Universe Sandbox 2 accurate representation of gravity

A NEO only impacts Earth if both bodies are in the orbit intersection zone at the same time, closely enough for gravity $(F \propto 1/r^2)$ to bring them in contact.
In a simulation, the timestep should also be short enough to detect a collision.
An object at a moderate relative speed of 10 km/s crosses Earth's diameter in about 20 minutes; a longer timestep could turn a hit into a miss.

If you're already modeling real asteroids, try some which have hit Earth:
2008 TC3,
2014 AA,
2018 LA.
The short observation arcs make their pre-impact orbits highly uncertain, so a few small bodies spread within $\pm\sigma$ of each orbital element value may improve your chances.
Alternatively you could set an object up for immediate collision with Earth at a relative speed of 12 to 20 km/s in any direction, and run the simulator backward to see how it would get there.

The asteroid risk is real, but we should not overstate it.
If you visit JPL CNEOS Sentry and use unconstrained settings, it shows only a few objects with more than 1 chance in 1000 to hit Earth in the next 100 years, and most of those are small.
Few NEO orbits are known precisely enough to make meaningful predictions beyond that.
The Palermo scale quantifies a given object's threat to Earth in the foreseeable future.


Orange light that rises quickly and moves of to the righ

The object is (most likely) Mars! It's been rising in the south as of late. When it will rise of course depends on your location...for me it rises pretty close to sunset. Planets appear in an essentially constant position relative to the background stars over the course of the night, and as with the stars, the movement you see is due to the rotation of the Earth. Over the course of weeks, however, you can see the planets change position relative to the stars due to their motion in their orbit around the Sun. We don't see this wandering of position with stars - at least, not in an amount or on a timescale we can appreciate - because they're so far away.

Keep an eye out for other bright objects in the sky that appear to move faster and (unless conditions are poor) don't really twinkle, especially near sunset before the stars properly come out - Mars probably isn't the only planet you can see! Jupiter, Saturn and Venus are often visible as well.

Check out this site if you want to see what planets are visible from your location and when.

https://www.timeanddate.com/astronomy/night/

ETA: Corrected explanation of planetary motion vs. motion of stars based on James' comment.


What are the chances of a star colliding with another during a galactic collision?

Let's see what we get from some back-of-the-envelope estimates.

Imagine throwing one star (e.g., the Sun) at the other galaxy. How likely is it we'll hit a star in the other galaxy? Well, it's basically proportional to how big a target each star in the other galaxy is (its cross-sectional area) compared to the size of the whole galaxy, multiplied by the total number of stars in the target galaxy.

Let's assume it's the Milky Way-Andromeda scenario, so each galaxy has about 100 billion stars, and each star is roughly the same size as the Sun (some are much larger, most are smaller). The actual target area for an individual star is a circle with twice the star's radius (we're counting one star just grazing the other as a collision). Let's also assume the stars are more or less evenly distributed in a circular disk. Since "100,000 light years" is a common (and not completely crazy) estimate of the Milky Way's size, that's a circle of radius = 50,000 light years (about $10^{16}$ meters).

So: 100 billion stars in the target galaxy, each with target radius $\sim 2 R_{\odot}$, gives us a total target area of $10^{11} \times \pi (2 R_{\odot})^{2} \approx 10^{30}$ m$^{2}$.

The area of the target galaxy is $\pi R_{gal}^{2} \approx 10^{42}$ m$^{2}$. So the chance of our Sun hitting a star in the other galaxy is $\approx 10^{30} / 10^{42} = 10^{-12}$ -- or about one in a trillion.

The odds of any star from our galaxy not hitting a star in the other galaxy would be $(1 - 10^{-12})^{10^{11}} \approx 0.90$.

So there's only about a 10% chance of one (or more) of the galaxy's 100 billion
stars hitting a star in the other galaxy. And the chances of any one particular star (like our Sun) hitting a star in the other galaxy is about one in a trillion.


How to calculate the time and distance of closest approach of a star?

This is not really an astronomy so much a basic Maths problem.

Put the Earth at the origin and the star at $\vec{r}(t)$ with a position $\vec{r_0}$ at $t=0$. Define your axes such that the star is in the z direction and its tangential motion components are along the x and y axes.

$$\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k},$$
where these components are the two proper motion velocities and the radial
 velocity.

Assume no acceleration, so that
$$\vec{r} = \vec{r_0} + \vec{v} t $$

Use Pythagoras to get the distance to the Earth using the components of $\vec{r}$ at any time $t$ and then find the time at which this is a minimum (and if you are unsure how to do that $\rightarrow$ Maths SE).


Could life survive on a planet orbiting a neutron star?

It is possible that there could be planets where terrestrial-type life could survive orbiting a neutron star, but they are likely extremely rare and transient.

The habitable zone where liquid water can persist around a star has a range $\approx 0.7\sqrt{L_*}$ to $1.4\sqrt{L_*}$ AU, where $L_*$ is the luminosity of the star measured in solar luminosities. The exact limits are debatable and depends somewhat on the planet's properties, but this is close enough for this answer. A typical neutron star has a low luminosity, so the habitable zone will be very close and narrow.

Neutron stars cool down very rapidly at first. A young, million-Kelvin neutron star has luminosity around 0.18 solar luminosities, but a 10 million year old neutron star is down to 0.000026, and after a billion years it will be down to a millionth of the sun. This rapid cooling means that the life zone moves inwards from around 0.3-0.6 AU for the young star to 0.003-0.007 for the 10 million year star to 0.0007-0.0014 AU for the old star.

This means that a planet orbiting inside the zone at some time will become too cold before too long unless it is very close to the star since the zone drift slows down over time. It is hence unlikely that life would naturally be able to emerge on a planet like this, since planets that are in the habitable zone early only get a few million years before they freeze. The close planets that get a long period with decent temperature also have a period of extremely hot temperature before that, when most likely all volatiles are boiled away.

(Given that such a system has experienced a supernova in its past, volatiles will have been reduced even further).

That said, I don't see a fundamental reason a close planet might not be terraformed with volatiles added from elsewhere. If we consider a planet orbiting a $L_*=10^{-6}L_\odot$ neutron star $10^{-3}$ AU away (that is, 149,597 km from the star) it could maintain water. It would have a period of 16 minutes. There would be some relativistic corrections to the orbit, but they are minor (we are $10^5$ Schwarzschild-radii away).  It would be tidally locked (since tidal locking happens on a timescale $\propto a^{6}$), but currently there are a fair number of atmosphere models suggesting such worlds could remain stable/ Living on the shady side would also prevent the damaging UV and X-rays, which would likely strip away the atmosphere over time. The main problem is that there would not be any geodynamic dynamo to power a protective magnetic field; this is a planet that really needs it.

It is worth noting that there is a very different system around a neutron star that in principle (but likely not in practice) may be habitable, and that would be far away from an accretion disk. A quiescent accretion disk may have a luminosity about 1/3 of the sun, but when going active this increases by a factor of 1,000-10,000. So here one could imagine a planet orbiting at distance, getting powered by the accretion disk light. This will likely be too unstable to actually work over long periods and have the same hard radiation problem as the close one, but on paper (or in a suitable work of fiction) it is not impossible.


How do we know that other planetary systems aren't composed of antimatter?

There are two major points about this, I think.

First, and ultimately most importantly, is that if we have matter here, and they have antimatter over there, then somewhere in between we must have a region that transitions from matter to antimatter.  Even if this region is located in the intergalactic medium, where densities are typically very low, collisions between matter and antimatter would happen often enough that there would be a tell-tale radiation signature from this region.  To date, no confirmed detections of such a region have been made.

Now for the second point, it is not actually true that matter and antimatter have the same observable physics, though the observable differences are hard to detect even within our own laboratories.

Electromagnetism (classical or quantum), gravity, and the strong force all obey C-symmetry, and so you would not see any observable differences between matter and antimatter with these forces.

However, C-symmetry is not preserved in the weak interaction/weak force.  It is also known that parity symmetry can be violated in weak interactions.  Parity symmetry, loosely, means reflecting all particles in a mirror, and violation of it means that something has "handedness": you can tell your left hand from a right hand, and the mirror makes your left look like a right.  Remarkably, there is reason to believe that the weak force also violates the combined CP symmetry.

So the weak force introduces this sort of handedness into particle physics.  It does so with the neutrinos and anti-neutrinos in particular.

Now this handedness of neutrinos is something that could, in principle, let us detect that a mass is composed of antimatter.  Beta decay is a result of the weak force and emits neutrinos or anti-neutrinos, depending on whether the original atom is matter or anti-matter.  So theoretically we might tell a planet is antimatter if it has some radioactive material undergoing beta decays.

Except that will actually be pretty impossible.  The scale of even an entire Jupiter's worth of radioactive mass would just be too small a source of neutrino emissions.  Current neutrino detectors confirm on the order of a handful of neutrinos that probably originated from outside of our solar system every decade.  And those aren't really looking for neutrinos from any particular area; they'll look for any neutrino coming in from anywhere in the universe (well, where you are on the earth matters a bit).

So our current technology basically isn't good enough to detect these weak sector asymmetries from sources outside of our solar system.

Otherwise, assuming some improvements, a lot of time, and/or one heck of a lot of luck, perhaps our best shot at confirming a large source of antimatter comes from supernovae.  Supernovae are very, very bright (they will outshine the entire galaxy they occur in), and therefore very noticeable.  They also emit tremendous numbers of neutrinos very quickly.  So if we chance to notice a supernova, and if we luck into our neutrino detectors having enough neutrino detections at the right time to correlate to that supernova, then if somehow that was an anti-matter star we might be able to tell due to detecting the difference in the neutrinos.

But that's really unlikely.  We're substantially more likely to detect the energy signature of annihilation events at the matter/anti-matter interface, as mentioned in the first point.

See also this Q&A, which proposes using polarized light as a means of detection.


Finding the lens diameter of a telescope by magnitude

The unaided eye can typically see mag 6 objects. With your telescope, you can see an additional 8 magnitudes. This requires a factor of $100^{8/5}$ of additional light gathering power (since 5 magnitudes equals 100 times the brightness). I.e. about 1580 times larger aperture than the pupil of the eye.  Assuming the pupil has a diameter of 7mm, then the diameter of your telescope would have to be $7\times 1580^{1/2}$, which is about 280 mm.

This is fairly close I think - perhaps a bit on the high side but not much.


How to explain this trajectory of the moon taken from the same spot on Earth over 28 days?

This is a digitally enhanced composite set of images of the Moon, with successive images taken 24 hours and 40 minutes apart. There's also a long gap in time between the left half and right half of the composite. The creator of the photograph took artistic license to enlarge the Moon; she thought that that made the composite image look better. Images of the new Moon / nearly new Moon are also enhanced.
Had consecutive images been taken 24 hours and 51 minutes apart the resulting composite would form a more or less closed curve. Successive images were instead taken 24 hours and 40 minutes apart so as to intentionally create a non-closed curve. The first image is at the top right while the last image is at bottom left.


At what wavelengths can black hole Sagittarius A* be observed from Earth?

From Genzel et al. (2010), here's part of Fig. 7.7.1:

This is part of the spectral energy distribution of Sagittarius A*, a flot of $\nu$ (frequency) vs. $\nu L_{\nu}$ (frequency times luminosity). For comparison, visible light is in wavelengths from $\sim4\times10^{14}\text{ Hz}$ to $\sim8\times10^{14}\text{ Hz}$, which happens to be around the bottom of the trough of nonthermal electron emission. Right away, this makes submillimeter and millimeter wavelengths good candidates, leading to studies using Very Long Baseline Interferometry. Likewise, infrared emissions are a good target, and so the Spitzer Space Telescope, for instance, has been used. X-ray flares of up to $\sim10^{36}\text{ erg/s}$ also occur from time to time,1 so that part of the spectrum is sometimes used to observe this activity. Finally, of course, Sagittarius A* is a very strong radio source, and it was initially observed in radio wavelengths (and it still is!).

As Fish & Doeleman 2010 write,2

  Interstellar scattering, which varies as $\lambda^2$, dominates over intrinsic source structure at longer wavelengths, and the emission from Sgr A* transitions from optically thick to optically thin near $\lambda = 1\text{ mm}$ (Doeleman et al. 2001).

This means that the visible part of the spectrum is dimmed even more. Combine that with relatively low emission at those wavelengths, and you have a pretty poor target for optical telescopes, and a pretty good target (all other factors considering) at other wavelengths, especially near the emission peak.

1 Flares also are visible in other wavelengths, but I mention X-rays here because Sagittarius A* is typically much less luminous in X-rays.

2 I believe they mean $\lambda^{-2}$. Power laws for scattering in the area vary, but they are generally given between as $\lambda^{-1.5}$ and $\lambda^{-2}$. At any rate, the index may differ over different regimes.


What's the difference between GSR and Grf

Short Answer
They're the same.
Long Answer
Local Standard of Rest (LSR)
Let me start out by defining the dynamical Local Standard of Rest (dynamical LSR or also LSRD). This definition is taken straight from An Introduction to Modern Astrophysics, 2nd Ed. by Carrol & Ostlie, page 903 (emphasis theirs).

To investigate the motion of the Sun and other stars in the solar neighborhood, we first define the dynamical local standard of rest (dynamical LSR) to be a point that is instantaneously centered on the Sun and moving in a perfectly circular orbit along the solar circle about the Galactic center.
...
Note that once the LSR is chosen, the Sun immediately begins to drift away from it, implying that we would effectively need to redefine the reference point constantly. In reality this is not a significant problem because (fortunately) the 230-Myr orbital period of the LSR is very long compared to the time since modern telescopic observations began (and even longer compared to the lifetime of a typical research grant); consequently there has not been sufficient time for the effect to become noticeable.

Galactic Standard of Rest (GSR)
Now it is important to understand how the LSR coordinate frame is defined, because the Galactic Standard of Rest (GSR) is defined with respect to the LSR. You can read about it here, but in effect, the LSR is a non-inertial coordinate frame co-rotating with the rotation of the galaxy. The GSR is defined in the same way, but removes that co-rotation by subtracting out the rotational velocity of $220\:km/s$. In the LSR frame, another star's velocity would essentially be with respect to our Sun but in the GSR, the star's velocity would be with respect to a fixed point in the galaxy and be a summation of that star's motion with respect to our Sun and it's orbital motion around the galaxy.
Galactic Rest Frame (GRF)
The GRF is identical to the GSR, although I've heard GSR more often. In this paper, you can see they define the GRF to be

... the galactic-rest-frame (grf) proper motions ... which are those seen by an observer at the location of the Sun and at rest with respect to the Galactic center.

This is identical to the definition of the GSR above. Another paper similarly defines the GRF.

Removing the contribution of the motion of the Sun and of the Local Standard of Rest to the measured proper motion produces a Galactic rest-frame proper motion.


Does tidal locking also slowly reduce the orbiting body's axial angle?

In general, tidal forces will cause the obliquity to decrease for an initially prograde rotation, so unless something is acting to disrupt the situation you would expect a tidally-locked satellite to end up with a very small obliquity. This is the observed rotation state of most of the major satellites in the Solar System.
The Moon is something of a tricky case thanks to its location and its history of outward migration. The Moon likely formed close to Earth in a near-equatorial orbit. As it moved outwards it crossed the Laplace radius, which marks the region in which the Laplace plane (the plane about which the satellite's orbital plane precesses) transitions from being aligned with the planet's equator to being aligned with the planet's orbit. If the Earth was originally in a relatively high-obliquity state, this transition would have resulted in rapid changes in the Moon's orientation, and the Earth's obliquity becoming more similar to its present-day value.
The upshot of this is, even though tidal forces tend to damp the inclination with respect to the Laplace plane, the Moon's orbit is still inclined by a few degrees to the Laplace plane (likely as a result of picking up a fairly high inclination during the transition across the Laplace radius), and the nodal precession has a timescale of about 18.6 years. This timescale is more rapid than tidal forces can decrease the Moon's obliquity, so the Moon cannot evolve into the zero-obliquity state.


As a black hole passes through spacetime, does it affect it?

No, a black hole, or any other gravitationally attracting object for that matter, does not change spacetime permanently, but only affects spacetime around it. The easiest way to see this, is to view the BH in its own frame of reference. Here, the BH is stationary, and spacetime around it is described by the Schwarzschild metric. If a BH came by you and left a physical imprint on spacetime after it (like a wake), then from the BH's frame that would mean that you would fly by it, and it would start producing a wake in the direction that you moved. You don't have such an effect on BHs.

Wrong scenario of observer and black hole passing each other, as viewed from the observer's (left) and the black hole's (right) reference frames.


Time domain astronomy and fastest eclipsing binary ZTF J1539+5027 (+20 mag, 6.91 minutes): How to measure its minimum brightness?

Unless I've done my maths wrong, the period of total eclipse is about 18 seconds.
The CHIMERA camera at Mt Palomar, the instrument which followed up the discovery of this system, can take exposures at up to 8 full 1k$\times$1k frames/second and considerably higher if windowed on an object. There is no need for photon-counting equipment for such a slowly varying source. Indeed, the CHIMERA observations used 3 second exposures.
The problem in time resolution I suspect is just how faint the object is. So moving to photomultiplier technology, which would mean a hit in terms of efficiency, is not going to help here. The hot primary has $g = 20.38 \pm 0.05$. Given the the ratios of radii and temperatures in the discovery paper - $R_1/R_2 = 0.5$ and $T_1/T_2 > 4.9$ - then just a simple bolometric scaling suggests the flux from the secondary is $>144$ times smaller (or 5.4 mag).
It is quite challenging to accumulate much in the way of signal-to-noise on a magnitude $\sim 26$ object in $\sim 10$ seconds, even on the biggest telescopes on Earth. Probably the best bet would be to use a standard CCD camera on say the Gemini-N telescope (or Subaru) and just take a ten second exposure exactly at the predicted time of the minimum and then exposures either side (to confirm you were in the right place!). This can be repeated every 6.91 minutes for a whole night if necessary.
It doesn't look hopeful though. I used the Gemini-N integration time calculator with a $g=26$ source of spectral-type $\sim$A0V (ok for a white dwarf) observed in the darkest skies. It would take $215\times 10$s exposures to get a signal-to-noise ratio of 5, and the the noise is dominated by sky background at that magnitude. So basically unfeasible.
Determining the effective temperature of the unheated side of the secondary would be interesting because it might give you a constraints on its intrinsic luminosity, its cooling age, and hence on the evolutionary pathway that led to the less massive, but larger secondary helium white dwarf forming before the hotter primary C/O white dwarf.


After a supernovae explosion from stars whose mass is more than the Chandrasekhar limit, the remaining neutron star is so dense that it collapses into a one single singularity( A point where the density in infinite and is compressed into a point).the only way we can detect a black hole is by the entropy given by the black hole. so at the point of formation the black hole is not taking any matter into it.After the black hole is formed we will see any object going into it as frozen in spacetime


How much light does Jupiter project onto the surface of Ganymede?

How much? Well how accurately do you need it? How do you want it quantifying? And in what wavelength range?

Jupiter scatters a fraction of its incident sunlight. It also has its own luminosity (predominantly in the infrared).

A quick calculation:

The solar constant (flux at 1 au) is about 1370 W/m$^{2}$. Jupiter is situated about 5.2 au from the Sun (it varies by about +/- 5%) and has a radius of 70,000 km. The albedo is about 0.34.

Thus it receives about $7.8\times 10^{17}$ W from the Sun and radiates about $2.6\times 10^{17}$ W back into space. Assuming this is done more-or-less isotropically into a hemmisphere, then Ganymede, at a distance of 1,070,000 km from Jupiter, receives a flux of only 0.1 W/m$^2$ multiplied by the fraction of the sunlit hemisphere that can be seen.

This compares with the $\sim 50$ W/m$^2$ it receives from the Sun!

This surprising result (to me) puts into perspective all the simulations you see of things in orbit around Jupiter. The planet is still pretty faint compared to the Sun.

I'd be grateful if someone could double-check the sums!

[The intrinsic infrared luminosity of Jupiter is less than a fifth of what it receives from the Sun, so this would increase, but not double the received power.]


A question about calculating the radius of a comet

Here's my approach to solving this problem. You provide some of the initial steps, but I'll still go through them just for completeness.
Orbital Distance
We're told the period is $T=5.5\:\mathrm{years}$. This means we can immediately calculate the orbital distance (or more precisely, the semi-major axis, $a$). Since we're talking about a comet orbiting the Sun, we can simply use:
$$T^2 = a^3$$
where $T$ is the orbital period in units of $\mathrm{years}$ and $a$ is the semi-major axis in units of $\mathrm{AU}$. I find that $a = 3.116\:\mathrm{AU} = 4.66\times 10^{11}\:\mathrm{m}$. Good so far.
Volume of Evaporated Shell
We're told that the radius of the comet decreases by $\Delta R = 20\:\mathrm{cm}$. We can assume the comet is perfectly spherical and calculate the volume of the evaporated shell which should depend on both $R$, the radius of the comet, and $\Delta R$. This will be necessary since we need to know the total amount of ice that was evaporated. The volume of this shell is given by
$$V = \frac{4}{3}\pi(R^3-(R-\Delta R)^3)$$
$$V = \frac{4}{3}\pi(R^3-R^3+3R^2\Delta R-3R\Delta R^2 + \Delta R^3)$$
$$V = \frac{4}{3}\pi(3R^2\Delta R-3R\Delta R^2)$$
$$V = 4\pi R^2\Delta R\left(1-\frac{\Delta R}{R}\right)$$
Note that I've made a specific choice here. I've dropped the $\Delta R^3$ term from the second line. The reason being that it is a third order term and $\Delta R^3 \ll R\Delta R^2$. You could argue that I could drop the second order term, $R\Delta R^2$ since $R\Delta R^2 \ll R^2\Delta R$, but I'm choosing to keep this second order term around so we wind up with an $R$ to solve for in the final answer.
Power Input
The next step is to find the total energy input per second to this comet, e.g., the power. You've basically already defined this part. The flux on this comet is defined as
$$F = \frac{L_\odot}{4\pi a^2}$$
The power input is simply the flux times the area of the comet.
$$P_{\mathrm{in}} = FA = \frac{L_\odot}{4\pi a^2} \pi R^2 = \frac{1}{4}L_\odot \frac{R^2}{a^2}$$
Note that this assumes the comet is in a circular orbit and thus always at the orbital radius of $a$. If the comet's orbit had any sort of eccentricity, then $F$ would be a function of radius and you'd have a much harder time.
Evaporation Energy
Now we need to calculate the total energy necessary to evaporate the evaporated shell of volume $V$ from above. In order to evaporate a solid ice, you have to go through four stages of heating. First, you raise the ice's temperature to the melting point. The energy input for this is defined by the specific heat capacity for ice, $c_{\mathrm{ice}}$. Then you add energy to convert the ice into water. The energy input for this is defined by the latent heat of fusion, $\ell_f$. Now you can raise the temperature of the water until it reaches the next stage. This is defined by the specific heat of water, $c_{\mathrm{water}}$. Finally, you add energy to convert the water to gas, defined by the latent heat of vaporization, $\ell_v$.
All of these can be put together into a single equation.
$$E_{\mathrm{evap}} = c_{\mathrm{ice}}(m_{\mathrm{shell}}\Delta T_1) + \ell_fm_{\mathrm{shell}}+c_{\mathrm{water}}(m_{\mathrm{shell}}\Delta T_2) + \ell_vm_{\mathrm{shell}}$$
$$E_{\mathrm{evap}} = (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)m_{\mathrm{shell}}$$
Each one of the terms in this equation accounts for the energy input of one of the stages I described above. Note that $\Delta T_1$ is the temperature change from the initial temperature to the melting point ($273.15\:\mathrm{K}$). A reasonable temperature change might be $73.15\:\mathrm{K}$ (assuming an initial temperature of $200\:\mathrm{K}$), based on the temperature of comet 67P as determined by Rosetta. The $\Delta T_2$ is the temperature change from the melting point to the boiling point which is necessarily $100\:\mathrm{K}$.
You can look in a table somewhere and find that $c_{\mathrm{ice}} = 2.108\:\mathrm{kJ\:kg^{-1}\:K^{-1}}$ and $c_{\mathrm{water}} = 4.187\:\mathrm{kJ\:kg^{-1}\:K^{-1}}$.
Lastly, you need to define the mass of the evaporated shell, $m_{\mathrm{shell}}$. This is simply the volume already determined, multiplied by the density of ice/water. Technically those densities will be different, but they're close enough that we can just use $\rho_{shell} = 1000\:\mathrm{kg\:m^3}$. So ultimately we have:
$$E_{\mathrm{evap}} = (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)V \rho_{\mathrm{shell}}$$
For simplicity's sake, I'm going to define
$$\eta \equiv (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)$$
so that
$$E_{\mathrm{evap}} = 4\pi \eta \rho_{\mathrm{shell}} R^2\Delta R\left(1-\frac{\Delta R}{R}\right)$$
Putting it all together
We now know the total energy necessary to evaporate the comet's $20\:\mathrm{cm}$ shell as well as the energy input per second. We know that it receives this energy input per second for one orbital period of $5.5\:\mathrm{years}$ which means we can say:
$$T = \frac{E_{\mathrm{evap}}}{P_{\mathrm{in}}}$$
$$T = \frac{4\pi \eta \rho_{\mathrm{shell}} R^2\Delta R\left(1-\frac{\Delta R}{R}\right)}{\frac{1}{4}L_\odot \frac{R^2}{a^2}}$$
$$T = 8\pi \eta \rho_{\mathrm{shell}} \Delta R \frac{a^2}{L_\odot} \left(1-\frac{\Delta R}{R}\right)$$
Again, for simplicity, I'll define
$$\xi \equiv 8\pi \eta \rho_{\mathrm{shell}} \Delta R \frac{a^2}{L_\odot}$$
so that
$$T = \xi \left(1-\frac{\Delta R}{R}\right)$$
It should be pretty easy to see now that
$$\boxed{R = \frac{\Delta R}{1-T/\xi}}$$
The rest is just plugging everything in.


Distance to Proxima Centauri (Gaia VS New Horizons parallax program)

LORRI will be used in 4x4 mode, which yields 4-arcsec pixels.  The error in positions is unlikely to be better than 1%, or 40 mas, about 200x larger than Gaia's error.  NH has a baseline ~20x larger, but this means it still misses Gaia by ~10x.

The date was selected for New Moon to help Earth observers find the two targets.  There is no Earth analogue for LORRI, but it should be readily matched by amateur rigs.  The project will use professional ~0.5 - 1.0m telescopes to get Earth obs.  Simultaneous Earth obs are not strictly required, but the simultaneous times are offered to conduct the aesthetically purest demonstration of instantaneous parallaxes.


Strength of gravity during the big bang?

The truth is we do not have a working or widely accepted "theory of everything" that unifies gravitation with the other fundamental, and quantum, field theories. What we do have is strong evidence that the other field theories - electromagnetic, strong nuclear and weak nuclear - are unified at high energies (indeed the evidence is so strong that I doubt it is widely disputed at all).

We do have a theory of gravitation - general relativity - that appears to work well in "most" circumstances, though it also has a singularity (black holes) and there is a fundamental difference between it and the quantum theories about vacuum energy (the so-called vacuum catastrophe).

What we also know is that gravitation appears to much weaker than the other forces, though it can act over very large distances and so is fundamental in shaping the nature of our universe.

There are various arguments about how/why our universe is able to support life/is the way it is. Perhaps it is just one of many/an infinite number of universes that have a range of physical properties and we just happen to be in one that "works" for life - obviously there could be no life in a universe that had physical properties that didn't support life.

At another extreme perhaps we are all inside a computer simulation.

I'd suggest hunting down works by Max Tegmark or Brian Greene for more information.


According to my calculations Jupiter's moons do not follow Kepler's 3rd law - Why is that?

Kepler's third law is that $R^3/P^2$ is a constant. However it is not a universal constant; it depends on the mass of the body that is being orbited.
$$\frac{R^3}{P^2} \simeq \frac{GM}{4\pi^2},$$
where $M$ is the mass of the orbiting body (assuming that $M\gg$ the mass of the moons).

Depending on your level of sophistication, you could try plotting $R^3$ against $P^2$, or better, $\log R$ against $\log P$, to demonstrate consistency with Kepler's third law.


What is the largest object on which the Yarkovsky effect has been observed?

According to Vokrouhlicky et al. 2015 Yarkovsky forces can be measured for small bodies with diameters up to 30-40 km. The largest object they have in their list of Yarkovsky detections is 4179 Toutatis with a diameter of (only) 2.8 km. I am not aware if Yarkovsky forces have been measured on anything larger than asteroids.


What is the intensity distribution of visible light over the solar disk?
  What is the distribution of the visible spectrum over the solar disk

Hopefully I've answered all the different questions.

The core of the sun is extremely hot (27 million degrees Fahrenheit) compared to the surface (10,000 degrees Fahrenheit) and the atmosphere (36,000 to 2 million degrees Fahrenheit). Source: JPL NASA - "The Sun's Layers and Temperatures". To our eye the sun generally appears to have a single color and brightness due to dynamic range and color mixing.

The AM0 spectrum and intensity looks like this:

Each element emits a different multitude of wavelengths. The distribution across the surface, and out into the atmosphere, varies.

For example here is a set of images taken at different wavelengths:

That answers your question in this comment:

  "Ah thank you, that is exactly what I was looking for. But how do images like these get produced? Are they adjusting the brightness to be equal througout the image before releasing it? – Jhonny007".

Images like those are produced by placing filters over the imaging sensor.

Limb darkening is caused by the different temperatures of the different layers. The center of the disk appears bumpy due to granules. Using a greater dynamic range it looks like this:

This image is a composite of 25 separate images spanning the period of April 16, 2012, to April 15, 2013. It uses the SDO AIA wavelength of 171 angstroms and reveals the zones on the sun where active regions are most common during this part of the solar cycle.

A closeup of a sunspot (a colder region, thus darker) looks like this:

The website Schurs Solar Imaging, and webpage: "Image Gallery of Solar Phenomenon - White Light Features" has a great set of photos, and one clearly showing limb darkening.

Adjusting the contrast of an image from the University of Virginia's webpage: "ASTR 1230 (Majewski) Lecture Notes" creates an image that shows limb darkening.


The La2010 long-term ephemeris Laskar et al. (2011), which is based on the INPOP numerical ephemeris integrated for 1 Myr, is valid for 250 million years before the present day and a unknown distance into the future. The authors note that due to chaotic motion in the Solar System, the accuracy degrades significantly beyond -50 Myr and likely a similar time into the future. The ephemeris is intended for paleoclimate and similar studies and provides the classical orbital elements which will need to be transformed into an Earth position or an Earth-Sun distance.


Is Barnard 68 the only cloud so close to us that there are no stars between us and it?

Barnard 68 has a surface area of around 10 square arcminutes on the sky and is at an estimated distance of around 160 pc. The volume of space encompassed by a sightline to Barnard 68 is therefore about 1 pc$^{3}$.

The the density of stars in the solar neighbourhood (hydrogen burning stars with mass above about 0.1$M_{\odot}$) is about 0.1 pc$^{-3}$. So it is not surprising that no stars are seen in front of Barnard 68.

At this distance, the brightness of the faintest stars would be around $V \sim 20$ so ought to have been picked up in various observations that have been performed on this cloud. Such stars would be more obvious in the near-infrared, but then you pick up background stars through the cloud and you would need an estimate of the distance.

I'm not actually aware of any clear statement that there are no faint stars in the foreground, though obviously you can't see any in the widely available visible pictures.

There are lots more Bok globules known which appear to have very little in the way of foreground stars. As I outlined in the calculation above - if they are small and nearby, then that is to be expected. You could also look at the list of "dark nebulae".

Other catalogues include the Lynds catalogue of dark nebulae or the catalogue of southern dark clouds by Hartley et al. (1986).


When using adaptive optics, what is the shortest timescale of atmospheric changes in refractive index that astronomers have to deal with?

I apologise, I'm posting an answer because I can't comment.

What is the fastest abberation coming from atmospheric turbulence in any measurement setting (sun observation during day, night-time observations)?

It depends on the astronomical seeing conditions at the location of the telescope, and also varies quite a lot. It could be anywhere from $1-100~ms$.

Does any astronomer need correction in the visible spectrum beyond e.g. 100 Hz?

It depends on the what you're after. In solar physics, for example, observations are generally interested in resolving features on the Sun. You can also find polarimeters that operate beyond $100~Hz$, so an Adaptive Optics system that operates in the $kHz$ is a fairly standard requirement for most modern Solar telescopes. However, if you're only interested in collecting unresolved spectra (say, a star), you may not really benefit from having a really fast Adaptive Optics system.

Are fast systems practically limited by spatial resolution of the sensor?

I'm not sure what you mean by limited and sensor. Are you talking about the sampling of the wavefront by the wavefront sensor?

Is there any way to get a quantitative overview how widespread adaptive optics is in astronomy today and which setups are currently in use?

I think you'd find this reference helpful: https://link.springer.com/article/10.12942/lrsp-2011-2


Could science be lost if a phenomena is observed before predicted?

The question leaves me confused on how to interpret it.  As some of the comments have noted, observations have routinely predated a theory that (accurately) predicts them.  In astronomy and cosmology this has tended to be the main driving force behind seeking new theories.  Yet I sense a more subtle question concerning how scientists would view a theory that predicts everything so far, but nothing new.

Ultimately I think there is a confusion concerning the typical mantra for adopting a new theory as the main theory for physics/science/whatever: if a new theory doesn't make any new testable predictions, and so can only reproduce what existing theories already predict, then preference is given to simpler and existing theories.  This has nothing to say about a single theory.  It's strictly about multiple theories being compared to each other.  It does not imply that a theory, viewed entirely on its own, that makes no new predictions is undesirable.  It simply means if we're going to replace theory X with theory Y, then theory Y better predict more stuff than X does.  It doesn't care if X does or doesn't predict testable things that have yet to be proven/disproven.


What is the reason the Jovian planets are in descending order of size?

Since the initial formation of the solar system and after the initial mass formation of each of the planets let's see what what we can derive from the laws of physics.

Any mass that could further accumulate on any planet would have to come from 3 places:

In the nearby vicinity of the planet's orbit.
From outside the solar system and extra Pluto Kuiper belt objects.
From the Sun as solar wind.

We can surmise that in the first 200 million to half billion years in the life of the solar system, the planets would have grown in the most mass by (1) sweeping up the asteroids and debris in their orbit and vicinity.  In fact, Jupiter with its huge mass continues to pull away stray asteroids to a lesser extent to this day.

The occasional encounters with objects coming from (2) outside the solar system are too rare and far in between to be of significance.  The cosmic rays from external to the solar system are also not significant for mass gain.

That leaves (3) the solar wind.  Whether a planet gains or loses mass from the solar wind is a balance between the

(a)  Gravitational escape velocity of the planet, which is directly proportional to the mass of the planet

(b)  Average temperature of planet's surface and atmosphere

(c)  Average speed or kinetic energy of the particles from the solar wind

(d)  Intensity of the planet's magnetic field.

(e)  Intensity of the solar wind, which is inversely proportional to the square of the distance from the sun.

The (e) factor was and still is much too strong for the inner 4 planets to gain mass from the solar wind.  Being that the escape velocity would be too low to keep the atmosphere with the ongoing assault of the solar wind as is the case with Mercury and our moon.  We can thank the earth's magneto sphere for saving our atmosphere from the solar wind.

Now when we get to where Jupiter's orbit is, about 10AU, the intensity of the solar wind is one hundredth of what it is on the earth.  The solar wind that does reach Jupiter is decelerated and deflected towards the poles by Jupiter's very strong magnetic field .  And since the escape velocity is so high due to Jupiter's mass, most of the solar wind is absorbed into Jupiter resulting in mass gain from solar wind over time. Not even hydrogen can escape Jupiter's  gravitation.

The same net gain from solar wind applies to the next 3 gas giants Saturn, Uranus, and Neptune.  Since these are in order further away, the solar wind intensity correspondingly less.  All three still have a high gravitational escape velocity due to their size and the ones further out have a very low surface temperature so the gases don't have enough average energy to escape.

As we go further out from Jupiter the same dynamic of mass gain through captured solar wind plays out, albeit less and less due to the distance from the sun.

While the solar wind for purposes of planetary  mass gain may seem insignificant now, we must understand that this was not the case earlier on.
And over 5 billion years much mass can be gained in this manner.


Would a tidally locked planet be always increasing in temperature?

Hot objects radiate heat. The hotter something is, the faster it radiates heat.

The side of the planet that faces the sun would heat up until the rate at which it is absorbing heat from the star is equal to the rate at which it is radiating heat. Once it reaches this point it will be stable, and neither heat up more or cool down.

As a simple analogy, consider standing next to a heater or a fire. You will heat up, but you won't heat up forever. You will reach thermal equilibrium.


If a solar system were surrounded by a cloud of debris, is it possible for a planet's orbit to intersect it?

The solar system IS surrounded by a cloud of debris,left over from its formation. This debris is called the Oort Cloud,& consists of thousands of comet-like bodies,sometimes described as dirty snowballs. None of the known major planets passes through them,but a few years ago some astronomers seemed certain that  they had detected signs of a 9th major planet much further out than the others. They predicted it was only a matter of time before this planet would be discovered,but so far it hasn't been. If there is indeed such a planet & part of its orbit passes through the Oort Cloud,it would probably sweep a lane through it eventually. The planet would not be a nice place to live,as the likelihood is that it has a temperature close to that of liquid nitrogen,& is frequently struck by comets & other debris.


Pulsars: How do astronomers measure minute changes in period (~picoseconds per year)?

Let us suppose that the pulsar is spinning down at a uniform rate. So it has a period $P$ and a rate of change of period $dP/dt$ that is positive and constant (in practice there are also second, third, fourth etc. derivatives to worry about, but this doesn't change the principle of my answer).

Now let's assume you can measure the period very accurately - say you look at the pulsar today and measure its radio signals for a few hours, do a Fourier transform of the signal and get a nice big peak with a period of 0.1 seconds (for example).

With that period, you can "fold" the data to create an average pulse profile. This pulse profile can then be cross-correlated with subsequent measurements of the pulse to determine an offset between the predicted time of "phase zero" in the profile, calculated using the 0.1 s period, and the actual time of phase zero. This is often called an "O-C" curve or a residuals curve.

If you have the correct period and $dP/dt=0$, then the residuals will scatter randomly around zero with no trend as you perform later and later observations (see plot (a) from Lorimer & Kramer 2005, The Handbook of Pulsar Astronomy). If the initial period was in error, then the residuals would immediately begin to depart from zero on a linear trend.

If however, you have the period correct, but $dP/dt$ is positive, then the residuals curve will be in the form of a parabola (see plot (b)).

If you have second, third etc. derivatives in the period, then this will affect the shape of the residuals curve correspondingly.

The residuals curve is modelled to estimate the size of the derivatives of $P$. The reason that $dP/dt$ can be measured so precisely is that pulsars spin fast and have repeatable pulse shapes, so changes in the phase of the pulse quickly become apparent and can be tracked over many years.

Mathematically it works something like this. The phase $\phi(t)$ is given by
$$\phi(t) \simeq \phi_0 + 2\pi \frac{\Delta t}{nP} - \frac{2\pi}{2}\frac{(\Delta t)^2}{nP^2} \frac{dP}{dt} + ...,$$
where $\phi_0$ is an arbitrary phase zero, $\Delta t$ is the time between the first and last observation and $n$ is the integer number of full turns the pulsar has made during that time. If the period is approximately correct, then $n = int(\Delta t/P)$.

The "residual curve" would be given by
$$\phi_0 - 2\pi\frac{\Delta t}{nP} -\phi(t) \simeq  \frac{2\pi}{2}\frac{(\Delta t)^2}{nP^2} \frac{dP}{dt} + ...,$$

For example, if the period of a $P \sim 0.01$ second pulsar changed by a picosecond in a year, then there would be an accumulated residual of almost $10^{-4}$ seconds after 1 year of observation. Depending on how "sharp" the pulse is, then this shift of about 1% in the phase of the pulse might be detectable.

Perhaps needless to say, but there are a host of small effects and corrections to make in order to get this very high precision timing. You need to know exactly how the Earth is moving in its orbit. The proper motion of the pulsar on the sky also has an effect. These and more can be found in Lorimer and Kramer's book, but there is also a summary here.


How can red supergiants be cooler than dwarfs of the same spectral type?

The HR diagram has many forms. There is no one to one relationship between temperature, colour and spectral type that is true for stars of all surface gravities and metallicities.
Spectral type depends on temperature primarily but spectral features can also be weaker or stronger at lower surface gravity or with changes in metallicity.
For example, the Handbook of Space Astronomy and Astrophysics has two tables of spectral type Vs temperature relationships, one for dwarfs and one for giants. And indeed the M-type giants are cooler than M-type dwarfs of the same subclass by ~200K.


Short answer:

A tiny fraction of the gravitational potential energy released by the very rapid collapse of the inert iron core gets transferred to the outer layers and this is sufficient to power the observed explosion.

In more detail:

Consider the energetics of an idealised model star. It has a "core" of mass $M$ and initial radius $R_0$ and an outer envelope of mass $m$ and radius $r$.

Now suppose the core collapses to a much smaller radius $R \ll R_0$ on such a short timescale that it decouples from the envelope. The amount of gravitational potential energy released will be $\sim GM^2/R$.

A fraction of this released energy can be transferred to the envelope in the form of outward moving shocks and radiation. If the transferred energy exceeds the gravitational binding energy of the envelope $\sim Gm^2/r$ then the envelope can be blown into space.

In an exploding star (a type II core collapse supernovae) $R_0\sim 10^4$ km, $R\sim 10$ km and $r \sim 10^8$ km. The core mass is $M \sim 1.2M_{\odot}$ and  the envelope mass is $m \sim 10M_{\odot}$. The dense core is mostly made of iron and supported by electron degeneracy pressure. The star is said to have "run out of fuel" because fusion reactions with iron nuclei do not release significant amounts of energy.

The collapse is triggered because nuclear burning continues around the core and so the core mass is gradually increased and as it does so it gradually shrinks (a peculiarity of structures supported by degeneracy pressure), the density increases and then an instability is introduced either by electron capture reactions or photodisintegration of iron nuclei. Either way, electrons (which are what is providing the support for the core) are mopped up by protons to form neutrons and the core collapses on a free fall timescale of $\sim 1$ s!

The collapse is halted by the strong nuclear force and neutron degeneracy pressure. The core bounces; a shock wave travels outwards; most of the gravitational energy is stored in neutrinos and a fraction of this is transferred to the shock before the neutrinos escape, driving away the outer envelope. An excellent descriptive account of this and the previous paragraph can be read in Woosley & Janka (2005).

Putting in some numbers.
$$GM^2/R = 4\times 10^{46}\ {\rm J}$$
$$Gm^2/r = 3\times 10^{44}\ {\rm J}$$

So one only needs to transfer of order 1% of the collapsing core's released potential energy to the envelope in order to drive the supernova explosion. This is actually not yet understood in detail, though somehow supernovae find a way to do it.

A key point is that the rapid collapse takes place only in the core of the star. If the entire star collapsed as one, then most of the gravitational potential energy would escape as radiation and neutrinos and there would be insufficient energy even to reverse the collapse. In the core collapse model, most (90%+) of the released gravitational energy is lost as neutrinos, but what remains is still easily sufficient to unbind the uncollapsed envelope. The collapsed core remains bound and becomes either a neutron star or black hole.

A second way to cause a star (a white dwarf) to explode is a thermonuclear reaction. If the carbon and oxygen can be ignited in nuclear fusion reactions then enough energy is released to exceed the gravitational binding energy of the white dwarf. These are type Ia supernovae.


What starting circumstances have to be for "grand tack" hypothesis to work, and why they are lacking in other planet systems(as far as we know)?

We do not yet know whether our solar system is a common or rare outcome of the planetary process.

The statistical properties of exoplanetary systems are severely affected by observational biases. It is far easier to detect close-in giant planets; objects like Saturn, at Saturn-like distances from their host star, have not been found due to the limitations of the observations.

It is therefore far too early to say whether the Grand Tack model may or may not be applicable to a wide variety of exoplanetary systems. What we do know, is that something drastically different happened in the case of those systems that have a Jupiter-sized planet within an astronomical unit or so of their parent star; but these are found in only a few per cent of stars.

As an addendum, it is also the case, that despite taking 10+ years of data to establish them, there are now many cases of Jupiter-like planets in Jupiter-like orbits, though they are probably not that common. The overall conclusion at present is that the solar system does not look that unusual, although the lack of close-in planets larger than the Earth is notable (see Martin & Livio 2015).


Let's say you have a rocket near a Schwarzschild black hole. Near the horizon, the thrust required to keep stationary goes to infinity; in this specific sense, the gravitational force becomes infinite at the horizon. If the rocket is hovering above the horizon, then the difference in gravitational forces becomes arbitrarily large near the horizon. This can tear it apart. However, an object falling into the black hole will also gain a velocity arbitrarily close to the speed of light (as measured by a stationary observer as it passes) as it gets closer to the horizon, meaning it will be length-contracted. It shrinks to zero size. As a result, the difference in 'gravitational force' across it is more complicated, and in fact remains finite.

  Knowing that spaghettification is dependent on the size of the object, as well as the difference in gravity on both sides, a shrinking particle may eventually become so small that the steepness of the gravity well is no longer a problem.

A shrinking object tames the infinite 'gravitational force' and keeps the tidal forces finite. But the curvature is already defined in the limit of the size of the measurement probe tending to zero. (The static tidal forces correspond to six of the twenty independent curvature components, but let's not worry about that much.) So no, you can't 'get rid of curvature' by shrinking your particle/system/whatever. That's conceptually broken.

  Is this a known theory or idea, and where can I find papers on it?

You would have to have a much clearer idea before that can be answered reasonably. I've skipped all the way to the end of your question because I haven't understood anything more of what you're trying to say.


What's the bleeding-edge model for how Population III stars are born and evolve?

This is a more recent review http://arxiv.org/abs/1209.4688 from someone hoping to understand population III by observing their supernovae

Since CR7 was discovered, there's been a bunch of papers on whether it could be pop III - http://arxiv.org/abs/1506.07173 for example. Try searching for "CR7 population III".


Does the radius of the Universe correspond to its total entropy?

Holographic entropy bounds are upper bounds for the maximum amount of entropy a given region can have, rather than its current entropy--for example, the spherical region of space just large enough to enclose the Earth and its atmosphere would currently contain a much lower entropy than a spherical black hole of the same radius, whose entropy is given by the Bekenstein bound. The scholarpedia article on the Bekenstein bound mentions the issue of cosmological entropy bounds:

  Problems with bound (1) or the holographic bound are known to occur in extreme situations. Both fail when the gravitational potential is large (strong self-gravity), e.g. system already collapsed inside a black hole. In an infinite universe the holographic bound fails when applied to a sufficiently large region. In a closed (finite) universe the specification of R or bounding area becomes ambiguous (Bousso 1999, 2002). The covariant entropy bound corrects these and other shortcomings.

So apparently the "covariant entropy bound" (also known as Bousso's holographic bound) is the thing to look into when considering the maximum entropy of the universe (and I assume you just mean the observable universe here rather than whatever may lie beyond it). This set of lecture slides by Raphael Bousso, the physicist who originally proposed the covariant entropy bound, says that this bound is "conjectured to hold in arbitrary spacetimes, including cosmology." But it also says "If correct, origin must lie in quantum gravity", probably meaning one can't derive this fact in a non-conjectural way from basic laws of physics without having a complete theory of quantum gravity.

For an introduction to this stuff, I recommend this Scientific American article that discusses various holographic bonds, including the Bekenstein bound along with Bousso's conjectured more general bound (the article was written by Bekenstein himself). Though it is from 2003 so I'm not sure what kind of theoretical progress has been made since then. Anyway, here's what it says about Buosso's entropy bound:

  In 1999 Raphael Bousso, then at Stanford, proposed a modified holographic bound, which has since been found to work even in situations where the bounds we discussed earlier cannot be applied. Bousso’s formulation starts with any suitable 2-D surface; it may be closed like a sphere or open like a sheet of paper. One then imagines a brief burst of light issuing simultaneously and perpendicularly from all over one side of the surface. The only demand is that the imaginary light rays are converging to start with. Light emitted from the inner surface of a spherical shell, for instance, satisfies that requirement. One then considers the entropy of the matter and radiation that these imaginary rays traverse, up to the points where they start crossing. Bousso conjectured that this entropy cannot exceed the entropy represented by the initial surface—one quarter of its area, measured in Planck areas. This is a different way of tallying up the entropy than that used in the original holographic bound.

  Bousso’s bound refers not to the entropy of a region at one time but rather to the sum of entropies of locales at a variety of times: those that are “illuminated” by the light burst from the surface. Bousso’s bound subsumes other entropy bounds while avoiding their limitations. Both the universal entropy bound and the ’t Hooft-Susskind form of the holographic bound can be deduced from Bousso’s for any isolated system that is not evolving rapidly and whose gravitational field is not strong. When these conditions are overstepped—as for a collapsing sphere of matter already inside a black hole—these bounds eventually fail, whereas Bousso’s bound continues to hold. Bousso has also shown that his strategy can be used to locate the 2-D surfaces on which holograms of the world can be set up.

But again, this bound should be the upper limit on the entropy of a given region, not necessarily the actual entropy in that region, so this type of holographic bound shouldn't imply that when we increase the entropy of the universe, the universe needs to grow. This paper specifically discusses how most matter-containing regions in cosmology apart from black holes would fail to "saturate" the covariant entropy bound, meaning that the actual entropy in the region is smaller than the maximum possible according to the bound.


A rough analytical expression for the Milky Way's radial mass distribution?

For the purposes of this simple exercise, what would be an analytical expression that roughly matches the Milky Way's radial density profile, projected on to its equatorial plane?

The simplest example astronomer's work with is the radial density profile of a Single Isothermal Sphere (SIS). It is called this because it is spherically symmetric (and thus applicable to a 2D plane for your purposes) and all objects orbit with the same velocity (and thUs have the same "temperature", hence isothermal). The density profile takes the form:

$$\rho(r) = \frac{v^2}{4\pi Gr^2}$$

where $v$ is the rotational velocity. Note that you may see other formulations which use $\sigma_v$ rather than $v$. In this case they're using the velocity dispersion which is slightly different than the rotational velocity.

Other, more realistic density profiles have been found by running simulations of the Universe and matching functional equations to the density profiles of the resulting galaxies. Such popular results are the NFW profile and the Einasto profile.

The NFW profile is a two-parameter function, given by

$$\rho(r) = \frac{\rho_0}{\frac{r}{R_S}\Big(1+\frac{r}{R_S}\Big)^2}$$

where $\rho_0$ and $R_S$ and two, halo-dependent parameters.

The Einasto Profile is again a two-parameter model given by

$$\rho(r) \propto \exp(-Ar^\alpha)$$

where $A$ and $\alpha$ are configurable parameters.

  For spherically symmetric distributions, Newton's Shell theorem allows one to treat all mass inside a sphere defined by an orbit's radius as if it were at the center, and to ignore all mass in the shell outside of that radius. Is there anything like an analog to this for a radial distribution within a plane?

The Shell theorem for gravity does not extend to a 2D ring. However, I will say that when talking about the orbits of stars in galaxies, the mass of stars outside a star's orbit are generally considered negligible. The primary reason for this is that it is Dark Matter which comprises most of the mass of a galaxy and contributes the most to defining a star's orbit in a galaxy. The Dark Matter halo is often assumed to be spherically symmetric, in which case Newton's Shell theorem does apply and the mass you're concerned with in determining a star's orbit, is the mass of the Dark Matter halo interior to the star's orbit.


Can a change in the density of interstellar and intergalactic quantum fields affect the Redshift of photons?

No.

Or at least such an effect has never been observed, neither in the locality of the Earth or in light detected from distant sources.

If a photon has an interaction with a quantum field (such as an electromagnetic field) this causes a scattering. Scattering would cause a blurring and dimming of distant sources. This is not observed.

Such an effect, if it existed, should also be present on more local scales. But there is no redshift related to distance within our galaxy. Nor is there any evidence of any shift, even tiny, in the photons emitted by other local sources

Thus such a proposal would require an unknown interaction of light, that causes no scattering, only affects light from outside the local group of galaxies, but is otherwise linear with distance. And one would have to explain such phenomena as the slowing of time in distant galaxies, (which an expanding universe model understands as the relativistic time dilation)

So in short: No, a change in the density of interstellar and intergalactic quantum fields does not cause the observed red-shift of photons.


Why do Uranus and Neptune have more methane than Jupiter and Saturn?

Why does Uranus and Neptune have more methane than Jupiter and Saturn?

It's a combination of equations of state (EOS), serpentinization, and mixing (rotational and convective) that favors a preference for some reactions (and resulting compounds) over others.

See the references below.

  The giant planets are all mostly hydrogen and helium, but Uranus and Neptune have relatively large amounts of hydrogen compounds like methane (that's what gives them their color).

Jupiter and Saturn are gas giants, Uranus and Neptune are ice giants.

  My question is why did that happen? How did Uranus and Neptune get their methane? My impression is that all the gas giants were far enough out for methane to condense into ice, so how did Uranus and Neptune end up preferentially with methane?

See Wikipedia's "Extraterrestrial Atmosphere":

Graphs of escape velocity against surface temperature of some Solar System objects showing which gases are retained. The objects are drawn to scale, and their data points are at the black dots in the middle. Data is based on "Lecture 5: Overview of the Solar System, Matter in Thermodynamc Equilibrium" and "Stargazer's FAQ - How exactly are atmospheres held?".

Wikipedia says little about the atmosphere of these planets, and the least about Uranus and Neptune:

Atmosphere of Jupiter:

  "There are no methane clouds as the temperatures are too high for it to condense." - Source: "Jupiter's ammonia clouds — localized or ubiquitous?" (April 9 2004), by S.K.Atreya, A.S.Wong, K.H.Baines,  M.H.Wong, and T.C.Owen.

  Quotes from the paper:

  Page 502: "For the production of polycyclic aromatic hydrocarbons (PAHs), chemistry begins with the destruction of methane (CH$_4$) by solar UV photons at $\lambda \le$160 nm, ultimately leading to the formation of benzene ($c$-C$_6$H$_6$, or A$_1$) and other complex hydrocarbons (Fig. 3). In the polar auroral regions where energetic particles also break down methane, ion chemistry becomes dominant in the production of benzene and heavy hydrocarbons (Wong et al., 2003, and Fig. 3).".

Atmosphere of Saturn:

  "Ultraviolet radiation from the Sun causes methane photolysis in the upper atmosphere, leading to a series of hydrocarbon chemical reactions with the resulting products being carried downward by eddies and diffusion. This photochemical cycle is modulated by Saturn's annual seasonal cycle.". - Source: "Ethane, acetylene and propane distribution in Saturn's stratosphere from Cassini/CIRS limb observations" (Nov. 2008), by S. Guerlet, T. Fouchet, and B. Bézard.

  Quotes from the paper:

  Page 406: "3 Method

  We used a line-by-line radiative transfer model to calculate synthetic spectra. It included opacity from CH$_4$, CH$_3$D, C$_2$H$_6$, C$_2$H$_2$, C$_3$H$_8$, C$_3$H$_4, C$_4$H$_2 and collision-induced opacity from H2-He and H2-H2 . The atmospheric grid consisted in [of] 360 layers from 10 bar to 10−8 bar. It was coupled with an iterative inversion algorithm adapted from Conrath et al. (1998), in order to retrieve the atmospheric state (temperature, hydrocarbon vertical profiles) from the measured spectra.

  As a molecular emission intensity depends on both its abundance and temperature, we proceeded in two steps. First, we retrieved the temperature vertical profile from the methane ν4 emission band at 1305 m$^{−1}$ (assuming it is uniformly mixed with a vmr of 4.5 x10$^{−3}$ (Flasar et al. 2005)), providing information in the 1 mbar - 2 $\mu$bar region.

  ...

  Figure 1 shows an example of a comparison between synthetic and observed emission bands of ethane, acetylene and propane at two given pressure levels (all the different pressure levels probed by CIRS have not
  been plotted for the sake of clarity) and Fig. 3 the Corresponding retrieved profiles.".

What that means is that more complex compounds than methane are favored by the conditions, see comments above concerning "equations of state".

Atmosphere of Uranus and Neptune:

  "The  gaseous outer layers of the ice giants have several similarities to those of the gas giants. These include long-lived, high-speed equatorial winds, polar vortices, large-scale circulation patterns, and complex chemical processes driven by ultraviolet radiation from above and mixing with the lower atmosphere.

  Studying the ice giants' atmospheric pattern also gives insights into atmospheric physics. Their compositions promote different chemical processes and they receive far less sunlight in their distant orbits than any other planets in the Solar System (increasing the relevance of internal heating on weather patterns).".

NASA Factsheets - Atmospheric composition (by volume, uncertainty in parentheses):

Jupiter

Major: Molecular hydrogen (H$_2$) - 89.8% (2.0%); Helium (He) - 10.2% (2.0%)
Minor (ppm): Methane (CH$_4$) - 3000 (1000); Ammonia (NH$_3$) - 260 (40); Hydrogen Deuteride (HD) - 28 (10); Ethane (C$_2$H$_6$) - 5.8 (1.5); Water (H$_2$O) - 4 (varies with pressure)
Aerosols: Ammonia ice, water ice, ammonia hydrosulfide

Saturn

Major: Molecular hydrogen (H$_2$) - 96.3% (2.4%); Helium (He) - 3.25% (2.4%)
Minor (ppm): Methane (CH$_4$) - 4500 (2000); Ammonia (NH$_3$) - 125 (75); Hydrogen Deuteride (HD) - 110 (58); Ethane (C$_2$H$_6$) - 7 (1.5)
Aerosols: Ammonia ice, water ice, ammonia hydrosulfide

Uranus

Major: Molecular hydrogen (H$_2$) - 82.5% (3.3%); Helium (He) - 15.2% (3.3%) Methane (CH$_4$) - 2.3%
Minor (ppm): Hydrogen Deuteride (HD) - 148
Aerosols: Ammonia ice, water ice, ammonia hydrosulfide, methane ice(?)

Neptune

Major: Molecular hydrogen (H$_2$) - 80.0% (3.2%); Helium (He) - 19.0% (3.2%); Methane (CH$_4$) 1.5% (0.5%)
Minor (ppm): Hydrogen Deuteride (HD) - 192; Ethane (C$_2$H$_6$) - 1.5
Aerosols: Ammonia ice, water ice, ammonia hydrosulfide, methane ice(?)

Additional references:

"Methane in the Solar System" in English, (Bol. Soc. Geol. Mex [online]. 2015, vol.67, n.3, pp.377-385.), by Andrés Guzmán-Marmolejo and Antígona Segura.

"Abiotic Production of Methane in Terrestrial Planets" (Astrobiology. 2013 Jun; 13(6): 550–559), by Andrés Guzmán-Marmolejo, Antígona Segura, and Elva Escobar-Briones.

"Methane clathrates in the solar system" (Astrobiology. 2015 Apr;15(4):308-26), by Mousis O, Chassefière E, Holm NG, Bouquet A, Waite JH, et al.

NASA - "Scientists Model a Cornucopia of Earth-sized Planets" (Sept. 24 2007).


Finding the suns position in the sky anywhere at any time

There are many resources that will let you compute the position of the Sun in a coordinate system with its center at the center of the Earth. The answer will be an angle above or below the equator, an angle eastward from the vernal equinox (first point of Aries, and a distance. Books such as Astronomical Algorithms 2nd ed. by Jean Meeus have information on how to transform such a position to a horizon system centered at an observer on the surface of the Earth. The answer will be an angle from the north, an angle above the horizon, and the distance. Wikipedia has articles about software libraries that will do this kind of calculation, such as "SOFA (astronomy)" and "Naval Observatory Vector Astrometry Subroutines". These subroutines are quite complicated because they are meant to be suitable for the most critical work, with angles accurate to a few milliarcseconds. The Meeus formulas are much simpler.


Why does the moon sometimes appear giant and a orange red color near the horizon?

(Source, Wikipedia Commons)

The moon is generally called a "Harvest Moon" when it appears that way (i.e. large and red) in autumn, amongst a few other names.  There are other names that are associated with specific timeframes as well.  The colour is due to atmospheric scattering (Also known as Rayleigh scattering):

  may have noticed that they always occur when the Sun or Moon is close to the horizon. If you think about it, sunlight or moonlight must travel through the maximum amount of atmosphere to get to your eyes when the Sun or Moon is on the horizon (remember that that atmosphere is a sphere around the Earth). So, you expect more blue light to be scattered from Sunlight or Moonlight when the Sun or Moon is on the horizon than when it is, say, overhead; this makes the object look redder.

As to the size, that is commonly referred to as the "Moon Illusion", which may be a combination of many factors.  The most common explanation is that the frame of reference just tricks our brains.  Also, if you look straight up, the perceived distance is much smaller to our brains than the distance to the horizon.  We don't perceive the sky to be a hemispherical bowl over us, but rather a much more shallow bowl.  Just ask anyone to point to the halfway point between the horizon and zenith, and you will see that the angle tends to be closer to 30 degrees as opposed to the 45 it should be.

University of Wisconsin discussion on the Moon Illusion.

NASA discussion on moon illusion.

A graphical representation of this:

Dr. Phil Plait discusses the illusion in detail.


Scholarly work on the gravitational disruption of a rocky planet by a close pass with another planet?

Scenario: This looks specifically at a near-miss scenario with an Earth-like intruder passing by at a distance of 1,000,000 km at a speed of 100 km/s.  This is about three times the distance to the Moon and at about three times Earth’s orbital velocity. The 100 km/s seems a reasonable number given the relative velocities of our stellar neighbors, the solar infall/escape velocity, and the likely difference in directions of the Earth and intruder velocities.
First Sighting: It's a pure guess, but maybe the intruder will be first recognized when it’s brightness (apparent magnitude) becomes similar to that of Pluto.  The difference in absolute magnitude between the Earth and Pluto is (-3.99)-(-0.70) = -3.29.  This corresponds to a difference in brightness of 100^(3.29/5) = 20.7.  At that distance from the Earth/Sun, the brightness will go as the 4th power of distance, so we should first spot the intruder when it is 20.7^(1/4) = 2.13 times further away than Pluto or about 12.6 billion km.  Just approximating with a constant 100 km/s, this means we will get about 4 years of advance warning.
Night Sky: Assuming the intruder approaches from roughly the opposite direction of the Sun, it will be visible in the night sky with a magnitude -3.99+(1/5)log100(1/s^2/(1+s)^2 where s is the distance from Earth in AU.  At 0.6 AU (90,000,000 km) it will be roughly as bright as Venus and clearly visible in the night sky.  This will occur about 10 days before closest approach.
At its closest approach (1 million km), the intruder will appear about 40% larger diameter than the Moon and will be about 7 times brighter, both because of the larger apparent area (0.73/0.52)^2 and higher albedo (0.43/0.12).  The intruder will take 5.6 hours to travel the 2 million kilometers  across 90 degrees of the night sky while its apparent area goes from 50% to 100% and back to 50% of maximum.  A sight worth watching!
Orbital Disturbance (Earth):  Assuming the deflections of the Earth and the intruder are small, the delta_V can be calculated as (1/2)GMe/d^2 x (2d/v), where Me is the mass of the Earth or the intruder, d is the distance of closest approach and v is the velocity.  The first factor corresponds to the maximum acceleration of the Earth and the second factor to the effective time during which the acceleration is applied.  The first factor is 0.000199 m/s^2 and the second factor is 20,000 seconds giving a delta_V of 3.98 m/s. This is 0.00013 of the Earth’s orbital speed of 30 km/s.  Generally this would change the ellipticity of the orbit slightly and could also change the mean radius if the delta-V were mainly along the direction of the orbit.  The orbital energy could change by +/-0.00026 and the mean radius would change the same fraction.
Corresponding to a change in the Earth’s mean orbital radius, the insolation would change by double that fraction or +/-0.00052.  The temperature based on black body radiation would change by the fourth root of the insolation, i.e. +/-0.00013 or +/-0.013% or +/-0.04 degrees C.   These changes seem quite minor.
Orbital Disturbance (Moon):  The Moon would be more severely affected because its mass is small compared to the intruder, so the acceleration would be roughly twice as large.  Also the Moon could be on the ‘wrong’ side where it would be closer to the intruder by a factor of 2/3.  So the acceleration and the delta_V could be higher by a factor of 2/(2/3)^2 = 4.5.  The resulting delta_V of 18 m/s on the Moon’s orbital velocity of about 1000 m/s is more significant.  This could result in a 3.6% change in orbital energy and in radius and a change of 5.4% in period, changing the 28 day period by more  than a day.  Since tidal effects go as the cube of distance, presumably the lunar tidal ranges would increase (or decrease) by about 11%.
Transient tidal effect: Tidal forces are proportional to mass/distance^3.  Ignoring the constant factor, the peak tidal force from the Moon : Sun : Intruder are in the ratio 7.35E22/(3.84E5)^3 : 1.989E30/(1.48E8)^3 : 5.97E24/(1E6)^3 or 1.30 : 0.61 : 5.97.  So the maximum tidal force as the intruder swings by will be about three times more that the combined lunar and solar tidal force.  Presumably, during the several hours of closest approach, this will raise ocean tides about three times larger than the familiar large spring tides.  Coastal areas would experiences tides tens of feet higher than normal much as in a storm surge or tsunami. I suppose it is also possible that some latent earthquakes could also be released by such large tidal forces
(The last time I did a back-of-the-envelope calculation like this, I was wrong by a factor of 10^6.  But it certainly elicited some pretty quick responses!)


Why don't planets give off their own light?

All matter radiates (except if it's at absolute zero temperature), regardless of its composition (you got that of Mercury badly wrong). The most important form of radiation is the black-body radiation which only depends on the temperature of the material, but line emission and absorption may also be important (but depends on the composition and ionisation state of the material) and other emission and absorption processes.

Stars are hot enough (the Sun has surface temperator $\sim5700$K) for the black-body radiation to peak in the visible part of the electro-magnetic spectrum. As a star shines, it looses energy, i.e. it cools, reducing the gas pressure that stabelizes it against gravitational collapse. In stars, this energy loss is balanced by the energy production from thermonuclear fusion in the core (requiring temperatures $\sim10^9$K). The transport of this energy to the surface makes stars non-trivial.

In planets and brown dwarves, there is (by definition) no thermonuclear energy source. Therefore, these objects must shrink, which generates energy from gravity. However, they cannot shrink indefinitely, as ultimately quantum mechanics becomes important: Pauli's exclusion principle demands that the electrons cannot be arbitrarily closely packed. Thus, for brown dwarves and giant gas planets (but also white dwarves) further shrinking is halted at a radius comparable to that of Jupiter (Jupiter itself is still shrinking at a very small rate). These as well as all smaller objects then merely cool down very much like a piece of glowing coal.

The situation is a often more complicated by sources of energy. Planets, for example are
irradiated by their host star, which may dominate the energy gains at their surface (in addition, the Earth gains energy from nuclear fission in its core). The balance between this energy gain and the loss by black-body radiation determines the temperature of a planet.

The Earth, for example, radiates in the infrared. However, the radiation losses are also regulated by line absorption of that infrared in the higher atomsphere by so-called green-house gases, in particular CO$_2$.


How do or could radio telescopes contribute to exoplanet studies?

Radio telescopes are frequently used to observe the births of stars and their planetary systems. The longer wavelengths are able to penetrate from beneath the envelope of gas and dust that shrouds any attempts to view these events at optical or infrared wavelengths. Unfortunately, the smallest angular resolution of a telescope goes as $\lambda/D$, where $D$ is the telescope diameter and $\lambda$ the observation wavelength. As radio waves are $10^5$ to $10^{6}$ times longer than optical waves, then one needs enormous telescopes to get similar resolving power. This is attempted/accomplished by constructing arrays that can act as a single aperture.

Whilst interferometric techniques to gain spatial resolution are certainly possible, if one wants to directly image exoplanets then optical/IR is still the way to go because this is where most of the intrinsic light from an exoplanet is found.
It is possible that planets orbiting stars may emit radio waves in their own right, through charged particles accelerated in their magnetospheres. Jupiter, for instance, is a strong source of radio waves in our solar system, but would be very difficult to spot at the distances of exoplanets. Some searches have been attempted, but nothing yet found (e.g. George & Stevens 2007).

Of course transit observations do not require superb angular resolution, but they do require lots of signal and precision. Levels of precision that are routinely achieved by optical telescopes (especially in space - e.g. Kepler) are not possible with radio telescopes, and the parent stars themselves are weak, variable radio sources, which makes studying any transit signature very difficult. In fact the main use of transits has been the other way around - to search for intrinsic planetary emission by looking for a drop in flux when the planet is eclipsed by the star (e.g. Lecavelier Des Etangs et al. 2011).  Radio emission might be enhanced in hot Jupiters if the exoplanetary magnetic field interacts with the stellar magnetic field.
This paper does claim to have detected a weak exoplanetary signal from a hot Jupiter, identified by its disappearance when eclipsed by the parent star. However, the result was not repeatable.

The SETI programme targeted a set of Kepler stars with evidence for a planet (or planet candidate) in the "habitable zone". The initial searches for narrow-band radio emission (at 1.1-1.9 Hz) appears to have concentrated on those systems containing multiple planets.

The (null) results for $\sim 100$ system were published by Siemion et al. (2013). The observations were not sensitive enough to detect the sort of "radio chatter" emitted from our planet, but might have detected some sort of deliberate signalling beacon.

A slightly more imaginative approach is to observe Kepler multiple planet systems when two planets are in conjunction as seen from the Earth. There is then the chance of seeing communications from one planet to the other. (e.g. http://adsabs.harvard.edu/abs/2014ebi..conf..5.2K ).


In order to execute the compiled binaries, all libraries on which the executable depends must be in the correct location. In particular, the error specified in the question is due to the fact that my .moog folder was in ~/, my user home folder, rather than /, the root of my hard drive, as explained here.

To solve the problem:

cd

mv .moog /.moog

See this link for answer on how to get MOOG to find SM libraries for plotting.


What azimuth description systems are in use?

There are many indeed.

It may be specified from South westwards(0° to 360°), or from North eastwards and westwards(180° to -180°).

The most popular one, as used by Roy and Clarke in "Astronomy : Principles and Practices" is to measure the azimuth from North eastwards 0° to 360°. For a southern hemisphere observer, this system changes to from South eastwards 0° to 360°.

With so much confusion, how do professional astronomers go about it? They simply specify the system they are using in a given document. Doesn't cause much trouble since the differences in all of the systems is by an angle of 180° or 360°. It just causes sign changes in all formula.


Where does Jupiter's gravitational force come from? Why don't Jupiters gasses fly away?

if they don't have any solid mass inside them is an incorrect conclusion.

Some have 'ordinary' solid cores, and the pressures in the centers of gas giants are so high, that they are even 'suspected' of compressing hydrogen into a 'metallic' state. Quoting from the Planetary core Wikipedia page:

Gas giants also have cores, though the composition of these are still a matter of debate and range in possible composition from traditional stony/iron, to ice or to fluid metallic hydrogen.

For Jupiter especially:

Jupiter has an observed magnetic field generated within its core, indicating some metallic substance is present. [... ]
Jupiter has a rock and/or ice core ten-thirty times the mass of the Earth [...]
Thermal contraction/evolution models support the presence of metallic hydrogen within the core in large abundances (greater than Saturn).

The earth's core has an average density of about 12,000 kg per cubic meter, Jupiter's core density is estimated at 25,000 kg per cubic meter (link). Together with its much larger radius than the earth (70000 km versus 6370 km), you will have a pretty massive core with a mass of from 12 to 45 times the Earth's mass, so there's plenty of gravitational pull there.

More to read: The outer planets: Giant planets: Interiors


how do we know the universe is STILL expanding?

We don't know. This is a depressing thing that makes me, at least, feel so small. The only thing we might assume, that if they once were accelerating that fast, they probably will do it still (even faster), but of course we can't be sure if anything have happened.

It's like this with all the long distances in the universe. How do we know if Proxima Centauri (4.2 light years away) have died/exploded? We wait, 4.2 years.


What are the prime challenges in search for extraterrestrial intelligence?

We have no idea how much we’ve been able to accomplish so far in the search for extraterrestrial intelligence.
If there isn’t extraterrestrial intelligence then our progress is zero.  If there is extraterrestrial intelligence and we already found some, then we’ve made 100% progress!
I think the major obstacles for SETI are threefold: funding, stigma, and physics.
Funding:  This article is a couple years old, but it does a nice job of summing up historical low funding for SETI.  Most of the funding is private and on the order of millions to tens of millions of dollars a year.  That might sound like a lot, but ALMA cost $1.3b.  A few million a year isn’t enough to sustain a very robust program or buy, maintain, and staff an observatory.
Stigma: At least when I was working for the NRAO, SETI work wasn’t considered serious astronomy.  There was some smirking in the lunchroom from some of the astronomers towards the folks “looking for little green men.”  I think it’s more useful to try to improve telescopes and data processing techniques, but I don’t think anyone should be stigmatized for pursuing their intellectual curiosity. Stigma usually means less funding, less access to highly sought after observation hours on the best telescopes, and less people choosing to join the field.
Physics: Radio signals decay in power as the distance squared and even the nearest star, Proxima Centauri, is 4.3 light years or about 271,937 AU away from our solar system.  That means they would need an enormously powerful emitter and we would need a very sensitive detector to detect their signal.  See: how far away could we detect that Earth has life?.


When is Gamma Draconis closest to the zenith in London on April 4th

I read this as "Benenaim" (another name for Alkaid) too (not Etamin, which is another name for Gamma Draconis), and that star does indeed transit the meridian near midnight on April 4th (12:44am to be more precise). Note that the hour angle is 0 in the Stellarium screenshot below, and the star is only 35 minutes from the zenith position:


The planets formed from a disc of dust and gas. Since the orbiting the proto-sun formed into a disc the planets, which formed from this matter are in a disc too, but not a perfect disc. There was some scattering of the matter in the primordial disc, and so the planets are also slightly scattered. Interactions between the planets have also scattered the planets out from the disc too. The formation of the planets caused them to have small, but not exactly zero inclination

The Earth has the greatest inclination (except pluto) however, by convention, we measure inclination relative to the Earth, all the planets have a positive inclination.

So there is no special reason that the planets have slight inclinations. It is just what would be expected due to random effects during their formation.


I think your question is on topic, but @RhysW has linked a very helpful post in understanding why your question is a common misconception about the Big Bang.

No Center

There is no 'center' to the universe. At any point, a local observer will claim that they are at the center of the universe by the way galaxies move away from them. How can we possibly know this? The universe appears to be both homogeneous (has the same structure everywhere) and isotropic (there is no preferred direction). If these are indeed properties of the universe, then the expansion of the universe must be the same at all other locations (See: The cosmological principle).

How the Big Bang and Explosions Differ

Additionally, the Big Bang is different from an explosion in the following ways:

1) Particles involved in an explosion slow down eventually due to frictional forces. Think of fireworks (http://www.youtube.com/watch?v=qn_tkJDFG3s). Particles are moving the fastest at the instant of explosion, and slow monotonically with time. The expansion of the early universe does not follow this trend, though sometimes people use the word 'explosion' to describe the enormous volumetric increase (an increase by a factor of $\sim10^{76}$) which occurred between $10^{-36}- 10^{-32}$ seconds after the Big Bang, which is aptly named inflation.

2) An explosion implies the existence of space. For an explosion to take place, particles (whether we're talking about matter or light) must have space to explode into. Strictly speaking the inflation of the universe is an expansion of space-time coordinates, and so the word explosion cannot really apply since there was nothing for space-time to explode into.


When you say "particle" candidates, I assume you're excluding MACHOs and RAMBOs. MACHOs are "dark" objects at the stellar scale like black holes, neutron stars, brown dwarfs, etc. RAMBOs are clusters of similar dark objects. MACHOs and RAMBOs are made of primarily baryonic matter (everyday stuff like protons and neutrons — electrons are not baryons but they can be a part of this too).

This leaves us with the WIMPs and possible other particle candidates. I don't know the details of any experiments looking for these offhand (I'll see if I can find it out and update the answer in a few days), but the major particle candidates are:

Neutrinos: It's highly unlikely that these make up any significant chunk of the DM in our universe, but they deserve a mention.
Sterile neutrinos (I'm using the term in the specific sense here, not the general sense): This type of neutrino is right handed and only interacts gravitationally (note that there can be right handed active neutrinos, too). There is some preliminary evidence suggesting that these might exist.
Axions: These are hypothetical, but if they exist (and make up the majority of DM), we kill two birds with one stone by both explaining DM and resolving the lack f symmetry breaking in the strong interaction.
Gravitinos and the Lightest Supersymmetric Partner: Both of these require supersymmetry to hold, but they are major DM candidates.


Are telescope eyepieces ever made with a diopter adjustment (similar to DSLRs)?

Are there eyepieces with diopter adjustments?  The answer is both yes and no.
Diopters are not needed on telescopes for simple spherical myopia (near-sighted or far-sighted vision issues) but may be needed if a person suffers from cylindrical myopia (astigmatism) and wants to be able to use the telescope without wearing prescription eyeglasses.
For near-sighted & far-sighted vision issues there is generally no need of a diopter adjustment on a telescope as their is on an SLR or DSLR camera for spherical correction.  Also, cameras generally don't a way to compensate for astigmatism ... but it turns out you can get eyepiece adapters that compensate when using a telescope (more on that later).
Why?  To answer the question, it's important to first recognize what the diopter adjustment does on a DSLR camera.
On a DSLR, the "SLR" stands for "Single Lens Reflex" because the camera is designed to allow the photographer to look through the same lens that will be used to capture the photo (as opposed to viewfinder or rangefinder cameras... or even twin-lens reflex cameras where the photographer looks through a separate lens used to "approximate" the image that will be captured.
If you remove the lens and inspect the top of the mirror chamber, you'll notice that the mirror is shining light up onto a piece of frosted glass.  The glass will often have etched marks to serve as aids when taking the photograph.
The distance that light must travel through the lens, into the camera body, onto the reflex mirror and up onto the frosted glass (focusing screen) is actually the same as the distance through the camera and onto the imaging sensor (when the mirror swings clear to capture a photograph).
This means that when you look through the viewfinder, you're technically not looking "through the lens" ... you're actually looking at a back-lit projection screen (the frosted glass).
This means you could have an image which is precisely focused on the focusing screen (and also precisely focused for the imaging sensor) but appears out of focus due to eyesight issues.
The Diopter Adjustment allows you to adjust for near-sighted/far-sighted issues (spherical myopia).  To adjust, manually put the camera out of focus and point at something with no contrast (such as a blue sky outdoors or a plain white wall if indoors) and adjust the diopter until the etched markings on the focus screen are as sharp as possible.  Once you've done this, you can be confident that if the image projected onto the focus screen is well-focused, your eyes will also perceive it as being well-focused.  It also means you may not need to wear your prescription eyeglasses when using the camera.
Telescopes
With a telescope, you aren't looking at an image on a focusing screen (such as frosted glass)... you really are looking through the telescope.  When doing visual observing, there is no camera imaging sensor involved.
This means if you are near-sighted or far-sighted (both of which are spherical corrections) you can just adjust the telescope's focus to compensate.  There is no need for a diopter adjustment.
Spherical myopia means the curvature of your eye is spherical... but is either slightly too flat ... or slightly too round to focus accurately at certain distances (resulting in a person being near-sighted or far-sighted).
But there is also cylindrical myopia (aka astigmatism).  This means that instead of having uniform curvature (spherical curvature) on your eye, your eye's curvature is a bit more barrel-shaped where the curvature in one direction is stronger than the orthogonal direction.
TeleVue (and so far as I know ... only TeleVue offers such a thing) makes a product called Dioptrx.  This is an adapter that fits onto many TeleVue eyepieces (not all TeleVue eyepieces are compatible with Dioptrx).  It is designed for those who suffer from astigmatism.
You would need to know your corrective lens prescription to get the correct Dioptrx adapter (and since your right vs. left eyes will each have their own prescription, you need to decide which Dioptrx to order based on the eye you prefer to use when viewing through the telescope.
If you follow the link to the TeleVue Dioptrx page, you'll notice that the instructions offer helpful information on how to read your eyeglass prescription and how to identify the cylindrical correction needed for your eye.  But you'll also notice it doesn't matter if the correction needed is positive or negative and also the axis can be ignored.  This is because you can't just rotate the lenses in your eyeglasses ... they have to mounted in your eyeglasses frames in the correct orientation to work properly.  But you can rotate the telescope eyepiece.  This is why the positve/negative and axis values can all be ignored (TeleVue marks the mounting ring of the Dioptrx with letters ... so once you decide which axis offers the sharpest view, you note the letter at the top and just make sure to always rotate the eyepiece to the same orientation relative to your eyes when using it.
The Dioptrx astigmatism corrector allows you to use your telescope without needing to wear prescription eyeglasses.  If your astigmatism if very weak (the cylindrical correction is a low value ... such as 0.50 ... then Dioptrx may not be needed.  But if your correction is very high (e.g. 2.00 or more) then Dioptrx will probably be a big help.


Could iron rain upset tidally locked WASP76b?

The important thing to bear in mind is that WASP-76b is a gas giant (at the time of writing, the NASA Exoplanet Archive gives values of 0.92 Jupiter masses and 1.83 Jupiter radii, so this planet is much less dense than Jupiter is): there's no solid surface for the iron to build up on. Instead what would happen is that the iron would rain down until it reached depths at which the temperatures are sufficiently high for it to vaporise again. This would then allow it to be carried back to the dayside as iron vapour. Once it is back on the day side of the planet, the increased temperature would allow it to rise back up to higher altitudes, where the winds would then carry it back to the night side in an "iron cycle".

Superrotation is predicted to be a general feature of hot Jupiter atmospheres, though the detection of westward displacements of hotspots on planets like CoRoT-2b suggests that things may be more complex: see for example Debras et al. (2019) "Acceleration of superrotation in simulated hot Jupiter atmospheres" for some discussion of how this might occur.

There's also the question of the so-called "thermal tide", where the atmospheric heating affects the rotation. This effect is thought to be responsible for Venus avoiding tidal lock. On a gas giant there isn't a solid surface for the atmosphere to couple to, but in some cases this effect may contribute to at least the outer radiative zone of the planet being in non-synchronous rotation, see Auclair-Desrotour & Leconte (2018) "Semidiurnal thermal tides in asynchronously rotating hot Jupiters" for details.


Would being ejected from the Milky Way Galaxy have any major impact on life on Earth?

I can think of two ways this benefits life.

Less chance of a passing star disrupting our Oort Cloud and sending
deadly comets crashing into the Earth.
Less chance of nearby supernova destroying our ozone layer.

On the other hand, if our Solar System passes near a relativistic jet emanating from the central black hole of a galaxy, Earth will get hit with high speed particles that will disrupt the magnetosphere and ozone layer. Definitely not good for any life on Earth at that time.

Getting ejected from the galaxy also means there is less chance of future humans exploring the galaxy by hopping to nearby stars. We'd only have our solar system to explore and colonize. There would not be another star system within a hundred thousand light years.