The movement of the moon I was in Cuba from January 15th to the 22nd this year and there was a quarter moon. If I use a compass to explain the orientation of the quarter moon, it started in the north position around 4 pm. By 7 pm, it was in the east position, and by 11 pm, it was in the south position. Normally the quarter moon would start in the east and end in the east as it travels across the sky.

Could someone please explain this phenomenon?

Thanks!

I think there might have been a problem with your compass readings, or "in the north" really means just a little north of east.

Cuba is in the northern hemisphere at around 20 to 23°N (tropic of Cancer) and so the Sun can be directly over head and the Moon with an additional 5 degrees inclination with respect to the ecliptic can go several degrees North. (see this answer for a bit about the Moon's motion).

  it started in the north position around 4 pm.
  By 7pm, it was in the east position,
  and by 11 pm, it was in the south position.

I think there might have been something affecting your compass. The Magnetic declination there is about 6-8 degrees, that might have had a tiny effect, but you might look for problems like

iron or other ferromagnetic material,
mechanical compass from the wrong hemisphere sticking due to dip angle
or if it was an electronic compass, incorrect degaussing/calibration.

The other possibility is that by "in the north" you really mean only "slightly north of east" or "east-north-east".

I wrote a short script in Python for your trip, and the three lines in the plot represent January 15 (lowest, blue), 17, and 19 from Santiago de Cuba, the southernmost area of Cuba. The moon does get to the zenith and perhaps a tiny bit past it (towards North) but I don't think that would be visually noticeable.

The Large black dots represent 7PM local time, and each dot to the west is one more hour.

Some Python:

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Loader, Topos

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads        = 180/pi, pi/180

load  = Loader('~/Documents/fishing/SkyData')  # single instance for big files
ts    = load.timescale()
de421 = load('de421.bsp')
earth = de421['earth']
moon  = de421['moon']

Santiago_de_Cuba = earth + Topos(latitude_degrees  = 20.019833,
                                 longitude_degrees = -75.813917,
                                 elevation_m       = 10.)

hours = np.arange(16, 23.1, 0.5) + 5

days   = (15, 17, 19, 21)

altazs, lines, linez= [], [], []
for day in days:
    times = ts.utc(2019, 1, day, hours)
    alt, az, d = (Santiago_de_Cuba).at(times).observe(moon).apparent().altaz()
    alt, az    = [thing.degrees for thing in (alt, az)]
    alt[alt<0] = np.nan
    altazs.append((alt, az))
    r = (1 - alt/90.)
    theta = rads * az
    lines.append((r, theta))
    x, y = [r*f(theta) for f in (np.sin, np.cos)]
    linez.append((x, y))

if True:
    plt.figure()
    plt.subplot(2, 1, 1)
    for alt, az in altazs:
        plt.plot(hours, alt)
        plt.plot(hours[::2], alt[::2], 'ok')
    plt.subplot(2, 1, 2)
    for alt, az in altazs:
        plt.plot(hours, az)
        plt.plot(hours[::2], az[::2], 'ok')
    plt.show()

th = np.linspace(0, twopi, 201)
cth, sth = [f(th) for f in (np.cos, np.sin)]
if True:
    plt.figure()
    plt.plot(cth, sth, '-k', linewidth=1.5)
    plt.plot([0], [0], 'or', markersize=8)
    plt.plot([-0.1, 0.1], [ 0,   0  ], '-k')
    plt.plot([ 0,   0  ], [-0.1, 0.1], '-k')

    plt.text(-0.05,  0.85, 'N', fontsize=16)
    plt.text( 0.9,  -0.05, 'E', fontsize=16)
    plt.text(-0.95, -0.05, 'W', fontsize=16)
    plt.text(-0.05, -0.90, 'S', fontsize=16)

    for x, y in linez:
        linewidth= 1
        plt.plot(x, y)
        plt.plot(x[:-1:2], y[:-1:2], '.k')
        plt.plot(x[:1], y[:1], 'ok')
    plt.show()


Does the orbital variation in planetary gravity affect the Sun's corona Dimitris (see below) argues that the syzygies of the Earth and Venus and those of Mercury, Earth and Jupiter distort the Sun's corona, which in some way affects climate on the scale of hundreds of years.  Is there any evidence to support the idea that the planets do have such an effect on the sun, other than the correlation noted by Dimitris?

Planetary orbits’ effect to the Northern Hemisphere climate, from solar corona
formation to the Earth climate.

Poulos Dimitris

Abstract
The four planets that influence the most the solar surface through tidal forcing seem to affect the Earth climate. A simple two cosine model with periods 251 years, of the seasonality of the Earth – Venus syzygies, and 265.4 years, of the combined syzygies of Jupiter and Mercury with Earth when Earth is in synod with Venus, fits well the Northern Hemisphere temperatures of the last 1000 years as reconstructed by Jones et al (1998). Later reconstructions that give too much emphasis on multy-centenial variation are due to increased error. The physical mechanism proposed is that planetary gravitational forces derange the Solar Corona that in turn deranges the planetary geomagnetic field causing
temperature variations.

http://www.itia.ntua.gr/getfile/1486/1/documents/PoulosPaper.pdf

The gravitational effect of the planets on the Sun can be calculated very accurately. The acceleration of a planet of mass $M$ at a distance from the Sun of $r$ is
$$a = \frac{G M}{r^2}.$$
If you plug in the numbers, you'll find that the largest effect comes from Jupiter due to its enormous mass. The gravitational acceleration of Jupiter on the Sun is $2\times10^{-7}$ m/s$^2$. Even if all planets were aligned so to add up all their gravitational accelerations, the total would only be $2.8\times10^{-7}$ m/s$^2$.

However, the acceleration is not what is important; rather it is the tidal forces, which are proportional to $1/r^3$ instead on $1/r^2$. The tidal acceleration felt over a distance $\Delta r$ at the surface of the Sun is
$$a_{\mathrm{tidal}} \simeq \frac{2GM}{r^3} \Delta r.$$
Now I didn't really bother to read the paper in detail, but the guy doesn't seem to write what exactly in the Sun's atmosphere should be affected by the planets. But even if we say that it's all of Sun's corona, which has a thickness of $\Delta r \sim 10^6$ km, the combined effect of all eight planets standing on a line would be roughly $\mathbf{10^{-9}}$ m/s$\mathbf{^2}$, a minor effect compared to the difference in the Sun's own gravitational field over the same distance, which is 11 orders of magnitude higher.


Will JWST be as durable as the hubble telescope? JW Space Telescope is designed for a 5-10 year mission duration. Seeing as Hubble and other space missions have paved the way for JWST for reliability issues, It even has zero friction gyro's with 100+ years mean time before failure. Is it reasonable to think that JWST can last longer than Hubble?

JWST will orbit a Lagrange point in earth's shadow and will be 4 times further than the moon, around 10 days of travel away, and NASA states that JWST won't be serviceable. Why?

Why are JWST's fuel limitations so short? Are there no provisions for refueling? even a refueling craft? How many gallons of dinitrogen tetroxide ($N_2O_4$) as oxidizer and hydrazine ($N_2H_4$) will it contain?

Surely NASA are secretly hoping for a 20 year mission?

Hubble was in low earth orbit, and was always intended to be serviceable. In fact, the original plan for Hubble was to have the space shuttle carry it down from orbit and take it back up, but they decided that was too risky compared to servicing in orbit.

JWST, on the other hand, will be at the Earth/Sun L2 Lagrange point, like WMAP and Planck before it. The distance to that point is 0.01 astronomical units, which is about 1.5 million kilometers from Earth's center. The Moon's orbit around the Earth is about $1/4^{\mathrm{th}}$ of that. So, a manned servicing mission would be an unprecedented project (beyond Apollo), and an unmanned mission, also unprecedented, would face automation challenges from the 5 second light travel time (10 second round trip).

Because the L2 Lagrange point is only meta-stable (stable to offsets in two directions, unstable along the Earth-Sun line), JWST will have to expend fuel on station-keeping just to stay there. It needs to stay there because it can only be as cool as it needs to be by using the Earth as a sun-shade - those tennis court sized shades built in are actually to shade JWST from the thermal radiation of Earth & the moon.

Put simply, it would be cheaper to build and launch another JWST than to service the existing one.


What effects besides "mass defect" cause the alpha ladder beyond iron-56/nickel-56 to be endothermic? Many sources state that fusion beyond iron-56/nickel-56 (and certainly beyond nickel-62) is impossible due to them being among the most tightly bound nuclei. For example, in the Wikipedia article on the iron peak (https://en.wikipedia.org/wiki/Iron_peak), it is said that:

For elements lighter than iron on the periodic table, nuclear fusion releases energy. For iron, and for all of the heavier elements, nuclear fusion consumes energy.

However, when you actually compute the mass defect, the alpha ladder would be exothermic up to Tin.
$$ Q=[m(Ni_{28}^{56})+m(He_{2}^{4})-m(Zn_{30}^{60})]c^2 $$
$$ Q=[55.942132022u+4.00260325415u-59.941827035u]m_uc^2 $$
$$ Q \approx 2.709 MeV $$
$$$$
$$ Ni_{28}^{56} + He_{2}^{4} \rightarrow Zn_{30}^{60} (+2.709 MeV)$$
$$ Zn_{30}^{60} + He_{2}^{4} \rightarrow Ge_{32}^{64} (+2.587 MeV)$$
$$ Ge_{32}^{66} + He_{2}^{4} \rightarrow Se_{34}^{68} (+2.290 MeV)$$
$$ Se_{34}^{68} + He_{2}^{4} \rightarrow Kr_{36}^{72} (+2.151 MeV)$$
$$ Kr_{36}^{72} + He_{2}^{4} \rightarrow Sr_{38}^{76} (+2.728 MeV)$$
$$ Sr_{38}^{76} + He_{2}^{4} \rightarrow Zr_{40}^{80} (+3.698 MeV)$$
$$ Zr_{40}^{80} + He_{2}^{4} \rightarrow Mo_{42}^{84} (+2.714 MeV)$$
$$ Mo_{42}^{84} + He_{2}^{4} \rightarrow Ru_{44}^{88} (+2.267 MeV)$$
$$ Ru_{44}^{88} + He_{2}^{4} \rightarrow Pd_{46}^{92} (+2.276 MeV)$$
$$ Pd_{46}^{92} + He_{2}^{4} \rightarrow Cd_{48}^{96} (+3.030 MeV)$$
$$ Cd_{48}^{96} + He_{2}^{4} \rightarrow Sn_{50}^{100} (+3.101 MeV)$$
I ended my calculation here because I wasn't able to find the masses of other isotopes that would, theoretically, follow the chain. I understand that these are highly unstable and their fusion would need an immense amount of energy to overcome the Coulomb barrier. However, my point is that, according to the calculations above, once the barrier is overcome, the fusion would actually release energy, not consume it. So, is the notion of fusion beyond the iron peak elements being endothermic false or am I missing something?

There are plenty of misleading statements in Wikipedia and elsewhere on the internet about nucleosynthesis (I am busy searching to see if I have said something similar in the past!)
The reason that the alpha chain does not proceed significantly beyond $^{56}$Ni is that in order to overcome the Coulomb barrier the temperatures need to be so high that the iron-peak nuclei are disintegrated by photons at this temperatures.
I suppose the sense in which the endothermic statement is true is when considering a core made of nickel. In order to produce alpha particles you need to disintegrate some Ni nuclei. This process is highly endothermic and can't be balanced by subsequent fusion.
e.g (And this is a bit simplistic) Photodisintegration of a Ni nucleus into 14 alpha particles requires 88.62 MeV. Then 14 fusion reactions with Ni nuclei, producing Zinc, would give back just 37.9 MeV. In contrast disintegrating $^{52}$Fe into 13 alpha particles needs 80.5 MeV, but 13 fusion reactions of $^{52}$Fe to Ni yields $8.1\times 13= 105.3$ MeV.


do rockets follow a straight path but is looks curved because of space time? Does the path of a rocket or anything going straight up look curved because the mass of the earth is curving space-time?

If we assume the earth, the rocket and all the celestial bodies whose gravitational field reaches the earth to be massless then will the rocket follow truly a straight line, because there would be nothing curving the space-time?

This is partly space exploration and partly physics.

Rockets do not normally go straight up. They start off going up, to get into thinner air, then they steer to turn themselves eventually to a horizontal path, until they are going fast enough to stay in orbit.

In general relativity, an object that is free to move in the Earth's gravitational field will appear to accelerate towards the ground as it is following a path (called a geodesic) in curved spacetime. This is not specific to rockets; a cricket ball will follow a curved path when thrown due to the curvature of spacetime around the Earth. And if one is far  from any gravitational field, then a cricket ball will travel in a straight line.

To directly answer your question: The path of a rocket going straight up will not look curved, it will look like it is going straight up.


Why did the moon have a distinct reddish edge tonight? In Monterey, California, with a "half-moon" high in the sky at about 9 pm (2017-08-30), the moon had a distinct red outline on the outer edge of the NE quadrant; for awhile I saw it also on the outer edge of the SW quadrant.
I observed this phenomena from two locations, about a mile distant from each other.
I've never seen that before. What would cause it?
I took a picture of it with my camera, but you can't tell anything "out of the ordinary" from the image.
UPDATE
It's a clunky picture (I've been watching My Three Sons lately) - red not visible, fuzzy, and makes it like a full moon, but here it is, at any rate:

Smoke (fire) is not the only explication - Dust is also a good candidate. I've noticed a drop in "seeing quality" even when observing deep sky objects or planets trough a telescope due to large amounts of dust in the air.

  The second reason for a red Moon is if there’s some kind of particle in the air. A forest fire or volcanic eruption can fill the air with tiny particles that partially obscure light from the Sun and Moon. Once again, these particles tend to scatter blue and green light away, while permitting red light to pass through more easily. When you see a red moon, high up in the sky, it’s probably because there’s a large amount of dust in the air.

(source: Red Moon - Universe Today)


Is the Tombaugh Regio on Pluto visible from Earth? Is the Tombaugh Regio on Pluto discernably visible from Earth ground based or Earth orbiting telescopes? And more generally, can surface features seen in Hubble images be matched with images from New Horizons? I can imagine a match from the image compilation I attach, with TR being the bright triangle, but is that intuition valid?

Your First Question - Can it be done from the ground?

Unfortunately making maps that have this level of detail from the ground is not possible with current telescopes. There are 2 problems; first, it takes a lot of magnification to resolve Pluto and even more to resolve surface features. Second, from the ground you need more than just magnification, you need EXTREMELY stable skies or that will wash out the surface features. Getting just one super sharp image is not enough, you need several to ensure that you have high enough signal to noise to believe the surface features are real.

This is why Marc Buie, who used the High Resolution Channel on the Advanced Camera for Surveys on HST to create the image you put in your question. Unfortunately, even that level of resolution is not possible anymore as HRC is no longer functional. These kind of maps can be very hard to make.

There have been other maps of pluto, both from the ground and using HST. Here is a compilation of several using a compilation of ground based photometry, the Fine Optical Camera on HST and the ACS HRC. As you can see from the ground based photometry you can make a map, but it is very poor resolution compared to what HST can do.

There have been other HST press releases about those other images

Second Question - How well do the maps match?

There is a good blog post about how they match up from the museum of applied arts and sciences at the Sydney Observatory. But the bottom line is that they match up pretty well, but not perfectly. This is expected as we expected volitle transport on the surface so the surface features will change. As you can see in the image below, that there were a lot of changes from the map made in 1994 to the one made in 2003. So one would expect changes going out to 2015.

So, yes your intuition is valid, but Pluto's surface is dynamic so it will be changing on you!


In 2016, the summer solstice will coincide with a full moon. How often does this happen? I found that the solstices work on a 400 year cycle based on this. I can't find anything similar for the lunar cycle. Is there a formula I can use to calculate the date of the full moon?

Florin Andrei what evidence can you supply to charge the oldest scientific body on this planet (The Royal Society at Greenwich) to be a profit-driven media entity.

Bishop, if you are asking when the 2 specific points in time match, it is true that they have a 'astronomical' infrequency to look at. But if you look at what the common terms for the definition of a solstice and full moon occurring at the same time, then it is not too infrequent. Matching a 24 hour period (solstice day) against a 72 hour period (a full moon) then they occur with a much more reasonable frequency. There are lots of lunar cycle calculators around. The current cycle we are in has the same middle day of the full moon  period In 2004, 2007, 2015, 2026, 2034, 2042 and 2045. further dates can be found using this http://aa.usno.navy.mil.
Just chose any date, calculate the drift on your 400 year cycle then calculate how often that drift occurs at the same time as the 3 day phase of the full moon drift. There will be large periods when the match occurs rarely (The center point of each phase occurring furthest apart). As the centre points close on each other the Frequency will increase. The recurring pattern of all astronomic movements.


What was that Asteroid/Comet in the 90s? I am now in my mid 20s and I remember when I was 6-10 years old, there was a comet or asteroid that flew really close to earth. I remember being able to see the object very clearly as if it were the moon. I also live in the west coast of the United States. What was this object?

Was it Comet Hale-Bopp? It was discovered in 1995, but made a very close approach in 1997, earning it the nickname "The Great Comet of 1997".

Wikipedia states

  As it passed perihelion on April 1, 1997 the comet developed into a spectacular sight. It shone brighter than any star in the sky except Sirius, and its dust tail stretched 40–45 degrees across the sky. The comet was visible well before the sky got fully dark each night, and while many great comets are very close to the Sun as they pass perihelion, comet Hale–Bopp was visible all night to northern hemisphere observers.

Image courtesy of Wikipedia user LoopZilla under the Creative Commons Attribution-Share Alike 2.0 Austria license.

My second guess is Comet Hyakutake, which made an extremely close approach to Earth in 1996 - becoming "The Great Comet of 1996".


If Jupiter was placed in Mars' orbit, how would Jupiter appear from the naked eye? If Jupiter were as far from us as Mars, how big would it appear in Earth's sky? Would it appear huge, or would it be as small as a star?

Since the question is actually asking for a size of Jupiter as seen from Mars, there is a very simple equation that will give you a good idea.

$$\frac{\alpha}{360^{\circ}}=\frac{\ell}{2\pi r} $$
$\alpha$ is the angle that Jupiter would subtend in the sky, or the angular size, and $\ell$ is the actual diameter, or linear size, of Jupiter. $2\pi r$ is the circumference of Jupiter's orbit about Mars. Now I know what you're thinking, that this doesn't make any sense because of the vast differences of Jupiter's orbit when it is close compared to when it is at opposition. It'll work out...

The following are all approximations.

Mars orbits the Sun at $\sim1.53AU$ and Jupiter at $\sim5.2AU$. Because $1AU \approx 150Mkm$, The difference in their orbits are roughly $550.5Mkm$.

$$\alpha=\frac{140000km}{2\pi(550.5\times10^6km)}\times360^{\circ}=0.015^{\circ} $$

That is just the average angular size, at the average closest distance, but given that the only real tool to measure angular size with the naked eye (that isn't a sextant) is your hand, and your smallest digit (the pinky) at arms length marks out $1^{\circ}$ you'll never be able to tell the difference in the variation of size due to the change in distance.

Now answering your other questions directly:

Jupiter would look larger, but not enough to actually tell a difference in size.
As shown above, Jupiter would be much smaller than the moon appears to us from Earth, $0.5^{\circ}$.
No disk, as this is very much a point-like size as far as human eyes are concerned.


9th planet location? I've seen a number of news reports indicating there is likely a 9th planet in our Solar System, something with an orbital period of between 10k-20k years, that is 10 times Earth's mass. I haven't seen any real indication of where this object might be. If I had access to a sufficient telescope, would I be able to find this planet, and what way would I point a telescope to find it? How far is it likely to be, or is that not well known?

It's too dim to be seen during a normal survey during the majority of its
orbit.

Update:  Scientists at the University of Bern have modeled a hypothetical 10 Earth mass planet in the proposed orbit to estimate its detectability with more precision than my attempt below.

The takeaway is that NASAs WISE mission would have probably spotted a planet of at least 50 Earth masses in the proposed orbit and that none of our current surveys would have had a chance to find one below 20 earth masses in most of its orbit.  They put the planets temperature at 47K due to residual heat from formation; which would make is 1000x brighter in infrared than it is in visible light reflected from the sun.

It should however be within reach of the LSST once it is completed (first light 2019, normal operations beginning 2022); so the question should be resolved within a few more years even if its far enough from Batygin and Brown's proposed orbit that their search with the Subaru telescope comes out empty.

My original attempt to handwave an estimate of detectability is below.
The paper gives potential orbital parameters of $400-1500~\textrm{AU}$ for the semi major axis, and $200-300~\textrm{AU}$ for perihelion. Since the paper doesn't give a most-likely case for orbital parameters, I'm going to go with the extreme case that makes it most difficult to find. Taking the most eccentric possible values from that gives an orbit with a $1500~\textrm{AU}$ semi-major axis and a $200~\textrm{AU}$ perihelion has a $2800~\textrm{AU}$ aphelion.

To calculate the brightness of an object shining with reflected light, the proper scaling factor is not a $1/r^2$ falloff as could be naively assumed.  That is correct for an object radiating its own light; but not for one shining by reflected light; for that case the same $1/r^4$ scaling as in a radar return is appropriate. That this is the correct scaling factor to use can be sanity checked based on the fact that despite being similar in size, Neptune is $\sim 6x$ dimmer than Uranus despite being only $50\%$ farther away: $1/r^4$ scaling gives a $5x$ dimmer factor vs $2.25$ for $1/r^2$.

Using that gives a dimming of 2400x at $210~\textrm{AU}\;.$ That puts us down $8.5$ magnitudes down from Neptune at perihelion or $16.5$ magnitude.  $500~\textrm{AU}$ gets us to $20$th magnitude, while a $2800~\textrm{AU}$ aphelion dims reflected light down by nearly $20$ magnitudes to $28$ magnitude. That's equivalent to the faintest stars visible from an 8 meter telescope; making its non-discovery much less surprising.

This is something of a fuzzy boundary in both directions. Residual energy from formation/radioactive material in its core will be giving it some innate luminosity; at extreme distances this might be brighter than reflected light. I don't know how to estimate this. It's also possible that the extreme cold of the Oort Cloud may have frozen its atmosphere out. If that happened, its diameter would be much smaller and the reduction in reflecting surface could dim it another order of magnitude or two.

Not knowing what sort of adjustment to make here, I'm going to assume the two factors cancel out completely and leave the original assumptions that it reflects as much light as Neptune and reflective light is the dominant source of illumination for the remainder of my calculations.

For reference, data from NASA's WISE experiment has ruled out a Saturn-sized body within $10,000~\textrm{AU}$ of the sun.

It's also likely too faint to have been detected via proper motion; although if we can pin its orbit down tightly Hubble could confirm its motion.

Orbital eccentricity can be calculated as:

$$e = \frac{r_\textrm{max} - r_\textrm{min}}{2a}$$

Plugging in the numbers gives:

$$e = \frac{2800~\textrm{AU} - 200~\textrm{AU}}{2\cdot 1500~\textrm{AU}} = 0.867$$

Plugging $200~\textrm{AU}$ and $e = 0.867$ into a cometary orbit calculator gives a $58,000$ year orbit.

While that gives an average proper motion of $ 22~\textrm{arc-seconds/year}\;,$ because the orbit is highly eccentric its actual proper motion varies greatly, but it spends a majority of its time far from the sun where its values are at a minimum.

Kepler's laws tell us that the velocity at aphelion is given by:

$$v_a^2 = \frac{ 8.871 \times 10^8 }{ a }  \frac{ 1 - e }{ 1 + e }$$

where $v_a$ is the aphelion velocity in $\mathrm{m/s}\;,$ $a$ is the semi-major axis in $\mathrm{AU},$ and $e$ is orbital eccentricity.

$$v_a = \sqrt{\frac{ 8.871 \times 10^8 }{ 1500 } \cdot \frac{ 1 - 0.867 }{ 1 + 0.867 }} = 205~\mathrm{m/s}\;.$$

To calculate the proper motion we first need to convert the velocity into units of $\textrm{AU/year}:$

$$205 \mathrm{\frac{m}{s}}\;  \mathrm{\frac{3600 s}{1 h}} \cdot \mathrm{\frac{24 h}{1 d}} \cdot \mathrm{\frac{365 d}{1 y}} \cdot \mathrm{\frac{1\; AU}{1.5 \times 10^{11}m}} = 0.043~\mathrm{\frac{AU}{year}}$$

To get proper motion from this, create a triangle with a hypotenuse of $2800~\textrm{AU}$ and a short side of $0.043~\textrm{AU}$ and then use trigonometry to get the narrow angle.

$$\sin \theta = \frac{0.044}{2800}\\ \implies \theta = {8.799×10^{-4}}^\circ   = 3.17~\textrm{arc seconds}\;.$$

This is well within Hubble's angular resolution of $0.05~\textrm{arc seconds};$ so if we knew exactly where to look we could confirm its orbit even if its near its maximum distance from the sun. However its extreme faintness in most of its orbit means that its unlikely to have been found in any survey. If we're lucky and it's within $\sim 500~\textrm{AU},$ it would be bright enough to be seen by the ESA's GAIA spacecraft in which case we'll located it within the next few years. Unfortunately, it's more likely that all the GAIA data will do is to constrain its minimum distance slightly.

Its parallax movement would be much larger; however the challenge of actually seeing it in the first place would remain.


Is or might there be lightning on Io? As far as we know, Jupiter's moon Io has the most volcanic activity of any known celestial body. It has the most volcanoes that are active and they're erupting frequently. On Earth (the third planet from the Sun) there's lightning during volcanic eruptions and on Venus (the 2nd planet) there's lightning too, probably also during eruptions. Has any lightning ever been detected during an eruption on Io, or isn't it possible for lightning to occur during eruptions on Io for some reason?

I would bet against lightning due to the very thin SO2 atmosphere; under normal conditions the pressure is  0.3 to 3 nbar, with plumes reaching 5 to 40 nbar. That does not sound like it could convey lightening. But given that the environment also has a lot of ionization there might also be all sorts of plasma and vacuum breakdown discharges. In short, I suspect this could go either way, but if there were lightning it would be rather different than on Earth since it would be a near-vacuum discharge.
As far as I have been able to find, there are no empirical detections of lightning at Io. I suspect the radio measurements that could detect it would be swamped by the overall noise of the Jupiter-Io environment.


Magnetic fields of peculiar HgMn A type stars Do HgMn (peculiar A type) stars really possess global magnetic fields? See, for example, this paper by Hubrig et al. from 2012.

One thing that seems to be clear is that HgMn stars have only an extremely weak net longitudinal magnetic field component, if any. Shorlin et al. (2002) did an early survey of HgMn, Am, and Ap stars, and detected no longitudinal magnetic fields in the former, with a median 1$\sigma$ uncertainty of 39 Gauss. Makaganiuk et al. (2010) also found $B_z$ values of 0 in the stars they surveyed, with a higher precision - a 1$\sigma$ uncertainty of 0.81-10 Gauss, varying between stars. Other studies also yielded precisions of less than a few Gauss for some stars (see mentions by Makaguniak (2011)).

Some reports have found longitudinal values in the 10s to 100s of Gauss, but as Kochukhov notes, subsequent inveistigations have failed to confirm these findings, which have had extremely high uncertainties. One example is Hubrig et al. (2012), the paper you cite, which claimed to have found weak longitudinal and quadratic fields in several stars, including HD 65949. Kochukhov et al. (2013) then found no longitudinal fields on the star, to within a few Gauss, and Bagnulo et al. (2013) attributed to 2012 findings to instrument error, leading to flawed data.

Non-longitudinal magnetic fields have not been observed in much detail (small-scale longitudinal fields have not yet been ruled out, either, by large-scale global ones appear to be nonexistent), and complicated ones could still exist. Kochukhov et al. (2013) do say that they have ruled out large so-called tangled magnetic fields, but small-scaled ones are still possible, according to Hubrig.

One thing worth noting is that the vast majority of these studies, including the one you referenced, which has been disputed, are focused on B-type HgMn stars, in part because fewer A-type HgMn stars have been discovered.


How would we detect an Earth doppelganger planet? With our current technology, or technology available in the near future (up to 2025), how would we detect a planet exactly like our own, and how close would it have to be to be detectable?

Which methods thus far have proven the most effective for detecting Earth-like planets and what would they be able to reveal regarding our planet's atmosphere, orbital properties, and our species?

This question is interesting food for thought on both how easily an extraterrestrial civilization with comparable technology might find us, and for how easily we might find them.

For the purpose of the question we will assume that there is a planet with a civilisation with equivalent technological advancements that could exist anywhere.

Using current technology (and by that I mean experiments and telescopes that are available now) we have not detected an "Earth-like" planet and we would probably be unable to detect life on Earth even if observed from a distance of a few light years. Therefore there is currently no prospect of detecting life on an "Earth doppelganger". I elaborate below:

No planets like the Earth have yet been detected around another star. That is to say, none that have a similar mass, radius and orbit at 1 au (or close to it) from a solar-type star. With current technology, it is just out of reach. Therefore any directed search for life on an Earth-like planet wouldn't actually know where to start. If you can't detect the planet at all then there is absolutely no chance of looking at its atmospheric composition to look for biomarkers (e.g. oxygen along with a reducing gas like methane, or chlorofluorocarbons from an industrial civilisation - Lin et al. 2014). The only exoplanets for which atmospheric compositions have been (crudely and tentatively) measured are "hot Jupiters". - giant exoplanets orbiting very close to their parent stars.
A "blind" search could look for radio signatures and of course this is what SETI has been doing. If we are talking about detecting "Earth", then we must assume that we are not talking about deliberate beamed attempts at communication, and so must rely on detecting random radio "chatter" and accidental signals generated by our civilisation.
The SETI Phoenix project was the most advanced search for radio signals from other intelligent life. Quoting from Cullers et al. (2000): "Typical signals, as opposed to our strongest signals fall below the detection threshold of most surveys, even if the signal were to originate from the nearest star". Quoting from Tarter (2001): "At current levels of sensitivity, targeted microwave searches could detect the equivalent power of strong TV transmitters at a distance of 1 light year (within which there are no other stars)...". The equivocation in these statements is due to the fact that we do emit stronger beamed signals in certain well-defined directions, for example to conduct metrology in the solar system using radar. Such signals have been calculated to be observable over a thousand light years or more. But these signals are brief, beamed into an extremely narrow angle and unlikely to be repeated. You would have to be very lucky to be observing in the right direction at the right time if you were performing targeted searches.

Hence my assertion that with current methods and telescopes there is not much chance of success. But of course technology advances and in the next 10-20 years there may be better opportunities.

The first step in a directed search would be to find planets like Earth. The first major opportunity will be with the TESS spacecraft, launching in 2017, capable of detecting earth-sized planets around the brightest 500,000 stars. However, it's 2-year mission would limit the ability to detect an Earth-analogue. The best bet for finding other Earths will come later (2024 perhaps) with the launch of Plato, a six-year mission that again, studies the brightest stars. However, there is then a big leap forward required to perform studies of the atmospheres of these planets. Direct imaging and spectroscopy would probably require space-borne nulling interferometers; indirect observations of phase-effects and transmission spectroscopy through an exoplanet atmosphere does not require great angular resolution, just massive precision and collecting area. Spectroscopy of something the size of Earth around a normal star will probably require a bigger successor to the James Webb Space Telescope (JWST - launch  2018), or even more collecting area than will be provided by the E-ELT in the next decade. For example Snellen (2013) argues it would take 80-400 transits-worth of exposure time (i.e. 80-400 years!) to detect the biomarker signal of an Earth-analogue with the E-ELT!

It has been suggested that new radio telescope projects and technology like the Square Kilometre Array may be capable of serendipitously detecting radio "chatter" out to distances of 50 pc ($\sim 150$ light years) - see Loeb & Zaldarriaga (2007). This array, due to begin full operation some time after 2025 could also monitor a multitude of directions at once for beamed signals. A good overview of what might be possible in the near future is given by Tarter et al. (2009).


What shape are black holes theorized to be? I have always been curious about what black holes (or their event horizons) would look like if we could directly observe them. I notice that they tend to be depicted as flat holes in space but it seems to me that they would be spheres. Is there a scientific consensus on this. Or is it a vague question?

The event horizon would be spherical. What is usually depicted in drawings is the accretion disk, where all the debris gather before entering the event horizon.


Can our Sun become a black hole Does every star become a black hole? Is there any probability that our sun can become a black hole? If yes then is it on its way to become a black hole? what is the current state of sun as per the black hole life cycle? What will be the effect on all the planetary objects in the solar system if the sun turns into a black hole.

Sorry for so many questions but i cannot miss them as these are some questions on my mind.

No, the sun won't ever become a black hole.
The choice between the three fates of stars (white dwarf, neutron star, black hole) is entirely determined by the star's mass.
A star on the main sequence (like most stars, including our sun) is constantly in a balance between the inward pressure of gravity and the outward pressure of the energy generated by the hydrogen fusion that makes it "burn".1  This balance stays relatively stable until the star runs out of whatever its current fuel is - at that point, it stops burning, which means there's no longer outward pressure, which means it starts collapsing.  Depending on how much mass there is, it might get hot enough as it collapses to start fusing helium together. (If it's really massive, it might continue on to burn carbon, neon, oxygen, silicon, and finally iron, which can't be usefully fused.)
Regardless of what its final fuel is, eventually the star will reach a point where the collapse from gravity is insufficient to start burning the next fuel in line.  This is when the star "dies".
White dwarfs
If the star's remains2 mass less than 1.44 solar masses (the Chandrasekhar limit3), eventually gravity will collapse the star to the point where each atom is pushed right up against the next.  They can't collapse further, because the electrons can't overlap.  While white dwarfs do shed light, they do so because they are extremely hot and slowly cooling off, not because they're generating new energy.  Theoretically, a white dwarf will eventually dim until it becomes a black dwarf, although the universe isn't old enough for this to have happened yet.
Neutron stars
If the collapsing star is above the Chandraskhar limit, gravity is so strong that it can overcome the "electrons can't overlap" restriction.  At that point, all the electrons in the star will be pushed into combining with protons to form neutrons.  Eventually, the entire star will composed primarily of neutrons pushed right up next to each other.  The neutrons can't be pushed into occupying the same space, so the star eventually settles into being a single ball of pure neutrons.
Black holes
Black holes are the step beyond neutron stars, although they're worth discussing in a bit more detail.  Everything, in theory, has a Schwarzschild radius.  That's the radius where a ball of that mass would be so dense that light can't escape.  For example, the Schwarzschild radius for Earth is about 9mm.  However, for all masses smaller than somewhere between 2-3 times the mass of the sun, it's impossible to squeeze the matter small enough to get it inside that radius.  Even a neutron star isn't massive enough.
But a star that becomes a black hole is.  We don't actually know what happens to a star once it's become a black hole - the edges of the "hole" itself is simply the Schwarzschild radius - the point light can't escape.  From outside, it doesn't matter whether the matter collapsed to the point that the neutrons started overlapping, whether it stopped just inside the radius, or whether it continued collapsing until it broke all known physical laws.  The edges are still the same, because they're just a cutoff based on the escape velocity.

1  I'm ignoring the red giant phase here, since it's just a delay in the "run out of fuel" step.  Basically, the core is helium "ash", while the hydrogen fusion process takes place further and further out.  Once that runs out, you get a nova and the collapse continues.
2 Likewise, I'm ignoring the mass that stars shed in their various nova phases.  All given masses are based on the remnants left behind.
3 Every source I've found for Chandrasekhar mass, except Wikipedia, gives 1.44 or 1.4 solar masses (which are compatible).  Wikipedia gives 1.39, and gives at least one source to back that number.


Data for Sun's Orbit The planets approximately orbit the sun, but a more precise statement is that the planets and the sun all orbit around a common barycenter.

Because the sun is so massive, this barycenter is very close to the sun.  If we take the sun to be the origin, the barycenter travels in a complex, nonlinear trajectory (from Wikipedia):

Modeling this accurately is critically important to predicting celestial events.  I wrote a simulator to predict the 2012 transit of Venus, based on estimates of orbits and the assumption that the sun is at the origin.

The error introduced by these assumptions was enough to miss the transit by a solar diameter.

This data must exist, since the transit actually did happen.  Additionally, errors in the orbital data of the planets could have affected it.  Where can I find data on where the planets and the sun are and what trajectories they're moving in (relative to the solar system's barycenter)?

A well-known analytical planetary theory is VSOP.  The paper describes six versions, most of which are heliocentric.  One of them (E) is barycentric and has an extra (ninth) body: the Sun.  Hence, for VSOP you have the choice of using the Solar System's barycentre as the origin and computing the position of an extra body, or of computing heliocentric positions of the planets directly.

If you use a different theory, I would expect something similar: a heliocentric version and/or a barycentric version with equations for the Sun.


How do kepler orbits account for planetary migration? This article got me thinking if kepler orbits explain how a planet moves in a solar system how can a planet migrate as explained in the article. At some point don't the kepler orbits must need some other model to explain their positioning?

  "Jupiter would have moved right through the primordial asteroid belt, emptying it and then repopulating it with scrambled material from both the inner and outer solar system as Jupiter then reversed course and headed back towards the outer solar system," DeMeo and Carry write, stressing that further observations should nail down how faithfully this scenario represents reality.

Kepler orbits apply only to 2-body systems.

They apply only, if the masses aren't too large and the orbits aren't too close. Otherwise relativistic effects occur, as for the Sun - Mercury system.

Kepler orbits only apply, if the bodies are sufficiently spherical.

As soon as a third body comes into play, the system can become chaotic.

Too many conditions, especially for our early solar system.

Strange orbits like those around Lagrangian points, or horseshoe orbits can occur. Orbital resonance can stabilize as well as destablize orbits.


Are there any double stars that I can actually see orbit each other? If I had a nice amateur telescope†, are there any multiple star systems that I could observe over a few years or a few decades and actually see the movement of one or both of them over time?

My short human lifespan and limited telescope put heavy constraints on the orbital distance, brightness, and distance from the Sun, so I am guessing that if there are any at all, the number is probably small.

†arbitrarily defined as say 8-inch (20 cm) aperture, with a good set of eyepieces, and a sketch pad.

Vir (12h 42m, –01° 27′)

Probably Porrima, $\gamma$ Vir, is the best candidate for most observers in the Northern Hemisphere to see changes in a binary orbit, particularly using a small telescope. It is a pair of stars with similar size and visual magnitude, of about 3.6. Their orbital period is about 169 years, but the orbit is eccentric, e = 0.88. They are also relatively close at about 40 ly. Periapsis was in 2005, so the stars are now moving away from each other, but their rate of separation is decreasing. Separation at periapsis was about 0.4 arcsec, so would not have been resolved using a small telescope in 2005. By 2015 their separation was ~2.5 arcsec, and will increase to ~3 arcsec by 2020. I estimated a position angle change between 2015-2020 of ~7 degrees. These changes should be detectable with a 100-200 mm (4-8 inch) telescope.

Since most short period binaries are close together, with nearly circular orbits, and often more distant, they are very difficult or impossible to resolve with a small telescope.

Sirius B (06h 45m, −16° 43′)

As mentioned in @MichaelWalsby's answer, it is also possible to observe the orbit of the white dwarf binary companion to Sirius, the brightest star visible in the night sky. Sirius is only 8.6 ly away, and their orbit has a semi-major axis of about 7.5 arcsec, an eccectricity of e = 0.59, and a period of about 50 years. If this pair were similar in brightness, they would be an easy answer to this question. Unfortunately, Sirius B, or the Pup (as the companion is known to amateur astronomers), is ~10 magnitudes dimmer than Sirius, and usually lost in its glare. It takes a night with excellent seeing, i.e. a stable, non-turbulent atmosphere, especially since Sirius never gets much above 30 degrees elevation at the mid-northern latitude where I live. I have only seen the Pup 4 or 5 times (one view was probable but not certain) over almost 6 decades of observing, and I have never seen it in a telescope with aperture under 300 mm. I know other amateurs that have seen Sirius B in 150-200 mm telescopes, but mostly at lower latitudes. However, by seeing Sirius B at intervals separated by decades, I have observed its polar angle change.

I believe the separation of Sirius B is now over 10 arcsec, and still increasing slightly. So for the next couple of decades, observing it might be a bit easier. In recent winters I have  tried with telescopes from a 120 mm refractor to a 250 mm Dobsonian, and occasionally larger, but still have not seen it for several years. This Hubble photo of Sirius gives some idea why Sirius B is hard to observe in small telescopes.

Indirect methods

Also, many eclipsing binaries are often observed, and their light curves measured. By analyzing light curves, orbital elements can be estimated. However, these indirect orbital observations are  probably stretching the intent of the original question.


Why type Ia supernova reaches its peak absolute magnitude(~ -19.3mag) at around 15-20 days after the explosion? I'm not sure why type Ia supernova reaches its peak magnitude 15-20 days after the explosion. Is this luminosity generated only by the radioactive decay of nickel-56 to cobalt-56?(considering a half-life, I don't think the decay to iron-56 can be involved with it) Or is there any combination that produces the peak luminosity?

The early light curve of a Type Ia SN (i.e., the fact that the brightness increases to a peak, then declines) is the result of the combination of two things, both of which are decreasing over time: energy from the decay of Ni-56 to Co-56 (and Co-56 to Fe-56), and the opacity of the explosion as it expands.

Just after the actual explosion, the expanding fireball is dense and opaque. Only light from the very outermost layer escapes to the outside.[*] Since the Ni-56 created in the thermonuclear burning that caused the explosion is in the interior regions, the initially visible part of the expanding fireball has little or no Ni-56, and thus no energy from its decay.

As the fireball expands, it gets less dense, and thus less opaque, to the point where light from the layers heated by radioactive decay can actually escape. Over time, more and more of the inner part of the fireball becomes visible, and so we see more and more of the regions where Ni-56 (and its daughter product Co-56) is decaying. Roughly speaking, the peak of the light curve is when the opacity is low enough that all the light from the Ni-56 + Co-56 decay can escape.[**]

But since there are fewer and fewer radioactive nuclei as time goes by, the amount of energy produced by the decay decreases with time. This is why the light curve turns over: now we're seeing all the light produced by Ni-56 and Co-56 decay, but that just keeps getting fainter over time.

[*] At this point, the SN is too faint to be seen. Even the initial thermonuclear explosion is too faint to be seen under most circumstances.

[**] Ni-56 and Co-56 actually produce gamma rays (and energetic positrons) when they decay, so what "the light from Ni-56 + Co-56 decay" means is thermal radiation from the gas heated by the gamma rays. Also, as I understand it, what we're seeing is really the combination of energy from current decay and residual heat from previous radioactive decay.


Understanding Scatter in Data collected for HR Plot I was recently asked a question for a lab I am doing for an astronomy class where we collected data from the M44 cluster (through Google Sky) and used information about the Hyades cluster to estimate the age and distance of M44. Anyways, I was asked whether there was more scatter in my M44 data or the provided Hyades data. There was definitely more scatter (see below) and I'm not sure how to explain it. I believe it is because of these reasons:

There is measurement error.
And I might be picking up background/foreground stars This would definitely lead to scatter. If I measure a foreground star and assume it is as far as the M44 star cluster, it would appear brighter than if it were truly in the M44 cluster.

Am I on the right track or am I missing something? Any help would be great. Here is my data, the green is the collected M44 data.

So, scatter in observational measurements of clusters in the HR diagram is mainly, in my experience, caused by 3 things: observational errors, binaries and contamination.

As you mentioned, there are observational errors, which are contributed to by the instrument, the data reduction, the weather conditions, etc. This error usually gets worse as you get fainter toward the limits of the telescope and looks like a "widening" of the CMD (seen here notably in magnitudes > 19).

The next problem is contamination. When you point your telescope at an object like M44 or the Hyades, it's not obvious which stars belong to the cluster and which stars just happen to be in the field of view, so you will get contamination from normal milky way stars (which will not follow that isochronic relationship that cluster members will). This looks something like this.

The final problem that I can think of is a binary population. A "photometric binary" star is what happens when two stars overlap in an observation. This happens quite often in clusters, since there are "intrinsic binaries" which are stars that are very close to, and orbit each other. They are usually different masses, and hence different colors. This is usually seen as another line, next to the main isochrone line, that is a little redder than all of the other stars and would look something like this image.

Now, this scatter looks far too big to be measurement error alone, modern telescopes will have magnitude errors of much much less than a magnitude (think milli-magnitudes, depending on specifics -- I'm unfamiliar with Google Sky, but any telescope is going to do better than a magnitude error, for reference the magnitude system is based on eyesight "just-noticable-difference", so your eyes have an approximate error of one magnitude).

And since the scatter doesn't follow some other systematic relationship, I would discard the binaries idea.

The most reasonable explanation to me is, like you said, foreground contamination.

Figures:

https://www.researchgate.net/figure/231129963_fig2_Fig-4-A-B-vs-B-V-color-magnitude-diagram-of-all-9507-stars-in-the-WFPC2-image-of-M30

http://home.strw.leidenuniv.nl/~piladiez/

https://inspirehep.net/record/1217369/plots


Determining the area of Lagrange Points I have looked into Lagrange points a decent amount and I see many images that show the L4 and L5 locations as wide sweeping areas. Jupiter's Trojan asteroids are a good example of this. As well, L3 sometimes is shown inset of the Hill Sphere [but that is another question]. I assume that each point is an infinitely small point in space at the exact mathematical position (ignoring any other gravitational affects that might be there).

My question: is there a way to determine the "stable" region in each Lagrange location?
Particularly in the L4 and L5 regions.

Essentially, I would like to be able to know how much additional thrust a craft might need to stay at these locations depending on how far away they are from the center of the Lagrange point.

My question: is there a way to determine the "stable" region in each Lagrange location? Particularly in the L4 and L5 regions.

tl;dr: Yes but it's usually a fancy version of "trial and error".

We are sometimes wrongly told that this abominable zero-velocity potential diagram show areas that are "bound" or "unbound" to Lagrange points, but it does not and that's obvious, because while the surface dips near L1, L2 and L3 it is a maximum near L4 and L5 and yet we know that for some mass ratios these regions can collect and accumulate trojan asteroids.

  Source

The problem is that these are 2D plots of the pseudo-potential at one specific velocity. The pseudo-potential is given by

$$C = \omega^2 (x^2+y^2) + 2\left(\frac{\mu_1}{r_1} + \frac{\mu_2}{r_2} \right) - (v_x^2 + v_y^2 + v_z^2)$$ and these zero velocity surfaces which are contours when drawn in a 2D slice don't depict orbits at all. If you put a particle on any contour other than $C=0$ it will accelerate rather than follow one of these lines.

So instead what people do is to run numerical simulations with a whole range of starting position and velocity vectors and find out by a fancy version of "trial and error" which particles will end up back at the same position where they started with the same velocity vector they started with. Every time they find a closed, periodic orbit it helps map out the "stable region".

One example of a method to find regions of 4D or 6D phase space ($x, y, v_x, v_y $ or $x, y, z, v_x, v_y, v_z$) that have closed periodic orbits is the shooting method which can be made faster by use of the transition matrix.

In Figure 1 of F. Marzari et al. Origin and Evolution of Trojan Asteroids you can see how tempting it is to use a zero velocity contour to indicate where the region of stable orbits lay with out actually saying it. Instead the the words "are closely related to" are used.

  Fig. 1. The location of the five Lagrange equilibrium points in the circular-restricted three-body problem. The primary and secondary masses (the Sun and planet in our examples) are denoted by the large and small filled circles respectively. The selected zero velocity curves (see text) are closely related to the types of orbits that can occur in the system. The letters P, H, and T denote passing, horseshoe, and tadpole orbits respectively. Note that the two masses and each of the L4 and L5 points form an equilateral triangle.

If you want to go beyond just a circular restricted three-body model and to the real solar system, numerical simulations become much more complex because there is long term evolution of these orbits due to all of the devitaions of reality from the simple CR3BP model. See just for an example The resonant structure of Jupiter's Trojan asteroids – I. Long-term stability and diffusion


What is the first recorded reference to the Moon being a satellite of the Earth? I saw a list recently of the moons of the various planets and the discoverer(s) and date of discovery. Earth's Moon was listed as Unknown.

I know you don't 'discover' a huge disc in the sky. But still, what is the first recorded reference to the Moon being a satellite of the Earth? That someone knew it is smaller and orbiting? Celestial Sphere does not count.

From before the dawn of history people naturally assumed that the sky was a solid dome above the flat earth.  The dome was assumed to rotate once a day, so the stars were assumed to be lights attached to the dome.

Anyone who assumed that the sky dome was opaque had to assume that the sun and the wandering planets were nearer than the sky dome.  I think that it was usually assumed the sky dome was opaque and colored blue, thus making the sky seem blue when it reflected the light of the sun during the day, thus making the sun closer than the sky dome.

Those who kept records of astronomical observations soon noticed that the moon occulted or passed in front of stars, planets, and the sun (solar eclipses), and thus was closer.

Everyone assumed that the sun and the moon were tiny, until traveler's reports showed that they had the same apparent diameters every place that as visited, and so were about equally far away from every place that was visited.  Thus as the size of the known world grew larger and larger, the minimum possible size and distance of the moon and the sun grew larger and larger.

The idea of a spherical earth was proposed and gradually accepted by Greek philosophers between the sixth and third centuries BC.  Thus the earth became known as Earth, a sphere instead of a flat disc, and the dome of the sky became a hollow spherical shell around it.

In Hellenistic times the diameter of the Earth was calculated with reasonable accuracy, as well as the distance to the moon and thus its diameter.  So it became known that the moon was about a quarter the diameter of the Earth and over sixty Earth radii distant.

Because of solar eclipses, it had been known since prehistoric times that the moon was closer than the sun, and keepers of astronomical records of events when the moon occulted (passed in front of) stars and planets knew that the moon was closer than planets and stars.

Anyone who assumed geocentrism, that the Earth was the enter of the universe, naturally assumed that every object revolved around the Earth, including the moon, and that the moon was thus the closest satellite of the Earth.  A few ancient Greek philosphers supported the Heliocentric theory, that the Earth and the planets revolved around the sun.  Some of them could have believed that the moon also revolved around the sun, but as far as I know all heliocentric believers also had the moon revolve around the Earth.

By the last few centuries BC there were many educated persons who believed that the Earth, the moon, and the sun were giant balls of rock (and thus somewhat similar objects), that the sun was a giant burning rock, and that the stars and planets might also be giant burning rocks far, far away from the Earth and appearing as dots as seen from Earth.  Since everybody already since prehistoric times believed that the moon revolved around the Earth, by the last few centuries BC educated people in Western civilization believed that the moon was what we now call a natural satellite of the Earth, though some of them believed and some did not that the sun and planets were also natural satellites of the Earth.


Do planetary magnetic fields slow the movement of conductive extraplanetary moving objects via induction? A fixed magnetic field will induce an electrical current in a conductive object moving through the field.  The strength of the eddy currents will depend on the shape and composition of the conductor and also how fast it is moving.  The electrical current produces ohmic heating in the conductor, and some of the kinetic energy of the moving conductor will be turned to heat. This is how induction brakes work.

I wonder if this principle applies to fast moving conductors like asteroids or satellites that encounter planetary magnetic fields.  Are planetary magnetic fields strong enough to act as induction brakes on moving conductors?  A satellite experiencing induction braking would fall to a lower and lower orbit and eventually out of orbit.  Passing asteroids might have their course deviated more than would be expected just from gravity.

My search turned up only this abstract, which attempted to explain why meteorites are hot via this principle.

http://adsabs.harvard.edu/full/1946PA.....54..482R

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

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

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

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


Can a telescope be built to focus on a point source star? If a Kepler telescope were built to focus on one point source star at a time, would it provide more information, or did Kepler capture virtually every photon from any given star anyway?

In order to study known transit events more closely one star at a time, to possibly detect moons or atmosphere or rings by resolving the transit light curve's shape better, is a larger aperture necessary or would a smaller field of view help? What field of view would be optimal for observing a single point source?

A smaller field of view would not necessarily help. The data transmitted back to Earth by Kepler was already limited by telemetry constraints and the portions of the CCD that were read out were already focused on small windows around (many) selected stars. Having a field of view that is at least 10 arcsecond across would be sufficient. What is more important is to make sure your star images are well-sampled. This is a bit of a problem for Kepler. The CCD pixels are quite big (4 arcseconds) to give a big field of view, but this is similar to the angular resolution of the telescope. This means all the light can fall in basically one pixel, which is not ideal for precise photometry. If you were focusing on one star you would make the field of view smaller but have lots of CCD pixels across your star image. This improves the precision of the photometry considerably.

If you were to construct a similar telescope designed to look at one star at a time, then in principle you could have a much higher telemetry rate for that star. Hence, if you had enough photons and a low CCD readout noise, you could take much shorter exposures and see finer detail in the light curve transits - maybe moons, rings, starspots etc. The standard Kepler cadence was 30 minutes for most stars and one minute for some of the brightest stars - see this link. A more rapid cadence would allow a much better characterisation of the host star through asteroseismology (e.g. mass and radius), which in turn leads to better characterisation of its planets.

To take advantage of the higher telemetry rates, it would be advantageous to have a larger aperture telescope, simply to collect enough photons. The angular resolution of the telescope is not that important if investigating obviously single , bright sources. You would need an absolutely enormous telescope (100m+) to resolve the disks or planetary systems of even nearby stars, so this is not really feasible.

A larger telescope would also allow you to split the light by wavelength and investigate transits in a wavelength dependent way. This is really important for investigating atmospheric composition and chemistry. This is perhaps the only area where focusing on a single star might be useful - because you could design a slit spectrograph to monitor the transit spectrum. On the other hand, these days it is possible to design multi-object spectrographs using fibres or slit masks, so overall I can't see any real advantage to having a telescope that is capable of gathering data for only one star at a time.


How large could gravitational waves get and what effect could they have on us? The waves detected by LIGO were basically subatomic distortions of space-time. Could there be much greater distortions closer to a source. Would they become perceptible to a human in any way and what would it do to us?

Wave strain scales as $1/r$; the waves that were detected came from a source about a Gpc away and had a strain of $10^{-20}$.

To get a strain that would affect things at the 1% level, you would need to be $10^{18}$ times closer, or at about 30,000 km from the merging black holes.

The tidal force due to the black holes is roughly $GMmh/4r^3$, where $h$ is your height, $m$ your mass and $M\sim 60M_{\odot}$ the combined mass of the black holes. Assuming 1.8m height and 80kg mass, the tidal force stretching you is around 50N. So I think you would notice this more than the gravitational wave, and even more so if you were to get any closer...


How fast would Earth need to rotate to fling someone off due to centrifugal force? If Earth started spinning fast enough, would the centrifugal force eventually overcome Earth's gravity enough enough to fling someone (let's assume 75 kg) into orbit or into space? If so, how fast would it need to spin?

Being that I'm not a scientist, I'm hoping for a simple answer like, "one rotation per xx hours/mins/seconds needed to throw someone off Earth" or something to that effect. Complicated equations and/or jargon is fine for the sciencey types, of course. I just probably won't understand it too much.

If it would rotate fast enough to throw a person off, it would throw everything off - dirt, houses, cars, etc., as the mass of an object is irrelevant - the centrifugal force (pulling out) and the gravitational force (pulling in) are both proportional to an object's mass.

Basically the whole thing would explode out at the same time. To calculate we equate the force of gravity $mg$ with the centrifugal force $mr\omega^2$, where $\omega$ is the angular velocity in radians per second

$$F = mg = mr\omega^2 $$ $$\Rightarrow g = r\omega^2 $$ $$\Rightarrow \omega = \sqrt{g/r} \approx 0.0012/s.$$

Insert the radius of the earth $r$ and the gravitational constant $g$, take the square root, and you have the necessary angular speed $\omega$.

with $\omega = 2 \pi/T$, you get $T = 2 \pi/\omega \approx 1.5 \mathrm{hours}$ [I did that in my head, so it might be wrong].


Power requirement for transmitting communications further than 1 ly? The StarShot project from Breakthrough Initiatives aims to send a paper light spacecraft to alpha centauri using pressure from a laser located on earth at 20,000kms.

What are the options for a transmitter that can be received from 1-4 lyrs away, if the transmitter weighs the same as a few jellybeans (<10 grams)? What wavelength and power range can they work with for that distance using that weight of equipment?

It's possible to transmit an intelligible signal to a target roughly 4.3 light-years away using 0.1 milliwatts. The trick for that particular level of power is that both the sender and receiver have to be carefully positioned several hundred AU from a star, with those stars lined up between them, in order to take advantage of a gravitational lensing effect. If only one of the two radios can be positioned at a star's gravitational focus, then the necessary power rises, but can be under 40 watts, according to one of the diagrams reprinted at https://www.centauri-dreams.org/?p=10123 .


What units are used for the Stefan-Boltzmann law? I have a star with given temperature in Kelvin and radius in solar radii. I tried to calculate the luminosity of the star using Stefan Boltzmann's law, and got an absurd number (over 1 million). What am I doing wrong, and are there any units that I should use instead of Kelvin and solar units?

The Stefan-Boltzmann constant $\sigma$ is not a dimensionless quantity, it comes with units. So whatever units you use, you must ensure that the value you use for the Stefan-Boltzmann constant is consistent with them.
So using the value expressed in terms of SI units:
$$\sigma = 5.670\,374\,419\ldots \times 10^{-8}\,\rm W\,m^{-2}\,K^{-4}$$
you would either have to work with radius, luminosity and temperature in metres, watts and kelvins, or convert $\sigma$ to the units you are actually using.
For example, if you want to work in terms of solar radii and luminosities you have to account for the conversion factors $L_\odot = 3.828 \times 10^{26}\,\rm W$ and $R_\odot = 6.957 \times 10^8\,\rm m$, giving:
$$\sigma = 7.169\ldots \times 10^{-17}\ L_\odot\, R_\odot^{-2}\, \rm K^{-4}$$


What do we know about the lifecycle of the Milky Way (or any other spiral galaxy)? I know that the Milky Way will collide with Andromeda in the distant future but my knowledge about the lifecycle of a galaxy is very lacking.

I was surprised that Wikipedia is so sparse on information in this topic. So how can I visualize its lifecycle? What will happen in the future?

FYI: I just found this very informative video on the topic.

If we look at Newtonian physics, and how galaxies will interact, a central black hole should just be considered as a massive, dense object.

The Milky Way does not fall into its own central black hole, it orbits the centre of mass - as physics tells us it should.

When we get closer to Andromeda, the gravitational influence of Andromeda will act more strongly on us, and when we get really close, individual masses within each galaxy will have dramatic effects on each other, but as with any such system, the two black holes will not suck everything in.

If the two galaxies end up coalescing - which is not a given - the orbits of the stars and the black holes will be very complex. For billions of years the 2 black holes will orbit each other, getting closer as they shed energy, but during that time the stars around them will suffer many effects, including:

some will be expelled
some will hit the black holes
some will go nova
and so on

The best way to understand what will happen is to avoid thinking too deeply about black holes being weird, and treat them as dense masses. For most purposes this will help you model galactic collisions.

Have a look at this NASA simulation of the collision between the Milky Way and Andromeda:


How do we calculate the escape velocity of galaxies? How do we calculate the escape velocity of galaxies?
Do we have to know the total mass?
What is the escape velocity for Milky Way?

In short, the escape velocity from the region we find ourselves in the milky way is 544 Km/s +- 10%.

This number, and an extensive explanation of how it was calculated can be found on a paper called The RAVE Survey: Constraining the Local Galactic Escape Speed

Also important to consider is the fact that we already are at aproximately 220 Km/s because of the sun's orbital velocity, so the actual value needed to get there from earth is much lower.


Could dark matter particles be unstable? Is it possible that dark matter particles are unstable and the existing dark matter will decay in the long run and turn into hydrogen, for example? What would be possible mechanisms for such a decay?

Could some of the hydrogen we observe be "new" hydrogen, i.e. be the product of dark matter decay?

Yes, as mentioned elsewhere, it is possibly possible. Dark matter particles may be intrinsically unstable (though having long lifetimes, which are at least significantly longer than Hubble time).

Check for more info here:
http://arxiv.org/abs/1307.6434


Why is Ganymede's aurora only visible in ultraviolet while Earth's is bright green and red? GANYMEDE AURORAE

NASA's Hubble Space Telescope observed a pair of auroral belts encircling the Jovian moon Ganymede. The belts were observed in ultraviolet light by the Space Telescope Imaging Spectrograph and are colored blue in this illustration.

Why is Ganymede's aurora only visible in ultraviolet while Earth's is in the visible part of the spectrum?

Quoting from Nat.Geo. article (which has that same UV image),

  Then, in 2010 and 2011, Hubble took a close look at the moon. More
  specifically, it looked at the auroral bands ringing Ganymede’s poles.
  Hubble studied the auroras in the ultraviolet, but Saur said the
  shimmering lights would be visible to human eyes.

  “If somebody could be standing on Ganymede looking up into the night
  sky, it would appear as red aurora, to you,” he described.


Earth rising and setting from moon's perspective Why the earth is rising and setting seen from the moon,
when the moon is tidally locked?
Shouldn't the earth be always on the same spot because of the tidal lock, if observed from the moon?

You're quite right.  The Earth is (nearly) stationary in the Moon's sky.

(I say "nearly" because the Moon is in a slightly elliptical orbit around the Earth, but rotates perfectly smoothly. This means that the Earth's motion through the Lunar sky is
 a bit faster when the Moon is at perigee and a bit slower when it's at apogee. Because the Moon is tide-locked, on average they match perfectly, but during the course of the sidereal month the Earth appears to move back and forth by a few degrees.  The effect isn't huge, but it's easily observable with a telescope.  See the Wikipedia article on
"libration" for an OK discussion and a really great animated gif illustration of libration as viewed from Earth.)

Is it possible that you're reacting to the amazing movie from Apollo 8 showing Earthrise over the Lunar horizon? It's been int he news a lot given that this is the 50th anniversary of the flight.  If so, what you saw there was not from the Moon's surface, but from orbit around the Moon and the Earthrise happened as Apollo 8 came around from behind the moon (where the Earth is not visible in the sky) and the Earth first became visible around the edge of the Moon.


Are Symbiotic and Vampire Stars the same type of star or different? Are Vampire Stars and Symbiotic Stars the same type of star or is there a difference?

Vampire Stars siphon off hydrogen from its neighbour.  According to https://www.cfa.harvard.edu/~kenyon/ss/, Symbiotic Stars seem like the same description.

"In all systems, the hot component - a main sequence star, a white dwarf, or a neutron star - accretes material lost by the red giant." (Harvard)

SS Leporis has been described as Symbiotic Star and in some places as Vampire Stars.

As the title of the research paper linked to in the ESO press release is "An incisive look at the symbiotic star SS Leporis", it is safe to conclude that they are the same type of star. My suspicion is that, since symbiotic stars transfer and accrete mass via either Roche-lobe overflow or stellar wind overflow, the term "vampire star" was coined as part of the press release as a "cute" way to imply that one star is "feeding" on another.


Telescope showing high chromatic aberration Hi i have made a telescope with 1100 mm focal length objective lense and 21mm plano convex as eyepiece. Both these lenses are not achromatic. But when is looked through it , the images were completely blurry. Is this because I have not used achromatic lense.?? Can it be resolved if I used any coated eyepiece.? The diameter of objective is 7.5 cm and eyepiece is 1cm. Is this issue is because the size of eyepiece is so small.

Is this because I have not used achromatic lens?

An easy way to check for the effects of chromatic aberration will be to look at a narrow wavelength source. These days that's easy because the spread in wavelengths from single color LEDs is only 10 to 20 nm1 depending on the specific LED. If you can find something with red, green or yellow LEDs (not the white ones) can you check to see if those can be focused more clearly? All you need is one small LED to do this, maybe across a room. If your focal length is 1.1 meters and the LED is 10 meters away, you'll have to pull your eyepiece out by another 10 cm farther than you would for focusing at infinity.
1random example
Note also that chromatic aberration is stronger in the blue than in the red.
You can also estimate the effect with a bright white object with a sharp edge (like a white piece of paper) on a black or dark background. If the edges are blurry with strong color effects (blue and red in different places) then this may indeed be the problem.
If so, you might have to invest in an achromatic objective to get rid of this problem, or look only at things that are monochromatic (and in Astronomy there aren't many of those!)
In addition to making sure the eyepiece lens is centered on the objective lens' axis, you also have to make sure the eyepiece lens isn't tilted. Both of those will cause problems and the point-like LED spot will show this. It might look like a fuzzy comet coma aberration.

Can it be resolved if I used any coated eyepiece?

No, don't worry about that for now. Adding antireflection coatings to lenses reduces reflections which are a big problem when there are multiple lens elements with air gaps between them within the eyepiece or the objective. Those cause lens flare usually only a big problem if there are very bright, point-like objects in the field of view. But in your case each is a single element, so not an issue.

Does this smartphone photo show Mars just below the Sun?
Apparently two objects in Solar Eclipse image
Weird Blob near Polaris?

Is this issue is because the size of eyepiece is so small.

No, I think it's because of some combination of the eyepiece not being positioned properly and chromatic aberration, and a test on an LED might be one way to separate the two effects for diagnostic purposes. If the blurriness has a lot of color associated with it, especially around white objects on black backgrounds, this is your primary culprit I think.


Planetary tails: How many have we observed yet? Just recently I learned that Mercury has a sodium tail and it is actually well studied since long:

As a result, the planet has the appearance of a comet, with a tail that's been observed streaming nearly 3.5 million kilometres away from the planet.

This made me wonder about exoplanets with tails, like WASP-69b. How many (exo)planets are known to have a tail?
References

Why would a planet form a cometlike tail under conditions of weaker rather than stronger solar wind?

Yes, apart from WASP-69b, there has been reports of other exoplanets emitting some form of planetary tails. Let's start from the Wikipedia:

KIC 12557548 b is a small rocky planet, very close to its star, that is evaporating and leaving a trailing tail of cloud and dust like a comet. The dust could be ash erupting from volcanos and escaping due to the small planet's low surface-gravity, or it could be from metals that are vaporized by the high temperatures of being so close to the star with the metal vapor then condensing into dust.
In June 2015, scientists reported that the atmosphere of GJ 436 b was
evaporating, resulting in a giant cloud around the planet and, due to
radiation from the host star, a long trailing tail 14 million km (9
million mi) long.

More on the dust particles that is trailing from the exoplanets KIC 12557548 b as well as KOI-2700b:

The observed tail lengths are consistent with dust grains composed of
corundum (Al2O3) or iron-rich silicate minerals (e.g., fayalite,
Fe2SiO4). Pure iron and carbonaceous compositions are not favored. In
addition, we estimate dust mass loss rates of 1.7 ± 0.5 M⊕ Gyr-1 for
KIC 12557548b, and > 0.007 M⊕ Gyr-1 (1σ lower limit) for KOI-2700b.
Source: Dusty tails of evaporating exoplanets, I. Constraints on the dust composition, R. van Lieshout1, M. Min and C. Dominik, A&A, Volume 572, December 2014, DOI: 10.1051/0004-6361/201424876

Astronomers using NASA's Hubble Space Telescope have confirmed the
existence of a baked object that could be called a "cometary planet."
The gas giant planet, named HD 209458b, is orbiting so close to
its star that its heated atmosphere is escaping into space.
Source: https://www.nasa.gov/mission_pages/hubble/science/planet-tail.html

Astronomers have confirmed that the planet Gliese 436b seems to be
trailing a gigantic, comet-like cloud of hydrogen.

Source: https://skyandtelescope.org/astronomy-news/exoplanet-with-a-comet-tail-06302015234/

Disintegrating planets allow for the unique opportunity to study the
composition of the interiors of small, hot, rocky exoplanets because
the interior is evaporating and that material is condensing into dust,
which is being blown away and then transiting the star. Their transit
signal is dominated by dusty effluents forming a comet-like tail
trailing the host planet K2-22b, making these good candidates for
transmission spectroscopy.
Source: Eva H. L. Bodman et al 2018 AJ 156 173, DOI: 10.3847/1538-3881/aadc60


Ice filled craters on Mars How many craters on Mars have ice in them and which ones have permanent ice? I could only find Korolev and the unnamed one - strange for such a remarkable crater.
http://www.esa.int/Science_Exploration/Space_Science/Mars_Express/Water_ice_in_crater_at_Martian_north_pole
One would expect all such craters to be famous already.
Which such crater is closet to the equator?

Since no one has posted an answer, I'll turn my comment into one, even though it is not going to give you a quantitative answer.
The reason that I can't give you a number is because the answer is simply, "lots."  Any crater that is on the north or south polar layered deposits (N/SPLD) are going to have permanent ice in them, by definition, since they formed in ice.  Since the NPLD is much younger than the SPLD, it has many fewer craters.
The largest (off the top of my head) on the SPLD is McMurdo, roughly 30 km across.  Korolev is larger at about 81 km across, but it is not on the NPLD.  Korolev itself is large enough that it has its own mini-climate with positive feedback mechanisms to keep the ice there, but smaller craters can't do that.
To get permanent ice in craters not in the N/SPLD, you need to keep the temperatures cold enough and pressure high enough for the ice to remain throughout that hemisphere's summer.  Since the northern hemisphere's summer is when Mars is near aphelion, getting 45% less sunlight than during southern summer, craters can be closer to the equator and still meet that requirement.  Also, the northern hemisphere is lower in elevation than the southern, so you also get more atmospheric pressure which helps keep the ice from sublimating.  You're also going to want a high enough rim and/or deep enough crater cavity to guard against sunlight as much as possible, so larger craters (like Korolev) are again going to be preferable to small ones.  But, Korolev is the largest crater closest to the NPLD, so that's why it's be best known for this sort of phenomenon.
So, are there others?  Certainly.  But, they're smaller.  And, unless an impact crater is scientifically important and has had a name nominated for it through the International Astronomical Union, it is not going to be named, so very few craters on Mars are named.
To answer your question about which is the closest to the equator, it is not something I'm sure has ever really been looked at.  One would need good, repeat mosaics at different points in Mars' orbit and simply do a visual search.  If you do do that, happy hunting!


Can't see any detail on planets with my new Celestron NexStar 130 SLT Computerized Telescope Mars looks like a small white blob/circle and the same with all other planets. This is very frustrating. I've looked all over for a solution jumping from site to site and nothing. Please be different on this site.

Edit: Many people have marked this post as duplicate. Yes, I know, but the other post did not answer my question.

Mars is a small planet and currently far from Earth (just over 1 Astronomical Unit). Therefore, it currently appears to be very small -- smaller than 9 arcseconds. At that size, it will look like a small disk.

Based on this one observation, I think that no conclusion can be made about any possible problem.

Also, a 130 mm f/5 scope is rather short for planetary observing where you want high magnification. This scope is better for wide fields of view, such as the Pleiades or Double Cluster.


What process shaped Ultima Thule? When I see this picture, I'm trying to understand how these two elements merged together and have two hypothesis :

two balls of lava in fusion together
two solid rocks fused at their contact point

But it appears that the collision speed was not particularly fast so I'm not sure in the end.

What process likely formed Ultima Thule?

Ultima Thule is a contact binary comprised of two formerly separate solid objects that gently1 (probably <5 km per hour) came together.  If the objects had been molten when they came together they would have formed a spherical object.  Since the lobes are distinct the objects must have been solid.

As for "understand[ing] how these two elements merged together", Ultima and Thule are believed to be have been two objects that formed separately through accretion.  At some point the two (solid) objects collided at a very low velocity of "one or a few miles per hour"1.  Since the escape velocity at point of impact is about 16 miles per hour2, gravity alone would be sufficient to keep the two lobes together1.

1 This topic was discussed in the 2019-01-02 New Horizons press conference.
The presentation talking about formation mechanism starts here (the portion discussing approach speed is ~80 seconds later).  A discussion of how strongly bound the lobes are starts here.

A note about the video links above: The official JHU APL video linked to above is missing the first few minutes.  If the video is updated to include the complete press conference then the links above will point to the wrong times.  The first two links are to Jeff Moore's presentation and the third is him fielding a question from Bill Horowitz (CBS News).

2 To quote the source:

  Modeling it as two spheres, one 16 km across the other 12 km, with a density the same as water, ... [a]t a distance of 8 km from the center of mass, the escape velocity ... would be about ... 16 miles/hour.


Whats the probability of a hazardous impact for human life on Mars? Yesterday my sister and I have discussed the recent efforts of SpaceX to create a permanent colony on mars.

When coming to the several dangers of creating a habitat on a different planet, we got stuck on the probability of mars getting hit by an astronomical object thats large enough to wipe out life anywhere on the planet.
I assume the probability is higher than on Earth, due to Mars being closer to the Asteroid Belt, its size and thinner atmosphere. There are more potential objects to enter martian atmosphere and can be smaller than on Earth to create the same hazardous impact for life.

I have tried to research the numbers, but the only calculations I found were for specific asteroids and their probability to hit Mars.

On this StackExchange there are some questions concerningthe probability of an individual  (on earth) or a specific spot (on the moon) getting hit, but I found none for the hit of an object with a certain amount of impact (I hope thats the right word here).

So my question is:

What is the probability of an astronomical object hitting mars anywhere on its surface and kill all people currently living there?

Any impact destroying life-support systems and thus wiping out life on mars indirectly can also be included.

Assume that there is only one habitat, it is built directly on the surface and the first humans have arrived with the latest flight.

If there are any questions about the scope of this question, don't hesitate to ask.

We actually have a very good idea of this because the Mars Reconnaissance Orbiter has been orbiting Mars for over a decade. The MRO is, basically, a spy satellite around Mars and is continually taking high-resolution photos of the surface. It has revisted much of the surface, taking pictures multiple times over the years.

As a consequence, we have a very good idea of the rate of formation of craters on Mars. We have seen numerous craters that have formed between visits to a spot, but none are very large. The key thing is that we now have a very solid measurement of the flux of smaller objects onto Mars and can say with certainty, that that sort of impact is way down on the list of perils. (It's probably less than the odds of being killed by a tornado in the Midwest.)

Larger impacts are a bit harder since we haven't seen any, but we have two lines of evidence. First, space debris follows a distribution in size vs. frequency that we've measured, and it's not much different around Mars' orbit than it is around Earth's.

Second, the extensive photography we have of Mars' surface (which we now have mapped better than Earth's!) allows us to count craters and to make estimates of their age. We know that small craters a few yards to tens of yards across are much more common and we know the rate at which they hit, so astronomers count the number of small craters on the floors of larger craters and can make a very good estimate of the age of the larger crater.

From this we have developed measurements of the rate of cratering at all sizes, and, again, the hazard isn't that great.

I'd note that the risk of what on Earth would be civilization-killers is actually lower on Mars because of its lack of an atmosphere. The Chicxulub impact that killed the dinosaurs killed most of them through the superheated steam and air shockwave that went around the Earth followed by the years of Asteroid Winter that followed.  Mars' atmosphere isn't heavy enough to support either, so "all" you'd have to do is hunker down and wait out the fall of rock fragments... (Still, it's an event better watched from orbit than from the ground.)


How far apart are stars in a binary system? I'm wondering how far apart from one another the stars are in binary systems.
What is the distribution of the separation between binary stars? Are most of the binary stars very far apart like Sirius (8.2 to 31.5 AU)? Are there many stars in orbits even wider? Are there more pairs with grazing orbits?

Is there a database similar to exoplanets.org where I could see the distance between the pair of stars for all the binary systems we know of?

Distances ($a$) between binary stars vary wildly, from the order of the radius of the stars, to more than a light-year! The plot below (from here) shows a compilation of several surveys, with the color indicating the method by which they've been detected.

Separations are roughly normally distributed in $\log a$, peaking at a typical separation of tens of AU. The corresponding periods have median values of a few hundred years (e.g. Raghavan et al. 2010; Duquennoy & Mayor 1991).

As commented by Oddthinking, if stars are too close we are no longer able to resolve them visually (green bars). But we can still detect them spectroscopically (yellow bars): If we observe the blended spectral lines of two stars, we see the lines shift back and forth as the two stars orbit each other and their light is Doppler shifted.

On the other hand, if stars are too far from each other, their orbital periods of several thousand years makes it impractical to observe them orbit each other (we have only observed double stars for some 200 years); instead they're detected because they simply follow the same path through space — i.e. they have a common proper motion (blue bars).

As commented by PM2Ring, the fact that the two $x$ axes suggest a 1:1 correspondence between orbital radius and period seems odd — by Kepler's law, this also depends on mass. My random guess is that, because low-mass stars (which are most abundant) are poorly constrained (difficult to observe) and high-mass stars are rare, then most observed binaries are of the order of a Solar mass. If they're not too far from each other in mass, then there is an approximate correspondence.


Why do photos of Earth always have a black background? if we are in the milky way, and people are able to take photographs from Earth that look like this:

Then how come almost all of Earth's photos look like this?

In the first photo, we are seeing what is "outside Earth", and so if we actually went "outside Earth", why don't we see these gases?

Is it because of the Sun blinding our vision? Suppose a satellite was hovering over the part of Earth where its night time (very little sun interference). If it "turned around and faced the opposite direction from earth", it would see the void, but would it see all these gases, or just the black void.

Why?

The photo of the Milky Way is an exposure that was 30 seconds or longer in order to capture the stars. The photo of the Earth was an exposure that was probably 1/100 of a second. (It is daytime!) The stars will not appear in such a short exposure.


What constellation would the Sun be in from an exoplanet? Where would it be in the sky from Proxima B (planet near Proxima Centauri) for example?

The sun would appear roughly half-way between Capella and the W of Cassiopeia. The constellations are a human invention, and don't correspond to actual groups of stars. Since all the nearby stars would be in somewhat different positions the constellations would be mixed up, for example, Sirius would be close to Betelgeuse. You can't really talk about constellations except from the perspective of Earth, but you could say that the sun would be near some of the stars that form Perseus and Camelopardalis.

From more distant exoplanets the constellations would be completely mixed up, so there would be no sense in describing the sun's location in terms of a constellation. The sun would also be too dim to be seen with the naked eye from many exoplanets.


Instead of using flexible mirrors, could software be used for adaptive optics? Telescopes that use adaptive optics now use flexible mirrors to adjust for distortions in the atmosphere.  The mirrors are flexed in real time to account for the distortions.  I am wondering if it would be feasible to instead use fixed mirrors and use software to make the needed corrections to the image?

Short answer: no.

Why not?  because the mirror is placed at a position in the optical train such that it controls the wavefront phase and tilt of each subsection, "subaperture" of the incoming beam.  Once an optical system has formed an image, all phase information is lost.  There  are some software methods which can correct for simple aberrations such as astigmatism or coma, but those require a near-constant type of error across the input field.  The problem with atmospheric turbulence is that the errors are essentially of a localized and random nature.

Note that even a plenoptic camera which, by tricky use of sublenses at the focal plane, record enough information to change focus, cannot correct for general phase aberrations.


Calculate how long it takes for the sun to complete one orbit around the galaxy? We know the speed that the sun travels around the galaxy is $220\, {\rm km/s} = 0.00021 \, {\rm parsec/year}$. I'm unsure of how to find the period?

The time it takes for the Sun to complete one orbit around the center of the Galaxy is called a Galactic year. It is estimated to last 225 to 250 million terrestrial years.
Now if you know the speed of the Sun in the Galaxy and you know how far the Sun is from the center, you can estimate this period. Let's assume a circular orbit and ignore the (large) error bars, for the sake of simplicity. The Sun is roughly $8\,200$ parsecs away from the Galactic center. So the circumference of the Sun's orbit is $8\,200 \cdot 2 \cdot \pi = 52\,000 {\rm \,parsecs}$. You mention the speed of the Sun is $2.1\cdot 10^{-4} {\rm \,parsec/year}$. So to complete one orbit, it takes circumference/speed = $52\,000 / 2.1\cdot 10^{-4} = 245 \cdot   10^6 {\rm years}$, that is 245 million years. Pretty close to the actual value!


Is gravitational energy released when a body contracts? Kelvin stated that gravitational energy released due to contraction was responsible for the energy radiated in the stars. Although the theory was wrong, was the assumption practically right? If it's right, how does contraction take place?

It is correct. Contracting stars heat up due to their own contraction. This justifies the initial temperature raise in a star, before fusion begins in earnest.

Kelvin was not 100% wrong, it's just that fusion, over the life time of the star, makes much, much more energy than contraction. The collapse heats up the star at the very beginning, then fusion takes over and makes essentially all the heat you observe.

This is also true of gas giant planets. Even a tiny amount of contraction may raise their temperature measurably. E.g., Jupiter radiates more energy than it receives from the Sun - this is due to contraction, at a rate of 2 cm / year (diameter).

TLDR: Gravitational energy becomes heat.