Why was Neptune rather than Uranus chosen as an archetype? The largest exoplanets are typically called Jupiter-like planets, or hot/cold Jupiters. Since Jupiter is the largest most massive planet in our Solar system, it makes sense that it would be chosen as an analog. For intermediate-size exoplanets, Neptune is used as an analog (e.g. 1).
Neptune and Uranus are both roughly the same size and mass. So why is Neptune rather than Uranus used as the archetype of intermediate-mass planets?
Did someone call an exoplanet a Neptune-like planet, choosing Neptune over Uranus for no particular reason, and then everyone else followed suit? Or is there a fundamental difference between the two planets that makes Neptune a better analog?

NASA distinguishes four types of exoplanets: Gas Giant, Super-Earth, Neptune-like, and Terrestrial. The question asks why ice planets termed "Neptune-like" and not "Uranus-like".
Let's start with some statistics: What is curious, that (at least according to scholar.google.com), intially more icy exoplanets were termed Uranus like planet rather than Neptune like planet before eventually, the latter term started growing:
1980 to 1990

2 citations with Uranus like planet
0 citations with Neptune like planet

1990 to 1995

3 citations with Uranus like planet
5 citations with Neptune like planet

1995 to 2000

2 citations with Uranus like planet
4 citations with Neptune like planet

2000 to 2005

4 citations with Uranus like planet
12 citations with Neptune like planet

2005 to 2010

4 citations with Uranus like planet
43 citations with Neptune like planet

This supports the hypothesis that it was a "vote with the feet", rather than something coordinated, much like initial density fluctuations in a protoplanetary disk which might serve as the seed for a planet, or not.
@uhoh already suspected this in a comment, plus also mentions that the pronouncaction of Uranus might lead to unintended puns when talking about "planets like ...". It is sometimes these linguistic pecularities which make a brand successful and another one not.

I'm 100% sure that

it's because of the way Uranus sounds when said out loud, and
the decision was made in the collective unconscious, we'll never find a smoking gun, or a...

After all, there quite some similarities between Uranus and Neptune, see e.g. the concise comparison between the two planets at WolframAlpha, although there is also reasoning Why Neptune and Uranus are different.


Will intergalactic travel be impossible in roughly 15 billion years? Even if we (or aliens somewhere in the universe living at that time) somehow invent something that can go just short of the speed of light, will intergalactic travel still be impossible in roughly 15 billion years?  Another way to put it, will even the nearest galaxy be receding from us faster than the speed of light in roughly 15 billion years?  To me, that would seem to require dark energy to increase dramatically, but seems to be the conclusion after thinking on the following points (although the expansion of space itself probably alters this estimate significantly):

In some further thought on the premises I posed in this linked question, I came to the realization that the universe may be "rapidly" changing:
Is there a physical limit to how far we can go?

Edit: This is different than that question in that I am focused now on the future state of our universe (when - the focus of this question) vs. the location (where - which was the focus of the prior question).

Known facts:

There is a part of the universe, about 15 billion years away, that is receding beyond the speed of light, and hence we can never reach.
If we were to pick a direction, there are a finite number of galaxies between here and that 15 billion light year position, (and I therefore deduce at some point, we would not be able to reach the next galaxy.)
Space is fairly uniform, so similar conditions should be present on a large scale
It would take at least 15 billion years to reach that edge, not accounting for space expansion.

My thought / question with the above in mind, does this mean that if we were to keep traveling, it is inevitable that we will eventually pass the last galaxy we can reach, and then all galaxies would be out of reach (except possibly the last one)?  Does this mean that in 15 billion years (then adjust for the expansion of space) that we won't even be able to reach the nearest galaxies, even if we find a way to travel at the speed of light?

Yes, over time the accelerating expansion of the universe (assuming the $\Lambda$CDM model is right) will separate gravitationally bound clusters of galaxies from each other with exponentially growing distances. However, this will not happen in 15 billion years.

This excellent paper analyses the expansion in detail. What takes about 17 billion years is for the distances to increase by 2.71 times. They predict that typical clusters will grow isolated in about 120 billion years, with the Local Group becoming isolated in 175 billion years from the other nearby groups.

In this "era of isolation" the only reachable galaxies are the ones in one's local cluster. However, they will tend to merge over time into one big elliptical galaxy (like the Milky Way and Andromeda are likely to do in a few billion years). Extragalactic astronomy also becomes really boring since the remote clusters are redshifted so much they cannot be observed.

If one just travels outwards from Earth in some direction with a high velocity one will eventually find oneself in a last cluster beyond which there is nothing reachable.


Do far away stars that we observe today still exist? I don't know much about the astronomy or the universe but enjoy reading articles from time to time as they pop up. Just finished reading an article on CNN that discussed how a South African telescope has discovered hundreds of galaxies, some 200 million light years away.

This got me thinking: Are these galaxies and or some of their stars we see in telescopes still there? Let's say STAR X is confirmed to be 200 million light years away. What if it completely exploded 1 million years ago? Wouldn't we need to wait 199 million years before we could detect that?

Simple answer is yes. It will take light from an explosion that happened 1 mya at a distance of 200 Mly, about 199 million years to reach "us" - not that any of "us" will be around then. The speed of light is the fastest speed we can get information from anywhere in the Universe. There's nothing special in asking about what happens to that star after that light starts out on its journey here. The fact is, we can't know, until we are able to gather more light from that same source. So, that star would appear to explode if you were looking up at it 199 million years from now (keeping in mind that both the Solar System and the Milky Way are also moving through space, so the exact time it takes the light to get here will be different than the time it took the light we are now seeing from that star to have gotten here.)


How to deal with shifting CCD bias (zero exposure) levels? The median values of the bias frames coming from our teaching telescope shift up or down with each capture. For example:

import glob, numpy as np, astropy.io.fits as fio, matplotlib.pyplot as plt

data = np.zeros((10,255,765), dtype=np.int)
for i, f in enumerate(glob.glob('bias_*.fit')):
    data[i,:,:] = fio.getdata(f)

print(np.median(data, axis=(1,2)))

[ 106.  106.  108.  108.  110.  108.  105.  106.  107.  107.]

Now, during data reduction we average combine our bias frames. We also reject outliers (either minmax or sigmaclip). But if some of our frames are "shifted" higher than others, wouldn't this skew the averaging (and especially the outlier rejection) so that we might be introducing more noise during averaging?

To give you an idea of the nature of this fluctuation, I did a quick analysis: read 10 bias frames, then iterated through them, subtracting each frame from the median combination of all other frames. This gives me something like "noise frames", showing only how much each frame deviates from the median of the rest. Below is the code and the distributions of two of these noise frames.

noiseframes = np.zeros((10,255,765), dtype=np.int)
otherframes = np.ones(10, dtype=np.bool) # just a mask
for i in xrange(10):
    otherframes[:] = True
    otherframes[i] = False # exclude current frame from mask
    median_of_others = np.median(data[otherframes,:,:], axis=0)
    noiseframes[i,:,:] = data[i,:,:] - median_of_others

def plotdist(i):
    frame = noiseframes[i].flatten()
    x = np.arange(np.min(frame),np.max(frame)+1)
    y = np.bincount(frame-np.min(frame))
    plt.step(x, y, label=str(np.median(data[i,:,:])))

plotdist(4)
plotdist(3)
plt.xlabel('Difference from median [ADU]')
plt.legend()
plt.show()

All of the noise frames have near-perfect gaussian distributions with pretty much constant FWHM, it's just the central value that shifts around with each frame.

This leads me to a couple of conclusions: firstly, this shift truely is a shift in median value and not just a result of something like cosmic rays; secondly, the fluctuation is an additive effect and so not due to a change in gain due to a difference in voltage or something like that.

I also could not find any correlation with temperature or time since last readout, both of which I suspected for a moment.

Perhaps it's slight changes in readout time?

I couldn't find any mention of this kind of artefact in my own course notes. It's also not mentioned anywhere online that I can find.

Can any of you help me to a) identify the possible sources of such a fluctuation and/or b) provide a proven strategy for dealing with this during data reduction?

Looks like you've got some flaky electronics attached to your CCD if the bias level is changing by several units in consecutive exposures. However, all is not lost if your CCD has an overscan region.

This is normally several columns of "pixels" that appear appended to each row of your data, but which do not correspond to physical pixels on the CCD. They are actually dummy readouts to assess the actual bias level in each particular row.

The overscan region should be clearly visible in any real exposure of a light source, since the overscan region will be un-illuminated.

You can use the overscan in two ways. The conventional method is to overscan-correct your images before doing a bias subtraction. You should also do this to your bias frames before combining them into your "master bias" frame.

So, take your overscan region and either (i) calculate a median (or clipped mean) within it and subtract this value from every pixel in the image before use (including the bias frames); or (ii) fit something more complicated as a function of row number and then subtract this from your actual data on a row-by-row basis.

Method (ii) is required if you notice that there is any structure or drift/trend in the overscan values as a function of row number.

This link gives plenty of details, some pictures and a recipe.


Requirements for a satellite/planet to be tidally locked to a planet/star The Moon is tidally locked to the Earth, the Four Galilean are tidally locked as well, and the recently found planetary sistem TRAPPIST-1 has seven tidally locked planets, but Venus or Mercury are not. Why? Is there any conditions or analytic formula which says if a small body will be eventually tidally locked to the major body?

Wikipedia gives the formula

$$t_{\text{lock}} \approx \frac{\omega a^6 I Q}{3 G m_p^2 k_2 R^5}$$

where

$\omega$ is the initial spin rate expressed in radians per second,
$a$ is the semi-major axis of the motion of the satellite around the central body (given by the average of the periapsis and apoapsis distances),
$I\approx 0.4 m_s R^2$ is the moment of inertia of the satellite, where $m_s$ is the mass of the satellite and $R$ is the mean radius of the satellite,
$Q$ is the dissipation function of the satellite,
$G$ is the gravitational constant,
$m_p$ is the mass of the central body, and
$k_2$ is the tidal Love number of the satellite." source

Notice that the semi-major-axis is the power of 6. A small change in the orbital distance can have a very large effect on whether the body will become locked.

Also note the terms $Q$ (a measure of how elastic the body is, and hence how much energy is lost in distortion source) and $k_2$ (a measure of the rigidity of the body: how much it is distorted by tides source), which are both hard to measure. The 7 detected planets in the Trappist-1 system are all much closer to their star than any of the sun's planets and for reasonable values of $Q$ and $k_2$, the planets will be tidally locked within a few million years, much as the moons of Jupiter are tidally locked.

In our solar system, mercury is asynchronously locked with the sun (on a 3:2 resonance) The Earth is not locked: it is too far from the sun, and the moon has a big effect. Mars is much too far out, and Venus is odd.


How much of a molecular cloud can end up as "starstuff"? Stars form within molecular clouds. These clouds can be up to 6 millions solar masses. When the cloud collapses into stars, is it possible to know a rough figure for how much of this material actually turns into stars? 1%? 99%?

Do larger clouds tend to be more mass-efficient in star formation than smaller ones?

This quantity is referred to as star formation efficiency (SFE) by astronomers who study star formation and galaxy evolution.   Estimates can vary but typically are around a few percent.  In Sec. 4.1 of this paper Inoue et al. review  some estimates from the literature.  Those numbers are for the central regions of spiral galaxies, and the rate may be lower in the outer regions where the density is lower.
Note that even if the rate is a few percent in a given cloud for a given episode of star formation, the rest of that gas mass will be dispersed into the interstellar medium and eventually find its way back into another cloud to form stars later - star formation is ongoing in spiral galaxies.
Note that you will often see SFE expressed as star formation rate (solar masses per year converted to stars) divided by total gas mass.  That has units of (mass/yr) / mass, which reduces to units of 1/yr.  You can think of that as the fraction of gas that forms into stars per year, or alternatively, the inverse of that number gives you a timescale in years to deplete the gas entirely.


Old star vs New star We know that with the passage of time, hydrogen inside (near or at the core) a star is converted to helium and other heavy elements via fusion. We can then deduce that the old sun contains heavier elements in comparison to the new one. Does that mean that the mass of the old sun is more than the new sun?

Let me try to add some numbers to Steve's answer.

The Sun's luminosity is about $L_{\odot}=4\times10^{26}\text{ J/s}$. Now, if we assume that the majority of that energy comes from nuclear fusion, we have
$$L_{\odot}=\frac{dE}{dt}=\frac{d(mc^2)}{dt}=c^2\frac{dm}{dt}=c^2\dot{M}_{\text{nuc}}$$
Therefore, we can write the mass-loss due to nuclear fusion as $\dot{M}_{\text{nuc}}=L_{\odot}/c^2\sim4\times10^{9}\text{ kg/s}$.

Now, the Sun also loses mass via the solar wind. I believe the mass loss rate is a few times about $\dot{M}_{\text{wind}}\sim3\times10^{-14}M_{\odot}/\text{yr}\sim2\times10^9\text{ kg/s}$. Coronal mass ejections, solar flares and similar phenomena are often accounted for in this figure; at any rate, they don't add a significant amount.

This figure should increase as the Sun moves into the red giant (and especially asymptotic giant branch) phase of its life; it winds will increase to become much more significant that they are today. However, fusion will also increase, and the Sun's luminosity will increase by a factor of 1000 or 10000; therefore, the mass loss due to fusion will also increase. However, I think mass loss due to winds will increase more significantly, possibly by a factor of $\sim10^6$ - but I'll check that.

At any rate, an older star will have a lower mass than a younger star. For a Sun-like star, this would be hard to measure over the span of a human lifetime - remember that it only loses about $10^{-13}$ solar masses each year. Long-term records could make it possible, though, through various observations of the Sun and its effects. Centuries or millennia of data-gathering do tend to yield results.


How close to circular is the Earth's equator Earth's orbit is about 99% circular. How circular is Earth around the equator? I know it bulges around the equator and is a spheroid. I also know it is not smooth and has oceans and mountains and all that.

Is there a similar measure of eccentricity to describe how circular it is around the equator?

Is it stretched into an ellipse by the sun/moon, and if so by how much?

Just for comparison, does the Earth's equator deviate from a circle more than the Earths' orbit around the Sun deviates from a circle?

My answer won't be complete, from lack of time and resources, but I still wanted to share some interesting aspects here that could be helpful.

The difficulty in aswering this question revolves around the complex and irregular shapes involved here. Also, finding a "best-fit" ellipse for comparison is not as easy as it seems, because it depends on what and how you want to model. The actual shape of the equator is rather complex and irregular. Generally speaking, it is pretty much circular, but indeed topography and the geoid complicate matters. The Earth's movement in the solar system is not a perfect ellipse either, because of the gravitational interactions with other celestial bodies. Let's review the different irregularities involved here.

The Earth is usually modeled as an ellipsoid of revolution (oblate spheroid), and the equator as a circle. A good example of that is the WGS 1984 geodetic reference system used by the Global Satellite Navigation System. Of course, the equator is not a perfect circle, it has irregularities mainly because of topography, and even sea level itself is a little irregular too. We can approximate sea level with a geoid, for example, here is a map of the EGM2008, a geoid used with WGS 1984 to transform ellipsoidal heights to geoid heights:

Basically, this map shows the height of the geoid (the idealized sea level without the effects of tides and currents) with respect to the WGS84 reference ellipsoid of revolution (semi-major axis 6,378,137 m, semi-minor axis 6,356,752.314 m). The differences are mostly less than 100 meters, and are caused by the irregular distribution of mass inside the Earth itself.

Now, some studies show that the Earth's shape could be slightly better modeled by a triaxial ellipsoid, and one could try to model the equator as an ellipse, and the Earth as a triaxial ellipsoid, however, even with a best-fit triaxial, we would still need geoid corrections for the irregular mean sea level, let alone topography, and geodetic computations would be more complex on a triaxial. Other funny models and names have come up over time, like a pear-shaped (because of a slight bulge in mid southern latitudes) model as a best-fit shape. But if you look at the map above, good luck visually finding the pear shape in there, or other mathematically modelizable aspects of these bumps. We are talking about very subtle differences here, that do not necessarily need to be taken into account for most purposes when describing the general shape of the Earth.

So depending on how you consider the shape of the equator (i.e. by topography and ocean floor, or sea level) you will arrive at a shape that is mostly circular with irregular bumps along the way. There is no authoritative agreement that I know of about an eccentricity of the equator. For instance, This study proposes a flattening of about 70 meters for the equator's ellipse, This article on Encyclopaedia Britannica proposes 80 meters.

For the Earth's orbit, for the sake of this comparison, we can use an best-fit ellipse of 149.598 million km by 149.577 million km. Of course, that is only a idealized ellipse, the real movement of the Earth in the Solar System is more complex.

Finally, say we scale down and superimpose the Earth's orbit's ellipse on the equator to compare. The eccentricity of the Earth's orbit is 0.0167, the semi-major axis is 149.598 million km and semi-minor axis is 149.577 million km. Scaled down by a factor of 23,455 to the equator's size, this corresponds to a difference of about 900 meters in both orbit axes. So I think we can agree that a best-fit ellipse of sea level along the equator is more circular than the Earth's orbit.

However, topography-wise, there are bumps up to over 4,000 meters in the Andes, and the sea floor reaches 5000 meters below sea level in several places. So the "topographic surface" equator would, for one thing, definitely appear more bumpy than the Earth's orbit. Unfortunately, I haven't found a example or study showing what a best-fit ellipse of the equator (including topography and bathymetry) could look like, mainly because we tend to approximate sea level, not topography itself, but with more time, tools and data, it might be possible to work out the answer with a "topographic best-fit" ellipse.


Did the Late Heavy Bombardment Period Happen Because of a Stellar System Collision? I was thinking about the impending collision between M31 and the milky way galaxy, and I've heard that since a galaxy is mostly space most of the contents of both galaxies would be just fine. Could a similar effect be responsible for the late heavy bombardment era? Just instead of galaxies, stellar systems?

There are several potential explanation for the cause of the late heavy bombardment. One of them includes a passing star that disturbed the Oort cloud. A "stellar system collision" would have destroyed the solar system, though. While galaxies are mostly empty and a galactic collision won't involve that many actual collisions (if any), the gravitational interactions between the galaxies actually changes star orbits around the galactic centers. Similarly a "stellar system collision" would heavily change planetary orbits, even if there were no collisions at first.

Another hypothesis is a totally "interior" one: A 2:1 orbital resonance between Jupiter and Saturn changed the orbits of Neptune and Uranus, potentially even causing them to swap positions. Neptune's new orbit then affected many Kuiper belt objects, either sending them way out to form or join the Oort cloud, or sending them in as part of the late heavy bombardment.


How do we distinguish old craters from new ones on the Moon? The following image is the observation made by the Dual Frequency Synthetic Aperture Radar (DF-SAR) onboard the Chandrayaan-2 Orbiter:

Due to image size restriction, a lower resolution image is used here. In case you wish to see the original image, visit this ISRO Webpage.

In the above image, there are three craters marked Young (Fresh), Intermediate and Old, and we are able to see the difference clearly. This ISRO Webpage states that the difference is due to weathering processes. I can't understand how can such processes undergo in our Moon, where there is no appreciable amount of atmosphere. Is that because of Moonquakes?

It would be great if you could explain this.

Aside from the excellent points made in James K's answer, there are other ways to date craters.  For example, when the rays of one crater overlay those of another, we know that the former is younger than the latter.

We can also estimate ages of large craters by counting subsequent craters inside the crater floor.  Crater counting is one of the more common ways of estimating the age of lunar surface features.

We can also determine ages of some craters by looking at whether it was affected by known ancient lunar processes.

Then there is radiometric dating.  Large impacts spread materials across the surface of the moon, so the Apollo mission samples contained ejecta from many craters, some of which are identifiable by things like mineral composition unique to that part of the moon.

Once you have a good date for one crater, you can use that data to help determine the age of others.  For example, if the floor of a crater has X smaller craters  in it, and you know how old that crater is,  you can then use those counts to estimate ages of other craters with similar counts.   With enough varied data sets, you can generate some pretty good functions for estimating absolute ages.

For example,  have a look at the image of the Aristarchus Plateau below:

NASA (image by Lunar Reconnaissance Orbiter) [Public domain]

Notice the two craters at the bottom of the plateau.  The one on the left is the crater Herodotus, and the one on the right is Aristarchus.

Looking at the differences between the two of them, the first thing you notice is that Herodotus is filled in with the same lava that fills the basin around the plateau, but Aristarchus does not have that fill.  So right there we know that Herodotus pre-dates the event that filled in Oceanus Procellarum, while Aristarchus came later.  This makes Herodotus at least 1.2 billion years old, which is the current best estimate for the last time lava flowed on the moon.

Now, we can do better than that with sampling.  If we get a sample of the basalt inside Herodotus, we can use radiation dating to determine when it crystallized, and that would give us a very good upper limit of the age of the crater.  And in fact we did get such samples from Apollo, and that, plus crater counting, plus looking at other features around it with known ages, plus the amount of degradation in the crater walls and the coloration of the material, we can tell that Herodotus dates to the Imbrium era, or about 3.8 billion years ago.

As for Aristarchus crater, you can see lots of evidence that it is young.  For one thing, its interior and ejecta are incredibly bright - it's the brightest crater on the moon.  We know from radiological dating and crater counting that similarly bright craters are young.  The walls of the crater have minimal slumping, and the ejecta around the crater is still well defined.

Aristarchus is one of the 'Copernican' craters, so named because they all formed during the Copernican period, which starts 1.2 billion years ago and goes right up to the present.  Other large copernican craters include Copernicus, Tycho, and Kepler, among others.

Now look at this wider, high contrast image of the Tycho region.  You can see Copernicus and Kepler on the right, and Tycho lower center.  Note that the crater rays overlap each other, and by careful examination we can tell the order of the impacts.  That helps us further refine the relative ages of the craters,  and if we can narrow down the age of one with sampling or other techniques, it helps narrow down the ranges of the others.

Btw,  the order of creation from oldest to newest Copernicus, Kepler, Aristarchus, Tycho.

Here's a pretty good reference describing a number of techniques for dating features on the moon:

Relative Crater Age Determination

This is an iterative process - each time you narrow down the age of a crater, that can be used as data to further refine the ages of others,   But it's not a precise science - age estimates of various craters can still vary by hundreds of millions of years.  There is still a lot we don't know about how the Moon formed and evolved.


What are the best candidates for solar lensing of exoplanets, and what would the results be? If we were able to send a probe to 550AU from the Sun, which exoplanets could be examined, to what resolution, and what form would the examination take?  It is radio telescopy only, or would it be possible to optically see what the surface of an exoplanet looked like?

I've seen this question for the past few days and have shied away from it because it seemed too difficult, but here goes:

550AU is the focal point of the sun, so theoretically anything on the opposite side would be focused into a perfectly coherent image right? Unfortunately due the density of the sun (Mass to radius ratio specifically) only radio waves that just miss the sun would be focused to that point, all other shorter wavelengths don't diffract enough and simply 'collide' with the sun or form a lense ring at a greater distance. So radio-observation only. Explained at this article https://physics.stackexchange.com/a/25501

A star such as a neutron star however would make an excellent candidate for stellar lensing because it has a high density and low radius, meaning much shorter wavelength radiation is able to go around the star.

This wouldn't be particularly effective at discovering new exoplanets, only observing them in more detail.


Why does the Moon never set in Svalbard, Norway? I heard in a documentary that, in Svalbard (Spitsberg), Norwegian archipelago in the Arctic Ocean, the Moon never sets. Why? A drawing would certainly help.

You must have misheard it, or the documentary you watched wasn't presenting very precise information. It does set but it also stays on the night sky for several days during the polar winter (polar night) when the Moon if full. This is relatively simple to imagine, so I'll describe it;

So what's happening is that the Earth's axial tilt during the polar winters leans the whole Northern hemisphere towards the night side, away from the Sun. This tilt is big enough (~ 23.4°) that the night sky objects aligned with the Earth's equatorial plane stay visible relatively low on the horizon. With those regions being either relatively flat and/or with a view towards the sea, there's not many obstructions limiting the viewing angle, so the Moon (and analogous also the Sun during polar summers) stays "locked" low above the horizon. To help a bit with imagining this, here's an animation of the Earth's axial tilt, courtesy of Wikipedia:

If we imagine this animation of the Earth with the Sun in the distant left of the image, so during Northern hemisphere's winter (winter solstice to be precise), and the Moon to the distant right of the image (roughly 25 widths of the image away), so when it's either full or close to this lunar phase, it's not too difficult to appreciate that the northernmost polar regions have a direct line of sight of the Moon during Earth's full rotation on its axis, or a day. If you keep in mind that other celestials, including the Moon, are oblivious to the Earth's axial tilt (well, not exactly, but let's not nitpick about tidal effects that might take millions of years to make a difference), as the Moon moves farther in its orbit, in our case towards the viewer, this observation angle decreases further still and those northernmost latitudes hide to us for some part of the day. At lunar last quarter, it would be directly towards us relative to the image, so this direct line of sight relationship between the Earth and the Moon becomes reciprocal to how we're seeing places on the Earth on the animation.

Why when the Moon is full? Simply because that's when the Moon is also behind the Earth (but not in its shadow), so the relative angle between the observation point and the Moon would stay high enough to observe it. As it moves in lunar phase and in orbit around the Earth farther, this angle becomes lower and the Moon indeed does set also in the arctic region. For what is worth, this goes exactly the same for South pole, only with a half a year difference.

One other effect that plays a role here is the Earth's atmospheric refraction which also adds to the duration during which the Moon appears not to set. Meaning, that even when the Moon wouldn't be in direct line of sight, but only marginally so, it would still appear low on the skies due to optical effect (displacement) of the atmosphere. This effect would somewhat offset observing the Moon from lowlands with possibly shallower observation angle when compared to higher altitude observation points with less direct line of sight obstructions, due to denser atmosphere and thus higher refraction index.


Star formation around rotating black holes? Please excuse an amateur question.
While trying to think of anything but what was happening during a dental procedure my mind turned to a model of a star close to a rotating black hole and the effects on the drawn in matter.

While it is obvious such matter would be excited to high temperature could the combination of rotation and excitation be enough to induce a sustained fusion reaction?

If so would this produce enough energy to maintain a fusion 'ring' at the event horizon - essentially a doughnut star?

Would there be enough of a reaction to start producing lighter elements?

Pure curiosity generated by an attempt to distract myself

The accretion of material onto (into) black holes (and neutron stars) provides environments that are both very hot and (relatively) dense. Under these circumstances it is possible for nuclear fusion to occur, the question is whether this is significant, both energetically or as a means of producing new chemical elements (nucleosynthesis).

The answer to the first of these questions is relatively straightforward. As material falls towards the black hole, its angular momentum forces it to form an accretion disk. Viscous processes heat the disk and provide torques, cause the material to lose energy and angular momentum and eventually allow it to fall into the black hole. Much of the gravitational potential energy (GPE) gained as the material falls towards the black hole does end up heating the material.

The innermost stable circular orbit of a black hole is at 3 Schwarzschild radii $=6GM/c^2$, where $M$ is the black hole mass.  The GPE released for material of mass $m$ falling to this radius is $\sim GMmc^2/6GM = mc^2/6$.
i.e. fully one sixth of the rest mass energy of the material could be released as heat.

Compare this with nuclear fusion. The fusion of hydrogen into helium only releases 0.7% of the rest mass as energy that can heat the accretion disk.

So from the energetic point of view, fusion reactions are negligible, unless they can occur much further out in the disk

The question about nucleosynthesis yields is more complex. The more massive a black hole and the higher the accretion rate, then in general the higher the disk temperature and density and the higher the fusion rate. But it also depends on the details of the cooling processes that are possible and how much material is advected into the black hole. Hu & Peng (2008) present some models of accretion onto a 10 solar-mass black hole and suggest that it may be possible to produce certain rare isotopes by this mechanism.  Stellar-sized black holes probably need very substantially super-Eddington accretion rates to achieve the necessary temperatures to sustain nuclear fusion (i.e. much greater accretion rates than are possible by radiation-pressure opposed spherical accretion flows), according to Frankel (2016). Such rates are likely only in the cases where black holes disrupt a binary companion, rather than through a steady accretion flow.


Why would the Chang'e-4 lander find lunar far side temp. "colder than scientists expected", when the LRO has already been taking thermal readings? In January, the Chinese probe lander Chang'e-4 was was announced to have found temperatures dipping lower on the far side than expected ("Chinese rover finds lunar nights 'colder than expected'" by R. Jackson, Jan. 31, 2019. AFP via Phys.org):

  Temperatures on the moon's surface plummeted to minus 190 degrees celsius (-310 degrees Fahrenheit) during the probe's first lunar night, which "was colder than scientists expected," the China National Space Administration (CNSA) said.

This was said to be surprising, because near-side temperature measurements by U.S. missions showed less substantial dips. It was suggested that this was probably due to a characteristic difference between regolith on the two sides:

  They were lower than those recorded by previous US missions to the near side of the moon, Zhang He, executive director of the Chang'e-4 mission, told Xinhua news agency.

  "That's probably due to the difference in lunar soil composition between the two sides of the moon," he said.

But the U.S. surface missions were not the last thermal readings taken of the Moon, and the Chinese mission isn't the first mission to take thermal readings of the far side. Unless the thermal variability measured by Chang'e-4 is particular to a very localized area of regolith on which it took those lunar dawn readings (in which case it can't be said to be representative of a generalizable difference from near-side regolith), how would that not be known already, from the Infrared measurements that have been taken by the Lunar Reconnaissance Orbiter for the past decade,
 with enough resolution and sensitivity to distinguish the thermal variation of craters from surrounding regolith?

I would not attribute too much importance to the quote "colder than expected", because that always depends which expectations one had initially. Here the quote you are discussing:

Temperatures on the moon's surface plummeted to minus 190 degrees celsius (-310 degrees Fahrenheit) during the probe's first lunar night, which "was colder than scientists expected," the China National Space Administration (CNSA) said.

As you correctly state, it was scientific knowledge on Moon temperature could get very low. Tim Sharp from Space.com writes in an article What is the Temperature on the Moon? back in 2017:

The Diviner instrument on NASA's Lunar Reconnaissance Orbiter measured temperatures of minus 396 F (minus 238 C [${=\rm 35 K}$]) in craters at the southern pole and minus 413 F (minus 247 C [${\approx\rm 26 K}$]) in a crater at the northern pole.

The conversion to Kelvin was added by myself.
On the other hand, Ramesh B.Malla et al. state in their article Determination of temperature variation on lunar surface and subsurface for habitat analysis and design in 2014:

Aside from the dangerous radiation levels and hypervelocity micrometeoroid impacts, the equatorial temperature on the surface of the Moon can range from ${\rm 102.4\,K}$ to ${\rm 387.1\, K}$.

So if you take the 2nd article as expectation, the value reported by 嫦娥四号 (the Chinese lunar lander) of $-193.15^\circ {\rm C} \approx 80 {\rm K}$ is indeed lower than ${102.4 \rm K}$.
From my time as researcher I remember that the public relation office did indeed sometimes use a more striking formulation than what the researcher themselves would - that's a factor also worth considering.


Will all of the gas in the universe be converted into red dwarf stars? Red dwarfs have lifespans of trillions of years and are also the most abundant stars, making up 70% of the stars in the Milky Way. No red dwarf star has ever died and none will die for a very, very long time. Because it take so long for them to die, they're not releasing their hydrogen back into the interstellar medium to make new stars.

As more massive stars live and die over the course of years, those stars release their hydrogen into the interstellar medium, which can form new stars, 70% of which will be red dwarfs. Does this mean that, billions of years into the future, all of the matter will get locked within red dwarfs until they begin dying out trillions of years later?

No. The reason is that gas recycling only recycles only about 40-50% of the gas in a sun-like star, leaving the rest as a white dwarf that slowly cools off. Heavier stars are even less effective in converting their gas into a condensed core. So the process will tend to produce a fair fraction of heavier white dwarfs now in the early stages of the stelliferous era, followed by a slow emergence of lighter white dwarfs from red dwarfs in the rest of the era.

Adams and Laughlin estimate that the final mass fractions will be brown dwarfs (never fused) 9.7%, white dwarfs 88%, and 2.4% neutron stars and black holes. In the post-stelliferous era some of the brown dwarfs will merge and make a few extra red dwarfs .


How far can the Sun and the Moon get? I noticed that last June the Sun was on the tropic of Cancer and moon on Madagascar:

Can you explain how that works and if they can get even farther in latitude?

The Earth's equator is inclined to the plane of the solar system (called the ecliptic) by 23 degrees. In other words, the Sun's position, as seen from the Earth, varies between 23 degrees North (at the northern hemisphere summer solstice) and 23 degrees South (at northern hemisphere winter solstice), appearing to cross the Earth's equator twice (at the equinoxes).

The tropic lines mark the maximum apparent solar move north or south.

The Moon's orbit is inclined by about 5 degrees to the ecliptic so moves over a somewhat wider range of the sky as seen from the Earth. The Moon orbits the Earth every 28 days or so, so moves over that range each month.

At full Moon the Moon is opposite the Sun in the sky, so if the Sun was close to the tropic, then the Moon would be around the opposite tropic when full, and so on.


Can a pulsar also be a magnetar? I know this is a simplistic question, but I cannot find a straight answer...
Also, is it possible most or all neutron stars are magnetars and/or pulsars?  It's just hard to see all of their properties from our vantage point, etc.?

Magnetars are the most widely-accepted explanations for two observed classes of objects:

Soft gamma repeaters (SGRs)
Anomalous X-ray pulsars (AXPs)

Under the magnetar hypothesis, these classes of objects are explained as highly magnetised neutron stars, evolving from a soft gamma repeater (expected lifetime ~10000 years) to an anomalous X-ray pulsar (expected lifetime ~10000 years) as the rotation period lengthens and the magnetic field decays.
The AXPs are pulsars, and at least some SGRs have been identified with pulsars (e.g. Sgr 0526-26 = PSR B0525-26). So under the magnetar hypothesis, magnetars can indeed be pulsars.
It's worth noting that the magnetar hypothesis may not explain (all?) SGRs and AXPs: there does seem to be evidence that some of these objects do not have especially strong magnetic fields compared to pulsars, e.g. SGR 0418+5729 (Rea et al. 2010), and there have been suggestions that some of these objects may actually be white dwarfs (Lobato et al. 2016), perhaps somewhat similar to the white dwarf pulsar in AR Scorpii.


Is the structure of crystalized white dwarf related to nuclear pasta? There is news of an analysis of a large number of nearby white dwarfs revealing a "pile-up" at a certain point in their cooling which has been predicted to happen due to a phase change where nuclear material crystalizes. This is analogous to a mixture of ice and water remaining at 0C while it is in the process of freezing (or thawing).

What is this crystalline material like? Is it really a lattice of Carbon and Oxygen nuclei in some recognizable structure (cubic BCC... etc.)? Is it related in some way to nuclear pasta?

Nature: Core crystallization and pile-up in the cooling sequence of evolving white dwarfs

It's thought to be crystalline in the same sense that diamonds are a crystal. The coulomb interactions between the positively charged ions becomes much larger than their thermal energies and they settle into a crystal lattice.

This both releases latent heat and increases the thermal heat capacity of the ions (roughly doubles it). This causes a "pile up" of white dwarfs with interior temperatures at around the crystallisation phase transition. However, the pile-up is not too steep because the freezing temperature increases with density and white dwarf interiors are roughly isothermal. This means they freeze from the inside out.

The crystalline structure is expected to be bcc.

I think the only thing this has in common with nuclear pasta is that they are both phase transitions. Nuclear pasta is the aggregation of nucleons (neutrons and protons) into wacky shapes in order to minimise the total energy density of the gas. The temperature is not really an issue here and it is all to do with density.


How dark would the sun become if we traversed a Bok globule? Bart Bok gave his name to dust clouds of 2-50 solar masses, which by their darkness may be 100 times more frequent than telescopes can detect.

If the solar system traversed a Bok Globule, how dark would the sun become towards the surface of the Earth, as visible by humans?

Not noticeably darker.

Assuming such a globule has a mass of 50 solar masses and a diameter of 1 light year, that would make it's average density about $2.2\times 10^{-16}\,kg\,m^{-3}$ which is fairly close to not being there at all in human terms.

An imaginary tube of this stuff 1 AU tall and of area 1 meter squared would contain about $3\times 10^{-5}\,kg$ of mass.  That is an imaginary tube from Earth to the Sun.

That much matter won't block much light.

For comparison the density of air at ground level is about $1.2\,kg\,m^{-3}$.  So less than a millimeter of air at ground level blocks more sunlight than the dust cloud would.

Also


How does space time differ within a galaxy? Not the same question but is similar to other marked duplicate.
How much time dilation  does the center of a galaxy can exist and sustain human life from our point of view? What would a day equal to each others point of view looking at each other?

I'm not asking about moving celestial bodies moving from each other but the difference in the gravity wells from the center of our galaxy being generally denser then the edge of a galaxy.

This is another way of asking if I wasn't clear above the line. Not to invalidate any answers.

Point 1 being near the center of the galaxy and point 2 at the the most edge the galaxy. How much would the differences in the galaxies gravity well point 1 being deeper in the center compared to the point 2 the edge effect time?

The gravitational redshift factor and also the factor by which time dilation occurs in a gravitational field is approximately $GM/Rc^2$. Here, $R$ would be the radius from which light was emitted and $M$ is the amount of mass within radius $R$ (making the crude assumption of spherical symmetry).

Thus, for an external observer,  light emitted from a radius of 10 kpc, might have $M \sim 10^{11} M_{\odot}$ in the galaxy shown. The redshift/time dilation factor is only $5\times 10^{-7}$ and is basically negligible.

This could perhaps gets an order of magnitude bigger as one approaches the denser central bulge, but is still very small.

The only appreciable redshift and time dilation effects will come in the immediate surroundings of any supermassive black hole. By immediate, I mean within about a hundred Schwarzschild radii, where $R_s = 2GM/c^2$. In our Galaxy that would mean within a billion kilometres.


When graphing radial velocity measurements, whats the relationship between days and the orbital phase I graphed the radial velocity measurements for various planets, however, I am able to calculate the RV and the phase but whats the translation between phase and days. I want it to be days vs RV.

If the orbital period is $P$ (in units of days) and the time of periastron passage (a specific date when the stars are closest together) is $T_0$ (generally given as a Julian Date, or JD), then the conversion from orbital phase $\phi$ to JD (days) will be:
$$ JD = T_0 + \phi P $$
In other words, $\phi$ is just the fraction of the orbital period, so multiplying it by the period gives the number of days since periastron within the orbit.
If the orbit is circular, there is no specific periastron date so $T_0$ can be set to any arbitrary date, but is generally chosen to be a date when the radial velocity is zero (or at the systemic velocity if that is non-zero) and crossing from negative to positive radial velocities.


Eyepiece recommendation for Orion XT8 I am looking for some eyepiece recommendations for my Orion XT8. There are so many kinds out there, and I am not quite sure what to get as recommendations vary wildly. Here are some Q&As that hopefully can give some insight into what I am looking for. Any thoughts or suggestions would be much appreciated.
What kind of telescope do you have?
Orion XT8. Focal length 1200 mm, F/5.9.
I have had it for a little less than a year. Before that I had a 4 inch Orion tabletop reflector.
What kind of eyepieces do you have today?
I have a kit of eyepieces from Canada Telescopes, which seems to be a rebranded imported kit. I can't say much about the quality. It has 6, 9, 15 and 20 mm eyepieces. They have approx 68 deg FOV. The 20 mm is the only semi-usable one, the others aren't really sharp enough.
I also came across a used Celestron Luminos 15 mm. It is nice, but I was underwhelmed by the wider 82 deg FOV. I have heard that these eyepieces do better on other kinds of telescopes that Newtonian reflectors, but I don't know if it is true or not.
What kind of objects do you observe?
Previously it has mostly been the moon and planets. I have spent countless hours looking at Saturn and Jupiter, and recently also Mars that it has been in opposition.
Since a month or so ago, I decided to spend a year trying to observe and check off everything in the book "Turn Left at Orion". That is when the eyepiece problems began :-). The Pleiades requires something wider than I have, and some of the double stars I have been trying to see require higher quality eyepieces for the magnification (dark skies aside).
So, literally a little bit of everything right now.
What magnifications are you looking for?
Low (40x), medium (75x) and high (150x). This means about 30 mm, 16 mm and 8 mm.
Do you wear glasses?
Yes, but I can take them off while observing if that increases my options.
Do you plan to use the eyepieces with another telescope?
Sometime in the future, I'd like to upgrade to a 10 or 12 inch Newtonian. I have been toying with the idea of astrophotography, so perhaps something that would work with a SCT too. But that is much lower priority.
What is your budget?
So hard to answer.
If the recommendation would be for general purpose eyepieces that will help me during my year of visual observation with the XT8 only, then I would like something below 200 USD/eyepiece. Meaning, that if they would be of less value on a 12 inch scope looking at DSOs or a smaller SCT looking at planets, then I'd like to limit the spend.
If the recommendation would be for "forever eyepieces" that are great no matter if it is the XT8, a 12 inch dob or a SCT, then I am more flexible on price because I would know that I could hang on to them for a longer period of time. Perhaps 500 USD/eyepiece maximum (but that would hurt...).
I hope you this gives some valuable context! Thanks!

This is all subjective. Ask three different people about eyepieces and you'll get three different opinions. My answer is not the right answer, it's just my view as an amateur who has a case full of eyepieces.

The main difference between the less expensive and the more expensive eyepieces is in the quality of the view towards the edges.
Almost all eyepieces will give good views in the centre of the view. It's towards the edges that the view starts to degrade.
Wide field eyepieces need more lenses to provide a well-corrected view to your eye. But the more glass you add, the more surfaces you have which can reflect light back and away from your eyes, so this glass needs to be well-polished.
This is why the expensive eyepieces cost as much as they do. To provide a wide field that's well corrected at the edges needs lots of quality glass, and a good QA process.
Finally, there are definitely diminishing returns when paying more money. The difference between a €50 eyepiece and a €200 eyepiece will be more apparent than the difference between a €500 and a €1000 eyepiece.

I see three main groups of eyepieces:

€1 to €199
€200 to €399
€400+

The first group consists of a lot of different brand names all selling pretty much the same eyepieces.
Generally speaking, a factory in China makes a batch of eyepieces, and different brands buy these eyepieces and pay the manufacturer to put their brand names on the eyepieces they buy.
The brands who pay more money probably get to pick the better eyepieces from a production run.
Or take the case of Explore Scientific, who are partnered with Jinghua Optical Corporation (JOC), which suggests a closer relationship and explains why some of Explore Scientific's eyepieces (the 92°, 120°, and 3" 100°) aren't available under other names.
The eyepieces in this group are mostly all much of a muchness, but there are some gems to be found (the 8.8mm 82° offered by ES and Meade, or the Orion E-series 7-21mm zoom).
Also, it's perfectly possible to buy a €50 eyepiece and get lucky with a great manufacturing sample that gives eyepieces ten times its price a run for their money.

The middle group begins where I see a noticeable increase in optical quality.
The Baader Morpheus 76° eyepieces are at the lower end, and are an example of some fine "forever eyepieces", as you put it, which won't break the bank. (They're slightly more expensive in the USA than Europe, I believe, which might make the Pentax XW a better option if you're in the Americas)
APM have some good eyepieces in this price range, too, such as their 100° series (which many experienced observers have said are the second-best 100° eyepieces available today) and their 12.5mm 84° eyepiece.
Pentax's 70° XW range are well-regarded, with eyepieces from 3.5mm to 40mm.
Also Tele Vue have their "budget" Delite 62° eyepieces in this price range.

The expensive group contains the Tele Vue Delos 72°, Nagler 82, and Ethos 100° ranges.
Right at the top are the Tele Vue Apollo 11mm and the Nikon NAV HW with 102° FoV, which comes in only two focal lengths - 12.5mm and 17mm - but comes with a screw-in lens to give an extra focal length, allowing for 10mm and 14mm.

So, having said all of that, what would I recommend?
Low power I would recommend the APM UFF 30mm. Other options are the Pentax XW, Explore Scientific, and Tele Vue Nagler. (You could go up to 40mm if your pupil dilates to 6.5mm in the dark).
By the way, at low magnifications a FoV of ~70° is wide enough, so don't feel that you have to go for an expensive ~80° eyepiece here.
Medium power I would recommend the Baader Morpheus 17.5mm. Another option is the APM Hi-FW 12.5mm. (I have both of these, and am therefore biased towards them, but they are my most-used and favourite eyepieces)
There are many different options in this range. Anything from 10mm to 20mm would be good.
You might even want to wait a while before buying a medium power eyepiece. Get the low power and high power ones and use them for a while, and you'll get a much better idea of what you'll want out of a medium power eyepiece.
High power I would recommend the Baader Morpheus 6.5mm. Other options are the APM 100° 7mm, the Pentax XW 7mm, the aforementioned Meade/ES 8.8mm.
There are quite a few different options in this range, too.
(At high magnifications the main thing you'll want to consider is minimising "time to nudge": when observing a planet you want to be able to have it drift across the field of view for as long as possible before you have to nudge your telescope to bring it back into view.
For this you'll need a high FoV eyepiece - 82° is a comfortable minimum these days, now such a relatively high FoV is widely available at reasonable prices.)

This answer has become a lot longer than I thought it would. And I haven't even mentioned zoom eyepieces yet! (You could get a zoom. And a Barlow. And a low power eyepiece.)
Another option is to buy a set of SVBONY eyepieces and use them for a while, and gradually replace your most-used focal lengths with better quality versions.
But I'll wrap up. To conclude: don't take my word for it! Get some more points of view. Read some reviews. Spend some time with the astronomy.tools FoV calculator to plan the focal lengths you want to have in your eyepiece collection.
(And if you decide to go for the expensive eyepieces then look at the second hand market - people generally take good care of their expensive eyepieces, and some brands maintain good resale value)


How is interstellar gas density mapped from GAIA data? I found the image below in Space.com's article This 3D Color Map of 1.7 Billion Stars in the Milky Way Is the Best Ever Made.

The caption for this image reads:

  The Gaia spacecraft gathered observations for this all-sky view of the Milky Way and neighboring galaxies between July 2014 and May 2016, releasing the data on April 25, 2018. This image shows all the stars' colors and brightness (top), the total density of stars (middle) and the distribution of interstellar gas and dust across the galaxy (bottom). Credit: ESA/Gaia/DPAC

The first two maps reflect data obtained from analysis of GAIA's measurement of starlight as explained in this answer, but the third map is described as "the distribution of interstellar gas and dust across the galaxy."

As far as I understand, GAIA does not transmit all complete image data to Earth, but instead does a significant amount of image processing, object detection, classification and analysis, and other data reduction in order to extract positions and radial velocity measurements for point-like objects such as stars, though of course small solar system bodies such as comets and asteroids end up in the raw data as well.

Question: How is interstellar gas density obtained from GAIA data?

Two techniques immediately spring to mind.

For the stars you detect, you can compare their colours and luminosities (Gaia provides photometric colours and distances) with what you expect for a star of that type at that distance. The difference between what you expect and what you observe tells you the reddening and extinction caused by interstellar dust, integrated in a column towards that star. By going through this process for lots of stars in different directions you build up a 3D picture of the distribution of dust.

The measurements are mostly insensitive to gas - simple extrapolation between dust and gas is often assumed.

A second technique would be just to compare how many stars you can see and with what brightness distribution, with a model for the Galactic stellar population and dust distribution. This is an old-fashioned way of estimating dust extinction, works well locally, but I suspect not what is used in that map, for the simple reasons that Gaia also provides the distances to stars, and one of the purposes of the mission is to improve the Galactic model, not assume it.


Why doesn't the fusion process of the sun speed up? Am I correct in saying that the fusion process of the sun is constant, i.e. X amount of fusion happens per day, more or less?
Why does this not speed up, i.e. one fusion event creates energy for two fusion events, etc.?
Does every collision of atom cause a fusion event, or is the probability small for a fusion event to happen thus it's not a runaway reaction?
I have heard that the probability for fusion event to happen is only 1 in 1012 for every collision.

Am I correct in saying that the fusion process of the sun is constant, i.e. X amount of fusion happens per day, more or less?

Yes, at least over human timescales. You could reasonably expect the fusion rate within the sun to be the same today as a few thousand years ago or into the future, give or take some small fraction.

  Why does this not speed up, i.e. one fusion event creates energy for two fusion events, etc.?

The energy released by fusion is quickly distributed as thermal energy in the centre of the sun, and the temperature difference between surface (around 6000K) and centre (estimated 15 million K) drives an energy flow from hot to cold.

  Does every collision of atom cause a fusion event, or is the probability small for a fusion event to happen thus it's not a runaway reaction?

Fusion in the sun is not a runaway nuclear reaction (like a critical mass of uranium in a fission reaction).

It is possible in theory to have runaway fusion events, but the pressure and temperature for these to happen are not approached in the core of the sun. For stable stars like the sun, the forces and energy flows are in equilibrium - if the core grew slightly hotter, then the pressure would increase and the star expand slightly against the force of gravity to compensate. Interesting things happen when stars fall out of equilibrium and runaway fusion ignition can happen in some scenarios.

In addition, this equilibrium point moves during the lifetime of a star as its mix of elements changes due to fusion. This is predictable for many stars and forms the basis of the main sequence stars in the Hertzsprung-Russell diagram

  I have heard that the probability for fusion event to happen is only 1 in 10^12 for every collision

I don't know the accuracy of that, but it seems reasonable. The definition of "collision" becomes somewhat arbitrary in such a hot dense environment. If you only include approaches close enough to make the strong nuclear force dominate the interaction, the ratio could be higher.

Another fact that I found interesting in the same area is that the power density from fusion - i.e. the Watts per cubic metre of substance - in the sun is roughly the same as that found in a typical compost heap. It is a very different environment to the inside of a fusion reactor experiment or a fusion bomb, which have much higher power densities.


Why would there be no day/night cycle even though the firmament would still be rotating, according to Copernicus In chapter 6 of "The Revolutions of the Heavenly Spheres" where Copernicus maintains that the heavens are immense compared to the size of the Earth, the Edward Rosen translation of the book states this

  "The upshot of the argument, then, is the claim that the earth as a part of the
  celestial sphere shares in the same nature and movement so that, being close to the
  center, it has a slight motion. Therefore, being a body and not the center, it too will
  describe arcs like those of a celestial circle, though smaller, in the same time. The
  falsity of this contention is clearer than daylight. For it would always have to be
  noon in one place, and always midnight in another, so that the daily risings and
  settings could not take place, since the motion of the whole and the part would be
  one and inseparable."
  To add context. Copernicus is debating the classical point which assumes that "the center is at rest, and the nearer an object is to the center, the slower its motion." Copernicus in the previous paragraph states that this assertion could be translated into a sphere. Where the center are the poles, and the constellations around those poles (i.e. Ursa Minor, Draco, etc...) would be rotating slower as they are closer to those poles. Yet Copernicus refutes this by saying that if those constellations truly moved slower than the rest, they wouldn't be able to make a full 360° rotation in a year.

Copernicus then in the paragraph above, argues that the Earth too must orbit around the center in order to account for the "slowness" in the poles. However Copernicus further refutes this by stating that there would be no day/night cycle in this case.

To elaborate on my question asked in the title. Why would the Earth need to orbit in order to account for this "slowness." And why would there be no day/night cycle even though the firmament would be rotating and not the Earth along with it.

Find a spherical object and imagine it's a celestial sphere with the world at the middle... consider the daily rotation of the sphere around as a spinning top, and a yearly tilt through the constellations which is the same as an unstable spinning top which is about to tip over.

Now, rotate it and dont spin the days, and only spin the sphere for the yearly seasons.

How fast are all points on the sphere rotating as it tilts, same as a spinning top is not straight, and omitting the individual rotations?

All points on the sphere are going at the same speed, the top of an orange describes a 1cm axial tilt, and the sides of the sphere also describe a 1cm axial tilt... put the orange sideways and tilt it again, the rind is at the same speed as the side whichever way the orange is viewer. How to combine the slower star speed of rotation in 24 hours with the equal star speed of rotation through the seasons? The previous theory didn't give two speeds and axes of rotation.

So, the 7 heavens other than the stars can be described as an epicycle on a 2D, as a turntable and epicycle.

For the stars, the 2D epicycle is more complicated, because it has to account for two speeds of rotation, the constant one around the seasons and the one that is slower in the middle. The epicycle geometry breaks down in 3D as a sphere, and only a sphere can explain the two types of rotations in the stars, and the 2100 year one.

Copernicus said: A sphere cannot move as an epicycle. Take an orange and try to move it as an epicycle.


What soviet telescope is this? A friend has a vintage telescope with a broken lens, I asked him to send me its catalog so I can see if I can buy him a new one, he gave me what looks like a Russian catalog, it seems a soviet telescope, I can't read it, I hope you could tell me what it is. I want to search for its lens using English characters.

Possible search terms to use:

Vintage Russian CCCP Spotting Scope 3PT-460 20x
Komz Russian Military Telescope - 3PT-460 20x
ZRT 460 Russian Scope

Images of item:

Prices I found range from \$50 to less than \$100.


What is a possible software for simulating binary star systems? I would like to define both initial velocities and masses of components so that orbits may be either elliptical of hyperbolic. Is there any software to make such simulation?

Unfortunately, Binary Star Combined Solution Package is out of date and NASA online plotter supports only circular orbits.

Thanks.

MESA

It is designed specifically for stellar evolution. There are many modules that you can play around with, including binary interaction.


Right ascension, Equinox, Positions Help I am trying to answer the following question which I slightly understand but need a clearer explanation:
What are the times of year to observe an object at X hours of RA at visual wavelength?
I am trying to use the march equinox as a starting point for 0 RA and for a 12 hour RA object, it will begin to rise from the east as the sun sets west of course, around 6 months from the march equinox???
I appreciate an explanation and if you can link a map or a helpful visual as well. Thanks!

For an object at about RA 6h and close to the celestial equator, viewed from 37 degrees N, it becomes visible in the early morning on about July 20th, and it is is visible at some point during the night until about May 20th, when it sets shortly after the sun.  Between May 20th and July 20th it is not visible. (Determined Stellarium)
An example of such an object is Betelgeuse.
An object that is at 6h of right ascension will culminate at midnight (ie reach its highest point at midnight about on the winter solstice on Dec 21st. The sun has a right ascension of 6 hrs on the summer solstice on June 21st and a RA of 18 hours on the winter solstice.  So the best time to observe the object would be on the winter solstice.


Is Ceres a terrestrial-type (dwarf) planet? Does Ceres belong to the terrestrial-type celestial bodies?

I'm inclined to say no (and footnote, I realize Wikipedia isn't a good source for scientific proof as it's not always right, but I'm using it more to demonstrate a point than than use it as an authoritative definition).     Wikipedia:

  A terrestrial planet, telluric planet or rocky planet is a planet that
  is composed primarily of silicate rocks or metals.

and

  All terrestrial planets have approximately the same type of structure:
  a central metallic core, mostly iron, with a surrounding silicate
  mantle. The Moon is similar, but has a much smaller iron core. Io and
  Europa are also satellites that have internal structures similar to
  that of terrestrial planets.

and

  Dwarf planets, such as Ceres and Pluto, and large small Solar System
  bodies are similar to terrestrial planets in the fact that they do
  have a solid surface, but are, on average, composed of more icy
  materials (Ceres and Pluto have densities 2.17 and 1.87 g cm−3,
  respectively, and Haumea's density is similar to Pallas's 2.8 g cm−3).

So, in my opinion, half ice/half rock if a different category and not terrestrial.   There probably are planets that formed outside the frost line that are half rock-half ice in other solar systems, but I don't think I'd call them terrestrial.   We probably need a new word for them.

If Ceres was to drift closer to an Earth's orbit, it's ice would melt and it would have oceans (and an H20 rich atmosphere), at least, for a little while anyway until it lost it's atmosphere.   Planets with Ceres composition would likely be water worlds if they were warm enough.

Now, I can't speak for any the official answer, but that's my opinion.  If Ceres is Terrestrial and Haumea is more rocky than Ceres, then Ceres probably shouldn't be the only terrestrial dwarf planet in the solar-system.

edit.

Link provided that says Ceres is a terrestrial dwarf planet.

reasoning:

  Terrestrial planets have numerous similarities to plutoids (objects
  like Pluto), which also have a solid surface, but are composed of more
  icy materials.

Some of this is just semantics, but lets look at density, which is a pretty good measure of Water-Ice and other ices to Rock content.

Ceres 2.08 g/cm^3

and the other dwarfs, by size (not all of them are well measured so density isn't certain).

Eris 2.3
Pluto 1.88
Makemake ~2
Haumea  2.6-3.3
Quaoar ~2.2
Sedna (2.0?)

and I could go on.

Lets do a few moons, just for fun.

Io:  3.55  (extreme volcanism)
Europa:  3.0
Ganymede:  1.93
Calisto:  1.83
Titan:  1.88
Triton:  2.06
Enceladus:   1.61

Moons are easy, we don't call them dwarf planets even if they are larger than dwarf planets cause we call them moons. - definition averted.   :-)

But I have a hard time seeing why Ceres gets a different classification than other very similar objects which just happen to be further away.   Now, granted, Ceres, while it formed outside the frost line, it's now inside the frost line so surface ice on Ceres doesn't last.  It's got a rocky surface while most Kuiper belt objects have an icy surface.    I can see that, defining an object based on it's surface, but I still think a half rock/half ice object (of significant size, large enough to be round and meet dwarf planet criteria), should be called half rock-half ice or whatever that definition is.   I don't think it should be rocky/terrestrial if it's Ceres and icy if it's Pluto, even if that's how they look the surface, but that's just my opinion.

/// hope that wasn't too much of a rant.  :-)


Would Venus have any significant effects on Earth if its orbit were entirely within the habitable zone? If Venus was entirely inside the habitable zone, or life zone, would its proximity to the Earth provoke any remarkable changes to the Earth?

Changes on Earth:

Except for possible tides and possible eclipses, if it were near enough, no.

Changes on Earth's orbit:

Venus would establish some resonance with Earth's orbit, if it were near enough or in one of the exact possible spots. This will not have a major impact except on year's lenght and Sun illumination on Earth.


Where might a semi proficient amateur analyst participate in meaningful astronomical efforts I am a retired engineer that has an ongoing interest in space efforts.  In my youth I did work on the Apollo program but on propulsion and vehicle thermal control: not flight dynamics.

I have reviewed/studied some astrodynamics (Vallado, Bates, etc) and can run GMAT and tudat libraries from Delft University.  I have explored Systemic Console, tap, and other software for implying the existence of exoplanets from RV curves and light transit curves.  I can download data in FITS format from various astronomical databases.

Where might I direct my efforts in some meaningful way by participating in a project, collaborating with an investigator, or pursuing some research oriented endeavour on my own?   Could someone point me in a direction where I might do some meaningful work?  Just picking a star and starting a transit analysis does not seem to have any probability of leading anywhere.

Those sites where one views a light transit curve or tries to classify a galaxy type are not very interesting to me.  I feel I could investigate things more in depth than that.

Directions, Ideas, offers! Anyone.
I will post this in Space Exploration forum also.
Thanks,

Tom Kosvic

Those sites where one views a light transit curve or tries to classify a galaxy type are not very interesting to me. I feel I could investigate things more in depth than that.

There is a lot more to many citizen science projects than that! I'd recommend having a look through all the projects available as there may be one in an area you are more interested in : Wikipedia: list of citizen science projects

Zooniverse is a great collection of Astronomy related projects. They are not just toys or made for school children, they are real science projects where you will be contributing to important and current research. Members who show they have advanced skills often get given more credit in the results weighting algorithms, or upgraded to more difficult tasks. People have even spotted things professional scientists had missed, and been able to bring new things to the attention of the scientific community and be credited on published papers.

Amateur observational astronomers are also useful members of the scientific community. Some beautiful and intriguing images of object have been taken by skilled amateurs, who don't have the pressures of short observing times.

Are you focused on research or on generally aiding science?

As someone with a lifetime of experience, you would be invaluable as a volunteer at outreach and engagement activities. You could ask your local University or even city council if they are running any science outreach projects you could collaborate with. Again, this does not just mean school children, although that can be a part of it. Lobbying for science literacy in politics, organising science fairs, giving talks to interested groups, etc.

I will try to think of more research related examples later today. It would be helpful to know what country you are based in, as my experience may be very different to yours.

Later edit: If you are interested in exoplanets but want more autonomy of research than the citizen science projects afford, you may be interested to know that NASAs Exoplanet Archive is a publicly available archive of data from exoplanet missions along with the tools to work with the data. Happy planet hunting!


Is the angular size of the black hole in the movie "interstellar" completely overblown? I assume everybody is tired of reading questions derived from the movie Interstellar, I will try to keep this short and simple:
In general in movies, in order to have stunning visuals, celestial objects are displayed as if they were extremely close to the observers, having a enormous apparent angular size.
One striking example of this that has seems counter-intuitive, is the angular size displayed for the black hole "Gargantua" in the movie Interstellar. The characters in the movie are so close to it that it can be fully displayed in great detail (see attached image: Black hole seems to cover almost/over 90 degrees of the field of view..!)
Now, we can fully experience gravitational pulls from much smaller objects than a supermassive black hole in reality. The angular size of the sun and the moon are so small that we need telescopes to appreciate details, despite their gravitational effects dominating our everyday life.
It seems to me that a supermassive black hole would have such immense gravitational effects, that in order to be able to "freely" navigate around it (as done in the movie), observers would be so far away that the angular size of the black hole would be tiny (which would result in very non-spectacular cinematography).
The question is: Realistically, for a supermassive black hole ("with a mass of 100 million times of the sun"), what would be roughly its angular size when standing at a "safe" distance from it?

Assuming the spin parameter of the black hole is close to the critical value, it could be quite large.
The scenario of heating a planet orbiting close to a black hole via blue-shifted cosmic background radiation has been explored. The shadow of the black hole would take up a large fraction of the sky on such a planet, see for example figure 1 in Bakala et al. (2020). They find that for an Earth-like planet, the region where this can be achieved is located beyond the Roche limit (where the planet would be broken up) for black hole masses $M_{\rm BH} > 1.63\times10^8 M_\odot$. As shown in figure 6 of the paper, the bulk of this radiation will be in the ultraviolet, but there's probably enough visible light in the mix to make a bluish patch near the edge of the black hole shadow. Bear in mind that this would need to be a very isolated black hole, because stars would also be blue-shifted and therefore be extremely bright sources of high-energy radiation, which would be an extremely unpleasant environment.
The scenario in the movie includes an accretion disc, this scenario is considered in Schnittman (2019), who notes that for accretion rates low enough to produce the Sun-like colour of the radiation depicted in the movie, you'd have to be a long way out:

Regardless of the specific details—whether the planet is tidally locked or not, embedded
in the disk or not—it is clear that for any planet to be habitable while also receiving its
primary heating flux from a Novikov-Thorne type accretion disk, the planet either has to
be extremely far from the black hole, or the mass accretion rate has to be an infinitesimal
fraction of Eddington. This leads to an important inconsistency: The Novikov-Thorne model
is predicated on an optically thick, radiatively efficient disk. At accretion rates down near
$10^{-12} \dot{M}_{\rm Edd}$ the optical depth is unlikely to be large enough to allow the gas to cool, thus
leading to a very hot, low-density accretion flow.


Determining moonset time from Nautical Almanac I have recently learnt to determine moonrise and moonset times using the Nautical Almanac, as part of my maritime navigation course. I have a query regarding the determination of moonset times. I will use an arbitrary date and an arbitrary latitude to illustrate my confusion. On 3rd February 2020, at the equator, the Nautical Almanac tells me that the moon will set at 0041 hrs. Does this mean that the moon that rises on 3rd February 2020 sets at 0041 hrs on 4th February 2020 or does this mean that the moon that rose on 2nd February sets at 0041 hrs on 3rd February 2020?

It means that there is a moonset at that time on the third of February.

At the equator, it is certain that the moon that set at 0041, rose the previous day (Feb 2nd), at about midday.  (the moonset is about 50 minutes later each day, which means that there can be days with no moonset, but you can't have a day with two moonsets)

On days with no moonset the time of moonset in your almanac might be indicated with a time that is more than 23:59

For example:

Date of month
1    2    3    4
h m  h m  h m  h m
2052|2250|2500|0100

Here you see on the first day of the month the moon sets at 20:52, on the second it sets at 22:50, on the third day there is no moonset. The moon sets one hour after midnight. The moonset 25:00 and 01:00 refer to the same event.


How to form Copper from Calcium in a supernova explosion? What are the reasonable chain of reactions to form copper (63, 29) from ca(40, 20) during supernova explosion? And how it will happen? I do not understand the theory behind it. I thought that the reaction will end at Iron and now it does not make sense to me.

Copper is not thought to be primarily made in a supernova. It is thought to be mainly produced by the s-process of slow neutron capture onto iron-peak nuclei that already exist inside a star. These reactions are endothermic.

The source of the neutrons is still somewhat debated, it could either be from the decay of $^{13}$C in relatively low-mass asymptotic giant branch stars or is more likely from the decay of $^{22}$Ne in more massive evolved stars (e.g. Pignatari et al. 2010).

Below you can see a typical route for producing copper from $^{56}$Fe. The axes of the plot are neutron number on the x-axis and proton (atomic) number on the y-axis. Three neutron captures are followed by a beta decay, a neutron capture, a beta decay, then 3 more neutron captures followed by another beta decay to form $^{63}$Cu.

The net process is
$$ ^{56}{\rm Fe} + 7n \rightarrow \  ^{63}{\rm Cu} + 3e + 3\bar{\nu}_e$$

The copper is then distributed into the interstellar medium by a later supernova explosion in the same star.

If there are no "seed" iron-peak nuclei (e.g. in the first/second generation of stars), then copper can be inefficiently produced by explosive r-process neutron capture during supernovae explosions. However, this would not contribute very much to the copper we see on the Earth.


What is the naming convention for newly discovered objects? Often, a newly discovered star is identified by a seemingly random string of letters and numbers. I'm sure that there is some order to it, though. What is the naming convention for newly discovered astronomical objects?

I'm sure NASA doesn't ask the public, as something like this would happen:

The International Astronomical Union's Naming Astronomical Objects webpage is a comprehensive guide to how astronomical objects are named.

For example, for "Specifications concerning designations for astronomical radiation sources outside the solar system", the IAU states that the designation (or sequence of letters and numbers) follow a general form:

  Acronym  ^ Sequence ^ (Specifier)

Where the Acronym

  is a code (i.e., alphanumerical string of characters) that specifies the catalog or collection of sources.

The sequence

  The sequence (or numbering) is an alphanumeric string of characters, normally only numerical, that uniquely determines the source within a catalog or collection. It may be a sequence number within a catalog (e. g., HD^224801), a combination of fields, or it may be based on coordinates.

and finally, the Specifier

  The specifier is optional and allows one to indicate other source parameters. However, they are not required syntax and are enclosed in parentheses.


Can I do M.Sc in physics to become an astronomer after doing B.Tech in aeronautical engineering? I am a student of class 12 and I am studying in India. I want to be an astronomer and I know that I have to get a B.Sc, then M.Sc, and then Ph.D in astronomy. I have also started to learn the python programming language. I came to know that there are less opportunities for research in astronomy, since many organizations like ISRO not only employ scientists, but also employ engineers to make satellites and spacecrafts. I will try my best to become an astronomer, but to have life security (if i am unemployed), can I do aeronautical engineering so that I can even get employed in ISRO? After having a degree in engineering and after doing engineering can i do M.Sc in physics and then i start preparing to be an astronomer?

Let me tell you about myself a bit, I love to study science and I love to learn about stars (I have also watched documentaries on astronomy like "How the universe works" and "Cosmos"). I also love aircraft so aeronautical engineering would also be a job that I would love. And please tell me should I also do some extra courses which would enhance my skills as an astronomer?

It is good to see that you are interested in astrophysics at an young age.
No problem with you doing the clichéd BE that everyone in India is forced with, you can still get your PhD in Astrophysics.

Your 4 years of B.Tech aeronautical is going to be tough, so be prepared for it. In the mean time, keeping in mind your aim of becoming an astrophysicist, prepare yourself equipped with programming languages, especially Python/C++. The reason I say this is because you don't need too have much of an astronomy background to do M.Sc Astronomy, but you definitely need to have a good knowledge in a programming language.

There are websites like Course Era or EdX and plenty more on the internet that can help you with learning python/C++ in an interactive way. Go for it, and trust me it will help you a lot.

As a final year student doing my PhD in Astrophysics (hailing from India) let me tell you the information I have acquired over the years.

With regard to the Masters/PhD in Astronomy/Astrophysics, the procedure is quite different in different countries.

Procedure within India

There are plenty of Universities in India that offer you a Masters course in Astrophysics.

Almost all the details required are summarised neatly in THIS PAGE,
so I am not going to go through the list of things required.

Procedure for Europe

You would need:

All your original mark sheets attested by the college/university (Bachelors, Masters, etc)
Two/three recommendation letters, and the number of letters depends on the university you are applying to.
A completely filled application form.
IELTS scores (depends on country)
Statement of purpose, as to why you want to do the course.

This is pretty much all you need for Europe. For certain universities, you might be having a Skype interview, wherein they might ask you about the stuff you know.

Procedure for USA:

When it comes to the USA, it gets quite complicated!

You will need:

All your original mark sheets attested by the college/university (Bachelors, Masters, etc)
Two/three recommendation letters, and the number of letters depends on the university you are applying to. For most universities it will be online forms that your professor needs to fill after he/she receives the mail.
A completely filled application form. This is going to cost you somewhere from around 50-80 dollars depending on the university.
GRE scores (both general GRE and subject (Physics,Chemistry etc.) GRE) that you can send to the university directly online by paying money (25 Dollars for each university). Only a few universities accept you without a subject GRE.
IELTS/TOEFL scores. TOEFL scores can be sent online, but if you are taking IELTS you have to go to British council and have them send it.
Statement of purpose, as to why you want to do the course.

So it depends on where you want to apply. I would recommend you to apply to both, to increase your chances. For Europe, the chances of you getting through depends on the country you apply. For USA, the chances of you getting through solely depend on your GRE scores. It doesn't matter whether you are a gold medallist in your Bachelors, it all boils down to GRE. If you fail there, your entry into USA will be difficult.

One advantage of Europe is that, mostly, the PhD is only for 3/4 years, depending on the country. Whereas in the USA, it is mandatory for the student to write exams in the first year, teach/be a research assistant in the second year, and only in the third year start the PhD project.

To search for PhDs in Astrophysics depends on the country you are applying for. But as a start, you can have a look at http://jobregister.aas.org/. PhD calls, internship calls, post-doc positions, etc. related to astrophysics will be posted here during the start of every month.

With regard to your question

  can I do aeronautical engineering so that I can even get employed in ISRO?

Yes you can definitely do that. ISRO not only employs aeronautical engineers, but also people who have completed their BE in electrical and electronics and also computer science.

Have a look at THE CAREERS WEBSITE OF ISRO to get more information.

Good luck with your future!


Could S5-HVS1 conceivably still hold on to any planets? If the star (S5-HVS1) has been accelerated to 1,700 km/s by a close encounter with the super massive black hole at the centre of our galaxy (Sagittarius A*) what would happen to any planets orbiting this star? Would they have been stripped off by the massive acceleration or could any conceivably still be in orbit? Presumably any orbiting bodies would be significantly perturbed!? see https://www.sciencealert.com/our-bossy-black-hole-kicked-out-a-star-and-it-s-shooting-through-the-galaxy-insanely-fast

According to Ginsburg, Loeb & Wegner (2012) "Hypervelocity planets and transits around hypervelocity stars", planet-hosting hypervelocity stars could form from the disruption of a stellar binary by a massive black hole, and the probability that the planet remains bound to the hypervelocity star is fairly high for suitable initial conditions.

In their simulations, this required initial planetary separations of 0.02–0.05 AU. The star S5-HVS1 is described in Koposov et al. (2019) "The Great Escape: Discovery of a nearby 1700 km/s star ejected from the Milky Way by Sgr A*" as a 2.35 solar mass A-type star. Planets around similar stars include ultra-hot Jupiters such as KELT-9b, which may at first appear promising.

The initial binary separation is another matter. The systems Ginsburg et al. investigated had binary separations in the range 0.05–0.5 AU. For the S-type planetary systems considered (the planet orbiting one of the stars as a satellite), this is a much smaller separation than any known planet-hosting binary: the planets known in such systems are in circumbinary systems which the paper does not explore.

It's therefore unclear whether the kind of progenitor system required would be capable of forming planets in the first place.


Comparing orbits between a planet and Red Dwarf I have a MATLAB problem, so I have an earth like planet orbiting around a sun like star and then there is a red dwarf on an elliptical orbit around the star that passes close to the planet. I want to show that sometimes when the Red Dwarf is at it's closest approach the planet may not be there so it won't gain any heat from the red dwarf.

To do this I was told to compare the orbital periods of the two. So for the planet it's orbital period is just like the earth 365 days and the red dwarf's period is 1896.59 days.

I was thinking is there a way I could plot the orbits of these two and run it for say like 100 years and see the if there is points when the earth could miss the red dwarf? If this is wrong please let me know a better way of tackling this problem.
Below is an image of the orbit, with the black line being the red dwarf's orbit and the  blue line being the planet's orbit. The eccentricity of the Red Dwarf's orbit is 0.866 since a=3 Au and b=1.5 Au.

earth like planet orbiting around a sun like star (365 days)
red dwarf on an elliptical orbit around the star that passes close to the planet (1896.59 days, eccentricity 0.866)

If we work in AU and years we can use $GM_{Sun} = 4 \pi^2 (\text{AU}^3 / \text{year}^2)$ for the standard gravitational parameter of the Sun. If we also assume the red dwarf's mass is small enough that it can be ignored we can leave the Earth in it's circular 1 AU orbit and assume that the Sun doesn't move. Then we can check the relationship between period and semimajor axis as follows:
$$T = 2 \pi \sqrt{\frac{a^3}{GM}} = a^{3/2}$$
$$a = T^{2/3} \left(\frac{GM}{4 \pi^2}\right)^{1/3} = T^{2/3}$$
which gives $a=1$ AU for 1 year, and $(1896.59/365)^{2/3} \approx 3 $ AU for the red dwarf so this checks out.
However your drawing is off. The periapsis and apoapsis of the red dwarf will be $3(1-e)$ and $3(1+e)$ or 0.4 and 5.6 AU and this plus this comment makes it clear that your drawing is wrong.
If the ratio of the two periods were a rational number, say $T_2/T_1 = n_2/n_1$ where $n_1, n_2$ are integers, then their relative motion would be periodic with the synodic period
$$T_S = \frac{1}{\frac{1}{T_1} - \frac{1}{T_2}} = \frac{T_1 T_2}{T_2 - T_1} = \frac{n_1 n_2}{n_2 - n_1}$$
but the distance of closest approach would depend on phasing; the initial positions at time $t=0$.
However in your case you have $a_2/a_1 = 3$ and so $T_2/T_1 = 3 \sqrt{3}$ which is irrational and not expressed as the ratio of two integers. That means that if these two orbits are in the same plane and the two bodies do not interact gravitationally with each other, sooner or later they may collide.

Python:
import numpy as np
import matplotlib.pyplot as plt

twopi = 2 * np.pi
a1, e1 = 1.0, 0.0
a2, e2 = 3.0, 0.866

theta = np.linspace(0, twopi, 201)

r1 = a1 * (1. - e1**2) / (1. + e1 * np.cos(theta))
r2 = a2 * (1. - e2**2) / (1. + e2 * np.cos(theta))

x1, y1 = [-r1 * f(theta) for f in (np.cos, np.sin)] # minus to match plot in question
x2, y2 = [-r2 * f(theta) for f in (np.cos, np.sin)]

plt.figure()
plt.plot(x1, y1, '-b', linewidth=2)
plt.plot(x2, y2, '-k', linewidth=1)
plt.plot([0], [0], 'or')
plt.gca().set_aspect('equal')
plt.xlim(-3.2, 6.8)
plt.ylim(-4, 4)
plt.xlabel('Distance (AU)', fontsize=14)
plt.ylabel('Distance (AU)', fontsize=14)
plt.show()


What would night sky look like if Earth was made of antimatter Let's assume that Earth in its current state is suddenly replaced with the Earth completely identical, but made entirely of antimatter. Ignoring meteorites and asteroids which would cause mass destruction, extinction events and whatnot, what would the night sky look like as the anti-Earth swept up the space dust?

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.