Would a Phobos full impact make Mars temporary habitable?

The impact of Phobos, even in one piece (less likely), would be different from Chicxulub. Phobos is probably a little larger than the Chicxulub impactor, but much slower, and comes in almost perfectly tangential. The impact energy would be less than a 10th of the Chicxulub impact, and the energy would be distributed over a large region around the Martian equator. It wouldn't start volcanism, and it wouldn't warm Mars in a relevant way. It wouldn't make Mars more friendly to life by warming. The only noteworthy contribution to habitability would be additional organic matter which Phobos is presumed to contain.

Phobos will (probably) desintegrate due to tidal forces, before its fragments will collide with Mars. Any loose rocks, dust, regolith on the surface will separate from Phobos below the Roche limit of
$$d=1.26~R_M\cdot\left(\frac{\rho_M}{\rho_m}\right)^\frac{1}{3},$$
with $R_M=3389.5\mbox{ km}$ the radius of Mars, $\rho_M=3.9335 \mbox{ g/cm}^3$ the mean density of Mars, and $\rho_m=1.887\mbox{ g/cm}^3$ the mean density of Phobos. This results in a Roche limit of
$$d=1.26\cdot 3389.5\mbox{ km}\cdot\left(\frac{3.9335 \mbox{ g/cm}^3}{1.887\mbox{ g/cm}^3}\right)^\frac{1}{3}=5455.6\mbox{ km}.$$
That's for Phobos' center about 2059 km above the equatorial surface of Mars (taking 3396.2 km as equatorial radius.)
Therefore desintegration will start long before the impact. Loose rocks will collide with Phobos and unlock more material. Whether there will be a considerable solid remnant to finally cause a big (tangential) impact is difficult to predict.

The orbital speed of the fragments at the time of collision will be about
$$v_0=\frac{v_e}{\sqrt 2},$$
with $v_e=5.03\mbox{ km/s}$ the escape velocity of Mars.
Phobos' inclination of 1.093° (relative to Mars's equator) is almost neglectable for the calculation of the impact velocity.
Mars' equatorial rotation velocity is about 241.17 m/s.

The impact velocity will hence be about
$$\frac{5030\mbox{ m/s}}{\sqrt 2}-241.17 \mbox{ m/s}=3316\mbox{ m/s}.$$
Phobos' mass is about $1.0659\cdot 10^{16}\mbox{ kg}.$
This results in a kinetic impact energy of
$$\frac{1}{2}mv^2=\frac{1}{2}\cdot 1.0659\cdot 10^{16}\mbox{ kg}\cdot (3316\mbox{ m/s})^2=5.86\cdot 10^{22}\mbox{ J}.$$

The heat of fusion of water is about 334 kJ/kg, the heat of vaporization of water at 0°C is about 2504 kJ/kg, together 2838 kJ/kg are needed to evaporate water ice of 0°C.
Hence the impact energy of Phobos would be sufficient to evaporate
$$\frac{5.86\cdot 10^{22}\mbox{ J}}{2.838 \cdot 10^{6}\mbox{ J/kg}}=2.06\cdot 10^{16}\mbox{ kg}$$ of water ice of 0°C, corresponding to $2.06\cdot 10^{13}\mbox{ m}^3$ of liquid water.

Mars' surface area is about 144,798,500 km², or $1.447985\cdot 10^{14}\mbox{ m}^2$.
Equally distributed, we could get a layer of water ice corresponding to
$$\frac{2.06\cdot 10^{13}\mbox{ m}^3}{1.447985\cdot 10^{14}\mbox{ m}^2}=0.142\mbox{ m},$$
(correspoding to 14.2 kg/m² of liquid water) evaporated by the impact energy.

With a surface gravity of 3.711 m/s², this would correspond to
a pressure of $14.2 \mbox{ kg/m}^2\cdot 3.711 \mbox{ m/s}^2= 52.8 \mbox{ N/m}^2=0.528\mbox{ hPa}$.

The current atmospheric pressure on Mars is about 6.3 hPa, with diurnal, seasonal and regional changes.
Thus we could get a maximum of 10% short-term pressure increase in the Martian atmosphere by water vapor, in an ideal case.

At the end, an impact of Phobos, or its fragments, would result in a regional reworking of the Martian surface in the equatorial region, but not in a long-term global change.

Just melting water ice would lead to a global ocean with a depth of about 1.2 m. But this ocean would boil due to the low atmospheric pressure, and freeze out again. Same with local lakes due to the uneven topography.

Model calculatios with a release of carbon dioxide would cause some temporary greenhouse effect, but in the equatorial region a release of carbon dioxide is even more unlikely than a release of water vapor.


What would happen if you jumped out of a moving spacecraft?

If the spacecraft is not accelerating, the astronaut leaves the airlock and just floats outside, not moving relative to it. If the rocket is accelerating they "fall" down towards the stern from the perspective of people onboard; the astronaut would instead say the rocket is accelerating "upwards" and they are just floating.

In space, away from gravitational fields and in vacuum, only relative motion matters. You cannot tell whether the astronaut and spacecraft are moving at a high or low velocity compared to the rest of the universe just by looking at the two of them. So one can flip between the perspective that the craft is fixed and by turning on the engine it makes the astronaut (and everything in the craft) is accelerated, or the the perspective that the astronaut is floating in place and the rocket is speeding away. They are equivalent.

Astronauts leave spacecraft like the ISS all the time, and only experience free-fall despite the station moving at high speed around the Earth. As far as I know nobody has ever done a spacewalk when a craft is accelerating, for obvious safety and sanity reasons.


Why aren't neutron stars full of dark matter?

Yes, neutron stars might actually accumulate weakly interacting dark matter and this allows some observational constraints on its nature. Basically, the temperature and continued existence of neutron stars places bounds on the density and interaction cross-section of dark matter.

A dark matter particle that does not interact with matter will just have its trajectory bent by the gravity field of a heavy object, so most unbound particles will just swoop past on a hyperbolic trajectory. But as discussed in (Adams & Laughlin 1997), if there is some matter-dark matter interaction then the particle may scatter from a matter particle and now have less than escape velocity. This way white dwarfs and neutron stars would indeed accumulate dark matter in their cores. The rate of accumulation is proportional to $\rho v \sigma$ where $\rho$ is the dark matter density, $v$ the average relative velocity and $\sigma$ the cross section. Adams & Laughlin estimate that a white dwarf star would accumulate its own mass in $10^{25}$ years, but this is going to be depend on the cross section (if it is too small the dark matter will pass through) which is at present unknown.

Were this accumulation the only thing happening it would eventually make white dwarfs and later neutron stars implode. However, dark matter is plausibly a mix of particles and antiparticles that annihilate each other at a rate $\sim \rho^2$; in an enriched environment like a white dwarf core this would produce energy from emitted photons heating things up. Adams & Laughlin estimate the luminosity as to about $10^{-12}L_\odot$, which is imperceptible in the current era but would eventually keep white dwarfs at 63 K in the far future (until the dark matter halo runs out).

Other, more elaborate, calculations lead to accretion estimates that are higher. If the rate were high enough, then we would not see any cool dense objects - so white dwarf and neutron star cooling gives some bounds on the possible density and cross sections, albeit not very strict ones. For example, one model suggests that neutron stars would level out at 10,000 K. Cool star observations can also already rule out some dark matter models.

There are even some arguments that super-earth planets in dense dark matter halos might be heated significantly, although this may require unrealistically dense halos and big cross sections. The current heath flow of Earth does give some constraints on how strongly it can interact.

So neutron stars are not directly giving us dark matter detection, but they (and planets and white dwarfs) are giving us some information.


Hills' mechanism for making Hypervelocity Stars

Consider two unit masses with velocity $V$. Give them a small push apart so they have velocities $V+v$ and $V-v$ with $v\ll V$ . Their kinetic energies are then $$\frac{(V+v)^2}{2} \approx \frac{(V)^2}{2} +vV$$ and $$\frac{(V-v)^2}{2} \approx \frac{(V)^2}{2} -vV$$
We have transferred energy $\approx vV$ from one mass to the other with the expenditure of a much smaller  energy $v^2$.

Now suppose the masses are initially moving with some tiny velocity and then travel into a potential well where they have velocity $V$. If we push while they are in the well then one will have enough kinetic energy to escape the well plus $vV$, while the other will be trapped in the well.

This is the same principle as the Oberth effect, where the ejected reaction mass plays the role of one of the masses.


How to get the longitude/latitude from solar zenith/azimuth?

Well let's see:

Local Standard Time Meridian (LSTM)

The Local Standard Time Meridian (LSTM) is a reference meridian used for a particular time zone and is similar to the Prime Meridian, which is used for Greenwich Mean Time.
The (LSTM) is calculated according to the equation:
$$
LSTM = 15^{o}.\Delta T_{GMT}
$$
where $\Delta T_{GMT}$ is the difference of the Local Time (LT) from Greenwich Mean Time (GMT) in hours.

Equation of Time (EoT)

The equation of time (EoT) (in minutes) is an empirical equation that corrects for the eccentricity of the Earth's orbit and the Earth's axial tilt.
$$
EoT = 9.87 sin\left(2B\right) - 7.53cos\left(B\right) - 1.5 sin\left(B\right)
$$

Where $B = \frac{360}{365}\left(d - 81\right) $ in degrees and d is the number of days since the start of the year.

Time Correction Factor (TC)

The net Time Correction Factor (in minutes) accounts for the variation of the Local Solar Time (LST) within a given time zone due to the longitude variations within the time zone and also incorporates the EoT above.
$$
TC = 4\left(Longitude - LSTM\right) + EoT
$$
The factor of 4 minutes comes from the fact that the Earth rotates 1° every 4 minutes.

Local Solar Time (LST)

The Local Solar Time (LST) can be found by using the previous two corrections to adjust the local time (LT).
$$
LST = LT + \frac{TC}{60}
$$

Hour Angle (HRA)

The Hour Angle converts the local solar time (LST) into the number of degrees which the sun moves across the sky. By definition, the Hour Angle is 0° at solar noon. Since the Earth rotates 15° per hour, each hour away from solar noon corresponds to an angular motion of the sun in the sky of 15°. In the morning the hour angle is negative, in the afternoon the hour angle is positive.
$$
HRA = 15^{o}\left(LST - 12\right)
$$

Declination angle:

The declination angle denoted by $\delta$, varies seasonally due to the tilt of the Earth on its axis of rotation and the rotation of the Earth around the sun. If the Earth were not tilted on its axis of rotation, the declination would always be 0°. However, the Earth is tilted by 23.45° and the declination angle varies plus or minus this amount. Only at the spring and fall equinoxes is the declination angle equal to 0°.
$$
\delta = 23.45^{o} sin\left[\frac{360}{365}\left(d - 81\right)\right]
$$

where d is the day of the year with Jan 1 as d = 1.

Elevation angle:

The elevation angle is the angular height of the sun in the sky measured from the horizontal.
$$
α=sin^{−1}[sin\delta sin\phi + cos\delta cos\phi cos(HRA)]
$$
Where $\delta$ is the declination angle, $\phi$ is the local latitude and HRA is the Hour angle.

Azimuth Angle:

The azimuth angle is the compass direction from which the sunlight is coming. At solar noon, the sun is always directly south in the northern hemisphere and directly north in the southern hemisphere.
$$
Azimuth = cos^{-1}\left[\frac{sin\delta cos\phi - cos\delta sen\phi cos(HRA)}{cos\alpha}\right]
$$

Where $\delta$ is the declination angle, $\phi$ is the local latitude and HRA is the Hour angle.

Zenith Angle:

The zenith angle is the angle between the sun and the vertical. The zenith angle is similar to the elevation angle but it is measured from the vertical rather than from the horizontal, thus making the zenith
$$
Zenith = 90° - \alpha
$$

Where $\alpha$ is the elevation angle.

Note that your input parameters are going to be:

Longitude
$\Delta T_{GMT}$ is the difference of the Local Time (LT) from Greenwich Mean Time (GMT) in hours
LT local military time in hours
$\phi$ the local latitude
d the day of the year


What is the furthest object in the observable universe?

About 13.1 billion light years.

The z8 GND 5296 Galaxy mentioned in the article linked by TildalWave, mere 700 millions years old (after Big Bang) would be the mentioned distance from us for the light to reach us at indicated speed, considering the 13.798 bln years old universe. As we push technological barriers, we keep finding more distant objects.

The age of the universe is the hard limit in light years of distance of any observable object - there was nothing to emit light more than 13.798 bln years ago, or respectively, visible from more than 13.798 bln light years away. visible from distance that what was 13.798 bln light years 13.798 bln years ago - or, after adjusting for universe expansion over that time, currently 46.6 billion light years away.

Of course, as RhysW mentions, this limit changes over time as the universe ages and grows. The exact rate of growth is 74.2km/sec per megaparsec, or about 0.000000007% per year. I'd say, comparing to our life and current technological progress, an insignificant factor.


Yes, there are hot Jupiters around stars more massive than the Sun: in fact the first known hot Jupiter host star 51 Pegasi has a mass of around 1.11 solar masses.

As for a list of them: use the confirmed planets table at the NASA Exoplanet Archive. You just need to add in the relevant criteria you're interested in. The actual criteria for where to draw the dividing line seems to be roughly around Saturn-mass planets, so using Saturn's properties you could do the following:

Filter by mass of planet > 0.3 Jupiters, orbital period < 10 days, mass of star > 1 solar mass. At the time of writing this selects 275 planets.
Filter by radius of planet > 0.8 Jupiters, orbital period < 10 days, mass of star > 1 solar mass. This gets me 286 planets at the time of writing.

There is a handy "Plot Table" button that will allow you to construct a histogram from the selected planets. Here's an example from the planets selected by mass between 1 and 5 solar masses (there are a few outliers at even higher stellar masses, but I'm not sure how reliable those mass estimates are):

Note that the above does not represent the actual distribution of hot Jupiters with stellar mass: there are detection biases at play that make it harder to find planets around more massive stars, which need to be corrected for to estimate the true distribution.

You can also do similar queries at the Extrasolar Planets Encyclopaedia but the user interface isn't quite as friendly.


The only one who seems legit is the first date, October 38172. It appears somewhere in Transits, by Jean Meeus (1989), and some say that also in the SOLET webpage, but I can't find it either. Taking into account that, if true, it comes from a book from 1989 and probably is not very accurate, I wouldn't take that date as an exact thing.
The other dates I can't find anything.


The formula you gave is to find the hour angle of the star while setting, not the setting time itself. Suppose the hour angle of star is $HA\star$, and $RA=\alpha$,then the Local Sidereel Time is given by $LST=HA\star+\alpha$. At the time of setting, $HA\star=10^h1^m30.2^s$. Thus, $LST=13^h16^m30.2^s$at the time of setting.

Now, $RA\odot=6^h$ on $\text{June} 21^{st}$ and $HA_\odot =\arccos(\tan(23.43^{\circ}))=4^h17^m17^s$ at sunset. Thus the star sets with the sun after $\text{ June} 21^{st}$. Since $RA_\odot$ increases approximately by $2^h$ every month, we can assume that the star sets with the sen near $\text{July }21^{st}$. Let it set after $n$ days. At $\text{July }21^{st}$, $\delta_\odot = 20.09^{\circ}$ which gives us $HA_\odot=4^h34^m11^s$, and $RA_{\odot}=8^h9^m44^s$. The $HA_\odot$ at sunset is approximately the same for nearby dates. Since $\delta_\odot$ is also near the maximum, we can approximate $\Delta RA_\odot= \omega_\odot n$. Thus,

$$13^h16^m30.2^s=8^h9^m44^s + \omega_\odot n +  4^h34^m11^s$$
$$ \omega_\odot n =0.5433^h$$
$$n=8$$

At  $\text{July } 29^{th}$, $\delta_\odot=18.23^{\circ}$, $HA_\odot= 4^h43^m6^s$, $RA_\odot=8^h42^m12^s$ which gives us $LST=13^h25^m18^s$, which is a bit more than expected. Trial and error for few days before gives us that on $\text{July } 26^{th}$, the sun sets at LST $13^h15^m9^s$ which would be the most appropriate answer.

Thus, the sun sets along with the star at $\text{July } 26^{th}$


Converting ecliptical to equatorial coordinates

Convert commments to community wiki

This is a standard problem with using an arctan function (inverse tan function returns a value between -90 and +90). A lot of programming languages have two functions the simple one like you appear to have used and a more complex one that takes two parameters for when you are dividing (before the arctan function) as in this case. (This is described in further detail at (http://youtu.be/_JbkTK17MrY)

In Java you can use α=atan2((cos ε∗sin λ−sin ε∗tan β), cos λ) you may need to add 360 if the angle returned is negative.

An alternative is α=atan2(cosβsinλcosε−sinβsinε,cosβcosλ). This is equivalent in most cases but avoids an infinite term tanβtanβ at the ecliptic poles, costing one more trig function call.


Is Io's orbit or rotation affected by its volcanism?
The rotations of comets and asteroids are affected by outgassing volatilities. Io is very volcanically active. Does this give Io a varying orbit and a slow rotation?

Jupiter's moon Io has a lot of volcanoes.
A similar answer to your question is provided at Quora's question: "Could the eruptions of all the volcanoes on Earth simultaneously knock it off its rotation or change its orbit?".
There are two differences: Eruptions on Io can reach escape velocity (2.558 km/s), and matter is subject to gravity as it is propelled away from the moon allowing some to escape and some to return; so it is a source of wobbling and rotational speed changes.
"Jupiter's rocky moon Io is the most volcanically active world in the solar system, with hundreds of volcanoes, some erupting lava fountains dozens of miles (or kilometers) high. Io’s remarkable activity is the result of a tug-of-war between Jupiter's powerful gravity and smaller but precisely timed pulls from two neighboring moons that orbit farther from Jupiter – Europa and Ganymede.
...
Orbit and Rotation
Although Io always points the same side toward Jupiter in its orbit around the giant planet, the large moons Europa and Ganymede perturb Io's orbit into an irregularly elliptical one. Thus, in its widely varying distances from Jupiter, Io is subjected to tremendous tidal forces.
These forces cause Io's surface to bulge up and down (or in and out) by as much as 330 feet (100 meters). Compare these tides on Io's solid surface to the tides on Earth's oceans. On Earth, in the place where tides are highest, the difference between low and high tides is only 60 feet (18 meters), and this is for water, not solid ground.
Io's orbit, keeping it at more or less a cozy 262,000 miles (422,000 kilometers) from Jupiter, cuts across the planet's powerful magnetic lines of force, thus turning Io into a electric generator. Io can develop 400,000 volts across itself and create an electric current of 3 million amperes. This current takes the path of least resistance along Jupiter's magnetic field lines to the planet's surface, creating lightning in Jupiter's upper atmosphere.
Surface
The tidal forces generate a tremendous amount of heat within Io, keeping much of its subsurface crust in liquid form seeking any available escape route to the surface to relieve the pressure. Thus, the surface of Io is constantly renewing itself, filling in any impact craters with molten lava lakes and spreading smooth new floodplains of liquid rock. ...".
The two-body problem explains how multiple interacting objects affect the orbit of each other. While the orbital eccentricity of Io is quite low (0.0041) the orbits of three of the Galilean moons (Ganymede, Europa and Io, excluding Callisto) have a 1:2:4 resonance. The "pushing" will act in periodic repetition to have a cumulative effect on the motion.

Orbital resonances greatly enhance the mutual gravitational influence of the bodies, i.e. their ability to alter or constrain each other's orbits. In most cases, this results in an unstable interaction, in which the bodies exchange momentum and shift orbits until the resonance no longer exists. Under some circumstances, a resonant system can be stable and self-correcting, so that the bodies remain in resonance.
The small effect of the volcanoes is corrected by the much greater effect of resonance. The website space.com estimates Io's age at about 4.5 billion years old, the same age as Jupiter. If the orbit was unstable and changing it may well have been destroyed long ago.


M87 Black hole. Why can we see the blackness?

Rob Jeffries' answer is excellent, I just wanted to add this picture trying to explain the geometry. Here, I assume a non-rotating black hole (BH); for a rotating BH the exact numbers are slightly different.

The photon sphere

Photons move on straight lines, but in the heavily curved space around a BH these straight lines appear curved. Although the event horizon (EH) at a distance of $r = 2GM/c^2 \equiv r_\mathrm{S}$ (the Schwarzschild radius) from the BH marks the region from which no photons may escape if emitted radially, photons on partially tangential orbit will fall back out to a distance of $r = 1.5r_\mathrm{S}$, where photons traveling fully tangentially will stay on the photon sphere (although this is an unstable orbit).

The innermost stable orbit and the accretion disk

Ordinary matter will spiral inward out to twice this distance; hence, inside the innermost stable circular orbit (ISCO) at $r=3r_\mathrm{S}$, matter is pretty much bound to be absorbed. Outside this region matter may orbit, forming the accretion disk, but since friction between the particles will cause them to lose energy, they will slowly approach the ISCO, after which they will rapidly fall into the BH. Note that the M87 BH doesn't have a thin accretion disk like the one depicted in the movie Interstellar; rather a thick "cloud" surrounding most of the BH.

Photons emitted tangentially just outside the photon sphere will spiral around the BH many times, slowly increasing their distance, until eventually they escape at a projected distance of $\sqrt{27/4}r_\mathrm{S} \simeq 2.6r_\mathrm{S}$ from the BH (e.g. Frolov & Novikov 1998).

The shadow

Just as the path of light rays are curved around the BH, so are the sightlines from you toward the BH (you can think of sightlines as reversed photons). That means that all sightlines that are closer than (a projected distance of) $2.6r_\mathrm{S}$ to the BH will, eventually, end up on the EH, even if taking several orbits around the BH. These sightlines comprise the so-called shadow (Falcke et al.(2000); Event Horizon Telescope Collaboration et al.(2019a), ). On the other hand, along sightlines that a farther away, you see the radiation emitted from the matter falling into the BH, both in front of and behind the BH. And since the first sightlines that don't terminate at the EH circle the photon sphere many times, those sightlines are actually very long paths through matter shining its last light before being engulfed, and hence they look exceptionally bright (e.g. Event Horizon Telescope Collaboration et al.(2019b)). This bright ring just outside the shadow is called the photon ring, or the emission ring.

The drawing

The drawing below may help understand. All the red lines are sightlines toward the BH. Only the uppermost one just grazes the photon sphere (and the luminous matter behind). The rest terminate at the EH, and hence look black (except for luminous matter in front). Close to the center, you see the front of the EH; farther out you actually see the back of the EH; even farther out you again see the front of the EH, and so on ad infinitum until you reach the photon ring.

The observation

Despite the observational resolution being an astonishing $\sim25$ micro-arcseconds, the photon ring is smeared out over a larger region, resulting in the doughnut shape you've seen. That is, what you see in that image is not "the EH in front of an accretion disk", but rather "the EH seen from all sides at the same time and enlarged, with light emitted from the photons ring".

Unless the accretion disk is viewed exactly face-on, half of the accretion disk$^\dagger$ has a velocity component toward you, making it brighter than the other half through a special relativistic effect called beaming. This is seen in the southern part of the M87 BH.

The figure below (from Event Horizon Telescope Collaboration et al.(2019b)) shows, from left to right, the actual observation, a model where you see the rather sharp photon ring, and this model blurred to match the resolution of the observation.

$^\dagger$At least the material just before it plunges into the BH, which follows the rotation of the BH. Farther out, the rest of the accretion disk may in principle rotate the other way.


Understanding date of astronomical events

Precession of equinoxes means that the North pole of the Earth is changing its direction. This also means, that the dates of the seasons (seen from outside the Solar System (= sidereal)) are changing. Nirayana calculates the date of Makar Sankranti with adding years to the first Makar Sankranti, but sayana calculates the date of Makar Sankranti relative to stars (sidereal).
In a nutshell: nirayana is on the same date as the first event (first Makar Sankranti), but sayana is on the same event as the first Makar Sankranti (winter solstice). Yes, winter solstice was once on the 14 January.


Accretion disks - why are they disk-shaped, rather than spherical?

There's two different effects here, and they're both related to viscuous forces in the accreting matter.

First, if the infalling matter has some nonzero angular momentum vector $\mathbf{L}$, then consider a plane perpendicular to this vector. Due to its conservation, we can't get rid of rotation along this plane, but the component of angular momentum orthogonal to the plane is zero. As the accreting matter heats up and radiates away energy, that energy is taken from the gravitational potential, which is lessened by bringing the matter particles closer together. Thus, such interactions tend to flatten the matter into the plane orthogonal to its angular momentum.

But there's something more that happens for rotating black holes, as a combination of its gravitomagnetic field and viscous forces in the matter, that drives the matter to the plane of the black hole's rotation, regardless of its initial angular momentum. This is called the Bardeen–Petterson effect [1]:

  In an accretion disk, viscosity causes each ring of gas to spiral inward toward the hole with a radial speed $|u^r|\ll\text{(orbital velocity)} = (M/r)^{1/2}$. If this were the only effect of viscosity, the warping of the disk would consist of a differential precession. Successive rings, each of smaller radius than the preceding, would have interacting longer with the [gravitomagnetic] field and thus would have precessed through successively larger angles... . As a result the disk would be radially twisted, but it would not be driven into the equatorial plane. However, the precession changes the disk's shear $\sigma_{jk}$ and thus changes the viscous forces; and these altered forces drive the disk toward the equatorial plane of the black hole.

This is only significant closer to the black hole, at $r\lesssim 100M$. At larger radii, the disk shape is due to the direction of the angular momentum of the matter instead.

[1] Black Holes: The Membrane Paradigm, ed. by Kip S. Thorne, Richard H. Price, and Douglas A. Macdonald.


Is there a significant population of neutron stars outside the galactic plane?

I'll partly answer your question with the following plot from the ATNF pulsar database (which is easy to play around with). The plot shows the Galactic z coordinate of the pulsars (distance from the Galactic plane in kpc) versus their estimated age (from their spin down rates, in years).

The basic picture is that young pulsars are all near the Galactic plane, because they are born from high mass stars that are born, live and die near the plane and even if the neutron star got a kick, there has been insufficient time to move far from the plane. A typical kick might be 500-1000 km/s, roughly 0.5-1 kpc in a million years.

Once pulsars reach a million years old, then indeed some reach a kpc above/below the Galactic plane and then a few even older examples reach 10 kpc.
Most normal pulsars only show pulsar activity for 10-100 million years (the older ones on the plot are probably a different population of pulsars in binaries), so they aren't observed at much greater distances (though there will be old neutron stars out there).


What size objects can be created from collision of two asteroids?

Since planets are forming from asteroid collisions, I would say those collisions won't necessarily create small objects.

Imagine throwing a snowball against a window.

If you throw it lightly, some snow will stick to the window and increase the mass of it (at least until it melts away) and the rest of the snowball will be scattered everywhere around but won't stick to the window.

If you throw the snowball too hard, the window will break, resulting in even more tiny pieces. But still, some snow will stick to some pieces of glass.

That's exactly what happens on asteroid impacts and depends on the speed, mass and structure of each asteroid.

The size of those debris should reach from microscopic dust up to gigantic rocks with several kilometer wide diameter.


What are the differences between a Black Hole and a Supermassive Black Hole

Stellar mass black holes form from the collapse of massive stars at the end of their lives. You can then find them scattered throughout galaxies, just like you find massive stars. They typically have a mass a few times the mass of the sun.

Supermassive black holes are found at the centers of galaxies. They typically have mass of millions of Suns.

Recently they have started to discover Intermediate Mass Black Holes which blur the lines between a stellar black hole and supermassive black hole. The typically have a mass in the range 100 to one million solar masses.


I asked a somewhat similar question but just about the Earth.  Here and in my question there's some links that you might be interested in.   Jupiter for example is thought to have moved closer to the sun during the late heavy bombardment, then back outwards.   So I suspect that through the wormhole didn't get that point quite right.   Observations by the Kepler Spacecraft in the hunt for exoplanets suggests that a number of planets in other solar systems have moved closer to their suns over time.

In general, there's a few parts to this question.

On the sun losing mass:

Yes, it is, but only a little bit.  See Here.  Their math isn't complete as they don't factor in coronal mass ejections or take into account the sun's expected increase in luminosity, but the gist of their point remains true.   The sun is only losing a tiny percentage of it's mass.  The effect on planetary orbits is pretty small.

Sunlight also creates a pressure against the planets, but this effect, isn't very much due to the comparatively enormous mass of the planets compared to the light and coronal mass ejection material that hits the planet.

Tides

Tidal effects can work either way depending on the rate of rotation compared to the planet's orbital rotation.  If the star turns ahead of the planet, the tidal force should push the planet slowly outwards, like the Earth is currently doing to the Moon.   If the planet moves around the star faster than the star turns, the tidal effect is reversed and it slowly pulls the planet in.  At our Sun's current rate of rotation it's moving ahead of all the planets, so there should be a small tidal force that pushes the planets outwards, but very very slowly.

Orbital debris and effective loss of angular momentum during planet formation.

If the orbit is littered with dust and comets, asteroids and such, the sun can gain mass faster than it loses it and the planets can effectively slow down as they move through the dust or orbital debris with more oblique orbits.   This may be the explanation for hot Jupiters and so many observed planets very close to their suns in other solar systems.

Planets can affect each other.

See here and a curious possibility here.   Planet to planet effects are tiny unless they are in resonance where the effect can be amplified and become significant over time.

Hope that's clear enough.  I can tidy up and provide additional links for reference if needed.


Since the Solar system is a multi-body system (with $N>2$ bodies of significant mass), the orbits of its constituents are not exact Keplerian orbits.
To lowest order, each planet orbits the Sun (or rather the centre of mass of all interior planets) on a Keplerian orbit, but the interactions with the planets as well as the fact that the centre of mass of the interior is not fixed lead to deviations of the true orbit from this simplification. These deviations can be treated either numerically or via perturbation theory, but are non-trivial functions of time.
The same holds for the Sun: to lowest order one can neglect all planets but Jupiter (which is more then twice as massive as all the remaining planets combined), when the Sun follows an elliptic orbit with semi-major axis of about 0.005AU (smaller than that of Jupiter by their mass ratio). This is of the same order as the radius of the Sun, i.e. the barycentre of the Solar system hardly leaves the Sun. However, as above, the pull by all other planets lead to a deviation from this simple model. Again, these deviations are non-trivial.


If you notice carefully, you'll see that their speeds away from the sun are fairly similar (7-9 km/s), and the difference is only in their speeds away from the earth. This is because of a small component of earth's motion around the sun (which is at about 30 km/s) adding to Voyager 1's motion and subtracting from Voyager 2's motion.


Question about semi-major axis mean separation

The semi major axis $a$ of an orbit is the same as the arithmetic average of the minimum and maximum distance $r$ or the object from the central object in sense of $a = 0.5\cdot(r_{min} + r_{max})$; thus semi major axis is sometimes also called mean separation. As such you are given the sum of the semi major axis of two components.
A two-body problem can be reduced / simplified to a mass-less body circling an object of the combined mass - as such you can calculate the combined mass of both stars with the numbers you are given.


How bright will Sirius be when it's closest to the solar system?

From Sirius: Brightest Diamond in the Night Sky by Jay B. Holberg we have current distance of Sirius ~8.5ly, closest approach will be ~6.7ly. The intensity of of light from Sirius assuming constant luminosity scales as distance squared, so the intensity (power per unit area at the eye) at closest approach will be: $\left(8.5/6.7 \right)^2 \approx 1.6$ times that today, which is a change of magnitude from about $-1.4$ to about $ -1.6$


Was the Geocentric Model correct at all?

Ptolemy's epicyclic, geocentric model, in use until the Renaissance, was very accurate in terms of predicting the positions of planets and the times of eclipses. What it couldn't account for were things like the correlations between apparent size and phase of Venus, or to properly account for the variation in brightness of the planets.
Thus the reason for discarding the geocentric model was not really because it lacked precision, but that it failed to explain various other observational facts, especially after the development of telescopes.
No doubt you could tune the Ptolemaic system even further (more epicycles?) to iron out some of the small errors that were revealed by Tycho's positional measurements at the turn of the 16th century, which had a precision unavailable to Ptolemy. However, the advent of Kepler's laws and subsequent explanation by Newton, rendered the geocentric model obsolete.
As you can judge from (well written) articles like this one, geocentrism is actually quite hard to kill-off observationally, if you are prepared to accept that the universe is arranged "just so".


Where can I find good places to watch stars?

It would have been better to ask about "which conditions must have a good starsighting place".

It must be:

Dark
Far from cities
Not under main flight paths (this depends on which altitude is the object you want to observe)
High altitude (or any special place with low cloud formation and low turbulence)
Far from big water bodies that can cause local humidity


How are galaxies moving so much and why are they moving?

There are several sources of movement for galaxies.

Galaxies may rotate.  Galaxies are formed essentially from matter collapsing under gravity into "clumps".  This process isn't uniform due to gravity from other clumps of matter and so it is very rare that all the particles in a single clump "fall" directly into it's centre.  This creates rotation (Angular Momentum) and so the eventual galaxy will rotate.  Note that some galaxies show very little overall rotation, with the individual stars orbiting the centre of the galaxy in a multiple directions.

Galaxies interact with each other. Most galaxies reside within a group or cluster of galaxies, our galaxy is within a group called the Local Group with Andromeda and several ($\geq$52) dwarf galaxies.  These dwarf galaxies are usually part of a "satellite-subsystem" where they orbit a large galaxy under gravity.  These orbits are affected by the other local galaxies and occasionally some systems can be "expelled" due to the gravitational forces happening to speed the body up past the velocity required to escape the system.

The major galaxies in a group or cluster interact with each other, orbiting a centre of gravity.  This is due to the same mechanism as galaxy rotation.  This is usually a very dense area of galaxies, or a massive Brightest Cluster Galaxy (BCG).  In the local group, the two main galaxies (our Milky Way and the Andromeda Galaxy) are getting closer together and will collide in approximately 4 billion years.

The universe is expanding. The universe is expanding and so galaxies appear to be moving away from us in every direction.  We see this as redshift, a change in the wavelength of light as it reaches us due to the difference in speed between the origin an us.  This is not really anything to do with galaxies moving, and more to do with the fabric of the universe.  Whether this will continue, or whether the universe will stop expanding and collapse is a current active area of research.


Capability of observing transits with terrestrial telescopes of various sizes?

A bright, nearby star with a large exoplanet would be best. For example, the first star to have a transiting planet observed was HD 209458, in Pegasus. It has a magnitude of 7.65.  When the large "hot jupiter" transits the star it dims by a relative flux of 0.984. That corresponds to a change in magnitude of about 0.016. I.e. it changes from a magnitude 7.65 star to a 7.67 star.

If you know the radii you can calculate the relative flux during a transit by assuming the planet blocks all the light.

The radius of the planet HD209458b is about 100000km and the star 800000km, so the planet has an area 1/64 of the area of the star's disc, and as expected 1-1/64 = 0.984: which is in good agreement with the light curve.

That won't be noticeable with the naked eye, and is challenging for an amateur. A couple of links:

http://www.britastro.org/vss/ccd_photometry.htm general information on photometry. (summary: you need some kit)
http://brucegary.net/book_EOA/EOA.pdf (summary: it can be done, with a 14 inch telescope and Arizonan dark skies.)


How can I estimate how long will a main sequence star stay on the main sequence given its temperature?

The website on Main sequence stars fromr the Austalian national telescope facility lists star mass, temperature and life span:

Mass/MSun
Luminosity/LSun
T=Effective Temperature/K
Radius/RSun
t=Main sequence lifespan/yrs

0.10
$3×10^{-3}$
2,900
0.16
$2×10^{12}$

0.50
0.03
3,800
0.6
$2×10^{11}$

0.75
0.3
5,000
0.8
$3×10^{10}$

1.0
1
6,000
1.0
$1×10^{10}$

1.5
5
7,000
1.4
$2×10^9$

3
60
11,000
2.5
$2×10^8$

5
600
17,000
3.8
$7×10^7$

10
10,000
22,000
5.6
$2×10^7$

15
17,000
28,000
6.8
$1×10^7$

25
80,000
35,000
8.7
$7×10^6$

60
790,000
44,500
15
$3.4×10^6$

If we suspect a lifespan, $t$ is related to the temperature $T$ by a power relationship, $t=aT^k$, we can plot log(t) against log(T) and look for a linear relationship

That looks plausible, and gives a relationship of $t = 10^{28}\times T^{-4.7}$ (rounding to 2sf)
looking more closely, this tends to underestimate the life span of mid-range stars, but overestimate large and small stars. You can adapt for this by considering smaller and larger stars separately: For stars less than 10000K, a model $t = 10^{37}\times T^{-7.6}$ and for hotter stars $t = 10^{20}\times T^{-3.0}$ gives reasonable fit to the data.
This is, of course, a purely empirical fitting of a curve, and isn't based on an astrophysical model. (though I speculate that the change in gradient is related to the change from the proton-proton to the CNO cycle fusion of hydrogen.)


How long does it take Dawn to orbit Ceres?

As of March 6, 2015, Dawn has entered orbit around Ceres.  But it's really so far only "captured by Ceres' gravitational pull".  Then, it was still 61,000 km from Ceres.

It's slowly spiraling down into an orbit that JPL's Dawn Journal calls "RC3", which will be 13,500 km above Ceres.  This orbit will last 15 days per revolution.

  It will take about 15 days to complete a single orbital revolution at this altitude.

The first news link above indicates that Dawn will reach RC3 on April 23, 2015.

This picture from the Dawn Journal gives us a picture of Dawn's path: it's not completing any revolutions around Ceres yet; it's still approaching the RC3 orbit as of this writing (March 10, 2015).

So as of now, March 10, 2015, there is no orbital period yet.  But when RC3 is achieved, it will be a slow orbit -- 15 days.

There are other orbital altitudes planned past RC3.

"second orbital phase, survey orbit" - Altitude 4,400 km, Orbital period 3 days, 3 hours
"high altitude mapping orbit" - Altitude 1,470 km, Orbital period 19 hours
"low altitude mapping orbit" - Altitude 375 km, Orbital period 5.5 hours

According to Dawn's schedule, the "survey orbit" will only last 22 days before it spirals in to the "high altitude mapping orbit" in August 2015, although it's unclear exactly when the "survey orbit" will take place.  The "low altitude mapping orbit" will be in November 2015.  In between orbits, when spiraling down to the next scheduled orbit, Dawn's orbital period will slowly decrease.  At the last orbit, Dawn will be left in this orbit after its batteries and hydrazine fuel are exhausted.

Update (Oct 23, 2015)

Excerpt from latest news update from the Dawn mission (Sept 30, 2015):

  Dawn is currently orbiting Ceres at an altitude of 915 miles (1,470
  kilometers), and the spacecraft will image the entire surface of the
  dwarf planet up to six times in this phase of the mission. Each
  imaging cycle takes 11 days.

  Starting in October and continuing into December, Dawn will descend to
  its lowest and final orbit, an altitude of 230 miles (375 kilometers).
  The spacecraft will continue imaging Ceres and taking other data at
  higher resolutions than ever before at this last orbit. It will remain
  operational at least through mid-2016.


What kind of telescope should I get my husband?

Video Astronomy or as some refer to it as Live or Near Live Viewing is a growing field of interest that lies between traditional viewing with an eyepiece and Astro Photography.

Eyepieces are substituted with cameras and the viewing is done on a screen. In general the cameras are more sensitive than the eye and often reveal more detail and color.

That being said it is a field that some love and others hate. The individual needs to decide for themselves. There are extremists in both groups. An open mind helps but have a few grains of salt on hand just in case. :)

In general, the electronic viewers prefer telescopes with a faster f ratio than the eyepiece viewers. Because of the increased camera sensitivity, many claim smaller aperture telescopes can work as well as those with double their aperture. Better to find out what type of viewing is more interesting before buying any equipment.

My suggestion is to see what the electronic viewers are doing with their equipment.

Many electronic viewers broadcast their viewing sessions on the NightSkiesNetwork. Anyone can join in on these broadcasts at

https://www.nightskiesnetwork.ca/

Most broadcasts originate in North America but they could be from anywhere in the world. Check for clear weather in the Eastern time zone at nightfall in the East. If the weather looks clear take a look at the NightSkiesNetwork site then look again occasionally as nightfall passes across the continent. It is all weather dependent. In general you will find a broadcast sooner rather than later.

Most viewers of the NightSkiesNetwork started out with traditional eyepiece viewing and like astronomers everywhere are more than willing to give people advice whether it be for viewing with an eyepiece or a camera.

I use several Mallincam cameras, but there are other cameras available.

Google "Video Astronomy" or "Electronic Assisted Astronomy" for more information.

For what it is worth I viewed with eyepieces for fifty years before going electronic. In the ten years since going electronic I have had no desire to use my traditional eyepieces. Even when my eyesight was at its best and with telescopes up to 14" aperture I have never seen as much as I do now with an electronic eyepiece on my 5" and SCT's 10". That may make me sound like an extremist, if so be prepared to deal with a few more like me. :)


Calculating right ascension for two planets and their separation using python

The issue appears to be in your understanding of the separation function. You state that

  it calculates the separation (angle) between those 2 planets with respect to their "x" projection in a plane.

however, the documentation states

  The separation() function computes the angle in degrees between two bodies, as measured by their right ascension and declination. In general separation() can measure the angle between any pair of spherical coordinates, so long as the elements of each coordinate are spherical longitude (angle around the sphere) followed by spherical latitude (angle above or below its equator). Each pair should be provided as a two-item sequence like a tuple or list. Appropriate coordinate pairs include right ascension and declination; heliocentric longitude and latitude; azimuth and altitude; and even the geographic longitude and latitude of two locations on earth.

This means that this function is calculating the angular separation between the two objects, not just the separation in a particular plane. If you want to perform the same calculation (as the documentation says they're performing), it looks like you actually want the general angular distance formula.

$$\theta = \cos^{-1}\left[\sin(\delta_1)sin(\delta_2)+\cos(\delta_1)\cos(\delta_2)\cos(\alpha_1-\alpha_2)\right]$$

where $\delta\in[-\pi/2,\pi/2]$ is the declination, and $\alpha\in[0,2\pi]$ is the Right Ascension.


76mm reflector vs 70mm refractor? Both are at sale at $200

Personally, I would go for neither of those. $200 is a bit steep for such a small refractor or reflector telescope (I am assuming this is USD?)

For a budget of $200, I would be looking at some second hand telescopes, as you will likely be able to pick up a 90mm refractor or 130mm reflector second hand with that budget.

It also depends what you want to do with it. For simple observations of planets and the moon, it will be fine, but don't expect to see much detail with such a small aperture. The optics on these aren't fantastic either so DSO's will not look great, although some of the brighter ones will be viewable.

So my answer would be don't get either of them, shop around second hand places and most importantly, read reviews on telescopes before buying them. If you can, join an astronomy group in your area and have a look through their scopes and see how you get on.


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

$$D=a(t)D_0$$

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

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

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

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


How to calculate center (not focus) of planetary orbit?

I assume that the Sun is at the origin: (0,0).  Then we can look at this image of an ellipse:
The foci are at +-c, where c = ea, by the definition of eccentricity. To put the perigee on the negative y-axis, we can then say the center of the ellipse is at (0,ea).  Then, to account for the argument of periapsis, w, we need to rotate the ellipse (and ellipse center) counter-clockwise around the origin to get the coordinates for the ellipse center, (x,y) =(-easin(w),eacos(w)).  A better approximation would include the inclination, i, for a 2-D projection, but the inclination is small for Mercury so it may not matter for the purposes of your animation.  This diagram for the orbital elements should also help:


It doesn't bend it more. It is just that the matter of the star is now concentrated in a very compact region of space. So, previously you could get only to the surface of the star but now as the star is very compact, you can get much nearer and thus you will experience a greater gravitational force.

I had a similar discussion here. Check it out and maybe your question will be answered.


How exactly is the Initial Mass Function (IMF) calculated?

What is it?
An IMF, $\Phi(m)$, is defined such as $\Phi(m){\rm d}m$ gives the fractions of stars with a mass between $m - {\rm d}m/2$ and $m + {\rm d}m/2$, and with a normalized distribution
$$\int_{m_{\rm min}}^{m_{\rm max}}m\Phi(m){\rm d}m = 1\ M_{\odot}.$$
Note that these boundaries ($m_{\rm min}$ and $m_{\rm max}$) are ill-defined, but typically of the order of 0.1 $M_{\odot}$ and 100 $M_{\odot}$, respectively.
IMFs
The various IMF used are the following, with their main characteristics:

the Salpeter's IMF, that is a parametrization of the IMF by a simple power-law, of the form
$$\Phi(m){\rm d}m \propto m^{-\alpha}{\rm d}m;$$
the Miller & Scalo's IMF, that is a parametrization of the IMF by a log-normal distribution of the form
$$\xi\left(\log(m)\right) = A_0 + A_1 \log(m) + A_2\left(\log(m)\right)^2;$$
the Kroupa's IMF, that is a parametrization of the IMF by a broken power-law;
the Chabrier's IMF and Chabrier's system IMF, that is a combination of log-normal distribution (for low mass stars with masses less than 1 $M_{\odot}$) and and a power-law distribution (for larger masses). The difference between the IMF and the system IMF is to merge resolved objects into multiple systems to compute the magnitude of systems instead of individual stars.

Determination
As you see, all these IMF are parametrization, deduced from observations. In general, the observations used to infer these mass functions come from star clusters in our galaxy. All you need to do it is to find a mass-magnitude relationship to deduce, from an observed luminosity, a mass function. In general, the number density distribution per mess interval, ${\rm d}n/{\rm d}m$, takes the following form
$$\frac{{\rm d}n}{{\rm d}m}\left(m\right)_\tau = \left(\frac{{\rm d}n}{{\rm d}M_\lambda(m)}\right) \times \left(\frac{{\rm d}m}{{\rm d}M_\lambda(m)}\right)^{-1}_\tau,$$
for a given age $\tau$ and an observed magnitude $M_\lambda$. Then, it is just a matter of parametrization, but also of how well it can arise from a proper theory.
For this matter, the Chabrier's IMF is probably the one that is best back up by theoretical arguments. It relies on a gravo-turbulent theory, taking into account all the possible supports (thermal support, turbulent support and magnetic support) plus the dual nature of turbulence, that both favors star formation by compressing the gas, and impedes star formation, by dispersing the fluid. All the dirty details are given in Hennebelle & Chabrier (2008) and Hennebelle & Chabrier (2009), showing how you can analytically deduce an IMF from these theoretical considerations.
Applications
As far as I know, these IMF are more or less used for every type of population. However, you won't favor Salpeter's IMF if you have enough resolution to resolve low-mass objects, that are not at all well-taken into account with this IMF. You should also favor the Chabrier's system IMF in case of unresolved objects.
To know if all these IMFs are really well-suited to any kind of population is an open and difficult question (the so-called question of the universality of the IMF), in particular because you need to resolve individual stars in clearly identified clusters to deduce an IMF. There are some papers investigating the question (for example, you could have a look at Cappellari et al. (2012) for a recent discussion of the problem).


Exoplanet hunters ignored or ridiculed in the 1980s or 1990s?

Everyone expected exoplanets to exist - there was no reason to suppose that the conditions that existed when the sun formed were unique, and we knew that multiple star systems were common. Newton wrote "And if the fixed stars are the centres of similar systems, they will all be constructed according to a similar design"

Observations in the 80s were not ignored, they just lacked sufficient evidence. I.e, they were considered and properly rejected. Better observations have since confirmed some of the planets that were "discovered" in the 80s, but denied others. Scientific scepticism means that you don't accept something without evidence.

We knew that finding planets would be hard. The technology of sensitive ccd devices that could be used for constant monitoring of multiple stars didn't exist in a form that could be put on a space telescope, and there were other priorities for space science: solar system exploration.

Knowing that planets would be impossible to observe with contempory technology meant that "planet hunting" wasn't a thing. The first planet to be detected was around a pulsar, as the detection depends on changes in the timing of the pulsar, not on measurements of variations in light intensity during a transit, or doppler shift as the star is wobbled.


What does the orbit of multiple star systems look like?

In the case of Alpha Centauri A&B and Proxima Centauri,   A&B orbit each other at elliptical orbits, getting almost as close as the Sun is to Saturn and about as far as the Sun is to Neptune.   Proxima Centauri is about 1,000 times further out and it's a tiny compared to the inner 2, so for simulation purposes, it's essentially a 2 star system with a far distant 3rd orbiting the inner 2.    The 3rd star is so far and comparatively small that it has essentially zero effect on the simulation.

Source of Photo

Now, in general, most 3 or more star systems operate in similar ways where you can get a 1 or 2 stars operating as the center of mass and then you need to extend a fair bit of distance for the 3rd star.  3 stars all about the same distance and about the same size can be set up on a computer, but it's unlikely to happen in reality cause instabilities tend to grow.

The Sun-Earth-Moon is effectively a stable 3 body system and not subject to 3-body chaos because the Moon is safely within the stable region of the Earth's Hill Sphere and those 3 objects have clear cut hierarchy in terms of gravitational influence.

Pluto, Neptune& the Sun, for example, is probably long term unstable.   See here

This is Wikipedia's 3 body simulation, but this is probably not something you'd see in solar-systems because of the inherent instability.   In such a 3 body orbital system, one of the 3 bodies would likely either be ejected or 2 of the 3 bodies would collide, creating a more stable 2 body system.   It wouldn't have long term stability.

https://en.wikipedia.org/wiki/N-body_problem

You could potentially get a 3 body system where the 2 outer bodies are in Trojan points to each other (L4 or L5 only).   I'm not sure how common that is but I'd think it happens occasionally, provided the central star had significantly more mass than the 2 outer stars.

L4 and L5 Trojans


How early/late is the human race as intelligent life in the universe/milkyway?

I think this is a very subjective question depending on your point of view. If you assume that the history of the universe is 13.8 billion years old, then humans have only been around for 200,000 years which makes us very late at evolving. But of course the entire history of the universe hasn't been written yet!

I always find these sort of calendar/24hr clock representations very insightful too. If you look at this image below, modern "intelligent" humans have been around for the last 6 minutes of the year, if the age of the universe was 365 days long, so not very long at all!

That being said, I think there is another answer to your question, but becuase we don't know how long the universe/milky way will last for, its impossible to say or indeed ever know for how long intelligent life could actually evolve for.

I'm afraid i'm going to have to say somewhere in between too, as we simply cant see into the future.


Could a close passing star be captured by the Sun's gravity?

Unless the stars comes so close that they actually collide, two stars will not be able to catch each other gravitationally. The reason is energy conservation: As they approach each other, their potential energy is converted into kinetic energy, increasing their velocities. When they are closest, their velocities are at their highest, but since there's nothing to take out energy from the system, their kinetic energy will be converted back to potential energy, propelling them far apart again.

If a third star is present, however, this may be able to extract energy from the system, so that one star is slung out while the other two start orbiting each other.

Note however that the probability of two stars encountering each other extremely small, and that of three stars is close to zero.

In the above, I have ignored the presence of planets. In principle, a large planet could help, but since their masses are much smaller than the stars', their effect is quite small.


Can you see both Jupiter and Saturn (and their moons) in detail during their upcoming conjunction?

At their closest on the 21th of December (in the UK), they will be very close, only about 6 arc minutes.  That is in the field of view of most telescopes (about 1/5 of the diameter of the moon.
It will look like this (simulated)


Is there a paper on galaxy mergers in clusters of galaxies?

Galaxy mergers in clusters explain the large central galaxies in clusters.  In fact, the centers of clusters often host a special type of galaxy called a cD which is usually extremely massive because of galactic cannibalism.  Dynamical friction makes galaxies spiral down toward the center where they get eaten by the central galaxy and that in turn makes the central galaxy more massive and more efficient at this process.

In the rest of the cluster, the velocity dispersion is significantly higher than the escape velocities from individual galaxies.  So a) galaxies cannot gravitationally attract other galaxies because they go by too fast and b) even if there is a head on collision by sheer coincidence, they would pass through and not capture one another.  There may well be huge consequences to both galaxies, such as stellar rings and shells formed and tidal tails extracted, but the two remnants would separate and perhaps never meet again.

This is an oversimplification though because clusters often have subgroups of galaxies which have lower velocity dispersion wrt each other and these could merge much as galaxies in groups merge.  But the tidal field of the cluster actually works against this and so the rate would be lower.

There is a lot of literature on cD or central galaxies since it is one of the standard types of galaxies. Here is a recent one:

"Growth of brightest cluster galaxies via mergers since z=1", Claire Burke and Chris A. Collins, 2014, MNRAS, v. 434, p. 2856.
http://mnras.oxfordjournals.org/content/434/4/2856

A study of merger rates in the outer parts of clusters may take a major effort of going through the results of N-body simulations on your own and your answer may depend on which simulation you use.


How many planetary systems exist in our galaxy?

A good, but hard question to answer - as you stated, only an indication could be provided.  But, there is some information available from the number of exoplanetary systems found 'locally', according to the NASA web page How many solar systems are in our galaxy?:

  So far, astronomers have found more than 500 solar systems and are discovering new ones every year. Given how many they have found in our own neighborhood of the Milky Way galaxy, scientists estimate that there may be tens of billions of solar systems in our galaxy, perhaps even as many as 100 billion.

Another estimate from the article 160 Billion Alien Planets May Exist in Our Milky Way Galaxy (Hall, 2012), particularly with their quote from a scientist:

  "This statistical study tells us that planets around stars are the rule, rather than the exception," said study lead author Arnaud Cassan of the Paris Institute of Astrophysics. "From now on, we should see our galaxy populated not only with billions of bright stars, but imagine them surrounded by as many hidden extrasolar worlds."

An important caveat from the NASA site is that exoplanetary observations are still in its infancy - so over time, this estimate is likely to improve.


What specific parts of the Milky Way do we see when we look up?

But do we see any of the central bulge?

With the naked eye and a dark sky, the galactic center is brighter than other areas of the galactic plane.  Some of the light from that region does make it to earth to make that direction identifiable.

If you look at the ESO night sky composite from Brunier and Tapissier (which is approximately what someone could see with the naked eye given dark skies and good viewing), you can see that while the galactic center is brighter, other directions along the plane seem fairly similar.  That makes it very difficult to say what fraction of scattered glow is coming from one arm or another.  It all just blends together.

https://www.eso.org/public/news/eso0932/

  If all we can see are stars in our galaxy that are relatively close to us, then the Milky Way should look relatively uniform, right?

Not necessarily.  The visible part of the galaxy is relatively thin.  It doesn't have a fixed distance, but a commonly quoted one is 1000ly.  That means that even if light were magically cut off at 3000ly or so, it could appear non-uniform.


When we see half moon, why is it always the lower half?

Actually, you sometimes can see the Moon illuminated by the Sun from the top, and therein lies the answer to your question - during day. When looking at the Moon during night time, the side of the Earth where you're standing and looking at it is pointed away from the Sun, while this is reversed during the day. The illuminated side of the Moon is also facing the Sun, of course. So we have:

Night time on Earth: Side illuminated by the Sun is "below us" and that's where the Moon is illuminated from too.
Day time on Earth: Side illuminated by the Sun is "above us" and that's where the Moon is illuminated from too:

Blue Moon in the daytime sky (Photograph source: LPOD)

So, the direction from which the Moon is illuminated from is the same as the direction the Earth is, both illuminated by the same celestial body - our Sun. So, when you see the Sun high above the skies, and if the Moon is visible during daytime, it's going to be illuminated from the same direction from which you are, and during night time, when you'd expect the Sun "below where you're standing", it's illuminated from that direction.


Expansion again: where does the energy come from?

Conceptually there are several things going on here.

Where does energy conservation come from? In modern understanding, energy is the Noether charge of time translation symmetry, as found by Noether's first theorem.
 But in general relativity, the metric is dynamical, and so in general we simply don't have any time translation symmetry. Static spacetimes do, and there is also a form of energy conservation for spacetimes that regain time translation symmetric far away from the gravitating system (e.g. ADM energy of asymptotically flat spacetimes). But those are the exceptions, not the rule.

In other words, in general relativity we don't have any scalar notion of "energy" that's applicable globally. Vacuously, it is neither conserved nor violated.

But what about locally? In a local inertial frame, energy is exactly conserved, but the gravitational forces exactly vanish.

One thing you can do in the context of cosmology is look at the Friedmann equations as some sort of analogue of energy conservation, by making a balance between the terms describing cosmic expansion and the energy density, pressure, and cosmological constant. The Friedmann equations come from the components of the Einstein field equation connecting the Einstein curvature tensor and the stress-energy tensor: $G_{\mu\nu} = 8\pi T_{\mu\nu}$. According to this interpretation, Einstein curvature always exactly balances the stress-energy of the matter in spacetime. But this is just a restatement of a dynamical law, so it's not really a "conservation" law.

The Einstein field equation itself can be found from the Einstein-Hilbert action, and trying to apply Noether's second theorem simply shows that the covariant derivative of the stress-energy tensor vanishes: $\nabla_\nu T^{\mu\nu} = 0$, which is analogous to $\nabla\cdot\mathbf{B} = 0$ of electromagnetism: "there are no local sources or sinks of [stress-energy/magnetic field] anywhere." This is actually trivial, because the covariant derivative of Einstein curvature always vanishes (a theorem of geometry lacking any physics), so Noether's second theorem didn't tell us much more than we would have known otherwise.

Because the derivative is covariant rather than partial, many people don't consider this to be a true conservation law either. Certainly it gives no information as to "how much" energy there in spacetime--that's still undefined.

So we have the following issues:

There is no global energy conservation is general relativity, except for very special spacetimes, and the FRW family used for Big Bang models don't qualify.
In a local inertial frame, energy is exactly conserved, but there are no gravitational forces. (Local inertial frames only exist as first-order approximations anyway.)
One can interpret the Einstein field equation as Einstein curvature exactly balancing the stress-energy of matter, which is also motivated by interpreting the Friedmann equations of cosmology as a balance between cosmic expansion and local energy, pressure, and cosmological constant. However, this is actually a dynamical law.
The vanishing of the covariant derivative of stress-energy can be construed as an analogue of local energy conservation, although doing so is conceptually misleading.

Addendum: It's notable that there is yet another sense in which the total energy of the a spatially finite universe is exactly zero. Intuitively, one can try to measure the content inside some closed surface, and then expand that surface to try to enclose everything in the universe. However, for a closed universe, that surface will contract to a point, thus enclosing nothing (picture a circle around the north pole of the Earth, and expand it to try to enclose all of Earth's surface--it just contracts to a point at the south pole).

More formally, one can find a sequence of asymptotically flat universes (for which, again, the energy actually is defined) that approximate a spatially finite universe. In the limit in which the approximating universes "pinch off" and separate from the asymptotically flat region (thus becoming actually finite), the ADM energy goes to $0$.


Why don’t we feel the Earth rotating beneath us?

There are two different things going on, and they aren't too related (from what I can see). For the hot air balloon situation, you would think that if you hovered above where you are for 6 hours, the Earth could spin under you, and you would land back in a completely different place. Unfortunately, because the hot air balloon was on the Earth to begin with, it was already moving with the Earth. There are many reference frames in play, even though we feel like we are standing still. Everyone on Earth is standing still on the surface. However, the surface is rotating around the Earth's axis. The Earth's axis (and the Earth itself) orbits the Sun. The Sun is goes around our galaxy and our galaxy is travelling through intergalactic space.

So how does this relate to the hot air balloon situation? Well because the balloon was on the surface, it was already moving with the surface of the Earth. Remember how I said that the surface of the Earth rotates around the Earth's axis? Well, since the balloon was on the surface to begin with, it will also rotate with Earth's axis, just like the surface! What if we wanted to achieve that hover effect? I said earlier that the Earth orbits the Sun. In order to achieve that hover effect, we would need our hot air balloon to orbit the Sun without rotating around the Earth. You can't do this with a hot air balloon, since the atmosphere also happens to move with the Earth, and hot air balloons can't go into space. To achieve that hover effect, we would need some sort of spaceship with lots of fuel. If I had a hovering spaceship that didn't orbit with the Earth (this would, again, cost lots and lots of fuel), then yes, I could hover in the same spot and have the Earth rotate under me.

Now, you're probably wondering how I could achieve this with a spacecraft that was originally on the surface of the Earth. There wouldn't be much point in this hovering effect since you may as well just fly to your destination (like the theoretical Big Falcon Rocket), but if you wanted to do it, you would need to go above the Earth's atmosphere, use your boosters to go opposite of the rotation of the Earth (to cancel out your velocity), hover for some time, then use your boosters to go along with the rotation of the Earth (to return to rotation velocity), and land on Earth. Of course, you could skip the 2nd part by using heat shields and ramming into Earth's atmosphere like every other spacecraft we have, and no rocket would ever do this as it's just much more practical to orbit around the Earth (hovering costs fuel, orbits don't).

Ok, what about the sniper situation? The Earth rotate on its axis. Since a day is 24 hours long, a location at the equator goes around once in the same time it takes for someone in the arctic to go around once. However, the further north or south you are from the equator, the slower that part of the Earth has to rotate to complete 1 rotation in 24 hours. Think about spinning a ball. The equator of the ball whips around, but the top and bottom move much slower. It's the same thing. Lets say my sniper is at the equator. When the sniper shoots east or west, they do not need to correct for the Earth's rotation because everywhere along that latitude, the Earth is rotating at the same speed. However, if the sniper shoots north, the bullet will go to the east. That's because when the bullet was shot at the latitude closer to the equator (shot from the south), that spot of the Earth was moving faster than the spot of the Earth that the target was standing on. It's like saying that the spot I'm standing rotates at a speed of say... 1000 mph, but my target's spot rotates at say...995 mph. Since my bullet came from the 1000 mph part, it will obviously go out of the barrel at high speeds because of the gun, but it will also be rotating around the Earth's axis at 1000 mph. However, once my bullet starts approaching my target's latitude, since the Earth rotates slower, it will appear to deviate in the direction of the rotation (which is east, or to the right). Why? Well since my target's speed is 995 mph from rotation, you do 1000 - 995 = 5. That means that my bullet will have a net velocity of 5 mph to the right relative to my target. That means my bullet will miss if the target is far away enough. If you shoot from the equator to the south, the bullet will also go east, so as a result, it will deviate to the left. It's the same effect, but "upside down". This effect is called the Coriolis effect, and it's what gives hurricanes their power.

Finally, you asked "why don't we feel the Earth rotating beneath us?" This is because we are moving with the Earth. When you are on a train that is going at a constant speed of 50 mph, you do not feel it move (you may feel some bumps due to bumpy tracks). It is only when you accelerate or decelerate do you feel things move. When you are on the train, your speed is constant, so you do not feel anything. It's the same for the Earth, except you are rotating around the Earth's axis at a constant 1000 mph. Nothing has changed, except the speed is faster.

I probably explained this horribly so feel free to ask anything.


The scale height of gas in a disk (if it were in equilibrium) is roughly $kT/mg$, where $T$ is the temperature, $g$ is the gravitational field, $m$ the mean mass of agas particle, and $k$ the Boltzmann constant.

If we assume most of the mass is in a thin disk, then Gauss's law for gravitation tells us that that $g = 2\pi G \sigma$, where $\sigma$ is the mass per unit area in the disk. According to Rix & Bovy, $\sigma \simeq 70 M_{\odot}$ pc$^{-2}$ at the location of the Sun (http://arxiv.org/abs/1309.0809).

If we assume hydrogen gas, then the effective particle mass is that of a proton, and this means the gas scale height is
$$ H = 4300 \left(\frac{T}{10^6\ K}\right)\ pc$$

Thus gas hotter than a million degrees will have a very substantial scale height and is not expected to be confined to the Milky Way disk.


Why does the moon appear on cloudy night sky?

Clouds are obviously not completely opaque, and what you are seeing is a break in the clouds. It is of course possible for the clouds to cover the moon, however, often the moon can shine brightly through the clouds and illuminate the clouds in some nice way.

This video might help in a visual way: https://www.youtube.com/watch?v=aLd_yhIWE48


Astro comes from the greek άστρο which not only means star but has a more general meaning that includes any small dot in the night sky (see https://en.wiktionary.org/wiki/%CE%AC%CF%83%CF%84%CF%81%CE%BF#Greek for details)

In today's language it still has a quite broad meaning:

Oxford English dictionary (https://www.oed.com/view/Entry/12240?isAdvanced=false&result=2&rskey=LlaQB9&):

  Forming terms relating to stars, other celestial objects, or outer space.

Merriam Webster (https://www.merriam-webster.com/dictionary/astr-)

  : star : heavens : outer space : astronomical

Conclusion: Sheldon and Raj are right


Where exactly is the modification that first corrected the spherical aberration in Hubble's primary mirror?

The image provided is of COSTAR which replaced the High Speed Photometer and fed corrected light to the Faint Object Camera (FOC), Faint Object Spectrograph (FOS) and the Goddard High Resolution Spectrograph (GHRS). Since the WFPC2 was already going to replace WFPC1, the correction for the primary mirror's spherical aberration could be designed in (as it has been for all subsequent instruments).

The overall layout of the HST optics (the Optical Tube Assembly (OTA)) and the WFPC camera optics is shown below in a figure taken from the Rodgers and Vaughan 1993 SPIE paper:

Light enters from the top left and is focused as normal by the HST primary and secondary mirrors and then part of the beam is picked-off into the WFPC2 camera.

The WFPC2 optical configuration is shown in more detail below (taken from the instrument description section of the WFPC2 Instrument Handbook) . The text just above that states:

  The aberrated HST wavefront is corrected by introducing an equal but
  opposite error in each of the four Cassegrain relays. An image of the HST
  primary mirror is formed on the secondary mirrors in the Cassegrain relays.
  (The fold mirror in the PC channel has a small curvature to ensure this.)
  The spherical aberration from the telescope's primary mirror is corrected
  on these secondary mirrors, which are extremely aspheric.

indicating that the correction for HST's primary is done on the part labeled 'Secondary Mirror' in the diagram. Note that is the secondary mirror of the mini-Cassegrain telescope that is inside each of the 3 widefield and 1 planetary camera channels inside WFPC2 itself and not the main telescope secondary mirror which is off the diagram and part of the 'OTA' (Optical Telescope Assembly)


Fate of helium moving toward a singularity

I'll take the reference of an observer falling with the helium atom. At the Event horizon, the tidal force across a helium atom is still minute, $10^{-29}N$ much less than the Coulomb forces between the electrons and nucleus, still less than the strong forces inside the nucleus. So the atom will fall through the event horizon without damage. (The region around the black hole can be a pretty violent place, with lots of hard X rays from the superheated matter, which would likely have ionised the atom, but I'll suppose it somehow manages to avoid this)

As it falls past the event horizon it is doomed. Think of the singularity less of a place in the middle of the black hole, and more as a point in time. It is inevitable our He atom will reach that point. As it approaches it the tidal forces will grow larger, eventually sufficient pull the atom apart. To describe what happens now would require a good description of quantum gravity, which we don't have.

The is no physics that describes what happens "after" the singularity. Think of the singularity as a point in time, but for the atom it is also the end of time for the atom. It is possible that something beyond our current understanding prevents a singularity from forming, but that is speculation.

This is explained diagramatically in a Penrose diagram


Generate an uniform distribution on the sky

Random points on the surface of a sphere can be generated by allowing the azimuthal angle $\phi$ to take a uniformly distributed random value between 0 and $2\pi$. To convert this to RA in degrees you multiply by $180/\pi$. To convert to hours, minutes and seconds you divide the $\phi$ in degrees by 15, which gives the hours, divide the remainder by 60 which gives the minutes and then divide the remainder of that by 60 to give the seconds.

The declination angle $\delta$ is not uniformly distributed, because the area covered at an angle $\delta$ goes as $\cos \delta$.

Note the potential confusion here between declination and the polar angle $\theta$, which is conventionally measured down from the pole. The area of a strip of width $\Delta \theta$ at a given $\theta$ on a unit sphere is $$\Delta A = 2\pi  \sin \theta\ \Delta \theta = 2\pi  \cos (\pi/2 - \delta)\ \Delta \delta = 2\pi  \cos \delta\ \Delta \delta$$

To translate this into something we can use, we say that the probability of finding a star in a small range of declinations
$$ dP \propto \cos \delta\ d\delta,$$
so the cumulative probability of finding a value of $\delta$ between $-90^{\circ}$ and $\delta$ is
$$ P(\delta) = \frac{\int^{\delta}_{-90^{\circ}}\cos \delta^{\prime}\ d\delta^{\prime}}{\int^{+90^{\circ}}_{-90^{\circ}}\cos \delta^{\prime}\ d\delta^{\prime}},$$
where the denominator makes sure the proportionality is correctly normalised. From this, we obtain
$$P(\delta) = \frac{1}{2}(1 + \sin \delta) \tag*{(1)}$$

The procedure for then obtaining a random declination is to assign a uniform random number to the cumulative probability $P$ between 0 and 1. Then from equation (1) we have
$$ \sin \delta = 2P -1$$
$$\delta = \sin^{-1}(2P-1),$$
where $P$ is a random number between 0 and 1.


Could photons orbit a black hole if fired at the correct angle?

There are no stable photon orbits around black holes (see this answer).
Black holes do have an unstable photon orbit though. "Unstable" here means that any tiny deviation from this orbit, will case the photon to be either scattered to infinity or sent into the black hole. It should be emphasized that there such orbits are always circular an can exist at only one radius (1.5 times the Schwarzschild radius for a Schwarzschild black hole). There are no elliptic photon orbits. This orbit is often called the "light ring" (or sometimes "photonsphere")
As long as we treat the photon as a perfect massless point (test) particle, we can find a highly fine-tuned trajectory that will asymptote to the light-ring. Of course, in reality, a photon always needs to be a wave packet with a finite size, making this fine tuning impossible in practice.


Yes! I can do it for K2 and I'm guessing it should be similar.
You can see with fits.info(fits_file) that it has the 84 extensions (21 modules * 4 channels per module) you mention.
Then you can select a channel, e.g. 65, and prepare also the wcs coordinates so you get the RA and dec associated with the image for the axis.
n = 65

with fits.open(fits_file, mode = "readonly") as hdulist:
    wcs_info = WCS(hdulist[n].header)
    cal_image = hdulist[n].data
    header = hdulist[n].header

Then you can just plot:
plt.figure(figsize = (15,15))

fig = plt.figure(figsize=(12, 8))
ax = fig.add_subplot(1, 1, 1, projection=wcs_info)
im = ax.imshow(cal_image, vmin = np.percentile(cal_image,1),vmax = np.percentile(cal_image, 98), origin='lower')

ax.coords[0].set_ticklabel_position('l')
ax.coords[0].grid(color='white', ls='dotted')
ax.set_ylabel('Dec')
ax.coords[1].set_ticklabel_position('b')
ax.coords[1].grid(color='white', ls='dotted')
ax.set_xlabel('RA');


What is the nature of bright spots found on Uranus?

Near-infrared pictures from the 8.1-meter Gemini North Telescope in Hawaii discovered a bright spot on Uranus. The spot is likely a tall methane cloud that reached high enough to become visible due to sunlight being scattered by its icy particles. The Uranian cloud is similar to an anvil cloud, a type of cumulonimbus cloud that's associated with severe thunderstorms. The cloud is also at a lower latitude on Uranus that could mean the spot is a storm that has migrated south.

Nobody knows how were the storms created. Some has speculated that the planet's tilt is the reason for such storms.
As Uranus is tilted on its axis is tilted, it virtually rolls around the Sun on its side. For half the orbit's period, the northern hemisphere is directed towards the Sun and the other half, the southern hemisphere is turned towards the sunlight. At two points in its orbit, known as the equinoxes, the Sun shines equally on both hemispheres. At these equinoxes, storms appear to become more frequent in Uranus’s disturbed atmosphere. Stormy weather has been observed a number of times in recent years, thus creating a notably bright spot appearing in 2014.
Reference:

https://www.nationalgeographic.com/news/2011/11/111021-uranus-planet-new-spot-storm-methane-gemini-space-science/
https://www.skymania.com/wp/mystery-bright-spot-appeared-on-uranus/
https://www.planetary.org/articles/08221504-new-spots-uranus
https://skyandtelescope.org/astronomy-news/bright-spot-uranus-atmosphere-11252014/


Is a swimming pool analogy to building of black holes in the centers of galaxies valid?

The process of an eddy gathering matter to the centre of a pool is due to the motion of water in a vortex. Roughly, if water is rotating  counterclockwise then a secondary motion will be set up, rotating clockwise about an axis parallel with the tangential motion of the water. This tends to push debris towards the centre of the vortex. (The same effect causes deposition on the inside of meanders)

Space time doesn't work like this. There is very little spinning of spacetime around a galaxy, the amount of frame dragging is very small. And spacetime doesn't follow the Navier–Stokes equations, so there is no true vortex. The debris flows with the water, where as matter travels through spacetime. Your pool has a clearly defined "up", the galaxy doesn't.

So while the appearance is somewhat similar, the mechanism is different. Though we don't really understand the mechanism of formation and growth of supermassive black holes.


What would Earth be like without the Moon?

Destroying the moon is not easy. It's pretty big. Has quite a lot of gravity and is sitting in the Earth's gravity well. It gets hit by a lot of asteroids and doesn't get destroyed.

The moon has a binding energy of about $10^{29}$ joules. That is a lot of energy. Even if you got the largest asteroid (ceres) to collide with the moon you would not get close to destroying it.

You would need something like the full power of the sun for about 15 minutes.

If something massive enough to disrupt the moon were to hit it, the energy of that explosion, and the massive amount of moon rocks that would subsequently hit the Earth would probably be enough to melt the surface of the Earth. It would be the end of life on Earth.

There is another way to get rid of the moon. Get a wizard to magic it away. The effect of not having a moon would then be pretty minor in the short term (no tides) and fairly significant in the long term. Human life could continue as normal for a while.

The moon stabilizes the rotational axis of the Earth, without the moon the angle would vary more (over thousands of years) this might mean more extreme ice ages and perhaps ultimately a snowball Earth scenario.


The amount of metals locked up in planets is completely negligible compared to that of the interstellar medium (ISM).

First, in a typical stellar system, the mass fraction of planets is less than 1%, and of this by far the most mass will be in gaseous planets, rather than rocky planets. For instance, in the Solar System, the mass fraction of all planets is 0.1%, while rocky planets comprises only a fraction of $6\times10^{-6}$.

Second, typically only a small fraction of the gas in a galaxy is locked up in stars. For small galaxies — which are the most abundant — as little as ~5% of the gas is in stars (e.g. Schombert et al. 2001), although for large galaxies like the Milky Way, which are more efficient at converting gas into stars, the ISM-to-stellar gas fraction can be of order unity (e.g. Morokuma-Matsui & Baba 2015).

That is, the fraction of metals locked up in planets is of the order $10^{-5}\text{–}10^{-7}\!$, depending somewhat on the type of galaxy considered.

On the other hand, the fraction of metals locked up in dust is considerable, with typical values of roughly 1/3 of the metals in dust. Thus, if you want to know the metallicity of a galaxy by looking at emission or absorption lines from metals, you must either account for this factor (hoping that it's a universal factor which it may or may not be, see e.g. Zafar & Watson 2013), or measure metals that tend not to deplete to dust grains, e.g. zinc or sulfur.


Are Tharsis Montes and Hellas Basin a result of the same event?

You were not the first to have seen that the Hellas Basin and parts of the Tharsis Rise are roughly antipodal (Peterson 1978, Williams and Greeley 1994). The Tharsis Montes themselves are far too close to the equator to be considered antipodal to the Hellas Basin. The shield volcano Alba Mons however is almost exactly antipodal to the Hellas Basin.

This shield volcano is a bit to the north of the Tharsis Rise proper, and well to the north of the Tharsis Montes. Both Peterson and Williams suggested that the Hellas Basin impact might well have triggered the vulcanism that resulted in Alba Mons.

There's an issue with this hypothesis, which is that the apparent ages do not align. Improved remote observations of Mars has provided a tool for estimating ages of Mars features: crater density. Based on crater density, the Hellas Basin appears to be very old, at least 3.8 billion years old (Carr 2010). Alba Mons, while also quite old, isn't that old. It formed at least 200 million years after the Hellas Basin impact and perhaps as much as a billion years after (Ivanov 2006).

Bottom line: If the huge impact that formed the Hellas Basin did have any antipodal effects, those effects are buried under the magma that formed Alba Mons several hundred million years later.

References:

Carr, Michael H., and James W. Head III. "Geologic history of Mars." Earth and Planetary Science Letters 294.3-4 (2010): 185-203.

Ivanov, Mikhail A., and James W. Head. "Alba Patera, Mars: Topography, structure, and evolution of a unique late Hesperian–early Amazonian shield volcano." Journal of Geophysical Research: Planets 111.E9 (2006).

Peterson, J. E. "Antipodal effects of major basin-forming impacts on Mars." Lunar and Planetary Science Conference. Vol. 9. 1978.

Williams, David A., and Ronald Greeley. "Assessment of antipodal-impact terrains on Mars." Icarus 110.2 (1994): 196-202.


Is parallax a "technique to measure the star distances" or a "tiny shift in star's position"?

It is both - a small shift of the position of a star on the sky as we see it, and a means of estimating the distance to the star.

The apparent position (with respect to very distant objects like quasars) changes because our viewing point changes as the Earth moves around the Sun in its orbit. The amount by which the position changes is inversely proportional to is distance.


does the orientation of orbital plane of our solar system rotate with the galaxy

The orbits in our solar system add up to a certain total angular momentum, which is a vector so points in some direction.   To get that vector to point some other way requires torque, it won't happen just because the solar system is orbiting the galaxy. So the "precession" rate of the plane of our solar system is controlled by very different things than the orbital rate, so there would never be any reason for picture #1.

Picture #2 is what you get if there is no torque, so at least there's a reason for it.  The kind of torque you'd need would be a force that points up from the plane on one side of the orbit and down from the plane on the other side, but a force like that would depend on distance from the Sun so it would be different for different planets.  That would tend to make their orbital planes undergo nodal precession at different rates, so they would not be in the same plane. So either such torques have had no effect over the lifetime of the solar system, or else something else shepherds the planets into the same plane.  But it can't be the latter, because Neptune is in the plane but Pluto isn't.  So I think it must be picture #2.


How does the Milky Way look like above 66° North and below 66° South?

On midnight, right around this time of year, the Milky Way will be in the zenith. Here is an XEphem rendering for the north of finland (65th latitude) for yesterday midnight (the brown outline marks the Milky Way):

You can also see this on photographs by finnish photographer Tommy Eliassen. He has many examples of this on his website. I won't put any in here, because I guess they are copyrighted, and not free to use.

This is an animated 24 hour version of the above sky map, which shows the rotation of the milkyway around the zenith:


Why do astrologers use the “wrong” sign?

Precession of the equinox, caused by the cyclical change in the direction of the Earth's tilt, causes the position of the vernal equinox (the point where the path of the sun crosses the celestial equator) to move over time.  When the constellations were first described, the vernal equinox was in Aries, and so it is still called "The First Point of Aries".  Now it is in Pisces and moving towards Aquarius. Astrologers use "signs" named according to the position of the vernal equinox when the constellations were first described.  So if a planet is at the vernal equinox (and therefore in Pisces, close to Aquarius) an Astrologer in the Western tradition will say that planet is "in Aries".
So for an astrologer, "Aries" is not a pattern of stars in the sky nor the region of the sky surrounding those stars, but the strip in the sky between 0 and 30 degrees of longitude, even though that region no longer contains the stars of the constellation Aries.


In principle yes, but in practice no.

At the so-called photon sphere, gravity is exactly so strong that a photon on a tangential trajectory would stay in orbit. For a non-rotating black hole of mass $M$, the radius of the photon sphere is $3/2$ times the radius of the event horizon — i.e. the "surface" of the black hole — which itself is given by $r_s \equiv 2GM/c^2$, where $G$ and $c$ are the gravitational constant and the speed of light, respectively.

However, the orbits are unstable; any perturbation will cause a photon on the photon sphere to either escape or plunge into the black hole.


Does the mass of a collapsing star change?

This is an old question, but I'd like to answer it anyway.

  Does a star, when collapsing into a neutron star, or even a black hole, pick up any significant relativistic mass due to its rapid collapse rate?

No. Conservation of energy applies. When you drop a 100000kg asteroid from some high elevation above the Earth, some of the asteroid's gravitational potential energy, which is mass-energy, is converted into kinetic energy. The mass-equivalence of the system does not increase as the asteroid falls. Instead the asteroid's rest mass reduces as the relativistic mass associated with its kinetic energy increases*.

When the asteroid hits the Earth (BOOM!) some of the kinetic energy will get dissipated and radiated out into space, and then you're left with a mass deficit. The mass of the system is then less than it was. See Wikipedia: "This missing mass may be lost during the process of binding as energy in the form of heat or light, with the removed energy corresponding to the removed mass through Einstein's equation E = mc²".

The same applies to a collapsing star. There is no "pick up" of relativistic mass. Conservation of energy applies. But some of this energy is typically radiated away, so you're left with a reduced mass.

The same also applies to a brick. When you drop a brick, gravity converts some of its rest mass into kinetic energy. This gets dissipated, and you're left with a mass deficit. Then when you lift the brick you do work on it. You add energy to it, thereby increasing its mass. Then you're back where you started.

The same also applies to an electron.

* this also applies to the Earth, but to a much lesser extent, so we tend to forget about it.


Do planetary magnetic fields slow the movement of conductive extraplanetary moving objects via induction?

A fixed magnetic field will induce an electrical current in a conductive object moving through the field.

No. It will induce an EMF in one direction as described in HyperPhysics. It will not induce a current since there is no return path.

This is different from induction braking, where you have a magnet with a non-uniform field moving near a conductor. It is the changing magnetic field at points within the conductor that induces the EMF and causes currents.

Alternatively, we can consider how electromagnetic fields transforn according to Relativity. If an object is moving at constant velocity through a uniform B field then, in its rest frame, it will experience a combination of uniform E and B fields which would not produce a current.  (Any current would have to be due to the spacial rate of change of the field, which would be tiny on the scale of the Earth's field.)


Should we expect clues from a gamma ray burst causing massive extinctions on Earth?

A gamma ray burst could cause an extinction event. There is no evidence that this has ever happened to the Earth.

Gamma ray bursts are rare events, we don't know exactly how rare, but to cause a mass extinction a GRB would need to be near (6000 light years or less is near in this context) and pointed right at Earth. GRBs are rare enough for it to be possible that this has never happened. It is possible that one or two of the extinctions in the last billion years were due to GRBs

The mass extinction that is most likely to have been caused by a GRB is the Orodvician-Silurian Event, 444 million years ago. There is no positive evidence of a GRB, but it can't be excluded. The best that can be said is that the pattern of extinction is consistent with a GRB.

A gamma ray burst would leave few signs that could be used to prove it occurred. The radiation would strip the ozone layer, leaving the surface vulnerable to UV radiation. But it wouldn't leave a crater, or a geological layer.

Gamma ray bursts can occur when a hypernova occurs. There is no reason to think that the Earth is more vulnerable to a hypernova when it is in the galatic plane. The evidence of a 26 million year periodicity to mass extinctions is weak, and there is no evidence that closely ties this to the sun's orbit around the milky way. There is good reason to tie other extinctions to particular impacts, for example there is no need to hypothesise that a GRB occurred 65 million years ago, at the K-T mass extinction

Gamma ray bursts are one of the less likely causes of a mass extinction. Researcher David Thompson likens them to the danger of finding a polar bear in his closet. It could happen; it would be dangerous, but the chance of it happening is so low that it is not a risk worth worrying about.