What is the area of the Summer Triangle?

The area of a triangle enclosed by 3 stars on the celestial sphere, in square degrees, is given by:
$$ A = \frac{180}{\pi}\times E$$
Where E is the spherical excess and is equal to the sum of all angles of the triangle minus 180°. Problem is, we don't know the angles of the Summer Triangle and instead we have to use an alternate equation for the spherical excess given by:
$$ E = 4\tan^{-1}\left({\sqrt{\tan\left(\frac{s}{2}\right)\times\tan\left(\frac{s-a}{2}\right)\times\tan\left(\frac{s-b}{2}\right)\times\tan\left(\frac{s-c}{2}\right)}} \right) $$
Where a, b, and c are the angular distances between 2 stars and thus the sides of the Summer Triangle in degrees, and $s = \frac{a+b+c}{2}$.
I highly recommend you just go to Wolfram Alpha and just type "Angular distance between Star 1 and Star 2" and get all values for a, b, and c and then you could calculate the spherical excess and subsequently the area.
But if you insist on giving yourself a headache and finding the values yourself then check the next section. Otherwise skip to the bottom.

The angular distance between 2 stars on the celestial sphere is given by:
$$\cos^{-1} \left ( \sin(\text{D} _{1})\sin(\text{D} _{2})+\cos(\text{D} _{1})\cos(\text{D} _{2}) \cos(\text{R} _{1}-\text{R} _{2})\right ) $$
Where:

$\text{D}_{1}$ and $\text{D}_{2}$ are the declination values of both stars in degrees.

$\text{R}_{1}$ and $\text{R}_{2}$ are the right ascension values for both stars, also in degrees.

To convert a star's declination from DMS to degrees, divide the arcminutes by 60 and arcseconds by 3600. For example Vega's declination is 38° 47' 1.28". Converted to degrees only, that becomes:
$$ 38 + \frac{47}{60} + \frac{1.28}{3600} = 38.7836° $$
The exact same formula is used to convert RA to degrees, only difference is you multiply it by 15 at the end. Vega's right ascension is 18h 36m 56.33s. Converted to degrees only, that becomes:
$$ \left(18 + \frac{36}{60} + \frac{56.33}{3600}\right) \times 15 = 279.2347° $$
You could just spare yourself all this mess by directly getting the angular distances from Wolfram Alpha or you could calculate them yourself. I recommend the former.

Once you plug in all the numbers, you get the spherical excess $E = 7.296508$ and the area of the Summer Triangle:
$$ \bbox[5px,border:2px solid #C0A000]{A = 418.059 \text{deg}^2} $$
It would cover 1.0134% of the celestial sphere and be the 40th largest constellation in the sky, bigger than Capricornus but smaller than Aries.


What's the point with Hawking Radiation?

I cannot show this or prove it but the main idea, in simplest terms, is that a matter-antimatter pair, near the event horizon, will form. A member of this pair will decay back into the event horizon and be absorbed by the black hole. The other particle will decay out into the free space.


How can satellites be safe in the atmosphere when there are so many asteroids coming in?

Space debris is a big problem for space industry. Natural bodies like dust or asteroids are not the main challenge, although probes are frequently hit by fine dust. The dust causes tiny craters on the surface of the probes or can change rotation or orbit slightly. This can mostly be corrected, e.g. by wheels fueled from solar panels.

The most concerning objects are remnants of human-made objects circling around the earth. With each collition of two such objects clouds of fragments form, leading to ever more collitions. Those fragments are tracked by radar, and probes try to avoid collitions by adjusting their orbits. But sometimes they are hit and destroyed.

An orbit is the path along which a satellite travels around e.g. Earth.

A satellite must remain outside the lower parts of the atmosphere, otherwise it will be slowed down and fall to the ground. The lower limit is somewhere near 200 km above ground. A satellite travels with about 8 kilometers per second around Earth, when in a lower Earth orbit.

Many satellites are in the geo-stationary orbit, about 36,000 km above ground. There Earth's atmosphere is neglectably thin, only some minor magnetic field can be found there; that part of the atmosphere is called magnetosphere.

The further away a satellite orbits the Earth the slower it travels to compensate Earth's gravity, because Earth's gravity grows weaker with distance from the Earth.

At present the probability of collisions, which destroy satellites, is very low for a single satellite. Actually there are only a few documented cases: "only one major incident has occurred: the 2009 satellite collision between Iridium 33 and Cosmos 2251", of the above wikipedia article. And "China was widely condemned after their 2007 anti-satellite missile test, both for the military implications as well as the huge amount of debris it created" of the same Wikipedia article.

The central problem is the growing number of fragments to be expected for the future, as a consequence of more satellites in the orbit, and the expected chain reaction of fragmentation.

Details of surface degradation of satellites due to impacts of dust particles can be read e.g. here. Most of that dust ist human-made. Surface degradation has consequences, e.g. for solar panels.

See also a related question at StackExchange.

Regarding the second part of the question:
The dust particles collide with the satellites and evaporate from the impact heat. So the annihilation distance is more or less zero.

When entering the earth's atmosphere "they usually disintegrate at altitudes of 50 to 95 km " (Meteorid). That's at least 100 km below the lowest satellite orbits.


What wavelengths are used to search for distant solar system objects like KBOs and Oort cloud members?

Virtually all KBO searches use mosaic CCD cameras operating in the visible part of the spectrum, usually with a red filter like SDSS-r' (centered at approx. 622nm SDSS filters) on big (~4m telescopes). This is what we did with the OSSOS survey I was (lightly) involved with (http://www.ossos-survey.org/about.html). Although cold KBOs are brightest in the infrared, the background from the atmosphere is much higher which means you are less sensitive, unless you go to space (very expensive).

Also IR detectors are smaller and less sensitive than optical CCDs and much harder to make into mosaic cameras and keep cold (to increase sensitivity). This means you can't cover as much sky area per night which makes finding rare and faint KBOs much slower. A CCD and red filter is a good balance between brightness of the objects you are looking for, the sensitivity of your detector (CCDs get less sensitive as you move more to the red or blue of the green-red central part of the optical spectrum) and a low background from the atmosphere.


What is the distribution of organic compounds in the Milky Way Galaxy?

The answer depends on which you are talking about (you mentioned both organic compounds and organic matter, which are two completely different things).

Organic compounds we do not have a clear definition of, but is most agreed upon to contain carbon atoms, either on their own (C), bonded with at least one hydrogen atom (C-H), or bonded with at least one other carbon atom (C-C). Compounds such as Methane (C-H4), Ethanol (C2-H6-O), and  Glucose (C6-H12-O6) are textbook examples of organic compounds. Methane has been discovered on several objects within our solar system, including Mars, Venus, and Saturn. Methane has also been discovered on the exoplanet HD 189733b (though it is the only example we have of methane on an exoplanet), and often exists in interstellar clouds, showing that it is quite common throughout our galaxy.

Organic matter refers to matter composed of organic compounds, as part of an organism. While we are searching for organic matter as proof of extraterrestrial life, we have yet to discover any organic matter or clear evidence for such matter outside of our own planet.


"Esperance" rock on Mars formed by erosion or weathering? or is there any rock on Mars not formed by erosion?

Rocks on Mars, just like rocks on Earth, can be igneous (formed from the cooling magma or lava) Sedimentary (formed from sediments that were laid down by water or wind) or Metamorphic (sedimentary rocks that have been changed by exposure to heat and pressure) (there may be other categories, such as meteorites)

After they have formed, rocks are subject to erosion and weathering. These processes don't form the rock, but can shape it or move it. The process of weathering produces sand, silt and clay, that can later form new rocks.

The Esperance rock has a significant amount of "clay" (a sedimentary rock). The clay in Esperance was formed by the weathering of other rocks to very fine particles, these then gathered together in wet environment to form a clay. When the water evaporated, the clay hardened into a rock. The significance of Esperance is that the water under which it formed must have been chemically neutral, not strongly acidic.


How would a Fast Fourier Transform Telescope work without a mirror, dish or lenses?

From one perspective, the purpose of a telescope lens, or mirror is to delay the light arriving in the centre of the instrument a bit relative to the light arriving at the edges. A lens is thicker in the middle, and light travels more slowly in glass, the path to the centre of the mirror and back to the detector is longer, because of the curve of the mirror than the one to the edge, and so on. The result is that all the waves which hit any part of the lens or mirror from a point source end up arriving at the detector in phase so they add up, while waves from anywhere else cancel out.

A phased array antenna (common for short wavelength radio operations) achieves the same delay by capturing the signal at lots of points on the (flat) antenna and then artificially delaying it before combining it with the signals from the other points.

An interferometer (such as the Event Horizon telescope) does the same thing digitally. The full waveform of the signal (more or less) is recorded in multiple places and then they are combined in a large computer with various slightly different delays to separate waves coming from slightly different directions (I am over-simplifying greatly).

What this article proposes is mainly a way to do that for more antennae at less computing cost by placing the antennae in a specific layout and using a bit of clever maths (the FFT) to do all the combining in one go.


If suddenly "knocked" or perturbed from its orbit, would gravity eventually return the Earth to its original orbit?

There's a few parts to this question so there's more than one answer.
Earth gets knocked a little bit out of its orbit all the time by gravitational influence of other planets in our solar system. Jupiter and Venus are the primary two, but all the planets have some effect.  These are called orbital perturbations and they tend to alternate, not add up.  They're the causes of Earth's Milankovich cycles and the effect isn't negligible.  These variations in Earth's orbit, which take thousands of years to move back and forth can cause the formation and recession of glaciation, sometimes called ice ages.
These orbital perturbations, in addition to being cyclical, as a rule, don't effect Earth's semi-major axis which is very consistent, and may only change slightly as the Sun loses mass.
Orbital perturbations can to lead to much greater variations of eccentricity and axial tilt.   Mars, for example, undergoes much bigger variations than Earth, but it still undergoes a cyclical back and forth that leads to general stability within a range.   All eight known planets in our solar system are thought to be relatively long term stable.
The exception to this cyclical back and forth is if there's a resonance where the effect can grow over time.   The closest example of this in our current solar system is Jupiter and Mercury, where they're not in, but they're close to resonance and it's possible that Mercury will be tossed out of its orbit in a few billion years.  It's the most unstable planet in our solar system.
This article isn't published, but I still think it's a good summary of both the stability of orbits in our solar system and the possible (but perhaps unlikely) destabilization of Mercury in a few billion years.
Planetary migration is another means for tossing a smaller planet out of its orbit.  This is thought to be fairly common, based on observations of other solar systems and some uncertainty that gas giant planets could form close to their suns.  (there may be room for debate on that), but migrating planets is thought to be fairly common, if somewhat slow. It's never been observed but it can be modeled.  It's been suggested that when our solar system was young, Jupiter moved inwards, perhaps tossing Uranus and Neptune outwards and reducing the material available that would eventually become Mars and when migrating back outwards, leading to the late heavy bombardment which brought water to Earth. This is called the grand tack hypothesis.
An inwardly migrating Jupiter could certainly toss Earth about easily into a completely different orbit, but there's no evidence that suggests Jupiter is likely to migrate that much in the future.
A massive enough object from beyond the solar system could toss Earth into a different orbit and in doing so, not only change Earth's eccentricity and orbital plane, but change Earth's semi-major axis as well and that change would effectively be permanent.
A gravity assist of that kind could give Earth both a new eccentricity, a new orbital plane and a new semi major axis and perhaps a new set of Milankovich cycles, though I suspect over time, the orbit could re-circularize and the Milankovich cycles, largely reset to where they were, so there could be some correction not by the sun, but by the other planets orbiting the sun, but not everything would go back to where it was.   The Semi-major axis change would likely be permanent and there would also be the question of whether the new orbit was in near resonance with other planets which would lead to further changes over time.
A higher eccentricity could create stronger seasonal variations and perhaps trigger a new ice age, or maybe fix man made global warming.  :-)
The effects of a change in semi-major axis would change the length of a year, so, we'd need new calendars, and could warm or cool if it was big enough.
There would also be tidal concerns if a massive object passed close enough to significantly change Earth's orbit and perhaps even, pulling the Moon out of its stable orbit.
The good news is, space is big and quite empty and an object massive enough to change Earth's orbit passing close enough to do so is extremely unlikely.


Determining latitude and longitude in bad weather

Sunstones are believed to have been used by vikings to determine the direction to the Sun on cloudy days. That helps you point the ship in the right direction, and by experience you can estimate the side ways drift in the water. But the longitude will still be very uncertain without precision time keeping. The news paper article I linked to says that a Sunstone was find at a wreckage in the English Channel from as late as 1592.


I'm not an astrophotographer but I'd like to make sure that you know that there are (at least) two ways to put a camera on a telescope from an optical perspective.

The first is to keep the eyepiece on which produces a virtual image at infinity, and keep the camera lens on, focused at infinity.

The second is to remove both the eyepiece and the camera's lens, which leaves the camera's CCD sensor right on the primary focal plane of the telescope.

The first method is the quickest, and it seems to be called Afocal photography (I didn't know that!). If you are careful and the image is very bright, say for example the moon, then you can just set the camera focus to infinity, hold it up to the eyepiece, moving around sideways until the inevitable vignetting is at least centered, and click! You can try to manage the vignetting by moving the camera closer or farther from the eyepiece. Ideally you want to place the entrance pupil of your camera's lens on top of the exit pupil of your telescope's eyepiece.
I predict you will quickly move to the second method which is called Prime Focus Photography. Find an eyepiece-to-camera body adapter and then you are a bona fide astrophotographer!
If you want to take advantage of your very nice f/1.8 lens then lose the telescope and just photograph the milky way or meteor showers. This works best if your camera can be programmed to take a sequence of regularly spaced images, because you can merge them later with a computer with offsets so that the motion of the stars doesn't produce trails.
For more on all of these, search questions tagged photography here and astrophotography in Photography SE.

For completenes' sake only, I'll mention a third way, called Eyepiece projection. While a properly focused eyepiece produces a virtual image at infinity, if you move it out, away from the telescope, it will produce a real image out past the eyepiece. If you are looking at the Moon or something very bright you could put a piece of paper out a ways away from the eyepiece, say 6 to 20 cm, and with some practice produce an image on the paper. You can then imagine having your camera's sensor at that plane. This gives you more magnification compared to prime focus photography, but it may not be useful magnification!
Further reading:

The Prime-Focus Method
Eyepiece Projection
What is eyepiece projection and how do I do it?
Sky & Telescope's Astrophotography With a Digital Camera
Afocal method
Afocal Deep Sky Astrophotography where you can clearly see what vignetting looks like!


What was the the value of the Hubble constant at the time of the CMB's 'release' (i.e., 379,000 years after Big Bang)?

This answer to the question “Is the Hubble constant dependent on redshift?” gives the formula (a form of the Friedmann equation) for the Hubble parameter $H(z)$ as a function of redshift $z$:
$$ H(z)^2 = H_0^2 \left[ (1+z)^4 \Omega_r + (1+z)^3 \Omega_M + (1+z)^2 \Omega_k +  \Omega_\Lambda \right] $$
where the $\Omega$ terms are the fractional densities in radiation, matter, curvature, and dark energy, respectively.
Using that, plus the knowledge that the redshift of the CMB is $z \simeq 1100$, you can plug in values for the densities (I used WMAP values quoted here) and get that $H$ at the time of the CMB was about 22,000 times larger than the current value.
That answer also gives a graph of the value of the Hubble parameter as a function of time.


Is the Earth going to evolve towards Mars' fate or Venus' fate?

Either or neither. It's impossible to tell from the present.

If runaway climate change occurs, then yes, the conditions on Venus could be a potential analogue for the kind of environment on Earth due to the greenhouse effect.

Mars's atmosphere is assumed to have been much thicker in the past, otherwise it could not have sustained liquid water on the surface; it would have evaporated away. This water is required to explain the gorges and riverbeds we see on the Martian surface today. Researchers are still unsure where the atmosphere went, but the prime candidate is through top-loss to outer space, rather than sequestration through minerals on the surface (see here).

The likelihood of such an atmospheric escape process becoming dominant on Earth is small due to the Earth's greater mass and strong relative magnetic field, which prevents ion escape. In fact, the dominant loss process on the earth is sequestration. Some estimates put the reservoirs of sequestered carbon from original CO$_2$ at 250000 times the size of the existing atmosphere.

The alternative is that the Earth remains in its current steady state, or close to it, far in to the future. There are many mechanisms to enable such an equilibirum, such as oceanic sequestration of C0$_2$ highlighted above. There are also more contested theories such as the Gaia hypothesis. This proposes that the global biological ecosystem on the Earth is self regulating, helping to maintain life enabling conditions on the Earth. It's pretty much a dead-cert that life doesn't exist on either the Martian or Venusian surfaces, so we can accept that if the Gaia hypoethesis is true, the Earth will never reach these conditions.


Why can't the surfaces of stars be observed?

There are many different ways to get spatial information about the surface of a star besides direct imaging.

Direct imaging is difficult because the angular resolution available goes as $\lambda/D$.
For a 8-m telescope and light at 500 nm, one can resolve $6\times10^{-8}$ radians (assuming the blurring of the atmosphere can be overcome by adaptive optics or similar).

The nearest stars are a couple of parsecs away, so the smallest spatial scales that could be resolved are $\sim 2\times 3.1\times10^{16} \times 6\times10^{-8} =  3.7\times10^{9}$ m, or about 500 solar radii. Hence no surface features or even a disk could be resolved.

Of course you could use interferometric techniques to effectively increase the size of $D$ and measurements of the angular radii are now possible for many nearby stars or giant stars at larger distances.

Surface imaging is harder. Indirect techniques are much more common. These include doppler imaging and eclipse mapping. The former uses the fact that there is a relationship between the position of a bright/dark feature on a rotating star and the doppler shift of the light from that feature. By observing a time-series of spectra, the lumps and bumps in spectral lines can be inverted to produce a "doppler map" of the surface. The technique is usually limited to stars that are considerably more rapidly rotating than the Sun. There is a lot of ambiguity in the image reconstruction process - many surfaces could lead to the same observable sigature and clever statistical techniques (and even philosphies) have to be deployed to choose between them. Many stars have published "doppler maps" of their surfaces. Here is a typical example of such a study and below I show an example of a "doppler image" for the star II Peg (a K-type subgiant), from Gu et al. (2003), showing dark spotted regions. A typical resolution for such an image is about 10 degrees on the star.

Eclipse mapping, for which I can't easily locate a good link, uses the fact that a star/accretion disk is orbited by another star or planet that periodically eclipses it. What happens to the light from the system in and out of eclipse can be used to probe the surface of the eclipsed object. There are of course limitations to the spatial resolution that can be obtained, depending on the size of the eclipsing object, how long the eclipse takes and how wide the orbit is. But useful constraints can be made on the structure of accretion disks, sizes of starspots etc, although "maps" are not usually produced. A recent example using the transits of a planet to probe starspot structure is Roettenbacher et al. 2013.

Another possibility is rotational modulation. Features on the surface that rotate around are self-eclipsed by the star and produce modulation of the observed light. This can be used to try and estimate e.g. the size and location of starspots. Again there are many degeneracies and ambiguities, but this has become a growth industry since the delivery of thousands of extremely high quality light curves from the Kepler satellite.


How do astronomers convert radio signals received by their antenna to images?

There is no universal answer to this question. Instrumentation, data formats and software differ greatly between telescopes.

Most software packages in use are written by astronomers as open source and shared and modified widely within the community. But instead of listing popular packages, perhaps a description of the common way of plotting images is a better answer here.

Consider a very basic radio telescope: A dish that's movable in two axis with a simple radiometer (think "power meter for RF") as the receiver. During an observation, you record four-tuples of time, azimuth, elevation and signal strength. Plotting these is pretty straightforward: Convert the instrument coordinates to celestial coordinates, and plot the signal strength over these coordinates. You'll get a classical intensity distribution map (source). Those maps can be created using simple spreadsheet programs like Excel or custom code snippets.

Adding the capability to record spectra to your receiver makes the table you record your data in a bit more complex (since your last column is no longer a scalar value but a spectrum), and allows the creation of intensity-velocity-plots (source). From a software perspective, this is still a simple plot-points-in-2D excercise.

For anything more complex, especially array setups with multiple antennas and/or focal plane arrays, data reduction and plotting requirements are far more involved and highly depend on the actual setup. CASA and AIPS, mentioned in the comments above, are widely used packages, but there is a variety of custom code around.


Distance between Earth and where The Big Bang's happened?

The distance to where the big bang theoretically happened would be "0" (and this should be true for any point within the universe).  The entire universe was theoretically produced in the big bang, with every point in space being "inside" the big bang.  The whole universe is very special, it is amazing what is out there (including black holes, earth, stars, planets, nebulae, etc...).

Whether or not there is even a "center" to the big bang / the universe, however is another related interesting question.  Current models seem to indicate that there is no center.  It would be interesting if there is a true center of the universe, and would be interesting to know if there is anything special about the center if it exists.

The question of whether or not there is a center of the universe is addressed in the following links:

What is in the center of the universe?

http://www.spaceanswers.com/deep-space/is-there-a-centre-of-the-universe/


How is Mercury tidally locked if the ratio is not 1:1?

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


Can visible wavelength spectroscopy study an exoplanet's chemical composition directly?

This can be done by high-contrast imaging.

Currently it is only feasible for a very limited class of systems (e.g. HR 8799) since it is difficult to achieve the required angular resolution (0.1 arcsecs for exoplanets at a distance 10 parsec) with today's adaptive optics.

Future large telescopes such as the European Extremely Large Telescope are designed for direct imaging as well as visible and near-infrared spectroscopy of exoplanets.


Why is there a gap in this image of supernova discoveries?

The coordinate system in this image is RA and Dec. It is a coordinate system which uses the Earth's equator (projected onto the sky) as its midline.

The inverted U is the Milky Way. The Milky Way is full of dust and gas, and blocks our view of galaxies (and supernovae) behind it. There is enough dust in the plane of the galaxy to block our view in that direction. For example the galaxy IC 342 is one of the nearest galaxies, and would be brilliant if it were not close to the galactic plane. There may be other galaxies that are completely hidden.

Our galaxy's bulk not only hides supernovæ that are in other galaxies, it also hides most of the supernovæ that occur in the Milky Way


Does anyone have spectra on exoplanet HD 219134b? How could one know either way?

For an authoritative source of data on exoplanets, your best bet is the NASA Exoplanet Archive.  Their page for HD 219134b is here.
For planet spectra, the archive has a page for planets with transit spectroscopy, but searching that page for HD 219134 yields no results.  Note that transit spectroscopy is challenging, so there aren't very many planets to which this technique has been applied.
In general, if you want to search for literature data on a particular astronomical object (whether it's an exoplanet or not), two good bets are NASA ADS or Simbad.  For ADS, you would type object:"HD 219134b" in the search box.  For Simbad, you would do an object search for "HD 219134b", and then scroll down to the "References" section and click "Display".  The advantage of both of these types of searches over a general web search, or even a general database search with that string, is that different names for the same object are cross-indexed; for example if you searched one of those for "alpha Ori", you would also find papers that only used the name "Betelgeuse", and vice versa.


Calculate depth and duration of exoplanet's transit using Python, astropy or lightkurve

Calculation of both, depth and duration, is usually done not on the raw data but derived from a fit to the data.

In your last three lines of code you also calculate the average / medium over all data while you should calculate the uneclipsed mean or median flux only for the non-transit time (with using median it possibly has only a tiny influence, yet it might).

As a first and crude step, I'd de-noise the data by applying a floating average filter over the data; you will have to test for its width and see what gives you best results: you don't want to average out features, but you want to average-out noise.

The better approach is to not smooth but actually fit a physical model to the data which takes into account the typical light curve behaviour of a transit. For an implementation I can point you to pytransit (reference paper). (Are you sure you are not re-inventing the wheel?). See also this paper by Maxted and Gill for a comparison of a few algorithms.


What does "Active Weather Patterns" mean?

It means that Titan has weather (driven by methane rather than water) and that it's weather changes with the seasons. Cassini has been observing Titan for almost half of a Titan year, which is 29.457 times longer than is our year. Titan's weather patterns (where it's cloudy, where it rains) has visibly changed over this time, just as weather patterns here change from summer to winter / winter to summer.


How much longer is this (July 27th 2018) lunar eclipse than others?

The NASA has a lot of data on eclipses. I've pulled the table for the 21st century in Excel. There are 228 eclipses of which 85 are total. The durations below are for the totality phase.

The duration of today's eclipse is 103 minutes.
The average duration is 68.7 minutes.
The duration of the next total eclipse, January 21st, 2019 is 62 minutes.
The shortest eclipse this century, April 4th, 2015, is 4.7 minutes.
The second longest eclipse this century, June 26th, 2029, is 101.9 minutes.
The median duration, as requested by @FooBar is 71.7 minutes (slightly more than the average).

For what it's worth, here in the Netherlands it's probably also the warmest lunar eclipse ever; at sunset, when the moon rose, it was 31°C.


What (actually) is Jupiter doing to this year's Perseids meteor shower?

To put it in simple terms, a comet's debris trail or dust trail is like a ring of debris that generally follows the entire orbit of the comet, more concentrated close to the comet, more spread out further away, but it can be thought of, for simplicity, as an entire ring of debris, kind of, a miniature asteroid belt with a fixed elliptical path.   The picture below isn't perfect, cause it gives the impression that orbits are all the same when there's some spreading out, but it's the gist of it.  (Note, the debris field and comet in the picture below isn't Swift-Tuttle/Perseid)

The reason for this is that any debris that's knocked off the comet at a low velocity has pretty much the same orbit as the original comet, and over hundreds or thousands of orbits, it tends to fill up the entire comet's orbit with mostly tiny particles, some larger ones.  The debris field or dust trail from the comet generally follows in the comets orbit.

This picture is a bit more accurate as to what the Earth is passing through.

https://lintvwpri.files.wordpress.com/2016/08/perseid_base4.png?w=650&h=366

The more compact the dust trail or debris field, the more meteors, but the smaller the target the Earth has to hit, so odds are greater that the Earth will miss the field entirely.  The more spread out, the greater chance the Earth hits the debris field, but more spread out means we get fewer meteors.

Earth's orbit crosses the Perseid (eliptical ring of debris) or Swift-Tuttle's debris field about the same time every year because Earth's and Swift-Tuttle's orbits nearly intersect.

This year, because of the placement of Jupiter, the debris field (which we can kind of think of as a ring), got bent by Jupiter, and Earth is hitting closer to the center of the debris field which means, more meteors.   I can't say on which plane the field got bent, cause it depends on it's orbital relation to Jupiter and I couldn't find that, but it was the ring of debris that got bent, so Earth will hit it more directly this year than usual.

Because the ring of debris is pretty much permanent, Jupiter bends the ring every 11 years as it passes closest to it, and occasionally that bend works in earths' favor and gives us more meteors, so its a rare event that might not happen again for 11 years, it might be over 100 years before Jupiter does this for us again.

See article.

  Typically Earth just grazes by Swift-Tuttle’s debris field, but our
  planet will be even closer to the particle stream this year thanks to
  some help from our neighbor Jupiter. The gas giant occasionally gets
  close to the comet stream, and its immense gravity pushes the debris
  nearer to Earth.


What's the difference between optical telescopes and solar telescopes?

Solar telescopes are optical telescopes, and it is quite possible to use an optical telescope to observe the sun. The usual way is to use the telescope to project an image of the sun, by placing a card at the focus of the eyepiece.

A professional telescope is a complex piece of engineering, it is designed for a particular task, such as observing very faint objects. It isn't designed for solar observation, and would most likely be rapidly damaged if it were pointed at the sun.

Solar telescopes have plenty of light to observe. The technical challenges involve controlling that light, so that it doesn't damage the equipment, and filter it, so particular aspects of the sun can be observed. The engineering problems of a solar and "nocturnal" telescope are different, so you can't use the same professional telescope for both. Analogy: an F1 race car and a rally car are similar in some ways but have been designed to overcome different technical challenges. An F1 car would never win a rally, nor would a rally car win at F1.


When I add 289P/Blanpain in the solar system editor of Stellarium,
it too shows the the comet blazing away at apparent magnitude -1.4.
The problem must be upstream.
Planetarium software typically gets comet and asteroid data from the Minor Planet Center,
whose own ephemeris service gives a similar result under m1:

0289P
Date       UT      R.A. (J2000) Decl.    Delta     r     El.    Ph.   m1     Sky Motion
            h m s                                                            "/min    P.A.
2019 11 30 000000 22 29 37.8 -16 25 42   0.280   1.006   85.9  78.0  -0.2    1.25    023.6
2019 12 15 000000 22 46 29.4 -06 51 49   0.201   0.963   78.0  90.2  -1.1    2.39    021.8
2019 12 30 000000 23 17 53.0 +14 42 59   0.123   0.968   79.2  93.7  -2.2    6.04    019.2

However, MPEC 2019-W179 from the same organization lists some actual observations from late November and gives this ephemeris:

Date    TT    R. A. (2000) Decl.     Delta      r    Elong.  Phase   m1    m2
2019 11 30    22 29 37.7 -16 25 45   0.2802  1.0058    85.9    78.0  17.3
2019 12 15    22 46 29.3 -06 51 54   0.2013  0.9628    78.0    90.2  16.4
2019 12 30    23 17 52.7 +14 42 50   0.1232  0.9678    79.2    93.7  15.3

MPC has collected other observations since then.
The JPL Small Body Database record, based on the same observations, estimates a total absolute magnitude M1 = 20.8±1.
HORIZONS gives this ephemeris:

Date__(UT)__HR:MN     R.A._(ICRF/J2000.0)_DEC  T-mag  N-mag
2019-Nov-30 00:00     22 29 37.44 -16 25 42.6  18.05   n.a.
2019-Dec-15 00:00     22 46 28.88 -06 51 47.8  17.25   n.a.
2019-Dec-30 00:00     23 17 52.05 +14 43 16.0  16.19   n.a.

If I edit my .stellarium/data/ssystem_minor.ini

[289pblanpain]
absolute_magnitude = 2.5

and replace the faulty 2.5 with 20.8, Stellarium predicts a more realistic magnitude 16.9 for the comet today.
The MPEC ephemeris is consistent with M1 ≈ 20.0.

The MPC data file CometEls.txt lists a few other comets at absolute magnitude 2.5.
JPL estimates for those comets range from 4.8±0.8 to 10.7±1.


From my limited experience, being able to see the rings on Saturn is more or less the extent of how powerful eyepieces can get for a home telescope. To get better and larger images would require a better and probably larger telescope.


How long does it take for an asteroid to get to Earth?

The full orbital period of an object orbiting the sun is $$T=2\pi \sqrt{\frac{a^3}{GM_\odot}},$$ where $a$ is the semi-major axis of the orbit and $M_\odot$ the mass of the sun.

If the asteroid has apoapsis in the asteroid belt ($r_{max}= 3$ AU) and periapsis near Earth ($r_{min}=1$ AU) then the semimajor axis will be $a=(r_{max}+r_{min})/2=2$ AU and going in from the belt takes half a period, so the answer is $\pi\sqrt{(2 AU)^3/GM_\odot}=1.4140$ years. An asteroid coming from further out, say 4 AU, would take 1.9762 years. From 2 AU, 0.9184 years.

Of course, were one to just give an asteroid a big push it could move nearly arbitrarily fast (up to the lightspeed limit if we want to be absurd) along an Earth-intersecting orbit, but I suspect the question was about normal asteroids.

However, it is worth noting that long-periodic comets can move fast (tens of km/s). Such a comet (or an interstellar asteroid) would have a velocity on the order of $\sqrt{2GM_\odot/(1AU)}\approx 42.1285$ km/s if it came in from the outer edge of the solar system. That would cross the 2 AU from the asteroid belt to Earth in about 82 days.


If you looked at Earth while travelling at relativistic speeds, will you see sped-up activities?

It doesn't have to be a visible telescope, any electromagnetic radiation will act in the same way.

I'll give you the answer, but if you want the details you need to look at a description of the resolution to the so-called  twin paradox, for for example here.

Assume that the spaceship travels  outwards and back to Earth at the same speed, with a  brief period of acceleration in between. From the point of view of the observer on the spaceship, they will receive signals at at lower frequency on the outward half of the journey. The change in frequency to depends on how fast they go and is given by the relativistic Doppler effect formula.
$$f_{\rm out} = f \left(\frac{c - v}{c + v}\right)^{1/2} = \alpha f,$$
where $\alpha < 1$ and $f$ is the frequency of the light/signal on Earth. So the Earth would appear slowed down.

On the way back the observed frequency
$$f_{\rm in}= f \left(\frac{c + v}{c - v}\right)^{1/2} = \frac{f}{\alpha}$$ and the Earth would appear speed up.

According to the observer on the spaceship they spend half the travel time in each of these phases. The average frequency they receive is
$$\bar{f} = \frac{(f_{\rm out}+f_{\rm in})}{2} = \frac{f}{2} \left(\alpha + \frac{1}{\alpha}\right),$$ which is always $>f$.

This means that on average things on Earth appear speeded up, such that the Earth is older upon return, but what is seen from the spaceship is divided into distinct outbound and inbound phases.


What are the equations for an orbit with time as parameter?

I have done a lot more research into this and found an answer. It comes down to solving Kepler's Equation, and my main references are:
Kepler's equation and the Equation of Centre which gives an iterative method for solving the equation, and Determination of Position in Orbits which has the formulae for calculating orbital coordinates.

Here is the Processing sketch. If you are not familiar with Processing, it is a language developed for graphic designers, but much more versatile than just that, based on Java and hence C++. It expects two functions, one is setup(), called once at the start up of the sketch, and a second draw() that is repeated over and over until the sketch is stopped. In case you wonder: the whole is encased in a main() before it is compiled. For more details see the Processing home page.

int wsize = 400;
float rp = 9.0; // radius of planet symbol
float rs = 15.0; // --"-- sun
// Constants of orbital ellipse
float a = 150, b; // major and minor axes
float e = 0.5; // eccentricity
float halfa = a/2.0;
float c; // focal distance from centre
float E, E0; // eccentric anomaly, temporary values in iteration
float v; // true anomaly = polar coord angle
float M; /* mean anomaly which is effectively the ..
            .. time variable round the orbit*/
float r, x, y; // polar coord radius, rect coords
float k, cosE, cosv; // useful intermediate variables
float twopi = 2*PI;
float PIby2 = PI/2.0;
float Minc = twopi/250; // time proxy increment

void setup()
{ size(wsize,wsize);
  //e = sqrt(((a*a)-(b*b))/(a*a)); kept for reference
  b = a*sqrt(1-e*e);
  c = e*a;
  k = 1-e*e;
  M = 0; // Mean anomaly increases linearly with time
}

void draw()
{ background(255);
  translate(wsize/2,wsize/2); // move origin to centre of display window
  ellipse(0,0,2*a,2*b); // show orbit
  spot(c,0,rs,#FFF303); // draw sun
  translate(c,0); // move origin to focus
  // Calculate Eccentic Anomaly by iteration
  E = M; E0 = 1000; // make sure while starts
  while(abs(E-E0) > 0.000001)
  { E0 = M + e*sin(E);
    E = E0;
  }
  // calculate polar coordinates of planet
  cosE= cos(E);
  cosv = (cosE-e)/(1-e*cosE);
  v= acos(cosv);
  r = a*(1-e*e)/(1+e*cosv);
  // calculate x,y coordinates of planet
  x = r * cosv;
  y = -r*sin(v);// negative makes it go anti-clockwise
  if (M > PI){y = -y;} // show second half of orbit
  // println(v+" : "+x+" : "+y);
  // line(0,0,x,y);
  spot(x,y, rp, #03ADFF); // draw planet
  M += Minc;
  if(M > twopi){M = 0.0;}
}

void spot(float x, float y, float w, color tint)
{ fill(tint);
  ellipse(x,y,w,w);
  fill(255); // restore white background
}


Would it be possible to equip an asteroid to collect space junk in Earth's orbit?

This is plausible, and might even be a good idea if used right.

First off, NASA has been working on plans for an asteroid redirect mission, called ARM. While it remains to be seen whether this will be approved, and they plan to put it into lunar orbit, this is a hypothetical scenario anyways so I'll ignore that.

Now, putting an asteroid into earth orbit is a little bit of a difficult subject. Sure, you could do it, but messing up would net you a whole lot of problems. I presume it would have to go into high earth orbit if it goes into orbit at all.

This rules out using the asteroid directly like a satellite, but that would have been impractical even if it was sanctioned by everyone with enough influence to have an opinion on earth. From there the satellite would be best used as a base of operations, from which you could send drones to collect junk and keep refueling stations. I'd also include a telescope or two, but that's personal preference.

If you want a list of targets though, here would be a good place to look.


What does the Reddening line represents in this U-B vs B-V plot?

The reddening line is the locus that a star would move along in a colour-colour diagram as the interstellar reddening towards it increases.
i.e. It is an equation on the colour-colour plot of the form (for this colour-colour plot)
$$E(U-B) = f(E(B-V)) + c\ ,$$
where $E(U-B)$ is the colour excess in $U-B$ due to interstellar reddening, $E(B-V)$ is the colour excess in $B-V$ and $c$ is a constant chosen so that the locus would go through a star of the same spectral type but with no interstellar reddening.
Typically, for low reddening and a typical interstellar medium
$$ E(U-B) = 0.72 E(B-V) + 0.05 (E(B-V))^2\ ,$$
which I imagine is the line drawn on the plot you show.
To establish the reddening to a particular star you can follow it back from its observed position in the colour-colour plot, along a reddening line, until it intersects the main sequence locus (if it is a main sequence star). The intersection would give its unreddened, intrinsic colours and the x-axis increment through which you traced it back would give its $E(B-V)$.


Characteristics of the first planets in the Universe?

According to Mashian & Loeb (2016) (also in ArXiv), one possibility is that planet formation may have occurred around carbon-enhanced metal poor (CEMP) stars in the early universe. The paper focuses on CEMP-no stars as the most metal-poor CEMP stars mostly fall into this category.

CEMP-no stars are thought to form from material polluted by supernova ejecta from Population III stars. The main grain species that forms in these ejecta is amorphous carbon. From section 2 of the paper:

  Tailoring the
  SN explosion models to reproduce the observed elemental
  abundances of CEMP-no stars, they find that: (i) for all the
  progenitor models investigated, amorphous carbon (AC) is
  the only grain species that forms in significant amounts;
  this is a consequence of extensive fallback, which results in
  a distinct, carbon-dominated ejecta composition with negligible amounts of other metals, such as Mg, Si, and Al, that
  can enable the condensation of alternative grain types; (ii)
  the mass of carbon locked into AC grains increases when
  the ejecta composition is characterized by an initial mass
  of C greater than the O mass; this is particularly true in
  zero metallicity supernova progenitors, which undergo less
  mixing than their solar metallicity counterparts (Joggerst,
  Woosley, & Heger 2009)

In this case, the earliest planets may have been carbon planets. Carbon planets with bare graphite/amorphous carbon surfaces (the carbon planet equivalent of the Moon or Mercury) would be very dark. Atmospheres would modify this appearance. As the paper notes, hot carbon planet atmospheres are expected to have spectra dominated by carbon monoxide. Cooler carbon planets would likely support various kinds of hydrocarbon chemistry that would modify their appearance, possibly producing reddish tholins as a result of photochemistry depending on what else is around.

Mashian & Loeb predict that the maximum distance at which planets form would depend on the carbon abundance:

  We find that carbon planets may be orbiting iron-deficient stars with carbon abundances [C/H] ∼ -0.6, such as HE 2356-0410, as far out
  as ∼20 AU from their host star in the case where fcond = 1.
  Planets forming around stars with less carbon enhancement,
  i.e. HE 0107-5240 with [C/H] ∼ -1.6, are expected to have
  more compact orbits, with semi-major axes r < 2 AU. If the
  carbon condensation efficiency is only 10%, the expected orbits grow even more compact, with maximum semi-major
  axes of ∼5 and 0.5 AU, respectively.

So far this is just hypothetical: we don't know of any planets around CEMP stars, nor of any carbon planets but this may suggest that it is not completely pointless to search for them.


Is the Drake equation based on a logical fallacy?

To directly answer the question, the Drake Equation is not based on a "logical fallacy" (which means a different thing, philosophically). The logic is fine, the real problem is we don't have any meaningful data.

Regarding abiogenesis (the first formation of some simple life): as you correctly state, we know only that it happened once (here, presumably but that doesn't matter). However the argument that it happened quite quickly on our planet is meaningless statistically, because that's just a sample of one.

(If anyone thinks this answer is too waffly or off topic just say so and I'll delete it by the way; I think the question is a fair one but we just can't present any data, just reason.)


How big would the Earth look from the Moon?

There are a couple questions here. The ones you have asked are:

What would be the ratio between the respective diameters?
How would this ratio be calculated?

But I think you meant to ask:

How much larger does the Earth appear when standing on the Moon than the Moon appears when standing on the Earth? (and the calculation)

For the answer to #1 (and #2), it is the ratio between the numbers (of the Earth's diameter of 12,742 km and the Moon's diameter of 3,474 km) which is $Earth:Moon = 3.668:1$ as @pela's comment mentions.

For the answer to #3, it is the same! The more interesting part is why. We know that the distance from the Earth to the Moon is the same as the Moon to the Earth, so it follows that when we are on one of them looking at the other we are gazing across the same distance. Next, we know that an object (approximately) appears larger in the same proportion as it is larger, so an object that is twice as wide or tall will appear twice as wide or tall. Since the Earth is 3.668 times as wide as the Moon, it will appear that much larger when viewed from the same distance, and we've already discussed why it is the same distance. So, that's it really. It is the ratio of the diameters.


How do we get radio signals of the big bang?

The best estimates at the moment are that the Universe is "flat". By that we mean that a pair of parallel laser beams will travel as far as you like and they will always remain the same distance apart and parallel.

Ironically, the best evidence for this comes from detailed measurements of the cosmic microwave background by the WMAP and now Planck satellites.

The problem you are having is you (I think) are imagining that the big bang happened at a point in space 13.7 billion years ago, and that radiation has been travelling outwards from that point. This is incorrect. Every point in space that we see now was, about 13.7 billion years ago, part of the big bang. As the space in the Universe expanded, its contents cooled and due to the trapping of free electrons by positively charged protons to form hydrogen atoms, the light that was emitted by the hot gas everywhere in the Universe at a time about 400,000 years after the big-bang was able to travel unabsorbed and unscattered. This radiation has travelled (at the speed of light) in all directions.

Now, when we observed the microwave background, we are seeing light that has been travelling in a straight line for just short of 13.7 billion years. That light was in the near infrared and visible part of the spectrum when it was emitted, but because of the expansion of space, the wavelength has also been stretched by a factor of 1100 to the microwave/short-wave radio part of the spectrum.


Connection between Earth, Moon, Theia, and asteroid belt?

Let's start answering your question in reverse. Ceres could not be "the missing Theia", because of its shape. If an object hit the Earth at an angle (as is currently thought), it would be pretty deformed, if it managed to stay together. If it hit the Earth head on . . . Well, it would almost certainly not survive. So if it did survive the impact, it would be in pretty bad shape. Where would such an object go?

Let's investigate the idea that it went into the asteroid belt. That begs a question: Where is it today? Ceres is the largest object in the belt (and we've already ruled it out). Theia had a mass about that of Mars (see Canup (2012)). It's safe to say that if Theia survived with much of its mass intact and then disappeared into the asteroid belt, we should have observed it by now.

Now for your belt-formation theory. It is, admittedly a very cool idea. The mass of the asteroid belt is about 4% that of the Moon - so it could be the result of the planetary crash. And some theories say that the belt is, in fact, the remains of a planet. There is, of course, one problem - the different composition of the asteroids. There are three types, differentiated by their composition - C, S, and M. So it is very unlikely that they came from the same planet. Meier et al. (2014) show evidence that indicates Theia's composition would have been more like Earth, containing enstatite chondrites instead of carbonaceous chondrites.

Finally, let's look at your idea about the Earth having an orbit farther out. This one seems to have a good chance of being right. The early protoplanets in the Solar System were very - very - unstable in their orbits because of the frequency of collisions. So the Earth could have been in a different orbit early on, and then migrated inwards or outwards, though planet-planet scattering or perhaps gas disk migration.


What is wrong with this measurement of the synodic period of Mercury?

Mercury's orbit is highly eccentric: 0.21 according to Wikipedia. Therefore, the actual time between repeating occurrences will vary depending on the year. If you were to perform your calculations for many periods, the average should approach the value given by Stellarium.

The theoretical synodic period, using the sidereal period of Earth and Mercury, is 115.9 days.


Does Ganymede have a day/night cycle? If so, what does it look like?

Here's some pertinent facts:

Apparent magnitudes: Sun from Ganymede     = -23.11
                     Jupiter from Ganymede = -16.02 (full)
                     Difference            = - 7.09
                     Brightness difference = 2.512 E7.09 = 686 times brighter

Earth comparison:    Sun from Earth        = -26.74
                     Moon from Earth       = -12.74 (mean full)
                     Full moon/ Jupiter from Ganymede difference = 20.5 times

Ganymede facts:      Orbital period = 1 Ganymede day: 7.16 days
                     Orbital inclinations: .2 deg to Jupiter equator
                                          2.2 deg to ecliptic
                     Orbital eccentricity: .0011 (circular)
                     Orbital speed (ave.): 10.88 km/s

Jupiter facts:       Mean radius: about 70000 km

Summary:
If you are on the "far side" of Ganymede you won't see Jupiter (maybe a glow on the horizon from Ganymede's thin atmosphere). The sun will be 2.512 E3.63 = about 28 times dimmer than seen from Earth (which is still pretty darn bright). A day will last 7.16 Earth days.

The near side will be far more interesting. The sun will go through the same cycle at the same brightness as on the far side. However, Jupiter will "hover" in the sky (maybe a little wobble/ libration) and when full will be about 686 times dimmer than the sun in the sky. Jupiter will go through phases over the course of the 7.16 day cycle and be less bright of course as it approaches eclipse. The eclipse (by rough calculations) will last about three and a half hours. The planet will have an angular diameter of about 7.5 degrees (compare with 0.5 for sun/moon from Earth) and be about 20.5 times brighter than a full moon when Ganymede is between the sun and Jupiter.

A fun side note is that the sky won't be totally dark during eclipse. Light from the sun will reflect off aerosols and hazes in Jupiter's upper atmosphere making Jupiter "glow" in Ganymede's sky.


How long does it take to produce a star? Why does it take that long?

The time to "produce" a star actually depends on the mass of the star. Let's start with a solar type star.
For a typical solar type star, the formation time is about 10 million years (you can see this image on Wikimedia that gives an overview of the star formation process, adapted from Philippe André's figure in Observations of protostars and protostellar stages in The cold Universe, 1994). You can distinguish different epochs in the star formation process, that are the signature of different dominant physical processes. The very first stage of star formation is a gravitational collapse that leads to the formation of the protostar itself. The timescale for this collapse is the so-called free-fall time which depends only on the density of the object. When you get a central object in a hydrostatic equilibrium, things become more subtle: the core will contract adiabatically (without heat transfer) and when a temperature of about 2000 K is reached, dihydrogen dissociates (which is a highly endothermic reaction) which leads to a second phase of collapse, leading to the formation of the protostar itself. It takes about 1000 years to get to this stage, from the hydrostatic core.
The next stage, the protostellar phase, is mostly an accretion phase. It means that the timescale for this epoch is given by an accretion time, that varies with the accreted mass (which is quite low for a solar type star). It takes about 200 000 years to accrete 90% of the final mass of the star.
Then, since the star is still contracting, the temperature at its center is increasing; when a temperature of 1 million Kelvin is reached, the protostar starts to burn its deuterium. At this stage, the Kelvin-Helmholtz mechanism allows the protostar to contract and to radiate away its gravitational energy. The meaningful timescale is then the Kelvin-Helmholtz time (which varies as the square of the mass and the inverse of the radius and the luminosity), that is much longer than the previous timescales. Temperature continues to rise, up to 10 million Kelvin, when hydrogen eventually starts to burn, which is the birth certificate of a star. It takes around 10 million years to get to this point.
But, as I said, this scenario depends on the mass of the star. It is valid for Sun-like stars, but not quite for massive stars. It is much faster for massive stars, and the star formation process is quite different. In particular, the accretion rate is much higher, the radiation pressure of the protostar is insanely higher, and their interplay is not completely understood. However, there are some theroretical works that gives an estimate of about 100 000 years to form a massive stars (see for example works from McKee and Tan).


Was the cosmic microwave background (CMB) radiation originally gamma rays?

The cosmic microwave background is a result of an almost perfect blackbody emitter. That means the spectrum covers a broad wavelength range with a peak that is given by Wien's law:
$$\lambda_{peak} = \frac{2.9\times 10^{-3}}{T},$$
where the wavelength is in metres and temperature in Kelvin.

The microwave background is formed at the epoch of (re)combination, when protons and electrons combine to form hydrogen atoms and the Uinverse effectively becomes transparent to the radiation field within it. This occurs at approximately 3000 K. So at that time, the cosmic microwave background had a peak wavelength of 970 nm, which in the near infrared part of the spectrum.

The epoch of (re)combination occurred at a redshift of approximately $z=1100$. What this means is that since then, the wavelength of the photons has been stretched by a factor of $1+z = 1101$, so that $\lambda_{peak} \simeq 1$ mm, and this corresponds to a temperature of $T=2.7$ K. I.e It is near infrared photons emitted at the epoch of recombination that have their wavelengths stretched by the expansion of the universe, and are seen today as microwaves.

However, prior to the epoch of recombination, the Universe would still have been full of radiation, and as we go back in time towards the big bang, that radiation field becomes hotter and hotter, and the peak wavelengths get shorter and shorter.

If we look at models for the thermal history of the Universe, then at about 1 second after the big bang, the Universe would have been at about $10^{10}$ K and the radiation field would have been in equilibrium with the matter at this temperature. The formula above then gives $\lambda_{peak}=2.9\times10^{-13}$ m, which is definitely in the gamma-ray regime. But let me emphasize that these photons are emitted then absorbed many times and we do not see them now.

The "particles" that you are talking about are photons of light.


Convert orbital elements to positions and velocities

Is this what you're looking for? When you say you want something defined for an observer on the earth looking at the system, do you mean you wish to set the origin of the system at the earth? This can be done, but I don't really see why you would do it unless the earth was intrinsic to the system.


What is the current accepted theory as to why Venus has a slow retrograde rotation?

There seem to be a few, and none are accepted by the whole scientific community. The main ones:

Venus was struck by a large body during its early formation
The spin axis flipped, as can happen with a gyroscope
The spin slowed to a standstill and then reversed, caused by the sun's gravity, the dense atmosphere and friction between core and mantle

That final one seems to be the most recent, being proposed by Alexandre Correira and Jacques Laskar in 2001. Their research seems to imply that the conditions on Venus and its distance for the sun make a retrograde spin slightly more likely than a forward one.


If one galaxy were to collide with another, what would it look like to an immortal inhabitant of one of their planets?

In the vast, vast majority of cases, not much happens. Galaxies are a whole lot of empty space.

If the Sun was a soccer ball in a field in California, the nearest star of similar size, Alpha Centauri, would be another soccer ball all the way across the continent, in Greenland. This is a scaled view, where everything is proportional. Think of all that empty space in between.

So the galaxies would merge, or keep swirling around each other, but for the most part stars and their systems would just continue as they were. It is extremely unlikely that any given star would collide. That's true for most of the volume of the galaxies.

In the center, however, things are a little different. Most galaxies already have very massive black holes at the center. These may collide and merge, or orbit each other a very long time, or continue separately.

The formation of new black holes as a result of galactic collisions must be a very rare event. Even then, the entire galaxy would not "become a black hole"; rather, a small fraction of it would be added to an existing black hole or, in very rare cases, produce a new black hole.

For most observers in such galaxies, there would not be dramatic changes, even over very long time intervals. This is because, again, galaxies are mostly empty space, and because the stars you see with the naked eye are very close to you, within a few dozen to a few hundred light years, which is a small portion of the whole galaxy.

EDIT: As shown in the comments, it's more accurate to think of Alpha Centauri as not a single soccer ball, but instead a larger soccer ball and a volleyball separated by a few hundred meters - it's a double star. Then Proxima Centauri would be a pingpong ball hovering 100 km away from Alpha.


How often and over what period is Earth’s rotation averaged to compute UT1?

Conceptually, the average is over an average year. Practically, that's not what's done anymore.

Sidereal time is much more readily measured than is solar time. Radio astronomy can measure the position of remote quasars much, much more precisely than they can measure the position of the Sun. The Sun is a big fiery blob with a not quite-well defined surface. Quasars are pinpoints. Very long baseline interferometry makes the observations of those quasars extremely precise.

The equation of the equinoxes converts between apparent and mean sidereal time. A very simple relation converts between mean sidereal and mean solar time. This makes variations in UT1 observable on a rapid basis. Sub-arc second astronomy depends on these rapid updates. IERS Bulletin A is updated daily, with the data for the most recent week or so subject to subsequent small updates.

Update

Universal Time is not based on apparent solar time. Time is now effectively divorced from the the apparent revolution of the Earth with respect to the Sun. It took several thousands of years to make this divorce final. (And the divorce is still not quite final. There's the matter of leap seconds. There's a proposal to eliminate leap seconds that has been shelved/postponed multiple times since 2003. A vote is scheduled for 2015.)

The ancient Egyptians divided daytime into 12 equal parts and nighttime into 12 equal parts. Daytime and nighttime hours were of different lengths, and these lengths varied over the course of a year. Hipparchus (190 – 120 BCE) suggested dividing one day into 24 equal parts, the basis for our modern hour. Ptolemy, using information already garnered by ancient Babylonians, asked "what day"? The length of a solar day (local noon to local noon) varies over the course of a year thanks to what we now call the equation of time.

This was of mostly academic interest until Christopher Huygens' development of the pendulum clock in 1656. Pendulum clocks tick at a rate proportional to the mean solar day rather than the apparent solar day. Those early pendulum clocks weren't particularly accurate. They had to be reset on a regular basis. The equation of time was central in determining the time to which those clocks needed to be set.

By the mid 1800s it was becoming apparent that even a mean solar day was not that good a basis for defining time. The clock kept by the solar system didn't agree with a clock kept by the Earth's mean rotation. A clock based on the solar system was more accurate and more consistent than was a clock based on the Earth's rotation.

By 1900, the divorce between Earth rotation and time was well on its way to being finalized. The basis for time was switched from the Earth's rotation to the clock kept by the solar system, with the second based on the length of a year rather than the length of a day. About 150 years of observations of the length of the mean solar day went into that circa 1900 definition of the second.

In 1967, time was further divorced from the clock kept by the solar system. By that time, atomic clocks had become extremely accurate devices, better than observations of the solar system. The International Atomic Time second was defined to be consistent with that circa 1900 definition of a second, which in turn was defined to be consistent with that ~150 years of observations of the length of an apparent solar day, centered on about 1821.

That 1821-based definition of a second as 1/86,400 of a day is no longer correct. A day is now about 86400.002 seconds long, and it will grow longer with passing millennia. This leads to the need to insert leap seconds to keep UTC in synchronization with the Earth's mean rotation. There is considerable debate with regard to whether even that should be done. There is a serious proposal to abandon leap seconds. This is now scheduled for a vote in the upcoming year (2015).


How to calculate the integrated source flux density of a continuous source?

Maybe too late but probably he is referring to only take into account pixels with an intensity greater than 3*sigma. If not, you will be counting random noise as continuum. Obviously you will lose a lot of pixels, but those pixels doesn't carry any information at all.


What is the smallest possible planet with same sea level atmospheric pressure as Earth?

In the textbook treatment the pressure decreases exponentially with height as $P(z)=P(0)\exp(-z/H)$ where $H=kT/mg$ is the scale height of the atmosphere. This assumes that the gravity $g$ is constant with altitude, which is a decent approximation for planets with an atmospheric height much smaller than their radius. Since $H$ is about 8 km for Earth, this makes sense to use.

For an atmosphere with Earth-like temperature $T$ and mean molecular mass $m$ it would hence seem that for the tungsten planet that if we want $P(0)$ to be like on Earth $g$ must be equal to Earth's 9.82 m/s$^2$. So if we have a planet with $g=G\rho (4\pi r^3/3)/r^2$, we get $r=3g/4\pi G\rho$. For $\rho=19300$ kg/m$^3$ this gives $r=1820$ km.

The limit is met somewhere before white dwarf density. They have a density of about a billion kg/m$^3$ and a 35 meter radius white dwarf would have terrestrial surface gravity. At this point the above approximation has broken down (the scale height is bigger than the radius), and besides the low gravity would not be enough to hold it together. The real determinant of the limit will be the highest density material you can get that is stable at (near) zero pressure. For molecular matter tungsten is likely close to the limit, and we have reason to doubt degenerate white dwarf matter is up to the job.

Note that small but less dense bodies can have quite dense atmospheres. Titan has radius about 2550 km and a surface pressure 1.48 times Earth, despite lower surface gravity. The trick here is that the atmosphere is colder and slightly denser.


Will stars live longer if they consume more hydrogen?

Adding mass of similar composition to the Sun would effectively turn the Sun into a higher-mass main-sequence star.  That would not increase the gravitational pressure, it would lower it, because it would cause the Sun to expand to lower density.  It would also cause the Sun to fuse hydrogen faster, to the point that it would actually reduce the lifetime of the Sun, not increase it.  All this can be seen by consulting models of higher-mass main-sequence stars.

If it seems counterintuitive that adding mass would weaken the gravity, this is because the core must self-regulate its fusion rate to match the rate that light is escaping, and simply adding mass would upset that balance unless the Sun expanded to recover it.  Note that main-sequence star models do not particularly care about the history of how the mass got there, they are simply the only equilibrium models that have the appropriate mass and composition and are undergoing core fusion.


Sun, Moon: Relative Distance from Crescent?

This is exactly how the ancient Greeks attempted to estimate the relative distances to the sun and moon.

The difficulty is the angles involved are rather small

Aristarchus measured the Moon-Earth-Sun angle at half moon to be 87 degrees. (or 3 degrees less than a right angle) Which implies the Sun is about 19 times further than the moon. And from this (and an estimate of the sizes of the Earth and moon made at a lunar eclipe) he estimated the distance of the sun.

The actual angle is more like 89.85 degrees (which is very hard to distinguish from 90 degrees) (0.15 degrees less than a right angle).

So this method can be done, but it requires very exact measurement to succeed.


Is there a public record of planetary disks apart from ours?

A useful resource for protoplanetary discs is the recent DSHARP survey using the ALMA telescope. With the ALMA telescope the spatial resolution is good enough that orientation (i.e. position angle and inclination) can be derived from the images.
See here.


If Sunlike stars become a red giant and eventually a white dwarf, what do red dwarfs become?

A relevant paper here is Laughlin, Bodenheimer & Adams (1997) "The End of the Main Sequence". From the abstract:

We find that for masses $M_\ast < 0.25\  M_\odot$ stars remain fully
convective for a significant fraction of the duration of their evolution. The maintenance of full convection
precludes the development of large composition gradients and allows the entire star to build up a large
helium mass fraction. We find that stars with masses $M < 0.20\ M_\odot$ will never evolve through a red giant
stage. After becoming gradually brighter and bluer for trillions of years, these late M dwarfs of today
will develop radiative-conductive cores and mild nuclear shell sources; these stars then end their lives as
helium white dwarfs.

Section 3 of the paper provides a detailed description of the lifetime of a $0.1\ M_\odot$ star. A brief summary:

After approximately 2 Gyr of contraction, the star reaches the zero-age main sequence point with a temperature of 2228 K and a luminosity of $10^{-3.38}\ L_\odot$.

On the main sequence, the mass fraction of $^3 \rm He$ increases steadily over a trillion years. The completely convective nature of the star ensures that it is mixed throughout the structure of the star. The star slowly increases its temperature and luminosity.

The maximum mass fraction of 9.95% $^3 \rm He$ is reached at 1380 Gyr. After this, the mass fraction declines as the rate of consumption exceeds the rate of production.

Between 1500 and 4000 Gyr (the text appears to use values that are too small by a factor of 1000 judging by figure 1 and the statement of total lifetime at the start of §3.2) the star starts turning itself into $^4 \rm He$, with this isotope becoming the main component of the star (by mass) around 3050 Gyr.

By 5740 Gyr, the star develops a radiative core due to the helium mass fraction lowering the opacity. This causes a small amount of contraction of the star and a decrease in luminosity.

After the development of the radiative core, shell burning proceeds outward through the star, increasing the surface temperature to a maximum of 5807 K at 6144 Gyr. The luminosity at this point is about $10^{-2.3}\ L_\odot$. This star is called a "blue dwarf".

The star becomes cooler and less luminous. Shell burning continues during this time, eventually ending with the star having a hydrogen mass fraction of ~1%. The nuclear burning lifetime ends at 6281 Gyr, producing a helium white dwarf with temperature 1651 K and a luminosity of $10^{-5.287}\ L_\odot$.

A discussion of the appearance of the blue dwarf stage and how blue they actually are can be found in this question.
The $0.16 \le M_\ast / M_\odot \le 0.20$ range is transitional between the stars that become blue dwarfs and the stars that become red giants. From the paper:

In connection with their increased luminosity output, the
transitional stars in the mass range $0.16 \le M_\ast / M_\odot \le 0.20$
are able to produce increasingly larger expansions of the
overall stellar radius after the radiative hydrogen-exhausted
core has developed.

In the models calculated in the paper, the lowest mass object that unambiguously produced a red giant was $0.25\ M_\odot$, but as noted the transition region is not sharp. Nevertheless, this does mean that the higher-mass M dwarfs will eventually go through a red giant stage.


Deriving phase of a pulsating star from list of Julian date values of observing brightness

INT refers to the floor function. So $\mathrm{INT}(x)$ is the largest integer not exceeding $x$.

In this case, the function is used to get the fractional part of $\frac{\mathrm{HJD}-T_0}{P}$, i.e. the part after the decimal point. So if for example $\frac{\mathrm{HJD}-T_0}{P}=3.4$, then $\frac{\mathrm{HJD}-T_0}{P}-\mathrm{INT}\left(\frac{\mathrm{HJD}-T_0}{P}\right)=0.4$.


Actually, the stars and nebulae that make up the spiral arm are only temporarily part of that spiral arm. Spiral arms are more like sound waves where individual particles move around a more or less stationary position. (Look for instance at the animation of longitudinal waves from Dan Russel, the red dots move a bit to the left and to the right around a stationary position). Dust, gas and stars move towards or away from another just as longitudinal waves. Where the dust, gas and stars come close together (and where, therefore, the density increases), spiral arms can be seen as more stars are close together increasing the brightness at that position in the galaxy.

This effect is, furthermore, much increased because the increased density of dust and gas in the spiral arm causes protostars to form. The brightest stars burn up their energy so fast that they will cease to exist even before the longitudinal wave (the spiral arm) has passed. These very bright stars only exist for a small portion of their orbital period around the galaxy's centre, and only while they are in the spiral arm. The large majority of stars exist much longer, but are also much dimmer and contribute only little to the overal brightness of the galaxy.

This causes the spiral arms to be so much brighter than the rest of the disk, where also a lot of stars exist. But these can hardly be seen as they are much dimmer.

Of course, the stars do not revolve around a stable position in the galaxy (as the red dots in the wave animation) but follow their own orbits around the centre of the galaxy. Sometimes a bit faster, and sometimes a bit slower depending on the position relative to the spiral arms.

Because the spiral arms are waves, it does not matter that stars near the centre move faster than the stars at the edge. It just means that they will be part of the spiral arm for a shorter period of time.


How can you verify the theories presented on TV shows like "How the Universe Works" is current and/or valid?

Use other, more scientific sources.

You have no way to test them by some home measurements - science was tuned to pass even more sophisticated tests.

Popular media is often not very exact. They try to open the results to the public. For example, to find the geometry of the Kerr Black Holes was a decade long work for the best mathematicians of the time. The result is a relative simple formula, which is understable with not far more than high school math. Most in the popular media about the wormholes and white holes is based on it, or at least strongly related to it.

So, to check it, best if you try to find the formula and try to learn, what does it mean. It still doesn't need to understand the decade long work behind that, and also this is amazing - but only after you've understood it.


Once you are inside a black hole, every direction leads towards the singularity.  So for every particle, whether it is massless or not, would travel towards the singularity, and not escape.
Axions have not been proved to exist, but they would also not be able to escape.


While I'm lacking the formulae, the applicable surface area of a body would have to be great enough, and it's density low enough that it's gravity remains below that threshold. So a smaller body is much more likely. If you think about it, most dust from near the sun gets pushed out by the pressure.

There wouldn't be a specific point where you would get the matter buildup, it'd be more like the Roche limit where the gravity and density are the factors of the distance from the larger body.

Without running calculations I'm also unsure as to whether it would in fact be a stable arrangement but more like an L1 style Lagrange point, where it requires some energy every now and then otherwise it drifts.

Edits are welcome, I feel that I haven't used the best language available here and may have rambled.


If we could travel far enough would we be able to see a constelation in reverse?

If the stars are all similar distance from us, like the Pleiades, then yes, you could find a place from where the stars would appear in a similar pattern and brightness, but the pattern would be a mirror image.

You can't do it when the stars are significantly different distances because they wouldn't look the same.   Most constellations are a range of distances, which wouldn't look like a mirror image from any distance.

Take Canis Major for example.   Sirius is the brightest star in the sky but in large part because it's only 8 light years away.   The 2nd brightest star in that constellation, Epsilon Canis Majoris is about 430 light years away.   At that distance, Sirius might not even be visible to the human eye.  It certainly wouldn't be constellation worthy.


As it happens, I was writing code for the exact same transformation last night using this memo as a reference. I haven't yet tested my implementation but the document is quite clear about what each step is doing, which pleased me greatly.

You'll have to understand the terminology to some degree; "epoch" means a reference time for which you have the orbital elements, and "considered epoch" means the time at which you're evaluating the position, for example. Other points requiring caution: the equations inconsistently drop subscripts; the use of the Newton-Raphson solver isn't particularly clear (i.e. how many times should it run)? Make sure you get the units right also.


What is the tidal radius of the Sirius system?

The tidal radius for a star or collection of stars near the Sun and on a circular orbit (matches Sirius reasonably well) is given by (e.g. Pinfield et al. 1998)
$$R_T \simeq \left( \frac{GM}{2(A-B)^2} \right)^{1/3} ,$$
where $A$ and $B$ are the Oort constants and $M$ is the total mass.

Using the value of $A-B$ from Feast et al. (1997), this reduces to
$$R_T = 1.43 \left( \frac{M}{M_{\odot}} \right)^{1/3}\ \ {\rm pc}.$$

Using this formula, I get that the tidal radius for Fomalhaut is 2.01 pc (in reasonable agreement with whatever source you are using). As the mass of the Sirius system is only marginally larger then the tidal radius will be almost the same, since it depends on the cube root of mass.


How does gravity affect the wavelength of light?

A partial answer: The lower the gravitational potential the slower the clocks, see Gravitational time dilation. This means, as long as your flashlight is behind you in constant distance, it appears blue-shifted to you. With roles exchanged (light below you in constant distance) it appears red-shifted.
By the heterogeneity of the gravitational field the shift grows with the time, while falling in constant distance.

This is called the Einstein shift, which is not the same as the doppler effect. The doppler effect is caused by an objects velocity relative to the observer.


What were the challenges for the ancients to observe the orbit of the Moon (instead of Mars)?

Wasn't it pretty obvious to a careful astrologer a thousand years ago, that the Moon does not have a circular orbit and does not describe epi-cycles?

The ancient Greek model of the motion of planetary bodies remained unchallenged for almost two millennia, so obviously not.

Hipparchus' model did a fairly good job dealing with the elliptical motion of the Moon; it did even better with the planets. The Moon's motion is tough to model because of perturbations by the Sun, Venus, and Jupiter. Ptolemy discovered what would eventually be called evection, the largest of the perturbations caused by the Sun. There was one problem with Ptolemy's model: It had the Moon swinging in and out by a huge amount. If Ptolemy's model was correct, we would see the Moon changing in diameter by a factor of two over the course of a bit over half an orbit. Copernicus much later came up with a scheme that fixed this problem and still relied on those old concepts of deferents, equants, epicycles, etc.

While Newton pointed the way to describing the Moon's orbit, it wouldn't be until 200 years after Newton's death that a decent (one that matches observations) model of the Moon's orbit was developed.


First of all there could not be an eclipse on December 23rd because that is one day later than the full moon.

There can be a couple of reasons for the moon looking red or orange:

It is near the horizon, and being viewed through the atmosphere over a large city.  Air pollution (for example from cars, trucks, fires etc) push small sooty particles in to the air.  These particles scatter blue light more than red, so there is less blue light coming towards you directly from the moon.
Sometimes sand from deserts can get blown up into the atmosphere, and it can get carried a long way.  The sand has a similar effect to the soot.


I think it is absolutely safe to say that all neutron stars spin.

Conservation of angular momentum ensures that as they collapse from a massive stellar core the size (roughly) of the Earth, to something with a 10km radius, their angular velocity increases roughly as the square of the decrease in their radius (i.e. a factor of $\sim 4\times 10^5$. Thus, even if the stellar core had the slightest spin to begin with, then the neutron star will be spinning very fast indeed.

Young pulsars appears to be born with rotation periods that vary from about 0.01 s to perhaps a second or so. They then lose angular momentum as they get older.

The difficulty in answering your question without any equivocation is that once neutron stars spin down to periods of longer than 1-10 seconds or so (depending on the pulsar magnetic field strength - see picture below from Wang et al. 2011) then the pulsar mechanism switches off and the neutron star, to all intents and purposes, becomes invisible (the so-called "pulsar deathline", Zhang 2003) .

It is expected that the spin down will continue, even after the pulsar has switched off, but the spin down rate itself depends on the rotation speed, so the neutron star will never come to a halt, even after tens of billions of years.


What would night sky look like if Earth was made of antimatter

It's kind of a strange question but I'll give it a shot.

It depend, kind of obviously, on how much of the "space dust" hits the earth.   Estimates vary pretty widely on how much dust from space hits the Earth anyway, by this site, between 5 and 300 metric tons per day.

Lets start with the low end estimate, 5 metric tons (5,000 KG) per day.   Antimatter will meet with and turn into energy an equal amount of matter, so 10,000 KG per day turned into energy, presumably almost all gamma-rays, but a lot of that energy would reflect off the upper atmosphere, lets estimate 40% of it reaches the earth, so 4,000 KG of gamma ray energy hits Earth from Space every day.

4,000 KG, * C^2 (300,000,000^2) = 3.6*10^20 joules of energy per day.

The sun (I'll just borrow from this link rather than calculate).  1.6x10^22, so, the heat from the matter-anti matter interactions would be 2% of sunlight.   That would be enough to warm the earth measurably.  A 2% increase in energy form space could warm the Earth several degrees, maybe 10 degrees, maybe more.   The Earth would be measurably warmer, but that would be the least of our problems.

I think we can bring the number down a bit, cause you said no meteors, or asteroids, and even if you hadn't, larger objects like that in a matter-antimatter interaction, the contact would be so hot that most of the meteor wouldn't have a chance to evaporate but would simply be exploded away from the Earth, so we can round the number down quite a bit, like maybe a few hundred KG per day gets converted to gamma-rays and travels through the Earth's atmosphere and hits the Earth.   At this stage, the total heat the Earth gets is negligible enough at those levels, but would the sky glow visibly at night?   Honestly, I don't know, but maybe.

Gamma rays aren't visible, but as the rays pass through the atmosphere you might get enough visibly hot atmosphere/visible thermal radiation that you might see it.   I don't know how to begin to calculate that, but it seems possible.  The gamma ray effect would be similar to what would happen to the Earth if it was hit by a gamma ray burst (from a non lethal, but perhaps uncomfortably close distance).  Here's a description of what that would be like.   We'd lose much of our ozone layer and the chemical composition of our atmosphere would change.   over time, it could end much of the life on the Earth's land.  Life in the oceans could survive, but life on land would have a hard time.

That's not the only effect though.   As a few hundred tons of antimatter hits the upper atmosphere, the upper atmosphere would heat up and the Earth would lose it's atmosphere due to higher temperature much faster.   Over thousands, perhaps millions of years the Earth would lose much of it's atmosphere, which wouldn't be fun.

A final effect is that antimatter would change chemistry.   If an Oxygen atom is hit it becomes a Nitrogen, if a Nitrogen is hit it becomes a carbon (I think, unless the gamma ray energy of the evaporation causes further photo-disintigration), but you'd see chemical changes as a result, perhaps some of them toxic and some radioactive isotopes.   So, no ozone layer, gamma ray radiation and toxic chemicals forming in the upper atmosphere - not exactly sunshine and puppies.

Teeny-Tiny amounts of antimatter hitting the Earth isn't an issue, but in the amount you propose it would probably make the earth pretty inhospitable pretty fast.


Why can't supermassive black holes merge? (or can they?)

The main problem is angular momentum. In order for two gravitationally bound objects to merge (whether black holes, supermassive black holes, planets, stars, etc.), they must shed enough angular momentum for their orbital separation to become small enough.  Average orbital separation (semi-major axis) is determined entirely from the angular momentum of the orbit (at least in classical mechanics; I don't know if that is true for relativistic situations such as merging black holes when they get close together).  Removal of angular momentum requires interactions with other objects.

When two galaxies merge, their supermassive black holes both have angular momentum.  Through a phenomenon known as "dynamical friction," gravitational interactions with other stars sap the black holes of much of their angular momentum, until they are brought within a few parsecs or so of each other.  At this point, the black holes have flung out all of the stars that were in the region and there is (presumably) nothing left for dynamical friction to sap their angular momentum.  Once the black holes are close enough (I don't know off the top of my head how close), emission of gravitational waves will sap the orbiting pair of their remaining angular momentum and a merger becomes inevitable.

So to answer your question, the reason for supermassive black holes to be unable to merge is that they are too close together for there to be any remaining material (stars, gas, etc.) in the center of the galaxy to remove angular momentum from the orbiting pair, having already removed the material themselves, but are not close enough for emission of gravitational waves to remove angular momentum fast enough for their merger to occur any time soon (in an astronomical sense).


How common are exoplanets in distant orbits?

There are many planets known which have orbits longer than the longest exoplanet orbital periods found by Kepler. These planets were discovered using the "doppler wobble" or radial velocity technique. The plot below (a few months out of date now) shows many planets orbiting with similar periods to Mars and Jupiter. The red points we discovered by transits (including Kepler planets), whilst the green were discovered by the doppler technique.

The sensitivity of current observational doppler techniques (shown by the blue line) means that so long as we observe for a sufficiently long time (decades, in order to see an oscillation in the radial velocity), we will be able to detect Jupiter-like (or even a bit lower-mass) planets out to much longer periods.

The frequency of planets in wider orbits is also probed well by surveys for gravitational microlensing.

Single transit events cannot be securely identified as being caused by planets. I think there are a number of other possible causes, both astronomical and instrumental, that might also be to blame. Nevertheless I'm sure somebody will analyse these in some statistical way at some point to draw some inferences, but at the moment, the securest data are from the radial velocity technique. These show that the solar system is not necessarily unusual at all - there are plenty of stars that have Jupiter-sized objects in Jupiter-sized orbits.

According to the review by Howard et al. (2013), from radial velocity surveys, about 10 per cent of G/K stars have a 0.3-10 Jupiter-mass planet in orbit between 0.03 and 3 au from the parent star and around 7% have a Jupiter-sized planet orbiting between 3 and 20 au. This latter number probably has a significant error bar at the moment, but it is consistent with the information from microlensing surveys  (e.g. Gould et al. 2010). This latter paper has an interesting statistic in the abstract. they say that if all stars had a solar system like ours, they would have detected 18.2 "events" - 11.4 due to Jupiters,  6.4 due to Saturns and 0.5 Uranus/Neptunes; they would also have detected 6 "two-planet" events. In actual fact they detected 6 events, one of which was a two-planet event, and thus conclude that a broad-brush figure for the frequency of "solar system occurrence" is 1/6.


When astronomers "look into the past" are they accounting for the expansion of space?

When astronomers say something is seen "13 billion years in the past", this is (in most cases) a model-dependent statement, but a perfectly valid statement that tells us how long the light from an object has been travelling to us and which does take account of universal expansion.

What has actually been measured is a redshift, and to convert that into a "look-back" time requires an understanding (and a model) of the expansion history of the universe.

In the actual scientific research publication, as opposed to a press release, the details of what assumptions have been made will be stated. The current uncertainties in the cosmological parameters are small - they would not for example lead to uncertainties in the look-back time of even a few percent.

However, there is always a possibility that our understanding of cosmological expansion is wrong. In particular, the assumption that the cosmological constant is constant might be wrong, in which case there will be systematic shifts in estimated look-back times.


What can we expect to see with a telescope with a 70mm aperture and a 10mm eyepiece?

I would buy a good 2x barrow lens to go with that. She might see Saturn's rings as a haze. Good for looking at the moon. Maybe Mars. A good tripod is a must.  I would suggest a camera mount. A light weight camera.  Secure the front of the telescope with 3 rubber bands to the tripod. To reduce vibration. 2 seconds pause on the camera. For vibration to end. A telescope that size is a fair place to start on Photos. A 3 second exposure. For stars planets & such. She can get some good photo's. Also other photo's. 70mm is a good daytime telescope size. Make sure she never looks towards the sun with it. For night work you need a 90mm or above. Not into astronomy. But use a 90mm Vexin for photo work. Ships in the bay. Objects up in the mountains that can not be got close to in Asia. A camera makes it so much more fun. You can build a camera mount not hard to do.


Planet-timer "device" from 1970s or 1980s possibly by Edmund Scientific

I ended up finding this on page 14 of the Edmund Scientific Holiday Sale catalog, with sale ending on December 31, 1983:

except that it's "Planetimer" (and not "Planet-timer"), catalog number H9454.

A 1200dpi image somewhat shows how it's supposed to work:

My original question was going to be: could this device have worked?

I would also appreciate any other information on this product (including a better scan, ideally of the product itself, if anyone still has one).

For reference, here is the page on which the ad appears and the cover of the catalog in which I found it:


Higgs Bosons in the centre of Black holes or extremely large stars

No. That's not what a black hole is. A black hole is a vacuum solution to general relativity. In other words a black hole is just mass, without a "thing" left to be massive. The mass is collapsed to a singularity, and is surrounded by an event horizon.

If you pass over the event horizon, every path you can take in will lead you to the singularity. In this sense the singularity is more like a point in time than a point in space. You can't "see" the singularity, since it is always in your future (you can only see things that are in the past)

There is no "thing" at the singularity, There are no protons neutrons or higgs bosons, and any that are formed will find themselves at the singularity, and ceasing to exist in a short time.

So black holes are very very weird. They can't be easily understood my analogy. They are understood in terms of Einstein's mathematical theory of general relativity.


NASA has a very good interactive online map which also shows highways.  It looks like I-95 intersects with the path by a wildlife refuge.

A note on planning your trip: there will most likely be a very large number of people who also want to go out to see the eclipse.  So be prepared for lots of traffic.  Also, depending on where you plan on going, you will probably have to leave very early in order to get a good spot.


The moon: how to determine its size, speed and distance from earth using a simple telescope?

Finding the distance is tricky. To find the distance you need to carefully note the position of the moon (relative to the stars ) from two different places, *at the same time *.  The moon will appear to be in a slightly different position from the two viewpoints (an effect called paralax). That done, finding the distance, and size is a simple exercise in trigonometry, and Khan academy has a page which shows hoebto find the distance to the moon.

Greater accuracy could be obtained by measuring the time that a star is hidden by the moon, from two locations, and using that to determine the position of the moon to greater accuracy than possible by direct measurements.  Hipparchus apparently used a solar eclipse to obtain two positions atbthe same time needed for calculation of paralax.

Knowing the distance makes finding the diameter and speed simple to do, by observing the angular size, and using a little trig.  Working with photographs takenby the telescope is convenient but not strictly necessary.