ .$$


Historical astronomical lunar tables I hope this is the right place to ask this question, and I apologise if not.

I've come across a date written in an old book, but it is written using what seems to be a lunar/zodiac system.
Below is a picture of the date (ignore the text above it):

I'm trying to work out what this date would have been in our modern calendar, but I can't seem to find any tables or almanacs that go back that far.

Are there any publicly available tables that do go back to 1587? If not, how exactly could I go about calculating the answer for myself? I am a mathematician, so a bit of wacky maths should be within my capabilities.
This is the answer I'm probably expecting from this site: some kind of program/algorithm/formula to be able to work out on what date the moon would have been in Taurus in

If not, any ideas as to where best for me to ask/look next?

You can calculate this using the program "KAIROS", See here: http://www.raymondm.co.uk/
It will let you play through the variables: Is this a Julian or a Gregorian date? Is it using true or mean positions? And so forth.


Why do rocks on other solar system bodies that have an atmosphere seem to be flat? Images taken by landers on Titan and Venus and Mars show landscapes where rocks, to me at least, are surprisingly flat. Being used to walking around in forests with roundish meter sized boulders, I'd be very surprised to find myself in a landscape of flat rocks.

Their atmospheric density range between about 100 times, half and 1/100th of Earth's atmosphere. So that can't be the main explanation. Actually, I think that the landscape of the Moon is more Earth like than that of the atmospheric worlds out there.

Is this my impression of "flatness" a real phenomenon, and if so, why and how?

I would say that your initial observation is flawed, so the question is moot.

Huygens landing site, Titan:


Universal point for sky observation Is there a place on Earth, where polar axis lies on the horizont and heaven equator is in zenit? If it exists, can I, in theory, see all the constelations available worldwide this night?

With the axis of rotation of the earth matching the axis of rotation of the stars, and the equator of the earth matching the equator of the stars, any place along the equator would match what you have described. However due to closeness to the horizon, the polar-most stars wouldn't be visible. If you were able to get a hundred metres or so above the ground, with nothing in the surrounding landscape, you could see all of the stars in the sky over a year's time.


How bright would Alpha Centauri appear (in apparent Magnitude) if it were 1 light year away? I'm trying to determine what a star of comparable mass and luminosity to the Sun would look like from the surface of the Earth if it were 1 light year away - I chose Alpha Centauri since it is close to us, and similar to the Sun. Could a star like the Sun be that close without exerting significant influence over it, or would they have an orbital relationship? I'm trying to design a star system for a science fiction novel, and would like there to be a star close enough and bright enough to be visible during the day, and cast a shadow at night, but far enough away to provide no heat energy, and be free from the Sun's gravity.

The change in magnitude is given by:
$$m_2-m_1=-2.5\ \log_{10}(B)$$

Where B is the change in brightness.

Alpha Centauri is about 4.4ly so inverse square says it would be 19.4x brighter at 1ly, so a magnitude differential of -3.2.

Its apparent magnitude would therefore go from -0.27 to -3.47. This is brighter than Sirius (-1.46) but not close to the Sun (-26.7) nor Venus. Venus doesn't cast a shadow, so neither would Alpha Centauri at 1ly.

On gravity, Jupiter is roughly one thousandth the mass of Alpha Centauri but would be about 63,000x closer than the star at 1ly. It would therefore have a gravitational effect on the Sun of:
$$63\,000^2/1000=\mathrm{4\ million\ times}$$
that of Alpha Centauri at 1ly!

Sounds like this works for your story.

Data brightness sources: https://www.space.com/21640-star-luminosity-and-magnitude.html

Mass and distance sources: https://en.wikipedia.org


Can the moon appear to be in any position in the sky? It seems to me that the Moon has appeared in many locations - apparent locations - in the 'dome of the sky'. I know that the Sun follows an arc from east to west and that the arc is smaller in winter than in summer. Therefore the Sun cannot appear in every location in the 'dome of the sky'.
My question is: can the moon appear in any (every?) location in the 'dome of the sky'?

Like the sun, the moon, from our perspective on the surface of the Earth, rises in the East and sets in the West. However, it does not rise exactly due East and set exactly due West. If you were to track the position of the moonrise along the horizon over the course of a year, you would notice that is varies considerably. It cannot, however, be located everywhere in the night sky, even over the course of a year. To understand why, we need to understand some celestial geometry.

The ecliptic plane is the plane which a hypothetical line between the Earth and the sun sweeps out over the course of a year. In the image above, you can see
how the Earth's orbit exists within this plane.
Also notice that the line from the Earth's geographic south pole through its north pole (its rotational axis) is not perpendicular to this ecliptic plane. In fact it is 23.5 degrees off this perpendicular direction.
Notice that the Earth's rotational axis points in the same direction regardless of which side of the Sun the Earth is currently on. This rotational axis always points towards the star Polaris. This means that on different parts of the year, the sun will rise at varying points along the horizon.
The Earth's equatorial plane is a plane defined to extend out from the Earth's equator in all directions. This plane is inclined relative to the ecliptic plane by 23.5 degrees (the same amount as the rotational axis is tilted relative to the perpendicular direction of the ecliptic plane).

Now consider the Earth moon system. In order to determine how far North or South the moon can appear to rise from, you need to consider the angle between the Earth's equatorial plane and the moon's orbital plane. Since the Earth's rotational axis is inclined 23.5 degrees relative to the ecliptic, and the Moon's orbit is inclined 5.14 degrees relative to the ecliptic, the highest the Moon's orbit can be relative to the equatorial plane of the Earth is 28.64 degrees.

Thus, the Moon cannot be directly overhead of any point on the surface of the Earth if that point is 28.64 degrees above or (by an symmetric argument) below the equator.

Thus, the Moon cannot appear in every location in the sky for a given location on Earth. There are parts of the sky that the geometry of the Earth moon system simply will not permit the Moon to exist.


Why are solar systems stable and not chaotic? I am not trained in applied mathematics and am asking as a layperson.
It seems that the motions of the planets can be reliably predicted for thousands of years. Yet, if we consider a relatively simple system, the compound pendulum for example, its motion is highly unpredictable.
Chaotic pendulum - guess when it will stop flipping
The pendulum is highly constrained by its pivot points, therefore its components do not escape from the system. However, with a solar system, there is no rigid attachment and it seems intuitive that planets could randomly fly off into space as a result of chaotic orbits.
Question
Is our solar system inherently stable or unstable?
If unstable then how did it hang together long enough to form in the first place?
If stable, how can stability and predictability be consistent with many bodies moving relatively to one another?

The solar system is chaotic, but it is also stable!
The fixed and linkages between the bars of a double pendulum allow for very rapid energy transfer between the arms.  This makes the chaotic motion develop rapidly.
The interactions between planets are gravitational and much much weaker, moreover, the planets are heavier and it takes a lot more energy to deflect them. This means that in the medium to long-term it is possible to predict exactly where a planet will be.
In the very long-term the system is chaotic, but the gravitational linkage between planets is not strong enough to actually cause one of them to deviate very far from its orbit, unless a resonance develops.  So it is chaotic in the sense that we won't know which side of the sun Mars will be on.  But it is stable in the sense that we know that Mars will continue to be the planet between Earth and Jupiter (but perhaps slightly closer to Earth or slighty closer to Jupiter)
There is only one potential resonance: it is possible that in the very distant future, Mercury could develop a resonance with Jupiter, which could (in 3 to 4 billions years) eject it from the solar system.
So it is the very rigidity of the double pendulum that makes chaos develop so quickly.  Whereas the looseness of gravity means the motion of the planets can be predicted far into the future.